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# Subtracting Fractions with Whole Numbers
## Related Calculator: Fractions Calculator
Subtracting fractions with whole numbers doesn't differ much from adding fractions with whole numbers (just remember how to subtract integers correctly).
Indeed, suppose we want to subtract whole number m from fraction n/q.
It is known that whole number m can be represented as fraction m/1.
Now, m-n/q=m/1-n/q=(mq)/q-n/q=(mq-n)/q.
Formula for subtracting fractions with whole numbers: color(green)(m-n/q=(mq-n)/q).
Example 1. Find 3-6/7.
Let's solve it step-by-step:
3+6/7=3/1-6/7=(3*color(red)(7))/(1*color(red)(7))-6/7=21/7-6/7=15/7.
If you need mixed number, convert 15/7 into midex number: 15/7=2 1/7.
Answer: 15/7=2 1/7.
Next example.
Example 2. Find -9-13/8.
Let's use direct formula:
-9-13/8=(-9*8-13)/8=(-72-13)/8=-85/8.
Convert to mixed number if needed: -85/8=-10 5/8
Answer: -85/8=-10 5/8.
Next example.
Example 3. Find -9/4-3.
-9/4-3=-9/4-(3*color(red)(4))/(1*color(red)(4))=-9/4-12/4=(-9-12)/4=-21/4.
Convert to mixed fraction if needed: -21/4=-5 1/4.
Answer: -21/4=-5 1/4.
Now, it is time to practice.
Exercise 1. Find 2-6/7.
Answer: 8/7=1 1/7.
Next exercise.
Exercise 2. Find 29/5-9.
Answer: -16/5=-3 1/5.
Next exercise.
Exercise 3. Find 5-(-97/8).
Answer: 135/8=16 7/8.
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# Root Test
Why did you need to learn about nth roots and algebra when you were in algebra class? It was so you could figure out when series converge, of course!
## Root Test in Calculus
If you need to know if a series converges, but there is a power of $$n$$ in it, then the Root Test is generally the go-to test. It can tell you if a series is absolutely convergent or divergent. This is different from most tests which tell you whether a series converges or diverges, but doesn't say anything about absolutely convergence.
One of the limits you will frequently need to apply the Root Test is
$\lim\limits_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1,$
but why is that true. Showing that limit is actually equal to 1 uses the fact from properties of exponential functions and natural logs that
$e^{-\frac{\ln n}{n}} = \frac{1}{\sqrt[n]{n}}.$
Since the exponential function is continuous,
\begin{align} \lim\limits_{n \to \infty} e^{-\frac{\ln n}{n}} &= e^{-\lim\limits_{n \to \infty} \frac{\ln n}{n}} \\ &= e^{0} \\ &= 1, \end{align}
which gives you the desired result.
## Root Test for Series
First, let's state the Root Test.
Root Test: Let
$\sum\limits_{n=1}^{\infty} a_n$
be a series and define $$L$$ by
$L = \lim\limits_{n \to \infty} \left| a_n \right|^{\frac{1}{n}}= \lim\limits_{n \to \infty} \sqrt[n]{\left| a_n \right|} .$
Then the following hold:
1. If $$L < 1$$ then the series is absolutely convergent.
2. If $$L > 1$$ then the series diverges.
3. If $$L = 1$$ then the test is inconclusive.
Notice that, unlike many series tests, there is no requirement that the terms of the series be positive. However, it can be challenging to apply the Root Test unless there is a power of $$n$$ in the terms of the series. In the next section, you will see that the Root Test is also not very helpful if the series is conditionally convergent.
## Root Test and Conditional Convergence
Remember that if a series converges absolutely, then it is, in fact, convergent. So if the Root Test tells you that a series converges absolutely, then it also tells you that it converges. Unfortunately, it will not tell you if a conditionally convergent series actually converges.
In fact the Root Test often can't be used on conditionally convergent series. Take for example the conditionally convergent alternating harmonic series
$\sum\limits_{n \to \infty} \frac{(-1)^n}{n} .$
If you try to apply the Root Test, you get
\begin{align} L &= \lim\limits_{n \to \infty} \left| a_n \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left| \frac{(-1)^n}{n} \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left( \frac{1}{n} \right)^{\frac{1}{n}} \\ &= 1. \end{align}
So in fact the Root Test doesn't tell you anything about the series. Instead to tell that the alternating harmonic series converges you would need to use the Alternating Series Test. For more details on that test, see Alternating Series.
## Root Test Rules
The most significant rule about the Root Test is that it doesn't tell you anything if $$L = 1$$. In the previous section, you saw an example of a series that converges conditionally, but the Root Test couldn't tell you that because $$L = 1$$. Next, let's look at two more examples where the Root Test isn't helpful because $$L = 1$$.
If possible, use the Root Test to determine the convergence or divergence of the series
$\sum\limits_{n=1}^{\infty} \frac{1}{n^2}.$
This is a P-series with $$p = 2$$, so you already know it converges, and in fact it converges absolutely. But let's see what the Root Test gives you. If you take the limit,
\begin{align} L &= \lim\limits_{n \to \infty} \left| a_n \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left| \frac{1}{n^2} \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left( \frac{1}{n^2} \right)^{\frac{1}{n}} \\ &= 1. \end{align}
So in fact the Root Test is inconclusive with this series.
If possible, use the Root Test to determine the convergence or divergence of the series
$\sum\limits_{n=1}^{\infty} \frac{1}{n^2}.$
This is a P-series with $$p = 1$$, or in other words the harmonic series, so you already know it diverges. If you take the limit to try and apply the Root Test,
\begin{align} L &= \lim\limits_{n \to \infty} \left| a_n \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left| \frac{1}{n} \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left( \frac{1}{n} \right)^{\frac{1}{n}} \\ &= 1. \end{align}
So in fact the Root Test is inconclusive with this series.
## Root Test Examples
Let's look at a couple of examples where the Root Test is useful.
If possible, determine the convergence or divergence of the series
$\sum\limits_{n=1}^{\infty} \frac{5^n}{n^n}.$
You might be tempted to use the Ratio Test for this problem instead of the Root Test. But the $$n^n$$ in the denominator makes the Root Test a much better first attempt for looking at this series. Taking the limit,
\begin{align} L &= \lim\limits_{n \to \infty} \left| a_n \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left| \frac{5^n}{n^n} \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left( \frac{5^n}{n^n} \right)^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \frac{5}{n} \\ &= 0 . \end{align}
Since $$L <1$$, the Root Test tells you that this series is absolutely convergent.
If possible, determine the convergence or divergence of the series
$\sum\limits_{n=1}^{\infty} \frac{(-6)^n}{n}.$
Given the power of $$n$$ the Root Test is a good test to try for this series. Finding $$L$$ gives:
\begin{align} L &= \lim\limits_{n \to \infty} \left| a_n \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left| \frac{(-6)^n}{n} \right|^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \left( \frac{6^n}{n} \right)^{\frac{1}{n}} \\ &= \lim\limits_{n \to \infty} \frac{6}{n^{\frac{1}{n}}} \\ &= 6 . \end{align}
Since $$L > 1$$ the Root Test tells you that this series is divergent.
## Root Test - Key takeaways
• $\lim\limits_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1$
• Root Test: Let
$\sum\limits_{n=1}^{\infty} a_n$
be a series and define $$L$$ by
$L = \lim\limits_{n \to \infty} \left| a_n \right|^{\frac{1}{n}}= \lim\limits_{n \to \infty} \sqrt[n]{\left| a_n \right|} .$
Then the following hold:
1. If $$L < 1$$ then the series is absolutely convergent.
2. If $$L > 1$$ then the series diverges.
3. If $$L = 1$$ then the test is inconclusive.
The Root Test is used to tell if a series is absolutely convergent or divergent.
Take the limit of the absolute value of the nth root of the series as n goes to infinity. If that limit is less than one the series is absolutely convergent. If it is greater than one the series is divergent.
You don't solve a root test. It is a test to see if a series is absolutely convergent or divergent.
You use it to see if a series is absolutely convergent or divergent. It is good when there is a power of n in the terms of the series.
When the limit equals 1, the Root Test is inconclusive.
## Final Root Test Quiz
Question
Can the Root Test tell you if a series is conditionally convergent?
No. It can only tell you if a series is absolutely convergent or divergent, otherwise the test is inconclusive.
Show question
Question
Do you really need the absolute value signs when taking the limit in the Root Test?
Most definitely yes. Without them you might get that a limit doesn't exist and the Root Test doesn't apply, when in fact the limit does exist and you can apply the Root Test.
Show question
60%
of the users don't pass the Root Test quiz! Will you pass the quiz?
Start Quiz
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# Assume that T is a linear transformation. Find the standard matrix of T.
• $T:$ $\mathbb{R}^2$ → $\mathbb{R}^4$, $T(e_1)$ $= (3,1,3,1)$ $and$ $T (e_2)$ $= (-5,2,0,0),$ $where$ $e_1$ $= (1,0)$ $and$ $e_2$ $= (0,1)$
In this question, we have to find the standard matrix of the linear transformation $T$.
First, we should recall our concept of the standard matrix. The standard matrix has columns that are the images of the vector of standard basis.
$A = \left [\begin {matrix}1\\0\\0\\ \end {matrix} \right] B = \left [ \begin {matrix}0\\1\\0\\ \end {matrix}\right] C = \left [ \begin {matrix}0\\0\\1\\ \end {matrix} \right ]$
The transformation matrix is a matrix that changes the Cartesian system of a vector into a different vector with the help of matrix multiplication.
## Expert Answer
Transformation matrix $T$ of order $a \times b$ on multiplication with a vector $X$ of $b$ components represented as a column matrix transforms into another matrix $X’$.
A vector $X= ai + bj$ when multiplied with matrix $T$ $\left [ \begin {matrix} p&q\\r&s \\ \end {matrix} \right]$ is transformed into another vector $Y=a’i+ bj’$. Thus, a $2 \times 2$ transformation matrix can be shown as below,
$TX =Y$
$\left[\begin {matrix} p&q\\r&s \\ \end {matrix}\right] \times \left [ \begin {matrix}x\\y\\ \end {matrix} \right] =\left [\begin {matrix}x^\prime\\y^\prime\\ \end {matrix} \right ]$
There are different types of Transformation matrices such as stretching, rotation, and shearing. It is used in Dot and Cross Product of vectors and can also be used in finding the determinants.
Now applying the above concept on the given question, we know that the standard basis for $R^2$ is
$e_1=\left [\begin {matrix}1\\0\\ \end {matrix} \right ]$
and $e_2= \left [\begin {matrix}1\\0\\ \end {matrix} \right ]$
and we have
$T(e_1)= \left [ \begin {matrix}3\\1\\3\\1\\ \end {matrix} \right] T(e_2)= \left [ \begin {matrix}-5\\2\\0\\0\\ \end {matrix} \right ]$
To find the standard matrix of linear transformation $T$, let us suppose it is matrix $X$ and it can be written as:
$X = T(e_1) T(e_2)$
$X = \left [ \begin {matrix} \begin {matrix}3\\1\\3\\ \end {matrix}& \begin {matrix}-5\\2\\0\\ \end {matrix}\\1&0\\ \end {matrix} \right ]$
## Numerical Results
So the standard matrix for linear transformation $T$ is given as:
$X =\left [ \begin {matrix} \begin {matrix}3\\1\\3\\ \end {matrix}& \begin {matrix}-5\\2\\0\\ \end {matrix}\\1&0\\ \end {matrix} \right ]$
## Example
Find the new vector formed for the vector $6i+5j$, with the transformation matrix $\left[ \begin {matrix}2&3\\1&-1\\ \end{matrix} \right ]$
Given as:
Transformation matrix $T = \left [ \begin {matrix}2&3\\1&-1\\ \end {matrix} \right ]$
Given vector is written as,$A = \left [ \begin {matrix}6\\5\\ \end {matrix} \right ]$
We have to find the transformation matrix B represented as:
$B = TA$
Now putting the values in above equation, we get:
$B=TA=\left [ \begin {matrix}2&3\\1&-1\\\end {matrix} \right ]\times\left [ \begin {matrix}6\\5\\\end {matrix} \right ]$
$B=\left [\begin {matrix}2\times6+3\times(5)\\1\times6+(-1)\times5\\\end {matrix} \right ]$
$B=\left [\begin {matrix}27\\1\\ \end {matrix} \right ]$
so based on above matrix, our required transformation standard matrix will be:
$B = 27i+1j$
5/5 - (6 votes)
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Practise mathematics online - the new site of Mathefritz
Calculate pyramid - volume of a pyramid, formulas, 3D interactive
Calculate pyramid - The designations of a pyramid
The pyramid
Calculate pyramid
Before we start with calculations on a pyramid, let us first define what a pyramid is.
In very general terms, a pyramid is a geometric figure consisting of a base and triangular sides that meet at a point, the top of the pyramid. A typical pyramid that we know from Egypt, for example, has a square base, called the base, and triangular sides that run from the vertices of the base to the top.
Later on, we will also consider pyramids with a triangular base and with regular polygons such as a pentagon or an octagon.
Pyramid 3D model - interactive
Pyramid 3D Model
Explore the pyramid in our 3D model. Use the mouse or your fingers to view the pyramid from all perspectives.
The 3D model of the pyramid was created with Plotly.com
Also visit the Plotly page of Mathefritz.
Calculate the volume of a pyramid - The derivation
Volume of a pyramid - The approximation with stacked cuboids.
We decompose the pyramid into a number of stacked cuboids as shown in the 3D animation next to it. The smaller the height of the cuboids, the closer we get to the actual volume of the pyramid.
Try it out by choosing the number of levels yourself!
Derivation Calculate volume of a pyramid
We start with our example and 5 planes. Then we want to derive the volume for any number of n planes.
First, the key figures of the pyramid with a square base:
• We call the edge length: a
• We call the height: h
With 5 planes a cuboid has the height $$h_q=\frac{h}{5}$$
The half side length a is divided into 5 lengths of equal width.
The first (lowest) cuboid has the volume:
$$V_1 = a^2 \cdot \frac{h}{5}$$
The second cuboid has the volume:
$$V_2 = (\frac{4}{5} a)^2 \cdot \frac{h}{5}$$
The third cuboid has the volume:
$$V_3 = (\frac{3}{5} a)^2 \cdot \frac{h}{5}$$
The fourth cuboid has the volume:
$$V_4 = (\frac{2}{5} a)^2 \cdot \frac{h}{5}$$
The fifth cuboid has the volume:
$$V_5 = (\frac{1}{5} a)^2 \cdot \frac{h}{5}$$
If we summarize the elements as a sum and factor them out, we obtain as a first approximation for the volume:
$$V = a^2 \cdot \frac{1}{5} h \cdot (\frac{1}{25} + \frac{4}{25} + \frac{9}{25} + \frac{16}{25} + \frac{25}{25})=a^2 \cdot h \cdot \frac{11}{25} =0,44\cdot h\cdot a^2$$
The volume for 5 levels is larger than the actual volume, because all cuboids protrude slightly above the real pyramid. See picture.
Let's summarize the calculation with a summation formula:
We set n=5.
$$V = a^2 \cdot h \cdot \frac{1}{n^3}\cdot \sum_{i=1}^{n} i^2$$
Let's increase the number of layers as in our 3D animation for the pyramid:
n = 50 : $$V=\frac {1717}{5000}\cdot a^2 \cdot h = 0,34 \cdot a^2 \cdot h$$
n = 1000 : $$V= 0.3338\cdot a^2 \cdot h$$
Wir sehen, dass sich das Volumen immer mehr dem Wert:
$$V=\frac {1}{3}\cdot a^2 \cdot h$$ annähert.
At 1000 levels, we have more or less already reached the correct value.
Calculating the volume of a pyramid with the formula for 1000 levels with a calculator: Here a CASIO fx-991DE CW.
Pyramid in 3D - Where to find a pyramid in buildings?
Calculate pyramid in practice:
If you look for pyramids in nature, you will find many pyramids in structures.
Pyramid-shaped structures can be found in different cultures and at different times around the world. Here are some well-known examples:
1. The Egyptian pyramids: The pyramids of Giza in Egypt are probably the most famous examples of pyramidal structures. The largest and most famous pyramid is the Pyramid of Khufu, a burial place of the pharaoh Khufu.
2. The Central American pyramids: Numerous pyramid-shaped structures are found in the Mayan, Aztec regions. Examples are the pyramids of Chichén Itzá in Mexico or the pyramids of Tikal in Guatemala.
3. The Pyramids of Teotihuacán: In Mexico there is the archaeological site of Teotihuacán, which contains the remains of an ancient city. There are the Pyramid of the Sun and the Pyramid of the Moon, which are among the most impressive pyramid-shaped structures of the pre-Columbian era.
These are just a few examples of pyramid-shaped structures. However, there are also modern buildings inspired by the traditional pyramid shape, such as the glass pyramid in the courtyard of the Louvre in Paris or the Luxor Hotel in Las Vegas.
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# 5.4 Dividing polynomials (Page 2/6)
Page 2 / 6
Given a polynomial and a binomial, use long division to divide the polynomial by the binomial.
1. Set up the division problem.
2. Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor.
3. Multiply the answer by the divisor and write it below the like terms of the dividend.
4. Subtract the bottom binomial from the top binomial.
5. Bring down the next term of the dividend.
6. Repeat steps 2–5 until reaching the last term of the dividend.
7. If the remainder is non-zero, express as a fraction using the divisor as the denominator.
## Using long division to divide a second-degree polynomial
Divide $\text{\hspace{0.17em}}5{x}^{2}+3x-2\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}x+1.$
The quotient is $\text{\hspace{0.17em}}5x-2.\text{\hspace{0.17em}}$ The remainder is 0. We write the result as
$\frac{5{x}^{2}+3x-2}{x+1}=5x-2$
or
$5{x}^{2}+3x-2=\left(x+1\right)\left(5x-2\right)$
## Using long division to divide a third-degree polynomial
Divide $\text{\hspace{0.17em}}6{x}^{3}+11{x}^{2}-31x+15\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}3x-2.\text{\hspace{0.17em}}$
There is a remainder of 1. We can express the result as:
$\frac{6{x}^{3}+11{x}^{2}-31x+15}{3x-2}=2{x}^{2}+5x-7+\frac{1}{3x-2}$
Divide $\text{\hspace{0.17em}}16{x}^{3}-12{x}^{2}+20x-3\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}4x+5.\text{\hspace{0.17em}}$
$4{x}^{2}-8x+15-\frac{78}{4x+5}$
## Using synthetic division to divide polynomials
As we’ve seen, long division of polynomials can involve many steps and be quite cumbersome. Synthetic division is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is 1.
To illustrate the process, recall the example at the beginning of the section.
Divide $\text{\hspace{0.17em}}2{x}^{3}-3{x}^{2}+4x+5\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}x+2\text{\hspace{0.17em}}$ using the long division algorithm.
The final form of the process looked like this:
There is a lot of repetition in the table. If we don’t write the variables but, instead, line up their coefficients in columns under the division sign and also eliminate the partial products, we already have a simpler version of the entire problem.
Synthetic division carries this simplification even a few more steps. Collapse the table by moving each of the rows up to fill any vacant spots. Also, instead of dividing by 2, as we would in division of whole numbers, then multiplying and subtracting the middle product, we change the sign of the “divisor” to –2, multiply and add. The process starts by bringing down the leading coefficient.
We then multiply it by the “divisor” and add, repeating this process column by column, until there are no entries left. The bottom row represents the coefficients of the quotient; the last entry of the bottom row is the remainder. In this case, the quotient is $\text{\hspace{0.17em}}2{x}^{2}–7x+18\text{\hspace{0.17em}}$ and the remainder is $\text{\hspace{0.17em}}–31.\text{\hspace{0.17em}}$ The process will be made more clear in [link] .
## Synthetic division
Synthetic division is a shortcut that can be used when the divisor is a binomial in the form $\text{\hspace{0.17em}}x-k\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is a real number. In synthetic division , only the coefficients are used in the division process.
Given two polynomials, use synthetic division to divide.
1. Write $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ for the divisor.
2. Write the coefficients of the dividend.
3. Bring the lead coefficient down.
4. Multiply the lead coefficient by $\text{\hspace{0.17em}}k.\text{\hspace{0.17em}}$ Write the product in the next column.
5. Add the terms of the second column.
6. Multiply the result by $\text{\hspace{0.17em}}k.\text{\hspace{0.17em}}$ Write the product in the next column.
7. Repeat steps 5 and 6 for the remaining columns.
8. Use the bottom numbers to write the quotient. The number in the last column is the remainder and has degree 0, the next number from the right has degree 1, the next number from the right has degree 2, and so on.
what is the period of cos?
Patrick
if tan alpha + beta is equal to sin x + Y then prove that X square + Y square - 2 I got hyperbole 2 Beta + 1 is equal to zero
sin^4+sin^2=1, prove that tan^2-tan^4+1=0
what is the formula used for this question? "Jamal wants to save \$54,000 for a down payment on a home. How much will he need to invest in an account with 8.2% APR, compounding daily, in order to reach his goal in 5 years?"
i don't need help solving it I just need a memory jogger please.
Kuz
A = P(1 + r/n) ^rt
Dale
how to solve an expression when equal to zero
its a very simple
Kavita
gave your expression then i solve
Kavita
Hy guys, I have a problem when it comes on solving equations and expressions, can you help me 😭😭
Thuli
Tomorrow its an revision on factorising and Simplifying...
Thuli
ok sent the quiz
kurash
send
Kavita
Hi
Masum
What is the value of log-1
Masum
the value of log1=0
Kavita
Log(-1)
Masum
What is the value of i^i
Masum
log -1 is 1.36
kurash
No
Masum
no I m right
Kavita
No sister.
Masum
no I m right
Kavita
tan20°×tan30°×tan45°×tan50°×tan60°×tan70°
jaldi batao
Joju
Find the value of x between 0degree and 360 degree which satisfy the equation 3sinx =tanx
what is sine?
what is the standard form of 1
1×10^0
Akugry
Evalute exponential functions
30
Shani
The sides of a triangle are three consecutive natural number numbers and it's largest angle is twice the smallest one. determine the sides of a triangle
Will be with you shortly
Inkoom
3, 4, 5 principle from geo? sounds like a 90 and 2 45's to me that my answer
Neese
Gaurav
prove that [a+b, b+c, c+a]= 2[a b c]
can't prove
Akugry
i can prove [a+b+b+c+c+a]=2[a+b+c]
this is simple
Akugry
hi
Stormzy
x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0
x exposent4+4x exposent3+8x exposent2+4x+1=0
HERVE
How can I solve for a domain and a codomains in a given function?
ranges
EDWIN
Thank you I mean range sir.
Oliver
proof for set theory
don't you know?
Inkoom
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# Lesson 10
On or Off the Line?
Let’s interpret the meaning of points in a coordinate plane.
### Problem 1
1. Match the lines $$m$$ and $$n$$ to the statements they represent:
1. A set of points where the coordinates of each point have a sum of 2
2. A set of points where the $$y$$-coordinate of each point is 10 less than its $$x$$-coordinate
2. Match the labeled points on the graph to statements about their coordinates:
1. Two numbers with a sum of 2
2. Two numbers where the $$y$$-coordinate is 10 less than the $$x$$-coordinate
3. Two numbers with a sum of 2 and where the $$y$$-coordinate is 10 less than the $$x$$-coordinate
### Problem 2
Here is an equation: $$4x-4=4x+\text{__}$$. What could you write in the blank so the equation would be true for:
1. No values of $$x$$
2. All values of $$x$$
3. One value of $$x$$
(From Unit 4, Lesson 7.)
### Problem 3
Mai earns $7 per hour mowing her neighbors' lawns. She also earned$14 for hauling away bags of recyclables for some neighbors.
Priya babysits her neighbor’s children. The table shows the amount of money $$m$$ she earns in $$h$$ hours. Priya and Mai have agreed to go to the movies the weekend after they have earned the same amount of money for the same number of work hours.
$$h$$ $$m$$
1 $8.40 2$16.80
4 \$33.60
1. How many hours do they each have to work before they go to the movies?
2. How much will each of them have earned?
3. Explain where the solution can be seen in tables of values, graphs, and equations that represent Priya's and Mai's hourly earnings.
### Problem 4
For each equation, explain what you could do first to each side of the equation so that there would be no fractions. You do not have to solve the equations (unless you want more practice).
1. $$\dfrac{3x-4}{8} = \dfrac{x+2}{3}$$
2. $$\dfrac{3(2-r)}{4} = \dfrac{3+r}{6}$$
1. $$\dfrac{4p+3}{8} = \dfrac{p+2}{4}$$
2. $$\dfrac{2(a-7)}{15} = \dfrac{a+4}{6}$$
(From Unit 4, Lesson 6.)
### Problem 5
The owner of a new restaurant is ordering tables and chairs. He wants to have only tables for 2 and tables for 4. The total number of people that can be seated in the restaurant is 120.
1. Describe some possible combinations of 2-seat tables and 4-seat tables that will seat 120 customers. Explain how you found them.
2. Write an equation to represent the situation. What do the variables represent?
3. Create a graph to represent the situation.
4. What does the slope tell us about the situation?
5. Interpret the $$x$$ and $$y$$ intercepts in the situation.
(From Unit 3, Lesson 14.)
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Intermediate Algebra: Functions and Graphs
Section3.1Extraction of Roots
Subsection3.1.1Introduction
So far you have learned how to solve linear equations. In linear equations, the variable cannot have any exponent other than 1, and for this reason such equations are often called first-degree. Now we’ll consider second-degree equations, or quadratic equations. A quadratic equation includes the square of the variable.
A quadratic equation can be written in the standard form
\begin{equation*} ax^2+bx+c=0 \end{equation*}
where $$a,~b,~$$ and $$c$$ are constants, and $$a$$ is not zero.
Which of the following equations are quadratic?
1. $$3x+2x^2=1$$
• Yes
• No
2. $$4z^2-2z^3+2=0$$
• Yes
• No
3. $$36y-16=0$$
• Yes
• No
4. $$v^2=6v$$
• Yes
• No
$$\text{Yes}$$
$$\text{No}$$
$$\text{No}$$
$$\text{Yes}$$
Solution.
We would like to be able to solve quadratic equations, use them in applications, and graph quadratic equations in two variables. Let’s begin by considering some simple examples.
The simplest quadratic equation in two variables is
\begin{gather*} y=x^2 \end{gather*}
Its graph is not a straight line, but a curve called a parabola, shown in the figure. You can verify the table of values below for this parabola.
$$x$$ $$-3$$ $$-2$$ $$-1$$ $$0$$ $$1$$ $$2$$ $$3$$ $$y$$ $$9$$ $$4$$ $$1$$ $$0$$ $$1$$ $$4$$ $$9$$
Caution3.1.3.
Be careful when squaring negative numbers. To evaluate the square of a negative number on a calculator, we must enclose the number in parentheses. For example,
\begin{equation*} (-3)^2=(-3)(-3)=9,~~~~\text{but}~~~~-3^2=-3 \cdot 3=-9 \end{equation*}
Graph the parabola $$~y=x^2-4$$
Solution.
We make a table of values and plot the points. The graph is shown below.
$$x$$ $$-3$$ $$-2$$ $$-1$$ $$0$$ $$1$$ $$2$$ $$3$$ $$y$$ $$5$$ $$0$$ $$-3$$ $$-4$$ $$-3$$ $$0$$ $$5$$
Graph the parabola $$~y=4-x^2$$ on the same grid in the Example above.
Which of the following is the best match for the graph of the parabola $$~y=4-x^2$$ (using the same scale grid as in the Example)?
• (A)
• (B)
• (C)
• (D)
$$\text{(C)}$$
Solution.
A graph is also shown below.
How can we solve a quadratic equation? Consider the equation
\begin{equation*} x^2-4=5 \end{equation*}
First, we can solve it graphically. Look again at the graph of $$~y=x^2-4~$$ from Example 1.
We would like to find the $$x$$-values that make $$y=5\text{.}$$ The horizontal line $$y=5$$ intersects the graph at two points with $$y$$-coordinate 5, and their $$x$$-coordinates are the solutions of the equation. Thus, there are two solutions, namely $$3$$ and $$-3\text{.}$$
Algebraically, we solve the equation as follows.
• First, we isolate the variable. We add 4 to both sides, yielding $$~x^2=9\text{.}$$
• Because $$x$$ is squared in this equation, we perform the opposite operation, or take square roots, in order to solve for $$x\text{.}$$
\begin{align*} x^2 \amp = 9 \amp \amp \blert{\text{Take square roots of both sides.}}\\ x \amp = \pm \sqrt{9} = \pm 3 \amp \amp \blert{\text{Remember that every positive number}}\\ \amp \amp \amp \blert{\text{has two square roots.}} \end{align*}
The solutions are $$3$$ and $$-3\text{,}$$ as we saw on the graph.
Note3.1.6.
Notice that we have found two solutions for this quadratic equation, whereas linear equations have at most one solution. (Sometimes they have no solution at all.) We shall see that every quadratic equation has two solutions, which may be equal. The solutions may also be complex numbers, which we’ll study in Chapter 4.
Solve the equation
\begin{gather*} \frac{1}{2}x^2-7=8 \end{gather*}
graphically and algebraically.
Solution.
The figure shows the graph of $$~y=\dfrac{1}{2}x^2-7.$$
We would like to find the $$x$$-values that make $$y=8\text{.}$$ The horizontal line $$y=8$$ intersects the graph at two points with $$x$$-coordinates approximately $$5.5$$ and $$-5.5\text{.}$$ These are the solutions of the equation.
Algebraically, we solve the equation as follows.
• First, we isolate the variable. We add 7 to both sides, then multiply by 2, yielding $$~x^2=30\text{.}$$
• Because $$x$$ is squared in this equation, we perform the opposite operation, or take square roots, in order to solve for $$x\text{.}$$
\begin{align*} x^2 \amp = 30 \amp \amp \blert{\text{Take square roots of both sides.}}\\ x \amp = \pm \sqrt{30} \amp \amp \blert{\text{Remember that every positive number}}\\ \amp \amp \amp \blert{\text{has two square roots.}} \end{align*}
• We use a calculator to find that $$\sqrt{30}$$ is approximately 5.477, or about 5.5, as we saw on the graph.
Caution3.1.8.
It is important to make a distinction betweeen exact values and decimal approximations.
• For the example above, the exact solutions are $$\pm \sqrt{30}\text{.}$$
• The values from the calculator, $$\pm 5.477\text{,}$$ are decimal approximations to the solutions, rounded to thousandths.
Which solutions are exact values, and which are approximations?
1. $$x^2=40, \qquad\qquad x=\pm 6.32455532$$
• exact solutions
• approximations
2. $$t^2=\dfrac{81}{64}, \qquad\qquad t=\pm 1.125$$
• exact solutions
• approximations
3. $$w^2=50, \qquad\qquad w=\pm 5\sqrt{2}$$
• exact solutions
• approximations
4. $$b^2=(0.632)^2, \qquad b=\pm 0.632$$
• exact solutions
• approximations
$$\text{approximations}$$
$$\text{exact solutions}$$
$$\text{exact solutions}$$
$$\text{exact solutions}$$
Solution.
Only the solutions to (a) are approximations.
We can now solve quadratic equations of the form $$~ax^2+c = 0~\text{,}$$ by isolating $$x$$ on one side of the equation, and then taking the square root of each side. This method for solving quadratic equations is called extraction of roots.
Extraction of Roots.
To solve a quadratic equation of the form
\begin{equation*} ax^2+c=0 \end{equation*}
1. Isolate $$x$$ on one side of the equation.
2. Take the square root of each side.
Solve by extracting roots $$\quad\dfrac{3x^2-8}{4} = 10$$
$$x=$$ Enter solutions separated by a comma.
Hint.
First isolate $$x^2\text{.}$$ Then take the square root of both sides.
$$4, -4$$
Solution.
$$\pm 4$$
In the next Example, we compare the steps for evaluating a quadratic expression and for solving a quadratic equation.
Tux the cat falls off a tree branch 20 feet above the ground. His height $$t$$ seconds later is given by $$h=20-16t^2\text{.}$$
1. How high is Tux above the ground 0.5 second later?
2. How long does Tux have to get in position to land on his feet before he reaches the ground?
Solution.
1. We evaluate the formula for $$t=0.5\text{.}$$ We substitute $$\alert{0.5}$$ for $$t$$ into the formula, and simplify.
\begin{align*} h \amp= 20-16(\alert{0.5})^2 \amp \amp \blert{\text{Compute the power.}}\\ \amp = 20-16(0.25) \amp \amp \blert{\text{Multiply, then subtract.}}\\ \amp = 20-4=16 \end{align*}
Tux is 16 feet above the ground after 0.5 second. You can also use your calculator to simplify the expression for $$h$$ by entering
$$\qquad\qquad 20$$ - $$16$$ × $$0.5$$ x^2 ENTER
2. We would like to find the value of $$t$$ when Tux’s height, $$h\text{,}$$ is zero. We substitute $$h=\alert{0}$$ into the equation to obtain
To solve this equation we use extraction of roots. We first isolate $$t^2$$ on one side of the equation.
\begin{align*} 16t^2 \amp =20 \amp \amp \blert{\text{Divide by 16.}}\\ t^2 \amp = \dfrac{20}{16} = 1.25 \end{align*}
Next, we take the square root of both sides of the equation to find
\begin{equation*} t=\pm \sqrt{1.25} \approx \pm 1.118 \end{equation*}
Only the positive solution makes sense here, so Tux has approximately 1.12 seconds to be in position for landing.
A graph of the Tux’s height after $$t$$ seconds is shown below. The points corresponding to parts (a) and (b) are labeled.
Subsection3.1.4Geometric Formulas
The formulas for the volume and surface area of some everyday objects, such as cylinders and cones, involve quadratic expressions. We can use extraction of roots to solve problems involving these objects.
Formulas for Volume and Surface Area.
The volume of a can of soup is 582 cubic centimeters, and its height is 10.5 centimeters. What is the radius of the can, to the nearest tenth of a centimeter?
Solution.
The volume of a cylinder is given by the formula $$V=\pi r^2h\text{.}$$ We substitute $$\alert{582}$$ for $$V$$ and $$\alert{10.5}$$ for $$h\text{,}$$ then solve for $$r\text{.}$$
\begin{align*} \alert{582} \amp = \pi r^2(\alert{10.5}) \amp \amp \blert{\text{Divide both sides by}~ 10.5 \pi.}\\ 17.643 \amp \approx r^2 \amp \amp \blert{\text{Take square roots.}}\\ 4.200 \amp \approx r \end{align*}
The radius of the can is approximately 4.2 centimeters.
The glass pyramid at the Louvre in Paris has a square base, is 21.64 meters tall, and encloses a volume of 9049.68 cubic meters. Use the formula $$~ V=\dfrac{1}{3}s^2h~$$ to find the length of the base. Round your answer to hundredths.
$$\sqrt{\frac{3\cdot 9049.68}{21.64}}$$
Solution.
35.42 m
Subsection3.1.5Solving Formulas
Sometimes it is useful to solve a formula for one variable in terms of the others. You might want to know what radius you need to build cones of various fixed volumes. In that case, it is more efficient to solve the volume formula for $$r$$ in terms of $$v\text{.}$$
The formula $$~V=\dfrac{1}{3}\pi r^2h~$$ gives the volume of a cone in terms of its height and radius. Solve the formula for $$r$$ in terms of $$V$$ and $$h\text{.}$$
Solution.
Because the variable we want is squared, we use extraction of roots. First, we multiply both sides by 3 to clear the fraction.
\begin{align*} 3V \amp = \pi r^2h \amp \amp \blert{\text{Divide both sides by } \pi h.}\\ \frac{3V}{\pi h} \amp = r^2 \amp \amp \blert{\text{Take square roots.}}\\ \pm \sqrt{\frac{3V}{\pi h}} \amp = r \end{align*}
Because the radius of a cone must be a positive number, we use only the positive square root: $$\quad r = \sqrt{\dfrac{3V}{\pi h}}\text{.}$$
Find a formula for the radius of a circle in terms of its area, $$A\text{.}$$
$$r=$$
Hint.
Start with the formula for the area of a circle: $$A =$$
Solve for $$r$$ in terms of $$A\text{.}$$
$$\sqrt{\frac{A}{\pi }}$$
Solution.
$$r=\sqrt{\dfrac{A}{\pi}}$$
Match each quantity with the appropriate units.
1. Height of a cylinder
• I
• II
• III
• IV
2. Volume of a cone
• I
• II
• III
• IV
3. Surface area of a sphere
• I
• II
• III
• IV
4. Area of a triangle
• I
• II
• III
• IV
1. Square meters
2. Feet
3. Cubic centimeters
4. Kilograms
$$\text{II}$$
$$\text{III}$$
$$\text{I}$$
$$\text{I}$$
Solution.
1. II
2. III
3. I
4. I
Subsection3.1.6More Extraction of Roots
We can also use extraction of roots to solve quadratic equations of the form
\begin{equation*} a(x-p)^2=q \end{equation*}
We start by isolating the squared expression, $$~(x-p)^2\text{.}$$
Solve the equation $$~3(x-2)^2=48\text{.}$$
Solution.
First, we isolate the perfect square, $$~(x-2)^2\text{.}$$
\begin{align*} 3(x-2)^2 \amp = 48 \amp \amp \blert{\text{Divide both sides by 3.}}\\ (x-2)^2 \amp = 16 \amp \amp \blert{\text{Take the square root of each side.}}\\ x-2 \amp = \pm \sqrt{16} = \pm 4 \end{align*}
This gives us two equations for $$x\text{,}$$
\begin{align*} x-2 \amp = 4~~~~\text{or}~~~~x-2=-4 \amp \amp \blert{\text{Solve each equation.}}\\ x \amp = 6~~~~\text{or}~~~~x=-2 \end{align*}
The solutions are $$6$$ and $$-2\text{.}$$ You can check that both of these solutions satisfy the original equation.
Solve $$2(5x + 3)^2 = 38$$ by extracting roots.
1. Give your answers as exact values, separating the solutions with a comma.
Note: Enter “sqrt(2)” to get $$\sqrt{2}\text{,}$$ and take care to use parentheses appropriately.
Use the “Preview My Answers” button to see if you have entered valid syntax.
2. Find approximations for the solutions to two decimal places, separating the solutions with a comma.
$$\frac{-3+\sqrt{19}}{5}, \frac{-3-\sqrt{19}}{5}$$
$$0.27178, -1.47178$$
Solution.
1. $$\displaystyle x=\dfrac{-3\pm \sqrt{19}}{5}$$
2. $$x\approx -1.47$$ or $$x\approx 0.27$$
True or false.
1. The first step in extraction of roots is to take square roots.
• True
• False
2. The solutions of a quadratic equation are always of the form $$~\pm k\text{.}$$
• True
• False
3. Your calculator gives exact decimal values for square roots of integers.
• True
• False
4. The coefficients of a quadratic equation are called parabolas.
• True
• False
$$\text{False}$$
$$\text{False}$$
$$\text{False}$$
$$\text{False}$$
Solution.
1. False
2. False
3. False
4. False
Subsection3.1.7An Application: Compound Interest
Many savings accounts offer interest compounded annually: at the end of each year the interest earned is added to the principal, and the interest for the next year is computed on this larger sum of money. After $$n$$ years, the amount of money in the account is given by the formula
\begin{equation*} A=P(1+r)^n \end{equation*}
where $$P$$ is the original principal and $$r$$ is the interest rate, expressed as a decimal fraction.
Carmella invests $3000 in an account that pays an interest rate $$r$$ compounded annually. 1. Write an expression for the amount of money in Carmella’s account after two years. 2. What interest rate would be necessary for Carmella’s account to grow to$3500 in two years?
Solution.
1. We use the formula $$A=P(1+r)^n$$ with $$P=3000$$ and $$n=2\text{.}$$ Carmella’s account balance will be
\begin{equation*} A=3000(1+r)^2 \end{equation*}
2. We substitute $$\alert{3500}$$ for $$A$$ in the equation.
This is a quadratic equation in the variable $$r\text{,}$$ which we can solve by extraction of roots. First, we isolate the perfect square.
\begin{align*} 3500 \amp = 3000(1+r)^2 \amp \amp \blert{\text{Divide both sides by 3000.}}\\ 1.1\overline{6} \amp = (1+r)^2 \amp \amp \blert{\text{Take square roots.}}\\ \pm 1.0801 \amp \approx 1+r \amp \amp \blert{\text{Subtract 1 from both sides.}}\\ r \approx 0.0801 \amp \text{or}~~~~r \approx -2.0801 \end{align*}
Because the interest rate must be a positive number, we discard the negative solution. Carmella needs an account with interest rate $$r \approx 0.0801\text{,}$$ or over 8%, in order to have an account balance of $3500 in two years. The formula for compound interest also applies to calculating the effects of inflation. For instance, if there is a steady inflation rate of 4% per year, then in two years the price of an item that costs$100 now will be
\begin{align*} A \amp = P(1+r)^2\\ 100 \amp = (1+0.04)^2 = 108.16 \end{align*}
The average cost of dinner and a movie two years ago was $36. This year the average cost is$38.16. What was the rate of inflation over the past two years? (Round to two decimal places.)
$$100\sqrt{\frac{38.16}{36}}-100$$
Solution.
2.96%
Exercises3.1.8Problem Set 3.1
Warm Up
1.
Simplify.
1. $$\displaystyle 4-2\sqrt{64}$$
2. $$\displaystyle \dfrac{4-\sqrt{64}}{2}$$
3. $$\displaystyle \sqrt{9-4(-18)}$$
1. $$\displaystyle -12$$
2. $$\displaystyle -2$$
3. $$\displaystyle 9$$
2.
Give a decimal approximation rounded to thousandths.
1. $$\displaystyle 5\sqrt{3}$$
2. $$\displaystyle \dfrac{-2}{3}\sqrt{21}$$
3. $$\displaystyle -3+2\sqrt{6}$$
3.
Use the definition of square toot to simplify the expression.
1. $$\displaystyle \sqrt{29}(\sqrt{29})$$
2. $$\displaystyle (\sqrt{7})^2$$
3. $$\displaystyle \dfrac{6}{\sqrt{6}}$$
1. $$\displaystyle 29$$
2. $$\displaystyle 7$$
3. $$\displaystyle \sqrt{6}$$
4.
Solve. Remember that every positive number has two square roots.
1. $$\displaystyle 3x^2=147$$
2. $$\displaystyle 4x^2=25$$
3. $$\displaystyle 3x^2=15$$
1. $$\displaystyle \pm 7$$
2. $$\displaystyle \pm \dfrac{5}{2}$$
3. $$\displaystyle \pm \sqrt{5}$$
5.
1. Complete the table and graph $$~y=2x^2-5\text{.}$$
$$~x~$$ $$-3~$$ $$-2~$$ $$-1~$$ $$~0~$$ $$~1~$$ $$~2~$$ $$~3~$$ $$~y~$$
2. Use the graph to solve the equation $$~2x^2-5=7\text{.}$$ Show your work on the graph. How many solutions did you find?
3. Solve the equation $$~2x^2-5=7~$$ algebraically, by "undoing" each operation.
1. Two solutions. $$\approx \pm 2.5$$
2. $$\displaystyle \pm \sqrt{6}$$
6.
Use the Pythagorean theorem to find the unknown side.
Skills Practice
Exercise Group.
For problems 7–14, Solve by extracting roots. Give exact values for your answers.
7.
$$3x^2-9=0$$
$$\pm \sqrt{3}$$
8.
$$\dfrac{3x^2}{5}=6$$
9.
$$(2x-1)^2=16$$
$$\dfrac{5}{2},~\dfrac{-3}{2}$$
10.
$$4(x-1)^2=12$$
11.
$$(x-\dfrac{2}{3})^2=\dfrac{5}{9}$$
$$\dfrac{2}{3} \pm \dfrac{\sqrt{5}}{3}$$
12.
$$81(x+\dfrac{1}{3})^2=1$$
13.
$$3(8x-7)^2=24$$
$$\dfrac{7}{8},\pm \dfrac{\sqrt{8}}{8}$$
14.
$$2(5x-12)^2=48$$
Exercise Group.
For problems 15 and 16, solve by extracting roots. Round your answers to two decimal places.
15.
$$5x^2-97=3.2x^2-38$$
$$\pm 5.73$$
16.
$$17-\dfrac{x^2}{4}=43-x^2$$
Exercise Group.
For problems 17 and 18,
1. Use technology to graph the quadratic equation in the suggested window.
2. Use your graph to find two solutions for the equation in part (b).
3. Check your solutions algebraically, using mental arithmetic.
17.
1. $$y=3(x-4)^2$$
\begin{align*} \text{Xmin} \amp = -5 \amp \amp \text{Ymin} = -20\\ \text{Xmax} \amp = 15 \amp \amp \text{Ymax} = 130 \end{align*}
2. $$\displaystyle 3(x-4)^2 = 108$$
b. 10, -2
18.
1. $$y=\dfrac{1}{2}(x+3)^2$$
\begin{align*} \text{Xmin} \amp = -15 \amp \amp \text{Ymin} = -5\\ \text{Xmax} \amp = 5 \amp \amp \text{Ymax} = 15 \end{align*}
2. $$\displaystyle \dfrac{1}{2}(x+3)^2 = 8$$
Exercise Group.
For problems 19–22, solve the formula for the specified variable.
19.
$$F=\dfrac{mv^2}{r},~$$ for $$v$$
$$\pm \dfrac{Fr}{m}$$
20.
$$S=4 \pi r^2,~$$ for $$r$$
21.
$$L=\dfrac{8}{\pi^2}T^2,~$$ for $$T$$
$$\pm pi \sqrt{\dfrac{L}{8}}$$
22.
$$s= \dfrac{1}{2}gt^2,~$$ for $$t$$
Applications
Exercise Group.
For problems 23 and 24,
1. Make a sketch of the situation described, and label a right triangle.
2. Use the Pythagorean theorem to solve each problem.
23.
The size of a TV screen is the length of its diagonal. If the width of a 35-inch TV screen is 28 inches, what is its height?
21 in
24.
24 a 30-meter pine tree casts a shadow of 30 meters, how far is the tip of the shadow from the top of the tree?
27.
What size rectangle will fit inside a circle of radius 30 feet if the length of the rectangle must be three times its width?
19 ft by 57 ft
28.
A storage box for sweaters is constructed from a square sheet of cardboard measuring $$x$$ inches on a side. The volume of the box, in cubic inches, is
\begin{equation*} V=10(x-20)^2 \end{equation*}
If the box should have a volume of 1960 cubic inches, what size cardboard square is needed?
29.
A large bottle of shampoo is 20 centimeters tall and cylindrical in shape.
1. Write a formula for the volume of the bottle in terms of its radius.
2. Complete the table of values for the volume equation. If you cut the radius of the bottle in half, by what factor does the volume decrease?
$$~r~$$ $$~1~$$ $$~2~$$ $$~3~$$ $$~4~$$ $$~5~$$ $$~6~$$ $$~7~$$ $$~8~$$ $$~V~$$
3. What radius should the bottle have if it must hold 240 milliliters of shampoo? (A milliliter is equal to one cubic centimeter.)
4. Use your calculator to graph the volume equation. (Use the table to help you choose a suitable window.) Locate the point on the graph that corresponds to the bottle in part (c). Make a sketch of your graph, and lable the scales on the axes.
1. $$\displaystyle V=62.8r^2$$
2. $$\displaystyle \dfrac{1}{4}$$
3. 1.96 cm
30.
The area of a ring is given by the formula
\begin{gather*} A = \pi R^2 - \pi r^2 \end{gather*}
where $$R$$ is the radius of the outer circle, and $$r$$ is the radius of the inner circle.
1. Suppose the inner radius of the ring is kept fixed at $$r=4$$ centimeters, but the radius of the outer circle, $$R\text{,}$$ is allowed to vary. Find the area of the ring when the outer radius is 6 centimeters, 8 centimeters, and 12 centimeters.
2. Graph the area equation, with $$r=4\text{,}$$ in the window
\begin{align*} \text{Xmin} \amp = 0 \amp\amp \text{Xmax} = 14.1\\ \text{Ymin} \amp = 0 \amp\amp \text{Ymax} = 400 \end{align*}
3. Trace along the curve to the point $$(9.75, 248.38217)\text{.}$$ What do the coordinates of this point represent?
4. Use your graph to estimate the outer radius of the ring when its area is 100 square centimeters.
5. Write and solve an equation to answer part(d).
Exercise Group.
For Problems 31 and 32, solve for $$x$$ in terms of $$a\text{,}$$ $$b\text{,}$$ and $$c\text{.}$$
31.
1. $$\displaystyle \dfrac{ax^2}{b} = c$$
2. $$\displaystyle \dfrac{bx^2}{c} - a = 0$$
1. $$\displaystyle \pm \sqrt{\dfrac{bc}{a}}$$
2. $$\displaystyle \pm \sqrt{\dfrac{ac}{b}}$$
1. $$\displaystyle (x-a)^2 = 16$$
2. $$\displaystyle (ax+b)^2 = 9$$
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## Hitting Baseballs and Blocking Swords
###Learning to Fight
When I was an undergraduate student, I used to fight with the Society of Creative Anachronism (SCA). I remember when I was first learning how to strike with a rattan sword, my teacher told me that ideally, I should not hit my target with the tip of the sword, but rather about 1/3 of the way down. I didn’t think too much about this at the time; later I learned that baseball coaches often tell their players the exact same thing. You want to try to hit a baseball about 1/3 of the way down the bat. This was too much of a coincidence for me. Why is that the case?
###Setting up the Free Body Diagram
Take a look at the free body diagram of a sword below. Assume that the sword has a total length of $$L$$ and some mass $$M$$. In this simple problem, we will assume that the center of gravity and center of mass lie on the same point a distance $$L_M$$ from the bottom. The weight of the sword is $$F_G$$.
The fighter strikes or blocks another object at a distance of $$L_F$$ from the bottom of the sword. The force of the strike is $$F$$.
The fighter holds the sword at one end, exerting forces $$F_{Hx}$$ and $$F_{Hy}$$. The “H” signifies the force due to the fighter’s hands.
This is a two-dimensional problem involving the dynamics of a rigid body.
###Solving the Free Body Diagram
To solve this problem, we can sum the forces and the torques in the system.
