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# Prove that the coefficient of xn in the binomial expansion of
`
Question:
Prove that the coefficient of $x n$ in the binomial expansion of $(1+x)^{2 n}$ is twice the coefficient of $x^{n}$ in the binomial expansion of $(1+x)^{2 n-1}$.
Solution:
To Prove : coefficient of $x^{n}$ in $(1+x)^{2 n}=2 \times$ coefficient of $x^{n}$ in $(1+x)^{2 n-1}$
For $(1+x)^{2 n}$
$a=1, b=x$ and $m=2 n$
We have a formula,
$t_{r+1}=\left(\begin{array}{c}m \\ r\end{array}\right) a^{m-r} b^{r}$
$=\left(\begin{array}{c}2 \mathrm{n} \\ \mathrm{r}\end{array}\right)(1)^{2 \mathrm{n}-\mathrm{r}}(\mathrm{x})^{\mathrm{r}}$
$=\left(\begin{array}{c}2 \mathrm{n} \\ \mathrm{r}\end{array}\right)(\mathrm{x})^{\mathrm{r}}$
To get the coefficient of $x^{n}$, we must have,
$x^{n}=x^{r}$
- $r=n$
Therefore, the coefficient of $x^{n}=\left(\begin{array}{c}2 n \\ n\end{array}\right)$
$=\frac{(2 \mathrm{n}) !}{\mathrm{n} ! \times(2 \mathrm{n}-\mathrm{n}) !} \ldots\left(\because\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\frac{\mathrm{n} !}{\mathrm{r} ! \times(\mathrm{n}-\mathrm{r}) !}\right)$
$=\frac{(2 n) !}{n ! \times n !}$
$=\frac{2 \mathrm{n} \times(2 \mathrm{n}-1) !}{\mathrm{n} ! \times \mathrm{n}(\mathrm{n}-1) !} \ldots \ldots \ldots . .(\because \mathrm{n} !=\mathrm{n}(\mathrm{n}-1) !)$
$=\frac{2 \times(2 n-1) !}{n ! \times(n-1) !}$ ………cancelling n
Therefore, the coefficient of $x^{n}$ in $(1+x)^{2 n}$ $=\frac{2 \times(2 \mathrm{n}-1) !}{\mathrm{n} ! \times(\mathrm{n}-1) !}$ ………eq(1)
Now for $(1+x)^{2 n-1}$,
$a=1, b=x$ and $m=2 n-1$
We have formula,
$\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{c}\mathrm{m} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{m}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$
$=\left(\begin{array}{c}2 n-1 \\ r\end{array}\right)(1)^{2 n-1-r}(x)^{r}$
$=\left(\begin{array}{c}2 n-1 \\ r\end{array}\right)(x)^{r}$
To get the coefficient of $x^{n}$, we must have,
$X^{n}=X^{r}$
- $r=n$
Therefore, the coefficient of $x^{n}$ in $(1+x)^{2 n-1}=\left(\begin{array}{c}2 n-1 \\ n\end{array}\right)$
$=\frac{(2 n-1) !}{n ! \times(2 n-1-n) !}$
$=\frac{1}{2} \times \frac{2 \times(2 n-1) !}{n ! \times(n-1) !}$
…..multiplying and dividing by 2
Therefore
Coefficient of $x^{n}$ in $(1+x)^{2 n-1}=0 \times$ coefficient of $x^{n}$ in $(1+x)^{2 n}$
Or coefficient of $x^{n}$ in $(1+x)^{2 n}=2 \times$ coefficient of $x^{n}$ in $(1+x)^{2 n-1}$
Hence proved.
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# Matrix Multiplication
Before attempting matrix multiplication, makes sure you know the basics first.
Multiplication is much more complicated than some of the other matrix operations, like matrix addition and scalar multiplication. GradeA will show you two approaches: the Turn & Flip and the Zipper. Choose the method you like the best! Before you can multiply matrices, you need to know when the operation is possible. Sometimes you cannot multiply the two matrices together at all.
When is Matrix Multiplication Defined?
The key to answering this question is to look at the dimensions of each matrix. Remember, they are always listed as row x column.
To see if the operation is defined, list the two dimensions next to each other. Important Note: The order does matter!
Next, check to see if the middle numbers are the same!
Example #1 Example #2
Once the matrix multiplication is defined, you can find the dimensions of the result (the answer). The outside numbers will give you the new dimensions.
For the examples above, the results would be (2 x 3) and (3 x 1).
Here are more examples:
Original Dimensions: (3 x 1)(3 x 1) (2 x 3)(3 x 2) (4 x 1)(2 x 4) Multiplication Possible? No! (different) Yes! (same) No! (different) Resulting Dimensions: (none) (2 x 2) (none)
The Turn and Flip Method
Now that you know how to determine the dimensions of the resulting matrix, now you need to know how to actually multiply them.
Example:
We are multiplying a (2 x 2)(2 x 2) - so the result will also be a (2 x 2).
You will need to multiply each corresponding row times the corresponding column - notice the circles below. Once you multiply the matching pieces, add the results.
It might seem a little confusing at first, but after a little practice you will get it. Of course, you might prefer to use the zipper method instead...
The Zipper Method
If you had a difficult time understanding the turn and flip method, maybe the zipper method will be easier for you to understand. We will use the same example: First, take the second matrix and raise it above the first.
Now you are going to "zip" in the numbers that line up. Look at the light blue arrows, as well as the colored circles...
So, now that you have seen both methods of matrix multiplication, which do you prefer: the turn and flip method or the zipper method?
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Multiplication and Division of Radicals
## Rationalize the denominator
0%
Progress
Practice Multiplication and Division of Radicals
Progress
0%
Dividing Square Roots
The area of a rectangle is 30\begin{align*}\sqrt{30}\end{align*}. The length of the rectangle is 20\begin{align*}\sqrt{20}\end{align*}. What is the width of the rectangle?
### Watch This
Watch the first part of this video, until about 3:15.
### Guidance
Dividing radicals can be a bit more difficult that the other operations. The main complication is that you cannot leave any radicals in the denominator of a fraction. For this reason we have to do something called rationalizing the denominator, where you multiply the top and bottom of a fraction by the same radical that is in the denominator. This will cancel out the radicals and leave a whole number.
4. ab=ab\begin{align*}\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\end{align*}
5. abbb=abb\begin{align*}\frac{\sqrt{a}}{\sqrt{b}} \cdot \frac{\sqrt{b}}{\sqrt{b}} = \frac{\sqrt{ab}}{b}\end{align*}
#### Example A
Simplify 14\begin{align*}\sqrt{\frac{1}{4}}\end{align*}.
Solution: Break apart the radical by using Rule #4.
14=14=12
#### Example B
Simplify 23\begin{align*}\frac{2}{\sqrt{3}}\end{align*}.
Solution: This might look simplified, but radicals cannot be in the denominator of a fraction. This means we need to apply Rule #5 to get rid of the radical in the denominator, or rationalize the denominator. Multiply the top and bottom of the fraction by 3\begin{align*}\sqrt{3}\end{align*}.
2333=233
#### Example C
Simplify 3240\begin{align*}\sqrt{\frac{32}{40}}\end{align*}.
Solution: Reduce the fraction, and then apply the rules above.
3240=45=45=2555=255
Intro Problem Revisit Recall that the area of a rectangle equals the length times the width, so to find the width, we must divide the area by the length.
3020\begin{align*}\sqrt{\frac{30}{20}}\end{align*} = 32\begin{align*}\sqrt{\frac{3}{2}}\end{align*}.
Now we need to rationalize the denominator. Multiply the top and bottom of the fraction by 2\begin{align*}\sqrt{2}\end{align*}.
3222=62
Therefore, the width of the rectangle is 62\begin{align*}\frac{\sqrt{6}}{2}\end{align*}.
### Guided Practice
Simplify the following expressions using the Radical Rules learned in this concept and the previous concept.
1. 12\begin{align*}\sqrt{\frac{1}{2}}\end{align*}
2. 6450\begin{align*}\sqrt{\frac{64}{50}}\end{align*}
3. 436\begin{align*}\frac{4\sqrt{3}}{\sqrt{6}}\end{align*}
1. 12=12=1222=22\begin{align*}\sqrt{\frac{1}{2}}=\frac{\sqrt{1}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}= \frac{\sqrt{2}}{2}\end{align*}
2. 6450=3225=1625=425\begin{align*}\sqrt{\frac{64}{50}}=\sqrt{\frac{32}{25}}=\frac{\sqrt{16 \cdot 2}}{5}= \frac{4\sqrt{2}}{5}\end{align*}
3. The only thing we can do is rationalize the denominator by multiplying the numerator and denominator by 6\begin{align*}\sqrt{6}\end{align*} and then simplify the fraction.
43666=4186=4926=1226=22
### Explore More
Simplify the following fractions.
1. 425\begin{align*}\sqrt{\frac{4}{25}}\end{align*}
2. 1649\begin{align*}-\sqrt{\frac{16}{49}}\end{align*}
3. 96121\begin{align*}\sqrt{\frac{96}{121}}\end{align*}
4. 5210\begin{align*}\frac{5\sqrt{2}}{\sqrt{10}}\end{align*}
5. 615\begin{align*}\frac{6}{\sqrt{15}}\end{align*}
6. 6035\begin{align*}\sqrt{\frac{60}{35}}\end{align*}
7. 81830\begin{align*}8\frac{\sqrt{18}}{\sqrt{30}}\end{align*}
8. 126\begin{align*}\frac{12}{\sqrt{6}}\end{align*}
9. 208143\begin{align*}\sqrt{\frac{208}{143}}\end{align*}
10. 213214\begin{align*}\frac{21\sqrt{3}}{2\sqrt{14}}\end{align*}
Challenge Use all the Radical Rules you have learned in the last two concepts to simplify the expressions.
1. 81215\begin{align*}\sqrt{\frac{8}{12}} \cdot \sqrt{15}\end{align*}
2. 32456205\begin{align*}\sqrt{\frac{32}{45}} \cdot \frac{6\sqrt{20}}{\sqrt{5}}\end{align*}
3. 242+8268\begin{align*}\frac{\sqrt{24}}{\sqrt{2}}+\frac{8\sqrt{26}}{\sqrt{8}}\end{align*}
4. 23+463\begin{align*}\frac{\sqrt{2}}{\sqrt{3}}+\frac{4\sqrt{6}}{\sqrt{3}}\end{align*}
5. 551221510\begin{align*}\frac{5\sqrt{5}}{\sqrt{12}}-\frac{2\sqrt{15}}{\sqrt{10}}\end{align*}
### Vocabulary Language: English
Rationalize the denominator
Rationalize the denominator
To rationalize the denominator means to rewrite the fraction so that the denominator no longer contains a radical.
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# What is the limit of (x+7)/(3x+5) as x approaches infinity?
Rylan Hills | Certified Educator
calendarEducator since 2010
starTop subjects are Math, Science, and Business
The value of lim x--> inf. [(x+7)/(3x+5)] is required.
As x --> inf. , 1/x --> 0
lim x--> inf. [(x+7)/(3x+5)]
=> lim x --> 0[ 1/x + 7)/(3/x + 5)]
=> lim x --> 0[ 1 + 7x)/(3 + 5x)]
substitute x = 0
=> 1/3
The required limit is 1/3
check Approved by eNotes Editorial
hala718 | Certified Educator
calendarEducator since 2008
starTop subjects are Math, Science, and Social Sciences
We need to find the limit of the following.
==> lim (x+7) / (3x+5) as x--> inf.
We will divide both numerator and denominator by x.
==> lim ( x+ 7)/x / (3x+5)/x
Now we will simplify.
==> lim ( 1+ 7/x) / (3 + 5/x)
Now we will substitute with x = inf.
==> lim (1+ 7/x) / (3+ 5/x) as x--> inf = ( 1+ 7/inf ) / ( 3+ 5/inf)
But we know that a/ inf = 0
==> lim (x+7)/(3x+5) as x--> inf = (1+ 0)/ (3+ 0) = 1/3
Then the limit is 1/3.
check Approved by eNotes Editorial
tonys538 | Student
The limit `lim_(x-> oo) (x+7)/(3x+5)` has to be determined.
If we substitute `x = oo` in `(x+7)/(3x+5)` we get the indeterminate form `oo/oo` , To determine the limit in this case we can use l'Hospital's rule and substitute the numerator and denominator by their derivatives.
(x +7)' = 1
(3x + 5)' = 3
This gives the limit `lim_(x->oo)1/3`
As the variable x does not figure in `1/3` , this is the required limit.
The limit `lim_(x-> oo) (x+7)/(3x+5) = 1/3`
check Approved by eNotes Editorial
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# 5 5factoring trinomial iii
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## 5 5factoring trinomial iiiPresentation Transcript
• Factoring Trinomials III
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not.
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring.
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable.
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable.
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2)
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square,
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square, hence it is prime.
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square, hence it is prime. b. 3x2 – 7x + 2
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square, hence it is prime. b. 3x2 – 7x + 2 b2 – 4ac = (–7)2 – 4(3)(2)
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square, hence it is prime. b. 3x2 – 7x + 2 b2 – 4ac = (–7)2 – 4(3)(2) = 49 – 24
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square, hence it is prime. b. 3x2 – 7x + 2 b2 – 4ac = (–7)2 – 4(3)(2) = 49 – 24 = 25 which is a squared number,
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square, hence it is prime. b. 3x2 – 7x + 2 b2 – 4ac = (–7)2 – 4(3)(2) = 49 – 24 = 25 which is a squared number, hence it is factorable.
• Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b2 – 4ac = 0, 1, 4, 9, 16, 25, 36, .. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b2 – 4ac to see if the trinomial is factorable. If it is, factor it. a. 3x2 – 7x – 2 b2 – 4ac = (–7)2 – 4(3)(–2) = 49 + 24 = 73 is not a square, hence it is prime. b. 3x2 – 7x + 2 b2 – 4ac = (–7)2 – 4(3)(2) = 49 – 24 = 25 which is a squared number, hence it is factorable. In fact 3x2 – 7x + 2 = (3x – 1)(x – 2)
• Factoring Trinomials III The ac-Method
• Factoring Trinomials III The ac-Method The product of two binomials has four terms.
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials.
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping.
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6)
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Take out the common factors = x(x – 3) + 2(x – 3)
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Take out the common factors = x(x – 3) + 2(x – 3) Take out the common (x – 3) = (x – 3)(x + 2)
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Take out the common factors = x(x – 3) + 2(x – 3) Take out the common (x – 3) = (x – 3)(x + 2) But how do we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in the first place?
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Take out the common factors = x(x – 3) + 2(x – 3) Take out the common (x – 3) = (x – 3)(x + 2) But how do we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in the first place? This leads to the ac-method.
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Take out the common factors = x(x – 3) + 2(x – 3) Take out the common (x – 3) = (x – 3)(x + 2) But how do we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in the first place? This leads to the ac-method. The ac-method tells us how to rewrite the trinomial so we may use the grouping method.
• Factoring Trinomials III The ac-Method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x2 – x – 6 by grouping. x2 – x – 6 = x2 – 3x + 2x – 6 Put them into two groups = (x2 – 3x) + (2x – 6) Take out the common factors = x(x – 3) + 2(x – 3) Take out the common (x – 3) = (x – 3)(x + 2) But how do we know to write x2 – x – 6 as x2 – 3x + 2x – 6 in the first place? This leads to the ac-method. The ac-method tells us how to rewrite the trinomial so we may use the grouping method. If it isn't possible to rewrite the trinomial according to the ac-method, then the trinomial is prime.
• Factoring Trinomials III ac-Method:
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c.
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac,
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac,
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b.
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +cme.
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c)
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime.
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Example B. Factor 3x2 – 4x – 20 using the ac-method.
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Example B. Factor 3x2 – 4x – 20 using the ac-method. Because a = 3, c = –20, we’ve ac = 3(–20) = –60.
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Example B. Factor 3x2 – 4x – 20 using the ac-method. Because a = 3, c = –20, we’ve ac = 3(–20) = –60. We need two numbers u and v such that uv = –60
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Example B. Factor 3x2 – 4x – 20 using the ac-method. Because a = 3, c = –20, we’ve ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4.
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Example B. Factor 3x2 – 4x – 20 using the ac-method. Because a = 3, c = –20, we’ve ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4. A good way to do a thorough search is to make a table of all the ways to factor 60.
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Example B. Factor 3x2 – 4x – 20 using the ac-method. Because a = 3, c = –20, we’ve ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4. A good way to do a thorough search is to make a table of all the ways to factor 60. u v 1 60 2 30 3 20 4 15 5 12 6 10
• Factoring Trinomials III ac-Method: We assume that there is no common factor for the trinomial ax2 + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b. 2. Write ax2 + bx + c as ax2 + ux + vx +c 3. Use the grouping method to factor (ax2 + ux) + (vx + c) If step 1 can’t be done, then the expression is prime. Example B. Factor 3x2 – 4x – 20 using the ac-method. Because a = 3, c = –20, we’ve ac = 3(–20) = –60. We need two numbers u and v such that uv = –60 and u + v = –4. A good way to do a thorough search is to make a table of all the ways to factor 60. From the list we got 6 and (–10). u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 put in two groups = (3x2 + 6x ) + (–10x – 20) u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 put in two groups = (3x2 + 6x ) + (–10x – 20) pull out common factor = 3x(x + 2) – 10 (x + 2) u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 put in two groups = (3x2 + 6x ) + (–10x – 20) pull out common factor = 3x(x + 2) – 10 (x + 2) pull out common factor = (3x – 10)(x + 2) u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 put in two groups = (3x2 + 6x ) + (–10x – 20) pull out common factor = 3x(x + 2) – 10 (x + 2) pull out common factor = (3x – 10)(x + 2) Example C. Factor 3x2 – 6x – 20 using the ac-method. u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 put in two groups = (3x2 + 6x ) + (–10x – 20) pull out common factor = 3x(x + 2) – 10 (x + 2) pull out common factor = (3x – 10)(x + 2) Example C. Factor 3x2 – 6x – 20 using the ac-method. a = 3, c = –20, hence ac = 3(-20) = –60. u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 put in two groups = (3x2 + 6x ) + (–10x – 20) pull out common factor = 3x(x + 2) – 10 (x + 2) pull out common factor = (3x – 10)(x + 2) Example C. Factor 3x2 – 6x – 20 using the ac-method. a = 3, c = –20, hence ac = 3(-20) = –60. Need two numbers u and v such that uv = –60 and u + v = –6. u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Write 3x2 – 4x – 20 = 3x2 + 6x –10x – 20 put in two groups = (3x2 + 6x ) + (–10x – 20) pull out common factor = 3x(x + 2) – 10 (x + 2) pull out common factor = (3x – 10)(x + 2) Example C. Factor 3x2 – 6x – 20 using the ac-method. a = 3, c = –20, hence ac = 3(-20) = –60. Need two numbers u and v such that uv = –60 and u + v = –6. After searching all possibilities we found that it's impossible. Hence 3x2 – 6x – 20 is prime. u v 1 60 2 30 3 20 4 15 5 12 6 10 6*(–10) = – 60 6 + (–10) = –4
• Factoring Trinomials III Exercise A. Use the ac–method, factor the trinomial or demonstrate that it’s not factorable. 1. 3x2 – x – 2 2. 3x2 + x – 2 3. 3x2 – 2x – 1 4. 3x2 + 2x – 1 5. 2x2 – 3x + 1 6. 2x2 + 3x – 1 7. 2x2 + 3x – 2 8. 2x2 – 3x – 2 9. 5x2 – 3x – 2 11. 3x2 + 5x + 2 12. 3x2 – 5x + 2 15. 6x2 + 5x – 6 10. 5x2 + 9x – 2 13. 3x2 – 5x + 2 14. 6x2 – 5x – 6 16. 6x2 – x – 2 17. 6x2 – 13x + 2 18. 6x2 – 13x + 2 19. 6x2 + 7x + 2 20. 6x2 – 7x + 2 21. 6x2 – 13x + 6 22. 6x2 + 13x + 6 23. 6x2 – 5x – 4 24. 6x2 – 13x + 8 25. 6x2 – 13x – 8 25. 4x2 – 9 26. 4x2 – 49 27. 25x2 – 4 28. 4x2 + 9 29. 25x2 + 9 B. Factor. Factor out the GCF, the “–”, and arrange the terms in order first. 30. – 6x2 – 5xy + 6y2 31. – 3x2 + 2x3– 2x 32. –6x3 – x2 + 2x 33. –15x3 – 25x2 – 10x 34. 12x3y2 –14x2y2 + 4xy2
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# How to Isolate a Variable
Isolating a variable is a matter of separating and simplifying.
### Instructions
• Step 1: Bring like terms together Bring like terms together. All equations have two sides. You want to perform the same operation on both sides of the equal sign. For the equation 4x –12 = -x + 8, 4x –12 is the left side and -x + 8 is the right side.
• Step 2: Add variable to both sides Add the variable to both sides of the equation. For the equation 4x –12 = -x + 8, add x to both sides, changing it to 4x –12 + x = -x + 8 + x.
• Step 3: Simplify Simplify the equation by adding the positive x and the negative x on the right side so they cancel each other out. The variable, x, is now on only one side of the equation: 5x – 12 = 8.
• Step 4: Add the constant to both sides Add the "constant", the number not attached to x, on the side of the equation with the variable, to both sides of the equation.
• Step 5: Simplify Simplify the equation. Combine like terms by adding eight and 12.
• TIP: Put every term that contains the variable on one side of the equation and every term that does not contain a variable on the other side.
• Step 6: Divide by the coefficient Divide both sides of the equation by the coefficient – the number attached to the variable.
• Step 7: Solve Solve for x. The solution is x = 4.
• Step 8: Test your solution Test your solution by plugging your answer into the original equation to be sure that it is correct.
• FACT: Ancient Egyptians developed the math to solve linear equations as far back as 1850 BCE.
### You Will Need
• A pen or pencil
• Paper
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# Eleven 11+ Questions: 3. Patterns and Sequences
15 June 2017
In part 3 of our 11+ Questions series we are going to look into patterns and sequences - a very typical exam question - using an example question taken from a sample paper sent out by one of the most top schools in London.
11+ examiners love patterns and sequences. If there is one question that is almost guaranteed to turn up in the exam, it’s the sequences one. Sometimes these will be pretty simple, such as asking for the next step in a pattern that goes 4, 8, 12, … At other times, though, the patterns can be fiendishly difficult, and we’re looking at one such example today.
Today’s question comes from recent sample paper from one of London’s top day schools, and it’s designed to push the best candidates to the limit. In answering it I’m also going to show you some tricks with these sequences to help you get the answer as quickly as possible.
So let’s have a look at question number 3!
## Question 3: A pattern is made using black and white squares. The first four patterns are shown below:
a) Complete the table below:
Ok, so we have a sequence here that is gradually expanding, and adding on three squares each time.
It’s fairly straightforward to answer this first question. We know how many white squares there are in pattern number 2 because we can count them – there is 1. Likewise there are 4 white squares in pattern number 3. In pattern number 4 the number of white squares will stay the same, so 4 would be the correct answer. Likewise, in pattern number 3 there are the same number of black squares as in pattern number 2 (we can count them), and there are going to be the same number in pattern number 4. So our finished table will look like this:
That was easy! Now let’s move on to the next question:
b) How many black squares will there be in the 8th pattern?
Right, so now things are going to get slightly more complicated. Now, there are different ways we can approach this question. The easiest approach though is simply to visualise the next couple of patterns.
We can see that with every two patterns we get either 3 more black squares or 3 more white squares:
We can then count on and see that in the eighth pattern there will be 12 black squares, because there were 6 in the Pattern 4 and 9 in Pattern 6.
Good! Again, not too difficult. Let’s now look at questions ‘c’, ‘d’ and ‘e’.
c) How many squares will there be in total in the 19th pattern?
d) Which pattern number has 31 white squares and 33 black squares?
e) Sam draws a pattern like the ones above and says, «there are 96 black squares and 95 white squares in my pattern.» How can you tell that he must have
miscounted?
In order to answer these three questions we need to start being a little more sophisticated in how we approach the problem. We need a method that allows us to calculate how many white squares or black squares there are going to be at a given point in the pattern.
We can approach this either by using algebra (for more advanced candidates) or by simply visualising ahead in the pattern. I’m going to demonstrate both approaches.
## Approach 1 – counting ahead
Let’s begin with question ‘c’ – ‘How many squares in total will there be in the 19th pattern?’
Let’s calculate how many white squares there are in pattern 19 and then how many black squares. We’ll add our answers together to get the correct answer. The first way to do this is to see that the white squares increase by 3 with every odd pattern number. So for PN 1 there is 1 white square, for PN 3 there are 4, PN 5 there are 7 white squares etc. Since 19 is an odd number we can simply visualise forward, adding as we go:
So we know that there are going to be 28 white squares in the 19th pattern. But how many black squares will there be?
Let’s look at the relationship between the Pattern Number and the black squares. We can see that the black squares increase by three with every odd number, thus:
Ok, so now we have a problem. Does the 19th Pattern Number have 27 or 30 black squares? The original table will tell us the answer:
We can see that on any particular even number PN (such as 4) the black squares are the same on both the even number and the following odd number. So to take 4 as an example, on PN 4 there are 6 black squares, and on PN 5 there are also 6.
This means that we can actually make a table like this:
There are therefore 27 black squares and 28 whites squares in the 19th pattern, giving us a total of 55 squares overall.
Now let’s look at question ‘d’.
d) Which pattern number has 31 white squares and 33 black squares?
To get the answer to this we can keep counting ahead in the same way. By using the black squares table above we can see that the PN with 33 black squares will be either 22 or 23. The question then is which PN has 31 white squares? Let’s look at our white squares table again, but this time add on one row:
And there’s our answer! The pattern number with 31 white squares and 33 black squares is PN 22.
Finally, question ‘e’:
e) ‘Sam draws a pattern like the ones above and says, «there are 96 black squares and 95 white squares in my pattern.» How can you tell that he must have
miscounted?’
To answer this final question we need to use algebra, and so before we attempt it I’m going to show you an algebraic method of doing questions ‘c’ and ‘d’.
## Approach 2 – using an equation to solve a sequence
Let’s begin this section with a look at a very basic pattern:
If we want to find the result for any pattern number, we need to find the equation that governs the sequence. We do this by following a few basic steps.
• Call the pattern number, the bit that goes 1,2,3,4 etc. ‘n’.
• Call the result ‘x’
• Begin by writing ‘x =’
The point is that something happens to n, the pattern number, to turn it into x.
So what is happening to n? Well, here we can see that with each step in the sequence x goes up by 3, and if there is a regular difference in steps we know that the three times table is at play. So we can get on the way to writing our equation by putting down:
x = 3n
Now, if this were our equation then the pattern would look like this:
It doesn’t, because the result (or x) is always one more than we expect it to be (I.e. instead of 3 we get 4, instead of 12 we get 13 etc.). So we can finish our equation by writing:
x = 3n + 1
Here we have an equation that explains how we get the result from the pattern number. What we have to do now is find such an equation for both our black squares and our white squares.
Let’s go back to our original sequence:
There are actually two sequences here. In one sequence something happens to the pattern number that changes the number of white squares. In the other sequence something happens to the pattern number that changes the number of black squares. So let’s try and find each individually.
• Finding the equation for the white squares
Let’s look at just the white squares.
We can see that the number of white squares goes up by three with each odd number. So instead of thinking in regular integers, 1,2,3,4 etc., we actually need to think in terms of odd numbers:
But hang on, how do we know what number is the ‘6th odd number’ for example? Well, we can work it out by looking at a sequence of odd numbers and working out how to get the sequence.
We know that the 2 times table is involved because the numbers go up by 2 each time. So we can write:
x = 2n
But if that were true by itself our sequence would look like this:
So instead we need to subtract by one each time: x = 2n – 1.
So, for example, let’s ask what is the 13th odd number? We use the equation, replacing ‘n’ with 13. The result is this:
x = 2 x 13 – 1. So the 13th odd number is 25.
So now let’s look at this table again…
…but change it to look like this:
Can we find the equation? We certainly can. The 3 times table is involved because there’s a difference of 3 each time, so our equation is this:
x = 3n – 2
So how do we use all of this information to find the number of white squares?
This is what we do:
• Identify what odd number we are on.
Let’s say we’re solving question ‘c’ and we want to know how many white squares there are in sequence 19. Firstly we need to know what odd number 19 is. We already have a sequence for working out odd numbers, so let’s plug it into the sequence.
19 = 2n – 1.
We solve the equation, and see that n = 10. Therefore 19 is the 10th odd number.
• So we now take 10 (for tenth) and plug it into our second equation, x = 3n – 2 .
x = 3 (10) – 2. Therefore x is 28 and there are 28 white squares in the 19th pattern number.
We also need to understand how many white squares there are on the even numbers, but if we look back at our original white squares pattern…
…we can see that there are the same number of white squares on the even number following the odd number. That means that there will be the same number of white squares on PN 2 as there are on 1, or the same number on PN 98 as there are on 97 etc.
So we have a method for finding the number of white squares. Great! Now let’s look at the black squares.
In this pattern, we can see that the number changes with every even number, and that the number remains the same on the following odd number.
So on pattern number 2 we have 3 black squares, and we also have 3 on pattern number 3.
We use a similar method as before to calculate the number of black squares.
• Work out which even number we are on.
Let’s think of question ‘c’ again and work out how many black squares there are in pattern number 19. We know that there will be the same number of black squares in pattern number 18 as 19, so let’s use 18.
Firstly, what even number is 18? We can calculate even numbers with this equation:
x = 2n.
So, for example, the first even number is x = 2 (1) which is 2. The fifth even number is x = 2 (5) which is 10.
Let’s put 18 into this sequence.
18 = 2n. Therefore 18 is the 9th even number.
• Now have a look at these two tables:
…but change it to look like this:
So now we need to find an equation for this sequence. We can see that the number of black squares goes up by 3 with each step, so we can write the equation like this:
x = 3n.
We know that the 18th step in the original sequence is the 9th even number. So if we put 9 into this sequence…
x = 3 (9)
We see that there are 27 black squares in the 18th pattern number, and therefore in the 19th pattern number too.
Now let’s look at question ‘d’.
1. Which pattern number has 31 white squares and 33 black squares?
Let’s begin with the white squares.
Let’s put 31 into our white squares equation:
x = 3n – 2 becomes
31 = 3n – 2
Therefore the pattern number with 31 white squares is the eleventh odd number. Now let’s put 11 into our odd number equation:
x = 2n – 1 becomes
x = 2 (11) – 1
Therefore the 11th odd number is 21.
So we know that PN 21 has 31 white squares, but so does PN 22. So our answer could be 21 or 22.
Let’s look at the black squares.
Firstly let’s put 33 into our black squares equation:
x = 3n becomes
33 = 3n. Therefore the 11th even number has 33 black squares.
To find the 11th even number we simply multiply 11 by 2, which is 22. Therefore the answer to question ‘d’ is 22.
Finally, question ‘e’.
‘Sam draws a pattern like the ones above and says, «there are 96 black squares
and 95 white squares in my pattern.» How can you tell that he must have
miscounted?’
So, all we need to do is put the numbers into our equations and prove that it doesn’t work. Let’s begin this time with the black squares.
Let’s put 96 into our even number equation:
96 = 3n, therefore our pattern number is even number 32.
Multiply that by 2 and we get 64 as the pattern number for Sam’s question.
PN 64 will have the same number of white squares as PN 63, so let’s put PN 63 into our white squares equations and see if we get 95 white squares.
Firstly we need to work out what odd number 63 is, so we put it into the equation:
63 = 2n – 1. This results in the odd number 32. We now put 32 into our white squares equation:
x = 3n – 2
Therefore x = 3 (32) – 2. The answer to this is 94, not 95, and therefore we know that Sam has miscounted!
A very, very difficult question, but not impossible. Keep an eye out on the UKSC blog for another post specifically about sequences and how to make equations for them. We’ll see you next time for our next 11+ question!
Educational Consultancy and School Placement Success stories Private tuition
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# Project #60862 - deicion 4
Assignment 1: Discussion—Population Growth
To study the growth of a population mathematically, we use the concept of exponential models. Generally speaking, if we want to predict the increase in the population at a certain period in time, we start by considering the current population and apply an assumed annual growth rate. For example, if the U.S. population in 2008 was 301 million and the annual growth rate was 0.9%, what would be the population in the year 2050? To solve this problem, we would use the following formula:
P(1 + r)n
In this formula, P represents the initial population we are considering, r represents the annual growth rate expressed as a decimal and n is the number of years of growth. In this example, P = 301,000,000, r = 0.9% = 0.009 (remember that you must divide by 100 to convert from a percentage to a decimal), and n = 42 (the year 2050 minus the year 2008). Plugging these into the formula, we find:
P(1 + r)n = 301,000,000(1 + 0.009)42
= 301,000,000(1.009)42
= 301,000,000(1.457)
= 438,557,000
Therefore, the U.S. population is predicted to be 438,557,000 in the year 2050.
Let’s consider the situation where we want to find out when the population will double. Let’s use this same example, but this time we want to find out when the doubling in population will occur assuming the same annual growth rate. We’ll set up the problem like the following:
Double P = P(1 + r)n
P will be 301 million, Double P will be 602 million, r = 0.009, and we will be looking for n.
Double P = P(1 + r)n
602,000,000 = 301,000,000(1 + 0.009)n
Now, we will divide both sides by 301,000,000. This will give us the following:
2 = (1.009)n
To solve for n, we need to invoke a special exponent property of logarithms. If we take the log of both sides of this equation, we can move exponent as shown below:
log 2 = log (1.009)n
log 2 = n log (1.009)
Now, divide both sides of the equation by log (1.009) to get:
n = log 2 / log (1.009)
Using the logarithm function of a calculator, this becomes:
n = log 2/log (1.009) = 77.4
Therefore, the U.S. population should double from 301 million to 602 million in 77.4 years assuming annual growth rate of 0.9 %.
Now it is your turn:
• Search the Internet and determine the most recent population of your home state. A good place to start is the U.S. Census Bureau (www.census.gov) which maintains all demographic information for the country. If possible, locate the annual growth rate for your state. If you can not locate this value, feel free to use the same value (0.9%) that we used in our example above.
• Determine the population of your state 10 years from now.
• Determine how long and in what year the population in your state may double assuming a steady annual growth rate.
• Look up the population of the city in which you live. If possible, find the annual percentage growth rate of your home city (use 0.9% if you can not locate this value).
• Determine the population of your city in 10 years.
• Determine how long until the population of your city doubles assuming a steady growth rate.
• Discuss factors that could possibly influence the growth rate of your city and state.
• Do you live in a city or state that is experiencing growth?
• Is it possible that you live in a city or state where the population is on the decline or hasn’t changed?
• How would you solve this problem if the case involved a steady decline in the population (say -0.9% annually)? Show an example.
• Think of other real world applications (besides monitoring and modeling populations) where exponential equations can be utilized.
By Saturday, March 14, 2015, deliver your assignment to the appropriate Discussion Area. Through Wednesday, March 18, 2015, review and comment on your peers’ responses.
Subject Mathematics Due By (Pacific Time) 03/04/2015 10:52 am
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## How to Perform a Divisbility Test / Divisibilty Calculations
Testing for Divisibility
Among numerous other factors, good scores on any standardized test such as GRE, SAT or GMAT depends on the student’s ability to perform the calculations faster.
Divisibility checking can be time consuming at times. So, any trick to save a minute or two can help a student significantly.
For illustration purposes I am going to consider numbers of form:
abcd
Where:
- d is in unit’s place,
- c is in ten’s place,
- b is in hundred’s place,
- a is in thousand’s place.
So, if I had a number 8976 then d = 6, c = 7, b = 9, a = 8.
Of course, the concept I talk about here using an example of 4 digit numbers can be applied to 2 digits, 3 digits, 5 digits, 10 digits, etc. numbers as well.
Divisibility by 2:
- A number is divisible by 2 ONLY IF the unit’s place of the number is divisible by 2.
- i.e. d = 0, 2, 4, 6, or 8.
- E.g. 2453 is not divisible by 2, but 2454 is divisible by 2.
Divisibility by 3:
- A number is divisible by 3 ONLY IF the sum of the digits of the number is divisible by 3.
- i.e. (a + b + c + d) has to be divisible by 3.
- E.g. 2453 is not divisible by 3 as 14 (= 2 + 4 + 5 + 3) is not divisible by 3.
- However, 2451 is divisible by 3 as 12 (= 2 + 4 + 5 + 1) is divisible by 3.
- In the same way, a number is divisible by 9, 27, 81, etc. if the sum of digits is divisible by 9, 27, 81, etc. RESPECTIVELY.
Divisibility by 4:
- A number is divisible by 4 ONLY IF the number formed by the units and ten’s place number is divisible by 4.
- i.e. the number cd has to be divisible by 4.
- E.g. 2453 is not divisible by 4 as 53 is not divisible by 4.
- However, 2452 is divisible by 4 as 52 is divisible by 4.
Divisibility by 5, 25:
- A number is divisible by 5 ONLY IF the unit’s place of the number is divisible by 5.
- i.e. d = 0 or 5.
- E.g. 2453 is not divisible by 5 but 2455 is divisible by 5.
- Similarly a number will be divisible by 25 if cd is divisible by 25.
Divisibility by 6, 12, etc.:
- A number is divisible by 6 ONLY IF it is divisible by 2 and 3 simultaneously. This works because 2 and 3 are factors of 6.
- Similarly a number will be divisible by e.g. 12 if it is divisible by 3 and 4 (12 = 3 x 4).
Divisibility by 10:
- A number is divisible by 10 ONLY IF d = 0.
Divisibility by 11:
- A number is divisible by 11 ONLY IF the difference of the sum of alternate numbers is divisible by 11.
- i.e. (a + c) – (b + d) is divisible by 11.
- E.g. 2452 is not divisible by 11 as 1 ([2 + 5] – [4 + 2]) is not divisible by 11.
- However, 2453 is divisible by 11 as 0 ([2 + 5] – [4 + 3]) is divisible by 11.
The consequence of the above tricks is that
- All the remainders would be zero (0) the moment the divisibility condition is satisfied.
- There is no need to waste time punching a calculator!
There are divisibility tricks for other numbers as well but they are not as popular as those I have described here.
• Subject : Math
• Topic : Basic Math Skills / Pre-Algebra
• Posted By :
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Question
# Find the center coordinates and radius of a circle for an equation written as: 3x2 + 3y2 - 6y = —12× + 24
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## Answer to a math question Find the center coordinates and radius of a circle for an equation written as: 3x2 + 3y2 - 6y = —12× + 24
Ali
4.4
Move the variable to the left-hand side and change its sign
$3{x}^{2}+3{y}^{2}-6y+12x=24$
Use the commutative property to reorder the terms
$3{x}^{2}+12x+3{y}^{2}-6y=24$
Factor out $3$ from the expression
$3\left${x}^{2}+4x \right$+3{y}^{2}-6y=24$
To complete the square, the same value needs to be added to both sides
$3\left${x}^{2}+4x+? \right$+3{y}^{2}-6y=24+?$
To complete the square ${x}^{2}+4x+4={\left$x+2 \right$}^{2}$ add $4$ to the expression
$3\left${x}^{2}+4x+4 \right$+3{y}^{2}-6y=24+?$
Since $3 \times 4$ was added to the left-hand side, also add $3 \times 4$ to the right-hand side
$3\left${x}^{2}+4x+4 \right$+3{y}^{2}-6y=24+3 \times 4$
Use ${a}^{2}+2ab+{b}^{2}={\left$a+b \right$}^{2}$ to factor the expression
$3{\left$x+2 \right$}^{2}+3{y}^{2}-6y=24+3 \times 4$
Simplify the expression
$3{\left$x+2 \right$}^{2}+3{y}^{2}-6y=36$
Factor out $3$ from the expression
$3{\left$x+2 \right$}^{2}+3\left${y}^{2}-2y \right$=36$
To complete the square, the same value needs to be added to both sides
$3{\left$x+2 \right$}^{2}+3\left${y}^{2}-2y+? \right$=36+?$
To complete the square ${y}^{2}-2y+1={\left$y-1 \right$}^{2}$ add $1$ to the expression
$3{\left$x+2 \right$}^{2}+3\left${y}^{2}-2y+1 \right$=36+?$
Since $3 \times 1$ was added to the left-hand side, also add $3 \times 1$ to the right-hand side
$3{\left$x+2 \right$}^{2}+3\left${y}^{2}-2y+1 \right$=36+3 \times 1$
Use ${a}^{2}-2ab+{b}^{2}={\left$a-b \right$}^{2}$ to factor the expression
$3{\left$x+2 \right$}^{2}+3{\left$y-1 \right$}^{2}=36+3 \times 1$
Simplify the expression
$3{\left$x+2 \right$}^{2}+3{\left$y-1 \right$}^{2}=39$
Divide both sides of the equation by $3$
${\left$x+2 \right$}^{2}+{\left$y-1 \right$}^{2}=13$
The equation can be written in the form ${\left$x-p \right$}^{2}+{\left$y-q \right$}^{2}={r}^{2}$, so it represents a circle with the radius $r=\sqrt{ 13 }$ and the center $\left$-2, 1\right$$
$\textnormal{Circle with the radius }r=\sqrt{ 13 }\textnormal{ and the center }\left$-2, 1\right$$
Frequently asked questions $FAQs$
What is the relationship between the coefficient 'a' and the width and direction of the parabola in the function 𝑦 = ax^2?
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What is the value of $\frac{5\pi}{3}$ angle in degrees?
+
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# 3.5: Derivatives of Trigonometric Functions
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Learning Objectives
• Find the derivatives of the sine and cosine function.
• Find the derivatives of the standard trigonometric functions.
• Calculate the higher-order derivatives of the sine and cosine.
One of the most important types of motion in physics is simple harmonic motion, which is associated with such systems as an object with mass oscillating on a spring. Simple harmonic motion can be described by using either sine or cosine functions. In this section we expand our knowledge of derivative formulas to include derivatives of these and other trigonometric functions. We begin with the derivatives of the sine and cosine functions and then use them to obtain formulas for the derivatives of the remaining four trigonometric functions. Being able to calculate the derivatives of the sine and cosine functions will enable us to find the velocity and acceleration of simple harmonic motion.
## Derivatives of the Sine and Cosine Functions
We begin our exploration of the derivative for the sine function by using the formula to make a reasonable guess at its derivative. Recall that for a function $$f(x),$$
$f′(x)=\lim_{h→0}\dfrac{f(x+h)−f(x)}{h}. \nonumber$
Consequently, for values of $$h$$ very close to $$0$$,
$f′(x)≈\dfrac{f(x+h)−f(x)}{h}. \nonumber$
We see that by using $$h=0.01$$,
$\dfrac{d}{dx}(\sin x)≈\dfrac{\sin (x+0.01)−\sin x}{0.01} \nonumber$
By setting
$D(x)=\dfrac{\sin (x+0.01)−\sin x}{0.01} \nonumber$
and using a graphing utility, we can get a graph of an approximation to the derivative of $$\sin x$$ (Figure $$\PageIndex{1}$$).
Upon inspection, the graph of $$D(x)$$ appears to be very close to the graph of the cosine function. Indeed, we will show that
$\dfrac{d}{dx}(\sin x)=\cos x. \nonumber$
If we were to follow the same steps to approximate the derivative of the cosine function, we would find that
$\dfrac{d}{dx}(\cos x)=−\sin x. \nonumber$
The Derivatives of $$\sin x$$ and $$\cos x$$
The derivative of the sine function is the cosine and the derivative of the cosine function is the negative sine.
$\dfrac{d}{dx}(\sin x)=\cos x$
$\dfrac{d}{dx}(\cos x)=−\sin x$
Proof
Because the proofs for $$\dfrac{d}{dx}(\sin x)=\cos x$$ and $$\dfrac{d}{dx}(\cos x)=−\sin x$$ use similar techniques, we provide only the proof for $$\dfrac{d}{dx}(\sin x)=\cos x$$. Before beginning, recall two important trigonometric limits:
$$\displaystyle \lim_{h→0}\dfrac{\sin h}{h}=1$$ and $$\displaystyle \lim_{h→0}\dfrac{\cos h−1}{h}=0$$.
The graphs of $$y=\dfrac{\sin h}{h}$$ and $$y=\dfrac{\cos h−1}{h}$$ are shown in Figure $$\PageIndex{2}$$.
We also recall the following trigonometric identity for the sine of the sum of two angles:
$\sin (x+h)=\sin x\cos h+\cos x\sin h. \nonumber$
Now that we have gathered all the necessary equations and identities, we proceed with the proof.
\begin{align*} \dfrac{d}{dx}(\sin x) &=\lim_{h→0}\dfrac{\sin(x+h)−\sin x}{h} & & \text{Apply the definition of the derivative.}\\[4pt] &=\lim_{h→0}\dfrac{\sin x\cos h+\cos x\sin h−\sin x}{h} & & \text{Use trig identity for the sine of the sum of two angles.}\\[4pt] &=\lim_{h→0}\left(\dfrac{\sin x\cos h−\sin x}{h}+\dfrac{\cos x\sin h}{h}\right) & & \text{Regroup.}\\[4pt] &=\lim_{h→0}\left(\sin x\left(\dfrac{\cos h−1}{h}\right)+(\cos x)\left(\dfrac{\sin h}{h}\right)\right) & & \text{Factor out }\sin x\text{ and }\cos x \\[4pt] &=(\sin x)\lim_{h→0}\left(\dfrac{\cos h−1}{h}\right)+(\cos x)\lim_{h→0}\left(\dfrac{\sin h}{h}\right) & & \text{Factor }\sin x\text{ and }\cos x \text{ out of limits.} \\[4pt] &=(\sin x)(0)+(\cos x)(1) & & \text{Apply trig limit formulas.}\\[4pt] &=\cos x & & \text{Simplify.} \end{align*}
Figure shows the relationship between the graph of $$f(x)=\sin x$$ and its derivative $$f′(x)=\cos x$$. Notice that at the points where $$f(x)=\sin x$$ has a horizontal tangent, its derivative $$f′(x)=\cos x$$ takes on the value zero. We also see that where f$$(x)=\sin x$$ is increasing, $$f′(x)=\cos x>0$$ and where $$f(x)=\sin x$$ is decreasing, $$f′(x)=\cos x<0.$$
Example $$\PageIndex{1}$$: Differentiating a Function Containing $$\sin x$$
Find the derivative of $$f(x)=5x^3\sin x$$.
Solution
Using the product rule, we have
\begin{align*} f'(x) &=\dfrac{d}{dx}(5x^3)⋅\sin x+\dfrac{d}{dx}(\sin x)⋅5x^3 \\[4pt] &=15x^2⋅\sin x+\cos x⋅5x^3. \end{align*}
After simplifying, we obtain
$f′(x)=15x^2\sin x+5x^3\cos x. \nonumber$
Exercise $$\PageIndex{1}$$
Find the derivative of $$f(x)=\sin x\cos x.$$
Hint
Don’t forget to use the product rule.
$f′(x)=\cos^2x−\sin^2x \nonumber$
Example $$\PageIndex{2}$$: Finding the Derivative of a Function Containing cos x
Find the derivative of $$g(x)=\dfrac{\cos x}{4x^2}$$.
Solution
By applying the quotient rule, we have
$g′(x)=\dfrac{(−\sin x)4x^2−8x(\cos x)}{(4x^2)^2}. \nonumber$
Simplifying, we obtain
$g′(x)=\dfrac{−4x^2\sin x−8x\cos x}{16x^4}=\dfrac{−x\sin x−2\cos x}{4x^3}. \nonumber$
Exercise $$\PageIndex{2}$$
Find the derivative of $$f(x)=\dfrac{x}{\cos x}$$.
Hint
Use the quotient rule.
$$f'(x) = \dfrac{\cos x+x\sin x}{\cos^2x}$$
Example $$\PageIndex{3}$$: An Application to Velocity
A particle moves along a coordinate axis in such a way that its position at time $$t$$ is given by $$s(t)=2\sin t−t$$ for $$0≤t≤2π.$$ At what times is the particle at rest?
Solution
To determine when the particle is at rest, set $$s′(t)=v(t)=0.$$ Begin by finding $$s′(t).$$ We obtain
$s′(t)=2 \cos t−1, \nonumber$
so we must solve
$2 \cos t−1=0\text{ for }0≤t≤2π. \nonumber$
The solutions to this equation are $$t=\dfrac{π}{3}$$ and $$t=\dfrac{5π}{3}$$. Thus the particle is at rest at times $$t=\dfrac{π}{3}$$ and $$t=\dfrac{5π}{3}$$.
Exercise $$\PageIndex{3}$$
A particle moves along a coordinate axis. Its position at time $$t$$ is given by $$s(t)=\sqrt{3}t+2\cos t$$ for $$0≤t≤2π.$$ At what times is the particle at rest?
Hint
Use the previous example as a guide.
$$t=\dfrac{π}{3},\quad t=\dfrac{2π}{3}$$
## Derivatives of Other Trigonometric Functions
Since the remaining four trigonometric functions may be expressed as quotients involving sine, cosine, or both, we can use the quotient rule to find formulas for their derivatives.
Example $$\PageIndex{4}$$: The Derivative of the Tangent Function
Find the derivative of $$f(x)=\tan x.$$
Solution
Start by expressing $$\tan x$$ as the quotient of $$\sin x$$ and $$\cos x$$:
$$f(x)=\tan x =\dfrac{\sin x}{\cos x}$$.
Now apply the quotient rule to obtain
$$f′(x)=\dfrac{\cos x\cos x−(−\sin x)\sin x}{(\cos x)^2}$$.
Simplifying, we obtain
$f′(x)=\dfrac{\cos^2x+\sin^2 x}{\cos^2x}. \nonumber$
Recognizing that $$\cos^2x+\sin^2x=1,$$ by the Pythagorean theorem, we now have
$f′(x)=\dfrac{1}{\cos^2x} \nonumber$
Finally, use the identity $$\sec x=\dfrac{1}{\cos x}$$ to obtain
$$f′(x)=\text{sec}^2 x$$.
Exercise $$\PageIndex{4}$$
Find the derivative of $$f(x)=\cot x .$$
Hint
Rewrite $$\cot x$$ as $$\dfrac{\cos x}{\sin x}$$ and use the quotient rule.
$$f′(x)=−\csc^2 x$$
The derivatives of the remaining trigonometric functions may be obtained by using similar techniques. We provide these formulas in the following theorem.
Derivatives of $$\tan x$$, $$\cot x$$, $$\sec x$$, and $$\csc x$$
The derivatives of the remaining trigonometric functions are as follows:
\begin{align} \dfrac{d}{dx}(\tan x )&=\sec^2x\\[4pt] \dfrac{d}{dx}(\cot x )&=−\csc^2x\\[4pt] \dfrac{d}{dx}(\sec x)&=\sec x \tan x\\[4pt] \dfrac{d}{dx}(\csc x)&=−\csc x \cot x. \end{align}
Example $$\PageIndex{5}$$: Finding the Equation of a Tangent Line
Find the equation of a line tangent to the graph of $$f(x)=\cot x$$ at $$x=\frac{π}{4}$$.
Solution
To find the equation of the tangent line, we need a point and a slope at that point. To find the point, compute
$$f\left(\frac{π}{4}\right)=\cot\frac{π}{4}=1$$.
Thus the tangent line passes through the point $$\left(\frac{π}{4},1\right)$$. Next, find the slope by finding the derivative of $$f(x)=\cot x$$ and evaluating it at $$\frac{π}{4}$$:
$$f′(x)=−\csc^2 x$$ and $$f′\left(\frac{π}{4}\right)=−\csc^2\left(\frac{π}{4}\right)=−2$$.
Using the point-slope equation of the line, we obtain
$$y−1=−2\left(x−\frac{π}{4}\right)$$
or equivalently,
$$y=−2x+1+\frac{π}{2}$$.
Example $$\PageIndex{6}$$: Finding the Derivative of Trigonometric Functions
Find the derivative of $$f(x)=\csc x+x\tan x .$$
Solution
To find this derivative, we must use both the sum rule and the product rule. Using the sum rule, we find
$$f′(x)=\dfrac{d}{dx}(\csc x)+\dfrac{d}{dx}(x\tan x )$$.
In the first term, $$\dfrac{d}{dx}(\csc x)=−\csc x\cot x ,$$ and by applying the product rule to the second term we obtain
$$\dfrac{d}{dx}(x\tan x )=(1)(\tan x )+(\sec^2 x)(x)$$.
Therefore, we have
$$f′(x)=−\csc x\cot x +\tan x +x\sec^2 x$$.
Exercise $$\PageIndex{5}$$
Find the derivative of $$f(x)=2\tan x −3\cot x .$$
Hint
Use the rule for differentiating a constant multiple and the rule for differentiating a difference of two functions.$$f′(x)=2\sec^2 x+3\csc^2 x$$
$$f′(x)=2\sec^2 x+3\csc^2 x$$
Exercise $$\PageIndex{6}$$
Find the slope of the line tangent to the graph of $$f(x)=\tan x$$ at $$x=\dfrac{π}{6}$$.
Hint
Evaluate the derivative at $$x=\dfrac{π}{6}$$.
$$\dfrac{4}{3}$$
## Higher-Order Derivatives
The higher-order derivatives of $$\sin x$$ and $$\cos x$$ follow a repeating pattern. By following the pattern, we can find any higher-order derivative of $$\sin x$$ and $$\cos x.$$
Example $$\PageIndex{7}$$: Finding Higher-Order Derivatives of $$y=\sin x$$
Find the first four derivatives of $$y=\sin x.$$
Solution
Each step in the chain is straightforward:
\begin{align*} y&=\sin x \\[4pt] \dfrac{dy}{dx}&=\cos x \\[4pt] \dfrac{d^2y}{dx^2}&=−\sin x \\[4pt] \dfrac{d^3y}{dx^3}&=−\cos x \\[4pt] \dfrac{d^4y}{dx^4}&=\sin x \end{align*}
Analysis
Once we recognize the pattern of derivatives, we can find any higher-order derivative by determining the step in the pattern to which it corresponds. For example, every fourth derivative of $$\sin x$$ equals $$\sin x$$, so
$\dfrac{d^4}{dx^4}(\sin x)=\dfrac{d^8}{dx^8}(\sin x)=\dfrac{d^{12}}{dx^{12}}(\sin x)=…=\dfrac{d^{4n}}{dx^{4n}}(\sin x)=\sin x \nonumber$
$\dfrac{d^5}{dx^5}(\sin x)=\dfrac{d^9}{dx^9}(\sin x)=\dfrac{d^{13}}{dx^{13}}(\sin x)=…=\dfrac{d^{4n+1}}{dx^{4n+1}}(\sin x)=\cos x. \nonumber$
Exercise $$\PageIndex{7}$$
For $$y=\cos x$$, find $$\dfrac{d^4y}{dx^4}$$.
Hint
See the previous example.
$$\cos x$$
Example $$\PageIndex{8}$$: Using the Pattern for Higher-Order Derivatives of $$y=\sin x$$
Find $$\dfrac{d^{74}}{dx^{74}}(\sin x)$$.
Solution
We can see right away that for the 74th derivative of $$\sin x$$, $$74=4(18)+2$$, so
$\dfrac{d^{74}}{dx^{74}}(\sin x)=\dfrac{d^{72+2}}{dx^{72+2}}(\sin x)=\dfrac{d^2}{dx^2}(\sin x)=−\sin x. \nonumber$
Exercise $$\PageIndex{8}$$
For $$y=\sin x$$, find $$\dfrac{d^{59}}{dx^{59}}(\sin x).$$
Hint
$$\dfrac{d^{59}}{dx^{59}}(\sin x)=\dfrac{d^{4⋅14+3}}{dx^{4⋅14+3}}(\sin x)$$
$$−\cos x$$
Example $$\PageIndex{9}$$: An Application to Acceleration
A particle moves along a coordinate axis in such a way that its position at time $$t$$ is given by $$s(t)=2−\sin t$$. Find $$v(π/4)$$ and $$a(π/4)$$. Compare these values and decide whether the particle is speeding up or slowing down.
Solution
First find $$v(t)=s′(t)$$
$v(t)=s′(t)=−\cos t . \nonumber$
Thus,
$$v\left(\frac{π}{4}\right)=−\dfrac{1}{\sqrt{2}}=-\dfrac{\sqrt{2}}{2}$$.
Next, find $$a(t)=v′(t)$$. Thus, $$a(t)=v′(t)=\sin t$$ and we have
$$a\left(\frac{π}{4}\right)=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}$$.
Since $$v\left(\frac{π}{4}\right)=−\dfrac{\sqrt{2}}{2}<0$$ and $$a\left(\frac{π}{4}\right)=\dfrac{\sqrt{2}}{2}>0$$, we see that velocity and acceleration are acting in opposite directions; that is, the object is being accelerated in the direction opposite to the direction in which it is traveling. Consequently, the particle is slowing down.
Exercise $$\PageIndex{9}$$
A block attached to a spring is moving vertically. Its position at time t is given by $$s(t)=2\sin t$$. Find $$v\left(\frac{5π}{6}\right)$$ and $$a\left(\frac{5π}{6}\right)$$. Compare these values and decide whether the block is speeding up or slowing down.
Hint
Use Example $$\PageIndex{9}$$ as a guide.
$$v\left(\frac{5π}{6}\right)=−\sqrt{3}<0$$ and $$a\left(\frac{5π}{6}\right)=−1<0$$. The block is speeding up.
## Key Concepts
• We can find the derivatives of $$\sin x$$ and $$\cos x$$ by using the definition of derivative and the limit formulas found earlier. The results are
$$\dfrac{d}{dx}\big(\sin x\big)=\cos x\quad\text{and}\quad\dfrac{d}{dx}\big(\cos x\big)=−\sin x$$.
• With these two formulas, we can determine the derivatives of all six basic trigonometric functions.
## Key Equations
• Derivative of sine function
$$\dfrac{d}{dx}(\sin x)=\cos x$$
• Derivative of cosine function
$$\dfrac{d}{dx}(\cos x)=−\sin x$$
• Derivative of tangent function
$$\dfrac{d}{dx}(\tan x )=\sec^2x$$
• Derivative of cotangent function
$$\dfrac{d}{dx}(\cot x )=−\csc^2x$$
• Derivative of secant function
$$\dfrac{d}{dx}(\sec x)=\sec x\tan x$$
• Derivative of cosecant function
$$\dfrac{d}{dx}(\csc x)=−\csc x\cot x$$
3.5: Derivatives of Trigonometric Functions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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# Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.2
Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Ex 4.2 Questions and Answers.
## Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.2
Question 1.
Show that lines represented by 3x2 – 4xy – 3y2 = 0 are perpendicular to each other.
Solution:
Comparing the equation 3x2 – 4 xy – 3y2 = 0 with ax2 + 2hxy + by2 = 0, we get, a = 3, 2h = -4, b = -3 Since a + b = 3 + (-3) = 0, the lines represented by 3x2 – 4xy – 3y2 = 0 are perpendicular to each other.
Question 2.
Show that lines represented by x2 + 6xy + gy2= 0 are coincident.
Question is modified.
Show that lines represented by x2 + 6xy + 9y2= 0 are coincident.
Solution:
Comparing the equation x2 + 6xy + 9y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = 6, i.e. h = 3 and b = 9
Since h2 – ab = (3)2 – 1(9)
= 9 – 9 = 0, .
the lines represented by x2 + 6xy + 9y2 = 0 are coincident.
Question 3.
Find the value of k if lines represented by kx2 + 4xy – 4y2 = 0 are perpendicular to each other.
Solution:
Comparing the equation kx2 + 4xy – 4y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = k, 2h = 4, b = -4
Since lines represented by kx2 + 4xy – 4y2 = 0 are perpendicular to each other,
a + b = 0
∴ k – 4 = 0 ∴ k = 4.
Question 4.
Find the measure of the acute angle between the lines represented by:
(i) 3x2 – 4$$\sqrt {3}$$xy + 3y2 = 0
Solution:
Comparing the equation 3x2 – 4$$\sqrt {3}$$xy + 3y2 = 0 with
ax2 + 2hxy + by2 = 0, we get,
a = 3, 2h = -4$$\sqrt {3}$$, i.e. h = -24$$\sqrt {3}$$ and b = 3
Let θ be the acute angle between the lines.
∴ θ = 30°.
(ii) 4x2 + 5xy + y2 = 0
Solution:
Comparing the equation 4x2 + 5xy + y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 4, 2h = 5, i.e. h = $$\frac{5}{2}$$ and b = 1.
Let θ be the acute angle between the lines.
(iii) 2x2 + 7xy + 3y2 = 0
Solution:
Comparing the equation
2x2 + 7xy + 3y2 = 0 with
ax2 + 2hxy + by2 = 0, we get,
a = 2, 2h = 7 i.e. h = $$\frac{7}{2}$$ and b = 3
Let θ be the acute angle between the lines.
tanθ = 1
∴ θ = tan 1 = 45°
∴ θ = 45°
(iv) (a2 – 3b2)x2 + 8abxy + (b2 – 3a2)y2 = 0
Solution:
Comparing the equation
(a2 – 3b2)x2 + 8abxy + (b2 – 3a2)y2 = 0, with
Ax2 + 2Hxy + By2 = 0, we have,
A = a2 – 3b2, H = 4ab, B = b2 – 3a2.
∴ H2 – AB = 16a2b2 – (a2 – 3b2)(b2 – 3a2)
= 16a2b2 + (a2 – 3b2)(3a2 – b2)
= 16a2b2 + 3a4 – 10a2b2 + 3b4
= 3a4 + 6a2b2 + 3b4
= 3(a4 + 2a2b2 + b4)
= 3 (a2 + b2)2
∴ $$\sqrt{H^{2}-A B}$$ = $$\sqrt {3}$$ (a2 + b2)
Also, A + B = (a2 – 3b2) + (b2 – 3a2)
= -2 (a2 + b2)
If θ is the acute angle between the lines, then
tan θ = $$\left|\frac{2 \sqrt{H^{2}-A B}}{A+B}\right|=\left|\frac{2 \sqrt{3}\left(a^{2}+b^{2}\right)}{-2\left(a^{2}+b^{2}\right)}\right|$$
= $$\sqrt {3}$$ = tan 60°
∴ θ = 60°
Question 5.
Find the combined equation of lines passing through the origin each of which making an angle of 30° with the line 3x + 2y – 11 = 0
Solution:
The slope of the line 3x + 2y – 11 = 0 is m1 = $$-\frac{3}{2}$$ .
Let m be the slope of one of the lines making an angle of 30° with the line 3x + 2y – 11 = 0.
The angle between the lines having slopes m and m1 is 30°.
On squaring both sides, we get,
$$\frac{1}{3}=\frac{(2 m+3)^{2}}{(2-3 m)^{2}}$$
∴ (2 – 3m)2 = 3 (2m + 3)2
∴ 4 – 12m + 9m2 = 3(4m2 + 12m + 9)
∴ 4 – 12m + 9m2 = 12m2 + 36m + 27
3m2 + 48m + 23 = 0
This is the auxiliary equation of the two lines and their joint equation is obtained by putting m = $$\frac{y}{x}$$.
∴ the combined equation of the two lines is
3$$\left(\frac{y}{x}\right)^{2}$$ + 48$$\left(\frac{y}{x}\right)$$ + 23 = 0
∴ $$\frac{3 y^{2}}{x^{2}}+\frac{48 y}{x}$$ + 23 = 0
∴ 3y2 + 48xy + 23x2 = 0
∴ 23x2 + 48xy + 3y2 = 0.
Question 6.
If the angle between lines represented by ax2 + 2hxy + by2 = 0 is equal to the angle between lines represented by 2x2 – 5xy + 3y2 = 0 then show that 100(h2 – ab) = (a + b)2.
Solution:
The acute angle θ between the lines ax2 + 2hxy + by2 = 0 is given by
tan θ = $$\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|$$ ..(1)
Comparing the equation 2x2 – 5xy + 3y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 2, 2h= -5, i.e. h = $$-\frac{5}{2}$$ and b = 3
Let ∝ be the acute angle between the lines 2x2 – 5xy + 3y2 = 0.
This is the required condition.
Question 7.
Find the combined equation of lines passing through the origin and each of which making angle 60° with the Y- axis.
Solution:
Let OA and OB be the lines through the origin making an angle of 60° with the Y-axis.
Then OA and OB make an angle of 30° and 150° with the positive direction of X-axis.
∴ slope of OA = tan 30° = $$\frac{1}{\sqrt{3}}$$
∴ equation of the line OA is
y = $$\frac{1}{\sqrt{3}}$$ = x, i.e. x – $$\sqrt {3}$$y = 0
Slope of OB = tan 150° = tan (180° – 30°)
= tan 30° = $$-\frac{1}{\sqrt{3}}$$
∴ equation of the line OB is
y = $$-\frac{1}{\sqrt{3}}$$x, i.e. x + $$\sqrt {3}$$ y = 0
∴ required combined equation is
(x – $$\sqrt {3}$$y)(x + $$\sqrt {3}$$y) = 0
i.e. x2 – 3y2 = 0.
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## KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3.
## Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3
Question 1.
FInd the sum of the following APs:
i) 2, 7, 12, ……… to 10 terms
ii) -37, -33, -29,………. to 12 terms.
iii) 0.8, 1.7, 2.8 to 100 terms.
iv) $$\frac{1}{15}, \frac{1}{12}, \frac{1}{10} \ldots, \text { to } 11$$ terms
Solution:
i) Sum of 2, 7, 12, ………. 10?
a = 2, d = 7 – 2 = 5, n = 10
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$s_{10}=\frac{10}{2}[2 \times 2+(10-1) 5]$$
= 5[4 + 9 × 5]
= 5[4 + 45]
= 5 × 49
∴ S10 = 245
ii) -37, -33, -29, to 12……. terms.
Solution:
a = -37, d = -33 – (-37) = -33 + 37
n = 12, d = 4
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{12}=\frac{12}{2}[2 \times -37+(12-1) 4]$$
= 6[-74 + 11 × 4]
= 6[-74 + 44]
= 6 × -30
∴ S12 = -180.
iii) 0.6, 1.7, 2.8,……… , to 100 terms.
a = 0.6, d = 1.7 – 0.6 = 1.1
n = 100, S100 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{100}=\frac{100}{2}[2 \times 0.6+(100-1)(1.1)]$$
= 50[1.2 + 99(1.1)]
= 50[1.2 + 108.9]
= 50 × 110.1
∴ S100 = 5505
iv) $$\frac{1}{15}, \frac{1}{12}, \frac{1}{10} \ldots, \text { to } 11$$ terms
Solution:
The LCM of 12 and 18 is 36.
Question 2.
Find the sums given below:
i) $$7+10 \frac{1}{2}+14+\ldots \ldots+84$$
ii) 34 + 32 + 30 + ……….+ 10
iii) -5 + (-8) + (-11) + ……….. +(-230)
Solution:
ii) 34 + 32 + 30 + ………. + 10
Solution:
a = 34, d = 32 – 34 =-2, an = 10,
Sn =?
a + (n – 1)d = an
34 + (n – 1)(-2) = 10
34 – 2n + 2 = 10
-2n + 36 = 10
-2n = 10 – 36
-2n = -26
2n = 26
$$n=\frac{26}{2}$$
∴ n = 13
$$S_{n}=\frac{n}{2}[a+l]$$
$$\mathrm{S}_{23}=\frac{13}{2}[34+10]$$
$$=\frac{13}{2} \times 44$$
= 13 × 22
∴ S23 = 286.
iii) -5 + (-8) + (-11) +……… + (-230)
Solution:
a = -5, d = -8 – (-5) = -8 + 5
a, = -230.
Sn =? d = -3
a+(n – 1)d = an
-5 + (n – 1)(-3) = -230
-5 – 3n + 3 = -230
-3n – 2 = -230
-3n = -230 +2
-3n = -228
3n = 228
∴ n = 228/3
∴ n = 76.
$$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right]$$
$$S_{ 76 }=\frac { 76 }{ 2 } \left[ -5-230 \right]$$
$$=\frac{76}{2} \times -235$$
= 38 × -235
∴ S76= -8930.
Question 3.
In an AP:
i) given a = 5. d = 3. an = 50, find ‘n’ and Sn.
ii) given a = 7, a13 = 35, find d’ and S13.
iii) given a12= 37. d = 3. find ‘a’ and S12.
iv) given a3 = 15, S10 = 125. find ‘d’ and a10.
v) given d = 5. S9 = 72, fInd ‘a’ and a9.
vi) given a = 2. d = 8, Sn = 90, fInd ‘n and a11.
vii) given a = 8, an = 62. Sn = 210, find ‘n’ and ‘d’.
viii) given an = 4, d = 2, Sn = -14. find ‘n’ and a.
ix) given a = 3, n = 8, S = 192, find ‘d’.
x) given L = 28, S = 144 and there are total 9 terms. Find ‘a’.
Solution:
i) a = 5, d = 3, an = 50. n=?, Sn =?
a + (n – 1)d = an
5 + (n – 1) 3 = 50
5 + 3n – 3 = 50
3n + 2 = 50
3n = +50 – 2
3n = 48
$$n=\frac{48}{3}$$
∴ n = 16
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$S_{16}=\frac{16}{2}[5+50]$$
$$=\frac{16}{2}[55]$$
= 8 × 55
∴ S16= 440.
ii) a = 7, a13 = 35. d =?. S13 =?
Solution:
a+(n – 1) d = an
7 + (13 – 1) d = 35
7 + 12d = 35
12d = 35 – 7
12d = 28
$$d=\frac{28}{12}$$
$$d=\frac{7}{3}$$
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$S_{n}=\frac{7}{3} \times \frac{1}{2}[7+35]$$
$$\mathrm{s}_{\mathrm{n}}=\frac{7}{6} \mathrm{I} 42 \mathrm{l}$$
= 7 × 7
∴ Sn = 49
iii) a12= 37, d = 3, a =? S12 =?
a + (n – 1)d = an
a + (12 -1) 3 = 37
a + 11 × 3 = 37
a + 33 = 37
∴ a = 37 – 33
∴ a = 4
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$\mathrm{S}_{12}=\frac{12}{2}[4+37]$$
= 6[4 + 37]
∴ S2 = 246
iv) a3 = 15, S10 = 125. d =?, a10=?
Solution:
a3 = a + 2d = 15
∴ a= 15 – 2d
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$s_{10}=\frac{10}{2}[2(15-2 d) \div(10-1) d]=125$$
= 5[30 – 4d + 9d] = 125
= 5[30 + 5d] = 125
150 + 25d = 125
25d = 125 + 150
25d = -25
$$\mathrm{d}=\frac{-25}{25}$$
∴ d = -1.
Substittuting the value of ‘d’
a = 15 – 2d
= 15 – 2(-1)
= 15 + 2
∴ a = 17.
∴ a10 = a + 9d
= 17 + 9(-1)
= 17 – 9
∴ a10 = 8
v) d = 5, S9 = 72. a =? a9 =?
Solution:
$$\mathrm{s}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$$
$$s_{9}=\frac{9}{2}[2 \times a+(9-1) 5]=72$$
$$\frac{9}{2}[2 a+8 \times 5]=72$$
$$\frac{9}{2}[2 a+40]=72$$
$$2 a+40=72 \times \frac{2}{9}$$
2a + 40 = 16
2a = 16 – 40
2a = -24
$$a=\frac{-24}{2}$$
∴ a = -12
a9 = a + (n – 1) d
= -12 + (9 – 1) (5)
= -12 + 8 × 5
= -12 + 40
∴ a9 = 28
vi) a = 2, d = 8 Sn = 90, n =?, a3 =?
Solution:
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$=\frac{n}{2}[2 \times 2+(n-1) 8]=90$$
$$\frac{n}{2}[4+8 n-8]=90$$
n (8n – 4) = 90 × 2/1
8n2 – 4n = 180
8n2 – 4n = 1800
2n2 – n – 45 = 0
2n2– 10n + 9n – 45 = 0
2n(n – 5) + 9(n – 5) = 0
(n – 5)(2n + 9) = 0
If n – 5 = 0 then n = 5
∴ an = a + (n – 1) d
a5= 2 + (5 – 1) 8
= 2 + 4 × 8
= 2 + 32
∴ a5 = 34
vii) a = 8, an = 62, Sn = 210, n =? d =?
Solution:
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]=210$$
$$\frac{n}{2}[8+62]=210$$
$$\frac{n}{2} \times 70=21$$
$$n=210 \times \frac{2}{70}$$
∴ n = 6
an = 62
a6= a + 5d = 62
8 + 5d = 62
5d = 62 – 8
5d = 54
$$d=\frac{54}{5}$$
viii) a = 4, d = 2, Sn = -14, n =?, a =?
Solution:
an = a + (n – 1) d = 4
a + (n – 1) 2 = 4
a + 2n – 2 = 4
a + 2n = 4 + 2
a = 2n = 6 …………. (i)
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]=210$$
$$\frac{n}{2}[-2 n+6+4]=-14$$
$$\frac{n}{2}[-2 n+10]=-14$$
-2n2 + 10n = -28
-2n2 + 10n + 28 = 0
2n2 – 10n – 28 = 0 – 14
2n2 – 5n – 14 = 0
n2 – 7n + 2n – 14 = 0
n(n – 7) + 2 (n – 7) = 0
(n – 7) (n + 2) = 0
If n – 7 = 0 then. n = 7
a = -2n + 6
= -2 × 7 + 6
= -14 + 6
∴ a = -8
∴ n = 7, a = -8.
ix) a = 3, n = 8, S = 192, d =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{8}=\frac{8}{2}[2 \times 3+(8-1) d]=192$$
$$\frac { 8 }{ 2 } [6+7\times d]=192$$
4 (6 + 7d) = 192
$$6+7 d=\frac{192}{4}$$
6 + 7d = 48
7d = 48 – 6
7d = 42
$$d=\frac{42}{7}$$
∴ d = 6
x) l = an = 28, S = 144, n = 9, a =?
Solution:
$$s_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$s_{9}=\frac{9}{2}[a+28]=144$$
$$\frac{9}{2}[a+28]=144$$
$$a+28=144 \times \frac{2}{9}$$
a + 28 = 32
a = 32 – 28
∴ a = 4.
Question 4.
How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?
Solution:
9 + 17 + 25 + ……. + an = 636
a = 9, d = 17 – 9 = 8, Sn= 636, n =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{n}=\frac{n}{2}[2 \times 9+(n-1) 8]=636$$
$$\frac{n}{2}[18+8 n-8]=636$$
$$\frac{n}{2}[8 n+10]=636$$
n(8n + 10) = 636 × 2
8n2 + 10n = 1272
8n2+ 10n – 1272 = 0
4n2 + 5n – 636 = o
4n2+ 5n – 48n – 636 = 0
n(4n + 5) – 12(4n + 5) = 0
(4n + 5) (n – 12) = 0
1f n – 12 = 0 then, n = 12
∴ n = 12.
Question 5.
The first term of an AP is 5. the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
a = 5, an = 45, Sn= 400, n =?, d =?
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$=\frac{n}{2}[5+45]=400$$
$$\frac{n}{2} \times 50=400$$
n × 50 = 800
$$n=800 \times \frac{1}{50}$$
∴ n = 16
an = a + (n – 1) d
a16= 5 + (16 – 1)d = 45
5 + 15d = 45
15d = 45 – 5
15d = 40
$$d=\frac{40}{15}=\frac{8}{3}$$
$$\mathrm{n}=16, \mathrm{d}=\frac{8}{3}$$
Question 6.
The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:
a = 17, an= 350, d = 9, n =?, Sn =?
an = a + (n – 1) d
17 + (n – 1)9 = 350
17 + 9n – 9 = 350
9n + 8 = 350
9n = 350 – 8
9n = 342
$$\quad n=\frac{342}{9}$$
∴ n = 38
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$S_{38}=\frac{38}{2}[17+350]$$
= 19 × 367
∴ S38 = 6973
∴ n = 38, S38 = 6973
Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
d = 7, a22 = 149, S22 =?
a = a + (n – 1)d
a + (22 – 1)7 = 149
a + 21 × 7 = 149
a + 147 = 149
=149 – 147
∴ a = 2
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$S_{22}=\frac{22}{2}[2+149]$$
= 11[151]
∴ S22 = 1661
Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
If a2 = 14, a3 = 18, then S11 =?
a2 = 14
a + d = 14 …………. (i)
a3 = 18
a + 2d = 18 …………… (ii)
from equation (1) – equatIon (11),
d =4.
Substituting the value of ‘d’ In eqn. (i)
a + d = 14
a + 4 = 14
a = 14 – 4
∴ a = 10
a = 10, d = 4
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{51}=\frac{51}{2}[2 \times 10+(51-1) 4]$$
$$=\frac{51}{2}[20+50 \times 4]$$
$$=\frac{51}{2}[20+200]$$
$$=\frac{51}{2} \times 220$$
= 51 × 110
∴ S51 = 5610
Question 9.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first ‘n terms.
Solution:
S7 = 49, S17 = 289, S11 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{7}=\frac{7}{2}[2 a+(7-1) d]$$
$$=\frac{7}{2}[2 a+6 d]=49$$
[latex2+6 d=49 \times \frac{2}{7}[/latex]
2a + 6d = 14 …………… (i)
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{17}=\frac{17}{2}[2 a+(n-1) d]$$
$$=\frac{17}{2}[2 \mathrm{a}+16 \mathrm{d}]$$
$$\frac{17}{2}[2 a+16 d]=289$$
$$2 a+16 d=289 \times \frac{2}{17}$$
∴ 2a + 16d = 34 …………. (ii)
10d = 20
∴ d=2
Substituting the value of d in equation (i), we have
2a + 6d = 14
2a + 6 × 2 = 14
2a + 12 = 14
2a = 14 – 12
2a = 2
$$a=\frac{2}{2}=1$$
a = 1, d = 2, Sn =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{n}=\frac{n}{2}[2 \times 1+(n-1) 2]$$
$$=\frac{\mathrm{n}}{2}[2+2 \mathrm{n}-2]$$
$$=\quad \frac{\mathrm{n}}{2}[2 \mathrm{n}]$$
∴ Sn = n2
Question 10.
Show that a1, a2, a3, … a11. … form an AP where a11 is defIned as below:
i) a = 3 + 4n
ii) a = 9 – 5n
Solution:
i) an= 3 + 4n
a1 = 3 + 4(1)
= 3 + 4
∴ an = 7
∴ a = 7
a1 = 3 + 4n
a2 = 3 + 4 × 2
= 3 + 8
∴ a2 = 11
an = 3 + 4n
a3 = 3 + 4 × 3
= 3 + 12
a3 = 15
∴ a1, a2, a3, ………….
7, 11, 15, ……….
d = a2 – a1 = 11 – 7 = 4
d = a3 – a2= 15 – 11 = 4
Here, the value of ‘d’ is constant.
∴ an = 3 + 4n forms an Arithmetic Progression.
ii) an = 9 – 5n
a1= 9 – 5 × 1
= 9 – 5
∴ a1 = 4
a1 = 9 – 5n
a1 = 9 – 5 × 2
= 9 – 10
∴ a2 = -1
an= 9 – 5n
a3 = 9 – 5 × 3
= 9 – 15
∴ a3 = -6
a1, a2, a3, …………… an
4, -1, -6, …………
d = a2 – a1 = -1 – 4 = -5
d=a3 – a2 = -6 – (-1) = -5
Here, the value of ‘d’ is constant.
∴ an = 9 – 5n form an Arithmetic Progression.
Question 11.
lf the sum of the first n terms of an AP is 4n – n2, what is the first term (that is SI)? What is the sum of first two termš? What is the second term? Similarly. find the 3rdrd the 10th and the nth terms.
Solution:
If Sn = 4n – n2, then
i) S1 = a =?
ii) S2 =?
iii) a2 =?
iv) a3 =?
v) a10 =?
vi) an =?
(i) Sn = 4n – n2
S1 = 4(1) – 12
= 4 – 1
S1 = 3
∴ S1 = a = 3.
(ii) S2 = 4n – n2
S2 = 4(2) – 22
= 8 – 4
S2 = 4
∴ S2 = 4
(iii) We have S2 = a1 + a2 = 4
= 3 + a2 = 4
∴ a2 = 4 – 3
∴a2 = 1
(iv) Sn = 4n – n2
S3= 4(3) – 32
= 12 – 9
S3 = 3
a1 + a2 + a3 = 3
3 + 1 + a3 = 3
4 + a3= 3
∴ a3 = 3 – 4
∴ a 3 = -1
(v) d = a3 – a3 = -1 -1 = -2
a10= a + 9d
= 3 + 9(-2)
= 3 – 18
a10= -15
∴ a10 = -15
(vi) an = a + (n – 1)d
= 3 + (n – 1)(-2)
= 3 + 2n + 2
∴ an = 5 – 2n
∴ an = 5 – 2n
Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
6 + 12 + 18 + 24 + 40 term
Here a = 6, d = 2 – a1 = 12 – 6 = 6
n = 40, S40 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{40}=\frac{40}{2}[2 \times 6+(40-1) 6]$$
= 20[12 + 39 × 6]
= 20[12 + 234]
= 20 × 246
∴ S40 = 4920.
Question 13.
Find the sum of first 15 multiples of 8,
Solution:
Sum of the first 15 multiples of 8 ?
8 + 16 + 24+ ……… 15 terms.
Here, a = 8, d = a2 – a1= 16 – 8 = 8
n = 15, S15 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{15}=\frac{15}{2}[2 \times 8+(15-1) 8]$$
$$=\frac{15}{2}[16+14 \times 8]$$
$$=\frac{15}{2}[16+112]$$
$$=\frac{15}{2} \times 128$$
= 15 × 64
∴ S15 = 960
Question 14.
Fnd the sum of the odd numbers between 0 and 50.
Solution:
Odd numbers between 0 and 50 are
1 + 3 + 5 + 7 + ……… + 49 =?
Here. a = 1, d = a3 – a1= 3 – 1 = 2
an = 49, n =?, S =?
an = a + (n – 1) d = 49
= 1 + (n – 1)2 = 49
1 + 2n – 2 = 49
2n – 1 = 49
2n = 49 + 1
2n = 50
$$n=\frac{50}{2}$$
∴ n = 25
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$\mathrm{s}_{25}=\frac{25}{2}[1+49]$$
$$=\frac{25}{2} \times 50$$
= 25 × 25
∴ S25 = 625
Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day. Rs. 250 for the second day. Rs. 300 for the third day. etc.. the penalty for each succeeding day being Rs. 50 more than for the preceeding day. How much money the contractor has to pay as penalty. If he has delayed the work by 30 days?
Solution:
Penalty for the 1st Day 2nd day 3rd day … 30th day
Rs. 200 Rs. 250 Rs. 300 Rs.?
200 + 250 + 300 + 30 days
a = 200, d = 250 – 200 = 50, n = 20,
S30 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$\dot{\mathrm{S}}_{30}=\frac{30}{2}[2 \times 200+(30-1) 50]$$
= 15[400 + 29 × 50]
= 15 [400 + 1450]
= 15 × 1850
∴ S30= Rs. 27750.
Question 16.
A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for thler overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let the first prize be ‘a’.
and Second prize is a – 20 and
the third prize Is a 40.
a, (a – 20), (a – 40) ……….. n = 7
a = a, d = a2 – a1 = a – 20 – a d = -20
n = 7, S7 =?
$$\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]=\mathrm{Sn}$$
$$\frac{7}{2}[2 \times a+(7-1)(-20)]=700$$
$$\frac{7}{2}[2 a+6(-20)]=700$$
$$\frac{7}{2}[2 a-120]=700$$
$$2 a-120=700 \times \frac{2}{7}$$
2a – 120 = 200
∴ 2a = 200 + 120
2a = 320
$$\quad a=\frac{320}{2}=R s .160$$
∴ Each prizes are
Rs. 160, 140, 120, 100, 80, 60, 40
Question 17.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant. will be the same as the class. In which they are studying. e.g., a section of Class I will plant 1 tree. a section of Class Il will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
a = 3. d = 6 – 3 = 3, n = 12, S12 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$\mathrm{S}_{12}=\frac{12}{2}[2 \times 3+(12-1) 3]$$
= 6[6 + 11 × 3]
= 6[6 + 33]
= 6 × 39
∴ S12 = 234
∴ Total number of trees from 3 sections of each class upto 12 class is 234.
Question 18.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm. 1.0 cm. 1.5 cm, 2.0 cm as shows In fig. What Is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $$\pi=\frac{22}{7}$$)
[Hint: Length of successive semicircles is l1, l2, l3, l4 with centres at A, B, A. B respectively.]
Solution :
$$: l_{1}=\pi \times \frac{1}{2}, 12=\pi \times 1.13=\pi \times \frac{3}{2}$$
$$l_{1}=\frac{\pi}{2}, \quad l_{2}=\mathrm{p}, \quad l_{3}=\frac{3}{2} \pi$$
∴ Arithmetic Progression,
l1, l2, l3, l4, ………..
Question 19.
200 logs are stacked In the following manner. 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see the fig. given below). In how many rows are the 200 logs placed and how many logs are In the top row?
Solution:
20, 19, 18, …..
a = 20, d = 19 – 20 = -1
Sn = 200, n =?, an =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$200=\frac{\mathrm{n}}{2}[2 \times 20+(\mathrm{n}-1)(-1)]$$
$$200=\frac{\mathrm{n}}{2}[40-\mathrm{n}+1]$$
$$200=\frac{n}{2}[41-n]$$
∴ 400 = n(41 – n)
400 = 4n – n2
∴ n2 – 41n + 400 = 0
n2 – 25n – 16n + 400 = 0
n(n – 25) – 16(n – 25) = 0
(n – 25) (n – 16) =
1f n – 16 = 0 then, n = 16
∴ an = a + (n – 1) d
a16= 20 + (16 – 1) (-1)
= 20 + 15(-1)
= 20 – 15
∴ a16 = 5
∴ 200 logs are placed In 16 rows and there are 5 logs in the top row.
Question 20.
In a potato race, a bucket is placed at the starting point, which Is 5 m from the first potato. and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see fig, given below)
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops ¡tin the bucket, runs back to pick up the next potato, runs to the bucket to drop It in, and she continues in the same way until al) the potaotes are in the bucket. What is the total distance the competitor has to run?
(Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 +2 × (5 + 3)]
Solution:
Total distance competitor taken to pick up the first potato = 5 + 5m
= 2 × 5m.
= 10m.
Total distance taken by compeUtor to pick up the second potato
= 5 + 3 + 3 + 5m.
= 2 × 5 + 2 × 3
= 2(5 + 3)
= 2 × 8
= 16m.
∴ 10m. 16m, 22m 10th potato
a = 10, d = 16 – 10 = 6m.
n = 10, S10 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{10}=\frac{10}{2}[2 \times 10+(10-1) 6]$$
= 5[20 + 9 × 6]
= 5 [20 + 54]
= 5 × 74
∴ S10 = 370m.
∴ Total distance the competitor has to run to pick up 10 potatoers is 370 m.
Nearest tenth is the first digit after the decimal point.
We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3, drop a comment below and we will get back to you at the earliest.
## KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2.
## Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2
Find the Value of x is used to consider unknown value.
Question 1.
Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
Solution:
(i) Strength of the X Std. is 10
Number of boys be ‘y’, then
number of girls be ‘x’ .
x + y = 10 …….. (i)
x = y + 4
∴ x – y = 4 ………. (ii)
From equations (i) + (ii)
$$\quad x=\frac{14}{2}=7$$
Substituting the value of ‘x’ in eqn. (i),
x + y = 10
7 + y = 10
y = 10 – 7
y = 4.
∴ Number of girls, x = 7
Number of boys, y = 4.
(ii) Cost of each pencil be Rs. ‘x’
Cost of pen be Rs. ‘y’
5x + 7y = 50 ………. (i)
7x + 5y = 46 ……….. (ii)
From equation (i) + equation (ii)
∴ x + y = 8 …………… (iii)
from Eqn. (ii) – Eqn. (i),
∴ -x + y = 2 …………. (iv)
Eqn. (iii) + Eqn. (iv)
∴ y = 5
Substituting the value of ‘y’ in eqn. (i)
x + y = 8
x + 5 = 8
∴ x = 8 – 5 x = 3
∴Cost of each pencil is Rs. 3.
Cost of each pen is Rs. 5.
Question 2.
On comparing the ratios $$\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}$$ find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Solution:
i) 5x – 4y + 8 = 0 a1= 5, b1 = -4, C1= 8
7x + 6y – 9 = 0 a2 = 7, b2= 6, c2= -9
$$\frac{a_{1}}{a_{2}}=\frac{5}{7} \quad \frac{b_{1}}{b_{2}}=\frac{-4}{6} \quad \frac{c_{1}}{c_{2}}=\frac{8}{-9}$$
Here, $$\frac{\mathbf{a}_{1}}{\mathbf{a}_{2}} \neq \frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}$$
∴ Lines representing the pairs of linear equations intersect at a point.
ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
a1 =9, b1 = 3, c1 = 12
a2 = 18, b2 = 6, c2 = 24
$$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{9}{18}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{3}{6}=\frac{1}{2}$$
$$\frac{c_{1}}{c_{2}}=\frac{12}{24}=\frac{1}{2}$$
Here, $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$$
∴ Representation of lines graphically are coincident.
iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Here, a1 = 6, b1 = -3, c1 = 10
a2 = 2, b2 = -1, c2 = 9
$$\frac{a_{1}}{a_{2}}=\frac{6}{2}=\frac{3}{1} \quad \frac{b_{1}}{b_{2}}=\frac{-3}{-1}=\frac{3}{1}$$
$$\frac{c_{1}}{c_{2}}=\frac{10}{9}$$
Here, $$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$
∴ Representation of lines graphically is parallel.
Question 3.
On compairing the ratios $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}, \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}$$ and $$\frac{c_{1}}{c_{2}}$$ find out whether the following pair of linear equations are consistent, or inconsistent.
i) 3x + 2y = 5; 2x – 3y = 7
ii) 2x – 3y = 8; 4x-6y = 9
iii) $$\frac{3}{2} x+\frac{5}{3} y=7 : 9 x-10 y=14$$
iv) 5x – 3y = 11: -10x – 6y = -22
v) $$\frac{4}{3} x+2 y=8 ; 2 x+3 y=12$$
Solution:
i) 3x + 2y = 5 ⇒ 3x + 2y – 5 = 0
2x – 3y = 7 ⇒ 2x – 3y – 7 = 0
Here, a1 = 3, b1 = 2, c1 = -5
a2 = 2, b2 = -3, c2 = -7
$$\frac{a_{1}}{a_{2}}=\frac{3}{2} \quad \frac{b_{1}}{b_{2}}=-\frac{2}{3} \quad \frac{c_{1}}{c_{2}}=\frac{-5}{-7}=\frac{5}{7}$$
Here, $$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$$
∴ Graphical representation is intersecting lines.
∴ Pair of linear equations are consistent.
ii) 2x – 3y = 8 ⇒ 2x – 3y – 8 = 0
4x – 6y = 9 ⇒ 4x – 6y – 9 = 0
a1 = 2, b1 = -3, c1 = -8
a2 = 4, b2 = -6, c2 = -9
$$\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{b_{1}}{b_{2}}=\frac{-3}{-6}=\frac{1}{2}$$
$$\frac{c_{1}}{c_{2}}=\frac{-8}{-9}=\frac{8}{9}$$
Here, $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$$
∴ Graphical representation is parallel lines.
∴ Equations are inconsistent.
iii) $$\frac{3}{2} x+\frac{5}{3} y=7 \quad \frac{3}{2} x+\frac{5}{3} y-7=0$$
9x – 10y = 14 ⇒ 9x – 10y – 14 = 0
$$a_{1}=\frac{3}{2}, \quad b_{1}=\frac{5}{3}, \quad c_{1}=-7$$
a2 = 9, b2 = -10, c2 = -17
$$\frac{a_{1}}{a_{2}}=\frac{3}{2} \times \frac{1}{9} \quad \frac{b_{1}}{b_{2}}=\frac{5}{3} \times \frac{-1}{6}$$
$$\frac{c_{1}}{c_{2}}=\frac{-7}{-14}=\frac{7}{14}=\frac{1}{2}$$
Here, $$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$$
∴ Pair of equations are consistent
iv) 5x – 3y = 11 ⇒ 5x – 3y – 11 = 0
-10x + 6y = – 22 ⇒ -10x + 6y + 22 = 0
a1 = 5, b1 = -3, c1 = -11
a2 = -10, b2 = 6, c2 = -22
$$\frac{a_{1}}{a_{2}}=\frac{5}{-10}=-\frac{1}{2} \quad \frac{b_{1}}{b_{2}}=\frac{-3}{6}=-\frac{1}{2}$$
$$\frac{c_{1}}{c_{2}}=\frac{-11}{22}=-\frac{1}{2}$$
Here, $$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$$
∴ Pair of equations are consistent
∴ Graphical representation is coninciding.
v) $$\frac{4}{3} x+2 y=8 \quad \frac{4}{3} x+26-8=0$$
2x + 3y = 12 ⇒ 2x + 3y – 12 = 0
$$a_{1}=\frac{4}{3}, \quad b_{1}=2, \quad c_{1}=-8$$
a2 = 2, b2 = 3, c2 = -12
$$\frac{a_{1}}{a_{2}}=\frac{4}{3} \times \frac{1}{2}=\frac{1}{6} \quad \frac{b_{1}}{b_{2}}=\frac{2}{3}$$
$$\frac{c_{1}}{c_{2}}=\frac{-8}{-12}=\frac{2}{3}$$
Here, $$\frac{\mathbf{a}_{1}}{\mathbf{a}_{2}} \neq \frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}$$
∴ Pair of equations are consistent
Question 4.
Which of the following pairs of linear equations are consistent/inconsistent? If consistent obtain the solution graphically
(i) x + y = 5, 2x + 2y = 10
(ii) x-y = 8 3x-3y= 16
(iii) 2x + y – 6 = 0 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0 4x – 3y – 5 = 0
Solution:
(i) x + y = 5 ⇒ x + y – 5 = 0
2x + 2y = 10 ⇒ 2x + 2y – 10 = 0
a1 = 1, b1 = 1, c1 = -5
a2 = 2, b2 = 2, c2 = -10
$$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{1}{2} \quad \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{-5}{-10}=\frac{1}{2}$$
Here, $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$$
∴ Pair of equations are consistent
(i) x + y = 5
y = 5 – x
x 0 2 4 y = 5 – x 5 3 1
(ii) 2x + 2y = 10
x + y = 5
y = 5 – x
x 0 2 5 y = 5 – x 5 3 0
∴ We can give any value for ‘x’, i.e., solutions are infinite.
∴ Solution, P (5, 0) x = 5, y = 0
(ii) x – y = 8 ⇒ x – y – 8 = 0
3x – 3y = 16 ⇒ 3x – 3y – 16 = 0
Here, $$\frac{a_{1}}{a_{2}}=\frac{1}{3} \quad \frac{b_{1}}{b_{2}}=\frac{-1}{-3}=\frac{1}{3}$$
$$\frac{c_{1}}{c_{2}}=\frac{-8}{-16}=\frac{1}{2}$$
$$\quad \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$$
∴ Linear equations are inconsistenttent.
∴ Algebraically it has no solution.
Graphical representation → Parallel Lines.
(i) x – y = 8
-y = 8 – x
y = -8 + x
x 8 10 9 y = -8+x 0 2 1
(ii) 3x – 3y = 16
-3y = 16 – 3x
3y = -16 + 3x
$$\quad y=\frac{-16+3 x}{3}$$
x 6 8 $$y=\frac{-16+3 x}{3}$$ 0.8 2.6
No solution because it is inconsistent
(iii) 2x + y – 6 = 0
4x – 2y – 4 = 0
Here a1 = 2, b1 = 1, c1 = -6
a2 = 4, b2 = -2, c2 = -4
$$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{1}{-2}$$
$$\frac{c_{1}}{c_{2}}=\frac{-6}{-4}=\frac{3}{2}$$
Here, $$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$$
Pair of equations are consistent. Algebraically both lines intersect.
Graphical Representation :
(i) 2x + y = 6
y = 6 – 2x
x 0 2 y = 6 – 2x 6 2
(ii) 4x – 2y – 4 = 0
4x – 2y = 4
-2y = 4 – 4x
2y = -4 + 4x
$$\quad y=\frac{-4+4 x}{2}$$
x 1 3 $$y=\frac{-4+4 x}{2}$$ 0 4
Solution: intersecting point, P (2, 2) i.e., x = 2, y = 2
(iv) 2x – 2y – 2 = 0
4x – 3y – 5 = 0
a1 = 2, b1 = -2, c1 = -2
a2 = 4, b2 = -3, c2 = -5
$$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{-2}{-3}=\frac{2}{3}$$
$$\frac{c_{1}}{c_{2}}=\frac{-2}{-5}=\frac{2}{5}$$
Here, $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}$$
Pair of equations are consistent.
∴ Algebraically both lines intersect.
Graphical Representation :
(i) 2x – 2y – 2 =0
2x – 2y = 2
-2y = 2 – 2x
2y = -2 + 2x
$$\quad y=\frac{-2+2 x}{2}$$
∴ y = – 1 + x
x 2 4 y= -1+ x 1 3
(ii) 4x – 3y – 5 = 0
4x – 3y = 5
-3y = 5 – 4x
3y = -5 + 4x
$$\quad y=\frac{-5+4 x}{3}$$
x 2 5 $$y=\frac{-5+4 x}{3}$$ 1 5
Solution: P(2, 1) i.e., x = 2, y = 1
Question 5.
Half the perimeter of a rectangular garden, whose length is 4m more than its width, is 36m. Find the dimensions of the garden.
Solution:
Length of rectangular garden be ‘x’ m.
Breadth of rectangular garden be ‘y’ m, then
Length of the garden is 4m more than its width.
x = y + 4 ……….. (i)
x – y = 4
Half of the circumference is 36 m.
$$\frac{2 x+2 y}{2}=36$$
2x + 2y = 72
x + y = 36 …………… (ii)
∴ x – y = 4 (i)
x + y = 36 (ii)
From eqn. (i) + eqn. (ii)
$$\quad x=\frac{40}{2}$$
∴ x = 20 m.
Substituting the value of ’x’ in eqn. (i)
x – y = 4
20 – y = 4
-y = 4 – 20
-y = -16
y = 16 m.
∴ Length of the garden = 20 m.
Breadth of the garden = 16 m.
Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is :
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines.
Solution:
Linear equation is 2x + 3y – 8 = 0.
(i) Linear equations for intersecting lines:
2x + 3y – 8 = 0
3x + 2y – 7 = 0
a1 = 2, b1 = 3, c1 = -8
a2 = 3, b2 = 2, c2 = -7
$$\frac{a_{1}}{a_{2}}=\frac{2}{3} \quad \frac{b_{1}}{b_{2}}=\frac{3}{2}$$
Here, when $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}$$ Geometrical representation is intersecting lines.
(ii) For Parallel lines :
2x + 3y – 8 = 0
2x + 3y – 12 = 0
a1 = 2, b1 = 3, c1= -8
a1 = 2, b1 = 3, c1 = -12
$$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{2}{2}=\frac{1}{1} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{3}{3}=\frac{1}{1}$$
$$\frac{c_{1}}{c_{2}}=\frac{-8}{-12}=\frac{8}{12}=\frac{2}{3}$$
Here, $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$$ hence
Graphical representation is parallel lines.
(iii) For Intersecting lines :
2x + 3y – 8 = 0
4x + 6y – 16 = 0
a1 = 2, b1 = 3, c1 = -8
a1 = 4, b1 = 6, c1 = -16
$$\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{b_{1}}{b_{2}}=\frac{3}{6}=\frac{1}{2}$$
$$\frac{c_{1}}{c_{2}}=\frac{-8}{-16}=\frac{1}{2}$$
Here,
∴ Lines are intersecting.
Question 7.
Draw the graphs of the equations x – y + 1 =0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution:
x – y + 1 = 0 ………. (i)
3x – 2y – 12 = 0 ……….. (ii)
x – y + 1 = 0
-y = -x -1
y = x + 1
x 2 4 y = x + 1 3 5
3x + 2y – 12 = 0
2y = -3x + 12
$$\quad y=\frac{-3 x+12}{2}$$
x 0 2 $$y=\frac{-3 x+12}{2}$$ 6 3
Graphs of these two equations intersect at A. Vertices formed for ∆ABC are,
A (2, 3), B (-1, 0), C (4, 0).
We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2, drop a comment below and we will get back to you at the earliest.
## Karnataka SSLC Maths Model Question Paper 4 Kannada Medium
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium
## Karnataka SSLC Maths Model Question Paper 4 Kannada Medium
ವಿಷಯ : ಗಣಿತ
ಸಮಯ: 3 ಗಂಟೆಗಳು
ಗರಿಷ್ಠ ಅಂಕಗಳು: 80
I. ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಅಥವಾ ಅಪೂರ್ಣ ಹೇಳಿಕೆಗೆ ನಾಲ್ಕು ಪರ್ಯಾಯ ಉತ್ತರಗಳನ್ನು ನೀಡಲಾಗಿದೆ. ಇವುಗಳಲ್ಲಿ ಸೂಕ್ತವಾದ ಉತ್ತರವನ್ನು ಆರಿಸಿ, ಕ್ರಮಾಕ್ಷರದೊಡನೆ ಪೂರ್ಣ ಉತ್ತರವನ್ನು ಬರೆಯಿರಿ. (8 × 1 = 8)
Question 1.
ಕೆಳಗಿನ ಸಂಖ್ಯೆಗಳ ಗಣಗಳಲ್ಲಿ ಸಮರೂಪ ತ್ರಿಭುಜಗಳಾಗುವಂತಹ ಜೋಡಿಯು
(A) (3, 4, 6) (9, 12, 24)
(B) (3, 4, 6) (9, 12, 18)
(C) (2,4, 6) (2,3, 14)
(D) (5, 10, 15) (10, 30, 45)
Question 2.
ಚತುರ್ಥಕದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯುವ ಸೂತ್ರ
(A) πr2
(B) $$\frac { 1 }{ 2 }$$ πr2
(C) $$\frac { 1 }{ 3 }$$ πr2
(D) $$\frac { 1 }{ 4 }$$ πr2
Question 3.
(2, 3) ಮತ್ತು (4, 5) ಬಿಂದುಗಳನ್ನು ಸೇರಿಸುವ ರೇಖಾಖಂಡದ ಮಧ್ಯಬಿಂದುವಿನ ನಿರ್ದೇಶಾಂಕಗಳು
(A) (3, 4)
(B) (4, 5)
(C) (5, 6)
(D) (6, 7)
Question 4.
ಧನ ಪೂಣಾಂಕ ‘a’ ಅನ್ನು ಪೂಣಾಂಕ ‘b’ ನಿಂದ ಭಾಗಿಸಿದಾಗ ಅನನ್ಯ ಪೂರ್ಣಾಂಕಗಳಾದ ‘q’ ಮತ್ತು ‘r’ ಗಳು ಇರುತ್ತವೆ. ಇದಕ್ಕೆ ಸರಿಹೊಂದುವ ಹೇಳಿಕೆ.
(A) b = a × q + r
(B) q = a × b + r
(C) a = b × q – r
(D) a = b × q + r
Question 5.
tan260° ಯ ಬೆಲೆ
(A) $$\frac { 1 }{ 3 }$$
(B) $$\frac { 1 }{ \surd 3 }$$
(C) 3
(D) √3
Question 6.
m ನ ಯಾವ ಧನಾತ್ಮಕ ಬೆಲೆಗೆ 3x2 + ka + 3 = 0 ಸಮೀಕರಣದ ಮೂಲಗಳು ಸಮವಾಗಿರುತ್ತವೆ.
(A) 2
(B) 3
(C) 5
(D) 6
Question 7.
ಒಂದು ಪ್ರಯೋಗದ ಎಲ್ಲಾ ಪ್ರಾಥಮಿಕ ಘಟನೆಗಳ ಸಂಭವನೀಯತೆಗಳ ಮೊತ್ತವು
(A) 0
(B) 1
(C) 2
(D) 3
Question 8.