The forces in the y-direction must sum to zero if we want to keep the sword upright. We know that $$\vec{F}_G = -Mg\hat{y}$$. Thus:
$$$\sum F_y = 0 \Rightarrow$$$ \begin{align} \vec{F}_{Hy} + \vec{F}_G &= 0 \\[0.1cm] F_{Hy} - Mg &= 0 \\[0.1cm] F_{Hy} &= Mg \\[0.1cm] \end{align}
Since we assume that the incident force $$F$$ is directed entirely in the $$\hat{x}$$-direction, this just means that we need to exert a force equal to the weight of the sword to keep it level.
The sum of the forces in the $$\hat{x}$$-direction is given by:
$$$\sum F_x = ma_x \Rightarrow$$$ $$$\vec{F}_{Hx} + \vec{F} = M \vec{a}_G$$$ $$$\vec{a}_G = \frac{\vec{F}_{Hx} + \vec{F}}{M}$$$
Here, $$\vec{a_G}$$ is the acceleration of the sword at its center of gravity. In order to solve this expression further, we need to look at the torques on the sword. We will take our pivot to be the point where the fighter holds the sword. Then, the sum of the torques is equal to the moment of inertia of the sword multiplied by its angular acceleration:
$$$\sum \tau = I \alpha$$$ \begin{align} F L_F &= I \alpha \\[0.1cm] &= I \frac{a_G}{L_G} \end{align}
As we see above, we can replace the angular acceleration of the sword by the linear acceleration of the center of mass of the sword.
For now, let’s assume that the sword can be approximated as a stick with uniform mass density in order to calculate the moment of inertia. Obviously, if you want to do a more realistic calculation, you should carefully calculate the value for I. Using the parallel axis theorem:
\begin{align} F L_F &= I \frac{a_G}{L_G} \\[0.1cm] &= \left ( \frac{1}{12}ML^2 + ML_G^2 \right )\frac{a_G}{L_G} \\[0.1cm] &= \left ( \frac{1}{12}ML^2 + ML_G^2 \right ) \frac{1}{L_G}\frac{|\vec{F}_{Hx} + \vec{F}|}{M} \end{align}
Now, in the best case scenario, you would feel no recoil on the sword when you strike or block. Thus, we will set $$\vec{F}_{Hx} = 0$$:
$$$F L_F = \left ( \frac{1}{12}ML^2 + ML_G^2\right ) \frac{1}{L_G}\frac{F}{M}$$$ \begin{align} L_F &= \left ( \frac{1}{12} \frac{L^2}{L_G} + L_G \right ) \\[0.1cm] &= \frac{L^2 + 12L_G^2}{12L_G} \\[0.1cm] \end{align}
So here we have it! The value $$L_F$$ is the “sweet spot” that we are looking for. If the fighter manages to aim the input force at some distance $$L_F$$ from the bottom of the sword, she theoretically feel no force on her hands; she can keep swinging. If she is slightly above or below this point, she will feel a balancing force either forward or backwards.
Let’s look at the case where the center of mass ($$L_G$$) is exactly halfway up the length of the sword ($$L$$):
\begin{align} L_F &= \frac{L^2 + 12L_G^2}{12L_G} \\[0.1cm] &= \frac{L^2 + 12 \left ( \frac{L}{2} \right )^2}{12 \left ( \frac{L}{2} \right )} \\[0.1cm] &= \frac{4L^2}{6L} \\[0.1cm] &= \frac{2}{3}L \end{align}
As we see, the sweet spot is 1/3 of the way from the top of the bat (2/3 of the way from the bottom).
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# Multiplying decimals
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0/1
##### Intros
###### Lessons
1. How to multiply decimals?
0/6
##### Examples
###### Lessons
1. Complete each multiplication statement.
1. 0.57 x 100 =
2. 5.43 x 0.1 =
3. 8.71 x 3.6 =
4. 32.8 x 0.02 =
2. Hannah's favorite soft drink is coke. Her mother gives her $4.50 to spend on coke one afternoon. She goes to the local grocery store and buys four bottles of coke. Each bottle costs$1.05. Does she have enough money to make this purchase?
1. Every Saturday morning Alan's bakery sells cinnamon buns for a special price of \$2.50. If Barb's two sisters go to the bakery every Saturday morning for two years, how much will they spend? Assume they each purchase one bun per visit. Round your answer to the nearest hundredth.
0%
##### Practice
###### Topic Notes
Previously, we learned how to add and subtract decimal numbers. In this section, we will learn how to multiply decimal numbers. As learned in previous section, when adding and subtracting decimal numbers, the decimal points must be lined up. In contrast, when multiplying decimal numbers, it is not important that the decimal points be lined up. Instead, it is important to line up the digits in the lowest place values of both numbers. In order to figure out where to place the decimal point in the answer, we must count how many digits total, between the two numbers being multiplied together, are behind the decimal points.
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# Unit 1.7
Online Instructor for Everett Public Schools à Everett School District
8 Sep 2014
1 sur 22
### Unit 1.7
• 1. 1.7 Modeling with Functions Copyright © 2011 Pearson, Inc.
• 2. What you’ll learn about Functions from Formulas Functions from Graphs Functions from Verbal Descriptions Functions from Data … and why Using a function to model a variable under observation in terms of another variable often allows one to make predictions in practical situations, such as predicting the future growth of a business based on data. Copyright © 2011 Pearson, Inc. Slide 1.7 - 2
• 3. Example A Maximum Value Problem A square of side x inches is cut out of each corner of an 8 in. by 15 in. piece of cardboard and the sides are folded up to form an open-topped box. (a) Write the volume V as a function of x. (b) Find the domain of V as a function of x. (c) Graph V as a function of x over the domain found in part (b) and use the maximum finder on your grapher to determine the maximum volume such a box can hold. (d) How big should the cut-out squares be in order to produce the box of maximum volume? Copyright © 2011 Pearson, Inc. Slide 1.7 - 3
• 4. Solution A square of side x inches is cut out of each corner of an 8 in. by 15 in. piece of cardboard and the sides are folded up to form an open-topped box. (a) Write the volume V as a function of x. 15 2x V x8 2x15 2x x 8 2x Copyright © 2011 Pearson, Inc. Slide 1.7 - 4
• 5. Solution A square of side x inches is cut out of each corner of an 8 in. by 15 in. piece of cardboard and the sides are folded up to form an open-topped box. (b) Find the domain of V as a function of x. V x8 2x15 2x The depth of x must be nonnegative, as must the side length and width. The domain is [0,4] where the endpoints give a box with no volume. Copyright © 2011 Pearson, Inc. Slide 1.7 - 5
• 6. Solution V x8 2x15 2x (c) Graph V as a function of x over the domain found in part (b) and use the maximum finder on your grapher to determine the maximum volume such a box can hold. The maximum occurs at the point (5/3, 90.74). The maximum volume is about 90.74 in.3. Copyright © 2011 Pearson, Inc. Slide 1.7 - 6
• 7. Solution A square of side x inches is cut out of each corner of an 8 in. by 15 in. piece of cardboard and the sides are folded up to form an open-topped box. (d) How big should the cut-out squares be in order to produce the box of maximum volume? Each square should have sides of one-and-two-thirds inches. Copyright © 2011 Pearson, Inc. Slide 1.7 - 7
• 8. Example Finding the Model and Solving Grain is leaking through a hole in a storage bin at a constant rate of 5 cubic inches per minute. The grain forms a cone-shaped pile on the ground below. As it grows, the height of the cone always remains equal to its radius. If the cone is one foot tall now, how tall will it be in one hour? Copyright © 2011 Pearson, Inc. Slide 1.7 - 8
• 9. Solution The volume of a cone is V 1 / 3 r2h. Since the height always equals the radius,V 1 / 3 h3 . When h 12 inches, the volume will be V (1 / 3) 12 576 in.3 . One hour later, the volume will have grown by (60 min)(5 in.3 / min) 300 in3 . The total volume will be 576 300 in3 . 1 / 3 h3 576 300 h3 3576 300 h 3576 300 3 12.63 inches Copyright © 2011 Pearson, Inc. Slide 1.7 - 9
• 10. Constructing a Function from Data Given a set of data points of the form (x, y), to construct a formula that approximates y as a function of x: 1. Make a scatter plot of the data points. The points do not need to pass the vertical line test. 2. Determine from the shape of the plot whether the points seem to follow the graph of a familiar type of function (line, parabola, cubic, sine curve, etc.). 3. Transform a basic function of that type to fit the points as closely as possible. Copyright © 2011 Pearson, Inc. Slide 1.7 - 10
• 11. Example Curve-Fitting with Technology The table shows that the number of patent applications in the United States increased from 1993 to 2003. Find both a linear and a quadratic regression model for this data. Which appears to be the better model of the data? Copyright © 2011 Pearson, Inc. Slide 1.7 - 11
• 12. Solution Use a grapher to compute the linear and quadratic regression, using x = 0 for 1993, x = 1 for 1994, … The linear regression model is y = 19.23x +1 57.84. The quadratic regression model is y = 0.7894x2 + 9.7573x + 178.36. The quadratic regression equation appears to model the data better than the linear regression equation. Copyright © 2011 Pearson, Inc. Slide 1.7 - 12
• 13. Functions Copyright © 2011 Pearson, Inc. Slide 1.7 - 13
• 14. Functions (cont’d) Copyright © 2011 Pearson, Inc. Slide 1.7 - 14
• 15. Quick Review Solve the given formula for the given variable. 1. Area of a Triangle Solve for b : A 1 2 bh 2. Volume of a Right Circular Cylinder Solve for h : V 1 3 r2h 3. Volume of a Sphere Solve for r : V 4 3 r3 4. Surface Area of a Sphere Solve for r : A 4 r2 5. Simple Interest Solve for P : I Prt Copyright © 2011 Pearson, Inc. Slide 1.7 - 15
• 16. Quick Review Solutions Solve the given formula for the given variable. 1. Solve for b : A 1 2 bh b 2A h 2. Solve for h : V 1 3 r2h h 3V r2 3. Solve for r : V 4 3 r3 r 3V 4 3 4. Solve for r : A 4 r2 r A 4 5. Solve for P : I Prt P I rt Copyright © 2011 Pearson, Inc. Slide 1.7 - 16
• 17. Chapter Test Find the (a) domain and (b) range of the function. 1. h(x) (x 2)2 5 2. k(x) 1 9 x2 3. Is the following function continuous at x 0? f (x) 2x 3 if x 0 3 x2 if x 0 4. Find all (a) vertical asymptotes and (b) horizontal asymptotes of the function y 3x x 4 . Copyright © 2011 Pearson, Inc. Slide 1.7 - 17
• 18. Chapter Test 5. State the interval(s) on which y x3 6 is increasing. 6. Tell whether the function is bounded above, bounded below or bounded. g(x) 6x x2 1 7. Use a grapher to find all (a) relative maximum values and (b) relative minimum values. y x3 3x 8. State whether the function is even, odd, or neither. y 3x2 4 x Copyright © 2011 Pearson, Inc. Slide 1.7 - 18
• 19. Chapter Test 9. Find a formula for f -1. f (x) 6 x 4 10. Find an expression for f ogx given f (x) x and g(x) x2 4. Copyright © 2011 Pearson, Inc. Slide 1.7 - 19
• 20. Chapter Test Solutions Find the (a) domain and (b) range of the function. 1. h(x) (x 2)2 5 (a) , (b) [5,) 2. k(x) 1 9 x2 (a) 3, 3 (b) [1/3,) 3. Is the following function continuous at x 0? f (x) 2x 3 if x 0 3 x2 if x 0 yes 4. Find all (a) vertical asymptotes and (b) horizontal asymptotes of the function y 3x x 4 . (a) x 4 (b) y 3 Copyright © 2011 Pearson, Inc. Slide 1.7 - 20
• 21. Chapter Test Solutions 5. State the interval(s) on which y x3 6 is increasing. , 6. Tell whether the function is bounded above, bounded below or bounded. g(x) 6x x2 1 bounded 7. Use a grapher to find all (a) relative maximum values and (b) relative minimum values. y x3 3x a 2 b 2 8. State whether the function is even, odd, or neither. y 3x2 4 x even Copyright © 2011 Pearson, Inc. Slide 1.7 - 21
• 22. Chapter Test 9. Find a formula for f -1. f (x) 6 x 4 f 1 x 6 x 4 10. Find an expression for f ogx given f (x) x and g(x) x2 4. x2 4 Copyright © 2011 Pearson, Inc. Slide 1.7 - 22
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# Integration Help
• roam
#### roam
1. Homework Statement
Find the antiderivative:
$$\int\frac{2x^3-5x^2+5x-12}{(x-1)^2(x^2+4)}$$
## Homework Equations
3. The Attempt at a Solution
Using Integration by Partial Fractions:
$$\int\frac{2x^3-5x^2+5x-12}{(x-1)^2(x^2+4)} = \frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{Cx+D}{(x^2+4)}$$
2x3-5x2+5x-12 = A(x-1)(x2+4)+B(x2+4)+(Cx+D)(x-1)2
Multiplying out:
= -4A+4Ax-Ax2+Ax3+Bx2+4B+Cx+D-2Cx2-2Dx-Cx3+Dx2
Collecting like terms:
2x3-5x2+5x-12 = x3(A-C)+x2(B-2C+D-A)+x(4A+C-2D) + (D+4B)
Equating corresponding coefficients gives
A-C = 2
B-2C+D-A = -5
4A+C-2D =5
D+4B-4A=-12
in the matrix form:
1, 0, -1, 0, 2
-1, 1, -2, 1, -5
4, 0, 1, -2, 5
-4, 4, 0, 1, -12
I used MATLAB to get the reduced-row echelon form:
rref=
1, 0, 0, 0, 27
0, 1, 0, 0, 8
0, 0, 1, 0, 25
0, 0, 0, 1, 64
Therefore, A=27, B=8, C= 25 and D=64 (?)
$$\int \frac{27}{(x-1)}+ \int \frac{8}{(x-1)^2}+ \int \frac{25x+64}{(x^2+4)}$$
= 27 Log(-1 + x)-8/(x-1)+32arctan(x/2)+25/2log(4+x2)
I'm not sure if this is the correct answer to this problem, because I tried solving it using mathematica and I got:
$$\frac{2}{(x-1)} + log(x-1) + (1/2) log (4+x^2)$$
I really appreciate it if someone could show me my mistakes. Thanks!
Last edited:
$$2x^3-5x^2+5x-12 = A(x-1)(x^2+4)+B(x^2+4)+(Cx+D)(x-1)^2$$
is true for all values of x. Put x=1
you'd get -10=5B =>B=-2. So I think you did something wrong in multiplying out.
You got some of your coefficients wrong when you expanded the right side of your equation. After collecting like terms I got these coefficients:
x^3: A + C (you have A - C)
x^2: -A + B - 2C + D (same)
x: 4A + C - 2D (same)
1: -4A + 4B + D (you have 4B + D at first, but apparently added in the -4A term in later work)
I didn't work this through, but I did find that B = -2.
Thanks for your input guys! Changing "-1" to "1" made all the difference!
I row-reduced the new matrix and got:
A=1, B=-2, C=1 and D=0
And eventually I got the right answer. Thanks very much for spotting my error.
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# Vector math¶
## Introduction¶
This tutorial is a short and practical introduction to linear algebra as it applies to game development. Linear algebra is the study of vectors and their uses. Vectors have many applications in both 2D and 3D development and Godot uses them extensively. Developing a good understanding of vector math is essential to becoming a strong game developer.
Note
This tutorial is not a formal textbook on linear algebra. We will only be looking at how it is applied to game development. For a broader look at the mathematics, see https://www.khanacademy.org/math/linear-algebra
## Coordinate systems (2D)¶
In 2D space, coordinates are defined using a horizontal axis (`x`) and a vertical axis (`y`). A particular position in 2D space is written as a pair of values such as `(4, 3)`.
Note
If you're new to computer graphics, it might seem odd that the positive `y` axis points downwards instead of upwards, as you probably learned in math class. However, this is common in most computer graphics applications.
Any position in the 2D plane can be identified by a pair of numbers in this way. However, we can also think of the position `(4, 3)` as an offset from the `(0, 0)` point, or origin. Draw an arrow pointing from the origin to the point:
This is a vector. A vector represents a lot of useful information. As well as telling us that the point is at `(4, 3)`, we can also think of it as an angle `θ` and a length (or magnitude) `m`. In this case, the arrow is a position vector - it denotes a position in space, relative to the origin.
A very important point to consider about vectors is that they only represent relative direction and magnitude. There is no concept of a vector's position. The following two vectors are identical:
Both vectors represent a point 4 units to the right and 3 units below some starting point. It does not matter where on the plane you draw the vector, it always represents a relative direction and magnitude.
## Vector operations¶
You can use either method (x and y coordinates or angle and magnitude) to refer to a vector, but for convenience, programmers typically use the coordinate notation. For example, in Godot, the origin is the top-left corner of the screen, so to place a 2D node named `Node2D` 400 pixels to the right and 300 pixels down, use the following code:
```\$Node2D.position = Vector2(400, 300)
```
Godot supports both Vector2 and Vector3 for 2D and 3D usage, respectively. The same mathematical rules discussed in this article apply to both types.
### Member access¶
The individual components of the vector can be accessed directly by name.
```# create a vector with coordinates (2, 5)
var a = Vector2(2, 5)
# create a vector and assign x and y manually
var b = Vector2()
b.x = 3
b.y = 1
```
```var c = a + b # (2, 5) + (3, 1) = (5, 6)
```
We can also see this visually by adding the second vector at the end of the first:
Note that adding `a + b` gives the same result as `b + a`.
### Scalar multiplication¶
Note
Vectors represent both direction and magnitude. A value representing only magnitude is called a scalar.
A vector can be multiplied by a scalar:
```var c = a * 2 # (2, 5) * 2 = (4, 10)
var d = b / 3 # (3, 6) / 3 = (1, 2)
```
Note
Multiplying a vector by a scalar does not change its direction, only its magnitude. This is how you scale a vector.
## Practical applications¶
Let's look at two common uses for vector addition and subtraction.
### Movement¶
A vector can represent any quantity with a magnitude and direction. Typical examples are: position, velocity, acceleration, and force. In this image, the spaceship at step 1 has a position vector of `(1,3)` and a velocity vector of `(2,1)`. The velocity vector represents how far the ship moves each step. We can find the position for step 2 by adding the velocity to the current position.
Tip
Velocity measures the change in position per unit of time. The new position is found by adding velocity to the previous position.
### Pointing toward a target¶
In this scenario, you have a tank that wishes to point its turret at a robot. Subtracting the tank's position from the robot's position gives the vector pointing from the tank to the robot.
Tip
To find a vector pointing from `A` to `B` use `B - A`.
## Unit vectors¶
A vector with magnitude of `1` is called a unit vector. They are also sometimes referred to as direction vectors or normals. Unit vectors are helpful when you need to keep track of a direction.
### Normalization¶
Normalizing a vector means reducing its length to `1` while preserving its direction. This is done by dividing each of its components by its magnitude. Because this is such a common operation, `Vector2` and `Vector3` provide a method for normalizing:
```a = a.normalized()
```
Warning
Because normalization involves dividing by the vector's length, you cannot normalize a vector of length `0`. Attempting to do so will result in an error.
### Reflection¶
A common use of unit vectors is to indicate normals. Normal vectors are unit vectors aligned perpendicularly to a surface, defining its direction. They are commonly used for lighting, collisions, and other operations involving surfaces.
For example, imagine we have a moving ball that we want to bounce off a wall or other object:
The surface normal has a value of `(0, -1)` because this is a horizontal surface. When the ball collides, we take its remaining motion (the amount left over when it hits the surface) and reflect it using the normal. In Godot, the Vector2 class has a `bounce()` method to handle this. Here is a GDScript example of the diagram above using a KinematicBody2D:
```# object "collision" contains information about the collision
var collision = move_and_collide(velocity * delta)
if collision:
var reflect = collision.remainder.bounce(collision.normal)
velocity = velocity.bounce(collision.normal)
move_and_collide(reflect)
```
## Dot product¶
The dot product is one of the most important concepts in vector math, but is often misunderstood. Dot product is an operation on two vectors that returns a scalar. Unlike a vector, which contains both magnitude and direction, a scalar value has only magnitude.
The formula for dot product takes two common forms:
and
However, in most cases it is easiest to use the built-in method. Note that the order of the two vectors does not matter:
```var c = a.dot(b)
var d = b.dot(a) # These are equivalent.
```
The dot product is most useful when used with unit vectors, making the first formula reduce to just `cosθ`. This means we can use the dot product to tell us something about the angle between two vectors:
When using unit vectors, the result will always be between `-1` (180°) and `1` (0°).
### Facing¶
We can use this fact to detect whether an object is facing toward another object. In the diagram below, the player `P` is trying to avoid the zombies `A` and `B`. Assuming a zombie's field of view is 180°, can they see the player?
The green arrows `fA` and `fB` are unit vectors representing the zombies' facing directions and the blue semicircle represents its field of view. For zombie `A`, we find the direction vector `AP` pointing to the player using `P - A` and normalize it, however, Godot has a helper method to do this called `direction_to`. If the angle between this vector and the facing vector is less than 90°, then the zombie can see the player.
In code it would look like this:
```var AP = A.direction_to(P)
if AP.dot(fA) > 0:
print("A sees P!")
```
## Cross product¶
Like the dot product, the cross product is an operation on two vectors. However, the result of the cross product is a vector with a direction that is perpendicular to both. Its magnitude depends on their relative angle. If two vectors are parallel, the result of their cross product will be a null vector.
The cross product is calculated like this:
```var c = Vector3()
c.x = (a.y * b.z) - (a.z * b.y)
c.y = (a.z * b.x) - (a.x * b.z)
c.z = (a.x * b.y) - (a.y * b.x)
```
With Godot, you can use the built-in method:
```var c = a.cross(b)
```
Note
In the cross product, order matters. `a.cross(b)` does not give the same result as `b.cross(a)`. The resulting vectors point in opposite directions.
### Calculating normals¶
One common use of cross products is to find the surface normal of a plane or surface in 3D space. If we have the triangle `ABC` we can use vector subtraction to find two edges `AB` and `AC`. Using the cross product, `AB x AC` produces a vector perpendicular to both: the surface normal.
Here is a function to calculate a triangle's normal:
```func get_triangle_normal(a, b, c):
# find the surface normal given 3 vertices
var side1 = b - a
var side2 = c - a
var normal = side1.cross(side2)
return normal
```
### Pointing to a target¶
In the dot product section above, we saw how it could be used to find the angle between two vectors. However, in 3D, this is not enough information. We also need to know what axis to rotate around. We can find that by calculating the cross product of the current facing direction and the target direction. The resulting perpendicular vector is the axis of rotation.
|
# Unlocking the Power of Reverse Polish Notation (RPN) with Stacks
Mar 5, 2024
Unlocking the Power of Reverse Polish Notation (RPN) with Stacks
## Introduction: The Mathematical Puzzle
In the world of mathematics and computer science, expressing and evaluating mathematical expressions is akin to solving a puzzle. Traditional methods, like infix notation, often require us to navigate through complex rules of precedence and parentheses. However, Reverse Polish Notation (RPN) emerges as a simple and efficient alternative, where every operation follows its operands directly. But how does RPN achieve this efficiency? The answer lies in its clever utilization of stacks.
## Understanding RPN: A New Perspective
RPN, also known as postfix notation, flips the script on traditional mathematical notation. Instead of placing operators between operands, as seen in infix notation, RPN puts operators after their operands. This eliminates the need for parentheses to denote the order of operations, making expressions clearer and easier to evaluate.
## The Stack: A Pillar of RPN Evaluation
At the heart of RPN lies the stack data structure. Think of a stack as a pile of plates: you can only add or remove plates from the top. Similarly, in RPN evaluation, we use a stack to keep track of numbers and intermediate results.
## Step-by-Step Evaluation: A Visual Journey
Let’s walk through the evaluation of an RPN expression step-by-step:
2. Scan each token (operand or operator) from left to right.
3. If the token is an operand, push it onto the stack.
4. If the token is an operator, pop the necessary operands from the stack, perform the operation, and push the result back onto the stack.
5. Continue this process until the entire expression is evaluated, leaving the final result on the stack.
## Illustrating with an Example
Consider the RPN expression: “3 4 5 *”
1. Push “3” onto the stack.
2. Push “4” onto the stack.
3. Encounter “*”: Pop “4” and “3”, perform multiplication, and push “12” onto the stack.
4. Push “5” onto the stack.
5. Encounter “*”: Pop “5” and “12”, perform multiplication, and push “60” onto the stack.
6. Expression evaluation is complete, with “60” remaining on the stack as the final result.
## Comparing RPN with Other Notations
While infix notation is familiar, it often requires complex rules and parentheses. Postfix notation shares similarities with RPN but lacks the efficiency of stack-based evaluation. RPN’s simplicity and reliance on stacks make it a clear winner in terms of computational efficiency and ease of understanding.
## Conclusion: Unleashing the Potential of RPN
Reverse Polish Notation (RPN) offers a fresh perspective on expressing and evaluating mathematical expressions. By leveraging the stack data structure, RPN simplifies the evaluation process, eliminating the need for complex precedence rules and parentheses. Its efficiency and clarity make it a valuable tool in various fields, from computer science to mathematics. So, next time you encounter a mathematical puzzle, consider embracing the simplicity and power of RPN.
#### By Anshul Pal
Hey there, I'm Anshul Pal, a tech blogger and Computer Science graduate. I'm passionate about exploring tech-related topics and sharing the knowledge I've acquired. With two years of industry expertise in blogging and content writing, I'm also the co-founder of HVM Smart Solution. Thanks for reading my blog – Happy Learning!
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## What is tangent plane?
Well tangent planes to a surface are planes that just touch the surface at the point and are “parallel” to the surface at the point. Since the tangent plane and the surface touch at (x0,y0) ( x 0 , y 0 ) the following point will be on both the surface and the plane.
## Is linear approximation the same as tangent plane?
The function L is called the linearization of f at (1, 1). f(x, y) ≈ 4x + 2y – 3 is called the linear approximation or tangent plane approximation of f at (1, 1). However, if we take a point farther away from (1, 1), such as (2, 3), we no longer get a good approximation.
## How do you find the gradient of a tangent plane?
Use gradients and level surfaces to find the normal to the tangent plane of the graph of z = f(x, y) at P = (x0,y0,z0). w = f(x, y) – z. The graph of z = f(x, y) is just the level surface w = 0. We compute the normal to the surface to be vw = .
## What’s the tangent?
Tangent (tan) function – Trigonometry. (See also Tangent to a circle). In a right triangle, the tangent of an angle is the length of the opposite side divided by the length of the adjacent side. The tangent function, along with sine and cosine, is one of the three most common trigonometric functions.
## How do you find the normal line of a tangent line?
The normal line is defined as the line that is perpendicular to the tangent line at the point of tangency. Because the slopes of perpendicular lines (neither of which is vertical) are negative reciprocals of one another, the slope of the normal line to the graph of f(x) is −1/ f′(x).
## What is tangent plane in GD&T?
The tangent plane modifier means that the form error, flatness and straightness, of the surface is ignored. Therefore, it is necessary to provide some form control. In this case, a flatness control with a value larger than the profile tolerance would be appropriate.
## How do you find a tangent vector to a surface?
Directional derivatives are one way to find a tangent vector to a surface. A tangent vector to a surface has a slope (rise in z over run in xy) equal to the directional derivative of the surface height z(x,y). To find a tangent vector, choose a,b,c so that this equality holds.
## What is the equation of plane?
If we know the normal vector of a plane and a point passing through the plane, the equation of the plane is established. a ( x − x 1 ) + b ( y − y 1 ) + c ( z − z 1 ) = 0.
## At what point is the tangent plane parallel to the plane?
So the point where the tangent plane is parallel to the plane x + 2y + 4z = 1 is at (-1/4,-1,-1).
## What is the equation for linear approximation?
since ο(Δx) corresponds to the term of the second and higher order of smallness with respect to Δx. Thus, we can use the following formula for approximate calculations: f(x)≈L(x)=f(a)+f′(a)(x−a). where the function L(x) is called the linear approximation or linearization of f(x) at x=a.
## What is a gradient in math?
The Gradient (also called Slope) of a straight line shows how steep a straight line is.
## Is gradient and tangent the same?
In short, gradients are slopes (or ratios of “rise over run”), whereas tangents are lines that touch curves without crossing them at the tangency point (except at points of inflection).
### Releated
#### Equation of vertical line
How do you write an equation for a vertical and horizontal line? Horizontal lines go left and right and are in the form of y = b where b represents the y intercept. Vertical lines go up and down and are in the form of x = a where a represents the shared x coordinate […]
#### Bernoulli’s equation example
What does Bernoulli’s equation State? Bernoulli’s principle states the following, Bernoulli’s principle: Within a horizontal flow of fluid, points of higher fluid speed will have less pressure than points of slower fluid speed. Why is Bernoulli’s equation used? The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one […]
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# Math in Focus Grade 6 Chapter 9 Review Test Answer Key
Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 9 Review Test to score better marks in the exam.
## Math in Focus Grade 6 Course 1 B Chapter 9 Review Test Answer Key
Concepts and Skills
Use the coordinate plane below.
Question 1.
Give the coordinates of points A, B, C, D, and E.
Point A:
The first coordinate represents distance along negative x axis which is 4 and the second coordinate represents distance along negative y axis which is 3.
So, the point A will be (-4,-3)
Point B:
The first coordinate represents distance along x axis which is 0 and the second coordinate represents distance along negative y axis which is 6.
So, the point B will be (0,-6)
Point C:
The first coordinate represents distance along x axis which is 2 and the second coordinate represents distance along negative y axis which is 4.
So, the point C will be (2,-4)
Point D:
The first coordinate represents distance along x axis which is 6 and the second coordinate represents distance along y axis which is 3.
So, the point D will be (6,3)
Point E:
The first coordinate represents distance along the negative x axis which is 2 and the second coordinate represents distance along y axis which is 3.
So, the point E will be (-2,3)
Use graph paper. Plot the points on a coordinate plane. In which quadrant is each point located?
Question 2.
A (3, 5), B (-2, 0), C (7, -2), D (0, -5), and E (-3, -8)
Point A (3, 5): Since both the x and y coordinate are positive, the point A will be located in quadrant I.
Point B (-2, 0): Since the x coordinate is negative and y coordinate is origin, the point B will not lie in any quadrant. It will be located on x axis.
Point C (7, -2): Since the x coordinate is positive and y coordinate is negative, the point C will lie in quadrant IV.
Point D (0, -5): Since the x coordinate is zero and y coordinate is negative, the point D will not lie in any quadrant. It will be located on y axis.
Point E (-3, -8): Since both the x and y coordinate are negative, the point A will be located in quadrant III.
Use graph paper. Points A and B are reflections of each other about the x-axis. Give the coordinates of point B if the coordinates of point A are the following:
Question 3.
(3, 6)
Given that the points A and B are reflections of each other about the x-axis.
Given that B is reflection of point A in the x-axis and point A is (3, 6).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x coordinate will be 3 and y coordinate will be -6.
Therefore, point B will be (3,-6).
Question 4.
(-6, 2)
Given that B is reflection of point A in the x-axis and point A is (-6, 2).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x coordinate will be 3 and y coordinate will be -2.
Therefore, point B will be (-6,-2).
Question 5.
(5, -4)
Given that B is reflection of point A in the x-axis and point A is (5, -4).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x coordinate will be 5 and y coordinate will be 4.
Therefore, point B will be (5,4).
Question 6.
(-3, -5)
Given that B is reflection of point A in the x-axis and point A is (-3, -5).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x coordinate will be -3 and y coordinate will be 5.
Therefore, point B will be (-3,5).
Use graph paper. Points C and D are reflections of each other about the y-axis. Give the coordinates of point D if the coordinates of point C are the following:
Question 7.
(3, 6)
Given that the points C and D are reflections of each other about the y-axis.
Given that C is reflection of point D in the y-axis and point C is (3, 6).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be -3 and y coordinate will be 6.
Therefore, point D will be (-3,6).
Question 8.
(-6, 2)
Given that B is reflection of point A in the y-axis and point C is (-6, 2).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be 6 and y coordinate will be 2.
Therefore, point D will be (6,2).
Question 9.
(5, -4)
Given that B is reflection of point A in the y-axis and point C is (5, -4).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be -5 and y coordinate will be -4.
Therefore, point D will be (-5,-4).
Question 10.
(-3, -5)
Given that B is reflection of point A in the y-axis and point C is (-3, -5).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be 3 and y coordinate will be -5.
Therefore, point D will be (3,-5).
Use graph paper. For each exercise, plot the given points on a coordinate plane. Then connect the points in order with line segments to form a closed figure. Name each figure formed.
Question 11.
A (2, -4), B (2, 4), C (-6, 4), and D (-6, -4)
Plot the points A,B,C and D on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has all equal sides. Therefore, it can be a square.
Question 12.
E (0, -2), F (-3, 1), G(-5, -1), and H (-2, -4)
Plot the points E,F,G and H on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has all four sides different. Therefore, it can be a quadrilateral.
Question 13.
J (0, 1), K(1, 4), and L(-4, 3)
Plot the points J,K and L on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has all three sides different. Therefore, it can be a triangle.
Question 14.
M (6, 5), N (3, 5), P (3, -3), and Q (6, -3)
Plot the points M,N,P and Q on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has opposite sides equal. Therefore, it can be a rectangle.
Question 15.
A (6, -3), B (4, 2), C (-1, 2), and D (0, -3)
Plot the points A,B,C and D on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has a pair of opposite sides parallel and equal. Therefore, it can be a parallelogram.
Question 16.
E (-1, 6), F (-3, 3), G (3, 3), and H (5, 6)
Plot the points E,F,G and H on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has all different sides. Therefore, it can be a quadrilateral.
Question 17.
J(6, 1), K(8, -2), L (2, -2), and M (0, 1)
Plot the points J,K,L and M on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has a pair of opposite sides parallel and equal. Therefore, it can be a parallelogram.
Question 18.
P (2, 7), Q (-1, 7), R (-5, 4), and S (4, 4)
Plot the points P,Q,R and S on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has a pair of opposite sides parallel and equal. Therefore, it can be a trapezium.
Question 19.
T (-2, 1), U (-6, 1), V(-6, -3), and W(-2, -3)
Plot the points T,U,V and W on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has opposite sides equal. Therefore, it can be a rectangle.
Use graph paper. Plot the points on a coordinate plane and answer the question.
Question 20.
a) Plot points A (1, -1) and B (7, -1) on a coordinate plane. Connect the two points to form a line segment.
Point A:
The first coordinate represents distance along positive x axis which is 1 and the second coordinate represents distance along negative y axis which is 1.
So, the point A will be (1,-1)
Point B:
The first coordinate represents distance along positive x axis which is 7 and the second coordinate represents distance along negative y axis which is -1.
So, the point A will be (7,-1)
b) Point C lies above $$\overline{\mathrm{AB}}$$, and is 2 units away from the x-axis. If triangle ABC is an isosceles triangle with base $$\overline{\mathrm{AB}}$$, find the coordinates of point C.
Point C lies above $$\overline{\mathrm{AB}}$$, and is 2 units away from the x-axis. If triangle ABC is an isosceles triangle with base $$\overline{\mathrm{AB}}$$
Since it is an isosceles triangle, two sides will be equal. When the line segment is joined through midpoint, the formed line segments will be equal in length.
Midpoint of AB line segment is 3 units, so midpoint will be (4,2).
Thus, coordinates of point C will be (4,2)
c) Points D and E lie below $$\overline{\mathrm{AB}}$$ such that ABDE is a rectangle. If BD is 5 units, find the coordinates of points D and E.
Given that BD is 5 units.
Points D and E lie below $$\overline{\mathrm{AB}}$$ such that ABDE is a rectangle.
Start from point B and move 5 units above and mark it. Thus, the required point D will be formed.
Since it is a rectangle, opposite sides will be equal. Therefore, AB will be equal to DE, which is 6 units.
Start from point D and move 6 units towrads left and mark it is as E.
Thus, the formed D point will be (7,4) and E will be (1,4).
Use graph paper. Plot each pair of points on a coordinate plane. Connect the points to form a line segment and find its length.
Question 21.
A (-1,0) and B (8, 0)
To plot point A (-1,0):
Here, the x-coordinate of A is -1 and the y-coordinate is 0. Start at the Origin. As the x coordinate is negative, move 1 units on the negative x-axis and point it.
Thus, the required point (-1,0) is marked.
To plot point B (8, 0):
Here, the x-coordinate of A is 8 and the y-coordinate is 0. Start at the Origin. As the x coordinate is positive, move 8 units along the positive x-axis and point it.
Thus, the required point B (8, 0) is marked.
After connecting the points to form a line segment, the length will be 9 units.
Question 22.
C (-2, 4) and D (6, 4)
To plot point C (-2, 4):
Here, the x-coordinate is -2 and the y-coordinate is 4. Start at the Origin. As the x coordinate is negative, move 2 units on the negative x-axis and 4 units on the positive y -axis.
Thus, the required point C (-2,4) is marked.
To plot point D (6, 4):
Here, the x-coordinate is 6 and the y-coordinate is 4. Start at the Origin. As the x coordinate is positive, move 4 units along the positive x-axis and 4 units along positive y -axis.
Thus, the required point D (6, 4) is marked.
After connecting the points to form a line segment, the length will be 8 units.
Question 23.
E (-6, -2) and F(-6, -6)
To plot point E (-6, -2):
Here, the x-coordinate is 6 and the y-coordinate is-2. Start at the Origin. As the x coordinate is negative, move 6 units on the negative x-axis and 2 units on the negative y -axis.
Thus, the required point C (-2,4) is marked.
To plot point F(-6, -6):
Here, the x-coordinate is -6 and the y-coordinate is -6. Start at the Origin. As the x coordinate is negative, move 6 units along the negative x-axis and 6 units along negative y -axis.
Thus, the required point D (-6, -6) is marked.
After connecting the points to form a line segment, the length will be 4 units.
Question 24.
G (-5, -4) and H (2, -4)
To plot point G (-5, -4):
Here, the x-coordinate is -5 and the y-coordinate is-4. Start at the Origin. As the x coordinate is negative, move 5 units on the negative x-axis and 4 units on the negative y -axis.
Thus, the required point G (-5, -4) is marked.
To plot point H (2, -4):
Here, the x-coordinate is 2 and the y-coordinate is -4. Start at the Origin. As the x coordinate is positive, move 4 units along the positive x-axis and 4 units along negative y -axis.
Thus, the required point H (2, -4) is marked.
After connecting the points to form a line segment, the length will be 7 units.
Question 25.
J(0, -3) and K(0, -8)
To plot point J(0, -3):
Here, the x-coordinate is 0 and the y-coordinate is-3. Start at the Origin. As the x coordinate is origin, move 3 units on the negative y -axis.
Thus, the required point J(0, -3) is marked.
To plot point K(0, -8):
Here, the x-coordinate is 0 and the y-coordinate is-8. Start at the Origin. As the x coordinate is origin, move 8 units on the negative y -axis.
Thus, the required point K(0, -8) is marked.
After connecting the points to form a line segment, the length will be 5 units.
Question 26.
M (5, 2) and N (5, -5)
To plot point M (5, 2):
Here, the x-coordinate is 5 and the y-coordinate is 2. Start at the Origin. As the x and y coordinates are positive , move 5 units on the positive x-axis and 2 units on the positive y -axis.
Thus, the required point M (5, 2) is marked.
To plot point N (5, -5):
Here, the x-coordinate is 5 and the y-coordinate is 5. Start at the Origin. Move 5 units on the positive x-axis and 2 units on the negative y -axis.
Thus, the required point N (5, -5) is marked.
After connecting the points to form a line segment, the length will be 7 units.
Problem Solving
The diagram shows the plan of a room. The side length of each grid square is 10 feet. Use the diagram to answer questions 27 to 28.
Question 27.
The eight corners of the room are labeled points A to H. Give the coordinates of each of these corners.
The first coordinate represents distance along negative x axis which is 40 and the second coordinate represents distance along y axis which is 100.
The coordinates of A will be (-40,100)
Similarly,The coordinates of B will be (-40,20)
The coordinates of C will be (-60,20)
The coordinates of D will be (-60,-40)
The coordinates of E will be (60,40)
The coordinates of F will be (60,20)
The coordinates of G will be (40,20)
The coordinates of H will be (40,100)
Question 28.
The entrance of the room is situated along $$\overline{\mathrm{AH}}$$. What is the shortest possible distance in feet between the entrance and $$\overline{\mathrm{DE}}$$ of the room?
The entrance of the room is situated along $$\overline{\mathrm{AH}}$$.
The shortest distance between the the entrance and $$\overline{\mathrm{DE}}$$ of the room will be the straight route from AH to DE.
The distance between AH to DE is 14 square grids.
Each square grid equals to 10 feet.
Thus, the distance between AH to DE will be 14×10 = 140 ft.
Question 29.
Diana walks across the room from point B to point G, and then walks from point G to point H. Find the total distance, in feet, that Diana walks.
Given that Diana walks across the room from point B to point G, and then walks from point G to point H.
Distance between B and G will be 8 square grids, 8×10=80 ft.
Distance between G to H will be 8 square grids, 8×10=80 ft.
The total distance will be BG+GH = 80+80 = 160 ft
Question 30.
Calculate the floor area of the room in square feet.
As the complete floor area is in irregular shape, we will find the area in parts.
To calculate the floor area, we will find the area of rectangle AHGB and area of CFED
Area of rectangle AHGB:
Length of AB will be 8 square grids, 8×10=80 ft.
Length of AH will be 8 square grids, 8×10=80 ft.
Area of the rectangle = length×width
= 80×80
= 6400 sq.ft
Area of rectangle CFED:
Length of CD will be 6 square grids, 6×10=60 ft.
Length of DE will be 12 square grids, 12×10=120 ft
Area of the rectangle = length×width
= 60×120
= 7200 sq.ft
Thus, the total area of floor will be 6400+7200 = 13600 sq.ft
Use graph paper. Solve.
Question 31.
An athlete took part in a race. The distance the athlete ran, v meters, after t minutes, is given by v = 300t. Graph the relationship between t and v. Use 2 units on the horizontal axis to represent 1 minute and 1 unit on the vertical axis to represent 150 meters.
a) What type of graph is it?
It is a straight line graph.This is also called a linear graph.
b) What is the distance the athlete ran in 3.5 minutes?
An athlete took part in a race. The distance the athlete ran, v meters, after t minutes, is given by v = 300t.
Here, t=3.5 min
v = 300t
v = 300×3.5
= 1050 m
Thus the distance the athlete ran in 3.5 min will be 1050 m.
c) What is the average speed of the athlete?
The average speed of the athlete will be,
Average speed = $$\frac{total distance}{total time}$$
Total distance will be 0+300+600+900+1200 = 3000 m
Total time will be 0+1+2+3+4 = 10 min
= $$\frac{3000}{10}$$
= 300 meters per min
d) Assuming the athlete runs at a constant speed, what is the distance she will run in 8 minutes?
The distance the athlete ran, v meters, after t minutes, is given by v = 300t.
The distance she will run in 8 min will be, 300×8=2400 m
e) Name the dependent and independent variables.
From the graph, we can say v is the dependent variable and t is the independent variable.
Question 32.
A truck uses 1 gallon of diesel for every 12 miles traveled, The amount of diesel left in the gas tank, r gallons, after traveling s miles is given by s = 300 – 12r. Copy and complete the table. Graph the relationship between r and s. Use 1 unit on the horizontal axis to represent 1 gallon and 1 unit on the vertical axis to represent 12 miles. Start your horizontal axis at 17 gallons.
a)
A truck uses 1 gallon of diesel for every 12 miles traveled, The amount of diesel left in the gas tank, r gallons, after traveling s miles is given by s = 300 – 12r.
Here s=24,
24 = 300 – 12r
24+12r = 300-12r+12r
24+12r = 300
24+12r-24 = 300-24
12r = 276
r = 276÷12 = 23 gallons
Here r=17,
s = 300 – 12×17
= 300-204
= 96 miles
b) How many gallons of diesel are left after the truck has traveled 60 miles?
If 60 miles is travelled,
s = 300 – 12r
60 = 300 – 12r
60 + 12r = 300 – 12r + 12r
60 + 12r = 300
60 + 12r – 60 = 300-60
12r = 240
12r÷12 = 240÷12
r = 20
Thus, 20 gallons of diesel are left after the truck has traveled 60 miles
c) After the truck has traveled for 72 miles, how much farther can the truck travel before it runs out of diesel?
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Lesson 2: Multiplying decimals strategies
# Estimating decimal multiplication
This video is all about using estimation to simplify the process of multiplying decimals. It emphasizes the idea that rounding numbers to their nearest whole can make multiplication easier and quicker, especially when you're trying to do it in your head!
## Want to join the conversation?
• Hi this video helped me enough but I am still confused how to do your estimate problems can you tell me some more information?
• You round your numbers to the nearest tenths, hundredths,and ones, then you do your old buddy thing: Multiply
• So basically rounding is just the art of makeing this simple for example if i were bakeing cookies and i wanted to sell them, and someone was like hey, quick i gotta go i have a \$7 an 56 cents. evry cookie was 1 dollar. you would look at the cents and see if 56 cents is closer to 100 or 0 it is closer to 100 so i would be like you can get about 7-8. So this is a basic representation of rounding decimals.
• whats 450x24 and how do you do it?
• 450*24=10800
I got this answer by using the normal algorithm method.
Step 1: You jot down the numerals like so:
450
24
----
----
(You might wonder what the dotted lines under the equation are for, it's for writing down the answer)
Step 2: You multiply the units digits in both numbers like so:
0*4=0
(So after step 2 the equation would be like:
450
24
----
0
----
Step 3: You multiply the tens digit of the first numeral with the units digit of the second numeral.
Step 4: I guess you've caught the sequence by now. All you have to do this time is multiply the hundreds digit of the first numeral by the units digit of the second numeral.
Your equation would look like this so far:
450
24
----
1800
----
You might be wondering about the tens digit in the second numeral because we haven't even used it once yet right?
Don't worry I'll teach you in the next few steps:
(Hot Tip: Write down a plus sign underneath the units digit of your new answer so far)
Step 5: Now we come back to the units digit of the first numeral and multiply that by the tens digit of the first numeral.
Step 6: And the pattern starts all along again, but this time with the tens digit of the second numeral. This time you multiply the tens digit of the first numeral by the tens digit of the second numeral.
Step 7: Again, you multiply the hundreds digit of the first numeral by the tens digit of the second numeral.
Your equation will look like this so far:
450
24
----
1800
900+
----
Step 8: It's pretty obvious by now that all we have to do now is add the new answers we have acquired.
The new equation will look like this:
450
24
----
1800
900+
------
10800
------
Hope this helps :)
• I don't understand this whole video. Can someone help me?
• ok what he is doing is estimating so for EX 199 x 7.8 it is approximately equal to 1600 because 199 is equal to 200 and 7.8 is equal to 8.0 or just 8 because there's a zero next to it not a 1,2,3,4or any other number. Then multiply the 8 and the 200 you would get 1600
• Hi Sal, is there a specific strategy that will help solve multiplication with decimals for the exact sum?
• There is a few strategies. One is to use long multiplication, and just make sure that you have the correct amount of decimals. Example would be 0.25 x 1.23, you would first do 3 times 25, gives 75, then 20 times 25 for 500. and then 100 times 25 for 2500. Then add up for 3075, then you say first number had 2 decimal spaces, second also had 2, 2 plus 2 is 4, so you have 4 decimal places and get 0.3075.
Of course, this isn't the only way, you could also turn one or both decimals into fractions. For the example you could say 123/100 * 1/4 then multiply fractions to get 123/400. If you wanted to then turn that fraction into a decimal to get your answer you once again have different options but I would do long division.
• hi i need help at
• Ok, so is turning a fraction into a decimal and estimating then turning it into a fraction an option? Can you do it the other way around.
• so 99.87 rounds to 100 and 19 rounds to 20 so all that is left is to multiply 100x20
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# Ex.14.3 Q2 Statistics Solution - NCERT Maths Class 10
Go back to 'Ex.14.3'
## Question
If the median of the distribution given below is $$28.5$$, find the values of $$x$$ and $$y.$$
Class interval Frequency $$0 – 10$$ $$5$$ $$10 – 20$$ $$x$$ $$20 – 30$$ $$20$$ $$30 – 40$$ $$15$$ $$40 – 50$$ $$y$$ $$50 – 60$$ $$5$$ Total $$60$$
Video Solution
Statistics
Ex 14.3 | Question 2
## Text Solution
What is known?
The median of the distribution is $$28.5$$
What is the unknown?
The values of $$x$$ and $$y.$$
Reasoning:
Median Class is the class having Cumulative frequency $$(cf)$$ just greater than $$\frac {n}2$$
Median $$= l + \left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times h$$
Class size,$$h$$
Number of observations,$$n$$
Lower limit of median class,$$l$$
Frequency of median class,$$f$$
Cumulative frequency of class preceding median class,$$cf$$
Steps:
The cumulative frequency for the given data is calculated as follows.
Class interval Frequency Frequency $$0 – 10$$ $$5$$ $$5$$ $$20 – 30$$ $$x$$ $$5 + x$$ $$30 – 40$$ $$20$$ $$25 + x$$ $$40 – 50$$ $$15$$ $$40 + x$$ $$40 – 50$$ $$y$$ $$40 + x + y$$ $$50 – 60$$ $$5$$ $$45 +x+y$$ $$n=$$ $$60$$
From the table, it can be observed that
$$n =60$$ $$\Rightarrow \frac{n}{2}=30$$
\begin{align} \\ {45+x+y}&={60} \\ {x+y} &={15}.............(i) \end{align}
Median of the data is given as $$28.5$$ which lies in interval $$20 - 30.$$
Therefore, median class $$= 20 - 30$$
Class size ($$h$$) $$= 10$$
Lower limit of median class ($$l$$$$=20$$
Frequency of median class ($$f$$$$=20$$
Cumulative frequency of class preceding the median class, ($$cf$$$$=5 + x$$
\begin{align}{\text { Median }}&\!=\!{l\!+\!\left(\frac{\frac{n}{2}\!-\!c f}{f}\right)\!\times\!h} \\ {28.5}&\!=\!{20\!+\!\!\left(\frac{\frac{60}{2}-(5+x)}{20}\right) \!\!\times\!\!10} \\ {8.5}&={\left(\frac{25-x}{2}\right)} \\ {17}&={25-x} \\ {x}&={8}\end{align}
Putting $$x=8$$ equation, (i)
\begin{align} {8+y}&={15} \\ {y}&={7}\end{align}
Hence, the values of $$x$$ and $$y$$ are $$8$$ and $$7$$ respectively.
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# RD Sharma Solutions -Ex-22.1 (Part - 1), Tabular Representation Of Statistical Data, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9
## Class 9: RD Sharma Solutions -Ex-22.1 (Part - 1), Tabular Representation Of Statistical Data, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9
The document RD Sharma Solutions -Ex-22.1 (Part - 1), Tabular Representation Of Statistical Data, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9 is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
All you need of Class 9 at this link: Class 9
Q1. What do you understand by the word “statistics” in:
(a)Singular form (B) Plural form
Solution 1:
The word statistics is used in both its singular and plural senses.
(a) In singular sense, statistics may be defined as the science of collection, presentation, analysis and interpretation of numerical data.
(b) In plural sense, statistics means numerical facts or observations collected with definite purpose.
For example: Income and expenditure of persons in a particular locality,number of persons in a particular locality ,number of males and females in a particular town are statistics.
Q2.Describe some fundamental characteristics of statistics.
Solution 2:
Fundamental characteristics of statistics
(1) A single observation does not form statistics. Statistics are a sum total of observations.
(2) Statistics are expressed quantitatively not qualitatively.
(3) Statistics are collected with definite purpose.
(4) Statistics in an experiment are comparable and can be classified into groups.
Q3. What are (1) primary data (2) secondary data? Which of the two—the primary or the secondary data—is more reliable and why?
Solution 3:
The word data means information statistical data and are of two types:
(1) Primary data
When an investigator collects data himself with a definite plan or design in hs or her mind is called primary data.
(2)Secondary data
Data which are not originally collected rather obtained from published or unpublished sources are called secondary data.
Secondary data are collected by an individual or an institution for some purpose and are used by someone else in another context.
Primary data are reliable and relevant because they are original in character and are collected by some individuals or by research bodies.
Q4. Why do we group data?
Solution 4: The data obtained in original form are called raw data .Raw data does not give any useful information and is rather confusing to mind. Data is grouped so that it becomes understandable and can be interpreted. According to various characteristics groups are formed by us. After grouping the data, we are in position to make calculations of certain values which will help us in describing and analyzing the data.
Q5. Explain the meaning of the following terms:
(1)Variate
(2)Class-integral
(3)Class-size
(4)Class-mark
(5) Frequency
(6)Class limits
(7)True class limits
Solution 5:
(1) Variate
Any character that can vary from one individual to another is called variate.
(2)Class interval
In the data each group into which raw data is considered is called a class-interval.
(3)Class-size
The difference between the true upper limit and lower limit is called the class size of that class size.
(4)Class-mark
The middle value of the class is called as the class mark.
Class mark =
(5)Frequency
The number of observations corresponding to class is called its frequency.
(6)Class limits
Each class is bounded by two figures ,called the class limits .The figures on the left side of the classes are called lower limits while figures on the right side are called upper limits.
(7)True class limits
If classes are inclusive.Eg:15-19,20-24,25-29….
Then, true lower limit of class = upper limit of class-0.5
Eg:- True limits of the class is 15-19 are 14.5 and 19.5
But if classes are exclusive like 10-20,20-30,30-40…
Here class limits and true class limits are the same.
Q6.The ages of ten students of a group are given below.The ages have been recorded in years and months:
8-6,9-0,8-4,9-3,7-8,8-11,8-7,9-2,7-10,8-8
(i)What is the lowest age?
(ii)What is the highest age?
(iii)Determine the range?
Solution 6:
The ages of ten students of a group are given below
8-6,9-0,8-4,9-3,7-8,8-11,8-7,9-2,7-10,8-8.
(1)Lowest age is 7 years 8months
(2)Highest age is 9 years,3 months
(3)Range = Highest age-lowest age = 9 year,3 months,7 years,8 months = 1 year 7 months
Q7.The monthly pocket money of six friends is given below:
Rs 45,Rs 30, Rs 40, Rs 25, Rs 45.
(i)What is the lowest pocket money?
(ii)What is the highest pocket money?
(iii)What is the range?
(iv)Arrange the amounts of pocket money in ascending order
Solution 7:
The monthly pocket money of six friends is given below:
Rs45,Rs 30, Rs 40, Rs 25, Rs 45.
(1) Highest pocket money = Rs 50
(2) Lowest pocket money = Rs 25
(3) Range = 50-25 = 25
(4) The amounts of pocket money in an ascending order is: Rs 25,Rs 30, Rs 40, Rs 45, Rs 45,Rs 50.
Q8.Write the class -size in each of the following:
1)0-4, 5-9, 10-14
(2) 10-19, 20-29, 30-39
(3) 100-120, 120-140, 160-180
(4) 0-0.25, 0.25-00.50, 0.50-0.75
(5) 5-5.01, 5.01-5.02, 5.02-5.03.
Solution 8:
(1) 0-4 ,5-9, 10-14
True class limits are 0.5-4.5,4.5-9.5,9.5-14.5
Therefore class size = 14.5-9.5 = 5
(2) 10-19 , 20-29 , 30-39
True class limitsà 19.5-19.5,19.5-29.5,29.5-29.5
Class size = 39.5-29.5 = 10
(3) 100-120 , 120-140, 160-180
Here the class limits and true class limits are the same
Therefore class size = 120-100 = 20
(4) 0-0.25,0.25-00.50,0.50-0.75
Here the class limits and true class limits are the same
Therefore class size = 0.25-0 = 0.25
(5) 5-5.01,5.01-5.02,5.02-5.03.
Here the class limits and true class limits are the same
Therefore class size = 5.01-5.0 = 0.01.
Q9. The final marks in mathematics of 30 students are as follows:
53,61,48,60,78,68,55,100,67,0,75,88,77,37,84,58,60,48,62,56,44,58,52,64,98,59,70,39,50,60.
(i)Arrange these marks in ascending order,30 to 39 one group,40 to 49 second group etc.
Now answer the following:
(ii)What is the lowest score?
(iii)What is the highest score?
(iv)What is the range?
(v)If 40 is the pass mark how many failed?
(vi)How many have scored 75 or more?
(vii)Which observations between 50 and 60 have not actually appeared?
(viii) How many have scored less than 50?
Solution 9:
The final marks in mathematics of 30 students are as follows:
53,61,48,60,78,68,55,100,67,0,75,88,77,37,84,58,60,48,62,56,44,58,52,64,98,59,70,39,50,60.
(1)
Group Class Observations I 30-39 37,39 II 40-49 44,48,48 III 50-59 50,52,53,55,56,58,58,59 IV 60-69 60,60,60,61,62,64,67,68 V 70-79 70,55,77,78 VI 80-89 84,88 VII 90-99 90,98 VIII 100-109 100
(2) Highest score = 100
(3)Lowest score = 37
(4)Range = 100-37 = 63
(5)If 40 is the passing marks,2 students have failed
(6) 8 students have scored 75 or more
(7)Observation 51, 54, 57 between 50 and 60 has not actually appeared.
(8) 5 students have scored less than 50
Q10. The weights of new born babies are as follows: 2.3,2.2,2.1,2.7,2.6,2.5,3.0,2.8,2.8,2.9,3.1,2.5,2.8,2.7,2.9,2.4.
(i)Rearrange the weights in descending order.
(ii)What is the highest weight?
(iii)What is the lowest weight?
(iv)Determine the range?
(v)How many babies were born on that day?
(vi)How many babies weigh below 2.5 Kg?
(vii) How many babies weigh more than 2.8 Kg?
(viii)How many babies weigh 2.8 Kg?
Solution 10:
The weights of new born babies(in kg) are as follows
2.3,2.2,2.1,2.7,2.6,2.5,3.0,2.8,2.8,2.9,3.1,2.5,2.8,2.7,2.9,2.4.
(1) The weights in descending order
3.1,3.0,2.9,2.9,2.8,2.8,2.7,2.7,2.6,2.5,2.5,2.4,2.3,2.2,2.1.
(2) The highest weight = 3.1 Kg
(3) The lowest weight = 2.1 Kg
(4) Range = 3.1-2.1 = 1.0 Kg
(5) 15 babies were born on that particular day.
(6)4 babies weight below 2.5 Kg.
(7) Weight more than 2.8 Kg are 4 babies.
(8) Weightà2 babies
Q11. The number of runs scored by a cricket player in 25 innings is as follows : 26,35,94,48,82,105,53,0,39,42,71,0,64,15,34,15,34,6,71,0,64,15,34,15,34,67,0,42,124, 84,54,48,139,64,47
(i)Rearrange these runs in ascending order.
(ii)Determine the player, is highest score.
(iii)How many times did the player not score a run?
(iv)How many centuries did he score?
(v)How many times did he score more than 50 runs?
Solution 11:
The numbers of runs scored by a player in 25 innings are
26,35,94,48,82,105,53,0,39,42,71,0,64,15,34,15,34,6,71,0,64,15,34,15,34,67,0,42,124,84,54,48,139,64,47.
(i) Runs in an ascending order are
0,0,0,0,6,15,15,15,15,26,34,34,34,34,35,39,42,42,47,48,48,53,54,64,64,64,67,71,71,82,90,124,139.
(ii) The highest number = 139
(iii) The player did not score any runs 3 times
(iv) He scored 3 centuries.
(v) He scored more than 50 runs 12 times.
Q12. The class size of distribution is 25 and the first class-interval is 200-224. There are seven class-intervals.
(i)Write the class-intervals.
(ii)Write the class marks of each interval.
Solution 12:
Given
Class size = 25
First class interval = 200-224
(i)Seven class interval are:
200-240,225-249,250-274,275-299,300-324,325-349,350-374.
(ii) Class mark 200-224 =
= 212
Class mark 225-249 =
= 237
Class mark 250-274 =
= 287
Class mark 300-324 =
= 312
Class mark 325-349 =
= 337
Class mark 350-374 =
= 362
The document RD Sharma Solutions -Ex-22.1 (Part - 1), Tabular Representation Of Statistical Data, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9 is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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## RD Sharma Solutions for Class 9 Mathematics
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# 16/29 as a Decimal
16/29 as a decimal expansion provides the detailed information about what is 16/29 in decimal form, and the answer with steps help students to easily understand how it is being calculated.
16/29 as a Decimal Expansion
16/29 = 0.5517
Hence,
16/29 as a decimal equals to 0.5517
where,
16/29 is a given fraction,
The decimal expansion of 16/29 is 0.5517
16/29 as a Mixed Number
The given fraction 16/29 can't be represented as a mixed number since the numerator 16 of the given fraction is smaller than the denominator 29.
For values other than 16/29, use this below tool:
## How-to: 16/29 as a Decimal
The below work with steps provide the detailed information about how to convert fraction 16/29 as a decimal equivalent.
step 1 Address the input parameters and observe what to be found:
Input values:
The fraction = 16/29
What to be found:
Find the decimal expansion of fraction 16/29.
step 2 The numerator is smaller than the denominator of the fraction 16/29, so find how many 10s should be multiplied with both numerator and denominator. Since the number of digits are equal in both numerator and denominator, multiply 100 with both numerator and denominator of 16/29.
= 16/29 x 100/100
step 3 Rearrange the fraction as like the below:
= 100 x 16/29 x 1/100
= 1600/29 x 1/100
step 4 Simplify the above expression further:
= 1600/29 x 1/100
= 1595 + 5/29 x 1/100
= ( 1595/29 + 5/29) x 1/100
= (55 + 5/29 ) x 1/100
step 5 Repeat the step 2 to find the decimal equivalent for the fraction 5/29 in the above expression:
= (55 + 5/29 ) x 1/100
= (55 + 0.172) x 1/100
step 6 Simplify the above expression further:
= (55 + 0.172) x 1/100
= 55.172 x 1/100
= 55.172/100
16/29= 0.55172
Hence,
16/29 as a decimal is 0.55172
## Quick Reference Table: 16/29 as a Decimal
The below quick reference table for the nearest values of fraction 16/29 provides the equivalent value in decimal form.
Fractionas a Decimal
19/290.6552
18/290.6207
17/290.5862
16/290.5517
15/290.5172
16/260.61538
16/270.59259
16/280.57143
16/300.53333
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Q.3 Solve the following equations by trial and error method: (i) 5p + 2 = 17 (ii) 3m – 14 = 4
(i) 5p + 2 = 17
For p = 1, LHS
= 5 × 1 + 2 = 5 + 2 = 7 $$\neq$$RHS=17
For p = 2, LHS = 5 × 2 + 2
= 10 + 2 = 12 $$\neq$$ RHS=17
For p = 3, LHS = 5 × 3 + 2
= 15 + 2 = 17 = RHS
Since the given equation is satisfied for p = 3 Thus, p = 3 is the required solution.
(ii) 3m – 14 = 4
For m = 1, LHS = 3 × 1 – 14
= 3 – 14 = -11 $$\neq$$ RHS = 4
For m = 2, LHS = 3 × 2 – 14 = 6 – 14 = -8 $$\neq$$ RHS= 4
For m = 3, LHS = 3 × 3 – 14 = 9 – 14 = -5 $$\neq$$ RHS =4
Form m = 4, LHS = 3 × 4 – 14 = 12 – 14 = -2 $$\neq$$ RHS= 4
For m = 5, LHS = 3 × 5 – 14 = 15 – 14 = -1 $$\neq$$ RHS =4
For m = 6, LHS = 3 × 6 – 14 = 18 – 14 = 4 =RHS
Since, the given equation is satisfied for m = 6. Thus, m = 6 is the required solution.
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How to Calculate the Square Root of Any Number by Hand
Calculating square roots by hand is not even what a practising mathematician would call a leisurely pastime. The procedure is hardly, if ever, taught in school in this age of computers. It is, however, quite possible to determine the square root of a number to any degree of accuracy for which a person is willing to stretch the work, and the mathematics involved in the procedure is not complex. A brief glimpse of a hand-calculated square root problem appears very much like a long division problem, and some aspects of the procedure are similar.
Step One - The Set-up
To illustrate these steps, we will find the square root of 1,234.56.
Write the number for which the square root is to be calculated under a square root sign (√), elongated if desired to simulate the appearance of long division. Locate and mark the decimal point of this number. Above the radical sign1, place another decimal point directly above the decimal point of the original number. This second point will become the decimal point in the square root of the original number. Note that since the handwritten work will look very similar to a long division problem, care should be given in allowing for space to write products, take differences, and carry down digits just as in long division.
On both sides of the decimal point in the original number, use some sort of marking system to pair off adjacent digits of the number. On the whole number side (the left side of the decimal point), cease pairing off digits when there are no digits or one digit left at the front of the number. On the fractional side (the right side of the decimal point), pair off digits using zeroes as placeholders if necessary. The digits in our example are paired off, with apostrophes as separators.
Each pairing of digits on the fractional (right) side will constitute one digit of accuracy for the final square root calculation, so if exceptional accuracy is desired, leave plenty of room on the right side of the paper for additional work. In the above example as it is presented, one would be able to calculate the square root of 1234.56 to four decimal places beyond the decimal point2. As many additional pairs of zeroes could be added to the end of the number as one wished to calculate.
Step Two - First Calculation
The calculation begins. Look at the very first pairing at the front of the number, similar in a sense to the way one looks at a number when beginning long division. This first pairing will either consist of one or two digits - for procedural purposes, it makes no difference. In our example, the first pairing is '12'. Determine the largest perfect square number (1, 4, 9, 16, 25, 36, 49, 64, and 81 will be your only choices for this first calculation) which is less than or equal to the value of the first pairing. In this case, the largest perfect square less than 12 is 9. Write this square number underneath the first pairing in preparation for subtraction, and write the square root of this perfect square above the radical sign, directly above the first pairing.
Subtract the perfect square from the value of the first pairing and write the difference below (12 - 9 = 3). Carry the next pairing of digits in the original number (34) down to the difference that was just written, and let the number comprising all these digits (334) be called D, for the sake of later referral.
So far, the look of the problem should seem very much like a long division problem as promised. It helps to conform to this style of presentation, since keeping track of differences and products will be as vital here as in division.
Step Three - The Trial-by-Error Repeating Step
The mathematics involved thus far has been relatively simple. The first spot of good news is that it only gets slightly more difficult in this third step. The second spot of good news is that with this somewhat more lengthy step, the procedure will begin to repeat indefinitely, until the degree of accuracy desired is finally obtained.
Multiply the number above the square root sign by 2, treating it as a whole number. (Ignore where the decimal might reside.) In our example, we have 3 above the radical, so we multiply it by 2 to obtain 6. Never mind that the decimal has not yet been reached. Place this product to the left of the number D. Leave space at the end of this product for one additional digit - our number is 6, so we write '6_'.
A digit between 0 and 9 must be chosen, at random if one wishes. Let's call it X. This digit will be written in 2 places: once above the square root sign, directly over the pairing that was just carried down (above the 34), and once in the blank space at the end of the product that was just calculated (after the 6). Let the letter Y denote the number thus created. In our example, if '7' is chosen for X, then Y will be 67.
Multiply X and Y. What is being sought is the largest possible value of X for which the product of X and Y will still be less than D. Choosing X to be as large as possible is purely a matter of trial and error. A glimmer of hope: there are only 10 choices for X, so hang in there! We can see that 7 is too big, because 7 * 67 = 469, which is greater than D (334). A little experimentation will show that the largest X we can choose is 5, making Y=65. (5 * 65 = 325 < 334)
Once the correct value of X is found and the product of X and Y is calculated, this product is written underneath D, and subtracted to obtain a new difference. Then next pair of digits from the original number are then carried down to obtain a new D.
Now Step Three repeats, until the degree of accuracy desired is obtained. To continue our example, the number we write to the left of the new D is 35 * 2, which equals 70. We leave a space at the end and write '70_'. The X that will fill in the blank is 1, because 1 * 701 < 956, while any other choice of X is too big. The product of 1 and 701 is written under D and subtracted, and another pairing is brought down to obtain the next D.
Worked out through all four decimal places, our example should look like this:
So, the square root of 1234.56, to four decimal places, is 35.1363. We can be sure about three of those decimal places, while the last '3' might round up, depending on the next digit.
More Examples
Here is a number (13,234.251) where the first pairing consists of a single digit.
Here is a well-known square root:
Finally, an example that comes out to a whole number:
1 Another name for the square root sign.
2 Only three of those digits will be significant, and the fourth may be used for rounding.
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Entry Data
Entry ID: A827453 (Edited)
Written and Researched by:
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Edited by:
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Date: 30 October 2002
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# Implicit Differentiation
Implicit differentiation is the main type of differential calculus. It is widely used to find the derivative of one variable with respect to another variable. The derivative is a major topic of calculus used to calculate the rate of change in the given function.
The other type of derivative is explicit differentiation. Both types are used to solve the problems of differential calculus. In this post, we will discuss the definition, methods, and examples of implicit differentiation.
Index
## What is implicit differentiation?
Implicit differentiation is a branch of differentiation in which you can calculate the derivative of an equation. In this type of derivative, two variables are used like x and y. These variables behave as one is the function of the other and you have to calculate dy/dx of the given function.
In implicit differentiation, the term y with respect to x is not considered as constant. The derivative of y2 in the implicit differentiation must be 2y dy/dx. All the formulas and rules remain the same in this type of differentiation.
The implicit differentiation can be defined as calculating the derivative of y with respect to x without solving the given equation for y. The main purpose of this type of differentiation is to calculate the dy/dx term of the given equation. More simply, we can say that implicit differentiation is used to calculate the differential of the inverse function.
### Rules of implicit differentiation
Rules of differentiation remains unchanged for implicit differentiation.
Sum rule
d/dx (u + v) = d/dx (u) + d/dx (v)
Difference rule
d/dx (u – v) = d/dx (u) – d/dx (v)
Product rule
d/dx (u * v) = u d/dx (v) + v d/dx (u)
Quotient rule
d/dx (u / v) = 1/v2 [v d/dx (u) – u d/dx (v)]
Power rule
d/dx (un) = n (u)n-1 d/dx (u)
By using rules and formulas of differentiation, we can easily find the implicit differentiation of the given equation.
## How to calculate the implicit differentiation?
Implicit differentiation of the given function can be calculated by treating x and y as variables and applying the rules of differentiation. You can also use an implicit differentiation calculator for getting the step-by-step solution of the given problems. In just a single click, you can calculate the result. You have to put the question into the input box like:
Then click the “calculate” button the result will show below the calculate button in a couple of seconds.
By pressing show steps, you can see the step-by-step solution of the given problem.
Example 1
Calculate the derivative of the given function with respect to x, 9x2y + 17xy2 – 19xy3 = 3x5 / 3x2 + 17x + 2y + 3?
Solution
Step 1: First of all, write the given equation.
9x2y + 17xy2 – 2y3 = 3x5 / 3x2 + 17x + 2y + 3
Step 2: Now apply the differential operator on both side in the given equation.
d/dx (9x2y + 17xy2 – 2y3) = d/dx (3x5 / 3x2 + 17x + 2y + 3)
Step 3: Apply the difference, product, sum, and quotient rules on the above equation.
d/dx (9x2y) + d/dx (17xy2) – d/dx (2y3) = d/dx (3x5 / 3x2) + d/dx (17x) + d/dx (2y) + d/dx (3)
9x2 d/dx (y) + y d/dx (9x2) + x d/dx (17y2) + y2 d/dx 17x – d/dx (2y3) = 1/(3x2)2 [3x2 * d/dx (3x5) – 3x5 d/dx (3x2)] + d/dx (17x) + d/dx (2y) + d/dx (3)
Step 4: Now apply the constant and power rules on the above equation.
9x2 d/dx (y) + 9y d/dx (x2) + 17x d/dx (y2) + 17y2 d/dx (x) – 2d/dx (y3) = 1/(3x2)2 [3x2 * 3d/dx (x5) – 3x5 3d/dx (x2)] + 17d/dx (x) + 2d/dx (y) + 0
9x2 dy/dx + 9y (2x2-1) + 17x (2y2-1) dy/dx + 17y2 (x1-1) – 2 (3y3-1) = 1/(9x4) [9x2 * (5x5-1) – 9x5 (2x2-1)] + 17 (x1-1) + 2dy/dx
9x2 dy/dx + 18xy + 34xy dy/dx + 17y2 – 6y2 = 1/(9x4) [9x2 * (5x4) – 9x5 (2x)] + 17 + 2dy/dx
9x2 dy/dx + 18xy + 34xy dy/dx + 17y2 – 6y2 dy/dx = 1/(9x4) [45x6 – 18x6] + 17 + 2dy/dx
9x2 dy/dx + 18xy + 34xy dy/dx + 17y2 – 6y2 dy/dx = 1/(9x4) [27x6] + 17 + 2dy/dx
9x2 dy/dx + 18xy + 34xy dy/dx + 17y2 – 6y2 dy/dx = 3x2 + 17 + 2dy/dx
Step 5: Take the dy/dx term on one side and take it as common.
9x2 dy/dx + 34xy dy/dx – 6y2 dy/dx – 2dy/dx = 3x2 + 17 – 18xy – 17y2
(9x2 + 34xy – 6y2– 2) dy/dx = 3x2 + 17 – 18xy – 17y2
dy/dx = (3x2 + 17 – 18xy – 17y2) / (9x2 + 34xy – 6y2– 2)
Example 2
Calculate the derivative of the given function with respect to x, 3xy2 + 4x2y – 13y = 3x5 * 19y2 + 34x + 2?
Solution
Step 1: First of all, write the given equation.
3xy2 + 4x2y – 13y = 3x5 * 19y2 + 34x + 2
Step 2: Now apply the differential operator on both side in the given equation.
d/dx (3xy2 + 4x2y – 13y) = d/dx (3x5 * 19y2 + 34x + 2)
Step 3: Apply the difference, product, sum, and quotient rules on the above equation.
d/dx (3xy2) + d/dx (4x2y) – d/dx (13y) = d/dx (3x5 * 19y2) + d/dx (34x) + d/dx (2)
3x d/dx (y2) + y2 d/dx (3x) + 4x2 d/dx (y) + y d/dx (4x2) – d/dx (13y) = 19y2 d/dx (3x5) + 3x5 d/dx (19y2) + d/dx (34x) + d/dx (2)
Step 4: Now apply the constant and power rules on the above equation.
3x (2y2-1) dy/dx + y2 (3(1)) + 4x2 dy/dx + y ((4 * 2) x2-1) – 13 dy /dx = 19y2 ((3 * 5) x5-1) + 3x5 ((19 * 2) y2-1) dy/dx + (34(1)) + 0
3x (2y dy/dx) + y2 (3) + 4x2 dy/dx + y (8x) – 13 dy /dx = 19y2 (15x4) + 3x5 (38y) dy/dx + (34(1))
6xy dy/dx + 3y2 + 4x2 dy/dx + 8xy – 13 dy /dx = 285x4y2 + 114x5y dy/dx + 34
Step 5: Take the dy/dx term on one side and take it as common.
6xy dy/dx + 4x2 dy/dx– 13 dy /dx – 114x5y dy/dx = 285x4y2 + 34 – 8xy – 3y2
(6xy + 4x2 – 13 – 114x5y) dy/dx = 285x4y2 + 34 – 8xy – 3y2
dy/dx = (285x4y2 + 34 – 8xy – 3y2) / (6xy + 4x2 – 13 – 114x5y)
## Conclusion
Implicit differentiation is used to find the values of dy/dx by applying rules to the given equation without solving the terms of y. You can easily solve the problems of implicit differentiation, once you grab all the basics of implicit differentiation.
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# 10. Vector Calculus
## Rate of Change of Variable Vectors
Rate of Change of a Position Vector in Elliptical Motion
We saw in Variable Vectors how vectors can vary with time. In this section, we learn how to find the rate of change of such varying vectors.
To find the time rate of change of a vector, we simply differentiate each component.
### Example 1
Let's consider the 2-dimensional force vector example from before:
F = (3t2 + 5) i + 4t j
The time rate of change of this vector is given by the derivative with respect to t of each component.
(dbb{text(F)})/dt=6t\ bb{text(i)}+4\ bb{text(j)}
At time t = 5, the rate of change of the vector F is the vector 30 i + 4 j. The units will be N/s.
### Example 2
The 3-D acceleration vector we met earlier in Example 2, Variable Vectors was given by
A = (5t) i + (2t + 3) j + (t2 + 10) k
The rate of change of this vector is given by differentiating each term, as follows:
(dbb{text(A)})/dt=5\ bb{text(i)}+2\ bb{text(j)} +2t\ bb{text(k)}
The units will be m/s3.
At the specific time t = 4 s, the rate of change of the vector will be
5 i + 2 j + 8 k m/s3
## Rate of Change of a Position Vector in Elliptical Motion
Next we consider the case when the terminal point of a vector is moving in an ellipse.
### Example 3
The following vector (units in m) follows such a pattern at time t (units in s):
P = (3 cos t) i + (sin t) j
This expression is based on the expression for a circle , P = (cos t) i + (sin t) j. The 3 in the i term stretches the circle into an ellipse. Of course, the time t is measured in radians, not degrees.
Following is the graph of the motion of the terminal point of the ellipse.
The terminal point starts at t = 0 s at the position (3, 0) and proceeds in an anti-clockwise direction. Its position at various times is indicated on the graph.
This idea of a curve being generated as a point moves in time is the same concept as Parametric Equations that we came across before.
The resulting vector has initial point at the origin as above. The vector for the cases t = 0 s (magnitude 3 m, direction horizontal, to the right), t = 1 s and t = 2 s are shown below:
To find the time rate of change of the position vector in elliptical motion, we differentiate the terms as we did earlier.
Since
P = 3 cos t i + sin t j
then
(dbb{text(P)})/dt=-3\ sint\ bb{text(i)}+cost\ bb{text(j)}
Note that the magnitude of the vector changes as time goes on because the terminal point is moving around the ellipse.
To find the rate of change at particular times, we substitute in values of t.
At t = 0 s,
rate of change = 0 i + j = j
Considering our diagram above, this answer makes sense. When the terminal point starts to move, it is in the vertical direction only (no horizontal component is present).
At t = 1 s, we expect a negative horizontal component and a positive vertical component (the terminal point is moving up and to the left at t = 1 s) . Substituting t = 1 (in radians, of course) gives us:
rate of change = (-3 sin 1) i + (cos 1) j = -2.52 i + 0.54 j
The x-value is negative and the y-value is positive, as expected.
At t = 6 s, we expect a positive horizontal component (the terminal point is moving to the right) and a positive vertical component (the terminal point is moving up).
rate of change = (-3 sin 6) i + (cos 6) j = 0.84 i + 0.96 j
Both components are positive, as expected.
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Thursday, December 14, 2006
Andrea's Percentage Growing Post =)
Question 1:
What is a good definition of percent? You should use words symbols, pictures and numerical examples in your definition.
A percent is a ratio whose second term is 100. Percent means parts per hundred. The word percent comes from the Latin phrase per centum, which means per hundred. In mathematics, we use the symbol % for percent.
---
Question 2:
How are three fifths (3/5), 3:2, 60% and 0.6 all the same? Use pictures and words to show your answer.
They are all the same because...
» If you express 3/5 into a decimal, the answer will be 0.6. How? We divide 3 to 5 to get 0.6.
» If you express 3/5 into a ratio, the answer will be 3:2. How? The numerator becomes the first part of the ratio, and the denominator is being subtracted to the numerator (5-3 = 2).
» If you express 0.6 into a percent, it will be 60%. How? To change a decimal to a percent, we multiply by 100. This is the same as moving the decimal point two places to the right. To change the decimal 0.6 to a percent, we move the decimal point two places to the right and get 60%.
So 3/5, 3:2, 60% and 0.6 are all equivalents!
---
Question 3:
Show three different ways to find 35% of 80.
a.)
35% of 80
80 Divided By 100 = 0.8 or 1%
35% x 1% = 35%
35% x 0.8 = 28
35/100 = 28/80
b.)
100% - 35% = 65%
65% = 0.65
0.65 x 80 = 52
80 - 52 = 28
c.)
35% = 0.35
80 Divided By 0.35
= 28
BubbleShare
---
Question 4:
Check this out. It talks about...
» How to change a percent into a decimal
» How to change a decimal into a percent
» And how to change a percent into a fraction.
The post shows pictures-slash-diagrams to make us understand more.
---
Question 5:
The principal announced that 50% of the children in Ms. Stanzi's class met their reading goal for the month and that 55% of the children in Ms. Lowrey's class met their reading goal for the month. Ms. Stanzi said that a greater number of her students met their reading goal. Could Ms. Stanzi be correct? Why or Why not.
After reading the question, I came up with two scenarios:
a.) Ms. Stanzi could be right if she has more students than Ms. Lowrey. No matter how many students (in percent) Ms. Lowrey has, Ms. Stanzi's class would still have greater number of students that met their reading goals.
b.) On the other hand, Ms. Stanzi might not be right if she has the same number of students as Ms. Lowrey's. Then Ms. Lowrey's class would have the greater number of students because 55% of those have done their reading goals, according to the principal.
---
Question 6:
Use a hundred grid (unit square) to illustrate the following questions. Once you have explained and illustrated what the question means, solve it.
a) 16 is 40% of what number?
BubbleShare
I got 40 as an answer because:
First, I changed 40% into a decimal. The answer is 0.4 (I removed the zero at the end). Secondly, I divided 16 to 0.4, which gave me the answer 40.
b) What is 120% of 30?
We have to fill one whole grid and another 20 little squares on another grid to make 120. One grid consists a hundred teeny, weeny, itsy, bitsy little sqaures.
We all know that one grid is equal to 30.
120% of 30
30 Divided By 100 = 0.3 (is equal to 1%)
120% x 1% = 120%
120% x 0.3 = 36
BubbleShare
Therefore, 120% of 30 is 36.
---
Question 7:
Marked other assignments and left mark as comments. (You must leave links in your growing post on the posts that you have marked.) There may only be two marks per student. If that student already has 2 marks you must choose another. ;)
[Mark as many as you want..]
I chose to mark Boeun's, Rivka's, Jorel's, Camille's and Michael's Growing Post.
Now I am marking my own growing post:
Setup
5/5
Comments: I have the right title and I think I chose colours that made the questions different from my answers.
Question 1
10/10
Comments: I think I did pretty good on this part since it was my first time to use the software, Gliffy. Plus, I added a written (I mean typed) definition of percent.
Question 2
10/10
Comments: I'd give myself a 10 because I spent time (and effort) to make the picture look nice and I gave a pretty good explanation about the equivalents.
Question 3
9/10
Comments: I think I did well on the Bubbleshare (and since it's my first time to use such a software). But I think I need to give more details.
Question 4
5/5
Comments: Good choice of site. And I think I did my best on the explaining part to make it longer, but that's all I can say about that site.
Question 5
10/10
Comments: I really, really exerted effort on finding the right words that fits in this question. And I think I explained it well.
Question 6
14/15
Comments: I made a diagram to make it understandable. And I did my best explaining on how I found the answer. But still, I think I am lack of details.
Overall Mark
63/65
Overall Comment for the Growing Post: I gave myself a pretty good score because I did a lot of work and I spent lots of time. I did everything I can in explaining and exerted lots of effort in doing the diagrams in Bubbleshare and Gliffy. But I promise to improve my explanations next time.
---
Thank you for giving me a chance to answer this growing post. I learned a lot. I hope you learned something, too. Feel free to comment! :)
Mr. H said...
Andrea you have been part of our class for under a week. This is amazing. Thank you for putting so much effort into your work.
This is great
Harbeck
Andrea =) said...
Thanks Mr. H for the encouraging comment.
I am doing the best I can. :D
Thanks again! =)
-Andrea
boeun said...
Wow, i really liked your grwoing post Andrea. I feel like you are ahead of me!!:P
Keep up the good works!!
charmaine . said...
greaaat post andrea ! (:
Andrea =) said...
hahaha thanks boeun && charmaine. i am psyched. lol. :D
boeun said...
This comment has been removed by the author.
boeun said...
This comment has been removed by the author.
Andrea =) said...
This comment has been removed by the author.
Andrea =) said...
thanks a lot boeun for ur helful comment. i'll go change my font colors now. lol. and I did make my diagram in question no.1 in gliffy. XD =)
boeun said...
Setup
5/5
Question 1
8/10
Question 2
8/10
Comments: I don't know if your picture is Gliffy or not because I can't see on my computer. I'll change your mark later if the picture was from Gliffy or in Bubbleshare.
Question 3
9/10
Question 4
4/5
Comments: I would tell more about how they solved problems or what they could improve on.
Question 5
8/10
Question 6
14/15
Overall Mark
59/65
Overall Comment for the Growing Post:
Wow! You are catching up so fast! I think you should explain more about your answer. For example, how you got that answer and why do you think that is the right answer. Keep up the good work!!
Andrea =) said...
ok boeun thanks for marking my post. awwww lol. and, it really is from GLIFFY!!! promise!! =) lol thanks agen ! happy holidays!!
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# One-Sided Limits
When crossing the street, you look to the left and right to ensure there are no problems before walking across. When finding the limits of the function, you do the same thing, look to the left and right!
## One-Sided Limits in Calculus
Generally, we call these the limit from the left or the limit from the right because you are specifically looking to the left or right of a specific point.
For a review of the definition of the limit of a function, see Limits of a Function.
## Definition of One-sided Limits
How can we formally define "one-sided limits" in calculus? Let's have a look!
We say that is the left limit of a function at if we can get as close to as we want by taking on the left side of , and close to but not equal to . It is written
.
This is also called the limit from the left of a function. You can also look at the limit from the right of a function.
We say that is the right limit of a function at if we can get as close to as we want by taking on the right side of , and close to but not equal to . It is written
.
The nice thing is that if the limit of the function exists, then both the limit from the left and right exist and are the same.
Each of the results below follow from just the definition of the limit, the left limit, and the right limit. They are an immediate consequence of the definitions and so don't require a fancy proof.
1. Suppose that
where is a real number. Then
and .
2. Similarly, if
then
.
Note this gives you a handy way to tell if the limit doesn't exist, just by using the contrapositive of part 2.
3. If
then
does not exist.
You can read as " approaching from the right", and as " approaching from the left".
## Finding One-Sided Limits
So how do you figure out what a function's left or right limit is? You can determine one-sided limits by looking at:
1. The graph of a function, OR
2. A table of function values
So let's look at a specific example.
Using the function
,
find
and .
What does this tell you about
?
First, let's look at the limit from the left. In the graph below, you can see the function, a table of function values that are getting closer to from the left-hand side, and the points in the table plotted on the graph.
Limit from the left using a graph and table | StudySmarter Original
As you can see from the graph above, as , all of the function values are equal to . Therefore
.
Now instead, let's look at the limit from the right. In the graph below, you can see the function, a table of function values that are getting closer to from the right-hand side, and the points in the table plotted on the graph.
Limit of a function from the right using a graph and table | StudySmarter Original
As you can see from the graph above, as , all of the function values are equal to . Therefore
.
Finally, since you know that
,
you also know that
does not exist.
## Examples of One-Sided Limits
Let's look at more examples of determining one-sided limits.
Consider the function
.
Find the limits from the left and right of .
Rather than thinking about this function as one with an absolute value, it can help to think about possible values for . Let's look at the 3 possible cases here:
1. When , this function is not defined.
2. When is negative, .
3. When is positive, .
So you can instead think of this as the piecewise-defined function:
.
This is very similar to the previous example. In fact
and
For the function in the picture below, determine the following (if it exists):
1. , , and
2. the limit from the left at , , , and
3. the limit from the right at , , , and
4. the limit at , , , and
Finding One-Sided limits from a graph | StudySmarter Original
1. This part is just looking for the function values at these points. So looking at the graph, , , and .
2. Remember that when you are finding the limit from the left, you only look at points on the graph that are to the left of the point you care about. So using the graph,
, , , and .
3. When you are finding the limit from the right, you only look at points on the graph that are to the right of the point you care about. So using the graph,
, , , and .
4. The limit will exist only in cases where the limit from the left and the limit from the right are the same. Otherwise, the limit doesn't exist. Looking at the information in parts 2 and 3 above, that means that the limit exists at and at . You can also say that
and that .
Notice that the fact that the limit exists is independent of the actual function value at the point, or even if the function is defined there.
In addition, the limit does not exist at and .
## One-Sided Limits and Vertical Asymptotes
One question that still needs to be answered. How do we evaluate the left and right limits of a function at a vertical asymptote? The process for finding the limits from the left and right when there is a vertical asymptote is exactly the same as at any other point. Let's look at an example.
Consider the function
.
Find
and .
First, let's think about the limit from the left. Look at the graph and table below.
Limit from the left at a vertical asymptote | StudySmarter Original
As you can see from both the graph and table, as you take values that get closer and closer to from the left, the function values become further and further away from the axis, and are all negative. So you would say that in fact there is no number that is the limit from the left. When this happens you can say that "the limit from the left diverges to negative infinity", and write it as
.
This may seem odd given that limits usually have to be numbers, but the notation is just saying that the function values to the left at zero, but close to zero, can be as large a negative value as you want them to be.
When we say the limit equals , it is just another way of saying the limit does not exist, just being a bit more specific!
Now let's think about the limit from the right. Look at a graph and table below.
The limit from the right of a vertical asymptote | StudySmarter Original
As you can see from both the graph and table, as you take values that get closer and closer to from the right, the function values become further and further away from the axis, and are all positive. So you would say that in fact there is no number that is the limit from the right. When this happens you can say that "the limit from the right diverges to infinity", and write it as
.
This may seem odd given that limits usually have to be numbers, but the notation is just saying that the function values to the left at zero, but close to zero, can be as large a positive number as you want them to be.
If instead of vertical asymptotes you are interested in the limit as , also known as limits at infinity, see Infinite Limits
There may be cases where the limit from one side exists, but does not exist from the other side. We see this in the example below.
For the function in the graph below, find
and .
Limit from one side exists but limit from other side doesn't | StudySmarter Original
From the picture above, we see that to the left of the function values get closer and closer to as . That means
.
However, if you look at values to the right of , the function values get larger and larger as . That means
.
Looking at the examples above, you can draw some helpful conclusions:
1. If
or if
then the function has a vertical asymptote at .
2. If the function has a vertical asymptote at then either
or .
## One-Sided Limits - Key takeaways
• We say that is the left limit of a function at if we can get as close to as we want by taking on the left side of , and close to but not equal to . It is written
.
• We say that is the right limit of a function at if we can get as close to as we want by taking on the right side of , and close to but not equal to . It is written
.
• Suppose that where is a real number. Then and .
• If
then
.
• If
then
does not exist.
• If
or if
then the function has a vertical asymptote at .
• If the function has a vertical asymptote at then either
or .
You can use a graph, a table of function values, or the properties of limits.
It is when you look at x values which are only to the left or right of the point you care about, not at both at the same time.
Graph the function near the point you care about. Then only look to the left or right of the point, depending on if you are looking for the limit from the left or the limit from the right.
If you can show that the limit from the left at a point is not the same as the limit from the right at that point, then you know that the limit of the function at that point doesn't exist.
You can do it algebraically for some functions, or using the properties of limits, or theorems like the Squeeze Theorem.
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# How to Turn A Graph into An Equation?
TopLine equation represents the relation between two variables. This equation is called as linear equation. In this equation value of one variable depends upon another. Let us see how to turn a graph into an equation.
First of all we are requiring a graph. Now we will determine the starting and ending points of line from graph. As we get the coordinates of line we can convert it into an equation with help of a graph. Let us see this process to get the line equation step by step.
Step 1: First of all we require starting points say (x1, y1) and ending points (x2, y2) of line. Now calculate the Slope of line. We have a formula to get the slope of line:
m = (y2 – y1) / (x2 – x1),
Assume that we have the coordinates (2, 0) and (-1, 3) then slope of line will be:
m = (3 - 0) / (-1 - 2) = -1,
Step 2: Now we will calculate the y- intercept. We will multiply the slope of line with starting 'x' point and subtract it from the starting 'y' point as:
y intercept = 0 – (-1) * 2 = 2.
Step 3: Now as we know that equation of line is y = mx + c. Here 'm' is the slope of line and 'c' is the y- intercept of line. We have already calculated 'm' and 'c' of line. Now we will place these values in line equation.
Y = mx + c,
Y = -x + 2,
Step 4: If we want to verify the equation we can put the coordinates in equation and verify it.
0 = -2 + 2 = 0 and,
3 = 1 + 3 = 3.
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# Math Accelerated Chapter 8: Equations and Inequalities
Math Accelerated Chapter 8 provides students with an introduction to equations and inequalities. Through this chapter, students explore concepts such as solving equations, graphing inequalities, and interpreting solutions. In order to gain a better understanding of equations and inequalities, it is important for students to be familiar with the various types of equations and the various methods of solving them.
## Understanding Linear Equations
Linear equations are equations that contain two or more variables and can be represented graphically by a line. This type of equation is very important in mathematics, as it is used to describe relationships between two or more variables. Linear equations can be solved by using a variety of techniques, such as substitution, elimination, and graphing. Students should be familiar with these techniques in order to solve linear equations efficiently.
## Solving Systems of Equations
A system of equations is a set of two or more equations that must be solved together. Students need to understand how to solve these equations in order to find the solution of the system. There are several methods for solving systems of equations, such as substitution, elimination, and graphing. Each method has advantages and disadvantages, so it is important for students to understand when to use each method.
## Graphing Inequalities
Inequalities are mathematical statements that involve two or more variables. In order to solve inequalities, students need to be able to graph them. Through graphing, students can visualize the solution set of an inequality and can also determine if the inequality is true or false. There are several methods for graphing inequalities, including shading and graphing on a number line.
## Interpreting Solutions
Once students have solved an equation or inequality, it is important for them to be able to interpret the solution. This involves understanding the meaning of the solution and being able to use it to answer questions related to the original equation or inequality. Students need to understand how to interpret the solutions of equations and inequalities in order to use them in real-world applications.
##
Math Accelerated Chapter 8 introduces students to equations and inequalities. Through this chapter, students learn about different types of equations, how to solve them, how to graph inequalities, and how to interpret solutions. Understanding equations and inequalities is an important part of mathematics, and this chapter provides students with the necessary tools to do so.
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# Types of Triangles
In mathematics, a triangle is one of the basic shapes used for educational purposes. In simple terms, a triangle refers to a polygon or a particular structure that has three sides. However, it can be of different shapes, and hence there are different types of triangles.
Here, we are discussing various types of triangles along with their definitions. Before discussing triangle types, let us first understand the definition of a triangle, including the general properties:
## What is the Triangle?
As the name suggests, a triangle refers to a 2D (two-dimensional) shape closed by joining the three sides. In other words, a triangle is defined as a polygon containing three angles, three corners, and three vertices connected to create a closed shape.
The term 'triangle' is derived from the Latin word 'triangulus' that means 'three-cornered'. A triangle sign (∆) is used for denoting a triangle while solving or calculating relevant questions.
The following diagrams, A and B, are the basic examples of triangles:
### Properties of Triangle
A triangle has many properties. Some of the common properties of a triangle are listed below:
• Every triangle includes three angles, three sides, and three vertices.
• A triangle is a two-dimensional polygon.
• The addition of any two sides' lengths in a triangle is always more than the third side's length.
• The perimeter of a triangle is equal to the addition of the length of the three sides.
• The product of the triangle base and the height gives the area of a triangle.
## Types of Triangles
In particular, triangles are shapes that join three sides and three corresponding angles. However, there are generally six types of triangles, and each type has a specific name and characteristics. The type of triangle is defined based on the length of its sizes and different angles. Therefore, triangles are mainly divided into the following two types:
• Triangles based on sides
• Triangles based on angles
Let us discuss each type in detail:
## Classification of Triangles based on their Sides
According to the length of the triangles' sizes, they are classified into the following three types:
• Scalenes
• Isosceles
• Equilaterals
Let us now understand each type:
### Scalene Triangle
A scalene triangle refers to a triangle in which all sides are of different lengths. In such triangles, the length of each side is not equal to any other side. Also, there will be different interior angles in this type of triangle.
The following diagram is an example of the scalene triangle, where all side lengths and interior angles seem to be unequal.
### Isosceles Triangle
An isosceles triangle refers to a type of triangle in which the lengths of any two sides are equal. Additionally, the angles that lie in the opposite of the equal sides are of equal lengths. This means a triangle typically contains two equal sides and two equal angles.
The following diagram is an example of the isosceles triangle where two side lengths and opposite angles are equal.
### Equilateral Triangle
An equilateral triangle refers to a type of triangle that has all equal sides. Additionally, the three interior angles are also equal. In this triangle, each interior angle is equal to 60 degrees. Because all three angles of an equilateral triangle are equal, this triangle type is also called an equilateral triangle.
It is important to note that an equilateral triangle's lengths do not relate or depend on the interior angles and vice versa.
The following diagram is an example of the equilateral triangle where all three sides are of equal lengths, and all three interior angles have a measure of 60 degrees, i.e., AB = BC = CA and ∠ ABC =∠ ACB = ∠ BAC.
## Classification of Triangles based on their Angles
According to the interior angles of the triangles, they are classified into the following three types:
• Acute-angled Triangle
• Obtuse-angled Triangle
• Right-angled Triangle
Let us now understand each type separately:
### Acute-angled Triangle
A triangle that contains all three interior angles as acute is known as an acute-angled triangle or acute triangle. In simple terms, an acute triangle is a triangle with each interior angle less than 90 degrees that include all three angles.
The following diagram is an example of an acute-angled triangle in which all interior angles (a, b, and c) are below 90 degrees.
### Obtuse-angled Triangle
A triangle with an obtuse angle at one of the three interior angles is defined as an obtuse-angled triangle (also called an obtuse triangle). In other words, the obtuse triangle is the triangle in which one of the three interior angles is above 90 degrees.
The following diagram is an example of the obtuse-angled triangle where one angle (a) is more than 90 degrees (obtuse) while the others (b and c) are less than 90 degrees (acute).
### Right-angled Triangle
The right-angled triangle refers to the triangle with one of three interior angles equal to 90 degrees. Additionally, the side that lies at the opposite of the right angle (90-degree angle) is the longest side of all three sides. This longest side is referred to as the 'hypotenuse'.
It should be noted that the right-angled triangle is different from the isosceles triangle. A right-angled triangle always has one of the three interior angles equal to 90 degrees, while the isosceles triangle has no such condition.
The following diagram is an example of the right-angled triangle where one angle (a) among all three angles is 90 degrees while the other two angles (b and c) are acute.
## Examples of Triangles
We all know that triangles are one of the most important shapes in trigonometry and geometric study. Apart from this, triangles also have many general-life applications. Some of the real-life examples of triangles are as follows:
• When pyramids are drawn on paper or seen from a distance, they form like a triangle shape.
• The Eiffel Tower forms a triangular shape in photos and drawn structures.
• The trusses of roofs and bridges are generally constructed in a triangle shape because they are considered the strongest type of shape.
• Sandwiches are the most common real-life examples of triangles.
• The Bermuda Triangle is a very popular triangular region located in the Atlantic Ocean. Any aircraft or ship passing through this area is swallowed. It is a mysterious triangular area.
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06 - Let's recap trigonometry - Class 10 - Maths Class 10 Notes | EduRev
Class 10 : 06 - Let's recap trigonometry - Class 10 - Maths Class 10 Notes | EduRev
``` Page 1
INTRODUCTION TO TRIGONOMETRY
Trigonometry: Trigonometry is a branch of mathematics that studies the
relationships between the sides and angles of triangles.
Trigonometric Ratios: The ratios of the sides of a right triangle are called
trigonometric ratios. Sine (sin), cosine (cos), and tangent (tan) are the three
common trigonometric ratios, cosecant (cosec), secant (sec) and Cotangent (cot)
are the reciprocal of the ratios sin, cos and tan respectively. These are defined for
acute angle A in right angled triangle ABC below:
Sine of ?A =
Cosine of ?A =
Tangent of ?A=
Cosecant of ?A=
Secant of ?A =
Cotangent of ?A =
C
B
A
Side
opposite
to angle
A
Side
to angle
A
Hypotenuse
Hypotenuse
Side opposite to angle A BC
AC
=
Hypotenuse
Side adjacent to angle A AB
AC
=
Side opposite to angle A BC
AB
=
Hypotenuse
Side opposite to angle A
AC
BC
=
Hypotenuse
AC
AB
=
Side opposite to angle A
AB
BC
=
Page 2
INTRODUCTION TO TRIGONOMETRY
Trigonometry: Trigonometry is a branch of mathematics that studies the
relationships between the sides and angles of triangles.