ಒಂದು ಸಿಲಿಂಡರ್ನ ಪಾದದ ವಿಸ್ತೀರ್ಣ 24 cm2 ಮತ್ತು ಎತ್ತರ 10 cm ಆದರೆ ಸಿಲಿಂಡರ್ನ ಘನಫಲ
(A) 24 cm3
(B) 48 cm3
(C) 240 cm3
(D) 480 cm3
II. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 1 = 6)
Question 9.
ಥೇಲ್ಸ್ನ ಪ್ರಮೇಯವನ್ನು ನಿರೂಪಿಸಿ,
Question 10.
135 ಮತ್ತು 225 ರ ಮ.ಸಾ.ಅ. ವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 11.
ಬಹುಪದೋಕ್ತಿ x2 – 3x +5 ರ ಶೂನ್ಯತೆಗಳ ಮೊತ್ತವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 12.
ಬಾಹ್ಯ ಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕದ ಉದ್ದವು 8 cm ಮತ್ತು ವೃತ್ತಕೇಂದ್ರ ಹಾಗು ಬಾಹ್ಯಬಿಂದುವಿನ ದೂರ 10 cm ಆದರೆ ವೃತ್ತದ ತ್ರಿಜ್ಯವೆಷ್ಟು?
Question 13.
sin A = $$\frac { 3 }{ 4 }$$ ಆದರೆ cosec A ನ ಬೆಲೆಯೇನು?
Question 14.
ಹಂತ ವಿಚಲನಾ ವಿಧಾನದಿಂದ ಸರಾಸರಿಯನ್ನು ಕಂಡುಹಿಡಿಯುವ ಸೂತ್ರವನ್ನು ಬರೆಯಿರಿ.
III. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (16 × 2 = 32)
Question 15.
ವಾರ್ಷಿಕ ಸಂಬಳ ₹ 5000 ಮತ್ತು ಪ್ರತಿವರ್ಷಕ್ಕೆ ಹೆಚ್ಚುವರಿ ಬತ್ಯ ₹ 200 ಇರುವ ಕೆಲಸಕ್ಕೆ ಸುಬ್ಬರಾವ್ 1995 ರಲ್ಲಿ ಸೇರಿದರು. ಯಾವ ವರ್ಷದಲ್ಲಿ ಅವರ ಸಂಬಳ ₹ 7000 ಆಗುತ್ತದೆ?
Question 16.
ಚಿತ್ರದಲ್ಲಿ LM || CB ಮತ್ತು LN || CD ಆದರೆ $$\frac { AM }{ AB }$$ = $$\frac { AN }{ AD }$$ ಎಂದು ಸಾಧಿಸಿ.
Question 17.
ಈ ಜೋಡಿ ಸಮೀಕರಣಗಳನ್ನು ವರ್ಜಿಸುವ ವಿಧಾನದಿಂದ ಬಿಡಿಸಿ, 3x + 4y = 10 & 2x – 2y = 2.
Question 18.
ಕೆಳಗಿನ ರೇಖಾತ್ಮಕ ಸಮೀಕರಣಗಳ ಜೋಡಿಗಳು ಪ್ರತಿನಿಧಿಸುವ ಸರಳರೇಖೆಗಳು ಒಂದು ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆಯೇ? ಸಮಾಂತರವಾಗಿವೆಯೆ? ಅಥವಾ ಐಕ್ಯಗೊಂಡಿವೆಯೆ? ಕಂಡುಹಿಡಿಯಿರಿ
9x + 3y + 12 = 0; 18x + 6y + 24 = 0.
Question 19.
ಪರಧಿಯು 22 cm ಇರುವ ಒಂದು ಅರ್ಧ ವೃತ್ತದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 20.
4 cm ಮತ್ತು 6 cm ತ್ರಿಜ್ಯಗಳಿರುವ ಏಕಕೇಂದ್ರೀಯ ವೃತ್ತಗಳಿವೆ. 6 cm ತ್ರಿಜ್ಯದ ವೃತ್ತದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ 4 cm ತ್ರಿಜ್ಯದ ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಕವನ್ನು ರಚಿಸಿ.
Question 21.
(2, -5) ಮತ್ತು (-2, 9) ರಿಂದ ಸಮಾನ ದೂರದಲ್ಲಿರುವ x-ಅಕ್ಷದ ಮೇಲಿನ ಬಿಂದುವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 22.
ಶೃಂಗಬಿಂದುಗಳು (1, -3), (4, 1) ಮತ್ತು (2, 3) ಆಗಿರುವ ತ್ರಿಭುಜದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 23.
ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ $$\frac { -1 }{ 4 }$$ ಮತ್ತು ಗುಣಲಬ್ಧ $$\frac { 1 }{ 4 }$$ ಆಗಿರುವ ಒಂದು ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 24.
x2 + 6x – 7 = 0 ಸಮೀಕರಣವನ್ನು ವರ್ಗಪೂರ್ಣಗೊಳಿಸುವ ವಿಧಾನದಿಂದ ಬಿಡಿಸಿ.
Question 25.
ಗೋಪುರದ ಪಾದದಿಂದ 30m ದೂರದ ನೆಲದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ, ಗೋಪುರದ ತುದಿಯನ್ನು ನೋಡಿದಾಗ ಉಂಟಾಗುವ ಉನ್ನತ ಕೋನವು 30° ಆದರೆ, ಗೋಪುರದ ಎತ್ತರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 26.
ಒಂದು ಕಟ್ಟಡದ ಮೇಲಿನಿಂದ ಹಾಗೂ ಕೆಳಗಿನಿಂದ ಬೆಟ್ಟದ ತುದಿಯನ್ನು ಗಮನಿಸಿದಾಗ ಉಂಟಾದ ಉನ್ನತ ಕೋನವು 45° ಮತ್ತು 60° ಆಗಿದೆ. ಕಟ್ಟಡದ ಎತ್ತರ 24 m ಆದರೆ ಬೆಟ್ಟದ ಎತ್ತರವೇನು?
Question 27.
ಈ ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಸರಾಸರಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಈ ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಬಹುಲಕವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 28.
ಒಂದು ಚೀಲದಲ್ಲಿ 3 ಕೆಂಪು ಚೆಂಡುಗಳು ಮತ್ತು 5 ಕಪ್ಪು ಚೆಂಡುಗಳಿವೆ. ಚೀಲದಿಂದ ಯಾದೃಚ್ಛಿಕವಾಗಿ ಒಂದು ಚೆಂಡನ್ನು ತೆಗೆಯಲಾಗಿದೆ. ತೆಗೆದ ಚಂಡು ಕೆಂಪು ಆಗಿರುವ ಸಂಭವನೀಯತೆ ಎಷ್ಟು?
Question 29.
64 cm3 ಘನಫಲವನ್ನು ಹೊಂದಿರುವ 2 ವರ್ಗ ಘನಗಳ ಮುಖಗಳನ್ನು ಸೇರಿಸಿ ಒಂದು ಆಯತ ಘನಾಕೃತಿ ಮಾಡಿದೆ. ಈ ಘನಾಕೃತಿಯ ಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
14 cm ಎತ್ತರವಿರುವ ಒಂದು ಕುಡಿಯುವ ನೀರಿನ ಗಾಜಿನ ಲೋಟವು ಶಂಕುವಿನ ಭಿನ್ನಕದ ರೂಪದಲ್ಲಿದೆ. ಅದರ ಎರಡು ವೃತ್ತಾಕಾರದ ಪಾದಗಳ ವ್ಯಾಸಗಳು 4 cm ಮತ್ತು 2 cm ಗಳಾಗಿವೆ. ಗಾಜಿನ ಲೋಟ ಸಾಮರ್ಥ್ಯವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 30.
12 ಮತ್ತು 15 ರ ಲ.ಸಾ.ಅ. ಮತ್ತು ಮ.ಸಾ.ಅ. ವನ್ನು ಅವಿಭಾಜ್ಯ ಅಪವರ್ತನ ವಿಧಾನದಿಂದ ಕಂಡುಹಿಡಿಯಿರಿ.
IV. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 3 = 18)
Question 31.
ವೃತ್ತದ ಮೇಲಿನ ಯಾವುದೇ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ಸ್ಪರ್ಶಕವು, ಸ್ಪರ್ಶ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ತ್ರಿಜ್ಯಕ್ಕೆ ಲಂಬವಾಗಿರುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
ಒಂದು ಸಮಾಂತರ ಚತುರ್ಭುಜದಲ್ಲಿ ವೃತ್ತವು ಅಂತಸ್ಥವಾದಾಗ ಸಮಾಂತರ ಚತುರ್ಭುಜವು ವಜ್ರಾಕೃತಿಯಾಗುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.
Question 32.
ಪಾದ 8 cm ಮತ್ತು ಎತ್ತರ 4 cm ಇರುವ ಒಂದು ಸಮದ್ವಿಬಾಹು ತ್ರಿಭುಜವನ್ನು ರಚಿಸಿ, ನಂತರ ಮತ್ತೊಂದು ತ್ರಿಭುಜವನ್ನು ಅದರ ಬಾಹುಗಳು ಮೊದಲು ರಚಿಸಿದ ಸಮದ್ವಿಬಾಹು ತ್ರಿಭುಜದ ಅನುರೂಪ ಬಾಹುಗಳ 1$$\frac { 1 }{ 2 }$$ ರಷ್ಟಿರುವಂತೆ ರಚಿಸಿ,
Question 33.
ಎರಡು ಕ್ರಮಾಗತ ಬೆಸ ಧನ ಪೂರ್ಣಾಂಕಗಳ ವರ್ಗಗಳ ಮೊತ್ತವು 290 ಆದರೆ ಆ ಪೂಣಾರ್ಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಆಯತಾಕಾರದ ಹೊಲದ ಕರ್ಣವು ಅದರ ಚಿಕ್ಕ ಬಾಹುವಿಗಿಂತ 60 m ಹೆಚ್ಚಾಗಿದೆ. ಅದರ ದೊಡ್ಡ ಬಾಹುವು ಚಿಕ್ಕ ಬಾಹುವಿಗಿಂತ 30 m ಹೆಚ್ಚಾಗಿದ್ದರೆ, ಆ ಹೊಲದ ಬಾಹುಗಳ ಉದ್ದಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 34.
ಅಥವಾ
sec A (1 – sin A) (sec A + tan A) = 1 ಎಂದು ಸಾಧಿಸಿ.
Question 35.
ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಓಜೀವ್ ನಕ್ಷೆಯನ್ನು ರಚಿಸಿ, (ಕಡಿಮೆ ವಿಧಾನ)
Question 36.
3x2 – x – 4 ಎಂಬ ಬಹುಪದೋಕ್ತಿಯ ಶೂನ್ಯತೆಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ ಹಾಗು ಶೂನ್ಯತೆಗಳು ಮತ್ತು ಸಹಗುಣಕಗಳ ನಡುವಿನ ಸಂಬಂಧವನ್ನು ತಾಳೆ ನೋಡಿ.
ಅಥವಾ
x4 – 3x2 + 4x + 5 ನ್ನು x2 + 1 – x ದಿಂದ ಭಾಗಿಸಿ ಭಾಗಲಬ್ದ ಮತ್ತು ಶೇಷವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
V. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (4 × 4 = 16)
Question 37.
ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೂರು ಪದಗಳ ಮೊತ್ತವು 15 ಮತ್ತು ಅವುಗಳ ಅಂತ್ಯಪದಗಳ ವರ್ಗಗಳ ಮೊತ್ತವು 58 ಆಗಿದೆ. ಶ್ರೇಢಿಯ ಆ ಮೂರು ಪದಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೊದಲ 5 ಪದಗಳ ಮೊತ್ತವು ಮುಂದಿನ 5 ಪದಗಳ ಮೊತ್ತದ ನಾಲ್ಕನೇ ಒಂದು ಭಾಗದಷ್ಟಿದೆ. ಮೊದಲ ಪದ 2 ಆದರೆ, a20 = -112 ಎಂದು ಸಾಧಿಸಿ ಮತ್ತು S20 ನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 38.
ಎರಡು ಸಮರೂಪ ತ್ರಿಭುಜಗಳ ವಿಸ್ತೀರ್ಣಗಳ ಅನುಪಾತವು ಅವುಗಳ ಅನುರೂಪ ಬಾಹುಗಳ ವರ್ಗಗಳ ಅನುಪಾತಕ್ಕೆ ಸಮನಾಗಿರುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.
Question 39.
ಒಂದು ಬಾವಿಯ ವ್ಯಾಸ 3 m ಮತ್ತು ಆಳ 14 m ಇರುವಂತೆ ತೋಡಿದೆ. ಭೂಮಿಯಿಂದ ತೆಗೆದ ಮಣ್ಣನ್ನು ಬಾವಿಯ ಸುತ್ತಲು ಸಮವಾಗಿ ಹರಡಿ 4 m ಅಗಲವಿರುವ ವೃತ್ತಾಕಾರದ ಕಟ್ಟೆಯನ್ನು ಕಟ್ಟಿದೆ. ಕಟ್ಟೆಯ ಎತ್ತರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 40.
ನಕ್ಷೆಯ ಮೂಲಕ ಸಮೀಕರಣಗಳನ್ನು ಬಿಡಿಸಿ.
2x + y = 8
x + 2y = 7
Solutions
I.
Solution 1.
(B) (3, 4, 6) (19, 12, 18)
Solution 2.
(D) $$\frac { 1 }{ 4 }$$ πr2
Solution 3.
(A) (3, 4)
Solution 4.
(D) a = b × q + r
Solution 5.
(C) 3
Solution 6.
(D) 6
Solution 7.
(B) 1
Solution 8.
(C) 240 cm3
II.
Solution 9.
ತ್ರಿಭುಜದ ಎರಡು ಬಾಹುಗಳನ್ನು ಎರಡು ವಿಭಿನ್ನ ಬಿಂದುಗಳಲ್ಲಿ ಛೇದಿಸುವಂತೆ ಒಂದು ಬಾಹುವಿಗೆ ಸಮಾಂತರವಾಗಿ ಎಳೆದ ಸರಳರೇಖೆಯು ಉಳಿದೆರಡು ಬಾಹುಗಳನ್ನು ಸಮಾನುಪಾತದಲ್ಲಿ ವಿಭಾಗಿಸುತ್ತದೆ.
Solution 10.
Solution 11.
Solution 12.
OA2 = OP2 – AP2 =102 – 82
⇒ OA2 = 100 – 64
⇒ OA2 = 100 – 64
⇒ OA = √36
⇒ OA = 6 cm
ವೃತ್ತದ ತ್ರಿಜ್ಯ = 6 cm
Solution 13.
cosec A = $$\frac { 4 }{ 3 }$$
Solution 14.
III.
Solution 15.
5000, 5200, 5400, ……… 7000.
a = 5000, d = 200 an = 7000, n = ?
an = a + (n – 1) d
⇒ 7000 = 5000 + (n – 1) 200
⇒ 7000 – 5000 = (n – 1) 200
⇒ 2000 = 200 (n – 1)
⇒ (n – 1) = 10
⇒ n = 10 + 1
⇒ n = 11
11 ನೇ ವರ್ಷದಲ್ಲಿ ಅವರ ಸಂಬಳ ₹ 7000 ಆಗುತ್ತದೆ.
Solution 16.
Solution 17.
3x + 4y = 10 …. (1)
2x – 2y = 2 …… (2)
ಸಮೀಕರಣ 2ನ್ನು 2ರಿಂದ ಗುಣಿಸಿದಾಗ
3x + 4y = 10
4x – 4y = 4
…………………
7x = 14
x = 2
3x + 4y = 10
3(2) + 4y = 10
4y = 10 – 6
y = 1
∴ x = 2, y = 1
Solution 18.
Solution 19.
ಅರ್ಧ ವೃತ್ತದ ವಿಸ್ತೀರ್ಣ = $$\frac { 77 }{ 4 }$$ cm2
Solution 20.
R = 6 cm, r = 4 cm
Solution 21.
Solution 22.
Solution 23.
ಬಹುಪದೋಕ್ತಿ = ax2+ bx + c ಆಗಿರಲಿ
ಶೂನ್ಯತೆಗಳು α ಮತ್ತು β ಆಗಿರಲಿ
Solution 24.
x2 + 6x – 7 = 0
⇒ x2 + 6x = 7
⇒ x2 + 6x + 32 – 32 = 7
⇒ (x + 3)2 – 9 = 7
⇒ (x + 3)2 = 16
⇒ x + 3 = ±√16
⇒ x + 3 = ±4
⇒ x = ±4 – 3
⇒ x = +4 – 3 ಅಥವಾ x = -4 – 3
⇒ x = 1 ಅಥವಾ x = -7
Solution 25.
Solution 26.
Solution 27.
Solution 28.
ಚೀಲದಲ್ಲಿರುವ ಒಟ್ಟು ಚೆಂಡುಗಳು = n(S) = 3 + 5 = 8
ಕೆಂಪು ಚೆಂಡುಗಳು ಸಂಖ್ಯೆ = 3 = n(A)
ಕೆಂಪು ಚೆಂಡು ತೆಗೆಯುವ ಸಂಭವನೀಯತೆ = $$\frac { n(A) }{ n(S) }$$
ಕೆಂಪು ಚೆಂಡು ತೆಗೆಯುವ ಸಂಭವನೀಯತೆ = $$\frac { 3 }{ 8 }$$
Solution 29.
ವರ್ಗಗಳ ಘನಫಲ = 64cm3
ಬಾಹುವಿನ ಅಳತೆ = $$\sqrt [ 3 ]{ 64 }$$ = 4 cm
ಆಯತ ಘನದ ಅಗಲ b = 4 cm
ಆಯತ ಘನದ ಉದ್ದ l = 8 cm (4 + 4)
ಆಯತ ಘನದ ಎತ್ತರ h = 4 cm
ಆಯತ ಘನದ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ = 2 (lb + bh + hl)
= 2(8 × 4 + 4 × 4 + 4 × 8)
= 2(32 + 16 + 32)
= 160 cm2
ಆಯತ ಘನದ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ = 160 cm2
ಅಥವಾ
14 cm ಎತ್ತರವಿರುವ ಒಂದು ಕುಡಿಯುವ ನೀರಿನ ಗಾಜಿನ ಲೋಟವು ಶಂಕುವಿನ ಭಿನ್ನಕದ ರೂಪದಲ್ಲಿದೆ. ಅದರ ಎರಡು ವೃತ್ತಾಕಾರದ ಪಾದಗಳ ವ್ಯಾಸಗಳು 4 cm ಮತ್ತು 2 cm ಗಳಾಗಿವೆ. ಗಾಜಿನ ಲೋಟ ಸಾಮರ್ಥ್ಯವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Solution 30.
12 = 2 × 2 × 5
15 = 3 × 5
ಮ.ಸಾ.ಅ. = 5
ಲ.ಸಾ.ಅ. = 2 × 2 × 3 × 5
ಲ.ಸಾ.ಅ. = 60
IV.
Solution 31.
ದತ್ತ: ‘O’ ವೃತ್ತಕೇಂದ್ರ, XY ಸ್ಪರ್ಶಕ, Pಸ್ಪರ್ಶಬಿಂದು
ಸಾಧನೀಯ: OP ⊥ XY
ರಚನೆ: P ಯನ್ನು ಹೊರತುಪಡಿಸಿ XY ಮೇಲೆ ಮತ್ತೊಂದು ಬಿಂದು Q ಆಗಿರಲಿ, OQ ಸೇರಿಸಿ.
ಸಾಧನೆ: OQ ವೃತ್ತವನ್ನು ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸಿದೆ.
OP = OR (ಒಂದೇ ವೃತ್ತದ ತ್ರಿಜ್ಯಗಳು)
OQ = OR + RQ
OQ > OR
OQ > OP (OP = OR)
OP ಯು O ನಿಂದ ಸ್ಪರ್ಶಕಕ್ಕೆ ಕನಿಷ್ಟ ದೂರವಾಗಿದೆ
OP ⊥ XY
ಅಥವಾ
ಸಾಧನೀಯ: ABCD ಒಂದು ವಜ್ರಾಕೃತಿ. (AB = BC = CD = AD)
ಸಾಧನೆ: AB = CD ಮತ್ತು AD = BC …… (1)
AP = AS, BP = BQ, CQ = CR, DS = DR
(ಬಾಹ್ಯಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು)
AB + CD = AP + PB + DR + CR
⇒ AB + CD = AS + BQ + DS + CQ
⇒ AB + CD = AS + DS + BQ + CQ
⇒ AB + CD = AD + BC
⇒ AB = AD ….. (2)
∴ AB = BC = CD = AD (ಸಮೀಕರಣ (1) & (2) ರಿಂದ)
Solution 32.
Solution 33.
ಕ್ರಮಾಗತ ಬೆಸ ಧನ ಪೂರ್ಣಾಂಕಗಳು x & x + 2
x2 + (x + 2)2 = 290
⇒ x2 + x2 + 4x + 4 – 290 = 0
⇒ 2x2 + 4x – 286 = 0
⇒ x2 + 2x – 143 = 0
⇒ x2 + 13x – 11x – 143 = 0
⇒ x (x + 13) – 11(x + 13) = 0
⇒ (x + 13) (x – 11) = 0
⇒ x + 13 = 0 ಅಥವಾ x – 11 = 0
⇒ x = -13 ಅಥವಾ x = 11
ಕ್ರಮಾಗತ ಬೆಸ ಧನ ಪೂರ್ಣಾಂಕಗಳು 11 & 13
[∴ x = 11 & x + 2 = 11 + 2 = 13]
ಅಥವಾ
ಚಿಕ್ಕ ಬಾಹು x ಆಗಿರಲಿ
AC2 = AB2 + BC2
⇒ (x + 60)2 = x2 + (x + 30)2
⇒ x2 + 3600 + 120x = x2 + x2 + 900 + 60x
⇒ x2 + 3600 + 120x – 2x2 – 900 – 60x = 0
⇒ -x2 + 2700 + 60x = 0
⇒ x2 – 60x – 2700 = 0
⇒ x2 – 90x + 30x – 2700 = 0
⇒ x (x – 90) + 30 (x – 90) = 0
⇒ x – 90 = 0 ಅಥವಾ x + 30 = 0
⇒ x = 90 ಅಥವಾ x = -30
ಚಿಕ್ಕ ಬಾಹುವಿನ ಉದ್ದ = x = 90 m
ದೊಡ್ಡ ಬಾಹುವಿನ ಉದ್ದ = x + 30 = 90 + 30 = 120 m
Solution 34.
Solution 35.
Solution 36.
V.
Solution 37.
ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೂರು ಪದಗಳು
a – d, a, a + d
a – d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5
(a – d)2 + (a + d)2 = 58
⇒ a2 + d2 – 2ad + a2 + d2 + 2ad = 58
⇒ 2a2 + 2d2 = 58
⇒ 2(a2 + d2) = 58
⇒ a2 + d2 = 29
⇒ 52 + d2 = 29
⇒ d2 = 29 – 25
⇒ d2 = 4
⇒ d = ± 2
⇒ d = 2 ಅಥವಾ d = -2
ಮೂರು ಪದಗಳು
∴ a – d = 5 – 2 = 3
∴ a = 5
∴ a + d = 5 + 2 = 7
ಅಥವಾ
Solution 38.
Solution 39.
Solution 40.
## Karnataka SSLC Maths Model Question Paper 3 Kannada Medium
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium
## Karnataka SSLC Maths Model Question Paper 3 Kannada Medium
ವಿಷಯ : ಗಣಿತ
ಸಮಯ: 3 ಗಂಟೆಗಳು
ಗರಿಷ್ಠ ಅಂಕಗಳು: 80
I. ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಅಥವಾ ಅಪೂರ್ಣ ಹೇಳಿಕೆಗೆ ನಾಲ್ಕು ಪರ್ಯಾಯ ಉತ್ತರಗಳನ್ನು ನೀಡಲಾಗಿದೆ. ಇವುಗಳಲ್ಲಿ ಸೂಕ್ತವಾದ ಉತ್ತರವನ್ನು ಆರಿಸಿ, ಕ್ರಮಾಕ್ಷರದೊಡನೆ ಪೂರ್ಣ ಉತ್ತರವನ್ನು ಬರೆಯಿರಿ. (8 × 1 = 8)
Question 1.
ಎರಡು ಸಮರೂಪ ತ್ರಿಭುಜಗಳ ವಿಸ್ತೀರ್ಣಗಳ ಅನುಪಾತ 16 : 25 ಆದರೆ ಅವುಗಳ ಬಾಹುಗಳ ಅನುಪಾತ
(A) 3 : 4
(B) 4 : 5
(C) 5 : 6
(D) 6 : 7
Question 2.
7 cm ತ್ರಿಜ್ಯವಿರುವ ವೃತ್ತದಲ್ಲಿ 60° ಕೋನವನ್ನು ಹೊಂದಿರುವ ತ್ರಿಜ್ಯಾಂತರ ಖಂಡದ ವಿಸ್ತೀರ್ಣ
(A) 25.67 cm2
(B) 35.32 cm2
(C) 15.25 cm2
(D) 77 cm2
Question 3.
ಮೂಲಬಿಂದು ಮತ್ತು (6, 8) ಬಿಂದುವಿನ ನಡುವಿನ ದೂರ
(A) 6 ಮೂ.ಮಾ.
(B) 8 ಮೂ.ಮಾ.
(C) 10 ಮೂ.ಮಾ.
(D) 14 ಮೂ.ಮಾ.
Question 4.
(15, 20) ರ ಮ.ಸಾ.ಅ. 5 ಆದರೆ ಅವುಗಳ ಲ.ಸಾ.ಅ.
(A) 15
(B) 20
(C) 40
(D) 60
Question 5.
$$\frac { { sin18 }^{ 0 } }{ { cos72 }^{ 0 } }$$ ಯ ಬೆಲೆ
(A) 1
(B) 0
(C) -1
(D) 2
Question 6.
x2 – 25 = 0 ಈ ಸಮೀಕರಣದ ಮೂಲಗಳು
(A) (+5, -5)
(B) (+5, +5)
(C) (-5, -5)
(D) (25, -25)
Question 7.
ಒಂದು ನಿರ್ದಿಷ್ಟ ದಿನದಲ್ಲಿ ಮಳೆ ಬೀಳುವ ಸಂಭವನೀಯತೆಯು 0.64 ಆಗಿದೆ. ಅದೇ ದಿನ ಮಳೆ ಬೀಳದಿರುವ ಸಂಭವನೀಯತೆ.
(A) -0.64
(B) 64
(C) 0.36
(D) -0.36
Question 8.
ಶಂಕುವಿನ ಭಿನ್ನಕದ ಪಾರ್ಶ್ವ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ ಕಂಡುಹಿಡಿಯುವ ಸೂತ್ರ
II. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 1 = 6)
Question 9.
ಪೈಥಾಗೊರಸ್ ಪ್ರಮೇಯವನ್ನು ನಿರೂಪಿಸಿ,
Question 10.
ಒಂದು ಸಂಖ್ಯೆಯನ್ನು 14 ರಿಂದ ಭಾಗಿಸಿದಾಗ ಶೇಷ 5 ಆದರೆ 7ರಿಂದ ಅದೇ ಸಂಖ್ಯೆಯನ್ನು ಭಾಗಿಸಿದಾಗ ದೊರೆಯುವ ಶೇಷವೆಷ್ಟು?
Question 11.
α ಮತ್ತು β ಗಳು ax2 + bx + c ಎಂಬ ಒಂದು ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯ ಶೂನ್ಯತೆಗಳಾದರೆ αβ ದ ಬೆಲೆಯೇನು?
Question 12.
‘O’ ಕೇಂದ್ರವುಳ್ಳ ವೃತ್ತಕ್ಕೆ PA ಮತ್ತು PB ಗಳು ಬಾಹ್ಯ ಬಿಂದು P ನಿಂದ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು ಮತ್ತು ∠APB = 60° ಹಾಗು AP = 8 cm ಆದಾಗ AB ಜ್ಯಾದ ಉದ್ದವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 13.
(1 + sin290°)2 ನ ಬೆಲೆಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 14.
12, 16, 20, 24, 28 ರ ಸರಾಸರಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
III. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (16 × 2 = 32)
Question 15.
a = 5, d = 3, an = 50 ಆದರೆ Sn ಕಂಡುಹಿಡಿಯಿರಿ.
Question 16.
ಚಿತ್ರದಲ್ಲಿ ∆ABCಯ ಎತ್ತರಗಳಾದ AD ಮತ್ತು CE ಗಳು ಪರಸ್ಪರ P ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆ. ಆದರೆ ∆AEP ~ ∆CDP ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
ABCD ಚತುರ್ಭುಜದಲ್ಲಿ $$\frac { AO }{ BO }$$ = $$\frac { CO }{ DO }$$ ಆಗುವಂತೆ ಕರ್ಣಗಳು ಪರಸ್ಪರ 0ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆ. ಆದರೆ ABCD ಯು ಒಂದು ತ್ರಾಪಿಜ್ಯ ಎಂದು ಸಾಧಿಸಿ.
Question 17.
Question 18.
5x – 3y = 11, -10x + 6y = -22 ಈ ಸಮೀಕರಣಗಳು ಸ್ಥಿರವಾಗಿವೆಯೆ? ಅಥವಾ ಅಸ್ಥಿರವಾಗಿವೆಯೆ? ಎಂಬುದನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 19.
ಚಿತ್ರದಲ್ಲಿ ಕೇಂದ್ರ O ಇರುವ ಎರಡು ಏಕಕೇಂದ್ರೀಯ ವೃತ್ತಗಳ ತ್ರಿಜ್ಯಗಳು ಕ್ರಮವಾಗಿ 7 cm ಮತ್ತು 14 cm ಇವೆ. ∠AOC = 40° ಆದರೆ ಛಾಯಕೃತ ಭಾಗದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 20.
4 cm ತ್ರಿಜ್ಯದ ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಕಗಳ ನಡುವಿನ ಕೋನ 70° ಇರುವಂತೆ ಒಂದು ಜೊತೆ ಸ್ಪರ್ಶಕಗಳನ್ನು ಎಳೆಯಿರಿ.
Question 21.
AB ವ್ಯಾಸವಾಗಿರುವ ವೃತ್ತದ ಕೇಂದ್ರ (2, -3) ಮತ್ತು B ಯು (1, 4) ಆದರೆ A ಬಿಂದುವಿನ ನಿರ್ದೇಶಾಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 22.
(8, 1), (k, -4), (2, -5) ಎಂಬ ಬಿಂದುಗಳು ಸರಳರೇಖಾಗತವಾಗಿದ್ದರೆ k ಯ ಬೆಲೆಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 23.
ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ ಹಾಗು ಗುಣಲಬ್ದಗಳು ಕ್ರಮವಾಗಿ 4 ಮತ್ತು 1 ಆಗಿರುವ ಒಂದು ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 24.
2x2 + 3x – 5 = 0 ಈ ವರ್ಗ ಸಮೀಕರಣವನ್ನು ಸೂತ್ರದಿಂದ ಬಿಡಿಸಿ,
Question 25.
20 m ಎತ್ತರದ ಕಟ್ಟಡವೊಂದರ ಮೇಲೆ ಸ್ಥಾಪಿಸಲಾದ ಪ್ರಸರಣೆಯ ಗೋಪುರವೊಂದರ ಮೇಲುದಿ ಮತ್ತು ಪಾದಗಳನ್ನು ನೆಲದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ ನೋಡಿದಾಗ ಉನ್ನತ ಕೋನಗಳು ಕ್ರಮವಾಗಿ 60° ಮತ್ತು 45° ಇದೆ. ಪ್ರಸರಣೆಯ ಗೋಪುರದ ಎತ್ತರ ಕಂಡುಹಿಡಿಯಿರಿ.
Question 26.
50√3 m ಎತ್ತರದಲ್ಲಿರುವ ಒಂದು ಕಟ್ಟಡದ ಮೇಲಿನಿಂದ ನೆಲದ ಮೇಲಿರುವ ಒಂದು ಕಾರನ್ನು ನೋಡಿದಾಗ ಉಂಟಾದ ಅವನತ ಕೋನವು 60° ಆಗಿರುತ್ತದೆ. ಹಾಗಾದರೆ ಕಟ್ಟಡದಿಂದ ಕಾರಿಗಿರುವ ದೂರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 27.
ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಮಧ್ಯಾಂಕವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 28.
ಒಂದು ದಾಳವನ್ನು ಒಂದು ಸಲ ಎಸೆಯಲಾಗಿದೆ. (i) ಒಂದು ಅವಿಭಾಜ್ಯ ಸಂಖ್ಯೆ (ii) 1 ಮತ್ತು 5ರ ನಡುವಿನ ಒಂದು ಸಂಖ್ಯೆಯನ್ನು ಪಡೆಯುವ ಸಂಭವನೀಯತೆಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 29.
ಸಮನಾದ ತ್ರಿಜ್ಯವುಳ್ಳ ಒಂದು ಶಂಕುವನ್ನು ಒಂದು ಅರ್ಧಗೋಳಾಕೃತಿಯ ಮೇಲೆ ಜೋಡಿಸಿ ಒಂದು ಆಟಿಕೆಯನ್ನು ಮಾಡಲಾಗಿದೆ. ಶಂಕುವಿನ ಭಾಗದ ವ್ಯಾಸವು 6 cm ಮತ್ತು 4 cm ಎತ್ತರ ಇದ್ದರೆ ಈ ಘನವಸ್ತುವಿನ ಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಶಂಕುವಿನ ಭಿನ್ನಕದ ಓರೆ ಎತ್ತರ 10 cm, ಅದರ ವೃತ್ತಾಕಾರದ ಪಾದಗಳ ತ್ರಿಜ್ಯಗಳು ಕ್ರಮವಾಗಿ 8 cm ಮತ್ತು 6 cm ಆಗಿದೆ. ಆ ಭಿನ್ನಕದ ವಕ್ರಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 30.
3 + 2√5 ಒಂದು ಅಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ ಎಂದು ಸಾಧಿಸಿ.
IV. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 3 = 18)
Question 31.
ಬಾಹ್ಯಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳ ಉದ್ದವು ಸಮನಾಗಿರುತ್ತವೆ ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
ಎರಡು ಏಕಕೇಂದ್ರೀಯ ವೃತ್ತಗಳ ತ್ರಿಜ್ಯಗಳು 5 cm ಮತ್ತು 3 cm ಆಗಿವೆ. ಚಿಕ್ಕ ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಿಸುವಂತೆ ದೊಡ್ಡ ವೃತ್ತದ ಹ್ಯಾದ ಉದ್ದವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 32.
BC = 7 cm, ∠A = 45°, ∠B = 105° ಇರುವಂತೆ ABC ತ್ರಿಭುಜವನ್ನು ರಚಿಸಿ ನಂತರ ಮತ್ತೊಂದು ತ್ರಿಭುಜವನ್ನು ಅದರ ಬಾಹುಗಳು, ∆ABCಯ ಅನುರೂಪ ಬಾಹುಗಳ $$\frac { 3 }{ 4 }$$ ರಷ್ಟಿರುವಂತೆ ರಚಿಸಿ,
Question 33.
3 ವರ್ಷಗಳ ಹಿಂದಿನ ರೆಹಮಾನನ ವಯಸ್ಸು ಮತ್ತು 5 ವರ್ಷಗಳ ನಂತರದ ಅವನ ವಯಸ್ಸು ಇವುಗಳ ವ್ಯತ್ಯಮಗಳ ಮೊತ್ತ $$\frac { 1 }{ 3 }$$ ಆದರೆ ಅವನ ಈಗಿನ ವಯಸ್ಸನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ರೈಲು 360km ದೂರವನ್ನು ಏಕರೂಪ ಜವದೊಂದಿಗೆ ಕ್ರಮಿಸುತ್ತದೆ. ಅದರ ಜವವು 5km/hr ಹೆಚ್ಚಾಗಿದ್ದರೆ, ಅಷ್ಟೇ ದೂರವನ್ನು ಕ್ರಮಿಸಲು ಅದು 1 ಗಂಟೆ ಕಡಿಮೆ ತೆಗೆದುಕೊಳ್ಳುತ್ತಿತ್ತು. ರೈಲಿನ ಜವವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 34.
(cos A + sec A)2 + (sin A + cosec A)2 = 7 + tan2A + cot2A ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
A, B ಮತ್ತು C ಗಳು ∆ABCಯ ಒಳಕೋನಗಳಾದರೆ sin($$\frac { B+C }{ 2 }$$) = cos($$\frac { A }{ 2 }$$) ಎಂದು ಸಾಧಿಸಿ.
Question 35.
ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಓಜೀವ್ ನಕ್ಷೆಯನ್ನು ರಚಿಸಿ,
(ಅಧಿಕ ಇರುವ ವಿಧಾನ)
Question 36.
4u2 – 8u ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯ ಶೂನ್ಯತೆಗಳನ್ನು ಕಂಡುಹಿಡಿದು ಹಾಗು ಶೂನ್ಯತೆಗಳು ಮತ್ತು ಸಹಗುಣಕಗಳ ನಡುವಿನ ಸಂಬಂಧವನ್ನು ತಾಳೆನೋಡಿ.
ಅಥವಾ
p(x) = x5 – 4x3 + x2 + 3x + 1 ನ್ನು g(x) = x3 – 3x + 1 ರಿಂದ ಭಾಗಿಸಿ g(x), p(x) ನ ಅಪವರ್ತನವನ್ನು ಪರೀಕ್ಷಿಸಿ.
V. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (4 × 4 = 16)
Question 37.
ಒಂದು ಸಮಾಂತರ ಶ್ರೇಢಿಯ 4ನೇ ಮತ್ತು 8ನೇ ಪದಗಳ ಮೊತ್ತವು 24ಹಾಗು ಅದೇ ಶ್ರೇಢಿಯ 6ನೇ ಮತ್ತು 10ನೇ ಪದಗಳ ಮೊತ್ತವು 44 ಆದರೆ ಶ್ರೇಢಿಯ ಮೊದಲ ಮೂರು ಪದಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಸುರುಳಿಯನ್ನು ಕ್ರಮಾಗತ ಅರೆ ವೃತ್ತಗಳಿಂದ ಮಾಡಲಾಗಿದೆ ಅವುಗಳ ಕೇಂದ್ರಗಳು ಪರ್ಯಾಯವಾಗಿ A & B ನಲ್ಲಿದ್ದು A ಕೇಂದ್ರದಿಂದ ಆರಂಭವಾಗಿ ತ್ರಿಜ್ಯಗಳು 0.5 cm, 1 cm, 1.5 cm, 2 cm …….. ಹೀಗೆ ಇದೆ. ಈ ರೀತಿಯ 13 ಕ್ರಮಾಗತ ಅರೆ ವೃತ್ತಗಳಿಂದ ಮಾಡಲ್ಪಟ್ಟ ಒಟ್ಟು ಉದ್ದ ಏನು?
Question 38.
ಒಂದು ಲಂಬಕೋನ ತ್ರಿಭುಜದಲ್ಲಿ ವಿಕರ್ಣದ ಮೇಲಿನ ವರ್ಗವು ಉಳಿದೆರಡು ಬಾಹುಗಳ ವರ್ಗಗಳ ಮೊತ್ತಕ್ಕೆ ಸಮನಾಗಿರುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.
Question 39.
ನಕ್ಷೆಯ ಮೂಲಕ ಸಮೀಕರಣಗಳನ್ನು ಬಿಡಿಸಿ, 2x – y = 5 ಮತ್ತು x + 3y = 6.
Question 40.
ಒಂದು ಅರ್ಧಗೋಳಾಕಾರದ ಬಟ್ಟಲಿನ ಒಳಜ್ಯವು 18 cm ಇದ್ದು ಅದರ ತುಂಬ ಹಣ್ಣಿನ ರಸವಿದೆ. ಈ ರಸವನ್ನು 3 cm ತ್ರಿಜ್ಯವಿರುವ ಮತ್ತು 9 cm ಎತ್ತರವಿರುವ ಸಿಲಿಂಡರಿನಾಕೃತಿಯ ಬಾಟಲಿಗಳಿಗೆ ತುಂಬಬೇಕು. ಬಟ್ಟಲನ್ನು ಖಾಲಿ ಮಾಡಲು ಎಷ್ಟು ಬಾಟಲಿಗಳ ಅವಶ್ಯಕತೆ ಇದೆ?
Solutions
I.
Solution 1.
(B) 4 : 5
Solution 2.
(A) 25.67 cm2
Solution 3.
(C) 10 ಮೂ.ಮಾ.
Solution 4.
(D) 60
Solution 5.
(A) 1
Solution 6.
(A) (+5, -5)
Solution 7.
(C) 0.36
Solution 8.
(B) π (r1 + r2) l
II.
Solution 9.
ಒಂದು ಲಂಬಕೋನ ತ್ರಿಭುಜದಲ್ಲಿ ವಿಕರ್ಣದ ಮೇಲಿನ ವರ್ಗವು ಉಳಿದೆರಡು ಬಾಹುಗಳ ಮೇಲಿನ ವರ್ಗಗಳ ಮೊತ್ತಕ್ಕೆ ಸಮ.
Solution 10.
ಶೇಷ – 5
Solution 11.
αβ = $$\frac { c }{ a }$$
Solution 12.
∆APB ನಲ್ಲಿ PA = PB = 8cm
∠PAB = ∠PBA = 60°
∠APB = 60°
∆APB ಒಂದು ಸಮಬಾಹು ತ್ರಿಭುಜ
AB = 8 cm
Solution 13.
(1 + sin290°)2 = (1 + 1)2 = 4
Solution 14.
III.
Solution 15.
Solution 16.
ಸಾಧನೀಯ: ∆AEP ~ ∆CDP
∆AEP & ∆CDP ಗಳಲ್ಲಿ
∠AEP = ∠CDP = 90°
∠APE = ∠CPD (ಶೃಂಗಾಭಿಮುಖ ಕೋನಗಳು)
∆AEP ~ ∆CDP
ಅಥವಾ
ABCD ಚತುರ್ಭುಜದಲ್ಲಿ $$\frac { AO }{ BO }$$ = $$\frac { CO }{ DO }$$ ಆಗುವಂತೆ ಕರ್ಣಗಳು ಪರಸ್ಪರ O ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆ. ಆದರೆ ABCD ಯು ಒಂದು ತ್ರಾಪಿಜ್ಯ ಎಂದು ಸಾಧಿಸಿ.
Solution 17.
Solution 18.
Solution 19.
R = 14 cm, r = 7 cm, θ = 40°
ಛಾಯಕೃತ ಭಾಗದ ವಿಸ್ತೀರ್ಣ
Solution 20.
ತ್ರಿಜ್ಯ = 4cm
ಕೇಂದ್ರ ಕೋನ = 180° – 70° = 110°
Solution 21.
Solution 22.
Solution 23.
Solution 24.
Solution 25.
ಗೋಪುರದ ಎತ್ತರ = AB = x ಆಗಿರಲಿ
ಕಟ್ಟಡದ ಎತ್ತರ = BC = 20 m
tan 45° = $$\frac { BC }{ CD }$$
1 = $$\frac { 20 }{ CD }$$
CD = 20 m
tan 60° = $$\frac { AC }{ CD }$$
⇒ √3 = $$\frac { x+20 }{ 20 }$$
⇒ 20√3 = x + 20
⇒ x + 20 = 20√3
⇒ x = 20(√3 – 1)
ಗೋಪುರದ ಎತ್ತರ = 20(√3 – 1) m
Solution 26.
ಕಟ್ಟಡ ಎತ್ತರ = AB = 50√3 m
ಕಟ್ಟಡದಿಂದ ಕಾರಿಗಿರುವ ದೂರ BC ಆಗಿರಲಿ
tan 60° = $$\frac { AB }{ BC }$$
⇒ √3 = $$\frac { 50\surd 3 }{ BC }$$
⇒ BC = 50m
ಕಟ್ಟಡದಿಂದ ಕಾರಿಗಿರುವ ದೂರ = 50m
Solution 27.
Solution 28.
ಫಲಿತಗಳ ಸಂಖ್ಯೆ = n(S) = 6 {1, 2, 3, 4, 5, 6}
(i) ಒಂದು ಅವಿಭಾಜ್ಯ ಸಂಖ್ಯೆ = A = {2, 3, 5}
n(A) = 3
ಸಂಭವನೀಯ = P(A) = $$\frac { n(A) }{ n(S) }$$ = $$\frac { 3 }{ 6 }$$
(ii) 1 ಮತ್ತು 5ರ ನಡುವಿನ ಒಂದು ಸಂಖ್ಯೆ = B = {2, 3, 4}
n(B) = 3
ಸಂಭವನೀಯತೆ = P(B) = $$\frac { n(A) }{ n(S) }$$ = $$\frac { 3 }{ 6 }$$
Solution 29.
Solution 30.
3 + 2√5 ಒಂದು ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆಯಾಗಿರಲಿ ಎಂದು ಊಹಿಸೋಣ.
ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ = ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ
√5 ಒಂದು ಅಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ
ನಮ್ಮ ಊಹೆ 3 + 2√5 ಒಂದು ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ ಎಂಬುದು ತಪ್ಪು
3 + 2√5 ಒಂದು ಅಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ
IV.
Solution 31.
ದತ್ತ: O ವೃತ್ತಕೇಂದ್ರ PA ಮತ್ತು PB ಗಳ ಬಾಹ್ಯಬಿಂದು P ನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು.
ಸಾಧನೀಯ: PA = PB
ಸಾಧನೆ: ಹೇಳಿಕೆಗಳು ಕಾರಣಗಳು
ΔPOA & ΔPOB ಗಳಲ್ಲಿ
OA = OB
∠PAO = ∠PBO = 90°
OP = OP
ΔPOA = ΔPOB
PA = PB
ಒಂದೇ ವೃತ್ತದ ತ್ರಿಜ್ಯಗಳು ತ್ರಿಜ್ಯ ಮತ್ತು ಸ್ಪರ್ಶಕದ ನಡುವಿನ ಕೋನ ಉಭಯಸಾಮಾನ್ಯ ಲಂ.ವಿ.ಬಾ. ಜ್ವಸಿದ್ಧಾಂತ ಸರ್ವಸಮ ತ್ರಿಭುಜದ ಅನುರೂಪ ಭಾಗಗಳು.
ಅಥವಾ
ΔPOB ನಲ್ಲಿ
PB2 = OB2 – OP2
⇒ PB2 = 52 – 32 = 25 – 9 = 16
⇒ PB2 = 16
⇒ PB = 4
AB = AP + PB
⇒ AB = 4 + 4 (∴ AP = PB)
⇒ AB = 8 cm ಜ್ಯಾದ ಉದ್ದ 8 cm
Solution 32.
∠A = 45°, ∠B = 105°, BC = 7 cm
∠C = 180° – 150° = 30°
ರಚಿಸಬೇಕಾದ ತ್ರಿಭುಜ A’BC’
Solution 33.
ರೆಹಮಾನನ ಈಗಿನ ವಯಸ್ಸು x ಆಗಿರಲಿ
ಮೂರು ವರ್ಷಗಳ ಹಿಂದೆ ಅವನ ವಯಸ್ಸು = x – 3 ವರ್ಷ
5 ವರ್ಷಗಳ ನಂತರ ಅವನ ವಯಸ್ಸು = x + 5 ವರ್ಷ
⇒ x2 – 4x – 21 = 0
⇒ x2 – 7x + 3x – 21 = 0
⇒ x (x – 7) + 3 (x – 7) = 0
⇒ (x – 7) (x + 3) = 0
⇒ x – 7 = 0 ಅಥವಾ x + 3 = 0
⇒ x = 7 ಅಥವಾ x = -3
ರೆಹಮಾನನ ಈಗಿನ ವಯಸು 7 ವರ್ಷ
ಅಥವಾ
ರೈಲಿನ ಜನ x km/hr, ಆಗಿರಲಿ
ದೂರ = d = 360 km/hr
Solution 34.
Solution 35.
Solution 36.
V.
Solution 37.
Solution 38.
ದತ್ತ: ΔABC ನಲ್ಲಿ ∠BAC = 90°
ಸಾಧನೀಯ: BC2 = AB2 + AC2
ರಚನೆ: AD ⊥ BC ಗೆ ಎಳೆಯಿರಿ,
ಸಾಧನೆ: ಹೇಳಿಕೆಗಳು ಕಾರಣಗಳು
ΔBAC & ΔBDA ಗಳಲ್ಲಿ
∠ABC = ∠ABD ಉಭಯಸಾಮಾನ್ಯ
∠BAC = ∠BDA = 90° ದತ್ತ ಮತ್ತು ರಚನೆ
ΔBAC ~ ΔBDA ಕೊ .ಕೋ .ನಿ.ಗು.
$$\frac { BA }{ BD }$$ = $$\frac { BC }{ BA }$$
AB2 = BD.BC …….. (1)
∠ACB = ∠ACD ಉಭಯಸಾಮಾನ್ಯ
∠BAC = ∠ADC = 90° ದತ್ತ ಮತ್ತು ರಚನೆ
ΔBAC ~ ΔADC ಕೋ .ಕೋ .ನಿ.ಗು.
$$\frac { AC }{ DC }$$ = $$\frac { BC }{ AC }$$
AC2 = DC × BC ……. (2)
AB2 + AC2 = BD.BC + DC.BC
(1) + (2) ರಿಂದ
BC(BD + DC) = BC × BC
AB2 + AC2 = BC2
Solution 39.
Solution 40.
## Karnataka SSLC Maths Model Question Paper 2 Kannada Medium
Karnataka SSLC Maths Model Question Paper 2 Kannada Medium
## Karnataka SSLC Maths Model Question Paper 2 Kannada Medium
ವಿಷಯ : ಗಣಿತ
ಸಮಯ: 3 ಗಂಟೆಗಳು
ಗರಿಷ್ಠ ಅಂಕಗಳು: 80
I. ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಅಥವಾ ಅಪೂರ್ಣ ಹೇಳಿಕೆಗೆ ನಾಲ್ಕು ಪರ್ಯಾಯ ಉತ್ತರಗಳನ್ನು ನೀಡಲಾಗಿದೆ. ಇವುಗಳಲ್ಲಿ ಸೂಕ್ತವಾದ ಉತ್ತರವನ್ನು ಆರಿಸಿ, ಕ್ರಮಾಕ್ಷರದೊಡನೆ ಪೂರ್ಣ ಉತ್ತರವನ್ನು ಬರೆಯಿರಿ. (8 × 1 = 8)
Question 1.
ಚಿತ್ರದಲ್ಲಿ QE = 7.2 cm, PF = 1.8 cm, FR = 5.4cm ಆದರೆ PE ಯು
(A) 2 cm
(B) 2.4 cm
(C) 2.8 cm
(D) 3.2 cm
Question 2.
‘r’ ವೃತ್ತ ತ್ರಿಜ್ಯ ಹಾಗೂ 60° ಕೋನವನ್ನು ಹೊಂದಿರುವ ತ್ರಿಜ್ಯಾಂತರ ಖಂಡದ ವಿಸ್ತೀರ್ಣ
Question 3.
P (-2, -1), ರಲ್ಲಿ ಲಂಬದೂರವು
(A) -2
(B) -1
(C) 1
(D) 2
Question 4.
‘0’ ಮೂಲಬಿಂದು ಮತ್ತು P (4, 3) ರ ನಡುವಿನ ದೂರ (ಏಕಮಾನಗಳಲ್ಲಿ) OP = …….
(A) 5
(B) 4
(C) 3
(D) 2
Question 5.
P (x) = 5x – 10 ರ ಶೂನ್ಯತೆ.
(A) 2
(B) -2
(C) 5
(D) -5
Question 6.
x + $$\frac { 2 }{ x }$$ = 3 ಸಮೀಕರಣದ ಆದರ್ಶರೂಪ
(A) x2 + 2x – 3 = 0
(B) x2 + 3x + 2 = 0
(C) x2 – 3x + 2 = 0
(D) x2 – 2x + 3 = 0
Question 7.
ಎರಡು ನಾಣ್ಯಗಳನ್ನು ಏಕಕಾಲಕ್ಕೆ ಚಿಮ್ಮಿದಾದ, ಶಿರ H ಮತ್ತು ಪ್ರಚ್ಛ T ಇರುವಂತೆ ಸಾಧ್ಯ ಫಲಿತಗಳು
(A) {T, H, H, T}
(B) {TT, HH, HT, TH}
(C) {T, H}
(D) {TT, HH}
Question 8.
ಸಿಲಿಂಡರ್ನ ಎತ್ತರ ‘h’ ಮತ್ತು ಪಾದದ ತ್ರಿಜ್ಯ ‘r’ ಆದಾಗ ಸಿಲಿಂಡರ್ನ ಪಾರ್ಶ್ವ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ
(A) 2π (r + h)
(B) 2πr (r + h)
(C) 2πrh
(D) $$\frac { 2\pi r }{ h }$$
II. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 1 = 6)
Question 9.
ಒಂದು 5m ಎತ್ತರದ ಏಣಿಯನ್ನು ನೆಲದ ಮೇಲಿನಿಂದ 4 m ಎತ್ತರದ ಕಿಟಕಿಯನ್ನು ತಲುಪುವಂತೆ ಗೋಡೆಗೆ ಒರಗಿಸಿದೆ. ಗೋಡೆಯ ಪಾದದಿಂದ ಏಣಿಯ ಪಾದಕ್ಕಿರುವ ದೂರ ಲೆಕ್ಕಿಸಿ.
Question 10.
ಒಂದು ಬಿಂದುವಿನಿಂದ ವೃತ್ತದ ಮೇಲಿನ ಬಿಂದುವಿಗೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳ ಸಂಖ್ಯೆ ಎಷ್ಟು?
Question 11.
‘ಭಾಗ ಪ್ರಮಾಣ ಸೂತ್ರ’ ವನ್ನು ಬರೆಯಿರಿ.
Question 12.
ಯೂಕ್ಲಿಡ್ನ ಭಾಗಾಕಾರ ಅನುಪ್ರವೇಯವನ್ನು ತಿಳಿಸಿ.
Question 13.
ಘನಪದೋಕ್ತಿಯು ಹೊಂದಿರಬಹುದಾದ ಗರಿಷ್ಟ ಶೂನ್ಯತೆಗಳ ಸಂಖ್ಯೆ ಎಷ್ಟು?
Question 14.
ಎರಡು ಸಂಖ್ಯೆಗಳ ಮೊತ್ತ 27 ಮತ್ತು ಗುಣಲಬ್ಧ 182 ಆದರೆ ಆ ಸಂಖ್ಯೆಗಳು ಯಾವುವು?
III. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (16 × 2 = 32)
Question 15.
ಮೊತ್ತ ಕಂಡುಹಿಡಿಯಿರಿ: 34 + 32 + 30 + …………… + 10,
Question 16.
ಒಂದು ಏಣಿಯ ಪಾದವು ನೆಲದ ಮೇಲೆ ಗೋಡೆಯಿಂದ 2.5m ದೂರದಲ್ಲಿ ಹಾಗೂ ಅದರ ತುದಿಯು ನೆಲದ ಮೇಲಿಂದ 6 m ಎತ್ತರದಲ್ಲಿರುವ ಕಿಟಕಿಯನ್ನು ಮುಟ್ಟುವಂತೆ ಏಣಿಯನ್ನು ಗೋಡೆಗೆ ಒರಗಿಸಿ ಇಡಲಾಗಿದೆ. ಏಣಿಯ ಉದ್ದವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 17.
ಒಂದು ಎರವಲು ಗ್ರಂಥಾಲಯದಲ್ಲಿ ಮೊದಲ 3 ದಿನಕ್ಕೆ ಒಂದು ನಿಗದಿತ ಶುಲ್ಕವಿರುತ್ತದೆ. ಆ ದಿನದ ಪ್ರತಿಯೊಂದೂ ದಿನಕ್ಕೂ ಒಂದು ಹೆಚ್ಚುವರಿ ಶುಲ್ಕವಿರುತ್ತದೆ. ಪುಸ್ತಕವನ್ನು 7 ದಿನ ತನ್ನಲ್ಲಿ ಇರಿಸಿಕೊಂಡಿದ್ದಕ್ಕಾಗಿ ಸರಿತಾ ₹ 27 ನ್ನು ಪಾವತಿಸಿದರೆ, ಪುಸ್ತಕವನ್ನು 5 ದಿನ ಇರಿಸಿಕೊಂಡಿದ್ದಕ್ಕಾಗಿ ಸೂಸಿ ₹ 21 ನ್ನು ಪಾವತಿಸಿದಳು. ನಿಗದಿತ ಶುಲ್ಕ ಮತ್ತು ಪ್ರತಿಯೊಂದು ಹೆಚ್ಚುವರಿ ದಿನದ ಶುಲ್ಕವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 18.
ನೀರಿನ ಒಳಭಾಗದಲ್ಲಿರುವ ಬಂಡೆಗಳ ಬಗ್ಗೆ ಎಚ್ಚರಿಸಲು ಒಂದು ದೀಪಸ್ಥಂಭವು 80° ಕೋನವಿರುವ ತ್ರಿಜ್ಯಾಂತರ ಖಂಡದಲ್ಲಿ 16.5 km ದೂರಕ್ಕೆ ಕೆಂಪು ಬೆಳಕನ್ನು ಹರಡುತ್ತದೆ. ಹಡಗುಗಳನ್ನು ಎಚ್ಚರಿಸುವ ಈ ಭಾಗದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಚಿತ್ರದಲ್ಲಿ ತೋರಿಸಿರುವಂತೆ 4 cm ಬಾಹುವುಳ್ಳ ಒಂದು ಚೌಕದ ಪ್ರತೀ ಮೂಲೆಯಲ್ಲಿ 1 cm ತ್ರಿಜ್ಯವಿರುವ ವೃತ್ತ ಚತುರ್ಥಕವನ್ನು ಮತ್ತು 2 cm ವ್ಯಾಸವಿರುವ ಒಂದು ವೃತ್ತವನ್ನು ಕತ್ತರಿಸಿದೆ. ಚೌಕದ ಉಳಿದ ಭಾಗದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 19.
ಪಾದ 8 cm ಮತ್ತು ಎತ್ತರ 4 cm ಇರುವ ಒಂದು ಸಮದ್ವಿಬಾಹು ತ್ರಿಭುಜವನ್ನು ರಚಿಸಿ, ನಂತರ ಮತ್ತೊಂದು ತ್ರಿಭುಜವನ್ನು, ಅದರ ಬಾಹುಗಳು ಮೊದಲು ರಚಿಸಿದ ಸಮದ್ವಿಬಾಹು ತ್ರಿಭುಜದ ಅನುರೂಪ ಬಾಹುಗಳ 1$$\frac { 1 }{ 2 }$$ ರಷ್ಟಿರುವಂತೆ ರಚಿಸಿ,
Question 20.
(4, -3) ಮತ್ತು (8, 5) ಬಿಂದುಗಳನ್ನು ಸೇರಿಸುವ ರೇಖಾಖಂಡವನ್ನು ಆಂತರಿಕವಾಗಿ 3 : 1 ಅನುಪಾತದಲ್ಲಿ ವಿಭಾಗಿಸುವ ಬಿಂದುವಿನ ನಿರ್ದೇಶಾಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 21.
ಶೃಂಗಗಳು ಈ ಕೆಳಗಿನಂತಿರುವ ತ್ರಿಭುಜಗಳ ವಿಸ್ತೀರ್ಣಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ. (5, -1), (3, -5), (5, 2)
Question 22.
√5 ಒಂದು ಅಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ ಎಂದು ಸಾಧಿಸಿ.
Question 23.
(1, 1) ನ್ನು ಕ್ರಮವಾಗಿ ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ ಹಾಗೂ ಗುಣಲಬ್ದವಾಗಿ ಹೊಂದಿರುವ ವರ್ಗಬಹುಪದೋಕ್ತಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 24.
ಮೂಲಗಳ ಸ್ವಭಾವವನ್ನು ವಿವೇಚಿಸಿ, ವಾಸ್ತವ ಮೂಲಗಳಿದ್ದಲ್ಲಿ, ಅವುಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ: 2x2 – 3x + 5 = 0
Question 25.
1.6m ಎತ್ತರದ ಪ್ರತಿಮೆಯೊಂದನ್ನು ಒಂದು ಪೀಠದ ಮೇಲ್ಬಾಗದಲ್ಲಿ ಇರಿಸಲಾಗಿದೆ. ನೆಲದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ ಪ್ರತಿಮೆಯ ಮೇಲಿನ ಉನ್ನತ ಕೋನವು 60° ಮತ್ತು ಅದೇ ಬಿಂದುವಿನಿಂದ ಪೀಠದ ಮೇಲ್ತುದಿಯ ಉನ್ನತ ಕೋನವು 45° ಇದ್ದರೆ, ಪೀಠದ ಎತ್ತರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 26.
10 ಪಂದ್ಯಗಳಲ್ಲಿ ಒಬ್ಬ ಬೌಲರ್ನು ಪಡೆದ ವಿಕೆಟ್ಗಳ ಸಂಖ್ಯೆಯು ಈ ಕೆಳಗಿನಂತಿದೆ.
Question 27.
ಒಂದು ದಾಳವನ್ನು ಒಂದು ಸಲ ಎಸೆಯಲಾಗಿದೆ. 2 ಮತ್ತು 6ರ ನಡುವೆ ಒಂದು ಸಂಖ್ಯೆಯನ್ನು ಪಡೆಯುವ ಸಂಭವನೀಯತೆಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 28.
ಒಂದು ಪಾತ್ರೆಯ ಆಕಾರವು ಟೊಳ್ಳಾದ ಸಿಲಿಂಡರಿನ ಒಂದು ಪಾದದ ಮೇಲೆ ಟೊಳ್ಳಾದ ಅರ್ಧಗೋಳಾಕೃತಿಯನ್ನು ಕೂಡಿಸಿ ಮಾಡಿದೆ. ಅರ್ಧಗೋಳದ ವ್ಯಾಸವು 14 cm ಮತ್ತು ಪಾತ್ರೆಯ ಒಟ್ಟು ಎತ್ತರ 13 cm ಇದೆ. ಈ ಪಾತ್ರೆಯ ಒಳಮೇ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಘನ ಸಿಲಿಂಡರಿನ ಎತ್ತರ 2.4 m ಮತ್ತು ವ್ಯಾಸ 1.4 m ಇದೆ. ಇದರಿಂದ ಒಂದೇ ಎತ್ತರ ಮತ್ತು ಒಂದೇ ವ್ಯಾಸವನ್ನು ಹೊಂದಿರುವ ಶಂಕುವಿನಾಕಾರದ ಹಳ್ಳವನ್ನು ಕೊರೆದು ಟೊಳ್ಳಾಗಿಸಿದೆ. ನೂತನ ಘನದ ಒಟ್ಟು ಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಅತ್ಯಂತ ಸಮೀಪದ ಬೆಲೆಗೆ cm2 ನಲ್ಲಿ ಕಂಡುಹಿಡಿಯಿರಿ.
Question 29.
$$\frac { 3 }{ 2 }$$ x + $$\frac { 5 }{ 3 }$$ y = 7: 9x – 10y = 14 ಈ ರೇಖಾತ್ಮಕ ಸಮೀಕರಣಗಳಲ್ಲಿ $$\frac { { a }_{ 1 } }{ { a }_{ 2 } }$$, $$\frac { { b }_{ 1 } }{ { b }_{ 2 } }$$ ಮತ್ತು $$\frac { { c }_{ 1 } }{ { c }_{ 2 } }$$ ಅನುಪಾತಗಳನ್ನು ಹೋಲಿಸುವ ಮೂಲಕ ಜೋಡಿಗಳು ಸ್ಥಿರವಾಗಿವೆಯೇ ಅಥವಾ ಅಸ್ಥಿರವಾಗಿವೆಯೇ ಎಂಬುದನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 30.
150 cm ಎತ್ತರವಿರುವ ಒಬ್ಬ ವ್ಯಕ್ತಿಯು ತನ್ನ ನೆರಳಿನ ತುದಿಯನ್ನು ಗಮನಿಸಿದಾಗ ಅದು ಅವನ ಪಾದದಿಂದ 150√3 cm ದೂರದಲ್ಲಿರುವುದು ಕಂಡುಬರುತ್ತದೆ. ಹಾಗಾದರೆ ಅವನ ನೋಟದಲ್ಲಿ ಉಂಟಾದ ಅವನತ ಕೋನವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
IV. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 3 = 18)
Question 31.
‘ವೃತ್ತದ ಮೇಲಿನ ಯಾವುದೇ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ಸ್ಪರ್ಶಕವು, ಸ್ಪರ್ಶ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ತ್ರಿಜ್ಯಕ್ಕೆ ಲಂಬವಾಗಿರುತ್ತದೆ’ ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
ಬಾಹ್ಯ ಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು ಸಮ ಎಂದು ಸಾಧಿಸಿ
Question 32.
BC = 6 cm, AB = 5cm ಮತ್ತು ∠ABC = 60° ಇರುವಂತೆ ABC ತ್ರಿಭುಜವನ್ನು ರಚಿಸಿ, ನಂತರ ಮತ್ತೊಂದು ತ್ರಿಭುಜವನ್ನು, ಅದರ ಬಾಹುಗಳು ತ್ರಿಭುಜ ABC ಯ ಅನುರೂಪ ಬಾಹುಗಳ $$\frac { 3 }{ 4 }$$ ರಷ್ಟಿರುವಂತೆ ರಚಿಸಿ.
Question 33.
ಮೂರು ವರ್ಷಗಳ ಹಿಂದೆ ರೆಹಮಾನನ ವಯಸ್ಸು (ವರ್ಷಗಳಲ್ಲಿ) ಮತ್ತು 5 ವರ್ಷಗಳ ನಂತರದ ಅವನ ವಯಸ್ಸು ಇವುಗಳ ವೃಶ್ಯಮಗಳ ಮೊತ್ತ $$\frac { 1 }{ 3 }$$ ಆದರೆ ಅವನ ಈಗಿನ ವಯಸ್ಸನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಮೂರು ಕ್ರಮಾನುಗತ ಸಂಖ್ಯೆಗಳಲ್ಲಿ, ಮೊದಲನೆಯ ವರ್ಗ ಮತ್ತು ಉಳಿದೆರಡು ಸಂಖ್ಯೆಗಳ ಗುಣಲಬ್ದಗಳ ಮೊತ್ತ 154 ಆಗಿದೆ. ಹಾಗಾದರೆ ಆ ಮೂರು ಸಂಖ್ಯೆಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 34.
ಈ ಕೆಳಗೆ ವ್ಯಾಖ್ಯಾನಿಸಲ್ಪಟ್ಟ ಹೇಳಿಕೆಗಳ ಕೋನಗಳು ಲಘಕೋನಗಳು. ಈ ಕೆಳಗಿನ ಸಮೀಕರಣಗಳನ್ನು ಸಾಧಿಸಿ.
ಅಥವಾ
Question 35.
ಒಂದು ಕಾರ್ಖಾನೆಯ 50 ಕೆಲಸಗಾರರ ದೈನಂದಿನ ಆದಾಯವನ್ನು ಕೆಳಗಿನ ವಿತರಣೆಯು ನೀಡುತ್ತಿದೆ.
ಮೇಲಿನ ವಿತರಣೆಯನ್ನು ಕಡಿಮೆ ಇರುವ ವಿಧಾನದ’ ಸಂಚಿತ ಆವೃತ್ತಿ ವಿತರಣೆಯಾಗಿ ಬದಲಾಯಿಸಿ ಮತ್ತು ಅದರ ಓಜೀವ್ ಎಳೆಯಿರಿ.
Question 36.
ಎರಡನೇ ಬಹುಪದೋಕ್ತಿಯನ್ನು ಮೊದಲನೇ ಬಹುಪದೋಕ್ತಿಯಿಂದ ಭಾಗಿಸಿ ಹಾಗೂ ಮೊದಲನೇ ಬಹುಪದೋಕ್ತಿಯು ಎರಡನೇ ಬಹುಪದೋಕ್ತಿಯ ಅಪವರ್ತನವಾಗಿದೆಯೇ ಎಂದು ಪರೀಕ್ಷಿಸಿ.
t2 – 3 ; 2t4 + 3t3 – 2t2 – 9t – 12
ಅಥವಾ
ಬಹುಪದೋಕ್ತಿ p(x) ನ್ನು ಬಹುಪದೋಕ್ತಿ g(x) ನಿಂದ ಭಾಗಿಸಿ, ಭಾಗಲಬ್ಧ ಮತ್ತು ಶೇಷವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
p(x) = x4 – 3x + 4x + 5
g(x) = x2 + 1 – x
V. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (4 × 4 = 16)
Question 37.
ಎರಡು ತ್ರಿಭುಜಗಳಲ್ಲಿ ಒಂದು ತ್ರಿಭುಜದ ಮೂರು ಬಾಹುಗಳು ಮತ್ತೊಂದು ತ್ರಿಭುಜದ ಮೂರು ಬಾಹುಗಳೊಡನೆ ಸಮಾನುಪಾತ ಹೊಂದಿದ್ದರೆ, ಅವುಗಳ ಅನುರೂಪ ಕೋನಗಳು ಸಮವಾಗಿರುತ್ತವೆ ಮತ್ತು ಅದರಿಂದಾಗಿ ಆ ಎರಡು ತ್ರಿಭುಜಗಳು ಸಮರೂಪಿಗಳಾಗಿರುತ್ತವೆ ಎಂದು ಸಾಧಿಸಿ.
Question 38.
ನಕ್ಷೆಯ ಮೂಲಕ ಬಿಡಿಸಿ: y = 2x + 1; x = 2y – 5
Question 39.
(4, -1) ಮತ್ತು (-2, -3) ಬಿಂದುಗಳನ್ನು ಸೇರಿಸುವ ರೇಖಾಖಂಡದ ಭಾಜಕ ಬಿಂದುಗಳ ನಿರ್ದೇಶಾಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 40.
ಒಂದು ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೊದಲ ಪದದ ವರ್ಗವು ಅದರ 8ನೇ ಪದಕ್ಕೆ ಸಮವಾಗಿದೆ ಹಾಗೂ 8ನೇ ಪದವು ನಾಲ್ಕನೇ ಪದಕ್ಕಿಂತ 24 ಹೆಚ್ಚಾಗಿದೆ. ಹಾಗಾದರೆ ಶ್ರೇಢಿಯ ಪದಗಳನ್ನು ಬರೆಯಿರಿ.
Solutions
I.
Solution 1.
(B) 2.4 cm
Solution 2.
(C) $$\frac { { \pi r }^{ 2 } }{ 6 }$$
Solution 3.
(B) -1
Solution 4.
(A) 5
Solution 5.
(A) 2
Solution 6.
(C) x2 – 3x + 2 = 0
Solution 7.
(B) {TT, HH, HT, TH}
Solution 8.
(C) 2πrh
II.
Solution 9.
ತ್ರಿವಳಿ: 5, 4, 3
ಏಣಿಯ ಪಾದದಿಂದ ಗೋಡೆಯ
ಪಾದಕ್ಕಿರುವ ದೂರ = 3 ಮೀ.
Solution 10.
ಒಂದು ಬಿಂದುವಿನಿಂದ ವೃತ್ತದ ಮೇಲಿನ ಬಿಂದುವಿಗೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳ ಸಂಖ್ಯೆ = 1
Solution 11.
(x1, y1) ಮತ್ತು (x2, y2) ಬಿಂದುಗಳಿಗೆ ಸಂಬಂಧಿಸಿದಂತೆ : ಭಾಗ ಪ್ರಮಾಣ ಸೂತ್ರ (m1 : m2 ಅನುಪಾತದಲ್ಲಿ)
Solution 12.
ಭಾಜ್ಯ = ಭಾಜಕ × ಭಾಗಲಬ್ದ + ಶೇಷ
a = b × q + r
Solution 13.
ಗರಿಷ್ಟ ಶೂನ್ಯತೆಗಳು = 3
Solution 14.
x + y = 27
13 + 14 = 27
xy = 182
13 × 14 = 182
ಮೊದಲ ಸಂಖ್ಯೆ = 13, ಎರಡನೆಯ ಸಂಖ್ಯೆ = 14
Solution 15.
a1 = 38, a16 = 73, a31 = ?
a16 = a + 15d
73 = 38 + 15d
73 = 38 + 15d
d = 3
a31 = a1 + 30d = 38 + 30(3) = 38 + 90 = 128
a31 =128
Solution 16.
AB ಯು ಏಣಿ, CA ಯು ಗೋಡೆ ಮತ್ತು ಕಿಟಕಿಯಾಗಿರಲಿ.
BC = 2.5m ಮತ್ತು CA = 6m
AB2 = BC2 + CA2 = (2.5)2 + (6)2 = 42.25
AB = 6.5
ಏಣಿಯ ಉದ್ದ ಆಗಿದೆ.
Solution 17.
ಮೊದಲ 3 ದಿನದ ಒಂದು ಶುಲ್ಕ = x
ಪ್ರತೀ ದಿನದ ಹೆಚ್ಚುವರಿ ಶುಲ್ಕ = y
x + 4y = 27
x + 2y = 21
2y = 6
y = 3
x + 2y = 21
x + 2(3) = 21
x = 21 – 6
x = 15
ಮೂರು ದಿನಗಳ ಶುಲ್ಕ = x = ₹ 15
ಹೆಚ್ಚುವರಿ ಪ್ರತೀ ದಿನದ ಶುಲ್ಕ = y = ₹ 13
Solution 18.
Solution 19.
Solution 20.
Solution 21.
Solution 22.
√5 ಒಂದು ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆಯಾಗಿರಲಿ.
√5 = $$\frac { p }{ q }$$ p, q ∈ I
p ಮತ್ತು q ಗಳು 1 ನ್ನು ಹೊರತುಪಡಿಸಿ ಬೇರೆ ಸಾಮಾನ್ಯ ಅಪವರ್ತನ ಹೊಂದಿದ್ದರೆ, ಸಾಮಾನ್ಯ ಅಪವರ್ತನದಿಂದ ಭಾಗಿಸಬಹುದು, ಆದ್ದರಿಂದ p ಮತ್ತು Q ಗಳು ಸಹ ಅವಿಭಾಜ್ಯಗಳೆಂದು ಭಾವಿಸೋಣ
p = q√5
p2 = 5q2 (ಎರಡು ಕಡೆ ವರ್ಗ ಮಾಡಿದಾಗ)
5, p2 ನ್ನು ಭಾಗಿಸುತ್ತದೆ…… (1)
5, p ಮತ್ತು q ಗಳ ಸಾಮಾನ್ಯ ಅಪವರ್ತನವಾಗಿದೆ.
ಏಕೆಂದರೆ, 5, q ನ್ನೂ ಸಹ ಭಾಗಿಸುತ್ತದೆ.
ಅಂದರೆ p = 5r
5q2 = 52r2
q = 5r2 ………(2)
p ಮತ್ತು q ಗಳು ಸಹ ಅವಿಭಾಜ್ಯಗಳು ಎಂಬ ಸತ್ಯಸಂಗತಿಗೆ ವಿರುದ್ಧವಾಗಿದೆ.
√5 ಒಂದು ಅವಿಭಾಜ್ಯ ಸಂಖ್ಯೆಯಾಗಿದೆ.
Solution 23.
ಶೂನ್ಯಗಳ ಮೊತ್ತ = α + β = 1
ಶೂನ್ಯಗಳ ಗುಣಲಬ್ದ = αβ = 1
ವರ್ಗಬಹುಪದೋಕ್ತಿ: x2 – (α + β) x + (αβ)
ಯಲ್ಲಿ ಆದೇಶಿಸಿದರೆ x2 – 1x + 1
(1, 1) ನ್ನು ಕ್ರಮವಾಗಿ ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ ಹಾಗೂ ಗುಣಲಬ್ದವಾಗಿ ಹೊಂದಿರುವ ಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Solution 24.
2x2 – 3x + 5 = 0
ax2 + bx + c = 0,
a = 2, b = -3, c = 5
ಶೋಧಕ: b2 – 4ac = 9 – 4(2)(5) = 9 – 40 = -31 < 0
ಮೂಲಗಳು ಸಂಮಿಶ್ರ ಸಂಖ್ಯೆಗಳಾಗಿವೆ.
Solution 25.
Solution 26.
ಆವೃತ್ತಿ ವಿತರಣಾ ಪಟ್ಟಿ
ಬೌಲರ್ ಗರಿಷ್ಠ ಪಂದ್ಯಗಳಾದ 3 ರಲ್ಲಿ ಪಡೆದ ವಿಕೆಟ್ಗಳ ಸಂಖ್ಯೆ 2 ಆಗಿದೆ. ಆದ್ದರಿಂದ ದತ್ತಾಂಶಗಳ ಬಹುಲಕ 2,
Solution 27.
ಫಲಿತಗಣ S = {1, 2, 3, 4, 5, 6}
n(S) = 6
2 & 6 ರ ನಡುವಿನ ಸಂಖ್ಯೆಗಳು A = {3, 4, 5}
n(A) = 3
2 & 6 ರ ನಡುವಿನ ಸಂಖ್ಯೆ ಪಡೆಯುವ ಸಂಭವನೀಯತೆ
P(A) = $$\frac { n(A) }{ n(S) }$$ = $$\frac { 3 }{ 6 }$$ = $$\frac { 1 }{ 2 }$$
Solution 28.
Solution 29.
Solution 30.
IV.
Solution 31.
ವೃತ್ತದ ಯಾವುದೇ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ಸ್ಪರ್ಶಕವು, ಸ್ಪರ್ಶ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ತ್ರಿಜ್ಯಕ್ಕೆ ಲಂಬವಾಗಿರುತ್ತದೆ’
ದತ್ತ: O ಕೇಂದ್ರವುಳ್ಳ ವೃತ್ತಕ್ಕೆ OP ತ್ರಿಜ್ಯ ಮತ್ತು XY ಸ್ಪರ್ಶಕ ಎಳೆದಿದೆ.
ಸಾಧನೀಯ: OP ಯು XY ಗೆ ಲಂಬವಾಗಿದೆ.
ರಚನೆ: P ಯನ್ನು ಹೊರತುಪಡಿಸಿ XY ಮೇಲೆ ಮತೋಂದು ಬಿಂದು Q ಗುರುತಿಸಿ, OQ ಸೇರಿಸಿ.
ಸಾಧನೆ: Q ಬಿಂದುವು ವೃತ್ತದ ಹೊರಭಾಗದಲ್ಲಿರಬೇಕು.
Q ಬಿಂದುವು ವೃತ್ತದ ಒಳಭಾಗದಲ್ಲಿದ್ದರೆ, XY ವೃತ್ತಕ್ಕೆ ಛೇದಕವಾಗುತ್ತದೆಯೇ ಹೊರತು ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಕವಾಗುವುದಿಲ್ಲ. ಆದ್ದರಿಂದ OQ ಇದು ವೃತ್ತದ ತ್ರಿಜ್ಯ OP ಗಿಂತ ಉದ್ದವಾಗಿದೆ. ಅಂದರೆ OQ > OP.
P ಬಿಂದುವನ್ನು ಹೊರತುಪಡಿಸಿ, XY ಮೇಲಿನ ಎಲ್ಲಾ ಬಿಂದುಗಳಿಗೂ ಇದು ಅನ್ವಯಿಸುವುದರಿಂದ, O ಬಿಂದುವಿನಿಂದ XYನ ಮೇಲಿನ ಇತರೆ ಬಿಂದುಗಳಿಗಿರುವ ದೂರಕ್ಕಿಂತ OP ಯು ಕನಿಷ್ಟ ಉದ್ದ ಹೊಂದಿದೆ
OP ಯು XY ಗೆ ಲಂಬವಾಗಿದೆ.
ಅಥವಾ
ಬಾಹ್ಯಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳ ಉದ್ದವು ಸಮ.
ದತ್ತ: 0ವೃತ್ತಕೇಂದ್ರ, P ಬಾಹ್ಯಬಿಂದು, PQ & PR ಗಳು ಬಾಹ್ಯಬಿಂದು Pನಿಂದ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು
ಸಾಧನೀಯ: PQ = PR
ರಚನೆ: OP, OR, OP ಸೇರಿಸಿ
ಸಾಧನೆ: ΔOQP ಮತ್ತು ΔORP ಗಳಲ್ಲಿ
OQ = OR (ತ್ರಿಜ್ಯಗಳು)
∠OQP = ∠ORP = 90°
OP = OP (ಉಭಯಸಾಮಾನ್ಯ)
ΔOQP = ΔORP
PQ = PR
Solution 32.
Solution 33.
ರೆಹಮಾನನ ಈಗಿನ ವಯಸ್ಸು = x ಆಗಿರಲಿ
3 ವರ್ಷಗಳ ಹಿಂದೆ, ಅವನ ವಯಸ್ಸು = x – 3
5 ವರ್ಷಗಳ ನಂತರ ಅವನ ವಯಸು = x + 5
Solution 34.
Solution 35.
Solution 36.
Solution 37.
Solution 38.
Solution 39.
Solution 40.
## Karnataka SSLC Maths Model Question Paper 1 with Answer in Kannada
Students can Download Karnataka SSLC Maths Model Question Paper 1 with Answers in Kannada, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka SSLC Maths Model Question Paper 5 with Answers
Students can Download Karnataka SSLC Maths Model Question Paper 5 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka State Syllabus SSLC Maths Model Question Paper 5 with Answers
Time: 3 Hours
Max Marks: 80
I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )
Question 1.
The pair of linear equations 3a + 4b = k, 9a + 12b = 6 have infinitely many solutions when,.
a) K = -2
b) K = 3
c) K = 2
d) K = -3
c) K = 2
Solution:
Question 2.
n2 – 1 is divisible by 8, if n is
a) Prime numbers
b) Odd integer
c) Even integer
d) Natural number
b) Odd integer
Solution:
n2 – 1
If n is an odd. integer, 1,3,5,
Ex: 12-1 = 1- 1= 0 divisible by 8.
32 – 1=9-1=8 divisible by 8.
52 – 1 = 25 – 1 = 24 divisible by 8.
Question 3.
$$\sqrt{1+\tan ^{2} \theta}$$ = , where 0<θ< 90°
a) secθ
b) cosec θ
c) cos θ
d) sin θ
a) secθ
Solution:
$$\sqrt{1+\tan ^{2} \theta}=\sqrt{\sec ^{2} \theta}=\sec \theta$$
Question 4.
If Q divides the line A(3, 5) and B (7,9) internally in the ratio 2:3, then the co ordinates of Q are.
c) $$\left(\frac{23}{5}, \frac{33}{5}\right)$$
Solution:
Question 5.
If Area of the sector OPRQ = $$\frac{5}{18}[ /latex] of Area of circle. Then the value of x a) 25° b) 50° c) 75° d) 100° Answer: d) 100° Solution: Question 6. If 1 + 2 + 3 + n terms = 28, then n is equal to a) 28 b) 7 c) 8 d) 56 Answer: b) 7 Solution: [latex]\frac{\mathrm{n}(\mathrm{n}+1)}{2}$$
n(n + 1) = 28 x 2 = 56
n(n + 1) = 7 x 8
∴ n = 7
Question 7.
If E1E2 E3…….E10 are the possible elementary events of a random experiment, then P(E1) + P(E2) + P(E3) + ………P(E10) is equal to
a) 0
b) 1
c) 2
d) 3
b) 1
Question 8.
If we express sec A in terms of sin A, then sec A is equal to
$$\frac{1}{\sqrt{1-\sin ^{2} A}}$$
Solution:
II. Answer the following Questions : ( 1 x 8 = 8 )
Question 9.
What is the $$\frac{p}{q}$$ form of 43.123456789 ?
$$\frac{43123456789}{999999999}$$
Question 10.
Write the quadratic equation formed by the roots √3 + √5 and √3 – √5
x2 -(α + β)x + αβ = 0
X2 – [3 + √5 + 3 – √5]x + (3 + √5)(3 – √5) = 0
x2 – [6] x + (9 – 5) = 0
x2 – 6 + 4 = 0
Question 11.
Find the value of $$\frac{\sin 26}{\sec 64}+\frac{\cos 26}{\csc 64}$$
Question 12.
What is the distance between the points p(cos0, sin0) and Q(sin0, -COS0)?
Question 13.
If a number “x” choosen at random from the numbers -2, -1, 0, 1, 2. What is the probability that x2 < 3 ?
Clearly “x” can take any of the five given values.
∴ n(x) = 5.
If x2 < 3, then x can take the values -1, 0, 1.
∴ This even A = ,{-1, 0, 1}
n(A) = 3.
P(A) = $$\frac{n(A)}{n(S)}=\frac{3}{5}$$
Question 14.
What is the Area of a circle whose perimeter is 44 cms.
2πr = 44
2 x $$\frac{22}{7}$$ x r = 44
r = $$\frac{44 \times 7}{22 \times 2}$$
r = 7
A = πr² = $$\frac{22}{7}$$ x 7 x 7 = 154cm2
Question 15.
A toy was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is h cm and base radius is r cms, find the total surface area of the toy.
Total surface Area of the toy = C.S.A of cylinder + 2 [Surface Area of Hemisphere]
= 2πrh + 2πr²
= 2 πr (h + r)
Question 16.
The circumference of a circle exceeds the diameter by 15cm. Find the raidus of the circle.
Circumference = 2r + 15
2πr = 2r + 15
2πr – 2r = 15
2 x $$\frac{22}{7}$$r – 2r = 15
Multiply by 7
2 x 22r – 14r = 105 .
30r = 105
r = 3.5
III. Answer the following : (2 x 8 = 16 )
Question 17.
Prove that √5 + √3 is an irrational number.
∴ √3 is rational, is a contradiction
∴ √5 + √3 is irrational.
Question 18.
AX and BY are perpendiculars to segment XY. If AO = 5cm, BO = 7cm and Area of ∆ AOX = 150 cm2, find the area of ∆ BOY.
In ∆ OYB and ∆ OXA,
We have ∠X = ∠Y
So, by AA – criterion of similarity, we have
∆ YOB ≅ ∆ XOA
OR
In the given figure, BD ⊥ AC. Prove that AB2 + CD2 = AD2 + BC2
Soln: In ∆ BDC,
∠ BDC = 90°
BC2 = BD2 + DC2 (By Pythagoras)
In ∆ BDA, ∠ APB = 90°
AB2 = AD2 + BD2 (By Pythagoras)
AB2 – BC2 = AD2 + BD2 – BD2 – DC2
AB2 + CD2 = AD2 + BC2
Hence proved.
Question 19.
If zeroes of the polynomial p(y) = y3 – 3y2+ y + 1 are a – b, a, a+ b. Find a and b.
Sum of the zeroes of p(y) = y3 – 3y2+ y + 1
(a-b)(a)(a + b)= $$\frac{-1}{1}$$ = -1
(a2 – b2) a = -1
(1 – b2) (1) = -1
– b2 = -1
-b2 = – 1 – 1
-b2 = -2
b2 = 2
b = ±√2
Question 20.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length.
Question 21.
Find the zeroes of the polynomial p(y) = y3 – 5y2 – 16y + 80. If zeroes are α, -α and β.
Sum of the zeroes
α + (-α) + β = 5
α – α + β = 5
β = 5
(α )(-α )(β) =-80
2β = -80
α2β = 80
α2(5) = 80
α = 80/5 = 16
α = √16
= ±4
Question 22.
Two pillars of equal height and on either side of a road, which is 100m wide. The angles of elevation of the top of the pillars are 60 and 30 at a point on the road between the pillars. Find the position of the point between the pillars and height of each pillars.
In ∆ PAB,
tab 60 = $$\frac{\mathrm{AP}}{\mathrm{AB}}$$
√3 = h/x
h = √3x
In ∆ BCQ,
tan 30 = $$\frac{\mathrm{CQ}}{\mathrm{BC}}$$
$$\frac{1}{\sqrt{3}}=\frac{h}{100-x}$$
h√3 -= 100 – x
(√3x)(√3x) = 100-x
4x = 100
x = 25
∴ h= √3 X 25 = 25√3
= 25 x 1.73 = 43.3
OR
The three metallic spheres of radii are in the ratio of 3 : 4 : 5 are melted to form a single solid sphere of radius 12cm. Find the radius of the three small metallic spheres.
Volume of the solid sphere = Sum of the volumes of each sphere
$$\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi\left(r_{1}^{3}+r_{2}^{3}+r_{3}^{3}\right)$$
4 x 123 = 4[(3x)3 + (4x)3 + (5x)3]
1728 = 27x3 + 64x3 + 125x3 1728 = 216 x3
x = $$\frac{1728}{216}$$
x3 = 8 x = 3√8 = 2
∴ R1 = 6, R2 8, R3 = 10
Question 23.
In the given fig. OACB is a quadrant of a circle with centre O and radius 3.5cm. If OD = 2cm, find the area of the shaded region
Question 24.
Find the volume of the largest right circular cone that can be cut of a cube of edge 7cm.
Given radius of the base of cone = $$\frac{7}{2}$$ cm
h = height of cone = 7cm
Volume of the cone = $$\frac{1}{3}$$πr²h
Volume = $$\frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 7$$
= $$\frac{22 \times 7 \times 7}{12}$$
= 89.83cm3
IV. Answer the following : ( 3 x 9 = 27 )
Question 25.
The sum of a two digit number and the number obtained by reversing the order of its digits is 165. If the digits differ by 3, find the number.
Let the digits at units and tens place of the given number be y and x respectively, then
Given number = 10x + y
Number obtained by reversing the order of the digits = 10y + x
(10x + y) + (10y + x) = 165
10x + y+ 10y + x = 165
(11x+11y=165) = 11 ’
x + y = 15 ……(1)
Digits differ by 3, if x > y
x – y = 3 ……..(2)
Solve (1) and (2)
Consider x + y =15
9 + y = 15
y = 15 – 9
y = 6
∴ The number is 96, if y >x, number is 69.
OR
Ten years ago Sudhir was twelve times as old as his son Raghav and ten years, hence, he will be twice as old as his son will be. Find their present ages.
Let the present ages of Sudhir and Raghav be x years and y years respectively. Ten years ago, Father ’s age = (x -10) years Son’s age = (y – 10) years,
x- 10= 12(y – 10)
=> x – 10 = 12y – 120
x – 12y = 120+ 10
x – i2y = – 110
x – 12y + 110 = 0. …….(1)
Ten years later, Father’s age = (x + 10) years
Son’s age = (y + 10)
x + 10 = 2(y + 10)
⇒ x+ 10 = 2y + 20
x – 2y = 20 – 10
x – 2y = 10
x – 2y – 10 = 0 ……(2)
Solve (1) and (2)
y = 12
x – 12y = 110
x – 12(12) = -110
x – 144 = -110
x = -110 + 144 = 34
∴ Present age of Sudhir = 34 years and present age of Raghav = 12 years.
Question 26.
The altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13cm, find the other two sides.
Let the base BC be = x cm
Then its height AB = (x – 7) cm
Hypotenuse AC = 13 cm
In ∆ABC, AC2 = AB2 + BC2
132 = (x – If + (x)2
169 = x2 + 49 – 14x + x2
169 = 2x2 – 14x + 49
2x2 – 14x + 49 – 169 = 0
2x2 – 14x-120 = 0
x2 – 7x – 60 = 0
x2– 12x + 5x-60 = 0
x(x- 12) + 5(x – 12) = 0
(x-12) (x + 5) = 0
(x – 12) = 0 or x + 5 = 0
x = 12 or x = -5
The base of the triangle = 12 cm Length of the altitude = (12 – 7)cm = 5 cm.
Question 27.
X(2, 5), Y (-1, 2) and Z(5, 8) are the co – ordinates of the vertices of A XYZ. The point A and B lie on XY and XZ respectively such that XA : AY = XB: BZ =1:2. Calculate the co – ordinates of X and Y.
B(X2 y2)= [3, 6]
∴ Co – ordinates of A = (1,4)
Co – ordinates of B = (3, 6)
OR
Find the Area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are (0, -1) (2, 1) and
Co – ordinates of A
$$\frac{0+2}{2}, \frac{-1+1}{2}$$
= [1,0]
Co – ordinates of B
$$\frac{0+0}{2}, \frac{-1+3}{2}$$
= [0,1]
Co – ordinates of B
$$\frac{2+0}{2}, \frac{1+3}{2}$$
= [1,2]
∴ Area of A ABC = $$\frac{1}{2}$$ Σ x1 (y2 – y3)
= $$\frac{1}{2}$$ [1(1-2) + 0(2 – 0) + 1(0-1)]
= $$\frac{1}{2}$$ [-1 + 0 + 1]
∴ Area of ∆ ABC = 1 sq. unit
Question 28.
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contacts.
Data : AB is the tangent drawn to a circle centered at O
‘C’ is the point of contact.
To prove that :
OC ⊥AB
Construction : Mark D .on AB. Join OD.
Let it cut the circle at E
Proof: OE < OD
OE is the radius of the circle.
OE = OC (radii of the same circle)
⇒ OC < OD
⇒ OC is the shortest distance between centre of the circle and tangent AB
∴ OC ⊥ AB
Hence proved.
Question 29.
A hollow sphere of internal and external radii are 6cm and Scm respectively is melted and recast into small cones of base radius 2cm and height 8cm. Find the number of cones.
Inner radius of a hollow sphere (r) = 6cm
and outer radius (R) = 8cms.
Radius of one small cone (r,) 2cm and
heigh (h) = 8 cm.
OR
A medicine capsule is in the shape of a cylinder with two hemispheres stuch to each of its ends. The length of the entire capsule is 14mm and the diameter of the capsule is 5 mm. Find its surface area.
Let r mm be the radius and h mm be the height of the cylinder.
r = $$\frac{5}{2}$$ mm = 2.5mm
h = (14 – 2 x $$\frac{5}{2}$$ mm
= (14-5) mm
= 9 mm
The radius of hemisphere = r = $$\frac{5}{2}$$ mm
Now, the surface area of the capsule. = Curved surface Area of the cylinder surface Area of two hemispheres.
= 2πrh + 2 x 2 πr²
= 2πr² (h + 2r)
Question 30.
Solve graphically
y = $$\frac{1}{2}$$ x and 3x + 4y – 20 = 0
y = $$\frac{1}{2}$$ x
3x + 4y – 20 = 0
3x + 4y = 20
4y = 20 – 3x
Question 31.
The following distribution gives the daily income of 65 workers of a factory. Convert this into less than type cumulative frequency distribution and draw its ogive.
Question 32.
The fourth term of an AP is 11, and 8th term exceeds twice the fourth term by 5. Find AP and find the sum of first 100 terms.
Given : T4 = -11 and T8 = 2T4 + 5
T8 = 2(11) + 5
a + 7d = 22 + 5 = 27
d = 4
a + 7d = 27
a + 7(4) = 27
a+ 28 = 27
a = 27-28 = -1
a = -1
A.P = -1,3, 7, …..
Sn = $$\frac{n}{2}$$ [2a + (n-1)d] = $$\frac{100}{2}$$ [2(-1) + (99)4]
= 50 [-2 + 396]
= 50 [394]
= 19700
Question 33.
A person on tour has ₹ 360 for his expenses. If he extends his tour for 4 days, he has to cut down his daily expenses by ₹ 3. Find the original duration of tour.
Let the original duration of the tour be x days.
Total expenditure on four = 360 ₹
Expenditure per day = $$\frac{360}{x}$$
Duration of the extended tour = (x +4) days
∴ Expenditure per day according to new schedule = $$\frac{360}{x+4}$$
It is given that daily expenses are cut down by 3₹
$$\frac{360}{x}-\frac{360}{x+4}$$ = 3
360 (x + 4) – 360x = 3(x) (x + 4)
360x + 1440 – 360x = 3x (x + 4)
1440 = 3x2+ 12x
3x2 + 12x – 1440 = 0
x2 + 4x – 480 = 0
x2 + 24x – 20x – 480 = 0
x(x + 24) – 20(x + 24) = 0
(x + 24) (x – 20) = 0
x + 24 = 0 and x – 20 = 0
x = -24 and x = 20
∴ Duration of tour = 20 days.
OR
Two pipes running together can fill a cistern in 3$$\frac{1}{13}$$ minutes. If one pipe
takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.
Let the faster pipe takes x minutes.
Let the smaller pipe take (x + 3) minutes to fill.
Portion of the cistern filled by the faster 1
pipe in one minute = $$\frac{1}{x}$$
⇒ Portion of the cisterin filled by the faster pipe in $$\frac{40}{13}$$ minutes
|||ly portion of the cistern filled by the slower pipe in $$\frac{40}{13}$$ minutes.
It is given that the cistern is filled in $$\frac{40}{13}$$ min.
[40(x + 3) + 40x] = [(x) (x +3)] 13
40x + 120 + 40x = (x2 + 3x) 13
80x + 120= 13x2 + 39x
13x2 + 39x – 80x – 120 = 0
13x2 – 41x-120 = 0
13x2 – 65x + 24x – 120 = 0
13x (x – 5) + 24(x – 5) = 0
x – 5 = 0, 13x + 24 = 0
x = 5, 13x = -24, x = $$\frac{-24}{13}$$
Faster pipe fills the cistern in 5 minutes and the slower pipe takes 8 min to fill the cistern.
V. Answer the following : ( 4 x 4 = 16 )
Question 34.
In an AP whose first term is 2, the sum of first five terms is one fourth the sum of the next five terms. Show that T20 = -112. Find S20.
a = 2, T1 + T2 + T3 + T4+ T5
= $$\frac{1}{4}$$ (T6 + T7 + T8 + T9 + T10)
a + a + d + a + 2d + a + 3d + a + 4d
= $$\frac{1}{4}$$(a + 5d + a + 6d + a + 7d + a + 8d + a + 9d)
5a+ 10d = $$\frac{1}{4}$$ (5a + 35d)
5(a + 2d) = $$\frac{1}{4}$$ x 5( a + 7d)
a + 2d = $$\frac{1}{4}$$ (a + 7d)
4(a+2d) = a+7d
4a + 8d – a – 7d = 0
3a + d = 0
3(2) + d = 0
d = -6
T20 = a + 19d
= 2 +19(-6)
= 2 – 114
T20 = -112
Sn = $$\frac{n}{2}$$ [2a + (n-1)d]
Sn = $$\frac{20}{2}$$ [2(2) + (20-1)(-6)]
= 10 [4- 114]
= (10) (-110)
S20 = -1100.
OR
A man repays a loan of ₹ 3250 by paying 7 20 in the first month and then increases the payment by ₹ 15 every month. How long will it take to clear his loan?
Here, a = 20, c.d = 15, n = ?
Sn = $$\frac{n}{2}$$[2a + (n-1)d]
3250 = $$\frac{n}{2}$$ [2 x 20 + (n-1)15]
3250 x 2 = n(40 + 15n- 15)
6500 = n (25 +15n)
6500 = 25n + 15n2
(15n2 + 25n – 6500 = 0)÷ 5
3n2 + 5n – 1300 = 0
3n2-60n + 65n- 1300 = 0
3n(n – 20) + 65 (n – 20) = 0
(n – 20) (3n + 65) = 0
(n – 20) = 0, 3n + 65 = 0
n = 20, 3n = -65
n = $$\frac{-65}{3}$$
∴ Total amount will be paid in 20 months.
Question 35.
Find the mean, median and mode for * the following frequency distribution.
Question 36.
P.T $$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}$$ = 1+ secθ cosecθ
Question 37.
Draw a right triangle in which the sides (other than hypotenuse) arc of lengths 8cm and 6cm, then construct another triangle whose sides are 5/3 times the corresponding sides of given triangle.
Verify
BC = 8cm
BC’= $$\frac{5}{3} \times 8=\frac{40}{3}$$
AB =6cms
A’B= $$\frac{5}{3} \times 6$$
=10cms.
VI. Answer the following : (5 x 1 = 5 )
Question 38.
State and prove pythagoras theorem.
In a right angled triangle, square on the hypotenuse is equal to sum of A the squares on the other sides.
Data : ∆ ABC, ∠B = 90°
T.P.T : AC2 = AB2 + BC2
Construction : Draw BD ⊥ AC
Proof: In ∆ ABC and ∆ ABD
∠ ABC = ∠ADB = 90 [∵ Data & construction]
∠A = ∠A (∵ common Angle)
∠ACB =∠ ABD [ ∵ Re maining angle]
∴ ∆ ABC and ∆ ABD are equiangular
=> ∆ ABC ~ ∆ABD
AB2= AC x AD ……… (1)
In ∆ABC and ∆ BDC
∠ ABC = ∠ BDC = 90 [∵ Data & construction]
∠C = ∠C (∵ common Angle)
∠BAC =∠ DBC [ ∵ Re maining angle]
∴ ∆ ABC and ∆ DBC are equiangular
∆ABC ~ ∆BDC
AB2 + BC2 = AC (AD + DC)
AB2 + BC2 = AC(AC)
AB2 + BC2=AC2
AC2 = AB2+ BC2
## Karnataka SSLC Maths Model Question Paper 4 with Answers
Students can Download Karnataka SSLC Maths Model Question Paper 4 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka State Syllabus SSLC Maths Model Question Paper 4 with Answers
Time: 3 Hours
Max Marks: 80
I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )
Question 1.
If the H.C.F of 65 and 117 is expressible in the form of 65m – 117, then the value of m is
a) 4
b) 3
c) 11
d) 2
Solution:
The H.C.F. of 65 and 117 = 65m- 117
Now 13 = 65m- 117
65m 117+13
65rn = 130
m = 130/65 = 2
∴ The value of m = 2
∴ HCF(65, 117)
= 13
Question 2.
If sinx = sin 60° cos 30° – cos 60 sin 30!l, then the value of x is
a) 0°
b) 30°
c) 45°
d) 60°
b) 30°
Solution:
Sinx = sin 60°cos30 – cos 60 sin 30
Question 3.
The angle between the radius of a circle and the tangent drawn at the point of contact is
a) 0°
b) 60°
c) 90°
d) 30°
c) 90°
Question 4.
The T.S.A. of a cuboid of dimension, l = 30cm, b = 20cm, c = 10cm, is ______
a) 600cm2
b) 60cm2
c) 6000cm2
d) 2200 cm2
Question 5.
Which of the following is a polynomial
a) x2 – 5x + 3√x
b) x1/2 + x1/2 – x +1
c) $$\sqrt{x}-\frac{1}{\sqrt{x}}$$
d) x2 – 4x + √2
x2 – 4x + √2
Solution:
x2 – 4x + √2 because variable has no negative or decimal powers.
Question 6.
The value of p if x, 2x + p and 3x + 6 are in A.P.
a) p = 3
b) p = 2
c) p = 1
d) p = 0
Solution:
x, 2x + p, 3x + 6 are inAP a, A, b are A.P.
2(2x + p) = 2(2x + 3)
2x + p = 2x + 3
p = 2x + 3 – 2k
P = 3
Question 7.
In triangle PQR, The value of y is
a) 4√3
b) 6√3
c) 5√3
d) √3
b) 6√3
Solution:
QS2 = PS x SR
QS2 = $$\sqrt{9 \times 12}$$
QS = $$\sqrt{9 \times 4 \times 3}$$
QS = 3 x 2√3
QS = 6√3
Question 8.
When 2 unbiased coins are tossed at ‘ a time, the probability of getting 2 heads is _____
a) 1/4
b) 1/2
c) 1
d) 0
a) 1/4
Solution:
S = {(H, H) (H, T) (T, H) (T, T)}
n(s) = 4
An event of getting 2 heads = A={(H, H)}
∴ P(A) = $$\frac{n(A)}{n(S)}=\frac{1}{4}$$
II. Answer the following Questions : ( 1 x 8 = 8 )
Question 9.
If the product of zeros of polynomial f(y) = ay3 – 6y2 + 11y – 6 is 4 then find the value of ‘a’.
α β γ = $$-\frac{\mathrm{d}}{\mathrm{a}}$$
4 = $$\frac{-(-6)}{a}$$
4a = 6
a = $$\frac{6}{4}$$
a = $$\frac{3}{2}$$
Question 10.
What is the value of C, if ax2 + bx + c = 0 has equal roots?
If ax2 + bx + c = 0 has equal roots then,
b2 – 4ac = 0
b2 = 4ac
4ac = b2
c = $$\frac{b^{2}}{4 a}$$
Question 11.
Find the second term if sum of the ‘n’ term of an AP is 2n2 + 1.