Trigonometric Ratios: The ratios of the sides of a right triangle are called
trigonometric ratios. Sine (sin), cosine (cos), and tangent (tan) are the three
common trigonometric ratios, cosecant (cosec), secant (sec) and Cotangent (cot)
are the reciprocal of the ratios sin, cos and tan respectively. These are defined for
acute angle A in right angled triangle ABC below:
Sine of ?A =
Cosine of ?A =
Tangent of ?A=
Cosecant of ?A=
Secant of ?A =
Cotangent of ?A =
C
B
A
Side
opposite
to angle
A
Side
to angle
A
Hypotenuse
Hypotenuse
Side opposite to angle A BC
AC
=
Hypotenuse
Side adjacent to angle A AB
AC
=
Side opposite to angle A BC
AB
=
Hypotenuse
Side opposite to angle A
AC
BC
=
Hypotenuse
AC
AB
=
Side opposite to angle A
AB
BC
=
INTRODUCTION TO TRIGONOMETRY
Also, observe that tan A = and cot A =
Trigonometric ratios of some specific angles: In this we will be calculating the
trigonometric ratios value for different angles such as 0°, 30°, 45°, 60° & 90° and
we will also develop the relationship between them.
Trigonometric Ratios of 45°: In a right angled triangle if one angle is of 45° then
the other angle is also 45°. ? ?A = ?C = 45° also AB = BC
The sides in this triangle are in the ratio 1:1: v2
Using the definition of trigonometric ratios, we have,
Sin 45° =
BC
AC
=
a
v2a
=
1
v2
Cosec 45° =
1
Sin 45°
= v2
Cos 45° =
AB
AC
=
a
v2a
=
1
v2
Sec 45° =
1
Cos 45°
= v2
Tan 45° =
BC
AB
=
a
a
= 1 Cot 45° =
1
Tan 45°
= 1
Trigonometric Ratios of 30° & 60° : In a right angled triangle if one angle is of
30° then the other angle is 60°. Side opposite to 30° is always half of the
hypotenuse; side opposite to 60° is always v3 times its adjacent side.
?The sides in this triangle are in the ratio 1:2: v3
Using the definition of trigonometric ratios, we have
v2a
a
a
Cos A
Sin A
Sin A
Cos A
45°
45°
C
B
A
a
v3a
a
2a
30°
60°
C
B
A
Page 3
INTRODUCTION TO TRIGONOMETRY
Trigonometry: Trigonometry is a branch of mathematics that studies the
relationships between the sides and angles of triangles.
Trigonometric Ratios: The ratios of the sides of a right triangle are called
trigonometric ratios. Sine (sin), cosine (cos), and tangent (tan) are the three
common trigonometric ratios, cosecant (cosec), secant (sec) and Cotangent (cot)
are the reciprocal of the ratios sin, cos and tan respectively. These are defined for
acute angle A in right angled triangle ABC below:
Sine of ?A =
Cosine of ?A =
Tangent of ?A=
Cosecant of ?A=
Secant of ?A =
Cotangent of ?A =
C
B
A
Side
opposite
to angle
A
Side
to angle
A
Hypotenuse
Hypotenuse
Side opposite to angle A BC
AC
=
Hypotenuse
Side adjacent to angle A AB
AC
=
Side opposite to angle A BC
AB
=
Hypotenuse
Side opposite to angle A
AC
BC
=
Hypotenuse
AC
AB
=
Side opposite to angle A
AB
BC
=
INTRODUCTION TO TRIGONOMETRY
Also, observe that tan A = and cot A =
Trigonometric ratios of some specific angles: In this we will be calculating the
trigonometric ratios value for different angles such as 0°, 30°, 45°, 60° & 90° and
we will also develop the relationship between them.
Trigonometric Ratios of 45°: In a right angled triangle if one angle is of 45° then
the other angle is also 45°. ? ?A = ?C = 45° also AB = BC
The sides in this triangle are in the ratio 1:1: v2
Using the definition of trigonometric ratios, we have,
Sin 45° =
BC
AC
=
a
v2a
=
1
v2
Cosec 45° =
1
Sin 45°
= v2
Cos 45° =
AB
AC
=
a
v2a
=
1
v2
Sec 45° =
1
Cos 45°
= v2
Tan 45° =
BC
AB
=
a
a
= 1 Cot 45° =
1
Tan 45°
= 1
Trigonometric Ratios of 30° & 60° : In a right angled triangle if one angle is of
30° then the other angle is 60°. Side opposite to 30° is always half of the
hypotenuse; side opposite to 60° is always v3 times its adjacent side.
?The sides in this triangle are in the ratio 1:2: v3
Using the definition of trigonometric ratios, we have
v2a
a
a
Cos A
Sin A
Sin A
Cos A
45°
45°
C
B
A
a
v3a
a
2a
30°
60°
C
B
A
INTRODUCTION TO TRIGONOMETRY
Sin 30° =
BC
AC
=
a
2a
=
1
2
Cosec 30° =
1
Sin 30°
= 2
Cos 30° =
AB
AC
=
v3a
2a
=
v3
2
Sec 30° =
1
Cos 30°
=
2
v3
Tan 30° =
BC
AB
=
a
v3a
=
1
v3
Cot 30° =
1
Tan 30°
= v3
Similarly,
Sin 60° =
AB
AC
=
v3a
2a
=
v3
2
Cosec 60° =
1
Sin 60°
=
2
v3
Cos 60° =
BC
AC
=
a
2a
=
1
2
Sec 60° =
1
Cos 60°
= 2
Tan 60° =
AB
BC
=
v3a
a
= v3 Cot 60° =
1
Tan 60°
=
1
v3
Trigonometric Ratios of 0° & 90° : if angle A is made smaller and smaller angle C
becomes larger and larger. When angle C becomes smaller, side BC also decreases
& finally when ?A becomes very close to 0°, AC will be almost the same as AB
and side BC gets very close to zero. Therefore the value of sin A is very close to 0
and cos A is very close to 1.
Thus we have,
Sin 0° = 0 Cosec 0° =
1
Sin 0°
, which is not defined
Cos 0° = 1 Sec 0° =
1
Cos 0°
= 1
Tan 0° =
Sin 0°
Cos 0°
=
0
1
= 0 Cot 60° =
1
Tan 0°
, which is again not defined.
Now, if angle A is made larger and larger angle C becomes smaller and smaller.
Therefore the length of side AB goes on decreasing. Point A gets closer to B.
Finally when ?A is very close to 90°, ?C becomes very close to 0°. Side AC almost
coincides with side BC so, sin A is very close to 1 and cos A is very close to 0.
C
B
A
Page 4
INTRODUCTION TO TRIGONOMETRY
Trigonometry: Trigonometry is a branch of mathematics that studies the
relationships between the sides and angles of triangles.
Trigonometric Ratios: The ratios of the sides of a right triangle are called
trigonometric ratios. Sine (sin), cosine (cos), and tangent (tan) are the three
common trigonometric ratios, cosecant (cosec), secant (sec) and Cotangent (cot)
are the reciprocal of the ratios sin, cos and tan respectively. These are defined for
acute angle A in right angled triangle ABC below:
Sine of ?A =
Cosine of ?A =
Tangent of ?A=
Cosecant of ?A=
Secant of ?A =
Cotangent of ?A =
C
B
A
Side
opposite
to angle
A
Side
to angle
A
Hypotenuse
Hypotenuse
Side opposite to angle A BC
AC
=
Hypotenuse
Side adjacent to angle A AB
AC
=
Side opposite to angle A BC
AB
=
Hypotenuse
Side opposite to angle A
AC
BC
=
Hypotenuse
AC
AB
=
Side opposite to angle A
AB
BC
=
INTRODUCTION TO TRIGONOMETRY
Also, observe that tan A = and cot A =
Trigonometric ratios of some specific angles: In this we will be calculating the
trigonometric ratios value for different angles such as 0°, 30°, 45°, 60° & 90° and
we will also develop the relationship between them.
Trigonometric Ratios of 45°: In a right angled triangle if one angle is of 45° then
the other angle is also 45°. ? ?A = ?C = 45° also AB = BC
The sides in this triangle are in the ratio 1:1: v2
Using the definition of trigonometric ratios, we have,
Sin 45° =
BC
AC
=
a
v2a
=
1
v2
Cosec 45° =
1
Sin 45°
= v2
Cos 45° =
AB
AC
=
a
v2a
=
1
v2
Sec 45° =
1
Cos 45°
= v2
Tan 45° =
BC
AB
=
a
a
= 1 Cot 45° =
1
Tan 45°
= 1
Trigonometric Ratios of 30° & 60° : In a right angled triangle if one angle is of
30° then the other angle is 60°. Side opposite to 30° is always half of the
hypotenuse; side opposite to 60° is always v3 times its adjacent side.
?The sides in this triangle are in the ratio 1:2: v3
Using the definition of trigonometric ratios, we have
v2a
a
a
Cos A
Sin A
Sin A
Cos A
45°
45°
C
B
A
a
v3a
a
2a
30°
60°
C
B
A
INTRODUCTION TO TRIGONOMETRY
Sin 30° =
BC
AC
=
a
2a
=
1
2
Cosec 30° =
1
Sin 30°
= 2
Cos 30° =
AB
AC
=
v3a
2a
=
v3
2
Sec 30° =
1
Cos 30°
=
2
v3
Tan 30° =
BC
AB
=
a
v3a
=
1
v3
Cot 30° =
1
Tan 30°
= v3
Similarly,
Sin 60° =
AB
AC
=
v3a
2a
=
v3
2
Cosec 60° =
1
Sin 60°
=
2
v3
Cos 60° =
BC
AC
=
a
2a
=
1
2
Sec 60° =
1
Cos 60°
= 2
Tan 60° =
AB
BC
=
v3a
a
= v3 Cot 60° =
1
Tan 60°
=
1
v3
Trigonometric Ratios of 0° & 90° : if angle A is made smaller and smaller angle C
becomes larger and larger. When angle C becomes smaller, side BC also decreases
& finally when ?A becomes very close to 0°, AC will be almost the same as AB
and side BC gets very close to zero. Therefore the value of sin A is very close to 0
and cos A is very close to 1.
Thus we have,
Sin 0° = 0 Cosec 0° =
1
Sin 0°
, which is not defined
Cos 0° = 1 Sec 0° =
1
Cos 0°
= 1
Tan 0° =
Sin 0°
Cos 0°
=
0
1
= 0 Cot 60° =
1
Tan 0°
, which is again not defined.
Now, if angle A is made larger and larger angle C becomes smaller and smaller.
Therefore the length of side AB goes on decreasing. Point A gets closer to B.
Finally when ?A is very close to 90°, ?C becomes very close to 0°. Side AC almost
coincides with side BC so, sin A is very close to 1 and cos A is very close to 0.
C
B
A
INTRODUCTION TO TRIGONOMETRY
So we define sin90° = 1 & cos90° = 0, similarly other trigonometric ratios can be
found.
Trigonometric Ratios of Complementary angles: If the sum of two angles is one
right angle or 90°, then one angle is said to be complementary of the other. Thus, ?°
and (90 - ?)° are complementary to each other.
sin (90°- A) = cos A ; cos (90°- A) = sin A
tan (90°- A) = cot A; cot (90°- A) = tan A
sec (90°- A) = csc A; csc (90°- A) = sec A
These relations are valid for all the values of A lying between 0° and 90
Trigonometric Identities: An equation is called an identity when it is true for all
the value of the variables involved. Similarly, an equation involving trigonometric
ratios of an angle is called a trigonometric identity, if it is true for all the values of
the angle(s) involved.
sin
2
???? + cos
2
???? = 1
1 + tan
2
? = sec
-1
????
1 + cot
2
? = cosec
-1
????
We obtain these identities by using Pythagoras theorem so these are also known as
Pythagorean identities.
?°
(90- ?)°
C
B
A
```
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Crash Course for Class 10 Maths by Let`s tute
88 videos|31 docs
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;
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# Statistics – Probability of Combined Events
I ended my last post showing the probability of picking a type of card from a standard deck of 52 cards. For example, if the event of interest, A, is picking a Jack, then the probability of picking a Jack from a shuffled deck of cards is
$P\left(A\right)=\frac{4}{52}=\frac{1}{13}$
because there are 4 ways to pick a Jack out of 52 cards. Now let’s consider probabilities of events like “picking a Jack or a Heart” or “a face card and a Heart”.
If we let events A be picking a Jack, B be picking a Heart, and C be picking a face card (Jack, Queen, or King), then the maths notation for these statements are
$P\left(A\cup B\right)=\mathrm{probability\ of\ picking\ a\ Jack\ or\ a\ Heart}$ $P\left(B\cap C\right)=\mathrm{probability\ of\ picking\ a\ face\ card\ and\ a\ Heart}$
The symbol “∪” stands for the union of two events, but in English, you can use the word “or”: AB = “A union B” or “A or B“. The symbol “∩” stands for the intersection of two events, but in English, you can use the word “and”: BC = “B intersection C” or “B and C“. These concepts are easily seen in a Venn diagram:
Circle A is the set of all Jacks and circle B is the set of all Hearts. Now the probability of picking a card from set A is 4/52. The probability of picking a card from set B is 13/52. You may be tempted so say that the probability of A or B is the sum of the two individual probabilities. But both of these probabilities include the Jack of Hearts so it is used twice. We have to subtract out this intersection of the two probabilities, so in maths notation:
$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$
This equation can be rearranged to show that the probability of the intersection of the two events is equal to the sum of the individual probabilities minus the probability of the union:
$P\left(A\cap B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cup B\right)$
These two equations are different forms of what is called the addition rule of probability.
So P(AB) = 4/52 +13/52 – 1/52 = 16/52, because P(AB) is the probability of a Jack and a Heart. Only one card satisfies this, the Jack of Hearts, so the probability of that is 1/52.
Now let’s define event D as picking a Diamond and consider the probability of picking a Heart and a Diamond, P(BD). This is clearly 0 as a card cannot be both suits. The associated Venn diagram looks like:
Events like this are called mutually exclusive, that is, you can pick one or the other, the picked card cannot be both. For mutually exclusive events:
$P\left(B\cup D\right)=P\left(B\right)+P\left(D\right)\ \mathrm{and}\ P\left(B\cap D\right) =0$
In my next post, I will discuss what is called conditional probabilities and explore the probability of picking a Jack given that the card is a Heart.
|
# How do you find the number of terms in a polynomial expansion?
## How do you find the number of terms in a polynomial expansion?
For a binomial (a+b)n, the number of terms is n + 1. For a trinomial (a+b+c)n, the number of terms is (n+1)(n+2)(2). For a multinomial (a+b+c+d)n, the number of terms is (n+1)(n+2)(n+3)(6).
How do you find the number of terms?
To find the number of terms in an arithmetic sequence, divide the common difference into the difference between the last and first terms, and then add 1.
### What is a term in a polynomial?
Polynomials are algebraic expressions that contain any number of terms combined by using addition or subtraction. A term is a number, a variable, or a product of a number and one or more variables with exponents. Like terms (same variable or variables raised to the same power) can be combined to simplify a polynomial.
What are terms in polynomial?
#### How many terms are there in the expansion of the binomial expansion?
Detailed Solution. Concept: In the binomial expansion of (a + b) n, there are total n + 1 terms.
How do you find the term in a binomial expansion?
How To: Given a binomial, write a specific term without fully expanding.
1. Determine the value of n according to the exponent.
2. Determine (r+1).
3. Determine r.
4. Replace r in the formula for the ( r + 1 ) t h \displaystyle \left(r+1\right)\text{th} (r+1)th term of the binomial expansion.
## How do you expand terms?
In order to expand and simplify an expression, we need to multiply out the brackets and then simplify the resulting expression by collecting the like terms. Expanding brackets (or multiplying out) is the process by which we remove brackets.
How do you know if an expansion is a polynomial?
Each expansion is a polynomial. There are some patterns to be noted. 1. There is one more term than the power of the exponent, n. That is, there are terms in the expansion of (a + b) n. 2. In each term, the sum of the exponents is n, the power to which the binomial is raised.
### How many terms are there in the expansion?
The patterns we just noted indicate that there are 7 terms in the expansion: a 6 + c 1 a 5 b + c 2 a 4 b 2 + c 3 a 3 b 3 + c 4 a 2 b 4 + c 5 ab 5 + b 6. How can we determine the value of each coefficient, c i?
How many terms are there in polynomial expansion in Galois field?
But what about in number of terms for polynomial expansion in Galois Field (or characteristic of 2, where addition is addition mod 2 or Xor) Show activity on this post. | { ( i, j) | c i, j ≠ 0 } |. and the number of terms is 2.
#### How to find the exponents of a binomial expansion using Pascal’s triangle?
Binomial Expansions Using Pascal’s Triangle 1. There is one more term than the power of the exponent, n. That is, there are terms in the expansion of (a + b) n. 2. In each term, the sum of the exponents is n, the power to which the binomial is raised. 3. The exponents of a start with n, the power of
|
# Law of Cosine Word Problems
• Jul 12th 2009, 05:01 PM
jordangiscool
Law of Cosine Word Problems
A plane flies 810 miles from A to B with a bearing of N 75 degrees E. Then it flies 648 miles from B to C with a bearing of N 32 degrees E. Find the straight-line distance and bearing from C to A
• Jul 13th 2009, 03:20 AM
sa-ri-ga-ma
Quote:
Originally Posted by jordangiscool
A plane flies 810 miles from A to B with a bearing of N 75 degrees E. Then it flies 648 miles from B to C with a bearing of N 32 degrees E. Find the straight-line distance and bearing from C to A
The angle between AB and AC is 180 - 75 + 32 degrees
• Jul 13th 2009, 06:08 AM
Soroban
Hello, jordangiscool!
Quote:
A plane flies 810 miles from A to B with a bearing of N 75° E.
Then it flies 648 miles from B to C with a bearing of N 32° E.
Find the straight-line distance and bearing from C to A
Code:
Q o C : * :32°* : * : * P 105° |* : o B : * : : * 75° : : 75° * : : * R A o
The plane flies from $\displaystyle A$ to $\displaystyle B\!:\;\;AB \,=\,810$
. . $\displaystyle \angle PAB \,=\, 75^o \,=\,\angle ABR \quad\Rightarrow\quad \angle ABQ \,=\,105^o$
Then it flies from $\displaystyle B$ to $\displaystyle C\!:\;\;BC \,=\,648$
. . $\displaystyle \angle QBC \,=\,32^o \quad\Rightarrow\quad \angle ABC \,=\,137^o$
Draw line segment $\displaystyle AC.$
We have .$\displaystyle \Delta ABC\!:\;\;AB = 810,\;BC = 648,\;\angle B = 137^o$
Law of Cosines: .$\displaystyle AC^2 \;=\;AB^2 + BC^2 - 2(AB)(BC)\cos B$
Got it?
|
Get Started by Finding a Local Center
# Mathnasium #MathTricks: Divisibility (Rule for 6’s)
Jan 22, 2024
Welcome to Mathnasium’s Math Tricks series. Today we are determining whether a number is divisible by 6.
A composite number is a number that has three or more distinct factors: 1, itself, and at least one other number. We can determine whether any number is divisible by a composite number by determining whether that number is divisible by a pair of the composite number’s factors.
In this case, 6 is a composite number because its factors are: 1, 2, 3, 6. Therefore, a number is divisible by 6 if …
it is divisible by 2 AND it is divisible by 3.
Follow the examples below to determine whether the numbers are divisible by 6.
##### Example 1: Is 236 divisible by 6?
Step 1: Determine if the number is divisible by 2.
A number is divisible by 2 if it is even. We can quickly tell that 236 is an even number because its ones digit, 6, is even. So, 236 is divisible by 2.
If 236 weren’t divisible by 2, then we could stop there and determine that 236 also wasn’t divisible by 6. Because it is divisible by 2, we can move on to the next step.
Step 2: Determine if the number is also divisible by 3.
A number is divisible by 3 if the sum of its digits is also divisible by 3. The sum of the digits of 236 is
2 + 3 + 6 = 11.
Since 11 is not divisible by 3, then 236 is not divisible by 3.
Answer: No. Since 236 is not divisible by 2 and 3, it is not divisible by 6.
##### Example 2: Is 1,512 divisible by 6?
Step 1: Determine if the number is divisible by 2.
A number is divisible by 2 if it is even. We can quickly tell that 1,512 is an even number because its ones digit, 2, is even. So, 1,512 is divisible by 2.
If 1,512 weren’t divisible by 2, then we could stop there and determine that 1,512 also wasn’t divisible by 6. Because it is divisible by 2, we can move on to the next step.
Step 2: Determine if the number is also divisible by 3.
A number is divisible by 3 if the sum of its digits is also divisible by 3. The sum of the digits of 1,512 is
1 + 5 + 1 + 2 = 9.
Since 9 is divisible by 3, then 1,512 is divisible by 3.
Answer: Yes. Since 1,512 is divisible by 2 and 3, it is divisible by 6.
Now, with this strategy, you are ready to use this Mathnasium Math Trick to determine if a number is divisible by 6. Click here for more practice problems, then check your answers here.
If you missed this, or any of our other Math Tricks videos, check them out on our YouTube channel!
Do you have a math trick you want to see? Submit your request at: https://bit.ly/MathnasiumMathTricks.
## OUR METHOD WORKS
Mathnasium meets your child where they are and helps them with the customized program they need, for any level of mathematics.
|
# 3.4e: Exercises - Polynomial Graphs
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
### A: Concepts
Exercise $$\PageIndex{A}$$
1) What is the difference between an $$x$$-intercept and a zero of a polynomial function $$f$$?
2) If a polynomial function of degree $$n$$ has $$n$$ distinct zeros, what do you know about the graph of the function?
3) What is the relationship between the degree of a polynomial function and the maximum number of turning points in its graph? .
4) Explain how the factored form of the polynomial helps us in graphing it.
5) If the graph of a polynomial just touches the $$x$$-axis and then changes direction, what can we conclude about the factored form of the polynomial?
1. The $$x$$-intercept is where the graph of the function crosses the $$x$$-axis, and the zero of the function is the input value for which $$f(x)=0$$.
3. The maximum number of turning points is one less than the degree of the polynomial.
5. There will be a factor raised to an even power.
### B: Multiplicity from an Equation
Exercise $$\PageIndex{B}$$
$$\bigstar$$ Find the zeros and give the multiplicity of each.
6) $$f(x)=(x+2)^3(x−3)^2$$ 7) $$f(x)=x^2(2x+3)^5(x−4)^2$$ 8) $$f(x)=x^3(x−1)^3(x+2)$$ 9) $$f(x)=x^2(x^2+4x+4)$$ 10) $$f(x)=(2x+1)^3(9x^2−6x+1)$$ 11) $$f(x)=(3x+2)^5(x^2−10x+25)$$ 12) $$f(x)=x(4x^2−12x+9)(x^2+8x+16)$$ 13) $$f(x)=x^6−x^5−2x^4$$ 14) $$f(x)=3x^4+6x^3+3x^2$$ 15) $$f(x)=4x^5−12x^4+9x^3$$ 16) $$f(x)=2x^4(x^3−4x^2+4x)$$ 17) $$f(x)=4x^4(9x^4−12x^3+4x^2)$$
7. $$0$$ and $$4$$ with multiplicity $$2$$, $$−\dfrac{3}{2}$$ with multiplicity $$5$$ 9. $$0$$ with multiplicity $$2$$, $$-2$$ with multiplicity $$2$$ 11. $$−\dfrac{2}{3}$$ with multiplicity $$5$$, $$5$$ with multiplicity $$2$$ 13. $$0$$ with multiplicity $$4$$, $$2$$ and $$-1$$ with multiplicity $$1$$ 15. $$\dfrac{3}{2}$$ with multiplicity $$2$$, $$0$$ with multiplicity $$3$$ 17. $$0$$ with multiplicity $$6$$, $$\dfrac{2}{3}$$ with multiplicity $$2$$
### C: Multiplicity from a Graph
Exercise $$\PageIndex{C}$$
$$\bigstar$$ Use the graph to identify zeros and multiplicity.
19) 20) 21) 22)
19. $$–4, –2, 1, 3$$ with multiplicity $$1$$ 21. $$–2, 3$$ each with multiplicity $$2$$
### D: Graph polynomials
Exercise $$\PageIndex{D}$$
$$\bigstar$$ Graph the polynomial functions. State the $$x$$- and $$y$$- intercepts, multiplicity, and end behavior.
24) $$f(x)=(x+3)^2(x−2)$$ 25) $$g(x)=(x+4)(x−1)^2$$ 26) $$h(x)=(x−1)^3(x+3)^2$$ 27) $$k(x)=(x−3)^3(x−2)^2$$ 28) $$m(x)=−2x(x−1)(x+3)$$ 29) $$n(x)=−3x(x+2)(x−4)$$ 30. $$a(x) = x(x + 2)^{2}$$ 31. $$g(x) = x(x + 2)^{3}$$ 32. $$f(x) = -2(x-2)^2(x+1)$$ 33. $$g(x) = (2x+1)^2(x-3)$$ 34. $$f(x) = x^{3}(x + 2)^{2}$$ 35. $$P(x) = (x - 1)(x - 2)(x - 3)(x - 4)$$ 36. $$q(x) = (x + 5)^{2}(x - 3)^{4}$$ 37. $$h(x) = x^2(x-2)^2(x+2)^2$$ 38. $$h(t) = (3-t)(t^2+1)$$ 39. $$Z(b) = b(42 - b^{2})$$
25. $$x$$-intercepts, $$(1, 0)$$ with multiplicity $$2$$, $$(–4, 0)$$ with multiplicity $$1$$, $$y$$- intercept $$(0, 4)$$ . As $$x→−∞$$, $$f(x)→−∞$$, as $$x→∞,$$ $$f(x)→∞$$. 27. $$x$$-intercepts $$(3,0)$$ with multiplicity $$3$$, $$(2,0)$$ with multiplicity $$2$$, $$y$$- intercept $$(0,–108).$$ As $$x→−∞,$$ $$f(x)→−∞$$, as $$x→∞,$$ $$f(x)→∞$$. 29. $$x$$-intercepts $$(0, 0),(–2, 0),(4, 0)$$ with multiplicity $$1$$, $$y$$-intercept (0, 0). As $$x→−∞,$$ $$f(x)→∞$$, as $$x→∞,$$ $$f(x)→−∞$$. 31. $$(-2,0)$$ multiplicity $$3$$, $$(0,0)$$ multiplicity 1, y-intercept $$(0 ,0 )$$, end behaviour: $$\nwarrow \dots \nearrow$$ 33. $$(-\frac{1}{2} ,0 )$$ multiplicity $$2$$, $$(3 ,0 )$$ multiplicity $$1$$, y-intercept $$(0 ,-3 )$$, end behaviour: $$\swarrow \dots \nearrow$$ 35. $$(1 ,0 )$$, $$(2 ,0 )$$, $$(3 ,0 )$$, $$(4 ,0 )$$ all multiplicity $$1$$, y-intercept $$(0 , 24)$$, end behaviour: $$\nwarrow \dots \nearrow$$ 37. $$(-2 ,0 )$$, $$(2 ,0 )$$, $$(0 ,0 )$$ all multiplicity $$2$$, y-intercept $$(0 ,0 )$$, end behaviour: $$\nwarrow \dots \nearrow$$ 39. $$(\sqrt{42} ,0 )$$, $$(-\sqrt{42} ,0 )$$, $$(0 ,0 )$$ all multiplicity $$1$$, y-intercept $$(0 ,0 )$$, end behaviour: $$\nwarrow \dots \searrow$$
$$\bigstar$$ Graph the polynomial functions. State the $$x$$- and $$y$$- intercepts, multiplicity, and end behavior.
41. $$f\left(x\right)=\left(x+3\right)^{2} (x-2)$$ 42. $$g\left(x\right)=\left(x+4\right)\left(x-1\right)^{2}$$ 43. $$h\left(x\right)=\left(x-1\right)^{3} \left(x+3\right)^{2}$$ 44. $$k\left(x\right)=\left(x-3\right)^{3} \left(x-2\right)^{2}$$ 45. $$m\left(x\right)=-2x\left(x-1\right)(x+3)$$ 46. $$n\left(x\right)=-3x\left(x+2\right)(x-4)$$ 47. $$f(x) = 9x - x^3$$ 48. $$f(x) =8 + x^3$$ 49. $$f(x) = x^4 - 25x^2$$ 50. $$f(x) =16 - x^4$$ 51. $$f(x) = -x^4 + 2x^3 + 8x^2$$ 52. $$f(x) =x^3+7x^2 -9x$$ 53. $$f(x) = 2x^3 + 12x^2 - 8x - 48$$ 54. $$f(x) = 4x^4 + 10x^3 - 4x^2 -10x$$
41. $$(-3 ,0 )$$ multiplicity $$2$$, $$(2 ,0 )$$ multiplicity $$1$$, y-intercept$$(0 ,-18 )$$, $$\swarrow \dots \nearrow$$ 43. $$( -3, 0)$$ multiplicity $$2$$, $$(1 , 0)$$ multiplicity $$3$$, y-intercept$$( 0, -9)$$, $$\swarrow \dots \nearrow$$ 45. $$(-3 , 0)$$, $$(0 , 0)$$, $$(1 , 0)$$ all multiplicity $$1$$, y-intercept$$(0 ,0 )$$, $$\nwarrow \dots \searrow$$ 47. $$(-3 , 0)$$, $$(0 , 0)$$, $$(3 , 0)$$ all multiplicity $$1$$, y-intercept $$(0 ,0 )$$, $$\nwarrow \dots \searrow$$ 49. $$( -5, 0)$$, $$( 5, 0)$$ both multiplicity $$1$$, $$(0 , 0)$$ multiplicity $$2$$, y-intercept $$(0 ,0 )$$, $$\nwarrow \dots \nearrow$$ 51. $$( -2, 0)$$, $$( 4, 0)$$ both multiplicity $$1$$, $$(0 , 0)$$ multiplicity $$2$$, y-intercept $$(0 ,0 )$$, $$\swarrow \dots \searrow$$ 53. $$(-6 , 0)$$, $$(-2 , 0)$$, $$(2 , 0)$$ all multiplicity $$1$$, y-intercept $$(0 , -48)$$, $$\swarrow \dots \nearrow$$
### E: Polynomial Degree from a Graph
Exercise $$\PageIndex{E}$$
$$\bigstar$$ Determine the least possible degree of the polynomial function shown.
61) 62) 63) 64) 65) 66) 67) 68)
61. $$3$$, $$\qquad$$ 63. $$5$$, $$\qquad$$ 65. $$3$$, $$\qquad$$ 67. $$5$$
### F: Construct an Equation from a graph
Exercise $$\PageIndex{F}$$
$$\bigstar$$ Use the graphs to write the formula for the polynomial function of least degree.
69) 70) 71) 72) 73. 74)
69. $$f(x) = -(x+3)(x+1)(x-3)$$ or $$f(x) = -\frac{2}{9}(x+3)(x+1)(x-3)$$
71. $$f(x) = (x+2)^2(x-3)$$ or $$f(x) = \frac{1}{4}(x+2)^2(x-3)$$
73. $$f(x) = -(x+3)(x+2)(x-2)(x-4)$$ or $$f(x) = -\frac{1}{24}(x+3)(x+2)(x-2)(x-4)$$
$$\bigstar$$ Use the graphs to write a formula for the polynomial function of least degree.
75) 76) 77) 78) 79) 80) 81) 82)
75. $$f(x)=(x−500)^2(x+200)$$ 77. $$f(x)=(x+300)^2(x-100)^3$$
79. $$f(x)=(x+3)(x-3)(x^2+10)$$ 81. $$f(x)=4x(x-5)(x-7)$$
$$\bigstar$$ Use the graphs to write a formula for the polynomial function of least degree.
83. 84(a). 84(b). 85. 86. 87. 88. 89. . 90.
83. $$y = \dfrac{1}{24} (x + 4) (x + 2) (x - 3)^2$$ 85. $$y = \dfrac{1}{12} (x + 2)^2 (x - 3)^2$$ 87. $$y = \dfrac{1}{6} (x + 3) (x + 2) (x - 1)^3$$ 89. $$y = -\dfrac{1}{16} (x + 3)(x + 1) (x - 2)^2 (x - 4)$$
### G: Construct an Equation from a Description
Exercise $$\PageIndex{G}$$
$$\bigstar$$ Use the information about the graph of a polynomial function to determine the function. Assume the leading coefficient is $$1$$ or $$-1$$. There may be more than one correct answer.
91) The $$y$$-intercept is $$(0,−4)$$. The $$x$$-intercepts are $$(−2,0), (2,0)$$. Degree is $$2$$. End behavior: $$\nwarrow \dots \nearrow$$
92) The $$y$$-intercept is $$(0,9)$$. The $$x$$-intercepts are $$(−3,0), (3,0)$$. Degree is $$2$$. End behavior: $$\swarrow \dots \searrow$$
93) The $$y$$-intercept is $$(0,0)$$. The $$x$$-intercepts are $$(0,0), (2,0)$$. Degree is $$3$$. End behavior: $$\swarrow \dots \searrow$$
94) The $$y$$-intercept is $$(0,1)$$. The x-intercept is $$(1,0)$$. Degree is $$3$$. End behavior: $$\nwarrow \dots \searrow$$
95) The $$y$$-intercept is $$(0,1)$$. There is no $$x$$-intercept. Degree is $$4$$. End behavior: $$\nwarrow \dots \nearrow$$
$$\bigstar$$ Use the given information about the polynomial graph to write the equation.
97) Degree $$3$$. Zeros at $$x=–2$$,$$x=1$$, and $$x=3$$. $$y$$-intercept at $$(0,–4)$$
98) Degree $$3$$. Zeros at $$x=–5$$, $$x=–2$$, and $$x=1$$. $$y$$-intercept at $$(0,6)$$
99) Degree $$5$$. Roots of multiplicity $$2$$ at $$x=3$$ and $$x=1$$. Root of multiplicity $$1$$ at $$x=–3$$.
$$\quad$$$$y$$-intercept at $$(0,9)$$
100) Degree $$4$$. Root of multiplicity $$2$$ at $$x=4$$. Roots of multiplicity $$1$$ at $$x=1$$ and $$x=–2$$.
$$\quad$$$$y$$-intercept at $$(0,–3)$$
101) Degree $$5$$. Double zero at $$x=1$$. Triple zero at $$x=3$$. Passes through the point $$(2,15)$$
102) Degree $$3$$. Zeros at $$x=4$$, $$x=3$$, and $$x=2$$. $$y$$-intercept at $$(0,−24)$$
103) Degree $$3$$. Zeros at $$x=−3$$, $$x=−2$$ and $$x=1$$. $$y$$-intercept at $$(0,12)$$
104) Degree $$5$$. Roots of multiplicity $$2$$ at $$x=−3$$ and $$x=2$$. Root of multiplicity $$1$$ at $$x=−2$$. $$y$$-intercept at $$(0, 4)$$.
105) Degree $$4$$. Roots of multiplicity $$2$$ at $$x=\dfrac{1}{2}$$. Roots of multiplicity $$1$$ at $$x=6$$ and $$x=−2$$. $$y$$-intercept at $$(0,18)$$
106) Double zero at $$x=−3$$. Triple zero at $$x=0$$. Passes through the point $$(1,32)$$.
107. Degree 3. Zeros at $$x$$ = -2, $$x$$ = 1, and $$x$$ = 3. Vertical intercept at (0, -4)
108. Degree 3. Zeros at $$x$$ = -5, $$x$$ = -2, and $$x$$ = 1. Vertical intercept at (0, 6)
109. Degree 5. Roots of multiplicity 2 at $$x$$ = 3 and $$x$$ = 1. Root of multiplicity 1 at $$x$$ = -3. Vertical intercept at (0, 9)
110. Degree 4. Root of multiplicity 2 at $$x$$ = 4. Roots of multiplicity 1 at $$x$$ = 1 and $$x$$ = -2.
$$\quad$$Vertical intercept at (0, -3)
111. Degree 5. Double zero at $$x$$ = 1. Triple zero at $$x$$ = 3. Passes through the point (2, 15)
112. Degree 5. Single zero at $$x$$ = -2 and $$x$$ = 3. Triple zero at $$x$$ = 1. Passes through the point (2, 4)
91. $$f(x)=x^2−4$$ 93. $$f(x)=x^3−4x^2+4x$$ 95. $$f(x)=x^4+1$$ 97. $$f(x)=−\dfrac{2}{3}(x+2)(x−1)(x−3)$$ 99. $$f(x)=\dfrac{1}{3}(x−3)^2(x−1)^2(x+3)$$ 101. $$f(x) =−15(x−1)^2(x−3)^3$$ 103. (f(x)=−2(x+3)(x+2)(x−1)\) 105. $$f(x)=−\dfrac{3}{2}(2x−1)^2(x−6)(x+2)$$ 107. $$y = -\dfrac{2}{3} (x + 2) (x - 1) (x - 3)$$ 109. $$y = \dfrac{1}{3} (x - 1)^2 (x - 3)^2 (x + 3)$$ 111. $$y = -15(x - 1)^2 (x - 3)^3$$
### H: Turning Points
Exercise $$\PageIndex{H}$$
$$\bigstar$$ Determine whether the graph of the function provided is a graph of a polynomial function. If so, determine the number of turning points and the least possible degree for the function.
115) 116) 117) 118) 119) 120) 121)
115. Yes. Number of turning points is $$2$$. Least possible degree is $$3$$.
117. Yes. Number of turning points is $$1$$. Least possible degree is $$2$$.
119. Yes. Number of turning points is $$0$$. Least possible degree is $$1$$.
212. Yes. Number of turning points is $$0$$. Least possible degree is $$1$$.
$$\star$$
3.4e: Exercises - Polynomial Graphs is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
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# Calculus 2 : Definition of Derivative
## Example Questions
← Previous 1 3 4 5 6 7
### Example Question #1 : Definition Of Derivative
Evaluate the limit using one of the definitions of a derivative.
Does not exist
Explanation:
Evaluating the limit directly will produce an indeterminant solution of .
The limit definition of a derivative is . However, the alternative form, , better suits the given limit.
Let and notice . It follows that .
Thus, the limit is
### Example Question #2 : Definition Of Derivative
Evaluate the limit using one of the definitions of a derivative.
Does not exist
Explanation:
Evaluating the derivative directly will produce an indeterminant solution of .
The limit definition of a derivative is . However, the alternative form, , better suits the given limit.
Let and notice . It follows that . Thus, the limit is .
### Example Question #1 : Definition Of Derivative
Suppose and are differentiable functions, and . Calculate the derivative of , at
Explanation:
Taking the derivative of involves the product rule, and the chain rule.
Substituting into both sides of the derivative we get
.
### Example Question #4 : Definition Of Derivative
Evaluate the limit
without using L'Hopital's rule.
Explanation:
If we recall the definition of a derivative of a function at a point , one of the definitions is
.
If we compare this definition to the limit
we see that that this is the limit definition of a derivative, so we need to find the function and the point at which we are evaluating the derivative at. It is easy to see that the function is and the point is . So finding the limit above is equivalent to finding .
We know that the derivative is , so we have
.
### Example Question #2 : Definition Of Derivative
Approximate the derivative if where .
Explanation:
Write the definition of the limit.
Substitute .
Since is approaching to zero, it would be best to evaluate when we assume that is progressively decreasing. Let's assume and check the pattern.
## Find f'(x):
Explanation:
Computation of the derivative requires the use of the Product Rule and Chain Rule.
The Product Rule is used in a scenario when one has two differentiable functions multiplied by each other:
This can be easily stated in words as: "First times the derivative of the second, plus the second times the derivative of the first."
In the problem statement, we are given:
is the "First" function, and is the "Second" function.
The "Second" function requires use of the Chain Rule.
When:
Applying these formulas results in:
Simplifying the terms inside the brackets results in:
We notice that there is a common term that can be factored out in the sets of equations on either side of the "+" sign. Let's factor these out, and make the equation look "cleaner".
Inside the brackets, it is possible to clean up the terms into one expanded function. Let us do this:
Simplifying this results in one of the answer choices:
### Example Question #1 : Derivatives
What is the value of the limit below?
Explanation:
Recall that one definition for the derivative of a function is .
This means that this question is asking us to find the value of the derivative of at .
Since
and , the value of the limit is .
### Example Question #8 : Definition Of Derivative
Explanation:
Evaluation of this integral requires use of the Product Rule. One must also need to recall the form of the derivative of .
Product Rule:
Applying these two rules results in:
This matches one of the answer choices.
### Example Question #9 : Definition Of Derivative
Use the definition of the derivative to solve for .
Explanation:
In order to find , we need to remember how to find by using the definition of derivative.
Definition of Derivative:
Now lets apply this to our problem.
Now lets expand the numerator.
We can simplify this to
Now factor out an h to get
We can simplify and then evaluate the limit.
### Example Question #10 : Definition Of Derivative
Use the definition of the derivative to solve for .
Explanation:
In order to find , we need to remember how to find by using the definition of derivative.
Definition of Derivative:
Now lets apply this to our problem.
Now lets expand the numerator.
We can simplify this to
Now factor out an h to get
We can simplify and then evaluate the limit.
← Previous 1 3 4 5 6 7
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# Fun with continued fractions
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26 Sep 2012 CPOL
Illustrates the calculation and the usefulness of continued fractions
## Introduction
Continued fractions are of great importance in many aspects, as they have many implementations for real problems where you want to describe something with an approximate fraction, or you simply want to replace a decimal or double number with a fraction.
## Background
As always in mathematics the history of continued fractions are quite complicated. The earliest traces of continued fractions appear as far back as 306 b.c. Other records have been found that show that the Indian mathematician Aryabhata used a continued fraction to solve a linear equation.
In the western hemisphere it did not appear until the 17th century, when William Brouncker and John Wallis started to formulate some form of continued fractions. The Dutch mathematician and astronomer, Christiaan Huygens made the first practical application of the theory in 1687, and wrote a paper explaining how to use convergents to achive the best rational approximations for gear ratios. These approximations enabled him to pick the gears with the best numbers of teeth as he was going to build a mechanical planetarium.
## Representation and calculations of continued fractions
How to write a continued fraction anyway, well let's take the simple quadratic equation:
x^2 - bx -1 = 0
Move things over and divide by X and you can rewrite it like this:
x = b + 1/x
We realize that we have a formula for X on the left, and we have an X on the right side. We now substitute the X on the right side with the formula on the right. Sounds complicated, but here is the result with one substitution:
x = b + 1/(b + 1/x)
We can do this an infinite number of times and get a continued fraction. This is an interesting formula if we plug in b = 1 we would get the golden ration. (to get this number you simply type in (sqrt(5)-1)/2 in the program submitted).
In the general representation of a continued fraction is given below, where x is a number (irrational or rational) and the coefficients a0, a1, and so on are all positive integers.
The continued fraction is quite complicated to write in this way, so they are usually represented as a series of the coefficients as given below:
Finite numbers would have a finite set of fractions, while irrational number (sqrt(2)) could be given a rational approximate representation if you adruptly stop it at one of the coefficients. This is called the n'th convergent of the continued fraction.
To calculate the numbers an you should simply do this:
a0 = Math.Floor(some decimal number)
we must also store the rest of the number in the temporary variable b:
b = some decimal number - Math.Floor(some decimal number)
a1 = 1/(b)
If we continue this, as shown in the code below, we would get all the expansions we desire:
Private Function GetSeriesExpantion(ByVal input As Decimal, ByVal NrFraction As Integer) As Integer
Dim intput As Decimal = input
Dim temp_dec As Integer
For i As Integer = 0 To NrFraction
temp_dec = CInt(Math.Floor(intput))
intput = intput - Math.Floor(intput)
If Not intput = 0 Then
intput = 1 / intput
Else
intput = 0
End If
Next
Return temp_dec
End Function
To get the resulting n'th convergent fraction I have designed an own fraction struct that can add a whole number to a fraction, and the result would be given below the an series in the program.
## Continued fractions in action
Why go to all of this trouble? Well as it turns out continued fractions produces a much better
approximation then the 10 digit system (for instance 314/100 in the case of Pi). You should also notice that the fraction approximation is better, the sooner you get a large number in the An series. As a direct result of this, the worst case scenario would be an expansion were all the numbers (in the An series) is 1.
An example were a series can be terminated at a hugh number is shown in the first program demo picture. The result was discovered by Ramanujan, who was an extraordenary natural talent in mathematics. He studied continued fractions and found, among other things, an approximate formula for pi written like this:
pi = (2143/22)^(1/4)
This is an accurate to the 8 digit after the decimal sign, and that really should be good enough for any practical use! To see why this approximation works so well, you should take a note of the example picture in the beginning of the article. It shows the continued expasion of the number pi^4 and as you can see the number a5 is over 16000, and if we cut the continued series expansion we would get the number in Ramanujan's approximate formula for pi.
To summarize the uses of continued fractions:
• The scale models are can be created with small errors using the convergent
• They convert rather quickly in comparison to other techniques.
• We've seen that they have many computationally desirable properties
• They represent all real numbers accurately
• They yield good approximations when truncated
• They're easy to compare
• They have no weird biases toward the number 10
Why don't we see them used more often? (I've seen other articles on the codeproject site that convert a decimal number to a fraction; this should also be done by using the continued fractions.)
## History
The article with all the explanations is based heavily on the reference given below, and my contribution is quite small, as I only made the program that gives you the possibility to calculate them easily.
The evaluator is simply the one I made previously for complex numbers, as I do not have the permission to publish the code with double taken from "Programming Microsoft Visual Basic .NET" (2003) - Francesco Balena (pages 505 - 509.). The evaluator should be replased with one that takes decimal numbers to get more accurate results.
## References
There is an excellent video by Professor John Barrow on Continued Fractions available on fora.tv for free, if you like to know more be sure to check it out. Some of the material from this article is from the video by him:
And some description from Barrow's web site:
Some other websites that are useful:
## Share
Engineer Norway
I hope that you like the stuff I have created and if you do wish to say thank you then a donation is always appreciated.
You can donate here[^].
First Prev Next
What do you mean by accurate to the 30,000 digit. Pascal Ganaye26-Sep-12 3:38 Pascal Ganaye 26-Sep-12 3:38
Re: What do you mean by accurate to the 30,000 digit. Kenneth Haugland26-Sep-12 3:50 Kenneth Haugland 26-Sep-12 3:50
Re: What do you mean by accurate to the 30,000 digit. Andreas Gieriet27-Sep-12 10:19 Andreas Gieriet 27-Sep-12 10:19
Re: What do you mean by accurate to the 30,000 digit. Kenneth Haugland27-Sep-12 10:58 Kenneth Haugland 27-Sep-12 10:58
Re: What do you mean by accurate to the 30,000 digit. Andreas Gieriet27-Sep-12 11:23 Andreas Gieriet 27-Sep-12 11:23
Re: What do you mean by accurate to the 30,000 digit. Kenneth Haugland26-Sep-12 4:07 Kenneth Haugland 26-Sep-12 4:07
Re: What do you mean by accurate to the 30,000 digit. Pascal Ganaye26-Sep-12 21:42 Pascal Ganaye 26-Sep-12 21:42
Re: What do you mean by accurate to the 30,000 digit. Kenneth Haugland26-Sep-12 21:46 Kenneth Haugland 26-Sep-12 21:46
My 5! Andreas Gieriet3-Sep-12 13:30 Andreas Gieriet 3-Sep-12 13:30
Re: My 5! Kenneth Haugland3-Sep-12 13:58 Kenneth Haugland 3-Sep-12 13:58
Thank you! woutercx21-Jul-12 11:01 woutercx 21-Jul-12 11:01
Re: Thank you! Kenneth Haugland21-Jul-12 11:22 Kenneth Haugland 21-Jul-12 11:22
Re: Thank you! Kenneth Haugland22-Jul-12 2:27 Kenneth Haugland 22-Jul-12 2:27
Re: Thank you! woutercx22-Jul-12 2:33 woutercx 22-Jul-12 2:33
Re: Thank you! Kenneth Haugland22-Jul-12 2:35 Kenneth Haugland 22-Jul-12 2:35
Re: Thank you! woutercx22-Jul-12 3:06 woutercx 22-Jul-12 3:06
Re: Thank you! Kenneth Haugland22-Jul-12 3:39 Kenneth Haugland 22-Jul-12 3:39
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# Difference between revisions of "1992 AIME Problems/Problem 4"
## Problem
In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below.
$$\begin{array}{c@{\hspace{8em}} c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}} c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}} c@{\hspace{6pt}}c@{\hspace{6pt}}c} \vspace{4pt} \text{Row 0: } & & & & & & & 1 & & & & & & \\\vspace{4pt} \text{Row 1: } & & & & & & 1 & & 1 & & & & & \\\vspace{4pt} \text{Row 2: } & & & & & 1 & & 2 & & 1 & & & & \\\vspace{4pt} \text{Row 3: } & & & & 1 & & 3 & & 3 & & 1 & & & \\\vspace{4pt} \text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\\vspace{4pt} \text{Row 5: } & & 1 & & 5 & &10& &10 & & 5 & & 1 & \\\vspace{4pt} \text{Row 6: } & 1 & & 6 & &15& &20& &15 & & 6 & & 1 \end{array}$$ In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio $3 :4 :5$?
## Solution 1
Consider what the ratio means. Since we know that they are consecutive terms, we can say $$\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.$$
Taking the first part, and using our expression for $n$ choose $k$, $$\frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}$$ $$\frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!}$$ $$\frac{1}{3(n-k+1)} = \frac{1}{4k}$$ $$n-k+1 = \frac{4k}{3}$$ $$n = \frac{7k}{3} - 1$$ $$\frac{3(n+1)}{7} = k$$ Then, we can use the second part of the equation. $$\frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!}$$ $$\frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!}$$ $$\frac{1}{4(n-k)} = \frac{1}{5(k+1)}$$ $$\frac{4(n-k)}{5} = k+1$$ $$\frac{4n}{5}-\frac{4k}{5} = k+1$$ $$\frac{4n}{5} = \frac{9k}{5} +1.$$ Since we know $k = \frac{3(n+1)}{7}$ we can plug this in, giving us $$\frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1$$ $$4n = 9\left(\frac{3(n+1)}{7}\right)+5$$ $$7(4n - 5) = 27n+27$$ $$28n - 35 = 27n+27$$ $$n = 62$$ We can also evaluate for $k$, and find that $k = \frac{3(62+1)}{7} = 27.$ Since we want $n$, however, our final answer is $\boxed{062.}$ ~LaTeX by ciceronii
## Solution 2
Call the row x, and the number from the leftmost side t. Call the first term in the ratio $N$. $N = \dbinom{x}{t}$. The next term is $N * \frac{x-t}{t+1}$, and the final term is $N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}$. Because we have the ratio, $N : N * \frac{x-t}{t+1} : N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}$ = $3:4:5$,
$\frac{x-t}{t+1} = \frac{4}{3}$ and $\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}$. Solve the equation to get get $t= 26$ and $x = \boxed{062}$.
-Solution and LaTeX by jackshi2006
1992 AIME (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
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# What are all the possible rational zeros for f(x)=3x^4-10x^3-24x^2-6x+5 and how do you find all zeros?
Sep 28, 2016
Use the rational root theorem and the remainder theorem to determine a zero.
The possible factors in $y = a {x}^{n} + b {x}^{n - 1} + \ldots + p$ are given by $\text{factors of p"/"factors of a}$.
In our example, these would be $\pm 1 , \pm \frac{5}{3} , \pm \frac{5}{1} \pm \frac{1}{3} \to \pm 1 , \pm \frac{5}{3} \pm 5 , \pm \frac{1}{3}$.
We now use the remainder theorem to determine the remainder after substituting these values into the function. If the remainder equals 0, we have a factor.
$f \left(- 1\right) = 3 {\left(- 1\right)}^{4} - 10 {\left(- 1\right)}^{3} - 24 {\left(- 1\right)}^{2} - 6 \left(- 1\right) + 5$
$f \left(- 1\right) = 3 \left(1\right) - 10 \left(- 1\right) - 24 \left(1\right) + 6 + 5$
$f \left(- 1\right) = 3 + 10 - 24 + 6 + 5$
$f \left(- 1\right) = 0$
Hence, $- 1$ is a factor. Next, we need to use synthetic division to divide $3 {x}^{4} - 10 {x}^{3} - 24 {x}^{2} - 6 x + 5$ by $x + 1$
$- 1 \text{_|"3" "-10" "-24" "-6" 5}$
$\text{ "-3" "13" "11" "-5}$
$\text{-----------------------------------------------------------------------------}$
$\text{ "3" "-13" "-11" "5" 0 }$
Hence, the quotient is $3 {x}^{3} - 13 {x}^{2} - 11 x + 5$.
We repeat the process to find that $- 1$ is once again a factor. We divide synthetically again, to obtain the result of $3 {x}^{2} - 16 x + 5$.
We can factor this as:
$3 {x}^{2} - 16 x + 5 = 3 {x}^{2} - 15 x - x + 5 = 3 x \left(x - 5\right) - 1 \left(x - 5\right) = \left(3 x - 1\right) \left(x - 5\right)$
Setting all of these factors to $0$, we have that this function has a root of $x = - 1$ of multiplicity $2$, and roots of $\frac{1}{3}$ and $5$ of multiplicity $1$.
Here is the graph of the function.
Hopefully this helps!
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Q.5) The following pictograph shows the transport used by the students to reach the school –
Each symbol = 12 Students
i) How many students come by Motorcycle?
[A] 30
[B] 54
[C] 48
ii) How many students come by Car ?
[A] 50
[B] 48
[C] 36
iii) How many students come by Bus or Scooter ?
[A] 36
[B] 37
[C] 30
iv) How many students come by Auto ?
[A] 36
[B] 54
[C] 28
v) Which transport is used by the least number of students ?
[A] Auto
[B] Scooter
[C] Bus
### Data handling Worksheets for Grade 3 Explanations
Q.1) Explanation – Data handling Worksheets for Grade 3
(i) Given that each symbol = 2 bananas
So, the number of bananas they ate on Tuesday = Number of Stars on Tuesday x 2
= ( 2 x 2 )
= 4
(ii) Given that each symbol = 2 bananas
So, the number of bananas they ate on Monday = Number of Stars on Monday x 2
= ( 2 x 3 )
= 6
(iii) Given that each symbol = 2 bananas
So, the number of bananas they ate on Wednesday = Number of Stars on Wednesday x 2
= ( 2 x 4 )
= 8
(iv) Given that each symbol = 2 bananas
So, the number of bananas they ate on Sunday = ( 2 x 1 ) = 2
The number of bananas they ate on Wednesday = ( 2 x 4 ) = 8
The number of bananas they ate on Thursday = ( 2 x 5 ) = 10
The total number of bananas they ate on Sunday, Wednesday and Thursday = 2 + 8 + 10 = 20
(v) Since there are maximum number of symbols on Thursday, maximum number of bananas were eaten on Thursday.
Q.2) Explanation – Data handling Worksheets for Grade 3
(i) Given that each symbol = ₹ 5
So, money earned by Sanjay on Tuesday = Number of symbols on Tuesday X 5 = ( 5 x 3 ) = ₹ 15
(ii) Given that each symbol = ₹ 5
So, money earned by Sanjay on Friday = Number of symbols on x 5 = ( 5 x 4 ) = ₹ 20
(iii) Since there are maximum numbers of symbols on Thursday, the maximum money is earned on Thursday.
(iv) Given that each symbol = ₹ 5
So, money earned by Sanjay on Thursday = ( 5 x 5 ) = ₹ 25
Money earned by Sanjay on Friday = ( 5 x 4 ) = ₹ 20
Money earned by Sanjay on Saturday = ( 5 x 1 ) = ₹ 5
Total amount of money earned by him on Thursday, Friday and Saturday is = ₹ 25 + ₹ 20 + ₹ 5 = ₹50
(v) Since there are minimum numbers of symbols on Saturday, the minimum money is earned on Saturday.
Q.3) Explanation – Data handling Worksheets for Grade 3
(i) Given that each symbol = 6 Students
So, the number of students who like Football = Number of symbols in Football X 6
= ( 6 x 3 )
= 18
(ii) Given that each symbol = 6 Students
So, the number of students who like Cricket = Number of symbols in Cricket x 6 = ( 6 x 5 ) = 30
(iii) Since, there are maximum numbers of symbols for Cricket, Cricket is liked the most by the students.
(iv) Since, there are minimum numbers of symbols for Tennis, Tennis is liked the least by the students.
(v) Given that each symbol = 6 Students
Number of students who like Badminton = ( 6 x 4 ) = 24
The number of students who like Hockey = ( 6 x 2 ) = 12
Total number of students who like Badminton or Hockey = 24 + 12 = 36
Q.4) Explanation – Data handling Worksheets for Grade 3
(i) Given that each symbol = 12 Students
So, the number of students who like Chocolate flavour = Number of symbols in Chocolate x 5 = ( 12 x 5 ) = 60
(ii) Given that each symbol = 12 Students
So, the number of students who like Chocolate marshmallow flavour = Number of symbols in marshmallow x 4 = ( 12 x 4 ) = 48
(iii) Since, there are maximum number symbols for Chocolate flavour, Chocolate flavour is liked the most by the students.
(iv) Since, there are minimum number of symbols for Vanilla flavour, Vanilla is liked the least by the students.
(v) Since, there are equal numbers of symbols for Strawberry and Chocolate marshmallow, Strawberry and Chocolate marshmallow flavours are liked equally by the students.
Q.5) Explanation – Data handling Worksheets for Grade 3
(i) Given that each symbol = 15 Students
So, the number of students who use Motorcycle to reach school = Number of symbols in X 4 = ( 12 x 4 ) = 48
(ii) Given that each symbol = 15 Students
So, the number of students who use Car to reach school = ( 12 x 3 ) = 36
(iii) Given that each symbol = 15 Students
So, the number of students who use Bus to reach school = ( 12 x 2 ) = 24
The number of students who use Scooter to reach school = ( 12 x 1 ) = 12|
The number of students who use Bus or Scooter to reach school = 24 + 13 = 37
(iv) Given that each symbol = 15 Students
So, the number of students who use Auto to reach school = ( 12 x 3 ) = 36
(v) Since, there are minimum numbers of symbols for Scooter, Scooter is used by the least number of students.
#### Maths Worksheet for Class 3
Q.1) Express the amount of money in given terms ?
3 rupees 85 paise = ₹ ?
a) ₹ 3.085
b) ₹ 3.85
Q.2) Convert:
3 rupees = ___ paise
a) 30 paise
b) 300 paise
Q.3) Convert:
400 paise = ___ rupees.
a) ₹ 4
b) ₹ 40
₹ 300 and ₹ 500
a) ₹ 900
b) ₹ 800
c) ₹ 1000
₹ 41.35 and ₹ 23.21
a) ₹ 64.56
b) ₹ 62.16
c) ₹ 68.16
Q.6) Subtract:
₹ 864 – ₹ 153
a) ₹ 723
b) ₹ 699
c) ₹ 711
Q.7) Subtract the following:
₹ 84.98 and ₹ 33.54
a) ₹ 53.56
b) ₹ 51.44
c) ₹ 55.04
### Money Worksheets Grade 3 Explanations
Q.1) Explanation – Money Worksheets Grade 3
Whenever, we are given both Rupees and paise, and we need to write them in rupees, we convert the given paisa into Rupees by adding a dot ( . ) after rupees value , to separate rupees and paise.
Paise are always written as a two-digit number.
3 rupees 85 paise = ₹ 3.85
Correct Answer – b) ₹ 3.85
Q.2) Explanation – Money Worksheets Grade 3
As we know that,
1 rupees = 100 paise To find 3 rupees in paise .
We have to multiply both sides by 3
3 rupees = ( 3 x 100 ) paise
3 rupees = 300 paise
Correct Answer – b) 300 paise
Q.3) Explanation – Money Worksheets Grade 3
As we know that,
₹ 1 = 100 paise
or 100 paise = ₹ 1
then,
1 paise = ₹ 1/100
To find 400 paise in rupees. . We have to multiply both sides by 400
( 400 x 1 ) paise = ₹ ( 400 x 1/100 )
400 paise = ₹ 4
Q.4) Explanation – Money Worksheets Grade 3
Since both the values contain only Rupees ( ₹ ), we would add the values
Hence, ₹ 300 + ₹ 500 = ₹ 800
Correct Answer – b) ₹ 800
Q.5) Explanation
Step I: Write Rupees ( ₹ ) and Paise ( p ) aligned in the rupees and paise columns.
Step II: Add Rupees ( ₹ ) and Paise ( p ) as we add numbers in Addition.
We would add the numbers in a normal way, as we do addition, starting from the right side, i.e paise side
Hence, the sum is ₹ 64.56
Correct Answer – a) ₹ 64.56
Q.6) Explanation –
Since both the values contain only Rupees ( ₹ ), we would subtract the values
Hence, ₹ 864 – ₹ 153 = ₹ 711
Correct Answer – c) ₹ 711
Q.7) Explanation – Money Worksheets Grade 3
Step I: Write Rupees ( ₹ ) and Paise ( p ) aligned in the rupees and paise columns.
Step II: Subtract Rupees ( ₹ ) and Paise ( p ) as we subtract numbers in Subtraction.
We would subtract the numbers in a normal way, as we do in subtraction, starting from the right side, i.e paise side.
Hence, the difference is ₹ 51.44
Correct Answer – b) ₹ 51.44
Maths Worksheet for Class 3
Q.1) Read the time in the clock and write it both in words and figures?
a) 4:30 or half past four
b) 5:15 or quarter past five
c) 5:30 or half past five
d) 4:15 or quarter past four
Q.2) Read the time in the clock and write it both in words and figures?
a) 6:45 or quarter to seven
b) 7:45 or quarter to eight
c) 8:15 or quarter past eight
d) 8:30 or half past eight
Q.3) How many days are there in a week?
a) 9
b) 12
c) 7
Q.4) How many months are there in a year?
a) 14
b) 12
c) 9
Q.5) The shortest month of the year is _______ .
a) April
b) May
c) February
Q.6) Which is the fifth month of the year?
a) April
b) June
c) May
Q.7) Which is the tenth month of the year?
a) September
b) October
c) August
Q.8) Which is the seventh month of the year?
a) August
b) June
c) July
Q.9) May comes after ______ .
a) April
b) June
c) July
Q.10) March comes before ______ .
a) May
b) April
c) February
Q.11) July comes before ______ .
a) June
b) September
c) August
Q.12) ______ month comes between March and May.
a) February
b) April
c) July
Q.13) A leap year comes once in every ____ years.
a) 3
b) 4
c) 2
Q.14) You are making waffles in the morning. Is it A.M. or P.M.?
a) A.M.
b) P.M.
Q.15) You are taking your dog for a walk in the afternoon. Is it A.M. or P.M.?
a) A.M.
b) P.M.
Q.16) When is dinner eaten?
a) A.M.
b) P.M.
Q.17) You are watching the evening news. Is it A.M. or P.M.?
a) A.M.
b) P.M.
Q.18) When does it get dark outside?
a) A.M.
b) P.M.
### Time Worksheets Grade 3 Explanations
Q.1) Explanation – Time Worksheets Grade 3
Since the hour hand is between 4 and 5, the number of hours is 4;
Minutes past the hour, tells us how many minutes have passed after completion of an hour
Since the minute hand is on 6, and each number shows 5 minutes, so the total minutes are (5 x 6) = 30
Since, 30 minutes are equal to half an hour, we can say that the time is 4:30 or half past four.
Q.2) Explanation – Time Worksheets Grade 3
8:45 or quarter to nine
Since the hour hand is between 6 and 7, the number of hours is 6;
Minutes past the hour, tells us how many minutes have passed after completion of an hour
Since the minute hand is on 9, and each number shows 5 minutes, so the total minutes are (5 x 9) = 4
Since, 15 minutes are equal to quarter an hour, we can say that the time is 6:45 or quarter to seven.
Q.3) Explanation: Time Worksheets Grade 3
There are 7 days in a week.
Q.4) Explanation: Time Worksheets Grade 3
There are 12 months in a year, which come in the following order : –
Q.5) Explanation:
February has 28 days in normal years or 29 days in leap year but other months have 30 or 31 days.
So, February is the shortest month of the year.
Q.6) Explanation:
There are 12 months in a year, which come in the following order : -
So, May is the fifth month of the year.
Q.7) Explanation: Time Worksheets Grade 3
There are 12 months in a year, which come in the following order : -
So, October is the tenth month of the year.
Q.8) Explanation:
There are 12 months in a year, which come in the following order : -
So, July is the seventh month of the year.
Q.9) Explanation: Time Worksheets Grade 3
There are 12 months in a year, which come in the following order : –
So, the month of May comes after April.
Q.10) Explanation:
There are 12 months in a year, which come in the following order : –
So, the month of March comes before April.
Q.11) Explanation: Time Worksheets Grade 3
There are 12 months in a year, which come in the following order : –
So, the month of July comes before August.
Q.12) Explanation:
The months come in the following order : –
So, April comes between March and May.
Q.13) Explanation:
A leap year comes in every 4 years.
Q.14) Explanation:
A.M. means before 12 noon.
P.M. means after 12 noon.
It is given that you are making waffles in the morning. Morning is before 12 noon. Hence, it is A.M.
Q.15) Explanation: Time Worksheets Grade 3
A.M. means before 12 noon.
P.M. means after 12 noon.
It is given that you are taking your dog for a walk in the afternoon. Afternoon is after 12 noon. Hence, it is P.M.
Q.16) Explanation:
A.M. means before 12 noon.
P.M. means after 12 noon.
Dinner is eaten at night. Night is after 12 noon. Hence, it is P.M.
Q.17) Explanation:
A.M. means before 12 noon.
P.M. means after 12 noon.
It is given that you are watching the evening news. Evening is after 12 noon. Hence, it is P.M.
Q.18) Explanation: Time Worksheets Grade 3
A.M. means before 12 noon.
P.M. means after 12 noon.
It gets dark outside in the evening. Evening is after 12 noon. Hence, it is P.M.
#### Maths Worksheet for Class 3
a) 467
b) 352
c) 527
a) 7698
b) 6967
c) 5769
a) 730
b) 793
c) 603
a) 4222
b) 5632
c) 5422
Adding the digits at ones place,
4 + 3 = 7
Adding the digits at tens place,
1 + 5 = 6
Adding the digits at hundreds place,
1 + 3 = 4
Hence, the sum of the given numbers is 467
Adding the digits at ones place,
5 + 2 = 7
Adding the digits at tens place,
3 + 3 = 6
Adding the digits at hundreds place,
5 + 4 = 9
Adding the digits at thousands place,
4 + 2 = 6
Hence, the sum of the given numbers is 6967
Adding the digits at ones place,
8 + 5 = 13
Since the number is more than 9, we carry over 1
Adding the digits at tens place and the carry of 1
0 + 9 + 1 = 10
Since the number is more than 9, we carry over 1
Adding the digits at hundreds place and the carry of 1
3 + 2 + 1 = 6
Hence, the sum of the given numbers is 603
Adding the digits at ones place,
7 + 5 = 12
Since the number is more than 9, we carry over 1
Adding the digits at tens place and the carry of 1
8 + 3 + 1 = 12
Since the number is more than 9, we carry over 1
Adding the digits at hundreds place and the carry of 1
9 + 2 + 1 = 12
Since the number is more than 9, we carry over 1
Adding the digits at thousands place and the carry of 1
2 + 1 + 1 = 4
Since the number is more than 9, we carry over 1
Hence,the sum of the given numbers is 4222
#### Maths Worksheet for Class 3
Q.1) The Fraction that represents purple color ice creams, as a part of the total ice creams would be?
a) $\cfrac{15}{11}$
b) $\cfrac{11}{15}$
Q.2) Write down the Numerator and Denominator of the Fraction:
$\cfrac{5}{6}$
a) Numerator 5 , Denominator 6
b) Numerator 6 , Denominator 5
Q.3) Write the following number as a Fraction ?
3 ÷ 4
a) $\cfrac{4}{3}$
b) $\cfrac{3}{4}$
Q.4) Find the missing number to make the two Fractions equivalent ?
$\cfrac{2}{5} = \cfrac{4}{?}$
a) 10
b) 15
Q.5) If Numerator of Fraction is 2 and Denominator of Fraction is 3 then, the Fraction is?
a) $\cfrac{3}{2}$
b) $\cfrac{2}{3}$
Q.6) Find:
$\cfrac{3}{4}$ of 8
a) 4
b) 6
c) 3
Q.7) Abhishek had 8 icecreams he ate$\cfrac { 2 }{ 4 }$ of them. The number of icecreams Abhishek ate is?
a) 6
b) 5
c) 4
Q.8) Compare$\cfrac { 1 }{ 6 }$ and $\cfrac { 7 }{ 6 }$
a) $\cfrac { 1 }{ 6 }$ >$\cfrac { 7 }{ 6 }$
b) $\cfrac { 1 }{ 6 }$ <$\cfrac { 7 }{ 6 }$
Q.9) Compare$\cfrac { 3 }{ 7 }$ and $\cfrac { 3 }{ 2 }$
a) $\cfrac { 3 }{ 7 }$ >$\cfrac { 3 }{ 2 }$
b) $\cfrac { 3 }{ 7 }$ <$\cfrac { 3 }{ 2 }$
Q.10) Find the sum:
$\cfrac{5}{4}$ and $\cfrac{3}{4}$
a) $\cfrac{6}{4}$
b) $\cfrac{8}{4}$
c) $\cfrac{7}{4}$
Q.11) Find the difference of :
$\cfrac{7}{6} - \cfrac{4}{6}$
a) $\cfrac{3}{6}$
b) $\cfrac{5}{6}$
c) $\cfrac{1}{6}$
### Fractions Worksheets Grade 3 Explanations
Q.1) Explanation –
As we know that a Fraction represents a part of a whole.
Fraction of purple ice creams = $\cfrac { Total\quad number\quad of\quad purple\quad ice\quad creams }{ Total\quad number\quad of\quad ice\quad creams }$
In the given figure,
Total number of purple ice creams = 11
Total number of ice creams = 15
Therefore,
Fraction for the purple ice creams = $\cfrac{11}{15}$
Correct Answer – b) $\cfrac{11}{15}$
Q.2) Explanation –
Numerator is the upper part (Number on Top) of the Fraction
So, the Numerator of $\cfrac{5}{6}$ is 5
Denominator is the lower number ( Number at Bottom) of the Fraction
So, the Denominator of $\cfrac{5}{6}$ is 6
Correct Answer – a) Numerator 5 , Denominator 6
Q.3) Explanation –
Any two non zero numbers a ÷ b
can be written as a Fraction$\cfrac{a}{b}$
i.e,$\cfrac{3}{4}$
Correct Answer – b) $\cfrac{3}{4}$
Q.4) Explanation –
Given,
$\cfrac{2}{5} = \cfrac{4}{?}$
In order to obtain an equivalent Fraction, we need to multiply the Numerator and Denominator of a given number , by same digit
To get 4 as Numerator, we have to multiply 2 by ( 4 ÷ 2 ) or 2
In order to get an equivalent Fraction, we have to multiply the Denominator by the same number, i.e, 2
i.e,
$\cfrac { 2\times 2 }{5\times 2} = \cfrac{4}{10}$
Hence, $\cfrac{2}{5} = \cfrac{4}{10}$
Q.5) Explanation –
In a Fraction,
Numerator is the upper part (Number on Top) of the Fraction.
Denominator is the lower number ( Number at Bottom) of the Fraction.
Here,
Numerator of the Fraction = 2
Denominator of the Fraction = 3
Hence the Fraction is = $\cfrac{2}{3}$
Correct Answer – b) $\cfrac{2}{3}$
Q.6) Explanation –
Step 1 : Divide the Whole Number by the denominator to get the Quotient.
8 ÷ 4 = 2
Step 2 : Multiply the Quotient by the Numerator.
2 x 3 = 6
Hence, $\cfrac{3}{4}$ of 8 = 6
Q.7) Explanation –
Abhishek ate
$\cfrac { 2 }{ 4 }$ of 8 icecreams
=$\cfrac { 2 }{ 4 }$ x 8 icecreams
=$\cfrac { 16 }{ 4 }$ icecreams
= 4 icecreams
Q.8) Explanation –
When Denominators of two or more Fractions are same, the Fraction with the higher Numerator is Higher
When we compare
$\cfrac { 1 }{ 6 }$ and$\cfrac { 7 }{ 6 }$
Denominator of Fractions are same i.e, 6
On Comparing the Numerators, we find
1 < 7
Hence,$\cfrac { 1 }{ 6 }$ < $\cfrac { 7 }{ 6 }$
Correct Answer – b) $\cfrac { 1 }{ 6 }$ <$\cfrac { 7 }{ 6 }$
Q.9) Explanation –
When Numerator of two or more Fractions are same, then the Fraction with the higher Denominator is Smaller.
When we compare
$\cfrac { 3 }{ 7 }$ and$\cfrac { 3 }{ 2 }$
Numerator of Fractions are same i.e, 3
On Comparing the Denominator, we find
7 > 2
Hence,$\cfrac { 3 }{ 7 }$ < $\cfrac { 3 }{ 2 }$
Correct Answer – b) $\cfrac { 3 }{ 7 }$ <$\cfrac { 3 }{ 2 }$
Q.10) Explanation –
$Sum\quad of\quad Like\quad Fraction\quad =\quad \cfrac { Sum\quad of\quad their\quad Numerators }{ Common\quad Denominator }$
= $\cfrac{5+3}{4}$
= $\cfrac{8}{4}$
Hence, the sum of $\cfrac{5}{4}$ and $\cfrac{3}{4}$ = $\cfrac{8}{4}$
Correct Answer – b) $\cfrac{8}{4}$
Q.11) Explanation –
$Difference\quad of\quad Like\quad Fraction\quad =\quad \cfrac { Difference\quad of\quad their\quad Numerators }{ Common\quad Denominator }$
= $\cfrac{7 - 4}{6}$
= $\cfrac{3}{6}$
Hence, the difference of $\cfrac{7}{6} and \cfrac{4}{6} = \cfrac{3}{6}$
Correct Answer – a) $\cfrac{3}{6}$
Maths Worksheet for Class 3
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# 2007 AMC 12B Problems/Problem 16
## Problem
Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?
$\mathrm{(A)}\ 15 \qquad \mathrm{(B)}\ 18 \qquad \mathrm{(C)}\ 27 \qquad \mathrm{(D)}\ 54 \qquad \mathrm{(E)}\ 81$
~ pi_is_3.14
## Solution 1
A tetrahedron has 4 sides. The ratio of the number of faces with each color must be one of the following:
$4:0:0$, $3:1:0$, $2:2:0$, or $2:1:1$
The first ratio yields $3$ appearances, one of each color.
The second ratio yields $3\cdot 2 = 6$ appearances, three choices for the first color, and two choices for the second.
The third ratio yields $\binom{3}{2} = 3$ appearances since the two colors are interchangeable.
The fourth ratio yields $3$ appearances. There are three choices for the first color, and since the second two colors are interchangeable, there is only one distinguishable pair that fits them.
The total is $3 + 6 + 3 + 3 = 15$ appearances $\Rightarrow \mathrm{(A)}$
## Solution 2
Every colouring can be represented in the form $(w,r,b)$, where $w$ is the number of white faces, $r$ is the number of red faces, and $b$ is the number of blue faces. Every distinguishable colouring pattern can be represented like this in exactly one way, and every ordered whole number triple with a total sum of 4 represents exactly one colouring pattern (if two tetrahedra have rearranged colours on their faces, it is always possible to rotate one so that it matches the other).
Therefore, the number of colourings is equal to the number of ways 3 distinguishable nonnegative integers can add to 4. If you have 6 cockroaches in a row, this number is equal to the number of ways to pick two of the cockroaches to eat for dinner (because the remaining cockroaches in between are separated in to three sections with a non-negative number of cockroaches each), which is $\binom{6}{2} = 15$
Alternative explanation to solution 2: A regular tetrahedron is the only platonic solid in which any of the faces is adjacent to all the other 3 faces. Hence we only need to think about the number of faces we can colour for each face. Let the number of faces coloured with red, blue and white be $r, b, w$ respectively. So we are solving for the number of solutions of the equation: $r + b + w = 4$ for nonnegative integers $r, b, w$. By Stars and Bars, we obtain the final answer which is $\binom{6}{2} = 15/(A)$
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Lesson Explainer: Converting Recurring Decimals to Fractions | Nagwa Lesson Explainer: Converting Recurring Decimals to Fractions | Nagwa
# Lesson Explainer: Converting Recurring Decimals to Fractions Mathematics • 7th Grade
In this explainer, we will learn how to convert a recurring decimal to a fraction or a mixed number.
A fraction, in everyday language, often means a tiny part. In math, a fraction compares a part to a whole and describes what we call a proportion. The denominator of the fraction is the number of equal shares, or “portions,” the whole is split into, while the numerator is the number of these shares that make the part we are considering.
But a fraction can also be understood as a quotient: it is simply dividing one quantity by another. Any rational number can be expressed in this way, and in this case a “fraction” is actually a number.
Let us take as an example.
Fraction: If is a fraction, it means that we are comparing a part made of 3 equal shares to the whole made of 4 such equal shares.
If we consider the whole to be the unit (1), then the part is 0.75.
Number: If is a number obtained by dividing 3 by 4, then it is 0.75.
We see that the rational number (3 divided by 4) is the same as (three-quarters) of 1. They can be both expressed as a decimal, 0.75.
It is important to remember the definition of a rational number.
### Definition: Rational Number
A rational number is a real number that can be expressed as a simple fraction (i.e., whose denominator and numerator are integers).
As we have seen above with , a rational number can be expressed as a decimal. We are going to learn here how to convert a decimal back to a fraction (of 1). For instance, consider 0.7. It is seven-tenths (i.e., ). 0.03 is three-hundredths (i.e., ). And note that 2 is simply .
We can see here that our decimal number system makes it very easy to convert a decimal back into a fraction with a power of ten as the denominator.
Consider, for instance, 1.372. This decimal can be split as
• 1 unit,
• 3 tenths,
• 7 hundredths,
• 2 thousandths.
The smallest bit in this partition of 1.372 is the two-thousandths. We will, therefore, express 1.372 as a number of thousandths:
• 1 unit is one thousand-thousandths, written as .
• 3 tenths are three hundred-thousandths, written as .
• 7 hundredths are seventy-thousandths, written as .
• 2 thousandths are written as .
Adding all the parts together, we get
We see that, in this method, the position of the last digit of a number determines the denominator of the fraction. A last digit in the third position after the decimal point means that it is in the thousandths column, so the simplest denominator to start with is 1 000. Then, to find the denominator, we can simply multiply our number by the value of the denominator:
So far, we have converted a decimal to a fraction. Now, we may want to express this fraction in its simplest form. We can do this in our example by simplifying by two twice: . This fraction cannot be simplified any further since 343 is a multiple of neither 5 nor 2, which are the only two prime factors of 250.
This fraction is a so-called improper fraction since its numerator is greater than its denominator. It can, therefore, be written as a mixed number:
We are going to look at some examples to check our understanding.
### Example 1: Converting a Simple Decimal to a Fraction
Convert 0.4 to a fraction.
In the decimal 0.4, the last digit, 4, is in the tenths column, meaning that 0.4 is 4 tenths. Hence, .
Now that we have converted 0.4 to a fraction, we need to write the fraction in its simplest form. As both 4 and 10 are multiples of 2, the fraction can be simplified by dividing by 2: .
The next example is slightly more complex.
### Example 2: Converting a Decimal with Three Decimal Places to a Fraction
Convert 0.268 to a fraction.
In the decimal 0.268, the last digit is 8 and it is at the third decimal place. This means that it is in the thousandths column. Therefore, 0.268 can be expressed as a whole number of thousandths: .
Now that we have converted 0.268 to a fraction, we need to write the fraction in its simplest form. The numbers 268 and 1 000 are both multiples of 4; therefore, by dividing both the numerator and denominator by 4, we find .
In the next example, we are going to look at how to express a decimal as a mixed number.
### Example 3: Converting a Decimal to a Mixed Number
The age of Earth is about 4.54 billion years. Write this as a mixed number in its simplest form.
The age of Earth is given here as a number of billion years, and it is given as a decimal, 4.54. We are asked to write 4.54 as a mixed number in its simplest form.
First, we convert 4.54 into a fraction: .
Then, we express this improper fraction as a mixed number: .
Finally, we find the simplest form of the fractional part. As both 54 and 100 are multiples of 2, we can divide both the numerator and denominator by 2: . There is no common factor between 27 and 50 except 1; therefore, this fraction is in its simplest form.
So far, we have looked at examples with terminating decimals. Now, we are going to learn how to convert repeating decimals to a fraction. A repeating decimal occurs when the division never ends. The bar notation is often used to indicate that a digit or group of digits is repeated. For instance, is written as , and is written as .
We know that , but how can we find, having only , that it can be written as ? We are going to learn a simple method based on the realization that subtracting two repeating decimals that have exactly the same decimal part repeating after the decimal point leads to an integer.
Let us start with the simple case of . For the sake of clarity, we are going to call this number . If we multiply by 10, we find another number with exactly the same repeating decimal: . And if we subtract from , we find which can be simplified to
From this, by dividing both sides of the equation by 9, the value of can be expressed as
Let us look at two more complex examples to deepen our understanding of the method.
### Example 4: Converting a Repeating Decimal to a Fraction
Express as a rational number in its simplest form.
Let be . To express as a fraction in its simplest form, we need first to identify the repeating part: it is the digits with the bar on the top, so it is the group of digits “75.” We have to multiply by a power of 10 so that the result is a number with the same repeating decimals (i.e., “75”). We see that , and so which can be simplified to From this, it follows that which can be simplified by dividing both the numerator and denominator by 3 to
In the next example, the repeating decimals do not start just after the decimal point.
### Example 5: Converting a Repeating Decimal to a Fraction
Convert to a fraction.
Let be the number . To express as a fraction in its simplest form, we need first to identify the repeating part: it is the digit with the bar on the top, “7.” And we notice that the repeating decimal does not start straight after the decimal point. In this case, we will have to multiply by two different powers of 10, so that both results are a number with the same repeating decimal starting straight after the decimal point.
The smallest power of 10 that will work is 100:
For the sake of simplicity, we can choose to multiply by the next power of 10, which is 1 000. We get
We can now subtract the smaller from the larger: which can be simplified to
From this, it follows that
This fraction cannot be simplified; it is therefore the final answer.
### Key Points
• A fraction can be understood as a quotient: it is simply dividing one quantity by another. Any rational number can be expressed in this way, and in this case a “fraction” is actually a number. If is a number obtained by dividing 3 by 4, then it is 0.75.
• We see that the rational number (3 divided by 4) is the same as (three-quarters) of 1. They can both be expressed as a decimal, 0.75.
• Converting a decimal to a fraction is finding the quotient that gives this decimal. To do this, we use the properties of our decimal system. A terminating decimal can always be expressed as a whole number divided by a power of 10. For instance, .
• To convert a repeating decimal, , to a fraction, the method consists of multiplying by two different powers of 10 so that we get two decimals with exactly the same repeating decimal part. They can be subtracted from each other to give a whole number. This allows us to find the quotient that gives , as illustrated in the following two examples.
• First example: . We take and then subtract from : which simplifies to and gives (by dividing both sides by 99) Note that, in this example, we could use and did not need to multiply it by a power of 10.
• Second example: . We take and , and subtract from : which simplifies to and gives (by dividing both sides by 900)
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# Word Problem Wednesday: Get Ready!
Apr 18, 2023
In today's world, maths skills can come in handy at any moment. Practicing this week's word problems will sharpen those skills and help your child be at the ready!
Find the word problem below that’s the right skill level for your child and, together, give it a try. Take your time working it out — no peeking! When you both feel you’ve found the answer, look below to check your solution against ours. Enjoy the fun maths practice, and make sure you check back for more!
Years 1-3:
Question: Liza sells 2 bunches of violets for 6 pennies. If each bunch has 6 violets, then how many violets cost a penny?
Years 3-6:
Question: Henry teaches language lessons. His most expensive lessons cost 60 pounds, and his least expensive lessons cost 1 shilling. If 20 shillings make 1 pound, then what is the range in language lesson prices? Give your answer in pounds and shillings.
Years 6-9:
Question: Clara, Freddy, Henry, and Liza are all sitting together on a sofa. Clara is sitting next to Henry, and Freddy is not. Clara and Freddy are both between two people, and Liza is not. Liza is to the left of Henry. From left to right, in what order are they sitting?
Have you worked out the answer to the word problem you chose? When you’re ready to check your work, look below to find our solutions.
Years 1-3:
Solution: If each bunch has 6 violets, then 2 bunches have 12 violets. If Liza sells 12 violets for 6 pennies, then she sells 2 for each penny.
Years 3-6:
Answer: 59 pounds and 19 shillings
Solution: We find the range by subtracting the smallest value from the largest value. So, the range in prices is 60 pounds – 1 shilling. Since there are 20 shillings to a pound, we borrow 20 shillings from 60 pounds to set up our subtraction equation: 59 pounds and 20 shillings – 1 shilling = 59 pounds and 19 shillings.
Years 6-9:
Solution: The first pieces of information we can use to eliminate seating options are who’s in between two people and who isn’t. We know that Clara and Freddy cannot be on the ends, and Liza must be on the far left or right. Liza cannot be on the far right because she is to the left of Henry. So, since Liza must be on the far left and Clara and Freddy are in the middle, Henry must be on the far right. Since Clara is next to Henry, she must be third, which means Freddy must be second.
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# Rectangle Riddle
Practices
Explaining and justifying
Noticing and using mathematical structure
Topics
• Measurement
• Multiplication
3
4
Use an App
Pattern Shapes
2UXP-8UU5
Geoboard
306X-9FT9
Number Frames
2J7E-5M62
What relationships can you find between area and perimeter?
Find as many rectangles as you can that have a perimeter that is 4 units more than the number of square units in the area.
How could you get started?
• Start by building a rectangle and recording the area and perimeter. What do you notice? How could you change the rectangle to try to get the perimeter and area measurements closer to the rectangle described?
• Think about one of the rectangles you made. How could you make a rectangle with the same area that has a longer perimeter? What about one with a shorter perimeter?
• Can you make two different rectangles that have the same area and the same perimeter?
In this task, students use their knowledge of area and perimeter of rectangles to find rectangles that match the description given (a perimeter that is 4 more than the area). Third and fourth grade students are developing an understanding of area as the number of square units that make up a two-dimensional shape and connect the concept of area to multiplication of two factors (side lengths). Third grade students understand perimeter as the length around a figure, and fourth grade students can use a general formula to calculate the perimeter of a rectangle.
Students will likely experiment with several areas with different lengths and widths that get them closer to the rectangle described. They may use what they know about factors and products to choose a number that could easily be arranged into one or more rectangles and then compare and contrast the perimeters and areas of the different arrangements.
Any rectangle arranged into 2 rows or columns will result in a perimeter with a numerical value that is 4 more than the numerical value of the area, so students may make and test conjectures as they notice this pattern developing.
Various apps may be used to represent the context of this problem.
• In the Pattern Shapes app, students may build a rectangle and calculate the perimeter and area by counting (as shown here) or by using formulas (as shown here). Once they have found one rectangle that fits the requirements, they should explore further to find others.
• In the Geoboard app, students might build a rectangle and calculate the perimeter and area by counting (as shown here) or by using formulas (as shown here).
• In the Number Frames app, students might experiment, using the “Choose a Frame” tool. This tool shows the area as the rectangle changes. Students might work inside this tool, comparing area with perimeter.
This activity gives students an opportunity to look for and express regularity in repeated reasoning. You can ask such questions as: How can you find the perimeter without counting each exterior side? How can you find the area without counting by 1s? Students who have not recognized or used formulas for finding area or perimeter will likely lean into the structures that they have observed when counting the total number of square units (area) or linear units around the rectangle (perimeter), such as skip-counting or noticing and operating with equal groups.
Additionally, students may begin to notice how the relationship between perimeter and area changes as the dimensions of the rectangle approach those of a square. You could ask: Did the perimeter increase or decrease when you made a rectangle with the same area, but with more rows? Fewer rows? As students make multiple rectangles that fit the requirements, ask: What similarities do you notice between the rectangles that fit the requirements? Encourage students to make and explore conjectures: Will that always be true? How do you know?
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# An Elegant Geometric Derivation
Jim Wilson
Let us take a triangle ABC with sides of length a, b, and c. I will follow the convention of indicating the length a opposite the angle A and similarly for the other two pairs. Figure 1 is our triangle so labeled. If you want to skip the preliminary details of getting ready for the proof, move ahead to the discussion of Figure 4.
If we add the incircle to our construction and indicated the points of tangency with the sides there are several relationships that can be easily shown. If they are not immediately clear, they can be easily proven. See Figure 2.
We have added several things to our image. These include the incircle with center at I, the points of tangency of the incircle with the sides of the triangle at D, E, and F, line segments connecting I to each of the vertices, and radii of the incircle indicated by dashed lines.
Observe that if we have three triangles with a common vertex at I that cover the area of our original triangle. Thus, if we could express the length of the radius r of the incircle in terms of side lengths a, b, and c, we could derive a formula for the area of the original triangle in terms of the lengths of its three sides.
The expression in parentheses is the semiperimeter of the triangle s and so Area = rs.
The semiperimeter can be expressed as the sum of 3 segments, each half the length of a side. It is convenient to locate these segments in Figure 2. From each vertex, the distances to the points of tangency on the adjacent sides are equal. We have illustrated that with colors of the respective line segments from each vertex. We have AE = AF, BF = BD, and CD = CE. If we construct a segment CG along AC extended and let CG = BF, then the length AG is the semiperimeter.
Now the three segments along AG (length s) have lengths s – a, s – b, and s – c. We need something more, probably some similar triangles, in order to find some expression for the radius r in terms of s, a, b, and c.
We are now going to suppress some of the added line segments for a moment and and some others. In particular, the construction of point G seems rather arbitrary. It would be nice if it related to something. In Figure 4, we have added the excircle that is tangent to side BC of the triangle. This excircle is externally tangent to side AC and side AB. The distances from A to each of the external tangent points on sides AC and AB are equal and that distance is s. Thus G is a point of tangency of the excircle. The excenter is H and HG is a radius of the excircle.
We now have two pairs of similar triangles HGA and IEA; HGC and CIE.
The first pair are similar because IE and HG are parallel. Therefore from the first pair we get
The second pair are similar. One way to show that the second pair are similar is to observe that HC and IC are exterior angle and interior angle bisectors from point C. Therefore they are perpendicular and thus the angles GCH and ICE are complementary. This leads to the congruence of angle GHC and ECI. The similarity gives
Now solve each for HG and set the two resulting expressions equal to give
Now,
Since Area = rs we have
Recognize the intermediate derivation was to find an expression for the radius r of the incircle in terms of the lengths of the sides. This is the underlying strategies for some other proofs.
Return
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Graphing Ordered Pairs We have graphed values on the number line in pre- algebra and in earlier chapters of algebra. In this lesson, you will learn how to find the domain and range from ordered pairs. So you are looking for the list where x+y = 2 (add the the two numbers in each ordered pair, is the answer 2? This first quadrant of the xy-plane exercise has novelty in spades for children in grade 4, grade 5, grade 6. Since the three points are on the graph of y = -x + 3, shown above, the ordered pairs are solutions to the equation. The numeric values in an ordered pair can be integers or fractions. This lesson helps students understand complex math concepts in an accessible way. Identifying Integers with the Coordinate Plane. Then identify the second number in the pair and move that number of spaces on the vertical axis y axis. Want to see the step-by-step answer? $9^m +3^m-2 = 2 p^n$ Where p is a prime number. Does the ordered pair ( $5,4$ ) lie on this line? An ordered pair is often used to represent a point on a coordinate plane or the solution to an equation in two variables. This is just a few minutes of a complete course. From here, take 4 steps towards the “y” axis (upwards). Explore this batch of pdf worksheets and dig deep into coordinate systems - systems that incorporate latitudes and longitudes in place of the x and y axes. Learn how to determine whether relations such as equations, graphs, ordered pairs, mapping and tables represent a function. A relation or a function is a set of ordered pairs. •Find the x and y intercepts of the equation. They should then use these new ordered pairs to create their initials on a new piece of graph paper. Where, x = abscissa, the distance measure of a point from the primary axis “x”, And, y = ordinate, the distance measure of a point from the secondary axis “y”. Write f (x) = 3x−5 f ( x) = 3 x - 5 as an equation. * The first number in the pair is the x-coordinate. Find Missing Ordered Pairs Using Polygons In this math lesson, students learn how to determine unknown ordered pairs by using characteristics of polygons. (See Examples.Example 1 Draw the graph of each of the following equations: a. y = 2 x + 1 Solution We need to find some ordered pairs from the solution This is what I've got done so far. I wrote the equation as … Recognize the four infinite regions created by the x and y axes on a coordinate plane as the four quadrants. What I want to do is search layer 2 for the correct x value and then layer 3 for the corresponding y value. Ordered pairs. Both represent two different points as shown below. An ordered pair is a composition of the x coordinate (abscissa) and the y coordinate (ordinate), having two values written in a fixed order within parentheses. Have the students find and label the x and y axes on the first page of their Coordinate Plane and Ordered Pairs Worksheet. To graph a point, enter an ordered pair with the x-coordinate and y-coordinate separated by a comma, e.g., . The set of all first coordinates of the ordered pairs is the … DOWNLOAD IMAGE. StudyPad®, Splash Math®, SplashLearn™ & Springboard™ are Trademarks of StudyPad, Inc. See Example 7. Not anymore! lets you practice your graph-reading skills by identifying the correct coordinates. p cannot be 2. Printable Worksheets @ www.mathworksheets4kids.com Functions - Ordered Pairs Sheet 1 Name : A) State whether each set of ordered pairs represents a function. A relation or a function is a set of ordered pairs. I need to find ordered pairs . That’s my midpoint. Start identifying the position of points with these pdf worksheets! Identify the x-value in the ordered pair and plug it into the equation. In these pdf worksheets, you not only graph line segments but also measure them by counting the unit squares along their lengths. Sort the array first and count the pairs by two indexes. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Find the ordered pairs for the x- and y-intercepts of the equation 2x − 6y = 24 and select the appropriate option below 0 votes B) The x-intercept is (0,2);The y-intercept is (-6,0). Coordinates are truly versatile! Don't you see coordinate planes steadily up their engagement quotient? In this lesson, you will learn how to find the domain and range from ordered pairs. surely in a non-ordered pair it is not significant what order the climate are seen in. f (x) = 3x − 5 f ( x) = 3 x - 5. Geometric shapes like circle, triangle, square, rectangle and polygons use the ordered pairs to represent the center, vertices and the length of the sides with coordinates. Write the point located at each ordered pair in part A, and represent the location of the given points using the ordered pairs in part B. We have 2 methods, graphing, draw your right triangle, then find ½ of each side and connect those back to see where that point is. The task is for children to first figure out the ordered pairs that denote the position of the objects and then to locate the items placed at the indicated ordered pairs. CAN'T COPY THE GRAPH Problem 62 Determine whether each graph is the graph of a function. Then, identify the second number in the pair and move that number of … The numeric values in an ordered pair can be … It helps to locate a point on the Cartesian plane for better visual comprehension. The ordered pair (6, 4) is different from the pair (4, 6). Copyright © 2020 Studypad Inc. All Rights Reserved, The first number in the ordered pair shows the distance from “x" axis which is 6, The second number in the ordered pair shows the distance from “y" axis which is 4. Ordered Pair Notation, Lean How the x and y values of ordered pair notation relate to a point's location on the x and y planes Locating objects is only part of the process! This is what i have tried. Count of Ordered Pairs (X, Y) satisfying the Equation 1/X + 1/Y = 1/N Last Updated: 19-08-2020 Given a positive integer N , the task is to find the number of ordered pairs (X, Y) where both X and Y are positive integers, such that they satisfy the equation 1/X + 1/Y = 1/N . Suitable for grades 4 - 5, What's The Point? To graph two objects, simply place a semicolon between the two commands, e.g., y=2x^2+1; y=3x-1. While any numbers can be used, smaller integers, like −1 − 1, 0 0, and 1 1, are mathematically easier to work with. And, the second number in the pair is the y-coordinate. Geometric shapes like circle, triangle, square, rectangle and polygons use the ordered pairs to represent the center, vertices and the length of the sides with coordinates. The order of the two numbers is important—(a, b) is different from (b, a) unless a equals b. Give a new lease of life to your practice by unveiling the mystery picture in this set of ordered pairs and coordinate planes worksheets! Https Www Matsuk12 Us Cms Lib Ak01000953 Centricity Domain 3674 Lesson 205 1 20writing 20linear 20equations 20from 20graphs 20and 20situations 20notes Pdf. B) State whether each set of ordered pairs … Now before we move on let us try and define an ordered triplet, or 3-tuple . When you move three units to the north, south, east, or west on a coordinate plane, where will you land? you are a god sent ;(($\endgroup$ – Terrence Matthews Oct 16 at 0:42 Check out a sample Q&A here. Ordered pairs are a fundamental part of graphing. Method 1 (Brute Force): Generate all possible pairs (i, j) and check if a particular ordered pair (i, j) is such that, (arr i, arr j) satisfies the given equation of the line y = mx + c, and i ≠ j.If the point is valid(a point is valid if the above condition is satisfied), increment the counter which stores the total number of valid points. Get all lessons & more subjects at: http://www.MathTutorDVD.com. Properly plotted and linked, a pair of distinct ordered pairs will give you a line segment. Function Check: Because the x values of 2 occur more than once, the set of ordered pairs you entered are not a function.Find Domain: The Domain is just the input values Domain = (2, -3, 4) Find … Let x=0 and solve for y: (0 is easy!) 2 young teachers die from COVID-19 complications See Examples 2 through 6. With everyday objects scattered all over, the coordinate plane is a learners' paradise. Are you wondering what on earth a coordinate plane is doing in a map territory? I also thought about using the "completing the square" method, but am not sure on how to use it in this case. As per the definition of ordered pair, the point P will be written as: To mark the point on the Cartesian plane, start from the origin. Beginning (0,0), identify the first number in the pair and move that number of spaces on the horizontal axis (x-axis). Have them draw a dot where they Plot the point “P” with coordinates 6, 4. Play it online, here. How To: Find the Slope Given 2 Ordered Pairs By daylightspool 3/17/10 9:05 AM 2/13/17 11:14 AM WonderHowTo Slope is the change in Y over the change in X. Find Three Ordered Pair Solutions. y = 3x− 5 y = 3 x - 5. Experts are waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes! The two indexes approach is similar to the one in 2-sum problem, which avoids the binary-search for N times. Burger Kings of tomorrow will focus on drive-thrus. Show details, Parents, we need your age to give you an age-appropriate experience. Choose any value for x x that is in the domain to plug into the equation. The two ordered pairs are equal only if u = x, v = y i.e. Plot the points using the indicated ordered pairs, consider these points as vertices, and connect them in the order to obtain regular polygons. Functions. (GRAPH CANT COPY) Elizabeth X. Winsor School. Our printable worksheets on ordered pairs and coordinate planes are ideal for 4th grade, 5th grade, 6th grade, and 7th grade students. Does having to draw shapes on a coordinate plane put you in a spot of bother? Want to see this answer and more? you are a god sent ;(($\endgroup$ – Terrence Matthews Oct 16 at 0:42 To ensure step-by-step progress, our exercises include tasks like locating objects, plotting points on quadrants, graphing line segments, drawing shapes, and more. This time they surprise you by navigating – locating a person or a vehicle on the move. The mathematician Rene Descartes and Pierre de Fermat invented analytic geometry in 16th century and Cartesian plane was designed, Ordered pair in modern math is widely used in the field of computing and programming languages. Graph the points $(-3,2)$ and $(1,3),$ and draw a line through them. Find all ordered pairs (m.n) of natural numbers that satisfy the equation. Consider two ordered pairs (u, v) and (x,y). I wrote the equation as $(3^m+2)(3^m-1) = 2p^n$ ($3^m-1$) is always even and ($3^m+2$) is always odd. I have an ordered pair matrix, A=[x y], and a layered matrix, N, with three layers. When you simplify, if the y-value you get is the same as the y-value in the ordered pair, then that ordered pair is indeed a solution to the equation. Alright, we've avoided this long enough! (u, v) = (x, y). The key is to know the ordered pairs are the addresses of the points. To figure out if an ordered pair is a solution to an equation, you could perform a test. I see now i had trouble because i was forcing my solutions to have x and y variables to find a specific value for the ordered pairs :(( thank you so much!!! In this set of printable worksheets, locate the point using its ordered pair, find the ordered pair of a point, plot the points, and draw shapes - all on the xy-plane. Give your sense of direction a big shot in the arm! The ordered pair (x, y) is not the same ordered pair as (y, x). Find three ordered pair solutions by completing the table. Example (0,2) ==> x=0, y=2 and x+y = 0+2 = 2 (yes, this pair is a solution) Check them all until you find one choice where all three pairs have x+y = 2. Points on the Quadrants, X-Axis, and Y-Axis. In mathematics, an ordered pair is a set of two numbers usually written in the form (a, b). surely in a non-ordered pair it is not significant what order the climate are seen in. 2 (0)+y=10 0+y=10 y=10 This ordered pair (x,y)= (0,10) An ordered pair is a pair of numbers, (x, y), written in a particular order. Graphing and Measuring the Vertical and Horizontal Line Segments. Does it lie on the quadrants or axes? Take students on a skill-upgrade tour! After having gone through the stuff given above, we hope that the students would have understood "How to Write a Relation as a Set of Ordered Pairs ". Naomi Osaka roars back to win 2nd U.S. Open title. Then do a substitution. To do algebra, we can't just be all about solving equations, we eventually have to graph some stuff as well. fullscreen. The concept of ordered pair is highly useful in data comprehension as well for word problems and statistics. In this case, the point lies on the y-axis. Algebra Q&A Library find ordered pairs. DOWNLOAD IMAGE. find ordered pairs. Fun facts The mathematician Rene Descartes and Pierre de Fermat invented analytic geometry in 16th century and … Find the ordered pairs of following equations; 3x-2y=-4 x+2y=12 Please show the steps and elaborate them. See Answer. Solution for How do I find ordered pairs and graph y=7-3x/4? The Complete K-5 Math Learning Program Built for Your Child. Solved Examples. Ordered Pair. Graphing Ordered Pairs on the Grid Have you ever savored four different, delicious treats all at one bite? Provide an x/y chart and show your work in finding 3 ordered pairs. In this set of printable worksheets, locate the point using its ordered pair, find the ordered pair of a point, plot the points, and Have you ever savored four different, delicious treats all at one bite? … Plot the points and draw the simple shapes on the grid accordingly. Ordered pairs are a crucial part of graphing, but you need to know how to identify the coordinates in an ordered pair if you're going to plot it on a coordinate plane. The point where the two lines meet at “0” is the origin. Where does the point x = 0 and y < 0 lie on the coordinate grid? After multiplying each of the ordered pairs by their chosen number, they will find the new values of the ordered pairs and record them in the last column of the Ordered Pairs Initials worksheet. y= 2 + 3/2 x Then do a substitution in the The order of the two numbers is important: (1, 2) is not equivalent to (2, 1) -- (1, 2)≠ (2, 1). Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Home > Math > Geometry>Graphing Ordered Pairs : Identify Ordered Pair Graphing Ordered Pairs : Identify Ordered Pair Geometry Geometry Worksheets The first number in the pair is the x-coordinate. Ordered pairs are also called 2-tuples. Question. Identifying the Quadrants Recognize the four infinite regions created by the x and y axes on a coordinate plane as the four quadrants. Sorry, we could not process your request. (GRAPH CANT COPY) Problem 31. Should you require advice on a polynomial as well as systems of linear equations, Sofsource.com is … Ordered pairs are commonly used to specify a location on a map or coordinate plane. That means my ordered pair is (0, ½). For example, find two ordered pairs that fall on the line y=−2x+3 y … Does the ordered pair $(3,5)$ lie on this line? Answer: 3 question Find the ordered pairs of the inverse function. Identify the x-value in the ordered pair and plug it into the equation. I see now i had trouble because i was forcing my solutions to have x and y variables to find a specific value for the ordered pairs :(( thank you so much!!! Let’s look at the rules of Midpoint! Explain that they cross at point 0 on the x-axis and 0 on the y-axis. An ordered pair is a pair of numbers in a specific order. Ordered pairs make up functions on a graph, and very often, you need to plot ordered pairs in order to see what the graph of a function looks like. We use cookies to give you a good experience as well as ad-measurement, not to personalise ads. Get moving with these printable worksheets for grade 5, grade 6, and grade 7. This ensemble of free ordered pairs and coordinate planes worksheets gets students on the go practicing, keeping pace with, and perfecting coordinate grids or Cartesian planes. case in point, with letters, (a, b) and (b, a) are 2 diverse ordered pairs, yet they are the comparable unordered pair (because of the fact order does not remember). To find ordered pairs using the slope-intercept form, plug in the values for x x and solve for the corresponding y y. Graphing In Algebra Ordered Pairs … The task is for children to first figure out the ordered pairs that denote the position of the objects and then to locate the items placed at the indicated ordered pairs. How To Find Ordered Pairs On A Graph Beginning 00 identify the first number in the pair and move that number of spaces on the horizontal axis x axis. This algebra video tutorial explains how to determine if an ordered pair is a solution to a system of equations. Ordered Pairs Cowboy Math Locate the Aliens Graphing Ordered Pairs Soccer Coordinates Billy Bug Stock the Shelves The Coordinate Plane Graph Mole X,Y Wizard What's the Point? Sofsource.com makes available essential advice on ordered pair solution equation calculator, intermediate algebra syllabus and geometry and other algebra topics. So p cannot be even i.e. We could define it by extending the definition above to include 3 sets with a third containing a, b, and c, but that would end up getting to convoluted as we extend the definition to n-tuples (tuples with any number of elements). 3x + y = 6 lets say x=0 3 * 0 = 0 y = 6 (intercept at 0,6) 3x + y = 6 lets say y=0 y*0= 0 3*2 = 6 x=2 (intercept at 2,0) Now, I do not know how to find the third ordered pair. It helps to locate a point on the Cartesian plane for better visual comprehension. In these word problems, children move from one point to the other and find the ordered pair for the end point. For example, (1, 2) and (- 4, 12) are ordered pairs. Ordered pair definition, two quantities written in such a way as to indicate that one quantity precedes or is to be considered before the other, as (3, 4) indicates the Cartesian coordinates of … Read the axes, identify the integer coordinates, and list all the points in each quadrant specifying its name and the ordered pairs. How To Find Ordered Pairs In Ymxb DOWNLOAD IMAGE. How To: Find coordinates (ordered pair) How To: Find the equation of a line given 2 points How To: Find a slope of a straight line with: Ax + By + C = 0 How To: Find the point slope form of a line equation How To: Find the equation of a perpendicular & parallel line Find three ordered pairs that solve the equation and draw the graph of each. The set of all first coordinates of the ordered pairs is the … Printable Worksheets @ www.mathworksheets4kids.com Functions - Ordered Pairs Sheet 1 Name : A) State whether each set of ordered pairs represents a function. The coordinate geometry uses ordered pairs to represent geometric figures and objects in an open space for visual comprehension. An ordered pair is a composition of the x coordinate (abscissa) and the y coordinate (ordinate), having two values written in a fixed order within parentheses. Use that graph to answer Problems 31-34. … In physics, a vector can be represented as an ordered pair, , where the first number is called the vector's x-component and the second number is called the vector's y-component. Take 6 steps towards the “x” axis (towards right) starting from the origin. As the name “ordered pair” suggests, the order in which values are written in a pair is very important. more ... Two numbers written in a certain order. Write at which quadrant each figure is; record the figures in each quadrant. Note: To figure out if an ordered pair is a solution to an equation, you could perform a test. So (12,5) is 12 units along, and 5 units up. check_circle Expert Answer. Read the positive and negative coordinates from the ordered pairs, determine the quadrant first, and then proceed to graph the points and the shapes at their respective sites. The x-axis and the y-axis of the coordinate grid are graduated in integers. If it is not, find ordered pairs that show a value of that is assigned more than one value of . Using Ordered Pairs to Represent Variables To comprehend it better, let’s take an example. Plot the points as usual, link them as directed, and solve the mystery! Find all ordered pairs of integers (x,y) such that $$\displaystyle x^2+2x+18=y^2$$ I tried square-rooting both sides so that the equation could be solved for y, but didn't have any luck. Encounter such questions headfirst with our free printable worksheets. y=\frac{1}{3} x 1. 2. the ordered pairs are points on the graph of the equation. Find all ordered pairs (m.n) of natural numbers that satisfy the equation $9^m +3^m-2 = 2 p^n$ Where p is a prime number. Algebra. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In the event you seek advice on trinomials or even a line, Mathmusic.org is without question the right place to pay a visit to! Usually written in parentheses like this: (12,5) Which can be used to show the position on a graph, where the "x" (horizontal) value is first, and the "y" (vertical) value is second. Ranging from locating stationary points to determining their position after movement in a specific direction, our pdfs have every opportunity for skills to leap to huge heights. Mathmusic.org offers both interesting and useful tips on ordered pair calculator, quadratic functions and equations in two variables and other math subjects. You need to solve each equation for x or y. In the Cartesian plane, we define a two-dimensional space with two perpendicular reference lines, namely x-axis and y-axis. This tutorial will introduce you to Find the values of (2x – 2, y – 1) = (x + 3, 4)? Choose 0 0 to substitute in for x x to find the ordered pair. This is what i have tried. Graph the function and its inverse. Then use the ordered pairs to graph the equation. In 2-sum problem, which avoids the binary-search for N times 3,5 ) $and draw a line them. Graph paper they cross at point 0 on the grid accordingly between the two commands,,... ” with coordinates 6, 4 and find the ordered pairs and coordinate worksheets. The one in 2-sum problem, which avoids the binary-search for N times the corresponding y.. Y axes on the y-axis of the equation of the inverse function: //www.MathTutorDVD.com line in pre- and. Pairs by two indexes approach is similar to the north, south east... Addresses of the points a value of that is assigned more than one value of value for x to! 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# How do you solve $\dfrac{5}{a}=\dfrac{30}{54}$ ?