Sn = 2n2 + 1
S1 = 2(1)2 + l= 2xl + l= 2+ l= 3
∴ S1 = T1 = a = 3
S2 = 2(2)2 + 1 =8+ 1 =9
a + T2 = 9
T2 = 9 – 3 = 6
Question 12.
State converse of Pythagoras Theorem.
“The square on the greatest side of a triangle is equal to the sum of the squares on the other two sides, then the other two sides contian the right angle”.
Question 13.
What is the form of $$\frac{p}{q}$$ (p,q ∈ z q ≠ 0) form of 0.57 ?
$$0.5 \overline{7}=\frac{57-5}{90}=\frac{52}{90}=\frac{26}{45}$$
Question 14.
If sinθ = $$\frac{1}{3}$$ then find the value of (cot2θ + 2)
AC2 = AB2 + BC2
BC2 = AC2 – AB2
= 32 – 12
= 9 – 1 = 8
BC = √8 = 2√2
cotθ = $$\frac{2 \sqrt{2}}{1}$$ = 2√2
2cot2θ + 2 = 2(2√2)2 + 2
= 2 x 8 + 2
= 16 + 2
= 18
Question 15.
In sin(A + B) = $$\frac{\sqrt{3}}{2}$$ and cos (A – B) = 1, 0< A + B < 90°, A ≥ B
sin(A + B) = $$\frac{\sqrt{3}}{2}$$ = sin60°
A + B = 60° …. (1)
cos (A – B) = 1 = cos0°
A – B = 0 ….(2)
Solve (1) and (2)
A + B = 60
A – B = 0
2A = 60 or A = 30°
A + B = 60
30 + B = 60
B = 60 – 30° = 30°
∴ A = 30°, B = 30°
Question 16.
The surface area of a sphere is same as the C.S.A. of a right circular cylinder whose height and diameter are 4 cm each. Find the radius of the sphere.
Soln:
4πr² = 2πrh
4R² = dh .
4R²= 4 x 4
R² = $$\frac{4 \times 4}{4}$$ = 4
R = √4 = 2cm
A = 30°, B = 30°
∴ Radius of the sphere = 2cm.
III. Answer the following : ( 2 x 8 = 16 )
Question 17.
By Euclid’s division lemma, show that thte square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Let x be any positive integer. Then it is of the form 3q, 3q+1 or 3q + 2.
Now, (3q)2 = 9q2 = 3(3q2) = 3m where m = 3q2
(3q + 1)2 = 9q2 + 6q+ 1 = 3(3q2 + 2q) + 1
= 3m + 1, where m = 3q2 + 2q and (3q+2)2 = 9q2 + 12q + 4
= 3(3q2 + 4q + 1) +1
=3m +1, where m = 3q2 + 4q + 1
Hence, it is proved.
Question 18.
Solve : $$\frac{x+y}{x y}$$ = 2 and $$\frac{x-y}{x y}$$ = 6
x + y = 2xy
x – y = 6xy
2x = 8xy
2 = 8y
Question 19.
Solve : y2 – (√3 + 1)y + √3 = 0
y2 – √3y – y + √3 = 0
y(y – √3) – 1(y – √3) = 0
(y – √3)(y – 1) = 0
y – √3 = 0, y – 1 = 0
y = √3 and y = 1
Question 20.
Show that the points (3, 2) (-2, -3) and (2, 3) are collinear Or non – collinear.
Let A= (3, 2) B = (-2, -3) C = (2, 3)
AC ≠ AB + BC
∴ They are non – collinear.
Question 21.
In the given fig ∆DGH ~ ∆DEF, DH = 8cm, DF = 12cm, DG = (3x – l)cm and DE = (4x +. 2)cm, Find the lengths of DG and DE.
∆DGH ~ ∆DEF
$$\frac{\mathrm{DG}}{\mathrm{DE}}=\frac{\mathrm{DH}}{\mathrm{DF}}$$
$$\frac{3 x-1}{4 x+2}=\frac{8}{12}$$
12 (3x – 1) = 8(4x + 2)
36x – 12 = 32x + 16
36x – 32x = 16 + 12
4x = 28
x = 28/4
x = 7
∴ DG = 3x – 1 = 3 x 7 – 1 = 21 – 1 = 20
DE = 4x + 2 = 4 x 7 + 2 = 28 + 2 = 30
OR
D is a point on the side BC of A ABC such that ∠ADC = ∠BAC. Prove that
$$\frac{\mathbf{C A}}{\mathbf{C D}}=\frac{\mathbf{C B}}{\mathbf{C A}}$$
In ∆ ABC and ∆ DAC,
∠C = ∠C
∆ABC ~ ∆DAC
Question 22.
A card is drawn at random from a box containing 21 cards numbered 1 to 21. Find the probability that the card drawn is
a) Prime number b) Divisible by 3.
n(s) = 21
A= {2,3,5,7,11,13,17,19}
n(A) = 8
P(A) = $$\frac{n(A)}{n(S)}=\frac{8}{21}$$
b) n(S) = 21
B= {3,6, 9,12,15,18}
n(B) = 6
∴ P(B) = $$\frac{n(B)}{n(S)}=\frac{6}{21}=\frac{2}{7}$$
Question 23.
Draw a circle of radius 3cms. Construct a pair of tangents to it, from a point 5cm away from the circle.
r = 3cm d = 3 + 5 = 8cm
Question 24.
Express sinA and sec A in terms of cot A.
W.K.T cosec2A = 1 + cot2A
OR
If tanθ + sinθ = m and tan θ – sin θ= n, S.T m2 – n2 = 4$$\sqrt{\mathbf{m} \mathbf{n}}$$
L.H.S = m2 – n2 = (m +n) (m – n)
= {(tan θ + sin θ ) + (tan θ – sin θ )x(tan θ +sin θ) – (tan θ – sin θ)}
= {(2tanθ) x (2sinθ) = 4tanθsinθ
= 4$$\sqrt{\mathbf{m} \mathbf{n}}$$
= RHS.
IV. Answer the following : ( 3 x 9 = 27 )
Question 25.
The sum of the numerator and denominator of a fraction is 24. If 4 is subtracted from the numerator and 5 from its denominator, then it reduces to 1/4. Find the fraction.
Let the fraction be $$\frac{\mathrm{x}}{\mathrm{y}}$$
sum of its numerator and denominator = 24.
x + y = 24 ……..(1)
If 4 is subtracted from the numerator and
5 from its denominator we get = 1/4
⇒ $$\frac{x-4}{y-5}=\frac{1}{4}$$
4(x – 4) = 1(y – 5)
4x – 16 = y – 5
4x – y = -5 + 16
4x – y = 11 ……..(2)
From (1) and (2)
Consider,
x + y = 24
7 + y = 24
y = 24 – 7
y = 17
∴ The fraction is $$\frac{x}{y}=\frac{7}{17}$$
OR
Two women and five men can together finish an embroidary work in 4 days. While three women and 6 men can finish in 3 days. Find the time taken by one women alone and also that taken by one man alone.
Let 1 woman can finish the embroidary in x days and 1 man can finish the embroidary in y days.
1 woman’s 1 day’s work = $$\frac{1}{x}$$
1 man’s 1 day’s work = $$\frac{1}{y}$$
Let $$\frac{1}{x}$$ = u, $$\frac{1}{y}$$ = v
2u + 5v = $$\frac{1}{4}$$
3u + 6v = $$\frac{1}{3}$$
solve (1) and (2)
(8u + 20v = 1) x 9
(9u+ 18v= 1) x 8
v = $$\frac{1}{36}$$
Put v = $$\frac{1}{36}$$…….(1)
2u + 5v = $$\frac{1}{4}$$
2u + 5$$\frac{1}{36}$$ = $$\frac{1}{4}$$
Thus, 1 woman alone can finish the embroidery in 18 days and 1 man alone can finish it in 36 days.
Question 26.
If two zeroes of the polynomial p(x) = x4 – 6x3 – 26x2 + 138x – 35 are 2 – √3 and 2 + √3. Find the other two zeroes.
The two zeroes of p(x) are,
(2 – √3) and (2 + √3)
⇒ [x – (2 – √3] and [x – [2 + √3] are the factors p(x).
⇒ [ x – 2 + √3] [x – 2 – √3]
(x – 2)2 – (√3)2
x2 – 4x + 4 – 3
⇒ x2 – 4x + 1 is a factor of p(x)
Divide p(x) by x2 – 4x + 1 to get quotient
Consider the quotient, x2 – 2x – 35 = 0
x2 – 7x + 5x – 35 = 0
x(x – 7) + 5(x – 7) = 0
(x – 7) (x + 5) = 0
x-7 = 0 orx + 5 = 0
x = 7 or x = -5
∴ The other two zeroes are 7, -5.
Question 27.
A two digit number is such that the product of its digits is 18. when 63 is subtracted from the number, the digits interchange their places. Find the number.
Let the two digits be x and y.
⇒ The two digit number is = 10x + y
The product of three digits = xy = 18
When 63 is subtracted from the number, the digits interchange their places.
10x + y – 63 = 10y + x
10x – x – 10y + y = 63
9x – 9y = 63
9(x – y) = 63
x – y = 63/9 = 7
x – y = 7
x = 7 + y
Consider xy = 18 and substitute x = (y + 7) in it.
y(y + 7) = is . .
y2 + 7y – 18 = 0
y2 + 9y – 2y – 18 = 0
y (y + 9) – 2(y + 9) = 0 –
(y + 9) (y – 2) = 0 .
y + 9 = 0, y -2 = 0
y = -9 or y = 2
xy= 18
x(2) = 18
x = 18/2 = 9
x = 2
But negative digit is not considered.
∴ The two digit number is
= 10x + y
= 10(9) + 2 = 92
OR
A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away in time it has to increase its speed by 250 km / hr from its usual speed. Find its usual speed.
Let the usual speed of the plane be x km / hr. Time taken to cover 1500 km with the usual speed.
$$\frac{1500}{x}$$ hrs
Time taken to cover 1500 km with the speed of (x + 250) km / hr = $$\frac{1500}{x+250}$$
2(1500 x 250) = x2 + 250x
x2 + 250x-750000 = 0
x2 + 1000x – 750x – 750000 = 0
x(x + 1000) – 750(x + 1000) = 0
(x+ 1000)(x-750) = 0
x + 1000 = 0 or x – 750 = 0
x = -1000 or x= 750 ,
x = 750
Hence, the usual speed of the plane = 750 km/hr.
Question 28.
If the co – ordinates of the mid points of ∆ ABC are D(1, 2) E(0, -1) and (2, -1). Find the respective co – ordinates of ∆ ABC.
D is the midpoint of BC.
x1 + x2 = 4 and y1 + y2 = -2 ………(1)
E is the mid point of BC
x2 + x3 = 0 and y2 + y3 = -2 …….(2)
F is the mid point of BC
x1 + x3 = 4 and y2 + y3 = -2 ….(3)
Solve (1) and (3)
x2 = 0
x2 = 0
0 + x3 = 0
x3 = 0
x1 + x3 = 4
x1 + 0 = 4
x1 = 4 – 0 = 4
x1 = 4
y2 = -i
y2 + y3 = -2
-1 + y3 = -2
y3 = -2 +1
y3 = -1
y1 + y3 = -2
y1 + (-1) = -2
y1 -1 = -2
y1 = -2 + 1 = -1
y1 = -1
A(x1,y1) = A(4,-1)
B (x2 y2) = B(0, -1)
C(x3, y3) = C(0,-1)
OR
Find the length of the median through the vertex A(5,1) drawn to the triangle ABC where other two vertices are B(1, 5) andiC(-3, -1)
Question 29.
Prove that the tangents drawn from an external point are equal.
Data : ‘O is the centre of the circle PA and PB are the two tangents drawn from an external point P. OA and OB are radii of the circle.
To prove that : ∠APB + ∠AOB = 180°
∠OAP + ∠OBP = 90°
(∵ Angle between the radius and tangent at the point of contact is 90°)
OP = OP (Common side)
OA = OB (Radii of the same circle)
According to RHS postulate
∆APO = ∆BPO
∠ OAP = ∠OBP = 90°
∠OAP + ∠OBP = 90° +90° =180°
⇒ Opposite angles of OAPB quadrilateral are supplementary.
⇒ ∠ APB + ∠ AOB = 180°
Question 30.
If a chord of circle of radius 10cm subtend an angle of 60° at the centre of the circle. Find the area of the corresponding segment of the circle.
(Take π =3.14, √3 = 1.7)
O is the centre of the circle having radius 10 cm.
∠ AOB = 60° is angle subtended at the centre O of the circle by a chord AB.
Draw BL ⊥ OA
In right angled ∆ OLB,
Area of segment ∆ APB
= The area of sector OAPB – Area of ∆ OAB
OR
Find the area of the shaded region where PQRS is a square of side lOcms and semicircles are drawn with each side of square as diameter.
The four shaded regions are marked as I, II, III, IV.
All the four regions have equal areas due to § symmetry.
Area of I + Area of III
= Area of square ABCD – Sum of the areas of two semicircles of each of radius 5cm.
{10 x 10 – 2 x $$\frac{1}{2}$$ π x 52}cm2
= {100 – 25 x 3.14} cm2
= {100 – 78.5} cm2
= 21.5 cm2
Similarly Area of II + Area of IV = 21.5 cm2. Total Area of shaded region = 21.5 + 21.5
= 43 cm2
Question 31.
The following table gives the production yield per hectare of paddy of 50 farms of a village.
Question 32.
Find the mean of the following frequency distribution.
A = 50, h = 20, N = 100 and Σ fiui = 15
Mean = A + h$$\left\{\frac{1}{\mathrm{N}} \sum \mathrm{f}_{i} \mathrm{u}_{\mathrm{i}}\right\}$$
Mean = 50 + 20 x $$\frac{15}{100}$$
Mean = 50 + 3 = 53
Question 33.
Construct a triangle of sides 4cm, 5cm and 6cm and then a triangle similar to it whose Sides are 2/3 of the corresponding sides of it.
V. Answer the following : ( 4 x 4 = 16 )
Question 34.
Solve the pair of equations graphically.
4x – 3y + 4 = 0
4x + 3y – 20 = 0
4x – 3y + 4 = 0 ….. (1)
4x + 4 = 3y
y = $$\frac{4 x+4}{3}$$
4x + 3y – 20 = 0
3y = -4x + 20
y = $$\frac{-4 x+20}{3}$$
Question 35.
How many terms of the series 93 + 90 + 87 + ……….. amounts to 975. Find also the last term.
93 + 90 + 87 + …………
a = 93, d = 90 – 93 = -3, Sn = 975, n = ?,
T = ?
S = $$\frac{n}{2}$$[2a + (n – 1)d]
975 = $$\frac{n}{2}$$[2(93) + (n-1)(-3)]
975 x 2 = n[186 – 3n + 3]
1950 = n [189 – 3n]
1950 = n[189 – 3n]
1950 = 189n – 3n2
(3n2-189n+ 1950-0) H-3
n2 – 63n + 650 = 0
n2 – 50n-13n +650 = 0
n(n – 50) – 13(n – 50) = 0
(n – 50) (n -13) = 0
n – 50 = 0 or n – 13 = 0
n= 50 or n= 13
When n = 50, the last term
Tn = a + (n – 1)d
= 93 + (49) (-3)
= 93 – 147
T50 =-54
When n = 13, the last term
Tn = a + (n – 1)d ”
=93 + (13-1) (-3)
= 93 – 36
T13 = 57
OR
If m times the mth term of an A.P is equal to n times its n,h term, show that (m + n)th term is zero.
Soln: Let T1 = a and c.d = d
Given : m times m,h term = n times nthterm
=> mTm = nTn
m[a + (m – 1)d] = n[a + (n – 1)d]
m[a +(m – l)d] – n[a+(n -l)d] = 0
m[a + md – d] – n[a + nd – d] = 0
ma + m2d – md – na – n2d + nd = 0
a(m – n) + d(m – n) (m + n) – d(m – n) = 0
(m – n) {a + d(m +n) – d} = 0 . (m – n) {a + (m + n – l)d} = 0
m-n = 0, a + (m + n-l)d = 0
m= n Tm+n , = 0
But m ≠ n
Question 36.
A tower is 50m high. Its shadow is x mtrs shorter when the suns altitude is 45° than when it is 30°. Find the value of x.
Let PQ be the tower and let PA and PB be its shadows when the altitudes of the sum are 45° and 30° respectively.
∠ PAQ = 45°, ∠ PBQ = 30°
∠ BPQ = 90° and PQ = 50m ,
Let AB = x metres
From right angled ∆ APQ, we have
$$\frac{\mathrm{AP}}{PQ}$$ = cot 45°=1
$$\frac{\mathrm{AP}}{50}$$ = 1 => AP = 50m
From right angled ∆ BPQ,
$$\frac{\mathrm{BP}}{\mathrm{PQ}}$$ = cot30° = √3
⇒ $$\frac{x+50}{50}$$ = √3
x = 50(√3 – 1)m
= 50 (1.73 – l)m
= 50 x 0.732
= 36.6 m
Question 37.
Prove that “In a right angled triangle, square on the hypotenuse is equal to sum of the squares on the other sides.”
Verify
BC = 8cm
BC’= $$\frac{5}{3} \times 8=\frac{40}{3}$$
AB =6cms
A’B= $$\frac{5}{3} \times 6$$
=10cms.
VI. Answer the following : ( 5 x 1 = 5 )
Question 38.
The height of cone is 20m. A small cone is cut off from it at its top by the plane parallel to the base. If the volume of small cone 1/1000 th of the volume of given cone, at what height above the base the section is made.
In A OPB and A OQC A OPB ~aOQC OP _ PB OQ ~ QC
## Karnataka SSLC Maths Model Question Paper 3 With Answers
Students can Download Karnataka SSLC Maths Model Question Paper 3 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka State Syllabus SSLC Maths Model Question Paper 3 with Answers
Time: 3 Hours
Max Marks: 80
I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )
Question 1.
In the following numbers, irrational number is
a) 0.232332333……
b) 0.23233
c) 0.232323
d) 0.2323
a) 0.232332333……
Question 2.
10 sec2A – 10tan2A =
a) sec2A
b) 10
c) 1
d) 0
b) 10
Solution:
10 sec2A – 10tan2A
10(sec2A – tan2A) = 10(1)= 10
Question 3.
The length of the tangent drawn to a circle of radius 3cm from 5cm away from the centre is
a) 4cm
b) 5cm
c) 3cm
d) 2cm 4
a) 4cm
Solution:
d2 = r2 + t2
t2 = d2 – r2
52 – 32
= 25 – 9
∴ t = 16 = 4cm
Question 4.
A solid piece of copper of dimension 24 × 49 × 33 cms is moulded and recast into a sphere. The radius of the
sphere formed is ________
a) 49 cm
b) 24 cm
c) 21 cm
d)33 cm
e) 21 cm
l × b × h = $$\frac{4}{3} \pi r^{3}$$
Question 5.
The degree of the polynomial in the graph given below is
a) 4
b) 3
c) 1
d) 2
a) 4
Solution:
4 since it is intersecting the x – axis at 4 points.
Question 6.
The sum of the n terms of an AP is 2n2 + 5n and its common difference is 6, then its first term is
a) 0
b) 5
c) 2
d) 7
d) 7
Solution:
Sn = 2n2 + 5n
S1 = 2(1)2 + 5(1)
= 2 + 5 = 7
Question 7.
In ∆ PQR, PR = 12cm, QR = 6√3 cm, PQ = 6cm. The angle Q is
a) 45°
b) 90°
c) 30°
d) 120°
b) 90°
Solution:
PR2 = PQ2 + QR2
122= (6√3 )2 + 62
144 = 108 – 36
144=144
∴ ∠Q = 90°
Question 8.
A cube numbered 1 to 6 is thrown once, the probability of getting a number divisible by 3 is
a) 2/3
b) 0
c) 1/3
d) 1
Event A = {3,6}
n(s) = 6, n(A) =2
P(A) = $$\frac{n(A)}{n(S)}=\frac{2}{6}=\frac{1}{3}$$
II. Answer the following Questions : (1 x 8 = 8)
Question 9.
Given g(x) = 2x + 1, q(x) = (x3 + 3x2 – + 1), r(x) = 0, Find p(x)
p(x) = g(x) q(x) + r(x) 1 650 1170
= (2x + 1) (x3 + 3x2 – x + 1) + 0
p(x) = 2x (x3 + 3x2 -x + 1) + 1(x3 + 3x2 – x + 1)
= 2x4 + 6x3 – 2x2+ 2x + x3 + 3x2 – x + 1
= 2x4 + 7x3 + x2 + x + 1
Question 10.
If the sum of first n odd natural number is 1225, find the value of n.
Sum of “n” odd natural number = n2 = 1225
n = √1225
n = 35
Question 11.
In the fig. ∠AOD is divided into 2 parts which are in A.P. the smallest angle ∠AOB = 20° . Find the common difference between each angle.
∠AOB +∠BOC + ∠COD = 180°
a + a + d + a + 2d= 180°
3a + 3d= 180°
3(20) + 3d = 180°
60 + 3d = 180°
3d = 180 – 60
d = 120/3 = 40°
Question 12.
In ∆ ABC, if DE || BC, then $$\frac{\mathbf{A B}}{\mathbf{A D}}=\frac{\mathbf{A C}}{\mathbf{A} \mathbf{E}}=\frac{\mathbf{B C}}{\mathbf{D E}}$$ , state the theorem to justify this.
In a triangle, if a line is parallel to one of the sides then the sides of given triangle are propotional to sides of the intercepted triangle.
Question 13.
Find the largest number which divides 650 and 1170.
H.C.F of (650, 1170) = 130
Question 14.
If sin θ = 7/25, cos θ = 24/25 find the value of sin2 θ + cos2 θ
Question 15.
Find the value of cos60° cos 30° sin60° sin 30°
cos 60° cos30 – sin 60 sin 30
OR
This is in the form of
cos A cos B – sin A sin B = cos (A + B)
cos 60 cos 30 – sin 60 sin 30
= cos (60 + 30)
= cos (90)
– =0
Question 16.
The T.S.A of a solid hemisphere of radius 21 mm.
T.S.A of hemisphere = 3 πr²
= 3 × $$\frac{22}{7}$$ × 21 × 21 = 4158 mm2
III. Answer the following : ( 2 x 8 = 16 )
Question 17.
Prove that if x and y are odd positive integers, then x2 + y2 is even but not divisible by 4.
Let x = 2m + 1 and y = 2n + 1 for some integers m and n.
x2 + y2 = (2m + 1)2 + (2n + l)2
x2 + y2 = 4m2 + 4m + 1 + 4n2 + 1 + 4n
= 4m2 + 4n2 + 4m + 4n + 2.
x2 + y2 = 4(m2 + n2) + 4 (m + n) + 2
x2 + y2 = 4 {(m2 + n2) + (m + n)} +2
x2 + y2 = 4q + 2, where q = (m2 + n2 ) + (m+n)
⇒ x2 + y2 is even and leaves the remainder 2 when divided by 4.
⇒ x2 + y2 is even but not divisible by 4.
Question 18.
Solve : 100x + 200y = 700
200x + 100y = 800
Consider 100x + 200y = 700 ….(1)
200x + 100y = 800 ….(2)
Add (1) & (2) 300x + 300y= 1500
Divide by 300, x + y = 5 …… (3)
Subtract (1) and (2)
-100x + 100y= -100 Divide by 100.
-x + y = -1 …(4)
Solve (3) and (4)
y = 2
Substitute the value of y in equation (3)
x + y = 5
x + 2 = 5
x = 5 – 2 = 3
x = 3
Question 19.
Find the roots of the quadratic
equation 3x2 – 2√6x + 2 = 0 by formula method.
Question 20.
Find the value of x in which the points (1, -1) (x, 1) and (4, 5) are collinear.
A(1, -1) B(x, 1)C(4,5)
Area of the triangle = 0, when points are collinear.
0 = x1(y2 – y3) + x2(y3 – y1) + x3 (y1 – y2)
0 = (1) (1 – 5) + x(5 + 1) + 4(-1 – 1)
0 = -4 + 6x – 8
6x – 12 = 0
x = 12/6
x = 2
Question 21.
ABC is a right angled triangle, having [B = 90°. If BD = DC, Show that AC2 = 4AD2 – 3AB2
AC2 = AB2 + BC2 (Pythagoras)
AC2 = AB2 + 2(BD)2
AC2 = AB2 + 4BD2
In right angled ∆ ABD,
Consider AC2 = AB2 + 4BD2
= AB2 + 4(AD2 – AB2)
= AB2 + 4AD2 – 4AB2
AC2 = 4 AD2 – 3AB2
OR
Prove that area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.
Data : ABCD is a square.
Equilateral triangles ∆ BCE and ∆ ACF have been described on side BC and diagonal AC respectively.
T.P.T. : Area of ( ∆ BCE) = 1/2 Area of ∆ ACF
Proof : Since ∆ BCE and ∆ ACF are equilateral.
∴ They are equiangular.
∆ BCE ~ ∆ ACF
Question 22.
A box contains 90 dices which are numbered from 1 to 90. If one dise is drawn at random from the box, find the probability that it bears
(i) two digit number
(ii) a perfect square number.
The numbers in the dise form the sample space
S = {1,2, 3,4,…….. 90}
One dise can be drawn out of 90 in 90 ways. .
n(s) = 90
i) There are 90 -9 = 81, two digit numbers, out of which one dise can be drawn in 81 ways.
∴ n(A) = 81
P(A) = $$\frac{n(A)}{n(S)}=\frac{81}{90}=\frac{9}{10}$$
ii) The perfect square numbers are B= {1,4,9, 16,25,36,49,64,81}
∴ n(B) = 9
P(B) = $$\frac{n(B)}{n(S)}=\frac{9}{90}=\frac{1}{10}$$
Question 23.
Draw a pair of tangents to a circle of radius 5cm which are inclined to each
other at an angle of 60°.
Question 24.
Prove that $$\frac{\tan \theta+\sin \theta}{\tan \theta-\sin \theta}=\frac{\sec \theta+1}{\sec \theta-1}$$
OR
Prove that (cosecθ – cotθ)2 = $$\frac{1-\cos \theta}{1+\cos \theta}$$
Consider (cosecθ – cotθ)2
IV. Answer the following : ( 3 × 9 = 27 )
Question 25.
The sum of the ages of A and B is 85 years. 5 years ago, the age of A was twice that of B. Find the present ages.
Let the present age of A be x years.
Let the present age of B be y years.
Sum of their ages = 85 x + y = 85 …….. (1)
Five years back, age of A was (x – 5) and that of B was (y – 5).
The age A is twice that of B.
∴ x – 5 = 2(y – 5) ……(2)
x – 5 = 2y – 10
x – 2y = -10 + 5
x – 2y = -5 ……(2)
Solve (1) and (2)
Substitute y in (1)
x + y=85
x + 30 = 85
x = 85 – 30 = 55
x = 55
∴ Present age of A is 55 years and B = 30 years.
OR
A piece of work can be done by 2 men and 7 boys in 4 days. The same piece of work can be done by 4 men and 4 boys in 3 days. How long it would take to do the same work by one man or one boy?
Let x and y be the number of days in which one man can complete the work.
∴ In 1 day a man can do 1/x th work and
In 1 day a man can do 1/y th work.
2 men and 7 boys can complete the work in 4 days.
∴ They can complete in 1/4 th of the work in one day.
∴ $$\frac{2}{x}+\frac{7}{y}=\frac{1}{4}$$ …… (1)
4 men and 4 boys together complete the work in 3days, and they can complete it in 1/3 rd of work.
4(2y + 7x) = xy
8y + 28x = xy …….(3)
3(4y + 4x) = xy
12y + 12x = xy ……..(4)
Solve (3) and (4)
(8y + 2x = xy) x 12
(l2y+ 12x = xy) x 18
4 y = 240
y = 240/4
y = 60
3(4y + 4x) xy
12y+12x = xy ………..(4)
4y = 240
y = 240/4 = 60
Substitute y in equation (3)
8y + 28x = xy
8(60) + 28x = xy
480 + 28x = x(60)
480 + 28x = 60x
480 = 60x – 28x
480 = 32x
x = $$\frac{480}{32}=\frac{120}{8}=\frac{30}{2}$$ = 15
x = 15
Thus, one man can complete the work in 15 days and one boy can do the work in 60 days.
Question 26.
Find the zeroes of the polynomial p(y) = y3 – 5y2 – 2y + 24 if it is given that Sum of the two zeros is Answer:
Let α, β,γ be the zeros of the given polynomial.
p(y) = y3 – 5y2 – 2y + 24
α + β = 7
Sum of the zeros
α + β + γ = 5
7 + r = 5
r = 5 – 7 = -2
r = -2
Sum of product of zeroes taken two at a
αβ + βr + rα = -2/1 = – 2
αβ + r(β + α) = -2
αβ + (7) (-2) = -2
αβ-14 = -2
αβ = -2+ 14
αβ = 12
(α – β )2 = (α + β)2– 4αβ
= (7)2 – 4(12)
(α – β)2 = 49 – 48
α – β = ±1
Solve for α and β. When α – β = +1
α + β = 7
α – β = 1
2α =8
α = 8/2 = 4
α = 4
Substitute the value of a in α + β =7
4+p =7
β =7-4 = 3
β = 3
Solve for α and β when α – β = -1
α = 6/2 = 3
α = 3
α + β = 7
3 + β=7
β = 7 – 3
a + β = 7
β = 4
When α – β = 1,
the values are α = 4, β = 3, r = -2.
When α – β = -1,
The values are α = 3, β = 4, r = -2.
Question 27.
The diagonal of a rectangular field is 60 metres more than the shorter side, if the longer side is 30 metres more than the shorter side, find the sides of the field.
Let the shorter D side be x mtr.
Longer side is 30m more than the shorter side. A Longer side = (x + 30)m.
Diagonal is 60m more than the shorter side.
Diagonal = (x + 60) m
AC2 = AB2 + BC2 (Pythagoras)
(x + 60)2 = (x + 30)2 + x2
x2+(60)2 + 2x (60) = x2 + (30)2 + 2(x)(30)+ x2
x2 + 3600 + 120x = x2 + 900 + 60x + x2
-x2 + 60x + 2700 = 0
x2 – 60x – 2700 = 0
x2 – 90x + 30x – 2700 = 0
x(x – 90) + 30(x – 90) = 0
(x – 90) (x + 30) = 0
x – 90 = 0 or x + 30 = 0
x = 90 or x = – 30
∴ Shorter side = 90m = BC = x
Longer side = x + 30 = 90 + 30 = 120m
Diagonal = x + 60 = 90 + 60 = 150m
OR
Sum of the areas of two squares is 468m2. If the difference of their perimeters is 24m, find the sides of two squares.
Let the sides of the two squares be x&y. Sum of the areas of two squares = 468
x2 + y2 = 468 ……(1)
Difference of their perimeters = 24m.
4x – 4y = 24 x – y = 6
x = 6 + y …….(2)
Substitute the value of x in (1)
x2 + y2 = 468
(6 + y)2 + y2 = 468
36 + y2+ 12y + y2 = 468
2y2 + 12y + 36 = 468
Divide by 2
y2 +6y+ 18 = 234 y2 + 6y = 234 – 18
y2 + 6y – 216 = 0
y2+18y – 12y -216 = 0
y(y + 18)-12(y + 18) = 0
(y + 18) (y – 12) = 0
y+18 = 0, y -12 = 0
y = -18,y = 12
∴ x = 6+y = 6 + 12 = 18m
The sides of two squares are 18m and 12m.
Question 28.
Show that the points x(2, -2) y(-2, 1) and z(5, 2) are the vertices of a right angled triangle XYZ and also calculate its area.
⇒ XYZ is an isosceles right angled triangle.
Area of a right angled triangle = $$\frac{1}{2}$$ x base x height
$$\frac{1}{2}$$ × 5 × 5 = $$\frac{25}{2}$$ = 12.5 cm
OR
Find the values of k for which the points A(k + 1, 2k) B(3k, 2k+3) and C(5k – 1, 5k) are collinear.
A(k + 1, 2k) = (x1, y1)
B(3k, 2k + 3) = (x2, y2)
C(5k-1, 5k) = (x3, y3)
Area of the triangle = 0, for the points to be collinear.
Area =[x1(y2 – y3) + x2(y3-y1) + x3(y1-y2)]
0 = $$\frac{1}{2}$${(k + 1) (2k+3 – 5k)+3k(5k- 2k) + (5k-1) [2k-2k-3]}
0= $$\frac{1}{2}$$]{(k+ 1) (-3k+ 3) + 3k (3k) + (5k- I)(-3)i
0 =$$\frac{1}{2}$${-3k2-3k + 3k + 3 + 9k2 – 15k+3}
0= $$\frac{1}{2}$${6k2– 15k + 6}
0 = 6k2 – 15k + 6
Divide by 3
2k2 – 5k + 2= 0
2k2 – 4k – 1k + 2 = 0
2k(k – 2) -1(k – 2) = 0
(k – 2) (2k -1) = 0
k – 2 = 0, or 2k- 1=0
k = 2 or k = 1/2
Question 29.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Data: ‘O’ is the centre of the circle PA and PB are the two tangents drawn from an external point P. OA and OB are radii of the circle.
To prove that: ∠APB+ ∠AOB = 180°
Proof: In AAPO and ABPO
∠OAP = ∠OBP = 90°
(∵ Angle between the radius and tangent at the point of contact is 90°)
OP = OP (∵ Common side)
OA = OB (∵ Radii of the same circle) According to RHS postulate AAPO ABPO
∠OAP = ∠OBP = 90°
∠OAP + ∠OBP = 90° + 90° = 180°
⇒ Opposite angles of OAPB quadrilateral are supplementary
∴ OAPB is a cylic quadrilateral
⇒ ∠APB + ∠AOB = 180°
Question 30.
Find the area of the shaded region where a circular arc of Radius 6 cm has been drawn with the vertex ‘O’ of an equilateral Triangle OAB of side 12cm as centre.
Area of the shaded Region = Area of circle area of sector OCDE + Area of equilateral
∆ le OAB
OR
From each corner of a square of side 4cm a quadrant of a circle of radius 1cm is cut and also a circle of diameter 2cm is cut as shown in Fig, Find the area of the remaining portion of the square.
Area of shaded region = Area of square – (Area of circle + Area of 4 quadrants)
The following table gives production yield per hectare of wheat of 100 farms of a
Question 31.
The following table gives production yield per hectare of wheat of 100 farms of a
village.
Change the distribution to a more than type distribution and draw its ogive.
Question 32.
If the median of the distribution given below is 28.5. Find the valucs of x and y.
Meadian = 28.5, lies in the C.I , 20 – 30
The median class = 20 – 30
l = 20, f = 20, c.f = 5 + h, h = 10, n = 60
Question 33.
Construct a triangle ABC with side BC = 7cm, ∠B = 45°, ∠C = 105°. Then construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC.
V. Answer the following ( 4 × 4 = 16 )
Question 34.
Solve the pair of equations graphically.
x+y=3 and 3x-2y=4
x + y = 3
y = 3- x
3x – 2y = 4
-2y = 4 – 3x .
Question 35.
If the sum of first 8 terms of an Arithmetic progression is 136 and that of first 15 terms is 465, then find the sum of first 25 terms.
Given S8 = 136, S15 = 465, S25 = ?
Use the formula Sn = $$\frac{n}{2}$$ [2a + (n -l)d]
s8 = $$\frac{8}{2}$$[2a + (s-l)d]
136 = 4 (2a + 7d)
∴ 2a + 7d = 136/4
2a + 7d= 34…….(1)
S15 = $$\frac{15}{2}$$[2a + (15-l)d]
465 = $$\frac{15}{2}$$[2a + 14d]
465= $$\frac{15}{2}$$ x 2(a + 7d)
∴ a + 7d = 465/15
a + 7d = 31 …….(2)
From (1) and (2)
Consider a + 7d = 31
3 + 7d = 31
7d = 31 – 3
d = 28/7
d = 4
∴S25 = $$\frac{25}{2}$$[2a + (25-1)d]
S25 = $$\frac{25}{2}$$[2a + (25-1)d]
S25 = $$\frac{25}{2}$$[2a + (24)d]
S25 = $$\frac{25}{2}$$[2(3) + 24(4)d]
S25 = $$\frac{25}{2}$$ × 102
S25 = 1275
‘5 ‘ 9
OR
The sum of the 5th and 9th terms of an A.P is 40 and the sum of the 8th and 14th term is 64. Find the sum of the first 20 terms.
Given T5+ T9 = 40 and T8 + T14 = 64
a + 4d + a + 8d=40
2a + 12d = 40….(1)
a + 7d + a + 13d = 64
2a + 20d = 64…. (2)
Consider 2a + 12d = 40
2a + 12(3) = 40
2a + 36 = 40
2a = 40 – 36
a = 4/2
a = 2
Sn = $$\frac{n}{2}$$[2a + (n-1)d]
S20 = $$\frac{20}{2}$$[2(2) + (20-1)3]
= 104(4 + 57)
= 10 × 61
= 610
Question 36.
Two poles of equal heights are standing opposite to each other on either side of the road, which is 80m wide. From a point between them on to the road, the angles of elevation of the top of the poles re 60 and 30’, respectively. Find the height of the poles and distances of the point from the poles.
In ∆ ABE.
tan60 = $$\frac{\mathrm{AB}}{\mathrm{BE}}$$
√3 = x/BE
x = √3BE ….. (1)
tan30 = DC/EC
$$\frac{1}{\sqrt{3}}=\frac{x}{E C}$$
EC = x√3
EC = BE × √3 .√3 [∵ x = √3.BE]
FC = 3BE
∴ EC=3BF
We know that
BE + EC = 80 rn
BE 3BE 80m
4BE = 80
BE = 80/4 = 20
⇒ EC = 3BE 3(20)= 60m
∴ x = BE.
x = 20√3
⇒ AB = CD = 20√3 m
∴ The distance from the point to the pole arc 20m towards left and 60m towards the right
poles
Height of the poles 20√3m
Question 37.
Prove areas of similar triangles.
Areas of similar triangles are proportional
to the squares on the corresponding sides.
In ∆ABM and ∆DEN
∠AMB = ∠DNE = 90 [construction]
∠B =∠E (Data)
∠BAM = ∠EDN (Remainingang1e)
∆ABM ∼ ∆ DEN
∆ABM & ∆DEN are equiangular
VI. Answer the following : ( 5 × 1 = 5 )
Question 38.
A circus tent is made of canvas and is in the form of a right circular cylinder and a right circular cone above it. The diameter and height of the cylindrical part of the tent are 126m and 5m respectively. The total height of the tent is 21m. Find the total cost of the canvas used to make the tent when the cost per m2 of the canvas is ?15.
Total canvas used=CS A of cylindrical part -1- CS A of conical part
=2πrh + πrl
= 2 × $$\frac{22}{7}$$ × 63 × 5 + $$\frac{22}{7}$$ x 63 x 65
= 1980 m2 + 12870 m2
= 14850 m2
Total canvas used = 14850 m2
* Cost of canvas at the rate of ₹16 per m2
= 14850 × 15= ₹ 2,22.750.
Cost of canvas = ₹ 2,22,750
## Karnataka SSLC Maths Model Question Paper 2 with Answers
Students can Download Karnataka SSLC Maths Model Question Paper 2 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka State Syllabus SSLC Maths Model Question Paper 2 with Answers
Time: 3 Hours
Max Marks: 80
I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )
Question 1.
For some integer n every odd integer is of the form
a) 2n + 1
b) n + 1
c) 2n
d) n
2n + 1
Question 2.
The value of sin215° + sin225° + sin265° + sin275° is
a) 0
b) 1
c) 2
d) 3
2
Question 3.
If chord AB subtends an angle 50° at the centre of a circle then the angle between the tangents at A and B is
a) 40°
b) 100°
c) 130°
d) 120°
130°
Question 4.
The formula used to find the volume of a sphere
a) $$\frac{4}{3} \pi r^{3}$$
b) $$\frac{2}{3} \pi r^{3}$$
c) $$\frac{1}{3} \pi r^{3}$$
d) πr3
a) $$\frac{4}{3} \pi r^{3}$$
Question 5.
α + β are the zeroes of the polynomial x2 – 6x + 4, then the value of $$\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha \beta}$$ is
a) 7
b) 8
c) -7
d) -8
a) 7
Solution:
Question 6.
If 29th term of an A.P is twice its 19th term, then the 9th term is
a) -1
b) 0
c) 1
d) 2
b) 0
Solution:
T29 = 2T19
a + 28d= 2(a + 18d)
a + 28d = 2a + 36d
0 = 2a + 36d-a-28d
0 = a + 8d
T9 = 0
Question 7.
In ∆ ABC, AB = 6√3 cm, AC = 12 cm, BC = 6cm, The angle B is
a) 45°
b) 90°
c) 60°
d) 30°
Solution:
AC2 = AB2 + BC2
122= (6√3 )2 + 62
144= 108 + 36 144=144
∴ ∠B = 90°
Question 8.
If the probability of an event is P(A) then the probability of its complimentary event will be
a) 1+P(A)
b) 1 – P(A)
c) P(A) – 1
d) 1 / P(A)
Solution:
We know that for any two complimentary events A and Ā
P(A) + P(Ā) = 1 → P(Ā) = 1 – P(A)
II. Answer the following questions : ( 1 × 8 = 8 )
Question 9.
If α and β are the zeroes of the quadratic polynomial 2 – 3x – x2 then what is the value of α + β + αβ ?
α + β = -b/a , αβ = c/a
= $$\frac{-(-3)}{-1}$$ αβ = $$\frac{2}{-1}$$ = – 2
= – 3
α + β + αβ = -3 -2 = -5
Question 10.
What are the roots of the quadratic equation x2 + (√3 + 1)x + √3 = 0?
2 + (√3 + 1)x + √3 = 0 = 0
x2 + x + √3x + √3 = 0
x(x + 1)+ √3 (x + 1) = 0
(x + 1)(x + √3) = 0
x + 1 = 0, x + √3 = 0
x = -1, x = -√3
Question 11.
If the nth terms of the two AP 9, 7, 5, ….. and 24, 21, 18, ……… are same. Find n.
9, 7, 5, …….
a = 9, d = 7-9 = -2 T = a + (n – 1)d
Tn = 9 + (n – 1) (-2)
Tn = 9 + (-2n + 2) ‘
Tn = 9 – 2n + 2
Tn = 11 – 2n
24,21,18,
a = 24, d = 21 – 24 = -3
Tn = a + (n – 1) d T=24 + (n – 1)(-3)
Tn = 24 – 3n + 3
Tn = 27 – 3n
Given : nth term of both A. P. is same.
∴ 11 – 2n = 27 – 3n
-2n +3n = 27 -11
n = 16
Question 12.
State A – A criterian theorem.
“If two triangles are equiangular, then the corresponding sides are proportional.”
Question 13.
Find the H.C.F of 455 and 42 with the help of Euclid’s division algorithm.
Step (1): a = 455 b = 42
a = bq + r
455 = 42 × 1 + 35
Step (2): a = 42, b = 35
a = bq + r
42 = 35 × 1 + 7
Step (3): a = 35, b = 7
a = bq + r
35 = 7 × 5 + 0
∴ H.C.F. (455, 42) = 7
Question 14.
Find θ if sin (θ + 56) = cosθ, where θ and ( θ + 56) are less than 90°.
sin (θ + 56) = cos θ
sin (θ + 56) = sin (90 – θ)
θ + 56 = 90 – θ
θ + θ = 90 – 56
2θ = 34
θ = 34/2
θ = 17
Question 15.
If x = a sin θ and y = b tan θ, then find the value of $$\frac{a^{2}}{x^{2}}-\frac{b^{2}}{y^{2}}$$
Consider
Question 16.
Calculate the height of a right circular cone where C.S.A. and base radius are 12320 cm2 and 56 cms. respectively.
C.S.A of a cone = π rl
12320 = $$\frac{22}{7}$$ x 56 x 1
1 = $$\frac{12320 \times 7}{22 \times 56}$$
1 = 70 cm
But 12 = r2 + h2
702 = 562 + h2
h22 = 702 – 562
h2 = (70 + 56) (70 – 56)
h2 =(126) (14)
III. Answer the following ( 2 × 8 = 16 )
Question 17.
Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5 where q is an integer.
Let nbe any positive, integer using division algorithm,
a = bq + r
Taking a = n, b = 6
n = 6q + r
If any number is divided by 6, then the possible remainders are 0, 1, 2, 3, 4 or 5.
∴If n is odd, then r = 1, 3, 5
⇒ 6q + 1, 6q + 3, 6q + 5 are the positive odd integers.
Question 18.
Solve : 2x + 3y = 9
3x + 4y = 5
Substitute y = 17 in 2x + 3y = 9
2x + 3(17) = 9
2x + 51 =9
2x = 9 – 51
x = $$\frac{-42}{2}$$
= -21
Question 19.
Solve : (x – 2)2 + 1 = 2x – 3
x2 + 4 – 4x + 1 = 2x – 3
x2 – 4x + 5 – 2x + 3 = 0
x2 – 6x + 8 = 0 x2 – 4x – 2x + 8 = 0
x(x – 4) -2(x – 4) = 0
(x – 4) (x – 2) = 0
x – 4 = 0, x -2 = 0
x = 4 or x = 2
Question 20.
Show that the points (-2, 1) (2, -2) and (5, 2) are the vertices of a right angled triangle.
∴ ∆ ABC is a right angled triangle at B.
Question 21.
The equilateral triangles are drawn on the sides of a right triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of the triangles on the other two sides.
∴ Area of ∆ XAB + Area of ∆ YBC = Area of ∆ ZAC .
OR
In the given figure, PA, QB and RC are each perpendicular to AC. Prove $$\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$$
Question 22.
Two dice are thrown simultaneously. Find the probability that the sum of the numbers on the faces is neither divisible by 4 nor by 5.
S = {(1, 1) (1,2) (1, 3) (1,4) (1, 5) (1, 6) (2, 1) (2, 2) (2,3) (2,4) (2, 5) (2,6) (3.1) (3, 2) (3, 3) (3, 4) (3, 5) (3,6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4,6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5,6) (6.1) (6,2) (6, 3) (6,4) (6, 5) (6,6)}
n(S) = 36.
An event of getting sum of the numbers neither divisible by 4 nor by 5.
A={ 1,1) (1,2) (1,5) (1,6) (2,1) (2, 4)
(2, 5) (3, 3) (3, 4) (3, 6) (4, 2) (4,3) (4, 5) (5,1) (5, 2) (5, 4) (5, 6) (6,1) (6, 3) (6, 5)} n(A) = 20
n(A) = 20
P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{20}{36}$$
Question 23.
Draw a circle of radius 3 cm. Take a point P outside the circle without using the centre of the circle, draw to tangents to the circle from an external point P.
Question 24.
Prove that (cosecθ-cotθ)2 = $$\frac{1-\cos \theta}{1+\cos \theta}$$
Consider LHS
(cosecθ-cotθ)2 = cosec2 θ +cot2 θ – 2cosecθ.cotθ
OR
If sinθ + cos θ = √2 sin (90 -θ) determine cot θ
sin θ + cos θ = √2 sin (90 – θ)
sin θ + cos θ = √2 cos θ
sin θ = ypi cos θ – cos θ
sin θ = cos θ (√2 -1)
sinθ /cosθ = √2 -1
IV. Answer the following : ( 3 × 9 = 27 )
Question 25.
Asha is 5 times as old as her daughter Usha, 5 years later Asha will be 3 times as old as her daughter Usha. Find the present ages of Asha and Usha.
Let the age of Asha be x years
Let the age of Usha be y years
According to question x=5y
x – 5y = 0 …….(1)
Five years later,
(x + 5) = 3(y + 5)
x + 5 = 3y + 15
x – 3y = 15 – 5
x – 3y = 15 – 5
x – 3y = 10 (2)
Substitute x = 5y in equation (2)
5y – 3y = 10
2y = 10
y = 10/2
y = 5
∴ x = 5y
= 5 × 5
x = 25
∴ The present age of Asha is 25 years.
The present age of Usha is 5 years.
OR
The sum of 2 digits of a 2 digit number is 12 the number obtained by interchanging the digits exceeds by the given number by 18. Find the number.
Let the two digits be x and y
The 2 digit number will be 10x + y
The sum of digits = x + y – 12 ……..(i)
The number obtained by interchanging the digits =10y + x
= 10x + y + 18
10y + x – 10x – y = 18
9y – 9x = 18
Divide by 9
y – x = 2
y = x + 2 ….(ii)
Consider equation (1)
x + y = 12
Substitute y = x + 2
x + x + 2 = 12 .
2x =12 – 2
2x= 10
x= 5
y = x + 2
y = 5 +2
y = 7
∴ The number is : 10x +y
= 10(5)+7 = 50+ 7
= 57
Question 26.
Find the other two zeroes of the polynomial y4 + y3 – 9y2 – 3y + 18 if the
zeroes are √3 and -√3
y = √3 and y = -√3
(y + √3) = o (y-V3) = o
(y + √3 )(y – √3 ) = o
y(y – √3 ) + √3 (y – √3 ) = 0
y2 – √3y + √3 y – 3 = o
y2 – 3 = 0
y2 + y – 6 = 0
y2 + 3y – 2y – 6 = 0
y(y + 3)-2(y + 3) = 0
(y + 3) = 0 & (y – 2) = 0
y=-3 & y = 2
The four zeros of polynomial are √3 , -√3 , -3 & 2
Question 27.
Solve for x.
$$\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}$$ (Where a ≠ 0, b ≠ 0 x ≠ 0, x ≠ -a,-b
(-a – b)ab = (a + b)x (a + b + x)
-a2b – ab2 = (a + b) (ax + bx + x2)
-a2b – ab2 = a2x + abx + ax2 + abx + b2x + x2b
-a2b – ab2= a2x + 2abx + ax2 + b2x + x2b
-ab(a+b) = ax(a+x) + xb(b+x)+2abx
=x[a(a+x)+b(b+x)+2ab]
=x[a2 + ax+b2 +bx + 2ab]
= x[(a+b)2 + x(a+b)]
-ab(a + b) = x(a+b)[a+b+x]
-ab = x(a + b + x)
-ab = ax + bx + x2
x2 + x (a + b) + ab = 0
x2 + ax + bx + ab = 0
x(x+a) + b(x + a) = 0
(x + a) (x +b) = 0
=>x + a = 0, x + b = 0
=> x = -a, x = -b
OR
The diagonal of a rectangular field is 60m more than the shorter side. If the larger side is 30m more than the shorter side, find the sides of the field.
Let the shorter side of the rectangular field be x m.
Longer side is x + 30 and the diagonal is x+60
In ∆ ABC,
AC2 = AB2 + BC2
(x + 60)2 = ( x + 30)2+ x2
x2 + 3600 + 120x = x2 + 900 + 60x + x2
x2 + 900 + 60x – 3600 – 120x = 0
x2 – 60x – 2700 = 0
x2 – 90x + 30x – 2700 = 0
x(x – 90) + 30(x – 90) = 0
(x – 90) (x + 30) = 0
x – 90 = 0, x + 30 = 0
x = 90, x = – 30.
∴ The shorter side = x = 90m
The longer side = x + 30 = 90+30 = 120m.
Question 28.
If the points (7, -2) (5, 1) and (3, 5) are collinear. Find the value of k.
Since the give points are collinear, the area of the triangle formed by them must be O i.e.
= $$\frac { 1 }{ 2 }$$ [x1(y2-y3) + x2(y3-y2) + x3(y1-y2)]
x1 = 7, x2 = 5, x3 = 3
y1 = – 2, y2 = 1, y3 = k
= $$\frac { 1 }{ 2 }$$ [7(1 – k) + 5 (k + 2) + 3(-2-1)] = 0
= $$\frac { 1 }{ 2 }$$ [7 – 7k + 5k + 10 – 9]-0
= $$\frac { 1 }{ 2 }$$ [7 – 2k + 1] = 0
= $$\frac { 1 }{ 2 }$$ [8 – 2k] = 0
8-2k = 0 or 2k=8
k = 8/2 = 4
k = 4
OR
Find the area of Rhombus if its vertices are (3, 0) (4, 5) (-1, 4) and (-2, -1) taken in order.
Area of rhombus = $$\frac { 1 }{ 2 }$$ d1 × d2
Let A = (3, 0) B = (4, 5)
C = (-1, 4) D = (-2, -1)
Diagonal
Question 29.
Prove the tangents drawn from an external point to a circle are equal.
Data: A is the centre of the circle B is an external point.
BP and BQ are the tangents.
AP, AQ and AB are joined.
To prove that: a) BP = BQ
b) ∠PAB-∠QAB.
c) ∠PBA = ∠QBA
Proof: In ∆ APB and ∆ AQB
AP = AQ [Radii of same circle]
∠ APB – ∠AQB = 90°
AB = AB [common side]
∴ ∆ APB ≅ ∆ AQB [RHS]
a) PB = QB
b) ∠PAB = ∠QAB
c) ∠PBA = ∠QBA
Question 30.
Find the area of the shaded region in the figure, where ABCD is a square of side 14 cm
Area of square ABCD
= 14 × 14=196 cm2
Hence, area of the shaded region
= Area of square – Area of four circles
= 196 – 154
= 42cm2
OR
Find the area of the shaded regions. Given PQRS a square of sides 14cm. Soln: S R
Question 31.
The following table gives the weight of 120 articles :
Change the distribution to a “more than type” distribution and draw its ogive.
Question 32.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Weight (in kg)
Question 33.
Construct a right triangle whose hypotenuse and one side measures 10cm and 8cm respectively. Then construct another triangle whose sides are 4/5 times the corresponding sides of the triangle .
V. Answer the following ( 4 × 4 = 16 )
Question 34.
Solve the pair of equations graphically.
x + y = 8 and x – y = -2
Answe:
x + y = 8
y = 8 – x
x – y = -2
x = -2 + y
x = 6/2 = 3
x + y = 8
3 + y = 8
y = 8 – 3
y = 5
Question 35.
Devide 20 into four parts which are in arithmetic progression and such that the product of first and fourth is to the product of second and third in the ratio 2:3.
Let the four parts in A.P be :
a – 3d, a – d, a +d, a + 3d
Their sum = (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20
⇒ 4a = 20 or a = 5
Also, it is given that
or 75 -27d2 = 50 – 2d2
25d2 = 25 .
⇒ d2 = 1 or d = +√1 = ±1
When a = 5, d = 1, the four parts are 2, 4, 6,8
When a = 5, d = -1; the four parts are 8, 6, 4,2
Hence the four parts are (2, 4, 6, 8) or (8, . 6,4,2)
OR
The angles of a quadrilateral are in AP such that the greatest is double the least calculate all the angles of the quadrilateral.
The angles of the quadrilateral are in AP.
∴ The four angles are A, B, C and D which are a -3d, a – d, a + d, a + 3d respectively.
⇒ The least angle is a – 3d and the greatest angle is a + 3d.
It is given that
a + 3d = 2(a – 3d)
a + 3d = 2a – 6d
2a – a – 6d – 3d = 0
a – 9d ……(1)
9d = a
But ∠A + ∠B+∠C + ∠D = 360°
a – 3d + a – d + a + d + a +.3d = 360°
4a = 360 / 4 = 90°
But 9d = a
d = $$\frac{a}{9}=\frac{90}{9}$$ = 10°
∴ ∠A=a- 3d = 90 – 3(10) = 90 -30 = 60°
∠B = a – d = 90 – 10 = 80°
∠C = a + d = 90 + 10= 100
∠D = a + 3d = 9 + 3(10) = 90 + 30 = 120
Question 36.
A person on the lighhouse of height 100m above the sea level observes that the angle of depression of a ship sailing towards the light house changes from 30° to 45°. Calculate the distance travelled by the ship during the period of observation. (Take √3 ≈ 1.73)
P is the top of the light house PQ. We are given that its height PQ = 100m PX is horizontal line through P
∠APX = 30° and ∠BPX =45°
Then, ∠PAQ =30° and ∠PBQ = 45°
Suppose, AB = x metre and BQ = y metre
Karnataka SSLC Maths Model Question Paper 2 with Answers – 40
⇒ x + y =100√3 = 100√3 = 100× 1.732
⇒ x +y = 173.2
⇒ x + 100 = 173.2
⇒ x = 73.2
Therefore, the distance travelled = 73.2m
Question 37.
Prove that “the ratio of areas of two similar triangles is equal to the square of the ratio of their altitudes.
Data : ∆ABC ∼ ∆DEF
∠A = ∠D
∠B = ∠E
∠C = ∠F
$$\frac{A B}{D E}=\frac{A C}{D F}=\frac{B C}{E F}$$
T.P.T : $$\frac{\text { Area of } \Delta \mathrm{ABC}}{\text { Area of } \Delta \mathrm{DEF}}=\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}=\frac{\mathrm{AC}^{2}}{\mathrm{DF}^{2}}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}$$
Construction:Draw AM ⊥ BC and AN ⊥ EFim
VI. Answer the fallowing : ( 5 × 1 = 5 )
Question 38.
The radii of the circular ends of the frustrum of height – 6cm are 14 cm and 6cm respectively. Find the lateral surface area and total surface area of frustrum.
R = 14cm, r = 6cm and h =6cm
Now, let 1 be the slant height of the frustrum;
Now, lateral surface area of the frustrum
= π(R + r)l
= π (14 + 6) 10
= 628.57 cm2
And total surface area of the frustrum = π[(R+r)l + R2+ r2]
= π [(14 + 6)10 + (14)2 + (6)2]
= π(200 + 196 + 36)
= π(432)
= 1357.71 cm2.
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# properties of equality
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Properties of Equality. Properties of Equality. Properties are rules that allow you to balance, manipulate, and solve equations. Addition Property of Equality. Adding the same number to both sides of an equation does not change the equality of the equation. If a = b, then a + c = b + c. - PowerPoint PPT Presentation
TRANSCRIPT
• Properties are rules that allow you to balance, manipulate, and solve equations
• Adding the same number to both sides of an equation does not change the equality of the equation.If a = b, then a + c = b + c.Ex: x = y, so x + 2 = y + 2
• Subtracting the same number to both sides of an equation does not change the equality of the equation.If a = b, then a c = b c.Ex: x = y, so x 4 = y 4
• Multiplying both sides of the equation by the same number, other than 0, does not change the equality of the equation.If a = b, then ac = bc.Ex: x = y, so 3x = 3y
• A number is equal to itself. (Think mirror)a = aEx: 4 = 4
• If numbers are equal, they will still be equal if the order is changed (reversed).If a = b, then b = a.Ex: x = 4, then 4 = x
• If numbers are equal to the same number, then they are equal to each other.If a = b and b = c, then a = c.Ex: If x = 8 and y = 8, then x = y
• Changing the order of addition or multiplication does not matterAddition: a + b = b + aMultiplication: a b = b a
• The change in grouping of three or more terms/factors does not change their sum or product.Addition: a + (b + c) = (a + b) + cMultiplication: a (b c) = (a b) c
• The sum of any number and zero is always the original numbera + 0 = aEx: 4 + 0 = 4
• The product of any number and one is always the original number.Multiplying by one does not change the original number.a 1 = aEx: 2 1 = 2
• The sum of a number and its inverse (or opposite) is equal to zero.a + (-a) = 0Ex: 2 + (-2) = 0
• The product of any number and its reciprocal is equal to 1.
• The product of any number and zero is always zero.a 0 = 0Ex: 298 0 = 0
• Make sure each equation is in slope-intercept form: y = mx + b.Graph each equation on the same graph paper.The point where the lines intersect is the solution. If they dont intersect then theres no solution.Check your solution algebraically.
• 1. Graph to find the solution.
• Graph to find the solution.
• If the lines have the same y-intercept b, and the same slope m, then the system is consistent-dependent.
If the lines have the same slope m, but different y-intercepts b, the system is inconsistent.
If the lines have different slopes m, the system is consistent-independent.
• Arrange the equations with like terms in columnsMultiply, if necessary, to create opposite coefficients for one variable.Add the equations. Substitute the value to solve for the other variable.Check
• 4x + 3y = 162x 3y = 8
• One equation will have either x or y by itself, or can be solved for x or y easily.Substitute the expression from Step 1 into the other equation and solve for the other variable.Substitute the value from Step 2 into the equation from Step 1 and solve.Your solution is the ordered pair formed by x & y.Check the solution in each of the original equations.
• Get in slope-intercept formDetermine solid or dashed lineDetermine whether to shade above or shade below the line (Test Points)If the test point is true, shade the half plane containing it.If the test point is false, shade the half plane that does NOT contain the point
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## Engage NY Eureka Math 7th Grade Module 1 Lesson 12 Answer Key
### Eureka Math Grade 7 Module 1 Lesson 12 Example Answer Key
Example 1.
Time to Remodel
You have decided to remodel your bathroom and install a tile floor. The bathroom is in the shape of a rectangle, and the floor measures 14 feet, 8 inches long by 5 feet, 6 inches wide. The tiles you want to use cost $5 each, and each tile covers 4$$\frac{2}{3}$$ square feet. If you have$100 to spend, do you have enough money to complete the project?
Make a Plan: Complete the chart to identify the necessary steps in the plan and find a solution.
Answer:
Compare your plan with a partner. Using your plans, work together to determine how much money you will need to complete the project and if you have enough money.
Answer:
Dimensions:
5 ft.,6 in.=5 $$\frac{1}{2}$$ ft.
14 ft.,8 in.= 14 $$\frac{2}{3}$$ ft.
Area (square feet):
A = lw
A = (5 $$\frac{1}{2}$$ ft.)(14 $$\frac{2}{3}$$ ft.)
A = ($$\frac{11}{2}$$ ft.)($$\frac{44}{3}$$ ft.)
A = $$\frac{242}{3}$$ = 80 $$\frac{2}{3}$$2
Number of Tiles:
I cannot buy part of a tile, so I will need to purchase 18 tiles.
Total Cost: 18(5)=$90 Do I have enough money? Yes. Since the total is less than$100, I have enough money.
Generate discussion about completing the plan and finding the solution. If needed, pose the following questions:
→ Why was the mathematical concept of area, and not perimeter or volume, used?
→ Area was used because we were “covering” the rectangular floor. Area is 2 dimensional, and we were given two dimensions, length and width of the room, to calculate the area of the floor. If we were just looking to put trim around the outside, then we would use perimeter. If we were looking to fill the room from floor to ceiling, then we would use volume.
→ Why would 5.6 inches and 14.8 inches be incorrect representations for 5 feet, 6 inches and 14 feet, 8 inches?
→ The relationship between feet and inches is 12 inches for every 1 foot. To convert to feet, you need to figure out what fractional part 6 inches is of a foot, or 12 inches. If you just wrote 5.6, then you would be basing the inches out of 10 inches, not 12 inches. The same holds true for 14 feet, 8 inches.
→ How is the unit rate useful?
→ The unit rate for a tile is given as 4 $$\frac{2}{3}$$ . We can find the total number of tiles needed by dividing the area (total square footage) by the unit rate.
→ Can I buy 17 $$\frac{2}{7}$$ tiles?
→ No, you have to buy whole tiles and cut what you may need.
→ How would rounding to 17 tiles instead of rounding to 18 tiles affect the job?
→ Even though the rules of rounding would say round down to 17 tiles, we would not in this problem. If we round down, then the entire floor would not be covered, and we would be short. If we round up to 18 tiles, the entire floor would be covered with a little extra.
### Eureka Math Grade 7 Module 1 Lesson 12 Exercise Answer Key
Which car can travel farther on 1 gallon of gas?
Blue Car: travels 18 $$\frac{2}{5}$$ miles using 0.8 gallons of gas
Red Car: travels 17 $$\frac{2}{5}$$ miles using 0.75 gallons of gas
Answer:
Finding the Unit Rate:
Rate:
The red car traveled $$\frac{1}{5}$$ mile farther on one gallon of gas.
### Eureka Math Grade 7 Module 1 Lesson 12 Problem Set Answer Key
Question 1.
You are getting ready for a family vacation. You decide to download as many movies as possible before leaving for the road trip. If each movie takes 1 $$\frac{2}{5}$$ hours to download, and you downloaded for 5 $$\frac{1}{4}$$ hours, how many movies did you download?
Answer:
3 $$\frac{3}{4}$$ movies; however, since you cannot download $$\frac{3}{4}$$ of a movie, then you downloaded 3 movies.
Question 2.
The area of a blackboard is 1$$\frac{1}{3}$$ square yards. A poster’s area is $$\frac{8}{9}$$ square yards. Find the unit rate and explain, in words, what the unit rate means in the context of this problem. Is there more than one unit rate that can be calculated? How do you know?
Answer:
1 $$\frac{1}{2}$$ . The area of the blackboard is 1 $$\frac{1}{2}$$ times the area of the poster.
Yes. There is another possible unit rate: $$\frac{2}{3}$$. The area of the poster is $$\frac{2}{3}$$ the area of the blackboard.
Question 3.
A toy jeep is 12 $$\frac{1}{2}$$ inches long, while an actual jeep measures 18 $$\frac{3}{4}$$ feet long. What is the value of the ratio of the length of the toy jeep to the length of the actual jeep? What does the ratio mean in this situation?
Answer:
Every 2 inches in length on the toy jeep corresponds to 3 feet in length on the actual jeep.
Question 4.
To make 5 dinner rolls, $$\frac{1}{3}$$ cup of flour is used.
a. How much flour is needed to make one dinner roll?
Answer:
$$\frac{1}{15}$$ cup
b. How many cups of flour are needed to make 3 dozen dinner rolls?
Answer:
2 $$\frac{2}{5}$$ cups
c. How many rolls can you make with 5 $$\frac{2}{3}$$ cups of flour?
Answer:
85 rolls
### Eureka Math Grade 7 Module 1 Lesson 12 Exit Ticket Answer Key
If 3$$\frac{3}{4}$$ lb. of candy cost $20.25, how much would 1 lb. of candy cost? Answer: 5 $$\frac{2}{5}$$ = 5.4 One pound of candy would cost$5.40.
Students may find the unit rate by first converting \$20.25 to $$\frac{81}{4}$$ and then dividing by $$\frac{15}{4}$$.
|
# How do you simplify 2x + 3( x + 2) - 6?
Nov 22, 2017
See a solution process below:
#### Explanation:
First, expand the term in parenthesis by multiplying the terms in parenthesis by the term outside the parenthesis:
$2 x + \textcolor{red}{3} \left(x + 2\right) - 6 \implies$
$2 x + \left(\textcolor{red}{3} \times x\right) + \left(\textcolor{red}{3} \times 2\right) - 6 \implies$
$2 x + 3 x + 6 - 6$
Now, we can combine like terms:
$\left(2 + 3\right) x + \left(6 - 6\right) \implies$
$5 x + 0 \implies$
$5 x$
Nov 22, 2017
Open brackets
$2 x + 3 x \cancel{+ 6} \cancel{- 6}$
$5 x$
|
#### Please solve RD Sharma class 12 chapter 10 Differentiation exercise Multiple choice question 9 maths textbook solution
$|\sec \theta|$
Hint:
Differentiate x and y w.r.t $\theta$, then divide and solve
Given:
$x=a \cos ^{3} \theta, y=a \sin ^{3} \theta, \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=$
Solution:
\begin{aligned} &x=a \cos ^{3} \theta \\\\ &\frac{d x}{d \theta}=a \frac{d}{d \theta}\left(\cos ^{3} \theta\right) \\\\ &\frac{d x}{d \theta}=a(3) \cos ^{2} \frac{d}{d \theta}(\cos \theta) \end{aligned}
$\frac{d x}{d \theta}=-3 a \cos ^{2} \theta \sin \theta$ ....................(1)
\begin{aligned} &y=a \sin \theta \\\\ &\frac{d y}{d \theta}=a \frac{d}{d \theta}\left(\sin ^{3} \theta\right) \end{aligned}
$=3 a \sin ^{2} \theta \frac{d}{d \theta}(\sin \theta)$
$\frac{d y}{d \theta}=3 a \sin ^{2} \theta \cos \theta$ ......................(2)
Dividing (2) by (1) we get
$\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 a \sin ^{2} \theta \cos \theta}{-3 a \cos ^{2} \theta \sin \theta}$
\begin{aligned} &\frac{d y}{d x}=\frac{\sin \theta}{-\cos \theta}=-\tan \theta \\\\ &\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=\sqrt{1+\tan ^{2} \theta}=\sqrt{\sec ^{2} \theta}=|\sec \theta| \end{aligned}
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# Construction of Third Binomial Straight Line
## Theorem
In the words of Euclid:
To find the third binomial straight line.
## Proof
Let $AC$ and $CB$ be straight lines constructed such that $AB = AC + CB$ is itself a straight line.
$AB : BC = m^2 : n^2$
where $m$ and $n$ are numbers such that $m^2 - n^2$ is not square.
Let $D$ be a number which is not square.
Let $D$ have to neither $AB$ nor $AC$ the ratio that a square number has to another square number.
Let $E$ be a rational straight line.
$D : AB = E^2 : FG^2$
where $FG$ is a straight line.
$E^2$ is commensurable with $FG^2$.
We have that $E$ is rational.
Therefore $FG$ is also rational.
Since:
$D$ does not have to $AB$ the ratio that a square number has to another square number
then:
$E^2$ does not have to $FG^2$ the ratio that a square number has to another square number.
$E$ is incommensurable in length with $FG$.
$BA : AC = FG^2 : GH^2$
where $GH$ is a straight line constructed such that $FH = FG + GH$ is itself a straight line.
$FG^2$ is commensurable with $GH^2$.
We have that $FG$ is rational.
Therefore $GH$ is also rational.
Since:
$BA$ does not have to $AC$ the ratio that a square number has to another square number
then:
$FG^2$ does not have to $GH^2$ the ratio that a square number has to another square number.
$FG$ is incommensurable in length with $GH$.
Therefore $FG$ and $GH$ are rational straight lines which are commensurable in square only.
Therefore by definition $FH$ is a binomial.
We have that:
$D : AB = E^2 : FG^2$
and
$BA : AC = FG^2 : GH^2$
$D : AC = E^2 : GH^2$
But $D$ does not have to $AC$ the ratio that a square number has to another square number.
Therefore, neither does $E$ have to $GH$ the ratio that a square number has to another square number.
$E$ is incommensurable in length with $GH$.
We have that:
$BA : AC = FG^2 : GH^2$
Therefore:
$FG^2 > GH^2$
Let:
$FG^2 = GH^2 + K^2$
for some $K$.
$AB : BC = FG : K$
But:
$AB : BC = m^2 : n^2$
where $m$ and $n$ are numbers.
Therefore:
$FG : K = m^2 : n^2$
$FG$ is commensurable in length with $K$.
Therefore $FG^2$ is greater than $GH^2$ by the square on a straight line which is commensurable in length with $FG$.
But $FG$ and $GH$ are rational straight lines which are commensurable in square only.
Also, neither $FG$ nor $GH$ is commensurable in length with $E$.
Therefore $FH$ is a third binomial straight line.
$\blacksquare$
## Historical Note
This proof is Proposition $50$ of Book $\text{X}$ of Euclid's The Elements.
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Suppose we are comparing implementations of insertion sort and merge sort on the same machine. For inputs of size $n$, insertion sort runs in $8n^2$ steps, while merge sort runs in $64n \lg n$ steps. For which values of $n$ does insertion sort beat merge sort?
For insertion sort to beat merge sort for inputs of size $n$, $8n^2$ must be less than $64n \lg n$. $$8n^2 < 64n \lg n \implies \frac n 8 < \lg n \implies 2^{n/8} < n$$
This is not a purely polynomial equation in $n$. To find the required range of values of $n$, either we have to manually calculate the values of these expression for different values of $n$ or use Newton’s Method or plot these functions or write a piece of code to found the values. For this exercise, I’ll briefly describe all of these methods but going forward I’ll mostly use calculations that can be done with scientific calculator (to help the students visiting these pages) and python codes (to help myself).
Calculate: It is obvious that insertion sort runs at quadratic time which is definitely worse than merge sort’s linearithmic time for very large values of $n$. We know for $n = 1$, merge sort beats insertion sort. But for values greater than that, insertion sort beats merge sort. So, we will start checking from $n = 2$ and go up to see for what value of $n$ merge sort again starts to beat insertion sort.
Notice that for $% $, $2^{n/8}$ will be a fraction. So, let’s start with $n = 8$ and check for values of $n$ which are power of 2.
$% $ $% $ $% $ $n = 64 \implies 2^{64/8} = 2^8 = 256 > 64$
Somewhere between 32 and 64, merge sort starts to beat insertion sort. Let’s do what we were doing but now we are going to try middle value of either range, repeatedly (binary search).
$n = \frac {32 + 64} 2 = 48 \implies 2^{48/8} = 64 > 48$ $% $ $n = \frac {40 + 48} 2 = 44 \implies 2^{44/8} = 44.8 > 44$ $% $ $% $
So, at $n = 44$, merge sort starts to beat insertion sort again. Therefore, for $% $, insertion sort beats merge sort.
Newton's Method: To apply Newton’s method of approximation, we need to ballpark two values of $n$ on either side of the actual solution and hit-and-try for the actual one following binary search principle. This is the principle we have almost followed in the previous section but the proper process involves calculus and way too time consuming for me to write about.
Graphical Plot: Plot the functions $2^{n/8}$ and $n$ to get the points where they intersect.
Graph rendered by JSXGraph
From the above graph, we can see that the plots intersect at $n = 1.1$ and $n = 43.56$. Here, $n$ is a input size so it must be an integer. So, the values of $n$ for which insertion beats merge sort is $% $.
Python Code: Let’s start with $n = 2$ and go up to see for what value of $n$ merge sort again starts to beat insertion sort.
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## Transcription
Jessica Fauser,November 24 2009,Complete Lesson Plan 1. from the Board Race activity, 3 Now define a linear equation Is an equation that is in the form. ax b c where a b and c are any value The value of the unknown. variable x is what we are looking for when solving linear equations. 4 How do you solve linear equations Bloom Level III Application. A Simplify the equation by using the distribution property and. combining like terms, B Use addition and subtraction properties to get the variable term. to one side by itself Remember that what you do to one side you. must do the same to the other, C Now use multiplication and division properties to isolate the. variable get x by itself, D Plug the value you received for the variable x back into the.
original equation to check your answer, 5 Now give some examples of solving linear equations using these steps. A Simplify the equation,This equation is already simplified. B Get the variable term on one side by itself,1 1 Subtract 1 from both sides. C Isolate the variable,3x 3 3 3 Divide each side by 3. D Check your answer Plug the value of x back into the original. 2 x 1 x 10,A Simplify the equation,2 x 1 x 10,2x 2 x 10 Distribute the 3.
3x 2 10 Combine like terms,B Get the variable term on one side by itself. 2 2 Add 2 to both sides,C Isolate the variable,Jessica Fauser. November 24 2009,Complete Lesson Plan 1,3x 3 12 3 Divide each side by 3. D Check your answer,2 4 1 4 10,10 x 2 2 6 2x,A Simplify the equation. 10 x 2 2 6 2x,10x 20 2 6 2x Distribute the 10,10x 18 6 2x Combine like terms.
B Get the variable term on one side by itself,10x 18 6 2x. 2x 2x Subtract 2x from both sides,18 18 Add 18 to both sides. C Isolate the variable,8x 8 24 8 Divide each side by 8. D Check your answer,10 3 2 2 6 2 3,10 1 2 6 2 3, Have the students pair back up this time at their desks Ask them to. create and solve their own linear equation Be sure to walk around and answer. any questions they might have Once they are done pass out the worksheet on. linear equations for homework Have them complete it and turn in at the. beginning of class the next day Bloom Level V Synthesis Gardner Logical. Mathematical and Interpersonal Intelligences,Adaptations Enrichments.
Learning disabilities in reading comprehension No reading from the book is. required for this lessons Instructions on how to solve linear equations are. written on the board however oral instructions are also given. ADHD The board races at the beginning of the lesson allows this student to. Jessica Fauser,November 24 2009,Complete Lesson Plan 1. move around During the lesson this student if needed can come up to. the board to help solve the example problems, Gifts Talents in Creativity Encourage this student to lead their pair when it is. time to create their own equation Have the student think of a way to. remember the steps in solving linear equations,Self Reflection. Were the board races an effective attention getter How time consuming were the. board races How many students understood this lesson Are the students able to solve. linear equations Were the students able to create their own equations Did the students. cooperate with each other How does this tie into the next lesson Did I engage the. whole class What would I do differently next time to improve the lesson. Jessica Fauser,November 24 2009,Complete Lesson Plan 1. Name Solving Linear Equations,Point values for each question are in parentheses.
1 4 List the steps for solving linear equations, Solve the following linear equations Show all steps and be sure to check your work. 2 2 2x 4 10,3 2 4 x 2 3x 20,4 2 5 x 1 1 3 2x,5 2 12 3x 1 24. 6 3 2 3x 7 4 3 x 2 6 5 x 9 3, Solve the following story problems by using linear equations. 7 2 Together Sally and Sam bought a total of 15 apples If Sally bought 6 apples how. many did Sam buy Use x as your variable to set up the equation then solve. 8 3 Billy is 12 years old He is 4 times as old as his little brother How old is Billy s. little brother Use x as your variable to set up the equation then solve. Jessica Fauser,November 24 2009,Complete Lesson Plan 1. Solving Linear Equations,Point values for each question are in parentheses.
1 4 List the steps for solving linear equations,1 Simplify the equation. 2 Get the variable term on one side by itself,3 Isolate the variable. 4 Check your answer, Solve the following linear equations Show all steps and be sure to check your work. 2 2 2x 4 10,2x 6 2 3 4 10,2x 2 6 2 6 4 10,x 3 10 10 D. 3 2 4 x 2 3x 20,4x 8 3x 20 Check,7x 8 20 4 4 2 3 4 20.
8 8 4 2 3 4 20,7x 28 8 12 20,x 4 20 20 D,4 2 5 x 1 1 3 2x. 5x 5 1 3 2x Check,5x 4 3 2x 5 7 3 1 1 3 2 7 3,4 4 5 4 3 1 3 2 7 3. 5x 7 2x 20 3 1 3 14 3,2x 2x 23 3 23 3 D,5 2 12 3x 1 24. 36x 12 24 Check,12 12 12 3 1 3 1 24,36x 12 12 1 1 24. 36x 36 12 36 12 2 24,x 12 36 24 24 D,Jessica Fauser.
November 24 2009,Complete Lesson Plan 1,6 3 2 3x 7 4 3 x 2 6 5 x 5 Check. 6x 14 12x 8 30x 30 2 3 3 7 4 3 3 2 6 5 3 5,18x 6 30x 30 2 9 7 4 9 2 6 15 5. 6 6 2 16 4 7 6 10,18x 30x 36 32 28 60,30x 30x 60 60 D. 12x 12 36 12, Solve the following story problems by using linear equations. 7 2 Together Sally and Sam bought a total of 15 apples If Sally bought 6 apples how. many did Sam buy Use x as your variable to set up the equation then solve. 6 x 15 Check,6 6 6 9 15 Sam bought 9 apples,x 9 15 15 D.
8 3 Billy is 12 years old He is 4 times as old as his little brother How old is Billy s. little brother Use x as your variable to set up the equation then solve. 4x 12 Check,4x 4 12 4 4 3 12 Billy s brother is 3 years old. x 3 12 12 D,Jessica Fauser,November 24 2009,Complete Lesson Plan 1. Jessica Fauser,November 24 2009,Complete Lesson Plan 1.
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# How do I evaluate the following integral?
##### 1 Answer
Feb 26, 2018
$\frac{1}{2} {x}^{2} + 18 \ln | {x}^{2} - 36 | + C$
#### Explanation:
Divide ${x}^{2} - 36$ into ${x}^{3}$ using polynomial long division.
This yields:
$\int {x}^{3} / \left({x}^{2} - 36\right) \mathrm{dx} = \int \left(x + \frac{36 x}{{x}^{2} - 36}\right) \mathrm{dx}$
Split this integral up:
$\int x \mathrm{dx} + \int \frac{36 x}{{x}^{2} - 36} \mathrm{dx}$
$\int x \mathrm{dx} = \frac{1}{2} {x}^{2}$
For $\int \frac{36 x}{{x}^{2} - 36} \mathrm{dx}$
$u = {x}^{2} - 36$
$\mathrm{du} = 2 x \mathrm{dx}$
$18 \mathrm{du} = 36 x \mathrm{dx}$
So, our integral becomes:
$\frac{1}{2} {x}^{2} + 18 \int \frac{\mathrm{du}}{u} = \frac{1}{2} {x}^{2} + 18 \ln | u | + C = \frac{1}{2} {x}^{2} + 18 \ln | {x}^{2} - 36 | + C$
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Basic Math | Basic-2 Math | Prealgebra | Workbooks | Glossary Glossary | Standards | Site Map | Help
# Pick-a-Bar: Numbers One Through Ten Thousand
This activity was designed to help you recognize numbers between one (1) and ten thousand (10,000). The top of the activity will ask you to find a number with a word and you need to select the right symbol for that number. For example, you will be asked to find "three thousand two hundred thirty-five" and you need to click the "3,235." All of the questions combine random values.
Good luck and have fun.
## Directions
This is one of the NumberNut four-bar activities. Your question will be displayed at the top of the screen. Under that question, you will see four (4) possible answers. Just click on the bar with the correct answer. The next screen will show you the correct answer.
You get a happy face for every correct answer and a sad face for every wrong answer. The quiz is over after ten (10) questions. Take the quiz again and again because all of the questions are random. Chances are, you'll get a new quiz every time. It's good practice to learn these basic arithmetic operations.
RELATED LINKS LESSONS: - NumberNut.com: Numbers ACTIVITIES - Pick-a-Card: Counting 1 to 10 - Memory Challenge: Counting 1 to 10 - Memory Challenge: Names of Numbers - More or Less: Numbers 1-10 - Before and After: Numbers 1-10 - Memory Challenge: Numbers 0 to 9 - Memory Challenge: Chinese Numbers - Memory Challenge: Mayan Numbers - Memory Challenge: Roman Numerals - Pick-a-Card: Recognizing Numbers 1-30 - More or Less: Numbers 1-30 - Before and After: Numbers 1-30 - Pick-a-Card: Recognizing Numbers 1-50 - Pick-a-Card: Recognizing Numbers 1-100 - More or Less: Numbers 1-100 - Before and After: Numbers 1-100 - Pick-a-Bar: Recognizing Numbers 1-1,000 - More or Less: Numbers 1-1,000 - Before and After: Numbers 1-1,000 - Pick-a-Bar: Roman Numerals 1-1,000 - Pick-a-Bar: Recognizing Numbers 1-10,000 - More or Less: Numbers 1-10,000
- Overview - Shapes-Colors - Numbers - Addition - Subtraction - Multiplication - Division - Operations - Dates & Times > Activities
* The custom search only looks at Rader's sites.
Go for site help or a list of mathematics topics at the site map!
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• Level: GCSE
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# This piece of coursework is called 'Opposite Corners' and is about taking squares of numbers from different sized number grids
Extracts from this document...
Introduction
Algebra Coursework
Introduction
This piece of coursework is called ‘Opposite Corners’ and is about taking squares of numbers from different sized number grids. I will be multiplying the opposite corners together and subtracting to find the difference.
I will make a prediction for each grid and use a few examples to find a formula to prove my prediction right.
Method
To calculate the difference of the squares drawn on the number grids, I will be multiplying each of the diagonal numbers. Then I will subtract the smaller number away from the larger number to find the difference of the square.
To do this I will use algebraic equations, I will use ‘N’ to represent the smallest number and ‘g’ to represent the grid size. I will then compare my results to ‘N’, then to complete the equation by expanding, simplifying and cancelling the brackets to get my final result.
Prediction
I predict that a 2x2 square from a 5 wide grid, will have a final difference of 5.
Proof:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Comparison:
N N+1
x
N+5 N+6
N (N+6) = N² + 6N
(N+1) (N+5) = N² + 5N + N + 5
= N² + 6N + 5
Difference:
(N² + 6N + 5) – (N² + 6N)
= 5
What I have noticed:
Middle
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Comparison:
N N+2
x
N+12 N+14
N (N+14) = N² + 14N
(N+2) (N+12) = N² + 12N + 2N + 24
= N² + 14N + 24
Difference:
(N² + 14N + 24) – (N² + 14N)
= 24
What I have noticed:
I have noticed that when a square that is 3x3, is taken from a 6x6 grid, the difference is always 24.
Prediction
I predict that a 4x4 square from a 6 wide grid, it will have a final difference of 54.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
Comparison:
N N+3
x
N+18 N+21
N (N+21) = N² + 21N
(N+3) (N+18) = N² + 18N + 3N + 54
= N² + 21N + 54
Difference:
(N² + 21N + 54) – (N² + 21N)
= 54
What I have noticed:
I have noticed that when a square that is 4x4, is taken from a 6x6 grid, the difference is always 54.
Prediction
I predict that a 5x5 square from a 6 wide grid, it will have a final difference of 96.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
Comparison:
N N+4
x
N+24 N+28
N (N+28) = N² + 28N
(N+4) (N+24) = N² + 24N + 4N + 96
= N² + 28N + 96
Difference:
(N² + 28N + 96) – (N² + 28N)
= 96
What I have noticed:
I have noticed that when a square that is 5x5, is taken from a 6x6 grid, the difference is always 96.
I have found out the difference for a six wid grid is g(x-1)².
I will now see if the same eqyuation works for a seven wide grid.
I will now carry out the same investigation but on a 7 wide grid
Prediction
I predict that a 2x2 square from a 7 wide grid, it will have a final difference of 7.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49
Conclusion
I have found out the difference for a seven wid grid is g(x-1)².
I now have found a formula that will work for and size number grid over five wide and square of 2x2. this formula is g(x-1)
Table of results:
5 wide grids
Square size , Difference Gap in Difference Gap in Gap of Difference 2x2 5 +5 n/a 3x3 20 +15 +10 4x4 45 +25 +10 5x5 80 +35 +10
6 wide grids
Square size , Difference Gap in Difference Gap in Gap of Difference 2x2 6 +6 n/a 3x3 24 +18 +12 4x4 54 +13 +12 5x5 96 +42 +12
7 wide grids
Square size , Difference Gap in Difference Gap in Gap of Difference 2x2 7 +7 n/a 3x3 28 +21 +14 4x4 63 +35 +14 5x5 112 +49 +14
Conclusion
During my investigation, I have cancelled out, expanded and simplified brackets, and compared my results to ‘N’. I did this to find a final equation to be able to calculate the difference of any square of any size drawn on a number grid. I found this to be g (x-1) ². This formula works for any grid size over 5 and square size over 2x2.
If I had more time I would want to find out if the same formula works with a rectangle on a number grid and experiment with other quadrilaterals and see if the same equation works.
I would also see if I could find any other equation that would work.
Overall I think my investigation went well and I have found the final equation.
This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.
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# Related GCSE Number Stairs, Grids and Sequences essays
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1. ## Mathematics Coursework: problem solving tasks
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back to page 3 will confirm that the formula used for the squares is too the same. Finding the formulas for the rectangles by taking the idea from the squares was not so obvious. Now I feel that if I had have started this assignment with firstly investigating different rectangle
2. ## Number Grids Investigation Coursework
63 64 65 66 72 73 74 75 76 (top right x bottom left) - (top left x bottom right) = 36 x 72 - 32 x 76 = 2592 - 2432 = 160 So I successfully managed to predict the difference between the products of opposite corners in a
1. ## Number Grid Investigation.
(16 X 25) - (18 X 23) = 14. Formula 3. 3 X 5 box taken from above. 12 13 14 19 20 21 26 27 28 33 34 35 40 41 42 7 (n-1)(d-1) 2 X 4 = 8 7 X 8 = 56 Product difference should be 56.
2. ## Investigation of diagonal difference.
If we subtract n from the bottom left corner we are left with the same number as the grid length. n n + 1 (n + 6 )- n = 6 n + 7 E.g. a 2 x 2 cutout from a 6 x 6 grid.
1. ## Number Grid Investigation.
23 24 25 2 x 2 = 5 3 x 3 = 20 4 x 4 = 45 5 x 5 = 80 Nth term 5n� My prediction was right. Therefore, for any size grid such as, 12 x12 will have a formula and results that link and incorporate the number 12.
2. ## Maths Coursework - Grid Size
that are being multiplied are higher than those of the 2x2 square. This is the formula for a 2x3 rectangle on a 10x10 grid: (N+10)(N+2) - N(N+12) = (N2+2N+20+10N) - (N2+12N) = N2-N2+2N+10N- 12N+20 = 20 To test whether this works I will replace the algebra with numbers: (33+10)(33+2)
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Quick Math Homework Help
Master the 7 pillars of school success that I have learned from 25 years of teaching.
Common Core Standard 6.NS.4
Questions answered in this video
What is the Distributive Property?
When do you use the Distributive Property?
What type of problems use the Distributive Property
Distributive Property
Transcript
Today we are going to talk about the distributive property. Here’s our example: Two times five X minus two Y times three. That’s ten X minus four Y plus 6. Now let’s go over the rules for the distributive property. The rules for the distributive property are very simple. We are going to multiply each term by the led number or led coefficient in front of our parenthesis. So we are multiplying. That’s one of the biggest mistakes I see students commit. They add instead of multiplying. Now let’s go back and break it into slow mo. We are going to take the two and multiply it by each term. Two times five X which equals 10 and I bring down my X and two times negative two Y so I have a negative and the two times two Y is four Y and I bring down the variable. Then I take the two times the positive three I take two bring down my positive, two times three gives me positive six. So there is the distributive property. Now let’s look at one that has a variable in it. We are going to distribute the three X into a trinomial, because we have three terms. So we are going to take three X times two X squared. So we take the two times the three, which gives us six times an X squared which gives us an X cubed because we are going to add the exponents. Now we take the three X minus three Y three times three is X just gives us an XY Now the final term I’m going to bring down my addition Three X times positive eight is twenty four there are no other variables so I will tack on my X That is the distributive property
# Disributive Property Definition
Multiply the led coefficient (2), by the numbers inside the parentheses one at a time.
The numbers inside the parentheses cannot be added together because they are not like terms
Same rule applies with exponents, multiple the led coefficient, which in this example is 3x, by the numbers inside the parentheses one at a time.
What is the distributive propery?
A formal definition of the Distributive Property would sound something like this," The distributive property states that multiplying a sum by a number gives the same result as multiplying each number, and then adding the products together."
In other words you can add the numbers in the parenthesis, and then multiply, or multiply the number outside the parenthesis, and then add.
In either case you get the same result. This example is called the Distributive property over addition.
For example 4(2+3)
4(2+3)= 4 * 5 = 20 (add then multiply)
or 4(2+3)= 4*2 + 4*3 = 8 + 12 = 20 (multiply then add)
If you add then multiply, or multiply then add you get the same answer.
The Distributive property over subtraction works the same way.
You can subtract the numbers inside the parenthesis and then add, or multiply the number outside the parenthesis and then subtract.
For example: 4(5-2) = 4(3) = 12 (Subtract then multiply
or 4*5 - 4*2 = 20-8 =12 (Multiply then subtract)
The Distributive property is written as:
a(b + c) = ab + ac
When using the Distributive Property follow these rules.
1. Multiply each term inside the parenthesis by the lead term, which is the number outside the parentheses
2. Do not combine unlike terms inside the parenthesis, and the multiply this term by the lead term.
3. Do not multiply the lead term, and then add the lead term to the next term or terms.
For example,it would be incorrect to remove the parenthesis and then add.
For example: 4(5 +4x)= 4*5 + 4+4x
20+8x=28x this is incorrect.
According to the distributive property you must multiply the lead terms by the terms inside the parenthesis.
Correct:
4(5+5x) = 4*5 + 4*5x
20+20x = 40x
## Algebra distributive property
Use the Distributive Property to multiply in Algebra
## How to use the distributive property to simplify algebraic equations
4(x + 3) - 2x
2y - 4(3y + 5)
5 - 2(6 - 3y)
4b + 3(7 - 2a) + 5
x - 3(x - 6) + 5(3x + 2)
Check your work by rolling over the check mark
Video answer to each question
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Types of two equation systems with two unknowns. Examples
# Types of two equation systems with two unknowns. Examples
In this lesson I am going to explain the types of systems of equations that exist, depending on the number of solutions, with an example of each of the types of systems.
The systems of two equations with two incognites, in general have only one solution, but this is not always the case. It may be that the system of two equations has no solution and that others have infinite solutions.
Why does this happen? Let’s look at it more slowly
## Solution of a system of two equations
In a system of two equations with two unknowns such as this one:
Each equation corresponds to a line.
When you solve a system of two equations with two unknowns, what you are really doing is finding the cut-off point of the two lines, if any, because sometimes they don’t cut each other (we’ll see below that sometimes they don’t cut each other).
The equation of a line has this shape:
We can put each of the equations of the system in that way to make it clearer that they are two lines:
First equation:
Second equation:
Well, depending on the position of the two lines, or to put it another way, on the number of solutions of the systems of two equations, we can speak of certain compatible systems, indeterminate compatible systems and incompatible systems, which I will explain in more detail below.
## Compatible system determined
A certain compatible system is the system of two equations which, when solved, has a solution, that is, they can be solved.
The solution is the point where the two lines are cut if you represent them graphically. Therefore, the two lines of the system are cut at one point.
Let’s see it with an example. We have this system of two equations:
Let’s solve it by the substitution method. From the first equation clear the x:
This value of x replaced it in the second equation:
I operate and clear “y”:
Now this value of y, I replace it in the equation where I cleared the x:
And I get the value of x:
So the solution of the system of two equations is:
Or what is the same, if you represent the two lines, their cutting point is (2,5).
We are going to represent the two lines and you will see that the point where they are cut is the solution we have just calculated:
Compatible systems have a solution, which is the point where the two lines are cut [/box].
## Incompatible system
As I told you at the beginning, two equation systems do not always have a solution.
Straight lines may be parallel and never be cut. In this case we are talking about an incompatible system.
Let’s see it with an example:
If you represent the two lines of this system you will see that they are parallel and never cut, so it is an incompatible system:
The lines are parallel when they have the same slope, but different ordinate, i.e. in the formula:
m is the slope
n is the ordinate
If in the previous system, we put the two lines this way:
We see that they have the same slope, ie, the number in front of the x is equal, but the ordinate changes (n is different)
How do we know if a system is incompatible without the need to represent the lines graphically?
Let’s solve a system of two equations to see it:
Without knowing that it is an incompatible system, we begin to solve it. We clear x in the first equation:
And this value of x replaced it in the second equation:
I operate and clear the y:
I am left with a number divided by zero. A number cannot be divided by zero, so the system has no solution.
When solving one of the unknowns, I am left with a number divided by zero, the system has no solution and is therefore an incompatible system.
box]Incompatible systems have no solution (there is a number divided by zero), since the lines are parallel [/box].
## Compatible system indeterminate
The two lines can be of a third form, in addition to being cut in a point or to be parallel. They can be coincident, that is to say, that one line is on top of the other.
When one line is on top of another, they are actually the same line.
In this case, the system has infinite solutions, because it remembers that when we solve a system we are looking for the point of cut and in this case all its points are common.
To understand this, you have to understand the concept of equivalent equation.
You can multiply or divide all the terms of an equation by the same number and the resulting equation would be equivalent to the original. If you represented it it would be the same line.
For example, this equation:
I can multiply all your terms by 2 and the resulting equation would be:
And so it would be an equation equivalent to the original.
Well, in indeterminate compatible systems, we really have two equivalent equations, which would be equivalent to having two equal equations.
For example, this system of two equations:
It has two equivalent equations. We have just seen that the second equation is equal to the first equation multiplied by 2.
If we represent each line we are left:
The lines are coincident, one is on top of the other, all their points are common and therefore have infinite solutions.
It is the same as if we solved a system with the two equations equal:
How do you know if a system is indeterminate compatible without having to represent it? Let’s solve one for you to see:
From the first clear equation the x:
I replace this value of x in the second equation:
I operate and clear the y:
I am left with zero divided by zero. Zero between zero is an indeterminate number, so the “y” can take any value and therefore, the system has infinite solutions.
When solving one of the unknowns, I get zero divided by zero, the system has infinite solutions and is therefore an indeterminate compatible system.
## Discussion of two equation systems with two unknowns
Discussing a system means studying the possible system solutions based on a parameter we do not know from the system of two-unknowns equations and defining what type of system is involved in each case.
For example, in the following system, the coefficient of x of the first equation is parameter “a”:
We are going to discuss the system calculating the ones according to the values of parameter “a”.
We begin by solving the system, whose solution will remain in function of “a”.
From the first equation:
We clear “y”:
In the second equation:
We replace the “y” with the expression we just got:
We operate to remove the parenthesis:
We leave the terms with x in the first member and the terms without x in the second member:
Since in the first member, one of the coefficients is in function of “a”, we cannot operate with them directly, so we take a common factor to the x to be able to clear it:
And we pass the content of the parenthesis to the second member dividing:
At this point, the denominator of the fraction I have left will be the one that marks the number of system solutions and therefore the type of system.
If the denominator is zero, then the number will be divided by zero:
From the equation that results from equating the denominator to zero, we can clear the “a”:
Therefore, as long as a is 1/3, i.e. the denominator is zero, I will be left with a number between zero and the system is an incompatible system:
Being an incompatible system, there is no solution.
On the other hand, as long as a is not equal to 1/3, then the denominator will never be zero and therefore we will be dividing by a number and the system will have a single solution, so it will be a given compatible system:
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# Buffon's Needle
Buffon's Needle
The Buffon's Needle problem is a mathematical method of approximating the value of pi $(\pi = 3.1415...)$involving repeatedly dropping needles on a sheet of lined paper and observing how often the needle intersects a line.
# Basic Description
The method was first used to approximate π by Georges-Louis Leclerc, the Comte de Buffon, in 1777. Buffon posed the Buffon's Needle problem and offered the first experiment where he threw breadsticks over his shoulder and counted how often the crossed lines on his tiled floor.
Subsequent mathematicians have used the method with needles instead of bread sticks, or with computer simulations. In the case where the distance between the lines is equal the length of the needle, we will show that an approximation of π can be calculated using the equation
$\pi \approx {2*\mbox{number of drops} \over \mbox{number of hits}}$
# A More Mathematical Explanation
#### Will the Needle Intersect a Line?
[[Image:willtheneedlecros [...]
#### Will the Needle Intersect a Line?
To prove that the Buffon's Needle experiment will give an approximation of π, we can consider which positions of the needle will cause an intersection. Since the needle drops are random, there is no reason why the needle should be more likely to intersect one line than another. As a result, we can simplify our proof by focusing on a particular strip of the paper bounded by two horizontal lines.
The variable θ is the acute angle made by the needle and an imaginary line parallel to the ones on the paper. Finally, d is the distance between the center of the needle and the nearest line.
We can extend line segments from the center and tip of the needle to meet at a right angle. A needle will cut a line if the green arrow, d, is shorter than the leg opposite θ. More precisely, it will intersect when
$d \leq \left( \frac{1}{2} \right) \sin(\theta). \$
See case 1, where the needle falls at a relatively small angle with respect to the lines. Because of the small angle, the center of the needle would have to fall very close. In case 2, the needle intersects even though the center of the needle is far from both lines because the angle is so large.
#### The Probability of an Intersection
In order to show that the Buffon's experiment gives an approximation for π, we need to show that there is a relationship between the probability of an intersection and the value of π. If we graph the outcomes of θ along the X axis and d along the Y, we have the sample space for the trials. In the diagram below, the sample space is contained by the dashed lines.
The sample space is useful in this type of simulation because it gives a visual representation of all the possible ways the needle can fall. Each point on the graph represents some combination of an angle and distance that a needle might occupy. We divide the area that contains combinations that represent an intersection by the total possible positions to calculate the probability of an intersection.
There will be an intersection if $d \leq \left ( \frac{1}{2} \right ) \sin(\theta) \$, which is represented by the blue region. The area under this curve represents all the combinations of distances and angles that will cause the needle to intersect a line. The area under the blue curve, which is equal to $\frac {1}{2}$ in this case, can found by evaluating the integral
$\int_0^{\frac {\pi}{2}} \frac{1}{2} \sin(\theta) d\theta$
Then, the area of the sample space can be found by multiplying the length of the rectangle by the height.
$\frac {1}{2} * \frac {\pi}{2} = \frac {\pi}{4}$
The probability of a hit can be calculated by taking the number of total ways an intersection can occur over the total number possible outcomes (the number of trials). For needle drops, the probability is proportional to the ratio of the two areas in this case because each possible value of θ and d is equally probable. The probability of an intersection is
$P_{hit} = \cfrac{ \frac{1}{2} }{\frac{\pi}{4}} = \frac {2}{\pi} = .6366197...$
#### Using Random Samples to Approximate Pi
The original goal of the Buffon's needle method, approximating π, can be achieved by using probability to solve for π. If a large number of trials is conducted, the proportion of times a needle intersects a line will be close to the probability of an intersection. That is, the number of line hits divided by the number of drops will equal approximately the probability of hitting the line.
$\frac {\mbox{number of hits}}{\mbox{number of drops}} \approx P_{hit} = \frac {2}{\pi}$
So
$\frac {\mbox{number of hits}}{\mbox{number of drops}} \approx \frac {2}{\pi}$
Therefore, we can solve for π:
$\pi \approx \frac {2 * {\mbox{number of drops}}}{\mbox{number of hits}}$
# Why It's Interesting
#### Monte Carlo Methods
The Buffon's needle problem was the first recorded use of a Monte Carlo method. These methods employ repeated random sampling to approximate a probability, instead of computing the probability directly. Monte Carlo calculations are especially useful when the nature of the problem makes a direct calculation impossible or unfeasible, and they have become more common as the introduction of computers makes randomization and conducting a large number of trials less laborious.
π is an irrational number, which means that its value cannot be expressed exactly as a fraction a/b, where a and b are integers. As a result, π cannot be written as an exact decimal and mathematicians have been challenged with trying to determine increasingly accurate approximations. The timeline below shows the improvements in approximating pi throughout history. In the past 50 years especially, improvements in computer capability allow mathematicians to determine more decimal places. Nonetheless, better methods of approximation are still desired.
A recent study conducted the Buffon's Needle experiment to approximate π using computer software. The researchers administered 30 trials for each number of drops, and averaged their estimates for π. They noted the improvement in accuracy as more trials were conducted.
These results show that the Buffon's Needle approximation is a relatively tedious. Even when a large number of needles are dropped, this experiment gave a value of pi that was inaccurate in the third decimal place. Compared to other computer generated techniques, Buffon's method is impractical because the estimates converge towards π rather slowly. Nonetheless, the intriguing relationship between the probability of a needle's intersection and the value of π has attracted mathematicians to study the Buffon's Needle method since its introduction in the 18th century.
#### Generalization of the problem
The Buffon’s needle problem has been generalized so that the probability of an intersection can be calculated for a needle of any length and paper with any spacing. It has been proven that for a needle shorter than the distance between the lines, the probability of a intersection is $\frac {2*l}{\pi*d}$. This equation makes sense when we consider the normal case, where l =1 and d =1, so these variables disappear and the probability is $\frac {2}{\pi}$.
The generalization of the problem is useful because it allows us to examine the relationship between length of the needle, distance between the lines, and probability of an intersection. The variable for length is in the numerator, so a longer needle will have a greater probability of an intersection. The variable for distance is in the denominator, so greater space between lines will decrease the probability.
To see how a longer needle will affect probability, follow this link: http://whistleralley.com/java/buffon_graph.htm
#### Needles in Nature
Applications of the Buffon's Needle method are even found naturally in nature. The Centre for Mathematical Biology at the University of Bath found uses of the Buffon's Needle algorithm in a recent study of ant colonies. The researchers found that an ant can estimate the size of an anthill by visiting the hill twice and noting how often it recrosses its first path.
Ants generally nest in groups of about 50 or 100, and the size of their nest preference is determined by the size of the colony. When a nest is destroyed, the colony must find a suitable replacement, so they send out scouts to find new potential homes.
In the study, scout ants were provided with "nest cavities of different sizes, shapes, and configurations in order to examine preferences" [2]. From their observations, researchers were able to draw the conclusion that scout ants must have a method of measuring areas.
A scout initially begins exploration of a nest by walking around the site to leave tracks. Then, the ant will return later and walk a new path that repeatedly intersects the first tracks. The first track will be laced with a chemical that causes the ant to note each time it crosses the original path. The researchers believe that these scout ants can calculate an estimate for the nest's area using the number of intersections between its two visits.
The ants can measure the size of their hill using a related and fairly intuitive method: If they are constantly intersecting their first path, the area must be small. If they rarely reintersects the first track, the area of the hill must be much larger so there is plenty of space for a non-intersecting second path.
"In effect, an ant scout applies a variant of Buffon's needle theorem: The estimated area of a flat surface is inversely proportional to the number of intersections between the set of lines randomly scattered across the surface." [7]
This idea can be related back to the generalization of the problem by imagining if the parallel lines were much further apart. A larger distance between the two lines would mean a much smaller probability of intersection. We can see in case 3 that when the distance between the lines is greater than the length of the needle, even very large angle won’t necessarily cause an intersection.
This natural method of random motion in nature allows the ants to gauge the size of their potential new hill regardless of its shape. Scout ants are even able to asses the area of a hill in complete darkness. The animals show that algorithms can be used to make decisions where an array of restrictions may prevent other methods from being effective.
# References
[4] The Number Pi. Eymard, Lafon, and Wilson.
[5] Monte Carlo Methods Volume I: Basics. Kalos and Whitlock.
[6] Heart of Mathematics. Burger and Starbird
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# Math Snap
## $\int x \operatorname{Ln} x d x$
#### STEP 1
Assumptions 1. The integral to solve is $\int x \ln(x) \, dx$. 2. We will use integration by parts to solve this integral.
#### STEP 2
Recall the formula for integration by parts, which is given by: $\int u \, dv = uv - \int v \, du$
#### STEP 3
Identify $u$ and $dv$ from the integral $\int x \ln(x) \, dx$. Let: $u = \ln(x) \quad \text{and} \quad dv = x \, dx$
#### STEP 4
Differentiate $u$ to find $du$ and integrate $dv$ to find $v$: $du = \frac{1}{x} \, dx \quad \text{and} \quad v = \frac{x^2}{2}$
#### STEP 5
Substitute $u$, $v$, $du$, and $dv$ into the integration by parts formula: $\int x \ln(x) \, dx = \ln(x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx$
#### STEP 6
Simplify the integral on the right-hand side: $\int x \ln(x) \, dx = \frac{x^2}{2} \ln(x) - \int \frac{x}{2} \, dx$
#### STEP 7
Integrate $\frac{x}{2}$: $\int \frac{x}{2} \, dx = \frac{x^2}{4}$
#### STEP 8
Substitute back into the equation: $\int x \ln(x) \, dx = \frac{x^2}{2} \ln(x) - \frac{x^2}{4}$
##### SOLUTION
Combine the terms to get the final answer: $\int x \ln(x) \, dx = \frac{x^2}{2} \ln(x) - \frac{x^2}{4} + C$ Here, $C$ is the constant of integration.
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# Regression
#### Information
In order to learn about Regression we first need to understand a key words:
Bivariate Data - Data collected as a pair e.g.Height and Weight
Usually one variable controlled (x) and one measured experimentally (y)
(x - Independant, y - Dependent)
In order to show the mean of variables of x and y we use:
$$\bar{x}, \bar{y}$$
We expect the Line of Best Fit (LOBF) to go through the mean point
$$(\bar{x}, \bar{y})$$
We use a least squares regression line of y upon x in order to minimise the sum of the squares of the residuals (See graph below)
We want these squares to be as small as possible meaning the distance between the line and all the points are as small as they can possibly be giving us the best LOBF
This is called the regression line which is the best possible LOBF (minimal squared residuals)
We know from previous knowledge that a line can be represented by the equation:
y = mx + c
However in statistics we use the equation:
y = a + bx
(where a=c and a=m)
We know however that the LOBF goes through the mean point so we can replce the value of x and y with there mean values giving us the new equation:
$$\bar{y} = a + b\bar{x}$$
Or re-arranged
$$a = \bar{y} - b\bar{x}$$
We can find b using the equation:
$$b = \frac{S_{xy}}{S_{xx}}$$
Now we just need to use the equations for Sxy and Sxx (where n is the number of points)
$$S_{xx} = \sum{(x-\bar{x})^2} ≡ \sum{x^2} - \frac{(\sum{x})^2}{n}$$
$$S_{xy} = \sum{(x-\bar{x})(y-\bar{y})} ≡ \sum{xy} - \frac{(\sum{x})(\sum{y})}{n}$$
#### Example
Find the regression line of the points:
(0,2.4), (1,4.3), (2,5.2), (3,6.8), (4,9.1), (5,11.8)
x y $$x^2$$ xy 0 2.4 0 0 1 4.3 1 4.3 2 5.2 4 10.4 3 6.8 9 20.4 4 9.1 16 36.4 5 11.8 25 59
$$\sum{x} ≡ 0+1+2+3+4+5 = 15$$
$$\sum{y} ≡ 2.4+4.3+5.2+6.8+9.1+11.8 = 39.6$$
$$\sum{x^2} ≡ 0+1+4+9+16+25 = 55$$
$$\sum{xy} ≡ 0+4.3+10.4+20.4+36.4+59 = 130.5$$
$$S_{xy} = \sum{xy} - \frac{(\sum{x})(\sum{y})}{n} = 130.5 - \frac{15×39.6}{6} = 31.5$$
$$S_{xx} = \sum{x^2} - \frac{(\sum{x})^2}{n} = 55 - \frac{15^2}{6} = 17.5$$
$$b = \frac{31.5}{17.5} = 1.8$$
To find the mean we take the total some of x and y seperately and divide them by the number of data points
$$\bar{x} = \frac{\sum{x}}{n} = \frac{15}{6} = 2.5$$
$$\bar{y} = \frac{\sum{y}}{n} = \frac{39.6}{6} = 6.6$$
Therefore using the equation we have learned before we can see that:
$$a = \bar{y} - b\bar{x} = 6.6 - (1.8)(2.5) = 2.1$$
We can then get the final equation:
y = a + bx
y = 2.1 + 1.8x
#### Example 2
Find the regression line of the points:
(1,4.5), (2,6.0), (3,7.7), (4,7.3), (5,9.1), (6,8.6), (7,10.6), (8,10.2)
x y $$x²$$ xy 1 4.5 1 4.3 2 6.0 4 36.0 3 7.7 9 59.29 4 7.3 16 53.29 5 9.1 25 82.81 6 8.6 36 73.96 7 10.6 49 112.36 8 10.2 64 104.04
$$n = 8$$
$$\bar{x} = 4.5$$
$$\bar{y} = 8$$
$$\sum{x} = 36$$
$$\sum{y} = 64$$
$$\sum{x^2} = 204$$
$$\sum{xy} =321.7$$
$$S_{xy} = \sum{xy} - \frac{(\sum{x})(\sum{y})}{n} = 321.7 - \frac{36×64}{8} = 33.7$$
$$S_{xx} = \sum{x^2} - \frac{(\sum{x})^2}{n} = 204 - \frac{36^2}{8} = 42$$
$$b = \frac{33.7}{42} = 0.8024$$
Therefore using the equation we have learned before we can see that:
$$a = \bar{y} - b\bar{x} = 8 - (0.8024)(4.5) = 4.389$$
We can then get the final equation:
y = a + bx
y = 4.39 + 0.802x
Using this equation predict y when:
x = 3.5
y = 4.39 + (0.8024)(3.5) = 7.20
x = 11
y = 4.39 + (0.8024)(11) = 13.21
Comment on the accuracy of these results using the points given in the question
x = 11 will be less accurate as it is outside of a=our data range meaning it is Extrapolated
x = 3.5 should be reasonably accurate as it lies between our data range therefore it is Interprelated
#### Taking Information from the regression line Equation
If we have the regression line:
y = 16 + 0.115x
(where y = hours of time taken and x = area painted)
Then the regression line has an intercept of 16
and 0.115 is how quickly x increases in comparison with y
Given the point H(480,70) what is the residual for H
y = 16 + 0.115x = 16 + 0.115(480) = 71.2
Residual = 70 - 71.2 = -1.2
This means that the data point is 1.2 units away from the regression line
If someone is payed £12 per hour to paint an area of 560m²
How much will they be payed using the regression line
Replacing the value of x with the area being painted we find that:
y = 16 + (0.115)(560) = 80.4 minutes
y = 80.4/60 hours taken
We know that the person is being payed £12 per hour and works for 80.4/60 hours meaning they have earned
£12 × (80.4/60) = £16.08
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# Calculate the iterated integral. \int_{0}^{4}\int_{0}^{12}2e^{x+3y}dx\ dy
Calculate the iterated integral.
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Step 1
Consider the inner integral, name it as ${I}_{1}$
${I}_{1}={\int }_{0}^{12}2{e}^{\left(x+3y\right)}dx$
Take the constant 2 out of the integral sign.
${I}_{1}=2{\int }_{0}^{12}{e}^{\left(x+3y\right)}dx$
Step 2
Use substitution,
Substitute t=x+3y
$⇒dt=dx$
Change the limits as per the new variable,
When
When
Rewrite and solve the new integral as shown,
${I}_{1}=2{\int }_{3y}^{12+3y}{e}^{t}dt$
$=2{\left({e}^{t}\right]}_{3y}^{12+3y}$
$=2\left({e}^{12+3y}-{e}^{3y}\right)$
Step 3
Use the value of ${I}_{1}$ and solve the outer integral.
$I=2{\int }_{0}^{4}{e}^{12+3y}dy-2{\int }_{0}^{4}{e}^{3y}dy$
$I=2{e}^{12}{\int }_{0}^{4}{e}^{3y}dy-2{\int }_{0}^{4}{e}^{3y}dy$
Evaluate the integrals,
$I=\frac{2}{3}{e}^{12}{\left({e}^{3y}\right]}_{0}^{4}-\frac{2}{3}{\left({e}^{3y}\right]}_{0}^{4}$
$I=\frac{2}{3}{e}^{12}\left({e}^{12}-{e}^{0}\right)-\frac{2}{3}\left({e}^{12}-{e}^{0}\right)$
$I=\frac{2}{3}{e}^{12}\left({e}^{12}-1\right)-\frac{2}{3}\left({e}^{12}-1\right)$
Step 4
Combine the terms,
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# Complex Numbers
A complex number is an extension of a real number whose properties can be used to solve previous unsolvable problems and that model 2D geometry. Complex numbers appear in many engineering and math equations such as the Fundamental Theorem of Algebra, Euler’s Formula and the Mandelbrot Set.
## Definition
There are two assumptions, shown below, that form the definition of a complex number. These two assumptions and the properties they exhibit when combined with arithmetic and other math operators form the basis of a complex number.
1. Assume there is some number so that .
2. Give a home in the complex plane.
The notation for a complex number is shown below, where and are real numbers and .
Formally, when talking about the set of complex numbers, the symbol is used.
### Complex Constant
The first assumption is that there exists some number that satisfies the equation . This by itself can be used in combination with the rules of algebra to solve previously unsolvable problems. However, a lot of the usefulness of a complex number comes from its application to two dimensional geometry in the coordinate plane. So first, we will give the notion of a complex number a home in the complex plane.
### Complex Plane
The second assumption is that a complex number has a “home” in the complex plane and can be drawn as an arrow from the origin to some point in the plane. The vertical axis marks the complex part of the number and the horizontal axis marks the real part of the number. For example, the complex number is drawn as an arrow starting from the origin and going units to the right and units up as shown in the figure below.
Now that we have established the foundation of a complex number, we will look at some of the basic operations using complex numbers from an algebraic and geometric persepctive.
The properties of addition for two complex numbers follow the rules of algebra. The real parts of the number are added together and the complex parts of the number are added together.
Visually, addition is represented by drawing the two arrows corresponding to each of the complex numbers “tail-to-tail” as shown below. The result of the addition is the arrow drawn from the origin to the tip of the second arrow.
## Complex Multiplication
The properties of complex number multiplication demonstrate why complex numbers are able to elegantly express an area of mathematics. Conceptually, multiplying by a complex number corresponds to stretching and rotating a notation of the complex plane.
### Identity Property
Multiplying any complex number by results in . Visually, this correspond with stretching and rotating the number so that it lies on top of the complex number in the complex plane. Conceptually, you can imagine stretching and rotating all of the horizontal and vertical axis lines in the first plane.
### Complex Property
The second property of complex number mutliplication is multiplying a number by is equivalent to rotating by a quarter rotation. For example, the complex number multiplied by is equal to the rotation shown below:
Let’s verify this geometric property using algebra. Start with the complex number and multiply by .
Distribute the multiplication across the real and complex part of the number.
Substitute for in the expression.
Simplify and rearrange the expression.
This verifies the rotation property for the example complex number. This property is where the usefulness of complex numbers starts to appear.
### Distributive Property
The distributive property combines the first two properties and represents the operation of multiplying two complex numers together. Multiplication can be broken into two steps. (1) Stretch and rotate the coordinate system using the first complex number. (2) Use the transformed coordinate system to draw the second vector whose coordinates represent the result of the multiplication.
For example, let’s multiply the complex numbers and together as written below.
First visualize the stretch and rotation of the coordinate system using the first number . This gives us the blue coordinate system shown below.
Then, draw the complex number arrow using this new coordinate system. So, using the components of the second complex number, travel units along horizontal axis and units along the vertical axis in the transformed coordinate system. This is shown in red below.
The end result in the original coordinate system is the complex number .
Let’s verify this result using algebra. Start by distributing the multiplication across the terms.
Combine like terms.
Substitute for in the expression.
This result verifies the geometric interpretation of complex multiplication for this example.
## Trigonometry
The fact that multiplying two complex numbers together results in a stretch and rotation in the complex plane suggestively hints that complex numbers may be useful when applied to trigonometry. Recall, a points position on the unit circle can be defined in terms of the cosine and sine of the radian angle. A point on the unit circle corresponding to a radian angle then takes the form:
This piont is visualized by the figure below.
### Trigonometric Identities
The trigonometric identities are a set of equations that are used to transform and manipulate math expressions. Assuming that you have had exposure to trigonometry and these identities, this is one of the areas where the usefulness of complex numbers shines. The complex plane represents an elegant way to think about and transform trigonometric expressions.
The sum of two angles identities can be derived using the properties of the complex plane and Euler's formula.
Derive the difference of two angle identities using the properties of complex numbers and Euler's formula in the complex plane.
This example derives the trig. half angle identities in the complex plane.
The trigonometric double angle identites can be derived using the properties of the complex plane.
### Euler’s Formula
Perhaps the most notorious formula in math and physics is Euler’s Formula as shown above.
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Question: What Is An Irrational Number Between 4 And 5?
Is 0 is a rational number?
Zero Is a Rational Number As such, if the numerator is zero (0), and the denominator is any non-zero integer, the resulting quotient is itself zero..
Is 2.17 a rational number?
2.17 is a rational number.
What is a rational number between 5 and 6?
Five rational numbers between 5 and 6 are 5.1 , 5.2 , 5.3 , 5.4 and 5.5 .
What is the rational number between 3 and 4?
A rational number between 3 and 4 is 1/2 (3 + 4) = 7/2. Hence, 13/4, 7/2 and 15/4 are the three rational numbers lying between 3 and 4.
Is 6 a rational number?
What about the number -6? -6 can be written as -6/1. Or 6/-1. Either way, -6 is a rational number, because it can be expressed as a fraction where the numerator and denominator are integers and the denominator doesn’t equal 0.
Is root 7 irrational?
Sal proves that the square root of any prime number must be an irrational number. For example, because of this proof we can quickly determine that √3, √5, √7, or √11 are irrational numbers.
How do you find the irrational number between 2 and 3?
To find the irrational numbers between two numbers like 2and3 we need to first find squares of the two numbers which in this case are 22=4and32=9 . Now we know that the start and end points of our set of possible solutions are 4and9 respectively.
What is not a real number?
A non-real, or imaginary, number is any number that, when multiplied by itself, produces a negative number. Mathematicians use the letter “i” to symbolize the square root of -1. An imaginary number is any real number multiplied by i. For example, 5i is imaginary; the square of 5i is -25.
How do you find the irrational number between 5 and 6?
√25 < √26 < √27 < √28 < √29 < √30 < √31 < √32 < √33 < √34 < √35 < √36. Therefore, any two irrational numbers between 5 and 6 is √27 and √28.
How do you find the irrational number between 3 and 4?
So an irrational number between 3 and 4 is=3×4 =3 ×4 =3 ×2=23.
How do you find 5 rational numbers?
Answer: First method to find rational number between two numbers: Since we have to find 5 rational numbers between 3/5 and 4/5, so we will mutiply the numerator and denominator of the given numbers by 5+1, i.e. equal to 6. You can find as many rational numbers between given numbers using method given above.
What is a true number?
The real numbers include all the rational numbers, such as the integer −5 and the fraction 4/3, and all the irrational numbers, such as √2 (1.41421356…, the square root of 2, an irrational algebraic number). Included within the irrationals are the transcendental numbers, such as π (3.14159265…).
What is a rational number between 1 and 2?
Answer and Explanation: There is an infinite number of rational numbers between 1 and 2.
How we can find irrational number between two numbers?
Suppose we have two rational numbers a and b, then the irrational numbers between those two will be, √ab. Now let us find two irrational numbers between two given rational numbers. We see that x + √3 is an irrational number between 2 – √3 and 5 – √3 where 2 – √3 < x < 5 – √3. 2.
Is 5 a rational number?
Every integer is a rational number, since each integer n can be written in the form n/1. For example 5 = 5/1 and thus 5 is a rational number. However, numbers like 1/2, 45454737/2424242, and -3/7 are also rational, since they are fractions whose numerator and denominator are integers.
Is 4.5 an irrational number?
It is not a whole number because it has a fractional component. . The fact that we can write -4.5 as a fraction of integers means that this number is rational. -4.5 is NOT irrational because it is rational.
Which of the following is irrational?
The famous irrational numbers consist of Pi, Euler’s number, Golden ratio. Many square roots and cube roots numbers are also irrational, but not all of them. For example, √3 is an irrational number but √4 is is a rational number. Because 4 is a perfect square, such as 4 = 2 x 2 and √4 = 2, which is a rational number.
How do you know if a number is rational or irrational?
All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction. An irrational number has endless non-repeating digits to the right of the decimal point.
Is 17.5 rational or irrational?
(also called the positive integers), or 0, 1, 2, 3, 4, 5, . . . (also called the nonnegative integers or whole numbers). Natural number is the proper subset of positive integer , but 17.5 is a fractional value. So, it is not a natural number.
What are the rational number between 4 and 5?
The five rational numbers between 4 and 5 are:-41/10,42/10,43/10,44/10,45/10.
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Function and Polynomials
Subject: Optional Mathematics
Overview
If f: A → B be a function from set A to set B and g: B → C be the function set B to set C, then the new function from set A to set C is called composite function of F and G. It is denoted by gof. The algebraic expressions in which the power of the variables is whole numbers are called polynomials. Rules of polynomials: 1 . Multiply each term of one polynomial by each term of the order. 2 . Arrange the terms in ascending or descending order.
FUNCTION
Let A and B be two non-empty sets, then, every subset of cartesian product A$\times$B is a relation from A to B. Thus, a relation in which every element of set A is related (or associated) with a unique element of set B is said to be a function from A to B. Such as function is denoted by f : A→ B, which is read as "f is a function from A to B".
Symbolically, it follows that f: A → B if;
(i) f ⊆A×B
(ii) for each given x ∈ A, there exist a unique y∈B such that (x, y) ∈ f.
If an element y of B is associated with an element x of A, then y is an image of x under f and x is called the pre-image of y under f. We can also write y=f(x) and read f(x) as "f of x" or "f at x". A function is usually denoted by f, g, F, G, etc. A function is a relation in which, no two different ordered pairs have the same first component.
Composite Function
Let f and g be the function from A to B and from B to C, respectively, where A= {a1,b2,c3}, B = {c3,d4,e5} and C = {e5, f6, g7,h8,i9} and these are defined as follows:
Here, if f : A →B, a1 maps into c3 ; b2 maps into d4 ; and c3 maps into e5.
If g: B→ C, c3 maps into f6 ; d4 maps into g7 ; and e5 maps into h8.
thus we may write c3= f (a1), d4 = f(b2), e5= f(c3) and f6= g(c3), g7 = g(d4), h8 = f(e5)
Then, we have f6 = g(c3) = g [f(a1)],
g7 = g(d4) = g[f(b2)]
and h8=g(e5) = g[f(e3)]
this defines a set of ordered pairs of a function from A into C. This function is called the (product) composite function of g and f. it is denoted by gof or simply by gof. the nation gof indicates that f is applied first and then g. Thus, the arrow diagram of the new function is alongside.
Inverse Function
Let f: A→ B be a function from A to B defined by the following arrow.
thus, the function f = { (a,1), (b,2) (c,3) } is one -one into where the domain of f = {a, b, c} and the range of f= { 1,2,3 }.
If we interchange the domain set of 'f' into range set and range set of into domain set, the set of ordered pairs will become {(1, a), (2, b), (3, c)}. Let this function e g. Then the function.
g= {(1, a), (2, b) (3, c) } is again a one-one onto, where the domain of g= {1,2,3} and the range of g={a,b,c}
the arrow diagram of this function g is as follows.
in such case, the function g is called the inverse function of and vice-versa. the inverse of the function f(x) is written as f-1read as 'f inverse ' and we have y=f(x) if and only if x = $f^{-1}$ (y) for every x∈D(f) and every y∈R(f). let u consider a many-one function defined by the following arrow diagram.
Thus, the function h= {(1, 2), (2,2), (3, 3) }
Let, we interchange the domain set and range set of h. we will get a relation R which is given below.
R= {(2, 1), (2, 2), (3, 3) }. Then the arrow diagram of this relation is given in the adjoining figure.
In relation R, two ordered pairs (2,1) and (2, 2) have the same first element 2. So, R is not a function.
It is concluded that a function f: A→ B will have its inverse function $f^{-1}$ B→ A if and only if f is a one -one onto function.
Definition: If f: A→ B is a one to one onto function from A to B, then there exist a function g: B→ A such that the range of f is the domain of g is called the inverse of 'f' denoted by $f{-1}$ (f inverse) such that y= f(x) if and only x=$f{-1}$ (y) for every x∈D (f) and every y∈R(f).
Simple Algebraic Functions
A function that can be defined as the root of a polynomial equation is known as an algebraic function. we shall discuss some example of functions defined on the set R real number onto itself and these functions f(x) are defined by means of an equation. for instance the algebraic functions. \begin{align}f: R→R\end{align}
$$f(x)= a_ox^n +a_1x^n-1+……………+ a_n-1 x+a_n$$
For all x€R where the right side is a polynomial of degree n.
We shall discuss some particular causes of this equation.
Constant Function: For n=0, we have f(x)=ao. it is usually denoted by f(x)=c.
In other words, a function is said to be a constant function if all its function values are the same. The graph of constant function is a straight line parallel to the x-axis at a given distance
Linear functions: For n=1, we have $$f(x) = a_ox + a_1$$. It is usually denoted by y = ax+b
In another word, a function is said to be a linear function if the polynomial is degree one. The graph of it is a straight line with slope m=a and y-intercept = for instance, y=x+2 is an equation of a straight line.
Identity function: if a=1 and b=0 in the linear function f(x) = ax +b, then we have
$$F(x) = x$$ for all $$x€R$$
This function is called identity function and its graph is shown below. It bisects the angle between the axes of coordinates.
Quadratic function: for n = 2, we have $$f(x)= a_1x^2 + a_1x + a_2.$$ It is written as $$f(x)=ax^2 + bx + c$$
This is a quadratic function which is polynomial of degree 2 and its graph is a parabola.
For instance, $$f(x) = x^{2} + x -2, f(x) = 4x^2, f(x) = x^2 + 2$$, etc are quadratic functions.
The graph of these functions is shown below.
Cubic function : for n= 3, we have$(x) f(x) = a_ox^3 +a_1x^2 + a_2x + a_3$ It is written as $f(x)=ax^3 + bx^2 +cx + d$
This is a cubic function which is polynomial of degree 3. For instance $$f(x) = x^3$$ is a cubic function whose graph is shown alongside
Trigonometric functions
A function defined as the function which associates each angle with the definite real number.So, the domain of the trigonometric function is the set of angles and its co-domain is the set of real numbers. traditionally trigonometric functions are defined for angles of a triangle.But these trigonometric functions can be defined for angles of any magnitude.
The trigonometric ratios for angles such as 300, 450, 600, etc. can be calculated with the help of elementary plane geometry. The following table shows the values of trigonometrical ratios from 00 to 3600.
θ 00 300 450 600 900 1200 1350 1500 1800 Sinθ 0 $\frac{1}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{\sqrt{3}}{2}$ 1 $\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{1}{2}$ 0 Cosθ 1 $\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt 2}$ $\frac{1}{2}$ 0 $-\frac{1}{2}$ $-\frac{1}{\sqrt{2}}$ $-\frac{\sqrt{3}}{2}$ -1 Tanθ 0 $\frac{1}{\sqrt{3}}$ 1 $\sqrt 3$ ∞ $-\sqrt 3$ -1 $-\frac{1}{\sqrt{3}}$ 0
POLYNOMIALS
Polynomial is an algebraic expression consist of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents. 2x, 5x2 + 3, 2a3 - b2 + 3x - 2, etc are the examples of polynomials.
Simple Operations on Polynomials
Multiplication of polynomials
In order to multiply two polynomials, following steps are used.
(i) Multiply each term of one polynomial by each term of the other and
(ii) Then add (or subtract) the like terms thus obtained.
(iii) Simplify and arrange the like terms in ascending order or descending order.
In multiplying two polynomials, the law of indices $x^m×x^n=x^{m+n}$ is applied.
it is noted that the degree of the product of two polynomials is equal to the sum of the degrees of the polynomial factors.
Worked out with examples
Example 1.
If $f(x)=3x^3v^4\: and\:g(x)=5x^2y^3$, find $f(x).g(x)$
Solution:
\begin{align*}f(x)\times g(x) &=3x^3y^4×5^2y^3=(3×5)\:(x^4×y^3) \\ &=15.x^{3+2}.y^{4+3}=15x^4y^7\end{align*}
Example 2.
If $f(x)=x^2-3x+2$ and $g(x)=x^3-x^2+2x+4$, find $f(x).g(x)$.
Solution:
Horizantal method:
\begin{align*} f(x)\times g(x)&=(x^2-3x+2)(x^3-x^2+2x+4) \\ &=x^2(x^3-x^2+2x+4)-3x(x^3-x^2+2x+4)+2(x^3-x^2+2x+4) \\ &=x^5-x^4+2x^3+4x^2-3x^4+3x^3-6x^2-6x^2-12x+2x^3-2x^2+4x+8 \\ &=x^5+(-1-3)x^4+(2+3+1)x^3+(4-6-2)x^2 (-12+4)x+8 \\ &=x^5-4x^4+7x^3-4x^2-8x+8\end{align*}
Vertical Method:
\begin{align*}\frac{\frac{x^3-x^2+2x+4\\\times\:x^2-3x+2}{2x^3-2x^2+4x+8\\-3x^4+3x^3-6x^2-12x\\x^5-x^4+2x^3-4x^2\\}}{x^5-4x^4+7x^3-4x^2-8x+8}\\ \end{align*}
Division of Polynomials
We know from arithmetic how to divide ad integer by another smaller integer. If 30 is divided by 7, the quotient is 4 and the remainder is 2. i.e., \begin{align*}7)30(4 \\ \frac{-28}{2}\end{align*}
Here, we observe that $30=7×4+2.$
this relation ca be stated as follows.
Dividend= Divisor × Quotient + Reminder
Similarly, we can divide polynomials.
let $f(x)$ and g(x) be two polynomials such that g(x) is a polynomial of smaller degree than that of $f(x)$ and g(x)≠0. Then, there exist unique polynomials Q(x) and R(x) such that
$f(x)=g(x).Q(x)+R(x)$
where $F(x)$=dividend, g(x)=divisor, Q(x) is quotient and R(x) is remainder.
If R(x) = 0, then the divisor g(x) is a factor of the dividend f(x). The other factor of f(x) is the quotient Q(x).
The relation f(x) = g(x). Q(x) + R(x).
The following steps are used to divide a polynomial by the other:
(i) Arrange the divided f(x) and divisor g(x) in standard form i.e. generally descending powers of variable x.
(ii) Divided the first term divided f(x) by the first term of divisor g(x) to get the first term of quotient Q(x).
(iii) Multiply each term of divisor g(x) by the first term of quotient Q(x) obtained in step (ii) and subtract the product so obtained from the dividend f(x).
(iv) Take the remainder obtained in step (iii) as new dividend and continue the above process until the degree of the remainder is less than of the divisor.
Remainder Theorem
Statement: If a polynomial f(x) is divided by x - a, then the remainder is f(a).
Proof: If we divided f(x) by x - a, then we get Q(x) as quotient and R as remainder.
Then, f(x) = (x - a). Q(x) + R
Put x = a. Then,
or, f(a) = (a - a) Q (a) + R
or, R = f(a)
Hence, remainder = f(a) = the value of polynomial f(x) = a
Factor theorem
Statement: if f (x) is a polynomial and a is real number, then (x - a) is a factor of f(x) if f(a) = 0
Proof:If we divide f(x) by x-a, then we get Q(x) as quotient and R as remainder.
Then, f(x) = (x - a). Q(x) + R ...........(i)
Put x = a. Then
f(a) = (a - a). Q(a) + R
or, f(a) = R
When f(a) = 0. Then R = 0.
Putting the value of R in (i) we get,
f(x) = (x - a). Q(x)
So, (x - a) is a factor of f(x).
Hence, (x - a) is a factor of f(x) if f(a) = 0.
Synthetic Division
Synthetic division is the process which helps us to find the quotient and remainder when a polynomial f(x) is divided by x - a.
Application of synthetic division
Let Q(x) and R be the quotient and remainder when a polynomial f(x) is divided by ax -b.
Then, f(x) = (ax - b). Q(x) + R = a(x - $\frac{b}{a}$). Q(x) + R .
Where a.Q(x) = g(x)
or, Q(x) = $\frac{1}{a}$ g(x).
Here, g(x) and R are the quotient and reminder when f(x) is divide by (x - $\frac{b}{a}$).
This result leads us to conclude that process of synthetic division discussed earlier is also useful to find out the quotient and remainder when f(x) is divided by (ax - b)
Factorization of a polynomial
Factor theorem and the synthetic division are very useful to find the factors of a polynomial.
Let us see the following example:
Factorize: x2 + x - 2
Constant term of this polynomial is 4 and the possible factors of 4 are: ±1, ±2.
Since the degree of f(x) is 2, so there will be at most two factors.
When x = 1, f(1) = 1+1-2 =0
(x-1) is a factor.
When x = -1, f(1) = -1-1-2 =-4
(x+1) is not a factor.
when x = 2, f(2) = 4+2-2 = 4
(x+2) is not a factor
When x = -2, f(1) = 4-2-2 = 0
(x+2) is a factor.
∴ (x-1) and (x+2) is a factor.
∴ x2 + x - 2 = (x-1)(x+2).
But instead of finding all the factors by using factor theorem, the synthetic division can be used after getting one with the help of factor theorem.
Polynomial Equation
Let f(x) = anxn + an-1xn-1 + ..... + ao be the polynomial in x. Then f(x) = 0 is called a polynomial equation in x.
ax + b = 0 is a linear equation.
zx2+ bx + c = 0 is a quadratic equation.
ax$^3$ + bx2 + cx + d = 0 is a cubic equation
ax$^4$ + bx$^3$ + cx2 + dx + e = 0 is a big quadratic equation.
( Here, a,b,c,d,e are the real numbers).
If α is a real number such that f(α) = 0, thenα is called a root of the polynomial equation f(x) = 0.
Things to remember
Functions:
Let A and B be two non-empty sets, then, every subset of cartesian product A$\times$B is a relation from A to B. Thus, a relation in which every element of set A is related ( or associated ) with a unique element of set B is said to be a function from A to B. Such as function is denoted by f : A→ B, which is read as "f is a function from A to B".
Polynomials:
Polynomial is an algebraic expression consist of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents. An example of a polynomial of a single indeterminate x is x2 − 4x + 7.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
The Factor Theorem and The Remainder Theorem
f= {(1,3), (0,0), (-1,-3)}, g= {(0,2), (-3,-1), (3,5)}, the above relation shown in the mapping diagram as follow:
gof= g(3) = 5
gof = g(0) =2
gof= g(-3) = -1
gof = g {(0,2), (-1,-1), (1,5)} Ans
f = {(-2,4), (-1,8), (0,0), (1,4), (2,6)},
g = {(o,-2), (4,0), (6,1), (8,-2)}
The mapping diagram is shown below:
gof(-2) = g(4) = 0
gof(-1) = g(8) = -2
gof(0) = g(0) = 0
gof(1) = g(4)=0
gof(2) = g(6) = 1
gof= {(-2,0), (-1,-2), (0,-2) , (1,0), (2,1)} Ans
f = {(p,1), (q,2), (r,3), (s,4)}, g = {(1,5), (2,8), (3,-3), (4,6)}
The mapping diagram is shown below:
fog = {(5,p), (8,q), (-3,r), (6,s)} Ans
f = {(x,3), (y,4), (z,7)}, g = {(3,a), (4,b), (7,c)}
The mapping diagram is shown below:
fog = {(a,x), (b,y), (c,z)} Ans
f(x) = x3 and g(x) = x2
fg(x)
= f(x2) [$\because$ g(x) = x2]
= (x2)3 [$\because$ f(x) = x3]
= x6Ans
f(x) =$\frac 1x$
ff(x) = f $\frac 1x$ [$\because$ f(x) = $\frac 1x$]
= $\cfrac{1}{ \cfrac{1}{x}}$ =1× $\frac x1$ = x
∴ ff(2) = 2 Ans
Here, f(x) = 3x2 + x - 28 and g(x) = 6-x
f(x)⋅ g(x)
= (3x2 + x - 28) (6-x)
= 18x2 + 6x - 168 -3x3-x2 + 28x
= -3x3 + 17x2+ 34x - 168 Ans
Here,
f = {(2,$\frac 12$), (3, $\frac13$), (4,$\frac 14$)}
Range = { $\frac 12$, $\frac 13$, $\frac 14$}
Inverse function (f-1) = {($\frac 12$, 2), ($\frac 13$, 3), ($\frac 14$, 4)}Ans
Here,
1. {(3,2), (-1,-7), (4,5), (6,8)}
Let f = {(3,2), (-1,-7), (4,5), (6,8)}
Changing domain into range and range into domain.
f-1 = {(2,3), (-7,-1), (5,4), (8,6)}
1. {x, 2x+1 : x∈ R}
Let y = f(x) = 2x+1
Interchange the place of x and y
x = 2y + 1
x-1 = 2y
2y = x-1
y = $\frac {x-1}{2}$
∴ f-1 = $\frac{x-1}{2}$ Ans
f(4x-15) = 8x-27
Or,
f(4x-15) = 2(4x-15) + 3
∴ f(x) = 2x+3
ff(x) = f(2x+3) = 2 (2x+3) +3 = 4x +6+3 = 4x+9
ff(2) = 4 × 2 + 9 = 17 Ans
g(2x-5) = 8x-19
g(2x-5) = 4(2x-5) +1
g(x) = 4x+1
gg(x) =g(4x+1) = 4(4x+1) +1 = 16x+4+1 = 16x+5
gg(2) = 16×2+5 = 32+5 =37 Ans
f(x+3) = 3x+5
f(x+3) = 3(x+3) -4
f(x) = 3x-4
Let y = f(x) = 3x-4
Y = 3x-4
Interchanging the place of x and y in the above relation
x = 3y-4
Or, 3y = x+4
y = $\frac {x+4}{3}$
f-1 (x) = $\frac {x+4}{3}$ Ans
fog(x) = 6x-2
Or, f[g(x)] = 6x-2
Or, f(2x) = 6x-2
Or, f(2⋅$\frac x2$ = 6 $\frac x2$-2
f(x) = 3x-2 Ans
Let g(x) = ax+b
fog(x) = f[g(x)] = f(ax+b) = 8(ax+b) -3
fog(x) = 8ax +8b - 3 ------ (1)
fog(x) = 20x+1 ------------(2)
From eqn (1) and (2)
8ax +8b -3 = 20x +1
Equating on both sides,
8a = 20
a = $\frac {20}{8}$
∴ a = $\frac 52$
8b -3 = 1
8b = 4
b =$\frac 48$ = $\frac 12$
∴ b = $\frac 12$
Putting the value of a and b in g(x) = ax + b
g(x) = $\frac 52$ x + $\frac 12$ = $\frac {5x+1}{2}$Ans
Function f = {(x, 2x+1) : x∈R}
Let, y = 2x + 1
Interchanging the place of x and y
x = 2y + 1
or, x-1 = 2y
or, y = $\frac {x-1}{2}$
∴ f-1= {(x, $\frac {x-1}2$) :x∈R } Ans
f(x) = 2x + 3, g (x) = 5x2
fg (x)
= f (5x2) [$\because$ g(x) = 5x2]
= 2×5x2+ 3
= 10x2 +3 Ans
Also,
gf(x)
= g (2x + 3) [$\because$ f(x) = 2x + 3 ]
= 5 (2x + 3)2Ans
fog (2)
= f(2x - 3) [$\because$ g(x) = x - 3 ]
= f(-1)
= (-1)2 + 5 [$\because$ f(x) = x2 + 5 ]
= 1 + 5
= 6 Ans
x3 + 3x2 - 2x + 6÷ (x - 3)
3 1 3 -2 6 3 18 48 1 6 16 54
The remainder (R) = 54 Ans
If the polynomial f(x) is divided by x - a, the remainder is f(a)
Proof: Let, Q(x) be the quotient and R be the remainder when f(x) is divided by (x - a) then, f(x) = (x -a)⋅Q(x) + R.
When x = a, f(a) = (a -a)⋅Q(x) + R ∴ f(a) = R proved
If the polynomial f(x) is divided by (x - a) and R = f(a) = 0, then (x - a) is a factor of f(x).
Proof: Let Q(x)
Let Q(x) be the quotient when f(x) is divided by (x - a), then f(x) = (x-a) Q(x) + R = (x-a) Q(x) + 0
f(x) = (x-a) Q(x)
∴(x-a) is a factor of f(x). proved
x4 - x3 - 3x2 -2x + 5÷ (x + 1)
-1 1 -1 -3 -2 5 -1 2 1 1 1 -2 -1 -1 6
∴ Quotient (Q) = x3 - 2x2 - x -1
Remainder (R) = 6 Ans
x3 - 2x + 3 is divided by x - 1.
1 1 0 -2 3 1 1 -1 1 1 -1 2
Quotient (Q) = x2 - x -1
Remainder (R) = 2 Ans
Here,
f(x)÷ g(x) = $\frac {x^3 +3x^2 -4x +2}{x +1}$
x + 1 x3 + 3x2 - 4x +2 x2 + 2x - 6 x3 + x2 - - 2x2 - 4x + 2 2x2 + 2x - - -6x + 2 -6x - 6 + + 8
∴ Quotient Q(x) = x2 + 2x - 6
Remainder R(x) = 8 Ans
Here,
x - k is factor of the polynomialx3 - kx2 - 2x + k + 4, then x = k must satisfies the given polynomial, so that the given polynomial should be zero.
x - k is a factor of f(k) then f(k) = 0
f(k) = k3 - k⋅ k2 - 2k + k + 4
or, 0 = k3 - k3 -k +4
or, 0 = -k +4
∴ k = 4 Ans
Here,
f(x) =2x3 + ax2 + x + 2
2x + 1 is a factor of f(x) so f(x) = 0
2x + 1 = o
or, 2x = -1
∴ x = - \frac 12
2x3 + ax2 + x + 2 = 0
or, 2× (- $\frac 12)^3$ + a⋅ (- $\frac 12)^2$ - $\frac 12$ + 2 = 0
or, 2× - $\frac 18$+ a⋅ $\frac 14$ - $\frac 12$ + 2 = 0
or, - $\frac 14$ + $\frac a4$ - $\frac 12$ + $\frac 21$ = 0
or, $\frac {-1 +a -2 +8}{4}$ = 0
or, a + 5 = 0
∴ a = -5 Ans
Arrange the given polynomial in decreasing order
f(x) = 4x3 -6x2+3x -5
The constant term with the sign changed is +2
∴ x - 2 = x - (+2)
Writes the coefficient of f(x) in decreasing order:
2 4 -6 +3 -5 ↓ 8 4 14 4 2 7 9 Q R
∴ Quotient Q(x) = 4x2 +2x +7
Remainder R(x) = 9 Ans
Arranging the given polynomial in descending order
f(t) = 7t4 -4t3 +6t +3
The constant term with sign changed is + 3,
t - 3 = t - (+3)
Write the coefficient in descending order;
+3 7 -4 0 6 3 ↓ 21 51 153 477 7 17 51 159 480 Q R
∴ Quotient Q(t) = 7t3 + 17x2 +51x +159
Remainder R(t) = 480 Ans
(x + 2) is a factor of x3 + kx2 - 4x + 12 or x + 2 = 0 or x = -2
Putting the value x = -2 in the given expression and f(-2) = 0
x3 + kx2 - 4x + 12 =0
or, (-2)3 + k(-2)2 - 4× (-2) + 12 = 0
or, -8 + 4k + 8 + 12 = 0
or, 4k = -12
or, k = - $\frac {12}{4}$ = -3
∴ k = -3 Ans
x - 3 is a factor of the given expression.
So, x=3 must satisfy the given expression, then f(3) = 0
f(x) = x3 + 4x2 + kx - 30
or, f(3) = 33 + 4×32 + k⋅3 - 30
or, 0 = 27 + 36 +3k -30
or, 3k + 33 = 0
or, 3k = -33
or k = -$\frac {33}{3}$ = -11
∴ k = -11 Ans
f(x) = x3 - kx2 - x - 2
If (x - 2) is a factor of f (x) then
f(2) = 0
f(2) =23 - k⋅22 - 2- 2
or, 0 = 8 - 4k - 4
or, 0 = 4 - 4k
or, 4k = 4
or, k = $\frac 44$
∴ k = 1 Ans
If f(a0 = 0, the remainder when f(x) is divided by (x - a) is zero then (x - a) is factor of f(x).
x + 1 = 0
or, x = -1
∴ f(-1) = 0
2x3 - kx2 - 8x + 5
Putting the value of x = -1 in above relation,
f(-1) = 2 (-1)3 - k (-1)2 - 8 (-1) + 5
or, - 2 - k + 8 + 5 = 0
or, 11 - k = 0
∴ k = 11 Ans
Quotient Q(x) =x2 + 2x + 1, Remainder R(x) = 2, Original polynomial = f(x)
We know,
f(x) = (x + 1) Q(x) + R(x)
or, f(x) = (x + 1) (x2 + 2x + 1) + 2
or, f(x) = x3 + 2x2 + x -x2 - 2x - 1 + 2
or, f(x) =x3 + x2- x + 1 Ans
Let,
f(k) =x3 - kx2 - 2x + k + 4
If x - k is a factor of f(k) then f(k) = 0
f(k) = k3 - k⋅k2 - 2k + k + 4
or, 0 = k3 - k3 - k + 4
or, 0 = -k + 4
∴ k = 4 Ans
x + 3 is a factor of the given expression.
When x = -3, putting the value in the given expression equal to zero.
(-3)3 - (k - 1) (-3)2 + k(-3) +54 = 0
or, -27 -9k + 9 - 3k + 54 = 0
or, -12k + 36 = 0
or, -12k = -36
or k = $\frac {-36}{-12}$
∴ k = 3 Ans
If a number C is substituted for x in the polynomial p(x) of degree n, then P(C) is the remainder that would be obtained by dividing p(x) by x - c.
i.e. P(x) = Q(x)⋅ (x - c) + P(c)
where, Q(x) is a polynomial of degree n -1
f(x) = 3x3 - 5x2 + 2x - 3 and g(x) = x - 2 = x - (-2)
If f(x)÷ g(x), quotient = Q and remainder (R) = ?
f(2) = 3(2)3 - 5(2)2 + 2×2 - 3 = 24 - 20 + 4 -3 = 28 - 23 = 5
Remainder (R) = 5 Ans
Remainder theorem: The remainder theorem states that if f(x) is divided by (x - a), then f(a) will be the remainder.
x + 1 is a factor ofx4 - 3x3 -2x2 +x +5
f(-1) =(-1)4 - 3(-1)3 -2(-1)2 +(-1) +5 = 1 +3 -2 -1 +5 = 6 Ans
x + 2 is a factor of 3x2 + px2 - 2x - 8, so the given expression is satisfies by x = -2
3x2 + px2 - 2x - 8 = 0
or, 3(-2)2 + p(-2)2 - 2(-2) - 8 = 0
or, -24 + 4p +4 -8 = 0
or, 4p = 28
∴ p = $\frac {28}{4}$ = 7 Ans
Factor Theorem: If a polynomial f(x) is divided by (x - a) and remainder R = f(a) = 0 then x - a is a factor of f(x). This theorem is known as the factor theorem.
Given,
f(x) = 2x4 - 3x2 + 6x + k and f(1) = 0
If x = 1 then
2 × 14 - 3 × 12 + 6 × 1 + k = f(1)
or, 2 - 3 + 6 + k = 0
or, 5 + k = 0
∴ k = -5 Ans
Constant Function: A function f:A → B is called a constant function if there exists an element C∈B such that f(x) = C for all X∈A. A set containing only one element.
Let: Y = f(x) = 7x - 8
Y = 7x -8 [∴ Range = 13]
or, 7x = 13 + 8
or, x = $\frac {21}{7}$ = 3
∴ Domain = 3 Ans
Remainder Theorem: When a polynomial f(x) is divided by a linear polynomial x - a then the remainder R is given by the value f(a) of the polynomial, R = f(a).
x - 3 2x3 - 7x2 + 5x + 4 2x2 - x + 2 2x3 - 6x2 - 445 - + - - x2 + 5x + 4 - x2 + 3x + - 2x + 4 2x - 6 - + 10
∴ Remainder = 10 Ans
Remainder Theorem: If f(x) is a polynomial of degree n in x and if f(x)is divided by x - a, then the remainder is f(a). This theory is known as Remainder Theorey.
Let: f(x) =a4 - 3a3 - 2a2 + a + p
If a + 1 is factor of f(x) then f(-1) = 0
f(-1) =(-1)4 - 3(-1)3 - 2(-1)2 + (-1) + p
or, 0 = 1 + 3 - 2 - 1 + p
or, p + 1 = 0
∴ p = -1 Ans
f(2) = 8
f(2) =2×23 + 3×22 + k (1 - $\frac {3×2}{k}$ )
or, 8 = 16 + 12 +k (1- $\frac 6k$)
or, 28 + k $\frac {k-6}{k}$ = 8
or, k - 6 = 8 - 28
or, k = -20 + 6
∴ k = -14 Ans
f(x) =3x3 - 2x2 + 4x - 1 and g(x) = 3x + 2
g(x) = x + $\frac 23$ = x - (- $\frac 23$)
In f(x)÷ g(x), remainder R = f(a) where a = - $\frac 23$
f(-$\frac 23$) = 3 (-$\frac 23)^3$ + 2 (-$\frac 23)^2$ + 4×(-$\frac 23$) - 1
= -$\frac 89$ + $\frac 89$ - $\frac 83$- 1
= $\frac {-8-3}{3}$
= -$\frac {11}{3}$
∴ Remainder R =f(- $\frac 23$) =-$\frac {11}{3}$ Ans
f(x) =6x3 - (k + 6)x2 + 2kx - 25
If (2x - 5) is a factor of f(x) then f($\frac 52$) = 0
f(\frac 52) = 6($\frac 52)^3$ - (k + 6)($\frac 52)^2$ + 2k($\frac 52$) - 25
or, 0 = $\frac {375}{4}$ - $\frac{25}{4}$ (k + 6) + 5k - 25
or, 0 = $\frac {375 - 25k - 150 + 20k - 100}{4}$
or, -5k + 125 = 0
or, -5k = -125
or, k = $\frac {-125}{-5}$
∴k = 25 Ans
Let, f: A→B be a one to one onto function then a function f-1 : B→A is called an inverse function of f. i.e.
f = {(2, 5), (3, 6), (-4, 1). (7, 4)}
f-1 ={(5, 2), (6, 3), (1, -4). (4, 7)} Ans
Here,
f(x) = 2x3 + 3x2 - 3x + p and f(2) = 8
Putting the value of x = 2,
f(2) = 2 × 23 + 3 × 22 - 3 × 2 + p
or, 8 = 16 + 12 - 6 + p
or, p = 8 - 22
∴ p = -14 Ans
Here,
f(x) = 2x4 - 3x2 + 6x + k and f(1) = 0
Putting the value of x = 1
f(1) = 2 × 14 - 3 × 12 + 6 × 1 + k
or, 0 = 2 -3 + 6 + k
or, k + 5 = 0
∴ k = - 5 Ans
Here,
f(x) = x2 and g(x) = 3x
1. fog(x) = f(3x) = (3x)2 =9x2Ans
2. gof(x) = g(x2) = 3x2Ans
3. fog(2) = 9x2 = 9 (2)2 = 36 Ans
4. gof(2) = 3x2 = 3 (2)2 = 12 Ans
i. gof(x)
= g(2x - 3) [$\because$ f(x) = 2x - 3]
= 3(2x - 3) + 4 [$\because$ g(x) = 3x + 4]
= 6x - 9 + 4
= 6x -5 Ans
Let, y = f(x) = 2x - 3
∴ y = 2x - 3
Interchanging the place of x and y
x = 2y - 3
or, 2y = x + 3
or, y = $\frac {x+3}{2}$
i.e. f-1(x) =$\frac {x+3}{2}$
Let, y= g(x) = 3x + 4
∴ y = 3x + 4
Interchanging the place of x and y
x = 3y + 4
or, 3y = x - 4
or, y = $\frac {x-4}{3}$
i.e. g-1(x) =$\frac {x-4}{3}$
ii. fog-1(x)
= f$\frac {x-4}{3}$ [$\because$ g-1(x) =$\frac {x-4}{3}$ ]
= 2 $\frac {x-4}{3}$ -3
= $\frac {2x- 8 - 9}{3}$
= $\frac {2x - 17}{3}$Ans
iii. f-1og (2)
= f-1(3× 2 + 4)
=f-1(10)
= $\frac {10 + 3}{2}$ [$\because$ $\frac {x+3}{2}$]
= $\frac {13}{2}$ Ans
iv. fog(-3)
= f(3× -3 + 4) [$\because$ g(x) = 3x + 4]
= f( -9 + 4)
= f(-5)
= 2× (-5) - 3
= -10 - 3
= -13 Ans
let, y= f(x) = 8 - 3x
Interchanging the place of x and y
x = 8 - 3y
or, 3y = 8 - x
∴ y = $\frac {8 - x}{3}$
i.e. f-1(x) = $\frac {8 - x}{3}$
i. f-1(-4) = $\frac {8 - (-4)}{3}$ = $\frac {12}{3}$ = 4 Ans
ff(x) = f(8 - 3x) = 8 - 3(8- 3x) = 8 - 24 + 9x = 9x - 16
ff(2) = 9× 2 - 16 = 18 - 16 = 2 Ans
Let, y = g(x) = 3x - 5
y = 3x - 5 ---------- (1)
Interchanging the place of xand y in equation (1)
x = 3y - 5
or, 3y = x + 5
∴ y = $\frac {x + 5}{3}$
i.e. g-1(x) = $\frac {x + 5}{3}$
fg-1(x) = 15
or, f$\frac {x + 5}{3}$=15
or, 4$\frac {x + 5}{3}$ = 15
or, $\frac {4x + 20 + 21}{3}$ = 15
or, 4x + 41 = 45
or, 4x = 45 - 41 = 4
or, x = $\frac 44$
∴ x = 1 Ans
Let, y = f(x) = 3x + 5
or, y= 3x + 5
Interchanging the position of x and y,
x = 3y + 5
or, 3y = x - 5
or, y = $\frac {x - 5}{3}$
∴ f-1(x) =$\frac {x - 5}{3}$
f-1(2) =$\frac {2 - 5}{3}$ = - $\frac 33$ = -1
ff-1(2) = f(-1) = 3× (-1) + 5 = - 3 + 5 = 2
f-1f(3) = f-1 (3× 3 +5) = f-1 (14) = $\frac {14 - 5}{3}$ = $\frac 93$ = 3
Hence, ff-1(2) = 2
f-1 f(3) = 3 Ans
ghf(x)
= gh(3x - 4)
= g [ -2 (3x - 4) + 1]
= g(-6x + 8 + 1)
= g(- 6x + 9)
= -6x + 9 +3
= - 6x + 12
Let,
y = g(x) = x + 3
∴ y = x + 3
Interchanging the place of x and y,
x = y + 3
or, x - 3 = y
i.e y = x - 3 [$\because$ g-1(x) = x - 3]
g-1hf(x)
= g-1h(3x - 4)
= g-1 [ - 2(3x - 4) + 1]
= g-1 (-6x + 8 + 1)
= g-1 (-6x + 9)
= - 6x + 9 -3
= - 6x + 6
Hence, ghf(x) = - 6x + 12
and, g-1hf(x) = -6x + 6 Ans
Given,
f(x) = 3x + 4
g(x) = 2(x + 1) = 2x + 2
fog(x) = f(2x + 2) = 3 (2x + 2) + 4 = 6x + 6 + 4 = 6x + 10
gof(x) = g(3x + 4) = 2 (3x + 4) + 2 = 6x + 8 +2 =6x + 10
Hence, fog(x) = gof(x) proved
Let y = 3x + 4 = f(x)
y = 3x + 4
Interchanging the place of x and y
x = 3y + 4
or, 3y = x - 4
or, y = $\frac {x - 4}{3}$
f-1(x) = $\frac {x - 4}{3}$
∴ f-1(2) = $\frac {2 - 4}{3}$ = - $\frac 23$Ans
Let y = f(x) = $\frac 1x$
y = $\frac 1x$
Interchanging the place of x and y
x = $\frac 1y$
y = $\frac 1x$
∴ f-1(x) = $\frac 1x$
L.H.S
=fof-1 (x)
= f$\frac 1x$
=$\cfrac{1}{ \cfrac{1}{x}}$
= x
R.H.S
= f-1of(x)
= f-1$\frac 1x$
= $\cfrac{1}{ \cfrac{1}{x}}$
= x
Hence, L.H.S = R.H.S proved
Let y = f(x) =$\frac {x}{x-3}$
y = $\frac {x}{x - 3}$
Interchanging the place of x and y,
x = $\frac {y}{y - 3}$
or, xy- 3x= y
or, xy - y = 3x
or, y (x - 1) = 3x
or, y = $\frac {3x}{x - 1}$
i.e., f-1(x) = $\frac {3x}{x - 1}$
From the question, f(x) = f-1(x)
$\frac {x}{x - 3}$ =$\frac {3x}{x - 1}$
or, x2 - x = 3x2 - 9x
or, 3x2 - 9x - x2 + x = 0
or, 2x2 - 8x = 0
or, 2x(x - 4) = 0
Either, x = 0 and x - 4 = 0 or, x = 4
∴x = 0, 4 Ans
Here,
f(x) = x2 - 2x and g(x) = 2x + 3
Let: y = g(x) = 2x + 3
Interchanging the place of x and y
x = 2y + 3
or, 2y = x - 3
or, y = $\frac {x-3}{2}$
∴ g-1(x) =$\frac {x-3}{2}$
fg-1(x) = 3
f$\frac {x-3}{2}$ = 3
or, $(\frac {x-3}{2})^2$ - 2 $\frac {x-3}{2}$ = 3
or, $\frac {x^2 - 6x + 9}{4}$ -$\frac {x-3}{2}$ = 3
or, $\frac {x^2 - 6x + 9 - 4x + 12}{4}$ = 3
or, x2 - 10x + 21 =12
or, x2 -10x + 21 - 12 = 0
or, x2 - 10x + 9 = 0
or, x2 - 9x - x + 9 = 0
or, x(x - 9) -1 (x - 9)= 0
or, (x - 9) (x - 1) = 0
Either, x - 9 = 0 i.e. x = 9
or, x - 1 = 0 i.e. x = 1
∴ x = 9 and 1 Ans
Let: y =g(x) = $\frac {1}{1-x}$
y = $\frac {1}{1-x}$
Interchanging the place of x and y
x =$\frac {1}{1-y}$
or, 1 - y = $\frac 1x$
or, y = 1 - $\frac 1x$ = $\frac {x - 1}{x}$
g-1(x) = $\frac {x - 1}{x}$
g-1$\frac 12$
= $\cfrac{(\frac{1}{2})-1}{ \cfrac{1}{2}}$
= $\frac {1-2}{2}$× $\frac 21$
= -1 Ans
Again, fg(x) = f$\frac {1}{1-x}$ = 1 + $\frac {2×1}{1-x}$
fg(x) = f$\frac {1}{1-x}$ = 1 + $\frac {2×1}{1-x}$
= f$\frac {1}{1-x}$ = 1 + $\frac {2×1}{1-x}$
= 1 + $\frac {2×1}{1-x}$
= $\frac {1 - x + 2}{1 - x}$
= $\frac {3 - x}{1 - x}$
fg(-1)
= $\frac {3 - (-1)}{1 - (-1)}$
= $\frac {3 + 1}{1 + 1}$
= $\frac 42$
= 2 Ans
Let: y = f(x) = $\frac {2x + 3}{x + 2}$
Interchanging the position of x and y
x= $\frac {2y + 3}{y + 2}$
or, xy+ 2x - 2y = 3
or, y(x - 2) = 3 - 2x
∴y= $\frac {3 - 2x}{x - 2}$
1. f-1(x) = $\frac {3 - 2x}{x - 2}$
2. f-1(1) = $\frac {3 - 2(1)}{1 - 2}$ = -$\frac 11$ = -1
3. fg(x) = f(x - 2) =$\frac {2(x-2) + 3}{(x-2) + 2}$ =$\frac {2x - 4 + 3}{x}$ = $\frac {2x-1}{x}$
4. fg(1) = $\frac {2×1-1}{1}$ = $\frac {2-1}{1}$Ans
Let, y= f(x) =$\frac {3x + 11}{x - 3}$
Interchanging the value of x and y,
x =$\frac {3y + 11}{y - 3}$
or, xy - 3x = 3y + 11
or, xy - 3y = 3x + 11
or, y(x - 3) = 3x + 11
or, y = $\frac {3x + 11}{x - 3}$
∴ f-1(x) = $\frac {3x + 11}{x - 3}$ Ans
Again,
Let: y = g(x) = $\frac {x - 3}{2}$
y =$\frac {x - 3}{2}$
Interchanging the place of x and y,
x =$\frac {y - 3}{2}$
or, 2x = y - 3
or, y = 2x + 3
∴ g-1(x) = 2x + 3
From the given question,
f(x) = g-1(x)
or,$\frac {3x + 11}{x - 3}$ = 2x + 3
or, 3x + 11 = 2x2 + 3x - 6x - 9
or, 2x2 - 3x - 9 - 3x - 11 = 0
or, 2x2 - 6x - 20 = 0
or, 2(x2 - 3x - 10) = 0
or, x2 - 5x + 2x - 10 = 0
or, x(x - 5) + 2(x - 5) = 0
or, (x - 5) (x + 2) = 0
Either, x - 5 = 0∴ x = 5
Or, x + 2 = 0∴ x = -2
Hence, x = 5 or -2 Ans
Here,
Let: y = g(x) = 3x - 5
y = 3x - 5 ------------------(1)
Interchanging the position of x and y in equation (1),
x = 3y - 5
or, 3y = x + 5
or, y = $\frac {x + 5}{3}$
∴ g-1(x) = $\frac {x + 5}{3}$
fg-1(x) = 15
or, f $\frac {x + 5}{3}$ = 15
or, 4 $\frac {x + 5}{3}$ + 7 = 15
or, $\frac {4x + 20}{3}$ + 7 = 15
or, $\frac {4x + 20 + 21}{3}$ = 15
or, 4x + 41 = 45
or, 4x = 45 - 41
or, 4x = 4
or, x = $\frac 44$
∴ x = 1 Ans
x3 - 4x2 + x + 6 = 0
(x - 2) is a factor ofx3 - 4x2 + x + 6
or, x3 - 2x2 - 2x2 + 4x - 3x + 6 = 0
or, x2(x - 2) - 2x(x - 2) - 3(x - 2) = 0
or, (x - 2) (x2 - 2x - 3) = 0
or, (x - 2) (x2 - 3x + x - 3) = 0
or, (x - 2) [x(x - 3) + 1(x - 3)] = 0
or, (x - 2) (x - 3) (x + 1) = 0
Either, x - 2 = 0 ∴ x = 2
Or, x - 3 = 0 ∴ x = 3
or, x + 1 = 0 ∴ x = - 1
Hence, x = 2, 3, -1 Ans
Rough:
x = 2
x3 - 4x2 + x + 6
= 23 - 4⋅ 22 + 2 +6
= 16 - 16
= 0
x - 2 is a factor of given expression
or, 2x3 - 4x2 + 7x2 - 14x + 3x - 6 = 0
or, 2x2 (x - 2) + 7x (x - 2) + 3 (x - 2) = 0
or, (x - 2) (2x2 + 7x + 3) = 0
or, (x - 2) (2x2 + 6x + x +3) = 0
or, (x - 2) [2x (x + 3) + 1 (x + 3)] = 0
or, (x - 2) (x + 3) (2x + 1) = 0
Either, x + 3 = 0∴ x = -3
Or, x - 2 = 0∴ x = 2
Or, 2x + 1 = 0∴ x = -$\frac 12$
∴ x = 2 , - 3 , -$\frac12$ Ans
Rough:
If x = 2
2x3 + 3x2 - 11x - 6
= 2(2)3 + 3(2)2 - 11(2) - 6
= 16 + 12 - 22 - 6
= 0
x = -1 is correct.
x + 1 is a factor of above equation.
or, 3x3 + 3x2 - 16x2 - 16x + 16x + 16 = 0
or, 3x2(x + 1) - 16x(x + 1) + 16(x + 1) = 0
or, (x + 1) (3x2 - 16x + 16) = 0
or, (x + 1) (3x2 - 12x - 4x + 16) = 0
or, (x + 1) [3x (x - 4) - 4 (x - 4)] = 0
or, (x + 1) (x - 4) (3x - 4) = 0
Either, x + 1 = 0 ∴ x = - 1
Or, x - 4 = 0 ∴ x = 4
Or, 3x - 4 = 0 ∴ x = $\frac 43$
∴ x = -1, 4 and $\frac 43$ Ans
Rough:
Put x = -1
or, 3(-1)3 - 13(-1)2 + 16 = 0
or, -3 -13 + 16 = 0
or, - 16 + 16 = 0
∴ 0 = 0
x + 2 is a factor of given equation.
x3 + 2x2 - 2x2 - 4x - 15x - 30 = 0
or, x2(x + 2) - 2x(x + 2) - 15(x + 2) = 0
or, (x + 2) (x2 - 2x - 15) = 0
or, (x + 2) (x2 - 5x + 3x - 15) = 0
or, (x + 2) [x(x - 5) + 3(x - 5)] = 0
or, (x + 2) (x - 5) (x + 3) = 0
Either, x + 2 = 0 ∴ x = -2
Or, x - 5 = 0 ∴ x = 5
Or, x + 3 = 0 ∴ x = -3
∴ x = -2, -3, 5 Ans
x3 - 3x - 2 = 0
x + 1 is a factor of given equation,
or, x3 + x2 - x2 - x - 2x - 2 = 0
or, x2 (x + 1) - x (x + 1) - 2 (x + 1) = 0
or, (x + 1) (x2 - x - 2) = 0
or, (x + 1) [x(x - 2) + 1 -(x - 2)] = 0
or, (x + 1) (x - 2) (x + 1) = 0
Either, x + 1 = 0 ∴ x = -1
Or, x - 2 = 0 ∴ x = 2
∴ x = -1 and 2 Ans
(x + 1) (x + 2) (x + 3) (x + 4) - 8 = 0
or, (x2 + x + 4x + 4) ( x2 + 2x + 3x + 6) - 8 = 0
or, (x2 + 5x + 4) (x2 + 5x + 6) - 8 = 0
Let: x2 + 5x = k
(k + 4) (k + 6) - 8 = 0
or, k2 + 6k + 4k + 24 - 8 = 0
or, k2 + 10k + 16 = 0
or, k2 + 8k + 2k + 16 = 0
or, k(x + 8) + 2(x + 8) = 0
or, (k + 2) (x + 8) = 0
Putting the value of k
(x2 + 5x + 8) (x2 + 5x + 2) = 0
Either, x2 + 5x + 8 = 0 --------------(1)
Or, x2 + 5x + 2 = 0 ------------------(2)
Taking eqn(1)
x =$\frac {-5 ± (\sqrt{5^{2}-4×1×8})}{2×1}$ =$\frac {-5 ± \sqrt{-7}}{2×1}$ (Impossible)
Taking eqn(2)
x =$\frac {-5 ± (\sqrt{5^{2}-4×1×2})}{2×1}$ =$\frac {-5 ± \sqrt{17}}{2×1}$ Ans
Here,
f(x) = x4 - x3 - 3x2 - 2x + 5
When x = 2 then,
2 1 -1 -3 -2 5 2 2 -2 -8 1 1 -1 -4 -3
∴ Quotient = x3 + x2 - x - 4 and f(x) = -3 Ans
When x = -1 then,
-1 1 -1 -3 -2 5 -1 2 1 1 1 -2 -1 -1 6
∴ Quotient = x3 - 2x2 - x - 1 and f(2) = 6 Ans
When x = 3 then,
3 1 -1 -3 -2 5 3 6 9 21 1 2 3 7 26
∴ Quotient = x3 + 2x2 + 3x + 7 and f(3) = 26 Ans
x - 2 = 0
x - 2 is a factor of given expression
or, x3 - x2 - 14x + 24 = 0
or, x3 - 2x2 + x2 - 2x - 12x + 24 = 0
or, x2(x - 2) + x(x- 2) - 12(x - 2) = 0
or, (x - 2) (x2 + x -12) = 0
or, (x - 2) (x2 + 4x - 3x - 12) = 0
or, (x - 2) [x(x + 4) - 3(x + 4)] = 0
or, (x - 2) (x + 4) (x -3) = 0
Either, x - 2 = 0 ∴x = 2
Or, x + 4 = 0 ∴ x = - 4
Or, x - 3 = 0 ∴ x = 3
∴ x = 2, 3, -4 Ans
Rough:
x = 2
i.e 23 - 22 - 14× 2 + 24
= 8 - 4 -28 + 24
= 0
2x3 - 3x2 - 3x + 2 = 0
or, 2x3 + 2 - 3x2 - 3x = 0
or, 2(x3 + 13) - 3x (x + 1) = 0
or, 2(x + 1) (x2 - x + 1) - 3x (x + 1) = 0
or, (x + 1)[2x2 - 2x + 2 - 3x] = 0
or, (x + 1) (2x2 - 5x + 2) = 0
or, (x + 1) (2x2 - 4x - x + 2) = 0
or, (x + 1) [2x(x - 2) - 1(x - 2)] = 0
or, (x + 1) (x - 2) (2x - 1) = 0
Either, x + 1 = 0 ∴x = -1
Or, x - 2 = 0 ∴x = 2
Or, 2x - 1 = 0 ∴ x = $\frac 12$
∴ x = -1, 2 , $\frac12$ Ans
|
# maths 2
?
• Created by: natalie
• Created on: 19-02-15 09:31
## discrete and continuos data
Discrete data is like a pair of shoes size 4 and 5 are ok but size 4.3 is no use, no in between numbers.
Continous data is like measurements. they have in between points 0.1 0.2
1 of 11
## Percentage
10% move one dp to the left
on a calculator number divided by 100 x 10%
2 of 11
## Simple Interest
Divide the interest by he amount of the loan (other money number) and then x by 100.
Lucy has aloan of £750 she pays interest of £25.50 per month. What is the rate of a simple interest.
£25.50 divide by £750 = 0.034
Then times 0.034 by 100
3 of 11
## Percentage increase and decrease
% of an amount divided by 100 x amount =
There is a sle in top shop and your favourite jeans are 56% off they cost £20 originally
eg 4% of 920
4 divide by 100 x by 920
WORKING OUT THE PERCENTAGE RATE
Numberdivide by Number x 100- So the other way around!
4 of 11
## Percentage Original Amount
If a dress cost £42 in the sale it was reduced by 15% what was the original price?
You need the whole information is 100%
Take the percentage you have been given away from 100 so in this example 100- 15% = 0.85
Then
Divide the price or number by your percentage
42 divided by 0.85 = £49.41
The amount should be bigger!
5 of 11
## Factorising expressions
2x+6 becomes 2(x+3)
2x-24 becomes 2(x-12)
6n-48 becomes 6(n-8)
3+3t becomes 3(1+t)
6 of 11
## Polygons
To find number of triangles in a ploygon shape - take 2 away from the number of sides in that shape
Sum of interior angles =
times the number of triangles by 180 degrees
7 of 11
## sums of each interior and exterior angle
Sum of each exterior angle=
To find the size of each exterior angle, we divide the sum
of the interior angles by the number of sides. Then take
that number from 180 degrees.
Sum of each interior angle=
We divide the sum of interior angles by the number of
angles in the shape.
8 of 11
## Pythagoras theory
To work out the longest side
Square both of the other side seperatley anf then add them together
Then find square root
To work out the shortest side
the point of the right angle points to the shortest side
so square both of the sides seperatley but take away from each other
Then find sqaure root
9 of 11
## Dividing Fractions and whole numbers
3 divide by 1/2
Make the whole number into a fraction with denominator as 1
3/1
the swap the other fraction around, so denominator becomes numerator
2/1
multiply your numbers then simplify if you can
3/1 x 2/1 = 6/1
Simplified =6
10 of 11
## Decimals to fractions, fractions to decimals
DECIMAL TO FRACTION:
0.6=6/10 0.4= 4/10 0.65= 65/100 0.875= 875/1000
FRACTIONS TO DECIMALS
Divide the numerator by the denominator (with a calculator) or multiply the fraction by 1.00 eg.
2/5x1.00= 2 x 1.00 = 2.00 = 0.40 or .04
5 x1.00 5
the 2.00 has been changed to 200 then dp added
11 of 11
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With help from the teacher, the student has partial success with level 2 and level 3 elements. Plot a point for your home at the origin in a coordinate plane. Step 7: To demonstrate how the Pythagorean Theorem can be used to find the length of the diagonal of a rectangle, draw a rectangle on a grid with the following coordinates: A (-2, 2), B (-2, 6), C (-5, 6), and D (-5, 2). Point out that instead of adding (6, 2) as the third coordinate, (3, 6) could also have been used to create the triangle. Oct 25, 2019 - Explore Katie Kilgore Widener's board "Pythagorean Theorem", followed by 381 people on Pinterest. Played 0 times. Share practice link. Gravity. Learn vocabulary, terms, and more with flashcards, games, and other study tools. If segments are at right angles, the theorem holds and the math works out. In this explainer, we will learn how to find the distance between two points on the coordinate plane using the Pythagorean theorem. Students should pair up and find the distance between the two points. Jan 14, 2021 - Explore All Things Maths's board "Pythagorean Theorem Activities", followed by 2215 people on Pinterest. Play. The formula for the distance between two points in two-dimensional Cartesian coordinate plane is based on the Pythagorean Theorem. 0. On the town map at right, each unit represents 1 kilometer. See more ideas about pythagorean theorem, theorems, pythagorean theorem activity. Presented By: Sasha Scott Jamya Thomas Jada Scandrett Trenilyas Tatum Write Formula: Let's Try It Together Substitute: Simplify: c= 7.62 Solve: Steps 1. The coordinate here is X is four, Y is six. PYTHAGOREAN THEOREM IN THE COORDINATE PLANE WORKSHEET. The Activity. As you can guess, the Pythagorean Theorem generalizes to any number of dimensions. Write. Step 8: To demonstrate how the Pythagorean Theorem can be used to find the distance between two points, draw points (3, 2) and (6, 6) on a grid. Objective. Mensuration formulas. Substitute 3. Plug a = 4 and b = 5 in (a2 + b2 = c2) to solve for c. Find the value of â41 using calculator and round to the nearest tenth. Lesson 27: Applying the Pythagorean Theorem on the Coordinate Plane 149 Duplicating any part of this book is prohibited by law. Applying the Pythagorean Theorem, 32 + 42 = segment AD2. In the Designing With Geometry unit, students will solve problems using the coordinate plane, scale factors, angles, congruence, transformations, and the Pythagorean theorem. Step 2:It is a common misconception that the length of the hypotenuse equals the number of grid intervals it passes through. 8th grade . Save. Explain your reasoning. Watch the video (Level 2: Pythagorean Theorem) Complete the Notes & Basic Practice Check the Key and Correct Mistakes 2. WHICH ONE DOESN’T BELONG? The figure shows a right triangle. Using Ordered Pairs on the Coordinate Plane. Check your answer for reasonableness. So 81 + x2 = 225, which can be rewritten as x2 = 225 - 81. How many kilometers does he jog? For example, in a right triangle where the one leg measures 9 meters and the hypotenuse measures 15 meters, write 92 + x2 = 152. Construction of triangles - I Construction of triangles - II. Use the town map to answer questions 8 and 9. 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For more information, download the comprehensive Standards Chart: Geometry printable. This is going to be comma six. In a 2 dimensional plane, the distance between points (X 1, Y 1) and (X 2, Y 2) is given by the Pythagorean theorem: d = (x 2 − x 1) 2 + (y 2 − y 1) 2 Check your answer for reasonableness. by quigs6464. Create a right triangle on a coordinate plane, given 2 points. \end{align} So rectangles which fit in a skew way into the coordinate grid are related to writing whole numbers as sums of squares in two different ways. Classwork Sketch a picture of each situation and find the missing length. Answer: The diagonal ET is about 10.63 units long. SWBAT apply the Pythagorean Theorem to find the distance between two points in a coordinate system. Show It! STUDY. Plug a = 4 and b = 2 in (a2 + b2 = c2) to solve for c. Find the value of â20 using calculator and round to the nearest tenth. The length of AD = 3 (subtracting the x-coordinates: -2 -5 = 3). 1. 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Pythagorean Theorem on a Coordinate plane Studies, courses, subjects, and textbooks for your search: Press Enter to view all search results () Press Enter to view all search results () Login Sell. 2. Use the Pythagorean theorem to determine the distance between the two points on the coordinate plane. Pythagorean theorem. Similar to Lesson 9, students discover ways that the Pythagorean Theorem can be a useful tool to determine measurements they could not find before. View not found. Find the approximate, length of the hypotenuse to the nearest tenth. Students who successfully answer the problems in the second activity use the structure of the coordinate plane to draw a right triangle, an example of looking for and making use of structure in the coordinate plane (MP7). Round to the nearest hundredth. 2.24. Explanation: Plot 3 ordered pairs to make a right triangle. TM ® & © 2016 Scholastic Inc. All Rights Reserved. A discrete function consists of isolated points. This Pythagorean Theorem Unit is easy-to-implement and scaffolded to support student success. Practice. Bell Ringer. Download the PDF from here, Designing With Geometry: Polygons on the Coordinate Plane. Step 3: Show an example of the Pythagorean Theorem on the board, labeling one leg 3 meters, the other 4 meters, and the hypotenuse 5 meters. In the plane, any two points (a, c) and (b, d) may be joined by a segment, and this segment is a diagonal of a unique rectangle with edges parallel to the coordinate axes.Because the base of this rectangle has length |a - b| and because the height of the rectangle is |c - d|, the Pythagorean theorem tells us that the length of the diagonal is given by ((a-b) 2 + (c-d) 2) 1/2. Universities. Test. Use the distance formula and the coordinates of points and to prove that the Pythagorean theorem is an alternative method for calculating the distance between points on a coordinate plane. … Let us consider (1 3, − 7) and (− 1 1, 3) in a coordinate system of origin (0, 0). In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. Sketch a picture of each situation and find the missing length. Flashcards. In your final answer, include all of your calculations. Identify the legs and the hypotenuse of a right triangle. 6.33. Plot these points on the coordinate plane at the right and connect them to draw the rectangle. Use Pythagorean Theorem to find Distance Between Two Points in the Coordinate Plane. That is, you can chain a bunch of triangles together and tally up the “outside” sections: You can imagine that each triangle is in its own dimension. Print; Share; Edit; Delete; Host a game. Pythagorean Theorem Using Coordinate Plane - Displaying top 8 worksheets found for this concept.. Discover how the Pythagorean Theorem describes the relationship between the lengths of the sides of a right triangle. Complete 2 of the following tasks IXL Practice Worksheets Creating O.1, O.2, (8th) At Least to 80 Score = _____ Level 2: Pythagorean Theorem Showing 2 Examples of using Pythagorean Theorem (1 finding hyp./1 finding leg) 3. This color-coding technique works well with students of all levels! The same method can be applied to find the distance between two points on the y-axis. Check for reasonableness by finding perfect squares close to 41. â41 is between â36 and â49, so 6 < â41 < 7. 1. 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Counting Spacing on the Coordinate Plane. In this worksheet, we will practice finding the distance between two points on the coordinate plane using the Pythagorean theorem. Common Core Standard 8.G.8. If a and b are legs and c is the hypotenuse, then. Some of the worksheets for this concept are Concept 15 pythagorean theorem, Using the pythagorean theorem, Pythagorean distances a, Length, Pythagorean theorem 1, The pythagorean theorem, Chapter 9 the pythagorean theorem, The pythagorean theorem date period. Pythagorean Theorem on the Coordinate Plane. The Pythagorean Theorem is used in this set of 12 task cards that has students finding the distances between zoo animals that have been placed at points on a coordinate plane. Lisa drives her car 46 km south and then 13 km east. Answer: The distance = 10 units. Step 4: Indicate that the Pythagorean Theorem can be used to find the missing side length of a right triangle when the other two side lengths are known. MP.1 - Make sense of problems and persevere in solving them. 8.G.8 - Apply the Pythagorean Theorem to find the distance between two points in a coordinate system. In this Designing With Geometry lesson, students will apply the Pythagorean Theorem to find the distance between two points in a coordinate system. Properties of triangle. Created by. Apply the Pythagorean Theorem to find missing lengths and to calculate distances between points on the coordinate plane. Draw a right triangle on a grid on the board, using the right-angle symbol to identify the right angle, and labeling the hypotenuse C, one leg A, and the other B. Edit. Because a and b are legs and c is hypotenuse, by Pythagorean Theorem, we have. The length of the horizontal leg is 2 units. Make a class set of the Party on the Patio: Applying the Pythagorean Theorem printable. Regular Math Lesson Sequence: Student choice either weekly quiz or 11 squared through 20 squared as a warm up as I passed back exit tickets ; QSSQ ; Exit ticket … Homework. This quiz is incomplete! The distance of BC = 6, which is obtained by subtracting the two x-coordinates (8 - 2). More precisely, the Pythagorean theorem implies, and is implied by, Euclid's Parallel (Fifth) Postulate. The distance of AC = 8, which is obtained by subtracting the two y-coordinates (10 - 2). 12.81 . This interactive exercise focuses on using the Pythagorean Theorem to calculate distance and plotting points on a Cartesian grid. Edit. It's going to have the same X coordinate … Using the Pythagorean Theorem, 82 + 62 = (segment AB)2. Live Game Live. Homework. You can use the Pythagorean Theorem to find the distance be- tween two points on the coordinate plane. Find the distance between each pair of points. Put your answers in the blanks provided. The length of the horizontal leg is 5 units. Plan your 60-minute lesson in Math or Pythagorean Theroem with helpful tips from Christa Lemily Use the Pythagorean Theorem to find the exact length of $$\overline { ET }$$. 0. Which set of numbers does not belong with the other three? [Note: (8, 10) would also work but this example uses (2, 2).] The Pythagorean Theorem states that the sum of the squared sides of a right triangle equals the length of the hypotenuse squared. The distance between two points is the length of the path connecting them. Start studying Using Pythagorean Theorem to find distance in the Coordinate Plane. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. Problem 1 : The figure shows a right triangle. The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, were the Pythagorean theorem to fail for some right triangle, then the plane in which this triangle is contained cannot be Euclidean. Write formula for Pythagorean Theorem 2. The Pythagorean Theorem can easily be used to calculate the straight-line distance between two points in the X-Y plane. Step 1: Review the properties of right triangles as necessary for your class. Since 6.4 is between 6 and 7, the answer is reasonable. •Students determine the distance between two points on a coordinate plane using the Pythagorean Theorem. Step 10: Checking for Understanding: Review answers as a class and respond to any questions. This Triangle Worksheet will produce exterior angle theorem problems. Using the Pythagorean theorem, find the distance between and . Q20: Let us consider (− 1 … a. The Pythagorean Theorem in the Coordinate Plane 7 units across 4 units up You begin with two points (usually given to you as ordered pairs only) You can quickly plot the two points on a graph and count the units (left/right and up/down) between them Use those distances to find the missing length 7² + 4² = c² so c = ⎷65 Rule of thumb: leave coordinate plane distances under … Remind them that the distance is not equal to the number of grid intervals between the two points. Make sure students explain their mathematical thinking. Ask the class how they can find the distance between the two points. by kbianchi_44454. Use the Pythagorean theorem to determine the distance between the two points on the coordinate plane. Step 9: Distribute graph paper to students. Pythagorean Theorem Word Problems Unit 6 – Geometry on the Coordinate Plane Name:_____ Date: _____ Show your work on another sheet of paper. Example finding distance with Pythagorean theorem. All you need to know are the x and y coordinates of any two points. Pythagorean Theorem 3d & Coordinate Plane DRAFT. For example, if both side A and side B measure 1 cm, then c2 = 12 + 12. GEOMETRY. a 2+b 2=c 2Write the Pythagorean Theorem. Show how the horizontal leg is 3 units in length by subtracting the x-coordinates of the original two points. Solve How To Find The A and B Variables on a The Pythagorean theorem (8th grade) Find distance between two points on the coordinate plane using the Pythagorean Theorem An updated version of this instructional video is available. Finding distance in the Coordinate Plane using the Pythagorean Theorem.-distance between two points Learn. LESSON 8: Applying the Pythagorean Theorem to Coordinate GeometryLESSON 9: Applying Distance to Perimeter and Area on Coordinate Plane LESSON 10: Unit Assessment. 3 years ago. Step 6: Finally, provide an example where the missing side isn't a whole number. For an interactive preview, click here!This deck gives students the opportunity to use Pythagorean Theorem with right triangles to find distance between two points on a coordinate plane.Deck contains 25 cards, of which 10 are randomly selected each time.Compatible with Google Classroom, SeeSaw, Powe Find study resources for. Example finding distance with Pythagorean theorem. Introduction to Pythagorean Theorem. Add to … Save. PLAY. Indicate that the formula for side lengths in a right triangle is a2 + b2 = c2. Print a copy of the Answer Key: Designing With Geometry printable for your use. Understand visual and algebraic proofs of the Pythagorean Theorem. Common Core Math Practices. Four comma six, and so the coordinate over here is going to have the same Y coordinate as this point. Find the distance between each pair of points. The Exterior Angle Theorem Worksheets. Check your answer for … If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. Point out that the diagonal creates two right triangles, one of which is ÎABD. In this lesson, students continue to apply the Pythagorean Theorem to find distances between points in the coordinate plane. Find the approximate length of the hypotenuse to the nearest tenth. Show your work . WRITING How can the Pythagorean Theorem be used to fi nd distances in a coordinate plane? Find the approximate length of the hypotenuse to the nearest tenth. Repeat with another example, such as a right triangle with sides of 5 meters, 12 meters, and 13 meters, where 52 + 122 = 132. Then show how the length of the vertical leg is 4 units by subtracting the y coordinates of the original two points. An 8-day CCSS-Aligned Pythagorean Theorem Unit – including the Pythagorean Theorem, the Pythagorean converse, Pythagorean theorem word problems, distance on a coordinate plane, and 3D applications of the Pythagorean theorem. Area and perimeter. The shortest path distance is a straight line. Some coordinate planes show straight lines with 2 p As previously mentioned, the Pythagorean Theorem is a mathematical equation that states that the square of the hypotenuse (the side opposite to the right angle triangle) is … Since â144 = 12, the length of the missing side is 12 meters. (4, 3), (1, –1) 2. 1. Solo Practice. Properties of parallelogram. Then draw a vertical line through one of the points and a horizontal line through the other point. 262 Chapter 6 Square Roots and the Pythagorean Theorem 1. At the moment this is an example of a discrete function. 28% average accuracy. Terms in this set (10) 9.84. To determine the distance between two points on the coordinate plane, begin by connecting the two points. Usually, these coordinates are written as ordered pairs in the form (x, y). Put your answers in the blanks provided. Add a third point at (6, 2) and note that the three points form a right triangle when connected. Thus, x2 = 144. Use this simple high school geometry lesson to extend your students' understanding of the future applications of the Pythagorean Theorem. Problem 2 : The figure shows a right triangle. Then determine the distance between the points using the Pythagorean Theorem. Pythagorean Theorem Word Problems Unit 6 – Geometry on the Coordinate Plane Name:_____ Date: _____ Show your work on another sheet of paper. You can choose a single variable or an algebraic expression for the unknown angle. Determine the distance between any two points on the coordinate plane. Find the approximate length of the hypotenuse to the nearest tenth. In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. (1, 3) 2. In this Pythagorean theorem: Distance Between Two Points on a Coordinate Plane worksheet, students will determine the distance between two given points on seven (7) different coordinate planes using the Pythagorean theorem, one example is provided. (2, 4) Now you can just plot the five ordered pairs in the coordinate plane. Pythagorean Theorem Concept 15: Pythagorean Theorem Pre Score DEADLINE: (C) Level 2 1. In this image we can find the lengths of sides a and b by counting the spaced on the coordinate plane. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Delete Quiz. More precisely, the Pythagorean theorem implies, and is implied by, Euclid's Parallel (Fifth) Postulate. Step 2: It is a common misconception that the length of the hypotenuse equals the number of grid intervals it passes through. For a triangle where one leg is 6 meters and the other is 8 meters, show how 62 + 82 = 100 and since â100 = 10, the length of the hypotenuse is 10 meters. 9. Negative five comma eight. Draw a right triangle on a grid on the board, using the right-angle symbol to identify the right angle, and labeling the hypotenuse C, one leg A, and the other B. Preparing Materials. The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, were the Pythagorean theorem to fail for some right triangle, then the plane in which this triangle is contained cannot be Euclidean. khan academy pythagorean theorem coordinate plane, 1 + 2 = 3. 0. Let a = 4 and b = 2 and c represent the length of the hypotenuse. The coordinate of this point up here is negative five comma eight. Simplify 4. 2 + 2 = 4. Therefore, c2 = 2, so c = â2. Live Game Live. To play this quiz, please finish editing it. Find the distance between each pair of points. In your final answer, include all of your calculations. Mathematics. 2 minutes ago. Practice. Video: Khan Academy ... Pythagorean Theorem - Distance on the Coordinate Plane. 3, 6, 8 6, 8, 10 5, 12, 13 7, 24, 25 Find the perimeter of the fi gure. We can begin by recalling the Pythagorean theorem, which relates the length of the longest side of a right triangle to the lengths of the other two sides. The Pythagorean theorem, applied to these two triangles gives \begin{align} 125 &= 10^2 + 5^2\\ 125 &= 11^2 + 2^2. Students should construct a right triangle by adding point C at (2, 2). Step 12: Checking for Understanding: Review the answers to the Party on the Patio: Applying the Pythagorean Theorem printable, which are provided on page 1 of the Answer Key: Designing With Geometry printable. Guided Notes, Color Coded using Highlighters! 7th grade . Volume. 42+ 52=c 2Substitute 4 for aand 5 for b. 0. This quiz is incomplete! Regular Math Objective: Apply the Pythagorean theorem in the coordinate plane Regular Math Standards: 8.G.8 Apply the Pythagorean Theorem to find the distance between two points in a coordinate system. Construction of triangles - III. To play this quiz, please finish editing it. Since 4.5 is between 4 and 5, the answer is reasonable. Watch the video (Level 2: Pythagorean Theorem) Complete the Notes & Basic Practice Check the Key and Correct Mistakes 2. This can be demonstrated to be false by taking a square piece of paper… Maksim jogs directly from City Hall to the public library . 1. An application of the Pythagorean theorem allows you to calculate the length of a diagonal of a rectangle, the distance between two points on the coordinate plane and the height that a ladder can reach as it leans against a wall. Construction of angles - I Course 3 • Chapter 5 Triangles and the Pythagorean Theorem Chapter 5-Lesson 5 Homework Practice Distance on the Coordinate Plane Graph each pair of ordered pairs. 9 + 16 = 25, so AD = â25 which is 5. Start studying Pythagorean Theorem on the Coordinate Plane. Ask how the length of segments AC or BD could be found. Lesson Plans and Worksheets for Grade 8 Lesson Plans and Worksheets for all Grades More Lessons for Grade 8 Common Core For Grade 8 Examples, videos, and solutions to help Grade 8 students learn how to determine the distance between two points on a coordinate plane using the Pythagorean Theorem. tmarti6 TEACHER. Check for reasonableness by finding perfect squares close to 20. â20 is between â16 and â25, so 4 < â20 < 5. Q1: Find the distance between the point ( − 2 , … Real world application of the Pythagorean Theorem . Sum of the angle in a triangle is 180 degree. Step 5: Further demonstrate how the Pythagorean Theorem can be used to find the missing leg length of a right triangle when the lengths of the other leg and the hypotenuse are known. This can be demonstrated to be false by taking a square piece of paper, making a diagonal cut, and holding the hypotenuse of one of the right triangles formed by the cut next to one of the legs of the other right triangle. See more ideas about pythagorean theorem, theorems, middle school math. Solo Practice. Spell. Pythagorean Theorem Using Coordinate Plane - Displaying top 8 worksheets found for this concept.. Write the following two points on a grid on the board: A (2, 10) and B (8, 2). Edit. Share practice link. Side B has a length of 3 units, vertically parallel to the y-axis. Delete … This quiz … In this image we can can find the … Point out that these two right triangles are congruent because they have the same side lengths and angle measurements. 8. Let a = 4 and b = 5 and c represent the length of the hypotenuse. Finish Editing. Then plot points for the locations of the park and the library to form a right triangle. Find the distance between each pair of points. Match. Find the approximate length of the hypotenuse to the nearest tenth. Some of the worksheets for this concept are Concept 15 pythagorean theorem, Using the pythagorean theorem, Pythagorean distances a, Length, Pythagorean theorem 1, The pythagorean theorem, Chapter 9 the pythagorean theorem, The pythagorean theorem date … This Pythagorean Theorem implies, and other study tools by, Euclid 's Parallel ( Fifth Postulate... For more information, download the comprehensive Standards Chart: Geometry printable triangles real-world., Pythagorean Theorem 1 2 points 46 km south and then 13 km.... 82 + 62 = ( segment AB = 4 ( subtracting the x-coordinates: -5... Angle measurements expression for the locations of the vertical leg is 5 passes through = 6 2... In solving them choose a single variable or an algebraic expression for the unknown.! The moment this is an example of a right triangle the student has partial success with Level 2 Pythagorean... Them to draw the rectangle 3 ). of AB = â100 which equals 10 the origin in a is... Be- tween two points on the coordinate plane = 12, the Theorem holds and the hypotenuse to nearest. Coordinate plane people on Pinterest how they can find the exact length of \ ( \overline { ET \! By counting the spaced on the coordinate plane is based on the coordinate.. It 's going to have the same y coordinate as this point Basic Practice check Key... Is an example of a discrete function is x is four, y.... These two right triangles are congruent because they have the same y coordinate as this point = c2 using! To form a right triangle for Understanding: Review answers as a class set of the hypotenuse squared, with... Can the Pythagorean Theorem to find the distance of BC = 6, )! ( Fifth ) Postulate is part of the path connecting them hypotenuse a... ® & © 2016 Scholastic Inc. all Rights Reserved the vertical leg 3! & © 2016 Scholastic Inc. all Rights Reserved hypotenuse of a discrete function help from the stuff above! Same y coordinate as this point, so 6 < â41 < 7 the park the. Counting the spaced on the coordinate plane units by subtracting the two points in two-dimensional Cartesian plane. Understand visual and algebraic proofs of the points and a horizontal line through the other three if side... Rewritten as x2 = 225, which is ÎABD the activity can be applied find... X and y coordinates of the path connecting them relationship between the lengths of sides a and are... Plotting points on the coordinate plane as necessary for your home at the pythagorean theorem on coordinate plane: Middle school.., given 2 points = 100, which means that segment AB ) 2 five ordered in! 2 and c is the hypotenuse 4.5 is between 6 and 7, the Theorem holds and math. To extend your students ' Understanding of the vertical leg is 2.... Coordinate … plot a point for your use + x2 = 225 -.. The distance between two points on the coordinate plane ) and note that diagonal... Patio: Applying the Pythagorean Theorem to find the missing side is n't a whole number each situation find... Activities '', followed by 2215 people on Pinterest between points on the coordinate.... Four, y ). by connecting the two points pythagorean theorem on coordinate plane for classwork or homework,! = 5 and c represent the length of the vertical leg is 2 units scaffolded support! Study tools because they have the same side lengths in right triangles in real-world and problems! All of your calculations hypotenuse squared a coordinate system algebraic proofs of the park and the math at right... Will produce exterior angle Theorem problems book is prohibited by law ) Now you can use Pythagorean. Points E and T to form a right triangle: khan academy... Pythagorean Theorem to the! Produce exterior angle Theorem problems for b same y coordinate as this point the student has partial success Level! Solve how to find the a and b = 5 and c the! Problems and persevere in solving them, followed by 2215 people on Pinterest right, each unit 1! { ET } \ ). is between â36 and â49, so AD = 3 distance and points... = 52 = â25 which is obtained by subtracting the two y-coordinates ( 10 2! Make sense of problems and persevere in solving them: Polygons on the Pythagorean Theorem to find distance in coordinate..., 82 + 62 = ( segment AB = 4 ( subtracting the y coordinates any! Path connecting them by connecting the two points on a Cartesian grid support success! Leg is 3 units in length by subtracting the two points on the coordinate?! The figure shows a right triangle directly from City Hall to the of... Be- tween two points on the town map at right, each unit represents kilometer. = 25, so c = â2 identify the legs and c is hypotenuse, by Pythagorean,! 'S going to have the same method can be applied to find distance between the two y-coordinates ( 10 2! Plane is based on the coordinate plane, 1 + 2 = 4 and 5, the length of path. All Things Maths 's board Pythagorean Theorem - distance on the coordinate plane: ( c ) Level 1... Plot these points on the coordinate plane set of the sides of a right triangle is a2 + =... Just plot the five ordered pairs in the coordinate plane the y coordinates of the original points! Or homework by 2215 people on Pinterest 15: Pythagorean Theorem ) Complete the Notes & Practice. ( 10 - 2 = 4 and b = 2, so AD = (... Apply the Pythagorean Theorem to find the missing side is n't a whole number 1: Review the properties right. About 10.63 units long ). if both side a has a length of the Theorem! In two and three dimensions 3 ordered pairs to make a class set of numbers does not with... This concept each unit represents 1 kilometer - Explore all Things Maths 's board Pythagorean Theorem Pre Score:... Numbers does not belong with the other three coordinate plane use this simple school. Displaying top 8 worksheets found for this concept unit represents 1 kilometer, will! The five ordered pairs in the coordinate plane Delete ; Host a game for … Pythagorean Theorem printable for or. Hall to the nearest tenth + 16 = 25, so AD = (. ), ( 1, –1 ) 2, 32 + 42 = 52 AD = which. Unknown angle x is four, y is six 64 + 36 = 100, which is obtained by the! For the unknown angle then plot points for the locations of the math out... Problem 1: the diagonal creates two right triangles, one of the and! Them to draw the rectangle pythagorean theorem on coordinate plane finish editing it â100 which equals 10 theorems Pythagorean. Of problems and persevere in solving them < â41 < 7 equal to the tenth! Lengths in a right triangle - 2 ) and note that the formula for distance. Has partial success with Level 2: Pythagorean Theorem implies, and more with flashcards, games, is. At right, each unit represents 1 kilometer be applied to find approximate. ; Edit ; Delete ; Host a game the points and a horizontal line through one of is! Up and find the approximate length of the hypotenuse to the nearest.. Can just plot the five ordered pairs to make a class set of does... Please finish editing it lengths in right triangles, one of which is 5 Understanding of the hypotenuse and coordinates... Y ). Level 3 elements in length by subtracting the x-coordinates the... If both side a has a length of the hypotenuse equals the number grid! Units in length by subtracting the x-coordinates of the hypotenuse equals the number of grid intervals it through. Math works out include all of your calculations © 2016 Scholastic Inc. all Rights.! Download the PDF from here, Designing with Geometry: Polygons on the coordinate.! Your use - Displaying top 8 worksheets found for this concept, then c2 2. Units, vertically Parallel to the nearest tenth MAT 266 at Alabama State University © 2016 Inc.... Missing lengths and to calculate distances between points on the coordinate plane Fifth ).. Be found side lengths and angle measurements is going to have the same y coordinate as point... Home at the Core: Middle school collection it 's going to have the x... Horizontal leg is 3 units in length by subtracting the x-coordinates of the Pythagorean Theorem to find a! + b2 = c2 2 units distance on the coordinate plane 149 Duplicating any of. The video ( Level 2: Pythagorean Theorem - pythagorean theorem on coordinate plane on the coordinate over here is going have. Bc = 6, 2 ). for your home at the origin in right... Identify the legs and c is the length of the original two points in form..., by Pythagorean Theorem to determine the distance between two points in a coordinate.. For aand 5 for b teacher, the student has partial success with Level 2: Theorem! Triangle is a2 + b2 = c2 ; in this case, 32 + 42 52! The angle in a coordinate plane - Displaying top 8 worksheets found for this concept connecting two... For … Pythagorean Theorem to calculate distance and plotting points on the town map at,... 2016 Scholastic Inc. all Rights Reserved if both side a and b = and! Ab ) 2 and y coordinates of any two points on a coordinate plane Practice.
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Quick Homework Help
# Reciprocal Transformation
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One important concept in the study of polynomials is the reciprocal transformation. What happens when we take the reciprocal transformation of a function, or one over the function Specifically, there are ways to create the graph of the reciprocal transformation of a function from the graph of the function itself. The reciprocal transformation is important in the definition of rational functions.
I want to take a look at a particular transformation called the "Reciprocal Transformation" given the graph of a function y=f of x what does the graph of y equals 1 over f of x look like? To figure that out I want to start with a simple example. Let's use the graph of y equals 1 minus a half x to graph y equals 1 over 1 minus a half x.
First thing I want to do is graph y equals 1 minus a half x and that's pretty easy alright this is going to be a line with y intercept 1 and slope negative one half. So it's gong to go down 1 over 2 and right away when you have 2 points for a line you can graph the line immediately so let's graph it. So that's our line, now how do we get points for this graph 1 over that? Well you can just take reciprocals right? For example this point has a y coordinate of 1, the reciprocal of 1 is 1 so the reciprocal graph will pass through this point. This point has a y coordinate of a half the reciprocal of that is 2 so the reciprocal graph will pass through this point. Let's pick a nice integer here we're going to have a y coordinate of 2, so the reciprocal will have a y coordinate of a half. Here we have a y coordinate of 3, the reciprocal is one third and you can kind of see what's going to happen, as the graph goes up to infinity the reciprocal goes down to zero. And we get this kind of shape, now what happens past here notice that the y values of my lines are getting close to zero.
Well let's take a table of values, I've got x 1 minus 0.5x and 1 over 1 minus 0.5x. Let's let x get closer and closer to 2 and see what happens as we get close to this point. So we've already done x equals 1, let's do 1.5 now half of 1.5 is 0.75 and 1 minus that if 0.25, the reciprocal of 0.25 is 4 okay so it's going up. If we wanted to we could plot that point, it would be up here 1.9 half of that is 0.95 1 minus 0.95 is 0.05 and the reciprocal of that is 20. You might be able to tell already that as these numbers get closer to 2 these numbers are going to get closer to 0, and these numbers are going up to infinity. So that's what out graph is going to do, as we get close to 2 this graph is going to move up to infinity. Alright and that means that we have a vertical asymptote that x equals 2. So let me draw that in, it's a vertical line, a vertical line that the graph is going to get closer and closer to as it moves up.
Now what happens on this side, again we can just plot some points like on the line at this x value which looks like 4, we're going to have negative 1 the reciprocal of -1 is -1 so my reciprocal graph will actually go through that point. Here the y value is negative a half the reciprocal of that is negative 2 so we'll go through this point and let's say here at 3 we have 3 halves, the reciprocal of that is two thirds and you can see that we're going to get a similar graph, a similar kind of graph down here as up here. Now is this going to happen? As x approaches 2 from the right are we going down to negative infinity? Let's check really quickly with some values, 2.5 half of that is 1.25, 1 minus 1.25 is negative 0.25 and the reciprocal of that is negative 4 so this is 1 over 1 minus 0.5 x.
What about 2.1 half of that is 1.05, 1 minus that is minus 0.05 and the reciprocal of that is negative 20. So yeah you could see that as this goes to 2, we're approaching 2 from the right, this is approaching zero and this is going to negative infinity. And that verifies that the graph should actually just go down, straight down to negative infinity, so we've got a vertical asymptote x equals 2 and you might also recognize that the x axis is a horizontal asymptote] right the graph is getting closer and close to it as x goes to infinity or as x goes to the negative infinity.
So this purple graph is a graph of my reciprocal function, y equals 1 over 1 minus 0.5x and the red graph is the graph of my original line y equals 1 minus 0.5x.
## Find Videos Using Your Textbook
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The function f of x is graphed. What is its domain? So the way it’s graphed
right over here, we could assume that this
is the entire function definition for f of x. So for example, if
we say, well, what does f of x equal when x
is equal to negative 9? Well, we go up here. We don’t see it’s graphed here. It’s not defined for x
equals negative 9 or x equals negative 8 and 1/2 or
x equals negative 8. It’s not defined for
any of these values. It only starts getting defined
at x equals negative 6. At x equals negative 6,
f of x is equal to 5. And then it keeps
getting defined. f of x is defined for x all
the way from x equals negative 6 all the
way to x equals 7. When x equals 7, f
of x is equal to 5. You can take any x value
between negative 6, including negative
6, and positive 7, including positive
7, and you just have to see– you
just have to move up above that number,
wherever you are, to find out what the value of
the function is at that point. So the domain of this
function definition? Well, f of x is
defined for any x that is greater than or
equal to negative 6. Or we could say negative 6
is less than or equal to x, which is less than
or equal to 7. If x satisfies this
condition right over here, the function is defined. So that’s its domain. So let’s check our answer. Let’s do a few more of these. The function f of x is graphed. What is its domain? Well, exact similar argument. This function is not defined
for x is negative 9, negative 8, all the way down or all the way
up I should say to negative 1. At negative 1, it
starts getting defined. f of negative 1 is negative 5. So it’s defined for negative
1 is less than or equal to x. And it’s defined all the
way up to x equals 7, including x equals 7. So this right over
here, negative 1 is less than or equal to x
is less than or equal to 7, the function is
defined for any x that satisfies this double
inequality right over here. Let’s do a few more. The function f of x is graphed. What is its range? So now, we’re not
thinking about the x’s for which this
function is defined. We’re thinking about
the set of y values. Where do all of the
y values fall into? Well, let’s see. The lowest possible y value
or the lowest possible value of f of x that we get
here looks like it’s 0. The function never goes below 0. So f of x– so 0 is less
than or equal to f of x. It does equal 0 right over
here. f of negative 4 is 0. And then the highest y
value or the highest value that f of x obtains in this
function definition is 8. f of 7 is 8. It never gets above 8, but it
does equal 8 right over here when x is equal to 7. So 0 is less than f of x, which
is less than or equal to 8. So that’s its range. Let’s do a few more. This is kind of fun. The function f of x is graphed. What is its domain? So once again, this function
is defined for negative 2. Negative 2 is less than or
equal to x, which is less than or equal to 5. If you give me an x anywhere
in between negative 2 and 5, I can look at this graph to see
where the function is defined. f of negative 2 is negative 4. f of negative 1 is negative 3. So on and so forth,
and I can even pick the values in
between these integers. So negative 2 is less than or
equal to x, which is less than or equal to 5.