Last updated date: 22nd Feb 2024
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Hint: In this question we need to simplify the given expression $\dfrac{5}{a}=\dfrac{30}{54}$ . For answering and finding the value of $a$ for that we will perform the transformations and simplify this by performing basic arithmetic simplifications and reduce it and get the value.
Complete step-by-step solution:
Now considering from the question we need to find the value of $a$ in the given expression $\dfrac{5}{a}=\dfrac{30}{54}$ .
For that we will find the value of $a$ . For that we will perform the transformations and simplify this by performing basic arithmetic simplifications and reduce it and get the value.
We will shift the variable $a$ from left hand side to right hand side after doing that we will have $\Rightarrow 5=\dfrac{30}{54}a$ .
Now we will shift the number $54$ from right hand side to left hand side after doing that we will have $\Rightarrow 5\times 54=30a$ .
Now we will shift the number $30$ from right hand side to left hand side after doing that we will have $\Rightarrow \dfrac{5\times 54}{30}=a$ .
Now we will further simplify the expression by performing simple basic arithmetic calculations like multiplication and division.
After doing that we will have $\Rightarrow \dfrac{54}{6}=a$ .
For further simplification we will perform the division we will have $\Rightarrow a=9$ .
Therefore we can conclude that the value of $a$ in the given expression $\dfrac{5}{a}=\dfrac{30}{54}$ is $9$.
Note: Here in this question we should be sure with the calculations and concepts during the process of solving the question. Similarly we can solve any expression. For example there is an expression like $\dfrac{3}{p}=\dfrac{15}{20}$ after solving this expression we will have $\Rightarrow 3=\dfrac{15}{20}p\Rightarrow \dfrac{3\times 20}{15}=p\Rightarrow p=4$ .
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# Algebra
Algebra is an ancient form of mathematical analytical methodology and is one of the most fundamental in our modern practice of analysis.
## Digits
Numbers are made of digits. Here are their names:
0 - zero
1 - one
2 - two
3 - three
4 - four
5 - five
6 - six
7 - seven
8 - eight
9 - nine
## Rules of arithmetic and algebra
The following laws are true for all ${\displaystyle a,b,c}$ whether these are numbers, variables, functions, or more complex expressions involving numbers, variable and/or functions.
• Commutative Law: ${\displaystyle a+b=b+a}$ .
• Associative Law: ${\displaystyle (a+b)+c=a+(b+c)}$ .
• Additive Identity: ${\displaystyle a+0=a}$ .
• Additive Inverse: ${\displaystyle a+(-a)=0}$ .
### Subtraction
• Definition: ${\displaystyle a-b=a+(-b)}$ .
### Multiplication
• Commutative law: ${\displaystyle a\times b=b\times a}$ .
• Associative law: ${\displaystyle (a\times b)\times c=a\times (b\times c)}$ .
• Multiplicative identity: ${\displaystyle a\times 1=a}$ .
• Multiplicative inverse: ${\displaystyle a\times {\frac {1}{a}}=1}$ , whenever ${\displaystyle a\neq 0}$
• Distributive law: ${\displaystyle a\times (b+c)=(a\times b)+(a\times c)}$ .
### Division
• Definition: ${\displaystyle {\frac {a}{b}}=a\times {\frac {1}{b}}}$ , whenever ${\displaystyle b\neq 0}$ .
Let's look at an example to see how these rules are used in practice.
${\displaystyle {\frac {(x+2)(x+3)}{x+3}}}$ ${\displaystyle =\left[(x+2)\times (x+3)\right]\times \left({\frac {1}{x+3}}\right)}$ (from the definition of division) ${\displaystyle =(x+2)\times \left[(x+3)\times \left({\frac {1}{x+3}}\right)\right]}$ (from the associative law of multiplication) ${\displaystyle =((x+2)\times (1)),\qquad x\neq -3}$ (from multiplicative inverse) ${\displaystyle =x+2,\qquad x\neq -3}$ (from multiplicative identity)
Of course, the above is much longer than simply cancelling ${\displaystyle x+3}$ out in both the numerator and denominator. But, when you are cancelling, you are really just doing the above steps, so it is important to know what the rules are so as to know when you are allowed to cancel. Occasionally people do the following, for instance, which is incorrect:
${\displaystyle {\frac {2\times (x+2)}{2}}={\frac {2}{2}}\times {\frac {x+2}{2}}=1\times {\frac {x+2}{2}}={\frac {x+2}{2}}}$ .
The correct simplification is
${\displaystyle {\frac {2\times (x+2)}{2}}=\left(2\times {\frac {1}{2}}\right)\times (x+2)=1\times (x+2)=x+2}$ ,
where the number ${\displaystyle 2}$ cancels out in both the numerator and the denominator.
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# 2013 AMC 10A Problems/Problem 12
## Problem
In $\triangle ABC$, $AB=AC=28$ and $BC=20$. Points $D,E,$ and $F$ are on sides $\overline{AB}$, $\overline{BC}$, and $\overline{AC}$, respectively, such that $\overline{DE}$ and $\overline{EF}$ are parallel to $\overline{AC}$ and $\overline{AB}$, respectively. What is the perimeter of parallelogram $ADEF$?
$[asy] size(180); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); real r=5/7; pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r); pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y)); pair E=extension(D,bottom,B,C); pair top=(E.x+D.x,E.y+D.y); pair F=extension(E,top,A,C); draw(A--B--C--cycle^^D--E--F); dot(A^^B^^C^^D^^E^^F); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,W); label("E",E,S); label("F",F,dir(0)); [/asy]$
$\textbf{(A) }48\qquad \textbf{(B) }52\qquad \textbf{(C) }56\qquad \textbf{(D) }60\qquad \textbf{(E) }72\qquad$
## Solution 1
Note that because $\overline{DE}$ and $\overline{EF}$ are parallel to the sides of $\triangle ABC$, the internal triangles $\triangle BDE$ and $\triangle EFC$ are similar to $\triangle ABC$, and are therefore also isosceles triangles.
It follows that $BD = DE$. Thus, $AD + DE = AD + DB = AB = 28$.
The opposite sides of parallelograms are equal (you can prove this fact simply by drawing the diagonal of the parallelogram and proving that the two resulting triangles are congruent by SSS), so the perimeter is $2 \times (AD + DE) = 56\implies \boxed{\textbf{(C)}}$.
## Solution 2
We can set $AD=0$, by fakesolving, we get $56\implies \boxed{\textbf{(C)}}$.
## Solution 3
Drawing the diagram with a ruler and compass (and scaling back by x4), we can draw approximate parallel lines. This yields about 14, but we need to multiply by 4 to get 56, or $\boxed{(C)}$.
~savannahsolver
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# The Basics of Addition: A Guide for Elementary School Students
The Basics of Addition: A Guide for Elementary School Students
Addition is a fundamental mathematical concept that is taught to elementary school students. It is one of the building blocks for more advanced math concepts and helps students develop their problem-solving skills. This guide will provide an overview of the basics of addition, including examples and frequently asked questions.
Introduction
Addition is a mathematical operation that involves combining two or more numbers together to find the total. It is one of the four basic operations of arithmetic, along with subtraction, multiplication, and division. Addition is an important concept to learn in elementary school because it is used as the foundation for more advanced math concepts.
Body
Addition is usually taught to students in the form of a number sentence. A number sentence is a mathematical equation that uses numbers and symbols to represent a mathematical operation. For example, the number sentence 5 + 3 = 8 is an addition equation. The first number (5) is called the addend, the second number (3) is called the augend, and the result (8) is called the sum.
In addition to number sentences, students also learn to use addition in word problems. Word problems are mathematical problems that are written in a sentence or paragraph form. They require students to read the problem and then use their addition skills to solve it. For example, a word problem might say: “John has 5 apples and his friend has 3 apples. How many apples do they have altogether?” In this case, the student would need to use addition to find the answer (8 apples).
Examples
Here are some examples of addition equations and word problems:
Equation: 7 + 3 = 10
Word Problem: Mary has 6 oranges and her brother has 4 oranges. How many oranges do they have altogether?
Equation: 5 + 8 = 13
Word Problem: John has 7 balloons and his friend has 3 balloons. How many balloons do they have altogether?
Equation: 10 + 5 = 15
Word Problem: Sarah has 8 books and her sister has 5 books. How many books do they have altogether?
FAQ Section
Q: What is the difference between an addend and an augend?
A: The addend is the first number in an addition equation. The augend is the second number in an addition equation.
Q: How can I remember the order of operations in an addition equation?
A: The order of operations in an addition equation is always the same: addend + augend = sum.
Q: What is the difference between addition and subtraction?
A: Addition is the process of combining two or more numbers together to find the total. Subtraction is the process of taking away one number from another to find the difference.
Summary
Addition is a fundamental mathematical concept that is taught to elementary school students. It is one of the four basic operations of arithmetic, along with subtraction, multiplication, and division. Addition is usually taught to students in the form of a number sentence or a word problem. The addend is the first number in an addition equation, the augend is the second number, and the sum is the result.
Conclusion
Addition is an important concept to learn in elementary school because it is used as the foundation for more advanced math concepts. With the help of this guide, students should now have a better understanding of the basics of addition and be able to solve equations and word problems with ease.
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# What is the Solution Set of the Quadratic Inequality
What is the Solution Set of the Quadratic Inequality
In algebra, solving a quadratic inequality is very like to solving a quadratic equation. The departure is that with quadratic equations, yous set the expressions equal to aught, but with inequalities, you’re interested in what’s on either side of the naught (positives and negatives).
1. Movement all the terms to one side of the inequality sign.
2. Cistron, if possible.
3. Decide all zeros (roots, or solutions).
Zeros are the values of the variable that make each factored expression equal to zero.
4. Put the zeros in order on a number line.
5. Create a sign line to show where the expression in the inequality is positive or negative.
A sign line shows the signs of the different factors in each interval. If the expression is factored, show the signs of the private factors.
6. Make up one’s mind the solution, writing information technology in inequality notation or interval notation.
Hither’due south an example: Solve the inequality
x
2
10
> 12.
Here, y’all need to determine what values of
x
y’all can square so that when you subtract the original number, your answer will exist bigger than 12.
1. Subtract 12 from each side of the inequality
ten
2
x
> 12 to move all the terms to one side.
You lot cease up with
10
2
x
– 12 > 0.
2. Factoring on the left side of the inequality, you lot get (10
– 4)(x
+ 3) > 0.
3. Determine that all the zeroes for the inequality are
x
= 4 and
x
= –3.
4. Put the zeros in order on a number line, shown in the following effigy.
5. Create a sign line to show the signs of the different factors in each interval.
Between –iii and 4, try letting
ten
= 0 (you tin utilize any number between –3 and iv). When
x
= 0, the factor (ten
– 4) is negative, and the gene (x
+ 3) is positive. Put those signs on the sign line to represent to the factors. Practice the same for the interval of numbers to the left of –3 and to the correct of iv (run across the side by side analogy).
Go along in heed that the
10
values in each interval are really random choices (as you lot can see from the choices in this example of
x
= –5 and
x
= 10). Whatsoever number in each of the intervals gives y’all the same positive or negative value to the gene.
6. To determine the solution, await at the signs of the factors; you want the expression to exist positive, corresponding to the inequality greater than zero.
The interval to the left of –3 has a negative times a negative, which is positive. So, any number to the left of –iii works. You tin can write that office of the solution as
10
< –3 or, in interval notation,
The interval to the correct of 4 has a positive times a positive, which is positive. So,
10
> four is a solution; you can write it as
The interval between –3 and 4 is always negative; yous have a negative times a positive. The complete solution lists both intervals that have working values in the inequality.
The solution of the inequality
x
two
x
> 12, therefore, is
ten
< –3 or
x
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and write it as
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# Reasoning - Direction Sense
These are direction-based reasoning questions. We have to judge the exact direction and then have to answer accordingly. Questions are direction-based means that they give information upon a movement of a person in certain direction for a particular distance. We have to judge the direction and distance of the person from the starting point.
The questions on logical sense test or direction sense test are designed to examine the candidate’s ability to sense direction by using pen and paper. In these types of questions normally there is data given about a person who is travelling in different directions and finally we have to calculate either the total distance travelled by the person or distance between final and initial position or the direction of the person from his initial position.
Questions on direction sense test are simpler than other questions that come in reasoning part and if a candidate has the right-hand side trick to solve it then he/she can score maximum in this section. To know the right-hand side trick first right-hand side knowledge should be there about the directions on paper. Confusion may be created in the questions as it is frequently asked about simultaneous left-hand side and right-hand side turns to a specific direction or movement in different degrees. So students are, therefore, advised to use the diagram as given below for clarification about the directions on paper.
The top of the paper is always referred as the north direction and right-hand side of the paper is always east direction. The opposite of north or the bottom side of the paper is south and left-hand side of the paper is referred as west direction. The direction between north and east is north-east direction, south and east is south-east direction and so on.
Question 1 − Ram started from his home and moves in eastward direction. After walking some distance he turned to his right-hand side and walked, after walking some distance he turned to his left-hand side and walked. After some time he has taken two simultaneous left-hand side ward movement and after walking some distance he again taken two left-hand side movements simultaneously. Now he is facing to which direction. All the turns are 90 degrees.
Options
A - East
B - West
C - North
D - South
Explanation − First Ram started walking to east then he took a right-hand side turn. Now he is facing south. After walking some distance, he turned left-hand side so after turning now he is facing east. After this, he has taken two left-hand side movement simultaneously so now he is facing towards west direction. And finally after walking some distance he has again moved to his left-hand side twice so finally after his all movements, he is currently facing towards east direction.
Question 2 − Krishna is going to a local market place for grocery. He has taken his cycle and from home started moving to north. He went for 10 km. After that he turned 90 degrees to his left-hand side and cycled for 6 km. After that he again moved 90 degree to his left-hand side and cycled for 18 km. Now he has reached the market place. What is distance between Krishna’s home and market place and he is in which direction from his home?
Options
A - North-east, 15 km
B - North-west, 12 km
C - South-west, 10 km
D - South-east, 10 km
Explanation − The distance between Krishna’s house and the market place can be found out by using Pythagoras theorem 82 + 62 = 102. The direction in which Krishna is standing is south-west direction from his home. First he cycled towards north for 10 km and he turned towards his left-hand side and after that he turned his left-hand side and walked for 18 km. so finally he is in south-west direction because by walking 18 km he has surpassed his 10 km. So the correct answer is option C.
Question 3 − Pirnakanti walks 500 m to east and reaches Lata’s house. From there she walks towards north for 1000 m to reach his gym. His gym is in which direction of his home and what is the total distance travelled by Pirnakanti?
Options
A - North, 1 km
B - West, 2 km
C - North-east, 1.5 km
D - North-west, 1.5 km
Explanation − First Pirnakanti has walked for 500 m and then he has walked for 1000 m, so in total he has walked for 1500 m or 1.5 km. Pirnakanti first walked to east direction and after some time she has turned left-hand side and walked for 1000m towards north direction. So now she is currently in north-east direction from her earlier position. Hence option C is correct.
Question 4 − Salim started to move in the direction of west in an open field and stopped after moving 5km. Now he turned to his left-hand side and moved 15 km. How much minimum distance he has to cover to reach his starting point?
Options
A - 25km
B - 20km
C - 30km
D - 15km
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# McGraw Hill Math Grade 4 Chapter 2 Lesson 5 Answer Key Subtracting Whole Numbers
Practice the questions of McGraw Hill Math Grade 4 Answer Key PDF Chapter 2 Lesson 5 Subtracting Whole Numbers to secure good marks & knowledge in the exams.
## McGraw-Hill Math Grade 4 Answer Key Chapter 2 Lesson 5 Subtracting Whole Numbers
Subtract
Question 1.
Explanation:
Difference between 762 and 328:
762 – 328 = 434.
Question 2.
Explanation:
Difference between 8,295 and 3,667:
8,295 – 3,667 = 4,628.
Question 3.
Explanation:
Difference between 87,471 and 53,743:
87,471 – 53,743 = 33,728.
Question 4.
Explanation:
Difference between 630,119 and 228,754:
630,119 – 228,754 = 401,365.
Question 5.
Explanation:
Difference between 926,138 and 67,880:
926,138 – 67,880 = 858,258.
Question 6.
Explanation:
Difference between 725,914 and 699,311:
725,914 – 699,311 = 26,603.
Question 7.
238,960 – 729 = ___________
Difference between 238,960 – 729 we get 238,231.
Explanation:
238,960 – 729 = 238,231.
Question 8.
46,008 — 19,725 = _____
Difference between 46,008 and 19,725 we get 26,283.
Explanation:
46,008 – 19,725 = 26,283.
Question 9.
52,937 — 8,658 = ___________
Difference between 52,937 and 8,658 we get 44,279.
Explanation:
52,937 – 8,658 = 44,279.
Question 10.
783,209 — 97,120 = ____
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# How do you solve for each of the variable 5L= 10L- 15D+ 5?
Jul 30, 2015
You take turns isolating them on one side of the equation.
#### Explanation:
Your starting equation looks like this
$5 L = 10 L - 15 D + 5$
To solve this equation for $L$, get all the terms that contain this variable on one side of the equation.
In your case, you can do this by adding $- 10 L$ to both sides of the equation
$5 L - 10 L = \textcolor{red}{\cancel{\textcolor{b l a c k}{10 L}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{10 L}}} - 15 D + 5$
$- 5 L = - 15 D + 5$
Now divide both sides of the equation by $- 5$ to get
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 5}}} L}{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 5}}}} = \frac{- 15 D}{- 5} + \frac{5}{- 5}$
$L = \textcolor{g r e e n}{3 D - 1}$
Now do the same for $D$. Notice that you can write
$5 L = 10 L + 5 - 15 D$, so you could add $- 10 L - 5$ to both sides of the equation to get
$5 L - 10 L - 5 = \textcolor{red}{\cancel{\textcolor{b l a c k}{- 10 L - 5}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{10 L + 5}}} - 15 D$
This is equivalent to
$- 15 D = - 5 L - 5$
Finally, divide both sides of the equation by $- 15$ to get
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 15}}} D}{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 15}}}} = \frac{- 5 L}{- 15} - \frac{5}{- 15}$
$D = \frac{1}{3} L + \frac{1}{3}$, or
$D = \textcolor{g r e e n}{\frac{1}{3} \left(L + 1\right)}$
|
# Difference between revisions of "2008 AIME I Problems/Problem 11"
## Problem
Consider sequences that consist entirely of $A$'s and $B$'s and that have the property that every run of consecutive $A$'s has even length, and every run of consecutive $B$'s has odd length. Examples of such sequences are $AA$, $B$, and $AABAA$, while $BBAB$ is not such a sequence. How many such sequences have length 14?
## Solution
### Solution 1
Let $a_n$ and $b_n$ denote, respectively, the number of sequences of length $n$ ending in $A$ and $B$. If a sequence ends in an $A$, then it must have been formed by appending two $A$s to the end of a string of length $n-2$. If a sequence ends in a $B,$ it must have either been formed by appending one $B$ to a string of length $n-1$ ending in an $A$, or by appending two $B$s to a string of length $n-2$ ending in a $B$. Thus, we have the recursions \begin{align*} a_n &= a_{n-2} + b_{n-2}\\ b_n &= a_{n-1} + b_{n-2} \end{align*} By counting, we find that $a_1 = 0, b_1 = 1, a_2 = 1, b_2 = 0$. $$\begin{array}{|r||r|r|||r||r|r|} \hline n & a_n & b_n & n & a_n & b_n\\ \hline 1&0&1& 8&6&10\\ 2&1&0& 9&11&11\\ 3&1&2& 10&16&21\\ 4&1&1& 11&22&27\\ 5&3&3& 12&37&43\\ 6&2&4& 13&49&64\\ 7&6&5& 14&80&92\\ \hline \end{array}$$ Therefore, the number of such strings of length $14$ is $a_{14} + b_{14} = \boxed{172}$.
### Solution 2
We replace "14" with "$2n$". We first note that we must have an even number of chunks of $B$'s, because of parity issues. We then note that every chunk of $B$'s except the last one must end in the sequence $BAA$, since we need at least two $A$'s to separate it from the next chunk of $B$'s. The last chunk of $B$'s must, of course, end with a $B$. Thus our sequence must look like this : $$\boxed{\quad A\text{'s} \quad} \boxed{\quad B\text{'s} \quad} BAA \boxed{\quad A\text{'s} \quad} \cdots \boxed{\quad B\text{'s} \quad}BAA \boxed{\quad A\text{'s} \quad} \boxed{\quad B\text{'s} \quad} B \boxed{\quad A\text{'s} \quad} ,$$ where each box holds an even number of letters (possibly zero).
If we want a sequence with $2k$ chunks of $B$'s, then we have $(2n - 6k + 2)$ letters with which to fill the boxes. Since each box must have an even number of letters, we may put the letters in the boxes in pairs. Then we have $(n - 3k + 1)$ pairs of letters to put into $4k + 1$ boxes. By a classic balls-and-bins argument, the number of ways to do this is $$\binom{n + k + 1}{4k} .$$ It follows that the total number of desirable sequences is $$\sum_k \binom{n + k + 1}{4k} .$$ For $n = 7$, this evaluates as $\binom{8}{0} + \binom{9}{4} + \binom{10}{8} = 1 + 126 + 45 = \boxed{172}$.
2008 AIME I (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
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# Tutorial on Slope of a Line
This is a tutorial on the slope of a line with examples and detailed solutions, and matched exercises also with solutions. You may wish to go through a tutorial on the concept of the slope of a line first.
## Examples with Solutions
### Example 1
Find the slope of the line passing through the pairs of points and describe the line as rising, falling, horizontal or vertical.
1. (2 , 1) , (4 , 5)
2. (-1 , 0) , (3 , -5)
3. (2 , 1) , (-3 , 1)
4. (-1 , 2) , (-1 ,- 5)
#### Solution to Example 1
1. The slope of the line is given by
m = ( 5 - 1 ) / (4 - 2) = 4 / 2 = 2
Since the slope is positive, the line rises as x increases.
2. The slope of the line is given by
m = ( -5 - 0 ) / ( 3 - (-1) ) = -5 / 4
Since the slope is negative, the line falls as x increases.
3. We first find the slope of the line
m = ( 1 - 1 ) / ( -3 - 2 ) = 0
Since the slope is equal to zero, the line is horizontal (parallel to the x axis).
4. The slope of the line is given by
m = ( -5 - 2 ) / ( -1 - (-1) )
Since ( -1 - (-1) ) = 0 and the division by 0 is not defined, the slope of the line is undefined and the line is vertical. (parallel to the y axis).
Matched Exercise to Example 1
Find the slope of the line passing through the pairs of points and describe the line as rising, falling, horizontal or vertical.
Solution
1. (3 , -1) , (3 , 5)
2. (-1 , 0) , (3 , 7)
3. (2 , 1) , (6 , 0)
4. (-5 , 2) , (9 , 2)
### Example 2
A line has a slope of -2 and passes through the point (2 , 5). Find another point A through which the line passes. (many possible answers)
#### Solution to Example 2
1. Let x1 and y1 be the x and y coordinates of point A. According to the definition of the slope
( y1 - 5 ) / (x1 - 2) = -2
2. We need to solve this equation in order to find x1 and y1. This equation has two unknowns and therefore has an infinite number of pairs of solutions. We chose x1 and then find y1. if x1 = -1, for example, the above equation becomes
( y1 - 5 ) / (-1 - 2) = -2
3. We obtain an equation in y1
( y1 - 5 ) / -3 = -2
4. Solve for y1 to obtain
y1 = 11
5. One possible answer is point A at
(-1 , 11)
6. Check that the two points give a slope of -2
(11 - 5 ) / (-1 - 2) = 6/-3 = -2
Matched Exercise to Example 2
A line has a slope of 5 and passes through the point (1 , -4). Find another point A through which the line passes. (many possible answers)
### Example 3
Are the lines L1 and L2 passing through the given pairs of points parallel, perpendicular or neither parallel nor perpendicular?
1. L1: (1 , 2) , (3 , 1)
L2: (0 , -1) , (2 , 0)
2. L1: (0 , 3) , (3 , 1)
L2: (-1 , 4) , (-7 , -5)
3. L1: (2 , -1) , (5 , -7)
L2: (0 , 0) , (-1 , 2)
4. L1: (1 , 0) , (2 , 0)
L2: (5 , -5) , (-10 , -5)
5. L1: (-2 , 5) , (-2 , 7)
L2: (5 , 1) , (5 , 13)
#### Solution Example 3
In what follows, m1 is the slope of line L1 and m2 is the slope of line L2.
1. Find the slope m1 of line L1 and the slope m2 of line L1
m1 = ( 1 - 2 ) / ( 3 - 1 ) = -1 / 2
m2 = ( 0 - (-1) ) / ( 2 - 0 ) = 1/2
The two slopes m1 and m2 are not equal and their products is not equal to -1. Hence the two lines are neither parallel nor perpendicular.
2. m1 = ( 1 - 3 ) / ( 3 - 0 ) = -2 / 3
m2 = ( -5 - 4 ) / ( -7 - (-1) ) = -9 / -6 = 3/2
The product of the two slopes m1*m2 = (-2 / 3)(3 / 2) = -1, the two lines are perpendicular.
3. m1 = ( -7 - (-1) ) / ( 5 - 2 ) = -6 / 3 = -2
m2 = ( 2 - 0 ) / ( -1 - 0 ) = -2
The two slopes are equal, the two lines are parallel.
4. m1 = ( 0 - 0 ) / ( 2 - 1 ) = 0 / 1 = 0
m2 = ( -5 - (-5) ) / ( -10 - 5 ) = 0 / -15 = 0
The two slopes are equal , the two lines are parallel. Also the two lines are horizontal
5. m1 = ( 7 - 5 ) / ( -2 - (-2) )
m2 = ( 13 - 1 ) / ( 5 - 5 )
The two slopes are both undefined since the denominators in both m1 and m2 are equal to zero. The two lines are vertical lines and therefore parallel.
Matched Exercise to Example 3
Are the lines L1 and L2 passing through the given pairs of points parallel, perpendicular or neither parallel nor perpendicular?
1. L1: (1 , 2) , (1 , 1)
L2: (-4 , -1) , (-4 , 0)
2. L1: (2 , 3) , (3 , 1)
L2: (1 , -2) , (7 , -5)
3. L1: (1 , -1) , (2 , -2)
L2: (0 , 0) , (1 , 1)
4. L1: (1 , 9) , (-2 , 9)
L2: (18 , -1) , (0 , -1)
Solution
### Example 4
Is it possible for two lines with negative slopes to be perpendicular?
#### Solution to Example 4
No. If both slopes are negative, their product can never be equal to -1.
Matched Exercise to Example 4
Is it possible for two lines with positive slopes to be perpendicular?
Solution
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# The value of $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 - \cos x} }}{x}$ is equal toA.$- \dfrac{1}{{\sqrt 2 }}$B.$\dfrac{1}{{\sqrt 2 }}$C.$0$D.Does not exist
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Hint: Here in this question, we have to determine the given limit of a function. To find this first we have to write the given function using a trigonometric double or half angle formula $\cos 2x = 1 - 2{\sin ^2}x$ then limit of a function $f$ Which is satisfies the condition left hand limit is equal to right hand limit (i.e., $LHL = RHL$) by applying a limit in to the function using the properties of limits, otherwise limits doesn’t exist
Complete step by step solution:
The limit of a function exists if and only if the left-hand limit is equal to the right-hand limit.
A left-hand limit means the limit of a function as it approaches from the left-hand side.
$\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = {l_1}$
A right-hand limit means the limit of a function as it approaches from the right-hand side.
$\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = {l_2}$
Consider the given limit function,
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 - \cos x} }}{x}$------(1)
By using a double of half angle formula: $\cos 2x = 1 - 2{\sin ^2}x \Rightarrow 1 - \cos x = 2{\sin ^2}\dfrac{x}{2}$, then
As we know, $\sqrt {1 - \cos x} = \left\{ {\begin{array}{*{20}{c}} { - \sqrt 2 \sin \dfrac{x}{2},\,\,\,\,\,\,\,x < 0} \\ {\,\,\sqrt 2 \sin \dfrac{x}{2},\,\,\,\,\,\,\,x \geqslant 0} \end{array}} \right.$
Now, find the left-hand limit to the function (1)
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sqrt {1 - \cos x} }}{x}$
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} - \dfrac{{\sqrt 2 \sin \dfrac{x}{2}}}{x}$
By applying a properties of limit function, we have
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - \sqrt 2 \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sin \dfrac{x}{2}}}{x}$
Multiply and divide by $\dfrac{1}{2}$, then
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - \sqrt 2 \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sin \dfrac{x}{2}}}{x} \times \dfrac{{\left( {\dfrac{1}{2}} \right)}}{{\left( {\dfrac{1}{2}} \right)}}$
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - \sqrt 2 \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\left( {\dfrac{1}{2}} \right)\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}$
Again, by the properties of limit, we have
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - \sqrt 2 \left( {\dfrac{1}{2}} \right)\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}$
As we know the standard limit $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$, then
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - \dfrac{{\sqrt 2 }}{2}$
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - \dfrac{1}{{\sqrt 2 }}$---------(2)
Now, find the right-hand limit to the function (1)
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sqrt {1 - \cos x} }}{x}$
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sqrt 2 \sin \dfrac{x}{2}}}{x}$
By applying a properties of limit function, we have
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \sqrt 2 \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sin \dfrac{x}{2}}}{x}$
Multiply and divide by $\dfrac{1}{2}$, then
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = - \sqrt 2 \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sin \dfrac{x}{2}}}{x} \times \dfrac{{\left( {\dfrac{1}{2}} \right)}}{{\left( {\dfrac{1}{2}} \right)}}$
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = - \sqrt 2 \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\left( {\dfrac{1}{2}} \right)\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}$
Again, by the properties of limit, we have
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = - \sqrt 2 \left( {\dfrac{1}{2}} \right)\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}$
As we know the standard limit $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$, then
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = - \dfrac{{\sqrt 2 }}{2}$
$\Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \dfrac{1}{{\sqrt 2 }}$---------(3)
Since, by (2) and (3)
$\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sqrt {1 - \cos x} }}{x} \ne \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sqrt {1 - \cos x} }}{x}$
$LHL \ne RHL$
Hence, limit does not exist
Therefore, option (D) is correct
So, the correct answer is “Option D”.
Note: Remember, the limit of any function exists between any two consecutive integers. And the product and quotient properties of limits are defined as:
The function $f\left( x \right)$ and $g\left( x \right)$ is are non-zero finite values, given that
$\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) \cdot g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) \cdot \mathop {\lim }\limits_{x \to a} g\left( x \right)$
$\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}$ and
Also $\mathop {\lim }\limits_{x \to a} k\,f\left( a \right) = k\mathop {\lim }\limits_{x \to a} f\left( a \right)$.
|
# Chapter Notes: Roman Numerals Notes | Study Mathematics for Class 4 - Class 4
## Class 4: Chapter Notes: Roman Numerals Notes | Study Mathematics for Class 4 - Class 4
The document Chapter Notes: Roman Numerals Notes | Study Mathematics for Class 4 - Class 4 is a part of the Class 4 Course Mathematics for Class 4.
All you need of Class 4 at this link: Class 4
What are Roman Numerals?
Roman Numerals was a system that was the way that the Roman's would write different numbers. Instead of using numbers.
• The Hindu-Arabic numeration system uses 10 digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 to write any other numeral.
• The Romans had seven basic symbols to write any number. These seven symbols are I, V, X, L, C, D, and M.
Basic Roman NumeralsThe Roman numerals with their corresponding Hindu-Arabic numerals are as follows:
• These basic Roman symbols are combined in a specific manner to form other Roman numerals.
• Roman numerals are used in some analogue clocks and watches, to name the classrooms in the schools, for the copyright date on films programmes, etc.
One to Twelve Roman Numbers written on a watch.
Try yourself:Which of the following options is the Hindu-Arabic form for the Roman numeral XXXIV?
Rules for Forming Roman Numerals
The following rules are used to write the numbers in the Roman system.
Rule Number 1
When a Roman numeral is repeated more than once, we add its value each time to get the number.
Examples:
• II = 1 + 1 = 2,
• XX = 10 + 10 = 20,
• XXX = 10 + 10 + 10 = 30
Edurev Tips:
• The same symbol is not repeated more than 3 times together.
• The symbols V, L and D are never repeated.
• There is no symbol for zero in the Roman system.
• Symbols I, X and V are used to write Roman numerals up to 39.
Try yourself:Which of the following options is not written correctly according to the Roman numerals rules?
Rule Number 2
When a smaller Roman numeral is written to the right of a greater Roman numeral, the value of the greater numeral is added to the smaller numeral.
Examples:
• VI = 5 + 1 = 6,
• XII = 10 + 1 + 1 = 12,
• XXXVI = 10 + 10 + 10 + 5 + 1 = 36
Rule Number 3
When a smaller Roman numeral is written to the left of a greater Roman numeral, it is always subtracted from the value of the greater numeral.
Examples:
• IV = 5 – 1 = 4,
• IX = 10 – 1 = 9
Edurev Tips:
• Symbol I can be subtracted only once.
• The symbols V, L and D are never subtracted.
Rule Number 4
When a smaller Roman numeral is placed between two greater Roman numerals, then its value is subtracted from the value of the greater numeral on its right.
Examples:
• XIX = 10 + (10 – 1) = 10 + 9 = 19
• XXIV = 10 + 10 + (5 – 1) = 20 + 4 = 24
Example 1: Write the following in Hindu-Arabic numerals.
(a) XII
(b) VII
(c) XV
(d) XIV
(a) XII = 10 + 1 + 1 = 12
(b) VII = 5 + 1 + 1 = 7
(c) XV = 10 + 5 = 15
(d) XIV = 10 + (5 – 1) = 14
Example 2: Write the following in Roman numerals.
(a) 37
(b) 16
(c) 28
(d) 34
(a) 37 = 30 + 7 = X + X + X + V + I + I = XXXVII
(b) 16 = 10 + 6 = X + V + I = XVI
(c) 28 = 20 + 8 = X + X + V + I + I + I = XXVIII
(d) 34 = 30 + 4 = X + X + X + IV = XXXIV
The document Chapter Notes: Roman Numerals Notes | Study Mathematics for Class 4 - Class 4 is a part of the Class 4 Course Mathematics for Class 4.
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3
Q:
# 9 men and 12 boys finish a job in 12 days, 12 men and 12 boys finish it in 10 days. 10 men and 10 boys shall finish it in how many days ?
A) 15 days B) 11 days C) 14 days D) 12 days
Explanation:
9M + 12B ----- 12 days ...........(1)
12M + 12B ------- 10 days........(2)
10M + 10B -------?
108M + 144B = 120M +120B
24B = 12M => 1M = 2B............(3)
From (1) & (3)
18B + 12B = 30B ---- 12 days
20B + 10B = 30B -----? => 12 days.
Q:
70000 a year is how much an hour?
A) 80 B) 8 C) 0.8 D) 0.08
Explanation:
Given for year = 70000
=> 365 days = 70000
=> 365 x 24 hours = 70000
=> 1 hour = ?
70000/365x24 = 7.990 = 8
0 95
Q:
A, B and C can do a piece of work in 72, 48 and 36 days respectively. For first p/2 days, A & B work together and for next ((p+6))/3days all three worked together. Remaining 125/3% of work is completed by D in 10 days. If C & D worked together for p day then, what portion of work will be remained?
A) 1/5 B) 1/6 C) 1/7 D) 1/8
Explanation:
Total work is given by L.C.M of 72, 48, 36
Total work = 144 units
Efficieny of A = 144/72 = 2 units/day
Efficieny of B = 144/48 = 3 units/day
Efficieny of C = 144/36 = 4 units/day
According to the given data,
2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100
3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54
p = 198/16.5
p = 12 days.
Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day
(C+D) in p days = (4 + 6) x 12 = 120 unit
Remained part of work = (144-120)/144 = 1/6.
4 545
Q:
10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?
A) 215 days B) 225 days C) 235 days D) 240 days
Explanation:
Given that
(10M + 15W) x 6 days = 1M x 100 days
=> 60M + 90W = 100M
=> 40M = 90W
=> 4M = 9W.
From the given data,
1M can do the work in 100 days
=> 4M can do the same work in 100/4= 25 days.
=> 9W can do the same work in 25 days.
=> 1W can do the same work in 25 x 9 = 225 days.
Hence, 1 woman can do the same work in 225 days.
7 903
Q:
A,B,C can complete a work in 15,20 and 30 respectively.They all work together for two days then A leave the work,B and C work some days and B leaves 2 days before completion of that work.how many days required to complete the whole work?
Given A,B,C can complete a work in 15,20 and 30 respectively.
The total work is given by the LCM of 15, 20, 30 i.e, 60.
A's 1 day work = 60/15 = 4 units
B's 1 day work = 60/20 = 3 units
C's 1 day work = 60/30 = 2 units
(A + B + C) worked for 2 days = (4 + 3 + 2) 2 = 18 units
Let B + C worked for x days = (3 + 2) x = 5x units
C worked for 2 days = 2 x 2 = 4 units
Then, 18 + 5x + 4 = 60
22 + 5x = 60
5x = 38
x = 7.6
Therefore, total number of days taken to complete the work = 2 + 7.6 + 2 = 11.6 = 11 3/5 days.
522
Q:
M, N and O can complete the work in 18, 36 and 54 days respectively. M started the work and worked for 8 days, then N and O joined him and they all worked together for some days. M left the job one day before completion of work. For how many days they all worked together?
A) 4 B) 5 C) 3 D) 6
Explanation:
Let M, N and O worked together for x days.
From the given data,
M alone worked for 8 days
M,N,O worked for x days
N, O worked for 1 day
But given that
M alone can complete the work in 18 days
N alone can complete the work in 36 days
O alone can complete the work in 54 days
The total work can be the LCM of 18, 6, 54 = 108 units
M's 1 day work = 108/18 = 6 units
N's 1 day work = 108/36 = 3 units
O's 1 day work = 108/54 = 2 units
Now, the equation is
8 x 6 + 11x + 5 x 1 = 108
48 + 11x + 5 = 108
11x = 103 - 48
11x = 55
x = 5 days.
Hence, all M,N and O together worked for 5 days.
2 506
Q:
P, Q, and R can do a job in 12 days together. If their efficiency of working be in the ratio 3 : 8 : 5, Find in what time Q can complete the same work alone?
A) 36 days B) 30 days C) 24 days D) 22 days
Explanation:
Given the ratio of efficiencies of P, Q & R are 3 : 8 : 5
Let the efficiencies of P, Q & R be 3x, 8x and 5x respectively
They can do work for 12 days.
=> Total work = 12 x 16x = 192x
Now, the required time taken by Q to complete the job alone = days.
5 649
Q:
5 men and 3 boys can together cultivate a 23 acre field in 4 days and 3 men and 2 boys together can cultivate a 7 acre field in 2 days. How many boys will be needed together with 7 men, if they cultivate 45 acre of field in 6 days.
A) 6 B) 4 C) 2 D) 3
Explanation:
Let work done by 1 man in i day be m
and Let work done by 1 boy in 1 day be b
From the given data,
4(5m + 3b) = 23
20m + 12b = 23....(1)
2(3m + 2b) = 7
6m + 4b = 7 ....(2)
By solving (1) & (2), we get
m = 1, b = 1/4
Let the number of required boys = n
6(7 1 + n x 1/4) = 45
=> n = 2.
5 641
Q:
The ratio of efficiencies of P, Q and R is 2 : 3 : 4. While P and R work on alternate days and Q work for all days. Now the work completed in total 10 days and the total amount they get is Rs. 1200. Find the amount of each person(respectively).
A) 200, 600, 400 B) 400, 600, 200 C) 600, 200, 400 D) 400, 200, 600
Explanation:
Ratio of efficiencies of P, Q and R = 2 : 3 : 4
From the given data,
Number of working days of P, Q, R = 5 : 10 : 5
Hence, ratio of amount of p, Q, R = 2x5 : 3x10 : 4x5 = 10 : 30 : 20
Amounts of P, Q, R = 200, 600 and 400.
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Related Articles
# Expected number of moves to reach the end of a board | Matrix Exponentiation
• Last Updated : 09 Jun, 2021
Given a linear board of length N numbered from 1 to N, the task is to find the expected number of moves required to reach the Nth cell of the board, if we start at cell numbered 1 and at each step we roll a cubical dice to decide the next move. Also, we cannot go outside the bounds of the board. Note that the expected number of moves can be fractional.
Examples:
Input: N = 8
Output:
p1 = (1 / 6) | 1-step -> 6 moves expected to reach the end
p2 = (1 / 6) | 2-steps -> 6 moves expected to reach the end
p3 = (1 / 6) | 3-steps -> 6 moves expected to reach the end
p4 = (1 / 6) | 4-steps -> 6 moves expected to reach the end
p5 = (1 / 6) | 5-steps -> 6 moves expected to reach the end
p6 = (1 / 6) | 6-steps -> 6 moves expected to reach the end
If we are 7 steps away, then we can end up at 1, 2, 3, 4, 5, 6 steps
away with equal probability i.e. (1 / 6).