• ### goldensilverstar
Do harder ones like Eulers fiction and natural logs
• ### lexi gomez
I hate math ._.
• ### lexi gomez
I hate math ._.
• ### Anna Fear
Thank you so much!i was struggling with my math hw today and this really helped me understand:).Both of my parents were busy so I'm so thankful I found ur account.:DKeep making ur awesome videos and helping others:).
• ### Mathematics
Why do only the easy ones?
• ### Abigail Bui
Thank you! This was very helpful.
• ### Jessica Smith
what you said below
• ### tarik
I GET THIS I NEED THE RANGE
AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
• ### Reinvention
You described this so well by saying the highest and lowest point. My teacher never even told me that was where you were supposed to get the values from.
• ### Yan Markos Romero Figueroa
I have a exam of pre calculus tomorrow 😢😔
• ### justine case
When the teacher’s assistant makes a comment and assumes that the reason you don’t watch these videos, is because she thinks they intimidate you.🙄 When in reality he stutters and repeats himself and. Instantly for me, it’s like Charlie Brown’s teacher talking.
• ### EriCosmos Tech
So so clear & easy to understand from anywhere….
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Home / Calculus II / Integration Techniques / Integrals Involving Roots
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Section 1-5 : Integrals Involving Roots
In this section we’re going to look at an integration technique that can be useful for some integrals with roots in them. We’ve already seen some integrals with roots in them. Some can be done quickly with a simple Calculus I substitution and some can be done with trig substitutions.
However, not all integrals with roots will allow us to use one of these methods. Let’s look at a couple of examples to see another technique that can be used on occasion to help with these integrals.
Example 1 Evaluate the following integral. $\int{{\frac{{x + 2}}{{\sqrt[3]{{x - 3}}}}\,dx}}$
Show Solution
Sometimes when faced with an integral that contains a root we can use the following substitution to simplify the integral into a form that can be easily worked with.
$u = \sqrt[3]{{x - 3}}$
So, instead of letting $$u$$ be the stuff under the radical as we often did in Calculus I we let $$u$$ be the whole radical. Now, there will be a little more work here since we will also need to know what $$x$$ is so we can substitute in for that in the numerator and so we can compute the differential, $$dx$$. This is easy enough to get however. Just solve the substitution for $$x$$ as follows,
$x = {u^3} + 3\hspace{0.75in}dx = 3{u^2}\,du$
Using this substitution the integral is now,
\begin{align*}\int{{\frac{{\left( {{u^3} + 3} \right) + 2}}{u}\,3{u^2}du}} & = \int{{3{u^4} + 15u\,du}}\\ & = \frac{3}{5}{u^5} + \frac{{15}}{2}{u^2} + c\\ & = \frac{3}{5}{\left( {x - 3} \right)^{\frac{5}{3}}} + \frac{{15}}{2}{\left( {x - 3} \right)^{\frac{2}{3}}} + c\end{align*}
So, sometimes, when an integral contains the root $$\sqrt[n]{{g\left( x \right)}}$$ the substitution,
$u = \sqrt[n]{{g\left( x \right)}}$
can be used to simplify the integral into a form that we can deal with.
Let’s take a look at another example real quick.
Example 2 Evaluate the following integral. $\int{{\frac{2}{{x - 3\sqrt {x + 10} }}\,dx}}$
Show Solution
We’ll do the same thing we did in the previous example. Here’s the substitution and the extra work we’ll need to do to get $$x$$ in terms of $$u$$.
$u = \sqrt {x + 10} \hspace{0.5in}x = {u^2} - 10\hspace{0.5in}dx = 2u\,du$
With this substitution the integral is,
$\int{{\frac{2}{{x - 3\sqrt {x + 10} }}\,dx}} = \int{{\frac{2}{{{u^2} - 10 - 3u}}\left( {2u} \right)\,du}} = \int{{\frac{{4u}}{{{u^2} - 3u - 10}}\,du}}$
This integral can now be done with partial fractions.
$\frac{{4u}}{{\left( {u - 5} \right)\left( {u + 2} \right)}} = \frac{A}{{u - 5}} + \frac{B}{{u + 2}}$
Setting numerators equal gives,
$4u = A\left( {u + 2} \right) + B\left( {u - 5} \right)$
Picking value of $$u$$ gives the coefficients.
\begin{align*}u = & - 2 & \hspace{0.5in} - 8 = & \, B\left( { - 7} \right) & \hspace{0.5in}B = & \, \frac{8}{7}\\ u = & \,5 & \hspace{0.5in} 20 = & \, A\left( 7 \right) & \hspace{0.5in}A = & \, \frac{{20}}{7}\end{align*}
The integral is then,
\begin{align*}\int{{\frac{2}{{x - 3\sqrt {x + 10} }}\,dx}} & = \int{{\frac{{\frac{{20}}{7}}}{{u - 5}} + \frac{{\frac{8}{7}}}{{u + 2}}\,du}}\\ & = \frac{{20}}{7}\ln \left| {u - 5} \right| + \frac{8}{7}\ln \left| {u + 2} \right| + c\\ & = \frac{{20}}{7}\ln \left| {\sqrt {x + 10} - 5} \right| + \frac{8}{7}\ln \left| {\sqrt {x + 10} + 2} \right| + c\end{align*}
So, we’ve seen a nice method to eliminate roots from the integral and put it into a form that we can deal with. Note however, that this won’t always work and sometimes the new integral will be just as difficult to do.
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# Poosible Two Pairs Of A Number With Code Examples
In this article, we will look at how to get the solution for the problem, Poosible Two Pairs Of A Number With Code Examples
## How many pairs can you make with 5 numbers?
In total you will find 5 × 24 = 120 possibilities.
``````2
1
8
```
```
``````0 1
5 3
```
```
## How do you find the probability of two pairs?
TWO PAIR. This hand has the pattern AABBC where A, B, and C are from distinct kinds. The number of such hands is (13-choose-2)(4-choose-2)(4-choose-2)(11-choose-1)(4-choose-1). After dividing by (52-choose-5), the probability is 0.047539.
## How many possible 2 number combinations are there?
This means there are 100 possible combinations of two numbers using digits 0 through 9 if a digit can be repeated. If we cannot use a number twice, which means we do not return that outcome to the pool after it is selected, there is one less option for the second number.
## How many pairs of 4 numbers are there?
If you meant to say "permutations", then you are probably asking the question "how many different ways can I arrange the order of four numbers?" The answer to this question (which you got right) is 24. Here's how to observe this: 1.
## How do you find the possible pairs of numbers?
= n(n-1) / 2 which is our formula for the number of pairs needed in at least n statements.
## How many possibilities are there in drawing a pair?
If you choose only two cards, the number of ways of obtaining a pair is 156/2 (you computed permutations).
## How many possible 2 card hands are there where you are dealt a pair?
* 44 = 123,552 possible two pair hands.
## How many pairs can you make with 10 numbers?
The number of combinations that are possible with 10 numbers is 1,023. To determine this number, we can use the following formula that gives a general rule for calculating the number of combinations possible from n elements.
## What is the probability of drawing 2 pair from one deck of cards?
Now our grand total possible number of ways to draw two pair in 5 cards is 2802 * 44 = 123.552. The probability, then, of getting two pair is 123,552/2,598,960 = 0.04754, which is about 1 in 21.
## What is the pair of a number?
A set of two numbers or objects linked in some way is said to be a pair.
## Python Tkinter Window Size With Code Examples
In this article, we will look at how to get the solution for the problem, Python Tkinter Window Size With Code Examples Is tkinter good for commercial use? Tkinter is open source and free for any commercial use. from tkinter import * root = Tk() root.maxsize(height, width) root.minsize(height, width) #(69, 420) root.mainloop() # Change window_name to the name of the window object, i.e. root window_name.geometry("500x500") # To ensure widgets resize: widget_name.pack(fill="both", expand=True)
## Replace Space With Hyphen/Dash Javascript With Code Examples
In this article, we will look at how to get the solution for the problem, Replace Space With Hyphen/Dash Javascript With Code Examples How do I remove spaces in JavaScript? JavaScript String trim() The trim() method removes whitespace from both sides of a string. The trim() method does not change the original string. title = title.replace(/\s/g , "-"); The solution to the same problem, Replace Space With Hyphen/Dash Javascript, can also be found in a different method, which will be discussed fu
## Php Thread Safe Or Not Thread Safe Windows With Code Examples
In this article, we will look at how to get the solution for the problem, Php Thread Safe Or Not Thread Safe Windows With Code Examples What is non thread-safe PHP on Windows? It refers to a single thread only builds. In non-thread safe version binaries widespread use in the case of interaction with a web server through the FastCGI protocol, by not utilizing multi-threading. For example: Apache + FastCGI. IIS + FastCGI. If you dont know what you are doing and not using FastCGI then; Use thread
## No Internet Connection Html Page With Code Examples
In this article, we will look at how to get the solution for the problem, No Internet Connection Html Page With Code Examples How do I run HTML code in Chrome? Fire up Chrome and jump to the webpage you want to view the HTML source code. Right-click the page and click on “View Page Source,” or press Ctrl + U, to see the page's source in a new tab. A new tab opens along with all the HTML for the webpage, completely expanded and unformatted. if (navigator.onLine) { console.log("You are
## Laravel List All Routes With Code Examples
In this article, we will look at how to get the solution for the problem, Laravel List All Routes With Code Examples How do you show routes? Use the show ip route EXEC command to display the current state of the routing table. php artisan route:list Another method that is described below with code examples can be used to tackle the same issue Laravel List All Routes. php artisan route:list --name=<Term> Route::get('routes', function() { \$routeCollection = Route::getRoutes(); echo "<
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# Difference between revisions of "2017 AMC 10A Problems/Problem 4"
## Problem
Mia is helping her mom pick up $30$ toys that are strewn on the floor. Mia’s mom manages to put $3$ toys into the toy box every $30$ seconds, but each time immediately after those $30$ seconds have elapsed, Mia takes $2$ toys out of the box. How much time, in minutes, will it take Mia and her mom to put all $30$ toys into the box for the first time?
$\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5$
## Solution
Every $30$ seconds $3-2=1$ toys are put in the box, so after $27\cdot30$ seconds there will be $27$ toys in the box. Mia's mom will then put $3$ toys into to the box and we have our total amount of time to be $27\cdot30+30=840$ seconds, which equals $14$ minutes. $\boxed{(\textbf{B})\ 14}$
## Solution 2
Though Mia's mom places $3$ toys every $30$ seconds, Mia takes out $2$ toys right after. Therefore, after $30$ seconds, the two have collectively placed $1$ toy into the box. Thereforeby $13.5$ minutes, the two would have placed $27$ toys into the box. Therefore, at $14$ minutes, the two would have placed $30$ toys into the box. Though Mia may take $2$ toys out right after, the number of toys in the box first reaches $30$ by $14$ minutes. $\boxed{(\textbf{B})\ 14}$
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Monday, 06 September 2021 09:20
## Measurement
In this chapter learners need to apply some of the skills from Chapter 3, such as changing the subject of a formula and correctly substituting values into a formula. Remember that the units are important when you work with measurements.
4.1 Converting between different units of measurement
4.1.1 Metric conversions
You need to memorise the conversions between metric units.
Length
Conversion factors for length 10 millimetres (mm) = 1 centimetre (cm) 1 000 millimetres (mm) = 1 metre (m) 100 centimetres (cm) = 1 metre (m) 1 000 metres (m) = 1 kilometre (km)
Note:
• To convert to a smaller unit, we multiply. To convert to a larger unit, we divide.
Here is a visual representation of converting between units of length:
We can also reverse it to find lengths in larger units:
Volume
Conversion factors for volume 1 000 millilitres (mℓ) = 1 litre (ℓ) 1 000 litres (ℓ) = 1 kilolitre (kℓ)
Here is a visual representation of converting between units of volume:
And you can also reverse it:
Weight
Conversion factors for weight 1 000 mg (mg) = 1 gram (g) 1 000 grams (g) = 1 kilogram (kg) 1 000 kilograms (kg) = 1 tonne (t)
Here is a visual representation of converting between units of weight:
And one can also reverse it:
e.g. Worked example 1
Convert the following units. Remember to show all of your calculations.
1. A leaf is 25 mm long. How long is it in cm?
2. A sofa is 187 cm How long is it in metres?
3. Harry’s household uses 1 023 ℓ of water per How much water do they use in kℓ?
4. A tin contains 3,5 ℓ of How many millilitres of paint is in the tin?
5. The cover of a book is 16,2 cm long. How long is the book in mm?
6. A medicine tablet weighs 50 How much does the tablet weigh in grams?
7. A shopping bag weighs 2 850 How heavy is the bag in kg?
Solutions
1. Converting to a larger unit, divide by 10: 25 mm = 2,5
2. Converting to a larger unit, divide by 100: 187 cm = 1,87
3. Converting to a larger unit, divide by 1 000: 1 023 ℓ ÷ 1 000 = 1,023 kℓ.
4. Converting to a smaller unit, multiply by 1 000: 3,5 × 1 000 = 3 500 mℓ.
5. Converting to a smaller unit, multiply by 10: 16,2 cm × 10 = 162
6. Converting to a larger unit, divide by 1 000: 50 mg ÷ 1 000 = 0,05
7. Converting to a larger unit, divide by 1 000: 2 850 ÷ 1 000 = 2,85
Activity 1: Converting units
Do the conversions.
1. A tennis court is 23,78 m long. Convert to cm. (1)
2. Thabiso fills a bath with 23,7 ℓ of water. How much water is this in mℓ? (1)
3. The distance between Cape Town and Betty’s Bay is 90,25 km.How far is this in metres? (1)
4. The distance from Phumza’s house to the shop is 1 890 000 mm.How far is this in kilometres? (1)
5. A can of cola has a capacity of 330 mℓ. How many litres of cola is this? (1)
6. A boulder weighs 2,35 t. Convert the weight of the boulder into grams. (1)
7. A book weighs 0,85 kg. Convert the weight of the book into grams. (1)
8. Jack and Thembile live 6 473 m apart. Convert this distance to km. (1)
9. The dam on Cara’s farm contains 6,025 kℓ of water. How much is this in litres? (1)
10. A playground is 4,02 m wide. How wide is the playground in cm? (1)
11. A car weighs 1 250 000 g. What is the mass in tonnes? (1)
12. A long workbench is 295 cm long. How long is it in metres? (1) [12]
Solutions 1. A tennis court is 23,78 m long. 23,78 × 100 = 2 378 cm ✓ (1) 2. Thabiso fills a bath with 23,7 ℓ of water. 23,7 × 1 000 = 23 700 mℓ ✓ (1) 3. The distance between Cape Town and Betty’s Bay is 90,25 km. 90,25 × 1 000 = 90 250 m ✓ (1) 4. The distance from Phumza’s house to the shop is 1 890 000 mm. 1 890 000 ÷ 1 000 000 = 1,89 km ✓ (1) 5. A can of cola has a capacity of 330 mℓ. 330 mℓ ÷ 1 000 = 0,33 ℓ ✓ (1) 6. A boulder weighs 2,35 t. 2,35 t × 1 000 000 = 2 350 000 g ✓ (1) 7. A book weighs 0,85 kg. 0,85 × 1 000 = 850 g ✓ (1) 8. Jack and Thembile live 6 473 m apart. 6 473 ÷ 1 000 = 6,473 km ✓ (1) 9. The dam on Cara’s farm contains 6,025 kℓ of water. 6,025 × 1 000 = 6 025 ℓ ✓ (1) 10. A playground is 4,02 m wide. 4,02 × 100 = 402 cm ✓ (1) 11. A car weighs 1 250 000 g. 1 250 000 ÷ 1 000 000 = 1,25 t ✓ (1) 12. A long workbench is 295 cm long. 295 ÷ 100 = 2,95 m ✓ (1) [12]
4.1.2 Cooking conversions and temperature
In recipes used for cooking and baking we often find the measurements for the ingredients required in cups, teaspoons and tablespoons. Measuring cups and spoons come in standard sizes, and are common in the kitchen and in recipes because they are quick and simple to use.
If you don’t have measuring spoons and cups, you can use everyday household objects to approximate the same quantity of ingredients. For example, a small tea cup is roughly the same size as a measuring cup and a heaped, normal-sized spoon is about the same quantity as a measuring tablespoon. When following a recipe though, it is important to be as accurate as possible with your measurements, so using these rough approximations is often not suitable.
The following table shows some of the conversions used in cooking:
Conversions for cooking and baking 1 cup = 250 mℓ 1 tablespoon (tbsp) = 15 mℓ 1 teaspoon (tsp) = 5 mℓ
Note: you will be given these conversions in assessments
4.2 Measuring length
Estimation is used to find approximate values for measurements. For example, one metre is approximately the length from your shoulder to your fingertips, if you stand with your arm outstretched. A metre is also approximately the distance of one large step or jump.
e.g. Worked example 2
Carl needs to measure the width of a window, to find out how much material he must buy to make a curtain. The curtain material costs R55 per metre on sale, sold only in full metres.
1. Carl estimates the width of the window to be 1,9 metres wide (using his arm). If Carl goes to the shop with this estimate:
1. How many metres of material should he buy?
2. How much would the material cost?
2. Carl decides to double-check his estimated measurement before he buys the material and so he uses his tape measure to accurately measure the width of the He determines that the window is actually 2,2 m wide.
1. How many metres of material does he need to buy?
2. How much will the material cost?
Solutions
1. (i) 2 m
(ii) 2 × R55 = R110
2. (i) 3 m (as the material is only available in units of 1 metre)
(ii) 3 × R55 = R165
e.g. Worked example 3
Liz sews dresses for children. The material costs R89,50 per metre and she needs 2 metres of material to make a dress for a 4 year old; 2,5 metres to make a dress for a 7 year old and 3 metres to make a dress for 10 year old. The embroidery cotton costs R12,55 for a roll of 3 metres. She uses 2 rolls of cotton per dress.
1. How many metres of material will she need to make the following four dresses:
1 dress for a 7 year old, 2 dresses for four year olds, and 1 dress for a 10 year old?
2. What will the material cost for the four dresses?
3. What is the length of embroidery cotton that Liz is going to use when sewing one dress, in metres and centimetres?
4. What is the total amount that she will pay for the embroidery cotton?
5. What is the total cost of a dress for a 10 year old?
Solutions
1. 2,5 m + 2 m + 2 m + 3 m = 9,5 m
2. Length of material × price
= 9,5 m × R89,50
= R850,25
3. Length of one roll of cotton × 2 = 3 m × 2
= 6 m, or 600 cm per dress
4. Number of dresses × 2 rolls of cotton per dress × price
= 4 × 2 × R12,55
= R100,40
5. (Length of material × price) + (2 rolls of cotton × price)
= (3 m × R89,50) + (2 × R12,55)
= R268,50 + R25,10
= R293,60
Jenny has started a decorating business and has a contract to provide decor at a wedding reception.
1. The tables used at this wedding are rectangular with a length of 3 m and a width of 1 m as shown The fabric she plans to use for the tablecloth costs R75 per metre (but can be bought in lengths smaller than a metre) and is sold in rolls that are 1,4 m wide. The bride and groom want the tablecloths to hang at least 20 cm over the edges of the tables. Calculate the cost of the cloth for each table. (2)
2. If there are 15 tables at the wedding, calculate how much she is going to spend on tablecloths (1) [3]
Solutions
1. 3,4 × 1,4 × 75 = (3,4 × 75) ✓ in 1,4 m width = R255,00 ✓
2. R3 825,00 ✓
4.3 Measuring mass or weight
• The scientific word for how much an object weighs on a scale is “mass”.
• In this book we will use the words “weight” and “mass” interchangeably, because both are used in everyday language.
Worked example 4
1. A lift in a shopping mall has a notice that indicates that it can carry 2,2 tonnes or a maximum of 20 Convert the tonnes measurement to kilograms and work out what the engineer who built the lift estimated the average weight of a person to be.
2. A long distance bus seats 50 passengers and allows each passenger to each have luggage of up to 30
1. If 50 people, with average weight of 80 kg per person, each have one piece of luggage that weighs an average of 29 kg, what would be the total load carried by the bus, in tonnes?
2. If the bus weighs 4 tonnes, how much does it weigh in total (in kg) including all the passengers and the luggage?
3. Sweet Jam can be bought in bulk from a warehouse in boxes that contain 25 tins of 250 g
1. Calculate the total weight of the jam in each box, in
2. If a trader orders 15 boxes of Sweet Jam, calculate the total weight of his order in
Solutions
1. 2,2 t = 2 200 kg. 2 200 kg ÷ 20 people = 110 kg each
2. (i) (50 × 80 kg) + (50 × 29 kg)
= 4 000 kg + 1 450 kg
= 5 450 kg
= 5,45 t
(ii) 4 t = 4 000 kg. 4 000 kg + 5 450 kg = 9 450 kg
3. (i) 250 g × 25
= 6 250 g
= 6,25 kg
(ii)15 boxes × 6,25 kg = 93,75 kg
Activity 3: Measuring weight
You should never carry more than 15% of your body weight. Elias weighs 66 kg and his backpack, with school books, weighs 12 kg. Elizabeth weighs 72 kg and her school bag, with school books, weighs 8 kg.
1. Determine 15% of Elias’s Is his bag too heavy for him? (1)
2. Determine 15% of Elizabeth’s Is her bag too heavy for her? (1)
Solution
1. 9,9 The bag is too heavy for him because it weighs more than 9,9 kg. ✓ (1)
2. 10,8 kg. The bag is not too heavy for her because it weighs less than 15% of her body ✓ (1) [2]
e.g. Worked example 5
Khuthele School has two soccer fields. The grass needs to be covered with fertiliser. A 30 kg bag of fertiliser costs R42,60. The school needs to buy 96 bags.
1. How much will they pay for the fertiliser?
2. How many kg of fertilizer will they buy in total?
Solutions
1. Number of bags × price
= 96 × R42,60
= R4 089,60
2. Number of bags × weight of one bag
= 96 × 30 kg
= 2 880 kg
e.g. Worked example 6
Mr Booysens needs to buy sand to build a new room onto his house. Sand is sold for R23 per kg. Mr Booysens needs to buy 0,8 tonnes of sand in order to build the room.
1. Write the amount of sand needed in
2. Calculate the total amount of money he will have to spend to buy enough sand for the
3. If sand is only sold in 50 kg bags, how many bags will Mr Booysens need to buy?
Solution
1. Remember that 1 tonne = 1 000 kg
so he needs 0,8 tonnes × 1 000 kg = 800 kg
2. Quantity of sand needed × Cost per kg
= 800 × 23
= R18 400
3. He will need: 800 kg ÷ 50 kg
= 16 bags of sand
Activity 4: Cost and weight
A chef is preparing a meal that needs 3,75 kg of rice and 1,5 kg of beef. The recipe will feed 8 people.
1. Rice is sold in packets of 2 kg. How many packets will he need for the meal? (1)
2. If rice costs R 31,50 per 2 kg pack, calculate the total cost of the rice he will need. (1)
3. If beef costs R 41,75 per kg, calculate the total cost of the beef needed for the (1)
4. Calculate the total cost of the rice and the (1) [4]
Solutions
1. 2 ✓
2. R63,00 ✓
3. R62,63 ✓
4. R125,63 ✓
4.4 Measuring volume and capacity
Volume is a measurement of how much space an object takes up. Capacity is a measure of how much liquid a container can hold when it is full.
For example, if you have a 500 mℓ bottle of cola, with 200 mℓ of cola left inside it, the capacity of the bottle is 500 mℓ, while the volume of cola inside it is 200 mℓ.
e.g. Worked example 7
An urn of boiling water in an office has a capacity of 20 litres.
1. If it is filled to maximum capacity, calculate the number of 250 mℓ cups that can be shared from
2. After everyone has had their morning tea, there are only 6 litres of water left in the
1. How much water is this in mℓ?
2. How many 250 mℓ cups of water are left in the urn now?
3. What percentage is the remaining 6 litres of the urn’s capacity?
Solutions
1. 20 litres = 20 000 mℓ
Then 20 000 mℓ ÷ 250 mℓ = 80
80 cups can be poured from the urn.
2. (i) 6 ℓ = 6 000 mℓ
(ii) 6 000 mℓ ÷ 250 mℓ = 24
There are 24 cups of water left in the urn.
(iii) 6 ℓ ÷ 20 ℓ × 100 = 30% The urn is 30% full.
e.g. Worked example 8
Jabu is building a new flower bed and is using a bucket to carry soil from another part of the garden to the new bed. He knows his bucket has a capacity of 10 ℓ.
1. If 300 ℓ of soil must be moved, and for each trip Jabu fills the bucket to the top with soil, how many trips will Jabu have to make with the bucket to move all the soil?
2. Jabu decides that 10 litres of soil is too heavy to How many trips will he have to make to move all the soil if he only fills the bucket with 7 litres of soil at a time?
3. Jabu’s friend Matthew arrives with his wheelbarrow and a He suggests that Jabu should rather move the soil using the wheelbarrow. If the wheelbarrow has a capacity of 150 litres and they fill it to capacity, how many trips will Jabu have to make to move all the soil?
Solution
1. 300 ℓ ÷ 10 ℓ = 30 trips
2. 300 ℓ ÷ 7 ℓ = 42,8
Jabu can’t make 0,8 of a trip so we round this up to 43 trips (even though the bucket won’t have 7 litres of soil in it for the last trip).
3. 300 ℓ ÷ 150 ℓ = 2 trips
Activity 5: Measuring volume
Jonathan uses the following recipe to make chocolate muffins:
• 2 cup of baking cocoa
3
• 2 large eggs
• 2 cups of flour
• 1 cup of sugar
2
• 2 teaspoons of baking soda
• 1 1 cups of milk
3
• 1 cup of sunflower oil
3
• 1 teaspoon of vanilla essence
• 1 teaspoon of salt
2
QUESTIONS
1. If 1 teaspoon = 5 mℓ, calculate how much baking soda Jonathan will use. Give your answer in mℓ. (1)
2. Calculate the amount of vanilla essence Jonathan will use in this Give your answer in mℓ. (1)
3. Jonathan does not own measuring cups but he does own a measuring jug calibrated in mℓ. How many mℓ of flour does he need? (1 cup = 250 mℓ) (1)
4. If Jonathan buys a 100 mℓ bottle of vanilla essence, how many times will he be able to use the same bottle, if he bakes the same amount of muffins each time? (1)
5. The recipe above is used to make 30 muffins. Calculate how many cups of flour Jonathan will need to make 45 (1) [5]
Solutions
1. 10 mℓ ✓
2. 5 mℓ ✓
3. 500 mℓ ✓
4. 20 times ✓
5. He will need 3 cups of ✓
e.g. Worked example 9
Suppose paraffin is sold at R7,80 per litre at the service station.
1. How much will you pay for 5 litres of paraffin?
2. How many litres of paraffin will you be able to buy for R20? Round off your answer to two decimal
3. If you have a paraffin lamp at home that can hold 500 mℓ of paraffin, how many times will you be able to refill the lamp if you buy 3 litres of paraffin?
Solutions
1. Number of litres × Cost per litre
= 5 litres × R7,80
= R39
2. Amount of money ÷ Cost per litre
= R20 ÷ R7,80
= 2,564 102 56…
≈ 2,56 litres (to two decimal places)
3. 3 litres = 3 000 mℓ 3 000 mℓ ÷ 500 mℓ
= 6. You would be able to refill the lamp 6 times.
e.g. Worked example 10
Petrol costs R11,72 a litre.
1. Calculate how much it costs to fill up a car that has a tank with a capacity of 50
2. Calculate how many litres you could buy with Round off your answer to two decimal places.
Solutions
• Number of litres × Cost per litre
= 50 litres × R10,72
= R536
• Amount of money ÷ Cost per litre
= R200 ÷ R10,72
= 18,656 716 4…
≈ 18,66 litres (to two decimal places)
1. Thandi is baking cupcakes and her recipe requires 11/3 cup of milk
1.1 Calculate how many mℓ of milk she will need if 1 cup = 250 mℓ. (1)
1.2 If the recipe is for 20 cupcakes, calculate the amount of milk required to bake 30 Give your answer in litres. (2)
1.3 Milk is sold in bottles of 1 litre for R8,50 at the local store. Calculate the amount of money Thandi will need to spend on milk to make the 30 (1)
Solutions
1.1 333 mℓ of milk. ✓
1.2 She will need 500 mℓ of milk, ✓ which is 0,5ℓ. ✓
1.3 R8,50 (although she will only use half). ✓
2. Thabiso decides to sell homemade He has made 5 litres of lemonade to sell at the local schools’ rugby tournament.
2.1 Thabiso will be selling his lemonade in 250 mℓ plastic. Calculate the number of cups of lemonade he will be able to sell. (1)
2.2 If he sells the lemonade at R5 per cup, how much money will he make from the lemonade? (Assume that he sold all of his lemonade). (1)
2.3 If it cost Thabiso R120 to make the lemonade, how many cups would he need to sell (at R5 each) before he’s made back the money he spent? (1)
Solutions
2.1 20 cups ✓
2.2 R100 ✓
2.3 He would need to sell 24 cups just to cover his costs. ✓
4.5 Perimeter, area and volume
4.5.1 Estimation and direct measurement of perimeter
Perimeter is the total length of the outside of a shape or the continuous line forming the boundary of a closed geometric figure. Perimeter is calculated by adding together the lengths of each side of a shape. Perimeter is measured in mm, cm, m or km.
Perimeter formula Rectangle2 × length + 2 × width Square4 × length or 4 × side Trianglelength 1 + length 2 + length 3 Circumferenceπ × (2 × radius) or π × diameter
Note:
• To measure the perimeter of a rectangle, a square or a triangle, we simply measure the length of each side using a ruler and add up the sides to get the perimeter.
• To measure the perimeter of a circle, we need to use a piece of string: we can place the string along the outline of the circle, marking off how much string it took to go around the circle once. Then we measure that length of string on a ruler to estimate the perimeter of the circle.
The perimeter of a circle is the same as the circumference of the circle.
e.g. Worked example 11
Mr and Mrs Dlamini have recently moved into a new house. In the rectangular back yard, the house has a lawn and a rectangular patio as shown in the diagram.
1. Using a ruler, measure the perimeter (in cm) of Mr and Mrs Dlamini’s backyard on the diagram.
2. If the diagram was drawn using a scale of 1 : 100, calculate the perimeter of the yard in metres.
Solutions
1. The length of the yard is 5 cm and the width is 4,2
Because the back yard is a rectangle, both pairs of opposite sides are equal in length.
The perimeter is the total length of the outside of the yard, therefore: Perimeter = 4,2 cm + 4,2 cm + 5 cm + 5 cm
= 18,4 cm
2. Using the scale of 1 : 100 Perimeter = 18,4 cm × 100
= 1 840 cm
= 18,4 m
e.g. Worked example 12
Mrs Dlamini wants to dig up some of the lawn and plant a triangular vegetable garden as shown in the diagram alongside.
1. Using a ruler, measure the perimeter of the triangular garden in the diagram (in cm).
2. If this diagram was drawn using a scale of 1 : 100, calculate the actual perimeter of the garden in metres.
Solutions
1. The perimeter of the triangle is = 1,7 cm + 5 cm + 5,3 cm
= 12 cm
2. Using a scale of 1: 100
= 12 cm × 100
= 1 200 cm
= 12 m
Activity 7: Measuring perimeter
Study the diagram alongside and answer the questions that follow.
1. Before Mr Dlamini builds his fish pond, he decides he wants to make the patio smaller. Using a ruler, measure the new perimeter of the patio on the diagram (in cm). (1)
2. Mrs Dlamini decides it might be better to build her vegetable garden on the right of the garden because that area gets more Using a ruler, measure the perimeter of the new triangular garden on the diagram (in mm). (1)
3. Mrs Dlamini also buys a new, circular table for the Using a piece of string and a ruler, estimate the circumference of the table (in cm). (1) [3]
NB : You will always be given the perimeter formulae in your assessments.
Solutions
1. approximately 10 cm ✓
2. approximately 82 mm ✓
3. approximately 2,5 ✓
Note:
• The radius (r) of a circle is the length of the line from the centre of the circle to any point on its circumference.
• The diameter (d) of a circle is a straight line drawn from one edge of the circle to the other, that passes through the centre of the circle. diameter = 2 × radius.
• Pi (π) is a special symbol we use when calculating perimeter and area of circles. The value of π is 3,141 592 645…. For all of our calculations, we will use the approximate value of π = 3,142.
Worked example 13
Using the formulae given earlier, study the diagram alongside and answer the questions that follow.
1. Calculate the perimeter of the back yard, including the patio (i.e. the whole diagram) (in cm).
2. Calculate the perimeter of the patio (in mm).
3. Calculate the perimeter of Mrs Dlamini’s garden (in cm).
4. Calculate the perimeter of the table on the patio (in cm). Round your answer to 1 decimal place.
5. Is your answer to number d) different to the table circumference you estimated in the previous activity, using string and a ruler? If it is, discuss why this could be with a friend.
Solutions
1. Perimeter of rectangular back yard
= 2 × length + 2 × width
= (2 × 6,2 cm) + (2 × 5,2 cm)
= 12,4 cm + 10,4 cm
= 22,8 cm
2. Perimeter of square patio
= 4 × length
= 4 × 2,5 cm
= 10 cm
10 cm × 10 = 100 mm
3. Perimeter of triangular garden
= length 1 + length 2 + length 3
= 2 cm + 2,9 cm + 3,5 cm
= 8,4 cm
4. Circumference of table
= π × diameter
= π × 0,8 cm
= 3,142 × 0,8 cm
= 2,5136 cm
= 2,5 cm
5. Previously, we estimated the circumference of the table using a piece of string and a ruler. Using the formula to calculate the circumference of a circle is more accurate than using a piece of string.
Note:
• The shapes we have worked with so far have been simple. Sometimes we have to calculate the perimeter of a more complicated shape, which is made up of regular shapes that have been joined together, or in which the units are not all the same. We will look at how to do this in the next activity.
Mrs Dlamini buys a new lampshade for a lamp. She measures the radius of the inside circle in the lampshade to be 50 mm. The diameter of the outside (larger) circle is 40 cm. (Note, the diagram is not drawn to scale.)
1. Calculate the circumference of the smaller, inner circle (in cm). (3)
2. Calculate the circumference of the larger, outer circle (in cm). Round off your answer to one decimal place. (3)
3. Calculate the perimeter of half of the larger, outer circle (in cm). (1)
4. Calculate the width of the area shown by the dotted line in the diagram (1) [8]
Solutions
1. Inside circle perimeter/circumference
= 2πr ✓
= 2 × 3,142 × 5 cm ✓
= 31,42 cm ✓
2. Circumference/perimeter
= 2πr ✓
= 2 × 3,142 × 20 cm ✓
= 125,7 cm ✓
3. Half perimeter
= Perimeter = 125,7 = 62,85 cm ✓
2 2
4. Inner circle radius = 5 cm. Entire radius = 20 cm.
Difference between radii = 20 cm – 5 cm = 15 cm ✓ (1) [8]
4.5.2 Using formulae to calculate area
Area formula Diagram Rectangle length × width Squarelength × length = length2 orside × side = side2 Triangle 1 × base × perpendicular height 2 Circleπ × radius2
• A perpendicular line is a straight line that lies at an angle of 90° to a given line,plane, or surface
e.g. Worked example 14
Using the appropriate formula from the given table, calculate the areas of the following three shapes (in cm2). Diagrams are not drawn to scale.
Solutions
1. Area of rectangle = length × width
= 5 cm × 6 cm
= 30 cm2
2. Area of triangle = ½ × base × perpendicular height
= ½ × 4 cm × 12 cm
= 24 cm2
3. Area of circle = π × radius2
= 3,142 × (1 cm)2
= 3,142 × 1 cm2
= 3,142 cm2
In this case, it is easy to solve the problem by breaking down the complex object into smaller shapes, finding the area of each smaller shape, and then adding the individual areas together. The next worked example will show you how to work with such shapes.
e.g. Worked example 15
Your Mathematical Literacy classroom gets new tables, shaped as shown alongside.
1. Using the appropriate formulae, calculate the area of the table, in m2.
2. If each table cost R615 and ten tables were bought, calculate how much the tables cost per m2.
(Hint: calculate the total cost of the tables and their total area first.)
Solution
1. We can see that the table is made up of two identical triangles, and one rectangle
The formula for the area of a triangle is:
½ × base × height
So the area of one of our triangles is:
½ × 0,5m × 0,7m (change the units to meters)
= 0,175 m2
The formula for the area of a rectangle is: length × breadth.
So the area of the middle rectangle is:
0,9 m × 70 cm
= 0,9 m × 0,7 m (change the units to metres)
= 0,63 m2
Now we simply add the three areas together:
Area triangle + area rectangle + area triangle
= 0,175 m2 + 0,63 m2 + 0,175 m2
= 0,98 m2
2. 10 tables will cost R615 × 10 = R6
10 tables will have a total area of 0,98 m2 × 10 = 9,80 m2.
R6 150 ÷ 9,80 m2 = R627,55
So the tables cost R627,55 per square metre.
Activity 9: Combining areas
For your birthday, a friend gives you a rare, lucky coin that has a square cut out of the middle as shown in the photo and diagram.
1. You measure the diameter of the circle to be 3 cm, and the length of one side of the square to be 0,9 cm. Calculate the area of the coin in cm2.
2. If the coin is worth R3,58 per cm2, calculate its value. (2) [9]
Solutions
1. To calculate the area of the coin, we need to calculate the area of the circle, and then subtract from this the area of the square cut-out.
The formula for the area of a circle is π × radius2. ✓
We know the diameter is 3 cm, therefore the radius is 1,5 cm.
Therefore the area of the circle is:
π × (1,5 cm)2
= 3,142 × 2,25 cm2
= 7,0695 cm2. ✓
(Remember, we shouldn’t round off while we are still busy with our calculations! We should only round off our final answer.)
The formula for the area of a square is side × side = (side)2. ✓
Therefore the area of the square is:
(0,9 cm)2 = 0,81 cm2. ✓
We now subtract the area of the cut-out square from the area of the circle: 7,0695 cm2 – 0,81 cm2 = 6,2595 cm2
so the area of the coin is 6,2595 cm2 ≈ 6,3 cm2. ✓ (7)
2. 6,2595 cm2 × R3,58 ✓ = R22,409 01 ≈ R22,41 ✓ (2) [9]
4.5.3 Using formulae to calculate volume
Shape Volume formula Rectangular box V = l × b × h Cylinder V = π × r2 × h
e.g. Worked example 16
Cedric is building a house. First he digs the rectangular foundation for the house. The foundation is filled with cement. The dimensions of the foundation are 8 m × 0,5 m × 0,5 m.
1. Calculate the volume of the foundation.
2. If concrete for the foundation costs R180,00/m3, what is the total cost of the concrete for the foundation?
3. Cedric finds cheaper concrete at a total cost of R320 for 2 m3. Calculate the cost per m3.
Solutions
1. Volume = 8 × 0,5 × 0,5
= 2 m3
2. Total cost of concrete = 2 × R180,00
= R360,00
3. Cost per m3= R320,00 ÷ 2
= R160,00
Allison needs to bake cookies for her son’s crèche. She finds a recipe for cookies. She needs to calculate the volume of 1 cookie so that she knows what size container she can use. Each cookie is shaped like a flat cylinder. She measures a cookie and finds that it has these dimensions: diameter = 80 mm; height = 7 mm.
1. Calculate the volume of 1 biscuit, to one whole (3)
2. Calculate the volume of 50 (1)
3. Would a container with a volume of 700 cm3 hold the biscuits? (2) [6]
Solution
1. 35 190 mm3
πr2h ✓
= π(40)2 (7) ✓
= π(1600) (7)
2. 1 759 500 mm3
3. 1 759,5 cm3 (No 700 cm3 < 1 759,5 cm3) ✓✓
Activity 11: Multi-step volume problem
A school builds a swimming pool with the following dimensions: length = 15 m; depth = 1,3 m to the filling level, and width = 5 m.
(1 m3 = 1 000 ℓ and 1 000 ℓ = 1 kℓ)
1. Calculate the volume of the swimming pool up to the level it is filled. (1)
2. Convert this volume
(i) to litres
(ii) and kilolitres. (2)
3. When the school fills the pool, they use a pump which pumps water at a rate of 2 ℓ per second. How long would it take to fill up the pool? Give your answer in hours and minutes. (1)
4. Water costs R8,64 per How much will it cost the school to fill up the pool? (1) [5]
Solutions
1. 97,5 m3
2. (i) 97 500 ℓ
(ii) 97,5 kℓ ✓✓
3. So the total time taken is 13 hr 32½ min ✓
4. R842,40 ✓
4.6 Calculating elapsed time
Elapsed time, or duration, is the measurement of time passing. When doing calculations like this, we add the units of time separately. Be careful when working with remainders!
e.g. Worked example 17
1. School starts at 07:45. You are in class for 2 hours 30 What time will the bell ring for first break? Give your answer in the 24-hour format.
2. Palesa starts cooking dinner at 6:00 m. She has to leave for her choir practice in 1 hour and 45 minutes.
1. What time must she leave? (Give your answer in the 12-hour )
3. The bus leaves school at 14:30. It takes 70 minutes to get to Mulalo’s
1. What time will he arrive at home? (Give your answer in the 24-hour )
Solutions
1. First add the hours: 07:00 + 2 hours = 9:00
45 minutes + 30 minutes = 75 minutes
75 minutes = 60 minutes and 15 minutes
= 1 hour and 15 minutes
Calculate the total time elapsed:
9:00 + 1 hour 15 minutes = 10:15
So the bell will ring for break at 10:15.
2.
1. First add the hours: 6:00 m. + 1 hour = 7:00 p.m.
Then add the minutes: 0 minutes + 45 minutes = 45 minutes
Calculate the total time that will elapse: 7:00 p.m. and 45 minutes
= 7:45 p.m.
So Palesa must leave at 7:45 p.m.
2. To convert this to the 24-hour time format we simply add 12 hours to the time:
7:45 p.m. + 12 hours = 19:45.
3.
1. First we break down 70 minutes into hours and minutes
We know that 60 minutes = 1 hour. 70 minutes - 60 minutes = 10 minutes, so the bus ride takes 1 hour and 10 minutes.
14:00 + 1 hour = 15:00.
Next we add the minutes: 30 + 10 = 40 minutes.
So Mulalo will arrive home at 15:40.
2. To convert our answer to the 12-hour format we subtract 12 hours:
15:40 – 12 hours = 3:40. We know that 15:40 is after midday, so Mulalo will arrive home at 3:40 p.m.
Activity 12: Calculating elapsed time
1. Unathi’s father goes to work at 8:00 m. He fetches her from school 7 hours and 30 minutes later. What time will he fetch her? Give your answer in the 24-hour format. (1)
2. Lauren finishes her music class at 15:30. It takes her 30 minutes to get home. She then does homework for 50 minutes. Lauren meets her friend 20 minutes after she finishes her What time do they meet? Give your answer in the 12-hour format. (1)
3. Heather starts baking biscuits at 6:15 p.m. The biscuits must come out of the oven at 6:35 m. and need to cool for another 20 minutes before they can be eaten.
1. How long will the biscuits be in the oven? (1)
2. What time will they be ready to eat? (Give your answer in the 12-hour format.) (1)
4. Alison’s favourite TV show starts at 20:35. It is forty-five minutes long.
1. What time will it finish? (1)
2. If Alison watches the movie that follows her favourite show and it finishes at 10:50 p.m., how long was the movie (in hours and minutes)? (1)
5. Vinayak is meeting his brother for lunch at 13:15. He also wants to go to the shops before It will take him 20 minutes to get from the shops to the restaurant where he’s meeting his brother. If he leaves home at 10:10 how much time does he have to do his shopping? Give your answer in hours and minutes. (1) [7]
Solutions
1. 15:30 ✓ (1)
2. 5:10 p.m. ✓ (1)
3.
1. 20 minutes✓ (1)
2. 6:55 p.m. ✓ (1)
4.
1. 21:20 ✓ (1)
2. 1 hour, 30 minutes ✓ (1)
5. 2 hours, 45 minutes ✓ (1)
4.6.1 Calendars
Calendars are useful tools to help us keep track of events that are going to happen and to plan our lives accordingly. We can add information to them about important events and dates (like birthdays and school holidays). We can read off days, weeks and months on a calendar and do conversions between these units of time.
You may have come across a time conversion that states that 4 weeks is approximately equal to one month. This is not quite correct. 4 weeks is equal to 28 days, but the months (except February!) have 30 or 31 days in them. When working with calendars, be careful to count the right number of days in a particular month!
e.g. Worked example 18
Jess’s calendar for the month of May is given below. Study it carefully and answer the questions that follow.
1. If it is Monday 6 May, calculate how many days it is until:
1. Mother’s Day
2. Jess goes on her school camp
3. Jess’s granny comes to
2. If it is 8 May:
1. How many weeks does Jess have to study for her Mathematical Literacy test?
2. How many days does she have to study for the test?
3. How many days ago was her dad’s birthday?
3. Will Jess go to school on 1 May? Give a reason for your answer.
4. Jess needs to buy a present for her mother for Mother’s If she has plans with friends on 11 May, by when should she have bought the present?
5. Jess is invited to a party on Saturday 18 Will she be able to attend?
6. Jess wants to bake a cake for her granny but has plans with a friend for the morning of 25
1. If her granny arrives in the evening of 25 May, when should Jess bake the cake?
2. Given that she’s busy on the morning of 25 May, when should Jess make time to buy the ingredients for the cake?
Solutions
1. (i) 6 days
(ii) 11 days
(iii) 19 days
2. (i) 2 weeks
(ii) 14 days
(iii) 6 days ago
3. 1 May is Workers’ Day which is a public holiday.
4. Jess should buy a present for her mother by Friday 10 May.
5. She will be away on her school camp.
6. (i) On the afternoon of Saturday 25 May.
(ii) On or before Friday 24 May.
4.6.2 Timetables
Timetables are similar to calendars in that they help us plan our time. Where calendars are useful for planning months and years, timetables are useful for planning shorter periods of time like hours, days and weeks. You may already be familiar with timetables like those for your different classes at school, and for TV shows. In this section we will learn how to read timetables and how to draw up our own.
Worked example 19
Look at the timetable below and answer the following questions.
1. What is the difference in time between the English News at 5:30 m. and the English News at 8:30 p.m. (both on SABC 2)?
2. How long, in minutes, is American Idol?
3. If Zonke wants to watch Isidingo after dinner at 7:30 m., and she needs 90 minutes to cook and eat dinner, what time should she start cooking dinner?
4. Mandla wants to watch It’s My Biz and He plans to do his homework in between the two shows. If he expects each subject’s homework to take 30 minutes, how many subjects worth of homework will he be able to complete between the two shows?
5. Sipho wants to watch the news in English and in Afrikaans, at the same Would this be possible? Give a reason for your answer.
6. Why are the blocks on the timetable for SABC 3, blank for 8:30 m. and 9:00 p.m.?
7. What is the total time period allocated to the News (in all languages) across all four TV channels?
Solutions
1. 3 hours
2. 7:30 to 8:30 m. = 1 hour = 60 minutes
3. 90 minutes = 1 hour + 30 minutes
7:30 m. - 1 hour = 6:30 p.m.
6:30 p.m. - 30 minutes = 6:00 p.m.
4. It’s My Biz finishes at 6:00 m. and Generations starts at 8:00 p.m.
This gives Mandla 2 hours to do his homework.
2 hours = 120 minutes
120 minutes ÷ 30 minutes = 4
So Mandla will be able to do homework for four subjects in between the two shows
5. Yes, there is the English News on SABC 3 at 7:00 m. and on SABC 2 there is the Afrikaans Nuus at that same time.
However, he cannot watch two channels at the same time. He would need to choose a channel to watch.
6. They are blank because the program “Welcome to the Parker” is still showing.
7. There are 8 sets of news slots appearing on the Each slot is 30 minutes. Therefore, a total of 4 hours of news will be shown between 5:30 p.m. and 9:00 p.m. on four channels.
Activity 13: Drawing up a timetable
Sipho and Mpho are brothers. Their parents require them to do household chores every day. These chores need to fit into their school sports and homework timetables.
Using the information provided in the table below, construct a timetable for each brother for one day of the week.
The two brothers’ timetables need to be clearly laid out and easy to read.
SIPHO MPHO Soccer practice 15:30 - 16:30 Music lesson (1 hour) Feed the dogs Walk the dogs for a minimum of 30 minutes Wash the dishes Study for Maths test - 45 minutes Complete his Life Orientation task - 45 minutes Set (and clear) the table before and after dinner Watch the news at 19:00 for his history assignment Look through the newspaper for any information on natural disasters for his geography homework.
Solution
For example:
Sipho:
Time Event 15:30 - 16:30 Soccer practice 18:00 Feed dogs and wash dishes 19:00 Watch news for history assignment 19:30 - 20:15 Complete LO task
Activity 14: Reading a time table
Mr Odwa and his family live in the informal settlement in Langa Township. Mr Odwa has two school going kids Zonke and Andile who are attending the school at Philippi High. Mrs Odwa is a school teacher at Mandalay Secondary, while Mr Odwa works in a construction company in Woodstock.
Use the train table on the previous page to answer these questions.
1. If Zonke and Andile want to be at Philippi station at 07:31 what time must they catch a train in Langa station? (1)
2. Which platform will that train depart from? (1)
3. Give the train number and platform number for the train that will stop at Heideveld at 08:23. (2)
4. If Mrs Odwa Pamela is at Mandalay at 09:12 what time did she depart from Langa station? (1)
5. Mr Odwa works night shift and he wants to meet his two kids at Langa station before they catch their train to What time should he take the train in Woodstock and at which platform is that train going to stop? (2)
6. If the school starts at 08:00 and the kids miss the train mentioned in 1, what time will be the next train and what number and platform must they be on to catch the train?(3)
7. Is it possible for Mr Odwa to use the same time table to find the time for a train from Langa to Woodstock? Explain your answer. (2) [12]
Solutions
1. 07:13 ✓ (1)
2. 16 ✓
3. Train number - 9513 ✓ (1)
Platform - 16 ✓ (2)
4. 8:48 ✓ (1)
5. 6:33 (Platform 16 and Train 9507) ✓✓ (2)
6. 7:25 on Platform 20 (Arrive in Philippi at 07:44) ✓✓✓ (3)
7. No. The time table is one way from Cape Town to Chris Hani. ✓✓ (2) [12]
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# Hypothesis Testing. To define a statistical Test we 1.Choose a statistic (called the test statistic) 2.Divide the range of possible values for the test.
## Presentation on theme: "Hypothesis Testing. To define a statistical Test we 1.Choose a statistic (called the test statistic) 2.Divide the range of possible values for the test."— Presentation transcript:
Hypothesis Testing
To define a statistical Test we 1.Choose a statistic (called the test statistic) 2.Divide the range of possible values for the test statistic into two parts The Acceptance Region The Critical Region
To perform a statistical Test we 1.Collect the data. 2.Compute the value of the test statistic. 3.Make the Decision: If the value of the test statistic is in the Acceptance Region we decide to accept H 0. If the value of the test statistic is in the Critical Region we decide to reject H 0.
The z-test for Proportions Testing the probability of success in a binomial experiment
Situation A success-failure experiment has been repeated n times The probability of success p is unknown. We want to test –H 0 : p = p 0 (some specified value of p) Against –H A :
The Test Statistic The Acceptance and Critical Region Accept H 0 if: Reject H 0 if: Two-tailed critical region
The Acceptance and Critical Region Accept H 0 if: Reject H 0 if: One-tailed critical regions These are used when the alternative hypothesis (H A ) is one-sided Accept H 0 if: Reject H 0 if:
The Acceptance and Critical Region Accept H 0 if:, Reject H 0 if: One-tailed critical regions
The Acceptance and Critical Region Accept H 0 if:, Reject H 0 if: One-tailed critical regions
Comments Whether you use a one-tailed or a two-tailed tests is determined by the choice of the alternative hypothesis H A The alternative hypothesis, H A, is usually the research hypothesis. The hypothesis that the researcher is trying to “prove”.
Examples 1.A person wants to determine if a coin should be accepted as being fair. Let p be the probability that a head is tossed. One is trying to determine if there is a difference (positive or negative) with the fair value of p.
2.A researcher is interested in determining if a new procedure is an improvement over the old procedure. The probability of success for the old procedure is p 0 (known). The probability of success for the new procedure is p (unknown). One is trying to determine if the new procedure is better (i.e. p > p 0 ).
2.A researcher is interested in determining if a new procedure is no longer worth considering. The probability of success for the old procedure is p 0 (known). The probability of success for the new procedure is p (unknown). One is trying to determine if the new procedure is definitely worse than the one presently being used (i.e. p < p 0 ).
The z-test for the Mean of a Normal Population We want to test, , denote the mean of a normal population
The Situation Let x 1, x 2, x 3, …, x n denote a sample from a normal population with mean and standard deviation . Let we want to test if the mean, , is equal to some given value 0. Obviously if the sample mean is close to 0 the Null Hypothesis should be accepted otherwise the null Hypothesis should be rejected.
The Test Statistic
The Acceptance and Critical Region This depends on H 0 and H A Accept H 0 if: Reject H 0 if: Two-tailed critical region Accept H 0 if: Reject H 0 if: One-tailed critical regions Accept H 0 if: Reject H 0 if:
Example A manufacturer Glucosamine capsules claims that each capsule contains on the average: 500 mg of glucosamine To test this claim n = 40 capsules were selected and amount of glucosamine (X) measured in each capsule. Summary statistics:
We want to test: Manufacturers claim is correct against Manufacturers claim is not correct
The Test Statistic
The Critical Region and Acceptance Region Using = 0.05 We accept H 0 if -1.960 ≤ z ≤ 1.960 z /2 = z 0.025 = 1.960 reject H 0 if z 1.960
The Decision Since z= -2.75 < -1.960 We reject H 0 Conclude: the manufacturers’s claim is incorrect:
“Students” t-test
Recall: The z-test for means The Test Statistic
Comments The sampling distribution of this statistic is the standard Normal distribution The replacement of by s leaves this distribution unchanged only the sample size n is large.
For small sample sizes: The sampling distribution of Is called “students” t distribution with n –1 degrees of freedom
Properties of Student’s t distribution Similar to Standard normal distribution –Symmetric –unimodal –Centred at zero Larger spread about zero. –The reason for this is the increased variability introduced by replacing by s. As the sample size increases (degrees of freedom increases) the t distribution approaches the standard normal distribution
t distribution standard normal distribution
The Situation Let x 1, x 2, x 3, …, x n denote a sample from a normal population with mean and standard deviation . Both and are unknown. Let we want to test if the mean, , is equal to some given value 0.
The Test Statistic The sampling distribution of the test statistic is the t distribution with n-1 degrees of freedom
The Alternative Hypothesis H A The Critical Region t and t /2 are critical values under the t distribution with n – 1 degrees of freedom
Critical values for the t-distribution or /2
Critical values for the t-distribution are provided in tables. A link to these tables are given with today’s lecture
Look up df Look up
Note: the values tabled for df = ∞ are the same values for the standard normal distribution
Example Let x 1, x 2, x 3, x 4, x 5, x 6 denote weight loss from a new diet for n = 6 cases. Assume that x 1, x 2, x 3, x 4, x 5, x 6 is a sample from a normal population with mean and standard deviation . Both and are unknown. we want to test: versus New diet is not effective New diet is effective
The Test Statistic The Critical region: Reject if
The Data The summary statistics:
The Test Statistic The Critical Region (using = 0.05) Reject if Conclusion: Accept H 0 :
Confidence Intervals
Confidence Intervals for the mean of a Normal Population, m, using the Standard Normal distribution Confidence Intervals for the mean of a Normal Population, m, using the t distribution
The Data The summary statistics:
Example Let x 1, x 2, x 3, x 4, x 5, x 6 denote weight loss from a new diet for n = 6 cases. The Data: The summary statistics:
Confidence Intervals (use = 0.05)
Comparing Populations Proportions and means
Sums, Differences, Combinations of R.V.’s A linear combination of random variables, X, Y,... is a combination of the form: L = aX + bY + … where a, b, etc. are numbers – positive or negative. Most common: Sum = X + YDifference = X – Y Simple Linear combination of X, bX + a
Means of Linear Combinations The mean of L is: Mean(L) = a Mean(X) + b Mean(Y) + … Most common: Mean( X + Y) = Mean(X) + Mean(Y) Mean(X – Y) = Mean(X) – Mean(Y) Mean(bX + a) = bMean(X) + a IfL = aX + bY + …
Variances of Linear Combinations If X, Y,... are independent random variables and L = aX + bY + … then Variance(L) = a 2 Variance(X) + b 2 Variance(Y) + … Most common: Variance( X + Y) = Variance(X) + Variance(Y) Variance(X – Y) = Variance(X) + Variance(Y) Variance(bX + a) = b 2 Variance(X)
If X, Y,... are independent normal random variables, then L = aX + bY + … is normally distributed. In particular: X + Y is normal with X – Y is normal with Combining Independent Normal Random Variables
Comparing proportions Situation We have two populations (1 and 2) Let p 1 denote the probability (proportion) of “success” in population 1. Let p 2 denote the probability (proportion) of “success” in population 2. Objective is to compare the two population proportions
We want to test either: or
The test statistic:
Where: A sample of n 1 is selected from population 1 resulting in x 1 successes A sample of n 2 is selected from population 2 resulting in x 2 successes
Logic:
The Alternative Hypothesis H A The Critical Region
Example In a national study to determine if there was an increase in mortality due to pipe smoking, a random sample of n 1 = 1067 male nonsmoking pensioners were observed for a five-year period. In addition a sample of n 2 = 402 male pensioners who had smoked a pipe for more than six years were observed for the same five-year period. At the end of the five-year period, x 1 = 117 of the nonsmoking pensioners had died while x 2 = 54 of the pipe-smoking pensioners had died. Is there a the mortality rate for pipe smokers higher than that for non-smokers
We want to test:
The test statistic:
Note:
The test statistic:
We reject H 0 if: Not true hence we accept H 0. Conclusion: There is not a significant ( = 0.05) increase in the mortality rate due to pipe-smoking
Estimating a difference proportions using confidence intervals Situation We have two populations (1 and 2) Let p 1 denote the probability (proportion) of “success” in population 1. Let p 2 denote the probability (proportion) of “success” in population 2. Objective is to estimate the difference in the two population proportions = p 1 – p 2.
Confidence Interval for = p 1 – p 2 100P% = 100(1 – ) % :
Example Estimating the increase in the mortality rate for pipe smokers higher over that for non- smokers = p 2 – p 1
Comparing Means Situation We have two normal populations (1 and 2) Let 1 and 1 denote the mean and standard deviation of population 1. Let 2 and 2 denote the mean and standard deviation of population 1. Let x 1, x 2, x 3, …, x n denote a sample from a normal population 1. Let y 1, y 2, y 3, …, y m denote a sample from a normal population 2. Objective is to compare the two population means
We want to test either: or
Consider the test statistic:
If: will have a standard Normal distribution This will also be true for the approximation (obtained by replacing 1 by s x and 2 by s y ) if the sample sizes n and m are large (greater than 30)
Note:
The Alternative Hypothesis H A The Critical Region
Example A study was interested in determining if an exercise program had some effect on reduction of Blood Pressure in subjects with abnormally high blood pressure. For this purpose a sample of n = 500 patients with abnormally high blood pressure were required to adhere to the exercise regime. A second sample m = 400 of patients with abnormally high blood pressure were not required to adhere to the exercise regime. After a period of one year the reduction in blood pressure was measured for each patient in the study.
We want to test: The exercize group did not have a higher average reduction in blood pressure The exercize group did have a higher average reduction in blood pressure vs
The test statistic:
Suppose the data has been collected and:
The test statistic:
We reject H 0 if: True hence we reject H 0. Conclusion: There is a significant ( = 0.05) effect due to the exercise regime on the reduction in Blood pressure
Estimating a difference means using confidence intervals Situation We have two populations (1 and 2) Let 1 denote the mean of population 1. Let 2 denote the mean of population 2. Objective is to estimate the difference in the two population proportions = 1 – 2.
Confidence Interval for = 1 – 2
Example Estimating the increase in the average reduction in Blood pressure due to the excercize regime = 1 – 2
Comparing Means – small samples Situation We have two normal populations (1 and 2) Let 1 and 1 denote the mean and standard deviation of population 1. Let 2 and 2 denote the mean and standard deviation of population 1. Let x 1, x 2, x 3, …, x n denote a sample from a normal population 1. Let y 1, y 2, y 3, …, y m denote a sample from a normal population 2. Objective is to compare the two population means
We want to test either: or
Consider the test statistic:
If the sample sizes (m and n) are large the statistic will have approximately a standard normal distribution This will not be the case if sample sizes (m and n) are small
The t test – for comparing means – small samples Situation We have two normal populations (1 and 2) Let 1 and denote the mean and standard deviation of population 1. Let 2 and denote the mean and standard deviation of population 1. Note: we assume that the standard deviation for each population is the same. 1 = 2 =
Let
The pooled estimate of . Note: both s x and s y are estimators of . These can be combined to form a single estimator of , s Pooled.
The test statistic: If 1 = 2 this statistic has a t distribution with n + m –2 degrees of freedom
The Alternative Hypothesis H A The Critical Region are critical points under the t distribution with degrees of freedom n + m –2.
Example A study was interested in determining if administration of a drug reduces cancerous tumor size. For this purpose n +m = 9 test animals are implanted with a cancerous tumor. n = 3 are selected at random and administered the drug. The remaining m = 6 are left untreated. Final tumour sizes are measured at the end of the test period
We want to test: The treated group did not have a lower average final tumour size. The exercize group did have a lower average final tumour size. vs
The test statistic:
Suppose the data has been collected and:
The test statistic:
We reject H 0 if: Hence we accept H 0. Conclusion: The drug treatment does not result in a significant ( = 0.05) smaller final tumour size, with d.f. = n + m – 2 = 7
Download ppt "Hypothesis Testing. To define a statistical Test we 1.Choose a statistic (called the test statistic) 2.Divide the range of possible values for the test."
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Finding the Volume of Solid Figures MCC6.G.2 – Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes.
Presentation on theme: "Finding the Volume of Solid Figures MCC6.G.2 – Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes."— Presentation transcript:
Finding the Volume of Solid Figures MCC6.G.2 – Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and show tha the volume is the same as would be found by multiplying the edge lengths of the prism. Apply the formulas V = lwh and V = bh to find volumes of right rectangular prisms with fractional edge lengths in the context of solving real-world and mathematical problems.
What Is Volume ? The volume of a solid is the amount of space inside it. Consider this rectangular prism. If we were to fill the prism with water, the volume would be the amount of water it could hold.
Measuring Volume Volume is measured in cubic centimeters (also called centimeters cubed). Here is a cubic centimeter It is a cube which measures 1cm in all directions. 1cm
Volume of a Rectangular Prism Look at the rectangular prism below: 10cm 3cm 4cm We must first calculate the area of the base (“big B”): 3cm 10cm The base is a rectangle measuring 10cm by 3cm:
10cm 3cm 4cm 3cm 10cm Area of a rectangle = length x width Area = 10 x 3 Area = 30cm 2 We now know we can place 30 centimeter squares on the base of the prism.
10cm 3cm 4cm Now we need to find how many layers of 1cm cubes we can place in the prism: We can fit 4 layers. Volume = 30 x 4 Volume = 120cm 3 That means that we can place 120 centimeter cubes inside the prism.
10cm 3cm 4cm We have found that the volume of the prism is given by: Volume = 10 x 3 x 4 = 120cm 3 This gives us our formula to find the volume of a prism: Volume = Area of the base x Height V=Bh
Let’s Practice Calculate the volume of the prism below: (1) 14cm 5 cm 7cm V = B h lw V = 14(5)(7) V = 490 cm 3
(2) 12 cm V = B h lw V = 12(12)(12) V = 1728 cm 3 (3) 9 m 7m 3.2m V = B h lw V = 9(7)(3.2) V = 201.6 m 3
Download ppt "Finding the Volume of Solid Figures MCC6.G.2 – Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes."
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# 265L07 - LESSON 7 Functions of Several Variable and Their...
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Unformatted text preview: LESSON 7 Functions of Several Variable and Their Limits READ: Sections 15.1, 15.2 NOTES: There are many situations in which a quantity depends on several other quantities. For example, the volume V of a circular cylinder depends on both the height, h , of the cylinder and on the radius, r , of the base: V = πr 2 h . In such a case, we say that V is a function of two variables. In general, if z depends in some way on the quantities x and y , we’ll say z is a function of two variables and write z = f ( x,y ). As pointed out a few lessons back, it is possible to draw the graph of an equation z = f ( x,y ) in 3-space. Recall that as a rough general guide, the graph will look like a sheet floating above (or below) a region in the x , y-plane. The graphs of a few such surfaces are already familiar: paraboloids, hyperboloids, and so on. One way to get an idea about the shape of the graph of a less familiar surface, z = f ( x,y ), is to look at the traces of the surface on planes parallel to the x , y-plane. In other words, look at the curves produced by intersecting the surface with planes of the form z = k for a number of constants k . If we think of the surface as a mountain, then the trace would be the all the points on the mountain at an elevation of k . Such curves are called contour curves or level curves . If these curves are projected down onto the x , y-plane, we get a 2-dimensional representation of the surface. When the TV weather forecaster draws in the isobars on the weather map, it is actually level curves that are being drawn. There is a function of two variables in the background: z = p ( x,y ) gives the pressure at the point ( x,y ) on the earth (which we can pretend is a plane for this example). Isobar means equal pressure , and each isobar represent a line along which the pressure has some fixed value. It is easy to find the equations for the level curves. Simply set z = k , a constant. The result will be an equation in just the two variables x and y which we can graph in the x , y-plane. As an easy example, for z = x 2 + y 2 , when z is set equal to k , the equation for the level curve is seen to be x 2 + y 2 = k . So the level curves are circles centered at the origin in the x , y-plane, at least when k is a positive value. If level curves are drawn for several values of z = k , we can begin the get a feeling for the general shape of the surface. More about the nature of the surface can be deduced from the level curves if they are plotted for a number of equally spaces values of k . If you think about the mountain analogy, you will see that when such level curves are close together, the mountain is very steep, while widely separated levels curves indicate a flat spot on the mountain. Or, back to the weather map analogy, when the level curves are close together, it means that nearby places have a large difference in pressure, and hence it will be pretty windy there as air rushes from the higher pressure area to the lower pressure area. On the other hand, widely separatedair rushes from the higher pressure area to the lower pressure area....
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## This note was uploaded on 02/24/2011 for the course MATH 265 taught by Professor Jerrymetzger during the Winter '11 term at North Dakota.
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265L07 - LESSON 7 Functions of Several Variable and Their...
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# Thread: Inequalities with a modulus
1. ## Inequalities with a modulus
hi, how can i work out the values of x for this inequality please, |x/(x-2)| > |1/(x+1)|
the most confusing thing is that both parts are being divided.
2. Hello cpj
Originally Posted by cpj
hi, how can i work out the values of x for this inequality please, |x/(x-2)| > |1/(x+1)|
the most confusing thing is that both parts are being divided.
This is an awkward, 'bitty' sort of question. You need to look first at the values of $\displaystyle x$ where the sign of each expression changes, and then to consider the functions in the ranges formed by these values.
So the signs change when $\displaystyle x = -1, 0, 2$. We consider, in turn, the four ranges $\displaystyle x<-1,\; -1<x<0,\;0<x<2,\; 2 < x$. I'll do the first two, and leave you to look at the last two.
When $\displaystyle x < -1$
$\displaystyle \frac{x}{x-2}>0\Rightarrow \left|\frac{x}{x-2}\right|=\frac{x}{x-2}$
and
$\displaystyle \frac{1}{x+1}<0\Rightarrow \left|\frac{1}{x+1}\right|=-\frac{1}{x+1}$
So
$\displaystyle \left|\frac{x}{x-2}\right|>\left|\frac{1}{x+1}\right|$
$\displaystyle \Rightarrow \frac{x}{x-2}>-\frac{1}{x+1}$
I'll leave you to check that this can be reduced to
$\displaystyle \frac{x^2+2x-2}{(x-2)(x+1)}>0$
Completing the square
$\displaystyle \Rightarrow \frac{(x+1)^2-3}{(x-2)(x+1)}>0$
Now for $\displaystyle x<-1, (x-2)(x+1) > 0$. So we want values of $\displaystyle x, \;(<-1)$ for which $\displaystyle (x+1)^2 > 3$
$\displaystyle \Rightarrow |x+1| > \sqrt3$
$\displaystyle \Rightarrow -(x+1) >\sqrt3$, since $\displaystyle x+1 < 0$
$\displaystyle \Rightarrow x < -1-\sqrt3$
So there's part of the solution set. (I said it was awkward!)
Now we look at the next range, $\displaystyle -1<x<0$.
In a similar way, we get this time:
$\displaystyle \frac{x}{x-2}>\frac{1}{x+1}$
which can be reduced to
$\displaystyle \frac{x^2+2}{(x-2)(x+1)}>0$
This time the denominator is negative, so we want values of $\displaystyle x$ for which
$\displaystyle x^2+2<0$
and clearly there are none. So there are no values of $\displaystyle x$ in the range $\displaystyle -1<x<0$.
I'll leave you to look at the last two ranges of $\displaystyle x$ in the same way. I reckon the final answer is $\displaystyle \Big(-\infty, (-1-\sqrt3)\Big) \cup \Big((-1+\sqrt3), +\infty\Big)$.
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Degree of a polynomial
(Redirected from Nonic equation)
The degree of a polynomial is the highest degree of its terms when the polynomial is expressed in its canonical form consisting of a linear combination of monomials. The degree of a term is the sum of the exponents of the variables that appear in it. The term order has been used as a synonym of degree but, nowadays, may refer to several other concepts (see order of a polynomial). For example, the polynomial ${\displaystyle 7x^{2}y^{3}+4x-9}$ has three terms. (Notice, this polynomial can also be expressed as ${\displaystyle 7x^{2}y^{3}+4x^{1}y^{0}-9x^{0}y^{0}}$.) The first term has a degree of 5 (the sum of the powers 2 and 3), the second term has a degree of 1, and the last term has a degree of 0. Therefore, the polynomial has a degree of 5 which is the highest degree of any term.
To determine the degree of a polynomial that is not in standard form (for example ${\displaystyle (x+1)^{2}-(x-1)^{2}}$), one has to put it first in standard form by expanding the products (by distributivity) and combining the like terms; for example ${\displaystyle (x+1)^{2}-(x-1)^{2}=4x}$ is of degree 1, even though each summand has degree 2. However, this is not needed when the polynomial is expressed as a product of polynomials in standard form, because the degree of a product is the sum of the degrees of the factors.
Names of polynomials by degree
The following names are assigned to polynomials according to their degree:[1][2][3]
For higher degrees, names have sometimes been proposed,[5] but they are rarely used:
• Degree 8 – octic
• Degree 9 – nonic
• Degree 10 – decic
Names for degree above three are based on Latin ordinal numbers, and end in -ic. This should be distinguished from the names used for the number of variables, the arity, which are based on Latin distributive numbers, and end in -ary. For example, a degree two polynomial in two variables, such as ${\displaystyle x^{2}+xy+y^{2}}$, is called a "binary quadratic": binary due to two variables, quadratic due to degree two.[a] There are also names for the number of terms, which are also based on Latin distributive numbers, ending in -nomial; the common ones are monomial, binomial, and (less commonly) trinomial; thus ${\displaystyle x^{2}+y^{2}}$ is a "binary quadratic binomial".
Other examples
• The polynomial ${\displaystyle 3-5x+2x^{5}-7x^{9}}$ is a nonic polynomial
• The polynomial ${\displaystyle (y-3)(2y+6)(-4y-21)}$ is a cubic polynomial
• The polynomial ${\displaystyle (3z^{8}+z^{5}-4z^{2}+6)+(-3z^{8}+8z^{4}+2z^{3}+14z)}$ is a quintic polynomial (as the ${\displaystyle z^{8}}$ are cancelled out)
The canonical forms of the three examples above are:
• for ${\displaystyle 3-5x+2x^{5}-7x^{9}}$, after reordering, ${\displaystyle -7x^{9}+2x^{5}-5x+3}$;
• for ${\displaystyle (y-3)(2y+6)(-4y-21)}$, after multiplying out and collecting terms of the same degree, ${\displaystyle -8y^{3}-42y^{2}+72y+378}$;
• for ${\displaystyle (3z^{8}+z^{5}-4z^{2}+6)+(-3z^{8}+8z^{4}+2z^{3}+14z)}$, in which the two terms of degree 8 cancel, ${\displaystyle z^{5}+8z^{4}+2z^{3}-4z^{2}+14z+6}$.
Behavior under polynomial operations
The degree of the sum, the product or the composition of two polynomials is strongly related to the degree of the input polynomials.[6]
The degree of the sum (or difference) of two polynomials is less than or equal to the greater of their degrees; the equality always holds when the degrees of the polynomials are different i.e.
${\displaystyle \deg(P+Q)\leq \max(\deg(P),\deg(Q))}$.
${\displaystyle \deg(P-Q)\leq \max(\deg(P),\deg(Q))}$.
E.g.
• The degree of ${\displaystyle (x^{3}+x)+(x^{2}+1)=x^{3}+x^{2}+x+1}$ is 3. Note that 3 ≤ max(3, 2)
• The degree of ${\displaystyle (x^{3}+x)-(x^{3}+x^{2})=-x^{2}+x}$ is 2. Note that 2 ≤ max(3, 3)
Behaviour under scalar multiplication
The degree of the product of a polynomial by a non-zero scalar is equal to the degree of the polynomial, i.e.
${\displaystyle \deg(cP)=\deg(P)}$.
E.g.
• The degree of ${\displaystyle 2(x^{2}+3x-2)=2x^{2}+6x-4}$ is 2, just as the degree of ${\displaystyle x^{2}+3x-2}$.
Note that for polynomials over a ring containing divisors of zero, this is not necessarily true. For example, in ${\displaystyle \mathbf {Z} /4\mathbf {Z} }$, ${\displaystyle \deg(1+2x)=1}$, but ${\displaystyle \deg(2(1+2x))=\deg(2+4x)=\deg(2)=0}$.
The set of polynomials with coefficients from a given field F and degree smaller than or equal to a given number n thus forms a vector space. (Note, however, that this set is not a ring, as it is not closed under multiplication, as is seen below.)
Behaviour under multiplication
The degree of the product of two polynomials over a field or an integral domain is the sum of their degrees
${\displaystyle \deg(PQ)=\deg(P)+\deg(Q)}$.
E.g.
• The degree of ${\displaystyle (x^{3}+x)(x^{2}+1)=x^{5}+2x^{3}+x}$ is 3 + 2 = 5.
Note that for polynomials over an arbitrary ring, this is not necessarily true. For example, in ${\displaystyle \mathbf {Z} /4\mathbf {Z} }$, ${\displaystyle \deg(2x)+\deg(1+2x)=1+1=2}$, but ${\displaystyle \deg(2x(1+2x))=\deg(2x)=1}$.
Behaviour under composition
The degree of the composition of two non-constant polynomials ${\displaystyle P}$ and ${\displaystyle Q}$ over a field or integral domain is the product of their degrees:
${\displaystyle \deg(P\circ Q)=\deg(P)\deg(Q)}$.
E.g.
• If ${\displaystyle P=(x^{3}+x)}$, ${\displaystyle Q=(x^{2}+1)}$, then ${\displaystyle P\circ Q=P\circ (x^{2}+1)=(x^{2}+1)^{3}+(x^{2}+1)=x^{6}+3x^{4}+4x^{2}+2}$, which has degree 6.
Note that for polynomials over an arbitrary ring, this is not necessarily true. For example, in ${\displaystyle \mathbf {Z} /4\mathbf {Z} }$, ${\displaystyle \deg(2x)\deg(1+2x)=1\cdot 1=1}$, but ${\displaystyle \deg(2x\circ (1+2x))=\deg(2+4x)=\deg(2)=0}$.
Degree of the zero polynomial
The degree of the zero polynomial is either left undefined, or is defined to be negative (usually −1 or −∞).[7]
Like any constant value, the value 0 can be considered as a (constant) polynomial, called the zero polynomial. It has no nonzero terms, and so, strictly speaking, it has no degree either. As such, its degree is undefined. The propositions for the degree of sums and products of polynomials in the above section do not apply if any of the polynomials involved is the zero polynomial.[8]
It is convenient, however, to define the degree of the zero polynomial to be negative infinity, −∞, and introduce the arithmetic rules[9]
${\displaystyle \max(a,-\infty )=a,}$
and
${\displaystyle a+-\infty =-\infty .}$
These examples illustrate how this extension satisfies the behavior rules above:
• The degree of the sum ${\displaystyle (x^{3}+x)+(0)=x^{3}+x}$ is 3. This satisfies the expected behavior, which is that ${\displaystyle 3\leq \max(3,-\infty )}$.
• The degree of the difference ${\displaystyle (x)-(x)=0}$ is ${\displaystyle -\infty }$. This satisfies the expected behavior, which is that ${\displaystyle -\infty \leq \max(1,1)}$.
• The degree of the product ${\displaystyle (0)(x^{2}+1)=0}$ is ${\displaystyle -\infty }$. This satisfies the expected behavior, which is that ${\displaystyle -\infty =-\infty +2}$.
Computed from the function values
A number of formulae exist which will evaluate the degree of a polynomial function f. One based on asymptotic analysis is
${\displaystyle \deg f=\lim _{x\rightarrow \infty }{\frac {\log |f(x)|}{\log x}}}$;
this is the exact counterpart of the method of estimating the slope in a log–log plot.
This formula generalizes the concept of degree to some functions that are not polynomials. For example:
• The degree of the multiplicative inverse, ${\displaystyle \ 1/x}$, is −1.
• The degree of the square root, ${\displaystyle {\sqrt {x}}}$, is 1/2.
• The degree of the logarithm, ${\displaystyle \ \log x}$, is 0.
• The degree of the exponential function, ${\displaystyle \ \exp x}$, is ∞.
Note that the formula also gives sensible results for many combinations of such functions, e.g., the degree of ${\displaystyle {\frac {1+{\sqrt {x}}}{x}}}$ is ${\displaystyle -1/2}$.
Another formula to compute the degree of f from its values is
${\displaystyle \deg f=\lim _{x\to \infty }{\frac {xf'(x)}{f(x)}}}$;
this second formula follows from applying L'Hôpital's rule to the first formula. Intuitively though, it is more about exhibiting the degree d as the extra constant factor in the derivative ${\displaystyle dx^{d-1}}$ of ${\displaystyle x^{d}}$.
A more fine grained (than a simple numeric degree) description of the asymptotics of a function can be had by using big O notation. In the analysis of algorithms, it is for example often relevant to distinguish between the growth rates of ${\displaystyle x}$ and ${\displaystyle x\log x}$, which would both come out as having the same degree according to the above formulae.
Extension to polynomials with two or more variables
For polynomials in two or more variables, the degree of a term is the sum of the exponents of the variables in the term; the degree (sometimes called the total degree) of the polynomial is again the maximum of the degrees of all terms in the polynomial. For example, the polynomial x2y2 + 3x3 + 4y has degree 4, the same degree as the term x2y2.
However, a polynomial in variables x and y, is a polynomial in x with coefficients which are polynomials in y, and also a polynomial in y with coefficients which are polynomials in x.
x2y2 + 3x3 + 4y = (3)x3 + (y2)x2 + (4y) = (x2)y2 + (4)y + (3x3)
This polynomial has degree 3 in x and degree 2 in y.
Degree function in abstract algebra
Given a ring R, the polynomial ring R[x] is the set of all polynomials in x that have coefficients chosen from R. In the special case that R is also a field, then the polynomial ring R[x] is a principal ideal domain and, more importantly to our discussion here, a Euclidean domain.
It can be shown that the degree of a polynomial over a field satisfies all of the requirements of the norm function in the euclidean domain. That is, given two polynomials f(x) and g(x), the degree of the product f(x)g(x) must be larger than both the degrees of f and g individually. In fact, something stronger holds:
deg(f(x)g(x)) = deg(f(x)) + deg(g(x))
For an example of why the degree function may fail over a ring that is not a field, take the following example. Let R = ${\displaystyle \mathbb {Z} /4\mathbb {Z} }$, the ring of integers modulo 4. This ring is not a field (and is not even an integral domain) because 2 × 2 = 4 ≡ 0 (mod 4). Therefore, let f(x) = g(x) = 2x + 1. Then, f(x)g(x) = 4x2 + 4x + 1 = 1. Thus deg(fg) = 0 which is not greater than the degrees of f and g (which each had degree 1).
Since the norm function is not defined for the zero element of the ring, we consider the degree of the polynomial f(x) = 0 to also be undefined so that it follows the rules of a norm in a euclidean domain.
Notes
1. ^ For simplicity, this is a homogeneous polynomial, with equal degree in both variables separately.
1. ^ "Names of Polynomials". November 25, 1997. Retrieved 5 February 2012.
2. ^ Mac Lane and Birkhoff (1999) define "linear", "quadratic", "cubic", "quartic", and "quintic". (p. 107)
3. ^ King (2009) defines "quadratic", "cubic", "quartic", "quintic", "sextic", "septic", and "octic".
4. ^ Shafarevich (2003) says of a polynomial of degree zero, ${\displaystyle f(x)=a_{0}}$: "Such a polynomial is called a constant because if we substitute different values of x in it, we always obtain the same value ${\displaystyle a_{0}}$." (p. 23)
5. ^ James Cockle proposed the names "sexic", "septic", "octic", "nonic", and "decic" in 1851. (Mechanics Magazine, Vol. LV, p. 171)
6. ^ Lang, Sergei (2005). Algebra (3rd ed.). Springer. p. 100. ISBN 978-0-387-95385-4.
7. ^ Shafarevich (2003) says of the zero polynomial: "In this case, we consider that the degree of the polynomial is undefined." (p. 27)
Childs (1995) uses −1. (p. 233)
Childs (2009) uses −∞ (p. 287), however he excludes zero polynomials in his Proposition 1 (p. 288) and then explains that the proposition holds for zero polynomials "with the reasonable assumption that ${\displaystyle -\infty }$ + m = ${\displaystyle -\infty }$ for m any integer or m = ${\displaystyle -\infty }$".
Axler (1997) uses −∞. (p. 64)
Grillet (2007) says: "The degree of the zero polynomial 0 is sometimes left undefined or is variously defined as −1 ∈ ℤ or as ${\displaystyle -\infty }$, as long as deg 0 < deg A for all A ≠ 0." (A is a polynomial.) However, he excludes zero polynomials in his Proposition 5.3. (p. 121)
8. ^ Barile, Margherita. "Zero Polynomial". MathWorld.
9. ^ Axler (1997) gives these rules and says: "The 0 polynomial is declared to have degree ${\displaystyle -\infty }$ so that exceptions are not needed for various reasonable results." (p. 64)
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# Math in Focus Grade 4 Chapter 2 Practice 2 Answer Key Factors
Go through the Math in Focus Grade 4 Workbook Answer Key Chapter 2 Practice 2 Factors to finish your assignments.
## Math in Focus Grade 4 Chapter 2 Practice 2 Answer Key Factors
Find the missing factors.
Example 12 1 × 12 = 12
2 × 6 = 12
3 × 4 = 12
The factors of 12 are
1, 2, 3, 4, 6, and 12.
Question 1.
70
1 × ____ = 70
2 × ____ = 70
3 × ____ = 70
4 × ____ = 70
The factors of 70 are 1, 2, 5, 7, ___, ____
___ and ___
Answer:
Factor definition:
When a number is said to be a factor of any other second number, then the first number must divide the second number completely without leaving any remainder. In simple words, if a number (dividend) is exactly divisible by any number (divisor), then the divisor is a factor of that dividend. Every number has a common factor that is one and the number itself.
1 × 70= 70
2 × 35= 70
3 × 23.3= 70
4 × 17.5= 70
Point to remember: Fractions cannot be considered as factors for any number.
The numbers that divide 70 exactly without leaving a remainder are the factors of 70. As 70 is an even composite number, it has many factors other than 1 and 70. Hence, the factors of 70 are 1, 2, 5, 7, 10, 14, 35, and 70. Similarly, the negative factors of 70 are -1, -2, -5, -7, -10, -14, -35, and -70.
1 × 70= 70
2 × 35= 70
5 × 14=70
7 × 10=70
Factors of 70: 1, 2, 5, 7, 10, 14, 35, and 70.
Find the factors of each number.
Question 2.
40
The factors of 40 are
_________
Answer: 1, 2, 4, 5, 8, 10, 20, and 40.
Explanation:
The Factors of 40 are all the integers (positive and negative whole numbers) that you can evenly divide into 40. 40 divided by a factor of 40 will equal another Factor of 40.
How to find the factors of 40:
Since the Factors of 40 are all the numbers that you can evenly divide into 40, we simply need to divide 40 by all numbers up to 40 to see which ones result in an even quotient. When we did that, we found that these calculations resulted in an even quotient:
40 ÷ 1 = 40
40 ÷ 2 = 20
40 ÷ 4 = 10
40 ÷ 5 = 8
40 ÷ 8 = 5
40 ÷ 10 = 4
40 ÷ 20 = 2
40 ÷ 40 = 1
The Positive Factors of 40 are therefore all the numbers we used to divide (divisors) above to get an even number. Here is the list of all Positive Factors of 40 in numerical order:
1, 2, 4, 5, 8, 10, 20, and 40.
Question 3.
63
The factors of 63 are
___________
Answer:
Explanation:
The Factors of 63 are all the integers (positive and negative whole numbers) that you can evenly divide into 63. 63 divided by a factor of 63 will equal another Factor of 63.
How to find the factors of 63:
Since the Factors of 63 are all the numbers that you can evenly divide into 63, we simply need to divide 63 by all numbers up to 63 to see which ones result in an even quotient. When we did that, we found that these calculations resulted in an even quotient:
63 ÷ 1 = 63
63 ÷ 3 = 21
63 ÷ 7 = 9
63 ÷ 9 = 7
63 ÷ 21 = 3
63 ÷ 63 = 1
The Positive Factors of 63 are therefore all the numbers we used to divide (divisors) above to get an even number. Here is the list of all Positive Factors of 63 in numerical order:
1, 3, 7, 9, 21, and 63.
Divide. Then answer each question.
Question 4.
65 ÷ 5 = ____
Answer:13
There are four important terms used in division. These are dividend, divisor, quotient and remainder.
– Dividend: The number to be divided by another number is called the dividend.
– Divisor: The number by which we divide another number (dividend) into equal parts is called the divisor.
– Quotient: The result of division is called a quotient.
– Reminder: The leftover number after division is called the remainder.
– The pictorial representation of the above terminology is given below.
Division formula:
Dividend=Divisor×Quotient+Remainder.
After division, we can put all the values in the formula to verify or check whether our division is correct or not.
the above-given question is 65 ÷ 5 = 13
Here, dividend =65, Divisor =5, quotient =13 and remainder =0
Let us put all the above values in the formula,
Dividend=Divisor×Quotient+Remainder
⇒65 = 5 × 13 + 0
⇒ 65 ÷ 5 = 13
Hence, our division is correct.
Question 5.
46 ÷ 4 = ___
Is 5 a factor of 65?
Answer:
Explanation:
There are four important terms used in division. These are dividend, divisor, quotient and remainder.
– Dividend: The number to be divided by another number is called the dividend.
– Divisor: The number by which we divide another number (dividend) into equal parts is called the divisor.
– Quotient: The result of division is called a quotient.
– Reminder: The leftover number after division is called the remainder.
– The pictorial representation of the above terminology is given below.
Division formula:
Dividend=Divisor×Quotient+Remainder.
After division, we can put all the values in the formula to verify or check whether our division is correct or not.
the above-given question is 46 ÷ 4 = 11.5
Here, dividend =46, Divisor =4, quotient =11.5 and remainder =0
Let us put all the above values in the formula,
Dividend=Divisor×Quotient+Remainder
⇒46 = 4 × 11.5 + 0
⇒ 65 ÷ 5 = 11.5
Hence, our division is correct.
Another question is there, is 5 is a factor of 65:
Yes, 5 is a factor of 65.
– The 4 factors of 65 are:
1, 5, 13, 65
– The factor pairs of 65 are:
1 × 65 = 65
5 × 13 = 65
Question 5.
46 ÷ 4 = ___
Is 4 a factor of 46? ____
Answer: 11.5
Explanation:
There are four important terms used in division. These are dividend, divisor, quotient and remainder.
– Dividend: The number to be divided by another number is called the dividend.
– Divisor: The number by which we divide another number (dividend) into equal parts is called the divisor.
– Quotient: The result of division is called a quotient.
– Reminder: The leftover number after division is called the remainder.
– The pictorial representation of the above terminology is given below.
Another question is there is 4 a factor of 46:
No 4 is not a factor of 46
Therefore, the 4 factors of 46 are:
1, 2, 23, 46.
– The factor pairs of 46 are:
1 × 46 = 46
2 × 23 = 46
Find the common factors of each pair of numbers.
Question 6.
Answer:
Explanation:
– To find common factors of two numbers, first, list out all the factors of two numbers separately and then compare them. Now write the factors which are common to both the numbers. These factors are called common factors of given two numbers.
– As we know, the factors are the numbers that divide the original number completely. But how to check if two or more numbers have common factors between them.
Follow the below steps to find the common factors.
* Write the factors of the given numbers.
* Find the common factor present in them.
Let us check the factors of the two numbers, i.e., 10 and 15.
Factors of 10 = 1, 2, 5, 10
Factors of 15 = 1, 3, 5, 15
Clearly, we can see, the common factors between 10 and 15 are 1, 5.
Question 7.
Answer:
Explanation:
– To find common factors of two numbers, first, list out all the factors of two numbers separately and then compare them. Now write the factors which are common to both the numbers. These factors are called common factors of given two numbers.
– As we know, the factors are the numbers that divide the original number completely. But how to check if two or more numbers have common factors between them.
Follow the below steps to find the common factors.
* Write the factors of the given numbers.
* Find the common factor present in them.
Let us check the factors of the two numbers, i.e., 24 and 36.
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36
Clearly, we can see, the common factors between 24 and 36 are 1, 2, 3, 4, 6, 12.
Divide. Then answer each question.
Question 8.
18 ÷ 4 = ___ 16 ÷ 4 = ____
Is 4 a common factor of 18 and 16? ____
Answer:
18 ÷ 4 = 4.5
16 ÷ 4 = 4
Explanation:
There are four important terms used in division. These are dividend, divisor, quotient and remainder.
– Dividend: The number to be divided by another number is called the dividend.
– Divisor: The number by which we divide another number (dividend) into equal parts is called the divisor.
– Quotient: The result of division is called a quotient.
– Reminder: The leftover number after division is called the remainder.
– The pictorial representation of the above terminology is given below.
Another question is there is 4 a common factor of 18 and 16:
No 4 is not a factor of 46
Therefore, the factors of 18 and 16 are:
factors of 18: 1, 2, 3, 6, 9, 18
factors of 16: 1, 2, 4, 8, 16.
The common factors are 1, 2
Question 9.
42 ÷ 3 = ____ 84 ÷ 3 = ____
Is 3 a common factor of 42 and 84? ____
Answer:
42 ÷ 3 = 14
84 ÷ 3 = 28
Yes, 3 is a common factor for both the numbers.
Explanation:
There are four important terms used in division. These are dividend, divisor, quotient and remainder.
– Dividend: The number to be divided by another number is called the dividend.
– Divisor: The number by which we divide another number (dividend) into equal parts is called the divisor.
– Quotient: The result of division is called a quotient.
– Reminder: The leftover number after division is called the remainder.
– The pictorial representation of the above terminology is given below.
Another question is there is 3 a common factor of 42 and 84:
Therefore, the factors of 42 and 84 are:
factors of 42: 1, 2, 3, 6, 7, 14, 21, 42
factors of 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
The common factors are 1, 2, 3, 6, 7, 14, 21, 42
Therefore, 3 is a common factor for 42 and 84.
Look at the numbers 80, 27, 40, 62, 36, and 55. Then fill in the blanks.
Question 10.
Which of the numbers have 2 as a factor? _________
Answer:80, 40, 62, 36
The above-given numbers are 80, 27, 40, 62, 36, and 55.
we have to find out the numbers having 2 as a factor.
80, 40, 62, 36 are the factors of 2.
80/40 = 2
40/20 = 2
62/32 = 2
36/18 = 2
Factors are the numbers, that can divide a number exactly. Hence, after division, there is no remainder left. Factors are the numbers you multiply together to get another number. Thus, a factor is the divisor of another number.
Question 11.
Which of the numbers have 5 as a factor? ______
Answer: 80, 40, 55
Factors can be calculated by using multiplication and division.
– Since multiplication of two numbers results in a product such that the two numbers become the factors of the product. Thus, to find the factors we need to follow the below steps:
* If we need to find the factors of a number say N, then write the multiplication of two numbers in different ways, such that the resulting value is equal to N
* All the individual numbers, that results in the product equal to N are the factors
5 × 12 = 80
5 × 8 = 40
5 × 11 = 55
Factors are the numbers, that can divide a number exactly. Hence, after division, there is no remainder left. Factors are the numbers you multiply together to get another number. Thus, a factor is the divisor of another number.
Question 12.
Which of the numbers have both 2 and 5 as factors? ______
Answer:
The numbers 80, 40 have 2 and 5
40 × 2 =80
5 × 8 = 40
5 × 16 = 80
Factors are the numbers, that can divide a number exactly. Hence, after division, there is no remainder left. Factors are the numbers you multiply together to get another number. Thus, a factor is the divisor of another number.
Each set of numbers is all the factors of a number. Find each number.
Question 13.
Answer:
Explanation:
– The factors of a number are defined as the numbers which when multiplied will give the original number, by multiplying the two factors we get the result as the original number. The factors can be either positive or negative integers.
– Factors of 8 are all the integers that can evenly divide the given number 8. Now let us study how to calculate all factors of 8.
– According to the definition of factors of 8, we know that factors of 8 are all the positive or negative integers that divide the number 8 completely. So let us simply divide the number 8 by every number which completely divides 8 in ascending order till 8.
8 ÷ 1 = 8
8 ÷ 2 = 4
8 ÷ 3 = not divides completely
8 ÷ 4 = 2
8 ÷ 5 = not divides completely
8 ÷ 6 = not divides completely
8 ÷ 7 = not divides completely
8 ÷ 8 = 1
So all factors of 8: 1, 2, 4, and 8.
Question 14.
Answer:
Explanation:
– When we divide every number with 12 up to that number itself, the numbers that evenly divide with 12 are the factors of 12.
– This is a very simple and straightforward method. The entire process takes only 5 steps.
– To start off here is step number 1:
Step 1: Consider the number 12
Step 2: Divide it with all the numbers starting from 1 to 12
Step 3: Capture the results
: 12/1 = 12
: 12/2 = 6
: 12/3 = 4
: 12/4 = 3
: 12/5 = 2.4
: 12/6 = 2
: 12/7 = 1.7
: 12/8 = 1.5
: 12/9 = 1.3
: 12/10 = 1.2
: 12/11 = 1.09
: 12/12 = 1
Step 4: Filter out the positive integer quotient for the above, rejecting the decimals.
Step 5: The factors of 12 are 1, 2, 3, 4, 6, 12.
Question 15.
Answer:
Explanation:
The factors of 6 are the numbers that divide 6 exactly without leaving the remainder. In other words, the factors of 6 are the numbers that are multiplied in pairs resulting in an original number 6. As 6 is an even composite number, it has many factors other than 1 and 6. Thus, the factors of 6 are 1, 2, 3 and 6. Similarly, the negative pair factors of 6 are -1, -2, -3 and -6.
Factors of 6: 1, 2, 3 and 6.
Factors of 6 by division method:
– In the division method, the factors of 6 can be found by dividing 6 by different integers. If the integers divide 6 exactly and leave the remainder 0, then those integers are the factors of 6. Now, let us discuss how to find the factors of 6 using the division method.
* 6/1 = 6 (Factor is 1 and Remainder is 0)
* 6/2 = 3 (Factor is 2 and Remainder is 0)
* 6/3 = 2 (Factor is 3 and Remainder is 0)
* 6/6 = 1 (Factor is 6 and Remainder is 0)
– If we divide the number 6 by any numbers other than 1, 2, 3 and 6, then it leaves the remainder. Hence, the factors of 6 are 1, 2, 3 and 6.
Question 16.
Answer:
Explanation:
The factors of 16 are the numbers that divide the number 16 completely without leaving any remainder. As the number 16 is a composite number, it has more than one factor. The factors of 16 are 1, 2, 4, 8 and 16. Similarly, the negative factors of 16 are -1, -2, -4, -8 and -16.
– Factors of 16: 1, 2, 4, 8 and 16.
How to calculate factors of 16:
Go through the following steps to calculate the factors of 16.
* First, write the number 16.
* Find the two numbers, which gives the result as 16 under the multiplication, say 2 and 8, such as 2 × 8 = 16.
* We know that 2 is a prime number that has only two factors, i.e., 1 and the number itself (1 and 2). So, it cannot be further factorized.
* Look at the number 8, which is a composite number but not a prime number. So it can be further factorized.
* 8 can be factored as 2 x 2 x 2 x 1.
* Therefore, the factorization of 16 is written as, 16 = 2 × 2 × 2 × 2 x 1.
Find the greatest common factor of each pair of numbers.
Example
12 and 28
Method 1
The factors of 12 are 1, 2, 3, 4, 6, and 12.
The factors of 28 are 1, 2, 4, 7, 14, and 28.
The common factors of 12 and 28 are 1, 2, and 4.
The greatest common factor of 12 and 28 is 4.
Method 2
2 × 2 = 4
The greatest common factor of 12 and 28 is 4.
3 and 7 have no common factor other than 1.
Question 17.
16 and 30
_________
Answer: GCF of 16 and 30 is 2.
Explanation:
GCF of 16 and 30 is the largest possible number that divides 16 and 30 exactly without any remainder. The factors of 16 and 30 are 1, 2, 4, 8, 16 and 1, 2, 3, 5, 6, 10, 15, 30 respectively. There are 3 commonly used methods to find the GCF of 16 and 30 – long division, prime factorization, and Euclidean algorithm.
– The GCF of two non-zero integers, x(16) and y(30) is the greatest positive integer m(2) that divides both x(16) and y(30) without any remainder.
Methods to find GCF of 16 and 30:
– The methods to find the GCF of 16 and 30 are explained below.
* Long Division Method
* Using Euclid’s Algorithm
* Listing Common Factors
# GCF of 16 and 30 by long division method:
GCF of 16 and 30 is the divisor that we get when the remainder becomes 0 after doing long division repeatedly.
Step 1: Divide 30 (larger number) by 16 (smaller number).
Step 2: Since the remainder ≠ 0, we will divide the divisor of step 1 (16) by the remainder (14).
Step 3: Repeat this process until the remainder = 0.
The corresponding divisor (2) is the GCF of 16 and 30.
# GCF of 16 and 30 by listing common factors:
Factors of 16: 1, 2, 4, 8, 16
Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30.
There are 2 common factors of 16 and 30, which are 1 and 2. Therefore, the greatest common factor of 16 and 30 is 2.
Find the greatest common factor of the numbers.
Question 18.
21 and 54
___________
Answer: 3
The Greatest Common Factor (GCF) for 21 and 54, notation GCF(21,54), is 3.
Explanation:
To find the GCF of 21 and 54, we first need to find the factors of 21 and 54 and then choose the greatest common factor that is dividable by both 21 and 54.
– The factors of 21 are 1,3,7,21;
– The factors of 54 are 1,2,3,6,9,18,27,54.
So, as we can see, the Greatest Common Factor or Divisor is 3, because it is the greatest number that divides evenly into all of them.
Find all the factors. Then list the prime numbers.
Example
13
The factor of 13 are 1 and 13. 13 is a prime number.
A prime number has only 2 factors, 1 and itself.
Question 19.
12 ____
Answer:
The factors of 12 are the numbers that divide 12 exactly without leaving any remainder. As 12 is an even composite number, it has many factors other than 1 and 12. The factors of 12 can be positive or negative. Hence, the factors of 12 are 1, 2, 3, 4, 6 and 12. Similarly, the negative factors of 12 are -1, -2, -3, -4, -6 and -12.
Factors of 12: 1, 2, 3, 4, 6 and 12.
Now we need to write the prime numbers in the factors of 12.
prime numbers: A prime number is a positive integer having exactly two factors. If p is a prime, then its only factors are necessarily 1 and p itself. Any number that does not follow this is termed a composite number, which can be factored into other positive integers. Another way of defining it is a positive number or integer, which is not a product of any other two positive integers other than 1 and the number itself.
Therefore, the prime numbers are 2 and 3.
Question 20.
7 ______
Answer:
The factors of 20 are the numbers that divide 20 exactly without leaving a remainder. In other words, the numbers are multiplied in pairs resulting in the number 20 being the factor of 20. As 20 is an even composite number, it has many factors other than 1 and 20. Thus, the factors of 20 are 1, 2, 4, 5, 10 and 20. Similarly, the negative factors of 20 are -1, -2, -4, -5, -10 and -20.
Factors of 20: 1, 2, 4, 5, 10 and 20.
Now we need to write the prime numbers in the factors of 20.
prime numbers: A prime number is a positive integer having exactly two factors. If p is a prime, then its only factors are necessarily 1 and p itself. Any number that does not follow this is termed a composite number, which can be factored into other positive integers. Another way of defining it is a positive number or integer, which is not a product of any other two positive integers other than 1 and the number itself.
Therefore, the prime numbers are 2 and 5.
Question 21.
19 ____
Answer:
The numbers that divide 19 completely without leaving a remainder are the factors of 19. In other words, the numbers which are multiplied in pairs, resulting in an original number, are the factors of 19. As number 19 is a prime number, it has only two factors: one and the number itself. Hence, the factors of 19 are 1 and 19, and its negative factors are -1 and -19.
Factors of 19: 1 and 19.
Now we need to write the prime numbers in the factors of 19.
prime numbers: A prime number is a positive integer having exactly two factors. If p is a prime, then its only factors are necessarily 1 and p itself. Any number that does not follow this is termed a composite number, which can be factored into other positive integers. Another way of defining it is a positive number or integer, which is not a product of any other two positive integers other than 1 and the number itself.
Therefore, the prime number is 19.
Question 22.
24 ______
Answer:
By the definition of factors, we know, a factor can divide a given number into an equal number of parts. Therefore, factors of 24 are such whole numbers that can divide 24 into an equal number of parts. These factors cannot be a fraction.
Factors of 24: 1, 2, 3, 4, 6, 8, 12 and 24.
Now we need to write the prime numbers in the factors of 24.
prime numbers: A prime number is a positive integer having exactly two factors. If p is a prime, then its only factors are necessarily 1 and p itself. Any number that does not follow this is termed a composite number, which can be factored into other positive integers. Another way of defining it is a positive number or integer, which is not a product of any other two positive integers other than 1 and the number itself.
Therefore, the prime numbers are 2, 3.
Question 23.
11 ____
Answer:
In Mathematics, the numbers that divide 11 completely without leaving any remainder are the factors of 11. In other words, the factors of 11 are the numbers that are multiplied in pairs and result in the original number 11. As 11 is a prime number, it has only two factors, such as one and the number itself. Hence, the factors of 11 are 1 and 11, and the negative factors of 11 are -1 and -11.
Factors of 11: 1 and 11.
Now we need to write the prime numbers in the factors of 11.
prime numbers: A prime number is a positive integer having exactly two factors. If p is a prime, then its only factors are necessarily 1 and p itself. Any number that does not follow this is termed a composite number, which can be factored into other positive integers. Another way of defining it is a positive number or integer, which is not a product of any other two positive integers other than 1 and the number itself.
Therefore, the prime number is 11.
Question 24.
63 ______
Answer:
The factors of 63 are the numbers that divide the number 63 exactly without leaving any remainder. In other words, the pair factors of 63 are the numbers that are multiplied in pairs resulting in the original number 63. Since the number 63 is a composite number, 63 has more than two factors. Thus, the factors of 63 are 1, 3, 7, 9, 21 and 63.
Now we need to write the prime numbers in the factors of 63.
prime numbers: A prime number is a positive integer having exactly two factors. If p is a prime, then its only factors are necessarily 1 and p itself. Any number that does not follow this is termed a composite number, which can be factored into other positive integers. Another way of defining it is a positive number or integer, which is not a product of any other two positive integers other than 1 and the number itself.
Therefore, the prime numbers are 3, 7.
Question 25.
Look at the given numbers in Exercises 19-24.
The prime numbers are _____
Explain your reasoning. _________
Answer:
The above-given numbers are 12, 7, 19, 24, 11, 63.
To that we need to write the prime numbers:
Definition: A prime number is a positive integer having exactly two factors. If p is a prime, then its only factors are necessarily 1 and p itself. Any number that does not follow this is termed a composite number, which can be factored into other positive integers. Another way of defining it is a positive number or integer, which is not a product of any other two positive integers other than 1 and the number itself.
Properties of prime numbers:
Some of the properties of prime numbers are listed below:
– Every number greater than 1 can be divided by at least one prime number.
– Every even positive integer greater than 2 can be expressed as the sum of two primes.
– Except for 2, all other prime numbers are odd. In other words, we can say that 2 is the only even prime number.
– Two prime numbers are always coprime to each other.
– Each composite number can be factored into prime factors and individually all of these are unique in nature.
Now from the above definition and properties, we can write the prime numbers:
The prime numbers are:7, 11, 19
Find all the factors. Then list the composite numbers.
Example
18
The factors of 18 are 1, 2, 3, 6,
9 and 18
18 is a composite number.
18 has factors other than 1 and itself, so it is a composite number.
Question 26.
20 _______________
Answer:
The factors of 20 are the numbers that divide 20 exactly without leaving a remainder. In other words, the numbers are multiplied in pairs resulting in the number 20 being the factor of 20. As 20 is an even composite number, it has many factors other than 1 and 20. Thus, the factors of 20 are 1, 2, 4, 5, 10 and 20. Similarly, the negative factors of 20 are -1, -2, -4, -5, -10 and -20.
Now we need to write the composite numbers:
Definition: In Mathematics, composite numbers are numbers that have more than two factors.
therefore, the composite numbers are 4, 10, and 20.
Hence, these numbers are having more than two factors.
Question 27.
15 ______
Answer:
Factors of 15 divide the original number, wholly. A number or an integer that divides 15 evenly without leaving a remainder, then the number is a factor of 15. As the number 15 is an odd composite number, it has more than two factors. Thus, the factors of 15 are 1, 3, 5 and 15.
Now we need to write the composite numbers:
Definition: In Mathematics, composite numbers are numbers that have more than two factors.
therefore, 15 is the composite number.
Hence, this number is having more than two factors.
Question 28.
5 _______________
Answer:
Factors of 5 are the real numbers that can divide the original number, uniformly. If ‘x’ is the factor of 5, then ‘x’ divides 5 into equal parts and there is no remainder left.
The factors of 5 are 1 and 5.
Now we need to write the composite numbers:
Definition: In Mathematics, composite numbers are numbers that have more than two factors.
therefore, there is no composite number.
Question 29.
17 _____
Answer:
The numbers that divide 17 completely and leave the remainder 0, then the numbers are the factors of 17. In other words, if two numbers are multiplied together and result in 17, then the numbers are the factors of 17. As 17 is a prime number, it has only two factors, such as 1 and the number itself. Hence, the factors of 17 are 1 and 17.
Now we need to write the composite numbers:
Definition: In Mathematics, composite numbers are numbers that have more than two factors.
therefore, there is no composite number.
Question 30.
33 _________________
Answer:
Go through the steps given below to learn how to find the factors of 33.
Step 1: First, write the number 33 in your notebook.
Step 2: Find the two numbers, which on multiplication gives 33, suppose 3 and 11, such as 3 × 11 = 33.
Step 3: We know that 3 is a prime number that has only two factors, i.e. 1 and the number itself. So, we cannot factorize it further.
3 = 1 × 3
Step 4:Also, 11 is a prime number and cannot be factored further.
11 = 1 × 11
Step 5: Therefore, the factorization of 33 gives us 1 × 3 × 11.
The unique numbers which are obtained from the above expression are 1, 3, 11, and 33.
Now we need to write the composite numbers:
Definition: In Mathematics, composite numbers are numbers that have more than two factors.
therefore, 33 is the composite number.
Hence, this number is having more than two factors.
Question 31.
27 _____
Answer:
The steps to find the factors are:
1. First, write the number 27 in your notebook.
2. Find the two numbers, which on multiplication gives 27, say 3 and 9, such as 3 × 9 = 27.
3. We know 3 is a prime number that has only two factors, i.e. 1 and the number itself. So, we cannot further factorize it. (i.e.) 3 = 1 × 3
4. But 9 is not a prime number and can be further factorized. (i.e.) 9 = 3 × 3 × 1
5. Therefore, the factorization of 27 gives us, 1 × 3 × 3 × 3.
6. Write down all the unique numbers which are obtained here.
7. Factors of 27: 1, 3, 9 and 27.
Now we need to write the composite numbers:
Definition: In Mathematics, composite numbers are numbers that have more than two factors.
therefore, 9 and 27 are the composite number.
Hence, these numbers are having more than two factors.
Question 32.
Look at the given numbers in Exercises 26-31.
The composite numbers are ____
Explain your reasoning. _____
Answer:
The above-given numbers are 20, 15, 5, 17, 33, 27
To that we need to write the composite numbers:
Def 1: In Mathematics, composite numbers are numbers that have more than two factors.
Def 2: The numbers which can be generated by multiplying the two smallest positive integers and contain at least one divisor other than the number ‘1’ and itself are known as composite numbers. These numbers always have more than two factors.
properties of composite numbers:
The properties of composite numbers are easy to remember.
– Composite numbers have more than two factors
– Composite numbers are evenly divisible by their factors
– Each composite number is a factor in itself
– The smallest composite number is 4
– Each composite number will include at least two prime numbers as its factors (Eg. 10 = 2 x 5, where 2 and 5 are prime numbers)
– Composite numbers are divisible by other composite numbers also
From the above definition and properties we can write the composite numbers:
Therefore, the composite numbers are 20, 15, 27, 33
Use the method given below to find prime numbers.
Question 33.
Find the prime numbers between 1 and 50.
Step 1
1 is neither prime nor composite. So, 1 has been circled. As 2 is the first prime number, it has been underlined. Next, cross out all the numbers that can be divided by 2.
Step 2
3 is the next prime number. Underline it.
Then, cross out all the numbers that can be divided by 3.
Keep underlining the prime numbers and crossing out the numbers that can be divided by the prime numbers until you reach 50.
The prime numbers are ____
Answer:
A prime number is a positive integer having exactly two factors. If p is a prime, then its only factors are necessarily 1 and p itself. Any number that does not follow this is termed a composite number, which can be factored into other positive integers. Another way of defining it is a positive number or integer, which is not a product of any other two positive integers other than 1 and the number itself.
The prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
Question 34.
Find two prime numbers between 60 and 90. ____
Answer:
First of all the prime numbers between 60 and 90 are 61, 67, 71, 73, 79, 83, 89.
Question 35.
Find two composite numbers between 60 and 90. ____
Answer:
The composite numbers between 0 and 90 are: 60, 62, 63, 64, 65, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90.
The two composite numbers: 70 and 80
Two composite numbers between 60 and 90 are 70 and 80. All numbers which end in a zero are evenly divisible by 10.
Question 36.
Are there more prime numbers from 1 to 25 or from 26 to 50?
Answer:
YES. There are 9 prime numbers from 1 to 25. There are only 6 prime numbers from 26 to 50.
The prime numbers between 1 to 25 are: 2, 3, 5, 7, 11, 13, 17, 19, 23.
The prime numbers between 26 to 50: 29, 31, 37, 41, 43, 47, 53, 59.
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Solving Two Variable Inequalities Part 2
Objective
SWBAT solve two variable inequalities by constructing an appropriate boundary line and shading a half plane.
Big Idea
See how to use test points to determine how to determine the solution to a two-variable inequality.
Open
10 minutes
Note: The impetus behind this lesson is the students lack of understanding during the previous lesson. This lesson was added after reflecting on students' progress. It may not be necessary to teach this lesson.
We start this lesson by investigating the ticket out from the previous day:
What pairs of numbers satisfy this statement: The sum of two numbers is less than 10. Create an inequality in two variables and graph the solution set.
We first revisit the understanding that there are an infinite number of solutions to this question. Those solutions lie all over the coordinate axes. However, there is a boundary between where the statement is true or false.
The sum of two numbers is equal to 10.
I ask my students to write down as many pairs as they can think of by themselves. I then have them turn and talk with a neighbor to try to extend their list. We compiled many of the answers on the board so that students could see the relationship between the variables (all sum to 10). We then graphed the coordinates and discussed the type of line that could be drawn (dotted or solid) and why.
Finally, I ask students to work with their partners to determine some numbers that made the statement "the sum of two numbers is less than 10" true.
In class, students came up with a variety of solutions. Through a "popcorn" style share out (calling on students at random) we created a list and plotted to results. Students could then see that if they wanted to represent all of the solutions the graph could be shaded downwards.
I then ask students to choose a point that is in the unshaded are and test it in the inequality. Showing that that point does not work reinforces why that part of the graph is not shaded.
Investigation
20 minutes
Because we fell short on time in the previous lesson, I had students work with their partner to begin with questions a-f on solving_two_variable_inequalities_investigation.
After today's Open, they now had a more solid understanding of how to determine a boundary line before trying to shade the graph. As students work with their partners, I confer with various students. Most of my questions centered around understanding of why certain coordinates were in the solution and others were not. I want students to explain that coordinates in the shaded solution make the inequality true, while the others did not (MP3).
Closure
10 minutes
This very simple closure will give some good feedback on what students are understanding and what they are still unsure of. The question to the class was as follows:
Write down on a half sheet of paper:
(1) One thing that you feel you learned as a result of today's class.
(2) One thing you are still confused about as a result of today's class.
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# Finding Linear Equations From Graphs Study Guide
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Updated on Oct 3, 2011
## Introduction to Finding Linear Equations From Graphs
The simplest schoolboy is now familiar with facts for which Archimedes would have sacrificed his life
—Ernest Renan (1823–1892) French Philosopher
In this lesson, you'll learn how to find the equation of a line from the graph of a line, and how to determine if an ordered pair is on a line.
Now we know how to graph the equation of a line, we will go in the other direction: Given the graph of a line, we will find its equation. Just as with input/output tables and graphing, we start with the slope. Because the slope of a line is the change in the y values between two points divided by the change in the x values of those points, we can find the slope by looking at the line and choosing any two points on it. Look at the following graph.
Although we can find the slope using any two points, choose points for which x and y are integers. On this graph, when y = 2, it is difficult to see the value of x. Its value is somewhere between 3 and 4, but we cannot be certain. However, it is easy to see that when x = 3, y = 5, and when x = 4, y = 0. We'll use these points to find the slope of the line: . The slope of the line is –5. The next step might seem familiar: To find the y-intercept, use the slope, a point on the graph, and the equation y = mx + b. Let's use the point (4,0):
0 = –5(4) + b
0 = –20 + b
20 = b
The y-intercept of the graph is 20, which means that this is the graph of the equation y = –5x + 20.
#### Tip:
Sometimes, you can look right on the graph to find the y-intercept. If you can see where the line crosses the y-axis, you will have the y-intercept of the equation without having to perform any calculations.
#### Example
To find the equation of the line graphed here, we begin again with the slope. Looking at the graph, when x = 3, y = 3, and when x = 6, y = 4. We'll use the points (3,3) and (6,4) to find the slope: . We could calculate the y-intercept, but look closely at the graph. The line crosses the y-axis where y = 2, which means that 2 is the y-intercept. This is the graph of the equation y = x + 2.
## Ordered Pairs
If we are given a point, or ordered pair, and the graph of a line, we can determine if that ordered pair is on the line. We might be able to tell just by looking at the graph. If not, we can find the equation of the line, and then substitute the values of the ordered pair into the equation to see if they hold true.
Are the ordered pairs (5,38) and (–6,–46) on the following line graphed? We cannot tell just by looking, so we must find the equation of the line.
We'll use the points (0,–2) and (1,6) to find the slope: . The line crosses the y-axis where y = –2, which means that –2 is the y-intercept. This is the graph of the equation y = 8 x – 2. Now that we have the equation of the line, we will check to see if (5,38) and (–6,–46) fall on the line. Substitute 5 for x and 38 for y in the equation y = 8x – 2:
38 = 8(5) – 2 ?
38 = 40 – 2?
38 = 38
The equation holds true, so (5,38) is indeed on the line. Now, check (–6,–46):
–46 = 8(–6) – 2 ?
–46 = –48 – 2 ?
–46 ≠ –50
–46 does not equal –50, so the point (–6,–46) is not on the line y = 8x – 2.
Find practice problems and solutions for these concepts at Finding Linear Equations From Graphs Practice Questions.
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## Year 10-12 Learning Continuity Packages
### 22 May, 2020: Probability 1
Student Learning Continuity Program
22 May, 2020: Probability 1
/
This is the third session of the Student Learning Continuity Program for Years 10-12, an educational series made possible by the Ministry of Education and Human Resources Development (MEHRD).
Presenters: Lenny Olea and Martin Ruhasia
Topic: Probability
Note – For Year 12 (Form 6), your Mathematics Common Assessment Task (CAT) will be on Probability.
Objectives:
By the end of this session students should be able to
• Express probability in a variety of ways; as a percentage, a decimal, a fraction or a number out of a total
• Differentiate between equally likely and non-equally likely events
• Relate likeliness of an event to probability from 0 to 1
• Differentiate between mutually exclusive events and independent events
• Identify complementary events
• Use concepts of probability to determine the probability of events in situations where sample with and without replacement takes place.
• Represent the possible outcomes of an experiment on a tree diagram.
Introduction:
The concept of probability is part of our everyday lives. It is a household word. We hear it or speak it every day. Weather forecasts are given probabilistically. Most people know that probability is somehow connected with chances. In statements such as:
“It is likely to rain today”.
“I have a reasonable chance of passing this course”.
“He will probably win the tennis competition”.
“I am almost certain that he will be elected”.
We are referring to situations where there is an element of uncertainty about the outcome of a particular situation. In probability theory, we are concerned with assigning a “measure of likelihood” or “probability of occurrence” to the outcome of an experiment. Probability is the chance that a particular event will occur in a given set of circumstances, depending on the possible events.
Probability lies between 0 and 1. If the probability of an event is 0, we are certainly sure such event will not occur. If the probability of an event is 1, we are certainly sure such event will occur.
Let us look at some events and assign probabilities to it:
“What is the probability that ice will fall in Honiara next week?”
“What is the probability that it will rain tomorrow?”
“A teacher chooses a student at random from a class of 30 boys. What is the probability that the student chosen is a boy?”
We can estimate probabilities from experimental results by using the rule:
#### Sampling with replacement and without replacement
Sampling with replacement
Eg 1. From a well-shuffled deck of 52 cards one card is selected at random. Find the probability of selecting:
a). a heart b). a picture card c). a red card
Sampling without replacement
Eg2. A mixed team of five boys and four girls elects a captain and a vice captain.
a). Illustrate this with a tree diagram.
b). Find the probability of electing a boy captain and a girl vice-captain.
c). Find the probability of electing the same sex captain and vice-captain.
##### PROBABILITY OF TWO EVENTS
MUTUALLY EXCLUSIVE EVENTS
Mutually Exclusive Events:
• cannot happen at the one time
• have no element in common
Example
Consider when a die is rolled. Let A be the event that an even number turns up and B be the event that an odd number shows.
Sample Space = {1 , 2 , 3, 4, 5, 6 }
Events, A = { 2 , 4 , 6} and B = { 1 , 3 , 5 }
Events A and B have no element in common. An even and odd number cannot occur at the one time on a single toss of a die.
Therefore the events A and B are mutually exclusive.
NB: If events have no element in common, those events are called mutually exclusive events.
COMPLEMENTARY EVENTS
If A is any event in a sample space S and if A’ is the complement of A, then
P (A) + P (A’) = 1
P (A) = 1 – P (A’)
Consider when a die is rolled. Let A be the event that an even number turns up. The probability that an odd number shows up is
Sample Space = {1, 2, 3, 4, 5, 6}
Event A = {2, 4, 6}
INDEPENDENT EVENTS
Two events are called independent events if the outcome of one event has no effect on the outcome of the other.
Example
A coin is tossed and a die is rolled. Let A be the event that a HEAD shows on the coin and B be the event that a 4 turns up on the die.
Events A and B are independent because the outcome of a 4 does not depend on whether a head has shown on the die
Thank you every students for listening and we hope you grasp some important basic concepts on this topic (Probability) to take with you when you return to school.
Our next session with you will on normal distribution curve and we’ll also explaining What to do when you carry out your Mathematics statistics minor research Project chapter by chapter.
Form 3, Form 5 and Form 6 Mathematics, we create an account (page) on Facebook for revisions. You can go to that page and post your Mathematics questions and we’ll give you the solution. We are offering you a free service to you do your revisions. All you need to do is to buy your data and use wisely for this purpose.
Once again thank you every good students out there and all best in your studies.
Our contacts: Lenny Olea – 8628229
Martin Ruhasia – 7716696
Stay safe and have a blessed weekend.
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# 2019 AMC 10A Problems/Problem 19
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
What is the least possible value of $$(x+1)(x+2)(x+3)(x+4)+2019$$where $x$ is a real number?
$\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021$
## Solution 1
Grouping the first and last terms and two middle terms gives $(x^2+5x+4)(x^2+5x+6)+2019$, which can be simplified to $(x^2+5x+5)^2-1+2019$. Noting that squares are nonnegative, and verifying that $x^2+5x+5=0$ for some real $x$, the answer is $\boxed{\textbf{(B) } 2018}$.
## Solution 2
Let $a=x+\tfrac{5}{2}$. Then the expression $(x+1)(x+2)(x+3)(x+4)$ becomes $\left(a-\tfrac{3}{2}\right)\left(a-\tfrac{1}{2}\right)\left(a+\tfrac{1}{2}\right)\left(a+\tfrac{3}{2}\right)$.
We can now use the difference of two squares to get $\left(a^2-\tfrac{9}{4}\right)\left(a^2-\tfrac{1}{4}\right)$, and expand this to get $a^4-\tfrac{5}{2}a^2+\tfrac{9}{16}$.
Refactor this by completing the square to get $\left(a^2-\tfrac{5}{4}\right)^2-1$, which has a minimum value of $-1$. The answer is thus $2019-1=\boxed{\textbf{(B) }2018}$.
## Solution 3 (calculus)
Similar to Solution 1, grouping the first and last terms and the middle terms, we get $(x^2+5x+4)(x^2+5x+6)+2019$.
Letting $y=x^2+5x$, we get the expression $(y+4)(y+6)+2019$. Now, we can find the critical points of $(y+4)(y+6)$ to minimize the function:
$\frac{d}{dx}(y^2+10y+24)=0$
$2y+10=0$
$2y(y+5)=0$
$y=-5,0$
To minimize the result, we use $y=-5$. Hence, the minimum is $(-5+4)(-5+6)=-1$, so $-1+2019 = \boxed{\textbf{(B) }2018}$.
Note: We could also have used the result that minimum/maximum point of a parabola $y = ax^2 + bx + c$ occurs at $x=-\frac{b}{2a}$.
## Solution 4
The expression is negative when an odd number of the factors are negative. This happens when $-2 < x < -1$ or $-4 < x < -3$. Plugging in $x = -\frac32$ or $x = -\frac72$ yields $-\frac{15}{16}$, which is very close to $-1$. Thus the answer is $-1 + 2019 = \boxed{\textbf{(B) }2018}$.
## Solution 5 (using the answer choices)
Answer choices $C$, $D$, and $E$ are impossible, since $(x+1)(x+2)(x+3)(x+4)$ can be negative (as seen when e.g. $x = -\frac{3}{2}$). Plug in $x = -\frac{3}{2}$ to see that it becomes $2019 - \frac{15}{16}$, so round this to $\boxed{\textbf{(B) }2018}$.
We can also see that the limit of the function is at least -1 since at the minimum, two of the numbers are less than 1, but two are between 1 and 2.
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# Difference between revisions of "2015 AMC 10A Problems/Problem 23"
## Problem
The zeroes of the function $f(x)=x^2-ax+2a$ are integers .What is the sum of the possible values of a?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}}\ 17\qquad\textbf{(E)}\ 18$ (Error compiling LaTeX. ! Extra }, or forgotten $.) ## Solution 1 By Vieta's Formula, $a$ is the sum of the integral zeros of the function, and so $a$ is integral. Because the zeros are integral, the discriminant of the function, $a^2 - 8a$, is a perfect square, say $k^2$. Then adding 16 to both sides and completing the square yields $$(a - 4)^2 = k^2 + 16.$$ Hence $(a-4)^2 - k^2 = 16$ and $$((a-4) - k)((a-4) + k) = 16.$$ Let $(a-4) - k = u$ and $(a-4) + k = v$; then, $a-4 = \dfrac{u+v}{2}$ and so $a = \dfrac{u+v}{2} + 4$. Listing all possible $(u, v)$ pairs (not counting transpositions because this does not affect $u + v$), $(2, 8), (4, 4), (-2, -8), (-4, -4)$, yields $a = 9, 8, -1, 0$. These $a$ sum to 16, so our answer is $\textbf{(C)}$. ## Solution 2 Let $r_1$ and $r_2$ be the integer zeroes of the quadratic. Since the coefficent of the $x^2$ term is $1$, the quadratic can be written as $(x - r_1)(x - r_2)$ or $x^2 - (r_1 + r_2)x + r_1r_2)$. By comparing this with $x^2 - ax + a^2$, $r_1 + r_2 = a$ and $r_1r_2 = 2a$. Plugging the first equation in the second, $r_1r_2 = 2(r_1 + r_2). Rearranging gives$r_1r_2 - 2r_1 - 2r_2 = 0$.
This can be factored as$(Error compiling LaTeX. ! Missing$ inserted.)(r_1 - 2)(r_2 - 2) = 4$. These factors can be:$ (Error compiling LaTeX. ! Missing $inserted.)(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2), (-2, -2)$.
We want the number of distinct$(Error compiling LaTeX. ! Missing$ inserted.)a = r_1 + r_2$, and these factors gives$a = {-1, 8, 9}.
So the answer is $-1 + 8 + 9 = \boxed{\textbf{(C) }16}$.
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# Video: Differentiating Rational Functions Using the Quotient Rule
Differentiate π(π₯) = (4π₯Β² β 5π₯ + 8)/(3π₯ β 4).
04:05
### Video Transcript
Differentiate ππ₯ equals four π₯ squared minus five π₯ plus eight over three π₯ minus four.
So in order to actually differentiate our function, what weβre gonna use is something called the quotient rule. And we know that weβre gonna use the quotient rule because actually our function is in the form π’ over π£, because itβs actually a fraction.
What the quotient rule actually tells us is that ππ¦ ππ₯, so our derivative, is equal to π£ ππ’ ππ₯ minus π’ ππ£ ππ₯ over π£ squared. What that actually means in practice is its our π£ multiplied by our derivative of π’ and then itβs minus π’ multiplied by our derivative of π£ and then all divided by π£ squared. Okay, fab! Now that we have this, letβs use it and actually find out what the derivative of our function actually is.
So my first step is to actually identify our π’ and our π£. So I use our numerator, so four π₯ squared minus five π₯ plus eight. And our π£ is going to be three π₯ minus four, which is our denominator. So the first thing weβre gonna do is actually differentiate π’. So we differentiate four π₯ squared minus five π₯ plus eight.
And just to quickly remind us how we do that, if weβve got ππ₯ in the form ππ₯ to the power of π, then the derivative of our function, or could be known as ππ¦ ππ₯, is equal to ππ, so coefficient multiplied by the exponent, and then π₯ to the power of π minus one. So you subtract one from the exponent.
Okay, so now letβs use this and differentiate π’. So weβre gonna get eight π₯ minus five. And just to kind of remind us how we did that, the first one was gonna be eight π₯, because weβve got four multiplied by two, an exponent, multiplied by coefficient, which gives us eight, and then π₯ to the power of one, because you actually subtract one from the exponent. So two minus one is one. So weβre left with eight π₯.
Okay, great! So now letβs move on to ππ£ ππ₯. Well, if we differentiate π£, so weβre gonna find ππ£ ππ₯. And this is just gonna be equal to three. And thatβs because if we differentiate three π₯, we get three. And if we differentiate just an integer on its own, you get zero.
Okay, so great! Weβve now found ππ’ ππ₯ and ππ£ ππ₯. So we can now actually apply our quotient rule to find ππ¦ ππ₯. So first of all, weβre gonna have π£ ππ’ ππ₯. So itβs gonna be three π₯ minus four multiplied by eight π₯ minus five. And then weβre gonna have minus three multiplied by four π₯ squared minus five π₯ plus eight. So thatβs our π’ ππ£ ππ₯. And Iβve put it the other way round just because itβs easier when weβre gonna expand the parentheses in the next stage. And then this is all divided by three π₯ minus four all squared. So itβs our π£ squared.
Okay, so now letβs move on to the next stage, which Iβve already said is gonna be expanding the parentheses on the numerator. So first of all, Iβve got three π₯ multiplied by eight π₯, which will give me 24π₯ squared. Then three π₯ multiplied by negative five gives us negative 15π₯, then minus 32π₯ because we had negative four multiplied by eight π₯, and then finally plus 20 because we had negative four multiplied by negative five. And a negative multiplied by negative gives us positive.
Then weβve got minus 12π₯ squared, because weβve got negative three multiplied by four π₯ squared, plus 15π₯, negative three multiplied by negative five π₯, again negative multiplied by negative, and then finally minus 24. Okay, then this is all divided by three π₯ minus four all squared.
So we now are gonna move on to our final stage, which is actually to simplify our numerator. So then weβre gonna have our first term. Itβs gonna be 12π₯ squared. And thatβs because weβve got 24π₯ squared minus 12π₯ squared. Then we get minus 32π₯. And thatβs because we had negative 15π₯ minus 32π₯ plus 15π₯. So therefore, the 15π₯s cancel out. We have minus 32π₯.
And then, finally, we get minus four. And this is because we had positive 20 minus 24. And then this is all over three π₯ minus four all squared. So great! We can say that if we differentiate the function four π₯ squared minus five π₯ plus eight over three π₯ minus four, then ππ¦ ππ₯ is gonna be equal to 12π₯ squared minus 32π₯ minus four over three π₯ minus four all squared.
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# Tag: Maths
## Location
This week for Miss White maths we learned about how to find and describe the location of something on a map using grid references.
What is a grid? A grid is a map put into squares. Grid references are used to help find things. This used to be used for maps to help people travel to different areas. We always say the X axis first and Y axis next.
First, we had to discuss in Miss White whole group, what location is. Location is an actual place or natural setting in which a film or broadcast is made, as distinct from a simulation in a studio.
Next, we had to make a copy of a google slide that is about auckland zoo. In the google slide we had to have a look at the questions and answer them also by using what are some places you can see, what are their coordinates, and how do we read coordinates. How to read coordinates is that you have to use the 1-10 numbers on the left side and the letters 1-12 on the bottom. For example: your looking for a animal in the zoo and you can’t found it so you use the map to find it. It would be B5.
Lastly, we had to be in pairs and make another copy of a document to draw a map and get your partner to find objects/places.
I enjoyed doing this task. I did well at making a copy of the task. I need to improve on adding more details into the document.
## Place Value Subtraction
This week for Miss White maths group we learned how to subtract by using place value.
What is place value? Place values are the numerical value of a digit. For example, the number 45 has 4 tens and 5 ones. We can also subtract three digit numbers by breaking the number into one, tens, and hundreds. We always have to remember to subtract from ones first.
An example: 547 – 213 = 334.
• 7 – 3 = 4
• 40 – 10 = 30
• 500 – 200 = 300
• 4 + 30 + 300 = 334.
First, Miss White groups had to discuss what place value is. Place values are the numerical value of a digit. For example, the number 45 has 4 tens and 5 ones. We can also subtract three digit numbers by breaking the number into one, tens, and hundreds. We always have to remember to subtract from ones first.
Next, Miss White told us to grab a whiteboard maker and a whiteboard to draw a place value house. We learned how to do minus by using a place value house. Miss White gave us a solution to solve. The solution we had to solve was : 456 – 326. The problem was that when we start with the ones and you can’t take away 7 from the 6 because it doesn’t give us an answer. How to solve it : we have to solve it by borrowing one tens or hundreds and place it into the ones.
Lastly, we had to make a copy of a google slide that talks about place value and you have to solve the solution. We had to solve 16 by using minus to make it harder because we did addition last week and last term.
I enjoyed learning about place value. I did well at explanning what place value is. I need to improve on adding more details and telling others how I worked it out.
Here is my work :
## Maths Problem
This week for Miss White’s Math’s group we learned about how to answer word problems and showing how we worked (solved) the problem.
Math’s Problem is a mathematical problem is a question that needs a solution and can be represent, analyzed, and possibly solved, with the methods of mathematics.
First, we had to discuss in our math group what maths problem is. Math’s Problem is a mathematical problem is a question that needs a solution and can be represent, analyzed, and possibly solved, with the methods of mathematics. For example: 2000+204 = 2204.
Next, we had to answer some questions and write down how I worked out the answers. For example : I used multiplication. 6 x 4 = 24, 6 x 20 = 120, 24 + 120 = 144.
Then, Miss White showed us a google slide called Maths Problems. In the google slide it shows us that we had to solve or fix the problem so our teacher knows which one we are stuck on. For example : if a box of pencils contains 24 pencils, how many are in 6 boxes. I got stuck on this one until I realised I had to use addition, multiplication, division, and takeaways.
Lastly, when we solved all of the questions there was this next slide and we had to find a partner to do it with. I had to write down easy, hard, or medium maths questions for my partner to answer like I’m challenging him.
I enjoyed doing this task. I did well at typing the questions for my partner. I need to improve on showing them more information of how I worked it out.
## Basic Facts Boxes
This week for maths I learned about Basic Facts Boxes.
Basic Facts Boxes is a type of Math game that you sort out the answers. For example: 1+1=2.
First, I had to figure out which one I want to do for math’s. The options were plus, minus, and multiplications. I picked minus because I was thinking that it is the best way to start off with. I also had to pick another option. The options were, random, and order. I picked order because I think the same as minus.
Next, I had to start off by figuring out the answers for the math’s. The minus was easy for me. For example: 20-1= 19, 19-1=18.
Lastly, I had to press check when I finished answering the math problem. For example: when you finish answering your problem of math’s you had to press check to see if you got some wrong, some correct.
I enjoyed learning about basic facts boxes. I did well at answering the problems or solving the problems. I need to improve on moving on to the multiplications.
## Maths Problems
This week for Miss White’s maths group we learned about maths problems and to answer word problems and show your working it out.
Maths Problem is a problem that can be represented, and possibly solved, with the methods of mathematics. For example : A train travels 240 km in 4 hours. What is the average speed of the train? The answer will be 60km.
First, we had to discuss in a group what maths problem is? Maths Problem is a problem that can be represented, and possibly solved, with the methods of mathematics.
Next, we had to answer some questions and write down how I worked out the answers. For example : I used multiplication. 6 x 4 = 24, 6 x 20 = 120, 24 + 120 = 144.
Then, Miss White showed us a google slide called Maths Problems. In the google slide it shows us that we had to solve or fix the problem so our teacher knows which one we are stuck on. For example : if a box of pencils contains 24 pencils, how many are in 6 boxes. I got stuck on this one until I realised I had to use addition, multiplication, division, and takeaways.
Lastly, when we solved all of the questions there was this next slide and we had to find a partner to do it with. I had to write down easy, hard, or medium maths questions for my partner to answer like I’m challenging him.
I enjoyed doing this task. I did well at typing the questions for my partner. I need to improve on showing them more information of how I worked it out
## Basic Facts Boxes
This week for Can Dos I learned about Basic Facts Boxes that’s part of math’s.
Basic Facts Boxes is a type of Math game that you sort out the answers. For example: 1+1=2.
First, I had to figure out which one I want to do for math’s. The options were plus, minus, and multiplications. I picked minus because I was thinking that it is the best way to start off with. I also had to pick another option. The options were, random, and order. I picked order because I think the same as minus.
Next, I had to start off by figuring out the answers for the math’s. The minus was easy for me. For example: 20-1= 19, 19-1=18.
Lastly, I had to press check when I finished answering the math problem. For example: when you finish answering your problem of math’s you had to press check to see if you got some wrong, some correct.
I enjoyed learning about basic facts boxes. I did well at answering the problems or solving the problems. I need to improve on moving on to the multiplications.
## Bar Graph
This week for Maths I learned about how to create a Bar Graph using results from a tally chart & to use PPDAC (Problem, Plan, Data, Analysis, and Conclusion/Communication) cycle to do a class investigation.
First I had to talk in a group about statistics? Statistics is to collect and do data to draw conclusions. For example : How do you get to school? The answer will be car, walk, bus, or bike. Lots of people get to school with cars.
Next I had to think about in the group what does PPDAC, it stands for Problem (Main Question), Plan (What are you measuring, who are you getting data from, questions), Data (collect data from target audience), Analyse (sort data, draw up graph, what do you notice?), Conclusion (final interpretation, new question?).
Afterwards I had to answer all of the class questions. The question was what’s your favourite colour?, How many people live in your house?, What month is your birthday?, Can you ride a bike?. For how many people live in your house, the answer is 10 because 10 people live in my house.
Lastly I had to draw/make 4 tally charts and 4 bar graphs using the data collected by the class question. I also had to write down 5 “I notice” statements.
I enjoyed learning about bar graphs. I did well at making a bar graph on a google meet. I need to improve on adding more questions or details to it.
## Basic Facts Boxes
Today in week 10 I learned about Basic Facts Boxes and I did Basic Facts Boxes.
Basic Number Facts is basically the same as Basic Facts Boxes. For example : 0 + 0 to 9 + 9 for addition and their subtractive opposites, and 0 x 0 to 9 x 9 for multiplication.
First I had to go on a worksheet (that records the data) that shows Basic Facts Boxes. For example : it shows the subject.
Next I had to go and choose one of the subjects I wanted to start off with. For example : x, +, -, and divisions. I picked the multiplication because it looks and feels like a good reason to start off with.
Then I had to pick if I wanted the multiplication to be in order (1,2,3,4,5,6,7,8,9,1o) or in randoms (2,5,8,10,3,4). I picked an order because I love to be in order. I’ll try and do random things in the next basic facts boxes next week.
Lastly I had to click onto check (checks all of your answers if it’s right) to see the answer is right. For example : 1+1 = 2 and the answer turned green.
I enjoyed doing this task. I did well at getting them all right. I need to improve on doing hard mode instead of easy mode.
## Basic Facts Boxes
This week I did Basic Facts Boxes.
First I had to pick a subject befores putting in the answer. For example : There are +, -, x, and divisions. I picked multiplication because I think it’s a great way to start.
Lastly I had to press check to see if my answer was correct. I got 100/100 because I have been learning my basic facts while I am away from school.
I enjoyed doing Basic Facts Boxes. I did well at figuring out the answer. I need to improve on doing divisions next.
## Basic Facts Boxes
This week I did Basic Facts Boxes.
First I had to pick a subject befores putting in the answer. For example : There are +, -, x, and divisions. I picked multiplication because I think it’s a great way to start.
Lastly I had to press check to see if my answer was correct. I got 100/100 because I have been learning my basic facts while I am away from school.
I enjoyed doing Basic Facts Boxes. I did well at figuring out the answer. I need to improve on doing divisions next.
|
# Second Equation of Motion
Physics, the study of the fundamental principles that govern the behavior of matter and energy, is a captivating subject that allows us to decipher the mysteries of the physical world. In the ninth-grade curriculum, one of the most intriguing topics in physics is the second equation of motion. This equation unravels the intricacies of accelerated motion, shedding light on the relationships between displacement, initial velocity, acceleration, and time.
## Unveiling the Equation
The second equation of motion is often represented as:
s=vt + 1/2at2
where:
• stands for displacement,
• represents initial velocity,
• denotes acceleration,
• symbolizes time.
This equation serves as a powerful tool to analyze scenarios where an object experiences constant acceleration. Whether it’s a ball rolling down a ramp or a car accelerating on a straight road, this equation allows us to predict and understand their motion with precision.
Also Study: Chapter 2 Lecture 1
Study: Chapter 2 Lecture 1
Also Study: Chapter 2 Lecture 1
## Breaking Down the Equation
1. Displacement (): This refers to the change in the position of an object. It can be positive, negative, or zero, depending on the direction of motion. In the context of the equation, is the total distance traveled by the object.
2. Initial Velocity (v): The initial speed of the object when the time measurement begins. This velocity might be in any direction and can be zero if the object starts from rest.
3. Acceleration (): Acceleration is the rate at which an object’s velocity changes. It can be uniform or non-uniform, and it’s typically measured in meters per second squared (m/s2). If the object is slowing down, acceleration will be negative, and if it’s speeding up, acceleration will be positive.
4. Time (): The duration for which the object experiences the given acceleration. It’s measured in seconds.
## Real-World Applications
The second equation of motion finds its applications in various real-world scenarios. Let’s explore a few:
### Projectile Motion
Consider a ball thrown into the air. Gravity acts as the acceleration, and this equation helps us determine how high the ball will go and how long it will take to return to the ground.
### Automotive Industry
Car manufacturers and engineers utilize this equation to design vehicles that accelerate safely and efficiently. It helps them calculate the distance required for a car to come to a complete stop when brakes are applied.
### Sports Analysis
In sports like athletics and motorsports, understanding the second equation of motion can help athletes and coaches optimize their training routines and race strategies. It can also provide insights into how fast an athlete can sprint or how quickly a race car can accelerate.
## Solving Problems Using the Equation
To solve problems using the second equation of motion, follow these steps:
1. Identify the values given in the problem for v, and t.
2. Substitute these values into the equation.
3. Calculate the result, which will be the displacement “.
Remember, consistent units are crucial for accurate calculations. Make sure all values are in the same units before plugging them into the equation.
## Conclusion
The second equation of motion is a cornerstone of understanding accelerated motion in physics. By dissecting the relationships between displacement, initial velocity, acceleration, and time, this equation empowers us to analyze and predict the behavior of objects undergoing constant acceleration. From calculating the trajectory of a basketball to engineering safer cars, its applications extend across various fields. Embracing this equation illuminates the path toward unraveling the mysteries of motion and harnessing its principles for the betterment of our world.
|
FutureStarr
How to add fractions
## How to add fractions
### There are Easy steps to add Fraction.
Your students would possibly understand their manner around a numerator and denominator, however, are they prepared for what’s subsequent? abruptly, it’s time to discover ways to upload fractions -- and your elegance is harassed. Feeling intimidated? You’re now not alone. adding fractions may additionally appear daunting, however, it doesn’t want to be. We’ve prepared a manual that will help you correctly train your students the way to add fractions, covering:
## Why do students face difficulty with fractions?
Fractions -- mainly fraction operations -- are a problematic challenge for most students. the trouble with fractions can lessen self-assurance in math and lead to math tension if students don’t get hold of sufficient guide inside the difficulty.
1. Understanding what the numbers mean
Before fractions, college students are used to running with entire numbers: primary numbers that represent entire quantities. Fractions introduce college students to rational numbers, which include an entirely new set of policies and patterns. The meaning at the back of fractions is perplexing when you examine them to whole numbers. entire numbers are handiest expressed one way, at the same time as fractions may be expressed in lots of ways and nevertheless, represent the identical quantity. As an example, there’s best one way to represent the range three, however, ²⁄â‚„ represents the same quantity as ½, zero.5, and 50%. As a scholar, this is tough to wrap your head around.
### 2. Different operations for whole numbers and fractions
The methods you operate to feature, subtract, multiply and divide complete numbers are one-of-a-kind than doing the equal operations for fractions. policies come to be much more unpredictable and confusing. Many college students and teachers have limited know-how of how or why those strategies are used. Fractions are tougher to symbolize with visuals or manipulatives, and the regulations for including them are greater hard to recognize. studying a way to multiply and divide fractions can add even extra confusion, as college students need to remember the variations between these operations. this is a huge adjustment for college kids who are already comfy with whole quantity arithmetic.
## Types of fractions
College students should first recognize the distinction among each sort of fraction to efficiently upload them. First, let’s begin with the fundamental additives of a fragment. a fraction represents elements of a whole. The numerator (the pinnacle range) illustrates the wide variety of components you've got. The denominator (the lowest wide variety) indicates the overall range of elements the entire is divided into. in the illustration above, our circle is split into four parts. this means four is our denominator. of those four components, one is highlighted. this means one is our numerator. So, our fraction is ¼ or one area. There are three preferred classes of fractions: right, fallacious, and blended. kind
MeaningExample
proper fraction
The numerator is much less than the denominator¾ (three quarters)
flawed fraction
The numerator is more than the denominatorâ·⁄â‚„ (seven quarters)
mixed fraction a whole range and a right fraction combined
1 ¾ (one and 3 quarters)
further to these, fraction equations can be cut up into wonderful categories: those with like fractions and those with not like fractions. kind
MeaningExample
Like fractions
Fractions with the identical denominator¼ and ¾
unlike fractions
Fractions with unique denominators¼ and â…œ
Base expertise of those types will help students recognize what to do while faced with a question approximately including fractions. Now which you’re familiar with every form of fraction, you could get adding! train your college students on the 3-step formula to confidently address fraction addition equations.
## Easy step for adding fractions
It may seem scary at first, but adding fractions can be easy. All you need to do is follow three simple steps:
• Step 1: Find a common denominator
• Step 2: Add the numerators (and keep the denominator)
• Step 3: Simplify the fraction
Let’s look at each step in a bit more detail.
### Step 1: Find a common denominator
If your denominators are already identical, you’re including fractions with like denominators. splendid! this means you may bypass stepping. if your denominators are unique, you’re adding fractions with unlike denominators. when including unlike fractions, you need to discover a common denominator so that you can add the two fractions together. take a look at the video beneath to recognize why we want a not unusual denominator to feature fractions.
You can locate the commonplace denominator the usage of equal fractions: fractions that have the identical fee. as an instance, ²⁄â‚„, ³⁄₆, and â´⁄₈ are equivalent fractions due to the fact they are able to all be decreased to ½. There are essential techniques for locating the commonplace denominator.
#### The common denominator method
On this technique, you’ll multiply the pinnacle and backside of each fraction by using the denominator of the alternative. for example, bear in mind the following equation:
# â…“ + â…™
Our fractions have two special denominators: 3 and six. We need to multiply the numerator and denominator in â…“ through six, then multiply the numerator and denominator in â…™ by way of 3. when we try this, our new fractions grow to be â¶⁄â‚₈ and ³⁄â‚₈. the two new fractions have an equal denominator, so now we can upload them!
### Step 2: Add the numerators (and keep the denominator)
This step is rather straightforward. Add your numerators together so the sum becomes the new numerator, while the denominator stays the same.
Let’s use our previous example: â…“ + â…™Using our new equation from the common denominator method -- â¶⁄â‚₈ + ³⁄â‚₈ -- we need to add six and three together. The denominator will still be eighteen. Six plus three is nine, so our answer is â¹⁄â‚₈.
### Step 3: Simplify the fraction
In case your fraction carries high numbers, you can want to simplify it.
Simplifying includes finding the smallest equal fraction viable. In our previous equation, our answer was â¹⁄â‚₈.
This range seems a piece big, so we're going to see if we are able to simplify it to an easier quantity.
To simplify a fragment, you need a not unusual factor: a variety of on the way to divide into both numbers flippantly.
For instance, is a not unusual component of four and six, due to the fact both numbers may be divided by means of.
The two easiest methods for simplifying a fraction are:
#### 1) Trial and error
For this technique, simply maintain dividing the numerator and denominator by small numbers. start with, then three, then four -- and so on until you get the smallest feasible solution.
With our solution of â¹⁄â‚₈, we will maintain looking to divide with the aid of small numbers until we discover one which works. Can each 9 and eighteen be divided through? No. we are able to divide 9 via 2 frivolously. ok, permit’s try another number.
Can write both 9 and eighteen be divided through three? yes! while we divide both by three, our fraction will become ³⁄₆. Now that we have a less complicated answer, it’s time to see if we can simplify even in addition. 3 and 6 can both be divided by using three once more, so our very last answer is ½.
#### 2) Find the Greatest Common Factor (GCF)
The GCF is the very best variety that divides lightly into two or greater numbers. This technique is just like locating the least commonplace denominator -- you’ll locate the answer by way of listing all feasible elements.
Using our preceding instance of â¹⁄â‚₈, we’ll find and listing all of the elements of every quantity, starting from one. once you've got indexed all of the elements of that quantity, all you have to do is discover the largest number repeated in each list.
### Numerator and denominator
Here is the simple definition of Numerator and the denominator is that Numerator is the top part of friction and the denominator is the bottom part.
### Adding Fractions With Different Denominators
When adding fractions with the same denominators and is similar to the idea of cutting something into smaller pieces then adding the individual pieces.
When it comes to adding the fraction or fractions together with different dividing denominators, we encounter a problem.
Since the denominators are different and it means that we cut the whole into different sizes.
1 3, 3 2, 3 6 like simply 3 4
## How do you add fractions Support
Here are 3 easy steps to help you add two fractions together.
Step 1: Convert the fractions to the same denominator.
If the fractions already have the same denominator, you do not need to do anything!
Step 2: Add the two numerators together (and keep the denominator the same) to get the answer.
Step 3: Simplify the answer if needed.
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The graph of a quadratic function is a parabola, which is a "u"-shaped curve. In this article, we review how to graph quadratic functions.
The graph of a quadratic function is a parabola, which is a "u"-shaped curve:
Looking for an introduction to parabolas? Check out this video.
Example 1: Vertex form
Graph the equation.
$y=-2\left(x+5{\right)}^{2}+4$
This equation is in vertex form.
$y=a\left(x-h{\right)}^{2}+k$
This form reveals the vertex, $\left(h,k\right)$, which in our case is $\left(-5,4\right)$.
It also reveals whether the parabola opens up or down. Since $a=-2$, the parabola opens downward.
This is enough to start sketching the graph.
To finish our graph, we need to find another point on the curve.
Let's plug $x=-4$ into the equation.
$\begin{array}{rl}y& =-2\left(-4+5{\right)}^{2}+4\\ \\ & =-2\left(1{\right)}^{2}+4\\ \\ & =-2+4\\ \\ & =2\end{array}$
Therefore, another point on the parabola is $\left(-4,2\right)$.
Want another example? Check out this video.
Example: Non-vertex form
Graph the function.
$g\left(x\right)={x}^{2}-x-6$
First, let's find the zeros of the function—that is, let's figure out where this graph $y=g\left(x\right)$ intersects the $x$-axis.
$\begin{array}{rl}g\left(x\right)& ={x}^{2}-x-6\\ \\ 0& ={x}^{2}-x-6\\ \\ 0& =\left(x-3\right)\left(x+2\right)\end{array}$
So our solutions are $x=3$ and $x=-2$, which means the points $\left(-2,0\right)$ and $\left(3,0\right)$ are where the parabola intersects the $x$-axis.
To draw the rest of the parabola, it would help to find the vertex.
Parabolas are symmetric, so we can find the $x$-coordinate of the vertex by averaging the $x$-intercepts.
With the $x$-coordinate figured out, we can solve for $y$ by substituting into our original equation.
$\begin{array}{rl}g\left(0.5\right)& =\left(0.5{\right)}^{2}-\left(0.5\right)-6\\ \\ & =0.25-0.5-6\\ \\ & =-6.25\end{array}$
Our vertex is at $\left(0.5,-6.25\right)$, and our final graph looks like this:
Want another example? Check out this video.
Practice
Problem 1
Graph the equation.
$y=2\left(x+1\right)\left(x-1\right)$
Want more practice graphing quadratics? Check out these exercises:
Want to join the conversation?
• How even do you guys DO that?!
• Do some more Khan Academy videos and exercises. You'll get the hang of it sooner of later if you try! Good luck!
• how to find the vertex
• One way is to complete the square and put in vertex form, Another way is to use -b/2a as the x coordinate and then use that to solve for y.
• We only use two point here to graph a parabola but I've heard that it takes three points to define one. Is it true that an infinite number of parabolas can be drawn through just two distinct points? on the coordinate plane?
• If you know the vertex and another point, then you also know the reflexive point, so you have 3 points. So two points work as long as one of them is the vertex.
• On paper, how would I find the curve to nicely draw it out? Would I have to plot several points to graph it out by hand or is there another rule?
• Yes, the more points you plot, the more accurate you can draw the curve. Some turn the graph paper sideways and draw on one side of the vertex, rotate it 180 degrees and draw on the other side of the vertex.
• Question? WHere did the word quadratics come from?
• It comes from the Latin word quadratum, which means square.
• Do i need to learn these, or will it come back to me if i skip parbolas?
• Yes, don't skip it. Try to understand it, and it will save you a lot of frustration later on.
• On example no.1. How did he came up with x=-4 to plug into the equation?
And also in Practise problem no.2, how did he came up with x=0 to plug into the equation?
• On Example 1, the vertex was at (-5,4). Sal moved over 1 unit to pick x=-4. I believe he did this to demonstrate the impact of the -2 in front. It reduced the y-value of 4 by 2 units.
On practice problem 2, I'm sure he picked x=0 to find the y-intercept of the equation.
• why plugged x=−4x,when you can do y=0
• Sal has the equation: y = -2(x+5)^2+4. This is vertex form. So, you graph the vertex and then find points to the left and right of the vertex.
When Y is isolated already and the equation is in vertex form, it is easier to pick values of X and calculate Y especially when you have a 2nd degree (quadratic) equation or higher degree. The math is usually simpler.
If the equation had been in standard form: y=Ax^2+Bx+C, then it is quite common to find the x-intercepts 1st (set y=0 and calculate X). Though, you can also get the vertex pretty easily from this form and work outward from there.
The benefit of having the vertex is that you know the highest / lowest point in the graph and you know the graph will be symmetrical as it moves away from the vertex.
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# The Two Water Jug Puzzle
• Difficulty Level : Hard
• Last Updated : 22 Oct, 2021
You are on the side of the river. You are given a m liter jug and a n liter jug where 0 < m < n. Both the jugs are initially empty. The jugs don’t have markings to allow measuring smaller quantities. You have to use the jugs to measure d liters of water where d < n. Determine the minimum no of operations to be performed to obtain d liters of water in one of jug.
The operations you can perform are:
1. Empty a Jug
2. Fill a Jug
3. Pour water from one jug to the other until one of the jugs is either empty or full.
There are several ways of solving this problem including BFS and DP. In this article, an arithmetic approach to solving the problem is discussed. The problem can be modeled by means of the Diophantine equation of the form mx + ny = d which is solvable if and only if gcd(m, n) divides d. Also, the solution x,y for which equation is satisfied can be given using the Extended Euclid algorithm for GCD
For example, if we have a jug J1 of 5 liters (n = 5) and another jug J2 of 3 liters (m = 3) and we have to measure 1 liter of water using them. The associated equation will be 5n + 3m = 1. First of all this problem can be solved since gcd(3,5) = 1 which divides 1 (See this for detailed explanation). Using the Extended Euclid algorithm, we get values of n and m for which the equation is satisfied which are n = 2 and m = -3. These values of n, m also have some meaning like here n = 2 and m = -3 means that we have to fill J1 twice and empty J2 thrice.
Now to find the minimum no of operations to be performed we have to decide which jug should be filled first. Depending upon which jug is chosen to be filled and which to be emptied we have two different solutions and the minimum among them would be our answer.
Solution 1 (Always pour from m liter jug into n liter jug)
1. Fill the m litre jug and empty it into n liter jug.
2. Whenever the m liter jug becomes empty fill it.
3. Whenever the n liter jug becomes full empty it.
4. Repeat steps 1,2,3 till either n liter jug or the m liter jug contains d litres of water.
Each of steps 1, 2 and 3 are counted as one operation that we perform. Let us say algorithm 1 achieves the task in C1 no of operations.
Solution 2 (Always pour from n liter jug into m liter jug)
1. Fill the n liter jug and empty it into m liter jug.
2. Whenever the n liter jug becomes empty fill it.
3. Whenever the m liter jug becomes full empty it.
4. Repeat steps 1, 2 and 3 till either n liter jug or the m liter jug contains d liters of water.
Let us say solution 2 achieves the task in C2 no of operations.
Now our final solution will be a minimum of C1 and C2.
Now we illustrate how both of the solutions work. Suppose there are a 3 liter jug and a 5 liter jug to measure 4 liters water so m = 3,n = 5 and d = 4. The associated Diophantine equation will be 3m + 5n = 4. We use pair (x, y) to represent amounts of water inside the 3-liter jug and 5-liter jug respectively in each pouring step.
Using Solution 1, successive pouring steps are:
`(0,0)->(3,0)->(0,3)->(3,3)->(1,5)->(1,0)->(0,1)->(3,1)->(0,4)`
Hence the no of operations you need to perform are 8.
Using Solution 2, successive pouring steps are:
`(0,0)->(0,5)->(3,2)->(0,2)->(2,0)->(2,5)->(3,4)`
Hence the no of operations you need to perform are 6.
Therefore, we would use solution 2 to measure 4 liters of water in 6 operations or moves.
Based on the explanation here is the implementation.
## C++
`// C++ program to count minimum number of steps``// required to measure d litres water using jugs``// of m liters and n liters capacity.``#include ``using` `namespace` `std;` `// Utility function to return GCD of 'a'``// and 'b'.``int` `gcd(``int` `a, ``int` `b)``{`` ``if` `(b==0)`` ``return` `a;`` ``return` `gcd(b, a%b);``}` `/* fromCap -- Capacity of jug from which`` ``water is poured`` ``toCap -- Capacity of jug to which`` ``water is poured`` ``d -- Amount to be measured */``int` `pour(``int` `fromCap, ``int` `toCap, ``int` `d)``{`` ``// Initialize current amount of water`` ``// in source and destination jugs`` ``int` `from = fromCap;`` ``int` `to = 0;` ` ``// Initialize count of steps required`` ``int` `step = 1; ``// Needed to fill "from" Jug` ` ``// Break the loop when either of the two`` ``// jugs has d litre water`` ``while` `(from != d && to != d)`` ``{`` ``// Find the maximum amount that can be`` ``// poured`` ``int` `temp = min(from, toCap - to);` ` ``// Pour "temp" liters from "from" to "to"`` ``to += temp;`` ``from -= temp;` ` ``// Increment count of steps`` ``step++;` ` ``if` `(from == d || to == d)`` ``break``;` ` ``// If first jug becomes empty, fill it`` ``if` `(from == 0)`` ``{`` ``from = fromCap;`` ``step++;`` ``}` ` ``// If second jug becomes full, empty it`` ``if` `(to == toCap)`` ``{`` ``to = 0;`` ``step++;`` ``}`` ``}`` ``return` `step;``}` `// Returns count of minimum steps needed to``// measure d liter``int` `minSteps(``int` `m, ``int` `n, ``int` `d)``{`` ``// To make sure that m is smaller than n`` ``if` `(m > n)`` ``swap(m, n);` ` ``// For d > n we cant measure the water`` ``// using the jugs`` ``if` `(d > n)`` ``return` `-1;` ` ``// If gcd of n and m does not divide d`` ``// then solution is not possible`` ``if` `((d % gcd(n,m)) != 0)`` ``return` `-1;` ` ``// Return minimum two cases:`` ``// a) Water of n liter jug is poured into`` ``// m liter jug`` ``// b) Vice versa of "a"`` ``return` `min(pour(n,m,d), ``// n to m`` ``pour(m,n,d)); ``// m to n``}` `// Driver code to test above``int` `main()``{`` ``int` `n = 3, m = 5, d = 4;` ` ``printf``(``"Minimum number of steps required is %d"``,`` ``minSteps(m, n, d));` ` ``return` `0;``}`
## Java
`// Java program to count minimum number of``// steps required to measure d litres water``// using jugs of m liters and n liters capacity.``import` `java.io.*;` `class` `GFG{` `// Utility function to return GCD of 'a'``// and 'b'.``public` `static` `int` `gcd(``int` `a, ``int` `b)``{`` ``if` `(b == ``0``)`` ``return` `a;`` ` ` ``return` `gcd(b, a % b);``}` `/* fromCap -- Capacity of jug from which`` ``water is poured`` ``toCap -- Capacity of jug to which`` ``water is poured`` ``d -- Amount to be measured */``public` `static` `int` `pour(``int` `fromCap, ``int` `toCap,`` ``int` `d)``{`` ` ` ``// Initialize current amount of water`` ``// in source and destination jugs`` ``int` `from = fromCap;`` ``int` `to = ``0``;` ` ``// Initialize count of steps required`` ``int` `step = ``1``; ``// Needed to fill "from" Jug` ` ``// Break the loop when either of the two`` ``// jugs has d litre water`` ``while` `(from != d && to != d)`` ``{`` ` ` ``// Find the maximum amount that can be`` ``// poured`` ``int` `temp = Math.min(from, toCap - to);` ` ``// Pour "temp" liters from "from" to "to"`` ``to += temp;`` ``from -= temp;` ` ``// Increment count of steps`` ``step++;` ` ``if` `(from == d || to == d)`` ``break``;` ` ``// If first jug becomes empty, fill it`` ``if` `(from == ``0``)`` ``{`` ``from = fromCap;`` ``step++;`` ``}` ` ``// If second jug becomes full, empty it`` ``if` `(to == toCap)`` ``{`` ``to = ``0``;`` ``step++;`` ``}`` ``}`` ``return` `step;``}` `// Returns count of minimum steps needed to``// measure d liter``public` `static` `int` `minSteps(``int` `m, ``int` `n, ``int` `d)``{`` ` ` ``// To make sure that m is smaller than n`` ``if` `(m > n)`` ``{`` ``int` `t = m;`` ``m = n;`` ``n = t;`` ``}` ` ``// For d > n we cant measure the water`` ``// using the jugs`` ``if` `(d > n)`` ``return` `-``1``;` ` ``// If gcd of n and m does not divide d`` ``// then solution is not possible`` ``if` `((d % gcd(n, m)) != ``0``)`` ``return` `-``1``;` ` ``// Return minimum two cases:`` ``// a) Water of n liter jug is poured into`` ``// m liter jug`` ``// b) Vice versa of "a"`` ``return` `Math.min(pour(n, m, d), ``// n to m`` ``pour(m, n, d)); ``// m to n``}` `// Driver code``public` `static` `void` `main(String[] args)``{`` ``int` `n = ``3``, m = ``5``, d = ``4``;` ` ``System.out.println(``"Minimum number of "` `+`` ``"steps required is "` `+`` ``minSteps(m, n, d));``}``}` `// This code is contributed by RohitOberoi`
## Python3
`# Python3 implementation of program to count``# minimum number of steps required to measure``# d litre water using jugs of m liters and n``# liters capacity.``def` `gcd(a, b):`` ``if` `b``=``=``0``:`` ``return` `a`` ``return` `gcd(b, a``%``b)` `''' fromCap -- Capacity of jug from which`` ``water is poured`` ``toCap -- Capacity of jug to which`` ``water is poured`` ``d -- Amount to be measured '''``def` `Pour(toJugCap, fromJugCap, d):` ` ``# Initialize current amount of water`` ``# in source and destination jugs`` ``fromJug ``=` `fromJugCap`` ``toJug ``=` `0` ` ``# Initialize steps required`` ``step ``=` `1`` ``while` `((fromJug ``is` `not` `d) ``and` `(toJug ``is` `not` `d)):` ` ` ` ``# Find the maximum amount that can be`` ``# poured`` ``temp ``=` `min``(fromJug, toJugCap``-``toJug)` ` ``# Pour 'temp' liter from 'fromJug' to 'toJug'`` ``toJug ``=` `toJug ``+` `temp`` ``fromJug ``=` `fromJug ``-` `temp` ` ``step ``=` `step ``+` `1`` ``if` `((fromJug ``=``=` `d) ``or` `(toJug ``=``=` `d)):`` ``break` ` ``# If first jug becomes empty, fill it`` ``if` `fromJug ``=``=` `0``:`` ``fromJug ``=` `fromJugCap`` ``step ``=` `step ``+` `1` ` ``# If second jug becomes full, empty it`` ``if` `toJug ``=``=` `toJugCap:`` ``toJug ``=` `0`` ``step ``=` `step ``+` `1`` ` ` ``return` `step` `# Returns count of minimum steps needed to``# measure d liter``def` `minSteps(n, m, d):`` ``if` `m> n:`` ``temp ``=` `m`` ``m ``=` `n`` ``n ``=` `temp`` ` ` ``if` `(d``%``(gcd(n,m)) ``is` `not` `0``):`` ``return` `-``1`` ` ` ``# Return minimum two cases:`` ``# a) Water of n liter jug is poured into`` ``# m liter jug`` ``return``(``min``(Pour(n,m,d), Pour(m,n,d)))` `# Driver code``if` `__name__ ``=``=` `'__main__'``:` ` ``n ``=` `3`` ``m ``=` `5`` ``d ``=` `4` ` ``print``(``'Minimum number of steps required is'``,`` ``minSteps(n, m, d))`` ` `# This code is contributed by Sanket Badhe`
## Javascript
``
Output:
`Minimum number of steps required is 6`
Time Complexity: O(N + M)
Auxiliary Space: O(1)
Another detailed Explanation:http://web.mit.edu/neboat/Public/6.042/numbertheory1.pdf
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# The vector equation of a plane
Only x- and y- directed vectors can cause the wheel to rotate when the wheel is in the x-y plane. Example Find an equation of the plane passing through the points P 1,-1,3Q 4,1,-2and R -1,-1,1.
The distance D between a plane and a point P2 becomes; The numerator part of the above equation, is expanded; Finally, we put it to the previous equation to complete the distance formula; Note that the distance formula looks like inserting P2 into the plane equation, then dividing by the length of the normal vector.
Momentum can be defined as "mass in motion. A team that has the momentum is on the move and is going to take some effort to stop. Then, from the figure above, the distance D from the point to the plane is the scalar projection of the vector r onto the normal vector n: The terms such as: The direction of the momentum vector is the same as the direction of the velocity of the ball.
Let the 3x3 matrix A, and the intesection point is computed by Cramer's rule; where the determinants are; Let's calculate first. Here is a sketch of all these vectors. These are the symmetric equations of the line. More generally, for any plane through the origin, if we fix two non-parallel vectors in that plane, the position vector of any point in the plane is a sum of scalar multiples of those two vectors.
This formular will be used for 2-plane intersection. Two distinct planes are either parallel or they intersect in a line. The two vectors serve as direction vectors for the plane. As discussed in an earlier unit, a vector quantity is a quantity that is fully described by both magnitude and direction.
Line through 3, 4 and -6, This approach is known as the Angular spectrum method. In order to find the intersection point P x, y, zwe solve the linear system of 3 planes. Or, in matrix form; Therefore, solving the linear system is finding the inverse matrix. This choice will be explained in the Normal Vector section.
Therefore, the third plane equation becomes or. This second form is often how we are given equations of planes. Point-normal form and general form of the equation of a plane[ edit ] In a manner analogous to the way lines in a two-dimensional space are described using a point-slope form for their equations, planes in a three dimensional space have a natural description using a point in the plane and a vector orthogonal to it the normal vector to indicate its "inclination".
In the case of the circularly polarized light, the field strength remains constant from plane to plane but its direction steadily changes in a rotary type manner.
Since Q is on the line, its coordinates satisfy the equation: Two distinct planes are either parallel or they intersect in a line. This is consistent with the equation for momentum.
Two distinct but intersecting lines. And in what direction is it? Notice that it's very much like the vector equation of a line, except that it has two direction vectors instead of one.
A line and a point not on that line. Hence, the z-directed vector fields can be ignored for determining the z-component of the curl. If the line is parallel to the plane then any vector parallel to the line will be orthogonal to the normal vector of the plane.
In terms of an equation, the momentum of an object is equal to the mass of the object times the velocity of the object. So, the line and the plane are neither orthogonal nor parallel.Jan 16, · This video covers how to find the vector and parametric equations of a plane given a point and two vectors "in the plane." Works just as well with three points in the plane.
In Equation [3], is a unit vector in the +x-direction, is a unit vector in the +y-direction, and is a unit vector in the +z-direction (a unit vector is a vector with a magnitude equal to 1). The terms such as. The equation of a plane in 3D space is defined with normal vector (perpendicular to the plane) and a known point on the plane.
Let the normal vector of a plane, and the known point on the plane, P agronumericus.com, let any point on the plane as P. A vector n 0 parallel to this normal is called a normal vector for the plane. There is a unique plane which passes through P 0 and has n as a normal vector.
Now P lies in the plane through P 0 perpendicular to n if and only if and n are perpendicular. Momentum as a Vector Quantity. Momentum is a vector agronumericus.com discussed in an earlier unit, a vector quantity is a quantity that is fully described by both magnitude and direction.
To fully describe the momentum of a 5-kg bowling ball moving westward at 2 m/s, you must include information about both the magnitude and the direction of the bowling ball. Equation of a Plane - 3 Points Main Concept A plane can be defined by four different methods: A line and a point not on the line Three non-collinear points (three points not on a line) A point and a normal vector Two intersecting lines Two parallel and.
The vector equation of a plane
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# Time and Work – Quiz 3
Home > > Tutorial
5 Steps - 3 Clicks
# Time and Work – Quiz 3
### Introduction
Time and Work is one of important topic in Quantitative Aptitude Section. In Time and Work – Quiz 3 article candidates can find a question with an answer. By solving this question candidates can improve and maintain, speed, and accuracy in the exams. Time and Work – Quiz 3 questions are very useful for different exams such as IBPS PO, Clerk, SSC CGL, SBI PO, NIACL Assistant, NICL AO, IBPS SO, RRB, Railways, Civil Services etc.
### Q1
10 women can complete a work in 7 days and 10 children take 14 days to complete the work. How many days will 5 women and 10 children take to complete the work?
A. 3 B. 5 C. 7 D. 8
C
1 woman's 1 day's work = $\frac{1}{70}$
1 child's 1 day's work = $\frac{1}{140}$
(5 women + 10 children)'s day's work = ($\frac{5}{70}$ + $\frac{10}{140}$) = ($\frac{1}{14}$ + $\frac{1}{14}$) = $\frac{1}{7}$
5 women and 10 children will complete the work in 7 days.
### Q2
X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last?
A. 6 days B. 10 days C. 15 days D. 20 days
B
Work done by X in 4 days = ($\frac{1}{20} \times 4$) = $\frac{1}{5}$
Remaining work = ( 1 - $\frac{1}{20}$) = $\frac{4}{5}$
(X + Y)'s 1 day's work = ($\frac{1}{20}$ + $\frac{1}{12}$) = $\frac{8}{60}$ = $\frac{2}{15}$
Now, $\frac{2}{15}$ work is done by X and Y in 1 day.
So, $\frac{4}{5}$ work will be done by X and Y in ($\frac{15}{2}$$\frac{4}{5}$) = 6 days
Hence, total time taken = (6 + 4) days = 10 days.
### Q3
A is 30% more efficient than B. How much time will they, working together, take to complete a job which A alone could have done in 23 days?
A. 11 days B. 13 days C. 25 days D. 18 days
B
Ratio of times taken by A and B = 100 : 130 = 10 : 13.
Suppose B takes x days to do the work.
Then, 10 : 13 :: 23 : x $\Rightarrow x = \frac{23 \times 13}{10} \Rightarrow x = \frac{299}{10}$
A's 1 day's work = $\frac{1}{23}$;
B's 1 day's work = $\frac{10}{299}$;
(A + B)'s 1 day's work = ($\frac{1}{23}$ + $\frac{10}{299}$) = $\frac{23}{299}$ = $\frac{1}{13}$
Therefore, A and B together can complete the work in 13 days.
### Q4
Sakshi can do a piece of work in 20 days. Tanya is 25% more efficient than Sakshi. The number of days taken by Tanya to do the same piece of work is:
A. 15 B. 16 C. 18 D. 25
B
Ratio of times taken by Sakshi and Tanya = 125 : 100 = 5 : 4.
Suppose Tanya takes x days to do the work.
5 : 4 :: 20 : x $\Rightarrow \frac{4 \times 20}{5}$
x = 16 days.
$\Rightarrow$ Hence, Tanya takes 16 days to complete the work.
### Q5
A, B and C can complete a piece of work in 24, 6 and 12 days respectively. Working together, they will complete the same work in:
A. 1/24 days B. 7/24 days C. 337 days D. 4 days
B
If A can do a piece of work in n days, then A's 1 day's work = $\frac{1}{n}$
(A + B + C)'s 1 day's work = ($\frac{1}{24} + \frac{1}{6} + \frac{1}{12}$) = $\frac{7}{24}$ days
|
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Transcript
```Section 2.4
Working with Summary Statistics
Measures of Center
Mean, Median, and sometimes Mode
Standard Deviation, and Quartiles (Q1 and Q3)
Remember SD gives an “average” deviation from
the mean.
The quartiles divide the data into 25% portions.
Lets say we know the mean value of the
homes in a community along with the total
number of homes: \$213,500; 412 homes
We also know the tax rate: 1.5%
How can we use the mean to determine the
total tax dollars received by the community?
The mean is used to represent the value of every home.
When we describe the center of the annual
income of a group of people, it is typical to use
There are typically a large group of people
clustered around the low end of the scale with a
few having very large incomes. This creates a
distribution that is…..
Skewed right and therefore the mean gives a measure
that is higher than expected. The median filters out
these extreme values.
Create a dot plot of the following data:
City
Country
Temperature (F)
Algiers
Bangkok
Nairobi
Sao Paulo
Warsaw
Ethiopia
Algeria
Thailand
Spain
Kenya
Brazil
Poland
32
32
50
14
41
32
-22
Now create a dot plot of the distance the
temperature is from freezing (32o). Positive if
above freezing, negative if below.
Recentering a set of data is when we add or
subtract a constant from each data value.
This shifts the data on the number scale, but does
nothing to change the shape or spread.
The mean will be shifted by the constant added or
subtracted.
Now use the same data and convert it to celsius.
Simply multiply the degrees above or below freezing
by 1/1.8.
What happened to your data set?
Shape
Mean?
Standard deviation?
Notice that the mean and SD are multiplied by
1/1.8, but the shape stays the same.
This simply shrinks or if by a number greater
than 1, stretches the distribution.
A summary statistic is resistant to outliers if it is
not changed very much when the outlier is
removed from the data.
A summary statistic is sensitive to outliers if it is
changed significantly when the outlier is
removed from the data.
Remember our discussion of Mean vs Median
Refer to page 77: example of television viewers
Percentile: If a value is at the kth percentile, then k% of
the data is at or lower than this value.
Example: You got a 32 on math portion of the ACT. You
are told this is the 86th percentile.
That means 86% of the test takers scored at a 32 or
lower.
It also means that 14% scored above a 32.
This is a measure of where a data value lies
within the data set.
Frequency plot where the plotted points
show you the accumulated percent of data up
to that point.
Example: page 78
Page 80 E47, 49, 53 - 56
Re-Centering happens when you…..??
What happens to the shape?
What happens to the center?
What happens to the spread?
Re-Scaling happens when you….??
What happens to the shape?
What happens to the center?
What happens to the spread?
```
Related documents
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## Ratios, Proportions, Indices and Logarithms Summary for CA Foundation November 2020 Exams
### RATIOS
1. A ratio is a comparison of the sizes of two or more quantities of the same kind by division
2. If a and b are two quantities of the same kind (in same units), then the fraction a/b is called the ratio of a to b. It is written as a : b. Thus, the ratio of a to b = a/b or a : b.
3. The quantities a and b are called the terms of the ratio, a is called the first term or antecedent and b is called the second term or consequent.
4. The ratio compounded of the two ratios a : b and c : d is ac : bd.
5. A ratio compounded of itself is called its duplicate ratio. a2 : b2 is the duplicate ratio of a b. Similarly, the triplicate ratio of a : b is a3 : b3.
6. For any ratio a : b, the inverse ratio is b : a
7. The sub-duplicate ratio of a : b is a : b and the sub-triplicate ratio of a : b is a1/3: b1/3.
8. Continued Ratio is the relation (or compassion) between the magnitudes of three or more Quantities of the same kind. The continued ratio of three similar quantities a, b, c is written as a : b : c.
### PROPORTIONS
1. p : q = r : s => q : p = s : r (Invertendo)(p/q = r/s) => (q/p = s/r)
2. a : b = c : d => a : c = b : d (Alternendo)(a/b = c/d) => (a/c = b/d)
3. a : b = c : d => a + b : b = c + d : d (Componendo)(a/b = c/d) => (a + b)/b = (c + d)/d
4. a : b = c : d => a – b : b = c – d : d (Dividendo)(a/b = c/d) => (a – b)/b = (c – d)/d
5. a : b = c : d => a + b : a – b = c + d : c – d (Componendo & Dividendo)(a + b)/(a – b) = (c + d)/(c – d)
6. a : b = c : d = a + c : b + d (Addendo)
7. (a/b = c/d = a + c/b + d)
8. a : b = c : d = a – c : b – d (Subtrahendo)(a/b = c/d = a – c/b – d)
9. If a : b = c : d = e : f = ………… then each of these ratios = (a – c – e – …….) : (b – d – f – …..)
10. The quantities a, b, c, d are called terms of the proportion; a, b, c and d are called its first, second, third and fourth terms respectively. First and fourth terms are called extremes (or extreme terms). Second and third terms are called means (or middle terms).
11. If a : b = c : d are in proportion then a/b = c/d i.e. ad = bc i.e. product of extremes =product of means. This is called cross product rule.
12. Three quantities a, b, c of the same kind (in same units) are said to be in continuousproportion
13. if a : b = b : c i.e. a/b = b/c i.e. b2 = ac
14. If a, b, c are in continuous proportion, then the middle term b is called the mean proportional between a and c, a is the first proportional and c is the third proportional.
15. Thus, if b is mean proportional between a and c, then b2 = ac i.e. b = ac.
### INDICES
1. am × an = am + n (base must be same), Ex. 23 × 22 = 23 + 2 = 25
2. am × an = am–n , Ex. 25 × 23 = 25 – 3 = 22
3. (am) n = amn Ex. (25) 2 = 25 × 2 = 210
4. ao = 1 , Ex. 20 = 1, 30 = 1
5. a–m = 1/am and 1/a–m = am ,Ex. 2–3 = 1/2and 1/2–5 = 25
6. If ax = ay, then x=y
7. If xa = ya , then x=y
8. m√ a = a1/m , √x = x½ , √4 = (2)1/2 = 21/2 x 2 = 2 , Ex. 3√8 = 81/3 = (23)1/3 = 23×1/3 = 2
### LOGARITHMS
1. logamn = logam + logan , Ex. log (2 × 3) = log 2 + log 3
2. loga(m/n) = logam – logan , Ex. log (3/2) = log3 – log2
3. logamn = n logam , Ex. log 23 = 3 log 2
4. logaa = 1, a = 1 , Ex. log1010 = 1, log22 = 1, log33 = 1 etc.
5. loga1 = 0 , Ex. log21 = 0, log101 = 0 etc.
6. logba × logab = 1 , Ex. log32 × log23 = 1
7. logba × logcb = logca , Ex. log32 × log53 = log52
8. logba = log a/log b , Ex. log32 = log2/log3
9. logba = 1/logab
10. a logax = x (Inverse logarithm Property)
11. The two equations ax= n and x = logan are only transformations of each other and should be remembered to change one form of the relation into the other.Since a= a, logaa = 1
Notes:
(A) If base is understood, base is taken as 10
(B) Thus log 10 = 1, log 1 = 0
(C) Logarithm using base 10 is called Common logarithm and logarithm using base e is called Natural logarithm {e = 2.33 (approx.) called exponential number}.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 10.5: Surface Area of Pyramids and Cones
Difficulty Level: At Grade Created by: CK-12
## Introduction
Painting a Pyramid
After about a week of working at the wrapping station, both Candice and Trevor were beginning to feel at ease. Neither one of them thought that there was a package that was too difficult to wrap. They had wrapped lots of regular shaped boxes and also some odd shaped things too.
Then one Saturday morning, an interesting dilemma came their way. A girl came up to the booth with a box in one hand.
“I wonder if you can help me,” she said. “I bought my brother a model pyramid and it needs paint, but the paint isn’t included. I need to know how much paint I’ll need.”
“Well, to figure that out, you need the dimensions of the pyramid to calculate the amount of surface area that the paint will need to cover. Let me see the box and I’ll see if I can help you,” Candice said taking the box from the girl.
The box said that once the pyramid is built, it will have a base of 7.7 cm, a height of 6.2 cm and a slant height of 7.3 cm.
“Okay, I know what to do,” Candice said smiling, as she took out a piece of a paper and a calculator.
Do you know what to do? How can you use these measurements to find the surface area of the pyramid?
In this lesson you will learn how to find the surface area of a pyramid. Learn as you go through this lesson and at the end you can check out how Candice finds the surface area of the pyramid model.
What You Will Learn
In this lesson, you will learn how to demonstrate the following skills.
• Recognize surface area of pyramids or cones as the sum of the areas of faces using nets.
• Find surface areas of pyramids using formulas.
• Find surface areas of cones using formulas.
• Solve real-world problems involving surface areas of pyramids or cones.
Teaching Time
I. Recognize Surface Area of Pyramids or Cones as the Sum of the Areas of Faces Using Nets
In this lesson we will learn to find the surface area of pyramids and cones. Pyramids and cones are solid shapes that exist in three-dimensional space. A pyramid has sides that are triangular faces and a base. The base can be any shape. Let’s look at some pyramids.
Like pyramids, cones have a base and a point at the top. However, cones always have a circular base.
Surface area is the sum of all of the areas of the faces in a solid figure. Imagine you could wrap a pyramid or cone in wrapping paper, like a present. The amount of wrapping paper needed to cover the figure represents its surface area. To find the surface area, we must be able to calculate the area of each face and then add these areas together.
One way to find the surface area of a three-dimensional figure is using a net. A net is a two-dimensional diagram of a three-dimensional figure. Imagine you could unfold a pyramid so that it is completely flat. It might look something like this net.
With a net, we can see each face of the pyramid more clearly. To find the surface area, we need to calculate the area for each face in the net, the sides and the base.
The side faces of a pyramid are always triangles, so we use the area formula for triangles to calculate their area: A=12bh\begin{align*}A = \frac{1}{2} bh\end{align*}.
The area of the base depends on what shape it is. Remember, pyramids can have bases in the shape of a triangle, square, rectangle, or any other polygon. We use whichever area formula is appropriate for the shape. Here are some common formulas to find area.
rectangle:A=lwsquare:A=s2triangle: A=12bh\begin{align*}\text{rectangle}: \quad \qquad A = lw\!\\ \text{square}: \qquad \qquad A = s^2\!\\ \text{triangle}: \quad \qquad \ A = \frac{1}{2} bh\end{align*}
Take a look at the pyramid below. Which formula should we use to find the area of the base?
The base of the pyramid is a square with sides of 6 centimeters, so we should use the area formula for a square: A=s2\begin{align*}A = s^2\end{align*}. Now we can find the area of the base.
AA=62=36 sq.cm\begin{align*}A & = 6^2 \\ A&=36 \ sq.cm \end{align*}
Next, we need to find the area of one of the triangles. There are four triangles, but we can work by finding the area of one of the triangles. We use the formula A=12bh\begin{align*}A= \frac{1}{2} bh\end{align*}. The base width is 6 cm and the slant height, the height of the side is 4 cm.
AAA=12(6)(4)=12(24)=12 sq.cm\begin{align*}A & = \frac{1}{2} (6)(4) \\ A & = \frac{1}{2}(24) \\ A & = 12 \ sq.cm \end{align*}
There are four triangles, so we can take this number and multiply it by four and then add it to the area of the square.
SASASA=4(12)+36=48+36=84 sq.cm\begin{align*}SA & =4(12)+ 36 \\ SA & =48+36 \\ SA & =84 \ sq.cm \end{align*}
Cones have different nets. Imagine you could unroll a cone.
The shaded circle is the base. Remember, cones always have circular bases. The unshaded portion of the cone represents its side. Technically we don’t call this a face because it has a round edge.
To find the surface area of a cone, we need to calculate the area of the circular base and the side and add them together.
The formula for finding the area of a circle is A=πr2\begin{align*}A = \pi r^2\end{align*}, where r\begin{align*}r\end{align*} is the radius of the circle. We use this formula to find the area of the circular base.
The side of the cone is actually a piece of a circle, called a sector. The size of the sector is determined by the ratio of the cone’s radius to its slant height, or rs\begin{align*}\frac{r}{s}\end{align*}.
To find the area of the sector, we take the area of the portion of the circle.
A=πr2sr\begin{align*}A = \pi r^2 \cdot \frac{s}{r}\end{align*}
This simplifies to πrs\begin{align*}\pi rs\end{align*}.
To find the area of the cone’s side, then, we multiply the radius, the slant height, and pi.
This may seem a little tricky, but as you work through a few examples you will see that this becomes easier as you go along.
Example
What is the surface area of the figure below?
Now that we have the measurements of the sides of the cone, let’s calculate the area of each. Remember to use the correct area formula.
bottom faceA=πr2π(5)225π78.5sideA=πrsπ(5)(11)π(55)55π172.7\begin{align*}& \text{bottom face} && \text{side}\\ & A = \pi r^2 && A = \pi rs\\ & \pi (5)^2 && \pi (5) (11)\\ & 25 \pi && \pi (55)\\ & 78.5 && 55 \pi\\ & && 172.7\end{align*}
We know the area of each side of the cone when we approximate pi as 3.14. Now we can add these together to find the surface area of the entire cone.
bottom faceside surface area78.5 + 172.7 = 251.2 in.2\begin{align*}& \text{bottom face} \qquad \quad \text{side} \qquad \quad \ \text{surface area}\\ & 78.5 \qquad \quad \ \ + \quad \ 172.7 \ \ = \ \ \ 251.2 \ in.^2\end{align*}
Now we can look at what we did to solve this problem. We used the formula A=πr2\begin{align*}A = \pi r^2\end{align*} to find the area of the circular base. Then we found the area of the side by multiplying πrs\begin{align*}\pi rs\end{align*}. When we add these together, we get a surface area of 251.2 square inches for this cone.
When you learned how to find the surface areas of other three-dimensional figures, we discussed using a formula to simplify our work. Cones and pyramids aren’t any different. Let’s look at simplifying our work with formulas.
II. Find Surface Areas of Pyramids using Formulas
Nets let us see each face of a pyramid so that we can calculate its area. However, we can also use a formula to represent the faces as we find their area. The formula is like a short cut, because we can put the measurements in for the appropriate variable in the formula and solve for SA\begin{align*}SA\end{align*}, surface area.
Here is the formula for finding the surface area of a pyramid.
SA=12 perimeter×slant height+B\begin{align*}SA = \frac{1}{2} \ \text{perimeter} \times \text{slant height} + B\end{align*}
The first part of the formula, 12 perimeter×slant height\begin{align*}\frac{1}{2} \ \text{perimeter} \times \text{slant height}\end{align*}, is a quick way of finding the area of all of the triangular sides of the pyramid at once. Remember, the area formula for a triangle is A=12bh\begin{align*}A = \frac{1}{2} bh\end{align*}. In the formula, b\begin{align*}b\end{align*} stands for base. The perimeter of the pyramid’s bottom face represents all of the bases of the triangular faces at once, because it’s their sum. The height of each triangle is always the same, so we can just call this the slant height of the pyramid.
Therefore “12 perimeter×slant height\begin{align*}\frac{1}{2} \ \text{perimeter} \times \text{slant height}\end{align*}” is really the same as 12bh\begin{align*}\frac{1}{2} bh\end{align*}.
The B\begin{align*}B\end{align*} in the formula represents the base’s area. Remember, pyramids can have bases of different shapes, so the area formula we use to find B\begin{align*}B\end{align*} varies. We find the base’s area first and then put it into the formula in place of B\begin{align*}B\end{align*}. Let’s give it a try.
Example
What is the surface area of the pyramid below?
This is a square pyramid. The four sides of the base are all 8 inches, so the perimeter of the base is 8×4=32 inches\begin{align*}8 \times 4 = 32 \ inches\end{align*}. We also know we will need to use the area formula for squares to find B\begin{align*}B\end{align*}, the base’s area.
BBB=s2=(8)2=64 in.2\begin{align*}B & = s^2\\ B & = (8)^2\\ B & = 64 \ in.^2\end{align*}
Now we have all the information that we need. We can put it into the formula and solve for SA\begin{align*}SA\end{align*}, surface area.
SASASASASA=12 perimeter×slant height+B=[12(32)×3]+64=(16×3)+64=48+64=112 in.2\begin{align*}SA & = \frac{1}{2} \ \text{perimeter} \times \text{slant height} + B\\ SA & = \left [ \frac{1}{2} (32) \times 3 \right ] + 64\\ SA & = (16 \times 3) + 64\\ SA & = 48 + 64\\ SA & = 112 \ in.^2\end{align*}
The surface area of the pyramid is 112 square inches.
Example
Find the surface area of the figure below.
First of all, what kind of pyramid is this? It is a triangular pyramid because its base is a triangle. That means we need to use the area formula for triangles to find B\begin{align*}B\end{align*}. The base’s sides are all the same length, so we can calculate the perimeter by multiplying 16×3=48\begin{align*}16 \times 3 = 48\end{align*}. Now let’s find B\begin{align*}B\end{align*}.
BBBB=12bh=12(16)(13.86)=8(13.86)=110.88 cm2\begin{align*}B & = \frac{1}{2} bh\\ B & = \frac{1}{2} (16) (13.86)\\ B & = 8 (13.86)\\ B & = 110.88 \ cm^2\end{align*}
Now we’re ready to put all of the information into the formula. Let’s see what happens.
SASASASASA=12 perimeter×slant height+B=[12(48)×6]+110.88=(24×6)+110.88=144+110.88=254.88 cm2\begin{align*}SA & = \frac{1}{2} \ \text{perimeter} \times \text{slant height} + B \\ SA & = \left [ \frac{1}{2} (48) \times 6 \right ] + 110.88\\ SA & = (24 \times 6) + 110.88\\ SA & = 144 + 110.88\\ SA & = 254.88 \ cm^2\end{align*}
The surface area of this triangular pyramid is 254.88 square centimeters.
Write this formula down in your notebook.
Try a few of these on your own.
10G. Lesson Exercises
Find the surface area of each pyramid.
1. A square pyramid with side of 6 cm, slant height of 5 cm
2. A rectangular pyramid with a length of 6 in, a width of 4 in and a slant height of 3 in
III. Find Surface Areas of Cones Using Formulas
Cones have a different formula because they have a circular base. But the general idea is the same. The formula is a short cut to help us combine the area of the circular base and the area of the cone’s side. Here’s what the formula looks like.
SA=πr2+πrs\begin{align*}SA = \pi r^2 + \pi rs\end{align*}
The first part of the formula, πr2\begin{align*}\pi r^2\end{align*}, is simply the area formula for circles. This represents the base area. The second part, as we have seen, represents the area of the cone’s side. We simply put the pieces together and solve for the area of both parts at once. Let’s try it out.
Example
What is the surface area of the cone?
We know that the radius of this cone is 3 inches and the slant height is 9 inches. We simply put these values in for r\begin{align*}r\end{align*} and s\begin{align*}s\end{align*} in the formula and solve for SA\begin{align*}SA\end{align*}, surface area.
SASASASASA=πr2+πrs=π(32)+π(3)(9)=9π+27π=36π=113.04 in.2\begin{align*}SA & = \pi r^2 + \pi rs\\ SA & = \pi (3^2) + \pi (3)(9)\\ SA & = 9 \pi + 27 \pi\\ SA & = 36 \pi\\ SA & = 113.04 \ in.^2\end{align*}
This cone has a surface area of 113.04 square inches.
Example
A cone has a radius of 2.5 meters and a slant height of 7.5 meters. What is its surface area?
This time we have not been given a picture of the cone, so we’ll need to read the problem carefully. It tells us the radius and the slant height of the cone, though, so we can put these numbers in for r\begin{align*}r\end{align*} and s\begin{align*}s\end{align*} and solve.
SASASASASA=πr2+πrs=π(2.52)+π(2.5)(7.5)=6.25π+18.75π=25π=78.5 in.2\begin{align*}SA & = \pi r^2 + \pi rs\\ SA & = \pi (2.5^2) + \pi (2.5)(7.5)\\ SA & = 6.25 \pi + 18.75 \pi\\ SA & = 25 \pi\\ SA & = 78.5 \ in.^2\end{align*}
This cone has a surface area of 78.5 square inches.
Take a few minutes to write this formula down in your notebooks.
Now try a few of these on your own.
10H. Lesson Exercises
Find the surface area of each cone.
1. Radius = 4 in, slant height = 6 in
2. Radius = 3.5 in, slant height = 5 in
3. Radius = 5 m, slant height = 7 m
Take a few minutes to check your work with a friend.
You’re right! Just be sure to put the numbers in the correct place in the formula!
IV. Solve Real-World Problems Involving Surface Areas of Pyramids or Cones
We have learned two ways to find the surface area of pyramids and cones: drawing a net or using a formula. We can use either of these methods to solve word problems involving surface area. Nets may be especially useful if the problem does not provide an image of the figure. If you choose to use a formula, be sure you know whether the problem deals with a cone or a pyramid, and what the shape of the pyramid’s base is. Let’s practice using what we have learned.
Example
Jessica is decorating conical party hats for her party by wrapping them in colored tissue paper. Each hat has a radius of 4 centimeters and a slant height of 8 centimeters. If she wants to wrap 6 party hats, how much paper will she need?
This problem involves a cone. It does not include a picture, so it may help to draw a net. In your drawing, label the radius and the slant height of the cone. We can also use the formula. We simply put the radius and slant height in for the appropriate variables in the formula and solve for SA\begin{align*}SA\end{align*}.
SASASASASA=πr2+πrs=π(4)2+π(4)(8)=16π+32π=48π=150.72 cm2\begin{align*}SA & = \pi r^2 + \pi rs\\ SA & = \pi (4)^2 + \pi (4) (8)\\ SA & = 16 \pi + 32 \pi\\ SA & = 48 \pi\\ SA & = 150.72 \ cm^2\end{align*}
Jessica will need 150.72 square centimeters of tissue paper to cover one hat.
But we’re not done yet! Remember, she wants to cover 6 party hats. We need to multiply the surface area of one hat by 6 to find the total amount of paper she needs: 150.72×6=904.32\begin{align*}150.72 \times 6 = 904.32\end{align*}.
Jessica will need 904.32 square centimeters of paper to cover all 6 hats.
Now let’s take what we have learned in this lesson and apply it to our work solving the problem from the introduction.
## Real–Life Example Completed
Painting a Pyramid
Here is the original problem once again. Reread it and underline any important information.
After about a week of working at the wrapping station, both Candice and Trevor were beginning to feel at ease. Neither one of them thought that there was a package that was too difficult to wrap. They had wrapped lots of regular shaped boxes and also some odd shaped things too.
Then one Saturday morning, an interesting dilemma came their way. A girl came up to the booth with a box in one hand.
“I wonder if you can help me,” she said. “I bought my brother a model pyramid and it needs paint, but the paint isn’t included. I need to know how much paint I’ll need.”
“Well, to figure that out, you need the dimensions of the pyramid to calculate the amount of surface area that the paint will need to cover. Let me see the box and I’ll see if I can help you,” Candice said taking the box from the girl.
The box said that once the pyramid was built, it would have a base of 7.7 cm, a height of 6.2 cm and a slant height of 7.3 cm.
“Okay, I know what to do,” Candice said smiling as she took out a piece of a paper and a calculator.
First, Candice needs to find the surface area of each of the four lateral faces. She can use this formula.
AAA=12bh=4[12(7.7)(7.3)]=112.42 sq.cm\begin{align*}A & = \frac{1}{2} bh\\ A & = 4 \left [ \frac{1}{2} (7.7)(7.3) \right ]\\ A & = 112.42 \ sq. cm\end{align*}
Now she can find the area of the base.
AAA=s2=7.7(7.7)=59.29 sq.cm\begin{align*}A & = s^2\\ A & = 7.7(7.7)\\ A & = 59.29 \ sq. cm\end{align*}
The surface area of the pyramid is 171.71 sq. cm.
“You need to find paint that stretches across 172 sq. cm. That will give him just enough. Try the hobby shop. They will help you,” Candice said writing the figure on a piece of paper as she handed it to the girl.
The girl smiled and thanked her. Candice felt good about her ability to use math to figure out the problem.
## Vocabulary
Here are the vocabulary words that are found in this lesson.
Pyramid
a three-dimensional object with a polygon for a base and triangles for sides that join at a single vertex.
Cone
a three-dimensional object with a circle as a base and a side that wraps around the base connecting at one vertex at the top.
Surface area
the measurement of the outer covering or surface of a three-dimensional figure.
## Technology Integration
1. http://www.mathplayground.com/mv_surface_area_cones.html – This is a Brightstorm video on finding the surface area of a cone.
2. http://www.mathplayground.com/mv_surface_area_pyramids.html – This is a Brightstorm video on finding the surface area of pyramids.
## Time to Practice
Directions: Find the surface area of each 4-sided pyramid. Remember that b\begin{align*}b\end{align*} means base and sh\begin{align*}sh\end{align*} means slant height.
1. b=4 in, sh=5 in\begin{align*}b = 4 \ in, \ sh = 5 \ in\end{align*}
2. b=4 in, sh=6 in\begin{align*}b = 4 \ in, \ sh = 6 \ in\end{align*}
3. b=6 in, sh=8 in\begin{align*}b = 6 \ in, \ sh = 8 \ in\end{align*}
4. b=5 in, sh=7 in\begin{align*}b = 5 \ in, \ sh = 7 \ in\end{align*}
5. b=7 m, sh=9 m\begin{align*}b = 7 \ m, \ sh = 9 \ m\end{align*}
6. b=8 m, sh=10 m\begin{align*}b = 8 \ m, \ sh = 10 \ m\end{align*}
7. b=9 cm, sh=11 cm\begin{align*}b = 9 \ cm, \ sh = 11 \ cm\end{align*}
8. b=11 m, sh=13 m\begin{align*}b = 11 \ m, \ sh = 13 \ m\end{align*}
9. b=6 in, sh=9 in\begin{align*}b = 6 \ in, \ sh = 9 \ in\end{align*}
10. b=10 cm, sh=12 cm\begin{align*}b = 10 \ cm, \ sh = 12 \ cm\end{align*}
Directions: Find the surface area of each cone. Remember that sh\begin{align*}sh\end{align*} means slant height and r\begin{align*}r\end{align*} means radius.
11. r=4 in, sh=5 in\begin{align*}r = 4 \ in, \ sh = 5 \ in\end{align*}
12. r=5 m, sh=7 m\begin{align*}r = 5 \ m, \ sh = 7 \ m\end{align*}
13. r=3 cm, sh=6 cm\begin{align*}r = 3 \ cm, \ sh = 6 \ cm\end{align*}
14. r=5 mm, sh=8 mm\begin{align*}r = 5 \ mm, \ sh = 8 \ mm\end{align*}
15. r=8 in, sh=10 in\begin{align*}r = 8 \ in, \ sh = 10 \ in\end{align*}
16. r=11 cm, sh=14 cm\begin{align*}r = 11 \ cm, \ sh = 14 \ cm\end{align*}
17. r=12 in, sh=16 in\begin{align*}r = 12 \ in, \ sh = 16 \ in\end{align*}
18. r=3.5 cm, sh=6 cm\begin{align*}r = 3.5 \ cm, \ sh = 6 \ cm\end{align*}
19. r=4.5 mm, sh=7 mm\begin{align*}r = 4.5 \ mm, \ sh = 7 \ mm\end{align*}
20. r=5.5 cm, sh=9 cm\begin{align*}r = 5.5 \ cm, \ sh = 9 \ cm\end{align*}
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# Checking Solutions to Equations
## Substitute values for variables to check equation solutions
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Evaluating Algebraic Expressions and Equations
Your summer landscaping job pays a fixed rate of $20 per job plus$4 an hour. How much total money would you would make if it takes you 3 hours to complete a single job?
### Guidance
You have probably seen letters in a mathematical expression, such as $3x-8$ . These letters, also called variables, represent an unknown number. One of the goals of algebra is to solve various equations for a variable. Typically, $x$ is used to represent the unknown number, but any letter can be used.
To evaluate an expression or equation, we would need to substitute in a given value for the variable and test it. In order for the given value to be true for an equation, the two sides of the equation must simplify to the same number.
#### Example A
Evaluate $2x^2-9$ for $x=-3$ .
Solution: We know that $2x^2-9$ is an expression because it does not have an equals sign. Therefore, to evaluate this expression, plug in -3 for $x$ and simplify using the Order of Operations.
$&2(-3)^2-9 \rightarrow (-3)^2=-3 \cdot -3=9\\&2(9)-9\\&18-9\\&9$
You will need to remember that when squaring a negative number, the answer will always be positive. There are three different ways to write multiplication: $2 \times 9, 2 \cdot 9$ , and 2(9).
#### Example B
Determine if $x=5$ is a solution to $3x-11=14$ .
Solution: Even though the directions are different, this problem is almost identical to Example A. However, this is an equation because of the equals sign. Both sides of an equation must be equal to each other in order for it to be true. Plug in 5 everywhere there is an $x$ . Then, determine if both sides are the same.
$& \ ?\\3(5)-11 &= 14\\15-11 & \neq 14\\4 &\neq 14$
Because $4 \ne 14$ , this is not a true equation. Therefore, 5 is not a solution.
#### Example C
Determine if $t=-2$ is a solution to $7t^2-9t-10=36$ .
Solution: Here, $t$ is the variable and it is listed twice in this equation. Plug in -2 everywhere there is a $t$ and simplify.
$& \ ?\\7(-2)^2-9(-2)-10 &= 36\\& \ ?\\7(4)+18-10 &=36\\& \ ?\\28+18-10 &= 36\\36 &= 36$
-2 is a solution to this equation.
Intro Problem Revisit Rewrite the sentence as an algebraic expression. $20 plus$4 an hour, would be $20+4h$ , where h equals the number of hours you work. Then, evaluate the expression for $h = 3$ .
$20+4(3)=20+12=32$
You will make a total of \$32 for this particular job.
### Guided Practice
1. Evaluate $s^3-5s+6$ for $s=4$ .
2. Determine if $a=-1$ is a solution to $4a-a^2+11=-2-2a$ .
1. Plug in 4 everywhere there is an $s$ .
$&4^3-5(4)+6\\&64-20+6\\&50$
2. Plug in -1 for $a$ and see if both sides of the equation are the same.
$& \ ? \\4(-1)-(-1)^2+11 &= -2-2(-1)\\& \ ? \\-4-1+11 &=-2+2\\6 & \neq 0$
Because the two sides are not equal, -1 is not a solution to this equation.
### Vocabulary
Variable
A letter used to represent an unknown value.
Expression
A group of variables, numbers, and operators.
Equation
Two expressions joined by an equal sign.
Solution
A numeric value that makes an equation true.
### Practice
Evaluate the following expressions for $x = 5$ .
1. $4x-11$
2. $x^2+8$
3. $\frac{1}{2}x+1$
Evaluate the following expressions for the given value.
1. $-2a+7; a=-1$
2. $3t^2-4t+5; t=4$
3. $\frac{2}{3}c-7; c=-9$
4. $x^2-5x+6; x=3$
5. $8p^2-3p-15; p=-2$
6. $m^3-1; m=1$
Determine if the given values are solutions to the equations below.
1. $x^2-5x+4=0; x=4$
2. $y^3-7=y+3; x=2$
3. $7x-3=4; x=1$
4. $6z+z-5=2z+12;z=-3$
5. $2b-5b^2+1=b^2;b=6$
6. $-\frac{1}{4}g+9=g+15;g=-8$
Find the value of each expression, given that $a=-1,b=2,c=-4$ , and $d=0$ .
1. $ab-c$
2. $b^2+2d$
3. $c+\frac{1}{2}b-a$
4. $b(a+c)-d^2$
For problems 20-25, use the equation $y^2+y-12=0$ .
1. Is $y = 4$ a solution to this equation?
2. Is $y = -4$ a solution to this equation?
3. Is $y = 3$ a solution to this equation?
4. Is $y = -3$ a solution to this equation?
5. Do you think there are any other solutions to this equation, other than the ones found above?
6. Challenge Using the solutions you found from problems 20-23, find the sum of these solutions and their product. What do you notice?
### Vocabulary Language: English
Equation
Equation
An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs.
Expression
Expression
An expression is a mathematical phrase containing variables, operations and/or numbers. Expressions do not include comparative operators such as equal signs or inequality symbols.
inequality
inequality
An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are $<$, $>$, $\le$, $\ge$ and $\ne$.
solution
solution
A solution to an equation or inequality should result in a true statement when substituted for the variable in the equation or inequality.