Look at the above simulation to understand better.
dp[N – 1] = dp[7]
= 1 + (dp[1] + dp[2] + dp[3] + dp[4] + dp[5] + dp[6]) / 6
= 1 + 6 = 7
Input: N = 10
Output: 7.36111
Approach: An approach based on dynamic programming has already been discussed in an earlier post. In this article, a more optimized method to solve this problem will be discussed. The idea is using a technique called Matrix-Exponentiation
Let’s define An as the expected number of moves to reach the end of a board of length N + 1
The recurrence relation will be:
An = 1 + (An-1 + An-2 + An-3 + An-4 + An-5 + An-6) / 6
Now, the recurrence relation needs to be converted in a suitable format.
An = 1 + (An-1 + An-2 + An-3 + An-4 + An-5 + An-6) / 6 (equation 1)
An-1 = 1 + (An-2 + An-3 + An-4 + An-5 + An-6 + An-7) / 6 (equation 2)
Substracting 1 with 2, we get An = 7 * (An-1) / 6 – (An-7) / 6
Matrix exponentiation technique can be applied here on the above recurrence relation.
Base will be {6, 6, 6, 6, 6, 6, 0} corresponding to {A6, A5, A4…, A0}
Multiplier will be
{7/6, 1, 0, 0, 0, 0, 0},
{0, 0, 1, 0, 0, 0, 0},
{0, 0, 0, 1, 0, 0, 0},
{0, 0, 0, 0, 1, 0, 0},
{0, 0, 0, 0, 0, 1, 0},
{0, 0, 0, 0, 0, 0, 1},
{-1/6, 0, 0, 0, 0, 0, 0}
To find the value:
• Find mul(N – 7)
• Find base * mul(N – 7).
• The first value of the 1 * 7 matrix will be the required answer.
Below is the implementation of the above approach:
## C++
`// C++ implementation of the approach``#include ``#define maxSize 50``using` `namespace` `std;` `// Function to multiply two 7 * 7 matrix``vector > matrix_product(`` ``vector > a,`` ``vector > b)``{`` ``vector > c(7);`` ``for` `(``int` `i = 0; i < 7; i++)`` ``c[i].resize(7, 0);` ` ``for` `(``int` `i = 0; i < 7; i++)`` ``for` `(``int` `j = 0; j < 7; j++)`` ``for` `(``int` `k = 0; k < 7; k++)`` ``c[i][j] += a[i][k] * b[k][j];`` ``return` `c;``}` `// Function to perform matrix exponentiation``vector > mul_expo(vector > mul, ``int` `p)``{` ` ``// 7 * 7 identity matrix`` ``vector > s = { { 1, 0, 0, 0, 0, 0, 0 },`` ``{ 0, 1, 0, 0, 0, 0, 0 },`` ``{ 0, 0, 1, 0, 0, 0, 0 },`` ``{ 0, 0, 0, 1, 0, 0, 0 },`` ``{ 0, 0, 0, 0, 1, 0, 0 },`` ``{ 0, 0, 0, 0, 0, 1, 0 },`` ``{ 0, 0, 0, 0, 0, 0, 1 } };` ` ``// Loop to find the power`` ``while` `(p != 1) {`` ``if` `(p % 2 == 1)`` ``s = matrix_product(s, mul);`` ``mul = matrix_product(mul, mul);`` ``p /= 2;`` ``}` ` ``return` `matrix_product(mul, s);``}` `// Function to return the required count``double` `expectedSteps(``int` `x)``{` ` ``// Base cases`` ``if` `(x == 0)`` ``return` `0;`` ``if` `(x <= 6)`` ``return` `6;` ` ``// Multiplier matrix`` ``vector > mul = { { (``double``)7 / 6, 1, 0, 0, 0, 0, 0 },`` ``{ 0, 0, 1, 0, 0, 0, 0 },`` ``{ 0, 0, 0, 1, 0, 0, 0 },`` ``{ 0, 0, 0, 0, 1, 0, 0 },`` ``{ 0, 0, 0, 0, 0, 1, 0 },`` ``{ 0, 0, 0, 0, 0, 0, 1 },`` ``{ (``double``)-1 / 6, 0, 0, 0, 0, 0, 0 } };` ` ``// Finding the required multiplier`` ``// i.e mul^(X-6)`` ``mul = mul_expo(mul, x - 6);` ` ``// Final answer`` ``return` `(mul[0][0]`` ``+ mul[1][0]`` ``+ mul[2][0]`` ``+ mul[3][0]`` ``+ mul[4][0]`` ``+ mul[5][0])`` ``* 6;``}` `// Driver code``int` `main()``{`` ``int` `n = 10;` ` ``cout << expectedSteps(n - 1);` ` ``return` `0;``}`
## Java
`// Java implementation of the approach``class` `GFG``{``static` `final` `int` `maxSize = ``50``;` `// Function to multiply two 7 * 7 matrix``static` `double` `[][] matrix_product(``double` `[][] a,`` ``double` `[][] b)``{`` ``double` `[][] c = ``new` `double``[``7``][``7``];` ` ``for` `(``int` `i = ``0``; i < ``7``; i++)`` ``for` `(``int` `j = ``0``; j < ``7``; j++)`` ``for` `(``int` `k = ``0``; k < ``7``; k++)`` ``c[i][j] += a[i][k] * b[k][j];`` ``return` `c;``}` `// Function to perform matrix exponentiation``static` `double` `[][] mul_expo(``double` `[][] mul, ``int` `p)``{` ` ``// 7 * 7 identity matrix`` ``double` `[][] s = {{ ``1``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0` `},`` ``{ ``0``, ``1``, ``0``, ``0``, ``0``, ``0``, ``0` `},`` ``{ ``0``, ``0``, ``1``, ``0``, ``0``, ``0``, ``0` `},`` ``{ ``0``, ``0``, ``0``, ``1``, ``0``, ``0``, ``0` `},`` ``{ ``0``, ``0``, ``0``, ``0``, ``1``, ``0``, ``0` `},`` ``{ ``0``, ``0``, ``0``, ``0``, ``0``, ``1``, ``0` `},`` ``{ ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``1` `}};` ` ``// Loop to find the power`` ``while` `(p != ``1``)`` ``{`` ``if` `(p % ``2` `== ``1``)`` ``s = matrix_product(s, mul);`` ``mul = matrix_product(mul, mul);`` ``p /= ``2``;`` ``}`` ``return` `matrix_product(mul, s);``}` `// Function to return the required count``static` `double` `expectedSteps(``int` `x)``{` ` ``// Base cases`` ``if` `(x == ``0``)`` ``return` `0``;`` ``if` `(x <= ``6``)`` ``return` `6``;` ` ``// Multiplier matrix`` ``double` `[][]mul = { { (``double``)``7` `/ ``6``, ``1``, ``0``, ``0``, ``0``, ``0``, ``0` `},`` ``{ ``0``, ``0``, ``1``, ``0``, ``0``, ``0``, ``0` `},`` ``{ ``0``, ``0``, ``0``, ``1``, ``0``, ``0``, ``0` `},`` ``{ ``0``, ``0``, ``0``, ``0``, ``1``, ``0``, ``0` `},`` ``{ ``0``, ``0``, ``0``, ``0``, ``0``, ``1``, ``0` `},`` ``{ ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``1` `},`` ``{(``double``) - ``1` `/ ``6``, ``0``, ``0``,`` ``0``, ``0``, ``0``, ``0` `}};` ` ``// Finding the required multiplier`` ``// i.e mul^(X-6)`` ``mul = mul_expo(mul, x - ``6``);` ` ``// Final answer`` ``return` `(mul[``0``][``0``] + mul[``1``][``0``] + mul[``2``][``0``] +`` ``mul[``3``][``0``] + mul[``4``][``0``] + mul[``5``][``0``]) * ``6``;``}` `// Driver code``public` `static` `void` `main(String[] args)``{`` ``int` `n = ``10``;` ` ``System.out.printf(``"%.5f"``,expectedSteps(n - ``1``));``}``}` `// This code is contributed by PrinciRaj1992`
## Python3
`# Python3 implementation of the approach``import` `numpy as np` `maxSize ``=` `50` `# Function to multiply two 7 * 7 matrix``def` `matrix_product(a, b) :`` ``c ``=` `np.zeros((``7``, ``7``));`` ` ` ``for` `i ``in` `range``(``7``) :`` ``for` `j ``in` `range``(``7``) :`` ``for` `k ``in` `range``(``7``) :`` ``c[i][j] ``+``=` `a[i][k] ``*` `b[k][j];`` ``return` `c` `# Function to perform matrix exponentiation``def` `mul_expo(mul, p) :` ` ``# 7 * 7 identity matrix`` ``s ``=` `[ [ ``1``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0` `],`` ``[ ``0``, ``1``, ``0``, ``0``, ``0``, ``0``, ``0` `],`` ``[ ``0``, ``0``, ``1``, ``0``, ``0``, ``0``, ``0` `],`` ``[ ``0``, ``0``, ``0``, ``1``, ``0``, ``0``, ``0` `],`` ``[ ``0``, ``0``, ``0``, ``0``, ``1``, ``0``, ``0` `],`` ``[ ``0``, ``0``, ``0``, ``0``, ``0``, ``1``, ``0` `],`` ``[ ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``1` `] ];` ` ``# Loop to find the power`` ``while` `(p !``=` `1``) :`` ``if` `(p ``%` `2` `=``=` `1``) :`` ``s ``=` `matrix_product(s, mul);`` ` ` ``mul ``=` `matrix_product(mul, mul);`` ``p ``/``/``=` `2``;` ` ``return` `matrix_product(mul, s);` `# Function to return the required count``def` `expectedSteps(x) :` ` ``# Base cases`` ``if` `(x ``=``=` `0``) :`` ``return` `0``;`` ` ` ``if` `(x <``=` `6``) :`` ``return` `6``;` ` ``# Multiplier matrix`` ``mul ``=` `[ [ ``7` `/` `6``, ``1``, ``0``, ``0``, ``0``, ``0``, ``0` `],`` ``[ ``0``, ``0``, ``1``, ``0``, ``0``, ``0``, ``0` `],`` ``[ ``0``, ``0``, ``0``, ``1``, ``0``, ``0``, ``0` `],`` ``[ ``0``, ``0``, ``0``, ``0``, ``1``, ``0``, ``0` `],`` ``[ ``0``, ``0``, ``0``, ``0``, ``0``, ``1``, ``0` `],`` ``[ ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``1` `],`` ``[ ``-``1` `/` `6``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0` `] ];` ` ``# Finding the required multiplier`` ``# i.e mul^(X-6)`` ``mul ``=` `mul_expo(mul, x ``-` `6``);` ` ``# Final answer`` ``return` `(mul[``0``][``0``] ``+` `mul[``1``][``0``] ``+` `mul[``2``][``0``] ``+`` ``mul[``3``][``0``] ``+` `mul[``4``][``0``] ``+` `mul[``5``][``0``]) ``*` `6``;` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` ` ``n ``=` `10``;` ` ``print``(expectedSteps(n ``-` `1``));`` ` `# This code is contributed by AnkitRai01`
## C#
`// C# implementation of the approach``using` `System;` `class` `GFG``{``static` `readonly` `int` `maxSize = 50;` `// Function to multiply two 7 * 7 matrix``static` `double` `[,] matrix_product(``double` `[,] a,`` ``double` `[,] b)``{`` ``double` `[,] c = ``new` `double``[7, 7];` ` ``for` `(``int` `i = 0; i < 7; i++)`` ``for` `(``int` `j = 0; j < 7; j++)`` ``for` `(``int` `k = 0; k < 7; k++)`` ``c[i, j] += a[i, k] * b[k, j];`` ``return` `c;``}` `// Function to perform matrix exponentiation``static` `double` `[,] mul_expo(``double` `[,] mul, ``int` `p)``{` ` ``// 7 * 7 identity matrix`` ``double` `[,] s = {{ 1, 0, 0, 0, 0, 0, 0 },`` ``{ 0, 1, 0, 0, 0, 0, 0 },`` ``{ 0, 0, 1, 0, 0, 0, 0 },`` ``{ 0, 0, 0, 1, 0, 0, 0 },`` ``{ 0, 0, 0, 0, 1, 0, 0 },`` ``{ 0, 0, 0, 0, 0, 1, 0 },`` ``{ 0, 0, 0, 0, 0, 0, 1 }};` ` ``// Loop to find the power`` ``while` `(p != 1)`` ``{`` ``if` `(p % 2 == 1)`` ``s = matrix_product(s, mul);`` ``mul = matrix_product(mul, mul);`` ``p /= 2;`` ``}`` ``return` `matrix_product(mul, s);``}` `// Function to return the required count``static` `double` `expectedSteps(``int` `x)``{` ` ``// Base cases`` ``if` `(x == 0)`` ``return` `0;`` ``if` `(x <= 6)`` ``return` `6;` ` ``// Multiplier matrix`` ``double` `[,]mul = {{(``double``)7 / 6, 1, 0, 0, 0, 0, 0 },`` ``{ 0, 0, 1, 0, 0, 0, 0 },`` ``{ 0, 0, 0, 1, 0, 0, 0 },`` ``{ 0, 0, 0, 0, 1, 0, 0 },`` ``{ 0, 0, 0, 0, 0, 1, 0 },`` ``{ 0, 0, 0, 0, 0, 0, 1 },`` ``{(``double``) - 1 / 6, 0, 0,`` ``0, 0, 0, 0 }};` ` ``// Finding the required multiplier`` ``// i.e mul^(X-6)`` ``mul = mul_expo(mul, x - 6);` ` ``// Final answer`` ``return` `(mul[0, 0] + mul[1, 0] + mul[2, 0] +`` ``mul[3, 0] + mul[4, 0] + mul[5, 0]) * 6;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{`` ``int` `n = 10;` ` ``Console.Write(``"{0:f5}"``, expectedSteps(n - 1));``}``}` `// This code is contributed by Rajput-Ji`
## Javascript
``
Output:
`7.36111`
Time Complexity of the above approach will be O(343 * log(N)) or simply O(log(N)).
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
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My Personal Notes arrow_drop_up
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# The Fourth Dimension
## Explorations
Begin learning about flatland and the fourth dimesion with:
## Introduction to Dimensions
Escher's Reptiles showing both 2 and 3 dimensional objects
Webster’s Dictionary gives a description of dimensions 1, 2 , 3, and 4. We will also give a description of the 0th dimension.
• Space of zero dimensions: A space that has no length breadth or thickness. An example is a point.
• Space of one dimension: A space that has length but no breadth or thickness; a straight or curved line.
• Space of two dimensions: A space which has length and breadth, but no thickness; a plane or curved surface.
• Space of three dimensions: A space which has length, breadth, and thickness; a solid.
• Space of four dimensions: A kind of extension, which is assumed to have length, breadth, thickness, and also a fourth dimension.
Spaces of five, six, or more dimensions can also be studied using mathematics.
You have seen examples of 0-, 1-, 2-, and 3-dimensional objects all your life. Some examples:
• A pin-prick can be thought of as 0-dimensional. To the naked eye it has no length, width or height.
• When you draw a line on a piece of paper, you are drawing a representation of a 1-dimensional object. We only measure one direction. We will ask for the length of a line-segment, but we would not ask questions about its width or height. We think of a line as simply not having any width or height.
• When you measure the area of a geometric object you are measuring something 2-dimensional. Think about how you would find the area of a rectangle. You would measure the length and the width and multiply the two, right? You measure the two dimensions, and so we think of the rectangle as 2-dimensional.
• Anything 3-dimensional will require 3 measurements. Hence the volume of a box is considered 3-dimensional. It has length, width, and height.
• Consider making an appointment with someone at a 10 story building. You will have to tell them where to meet and when. The location is 3-dimensional, because we need 3 coordinates to find a place in space: longitude, latitude and height. But this means that to determine our place and time of meeting we require 4 pieces of information: longitude, latitude, height AND time. This means that our appointment is something 4-dimensional.
This idea of space and time, appropriately named 4-dimensional space-time, was used by Einstein when he brought forth his theory of relativity.
A point (0-dim), a segment (1-dim), a square (2 dim) and a cube (3-dim)
In the image above we see a point, a line segment, a square and a cube. These shapes are 0-, 1-, 2-, and 3-dimansional respectively. One could ask: what is the next shape in that sequence? In other words, is there something like a 4-cube.
To construct a model of this 4-cube, also known as a tesseract, we can follow the procedure of creating higher dimensional shapes. Think about extending the shape we have into a dimension perpendicular to those we already have. For instance, starting with a line segment we can draw an exact copy of the segment in the plane and then connect the corresponding vertices. This creates a square.
Using two copies of a segment (1-dim) we construct a square (2-dim) by connecting the corresponding vertices
Similarly, connecting corresponding vertices on two copies of the square will result in a cube.
Given two cubes we can connect the corresponding vertices and construct a 4-dimensional cube or tesseract.
A 4-dimensional cube. The figure on the right shows the two cubes (yellow and blue) whose corresponding vertices have been connected.
## Escher and Dimension
Escher looked at the interplay between 3-dimensional objects and their 2-dimensional depictions. He used the play on dimensions to create several interesting prints. Some examples include:
A famous print by Escher showing the contrast between 2 and 3 dimensions is the print named drawing hands. The hands in the print are clearly 3-dimensional. The hands and the pencils are shown as existing in space. It gets more interesting when we move our eye to the wrists and the lower arm. Here Escher transitions to a 2-dimensional image. The underlying piece of paper is depicted as entirely flat.
Escher's tessellations are all 2-dimensional. He referred to them as "Regular Divisions of the Plane" (Regelmatige Vlakverdeelingen in Dutch) and they all depict patterns that decorate a nice flat, 2-dimensional surface.
Escher also studied regular 3-dimensional shapes. Some examples include:
First, there are the platonic solids. If we experiment with regular polygons and try to build 3-dimensional shapes, then there are only 3 regular polygons that can be used by themselves.
The triangle can be used in three different ways, while the square and the pentagon result in two more platonic solids:
• Four triangles will form a tetrahedron.
• Eight triangles will form a octahedron.
• Twenty triangles will form an icosahedron.
• Six squares will form a cube.
• Twelve regular pentagons will form a dodecahedron.
In "Stars" we see two chameleons inside a shape made up of three octahedra. In the background - floating around in space - are some platonic solids and several other geometric 3-dimensional figures. Wikipedia has a short article with links describing some of the more exotic geometric shapes that are shown in the background. [Stars (M. C. Escher)]
## Flatland
Flatland was written in the 19th century, and is both a satire on Victorian Society and an exploration of the mathematical notion of dimensions. We read this story to develop some ideas about how to think about the 4th dimension. Even though we do live in the 4-dimensional space-time, most people are not comfortable with the 4th dimension at first. There are two questions we are interested in. What would a 4-dimensional being look like if it interacted with us? What would our 3-dimensional world look like if someone moved us into the 4th dimension? One way to think through these questions is to first ask them with all the dimensions dropped down a bit. What would a 3-dimensional being look like if it interacted with 2-dimensional beings (i.e. Flatlanders)? Or, what would a 2-dimensional being look like if it interacted with a 1-dimensional being? What would the 2-dimensional world look like if someone moved a Flatlander into the 3rd dimension? Abbott answers all of these questions in the book Flatland.
The 3-dimensional sphere appeared one 2-dimensional slice at a time. The sphere would first appear as a dot, and then grow into ever increasingly large circles. After reaching its biggest circumference, the circles would shrink back down to a point again. But the important part here is that the flatlanders could only see a 2-dimensional cross-section. Their 2-dimensional eyes and brains were not used to thinking about or seeing 3-dimensional beings.
Similarly, we would expect to see 3-dimensional cross-sections of any 4-dimensional beings.
When A. Square (the main character in Flatland) traveled to Lineland, he could see all of their world at once. The King of Lineland at first doesn’t know who is talking to him, because he can’t see Mr. A. Square at all. Later in the story A. Square moves into Spaceland, and is able to look down upon his own world.
A. Square is able to see the interior and exterior of his house at the same time. He can also see his family moving about the house. If you look carefully at the picture, you would see that A. Square can also see inside his relatives. Similarly, if we were moved into the 4th dimension, we might be able to see our world all at once. We would see the interior and exterior of our houses simultaneously, and we would also be able to see all around people.
## The fourth dimension in art
Several well known artists have explored the concept of the fourth dimension, while the work of some others can certainly be interpreted as depicting the fourth dimension.
Crucifixion (Corpus Hypercubus) - 1954 A Propos of the Treatise on Cubic Form by Juan de Herrera - 1960
Salvador Dali explored the 4-dimensional cube in two of his works. In 1954 he painted Crucifixion (Corpus Hypercubus) depicting Jesus crucified on the net of a hypercube. Dali's wife Gala is shown as the woman appearing before the cross.
In the painting A Propos of the Treatise on Cubic Form by Juan de Herrera from 1960, Dali gives a nod to the 16th century Spanish architect, mathematician and geometer Juan de Herrera. The painting shows a tesseract: two cubes with the corresponding vertices connected. The cube on the outside shows the foundation stone of the St. John Apostle and Evangelist Church in Northern Spain on four of the sides. The cube on the interior is a normal cube, while the edges connecting the corresponding vertices spell out the name of Juan de Herrera.
Other artists have depicted movement in a way that suggests space-time: a 3-dimensional object moves over time.
Nude descending a staircase by Marcel Duchamp (1912) Dynamism of a Dog on a Leash by Giacomo Balla (1912) The Knife Grinder by Kazimir Malevich (1912)
Marcel Duchamp's Nude descending a staircaserepresents a "dynamic version of Facet Cubism". This work shows movement by "superimposing successive phases of movement on each other, as in multiple exposure photography". (quoted from History of Art by Janson - 3rd edition pg 692)
The painting shows a robot like figure walking down a set of stairs. This work was first displayed at the Armory Show in New York in 1913 and caused quite a scandal. The American public had not been exposed to cubism and found this work of art quite shocking.
The close-up of Giacomo Balla's painting shows movement in quite a bit of detail. We can see the legs of the dachshund churning away as he walks down the boulevard. We see the dog wagging its tail and we can see the ears moving as well. The gentle sway of the leash is displayed by showing a blur of movement. Next to the dog we see many copies of the woman's feet indicated her movement along the sidewalk.
Kazimir Malevich's painting of the knife grinder is very geometric in nature due to the use of the cubist style. All the components of the painting are depicted as geometric shapes. There is also a sense of movement in the painting. We can see both the feet and the hands moving back and forth as the man is sharpening the knife on the wheel. This painting is an interesting juxtaposition of the 2-dimensional treatment of the surfaces due to the use of the cubist style, while also portraying the 4-dimensional space-time suggested by the movement of solid shapes.
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# Reasoning to an interpretation before applying Bayes’ rule
What’s the point of Bayes’ rule? This web page by Eliezer S. Yudkowsky gives a long intuitive explanation (thanks to Keith Frankish for pointing to it). This blog post is an attempt at a slightly shorter version with a bit more maths, and a bit of rambling about interpretation.
The information in the example problem given there is as follows:
1. 1% of women at age forty who participate in routine screening have breast cancer.
2. 80% of women with breast cancer will get positive mammographies.
3. 9.6% of women without breast cancer will also get positive mammographies.
The task: A woman in this age group had a positive mammography in a routine screening. What is the probability that she actually has breast cancer?
The general problem solved by Bayes’ rule is that if you know the probability of if A, then B, how do you work out the probability of if B, then A? More precisely if you know P(B|A), what is P(A|B)?
Here B|A denotes the conditional event, a simultaenously easy and difficult concept. One way to think of it is as follows.
Consider a fair die with six sides. It’s thrown. What’s the probability of a six given that a side showing an even number lands upwards? (Van Frassen, 1976 used an example like this to explain the conditional event interpretation of the natural language if-then.) This is P(lands six|lands even). The idea is that you only consider cases where it’s showing an even number (2, 4, or 6). Assuming they’re all equally probable, then P(lands six|lands even) = 1/3.
Interpretation
The first stage of solving problems like that above is interpretating the problem in the language of the mathematical theory you want to use.
Let $C$ denote “has cancer”, $\neg C$ denote “does not have cancer”, $T$ denote “shows a positive test result”, and $\neg T$ denote “shows a negative test result”.
Let’s take each item of information individually.
1% of women at age forty who participate in routine screening have breast cancer.
There’s a mix of information here: a percentage of people (1%), from a particular sub-population (women, aged 40, who participate in routine screening), and a property they have. From the problem it is clear that the interpretation is supposed to be:
$P(C) = .01$
But one can imagine a more complicated formalisation, for instance if the population of interest contains women of many different ages, some, but not all, of whom were screened because they had some worry about their health.
Next sentence:
80% of women with breast cancer will get positive mammographies.
This is an instance of
X% of people with property A have property B
The intended interpretation is P(B|A) = X%, but this might not be obvious to all readers. Take some:
Some people with property A have property B
If this is interpreted as an existential quantifier, then it also follows that some people with property B have property A. The conditional event, B|A, is in general not reversable in this way, so would not be suitable for the interpretation of an existential “some”. Consider the following statement:
All people with property A have property B
This is not (in general) reversable. The percentage quantifier (used in the problem description) is also not reversible. So there’s quite a lot of trickiness involved in interpreting this innocent looking statement. Given some background knowledge (we know the article is about Bayes’ rule, and about conditional probabilities), the intended interpretation of the original information is:
$P(T|C) = .8$
The idea is that if we choose a person at random from the population of interest, who has cancer (i.e., we know for sure she has cancer), then the probability of her having a positive test result is .8.
Then similarly for the last sentence:
9.6% of women without breast cancer will also get positive mammographies.
The formalisation is:
$P(T|\neg C) = .096$
Here is the summary:
$P(C) = .01$
$P(T|C) = .8$
$P(T|\neg C) = .096$
Now the problem statement:
A woman in this age group had a positive mammography in a routine screening. What is the probability that she actually has breast cancer?
We have to infer $P(C|T)$. Note how this is a reversal of the conditional statements we encounted in the information given about the test.
Calculation
Now comes the calculation. A good place to start when thinking about conditional probability is the ratio formula for the probability of a condititional event:
$P(B|A) = \frac{P(A \& B) }{P(A)}$
Take an interpretatation of “If it is raining, then I have an umbrella” as the conditional event expression:
I have an umbrella | it is raining
The probability of this is the probability that I have an umbrella and it is raining, divided by the probability that it is raining.
This can easily be rewritten to
$P(A \& B) = P(B|A) P(A)$
So if you know the probability of rain, and the probability that I have an umbrella when it rains, then you can multiply them to infer the probability that it is raining and I have an umbrella.
One step towards Bayes’ rule begins with:
1. $P(B|A) = P(A \& B) / P(A)$
2. $P(A|B) = P(A \& B) / P(B)$ [$A \& B = B \& A$ in (this) probability theory, so it does not matter what order you write them]
From 2 we can infer $P(A \& B) = P(A|B)P(B)$, which slots into 1 to give
$P(B|A) = \frac{P(A|B) P(B)}{P(A)}$
Now use the same variables as in the original problem
$P(C|T) = \frac{P(T|C) P(C)}{P(T)}$
We can already fill in the numerator (top row) with $P(T|C) = .8$ and $P(C) = .01$, but not yet the denominator (bottom row).
Let’s work a bit further then. We can infer $P(T)$ as follows:
$P(T) = P(T \& C) + P(T \& \neg C)$
Which is easily calculated from the rewrite of the conditional probability above:
$P(T) = P(T|C) P(C) + P(T|\neg C) P(\neg C)$
One more thing: $P(\neg A) = 1 - P(A)$. So this gives:
$P(T) = P(T|C) P(C) + P(T|\neg C) P(\neg C)$
$= .8 \times .01 + .096 \times (1 - .01) = .10304$
Now we have everything we need:
$P(C|T) = \frac{.8 \times .01}{.10304} = .078$.
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# Setting Goals
Alignments to Content Standards: 6.NS.B.3
1. Seth wants to buy a new skateboard that costs \$167. He has \$88 in the bank. If he earns \$7.25 an hour pulling weeds, how many hours will Seth have to work to earn the rest of the money needed to buy the skateboard? 2. Seth wants to buy a helmet as well. A new helmet costs \$46.50. Seth thinks he can work 6 hours on Saturday to earn enough money to buy the helmet. Is he correct?
3. Seth’s third goal is to join some friends on a trip to see a skateboarding show. The cost of the trip is about \$350. How many hours will Seth need to work to afford the trip? ## IM Commentary The purpose of this task is for students to solve problems involving division of decimals in the real-world context of setting financial goals. The focus of the task is on modeling and understanding the concept of setting financial goals, so fluency with the computations will allow them to focus on other aspects of the task. This task is also good preparation for the study of ratios and proportional relationships in 6th and 7th grade and their lead-in to linear functions in 8th grade. This task is part of a set collaboratively developed with Money as You Learn, an initiative of the President’s Advisory Council on Financial Capability. Integrating essential financial literacy concepts into the teaching of the Common Core State Standards can strengthen teaching of the Common Core and expose students to knowledge and skills they need to become financially capable young adults. A mapping of essential personal finance concepts and skills against the Common Core State Standards as well as additional tasks and texts will be available at http://www.moneyasyougrow.org. ## Solution 1. 167 - 88 = 79, so Seth needs to make \$79. Since $$79 \div 7.25\approx 10.9$$ Seth will have to work about 11 hours to make enough money to buy the skateboard.
2. No, Seth is not correct. 6 x 7.25 = 43.5 which is not enough to buy the helmet; he needs \$3 more which will require a bit less than a half an hour more work. 3. Since$350 ÷ 7.25 \approx 48.3\$ Seth will have to work about 50 hours.
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Curvature at a Point on a Curve Examples 2
# Curvature at a Point on a Curve Examples 2
Recall that if $\vec{r}(t) = (x(t), y(t), z(t))$ is a vector-valued function that traces the smooth curve $C$, then the curvature of $C$ at the point $\vec{r}(t)$ can be given by either of the formulas:
• $\kappa (s) = \biggr \| \frac{d\hat{T}(s)}{ds} \biggr \| = \biggr \| \frac{d\hat{T}(s) / dt}{ds / dt} \biggr \|$ if we find an arc length parameterization of $\vec{r}(t)$ as $\vec{r}(s)$.
• $\kappa (t) = \frac{\| \hat{T'}(t) \|}{\| \vec{r'}(t) \|}$ if $\hat{T'}(t)$ is not too difficult to calculate.
• $\kappa (t) = \kappa (t) = \frac{\| \vec{r'}(t) \times \vec{r''}(t) \|}{\| \vec{r'}(t) \|^3}$.
## Example 1
Find the curvature of $\vec{r}(t) = (1 - t^3, t^3 + 2, -2t^3)$.
We will use the third formula to calculate $\kappa (t)$.
First we will calculate the first and second derivatives. Firstly, $\vec{r'}(t) = (-3t^2, 3t^2, -6t^2)$, and secondly, $\vec{r''}(t) = (-6t, 6t, -12t)$. We now need to take the cross product $\vec{r'}(t) \times \vec{r''}(t)$ as follows:
(1)
\begin{align} \quad \quad \vec{r'}(t) \times \vec{r''}(t) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ -3t^2 & 3t^2 & -6t^2\\ -6t & 6t & -12t \end{vmatrix} = \begin{vmatrix} 3t^2& -6t^2\\ 6t& -12t \end{vmatrix} \vec{i} - \begin{vmatrix} -3t^2& -6t^2\\ -6t & -12t\end{vmatrix} \vec{j} + \begin{vmatrix} -3t^2& 3t^2\\ -6t & 6t \end{vmatrix} \vec{k} \\ = (-36t^3 + 36t^3) \vec{i} - (36t^3 - 36t^3) \vec{j} + (-18t^3 - 18t^3) \vec{j} = (0, 0, 0) \end{align}
Since we have arrived at the zero-vector, we know that $\| \vec{r'}(t) \times \vec{r''}(t) \| = 0$ and so $\kappa (t) = 0$. This implies that $\vec{r}(t)$ represents a line in $\mathbb{R}^3$.
## Example 2
Find the curvature of the elliptical helix $\vec{r}(t) = (2 \cos t, 3 \sin t, t^2)$. Use this formula to find the curvature at $t = \pi$
First we will calculate the first and second derivatives. Firstly, $\vec{r'}(t) = (-2 \sin t, 3 \cos t, 2t)$, and secondly, $\vec{r''}(t) = (-2 \cos t, -3 \sin t, 2)$. Now we need to take the cross product $\vec{r'}(t) \times \vec{r''}(t)$ as follows:
(2)
\begin{align} \quad \quad \vec{r'}(t) \times \vec{r''}(t) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ -2 \sin t & 3 \cos t & 2t \\ -2 \cos t & -3 \sin t & 2 \end{vmatrix} = \begin{vmatrix} 3 \cos t & 2t \\ -3 \sin t & 2 \end{vmatrix} \vec{i} - \begin{vmatrix} -2 \sin t & 2t \\ -2 \cos t & 2 \end{vmatrix} \vec{j} + \begin{vmatrix} -2 \sin t & 3\cos t \\ -2 \cos t & -3 \sin t \end{vmatrix} \vec{k} \\ \quad \quad = (6 \cos t + 6t \sin t) \vec{i} - (-4 \sin t + 4t \cos t) \vec{j} + (6 \sin ^2 + 6 \cos ^2 t) \vec{k} = (6 \cos t + 6t \sin t, 4 \sin t - 4t \cos t, 6) \end{align}
Now we need to calculate the magnitude of $\vec{r'}(t) \times \vec{r''}(t)$, that is $\| \vec{r'}(t) \times \vec{r''}(t) \| = \sqrt{(6 \cos t + 6t \sin t)^2 + (4 \sin t - 4t \cos t)^2 + (6)^2} = \sqrt{(6 \cos t + 6t \sin t)^2 + (4 \sin t - 4t \cos t)^2 + 36}$.
Now we need to calculate $\| \vec{r'}(t) \| = \sqrt{(-2\sin t)^2 + (3 \cos t)^2 + (2t)^2} = \sqrt{4 \sin ^2 t + 9 \cos ^2 t + 4t^2} = \sqrt{5 \cos^2 t + 4t^2 + 4}$. Therefore $\| \vec{r'}(t) \|^3 = \left ( \sqrt{5 \cos^2 t + 4t^2 + 4} \right )^3$. Putting this all together and we have that:
(3)
\begin{align} \kappa (t) = \frac{\sqrt{(6 \cos t + 6t \sin t)^2 + (4 \sin t - 4t \cos t)^2 + 36}}{\left ( \sqrt{5 \cos^2 t + 4t^2 + 4} \right )^3} \end{align}
So for $t = \pi$ we have that $\kappa(\pi) = \frac{\sqrt{16\pi^2 +72}}{\left ( \sqrt{ 4 \pi ^2 + 9} \right)^3}$.
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# The Central Limit Theorem Part 1
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1 The Central Limit Theorem Part. Introduction: Let s pose the following question. Imagine you were to flip 400 coins. To each coin flip assign if the outcome is heads and 0 if the outcome is tails. Question: If we average all of those values (all 400 of those 0s and s) together, what s the probability that the result will be 0.57 or more? 2. Building an Understanding We could approach the question by starting simple and looking at one coin. We assign X to be the average number of heads for the one flip, but of course for one coin this is just the number of heads. The flip results are T and H, corresponding to 0 and respectively, so we have distribution as follows. Remember this is the distribution for the average number of heads. which corresponds to the histogram x 0 p(x) /2 /2 Now let s do two coins. We assign X 2 to be the average number of heads. The flip results are TT, TH, HT and HH, corresponding to 0, /2, /2 and respectively (remember, average the number of heads per pair of rolls), so we have distribution as follows. Remember this is the distribution for the average number of heads. x 0 /2 p(x) /4 /2 /4
2 Now let s do three coins. We assign X 3 to be the average number of heads. The flip results are TTT, TTH, THT, THH, HTT, HTH, HHT and HHH, corresponding to 0, /3, /3, 2/3, /3, 2/3, 2/3 and respectively, so we have distribution as follows. Remember this is the distribution for the average number of heads. x 0 /3 2/3 p(x) /8 3/4 3/8 /8 Almost last but not least, before we jump to our result, let s do four coins. We assign X 4 to be the average number of heads. The flip results are TTTT, TTTH, TTHT, TTHH, THTT, THTH, THHT, THHH, HTTT, HTTH, HTHT, HTHH, HHTT, HHTH, HHHT and HHHH, corresponding to 0, /4, /4, 2/4, /4, 2/4, 2/4, 3/4, /4, 2/4, 2/4, 3/4, 2/4, 3/4, 3/4 and respectively. so we have distribution as follows. Remember this is the distribution for the average number of heads. x 0 /4 2/4 3/4 4/4 p(x) /6 4/6 6/6 4/6 /6 What do you see? The probability distribution is becoming a normal distribution! The more coins we flip the closer it becomes! Just to convince you here s the eight coin histogram. We ve scaled it up vertically so you can see what s going on - the highest rectangle is actually only 70/256 = high.
3 3. The Central Limit Theorem: The CLT states that our observation is accurate. More rigorously it states that if we have an event with random variable X (think a single coin toss, counting heads) and if we repeat it n times and assign X n to be the average of the results (think n coin tosses with X n being the average number of heads) then as n gets larger the distribution approaches a normal distribution. What s fascinating about this is that it s true no matter what the probability distribution function for the single variable is. It could be coin tosses, dice rolls, exponential distributions, whatever. In the long term the average outcome approaches the normal distribution. This is useful for us because it says that if we are interested in the average outcome of a very large number of flips (or any random variable at all) we can get an approximation by looking at the corresponding normal distribution rather than by actually flipping lots of coins. For large n the probability distribution for the average outcome is approximately a normal distribution. Since a normal distribution is determined completely by the mean µ and standard deviation σ we just need to determine what those would be for the normal distribution corresponding to our very large n. The CLT gives us the result as: The corresponding normal distribution has µ = E(X). The corresponding normal distribution has σ = σ(x)/ n. Note: You shouldn t look at this and say It s obvious! because it s not. The CLT is not easy to prove - we are unable to prove it with the mathematics we currently know.
4 4. Application: Some examples: Example : We flip 400 coins, assigning for head and 0 for tail each time. What s the probability that the average outcome will be greater than or equal to 0.57? If X is the outcome for a single coin and X 400 is the average outcome of 400 coins then for X we have E(X) = /2() + /2(0) = /2 σ(x) = V ar(x) = /2( /2) 2 + /2(0 /2) 2 = /2 The CLT says that the corresponding approximating normal distribution has µ = /2 σ = (/2)/ 400 = In friendly terms 400-Coin Flip Average Outcome Distribution Normal Distribution with µ = /2 and σ = So our desired value is P(0.57 X 400 < ) but by the CLT we can this approximate with the normal distribution. We find the corresponding z-scores x = = z = x = 0.57 = z = ( )/0.025 = 2.8 and so P(0.57 X 400 < ) P(2.8 Z < ) = = 0.26% This means that the probability of getting an average outcome of over 0.57 from flipping 400 coins is approximately 0.26%.
5 Example 2: Suppose the random variable X corresponds to rolling a die with winnings and losings given as follows: Rolling a wins \$2, rolling a 2 or 3 wins \$7, rolling a 4 wins nothing and rolling a 5 or 6 loses \$6. Suppose we roll 2000 dice and average our winnings from each roll. What s the probability that we ll win under \$0.75? Here X is the random variable that gives the average of one roll. Of course the average for just one roll is the same as the winnings for that one roll and so E(X) = 2(/6) + 7(2/6) + 0(/6) 6(2/6) = 4/6 = 2/3 σ(x) = (2 2/3) 2 (/6) + (7 2/3) 2 (2/6) + (0 2/3) 2 (/6) + ( 6 2/3) 2 (2/6) = 257/9 Let X 2000 be the random variable corresponding to the average earnings for 2000 rolls. The CLT says that the corresponding approximating normal distribution has µ = 2/3 σ = 257/9/ 2000 = 257/8000 In friendly terms 2000-Dice Roll Average Earnings Distribution Normal Distribution with µ = 2/3 and σ = 257/8000 Our desired value is P( < X ) but by the CLT we can approximate with the normal distribution. We find the corresponding z-scores x = = z = x = 0.75 = z = (0.75 2/3)/ 257/ and so P( < X ) Z( < Z < 0.70) = 75.80% Note that this makes sense. If the average winnings per roll is \$2/3 \$0.66 then after a lot of rolls we ought to be pretty close to that. Having under \$0.75 on average is pretty likely!
6 Example 3: The lifespan of a bacterium is randomly distributed with mean 5 days and standard deviation 2 days. Suppose you have 60 bacteria in your lab. What s the probability that the average lifespan will be over 4.8 days? Let X be the random variable corresponding to the lifespan of an individual bacterium and let X 60 correspond to the average of the lifespans of the 60 bacteria. Since X is already given to be distributed with E(X) = 5 σ(x) = 2 the CLT says that the corresponding approximating normal distribution has µ = 5 σ = 2/ 60 In friendly terms 60-Bacteria Average Lifespan Distribution Normal Distribution with µ = 5 and σ = 2/ 60 Our desired value is P(4.8 X 2000 ) but by the CLT we can approximate with the normal distribution. We find the corresponding z-scores x = = z = x = 4.8 = z = (4.8 5)/(2/ 60).26 and so P(4.8 X 60 < ) P(.26 Z < ) = = 89.62% Note that this seems reasonable. If the expected lifespan of a single organism is 5 days and we have a bunch of them then very probably the average lifespan of all of them will be over 4.8 days.
7 Example 4: The lifespan of a bacterium is distributed with mean 5 days and standard deviation 2 days. Suppose you have 60 bacteria in your lab. Over what age can you expect 70% of the bacteria to live? We ve reversed the question here, giving the desired percentage and asking for the age. Initially nothing changes from the previous problem. Let X be the random variable corresponding to the lifespan of an individual bacterium and let X 60 correspond to the average of the lifespans of the 60 bacteria. Since X is already given to be distributed with E(X) = 5 σ(x) = 2 the CLT says that the corresponding approximating normal distribution has µ = 5 σ = 2/ 60 Now things change. We want to know M such that P(M X 000 < ) = 0.7 but by the CLT we can approximate with the normal distribution. We find the corresponding z-scores x = = z = x = M = z = (M 5)/(2/ 60) and so P(M X 000 < ) = P ( (M 5)/(2 ) 60) Z < We wish this probability to be as close to 0.7 as possible. Since this probability is the area to the right of (M 5)/(2 60) we want the area to the left to be as close to 0.3 as possible. We reverse-lookup on the table and get the closest value Thus (M 5)/(2/ 60) = 0.52 M 5 =.04/ 60 M = 5.04/ 60 M 4.92 and so 70% of the bacteria will live over 4.92 days. We should note that this is important if we re collecting bacteria to study and we want to collect a sufficient quantity so that enough will survive for our study.
8 Problems Approximate all final answers to four places beyond the decimal point. Any intermediate value which is awkward (annoying square roots for example) may be approximated to four places beyond the decimal point. Any value which needs to be looked up in the standard normal table should be approximated as necessary first. Use when appropriate. Any value which needs to be reverse-looked up in the standard normal table should be reverse-looked up to the closest value.. Suppose you flip 2500 coins, assigning to heads and 0 to tails. (a) Calculate E(X) and σ(x) for a single flip. (c) What is the probability of the average outcome being between 0.49 and 0.52? (d) What is the probability of the average outcome being over 0.525? 2. Suppose you flip 000 coins. Each head wins \$2 and each tail loses \$.50. (a) Calculate E(X) and σ(x) for a single flip. (c) What is the probability of your average earnings being \$0.27 or more? What would your total earnings be if this were the case? (d) What is the probability of your average earnings being a dime or less? 3. Suppose you roll 3600 dice. Rolling a,2 or 3 wins \$3. Rolling a 4 or 5 loses \$4. Rolling a 6 loses \$2. (a) Calculate E(X) and σ(x) for a single roll. (c) What is the probability of your average earnings being positive? (d) Your average earnings have 80% probability of being below what value? 4. A honeybee drone has an expected lifespan which is exponentially distributed with mean 45 days. You collect one thousand honeybee drones. (a) What is the standard deviation? (b) If you pick a random bee, what is the probability it will live over 47 days? Note: This has nothing to do with the CLT! (c) Calculate µ and σ for the corresponding approximating normal distribution. (d) What is the probability that the average lifespan of all the bees in your collection will be over 47 days? (e) Under what average age will 0% of your honeybees live?
9 5. A certain drug is to be rated either effective or ineffective. Suppose lab results indicate the drug is effective 75% of the time and ineffective 25% of the time. An effective drug increases the lifespan of a patient by 5 years while an ineffective drug causes a complication which decreases the lifespan of a patient by year. As part of a study you administer the drug to 0000 patients. (a) Calculate E(X) and σ(x) for a single patient. (c) What is the probablity that the average lifespan increase will be 3.55 or more years? (d) What gain in lifespan will 25% of your patients experience? 6. A certain event is a random variable with PDF given by f(x) = x for x e. (a) Find E(X) and σ(x) for this event. (c) If this event occurs 900 times what is the probability of the average outcome being below.72? (d) There is a 90% probability that the average outcome is above which value?
### MATH 10: Elementary Statistics and Probability Chapter 4: Discrete Random Variables
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Lesson Video: The Equilibrium of a Body on a Rough Horizontal Plane | Nagwa Lesson Video: The Equilibrium of a Body on a Rough Horizontal Plane | Nagwa
# Lesson Video: The Equilibrium of a Body on a Rough Horizontal Plane Mathematics • Third Year of Secondary School
## Join Nagwa Classes
In this video, we will learn how to solve problems involving the equilibrium of a body on a rough horizontal plane.
15:43
### Video Transcript
In this video, we will learn how to solve problems involving the equilibrium of a body on a rough horizontal plane. We will begin by looking at some key definitions and then consider how we can resolve vertically and horizontally to calculate unknowns. These will include the normal reaction force, the frictional force between the body and the plane, and the coefficient of friction π. We will also consider the angle of friction.
Letβs begin by considering a body resting on a rough horizontal surface as shown. If the body is in equilibrium, it has zero resultant force acting on it. Two forces act on the body: firstly, its weight, which is equal to the mass of the body multiplied by the acceleration due to gravity. Secondly, as a result of Newtonβs third law, we have the normal reaction force π. For a body on a horizontal surface, this acts vertically upwards. As there is zero resultant force, these two forces must be equal such that π is equal to π, which we can also write as π is equal to ππ.
If the body was on a smooth surface, any net force applied horizontally on the body would accelerate the body horizontally. However, on a rough surface, the body will only accelerate if the horizontal force applied to it has magnitude greater than the frictional force between the body and the surface. In this video, we will label this frictional force π
π. This frictional force acts in the opposite direction to the applied force π
.
In this video, we will consider problems where the body is on the point of moving and, as such, the frictional force will be at its maximum. At this stage, the body will still be in equilibrium. And when we resolve horizontally, the applied force π
will be equal to the frictional force π
π. The maximum frictional force is also called the limiting friction. And this limiting friction π
π between a body at rest and the surface it rests on satisfies the formula π
π is equal to ππ. And since π, the normal reaction force, is equal to ππ, the frictional force π
π is equal to πππ.
We will now look at an example where the maximum applied force on a body that remains in equilibrium is to be determined.
A body weighing 25.5 newtons rests on a rough horizontal plane. A horizontal force acts on the body, causing it to be on the point of moving. Given that the coefficient of friction between the body and the plane is three seventeenths, determine the magnitude of the force.
We can begin by sketching a diagram to model the situation. We know that any body resting on a horizontal plane will have a downward force equal to its weight. And this weight π will be equal to the mass of the body multiplied by the acceleration due to gravity. In this question, we are told the body weighs 25.5 newtons. So there will be a force acting vertically downwards equal to 25.5 newtons. Newtonβs third law tells us that there will be a normal reaction force acting in the opposite direction to this. Since the body is in equilibrium, we know that these two forces are equal. The normal reaction force π is equal to 25.5 newtons.
We are told that a horizontal force π
acts on the body. And as the plane is rough, there will be a frictional force acting between the body and the plane. The body is on the point of moving, which means that the frictional force will be at its maximum. This is known as the limiting friction such that π
π is equal to π multiplied by π, where π is the coefficient of friction between the body and the plane. In this question, we are told this is equal to three seventeenths. The frictional force π
π is therefore equal to three seventeenths multiplied by 25.5, which is equal to 4.5. The maximum frictional force is equal to 4.5 newtons.
Since the body is on the point of moving and is still in equilibrium, the horizontal forces must also be equal to one another. The applied force π
must be equal to the frictional force π
π. We can therefore conclude that the magnitude of the horizontal force acting on the body is 4.5 newtons.
In our next example, we need to calculate a resultant force of which the limiting friction is a component.
A body is resting on a rough horizontal plane. The coefficient of friction between the body and the plane is 0.2, and the limiting friction force that is acting on the body is 80 newtons. Given that π
is the resultant of the force of friction and the normal reaction force, find the magnitude of π
.
We will begin by sketching a diagram to model the situation. If the body is in limiting friction, it will have four forces acting on it. Firstly, acting vertically downwards, we have the weight force, which is equal to the mass of the body multiplied by the acceleration due to gravity. Acting vertically upwards, we have the normal reaction force π. We will have an applied force π
that is causing the body to be on the point of moving, and finally a frictional force π
π between the body and the plane acting in the opposite direction to this.
Whilst the weight and applied force are not mentioned in this question, we know that for a body at rest on a surface, the normal reaction force π must be equal to the weight of the body. The limiting friction is the maximum frictional force that the surface exerts on the body before the body starts to move. And we therefore know that the applied force π
must be equal to the frictional force π
π. This limiting friction is equal to π multiplied by π, where π is the coefficient of static friction between the surface and the body.
Letβs now substitute the values we are given in this question. We are told that π is equal to 0.2. We are also told that the limiting frictional force is equal to 80 newtons. Substituting these into π
π is equal to ππ, we have 80 is equal to 0.2 multiplied by π. We can then divide through by 0.2. π is equal to 80 divided by 0.2, which equals 400. The normal reaction force acting on the body is 400 newtons. Whilst it is not required in this question, we could then calculate the weight force and the applied force π
. These are equal to 400 newtons and 80 newtons, respectively.