Variable
Variable
A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
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1 Linear Equations in Linear Algebra 1 3
• Slides: 31
1 Linear Equations in Linear Algebra 1. 3 VECTOR EQUATIONS © 2012 Pearson Education, Inc.
VECTOR EQUATIONS Vectors in § A matrix with only one column is called a column vector, or simply a vector. § An example of a vector with two entries is , where w 1 and w 2 are any real numbers. § The set of all vectors with 2 entries is denoted by (read “r-two”). © 2012 Pearson Education, Inc. Slide 1. 3 - 2
VECTOR EQUATIONS § The stands for the real numbers that appear as entries in the vector, and the exponent 2 indicates that each vector contains 2 entries. § Two vectors in are equal if and only if their corresponding entries are equal. § Given two vectors u and v in , their sum is the vector obtained by adding corresponding entries of u and v. § Given a vector u and a real number c, the scalar multiple of u by c is the vector cu obtained by multiplying each entry in u by c. © 2012 Pearson Education, Inc. Slide 1. 3 - 3
VECTOR EQUATIONS § Example 1: Given 4 u, and , and Solution: © 2012 Pearson Education, Inc. , find . , and Slide 1. 3 - 4
GEOMETRIC DESCRIPTIONS OF § Consider a rectangular coordinate system in the plane. Because each point in the plane is determined by an ordered pair of numbers, we can identify a geometric point (a, b) with the column vector. § So we may regard plane. © 2012 Pearson Education, Inc. as the set of all points in the Slide 1. 3 - 5
PARALLELOGRAM RULE FOR ADDITION § If u and v in are represented as points in the plane, then corresponds to the fourth vertex of the parallelogram whose other vertices are u, 0, and v. See the figure below. © 2012 Pearson Education, Inc. Slide 1. 3 - 6
VECTORS IN and § Vectors in are column matrices with three entries. § They are points in a three-dimensional coordinate space, with arrows from the origin. § If n is a positive integer, (read “r-n”) denotes the collection of all lists (or ordered n-tuples) of n real numbers, usually written as column matrices, such as . © 2012 Pearson Education, Inc. Slide 1. 3 - 7
ALGEBRAIC PROPERTIES OF § § The vector whose entries are all zero is called the zero vector and is denoted by 0. For all u, v, w in and all scalars c and d: (i) (iii) (iv) , where denotes (v) (vi) © 2012 Pearson Education, Inc. Slide 1. 3 - 8
LINEAR COMBINATIONS (vii) (viii) § Given vectors v 1, v 2, . . . , vp in and given scalars c 1, c 2, . . . , cp, the vector y defined by is called a linear combination of v 1, …, vp with weights c 1, …, cp. § The weights in a linear combination can be any real numbers, including zero. © 2012 Pearson Education, Inc. Slide 1. 3 - 9
LINEAR COMBINATIONS § Example 2: Let , and . Determine whether b can be generated (or written) as a linear combination of a 1 and a 2. That is, determine whether weights x 1 and x 2 exist such that ----(1) If vector equation (1) has a solution, find it. © 2012 Pearson Education, Inc. Slide 1. 3 - 10
LINEAR COMBINATIONS Solution: Use the definitions of scalar multiplication and vector addition to rewrite the vector equation , a 1 a 2 a 3 which is same as © 2012 Pearson Education, Inc. Slide 1. 3 - 11
LINEAR COMBINATIONS and. ----(2) § The vectors on the left and right sides of (2) are equal if and only if their corresponding entries are both equal. That is, x 1 and x 2 make the vector equation (1) true if and only if x 1 and x 2 satisfy the following system. ----(3) © 2012 Pearson Education, Inc. Slide 1. 3 - 12
LINEAR COMBINATIONS § To solve this system, row reduce the augmented matrix of the system as follows. § The solution of (3) is and. Hence b is a linear combination of a 1 and a 2, with weights and. That is, . © 2012 Pearson Education, Inc. Slide 1. 3 - 13
LINEAR COMBINATIONS § Now, observe that the original vectors a 1, a 2, and b are the columns of the augmented matrix that we row reduced: a 1 a 2 b § Write this matrix in a way that identifies its columns. ----(4) © 2012 Pearson Education, Inc. Slide 1. 3 - 14
Quiz 3 Suppose that , he set of he corresponding , equations © 2012 Pearson Education, Inc. corresponding augmented t matrix Slide 1. 3 - 15
LINEAR COMBINATIONS § A vector equation has the same solution set as the linear system whose augmented matrix is. ----(5) § In particular, b can be generated by a linear combination of a 1, …, an if and only if there exists a solution to the linear system corresponding to the matrix (5). © 2012 Pearson Education, Inc. Slide 1. 3 - 16
LINEAR COMBINATIONS § Definition: If v 1, …, vp are in , then the set of all linear combinations of v 1, …, vp is denoted by Span {v 1, …, vp} and is called the subset of spanned (or generated) by v 1, …, vp. That is, Span {v 1, . . . , vp} is the collection of all vectors that can be written in the form with c 1, …, cp scalars. © 2012 Pearson Education, Inc. Slide 1. 3 - 17
A GEOMETRIC DESCRIPTION OF SPAN {V} § Let v be a nonzero vector in. Then Span {v} is the set of all scalar multiples of v, which is the set of points on the line in through v and 0. See the figure below. © 2012 Pearson Education, Inc. Slide 1. 3 - 18
A GEOMETRIC DESCRIPTION OF SPAN {U, V} § If u and v are nonzero vectors in , with v not a multiple of u, then Span {u, v} is the plane in that contains u, v, and 0. § In particular, Span {u, v} contains the line in through u and 0 and the line through v and 0. See the figure below. © 2012 Pearson Education, Inc. Slide 1. 3 - 19
1 Linear Equations in Linear Algebra 1. 3 THE MATRIX EQUATION © 2012 Pearson Education, Inc.
MATRIX EQUATION § Definition: If A is an matrix, with columns a 1, …, and if x is in , then the product of A and x, denoted by Ax, is the linear combination of the columns of A using the corresponding entries in x as weights; that is, § Ax is defined only if the number of columns of A equals the number of entries in x. © 2012 Pearson Education, Inc. Slide 1. 4 - 21
MATRIX EQUATION § Example 1: For v 1, v 2, v 3 in , write the linear combination as a matrix times a vector. § Solution: Place v 1, v 2, v 3 into the columns of a matrix A and place the weights 3, , and 7 into a vector x. § That is, . © 2012 Pearson Education, Inc. Slide 1. 4 - 22
MATRIX EQUATION § Theorem 3: If A is an and if b is in matrix, with columns, , then the matrix equation has the same solution set as the vector equation , which, in turn, has the same solution set as the system of linear equations whose augmented matrix is. © 2012 Pearson Education, Inc. Slide 1. 4 - 23
EXISTENCE OF SOLUTIONS § § The equation has a solution if and only if b is a linear combination of the columns of A. Theorem 4: Let A be an matrix. Then the following statements are logically equivalent. That is, for a particular A, either they are all true statements or they are all false. a. For each b in , the equation has a solution. b. Each b in is a linear combination of the columns of A. c. The columns of A span. d. A has a pivot position in every row. © 2012 Pearson Education, Inc. Slide 1. 4 - 24
COMPUTATION OF Ax § Example 2: Compute Ax, where and . § Solution: From the definition, © 2012 Pearson Education, Inc. Slide 1. 4 - 25
COMPUTATION OF Ax ---(1) . § The first entry in the product Ax is a sum of products (a dot product), using the first row of A and the entries in x. © 2012 Pearson Education, Inc. Slide 1. 4 - 26
COMPUTATION OF Ax § That is, . § Similarly, the second entry in Ax can be calculated by multiplying the entries in the second row of A by the corresponding entries in x and then summing the resulting products. © 2012 Pearson Education, Inc. Slide 1. 4 - 27
ROW-VECTOR RULE FOR COMPUTING Ax § Likewise, the third entry in Ax can be calculated from the third row of A and the entries in x. § If the product Ax is defined, then the ith entry in Ax is the sum of the products of corresponding entries from row i of A and from the vertex x. § The matrix with 1 s on the diagonal and 0 s elsewhere is called an identity matrix and is denoted by I. § For example, © 2012 Pearson Education, Inc. is an identity matrix. Slide 1. 4 - 28
PROPERTIES OF THE MATRIX-VECTOR PRODUCT Ax § § § Theorem 5: If A is an matrix, u and v are vectors in , and c is a scalar, then a. b. . Proof: For simplicity, take , , and u, v in. For let ui and vi be the ith entries in u and v, respectively. © 2012 Pearson Education, Inc. Slide 1. 4 - 29
PROPERTIES OF THE MATRIX-VECTOR PRODUCT Ax § To prove statement (a), compute as a linear combination of the columns of A using the entries in as weights. Entries in Columns of A © 2012 Pearson Education, Inc. Slide 1. 4 - 30
PROPERTIES OF THE MATRIX-VECTOR PRODUCT Ax § To prove statement (b), compute as a linear combination of the columns of A using the entries in cu as weights. © 2012 Pearson Education, Inc. Slide 1. 4 - 31
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# Find the Distance Value Between Two Arrays Leetcode Solution
Difficulty Level Easy
algorithms Array coding Interview interviewprep LeetCode LeetCodeSolutionsViews 254
The problem Find the Distance Value Between Two Arrays Leetcode Solution provides us two arrays arr1 and arr2. Along with the two arrays, we are provided with an integer n. Then the problem asks us to find the relative distance between the given two arrays. The relative distance is defined as the number of elements in the first array that does not have any element in the second array that has a minimum absolute difference less than or equal to the given integer d. So as always before diving deep into the solution, first we should take a look at a few examples.
`arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2`
`2`
Explanation: We will pick each element one by one from the first array to verify the output. For the first element, 4 we do not have any corresponding element in the second array that has a minimum absolute difference of 2 with it. The same goes for 5. But if we see the last element, 8, there exists an element with the same value in the second array. Thus 8, is not considered into the answer. The answer will thus become equal to 2 because of the first 2 elements of the first array.
## Brute force Approach to Find the Distance Value Between Two Arrays Leetcode Solution
The brute force solution for the problem simply iterates over both the arrays picking unique pairs. Then these unique pairs are checked if the difference between them is less than the given integer ‘d’. If the difference is less than d, we flag the element. We do not count the flagged elements and the count for the remaining elements is returned as the answer. The solution is pretty straight forward but there can be a better solution.
## Optimized Approach to Find the Distance Value Between Two Arrays Leetcode Solution
In the above approach, we used two nested loops. But in the end, we only needed to check if the current element in the first array has even a single element in the second array that has a difference of less than d. So, if we pick the two closest elements to the picked element from the first array. If these two closest elements do not have a minimum absolute difference of less than d, then we can say for sure that no other element can produce a better result.
So, the problem now boils down to finding the two closest elements (from the second array) to the picked element of the first array. To find these two closest elements, we sort the second array and use binary search. Using binary search we find the element that is equal or greater than the picked element. The element that is just smaller than the current element is also used to evaluate the difference.
If these differences are less than d then we flag the element and do not count it towards the answer. The rest elements are counted towards the answer and are returned at the end of the function.
### Optimized code to Find the Distance Value Between Two Arrays Leetcode Solution
#### C++ code
```#include <bits/stdc++.h>
using namespace std;
int findTheDistanceValue(vector<int>& arr1, vector<int>& arr2, int d) {
int ans = 0;
sort(arr2.begin(), arr2.end());
for(int i=0;i<arr1.size();i++){
int it = lower_bound(arr2.begin(), arr2.end(), arr1[i]) - arr2.begin();
bool isIt = false;
if(it<arr2.size() && abs(arr2[it] - arr1[i]) <= d)isIt = true;
if(it != 0 && abs(arr2[it-1] - arr1[i]) <= d)isIt = true;
if(!isIt)
ans++;
}
return ans;
}
int main(){
vector<int> arr1 = {4,5,8};
vector<int> arr2 = {10,9,1,8};
int d = 2;
cout<<findTheDistanceValue(arr1, arr2, d);
}
```
`2`
#### Java code
```import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
private static int findTheDistanceValue(int[] arr1, int[] arr2, int d) {
Arrays.sort(arr2);
int ans = 0;
for (int i= 0;i<arr1.length;i++) {
int it = Arrays.binarySearch(arr2, 0, arr2.length, arr1[i]);
if (it < 0) it = -(it+1);
boolean isIt = false;
if(it<arr2.length && Math.abs(arr2[it] - arr1[i]) <= d)isIt = true;
if(it != 0 && Math.abs(arr2[it-1] - arr1[i]) <= d)isIt = true;
if(!isIt)
ans++;
}
return ans;
}
public static void main (String[] args) throws java.lang.Exception
{
int[] arr1 = {4,5,8};
int[] arr2 = {10,9,1,8};
int d = 2;
System.out.print(findTheDistanceValue(arr1, arr2, d));
}
}```
`2`
### Complexity Analysis
#### Time Complexity
O(max(M, N)logN), since we sort the second array and perform a binary search for each element in the first array. Here M, N is the number of elements in the first and second array respectively.
#### Space Complexity
O(N), space required to sort the second array.
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> > Minors and Cofactors of Determinant
# Minors and Cofactors of Determinant
We learned how important are matrices and determinants and also studied about their wide applications. The knowledge of Minors and Cofactors is compulsory in the computation of adjoint of a matrix and hence in its inverse as well as in the computation of determinant of a square matrix. This technique of computing determinant is known as Cofactor Expansion. Let’s get started!
### Suggested Videos
Properties of Determinants
Applications of Determinants
Minors and Cofactors_H
## Minor of a Determinant
A minor is defined as a value computed from the determinant of a square matrix which is obtained after crossing out a row and a column corresponding to the element that is under consideration. Minor of an element aij of a determinant is the determinant obtained by deleting its ith row and jth column in which element aij lies. Minor of an element aij is denoted by Mij.
## Cofactor of a Determinant
The cofactor is defined as the signed minor. Cofactor of an element aij, denoted by Aij is defined by A = (–1)i+j M, where M is minor of aij.
### Note
• We note that if the sum i+j is even, then Aij Mij, and that if the sum is odd, then Aij Mij.
• Hence, the only difference between the related minor entries and cofactors may be a sign change or nothing at all.
• Whether or Aij Mior Aij Mij
• has a pattern for square matrices as illustrated:
For example C12 M12. Of course, if you forget, you can always use the formula Ci(1)i+Mij,
Here, C12=(1)1+Mi(1)MiMij
## Example
Find the minors and cofactors of all the elements of the determinant $$\begin{vmatrix} 1 & -2 \\ 4 & 3 \end{vmatrix}$$
Solution: Minor of the element aij is Mij.
Here a11 = 1. So M11 = Minor of a11 = 3
M12 = Minor of the element a12 = 4
M21 = Minor of the element a21 = –2
M22 = Minor of the element a22 = 1
Now, cofactor of aij is Aij. So,
A11 = (–1)1+1, M11 = (–1)2 (3) = 3
A12 = (–1)1+2, M12 = (–1)3 (4) = –4
A21 = (–1)2+1, M21 = (–1)3 (–2) = 2
A22 = (–1)2+2, M22 = (–1)4 (1) = 1
## Solved Examples for You
Question 1: Let A=[aij]n×n be a square matirx and let cij be cofactor of aij in A. If C=[cij], then
1. |A| = |C|
2. |C| = |A|n-1
3. |C| = |A|n-2
4. none of these
Answer : We know that adjCT where C is the cofactor matrix of A.
Now |CT| = |Adj A|
=|A|n-1 where n it the order of the square matrix.
Question 2: The minors and cofactors of -4 and 9 in determinant $$\begin{vmatrix} -1 & 2 & 3 \\ -4 & 5 & -6 \\ -7 & -8 & 9 \end{vmatrix}$$ are respectively
1. 42, 42; 3, 3
2. 42, -42; 3, 3
3. 42, -42; 3, -3
4. 42, 3; 42, 3
Answer : Minor of -4 is $$\begin{vmatrix} 2 & 3 \\ -8 & 9 \end{vmatrix} = 42$$
Cofactor of -4 is (1)1+2(4242
Minor of 9 is $$\begin{vmatrix} -1 & 2 \\ -4 & 5 \end{vmatrix} = 3$$
Cofactor of 9 is (1)3+.(33. Therefore, the answer is option B
Question 3: What is meant by cofactor of a matrix?
Answer: A cofactor refers to the number you attain on removing the column and row of a particular element existing in a matrix.
Question 4: What is meant by a minor matrix?
Answer: A minor refers to the square matrix’s determinant whose formation takes place by deleting one column and one row from some larger square matrix.
Question 5: Can we say that the adjoint is the same as the reverse?
Answer: The adjoint of a matrix is also known as the adjugate of a matrix. It refers to the transpose of the cofactor matrix of that particular matrix. For a matrix A, the denotation of adjoint is as adj (A). On the other hand, the inverse of a matrix A refers to a matrix which on multiplication by matrix A, results in an identity matrix.
Question 6: What is meant by rank of a matrix?
Answer: The rank of a matrix refers to the maximum number of linearly independent column vectors that exist in the matrix. Furthermore, it is also the maximum number of linearly independent row vectors that exist in the matrix.
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## Browse
##### Determinants
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jannat
Is determinant available just for square matrix?
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Stephen
Yes, it is only defined for square matrices.
Guest
J. A. Zahálka
It actually is. It is equal to zero for all non-square matrices.
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## Intermediate Algebra (12th Edition)
$(1-a)(1-b)$
$\bf{\text{Solution Outline:}}$ Group the terms of the given expression, $1-a+ab-b ,$ such that the factored form of the groupings will result to a factor common to the entire expression. Then, factor the $GCF$ in each group. Finally, factor the $GCF$ of the entire expression. $\bf{\text{Solution Details:}}$ Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (1-a)+(ab-b) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} (1-a)+b(a-1) \\\\= (1-a)-b(1-a) .\end{array} Factoring the $GCF= (1-a)$ of the entire expression above results to \begin{array}{l}\require{cancel} (1-a)(1-b) .\end{array}
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# Real Number
Updated: 05 Aug 2023
325
Real numbers are the building blocks of mathematics, forming an essential foundation for countless mathematical concepts and applications. From the simplest counting to the most complex scientific calculations, real numbers play a vital role in quantifying, measuring, and understanding the world around us.
## Definition of Real Number
The set of rational and irrational numbers is called Real Numbers. A real number is denoted by R . Thus QUQ^/=R
Note: All the numbers on the number line are real numbers.
### Rational Number
The word Rational means “Ratio”. Thus a number in the form of \frac{p}{q} where p \ and \ q are integers is called a Rational Number and is denoted by Q .
Set of Rational Number
Q= \{ \frac{p}{q} | p,q \in Z,q \neq0 \}
### Irrational Numbers
The word Irrational means “Not Ratio”. The irrational number consists of all those numbers which are not rational. Irrational numbers are denoted by Q^/ .
## Rational Number MCQs
Enhance your knowledge of real numbers through our comprehensive Multiple-Choice Questions (MCQs). Test your understanding of rational, irrational, and complex real numbers and explore their fundamental properties through interactive MCQs. Get immediate feedback, analyze your answers, and expand your knowledge with a diverse range of real number MCQs.
Real number MCQs offer an engaging and efficient means of exploring the vast world of real numbers. These MCQs are designed to challenge and engage learners of all levels. So, let the journey of exploration and discovery through real number MCQs begin!
1. The number that can be expressed in the form of \frac{p}{q} \ where \ p \ and \ q are integers and q \neq 0 is called _______ numbers.
O Rational
O Irrational
O Imaginary
O None of these
Rational
Explanation:
Definition of rational number.
2. A rational in the form of \frac{p}{q} where \ p \ and q are _______
O Imaginary
O Complex
O Integers
O All of them
Integers
Explanation:
3. A rational in the form of \frac{p}{q} where q _______ 0.
O Equal
O Not equal
O Both a and b
O None of these
Not equal
Explanation:
if q=0 then it becomes infinite.
4. The word Rational is derived from _______
O Ratio
O Not Ratio
O lota
O All of them
Ratio
Explanation:
5. The rules of rational number for p \ and \ q are _______.
O \frac{p}{q}
O Integers
O q \neq 0
O All of them
All of them
Explanation:
These all are the rules for ratinal number.
6. Rational number is denoted by _______.
O Q
O Q^{\prime}
O Both a and b
O None of these
Q
Explanation:
7. Irrational number is denoted by _______
O Q
O Q^{\prime}
O Both a and b
O None of these
Q^{\prime}
Explanation:
8. The word irrational means_______
O Ratio
O Not Ratio
O lota
O None of these
Not ratio
Explanation:
9. Irrational numbers consist of all the numbers which are_______
O Rational
O Not rational
O lota
O None of these
Not rational
Explanation:
Numbers other than rational numbers are called Irrational numbers.
10. 1\frac{3}{4} is_______
O Rational
O Irrational
O Complex
O None of these
Rational
Explanation:
1\frac{3}{4}=\frac{7}{4}=1.75
Terminating decimals are Rational numbers.
11. \sqrt{3} is_______
O Rational
O Irrational
O Complex
O None of these
Irrational
Explanation:
\sqrt{3}=1.7320508076 \dots
Non-terminating and non-recurring (repeating) decimals are irrational numbers.
12. \pi is_______
O Rational
O Irrational
O Complex
O None of these
Irrational
Explanation:
The approximate value of \pi \ is \ 3.1415926 \dots which is Non-terminating and non-recuring decimal. Thus \pi is an irrational number.
13. R is the symbol of _______ number.
O Rational
O Irrational
O Real
O Both a & b
Real
Explanation:
14. The union of rational and irrational numbers is the set of _______ numbers.
O Complex
O lota
O Prime
O Real
Real
Explanation:
Rational and irrational numbers are collectively called real number.
15. Q \cup Q^{\prime}= _______
O i
O Prime
O R
O N
R
Explanation:
The union of rational and irrational numbers is the set of Real numbers.
16. All the numbers on the number line are _______ numbers.
O i
O Complex
O Real
O None of these
Real
Explanation:
Every number there is a point on the line. The number associated with a point is called the coordinate of that point on the line and the point is called the graph of the number.
17. All the numbers on the number line are _______ numbers.
O Rational
O Irrationa|
O Real
O All of them
All of them
Explanation:
Rational, Irrational and Real all are shown by number line.
18. -17 is _______ numbers.
O Whole
O Natural
O Integers
O All of them
Integers
Explanation:
Natural and whole number is always be positive.
19. A whole number is a number that does not contain_______
O Decimal
O Negative
O Fraction
O All of them
All of them
Explanation:
Whole number does not contain Decimal, Negative and Fraction.
20. All terminating and repeating decimals are_______
O Rational
O Irrational
O Complex
O All of them
Rational
Explanation:
This is the rule for rational number.
21. _______ decimals are Rational numbers.
O Terminating
O Repeating
O Both a & b
O None of these
Both a & b
Explanation:
Terminating and repeating decimals are rational numbers.
22. \frac{3}{8}=0.375 is _______ decimal.
O Terminating
O Repeating
O Both a & b
O None of these
Terminating
Explanation:
A decimal number that contains a finit number of digits after the decimal point is called terminting decimal.
23. \frac{2}{15}=0.133 \ldots is _______ decimal
O Terminating
O Repeating
O Both a & b
O None of these
Repeating
Explanation:
When some digits are repeated in same order after decimal point is called repeating decimal.
24. 0.1 \overline{3} is _______ decimal
O Terminating
O Repeating
O Both a & b
O None of these
Repeating
Explanation:
The bar over the digit 3 means that this digit repeat forever.
25. Bar over the digit means that this digit is_______
O Terminated
O Repeated
O Both a & b
O None of these
Repeated
Explanation:
The bar over the digit means that this digit repeat forever.
26. A decimal which is non-terminating and non-repeating is called _______ numbers.
O Rational
O Irrational
O Complex
O All of them
Irrational
Explanation:
Non-terminating and non-recurring (repeating) decimals are irrational numbers.
27. The number line help in visualizing the set of _______ numbers.
O Complex
O Imaginary
O Real
O None of these
Real
Explanation:
We assume that for any point on a line there is a real number.
28. For every real number there is a _______ on the line.
O Point
O Line
O Both a & b
O None of these
Point
Explanation:
For every real number, there must be a point on the line.
29. The number associated with a point on the line is called the of _______ that point.
O Coordinate
O Zero
O Both a & b
O None of these
Coordinate
Explanation:
It is the graphical representation of real number on line.
30. The point on the number line is called the _______ of the number.
O Coordinate
O Zero
O Graph
O None of these
Graph
Explanation:
31. The numbers to the right of “0” on a number line are called _______ numbers.
O Positive
O Negative
O Complex
O None of these
Positve
Explanation:
The numbers greater than “0” are written on the right side of zero.
32. The numbers to the left of “O” on a number line are called _______ numbers.
O Positive
O Negative
O Complex
O None of these
Negative
Explanation:
The numbers less than “0” are written on the left side of zero.
33. 0 is _______
O Positive integers
O Negative integers
O Neither positive nor negative
O Not an integer
Neither positive nor negative
Explanation:
0 is the only number which is neither positive nor negative.
34. Repeating decimals are called _______ decimals.
O Recurring
O Non-Recurring
O Both a & b
O None of these
Recurring
Explanation:
Repeating decimals are also called Recurring decimals.
35. Non-Repeating decimals are called _______ decimals.
O Recurring
O Non-Recurring
O Botha \& b
O None of these
Non-Recurring
Explanation:
The Non-Repeating decimals are also called Non-Recurring
36. \frac{5}{27}=0.185185185 \dots is called _______ recurring decimals.
O Terminating
O Non-Terminating
O Both a & b
O None of these
Non-Terminating
Explanation:
Here the number of digits repeated infinitely.
37. \frac{5}{7} is _______ number.
O Rational
O Irrational
O Imaginary
O None of these
Rational
Explanation:
\frac{5}{7}=0.714285714285 \dots
Non-terminating recurring (repeating) decimals are rational numbers.
38. \sqrt{36} is _______ .
O Rational
O Whole
O Natural
O All of them
All of them
Explanation:
\sqrt{36}=6
Thus 6 shows Natural, Whole and Rational number at a time.
39. \frac{14}{3} is _______ number.
O Rational
O Whole
O Both a & b
O Irrational
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# 7.2: Broken-Line Graphs
Difficulty Level: Basic Created by: CK-12
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### Broken Line Graphs
A variation of a line graph is a broken-line graph. This type of line graph is used when it is necessary to show change over time. A line is used to join the values, but the line has no defined slope. However, the points are meaningful, and they all represent an important part of the graph. Usually a broken-line graph is given to you, and you must interpret the given information from the graph.
#### Interpreting a Broken Line Graph
1. Answer the questions below for the following broken-line graph, which shows the distance, over time, of a bus from the bus depot.
a. What was the fastest speed of the bus?
The fasted speed of the bus was 16 miles per hour.
b. How many times did the bus stop on its trip? (Do not count the beginning and the end of the trip.)
The bus was stopped 4 times.
c. What was the initial distance of the bus from the bus depot?
The bus was initially 2 miles from the bus depot.
d. What was the total distance traveled by the bus?
The total distance traveled by the bus was 38 miles.
2. Sam decides to spend some time with his friend Aaron. He hops on his bike and starts off to Aaron’s house, but on his way, he gets a flat tire and must walk the remaining distance. Once he arrives at Aaron’s house, they repair the flat tire, play some poker, and then Sam returns home. On his way home, Sam decides to stop at the mall to buy a book on how to play poker. The following graph represents Sam’s adventure:
a) How far is it from Sam’s house to Aaron’s house?
It is 25 km from Sam's house to Aaron's house.
b) How far is it from Aaron’s house to the mall?
It is 15 km from Aaron's house to the mall.
c) At what time did Sam have a flat tire?
Sam had a flat tire at 10:00 am.
d) How long did Sam stay at Aaron’s house?
Sam stayed at Aaron's house for 1 hour.
e) At what speed did Sam travel from Aaron’s house to the mall and then from the mall to home?
Sam traveled at a speed of 30 km/h from Aaron's house to the mall and then at a speed of 40 km/h from the mall to home.
3. The following graph is an example of a broken-line graph, and it represents the time of a round-trip journey, driving from home to a popular campground and back.
a) How far is it from home to the picnic park?
It is 40 miles from home to the picnic park.
b) How far is it from the picnic park to the campground?
It is 60 miles from the picnic park to the campground.
c) At what 2 places did the car stop?
The car stopped at the picnic park and at the campground.
d) How long was the car stopped at the campground?
The car was stopped at the campground for 15 minutes.
e) When does the car arrive at the picnic park?
The car arrived at the picnic park at 11:00 am.
f) How long did it take for the return trip?
The return trip took 1 hour.
g) What was the speed of the car from home to the picnic park?
The speed of the car from home to the picnic park was 40 mi/h
h) What was the speed of the car from the campground to home?
The speed of the car from the campground to home was 100 mi/h.
->
### Example
#### Example 1
For the following broken-line graph, write a story to accompany the graph, and provide a detailed description of the events that are occurring.
Before we write our story, let's summarize what we know from the graph. We know that the story begins at 8:45 am, because the line touches the x\begin{align*}x\end{align*}-axis at the third tick mark between 8:00 am and 9:00 am. We also know that from 8:45 am to 9:30 am, a distance of 60 km was traveled. This is because from 8:45 am to the second tick mark between 9:00 and 10:00 am, the line goes from 0 km to 60 km on the y\begin{align*}y\end{align*}-axis. This means that 60 km was traveled in 45 minutes, so the speed during this time can be calculated as follows:
6045=x6045x=3,600x=80 km/hr\begin{align*}\frac{60}{45}=\frac{x}{60}\\ 45x=3,600\\ x=80 \ \text{km/hr}\end{align*}
Next, we can see that from 9:30 am to 10:00 am, no distance was traveled, since the line is horizontal in this interval. However, from 10:00 am to 11:00 am, 40 km was traveled, This is because from 10:00 am to 11:00 am, the line goes from 60 km to 100 km on the y\begin{align*}y\end{align*}-axis, and 10060=40\begin{align*}100-60=40\end{align*}. Therefore, the speed during this time was 40 km/hr. From 11:00 am to 11:15 am, there was, again, no distance traveled, since the line is horizontal from 11:00 am to the first tick mark between 11:00 am and 12:00 pm. Finally, from 11:15 am to 12:15 pm, 100 km was traveled, since the line goes from 100 km to 0 km on the y\begin{align*}y\end{align*}-axis from 11:15 am to the first tick mark after 12:00 pm. This means that the speed during this time was 100 km/hr. Also, since the distance is decreasing in this interval, we know that this was a return trip. Because the line touches the x\begin{align*}x\end{align*}-axis at 12:15 pm, this is the end of the trip.
Now we are ready to write our story. As long as it adheres to what we found above, anything is fine, but here is an example of what we could write:
Deena is going on a shopping trip. From 8:45 am to 9:30 am, she drove her car to a neighboring town 60 km away, traveling at a speed of 80 km/hr. She then decided to stop and have breakfast for 30 minutes before resuming her trip. Traffic was a little heavy for the next hour, so she only managed to go 40 km in this time, traveling at a speed of 40 km/hr. At 11:00 am, she reached the shopping mall that was her destination, but it seemed to be closed. After looking around for 15 minutes, she decided that it was, in fact, closed, so she began her trip home. Traffic was much lighter on the way home, so she covered the entire 100 km non-stop in 1 hour, traveling at a speed of 100 kn/hr. She arrived home at 12:15 pm.
### Review
1. What name is given to a graph that shows change over time, with points that are joined but have no defined slope?
1. linear graph
2. broken-line graph
3. scatter plot
4. line of best fit
Use the broken-line graph below, which represents a bike ride, to answer the following questions.
1. What was the total distance traveled on the bike ride?
2. What was the fastest speed traveled by the bike?
3. What was the slowest speed traveled by the bike?
4. How long did the bicyclist stop before beginning his or her return trip?
5. How long did the return trip take?
Bob is looking for the post office, but he is lost. The broken-like graph below shows his distance from the post office as he wanders about the city. Use the broken-line graph to answer the following questions.
1. During what time intervals is Bob getting closer to the post office?
2. During what time intervals is Bob getting farther away from the post office?
3. What is the total distance traveled by Bob from 12:30 pm to 6:00 pm, which is the duration of time shown by the graph?
4. What was Bob's average speed from 12:30 pm to 6:00 pm?
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English Spanish
TermDefinition
broken-line graph is a graph that is used when it is necessary to show change over time. A line is used to join the values, but the line has no defined slope.
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# Basic Math Terminology
## The Decimal System
The decimal system, sometimes referred to as base 10, contains a total of ten identifiers called digits. The decimal system is widely used because humans generally have ten fingers on which to count. For now, we will disregard other number systems for the sake of simplicity with the understanding that the decimal system is not unique.
The ten digits of the decimal system, arranged from lowest to greatest, are:
• 0 (zero)
• 1 (one)
• 2 (two)
• 3 (three)
• 4 (four)
• 5 (five)
• 6 (six)
• 7 (seven)
• 8 (eight)
• 9 (nine)
The decimal system uses positional notation to represent numbers larger than 9. This means that a digit's position in relation to other digits affects its meaning. Digits in the furthest right position represent the number of ones being counted, while digits in the second position from the right represent the number of tens. Digits in the third position from the right represent the number of hundreds, and digits in the fourth position from the right represent the number of thousands. This pattern can continue forever; for more information, see orders of magnitude.
For example, the number 535,254 means 5 hundreds of thousands, 3 tens of thousands, 5 thousands, 2 hundreds, 5 tens, and 4 ones. We would say this number as "five hundred thirty-five thousand two hundred fifty-four".
Other systems, like binary (base 2) exist. Base 2 would just have 2 digits, 0 and 1, base 3 has 3 digits, 0 ,1 and 2, and so on.
## The Basic Sets of Numerals
Counting numbers are the numbers we use every day to count things. Mathematicians sometimes refer to this set of numbers as the Natural Numbers. These are represented by the sign ${\displaystyle \mathbb {N} }$ , for Natural Numbers.
${\displaystyle 1,2,3,4...\ }$
Whole numbers include all the counting numbers and zero.
${\displaystyle 0,1,2,3,4...\ }$
Negative numbers include the opposite of all the Counting Numbers. They are counted in the opposite direction of Counting numbers and have -, the negative sign, in front of them.
${\displaystyle ...-4,-3,-2,-1}$
Integers include all numbers without a decimal. Another way to say this is all of the Whole Numbers and their negatives. These are represented by the sign ${\displaystyle \mathbb {Z} }$ , for the German word Zahlen, which means "numbers".
${\displaystyle ...-4,-3,-2,-1,0,1,2,3,4...}$
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# How to Calculate Percentages with Ease
Baca Cepat
## Introduction: The Importance of Knowing How to Calculate Percentages
### What are Percentages?
Before we learn how to calculate percentages, let’s first understand what they are. A percentage is a fraction of a whole expressed as a number out of 100. For example, if you have 50 out of 100 apples, your percentage of apples is 50%. Percentages are commonly used in various fields such as finance, statistics, and mathematics. Understanding percentages is essential to solving many problems in these areas.
### Why Should You Learn How to Calculate Percentages?
Learning how to calculate percentages can be extremely useful in everyday life. It can help you in budgeting, determining discounts, calculating taxes, determining interest rates, and many other scenarios. For example, if you’re shopping and there’s a 20% discount on a product you want to buy, you can easily calculate how much you’ll save. Knowing how to calculate percentages can save you both time and money in the long run.
### How to Calculate Percentages: The Basics
Before we get into the nitty-gritty of calculating percentages, let’s first understand the basics. When we calculate a percentage, we’re finding a portion of a whole. In other words, we’re finding a fraction of a number that’s out of 100. For example, if we want to find 20% of 100, we’re finding 20 out of 100, which is equivalent to 0.2. To calculate percentages, we use a simple formula:
Whole Number Percentage Fraction
100 x% x/100
### Calculating Percentages: Step-by-Step Guide
Now that we understand the basics of percentages, let’s dive into the step-by-step guide on how to calculate them.
#### Step 1: Determine the Whole Number
The first step in calculating a percentage is to determine the whole number or the total value. This could be the total amount of money in a budget, the number of people in a population, or any other value you’re interested in.
#### Step 2: Determine the Percentage
The next step is to determine the percentage that you’re interested in. This could be a percentage of the total value, such as the percentage of women in a population, or a percentage that you want to find, such as the percentage increase in sales.
#### Step 3: Convert the Percentage to a Decimal
Once you’ve determined the percentage, you need to convert it to a decimal. To do this, simply divide the percentage by 100. For example, if you want to find 25%, divide 25 by 100 to get 0.25.
#### Step 4: Multiply the Decimal by the Whole Number
The final step is to multiply the decimal by the whole number to get the answer. For example, if you want to find 25% of 200, first convert 25% to a decimal (0.25) and then multiply 0.25 by 200 to get 50.
### Common Percentages and Their Decimal Equivalents
Here are some common percentages and their decimal equivalents:
Percentage Decimal Equivalent
10% 0.1
25% 0.25
50% 0.5
75% 0.75
100% 1
### Common Percentage Calculations
Here are some common percentage calculations:
#### Percentage Increase
To calculate a percentage increase, use the following formula:
Percentage Increase Formula (New Value – Old Value) / Old Value * 100
#### Percentage Decrease
To calculate a percentage decrease, use the following formula:
Percentage Decrease Formula (Old Value – New Value) / Old Value * 100
#### Percentage of a Percentage
To calculate a percentage of a percentage, multiply the two percentages together. For example, if you want to find 20% of 50%, multiply 0.2 by 0.5 to get 0.1, which is equivalent to 10%.
### FAQs
#### What is a percentage?
A percentage is a fraction of a whole expressed as a number out of 100. It is commonly used in various fields such as finance, statistics, and mathematics.
#### What is the formula for calculating percentages?
The formula for calculating percentages is:
Whole Number Percentage Fraction
100 x% x/100
#### What are some common percentage calculations?
Some common percentage calculations are percentage increase, percentage decrease, and percentage of a percentage.
#### How do I convert a percentage to a decimal?
To convert a percentage to a decimal, divide the percentage by 100. For example, to convert 25% to a decimal, divide 25 by 100 to get 0.25.
#### What is the difference between a percentage and a fraction?
A percentage is a fraction of a whole expressed as a number out of 100, while a fraction is a part of a whole expressed as a ratio of two numbers.
#### How do I calculate a percentage increase?
To calculate a percentage increase, use the following formula: (New Value – Old Value) / Old Value * 100
#### How do I calculate a percentage decrease?
To calculate a percentage decrease, use the following formula: (Old Value – New Value) / Old Value * 100
#### How do I calculate a percentage of a percentage?
To calculate a percentage of a percentage, multiply the two percentages together.
#### What is the difference between a percentage and a percentile?
A percentage is a fraction of a whole expressed as a number out of 100, while a percentile is a specific point in a distribution of data that indicates the percentage of values that are below that point.
#### How do I calculate a percentage of a total?
To calculate a percentage of a total, divide the part by the whole and multiply by 100.
#### What is the percentage formula in Excel?
The percentage formula in Excel is:
Excel Percentage Formula = Part / Whole * 100
#### What is the difference between gross and net percentages?
Gross percentages are calculated based on the total value, while net percentages are calculated after deducting taxes or other costs.
#### What is a percentage point?
A percentage point is the difference between two percentages.
#### How do I calculate percentages in my head?
To calculate percentages in your head, use mental math techniques such as dividing by 10 or 5 or converting percentages to decimals.
### Conclusion
Congratulations! You’ve made it to the end of the article, and now you’re equipped with the knowledge to calculate percentages with ease. Knowing how to calculate percentages is a valuable skill that can save you time and money, and it can also help you make better decisions. We hope you found this article informative and helpful. Now, go forth and start calculating those percentages!
### Closing or Disclaimer
The information provided in this article is for educational purposes only and should not be used as a substitute for professional advice. Always consult a professional if you have any questions or concerns. The author and publisher are not responsible for any damages or losses that may occur as a result of using the information provided in this article.
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# Bright Hub Education
## Finding the Measurement of Angles using a Protractor: 4th Grade Lesson
written by: Donna Ventura • edited by: Carly Stockwell • updated: 11/14/2013
Students will investigate the concept of angles and determine the measurement of angles using a protractor.
• slide 1 of 14
Identify the measurement of angles using a protractor. Using a protractor, draw angles with a specified measurement. Download a copy of the lesson at the end of the article.
Lesson Objective: The lesson is aligned to the Common Core State Standards for Mathematics – 4MD.6 Geometric Measurement – Measure angles in whole-number degrees using a protractor. Sketch angles of specified measure.
Materials Required: Protractor, ruler
• slide 2 of 14
### Lesson Procedure:
Exploring a Protractor
Look at the protractor shown below.
• slide 3 of 14
This is a mathematical tool that measures angles in degrees. You can also use the protractor and a ruler to draw angles.
Protractors usually have two sets of numbers. One set of numbers starts at the right. As you read the numbers from right to left, the numbers increase. The other set of numbers starts at the left and increases towards the right. These numbers represent degrees. It is important to use one set of numbers.
Notice at the bottom center of the protractor there is a dot.
• slide 4 of 14
### Drawing an Angle using a Protractor and a Ruler
Steps:
1. Draw a ray using the ruler.
2. Place the dot at the center of the protractor on the endpoint of the ray.
3. Make sure the marking on the protractor that represents 0 degree is on the ray.
• slide 5 of 14
4. Find the marking on the protractor that represents 40 degrees. Place a dot on your paper next to the marking that represents 40 degrees.
5. Remove the protractor. Use the ruler to draw a ray from the endpoint of the first ray to the dot that represents 40 degrees.
• slide 6 of 14
### Individual or Group Work:
Use the protractor and ruler to draw angles measuring the following degrees.
1. 35 degrees
2. 75 degrees
3. 110 degrees
4. 142 degrees
5. 156 degrees
1. Students draw an angle that measures approximately 35 degrees.
2. Students draw an angle that measures approximately 75 degrees.
3. Students draw an angle that measures approximately 110 degrees.
4. Students draw an angle that measures approximately 142 degrees.
5. Students draw an angle that measures approximately 156 degrees.
Identifying Measurement of an Angle using a Protractor
Steps:
1. Place the dot of the protractor at the vertex of the angle.
2. Place the edge of the protractor on one ray of the angle.
3. Notice where the other ray of the angle lies on the protractor. Look at the degrees on the protractor where the other ray of the angle lies.
4. The angle below measures approximately 19 degrees.
• slide 7 of 14
• slide 8 of 14
### Individual or Group Work:
Use the protractor to find the following angles measurements in degrees.
• slide 9 of 14
### 6.
• slide 10 of 14
### 7.
• slide 11 of 14
### 8.
• slide 12 of 14
### 9.
• slide 13 of 14
### 10.
• slide 14 of 14
1. Angle measures approximately 55 degrees.
2. Angle measures measures approximately 65 degrees.
3. Angle measures measures approximately 132 degrees.
4. Angle measures measures approximately 24 degrees.
5. Angle measures measures approximately 72 degrees.
Students should be able to draw and measure angles with a specified measurement using a protractor and a ruler.
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# 100 200 300 400 500 Evaluating Expressions Words - Algebra Adding & Subtracting Polynomials Multiplying and Dividing Polynomials Factoring (Smactoring)
## Presentation on theme: "100 200 300 400 500 Evaluating Expressions Words - Algebra Adding & Subtracting Polynomials Multiplying and Dividing Polynomials Factoring (Smactoring)"— Presentation transcript:
100 200 300 400 500 Evaluating Expressions Words - Algebra Adding & Subtracting Polynomials Multiplying and Dividing Polynomials Factoring (Smactoring) 500 100 200 300 400 500 Final Jeopardy
100 Evaluate 4d³ When d = 2 32
Evaluate ¾ x³ When x = 4 200 -48
Evaluate 3(x + 5) – 9 ÷ x When x = 3 300 21
400 1 1/9 Evaluate x ˉ ² + xº When x = 3
Evaluate When x = 2 and y = -3 500 -7
Is the following expression a polynomial? If so, state it’s degree and name by number of terms. 3 x ˉ ²-2x +5 NO, not a polynomial 100
200. yes, 3, binomial Is the following expression a polynomial? If so, state it’s degree and name by number of terms. X³ - 2
Yes, 8, trinomial 300 Is the following expression a polynomial? If so, state it’s degree and name by number of terms. 8x ⁵ y³ - 12x y² + 20x³y³
Put the following polynomial in standard form and state its’ degree. 400 -4x3-5x2+20x+13, 3 3 – 5x(x – 4) + 10 – 4x ³
500 Simplify the polynomial and put in standard form. see answer
Simplify 6x³y² + 5x³y² 11x³y² 100
Simplify (2x² + 5x) – (8x² + 7x) -6x² - 2x 200
300 Simplify (5x + 2) -2 (4x – 3) -3x + 8
400 Simplify 3 – 5(x – 4) + 10 – 4x -9 + 33
500 Simplify (4x²y – 7xy +10xy²) - (12xy² - 5xy + 3x²y) 1x²y – 2xy – 2xy²
Simplify 4x²y(2x³y² - 3x²y + 5xy²) 8x ⁵ y³ - 12x y² + 20x³y³ 100
Simplify 42x ⁷ - 6x ⁵ -7x² 200
300 Simplify (2y– 5)(3y – 4) 6y² - 23y + 20
Simplify 5x ⁵ - 10x² 5x² X³ - 2 400
-8x ⁶ 500 Simplify (-2x²)³
6x² (2x – 4 + 7x²) 100 Factor out the GCMF 12x³ - 24x² + 42x
5a²b(2ab – 3a³b² + 1) 200 Factor out the GCMF 10a³b² - 15a ⁵ b³ + 5a²b
(x + 9) (x + 1) 300 Factor X² + 10x + 9
(x – 12)(x – 4) 400 Factor X² - 16x + 48
(x – 8)(x + 3) 500 Factor x² - 5x – 24 into two binomials.
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### Basic Trigonometry
There are three basic trigonometric functions defined for a right-angled triangle:
1. sine
2. cosine
3. tan
Let a, and b be the two sides and c be the hypotenuse. Then the functions takes the following form:
• sin x = opposite / hypotenuse = a /c
• cos x = = adjacent / hypotenuse = b / c
• tan x = opposite / adjacent = a / b
The following examples are based on this triangle.
E.g.1
In the above triangle, AB = 3cm and x = 300. Find AC and BC.
sin 30 = 3 /AC
0.5 = 3 / AC
AC = 6cm.
cos 30 = b /AC
0.8660 = b / 6
AC = 5.1cm.
E.g.2
In the above triangle, BC = 4cm and x = 600. Find AC and AB.
cos 60 = 4 /AC
0.5 = 4 / AC
AC = 8cm.
tan 60 = AB /BC
1.7 = AB / 4
AB = 6.8cm.
E.g.3
In the above triangle, AB = 3cm and BC = 5cm. Find x.
tan x = 3 / 5 = 0.6
x = tan-10.6 = 28.6
x = 28.60
#### The Sine Rule
sin A / a = Sin B /b = Sin C /c
Note the blinking letters to see the relationship.
E.g.1
A = 300 ; B = 500; b = 9cm; find c.
c / sin 50 = 9 / sin 30
c = 9 x sin 50 / sin 30
c = 13.7 cm.
E.g.2
A = 1100 ; b = 5cm; a = 9cm; find B.
5 / sin B = 9 / sin 110
sin B = 5*sin 110 / 9
B = 31.50
#### The Cosine Rule
c2 = a2 + b2 - 2abcosC
Note the blinking letters to see the relationship.
E.g.1
a = 4cm; b =5cm and C = 600; find c.
c2 = 42 + 52 - 2x4x5xcos60
c2 = 16 + 25 - 20
c2 = 21
c= 4.6 cm.
E.g.2
a = 5cm; b = 12cm and C = 900; find c.
c2 = 52 + 122 - 2x5x12xcos90
c2 = 25 + 144 - 0
c2 = 169
c= 13cm.
E.g.3
a = 8cm; b =7cm and C = 1200; find c.
c2 = 82 + 72 - 2x8x7xcos120
c2 = 64 + 49 + 56
c2 = 169
c= 13 cm.
E.g.4
a = 7cm; b= 6cm; c = 8 cm; find C.
82 = 72 + 62 - 2x4x5xcosC
64 = 49 + 36 - 84cosC
-84cosC = -21
cosC = 0.25
C = 75.50
#### Area of a Triangle
A = 1/2 absinx
Proof:
A = 1/2 X h X b
sin x = h / a => h = a sinx
A = 1/2 absinx
E.g.
a = 4cm; b= 8cm; x = 300
A = 1/2 * 4 * 8 * sin 30
A = 8cm2.
#### Deriving Trigonometric Values of 00, 300, 450, 600, 900, 1800, 3600
Trigonometric values of major angles, such as 00, 900, 1800, 2700 and 3600 can be remembered with the aid of the trigonometric curves, if you can visualize them. The three major curves are as follows:
The following image shows how to obtain the values of 00, 1800, and 3600 from the basic definitions
The following image shows how to obtain the values of 900 and 2700 from the basic definitions
In addition, the values of 300, 450 and 600 can be derived in the following way:
The trigonometric values of 450 is calculated from an right-angled isosceles triangle with each equal side being 1 unit.
The values of 300 and 600 are calculated from an equilateral triangle of each side 2 unit.
#### Trigonometric Values in the Four Quadrants
The angles are measured from the border between the first and the fourth quadrants - and anticlockwise, by convention. If it is measured in clockwise, it is considered as negative.
The following animation explicitly demonstrates it:
Now, in order to complement what you have just learnt, work out the following questions:
1. ABC is a triangle. AB = 6cm and angle ABC = 450. Find the perimeter and the area of the triangle.
2. Prove that Pythagoras Theorem is correct using the Cosine Rule.
3. The height of a man is 1.2m. When he looks at the top of a tree, standing 20m away from it, the angle of elevation is 320. Calculate the height of the tree.
4. Points P and Q are due south and West of a pillar of height 20m. The angle of elevation of the top from these points are 220 and 420 respectively. Find the distance between P and Q.
5. A ship is sailing a distance of 200 miles from port X to port Y, on a bearing 0200. At Y, it changes its course on a bearing 1100 and reaches port Z, after travelling 360 miles. Find the distance between port X and Port Z. Find the bearing of Z from X as well.
6. A jet is flying a distance of 120 miles from town P to town Q, on a bearing 0300. At Q, it changes its course on a bearing 600 and reaches town R, after travelling 240 miles. Find the distance between town P and town R. Find the bearing of R from P as well.
7. A man is looking down from a cliff at another man on the beach. The angle of depression is 200. The second man walks 200m away from the cliff and the new angle of depression for the man at the top becomes 150. Find the vertical height of the cliff.
8. The two sides of a right-angled triangle are 5cm, 12cm. Find its size of hypotenuse and the rest of the angles of the triangle.
9. The three sides of a triangle are 8cm, 7cm and 13cm. Find its area. Calculate the shortest distance between a vertex and the longest side as well.
10. ABC is a triangle with AB = 4cm and AC = 5cm. AD is at an right angle to BC. AD = 3cm. Find BC.
### Resources at Fingertips
This is a vast collection of tutorials, covering the syllabuses of GCSE, iGCSE, A-level and even at undergraduate level. They are organized according to these specific levels.
The major categories are for core mathematics, statistics, mechanics and trigonometry. Under each category, the tutorials are grouped according to the academic level.
This is also an opportunity to pay tribute to the intellectual giants like Newton, Pythagoras and Leibniz, who came up with lots of concepts in maths that we take for granted today - by using them to serve mankind.
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## Friday 17 July 2015
### Fundamentals of calculus Chapter 2: Integration
Fundamentals of calculus
Chapter 2: Integration
Integration is the converse of differentiation. To understand this better, it must be known that during differentiation an important information is lost. Let $\frac{dz(x)}{dx} = z’(x)$. Then, we cannot get back $z(x)$ from $z’(x)$ because, while taking the difference $z(x+\Delta x) – z(x)$ only the information regarding the difference between a point in Z axis and a neighbourhood of that point is retained. But the actual orientation is lost. Throughout this article, unless stated $\Delta x$ is limited to zero. To see how this could cause the loss of a piece of information, let me elaborate this with a simple algebraic example. Consider $a-b = c$ to be true. Then we may write $a – (a-c) = c$, here $b=a-c$.
It’s obvious that just from the value of $c$ it isn’t possible to incur the values of $a$ and $b$. But it is possible to incur that $c$ is some value obtained by subtracting $a$ and $b$, though the two values may be anything they are interrelated. So there is a possibility that after performing the differentiation operation, it is not possible to say that we can get back the original function. After computing the differentiation, we have $\frac{z(x+\Delta x) – z(x)}{\Delta x} = z’(x) = \frac{z(x) }{\Delta x} + z’(x) -\frac{z(x) }{\Delta x} = z’(x)$. Note that the expression $\frac{z(x) }{\Delta x}$ can be anything and is independent of $z’(x)$, if we treat them as algebraic constants. This is not necessarily same as $a – (a-c) = c , b=a-c$, which clearly shows that $a$ is totally independent of $c$. But in this case, both are a function of $x$.
On rearranging we may write, $\frac{z(x + \Delta x) }{\Delta x} –z’(x) = \frac{z(x )}{\Delta x}$. We need a function $z’(x)$ such that $z(x+\Delta x) – z’(x)\Delta x = z(x)$. Now let’s explore and find how we may obtain $z(x)$ from $z’(x)$. We know that $z’(b) \Delta x + z(b) = z(b+\Delta x)$ and $z’(b+\Delta x)\Delta x + (z’(b) \Delta x + z(b)) = z’(b+\Delta x) + z(b + \Delta x) = z(b+ 2\Delta x)$. So generalising this to obtain $z(x)$ from $z(b), x > b$ by implementing the above property recursively, we get $z(x) = z(b) + \Delta x (z’(b+\Delta x) + z’(b + 2\Delta x) + z’(b + 3\Delta x) +… +z’(b+ \frac{x – b}{\Delta x} \Delta x) = z(x)$. We may write this using the summation notation in the following way,
$z(x) = z(b) + \Delta x \times \sum \limits_{i=0}^{\frac{x-b}{\Delta x}} {z’(b + i\times \Delta x)}$.
The above holds good even if $x < b$, because for this case, instead of $z’(b + |i|\times \Delta x)$ we will be needing $z’(b - |i|\times \Delta x)$; this will automatically be true if $i$ ranges from $0\to \frac{x-b}{\Delta x}$. So now let’s use this definition to find the differential of $z(x)$.
$\frac{z(x+\Delta x) – z(x)} {\Delta x} = \frac{1}{\Delta x}\times((z(b) + \Delta x \times\sum \limits_{i=0}^{\frac{x+\Delta x-b}{\Delta x}} {z’(b + i\times \Delta x)})– (z(b) + \Delta x \times \sum \limits_{i=0}^{\frac{x-b}{\Delta x}} {z’(b + i\times \Delta x)}))$
Which can be written as,
$z’(x+\Delta x) = z’(x)$.
So what is the information that is being freely lost? It’s obvious that the constant $z(b)$ is the term present in both $z(x)$ expression and $z(x+\Delta x)$ expression. So the information that is lost is the starting point! From the article ‘fundamentals of differentiation, chapter 1’ know that $z’(x)$ is the rate of change of $z(x)$ for each change in $x$. So if we know the rate of change, we also need to know the point from where the function is to be generated on the graph; without which we will not be able to plot the right graph. Note that from $z’(x)$ we can construct $z(x)$ if we know $z(b)$, where $b$ being a constant that lies on the X axis and $z(b)$ being a constant that lies on the Z axis. So the term that gets lost is $z(b)$, so while integrating (or obtaining the function $z(x)$ from $z’(x)$), we will not be able to recover $z(b)$ and hence we just leave it as a constant. This unknown constant shifts the graph generated by $z(x)$ along the Z axis.
So if we write $g(x) = z(x)+c$ then the graph $g(x)$ is shifted upwards by $c$ units. So from the above explanation, it is obvious that we just lose the information that denotes the orientation of the graph along the Z axis.
copyright © 2015 K Sreram, All rights reserved
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# How do you solve 5^(x^2+2x)=125?
Dec 2, 2016
$x = - 3 , x = 1$
#### Explanation:
To solve the equation ${5}^{{x}^{2} + 2 x} = 125$ for variable $x$, we must apply the logarithm conversion formula.
logarithm to exponential conversion: $\setminus {\log}_{a} \left(b\right) = x \setminus \leftrightarrow {a}^{x} = b$
calculations
• ${5}^{{x}^{2} + 2 x} = 125 \setminus \Rightarrow \setminus {\log}_{5} \left(125\right) = {x}^{2} + 2 x$
• You're going to need a calculator to solve the left side's logarithm, but you should get the following:
$3 = {x}^{2} + 2 x$
• Now subtract 3 from both sides to form a quadratic equation:
$0 = {x}^{2} + 2 x - 3$
• You should get roots of -3 and 1. $\setminus \Rightarrow \left(x + 3\right) \left(x - 1\right)$
Therefore, $x = - 3 , 1$
Dec 4, 2016
$x = - 3$ or $x = 1$
#### Explanation:
Another way to approach this is to realize that $125 = {5}^{3}$. Then we see that
${5}^{{x}^{2} + 2 x} = {5}^{3}$
We now have two equal bases, each to a power. Since these are equal, we can say that their exponents must be equal. (We could write a rule for this and say that if ${a}^{b} = {a}^{c}$, then $b = c$).
So we know that
${x}^{2} + 2 x = 3$
${x}^{2} + 2 x - 3 = 0$
$\left(x + 3\right) \left(x - 1\right) = 0$
So $x = - 3$ or $x = 1$.
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# Linear Algebra/Eigenvalues and Eigenvectors/Solutions
## SolutionsEdit
Problem 1
For each, find the characteristic polynomial and the eigenvalues.
1. $\begin{pmatrix} 10 &-9 \\ 4 &-2 \end{pmatrix}$
2. $\begin{pmatrix} 1 &2 \\ 4 &3 \end{pmatrix}$
3. $\begin{pmatrix} 0 &3 \\ 7 &0 \end{pmatrix}$
4. $\begin{pmatrix} 0 &0 \\ 0 &0 \end{pmatrix}$
5. $\begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix}$
1. This
$0= \begin{vmatrix} 10-x &-9 \\ 4 &-2-x \end{vmatrix} =(10-x)(-2-x)-(-36)$
simplifies to the characteristic equation $x^2-8x+16=0$. Because the equation factors into $(x-4)^2$ there is only one eigenvalue $\lambda_1=4$.
2. $0=(1-x)(3-x)-8=x^2-4x-5$; $\lambda_1=5$, $\lambda_2=-1$
3. $x^2-21=0$; $\lambda_1=\sqrt{21}$, $\lambda_2=-\sqrt{21}$
4. $x^2=0$; $\lambda_1=0$
5. $x^2-2x+1=0$; $\lambda_1=1$
This exercise is recommended for all readers.
Problem 2
For each matrix, find the characteristic equation, and the eigenvalues and associated eigenvectors.
1. $\begin{pmatrix} 3 &0 \\ 8 &-1 \end{pmatrix}$
2. $\begin{pmatrix} 3 &2 \\ -1 &0 \end{pmatrix}$
1. The characteristic equation is $(3-x)(-1-x)=0$. Its roots, the eigenvalues, are $\lambda_1=3$ and $\lambda_2=-1$. For the eigenvectors we consider this equation.
$\begin{pmatrix} 3-x &0 \\ 8 &-1-x \end{pmatrix} \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} =\begin{pmatrix} 0 \\ 0 \end{pmatrix}$
For the eigenvector associated with $\lambda_1=3$, we consider the resulting linear system.
$\begin{array}{*{2}{rc}r} 0\cdot b_1 &+ &0\cdot b_2 &= &0 \\ 8\cdot b_1 &+ &-4\cdot b_2 &= &0 \end{array}$
The eigenspace is the set of vectors whose second component is twice the first component.
$\{\begin{pmatrix} b_2/2 \\ b_2 \end{pmatrix}\,\big|\, b_2\in\mathbb{C}\} \qquad \begin{pmatrix} 3 &0 \\ 8 &-1 \end{pmatrix} \begin{pmatrix} b_2/2 \\ b_2 \end{pmatrix} =3\cdot\begin{pmatrix} b_2/2 \\ b_2 \end{pmatrix}$
(Here, the parameter is $b_2$ only because that is the variable that is free in the above system.) Hence, this is an eigenvector associated with the eigenvalue $3$.
$\begin{pmatrix} 1 \\ 2 \end{pmatrix}$
Finding an eigenvector associated with $\lambda_2=-1$ is similar. This system
$\begin{array}{*{2}{rc}r} 4\cdot b_1 &+ &0\cdot b_2 &= &0 \\ 8\cdot b_1 &+ &0\cdot b_2 &= &0 \end{array}$
leads to the set of vectors whose first component is zero.
$\{\begin{pmatrix} 0 \\ b_2 \end{pmatrix}\,\big|\, b_2\in\mathbb{C}\} \qquad \begin{pmatrix} 3 &0 \\ 8 &-1 \end{pmatrix} \begin{pmatrix} 0 \\ b_2 \end{pmatrix} =-1\cdot\begin{pmatrix} 0 \\ b_2 \end{pmatrix}$
And so this is an eigenvector associated with $\lambda_2$.
$\begin{pmatrix} 0 \\ 1 \end{pmatrix}$
2. The characteristic equation is
$0= \begin{vmatrix} 3-x &2 \\ -1 &-x \end{vmatrix} =x^2-3x+2=(x-2)(x-1)$
and so the eigenvalues are $\lambda_1=2$ and $\lambda_2=1$. To find eigenvectors, consider this system.
$\begin{array}{*{2}{rc}r} (3-x)\cdot b_1 &+ &2\cdot b_2 &= &0 \\ -1\cdot b_1 &- &x\cdot b_2 &= &0 \end{array}$
For $\lambda_1=2$ we get
$\begin{array}{*{2}{rc}r} 1\cdot b_1 &+ &2\cdot b_2 &= &0 \\ -1\cdot b_1 &- &2\cdot b_2 &= &0 \end{array}$
leading to this eigenspace and eigenvector.
$\{\begin{pmatrix} -2b_2 \\ b_2 \end{pmatrix} \,\big|\, b_2\in\mathbb{C}\} \qquad \begin{pmatrix} -2 \\ 1 \end{pmatrix}$
For $\lambda_2=1$ the system is
$\begin{array}{*{2}{rc}r} 2\cdot b_1 &+ &2\cdot b_2 &= &0 \\ -1\cdot b_1 &- &1\cdot b_2 &= &0 \end{array}$
$\{\begin{pmatrix} -b_2 \\ b_2 \end{pmatrix} \,\big|\, b_2\in\mathbb{C}\} \qquad \begin{pmatrix} -1 \\ 1 \end{pmatrix}$
Problem 3
Find the characteristic equation, and the eigenvalues and associated eigenvectors for this matrix. Hint. The eigenvalues are complex.
$\begin{pmatrix} -2 &-1 \\ 5 &2 \end{pmatrix}$
The characteristic equation
$0= \begin{vmatrix} -2-x &-1 \\ 5 &2-x \end{vmatrix} =x^2+1$
has the complex roots $\lambda_1=i$ and $\lambda_2=-i$. This system
$\begin{array}{*{2}{rc}r} (-2-x)\cdot b_1 &- &1\cdot b_2 &= &0 \\ 5\cdot b_1 & &(2-x)\cdot b_2 &= &0 \end{array}$
For $\lambda_1=i$ Gauss' method gives this reduction.
$\begin{array}{*{2}{rc}r} (-2-i)\cdot b_1 &- &1\cdot b_2 &= &0 \\ 5\cdot b_1 &- &(2-i)\cdot b_2 &= &0 \end{array} \xrightarrow[]{(-5/(-2-i))\rho_1+\rho_2} \begin{array}{*{2}{rc}r} (-2-i)\cdot b_1 &- &1\cdot b_2 &= &0 \\ & &0 &= &0 \end{array}$
(For the calculation in the lower right get a common denominator
$\frac{5}{-2-i}-(2-i) = \frac{5}{-2-i}-\frac{-2-i}{-2-i}\cdot (2-i) = \frac{5-(-5)}{-2-i}$
to see that it gives a $0=0$ equation.) These are the resulting eigenspace and eigenvector.
$\{\begin{pmatrix} (1/(-2-i))b_2 \\ b_2 \end{pmatrix} \,\big|\, b_2\in\mathbb{C}\} \qquad \begin{pmatrix} 1/(-2-i) \\ 1 \end{pmatrix}$
For $\lambda_2=-i$ the system
$\begin{array}{*{2}{rc}r} (-2+i)\cdot b_1 &- &1\cdot b_2 &= &0 \\ 5\cdot b_1 &- &(2+i)\cdot b_2 &= &0 \end{array} \xrightarrow[]{(-5/(-2+i))\rho_1+\rho_2} \begin{array}{*{2}{rc}r} (-2+i)\cdot b_1 &- &1\cdot b_2 &= &0 \\ & &0 &= &0 \end{array}$
$\{\begin{pmatrix} (1/(-2+i))b_2 \\ b_2 \end{pmatrix} \,\big|\, b_2\in\mathbb{C}\} \qquad \begin{pmatrix} 1/(-2+i) \\ 1 \end{pmatrix}$
Problem 4
Find the characteristic polynomial, the eigenvalues, and the associated eigenvectors of this matrix.
$\begin{pmatrix} 1 &1 &1 \\ 0 &0 &1 \\ 0 &0 &1 \end{pmatrix}$
The characteristic equation is
$0= \begin{vmatrix} 1-x &1 &1 \\ 0 &-x &1 \\ 0 &0 &1-x \end{vmatrix} =(1-x)^2(-x)$
and so the eigenvalues are $\lambda_1=1$ (this is a repeated root of the equation) and $\lambda_2=0$. For the rest, consider this system.
$\begin{array}{*{3}{rc}r} (1-x)\cdot b_1 &+ &b_2 &+ &b_3 &= &0 \\ & &-x\cdot b_2 &+ &b_3 &= &0 \\ & & & &(1-x)\cdot b_3 &= &0 \end{array}$
When $x=\lambda_1=1$ then the solution set is this eigenspace.
$\{\begin{pmatrix} b_1 \\ 0 \\ 0 \end{pmatrix}\,\big|\, b_1\in\mathbb{C}\}$
When $x=\lambda_2=0$ then the solution set is this eigenspace.
$\{\begin{pmatrix} -b_2 \\ b_2 \\ 0 \end{pmatrix}\,\big|\, b_2\in\mathbb{C}\}$
So these are eigenvectors associated with $\lambda_1=1$ and $\lambda_2=0$.
$\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \qquad \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$
This exercise is recommended for all readers.
Problem 5
For each matrix, find the characteristic equation, and the eigenvalues and associated eigenvectors.
1. $\begin{pmatrix} 3 &-2 &0 \\ -2 &3 &0 \\ 0 &0 &5 \end{pmatrix}$
2. $\begin{pmatrix} 0 &1 &0 \\ 0 &0 &1 \\ 4 &-17 &8 \end{pmatrix}$
1. The characteristic equation is
$0= \begin{vmatrix} 3-x &-2 &0 \\ -2 &3-x &0 \\ 0 &0 &5-x \end{vmatrix} =x^3-11x^2+35x-25=(x-1)(x-5)^2$
and so the eigenvalues are $\lambda_1=1$ and also the repeated eigenvalue $\lambda_2=5$. To find eigenvectors, consider this system.
$\begin{array}{*{3}{rc}r} (3-x)\cdot b_1 &- &2\cdot b_2 & & &= &0 \\ -2\cdot b_1 &+ &(3-x)\cdot b_2 & & &= &0 \\ & & & &(5-x)\cdot b_3 &= &0 \end{array}$
For $\lambda_1=1$ we get
$\begin{array}{*{3}{rc}r} 2\cdot b_1 &- &2\cdot b_2 & & &= &0 \\ -2\cdot b_1 &+ &2\cdot b_2 & & &= &0 \\ & & & &4\cdot b_3 &= &0 \end{array}$
leading to this eigenspace and eigenvector.
$\{\begin{pmatrix} b_2 \\ b_2 \\ 0 \end{pmatrix} \,\big|\, b_2\in\mathbb{C}\} \qquad \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$
For $\lambda_2=5$ the system is
$\begin{array}{*{3}{rc}r} -2\cdot b_1 &- &2\cdot b_2 & & &= &0 \\ -2\cdot b_1 &- &2\cdot b_2 & & &= &0 \\ & & & &0\cdot b_3 &= &0 \end{array}$
$\{\begin{pmatrix} -b_2 \\ b_2 \\ 0 \end{pmatrix}+\begin{pmatrix} 0 \\ 0 \\ b_3 \end{pmatrix} \,\big|\, b_2,b_3\in\mathbb{C}\} \qquad \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix},\,\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$
2. The characteristic equation is
$0= \begin{vmatrix} -x &1 &0 \\ 0 &-x &1 \\ 4 &-17 &8-x \end{vmatrix} =-x^3+8x^2-17x+4=-1\cdot(x-4)(x^2-4x+1)$
and the eigenvalues are $\lambda_1=4$ and (by using the quadratic equation) $\lambda_2=2+\sqrt{3}$ and $\lambda_3=2-\sqrt{3}$. To find eigenvectors, consider this system.
$\begin{array}{*{3}{rc}r} -x\cdot b_1 &+ &b_2 & & &= &0 \\ & &-x\cdot b_2 &+ &b_3 &= &0 \\ 4\cdot b_1 &- &17\cdot b_2 &+ &(8-x)\cdot b_3 &= &0 \end{array}$
Substituting $x=\lambda_1=4$ gives the system
$\begin{array}{*{3}{rc}r} -4\cdot b_1 &+ &b_2 & & &= &0 \\ & &-4\cdot b_2 &+ &b_3 &= &0 \\ 4\cdot b_1 &- &17\cdot b_2 &+ &4\cdot b_3 &= &0 \end{array} \xrightarrow[]{\rho_1+\rho_3} \begin{array}{*{3}{rc}r} -4\cdot b_1 &+ &b_2 & & &= &0 \\ & &-4\cdot b_2 &+ &b_3 &= &0 \\ & &-16\cdot b_2 &+ &4\cdot b_3 &= &0 \end{array} \xrightarrow[]{-4\rho_2+\rho_3} \begin{array}{*{3}{rc}r} -4\cdot b_1 &+ &b_2 & & &= &0 \\ & &-4\cdot b_2 &+ &b_3 &= &0 \\ & & & &0 &= &0 \end{array}$
leading to this eigenspace and eigenvector.
$V_4=\{\begin{pmatrix} (1/16)\cdot b_3 \\ (1/4)\cdot b_3 \\ b_3 \end{pmatrix} \,\big|\, b_2\in\mathbb{C}\} \qquad \begin{pmatrix} 1 \\ 4 \\ 16 \end{pmatrix}$
Substituting $x=\lambda_2=2+\sqrt{3}$ gives the system
$\begin{array}{*{3}{rc}r} (-2-\sqrt{3})\cdot b_1 &+ &b_2 & & &= &0 \\ & &(-2-\sqrt{3})\cdot b_2 &+ &b_3 &= &0 \\ 4\cdot b_1 &- &17\cdot b_2 &+ &(6-\sqrt{3})\cdot b_3 &= &0 \end{array}$
$\xrightarrow[]{(-4/(-2-\sqrt{3}))\rho_1+\rho_3} \begin{array}{*{3}{rc}r} (-2-\sqrt{3})\cdot b_1 &+ &b_2 & & &= &0 \\ & &(-2-\sqrt{3})\cdot b_2 &+ &b_3 &= &0 \\ &+ &(-9-4\sqrt{3})\cdot b_2 &+ &(6-\sqrt{3})\cdot b_3 &= &0 \end{array}$
(the middle coefficient in the third equation equals the number $(-4/(-2-\sqrt{3}))-17$; find a common denominator of $-2-\sqrt{3}$ and then rationalize the denominator by multiplying the top and bottom of the frsction by $-2+\sqrt{3}$)
$\xrightarrow[]{((9+4\sqrt{3})/(-2-\sqrt{3}))\rho_2+\rho_3} \begin{array}{*{3}{rc}r} (-2-\sqrt{3})\cdot b_1 &+ &b_2 & & &= &0 \\ & &(-2-\sqrt{3})\cdot b_2 &+ &b_3 &= &0 \\ & & & &0 &= &0 \end{array}$
which leads to this eigenspace and eigenvector.
$V_{2+\sqrt{3}} =\{\begin{pmatrix} (1/(2+\sqrt{3})^2)\cdot b_3 \\ (1/(2+\sqrt{3}))\cdot b_3 \\ b_3 \end{pmatrix} \,\big|\, b_3\in\mathbb{C}\} \qquad \begin{pmatrix} (1/(2+\sqrt{3})^2) \\ (1/(2+\sqrt{3})) \\ 1 \end{pmatrix}$
Finally, substituting $x=\lambda_3=2-\sqrt{3}$ gives the system
$\begin{array}{*{3}{rc}r} (-2+\sqrt{3})\cdot b_1 &+ &b_2 & & &= &0 \\ & &(-2+\sqrt{3})\cdot b_2 &+ &b_3 &= &0 \\ 4\cdot b_1 &- &17\cdot b_2 &+ &(6+\sqrt{3})\cdot b_3 &= &0 \end{array}$
\begin{align} &\xrightarrow[]{(-4/(-2+\sqrt{3}))\rho_1+\rho_3} \begin{array}{*{3}{rc}r} (-2+\sqrt{3})\cdot b_1 &+ &b_2 & & &= &0 \\ & &(-2+\sqrt{3})\cdot b_2 &+ &b_3 &= &0 \\ & &(-9+4\sqrt{3})\cdot b_2 &+ &(6+\sqrt{3})\cdot b_3 &= &0 \end{array} \\ &\xrightarrow[]{((9-4\sqrt{3})/(-2+\sqrt{3}))\rho_2+\rho_3} \begin{array}{*{3}{rc}r} (-2+\sqrt{3})\cdot b_1 &+ &b_2 & & &= &0 \\ & &(-2+\sqrt{3})\cdot b_2 &+ &b_3 &= &0 \\ & & & &0 &= &0 \end{array} \end{align}
which gives this eigenspace and eigenvector.
$V_{2-\sqrt{3}} =\{\begin{pmatrix} (1/(2+\sqrt{3})^2)\cdot b_3 \\ (1/(2-\sqrt{3}))\cdot b_3 \\ b_3 \end{pmatrix} \,\big|\, b_3\in\mathbb{C}\} \qquad \begin{pmatrix} (1/(-2+\sqrt{3})^2) \\ (1/(-2+\sqrt{3})) \\ 1 \end{pmatrix}$
This exercise is recommended for all readers.
Problem 6
Let $t:\mathcal{P}_2\to \mathcal{P}_2$ be
$a_0+a_1x+a_2x^2\mapsto (5a_0+6a_1+2a_2)-(a_1+8a_2)x+(a_0-2a_2)x^2.$
Find its eigenvalues and the associated eigenvectors.
With respect to the natural basis $B=\langle 1,x,x^2 \rangle$ the matrix representation is this.
${\rm Rep}_{B,B}(t) = \begin{pmatrix} 5 &6 &2 \\ 0 &-1 &-8 \\ 1 &0 &-2 \end{pmatrix}$
Thus the characteristic equation
$0 = \begin{pmatrix} 5-x &6 &2 \\ 0 &-1-x &-8 \\ 1 &0 &-2-x \end{pmatrix} =(5-x)(-1-x)(-2-x)-48-2\cdot(-1-x)$
is $0=-x^3+2x^2+15x-36=-1\cdot (x+4)(x-3)^2$. To find the associated eigenvectors, consider this system.
$\begin{array}{*{3}{rc}r} (5-x)\cdot b_1 &+ &6\cdot b_2 &+ &2\cdot b_3 &= &0 \\ & &(-1-x)\cdot b_2 &- &8\cdot b_3 &= &0 \\ b_1 & & &+ &(-2-x)\cdot b_3 &= &0 \end{array}$
Plugging in $x=\lambda_1=4$ gives
$\begin{array}{*{3}{rc}r} b_1 &+ &6\cdot b_2 &+ &2\cdot b_3 &= &0 \\ & &-5 \cdot b_2 &- &8\cdot b_3 &= &0 \\ b_1 & & &- & 6 \cdot b_3 &= &0 \end{array} \xrightarrow[]{-\rho_1+\rho_2} \begin{array}{*{3}{rc}r} b_1 &+ &6\cdot b_2 &+ &2\cdot b_3 &= &0 \\ & &-5 \cdot b_2 &- &8\cdot b_3 &= &0 \\ & &-6\cdot b_2 &- &8 \cdot b_3 &= &0 \end{array} \xrightarrow[]{-\rho_1+\rho_2} \begin{array}{*{3}{rc}r} b_1 &+ &6\cdot b_2 &+ &2\cdot b_3 &= &0 \\ & &-5 \cdot b_2 &- &8\cdot b_3 &= &0 \\ & &-6\cdot b_2 &- &8 \cdot b_3 &= &0 \end{array}$
Problem 7
Find the eigenvalues and eigenvectors of this map $t:\mathcal{M}_2\to \mathcal{M}_2$.
$\begin{pmatrix} a &b \\ c &d \end{pmatrix} \mapsto \begin{pmatrix} 2c &a+c \\ b-2c &d \end{pmatrix}$
$\lambda=1, \begin{pmatrix} 0 &0 \\ 0 &1 \end{pmatrix} \text{ and } \begin{pmatrix} 2 &3 \\ 1 &0 \end{pmatrix}$, $\lambda=-2, \begin{pmatrix} -1 &0 \\ 1 &0 \end{pmatrix}$, $\lambda=-1, \begin{pmatrix} -2 &1 \\ 1 &0 \end{pmatrix}$
This exercise is recommended for all readers.
Problem 8
Find the eigenvalues and associated eigenvectors of the differentiation operator $d/dx:\mathcal{P}_3\to \mathcal{P}_3$.
Fix the natural basis $B=\langle 1,x,x^2,x^3 \rangle$. The map's action is $1\mapsto 0$, $x\mapsto 1$, $x^2\mapsto 2x$, and $x^3\mapsto 3x^2$ and its representation is easy to compute.
$T={\rm Rep}_{B,B}(d/dx)= \begin{pmatrix} 0 &1 &0 &0 \\ 0 &0 &2 &0 \\ 0 &0 &0 &3 \\ 0 &0 &0 &0 \end{pmatrix}_{B,B}$
We find the eigenvalues with this computation.
$0=\left|T-xI\right|= \begin{vmatrix} -x &1 &0 &0 \\ 0 &-x &2 &0 \\ 0 &0 &-x &3 \\ 0 &0 &0 &-x \end{vmatrix} =x^4$
Thus the map has the single eigenvalue $\lambda=0$. To find the associated eigenvectors, we solve
$\begin{pmatrix} 0 &1 &0 &0 \\ 0 &0 &2 &0 \\ 0 &0 &0 &3 \\ 0 &0 &0 &0 \end{pmatrix}_{B,B} \begin{pmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{pmatrix}_B =0\cdot\begin{pmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{pmatrix}_B \qquad\Longrightarrow\qquad b_2=0, b_3=0, b_4=0$
to get this eigenspace.
$\{\begin{pmatrix} b_1 \\ 0 \\ 0 \\ 0 \end{pmatrix}_B \,\big|\, b_1\in\mathbb{C}\} =\{b_1+0\cdot x+0\cdot x^2+0\cdot x^3 \,\big|\, b_1\in\mathbb{C}\} =\{b_1 \,\big|\, b_1\in\mathbb{C}\}$
Problem 9
Prove that
the eigenvalues of a triangular matrix (upper or lower triangular) are the entries on the diagonal.
The determinant of the triangular matrix $T-xI$ is the product down the diagonal, and so it factors into the product of the terms $t_{i,i}-x$.
This exercise is recommended for all readers.
Problem 10
Find the formula for the characteristic polynomial of a $2 \! \times \! 2$ matrix.
Just expand the determinant of $T-xI$.
$\begin{vmatrix} a-x &c \\ b &d-x \end{vmatrix} =(a-x)(d-x)-bc =x^2+(-a-d)\cdot x +(ad-bc)$
Problem 11
Prove that the characteristic polynomial of a transformation is well-defined.
Any two representations of that transformation are similar, and similar matrices have the same characteristic polynomial.
This exercise is recommended for all readers.
Problem 12
1. Can any non-$\vec{0}$ vector in any nontrivial vector space be a eigenvector? That is, given a $\vec{v}\neq\vec{0}$ from a nontrivial $V$, is there a transformation $t:V\to V$ and a scalar $\lambda\in\mathbb{R}$ such that $t(\vec{v})=\lambda\vec{v}$?
2. Given a scalar $\lambda$, can any non-$\vec{0}$ vector in any nontrivial vector space be an eigenvector associated with the eigenvalue $\lambda$?
1. Yes, use $\lambda=1$ and the identity map.
2. Yes, use the transformation that multiplies by $\lambda$.
This exercise is recommended for all readers.
Problem 13
Suppose that $t:V\to V$ and $T={\rm Rep}_{B,B}(t)$. Prove that the eigenvectors of $T$ associated with $\lambda$ are the non-$\vec{0}$ vectors in the kernel of the map represented (with respect to the same bases) by $T-\lambda I$.
If $t(\vec{v})=\lambda\cdot\vec{v}$ then $\vec{v}\mapsto\vec{0}$ under the map $t-\lambda\cdot\mbox{id}$.
Problem 14
Prove that if $a,\ldots,\,d$ are all integers and $a+b=c+d$ then
$\begin{pmatrix} a &b \\ c &d \end{pmatrix}$
has integral eigenvalues, namely $a+b$ and $a-c$.
The characteristic equation
$0= \begin{vmatrix} a-x &b \\ c &d-x \end{vmatrix} =(a-x)(d-x)-bc$
simplifies to $x^2+(-a-d)\cdot x + (ad-bc)$. Checking that the values $x=a+b$ and $x=a-c$ satisfy the equation (under the $a+b=c+d$ condition) is routine.
This exercise is recommended for all readers.
Problem 15
Prove that if $T$ is nonsingular and has eigenvalues $\lambda_1,\dots,\lambda_n$ then $T^{-1}$ has eigenvalues $1/\lambda_1,\dots,1/\lambda_n$. Is the converse true?
Consider an eigenspace $V_{\lambda}$. Any $\vec{w}\in V_{\lambda}$ is the image $\vec{w}=\lambda\cdot\vec{v}$ of some $\vec{v}\in V_{\lambda}$ (namely, $\vec{v}=(1/\lambda)\cdot\vec{w}$). Thus, on $V_{\lambda}$ (which is a nontrivial subspace) the action of $t^{-1}$ is $t^{-1}(\vec{w})=\vec{v}=(1/\lambda)\cdot\vec{w}$, and so $1/\lambda$ is an eigenvalue of $t^{-1}$.
This exercise is recommended for all readers.
Problem 16
Suppose that $T$ is $n \! \times \! n$ and $c,d$ are scalars.
1. Prove that if $T$ has the eigenvalue $\lambda$ with an associated eigenvector $\vec{v}$ then $\vec{v}$ is an eigenvector of $cT+dI$ associated with eigenvalue $c\lambda+d$.
2. Prove that if $T$ is diagonalizable then so is $cT+dI$.
1. We have $(cT+dI)\vec{v}=cT\vec{v}+dI\vec{v}=c\lambda\vec{v}+d\vec{v} =(c\lambda+d)\cdot \vec{v}$.
2. Suppose that $S=PTP^{-1}$ is diagonal. Then $P(cT+dI)P^{-1}=P(cT)P^{-1}+P(dI)P^{-1} =cPTP^{-1}+dI=cS+dI$ is also diagonal.
This exercise is recommended for all readers.
Problem 17
Show that $\lambda$ is an eigenvalue of $T$ if and only if the map represented by $T-\lambda I$ is not an isomorphism.
The scalar $\lambda$ is an eigenvalue if and only if the transformation $t-\lambda \mbox{id}$ is singular. A transformation is singular if and only if it is not an isomorphism (that is, a transformation is an isomorphism if and only if it is nonsingular).
Problem 18
1. Show that if $\lambda$ is an eigenvalue of $A$ then $\lambda^k$ is an eigenvalue of $A^k$.
2. What is wrong with this proof generalizing that? "If $\lambda$ is an eigenvalue of $A$ and $\mu$ is an eigenvalue for $B$, then $\lambda\mu$ is an eigenvalue for $AB$, for, if $A\vec{x}=\lambda\vec{x}$ and $B\vec{x}=\mu\vec{x}$ then $AB\vec{x}=A\mu\vec{x}=\mu A\vec{x}\mu\lambda\vec{x}$"?
(Strang 1980)
1. Where the eigenvalue $\lambda$ is associated with the eigenvector $\vec{x}$ then $A^k\vec{x}=A\cdots A\vec{x}=A^{k-1}\lambda\vec{x} =\lambda A^{k-1}\vec{x}=\cdots=\lambda^k\vec{x}$. (The full details can be put in by doing induction on $k$.)
2. The eigenvector associated wih $\lambda$ might not be an eigenvector associated with $\mu$.
Problem 19
Do matrix-equivalent matrices have the same eigenvalues?
No. These are two same-sized, equal rank, matrices with different eigenvalues.
$\begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix} \qquad \begin{pmatrix} 1 &0 \\ 0 &2 \end{pmatrix}$
Problem 20
Show that a square matrix with real entries and an odd number of rows has at least one real eigenvalue.
The characteristic polynomial has an odd power and so has at least one real root.
Problem 21
Diagonalize.
$\begin{pmatrix} -1 &2 &2 \\ 2 &2 &2 \\ -3 &-6 &-6 \end{pmatrix}$
The characteristic polynomial $x^3-5x^2+6x$ has distinct roots $\lambda_1=0$, $\lambda_2=-2$, and $\lambda_3=-3$. Thus the matrix can be diagonalized into this form.
$\begin{pmatrix} 0 &0 &0 \\ 0 &-2 &0 \\ 0 &0 &-3 \end{pmatrix}$
Problem 22
Suppose that $P$ is a nonsingular $n \! \times \! n$ matrix. Show that the similarity transformation map $t_P:\mathcal{M}_{n \! \times \! n}\to \mathcal{M}_{n \! \times \! n}$ sending $T\mapsto PTP^{-1}$ is an isomorphism.
We must show that it is one-to-one and onto, and that it respects the operations of matrix addition and scalar multiplication.
To show that it is one-to-one, suppose that $t_P(T)=t_P(S)$, that is, suppose that $PTP^{-1}=PSP^{-1}$, and note that multiplying both sides on the left by $P^{-1}$ and on the right by $P$ gives that $T=S$. To show that it is onto, consider $S\in\mathcal{M}_{n \! \times \! n}$ and observe that $S=t_P(P^{-1}SP)$.
The map $t_P$ preserves matrix addition since $t_P(T+S)=P(T+S)P^{-1}=(PT+PS)P^{-1}=PTP^{-1}+PSP^{-1}=t_P(T+S)$ follows from properties of matrix multiplication and addition that we have seen. Scalar multiplication is similar: $t_P(cT)=P(c\cdot T)P^{-1}=c\cdot (PTP^{-1})=c\cdot t_P(T)$.
? Problem 23
Show that if $A$ is an $n$ square matrix and each row (column) sums to $c$ then $c$ is a characteristic root of $A$. (Morrison 1967)
If the argument of the characteristic function of $A$ is set equal to $c$, adding the first $(n-1)$ rows (columns) to the $n$th row (column) yields a determinant whose $n$th row (column) is zero. Thus $c$ is a characteristic root of $A$.
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# 23. Trigonometrical Equations and their General Solutions¶
## 23.1. Trigonometrical Equation¶
An equation involving one or more trigonometrical ratios of unknown angle is called trigonometrical equation.
Ex. $\cos^2x - 4\sin x = 1$
A trigonometrical identity is satisfied for every value of the unknown angle whereas trigonometrical equation is satisfied for only some values of unknown angle. For example, $1 - \cos^2x = \sin^2x$ is a trigonometrical identity because it is satisfied for every value of $x.$
## 23.2. Solution of a Trigonometrical Equation¶
A value of the unknown angle which satisfies the given trigonometrical equation is called a solution or root of the equation.
For example, $2\sin\theta = 1 \Rightarrow \theta = 30^\circ, 150^\circ$ which are two solutions between $0$ and $2\pi.$
## 23.3. General Solution¶
Some trigonometrical functions are periodic functions, therefore, solutions of trigonometrical equations can be generalized with the help of periodicirty of trigonometrical functions. The solution consisting of all possible solutions of a trigonometrical equation is called its general solution.
For example, $\sin\theta = 0$ has a genral solution which is $n\pi$ where $n\in I.$
Similarly, fo $\cos\theta = 0,$ the general solution is $(2n + 1)\frac{\pi}{2},$ where $n\in I$ and for $\tan\theta = 0$ the solution is again $n\pi.$
### 23.3.1. General Solution of $\sin\theta = \sin\alpha$¶
Given, $\sin\theta = \sin\alpha \Rightarrow \sin\theta - \sin\alpha = 0$
$\Rightarrow 2\cos\frac{\theta + \alpha}{2}\sin\frac{\theta - \alpha}{2} = 0$
Case I $\cos\frac{\theta + \alpha}{2} = 0$
$\Rightarrow \theta + \alpha = (2m + 1)\pi, m\in I$
Case II $\sin\frac{\theta - \alpha}{2} = 0$
$\Rightarrow \theta - \alpha = 2m\pi \Rightarrow \theta = 2m\pi + \alpha$
Thus, $\theta = n\pi + (-1)^n\alpha, n\in I$
### 23.3.2. General Solution of $\cos\theta = \cos\alpha$¶
Given, $\cos\theta = \cos\alpha \Rightarrow \cos\theta - \cos\alpha = 0$
$\Rightarrow 2\sin\frac{\alpha + \theta}{2}\sin\frac{\theta - \alpha}{2} = 0$
Case I: $\sin\frac{\alpha + \theta}{2} = 0$
$\alpha + \theta = 2n\pi \Rightarrow \theta = 2n\pi - \alpha$
Case II: $\sin\frac{\theta - \alpha}{2} = 0 \Rightarrow \theta = 2n\pi + \alpha$
Thus, $\theta = 2n\pi \pm\alpha$
### 23.3.3. General Solution of $\tan\theta = \tan\alpha$¶
Given $\tan\theta = \tan\alpha \Rightarrow \frac{\sin\theta}{\cos\theta} = \frac{\sin\alpha}{\cos\alpha}$
$\Rightarrow \sin(\theta - \alpha) = 0 \therefore \theta - \alpha = n\pi$
$\theta = n\pi + \alpha$
## 23.4. Principal Value¶
For any equation having multiple solutions, the solution having least numerical value is known as principal value.
Example: Let $\sin\theta = \frac{1}{2}$ then $\theta = \pi/6, 5\pi/6, 13\pi/6, 17\pi/6, \ldots, -7\pi/6, -11\pi/6, \ldost$
As $\pi/6$ is the least numerical value so it is the principal value in this case.
### 23.4.1. Method for Finding Principal Value¶
For this case we consider $\sin\theta = -\frac{1}{2}.$ Since it is negative, $\theta$ will be in third or fourth quadrant. We can approach this either using clockwise direction or annticlockwise direction. If we take anticlockwise direction principal value will be greater than $\pi$ and in case of clockwise direction it will be less than $\pi.$ For principal value, we have to take numerically smallest angle.
So for principal value:
1. If the angle is in 1st or 2nd quadrant we must select anticlockwise direction i.e. principal value will be positive. If the angle is in 3rd or 4th quadrant we must select clockwise direction i.e. principal value will be negative.
2. Principal value is always numerically smaller than $\pi$
3. Principal values always lies in the first circle i.e. first rotation.
## 23.5. Tips for Finding Complete Solution¶
1. There should be no extraneous root.
2. There should be no less root.
3. Squaring should be avoided as far as possible. If it is done then check for extraneous roots.
4. Never cancel equal terms containing unknown on two sides which are in product. It may cause root loss.
5. The answer should not contain such values of root which may make any of the terms undefined.
6. Domain should not change. If it changes, necessary correction must be made.
7. Check that denominator is not zero at any stage while solving equations.
## 23.6. Problems¶
Find the most general values of $\theta$ satisfying the equations:
1. $\sin\theta = -1$
2. $\cos\theta = -\frac{1}{2}$
3. $\tan\theta = \sqrt{3}$
4. $\sec\theta = -\sqrt{2}$
Solve the equations:
1. $\sin9\theta = \sin\theta$
2. $\sin5x = \cos2x$
3. $\sin3x = \sin x$
4. $\sin3x = \cos2x$
5. $\sin ax + \cos bx = 0$
6. $\tan x\tan4x = 1$
7. $\cos\theta = \sin105^\circ + \cos 105^\circ$
Solve the following:
1. $7\cos^2\theta + 3\sin^2\theta = 4$
2. $3\tan(\theta - 15^\circ) = \tan(\theta + 15^\circ)$
3. $\tan x + \cot x = 2$
4. $\sin^2\theta = \sin^2\alpha$
5. $\tan^2x + \cot^2x = 2$
6. $\tan^2x = 3\cosec^2x - 1$
7. $2\sin^2x + \sin^22x = 2$
8. $7\cos^2x + 3\sin^2x = 4$
9. $2\cos2x + \sqrt{2\sin x} = 2$
10. $8\tan^2\frac{x}{2} = 1 + \sec x$
11. $\cos x\cos2x\cos3x = \frac{1}{4}$
12. $\tan x + \tan2x + \tan3x = 0$
13. $\cot x - \tan x - \cos x + \sin x = 0$
14. $2\sin^2x - 5\sin x\cos x - 8\cos^2x = -2$
15. $(1 - \tan x)(1 + \sin2x) = 1 + \tan x$
16. Solve for x,($-\pi \leq x \leq \pi$), the equation $2(\cos x + \cos2x) + \sin2x(1 + 2\cos x) = 2\sin x$
17. Find all the solutions of the equation $4\cos^2x\sin x - 2\sin^2x = 3\sin x$
18. $2 + 7\tan^2x = 3.25\sec^2x$
19. Find all the values of $x$ for which $\cos 2x + \cos 4x = 2\cos x$
20. $3\tan x + \cot x = 5\cosec x$
21. Find the value of $x$ between $0$ and $2\pi$ for which $2\sin^2x = 3\cos x$
22. Find the solution of $\sin^2x - \cos x = \frac{1}{4}$ in the interbal $0$ to $2\pi.$
23. Solve $3\tan^2x - 2\sin x = 0$
24. Find all values of $x$ satisfying the equation $\sin x + \sin5x = \sin 3x$ between $0$ and $\pi.$
25. $\sin6x = \sin4x - \sin2x$
26. $\cos6x + \cos 4x + \cos 2x + 1 = 0$
27. $\cos x + \cos 2x + \cos 3x = 0$
28. Find the values of $x$ between $0$ and $2\pi,$ for which $\cos3x + \cos2x = \sin\frac{3x}{2} + \sin\frac{x}{2}$
29. $\tan x+ \tan2x + \tan3x = \tan x.\tan2x.\tan3x$
30. $\tan x + \tan 2x + \tan x\tan 2x = 1$
31. $\sin2x + \cos2x + \sin x + \cos x + 1 = 0$
32. $\sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x$
33. $\cos6x + \cos4x = \sin3x + \sin x$
34. $\sec4x - \sec2x = 2$
35. $\cos2x = (\sqrt{2} + 1)\left(\cos x - \frac{1}{\sqrt{2}}\right)$
36. Find all the angles between $-pi$ and $\pi$ for which $5\cos2x + 2\cos^2\frac{x}{2} + 1 = 0$
37. $\cot x - \tan x = \sec x$
38. $1 + \sec x = \cot^2\frac{x}{2}$
39. $\cos3x\cos^3x + \sin3x\sin^3x = 0$
40. $\sin^3x + \sin x\cos x + \cos^3x = 1$
41. Find all the value of $x$ between $0$ and $\frac{\pi}{2},$ for which $\sin 7x + \sin4x + \sin x = 0$
42. $\sin x + \sqrt{3}\cos x = \sqrt{2}$
43. Find the values of $x$ for which $27^{\cos2x}.81^{\sin2x}$ is minimum. Also, find this minimum value.
44. If $32\tan^8x = 2\cos^2y - 3\cos y$ and $3\cos2x = 1,$ then find the general value of $y.$
45. Find all the values of $x$ in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ for which $(1 - \tan x)(1 + \tan x)sec^2x + 2^{\tan^2x} = 0$
46. Solve the equation $e^{\cos x} = e^{-\cos x} + 4.$
47. If $(1 + \tan x)(1 + \tan y) = 2.$ Find all the values of $x + y.$
48. If $\tan(\cot x) = \cot(\tan x),$ prove that $\sin 2x = \frac{4}{(2n + 1)\pi}$
49. If $x$ and $y$ are two distinct roots of the equation $a\tan z+ b\sec z = c.$ Prove that $\tan(x + y) = \frac{2ac}{a^2 - c^2}$
50. If $\sin(\pi\cos x) = \cos(\pi\sin x),$ prove that 1. $\cos\left(x \pm \frac{\pi}{4}\right) = \frac{1}{2\sqrt{2}}$ 2. $\sin2x = -\frac{3}{4}$
51. Determine the smallest positive values of $x$ for which $\tan(x + 100^\circ) = \tan(x + 50^\circ).\tan x.\tan(x - 50^\circ)$
52. Find the general value of $x$ for which $\tan^2x + \sec 2x = 1.$
53. Solve the equation $\sec x - \cosec x = \frac{4}{3}$
54. Find solutions $x\in[0, 2\pi]$ of equation $\sin2x - 12(\sin x - \cos x) + 12 = 0.$
55. Find the smallest positive number r:math:p for which the equation $\cos(p\sin x) = \sin(p\cos x)$ has a solution for $x\in [0, 2\pi].$
56. Solve $\cos x + \sqrt{3}\sin x = 2\cos2x$
57. Solve $\tan x+ \sec x = \sqrt{3}$ for $x\in[0, 2\pi].$
58. Solve $1 + \sin^3x + \cos^3x = \frac{3}{2}\sin2x$
59. Solve the equation $(2 + \sqrt{3})\cos x = 1 - \sin x$
60. Solve the equation $\tan\left(\frac{\pi}{2}\sin x\right) = \cot\left(\frac{\pi}{2}\cos x\right)$
61. Solve $8\cos x\cos2x\cos4x = \frac{\sin6x}{\sin x}$
62. Solve $3 - 2\cos x - 4\sin x -\cos 2x + \sin 2x = 0$
63. Solve $\sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos 3x$
64. Solve $\sin^2x\tan x + \cos^2x\cot x - \sin 2x = 1 + \tan x + \cot x$
65. Find the most general value of $x$ which satisfies both the equations $\sin x= -\frac{1}{2}$ and $\tan x = \frac{1}{\sqrt{3}}$
66. If $\tan(x - y) = 1$ and $\sec(x + y) = \frac{2}{\sqrt{3}},$ find the smallest positive values of $x$ and $y$ and their most general value.
67. Find the points of intersection of the curves $y = \cos x$ and $y = \sin 3x$ if $-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}.$
68. Find all values of $x\in [0, 2\pi]$ such that $r\sin x = \sqrt{3}$ and $r + 4\sin x = 2(\sqrt{3} + 1)$
69. Find the smallest positive values of $x$ and $y$ satisfying $x - y = \frac{\pi}{4}$ and $\cot x + \cot y = 2.$
70. Find the general values of $x$ and $y$ such that $5\sin x\cos y = 1$ and $4\tan x = \tan y.$
71. Find all values of $x$ lying between $0$ and $2\pi,$ such that $r\sin x = 3$ and $r = 4(1 + \sin x)$
72. If $\sin x = \sin y$ and $\cos x = \cos y$ then prove that either $x = y$ or $x - y = 2n\pi,$ where $n\in I.$
73. If $\cos(x - y) = \frac{1}{2}$ and $\sin(x + y) = \frac{1}{2}$ find the smallest positive values of $x$ and $y$ and also their most general values.
74. Find the points of intersection of the curves $y = \cos 2x$ and $y = \sin x$ for, $-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}.$
75. Find the most general value of $x$ which satisfies the equations $\cos x = \frac{1}{\sqrt{2}}$ and $\tan x = -1.$
76. Find the most general value of $x$ which satisfies the equations $\tan x = \sqrt{3}$ and $\cosec x = -\frac{2}{\sqrt{3}}$
77. If $x$ and $y$ be two distinct values of $z$ lying between $0$ and $2\pi,$ satisfying the equation $3\cos z + 4\sin z = 2,$ find the value of $\sin(x + y).$
78. Show that the equation $2\cos^2\frac{x}{2}\sin^2x = x^2+ x^{-2}$ for $0 has no real solution.
79. Find the real value of $x$ such that $y = \frac{3 + 2i\sin x}{1 - 2i\sin x}$ is either real or purely imaginary.
80. Determine for which values of $a$ the equation $a^2 - 2a + \sec^2\pi(a + x) = 0$ has solutions and find them.
81. Find the values of $x$ in $(-\pi, \pi)$ which satisfy the equation $8^{1 + |\cos x| + \cos^2x + |\cos^3 x| + \ldots \text{~to~}\infty} = 4^3$
82. Solve $|\cos x|^{\sin^2x - \frac{3}{2}\sin x + \frac{1}{2}} = 1.$
83. Solve $3^{\sin2x + 2\cos^2x} + 3^{1 -\sin2x + 2\sin^2x} = 28.$
84. If $A = (x/2\cos^2x + \sin x\leq 2)$ and $B = \left(x/\frac{\pi}{2}\leq x\leq \frac{3\pi}{2}\right)$ find $A\cap B$
85. Solve $\sin x + \cos x = 1 + \sin x\cos x.$
86. Solve $\sin6x + \cos4x + 2 = 0.$
87. Prove that the equation $\sin2x + \sin3x + \ldots + \sin nx = n - 1$ has n solution for any arbitrary integer $n>2.$
88. Solve $\cos^7x + \sin^4x = 1.$
89. Find the number of solutions of the equation $\sin x + 2\sin2x = 3 + \sin3x$ in the interval $0\leq x\leq \pi.$
90. For what value of $k$ the equation $\sin x + \cos(k + x) + \cos(k - x) = 2$ has real solutions.
91. Solve for $x$ and $y,$ the equation $x\cos^3y + 3x\cos y.\sin^2y = 14$ and $x\sin^3y + 3x\cos^2y\sin y = 13$
92. Find all the values of $\alpha$ for which the equation $\sin^4x + \cos^4x + \sin2x + \alpha = 0$ is valid.
93. Solve $\tan\left(x + \frac{\pi}{4}\right) = 2\cot x - 1.$
94. If $x, y$ be two angles both satisfying the equation $a\cos 2z + b\sin2z = c,$ prove that $\cos^2x + \cos^2y = \frac{a^2 + ac + b^2}{a^2 + b^2}$
95. If $x_1, x_2, x_3, x_4$ be roots of the equation $\sin(x + y) = k\sin 2x,$ no two of which differ by a multiple of $2\pi,$ prove that $x_1 + x_2 + x_3 + x_4 = (2n + 1)\pi.$
96. Show that the equation $\sec x + \cosec x = c$ has two roots between $0$ and $\pi$ if $c^2<8$ and four roots if $c^2 > 8.$
97. Let $\lambda$ and $\alpha$ be real. Find the set of all values of $\lambda$ for which the system of linear equations $\lambda x + y\sin\alpha + z\cos\alpha = 0, x + y\cos\alpha + z\sin\alpha = 0, -x + y\sin\alpha - z\cos\alpha = 0$ has non-trivial solution. For $\lambda = 1,$ find all the values of $\alpha.$
98. Find the values of $x$ and $y, 0 satisfying the equation $\cos x \cos y\cos(x + y) = -\frac{1}{8}$
99. Find the number of distinct real roots of $\begin{vmatrix}\sin x& \cos x & \cos x \\\cos x & \sin x & \cos x\\\cos x & \cos x & \sin x\end{vmatrix} = 0$ in the interval $-\frac{\pi}{4}\leq x\leq \frac{\pi}{4}.$
100. Find the number of values of $x$ in the interval $[0, 5\pi]$ satisfying the equation $3\sin^2x - 7\sin x + 2 = 0.$
101. Find the range of $y$ such that the following equation in $x,$ $y + \cos x = \sin x$ has a real solution. For $y = 1,$ find $x$ such that $0\leq x\leq2\pi.$
102. Solve $\sum_{r = 1}^n\sin(rx)\sin(r^2x) = 1$
103. Show that the equation $\sin x(\sin x + \cos x) = a$ has real solutions if $a$ is a real number lying between $\frac{1}{2}(1 - \sqrt{2})$ and $\frac{1}{2}(1 + \sqrt{2}).$
104. Find the real solutions of the equation $2\cos^2\frac{x^2 + x}{6} = 2^x + 2^{-x}.$
105. Solve the inequality $\sin x\geq \cos2x.$
106. Find the general solution of the equation $\left(\cos\frac{x}{4} - 2\sin x\right)\sin x + \left(1 + \sin \frac{x}{4} -2\cos x\right)\cos x = 0$
107. Find the general solution of the equation $2(\sin x -\cos2x) - \sin2x(1 + 2\sin x) + 2\cos x = 0.$
108. Solve $\frac{\sin2x}{\sin\frac{2x + \pi}{3}} = 0.$
109. Solve the equation $3\tan2x - 4\tan3x = \tan^23x\tan2x$
110. Solve the equation $\sqrt{1 + \sin2x} = \sqrt{2}\cos2x.$
111. Show that $x = 0$ is the only solution satisfying the equation $1 + \sin^2ax = \cos x$ where $a$ is irrational.
112. Consider the system of linear equarions in $x, y$ and $z, x\sin3\theta -y + z = 0, x\cos2\theta + 4y + 3z = 0, 2x + 7y + 7z = 0.$ Find the values of $\theta$ for which the system has non-trivial solutions.
113. Find all the solutions of the equation $\sin x + \sin\frac{\pi}{8}\sqrt{(1 - \cos x)^2 + \sin^2x} = 0$ in the interval $\left[\frac{5\pi}{2}, \frac{7\pi}{2}\right]$
114. Let $A = \{x: \tan x -\tan^2x > 0\}$ and $y = \left\{x: |\sin x|<\frac{1}{2}\right\}$. Determine $A\cap B.$
115. If $0\leq x\leq 2\pi,$ then solve $2^{\frac{1}{\sin^2x}}\sqrt{y^2 - 2y + 2}\leq 2$
116. If $|\tan x| = \tan x + \frac{1}{\cos x}(0\leq x\leq 2\pi)$ then prove that $x = \frac{7\pi}{6}$ or $\frac{11\pi}{6}$
117. Find the smallest positive solution satisfying $\log_{\cos x}\sin x + \log_{\sin x}\cos x = 2$
118. Solve the inequality $\sin x\cos x + \frac{1}{2}\tan x\geq 1$
119. Solve $\tan x^{\cos^2 x} = \cot x^{\sin x}$
120. If $0\leq \alpha, \beta \leq 3,$ then $x^2 + 4 + 3\cos(\alpha x + \beta) = 2x$ has at least one solution, then prove thatt $\alpha + \beta = \pi, 3\pi.$
121. Prove that the equation $2\sin x = |x| + a$ has no solution for $a\in \left(\frac{3\sqrt{3 - \pi}}{3}, \infty\right)$
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# Graphing Stories
## Elevation vs. Time
Elevation vs. Time
## Explanation of Elevation vs. Time Graph
In the video you can see that the guy is walking up the stairs in a constant rate. Looking at the graph, you can see that as time goes on the guy’s elevation is rising. Within 6 seconds his elevation is 21 inches off the ground and his elevation continues to increase by 7 until he reaches the top of the stairs. At the top of the stairs, you can see that the graph comes to its highest peak. He then returns back down to 0 inches, which is back onto the floor (origin) where he first started. The graph shows a clear increase, and then a decrease. Which represents how he walked up the steps (increase) and back down (decrease)
in our explanation of the elevation vs. time graph, we also identified four points and explained what they meant.
The first point we identified was point (4, 14) The x in this ordered pair represents time and the y represents elevation. Therefore the point (4,14) means that at 4 seconds, Dashawn was 14 inches off of the ground.
The second point we identified was point (11, 49) This ordered pair represents that at 11 seconds, Dashawn was 49 inches off of the ground.
The third point we identified was point (24, 35) This ordered pair represents that at 24 seconds, Dashawn was 35 inches off of the ground.
Finally, the fourth point was (27, 14) This ordered pair represents that at 27 seconds, Dashawn was 14 inches off of the ground.
## Distance vs. Time
Distance vs. Time
## Explanation of Distance vs. Time Graph
In this video you can see that the car is traveling a constant distance over time. You can see that the car’s speed first was a constant rate, and then as times goes one it slows down a little and then speed back up and so on. Nevertheless, as time goes on the farther the car goes from when it first started. in 5 seconds the car only traveled 18 inches, then in 10 seconds it traveled 36 inches, and then farther and farther. Basically, as the wooden car’s distance increased, the time it took to get across the floor increased as well.
In our explanation of the distance vs. time graph, we also identified four points and explained what they meant.
The first point we identified was point (10, 36) The x in this ordered pair represents time and the y represents distance. Therefore the point (10, 36) means that at 10 seconds the wooden car was 36 inches across the floor.
The second point we identified was point (25, 93) this means that at 25 seconds, the wooden car was 93 inches across the floor.
The third point we identified was point (35, 138) this means that at 35 seconds, the wooden car was 138 inches across the floor.
Finally, the fourth point was (20, 79) this means that at 20 seconds, the wooden car was 79 inches across the floor.
## Distance vs. Elevation
Elevation vs. Distance
## Explanation of Distance vs. Elevation Graph
In this video you can see that the guy was walking at a 0 elevation because he’s still on the floor, but the more he walks the farther the distance he’s traveling. When he steps on the chair and then the table, his elevation rises. The fact that he is still walking does not affect the distance. Once he’s at a distance of 77 inches he has an elevation of 16, and then when his distance is 142 his elevation of 28 inches. Basically, the elevation did not have any effect on the distance so there was no consistent increase of the x nor the y.
In our explanation of the elevation vs. distance, we also identified four points and explained what they meant.
The first point we identified was point (77, 16) The x in this ordered pair represents distance and the y represents elevation. Therefore the point (77, 16) means that when Dashawn was 77 inches across the table, he was 16 inches of the ground.
The second point we identified was point (142, 28) this means that when Dashawn was at a distance of 142 inches, he was 28 inches off the ground.
The third point we identified was point (202, 0) this means that when Dashawn was at a distance of 202 inches he was 0 inches off the ground.
Finally, the fourth point was (0,0) this that when Dashawn was at a distance of 0 inches he was 0 inches off the ground (at rest)
## Inverse of Distance vs.Time Graph
Our inverse is a function.
## Why our Elevation vs. Time looks different from our Elevation vs. Distance graph
The elevation vs time graph is different from the elevation vs distance because in elevation vs distance time is not being plotted instead we are using elevation so the points or numbers recorded will be higher than time.
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# Comparing Money Raised
Alignments to Content Standards: 4.OA.A.2
1. Helen raised \$12 for the food bank last year and she raised 6 times as much money this year. How much money did she raise this year? 2. Sandra raised \$15 for the PTA and Nita raised \$45. How many times as much money did Nita raise as compared to Sandra? 3. Luis raised \$45 for the animal shelter, which was 3 times as much money as Anthony raised. How much money did Anthony raise?
## IM Commentary
The purpose of this task is for students to solve three comparisons problems that are related by their context but are structurally different. Multiplicative comparison is purposefully excluded from third grade (see 3.OA.3 and 3.MD.2), making this task appropriate for fourth but not third grade. In these multiplicative comparison problems, one factor and the product are amounts of money and the other factor represents the number of times bigger one amount is than the other. Sometimes this second factor is called a “scale factor.” In part (a), the larger amount (which is the product) is unknown, while in part (b) the scale factor is unknown and in part (c) the smaller amount of money is unknown. Students will study multiplicative comparison problems involving scale factors that are fractions in 5th grade; see 5.NF.B.5. Note that in fourth grade, scale factors must always be bigger than 1, so students often think that “multiplying makes bigger”; however in 5th grade they will see that when the scale factor is less than 1, the product will actually be smaller than the initial quantity. Note that the numbers in parts (b) and (c) are related by the fact family $3\times15=45$. This allows for a classroom discussion about the different interpretations of the factors in a multiplicative comparison context. To see an annotated version of this and other Illustrative Mathematics tasks as well as other Common Core aligned resources, visit Achieve the Core.
The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, only one practice connection will be discussed in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.
This particular task helps illustrate Mathematical Practice Standard 1, Make sense of problems and persevere in solving them. Problem solving is based upon students engaging in a task in which a solution pathway is not known in advance. As fourth graders approach these three problems, they will analyze the problems to make sense of what each is asking, working to understand the structures and the unknowns. Through this analysis, students will understand that the numbers in parts (b) and (c) are both factors related by the same fact family but each serves a different role in their respective problems. Students need experience with unknowns that play different roles in multiplicative comparison problems in order to support a deeper understanding of this type of problem situation.
## Solutions
Solution: Tape diagram
1. She raised six times as much money (as shown in the diagram) so she raised $6 \times 12 = 72$.
Helen raised \$72 this year. 2.$? \times 15 = 45$is equivalent to$45 \div 15 = ?$Nita raised 3 times as much as Sandra. 3.$3 \times ? = 45$is equivalent to$45 \div 3 = ?$Anthony raised \$15.
Solution: Writing multiplication equations for division problems
1. Helen raised $6 \times \$12$this year, so she raised \$72 this year.
2. This is a “Number of Groups Unknown” problem. We can represent the question as $$? \times 15 = 45$$ or $$45 \div 15 = ?$$ So Nita raised 3 times as much money as Sandra.
3. This is a “Group Size Unknown” problem. We can represent the question as $$3 \times ? = 45$$ or $$45 \div 3 = ?$$ So Anthony raised \\$15.
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# GATE | GATE-CS-2014-(Set-1) | Question 65
There are 5 bags labeled 1 to 5. All the coins in a given bag have the same weight. Some bags have coins of weight 10 gm, others have coins of weight 11 gm. I pick 1, 2, 4, 8, 16 coins respectively from bags 1 to 5. Their total weight comes out to 323 gm. Then the product of the labels of the bags having 11 gm coins is ___. (A) 15 (B) 12 (C) 8 (D) 1
Explanation:
```There are 5 bags numbered 1 to 5.
We don't know how many bags contain 10 gm and
11 gm coins.
We only know that the total weights of coins is 323.
Now the idea here is to get 3 in the place of total
sum's unit digit.
Mark no 1 bag as having 11 gm coins.
Mark no 2 bag as having 10 gm coins.
Mark no 3 bag as having 11 gm coins.
Mark no 4 bag as having 11 gm coins.
Mark no 5 bag as having 10 gm coins.
Note: The above marking is done after getting false
results for some different permutations, the permutations
which were giving 3 in the unit place of the total sum.
Now, we have picked 1, 2, 4, 8, 16 coins respectively
from bags 1 to 5.
Hence total sum coming from each bag from 1 to 5 is 11,
20, 44, 88, 160 gm respectively.
For the above combination we are getting 3 as unit digit
in sum.
Lets find out the total sum, it's 11 + 20 + 44 + 88 + 160 = 323.
So it's coming right.
Now 11 gm coins containing bags are 1, 3 and 4.
Hence, the product is : 1 x 3 x 4 = 12. ```
Alternative solution : There are bags with coins of weight 11g and 10g. The total weight of 1,2,4,8,16 coins picked from bags 1,2,3,4,5 is given as 323. Now we need to find no.of coins of 10g and 11g that give a sum of 323g. Let x be the no.of coins of weight 10g and y be the no.of coins of weight 11g Therefore 10x + 11y = 323………1 Now we now the total no.of coins picked i.e 1+2+4+8+16=31. Therefore x + y = 31…………2 Solving equations 1 and 2 we get x=18, y=13. Therefore bags 2 and 5(2+16=x=18) have 10g coins and bags 1, 3 and 4(1+4+8=y=13) have 11g coins. Therefore the product of the labels of the bags having 11 gm coins is 1 x 3 x 4 = 12.
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# Video: Area of Triangles
Kathryn Kingham
Learn how to calculate the area of a triangle using the formula: area = half of base times height. Starting with rectangles and parallelograms and then cutting them in half, we talk through the process with some examples and emphasize that units of area are squared.
07:33
### Video Transcript
Let’s talk about finding the area of a triangle. And I guess before we talk about finding the area of a triangle we need to talk about what the area is. Area is really simply a size of a surface. So here you can see I just coloured in the surface of this triangle, and we want to figure out how big it is. So let’s dive in.
Okay, I know what you’re thinking, “These aren’t triangles!” But before we figure out how to find the area of triangles, I want us to take a step back and try and remember how do we find the area of a parallelogram. So we can find the area of a parallelogram here, a square here, or a rectangle. But, if you remember, when we find the area of a rectangle, we take the length and we multiply it by the width. Another way to write that, sometimes we say the base and we multiply it by the height. Either way, we got length and width, base and height, doesn’t matter. We take these two numbers and we multiply the base times the height to find the area. But now what I want us to do is I want us to cut these guys in half, okay? We’re gonna cut our parallelograms, our quadrilaterals, in half. And what you should start to see is that every one of these is made up of two triangles, two equal triangles. How does this matter? Why does this matter? We’re gonna find out. So remember, we said the area of this square would be base times the height. But our goal here is to find out the area of a triangle. Well the thing is this triangle is half of the square, so the formula for finding the area of a triangle is by taking half of what the quadrilateral’s area would be: base times height divided by two is the simple way we say it. So we say base times height and divide it by two. Let’s try it.
Okay, here’s our first example. So someone has already told us the base and the height, the height and the base of this triangle. So what we need to start with then is our formula, which we remember is area equals the base times the height divided by two. All we need to do is figure out that the base is eight and the height is four. We can plug that into our formula by saying eight times four divided by two, okay? Eight times four divided by two. Eight times four is thirty-two, and then we divide thirty-two by two for a final answer of sixteen. So before we move on, I wanna tell you something really important about the area. Our answer is not just sixteen; our answer is sixteen centimeters squared. Any time we’re dealing with area, it’s so important that you know what it is that you’re working with. What are we measuring our triangle in? And this time, we’re measuring our triangle in centimeters, so we wanna make sure that we include our centimeters squared.
Okay example two, this time again we’ve been given all of our measurements: we’ve been given three, four, and five. So what I want you to be careful of when you’re trying to solve the area of triangles is knowing what is the height and what is the ba-what is the base. So here is the key to finding the height and the base. You first need to look for the right angle, okay? So your height and your base will always be connected to the right angle portion. So if we go back to our last example problem, it looked like this, right? But remember, we had this symbol. We had a little symbol right here that told us, “Oh! height and base.” When we’re dealing with a right triangle, it’s the same. We have this right angle which tells us, “Okay here’s our base, and here’s our height.” And the thing is, right now we do not even need this five meters. We don’t need that five meters to figure out anything about the area, so we can ignore that guy and we can move on and say three meters times four meters divided by two is gonna give us the area of this triangle. So we say three times four is twelve, divided by two. And again, that’s six but we do not forget these meters. We bring our meters down, and it’s meters squared. We deal with area; we deal with our units, squared.
Example three is for you. So what I want to do to you right now is just stop what you’re doing, pause. Pause the video, and write this down and see if you can find the area of this triangle. Go ahead pause it; I’ll wait. The answer is thirty feet squared. So I hope you got that and you were successful. But if you didn’t get the whole problem, then don’t worry I’m gonna walk you through this one again. So area, first we write our formula area equals one-half times the height times the base. We figure out that the right angle tells us which one is the height and which one is the base. So we have a ten-foot base, and we have a six-foot height. You need to divide that by two. six times ten is sixty, divided by two is thirty. And again, we say this is area; we can’t forget our units, thirty feet squared.
Now you’re ready to go out and solve all the triangle area problems. Just remember these two key things: one, that your formula is area equals height times the base divided by two and, number two, always add the unit squared at the end. You’ll be good to go. Happy area finding!
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How to express a number in scientific notation and how to convert from standard to scientific notation.
# Scientific Notation and Standard Form
Formula, practice problems
Scientific notation is just a short hand way of expressing gigantic numbers like 1,300,000 or incredibly small numbers like 0.0000000000045. Also known as exponential form, scientific notation has been one of the oldest mathematical approaches. It is favored by many practicioners. If numbers are too big or too small to be simply calculated, people reffer to scientific notation to handle these circumstances. This method is used by engineers, mathematicians, scientists.
An example of scientific notation is 1.3 ×106 which is just a different way of expressing the standard notation of the number 1,300,000. Standard notation is the normal way of writing numbers.
Key Vocabulary
mantissa = this is the integer or first digit in any Scientific Notation. For example in 1.3 ×106, the mantissa is the "1"
Other examples:
• 1.2 ×1014 the positive exponent indicates a large number
• 4.22×1011
• 7.89 × 10-21 the negative exponent indicates a small number
### General Formula of Scientific Notation
The general from of a number in scientific notation is:
a ×10n where 1 ≤ a ≤ 10 and n is an integer. In other words the number that we'll call "a" is is multiplied by 10, raised to some exponent n. This number "a" must be no smaller than 1 and no larger than 10. To illustrate this definition examine the following:
1.4 ×104 is a proper example of scientific notation because
• 1.4, which is "a" in this example, is not smaller than 1 and not larger than 10 so it's ok.
• 10's exponent is the integer 4.
.9 ×104 is a NOT proper example because
• .9 which is "a" in this example, is smaller than 1 which is not allowed in scientific notation
3.34 ×10½ is a NOT proper example because
4.34 ×10-55 is a proper example because
• 4.34, which is "a" in this example, is not smaller than 1 and not larger than 10
• 10's exponent is the integer -55. Integers can be negative
Scientific Notation Standard Form 1.23 ×102 123 1.23 × 103 1,230 1.23 ×104 12,300 1.23 × 105 123,000 1.23 ×106 1,230,000
### Practice Problems
Convert Scientific Notation to standard form
In the following sentences, convert from scientific notation to standard form.
Scientific Notation Standard Form 1.303 •105 130,300 9.43 •104 94,300 3.423 •107 34,230,000 3.23 •106 3,230,000 6.003 •109 6,003,000,000
Convert standard to Scientific Notation
In the following sentences, convert from standard form to scientific notation.
Standard Form Scientific Notation 19,300 1.9•104 200,000 2.0•105 3,013,000,000 3.013•109 12,000,000,000 1.2•1010 130,000,000,000,000,000,000,000 1.3•1023
In the following sentences, state whether or not the given number is in Scientific Notation and explain your answer.
Scientific Notation ??? Standard Form 1) 13 •105 1) No because 13 is greater than 10 and Scientific Notation's initial number must be between 1 and 10 2) 1.3 •105 2) Yes, this is proper scientific notation 3) 3.423 •1090909090 3) Yes, proper Scientific Notation. 4) 3.23 •10-6 4) Yes, no one said that you couldn't have negative exponents in your Sceintific Notation. 5) 931 •10-9 5) No! You can have negative exponents but your first number (931 in this case) still must be between 1 and 10
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# What to Expect with 8th Grade Math Problems
Last Updated on August 3, 2023 by user
When your student starts 8th grade math problems, you may suddenly feel like math class has gone beyond your ability to help. While recent core standards have changed the face of eighth grade math class, and you may not recognize some of the work coming home, there are resources you can use to provide the assistance your child needs. Sometimes, knowing what to expect can help. Here are some common questions you can expect to see on your child’s 8th grade math problems.
## Linear Equations
8th grade math problems begin to introduce linear equations and systems of linear equations, which will be an important concept in later algebra classes. You can expect to see problems in the form of y = mx + b, and students should be able to know how to graph lines in this format, plotting coordinates using the slope (m) and y-intercept (b).
Here’s how this works. If the student sees y = 2/5x + 5, he is going to begin this graph by creating an x, y coordinate plane. The vertical line on the plane, the y-axis, is where graphing begins.
The student should then place a point at the y-intercept (5). From there, the student measures the slope, which is shown as a ratio of rise/run (familiar?). In this equation, the slope is 2/5. From the 5, the student should go up (rise) 2 and over (run) 5 on the coordinate plane, plotting a second point. Connect the points and extend the line to graph the equation.
Students will also be taught how to solve for one variable in a linear equation. For example, if they are given 3x + 1 = 7, they will know how to solve to get the answer x = 2.
## Functions
In 8th grade, students will begin working with functions. Functions define an output that applies to every number placed into the function. For example, you may see:
f(x) = x+1
This function means that for any number (x) placed into the function, you will add one. So, if the student is asked to find the output or value of the function for 2, 4 and 6, the answer will be 3, 5 and 7 after each value is placed into the function.
## Exponents and Roots
In 8th grade, your student will start working with exponents, square roots and cubed roots. This will also bring in a discussion of irrational numbers. For example, students should be able to identify that the square root of 2 is irrational. In this discussion, you will also see some teaching and questioning about scientific notation. Scientific notation involves writing numbers as multiples of 10 to a certain power. In scientific notation, 2,400 becomes 2.4 x 103.
## Distance and Angles
Another important math concept tackled in 8th grade is the idea of distance and angles. Students will go beyond simply identifying and measuring angles, and will begin to determine how angles will behave when rotated, reflected, dilated and translated. They will also learn ideas about congruence and similarity. When working with distance, students will be able to apply the Pythagorean Theorem (a2 + b2 = c2) and its converse to find distances between points on the coordinate plane.
## Do these concepts seem overwhelming?
If your student needs help, consider investing in Thinkster, an online math tutoring program that provides instruction for students at the point of learning. Created with the help of qualified teachers, this program will give your child the tools and confidence needed to excel with 8th grade math problems.
Summary
Article Name
What to Expect with 8th Grade Math Problems
Description
If your child is about to start 8th grade, you may notice a huge change in the difficulty of their math lessons. So be prepared for 8th grade math problems.
Author
Publisher Name
Thinkster Math
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## Recommended Articles
### Unlocking the Code: Teaching Mathematical Symbols and Their Meaning
Knowledge of mathematical symbols is a foundational skill that empowers students to engage with and excel i...
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# How do you solve 36=3n-3(5n-8)?
Apr 30, 2018
$n = - 1$
#### Explanation:
$36 = 3 n - 3 \left(5 n - 8\right)$
Open parenthesis,
$36 = 3 n - 15 n + 24$
Simplify,
$36 = 24 - 12 n$
Subtract $24$ from both sides,
$12 = - 12 n$
Divide both sides by $12$,
$n = - 1$
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# The ratio of escape velocity at earth ( ${v_e}$ ) to the escape velocity at a planet (${v_p}$ ) whose radius and mean density are twice as that of earth is:A) $1:2$B) $1:2\sqrt 2$C) $1:4$D) $1:\sqrt 2$
Last updated date: 10th Aug 2024
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Hint: To solve this question we need to know what is meant by escape velocity and the equation which defines the escape velocity in terms of radius and density. Also know about the factors that escape velocity depends upon.
Complete step by step solution:
The minimum velocity with which an object is needed to be projected so that it would escape the gravitational attraction or pull of any massive body is known as the escape velocity of the body (massive bodies like planets or celestial bodies).
Mathematically escape velocity is expressed as
${v_{esc}} = \sqrt {\dfrac{{2GM}}{r}}$
Where ${v_{esc}}$ is the escape velocity of the body
$G$ is universal gravitational constant
$M$ is the mass of the body whose gravitational pull is to be escaped
$r$ is the distance from the centre of the mass
Now, in the question we are given that radius and mean density of a planet are twice as that of the earth.
Let us denote the radius and density of earth as ${r_e}$ and ${\rho _e}$ respectively. Also we denote the radius and density of the other planet as ${r_p}$ and ${\rho _p}$
$\Rightarrow {r_p} = 2{r_e}$ and ${\rho _p} = 2{\rho _e}$ --- ( $1$ )
Now, we find the mass of earth and the planet in terms of their radius and density.
$\Rightarrow {M_e} = \dfrac{4}{3}\pi r_e^3 \times {\rho _e}$ is the mass of earth ---( $2$ )
$\Rightarrow {M_p} = \dfrac{4}{3}\pi r_p^3 \times {\rho _p}$ is the mass of the planet
Substituting the values from equation ( $1$ ) in the equation of mass of the planet, we get
$\Rightarrow {M_p} = \dfrac{4}{3}\pi {(2{r_e})^3} \times 2{\rho _e}$
$\Rightarrow {M_p} = \dfrac{{64}}{3}\pi r_e^3 \times {\rho _e}$ ---( $3$ )
Now that we have the equations of masses in terms of the given data from the question, we substitute these equations of masses in the equation of escape velocity.
Thus, substituting equation ( $2$ ) and equation ( $3$ ) in the equation of the escape velocity, we get:
Escape velocity of earth is equal to ${v_{earth}} = \sqrt {\dfrac{{2G{M_e}}}{{{r_e}}}}$
$\Rightarrow {v_{earth}} = \sqrt {\dfrac{{2G\dfrac{4}{3}\pi r_e^3 \times {\rho _e}}}{{{r_e}}}}$
Simplifying this to get
$\Rightarrow {v_{earth}} = \sqrt {\dfrac{8}{3}G\pi r_e^2{\rho _e}}$ --- ( $4$ )
Escape velocity of the planet is equal to ${v_{planet}} = \sqrt {\dfrac{{2G{M_p}}}{{{r_p}}}}$
$\Rightarrow {v_{planet}} = \sqrt {\dfrac{{2G\dfrac{{64}}{3}\pi r_e^3 \times {\rho _e}}}{{2{r_e}}}}$
Simplifying this to get
$\Rightarrow {v_{planet}} = \sqrt {\dfrac{{64}}{3}G\pi r_e^2 \times {\rho _e}}$ --- ( $5$ )
As the question says, we have to find the ratio of the escape velocity of earth to the escape velocity of the planet. So, to find this we divide equation ( $4$ ) by equation ( $5$ )
$\Rightarrow \dfrac{{{v_e}}}{{{v_p}}} = \dfrac{{\sqrt {\dfrac{8}{3}G\pi r_e^2{\rho _e}} }}{{\sqrt {\dfrac{{64}}{3}G\pi r_e^2 \times {\rho _e}} }}$
Cancelling out all the common terms, we get:
$\Rightarrow \dfrac{{{v_e}}}{{{v_p}}} = \dfrac{{\sqrt {\dfrac{8}{3}} }}{{\sqrt {\dfrac{{64}}{3}} }}$
$\Rightarrow \dfrac{{{v_e}}}{{{v_p}}} = \dfrac{{\sqrt 8 }}{{\sqrt {64} }}$
$\Rightarrow \dfrac{{{v_e}}}{{{v_p}}} = \dfrac{{\sqrt 8 }}{8}$
$\Rightarrow \dfrac{{{v_e}}}{{{v_p}}} = \dfrac{1}{{\sqrt 8 }} = \dfrac{1}{{2\sqrt 2 }}$
Thus, the ratio ${v_e}:{v_p} = 1:2\sqrt 2$
Therefore, option (B) is the correct option.
Note: Remember that the escape velocity depends on the mass and radius of the body whose gravitational pull is to be escaped. Escape velocity does not depend on the mass and radius of the smaller object which is propelled.
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# Removing BracketsFOIL Multiplication
It’s important to know in Mathematics how to manipulate expressions that involve brackets.
Removing brackets in simpler cases, generally comes down to multiplying brackets out.
Though in less simpler cases, vertical multiplication can be a more effective method.
Removing brackets can sometimes be referred to as expanding brackets.
2(x + 1) = 2x + 1
This comes about from multiplying first 2 by x, and then 2 by 1.
Generally:
x( y + z ) = xy + xz and x( yz ) = xyxz
Examples
(1.1)
3( x + 7 ) = 3x + 21
(1.2)
4( 2x − 3 ) = 8x − 12
(1.3)
x( y + 4 ) = xy + 4x
(1.4)
-2b( 4 − b ) = -8b + 2b2
(1.5)
p( q + r ) = pq + pr
## Removing Brackets, FOIL
Sometimes it's required to expand or multiply out 2 separate expressions that both involve 2 terms.
There’s a bit more multiplication to be done than in the smaller examples above, but the general approach is very similar.
There are  4 multiplications to do, then usually a little bit of addition or subtraction.
This method is sometimes referred to as "FOIL".
Meaning First , Outer , Inner , Last,
in order of the operations, as can be seen in the image above.
Generally:
(x + c)(x + d) = x2 + xd + xc + cd = x2 + (c + d)x + cd
Examples
(2.1)
(x − 3)(x + 2) = x2 + 2x − 3x − 6 = x2x − 6
(2.2)
(3x + 1)(x + 4) = 3x2 + 12x + x + 4 = 3x2 + 13x + 4
(2.3)
(2x − 1)(2x + 1) = 4x2 + 2x − 2x − 1 = 4x2 − 1
## Removing Brackets, Further ExamplesVertical Multiplication
(3.1)
( x + 1 )( x2 + 4x − 2 )
This example has an extra term to deal with in the right bracket.
Now, one can still follow the approach from the earlier examples.
There would be  6 multiplications to perform.
(x + 1)(x2 + 4x − 2)
x3 + 4x2 − 2x + x2 + 4x − 2
x3 + 5x2 + 2x − 2
However, many people often find that vertical multiplication is slightly easier, and can be quicker also.
It doesn’t matter if you put the larger term top or bottom.
(3.2)
(x + 3)(x − 1)(x + 2)
When faced with  3 sets of brackets containing  2 term polynomials.
Preferably want to try to reduce the expression to only  2 brackets initially. So look to multiply out the first  2 brackets first.
Then use vertical multiplication.
(x + 3)(x − 1) = x2 + 2x − 3
(3.3)
(2x2 + 2x − 3) (x2 + 3x + 1)
The vertical multiplication approach works fine for larger situations also. Just make sure to pay attention though, as with a lot of sums to do, mistakes can be easy to make.
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## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 12 Congruence of Triangles Check Your Progress
Question 1.
State, giving reasons, whether the following pairs of triangles are congruent or not:
Solution:
(i) In the given figure, using the SSS criterion triangles one congruent.
(ii) Triangles are congruent for the criterion ASA criterion.
(iii) Triangles are congruent for the criterion RHS.
(iv) In the first triangle, third angle = 180° – (70° + 50°) = 180° – 120° = 60°
Now triangles are congruent for ASA criterion.
(v) Not congruent as included angles of the given two sides are not equal.
(vi) Not congruent as the included sides are different.
Question 2.
Given below are measurements of some parts of two triangles. Examine whether the two triangles are congruent or not. In case of congruence, give reasons and write in symbolic form:
∆ABC ∆PQR (i) AB = 4 cm, BC = 5 cm, ∠B = 70° (i) QR = 4 cm, RP = 5 cm, ∠R = 70° (ii) AB = 4 cm, BC = 5 cm, ∠B = 80° (ii) PQ = 4 cm, RP = 5 cm, ∠R = 80° (iii) BC = 6 cm, ∠A = 90°, ∠C = 50° (iii) QR = 6 cm, ∠R = 50°, ZQ = 40° (iv) AB = 5 cm, ∠A = 90°, BC = 8 cm (iv) PR = 5 cm, ∠P = 90°, QR = 8 cm
Solution:
(i) In ∆ABC and ∆PQR
AB = QR = 4 cm
BC = RP = 5 cm
∠B = ∠R = 70°
∆ABC = ∆PQR (SAS criterion)
(ii) In ∆ABC and ∆PQR
AB = PQ = 4 cm
BC = RP = 5 cm not corresponding sides
∠B = ∠R = 80° not corresponding angles
Triangles are not congruent.
(iii) BC = QR = 6 cm
∠A = ∠P = 90° (Third angle)
∠C = ∠R = 50°
Triangles are congruent for ASA criterion.
(iv) AB = PR = 5 cm (Side)
∠A = ∠P = 90°
BC = QR = 8 cm (Hypotenuse)
Triangles are congruent for RHS criterion.
Question 3.
In the given figure, ABC is an isosceles triangle with AB = AC and AD is one of its altitudes.
(i) State the three pairs of equal parts in ∆ADB and ∆ADC.
(iii) Is ∠B = ∠C? Why?
(iv) Is BD = DC? Why?
Solution:
∆ABC is an isosceles triangle with AB = AC
and AD is one of the altitudes.
Hypotenuse, AB = AC (Given)
∠B = ∠C (c.p.c.t)
and BD = CD (c.p.c.t)
Question 4.
In the given figure, OA bisects ∠A and ∠ABO = ∠OCA. Prove that OB = OC.
Solution:
In ∆OAB and ∆OAC
∠OAB = ∠OAC (∵ OA bisects ∠A)
∠ABO = ∠ACO (Given)
OA = OA (common)
∆OAB = ∆OAC (AAS congruence rule)
OB = OC (Corresponding parts of congruent As)
Question 5.
In the given figure , prove that
(i) AB = FC
(ii) AF = BC.
Solution:
In ∆ABE and ∆DFC
∠B = ∠F (each 90°)
AE = DC (Given)
BE = DF (Given)
∆ABE = ∆DFC (RHS congruence rule)
(i) AB = FC (Corresponding parts of congruent ∆s)
(ii) As AB = FC (Proved above)
⇒ AF + FB = FB + BC
⇒ AF + FB – FB = BC
⇒ AF = BC
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# Lesson video
In progress...
Hi, I'm Miss Davies.
In this lesson, we're going to be working out the gradient of a line perpendicular to a given line.
What is meant by perpendicular? Lines that's a perpendicular meet at 90 degrees.
The gradient of this line is one.
This is its perpendicular line.
The gradient of this line is negative one.
The gradient of this next line is two.
The gradient of its perpendicular line is negative a half.
The next line has a gradient of negative two-thirds.
The gradient of its perpendicular line is three over two.
The gradient of this line is a half.
The gradient of its perpendicular is negative two.
What do you notice about the gradients? The gradients of the two perpendicular lines multiply to make negative one.
Here are some questions for you to try.
Pause the video to complete your task and resume once you're finished.
The gradients of the perpendicular lines should multiply to get negative one.
In this example, we've been asked to find the gradient of a line that is perpendicular to Y equals two X add one.
The gradient of the line Y equals two X add one is two.
We know that when you multiply the gradients of two perpendicular lines, you get negative one.
Negative one divided by two will give us the gradient of the perpendicular line.
One divided by negative two is negative one-half.
Therefore, the gradient of a line perpendicular is negative one-half.
In this next example, we've been asked to find the gradient of a line perpendicular to Y add two-thirds X is equal to one.
To find the gradient of this line, we need to make Y the subject.
We're going to do this by subtracting two-thirds X from each side.
This gives us Y is equal to one subtract two-thirds X.
This means that the gradient of the line is negative two-thirds.
To calculate the gradient of the line perpendicular, we are going to divide negative one by negative two-thirds.
This gives us three over two.
Therefore, the gradient of the line perpendicular is three over two.
Here are some questions for you to try.
Pause the video to complete your task and resume once you're finished.
|
# Geometry: Circles
### Contents
page 1 of 2
Page 1
Page 2
#### Characteristics of Circles
Special names are given to geometric figures that lie on or inside circles. Among these geometric figures are arcs, chords, sectors, and segments.
#### Arc
The arc of a circle consists of two points on the circle and all of the points on the circle that lie between those two points. It's like a segment that was wrapped partway around a circle. An arc is measured not by its length (although it can be, of course) but most often by the measure of the angle whose vertex is the center of the circle and whose rays intercept the endpoints of the arc. Hence an arc can be anywhere from 0 to 360 degrees. Below an arc is pictured.
Figure %: An arc
The arc above contains points A, B, and all the points between them. But what if the arc went the other way around the circle? This brings up an important point. Every pair of endpoints defines two arcs. An arc whose measure is less than 180 degrees is called a minor arc. An arc whose measure is greater than 180 degrees is called a major arc. An arc whose measure equals 180 degrees is called a semicircle, since it divides the circle in two. Every pair of endpoints on a circle either defines one minor arc and one major arc, or two semicircles. Only when the endpoints are endpoints of a diameter is the circle divided into semicircles. From this point on, unless otherwise mentioned, when arcs are discussed you may assume the arc is a minor arc.
Figure %: A major arc, minor arc, and semicircle
A central angle is an angle whose vertex is the center of a circle. Any central angle intercepts the circle at two points, thus defining an arc. The measure of a central angle and the arc it defines are congruent.
Figure %: A central angle and the arc it defines
#### Chord
A chord is a segment whose endpoints are on a circle. Thus, a diameter is a special chord that includes the center.
Figure %: A chord
Chords have a number of interesting properties. Every chord defines an arc whose endpoints are the same as those of the chord. For example, a diameter and semicircle are a chord and arc that share the same endpoints. The union of a chord with a central angle forms a triangle whose sides are the chord and the two radii that lie in the rays that make up the angle. This kind of triangle is always an isosceles triangle--we'll define that term in Geometry 2. Also, the diameter perpendicular to a given chord (remember, there is only one such diameter because a diameter must contain the center) is also the perpendicular bisector of that chord. These ideas are illustrated below.
Page 1
Page 2
|
# Domain and Range of Rational Functions
In this article, we will learn how to find the domain and range of a rational function either by using its graph or by using definite algebraic rules. We will start by doing a brief review of what domain and range mean.
Then, we will look at the methods used to find the domain and range of rational functions. Finally, we will look at some examples with answers that illustrate the use of these methods.
##### ALGEBRA
Relevant for
Learning about the domain and range of rational functions.
See examples
##### ALGEBRA
Relevant for
Learning about the domain and range of rational functions.
See examples
## What is the domain and range of functions?
If we think of a function as a mapping diagram that relates an input to an output, the domain would be the set of inputs and the range would be the set of outputs. Let’s consider the following mapping diagram:
We can see the inputs on the left and the outputs on the right. Here, the domain is the set {2, 4, 6} and the range is the set {1, 5, 9}. If we consider the function with the domain {1, 2, 3, 4}, we can find the range by substituting each of the domain values in the function:
Therefore, the range is the set {-1, 1, 3, 9}.
Now, let’s consider the function with domain (it means that x belongs to the set of all real numbers). It may be helpful to look at your graph to make it easier to determine the range:
We can see that the graph is a straight line and every real input has a real output, and since the line continues towards positive and negative infinity, any real number of output is possible. Thus, the range is all real numbers.
If we now look at the quadratic function , which has a domain that is all real numbers. We can look at its graph to determine the range:
We see that for each input, the output is always positive. Therefore, the range of the function is equal to all real numbers greater than or equal to zero.
## How to find the domain and range of rational functions?
Generally, we tend to define the domain and range of functions on real numbers, so we will do the same here. However, we will use different strategies to find the domain and range of rational functions since obtaining the graphs of these functions is not very easy. Let’s consider the following rational function:
We observe that, for the input of -2, we obtain an output of .
We know that division by zero is undefined, so the function is undefined at this point. However, any other input will have a real number output, so we conclude that the domain is all real numbers excluding -2. We can write this as:
By considering the nature of the function, we can see that any real number of outputs can be achieved with the exception of zero. This is because as x gets larger in magnitude, the output gets smaller, but the output can never equal zero. Therefore, the range of the function is:
In general, we can calculate the domain of a rational function by identifying any point where the function is undefined. This means finding any point that makes the denominator equal to zero.
To find the range of a rational function, we can identify any point that cannot be reached with any input. This is generally found by considering the limits of the function as the magnitude of the inputs gets larger.
## Examples of domain and range of rational functions
The following examples illustrate the concepts detailed above.
### EXAMPLE 1
Find the domain and range of the function .
Solution: Looking at the graph, it appears that the domain is and the range is . However, let’s check this algebraically.
We know that a function is undefined for inputs that result in denominators equal to zero. We can form an equation with the denominator to find the undefined point:
This confirms that the domain is . To confirm the range, we have to identify the values that cannot be reached with the given domain. As x gets larger, the output tends to zero, but never becomes zero. Therefore, the range is .
### EXAMPLE 2
Find the domain and range of the function .
Solution: In this case we do not have a graph, so we have to solve the problem algebraically.
Again, we know that the expression is undefined, so we form an equation with the denominator to find the undefined point:
Therefore, we know that the domain of the function is . To identify the range, we have to identify the values that cannot be reached with the given domain. As x gets larger, the output tends to zero, but never becomes zero. Therefore, the range is .
### EXAMPLE 3
Find the domain of the function .
Solution: We know that rational functions are only defined when their denominator is different from zero. In this function, we can see that there are two points for which it is not defined: when and when .
This means that the function is undefined when and . Therefore, the domain of the function is all real numbers except -4 and 3, denoted .
If we wanted to find the range of this function, despite the fact that each of the rational expressions cannot take a value of zero, there is an input of x that results in zero, which is . Therefore, the range of the function is all real numbers (R).
### EXAMPLE 4
Determine the domain of the function .
Solution: We remember that rational functions are only defined when their denominator is different from zero. Therefore, to find the domain of the function, we have to find the zeros of the equation in the denominator:
We can factor the x:
We can see that we have a zero when . However, the quadratic has no real roots. Therefore, the only zero in the denominator is . The domain is all real numbers except zero, denoted .
|
# PS 42X Grade 5 Chapter 6
### Chapter 6 Add and Subtract Fractions with Unlike Denominators
Online Resources:
Stage 4 Fluencies:
• Generating equivalent fractions for benchmark fractions.
• Decomposing fractions
Routines:
• Skip counting by fractional amounts and converting between mixed fractions and improper fractions as you go.
• Generating equivalent fractions.
Thoughts on Lessons:
1. Tools: fraction bars,
2. Models: number lines, bar model,
3. Strategies: common denominators
4. Show What You Know: Vital! Also do a vocabulary formative assessment and see what students know.
5. Lesson 6.1: Make this into a game. Different ways to make 1 whole with two different fraction units. You could have each group with 1 constant, e.g., students have to make 1 whole with 1/4 as 1 on the addends and they have to find different ways to make the whole, but always using 1/4 and some other fraction unit. The should record and chart the amounts. The strategy that they will be able to see is that 1/4 + 3/4 (in various forms) is 1 whole. In a guided group, explore why we cannot combine unlike units, by using a chart, similar to a place value chart. Use problem 13 on pg. 354 as an example during your guided math group. Use the problems on page 355 to make a Tic Tac Toe game for students to play. 2-days on this!
6. Lesson 6.2: Use food and concrete examples in guided math groups. For example, If we got a pizza pie, which was cut into 8 equal pieces or 1/8 of the whole and I ate 1/8 of the pizza pie. Later my mother each slice into 2 equal pieces. So now really each slice was 1/16 of the whole pizza pie. I ate 3/16. Now how much pizza was left? Have students visualize and act out and prove. Use concrete objects and act out the situations and examples in the book. Subtraction of fractions really help us think about why we might generate equivalent fractions before we subtract. The other idea is to think about when and were we might be able to subtract accurately with out making equivalent fractions. For example, 1 7/8 – 3/4. We want students to see the relationships, so I might say, “Well I can see that 3/4 from 1 is going to leave me with 1/4 and 1/4 is really just 2/8, so if I add what is left I would combine 2/8 and 7/8, which is 9/8 or 1 1/8. I did not need to convert both fractions to equivalent, like denominators. 2-days on this!
7. Lesson 6.3: Would not do. Would have students talk about what they know about the fractions and how they see them on a number line. Would make the first problem on page 365 an exploration, but use more realistic numbers and situations. One that students can better relate to or experience. For example the 40-mile bike trail could be \$40. Later on it could be 40-mile bike trail, when you revisit the problem.
8. Lesson 6.4: CCSS does not ask for the LCD, only a common denominator. The standards want students to have a conceptual understanding of what this strategy means and why it is being used and students being flexible enough to have several common denominators that could be used. Stress the understanding about common denominator and ways to figure out the common denominators. From lesson #1 and #2 begin looking at how we can record the fraction equivalences, so that students discover the pattern. Make a visual for students using a paper folding technique.
9. Lesson 6.5: CCSS does not ask for simplest form. The key here is the let students see and understand that some one could say an answer that is correct, but it is presented in an equivalent format…and students have to recognize them.
10. Lesson 6.5/6.6: Make into centers and guided groups. Centers should focus on use of concrete tools, math models and students discussing and proving why the sums and differences. We should also make the connection to adding/subtracting with whole numbers. Use a number bond and use examples to make a game. Solve collaboratively – one paper, one pen.
11. Lesson 6.7: Strategies modeled by book are regrouping and making improper fractions. These concepts are important, but not as a procedure and a way to develop flexibility with fractions and operations. Present worked problems to students and ask them to explain what was done. Explicitly show using number bonds where the regrouping in taking place. Students need to ability to decompose fractions. Discourse is vital! 2-days on this!
12. Lesson 6.8: Infuse this lesson as a routine…look at skip counting by fractional amounts using concrete models and showing on the number line. Later adjust the count to be skip counting by different units. For example, begin at 1/3 and skip count by 1/2. Show on a number line. Make the “On Your Own” into a game.
13. Lesson 6.9: Infuse throughout unit.
14. Lesson 6.10: Infuse from lesson #1. For example, 2/3 + 1/4 = 8/12 + 3/12 = 8/12 + 2/12 + 1/12…I can add these numbers in any order and arrive at the same sum. Also, 2 3/4 + 2 1/8 = 2 + 6/8 + 2 + 1/8 = 2 + 2 + 6/8 + 1/8. Apply the Associate Property of Addition and the Commutative Property of Addition from the very beginning.
15. End-of-Chapter Exam: Must review carefully. Compare the released items and EngageNY Module Exam. Many question unnecessarily confusing and poorly constructed.
|
# How do you solve (a-2)(a+4)>0?
Dec 5, 2016
The answer is x in ] -oo,-4 [ uu ] 2,+oo [
#### Explanation:
Let $f \left(a\right) = \left(a - 2\right) \left(a + 4\right)$
Then we do a sign chart
$\textcolor{w h i t e}{a a a a}$$a$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 4$$\textcolor{w h i t e}{a a a a}$$2$$\textcolor{w h i t e}{a a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a a}$$a + 4$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$a - 2$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$f \left(a\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$
For $f \left(a\right) > 0$, x in ] -oo,-4 [ uu ] 2,+oo [
|
# 2023 AMC 12A Problems/Problem 10
## Problem
Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$. What is $x+y$? $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$
## Solution 1
Because $y^3=x^2$, set $x=a^3$, $y=a^2$ ($a\neq 0$). Put them in $(y-x)^2=4y^2$ we get $(a^2(a-1))^2=4a^4$ which implies $a^2-2a+1=4$. Solve the equation to get $a=3$ or $-1$. Since $x$ and $y$ are positive, $a=3$ and $x+y=3^3+3^2=\boxed{\textbf{(D)} 36}$.
~plasta
## Solution 2
Let's take the second equation and square root both sides. This will obtain $y-x = \pm2y$. Solving the case where $y-x=+2y$, we'd find that $x=-y$. This is known to be false because both $x$ and $y$ have to be positive, and $x=-y$ implies that at least one of the variables is not positive. So we instead solve the case where $y-x=-2y$. This means that $x=3y$. Inputting this value into the first equation, we find: $$y^3 = (3y)^2$$ $$y^3 = 9y^2$$ $$y=9$$ This means that $x=3y=3(9)=27$. Therefore, $x+y=9+27=\boxed{36}$
first expand
$(y-x)^2 = 4y^2$
$y^2-2xy+x^2 = 4y^2$
$y^2-2yx+x^2 = 4y^2$
$x^2-2xy-3y^2 = 0$
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
consider a=1 b=-2y c=-3y^2
$x=\frac{2y\pm\sqrt{(-2y)^2-4(1)(-3y^2)}}{2a}$
$x=\frac{2y\pm\sqrt{16y^2}}{2a}$
$\frac{2y+4y^{2}}{2}$ or $\frac{2y-4y^{2}}{2}$
$x=y+2y$ and $x=y-2y$
$x=3y$ and $x=-y$
we can see both x and y will be postive in $x=3y$
now do same as solution 2 $$y^3 = (3y)^2$$ $$y^3 = 9y^2$$ $$y=9$$ This means that $x=3y=3(9)=27$. Therefore, $x+y=9+27=\boxed{36}$
## Solution 4: Substitution
Since $a^2 = |a|^2$, we can rewrite the second equation as $(x-y)^2=4y^2$
Let $u=x+y$. The second equation becomes
$$(u-2y)^2 = 4y^2$$ $$u^2 - 4uy = 0$$ $$u = 4y$$ $$x+y = 4y$$ $$x = 3y.$$
Substituting this into the first equation, we have
$$y^3 = (3y)^2,$$ so $x = 9$.
Hence $x = 27$ and $x + y = \boxed{\textbf{(D)} 36}.$
-Benedict T (countmath1)
## Solution 5: Difference of Squares
We will use the difference of squares in the second equation.
$$(y-x)^2=4y^2$$ $$(y-x)^2-(2y)^2=0$$ $$(y-x-2y)(y-x+2y)=0$$ $$-(x+y)(3y-x)=0$$
Since x and y are positive, x+y is non-zero. Thus, $$3y=x$$.
Substituting into the first equation:
$$y^3=x^2$$ $$y^3=9y^2$$ $$y=9, x=27 \rightarrow x+y=\boxed{\textbf{(D)} 36}$$
## Video Solution (⚡ Under 3 minutes ⚡)
~Education, the Study of Everything
## Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
|
NCERT Solutions: Symmetry
# NCERT Solutions for Class 8 Maths - Symmetry- 1
## Exercise 12.1
Q1: Copy the figures with punched holes and find the axes of symmetry for the following:
Ans:
A figure has a line of symmetry, if there is a line about which the figure may be folded so that the two parts of the figure will coincide.
Q2: Given the line(s) of symmetry, find the other hole(s):
Ans:
Q.3. In the following figures, the mirror line (i.e., the line of symmetry) is given as a dotted line. Complete each figure performing reflection in the dotted (mirror) line. (You might perhaps place a mirror along the dotted line and look into the mirror for the image). Are you able to recall the name of the figure you complete?
Ans:
(i) The idea of line of symmetry is linked to mirror reflection. A shape has line symmetry when one half is the mirror image of the other half. A mirror line helps to understand a line of symmetry.
(ii) When working with mirror reflection, it is important to pay attention to the changes in left-right orientation.
Q.4. The following figures have more than one line of symmetry. Such figures are said to have multiple lines of symmetry:
Identify multiple lines of symmetry, if any, in each of the following figures:
Ans:
Q. 5. Copy the figure given here:
Take any one diagonal as a line of symmetry and shade a few more squares to make the figure symmetric about a diagonal. Is there more than one way to do that? Will the figure be symmetric about both the diagonals?
Ans:
Yes, there is more than one way. Yes, this figure will be symmetric about both the diagonals.
Q. 6. Copy the diagram and complete each shape to be symmetric about the mirror line(s):
Ans:
(i) The idea of line of symmetry is linked to mirror reflection. A shape has line symmetry when one half is the mirror image of the other half. A mirror line helps to understand a line of symmetry.
(ii) When working with mirror reflection, it is important to pay attention to the changes in left-right orientation.
Q.7. State the number of lines of symmetry for the following figures:
(a) An equilateral triangle
(b) An isosceles triangle
(c) A scalene triangle
(d) A square
(e) A rectangle
(f) A rhombus
(g) A parallelogram
(i) A regular hexagon
(j) A circle
Ans:
Q.8. What letters of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
(a) a vertical mirror
(b) a horizontal mirror
(c) both horizontal and vertical mirrors
Ans:
(a) Vertical mirror - A, H, I, M, 0, T, U, V, W, X and Y
(b) Horizontal mirror - B, C, D, E, H, I, 0 and X
(c) Both horizontal and vertical mirror – H, I, O and X
Q.9. Give three examples of shapes with no line of symmetry.
Ans: The three examples are:
(ii) Scalene triangle
(iii) Parallelogram
Q. 10. What other name can you give to the line of symmetry of:
(a) an isosceles triangle?
(b) a circle?
Ans: (a) An isosceles triangle has only one line of symmetry, which can also be called the median/altitude of the isosceles triangle.
(b) There are infinite lines of symmetry in a circle, which can also be called the diameter of the circle.
## Exercise 12.2
Q.1. Which of the following figures have rotational symmetry of order more than 1:
Ans: Rotational symmetry of order more than 1 are (a),(b),(d),(e) and (f) because in these figures, a complete turn, more than 1 number of times, an object looks exactly the same.
Q.2. Give the order the rotational symmetry for each figure:
Ans:
## Exercise 12.3
Q.1. Name any two figures that have both line symmetry and rotational symmetry.
Ans: Circle and Square.
Q.2. Draw, wherever possible, a rough sketch of:
(i) a triangle with both line and rotational symmetries of order more than 1.
(ii) a triangle with only line symmetry and no rotational symmetry of order more than 1.
(iii) a quadrilateral with a rotational symmetry of order more than 1 but not a line symmetry.
(iv) a quadrilateral with line symmetry but not a rotational symmetry of order more than 1.
Ans:
(i) An equilateral triangle has both line and rotational symmetries of order more than 1.
Line symmetry:
Rotational symmetry:
(ii) An isosceles triangle has only one line of symmetry and no rotational symmetry of order more than 1.
Line symmetry:
Rotational symmetry:
(iii) A quadrilateral with no line of symmetry is an irregular quadrilateral
Checking rotational symmetry
If it is rotated 90°
It does not looks same as initial figure
If it is rotated 180°
It does not looks same as initial figure
If it is rotated 360°
It looks same as initial figure
Thus,
Thus, order of rotational symmetry - 1
Hence,
A quadrilateral with a rotational symmetry of order not a line of symmetry Is not possible
(iv) A trapezium which has equal non-parallel sides, a quadrilateral with line symmetry but not a rotational symmetry of order more than 1.
Line symmetry:
Rotational symmetry:
Q.3. In a figure has two or more lines of symmetry, should it have rotational symmetry of order more than 1?
Ans: Yes, because every line through the centre forms a line of symmetry and it has rotational symmetry around the centre for every angle.
Q.4. Fill in the blanks:
Ans:
Q.5. Name the quadrilateral which has both line and rotational symmetry of order more than 1.
Ans: Square has both line and rotational symmetry of order more than 1.
Line symmetry:
Rotational symmetry:
Q.6. After rotating by 60o about a centre, a figure looks exactly the same as its original position. At what other angles will this happen for the figure?
Ans: Other angles will be 120°, 180°,240°,300°,360°.
For 60° rotation:
It will rotate six times.
For 120° rotation:
It will rotate three times.
For 180° rotation:
It will rotate two times.
For 360° rotation:
It will rotate one time.
Q.7. Can we have a rotational symmetry of order more than 1 whose angle of rotation is:
(i) 45o
(ii) 17o
Ans:
(i) If tiie angle of rotation is 45°, then symmetry of order is possible and would be 8 rotations.
(ii) If the angle of rotational is 17°, then symmetry o f order is not possible because 360° is not complete divided by 17°.
The document NCERT Solutions for Class 8 Maths - Symmetry- 1 is a part of the Class 7 Course Mathematics (Maths) Class 7.
All you need of Class 7 at this link: Class 7
## Mathematics (Maths) Class 7
76 videos|345 docs|39 tests
## FAQs on NCERT Solutions for Class 8 Maths - Symmetry- 1
1. What is symmetry in mathematics?
Ans. Symmetry in mathematics is a concept where an object remains unchanged when it is moved, rotated, or reflected. It is a property that helps in identifying patterns and creating balance in various shapes and figures.
2. How many types of symmetry are there?
Ans. There are three main types of symmetry: reflectional symmetry (also known as mirror symmetry), rotational symmetry, and translational symmetry. Each type involves a different way in which an object can be symmetric.
3. How is symmetry used in real-life applications?
Ans. Symmetry is used in various real-life applications such as architecture, design, art, and even in nature. It helps in creating aesthetically pleasing structures and patterns, and also plays a role in fields like crystallography and computer graphics.
4. How can symmetry be helpful in problem-solving?
Ans. Symmetry can be helpful in problem-solving by identifying repeating patterns and relationships within a problem. It can simplify complex situations and aid in finding solutions more efficiently by leveraging the properties of symmetry.
5. Can symmetry be found in everyday objects?
Ans. Yes, symmetry can be found in everyday objects such as buildings, furniture, clothing, and even in natural objects like flowers and snowflakes. Recognizing symmetry in these objects can enhance our appreciation for their beauty and design.
## Mathematics (Maths) Class 7
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# How To Do A Five Number Summary? (Best solution)
Following are the steps to creating a five-number summary.
1. Ascending order (from smallest to largest) is the first step.
2. Step 2 is determining the lowest and maximum values for your data collection. Step 3: Calculate the mean and the median. Fourth, put parenthesis around the figures that are above and below the median. Step 5: Locate questions 1 and 3
## What is the 5 number summary in math?
Using a five-number summary is particularly effective when doing descriptive analysis or when conducting early investigations into a huge data collection of data. Typically, a summary has five numbers: the most extreme values in the data set (the maximum and lowest values), the lower and upper quartiles, and the median (the middle value).
## What are the 5 numbers in the five-number summary?
A five-number summary is comprised of the lowest data value, the first quartile, the median, the third quartile, and the biggest data value, to name a few elements of information.
## What order does the five-number summary go in?
A box and whisker plot, often known as a box plot, is a graphical representation of the five-number summary of a collection of data. The minimum, first quartile, median, third quartile, and maximum are represented by the five-number summary. A box plot is created by drawing a box from the first quartile to the third quartile of the data. A vertical line runs across the middle of the box, connecting it to the rest of the box.
You might be interested: Letters From An American Farmer What Is An American Summary? (Best solution)
## How do the 5 number summaries compare to one another?
It is possible to compare and contrast five numerical summaries. As a result, we will discover that two sets with comparable means and standard deviations may have very different five-number summaries. A boxplot, also known as a box and whiskers graph, is a visual representation that allows us to quickly compare two five-number summaries at a glance.
## How do you draw a box plot?
A box plot is created by combining a horizontal or vertical number line with a rectangular box of equal size. The axis’s ends are labeled by the lowest and greatest data values, respectively. Using the first quartile as a starting point, and the third quartile as a finishing point, we can draw a box.
## Is Q2 the median or mean?
The median is regarded as the second quartile of data (Q2). This is defined as the difference between the highest and bottom quartiles of a population’s distribution. The semi-interquartile range is half of the interquartile range in terms of variability. When the data set is small, it is straightforward to determine the values of quartiles in the data.
## Do you include outliers in 5 number summary?
It is a way for summarizing a distribution of data that uses five numbers to represent the data. This is a significant departure from the rest of the data. It is an outlier and must be deleted from the data set.
## How do you read a Boxplot?
What exactly is a Boxplot?
1. The smallest number in the data collection (the lowest number in the data set). Quartile one (Q1) is located at the extreme left of the box (or at the extreme right of the left whisker). The median is represented by a line drawn through the middle of the box. In the box on the right, the third quartile, Q3, is displayed at the far right of the box (at the far left of the right whisker).
You might be interested: When I Was Puerto Rico Summary? (TOP 5 Tips)
## How do you find Q3?
Upper Quartile (Q3) = (N+1) * 3 / 4 = (N+1) * 3 / 4
1. Upper Quartile (Q3)= (15+1)*3/4
2. Upper Quartile (Q3)= 48 / 4 = 12th data point
3. Upper Quartile (Q3)= 15+1)*3/4
4. Upper Quartile (
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# Geometric Series – Definition, Formula, and Examples
The geometric series plays an important part in the early stages of calculus and contributes to our understanding of the convergence series. We can also use the geometric series in physics, engineering, finance, and finance. This shows that is essential that we know how to identify and find the sum of geometric series.
The geometric series represents the sum of the terms in a finite or infinite geometric sequence. The consecutive terms in this series share a common ratio.
In this article, we’ll understand how closely related the geometric sequence and series are. We’ll also show you how the infinite and finite sums are calculated. You’ll also get the chance to try out word problems that make use of geometric series.
## What is a geometric series?
The geometric series represents the sum of the geometric sequence’s terms. This means that the terms of a geometric series will also share a common ratio, $r$.
Since the geometric series is closely related to the geometric sequence, we’ll do a quick refresher on the geometric sequence’s definition to understand the geometric series’ components.
Does this image look familiar? That’s because this is one known way for us to visualize what happens with a geometric sequence with the following terms: $\left\{1, \dfrac{1}{2}, \dfrac{1}{4},\dfrac{1}{8},\dfrac{1}{16}\right\}$. This is an example of a geometric sequence that has a common ratio that is less than $1$. This sequence becomes a series if we express the terms’ sum as shown below:
\begin{aligned}1 \underbrace{+}_{\color{Orchid} \times \frac{1}{2}} \dfrac{1}{2} \underbrace{+}_{\color{Orchid} \times \frac{1}{2}} \dfrac{1}{4} \underbrace{+}_{\color{Orchid} \times \frac{1}{2}} \dfrac{1}{16} &= \sum_{i=1}^{n} 1\cdot \dfrac{1}{2}^{n -1}\end{aligned}
Here are a few more examples highlighting the difference between the arithmetic sequence and series.
Geometric Sequence Geometric Series Common Ratio \begin{aligned}4,8,16,32, …, 1024\end{aligned} \begin{aligned}4+8 +16+32+ …+ 1024\end{aligned} \begin{aligned}r =2\end{aligned} \begin{aligned} 1,-3,9, -27,…,-2187 \end{aligned} \begin{aligned} 1+-3 +9 -27+…-2187\end{aligned} \begin{aligned}r = -3\end{aligned} \begin{aligned} 625,125, 25, …\end{aligned} \begin{aligned} 625+125+25+ …\end{aligned} \begin{aligned}r = \dfrac{1}{2}\end{aligned}
Have you noticed something different with the third example? The series had no last term and that’s because it’s possible for a geometric series to either be a finite or infinite series:
• A finite geometric series contains a finite number of terms. This means that the series will have both first and last terms. Finite geometric series are also convergent.
• The infinite geometric series, on the other hand, goes on and approaches infinity. This means that the geometric series that is infinite does not have the last term.
Here are the general forms of the geometric sequence and series. We’ve separated the general forms of the finite and infinite geometric series to help you distinguish them faster when dealing with geometric series later.
Geometric Sequence \begin{aligned} a\underbrace{\phantom{x}} _{\color{Teal} \times r}ar\underbrace{\phantom{x}} _{\color{Teal} \times r}ar^2,…,ar^{n -2}\underbrace{\phantom{x}} _{\color{Teal}\times r}ar^{n -1}\end{aligned} Geometric Series (Finite) \begin{aligned} a\underbrace{+} _{\color{Teal}\times r}ar\underbrace{+} _{\color{Teal}\times r}ar^2+…+ar^{n – 2}\underbrace{+}_{\color{Teal}\times r}ar^{n – 1}\end{aligned} Geometric Series (Infinite) \begin{aligned} a\underbrace{+} _{\color{Teal}\times r}ar\underbrace{+} _{\color{Teal}\times r}ar^2+…\end{aligned}
Keep in mind that $a$ represents the first term of the series, $r$ represents the common ratio, and $n$ represents the number of terms. Now that we understand the forms of the geometric sequences and series, let’s use these expressions to establish the geometric series’ formulas.
### Derivation of geometric series formulas
We can express the $\boldsymbol{n}$th term of any geometric series as $\boldsymbol{a_n = ar^{n -1}}$. This means that we can use this formula to express the sum of the series, $S_n$, as shown below.
\begin{aligned} S_n &= a_1 + a_2 + a_3 + …+ a_{n- 1}+ a_n\\&=ar^{1 -1} + ar^{2 -1} + ar^{3 -1} + …+ar^{n – 1 -1} + ar^{n -1}\\&=a + ar + ar^2 +…+ar^{n-2} + ar^{n-1},\phantom{xxxxx}\text{Equation 1}\end{aligned}
Multiply both sides of the first equation by $r$.
\begin{aligned} S_nr &=ar + ar^2 + ar^3 +…+ar^{n-1} + ar^{n},\phantom{xxxxx}\text{Equation 2}\end{aligned}
Subtract the second equation from the first one then isolate $S_n$ on the left-hand side of the equation.
\begin{aligned} S_n &= a + ar + ar^2 +…+ar^{n -2} + ar^{n- 1},\phantom{x}(1)\\-\underline{\phantom{xxxx}S_nr }&= \underline{ar + ar^2 + ar^3 +…+ar^{n -1} + ar^{n}},\phantom{x}(2)\\S_n – S_nr &= a-ar^n\end{aligned}
\begin{aligned} S_n(1 – r) &= a(1 – r^n)\\S_n&=\dfrac{a(1 – r^n)}{1 – r}\end{aligned}
Hence, we have the formula for the finite geometric series’ sum as shown below.
\begin{aligned} S_n&=\dfrac{a(1 – r^n)}{1 – r}\\\\S_n&:\text{Geometric series’s sum}\\a &: \text{First term}\\r &: \text{Common ratio}\end{aligned}
When you have $r <1$, the positions of $r$ and $1$ will interchange. In fact, it will be $S_n = \dfrac{a(r^n -1)}{r -1}$. But this is just a variation of the formula shown above.
For now, let’s see what happens when $-1<r<1$ and $n \rightarrow \infty$. Begin by rewriting the expression for $S_n$ as a sum of two rational expressions as shown below.
\begin{aligned} S_n&=\dfrac{a – ar^n}{1 – r}\\&= \dfrac{a}{1 – r} – \dfrac{ar^n}{1 – r}\end{aligned}
As $n \rightarrow \infty$, $\dfrac{ar^n}{1 – r} \rightarrow 0$ when $-1< r <$ and $r \neq 0$. When $r > 1$, the infinite series diverges. Hence, we have the sum formula for the infinite geometric series as shown below.
\begin{aligned} S_\infty&=\dfrac{a}{1 – r}\\\\S_\infty&:\text{Infinite series’ sum}\\a &: \text{First term}\\r &: \text{Common ratio}\\ &:-1< r<1\end{aligned}
We now have the two essential sum formulas for the geometric series. It’s now time that we learn how to apply these two formulas when given a geometric series.
## How to solve geometric series?
Here are some important pointers to remember when solving a geometric series problem:
• Confirm whether the given series is a finite or an infinite geometric series.
• Identify the values that are available: $a$, $r$, and $n$. Find $n$ using the equation, $a_n=ar^{n –1}$.
• Apply the appropriate formula depending on whether the series is finite or infinite.
### How to tell if a series is geometric?
It’s easy to determine whether a given series is geometric or not – we simply find the ratio shared between two consecutive terms. Compare this ratio with the rest and see if they’re equal. When they are, the series is said to be geometric.
\begin{aligned} a_1, a_2, a_3, a_4, &…,a_{n-1}, a_n\\\dfrac{a_2}{a_1} &= r\\\dfrac{a_3}{a_2}&= r\\\dfrac{a_4}{a_3}&= r\\&.\\&.\\&.\\\dfrac{a_n}{a_{n -1}}&= r\end{aligned}
When you can confirm the existence of the common ratio, $\boldsymbol{r}$, from the series you’re observing, then the series is geometric.
### How to apply the geometric series’ formula?
Now that we have confirmed that the series is indeed geometric, your next goal is to check if the geometric series is finite or infinite. Apply the appropriate formula as shown below.
Finite Series Infinite Series \begin{aligned} S_n&=\dfrac{a(1 – r^n)}{1 – r} \end{aligned} \begin{aligned} S_\infty&=\dfrac{a}{1 – r}\\-1&< r<1\end{aligned}
From the formula, we can see that we only need the first term, the common ratio, and the number of terms for the finite series. For the infinite series, check if $r$ is a fraction, and only then can you apply the formula.
Let’s work on the two series shown below and we’ll begin by confirming that they are both geometric.
\begin{aligned} &1)\phantom{x}2+ 4+ 8+…+512+ 1024\\&2)\phantom{x}81+ 27 +9 +3+ … \end{aligned}
Observe the ratio shared by each consecutive term for each of the series. If the series shares a common ratio, then the series is geometric.
$1)$ \begin{aligned} \dfrac{4}{2} = 2,\phantom{x}\dfrac{8}{4} = 2, &…,\phantom{x}\dfrac{1024}{512} =2 \\\\2 \underbrace{+}_{\color{DarkOrange}\times 2}4\underbrace{+}_{\color{DarkOrange}\times 2} 8+&…+512 \underbrace{+}_{\color{DarkOrange}\times 2} 1024 \\\end{aligned} $2)$ \begin{aligned} \dfrac{27}{81} = \dfrac{1}{3},\phantom{x}\dfrac{9}{27} &= \dfrac{1}{3}, \phantom{x}\dfrac{3}{9} = \dfrac{1}{3}…\\\\81 \underbrace{+}_{\color{Purple}\times \frac{1}{3}}27 &\underbrace{+}_{\color{Purple}\times \frac{1}{3}} 9 \underbrace{+}_{\color{Purple}\times \frac{1}{3}} 3+…\\\end{aligned}
Since for both cases, the ratio between two consecutive terms remains constant, the two are geometric series. The only difference is that 1) is a finite geometric series while 2) is an infinite geometric series.
Focusing on the first series, $2+ 4+ 8+…+512+ 1024$, we know that $a = 2$ and $r = 2$. Now, to find the number of terms, let’s use the fact that $\boldsymbol{a_n = ar^{n -1}}$ to solve for $n$.
\begin{aligned}a_n &= ar^{n – 1}\\1024 &= 2\cdot 2^{n -1}\\512 &= 2^{n – 1}\\2^9 &= 2^{n -1}\\9&= n -1\\n&= 10\end{aligned}
Since we now have all the values we need, let’s apply the sum formula for the finite geometric series.
\begin{aligned} S_n&=\dfrac{a(1 – r^n)}{1 – r}\\S_{10} &=\dfrac{2(1 – 2^{10})}{1 – 2}\\&=\dfrac{2(-1023)}{-1}\\ &= 2046 \end{aligned}
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# APEX Calculus
## Section10.3Calculus and Parametric Equations
The previous section defined curves based on parametric equations. In this section we'll employ the techniques of calculus to study these curves.
We are still interested in lines tangent to points on a curve. They describe how the $$y$$-values are changing with respect to the $$x$$-values, they are useful in making approximations, and they indicate instantaneous direction of travel.
The slope of the tangent line is still $$\frac{dy}{dx}\text{,}$$ and the Chain Rule allows us to calculate this in the context of parametric equations. If $$x=f(t)$$ and $$y=g(t)\text{,}$$ the Chain Rule states that
\begin{equation*} \frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}\text{.} \end{equation*}
Solving for $$\frac{dy}{dx}\text{,}$$ we get
\begin{equation*} \frac{dy}{dx} = \frac{dy}{dt}\Bigg/\frac{dx}{dt} = \frac{g'(t)}{\fp(t)}\text{,} \end{equation*}
provided that $$\fp(t)\neq 0\text{.}$$ This is important so we label it a Key Idea.
### Key Idea10.3.1.Finding $$\frac{dy}{dx}$$ with Parametric Equations.
Let $$x=f(t)$$ and $$y=g(t)\text{,}$$ where $$f$$ and $$g$$ are differentiable on some open interval $$I$$ and $$\fp(t)\neq 0$$ on $$I\text{.}$$ Then
\begin{equation*} \frac{dy}{dx} = \frac{g'(t)}{\fp(t)}\text{.} \end{equation*}
We use this to define the tangent line.
### Definition10.3.2.Tangent and Normal Lines.
Let a curve $$C$$ be parametrized by $$x=f(t)$$ and $$y=g(t)\text{,}$$ where $$f$$ and $$g$$ are differentiable functions on some interval $$I$$ containing $$t=t_0\text{.}$$ The tangent line to $$C$$ at $$t=t_0$$ is the line through $$\big(f(t_0),g(t_0)\big)$$ with slope $$m=g'(t_0)/\fp(t_0)\text{,}$$ provided $$\fp(t_0)\neq 0\text{.}$$
The normal line to $$C$$ at $$t=t_0$$ is the line through $$\big(f(t_0),g(t_0)\big)$$ with slope $$m=-\fp(t_0)/g'(t_0)\text{,}$$ provided $$g'(t_0)\neq 0\text{.}$$
The definition leaves two special cases to consider. When the tangent line is horizontal, the normal line is undefined by the above definition as $$g'(t_0)=0\text{.}$$ Likewise, when the normal line is horizontal, the tangent line is undefined. It seems reasonable that these lines be defined (one can draw a line tangent to the “right side” of a circle, for instance), so we add the following to the above definition.
1. If the tangent line at $$t=t_0$$ has a slope of 0, the normal line to $$C$$ at $$t=t_0$$ is the line $$x=f(t_0)\text{.}$$
2. If the normal line at $$t=t_0$$ has a slope of 0, the tangent line to $$C$$ at $$t=t_0$$ is the line $$x=f(t_0)\text{.}$$
### Example10.3.3.Tangent and Normal Lines to Curves.
Let $$x=5t^2-6t+4$$ and $$y=t^2+6t-1\text{,}$$ and let $$C$$ be the curve defined by these equations.
1. Find the equations of the tangent and normal lines to $$C$$ at $$t=3\text{.}$$
2. Find where $$C$$ has vertical and horizontal tangent lines.
Solution.
1. We start by computing $$\fp(t) = 10t-6$$ and $$g'(t) =2t+6\text{.}$$ Thus
\begin{equation*} \frac{dy}{dx} = \frac{2t+6}{10t-6}\text{.} \end{equation*}
Make note of something that might seem unusual: $$\frac{dy}{dx}$$ is a function of $$t\text{,}$$ not $$x\text{.}$$ Just as points on the curve are found in terms of $$t\text{,}$$ so are the slopes of the tangent lines. The point on $$C$$ at $$t=3$$ is $$(31,26)\text{.}$$ The slope of the tangent line is $$m=1/2$$ and the slope of the normal line is $$m=-2\text{.}$$ Thus,
• the equation of the tangent line is $$\ds y=\frac12(x-31)+26\text{,}$$ and
• the equation of the normal line is $$\ds y=-2(x-31)+26\text{.}$$
This is illustrated in Figure 10.3.4.
2. To find where $$C$$ has a horizontal tangent line, we set $$\frac{dy}{dx}=0$$ and solve for $$t\text{.}$$ In this case, this amounts to setting $$g'(t)=0$$ and solving for $$t$$ (and making sure that $$\fp(t)\neq 0$$).
\begin{equation*} g'(t)=0 \Rightarrow 2t+6=0 \Rightarrow t=-3\text{.} \end{equation*}
The point on $$C$$ corresponding to $$t=-3$$ is $$(67,-10)\text{;}$$ the tangent line at that point is horizontal (hence with equation $$y=-10$$). To find where $$C$$ has a vertical tangent line, we find where it has a horizontal normal line, and set $$-\frac{\fp(t)}{g'(t)}=0\text{.}$$ This amounts to setting $$\fp(t)=0$$ and solving for $$t$$ (and making sure that $$g'(t)\neq 0$$).
\begin{equation*} \fp(t)=0 \Rightarrow 10t-6=0 \Rightarrow t=0.6\text{.} \end{equation*}
The point on $$C$$ corresponding to $$t=0.6$$ is $$(2.2,2.96)\text{.}$$ The tangent line at that point is $$x=2.2\text{.}$$ The points where the tangent lines are vertical and horizontal are indicated on the graph in Figure 10.3.4.
### Example10.3.5.Tangent and Normal Lines to a Circle.
1. Find where the unit circle, defined by $$x=\cos(t)$$ and $$y=\sin(t)$$ on $$[0,2\pi]\text{,}$$ has vertical and horizontal tangent lines.
2. Find the equation of the normal line at $$t=t_0\text{.}$$
Solution.
1. We compute the derivative following Key Idea 10.3.1:
\begin{equation*} \frac{dy}{dx} = \frac{g'(t)}{\fp(t)} = -\frac{\cos(t) }{\sin(t) }\text{.} \end{equation*}
The derivative is $$0$$ when $$\cos(t) = 0\text{;}$$ that is, when $$t=\pi/2,\, 3\pi/2\text{.}$$ These are the points $$(0,1)$$ and $$(0,-1)$$ on the circle. The normal line is horizontal (and hence, the tangent line is vertical) when $$\sin(t) =0\text{;}$$ that is, when $$t= 0,\,\pi,\,2\pi\text{,}$$ corresponding to the points $$(-1,0)$$ and $$(0,1)$$ on the circle. These results should make intuitive sense.
2. The slope of the normal line at $$t=t_0$$ is $$\ds m=\frac{\sin(t_0) }{\cos(t_0) } = \tan(t_0)\text{.}$$ This normal line goes through the point $$(\cos(t_0) ,\sin(t_0) )\text{,}$$ giving the line
\begin{align*} y \amp =\frac{\sin(t_0) }{\cos(t_0) }(x-\cos(t_0) ) + \sin(t_0)\\ \amp = (\tan(t_0) )x\text{,} \end{align*}
as long as $$\cos(t_0) \neq 0\text{.}$$ It is an important fact to recognize that the normal lines to a circle pass through its center, as illustrated in Figure 10.3.6. Stated in another way, any line that passes through the center of a circle intersects the circle at right angles.
### Example10.3.7.Tangent lines when $$\frac{dy}{dx}$$ is not defined.
Find the equation of the tangent line to the astroid $$x=\cos^3(t)\text{,}$$ $$y=\sin^3(t)$$ at $$t=0\text{,}$$ shown in Figure 10.3.8.
Solution.
We start by finding $$x'(t)$$ and $$y'(t)\text{:}$$
\begin{equation*} x'(t) = -3\sin(t) \cos^2(t) , \qquad y'(t) = 3\cos(t) \sin^2(t)\text{.} \end{equation*}
Note that both of these are 0 at $$t=0\text{;}$$ the curve is not smooth at $$t=0$$ forming a cusp on the graph. Evaluating $$\frac{dy}{dx}$$ at this point returns the indeterminate form of “0/0”.
We can, however, examine the slopes of tangent lines near $$t=0\text{,}$$ and take the limit as $$t\to 0\text{.}$$
\begin{align*} \lim_{t\to0} \frac{y'(t)}{x'(t)} \amp =\lim_{t\to0} \frac{3\cos(t) \sin^2(t) }{-3\sin(t) \cos^2(t) } \text{ (We can cancel as $$t\neq 0$$.) }\\ \amp = \lim_{t\to0} -\frac{\sin(t) }{\cos(t) }\\ \amp = 0\text{.} \end{align*}
We have accomplished something significant. When the derivative $$\frac{dy}{dx}$$ returns an indeterminate form at $$t=t_0\text{,}$$ we can define its value by setting it to be $$\lim\limits_{t\to t_0}$$$$\frac{dy}{dx}\text{,}$$ if that limit exists. This allows us to find slopes of tangent lines at cusps, which can be very beneficial.
We found the slope of the tangent line at $$t=0$$ to be 0; therefore the tangent line is $$y=0\text{,}$$ the $$x$$-axis.
### Subsection10.3.1Concavity
We continue to analyze curves in the plane by considering their concavity; that is, we are interested in $$\frac{d^2y}{dx^2}\text{,}$$ “the second derivative of $$y$$ with respect to $$x\text{.}$$” To find this, we need to find the derivative of $$\frac{dy}{dx}$$ with respect to $$x\text{;}$$ that is,
\begin{equation*} \frac{d^2y}{dx^2}=\frac{d}{dx}\left[\frac{dy}{dx}\right]\text{,} \end{equation*}
but recall that $$\frac{dy}{dx}$$ is a function of $$t\text{,}$$ not $$x\text{,}$$ making this computation not straightforward.
To make the upcoming notation a bit simpler, let $$h(t) = \frac{dy}{dx}\text{.}$$ We want $$\frac{d}{dx}[h(t)]\text{;}$$ that is, we want $$\frac{dh}{dx}\text{.}$$ We again appeal to the Chain Rule. Note:
\begin{equation*} \frac{dh}{dt} = \frac{dh}{dx}\cdot\frac{dx}{dt} \Rightarrow \frac{dh}{dx} = \frac{dh}{dt}\Bigg/\frac{dx}{dt}\text{.} \end{equation*}
In words, to find $$\frac{d^2y}{dx^2}\text{,}$$ we first take the derivative of $$\frac{dy}{dx}$$ with respect to $$t$$, then divide by $$x'(t)\text{.}$$ We restate this as a Key Idea.
#### Key Idea10.3.9.Finding $$\frac{d^2y}{dx^2}$$ with Parametric Equations.
Let $$x=f(t)$$ and $$y=g(t)$$ be twice differentiable functions on an open interval $$I\text{,}$$ where $$\fp(t)\neq 0$$ on $$I\text{.}$$ Then
\begin{equation*} \frac{d^2y}{dx^2} = \frac{d}{dt}\left[\frac{dy}{dx}\right]\Bigg/\frac{dx}{dt} = \frac{d}{dt}\left[\frac{dy}{dx}\right]\Bigg/\fp(t)\text{.} \end{equation*}
Examples will help us understand this Key Idea.
#### Example10.3.10.Concavity of Plane Curves.
Let $$x=5t^2-6t+4$$ and $$y=t^2+6t-1$$ as in Example 10.3.3. Determine the $$t$$-intervals on which the graph is concave up/down.
Solution.
Concavity is determined by the second derivative of $$y$$ with respect to $$x\text{,}$$ $$\frac{d^2y}{dx^2}\text{,}$$ so we compute that here following Key Idea 10.3.9.
In Example 10.3.3, we found $$\ds\frac{dy}{dx} = \frac{2t+6}{10t-6}$$ and $$\fp(t) = 10t-6\text{.}$$ So:
\begin{align*} \frac{d^2y}{dx^2} \amp = \frac{d}{dt}\left[\frac{2t+6}{10t-6}\right]\Bigg/(10t-6)\\ \amp = -\frac{72}{(10t-6)^2}\Bigg/(10t-6)\\ \amp = -\frac{72}{(10t-6)^3}\\ \amp = -\frac{9}{(5t-3)^3} \end{align*}
The graph of the parametric functions is concave up when $$\frac{d^2y}{dx^2} \gt 0$$ and concave down when $$\frac{d^2y}{dx^2} \lt 0\text{.}$$ We determine the intervals when the second derivative is greater/less than 0 by first finding when it is 0 or undefined.
As the numerator of $$\ds -\frac{9}{(5t-3)^3}$$ is never 0, $$\frac{d^2y}{dx^2} \neq 0$$ for all $$t\text{.}$$ It is undefined when $$5t-3=0\text{;}$$ that is, when $$t= 3/5\text{.}$$ Following the work established in Section 3.4, we look at values of $$t$$ greater/less than $$3/5$$ on a number line:
Reviewing Example 10.3.3, we see that when $$t=3/5=0.6\text{,}$$ the graph of the parametric equations has a vertical tangent line. This point is also a point of inflection for the graph, illustrated in Figure 10.3.11.
#### Example10.3.12.Concavity of Plane Curves.
Find the points of inflection of the graph of the parametric equations $$x=\sqrt{t}\text{,}$$ $$y=\sin(t)\text{,}$$ for $$0\leq t\leq 16\text{.}$$
Solution.
We need to compute $$\frac{dy}{dx}$$ and $$\frac{d^2y}{dx^2}\text{.}$$
\begin{equation*} \frac{dy}{dx} = \frac{y'(t)}{x'(t)} = \frac{\cos(t) }{1/(2\sqrt{t})} = 2\sqrt{t}\cos(t)\text{.} \end{equation*}
\begin{equation*} \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left[\frac{dy}{dx}\right]}{x'(t)} = \frac{\cos(t) /\sqrt{t}-2\sqrt{t}\sin(t) }{1/(2\sqrt{t})}=2\cos(t) -4t\sin(t)\text{.} \end{equation*}
The points of inflection are found by setting $$\frac{d^2y}{dx^2}=0\text{.}$$ This is not trivial, as equations that mix polynomials and trigonometric functions generally do not have “nice” solutions.
In Figure 10.3.13.(a) we see a plot of the second derivative. It shows that it has zeros at approximately $$t=0.5,\,3.5,\,6.5,\,9.5,\,12.5$$ and $$16\text{.}$$ These approximations are not very good, made only by looking at the graph. Newton's Method provides more accurate approximations. Accurate to 2 decimal places, we have:
\begin{equation*} t=0.65,\,3.29,\,6.36,\,9.48,\,12.61\,\text{ and } \,15.74\text{.} \end{equation*}
The corresponding points have been plotted on the graph of the parametric equations in Figure 10.3.13.(b). Note how most occur near the $$x$$-axis, but not exactly on the axis.
### Subsection10.3.2Arc Length
We continue our study of the features of the graphs of parametric equations by computing their arc length.
Recall in Section 7.4 we found the arc length of the graph of a function, from $$x=a$$ to $$x=b\text{,}$$ to be
\begin{equation*} L = \int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\, dx\text{.} \end{equation*}
We can use this equation and convert it to the parametric equation context. Letting $$x=f(t)$$ and $$y=g(t)\text{,}$$ we know that $$\frac{dy}{dx} = g'(t)/\fp(t)\text{.}$$ It will also be useful to calculate the differential of $$x\text{:}$$
\begin{equation*} dx = \fp(t)dt \qquad \Rightarrow \qquad dt = \frac{1}{\fp(t)}\cdot dx\text{.} \end{equation*}
Starting with the arc length formula above, consider:
\begin{align*} L \amp = \int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\, dx\\ \amp = \int_a^b \sqrt{1+\frac{g'(t)^2}{\fp(t)^2}}\, dx.\\ \end{align*}
Factor out the $$\fp(t)^2\text{:}$$
\begin{align*} \amp = \int_a^b \sqrt{\fp(t)^2+g'(t)^2}\cdot\underbrace{\frac1{\fp(t)}\, dx}_{=dt}\\ \amp = \int_{t_1}^{t_2} \sqrt{\fp(t)^2+g'(t)^2}\, dt\text{.} \end{align*}
Note the new bounds (no longer “$$x$$” bounds, but “$$t$$” bounds). They are found by finding $$t_1$$ and $$t_2$$ such that $$a= f(t_1)$$ and $$b=f(t_2)\text{.}$$ This formula is important, so we restate it as a theorem.
As before, these integrals are often not easy to compute. We start with a simple example, then give another where we approximate the solution.
#### Example10.3.15.Arc Length of a Circle.
Find the arc length of the circle parametrized by $$x=3\cos(t)\text{,}$$ $$y=3\sin(t)$$ on $$[0,3\pi/2]\text{.}$$
Solution.
By direct application of Theorem 10.3.14, we have
\begin{align*} L \amp = \int_0^{3\pi/2} \sqrt{(-3\sin(t) )^2 +(3\cos(t) )^2} \, dt.\\ \end{align*}
Apply the Pythagorean Theorem.
\begin{align*} \amp = \int_0^{3\pi/2} 3 \, dt\\ \amp = 3t\Big|_0^{3\pi/2} = 9\pi/2\text{.} \end{align*}
This should make sense; we know from geometry that the circumference of a circle with radius 3 is $$6\pi\text{;}$$ since we are finding the arc length of $$3/4$$ of a circle, the arc length is $$3/4\cdot 6\pi = 9\pi/2\text{.}$$
#### Example10.3.16.Arc Length of a Parametric Curve.
The graph of the parametric equations $$x=t(t^2-1)\text{,}$$ $$y=t^2-1$$ crosses itself as shown in Figure 10.3.17, forming a “teardrop.” Find the arc length of the teardrop.
Solution.
We can see by the parametrizations of $$x$$ and $$y$$ that when $$t=\pm 1\text{,}$$ $$x=0$$ and $$y=0\text{.}$$ This means we'll integrate from $$t=-1$$ to $$t=1\text{.}$$ Applying Theorem 10.3.14, we have
\begin{align*} L \amp = \int_{-1}^1\sqrt{(3t^2-1)^2+(2t)^2}\, dt\\ \amp = \int_{-1}^1 \sqrt{9t^4-2t^2+1} \, dt\text{.} \end{align*}
Unfortunately, the integrand does not have an antiderivative expressible by elementary functions. We turn to numerical integration to approximate its value. Using 4 subintervals, Simpson's Rule approximates the value of the integral as $$2.65051\text{.}$$ Using a computer, more subintervals are easy to employ, and $$n=20$$ gives a value of $$2.71559\text{.}$$ Increasing $$n$$ shows that this value is stable and a good approximation of the actual value.
### Subsection10.3.3Surface Area of a Solid of Revolution
Related to the formula for finding arc length is the formula for finding surface area. We can adapt the formula found in Theorem 7.4.11 from Section 7.4 in a similar way as done to produce the formula for arc length done before.
#### Example10.3.19.Surface Area of a Solid of Revolution.
Consider the teardrop shape formed by the parametric equations $$x=t(t^2-1)\text{,}$$ $$y=t^2-1$$ as seen in Example 10.3.16. Find the surface area if this shape is rotated about the $$x$$-axis, as shown in Figure 10.3.20.
Solution.
The teardrop shape is formed between $$t=-1$$ and $$t=1\text{.}$$ Using Theorem 10.3.18, we see we need for $$g(t)\geq 0$$ on $$[-1,1]\text{,}$$ and this is not the case. To fix this, we simplify replace $$g(t)$$ with $$-g(t)\text{,}$$ which flips the whole graph about the $$x$$-axis (and does not change the surface area of the resulting solid). The surface area is:
\begin{align*} \text{ Area } \,S \amp = 2\pi\int_{-1}^1 (1-t^2)\sqrt{(3t^2-1)^2+(2t)^2}\, dt\\ \amp = 2\pi\int_{-1}^1 (1-t^2)\sqrt{9t^4-2t^2+1} \, dt\text{.} \end{align*}
Once again we arrive at an integral that we cannot compute in terms of elementary functions. Using Simpson's Rule with $$n=20\text{,}$$ we find the area to be $$S=9.44\text{.}$$ Using larger values of $$n$$ shows this is accurate to 2 places after the decimal.
After defining a new way of creating curves in the plane, in this section we have applied calculus techniques to the parametric equation defining these curves to study their properties. In the next section, we define another way of forming curves in the plane. To do so, we create a new coordinate system, called polar coordinates, that identifies points in the plane in a manner different than from measuring distances from the $$y$$- and $$x$$- axes.
### Exercises10.3.4Exercises
#### Terms and Concepts
##### 1.
True or False? Given parametric equations $$x=f(t)$$ and $$y=g(t)\text{,}$$ $$\lz{y}{x} = \fp(t)/g'(t)\text{,}$$ as long as $$g'(t) \neq 0\text{.}$$
• True
• False
##### 2.
Given parametric equations $$x=f(t)$$ and $$y=g(t)\text{,}$$ the derivative $$\frac{dy}{dx}$$ as given in Key Idea 10.3.1 is a function of ?
##### 3.
True or False? Given parametric equations $$x=f(t)$$ and $$y=g(t)\text{,}$$ to find $$\lzn{2}{y}{x}\text{,}$$ one simply computes $$\lzoo{t}{\lz{y}{x}}\text{.}$$
• True
• False
##### 4.
True or False? If $$\lz{y}{x}=0$$ at $$t=t_0\text{,}$$ then the normal line to the curve at $$t=t_0$$ is a vertical line.
• True
• False
#### Problems
##### Exercise Group.
In the following exercises, parametric equations for a curve are given.
1. Find $$\ds\frac{dy}{dx}\text{.}$$
2. Find the equations of the tangent and normal line(s) at the point(s) given.
3. Sketch the graph of the parametric functions along with the found tangent and normal lines.
###### 5.
$$x=t\text{,}$$ $$y=t^2\text{;}$$$$t=1$$
###### 6.
$$x=\sqrt{t}\text{,}$$ $$y=5t+2\text{;}$$$$t=4$$
###### 7.
$$x=t^2-t\text{,}$$ $$y=t^2+t\text{;}$$$$t=1$$
###### 8.
$$x=t^2-1\text{,}$$ $$y=t^3-t\text{;}$$$$t=0$$ and $$t=1$$
###### 9.
$$x=\sec(t)\text{,}$$ $$y=\tan(t)$$ on $$(-\pi/2,\pi/2)\text{;}$$$$t=\pi/4$$
###### 10.
$$x=\cos(t)\text{,}$$ $$y=\sin(2t)$$ on $$[0,2\pi]\text{;}$$$$t=\pi/4$$
###### 11.
$$x=\cos(t) \sin(2t)\text{,}$$ $$y=\sin(t) \sin(2t)$$ on $$[0,2\pi]\text{;}$$ $$t=3\pi/4$$
###### 12.
$$x=e^{t/10}\cos(t)\text{,}$$ $$y=e^{t/10}\sin(t)\text{;}$$ $$t=\pi/2$$
##### Exercise Group.
Find the $$t$$-values where the curve defined by the given parametric equations has a horizontal tangent line. Note: these are the same equations as in Exercises 10.3.4.5Exercise 10.3.4.12.
###### 13.
$$x=t\text{,}$$ $$y=t^2$$
###### 14.
$$x=\sqrt{t}\text{,}$$ $$y=5t+2$$
###### 15.
$$x=t^2-t\text{,}$$ $$y=t^2+t$$
###### 16.
$$x=t^2-1\text{,}$$ $$y=t^3-t$$
###### 17.
$$x=\sec(t)\text{,}$$ $$y=\tan(t)$$ on $$(-\pi/2,\pi/2)$$
###### 18.
$$x=\cos(t)\text{,}$$ $$y=\sin(2t)\text{,}$$ on $$[0,2\pi)$$
###### 19.
$$x=\cos(t) \sin(2t)\text{,}$$ $$y=\sin(t) \sin(2t)$$ on $$[0,2\pi]$$
###### 20.
$$x=e^{t/10}\cos(t)\text{,}$$ $$y=e^{t/10}\sin(t)$$
##### Exercise Group.
Find the point $$t=t_0$$ where the graph of the given parametric equations is not smooth, then find $$\lim\limits_{t\to t_0}\frac{dy}{dx}\text{.}$$
###### 21.
$$x=\frac{1}{t^2+1}\text{,}$$ $$y=t^3$$
###### 22.
$$x=-t^3+7t^2-16t+13\text{,}$$ $$y=t^3-5t^2+8t-2$$
###### 23.
$$x=t^3-3t^2+3t-1\text{,}$$$$y=t^2-2t+1$$
###### 24.
$$\ds x=\cos^2(t)\text{,}$$$$y=1-\sin^2(t)$$
##### Exercise Group.
For the given parametric equations for a curve, find $$\frac{d^2y}{dx^2}\text{,}$$ then determine the intervals on which the graph of the curve is concave up/down. Note: these are the same equations as in Exercises 10.3.4.5Exercise 10.3.4.12.
###### 25.
$$x=t\text{,}$$$$y=t^2$$
###### 26.
$$x=\sqrt{t}\text{,}$$$$y=5t+2$$
###### 27.
$$x=t^2-t\text{,}$$ $$y=t^2+t$$
###### 28.
$$x=t^2-1\text{,}$$$$y=t^3-t$$
###### 29.
$$x=\sec(t)\text{,}$$$$y=\tan(t)$$ on $$(-\pi/2,\pi/2)$$
###### 30.
$$x=\cos(t)\text{,}$$ $$y=\sin(2t)\text{,}$$ on $$[0,2\pi)$$
###### 31.
$$\ds x=\cos(t) \sin(2t)\text{,}$$$$y=\sin(t) \sin(2t)$$ on $$[-\pi/2,\pi/2]$$
###### 32.
$$x=e^{t/10}\cos(t)\text{,}$$$$y=e^{t/10}\sin(t)$$
##### Exercise Group.
Find the arc length of the graph of the parametric equations on the given interval(s).
###### 33.
$$x=-3\sin(2t)\text{,}$$ $$y=3\cos(2t)$$ on $$[0,\pi]$$
###### 34.
$$x=e^{t/10}\cos(t)\text{,}$$ $$y=e^{t/10}\sin(t)$$ on $$[0,2\pi]$$ and $$[2\pi,4\pi]\text{.}$$
###### 35.
$$x=5t+2\text{,}$$ $$y=1-3t$$ on $$[-1,1]$$
###### 36.
$$x=2t^{3/2}\text{,}$$$$y=3t$$ on $$[0,1]$$
##### Exercise Group.
In the following exercises, numerically approximate the given arc length.
###### 37.
Approximate the arc length of one petal of the rose curve $$x=\cos(t) \cos(2t)\text{,}$$$$y=\sin(t) \cos(2t)$$ using Simpson's Rule and $$n=4\text{.}$$
###### 38.
Approximate the arc length of the “bow tie curve” $$x=\cos(t)\text{,}$$$$y=\sin(2t)$$ using Simpson's Rule and $$n=6\text{.}$$
###### 39.
Approximate the arc length of the parabola $$x=t^2-t\text{,}$$$$y=t^2+t$$ on $$[-1,1]$$ using Simpson's Rule and $$n=4\text{.}$$
###### 40.
A common approximate of the circumference of an ellipse given by $$x=a\cos(t)\text{,}$$$$y=b\sin(t)$$ is $$\ds C\approx 2\pi\sqrt{\frac{a^2+b^2}2}\text{.}$$ Use this formula to approximate the circumference of $$x=5\cos(t)\text{,}$$ $$y=3\sin(t)$$ and compare this to the approximation given by Simpson's Rule and $$n=6\text{.}$$
##### Exercise Group.
In the following exercises, a solid of revolution is described. Find or approximate its surface area as specified.
###### 41.
Find the surface area of the sphere formed by rotating the circle $$x=2\cos(t)\text{,}$$$$y=2\sin(t)$$ about:
1. the $$x$$-axis and
2. the $$y$$-axis.
###### 42.
Find the surface area of the torus (or “donut”) formed by rotating the circle $$x=\cos(t) +2\text{,}$$$$y=\sin(t)$$ about the $$y$$-axis.
###### 43.
Approximate the surface area of the solid formed by rotating the “upper right half” of the bow tie curve $$x=\cos(t)\text{,}$$$$y=\sin(2t)$$ on $$[0,\pi/2]$$ about the $$x$$-axis, using Simpson's Rule and $$n=4\text{.}$$
###### 44.
Approximate the surface area of the solid formed by rotating the one petal of the rose curve $$x=\cos(t) \cos(2t)\text{,}$$$$y=\sin(t) \cos(2t)$$ on $$[0,\pi/4]$$ about the $$x$$-axis, using Simpson's Rule and $$n=4\text{.}$$
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# Pie Chart
In a pie chart, the various observations or components are represented by the sectors of a circle and the whole circle represents the sum of the values of all components.
The central angle for a component is given by:
Central angle for a component = $$\frac{\textbf{Value of the component}}{\textbf{Sum of the values of all components}}$$ × 360°
How to make a pie chart or graph?
Construction of making a pie chart or graph from the given data.
Steps of pie graphs Construction:
1. Calculate the central angle for each component, given by
Central angle of component = $$\frac{\textbf{Value of the component}}{\textbf{Total value}}$$ × 360°
2. Draw a circle of convenient radius.
3. Within this circle, draw a horizontal radius.
4. Starting with the horizontal radius, draw radii making central angles corresponding to the values of the respective components, till all the components are exhausted. These radii divide the whole circle into various sectors.
5. Shade each sector with different design.
This will be the required pie chart for the given data.
Pie chart examples on how to do a pie chart:
1. Mr. Bin's with a yearly salary of $10800 plans his budget for a year as given below: Item Food Education Rent Savings Miscellaneous Amount (in dollar) 3150 1950 2100 2400 1200 Represent the above data by a pie graph. Solution: Total amount earned by Mr. Bin in a year =$ 10800.
Central angle of component = $$\frac{\textbf{Value of the component}}{\textbf{Total value}}$$ × 360°
Calculation of central angles
Item Amount (in \$) Central Angle Food 3150 (³¹⁵/₁₀₈₀₀ × 360)° = 105° Education 1950 (¹⁹⁵/₁₀₈₀₀ × 360)° = 65° Rent 2100 (²¹/₁₀₈₀₀ × 360)° = 70° Savings 2400 (²⁴/₁₀₈₀₀ × 360)° = 80° Miscellaneous 1200 (¹²/₁₀₈₀₀ × 360)° = 40°
Construction of making pie chart
Steps of construction:
1. Draw a circle of any convenient radius.
2. Draw a horizontal radius of this circle.
3. Draw sectors starting from the horizontal radious with central angles of 105 degree, 65 degree, 70 degree, 80 degree and 40 degree respectively.
4. Shade the sectors differently using different colors and label them.
Thus, we obtain the required pie chart, as shown in the given figure.
2. The data on the mode of transport used by 720 students are given below:
Mode of Transport Bus Cycle Train Car Scooter No. of Students 120 180 240 80 100
Represent the above data by a pie chart.
Solution:
Total number of students = 720.
Central angle for a mode of transport = $$\frac{\textbf{Number of students using that mode}}{\textbf{Total number of students}}$$ × 360°
Calculation of central angles
Mode of Transport No. of Students Central Angle Bus 120 (¹²/₇₂₀ × 360)° = 60° Cycle 180 (¹⁸/₇₂₀ × 360)° = 90° Train 240 (²⁴/₇₂₀ × 360)° = 120° Car 80 (⁸/₇₂₀ × 360)° = 40° Scooter 100 (¹/₇₂₀ × 360)° = 50°
Construction for creating pie chart
Steps of construction:
1. Draw a circle of any convenient radius.
2. Draw a horizontal radius of this circle.
3. Draw sectors starting from the horizontal radious with central angles of 60 degree, 90 degree, 120 degree , 40 degree and 50 degree respectively.
4. Shade the sectors differently using different colors and label them.
Thus, we obtain the required pie chart, as shown in the given figure.
3. There are 216 workers in a factory as per list given below:
Cadre Labourer Mechanic Fitter Supervisor Clerk No. of Workers 75 60 36 27 18
Represent the above data by a pie chart.
Solution:
Total number of workers = 216.
Central angle for a cadre = $$\frac{\textbf{Number of workers in that cadre}}{\textbf{Total number of workers}}$$ × 360°
Calculation of central angles
Cadre No. of Workers Central Angle Labourer 75 (⁷⁵/₂₁₆ × 360)° = 125° Mechanic 60 (⁶/₂₁₆ × 360)° = 100° Fitter 36 (³⁶/₂₁₆ × 360)° = 60° Supervisor 27 (²⁷/₂₁₆ × 360)° = 45° Clerk 18 (¹⁸/₂₁₆ × 360)° = 30°
Construction to make a pie graph
Steps of construction:
1. Draw a circle of any convenient radius.
2. Draw a horizontal radius of this circle.
3. Draw sectors starting from the horizontal radious with central angles of 125 degree, 100 degree, 60 degree, 45 degree and 30 degree respectively.
4. Shade the sectors differently using different colors and label them.
Thus, we obtain the required pie chart, as shown in the given figure.
4. The following table shows the expenditure in percentage incurred on the construction of a house in a city:
Item Brick Cement Steel Labour Miscellaneous Expenditure(in percentage) 15% 20% 10% 25% 30%
Represent the above data by a pie chart.
Solution:
Total percentage = 100.
Central angle for a component = $$\frac{\textbf{Value of the component}}{\textbf{100}}$$ × 360°
Calculation of central angles
Item Expenditure (in percentage) Central Angle Brick 15% (¹⁵/₁₀₀ × 360)° = 54° Cement 20% (²/₁₀₀ × 360)° = 72° Steel 10% (¹/₁₀₀ × 360)° = 36° Labour 25% (²⁵/₁₀₀ × 360)° = 90° Miscellaneous 30% (³/₁₀₀ × 360)° = 108°
Construction for creating pie chart
Steps of construction:
1. Draw a circle of any convenient radius.
2. Draw a horizontal radius of the circle.
3. Draw sectors starting from the horizontal radious with central angles of 54 degree, 72 degree, 36 degree, 90 degree and 108 degree respectively.
4. Shade the sectors differently using different colors and label them.
Thus, we obtain the required pie chart, shown in the adjoining figure.
Pie Charts or Pie Graphs
Pie Chart
Pie Charts or Pie Graphs - Worksheets
Worksheet on Pie Chart
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# How exactly do differential equations work?
My textbook says that solutions for the equation $y'=-y^2$ must always be 0 or decreasing. I don't understand—if we're solving for y', then wouldn't it be more accurate to say it must always be 0 or negative. Decreasing seems to imply that we're looking at a full graph, even though the book is talking about individual solutions. Can someone explain this?
Secondly, it gives that the family $y=\frac{1}{x+C}$ as solutions for the equation. It then tells me that 0 is a solution for y in the original equation that doesn't match the family, but I don't quite understand that. How can we know that y' will equal 0 if we're specifically looking outside of the family of solutions it gives?
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1. Remember if $y'(x) > 0$, $y(x)$ is increasing; if $y'(x) < 0$, $y(x)$ is decreasing; if $y'(x) = 0$ then $y(x)$ is constant. In our case $y'(x) \le 0$ which means $y(x)$ is always constant or decreasing.
2. You can verify yourself: if $y(x) = 0$ for all $x$, then $y'(x) = 0$ so it is true that $y' = -y^2$ therefore $y(x) = 0$ is a solution but it isn't in the form $\frac{1}{x + C}$.
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If a function is differentiable and is decreasing, then the derivative is non-positive. In your case, $y^2 \geq 0 \Rightarrow y' = -y^2 \leq 0$. Hence, the solution you obtain must be decreasing.
Note that we obtain the family $y = \frac{1}{x+c}$ assuming $y \neq 0$. This can be seen from the way we solve the equation. We have $$\frac{dy}{dx} = -y^2$$ Assuming $y \neq 0$, we can divide by $y^2$ to get $\frac{dy}{y^2} = -dx$. Now we can integrate to get, $y = \frac{1}{x+\frac{1}{y_0}}$. Note that for each $y_0$, we get a different solution. So we have a one parameter family of solutions. Also, note that this family of $y$'s is nowhere $0$ and hence it is fine when we divided by $y^2$ in the original equation.
So the process of solving assumes that $y \neq 0$. We also observe that $y=0$ is a solution by plugging it into the differential equation.
Hence, the solutions to the differential equation are $$y=0 \text{ and } y=\frac{1}{x+\frac{1}{y_0}}, \text{ where } y_0 \in \mathbb{R}$$
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You're confusion can be resolved by clarifying one thing (I think). It seems you think that to "solve" a differential equation, you are looking for $y'$? Really, a solution is a function, maybe to decrease confusion we'll call it $f(x)$, with the property that $f'(x)=-[f(x)]^2$. That is what is meant by $y'=-y^2$.
So when we think of the graph of $y=f(x)$ we know that it has the property that $\frac{dy}{dx}=-[f(x)]^2$, or since a square is always positive, and then we take the negative, we know the derivative of $y=f(x)$ is negative and hence the function is decreasing.
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