We need to calculate the magnitude of π
, where π
is the resultant of the force of friction and the normal reaction force. This will act in a direction shown on the diagram. We can calculate the magnitude by creating a right triangle. And using the Pythagorean theorem, we have π
squared is equal to 400 squared plus 80 squared. The right-hand side is equal to 166,400. We can then take the square root of both sides of this equation. And since the resultant force must be positive, we have π
is equal to 80 root 26. The magnitude of the resultant force is equal to 80 root 26 newtons. As a decimal answer, this is equal to 407.92 newtons to two decimal places.
This question leads us onto a key fact about any body on a rough surface. A body on a rough surface has an angle of friction. This is the angle between the normal reaction force and the resultant of the normal reaction force and the limiting frictional force. If we consider the following body on a rough horizontal surface, then if the body is in limiting friction and on the point of moving, there will be four forces acting on it in the horizontal and vertical directions. We have the normal reaction force π, the weight force π which is equal to ππ, the applied force π
, and the limiting frictional force π
π.
We recall that this limiting frictional force π
π is equal to π multiplied by π. And whilst it is not required here, it is worth recalling that when resolving vertically and horizontally, our forces are equal. Drawing on the force π
, which is the resultant of the normal reaction force and the frictional force and not an additional force, we see that the angle of friction π is as shown. We can sketch a right triangle with this resultant force π
together with the normal reaction force and the limiting frictional force. Replacing π
π with ππ, we can use the tangent ratio to calculate π.
We know that in any right triangle, the tan of angle π is equal to the opposite over the adjacent. In our diagram, tan π is equal to ππ over π. And dividing through by π, we have tan π is equal to π. The angle of friction for a body on a rough horizontal surface can therefore be determined by tan π equals π.
Letβs now look at a specific example where we need to calculate the angle of friction.
Given that the coefficient of friction between a body and a plane is root three over four, what is the angle of friction? Round your answer to the nearest minute if necessary.
We will begin by recalling what we mean by the angle of friction. The angle of friction is the angle between the normal reaction force on a body and the resultant of the normal reaction force and the limiting frictional force on the body. Since the limiting frictional force π
π is equal to ππ, and using our knowledge of right angle trigonometry, we know that the angle of friction π satisfies the equation tan π is equal to ππ over π, which is equal to π, where π is the coefficient of friction and in this question is equal to root three over four. The tan of angle π is therefore equal to root three over four.
Taking the inverse tangent of both sides of this equation gives us π is equal to the inverse tan or arctan of root three over four. Ensuring that our calculator is in degree mode, typing in the right-hand side gives us 23.4132 and so on. We are asked to give our answer to the nearest minute. We can do this by using the degrees, minutes, and seconds button on the calculator or by multiplying the decimal part of our answer by 60. This is equal to 24.79 and so on minutes, giving us an answer to the nearest minute of 23 degrees and 25 minutes. This is the angle of friction when the coefficient of friction between a body and a plane is root three over four.
We will now summarize the key points from this video. We saw in this video that the normal reaction force for a body on a horizontal surface has a magnitude equal to the weight of the body. The frictional force between a body and a rough surface acts in the opposite direction to the net force on the body and parallel to the plane. The maximum frictional force between a body and a rough surface is given by π
π is equal to ππ, where π is the normal reaction force on the body and π is the coefficient of static friction between the body and the surface. Finally, we saw that the angle of friction for a body on a rough horizontal surface is given by π, which is equal to the inverse tan of π, where once again π is the coefficient of static friction between the body and the surface.
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# 2004 AMC 10B Problems/Problem 23
## Problem
Each face of a cube is painted either red or blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?
$\mathrm{(A) \ } \frac{1}{4} \qquad \mathrm{(B) \ } \frac{5}{16} \qquad \mathrm{(C) \ } \frac{3}{8} \qquad \mathrm{(D) \ } \frac{7}{16} \qquad \mathrm{(E) \ } \frac{1}{2}$
## Solution
Label the six sides of the cube by numbers $1$ to $6$ as on a classic dice. Then the "four vertical faces" can be: $\{1,2,5,6\}$, $\{1,3,4,6\}$, or $\{2,3,4,5\}$.
Let $A$ be the set of colorings where $1,2,5,6$ are all of the same color, similarly let $B$ and $C$ be the sets of good colorings for the other two sets of faces.
There are $2^6=64$ possible colorings, and there are $|A\cup B\cup C|$ good colorings. Thus the result is $\frac{|A\cup B\cup C|}{64}$. We need to compute $|A\cup B\cup C|$.
Using the Principle of Inclusion-Exclusion we can write $$|A\cup B\cup C| = |A|+|B|+|C| - |A\cap B| - |A\cap C| - |B\cap C| + |A\cap B\cap C|$$
Clearly $|A|=|B|=|C|=2^3=8$, as we have two possibilities for the common color of the four vertical faces, and two possibilities for each of the horizontal faces.
What is $A\cap B$? The faces $1,2,5,6$ must have the same color, and at the same time faces $1,3,4,6$ must have the same color. It turns out that $A\cap B=A\cap C=B\cap C= A\cap B\cap C =$ the set containing just the two cubes where all six faces have the same color.
Therefore $|A\cup B\cup C| = 8+8+8-2-2-2+2 = 20$, and the result is $\frac{20}{64}=\boxed{\frac{5}{16}}$.
## Solution 3
Suppose we break the situation into cases that contain four vertical faces of the same color:
I. Two opposite sides of same color There are 3 ways to choose the two sides, and then two colors possible, so $3*2=6$
II. One face different from all the others There are 6 ways to choose this face, and 2 colors, so $6*2=12$
III. All faces same There are 2 colors, so two ways for all faces to be the same.
Adding them up, we have a total of 20 ways to have four vertical faces the same color. The are $2^6$ ways to color the cube, so the answer is $\frac{20}{64}=\boxed{\frac{5}{16}}$
2004 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 24 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 All AMC 10 Problems and Solutions
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# Scientific Notation
## Related calculator: Scientific Notation Calculator
In general, we need scientific notation, if we want to write very big or very small number more compactly.
For example, it is known that mass of the Earth is 5973600000000000000000000 kg. Look how long it is!
To make it shorter, we will use powers of 10. We can count number of zeros and write that ${5973600000000000000000000}={59736}\cdot{100000000000000000000}={59736}\times{{10}}^{{{20}}}$.
But this is not scientific notation, in powers of 10 note, we saw that we can represent number using different powers of ten.
So, how do we define normalized scientific notation?
Definition. Scientific notation is when there is only one digit to the left of decimal point, greater than 0.
Let's see on example, what does this mean.
Number 450 can be written in many ways: ${450}={45}\times{{10}}^{{1}}={4.5}\times{{10}}^{{2}}={0.45}\times{{10}}^{{3}}$.
First representation is not applicable, since there are two digits, third representation has one digit to the left of decimal point, but this digit is less than 1. Correct representation is second.
So, $450=4.5\times 10^2$.
Similarly, ${0.54}={54}\times{{10}}^{{-{2}}}={5.4}\times{{10}}^{{-{1}}}={0.054}\times{{10}}^{{1}}$.
So, $0.54=5.4\times 10^{-1}$.
In example with the Earth's mass, correct representation is ${5.9736}\times{{10}}^{{{24}}}$.
Note, that if an exponent is $0$, we just don't write it $\left({5}={5}\times{{10}}^{{0}}={5}\right)$.
Finally, let's do some exercises.
Exercise 1. Write 3562000 in scientific notation.
Answer: ${3562000}={3.562}\times{{10}}^{{{6}}}$.
Next exercise.
Exercise 2. Write -5678.94 in scientific notation.
Answer: $-{5678.94}=-{5.67894}\times{{10}}^{{3}}$.
Next exercise.
Exercise 3. Write ${0.035}$ in scientific notation.
Answer: ${0.035}={3.5}\times{{10}}^{{-{2}}}$.
Now, try to convert number in scientific notation back to the number itself.
Exercise 4. Write ${1.35}\times{{10}}^{{3}}$ as normal number.
Answer: ${1.35}\times{{10}}^{{3}}={1350}$.
Last one.
Exercise 5. Write $-{3.5}\times{{10}}^{{-{2}}}$ as normal number.
Answer: $-{3.5}\times{{10}}^{{-{2}}}=-{0.035}$.
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Mathematics
Easy
Question
# What is the value of
## 20
Hint:
### The given expression is [(x3 + 4 ÷ 4) × 4 ] + 16There is one variable expression. The variable is: ‘x’. We have to perform the operations between them and find the right answer.Let’s consider the inner bracket. In inner bracket there are two terms. x^3 and (4 ÷4)They are unlike terms as one is variable and second is constant. Unlike terms means they have different variables.The whole bracket is multiplied by 4 and then to the final value 16 is added.Let’s solve the expression step by step[(x3 + 4 ÷ 4) × 4 ] + 16We will have to follow the BODMAS rule. We have to follow the this order.B stands for bracket, O stands for order, D stands for division, M stands for multiplication, A stands for addition. And S stands for subtraction.So we will solve (4 ÷ 4) first[(x3 + 4 ÷ 4) × 4 ] + 16 = [( x3 + 1) × 4 ] + 16Now we will use distribute law here to distribute 4 for both the terms inside the bracket.Distributive property - The quantity outside the bracket, or in other words, quantity in multiplication with bracket is equally distributed to the terms inside the bracket. This property is used in the above equation. And this property is distributive property[(x2+ 4 ÷ 4) × 4 ] + 16 = ( x3 × 4 + 1 × 4)+ 16[(x3 + 4 ÷ 4) × 4 ] + 16 = ( 4x3 + 4)+ 16No further operation is possible for the bracket term as they are unlike terms.So we can write[(x3+ 4 ÷ 4) × 4 ] + 16 = 4x3+ 4 + 16[(x3+ 4 ÷ 4) × 4 ] + 16 = 4x3 + 20No further operation is possible as both are unlike terms.Therefore, the value of expression is 4x3 + 20So the option which is'4x3 + 20' is the right option.
We have to follow the proper order of operations, or we will get errors. We have to be careful about different types of variables.
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# Common Core: 8th Grade Math : Apply the Pythagorean Theorem to Find the Distance Between Two Points in a Coordinate System: CCSS.Math.Content.8.G.B.8
## Example Questions
### Example Question #1 : Apply The Pythagorean Theorem To Find The Distance Between Two Points In A Coordinate System: Ccss.Math.Content.8.G.B.8
Give the perimeter of the above parallelogram if .
Explanation:
By the Theorem:
, and
The perimeter of the parallelogram is
### Example Question #1 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem
In a rectangle, the width is 6 feet long and the length is 8 feet long. If a diagonal is drawn through the rectangle, from one corner to the other, how many feet long is that diagonal?
Explanation:
Given that a rectangle has all right angles, drawing a diagonal will create a right triangle the legs are each 6 feet and 8 feet.
We know that in a 3-4-5 right triangle, when the legs are 3 feet and 4 feet, the hypotenuse will be 5 feet.
Given that the legs of this triangle are twice as long as those in the 3-4-5 triangle, it follows that the hypotense will also be twice as long.
Thus, the diagonal in through the rectangle creates a 6-8-10 triangle. 10 is therefore the length of the diagonal.
### Example Question #81 : Geometry
If and , how long is side ?
Not enough information to solve
Explanation:
This problem is solved using the Pythagorean theorem . In this formula and are the legs of the right triangle while is the hypotenuse.
Using the labels of our triangle we have:
### Example Question #1 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem
Sam and John both start at the same point. Sam walks 30 feet north while John walks 40 feet west. How far apart are they at their new locations?
Explanation:
Sam and John have walked at right angles to each other, so the distance between them is the hypotenuse of a triangle. The distance can be found using the Pythagorean Theorem.
### Example Question #1 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem
Daria and Ashley start at the same spot and walk their two dogs to the park, taking different routes. Daria walks 1 mile north and then 1 mile east. Ashley walks her dog on a path going northeast that leads directly to the park. How much further does Daria walk than Ashley?
2 – √2 miles
2 + √2 miles
√2 miles
Cannot be determined
1 mile
2 – √2 miles
Explanation:
First let's calculate how far Daria walks. This is simply 1 mile north + 1 mile east = 2 miles. Now let's calculate how far Ashley walks. We can think of this problem using a right triangle. The two legs of the triangle are the 1 mile north and 1 mile east, and Ashley's distance is the diagonal. Using the Pythagorean Theorem we calculate the diagonal as √(12 + 12) = √2. So Daria walked 2 miles, and Ashley walked √2 miles. Therefore the difference is simply 2 – √2 miles.
### Example Question #11 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem
Max starts at Point A and travels 6 miles north to Point B and then 4 miles east to Point C. What is the shortest distance from Point A to Point C?
7 miles
5 miles
2√13 miles
10 miles
4√2 miles
2√13 miles
Explanation:
This can be solved with the Pythagorean Theorem.
62 + 42 = c2
52 = c2
c = √52 = 2√13
### Example Question #51 : Sat Mathematics
Angela drives 30 miles north and then 40 miles east. How far is she from where she began?
50 miles
35 miles
60 miles
45 miles
50 miles
Explanation:
By drawing Angela’s route, we can connect her end point and her start point with a straight line and will then have a right triangle. The Pythagorean theorem can be used to solve for how far she is from the starting point: a2+b2=c2, 302+402=c2, c=50. It can also be noted that Angela’s route represents a multiple of the 3-4-5 Pythagorean triple.
### Example Question #1 : Apply The Pythagorean Theorem To Find The Distance Between Two Points In A Coordinate System: Ccss.Math.Content.8.G.B.8
To get from his house to the hardware store, Bob must drive 3 miles to the east and then 4 miles to the north. If Bob was able to drive along a straight line directly connecting his house to the store, how far would he have to travel then?
15 miles
25 miles
5 miles
7 miles
9 miles
Explanation:
Since east and north directions are perpendicular, the possible routes Bob can take can be represented by a right triangle with sides a and b of length 3 miles and 5 miles, respectively. The hypotenuse c represents the straight line connecting his house to the store, and its length can be found using the Pythagorean theorem: c2 = 32+ 42 = 25. Since the square root of 25 is 5, the length of the hypotenuse is 5 miles.
### Example Question #11 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem
A park is designed to fit within the confines of a triangular lot in the middle of a city. The side that borders Elm street is 15 feet long. The side that borders Broad street is 23 feet long. Elm street and Broad street meet at a right angle. The third side of the park borders Popeye street, what is the length of the side of the park that borders Popeye street?
27.46 feet
22.5 feet
18.5 feet
16.05 feet
17.44 feet
27.46 feet
Explanation:
This question requires the use of Pythagorean Theorem. We are given the length of two sides of a triangle and asked to find the third. We are told that the two sides we are given meet at a right angle, this means that the missing side is the hypotenuse. So we use a+ b= c2, plugging in the two known lengths for a and b. This yields an answer of 27.46 feet.
### Example Question #61 : Triangles
Kathy and Jill are travelling from their home to the same destination. Kathy travels due east and then after travelling 6 miles turns and travels 8 miles due north. Jill travels directly from her home to the destination. How miles does Jill travel?
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# What ratio is equivalent to 2 5
So, 2:5 is equivalent to 6:15.
## How do you calculate equivalent ratios?
• Enter a Ratio into the equivalent ratio calculator, for example, you could enter 7:25
• Select the number of equivalent ratios that you would like to see in the table of results
• The equivalent ratio calculator will calculate as you type and produce a lis of equivalent ratios in a table below the calculator
More items…
## How to calculate the ratio between two numbers?
• We want to work out \$20 shared in the ratio of 1:3.
• Step 1 is to work out the total number of parts in the ratio.
• 1 + 3 = 4, so the ratio 1:3 contains 4 parts in total.
• Step 2 is to divide the amount by the total number of parts in the ratio.
• \$20 ÷ 4 = \$5.
• Each of the four parts of the ratio is worth \$5.
More items…
## What ratios are equal to 5 to 2?
The ratio of 5 to 2 is the same as 2.5 to 1. In general, you can reduce a ratio by dividing both numbers by the smallest of the two numbers (so 5 to 2 is the same as 5 / 2 to 2 / 2 or 2.5 to 1 ). , Retired Computer Center Tech (AUTODIN) – lifetime enjoyment of math.
## How to write equivalent ratios?
WRITE AN EQUIVALENT RATIO. To find ratios which are equivalent to a ratio, multiply the terms of the original ratio by the same number. That is, multiply the terms of the original ratio by 2, 3, 4 and so on. Example 1 : Give two equivalent ratios of 6 : 4.
## What ratio is 2 4 equivalent to?
ratio 1 to 2Note that the ratio 2 to 4 is said to be equivalent to the ratio 1 to 2, that is 2:4 = 1:2. Note also that a fraction is a number that stands for “part of something”, so although this ratio can be expressed as a fraction, in this case it does NOT represent “part of something”.
## What is the fraction of 2 5?
4/10Decimal and Fraction Conversion ChartFractionEquivalent Fractions3/46/824/321/52/108/402/54/1016/403/56/1024/4023 more rows
## What are the five equivalent fraction of 2 5?
Thus, five equivalent fractions of 2/5 are 4/10, 6/15, 8/20, 10/25 and 12/30.
## How do you find the equivalent ratio?
Thus, to find a ratio equivalent to another we have to multiply the two quantities, by the same number. Another way to find equivalent ratios is to convert the given ratio into fraction form and then multiply the numerator and denominator by the same number to get equivalent fractions.
## What is 2/5 as a number?
Answer: 2/5 as a decimal is 0.4.
## What is 2/5 as a whole number?
0.4What is 2/5ths as a whole number? Answer: The number is 2/5 which can also be written as 0.4.
## What is 2/5 equivalent to as a decimal?
0.4Note that since 2 cannot be divided by 5, you can add zeros to the dividend by placing a decimal point in the quotient. The decimal form of is 0.4. The long division method is the standard method to convert any fraction to decimal form.
## Which of the following fraction is nearest to 2 5?
Answer: The fractions equivalent to 2/5 are 4/10, 6/15, 8/20, etc.
## What is 2/5 as a percentage?
40%Fraction to percent conversion tableFractionPercent3/475%1/520%2/540%3/560%41 more rows
## What is a equivalent ratio?
Equivalent ratios have the same value. To determine whether two ratios are equivalent, write them as fractions. If the fractions are equal, the ratios are equivalent.
## What is a ratio equivalent to 2 3?
Answer: 4/6, 6/9, 8/12, 10/15 … are equivalent to 2/3. All those fractions obtained by multiplying both the numerator and denominator of 2/3 by the same number are equivalent to 2/3.
## How do you make an equivalent ratio table?
Ratio tables show equivalent ratios between two quantities. You can create a ratio table by multiplying (or dividing) both quantities in the ratio by the same value to find an equivalent ratio. A ratio table can help you find equivalent ratios in order to solve problems.
## What is the form of 2 5?
2/5 is 0.4 when converted to decimal form.
## What is 2/5 as a decimal?
0.4Note that since 2 cannot be divided by 5, you can add zeros to the dividend by placing a decimal point in the quotient. The decimal form of is 0.4. The long division method is the standard method to convert any fraction to decimal form.
## What is 2 and 2/5 as an improper fraction?
Mixed Number 2 25 in the improper fraction is 125 . 3. Where can I get detailed steps for converting 2 25 to improper fraction?
## Ratio Calculator
Related: Fraction Calculator What is Ratio? A ratio is a quantitative relationship between two numbers that describe how many times one value can contain another. Applications of ratios are fairly ubiquitous, and the concept of ratios is quite intuitive.
## Equivalent Ratios Calculator – RankUpturn
Equivalent ratios calculator that shows work to find the equivalent ratios for the given ratio with two or more numbers. The step-by-step calculation help parents to assist their kids studying 4th, 5th or 6th grade to verify the work and answers of writing equivalent ratios homework and assignment problems in pre-algebra or in ratios and proportional relationships (RP) of common core state …
## Ratio Calculator | Good Calculators
Our innovative ratio calculator, which is also launched online, can settle all your equivalent fractions, proportions and ratio
## Ratios & Proportions Calculator – Symbolab
Free Ratios & Proportions calculator – compare ratios, convert ratios to fractions and find unknowns step-by-step
## How to find equivalent ratios?
As we previously mentioned, Equivalent Ratios are two ratios that express the same relationship between numbers. The Equivalent Ratio Calculator provides a table of equivalent ratios that have the same relationship between each other and directly with the ratio you enter into the calculator. We will look at how to calculate equivalent ratios shortly, first lets look at how to use the free online equivalent ratio calculator: 1 Enter a Ratio into the equivalent ratio calculator, for example, you could enter 7:25 2 Select the number of equivalent ratios that you would like to see in the table of results 3 The equivalent ratio calculator will calculate as you type and produce a lis of equivalent ratios in a table below the calculator 4 [Optional] Print or email the Table of Equivalent Ratios for later use
## What is a ratio?
A ratio is a direct comparison of one number against another. A ratio calculator looks to define the relationship that compares between those two numbers
## What is the importance of ensuring the right ratio of students to teachers?
Education: ensuring the right ratio of students to teachers is key for effective learning. Class sizes in terms of the ratio of pupils to a teacher is a common ratio concern.
## Where are ratios used?
Ratios are used everywhere, from cooking with your favourite recipes to building housing, here are some common applications of ratios in everyday life:
## Is there a formula for equivalent ratios?
As equivalent ratios have the same value there is technically no equivalent ratio formula but the following equivalent ratio formula will help you with the manual math calculations.
## How to find equivalent fractions?
To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 25) by the same natural number, ie, multiply by 2, 3, 4, 5, 6
## What is 4 10?
The fraction 4 10 is equal to 2 5 when reduced to lowest terms. To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 2 5) by the same integer number, ie, multiply by 2, 3, 4, 5, 6 … and so on …
## Is 2 5 a fraction?
Important: 2 5 looks like a fraction, but it is actually an improper fraction.
## Can you convert fractions to decimals?
This Equivalent Fractions Table/Chart contains common practical fractions. You can easily convert from fraction to decimal, as well as, from fractions of inches to millimeters.
## What is a Ratio?
A contrast, which exists between two particular numbers, is defined as ratio. Our ratio calculator is developed to compute this contrast and find out the relationship between these numbers.
## Using the Ratio Calculator
Resort to the help of this amazing ratio calculator when you have you settle ratio/proportion problems and check equivalent fractions. Despite the fact that you cannot enter a ratio of 4/5 into this calculator, it accepts values such as 4:5, for example, 4/3 should be written as 4:3.
## How to Calculate Ratios
In order to keep numbers in direct relation you should first divide or multiply, which depends on your task, them in the ratio. Therefore, a ratio of 8/6 is an equivalent ratio of 4/3: in that particular ratio calculation, you should just multiply 4, as well as 3, by 2.
## What is Ratio?
A ratio is a quantitative relationship between two numbers that describe how many times one value can contain another. Applications of ratios are fairly ubiquitous, and the concept of ratios is quite intuitive. This could likely be demonstrated by giving a child half as many cookies as his sister. While the child may not be able to voice the injustice using ratios, the raucous protestations that would most likely ensue should make it immediately obvious that he is well aware he has received 1:2 as many cookies as his sister, conceptually, if not mathematically.
## What is aspect ratio?
The aspect ratio is the ratio of a geometric shape’s sizes in different dimensions. In the case of a rectangle, the aspect ratio is that of its width to its height. Although aspect ratios are widely used in applications such as tire sizing, paper sizing, and standard photographic print sizes, some of the most frequent uses of aspect ratios involve computer screen dimensions, mobile phone screens, and video sizes. As such, below is a list of typical computer screen/video resolutions and aspect ratios.
## How to find equivalent ratios?
As we previously mentioned, Equivalent Ratios are two ratios that express the same relationship between numbers. The Equivalent Ratio Calculator provides a table of equivalent ratios that have the same relationship between each other and directly with the ratio you enter into the calculator. We will look at how to calculate equivalent ratios shortly, first lets look at how to use the free online equivalent ratio calculator: 1 Enter a Ratio into the equivalent ratio calculator, for example, you could enter 7:25 2 Select the number of equivalent ratios that you would like to see in the table of results 3 The equivalent ratio calculator will calculate as you type and produce a lis of equivalent ratios in a table below the calculator 4 [Optional] Print or email the Table of Equivalent Ratios for later use
## What is a ratio?
A ratio is a direct comparison of one number against another. A ratio calculator looks to define the relationship that compares between those two numbers
## What is the importance of ensuring the right ratio of students to teachers?
Education: ensuring the right ratio of students to teachers is key for effective learning. Class sizes in terms of the ratio of pupils to a teacher is a common ratio concern.
## Where are ratios used?
Ratios are used everywhere, from cooking with your favourite recipes to building housing, here are some common applications of ratios in everyday life:
## Is there a formula for equivalent ratios?
As equivalent ratios have the same value there is technically no equivalent ratio formula but the following equivalent ratio formula will help you with the manual math calculations.
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# Divisor
In Mathematics, Divisor means a number which divides another number. It is a part of the division process. In division, there are four significant terms which are commonly used, such as Dividend, Divisor, Quotient and Remainder. You will learn the definitions of these terms in this article.
In Mathematics, we perform the most basic operations, which are addition, subtraction, division and multiplication. These arithmetic operations have been taught in our primary classes. Here, you will learn about the division factor, known as the divisor.
Dividend ÷ Divisor = Quotient
## Divisor Meaning
In division, we divide a number by any other number to get another number as a result. So, the number which is getting divided here is called the dividend. The number which divides a given number is the divisor. And the number which we get as a result is known as the quotient. The divisor which does not divide a number completely produces a number, which is referred to as remainder.
## Divisor Formula
The operation of division in the form of:
Dividend ÷ Divisor = Quotient
The above expression can also be written as:
Divisor = Dividend ÷ Quotient
Here, ‘÷’ is the symbol of division. But sometimes, it is also represented by the ‘/’ symbol, such as
Dividend / Divisor = Quotient
### Examples
1. In 22 ÷ 2 = 11, 22 is the dividend, 2 is the divisor and 11 is the quotient.
2. If, 45/5 = 9, then 5 is the divisor of 45, which divides number 45 into 9 equal parts.
3. 1 ÷ 2 = 0.5, the divisor 2 divides the number 1 into fraction.
4. In the below-given example, 5 is the divisor, 52 is the dividend, 10 is the quotient and 2 is the remainder.
### How to represent the Divisor?
We can represent the divisor in three different ways. These are given in the figure below:
### Key Points to Remember
• Divisor divides the number into parts.
• The divisor can divide the dividend either completely or partially. When divided completely, the remainder is zero and when divided partially, the remainder is a non-zero integer.
• Divisor could be a positive or negative number.
• A number that divides an integer exactly, leaving no remainder, is also termed as the divisor.
• The divisor 1 and -1 can divide every integer, present in the number line.
### Divisors of 18
We know that the numbers which divide the given number exactly are referred to as factors. These are also called divisors. From this, we can write the divisors of 18 as given below:
Factors of 18:
1 × 18 = 18
2 × 9 = 18
3 × 6 = 18
Factors are 1, 2 , 3, 6, 9 and 18.
Therefore, the number of divisors of 18 is 5, they are 1, 2, 3, 6, 9, and 18.
### Divisors of 26
The divisors of 26 are the numbers which divide it exactly without leaving any remainder.
Factors of 26:
1 × 26 = 26
2 × 13 = 26
Hence, the divisors of 26 are 1, 2, 13 and 26.
## Frequently Asked Questions on Divisor
### What is a divisor?
A divisor is an integer that divides another integer to produce a result. The number which is divided is called the dividend and the result obtained here is the quotient.
Dividend ÷ Divisor = Quotient
### Is a number a divisor of itself?
Yes, a number is a divisor of itself, which results in quotient 1. For example, 4 is divided by 2 and 4 itself.
4 ÷ 4 = 1
### Can a divisor be negative?
Divisors can be both positive and negative, but usually, we consider positive terms for divisors. For example, 4 is divisible by six divisors: -2, -1, -4, and 2, 1, 4. But only positive integers are considered here.
### What is the divisor in a fraction?
A fraction is represented in the form of p/q. So, p, which is the numerator here, is considered as dividend and q, which is the denominator here, is considered as the divisor.
### Is 1 a prime divisor?
No, it is not. To be a prime divisor a number has to meet the prime number condition, according to which there should be a minimum of two factors.
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# Poisson Distribution
A Poisson experiment is a statistical experiment that has the following properties:
• The experiment results in outcomes that can be classified as successes or failures.
• The average number of successes (μ) that occurs in a specified region is known.
• The probability that a success will occur is proportional to the size of the region.
• The probability that a success will occur in an extremely small region is virtually zero.
Note that the specified region could take many forms. For instance, it could be a length, an area, a volume, a period of time, etc.
## Notation
The following notation is helpful, when we talk about the Poisson distribution.
• e: A constant equal to approximately 2.71828. (Actually, e is the base of the natural logarithm system.)
• μ: The mean number of successes that occur in a specified region.
• x: The actual number of successes that occur in a specified region.
• P(x; μ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is μ.
## Poisson Distribution
A Poisson random variable is the number of successes that result from a Poisson experiment. The probability distribution of a Poisson random variable is called a Poisson distribution.
Given the mean number of successes (μ) that occur in a specified region, we can compute the Poisson probability based on the following formula:
Poisson Formula. Suppose we conduct a Poisson experiment, in which the average number of successes within a given region is μ. Then, the Poisson probability is:
P(x; μ) = (e) (μx) / x!
where x is the actual number of successes that result from the experiment, and e is approximately equal to 2.71828.
The Poisson distribution has the following properties:
• The mean of the distribution is equal to μ .
• The variance is also equal to μ .
Poisson Distribution Example
The average number of homes sold by the Acme Realty company is 2 homes per day. What is the probability that exactly 3 homes will be sold tomorrow?
Solution: This is a Poisson experiment in which we know the following:
• μ = 2; since 2 homes are sold per day, on average.
• x = 3; since we want to find the likelihood that 3 homes will be sold tomorrow.
• e = 2.71828; since e is a constant equal to approximately 2.71828.
We plug these values into the Poisson formula as follows:
P(x; μ) = (e) (μx) / x!
P(3; 2) = (2.71828-2) (23) / 3!
P(3; 2) = (0.13534) (8) / 6
P(3; 2) = 0.180
Thus, the probability of selling 3 homes tomorrow is 0.180 .
## Cumulative Poisson Probability
A cumulative Poisson probability refers to the probability that the Poisson random variable is greater than some specified lower limit and less than some specified upper limit.
Cumulative Poisson Example
Suppose the average number of lions seen on a 1-day safari is 5. What is the probability that tourists will see fewer than four lions on the next 1-day safari?
Solution: This is a Poisson experiment in which we know the following:
• μ = 5; since 5 lions are seen per safari, on average.
• x = 0, 1, 2, or 3; since we want to find the likelihood that tourists will see fewer than 4 lions; that is, we want the probability that they will see 0, 1, 2, or 3 lions.
• e = 2.71828; since e is a constant equal to approximately 2.71828.
To solve this problem, we need to find the probability that tourists will see 0, 1, 2, or 3 lions. Thus, we need to calculate the sum of four probabilities: P(0; 5) + P(1; 5) + P(2; 5) + P(3; 5). To compute this sum, we use the Poisson formula:
P(x < 3, 5) = P(0; 5) + P(1; 5) + P(2; 5) + P(3; 5)
P(x < 3, 5) = [ (e-5)(50) / 0! ] + [ (e-5)(51) / 1! ] + [ (e-5)(52) / 2! ] + [ (e-5)(53) / 3! ]
P(x < 3, 5) = [ (0.006738)(1) / 1 ] + [ (0.006738)(5) / 1 ] + [ (0.006738)(25) / 2 ] + [ (0.006738)(125) / 6 ]
P(x < 3, 5) = [ 0.0067 ] + [ 0.03369 ] + [ 0.084224 ] + [ 0.140375 ]
P(x < 3, 5) = 0.2650
Thus, the probability of seeing at no more than 3 lions is 0.2650.
## Poisson Calculator
Clearly, the Poisson formula requires many time-consuming computations. The Stat Trek Poisson Calculator can do this work for you - quickly, easily, and error-free. Use the Poisson Calculator to compute Poisson probabilities and cumulative Poisson probabilities. It can found in the Stat Trek main menu under the Stat Tools tab. Or you can tap the button below.
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# 9.4: Multiplication of Polynomials by Binomials
Difficulty Level: Basic Created by: CK-12
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Practice Multiplication of Polynomials by Binomials
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Suppose a factory needs to increase the number of units it outputs. Currently it has \begin{align*}w\end{align*} workers, and on average, each worker outputs \begin{align*}u\end{align*} units. If it increases the number of workers by 100 and makes changes to its processes so that each worker outputs 20 more units on average, how many total units will it output? What would you have to do to find the answer? After completing this Concept, you'll be able to multiply a polynomial by a binomial so that you can perform the operation required here.
### Watch This
Multimedia Link: For further help, visit http://www.purplemath.com/modules/polydefs.htm – Purplemath’s website – or watch this CK-12 Basic Algebra: Adding and Subtracting Polynomials
### Guidance
A binomial is a polynomial with two terms. The Distributive Property also applies for multiplying binomials. Let’s think of the first parentheses as one term. The Distributive Property says that the term in front of the parentheses multiplies with each term inside the parentheses separately. Then, we add the results of the products.
\begin{align*}(a+b)(c+d)=(a+b)\cdot c+(a+b)\cdot d\end{align*} Let’s rewrite this answer as \begin{align*}c\cdot (a+b)+d\cdot (a+b)\end{align*}.
We see that we can apply the Distributive Property on each of the parentheses in turn.
\begin{align*}c \cdot (a+b)+d\cdot (a+b)=c\cdot a+c \cdot b+d \cdot a+d \cdot b \ (\text{or} \ ca+cb+da+db)\end{align*}
What you should notice is that when multiplying any two polynomials, every term in one polynomial is multiplied by every term in the other polynomial.
#### Example A
Multiply and simplify \begin{align*}(2x+1)(x+3)\end{align*}.
Solution: We must multiply each term in the first polynomial with each term in the second polynomial. First, multiply the first term in the first parentheses by all the terms in the second parentheses.
Now we multiply the second term in the first parentheses by all terms in the second parentheses and add them to the previous terms.
Now we can simplify.
\begin{align*}(2x)(x)+(2x)(3)+(1)(x)+(1)(3) & = 2x^2+6x+x+3\\ & = 2x^2+7x+3\end{align*}
#### Example B
Multiply and simplify \begin{align*}(4x-5)(x^2+x-20)\end{align*}.
Solution:
Multiply the first term in the binomial by each term in the polynomial, and then multiply the second term in the monomial by each term in the polynomial: \begin{align*}(4x)(x^2)+(4x)(x)+(4x)(-20)+(-5)(x^2)+(-5)(x)+(-5)(-20)&=4x^3+4x^2-80x-5x^2-5x+100\\ & = 4x^3-x^2-85x+100\end{align*}
Solving Real-World Problems Using Multiplication of Polynomials
We can use multiplication to find the area and volume of geometric shapes. Look at these examples.
#### Example C
Find the area of the following figure.
Solution: We use the formula for the area of a rectangle: \begin{align*}\text{Area}=\text{length}\cdot\text{width}\end{align*}. For the big rectangle:
\begin{align*}\text{Length} & = B+3, \ \text{Width}=B+2\\ \text{Area} &= (B+3)(B+2)\\ & = B^2+2B+3B+6\\ & = B^2+5B+6\end{align*}
### Vocabulary
Binomial: A binomial is a polynomial with two terms.
The Distributive Property for Binomials: The Distributive Property says that the term in front of the parentheses multiplies with each term inside the parentheses separately. Then, we add the results of the products.
\begin{align*}(a+b)(c+d)=c\cdot (a+b)+d\cdot (a+b)=c\cdot a+c \cdot b+d \cdot a+d \cdot b \ (\text{or} \ ca+cb+da+db)\end{align*}
### Guided Practice
Find the volume of the following figure.
Solution:
\begin{align*}The \ volume \ of \ this \ shape & = (area \ of \ the \ base) \cdot (height).\\ \text{Area of the base} & = x(x+2)\\ & = x^2+2x\end{align*}
\begin{align*}Volume&=(area \ of \ base ) \times height\\ Volume&=(x^2+2x)(2x+1)\end{align*}
Now, multiply the two binomials together.
\begin{align*}Volume&=(x^2+2x)(2x+1)\\ &= x^2\cdot 2x+x^2\cdot 1+2x\cdot 2x+ 2x\cdot 1\\ &= 2x^3+x^2+2x^2+2x\\ &=2x^3+3x^2+2x\end{align*}
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Multiplication of Polynomials (9:49)
Multiply and simplify.
1. \begin{align*}(x-2)(x+3)\end{align*}
2. \begin{align*}(a+2)(2a)(a-3)\end{align*}
3. \begin{align*}(-4xy)(2x^4 yz^3 -y^4 z^9)\end{align*}
4. \begin{align*}(x-3)(x+2)\end{align*}
5. \begin{align*}(a^2+2)(3a^2-4)\end{align*}
6. \begin{align*}(7x-2)(9x-5)\end{align*}
7. \begin{align*}(2x-1)(2x^2-x+3)\end{align*}
8. \begin{align*}(3x+2)(9x^2-6x+4)\end{align*}
9. \begin{align*}(a^2+2a-3)(a^2-3a+4)\end{align*}
10. \begin{align*}(3m+1)(m-4)(m+5)\end{align*}
Find the areas of the following figures.
Find the volumes of the following figures.
Mixed Review
1. Simplify \begin{align*}5x(3x+5)+11(-7-x)\end{align*}.
2. Cal High School has grades nine through twelve. Of the school's student population, \begin{align*}\frac{1}{4}\end{align*} are freshmen, \begin{align*}\frac{2}{5}\end{align*} are sophomores, \begin{align*}\frac{1}{6}\end{align*} are juniors, and 130 are seniors. To the nearest whole person, how many students are in the sophomore class?
3. Kerrie is working at a toy store and must organize 12 bears on a shelf. In how many ways can this be done?
4. Find the slope between \begin{align*}\left ( \frac{3}{4},1 \right )\end{align*} and \begin{align*}\left ( \frac{3}{4}, -16 \right )\end{align*}.
5. If \begin{align*}1 \ lb=454 \ grams\end{align*}, how many kilograms does a 260-pound person weigh?
6. Solve for \begin{align*}v\end{align*}: \begin{align*}|16-v|=3\end{align*}.
7. Is \begin{align*}y=x^4+3x^2+2\end{align*} a function? Use the definition of a function to explain.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English Spanish
TermDefinition
binomial A binomial is a polynomial with two terms.
Distributive Property for Binomials To use the Distributive Property with two binomials, multiply each term in the first factor by each term in the second. $(a+b)(c+d)=c\cdot (a+b)+d\cdot (a+b)=c\cdot a+c \cdot b+d \cdot a+d \cdot b \ (\text{or} \ ca+cb+da+db)$
distributive property The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, $a(b + c) = ab + ac$.
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# Clocks and Calendars Questions and Answer PDF
By | January 7, 2022
Clocks and calendars questions are part of the data interpretation and logical reasoning section. Quantitative Aptitude questions on clocks and calendars pdf are useful for competitive exams such as SSC, RRB, CAT, FCI, NTPC, railway, bank, CWC, etc.
Content Index
## Clock and Calendar Reasoning Questions PDF
Clock and Calendar Questions PDF book in English & Hindi – Download
30+ Calendar and Clock Questions with Solution Free PDF For SSC, RRB, FCI Exams – Download Now
### How to solve clock and calendar questions?
• Go through the different formulas to derive the clock angles and dial speed of the clock. Then use interpretational abilities by matching the derived answer with the options.
• The clocks and calendars questions test your analytical and reasoning skills. You should keep looking for any hints in the question.
• Be familiar with the different types of calendar weeks and days.
• Read clocks and calendars study guides to check various question patterns and their solution strategies.
## MCQ Questions on Clocks and Calendars
Let’s attempt some objective questions of Clocks and Calendars reasoning.
Question 1: Meetu’s birthday is on Wednesday 8th March. On what day of the week will be Ritu’s Birthday in the same year if Ritu was born on 10th July?
(A) Monday
(B) Wednesday
(C) Friday
(D) Saturday
(A) Monday
Question 2: Number of times 29th day of the month occurs in 400 consecutive year is ________ .
(A) 4400
(B) 4497
(C) 4800
(D) 4600
(B) 4497
400 consecutive years contain 97 leap years.
∴ In 400 consecutive years February has 29 days 97 times and the remaining 11 months have 29th day 400 x 11 = 4400 times
∴ 29th day of the month occurs 4400 + 97 = 4497 times.
Question 3: On what dates of April 2001 did Wednesday fall?
(A) 2nd,9th,16th,23rd
(B) 4th,11th,18th,25th
(C) 12th,18th,27th,6th
(D) 1st,8th,15th,22nd
(B) 4th,11th,18th,25th
We shall find the day on 1st April 2001.
1st April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)
Odd days in 1600 years = 0
Odd days in 400 years = 0
Jan. Feb. March April
(31 + 28 + 31 + 1) = 91 days 0 odd days.
Total number of odd days = (0 + 0 + 0) = 0
On 1st April, 2001 it was Sunday.
In April, 2001 Wednesday falls on 4th, 11th, 18th and 25th.
If you want to attempt more questions then download the above pdf books on Clocks and Calendars Questions.
Thank you for visiting our site. Furthermore if you have any questions regarding this topic of reasoning then do comment below.
Category: Reasoning PDF
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# Properties of Fractional Division
Properties of fractional division are discussed here:
Property 1: When a fractional number is divided by 1, the quotient is the fractional number itself.
For Example:
1. 2/3 ÷ 1
= 2/3 × 1/1
= (2 × 1)/(3 × 1)
= 2/3
2. 11/16 ÷ 1
= 11/16 × 1/1
= (11 × 1)/(16 × 1)
= 11/16
3. 21/43 ÷ 1
= 21/43 × 1/1
= (21 × 1)/(43 × 1)
= 21/43
Property 2:When zero is divided by non zero fractional number, then the quotient is always zero.
For Example:
1. 0 ÷ 7/9
= 0 × 9/7
= 0
2. 0 ÷ 18/33
= 0 × 33/18
= 0
3. 0 ÷ 29/16
= 0 × 16/29
= 0
Property 3: When any non zero fractional number is divided by itself, then the quotient is 1.
For Example:
1. 19/21 ÷ 19/21
= 19/21 × 21/19
= 1
2. 144/11 ÷ 144/11
= 144/11 × 11/144
= 1
Property 4: The reciprocal of zero or multiplicative inverse does not exist. So a fractional number cannot be divided by zero (0).
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# how to do x-y=2 and4x-3y=11 doing substitution
Hi Ciera,
This can be solved in a couple of steps.
Do you know how to "isolate" a variable? You know, get a variable on one side by itself? This is step one. Choose one of the equations, doesn't matter which, and solve for one of the variables (doesn't matter which).
For Example, if you chose this equation: x - y = 2
You could solve for x by adding y to both sides. This would eliminate the y variable on the left side and add it to the right side... giving you this: x = 2 + y
Now you have isolated the variable, x.
Take your new equation, x = 2 + y and substitute this equation into the second equation.
Notice that the second equation has both x and y variables also: 4x - 3y = 11
You can replace (substitute) your first equation into the position of the 'x' in the second equation. Why? So that you will only have one variable in the second equation instead of two. Then you'll be able to solve the second equation and find out the value of y. It looks like this:
4x - 3y = 11
4(2 + y) - 3y = 11
Notice that every other part of the second equation remains the same. The only thing we changed was that we replaced the x variable with part of the first equation. Since x = 2 + y, we can substitute the x for 2 + y in the second equation.
Now we solve for y by following these steps: distribute, add like terms, and isolate the y variable. It looks like this:
Given: 4(2 + y) - 3y = 11
Distribute: 4(2) + 4y - 3y = 11
Multiply: 8 + 4y - 3y = 11
Combine like terms (4y - 3y = y): 8 + y = 11
Isolate the variable by subtracting the 8 from both sides: y = 11 - 8
Subtract: y = 3
Now that you know the value of y, you can substitute that value into the first equation to find the value of x.
x = 2 + y
x = 2 + 3
x = 5
The solution to the system of equations is x = 5 and y = 3
Written as a coordinate: (5, 3)
Substitution? Easy as pie! Think of it this way, when you substitute something, you're replacing it with something else, right? That's what substitution in math is about too. You replace a letter you don't know with something you do know, so you can make the problem easy to solve.
Step 1: Write out both equations.
Step 2: Make one of them equal to y, by moving things numbers around (use the opposites to move them from one side of the equation to the other.)
Step 3: Go to the second equation and plug in the part of the first equation that is after the equal sign.
Step 4: Solve the second equation by adding or subtracting like terms.
Step 5: Plug in that answer into the first equation and solve for the second unknown number (variable).
That's all there is to it...well, besides checking to make sure they both equal the numbers on the other side of the equations!
Let's try your example:
x-y=2 and4x-3y=11
x-y=2
-x from both sides of the equation.
You are now left with -y=2-x
Divide by -1 to get y alone.
Do the same thing to the other side. Divide by -1
When you divide by a negative 1, all the signs change to the opposite signs.
So, y = -2+x
plug that into the second equation. Wherever you see y in the second equation, replace it with the -2+x.
Then, solve the second equation.
4x-3y=11
4x-3(-2+x)=11
Use distribution to solve for the parenthesis part and you will have 4x+6-3x=11
Add like terms: 4x-3x=1x or just x
You will now be left with: x+6=11
Subtract 6 from both sides.
x+6-6=11-6
x=5
Plug that answer back into the first original equation and solve for y.
x-y=2
5-y=2
subtract 5 from both sides: 5-5-y=2-5
-y=-3
divide by -1 and you will have y = 3
Now, you just check to make sure it all works out, by plugging both answers back into both equations and seeing if they equal the totals already in the original equations.
For instance: x-y=2
5-3=2
2=2 check
Does it check? Yes, it does!
Do the same with the other one and presto...you're a math whiz! ;)
What is meant by substitution is that if A=B, then in any equation with A, every occurrence of A can be replaced by B. In this casewe can start by isolating x in the first equation:
x - y = 2
x = y + 2
Now in the second equation, we can replace every x with a y+2 (parentheses are important here) and solve for y:
4(y+2) - 3y = 11
4y + 8 - 3y = 11 (Used distributive property)
y + 8 = 11 (Collected like terms)
y = 3
Then solve for x:
x = y+2 = 3+2 = 5
So the solution to the system is x=5, y=3.
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# A car starts from rest and accelerates uniformly to a speed of 180kmh−1 in 10 s. The distance covered by the car in the time interval is
A
200 m
B
300 m
C
500 m
D
250 m
Video Solution
Text Solution
Generated By DoubtnutGPT
## To solve the problem step by step, we can follow these calculations:Step 1: Understand the given information- The car starts from rest, which means the initial velocity u=0 m/s.- The final velocity v=180 km/h.- The time t=10 seconds.Step 2: Convert the final velocity from km/h to m/sTo convert km/h to m/s, we use the conversion factor 518:v=180km/h×518=50m/sStep 3: Calculate the accelerationUsing the formula for acceleration a:a=v−utSince u=0:a=50m/s−010s=5010=5m/s2Step 4: Calculate the distance covered using the formulaWe can use the formula for distance s when starting from rest:s=ut+12at2Substituting u=0:s=0⋅t+12at2=12⋅5m/s2⋅(10s)2Calculating further:s=12⋅5⋅100=5002=250mFinal AnswerThe distance covered by the car in the time interval is 250 meters.---
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Updated on:7/8/2024
### Knowledge Check
• Question 1 - Select One
## If a car at rest, accelerates uniformly to a speed of 144km/h in 20s, it covers a distance of
A2880m
B1440m
C400m
D20m
• Question 2 - Select One
## If a car at rest accelerates uniformly to a speed of 144km/h 20 s, it covers a distance of
A400m
B1440m
C2880m
Dnone of these
• Question 3 - Select One
## If a sports car at rest accelerates uniformly to a speed of 144 km h−1 in 5 s, it covers a distance of________ m.
A100
B140
C60
D80
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# How many chocolates?
One Christmas night, 3 kids were sleeping in their room. Their father came and kept a bag full of chocolates in the center of room with a card that says: "Sweets for my three sweet kids."
1. At midnight, the first kid wakes up, and sees the bag of chocolates. He takes exactly one third of the chocolates in the bag, hides them under his pillow, and leaves the remaining chocolates in the bag.
2. After some time the second kid wakes up. He does the same with the chocolates that were left by the first kid.
3. Then, the third kid wakes up after some time and does the same as well.
Finally, at morning they all wake up, looked at each other and said "Look, we got chocolates!" They divided the chocolates equally again.
So, what is the least number of chocolates their father would have kept in the bag?
Hint:
Each time at night a kid divided the chocolates, there was no remainder. The same happened in morning, it all divided equally.
• No trickery? When each kid woke up, the previous kid was asleep? Commented Jun 10, 2016 at 11:22
Let's say $X$ is the number of chocolates.
This means.
$X mod 3 = 0$.
how the first kid divided them
$\frac{2}{3} * X mod 3 = 0$
the second kid had only $\frac{2}{3}$ of the total chocolates to divide.
He takes $\frac{1}{3}$ of $\frac{2}{3} * X$ so the last kid counts
$\frac{4}{9} * X$ chocolates.
This means $\frac{4}{9} * X mod 3 = 0$.
The third kid takes $\frac{1}{3}$ of $\frac{4}{9}$ of X...so $\frac{4}{27}$ of chocolates.
S0 on the last morning there are left ($\frac{4}{9} - \frac{4}{27}) * X = \frac{8}{27} * X$.
And this $\frac{8}{27} * X mod 3 = 0$.
This means $\frac{8}{27} * X = 3k$.
in order for x to be integer, $3K$ must be divisible by 8.
The smallest possible $k$ is 8, so $\frac{8}{27} * X = 24$.
Solving the equation we get X =
$81$.
How it all happened.
The fist kid wakes up:
Divides 81 chocolates into 3 piles of 27.
He hides his 27 and puts 54 back.
Second kid wakes up.
Divides 54 chocolates into 3 piles of 18.
puts his 18 under the pillow and puts back 36.
The last kid wakes up
Divides 36 chocolates into 3 piles of 12.
puts his 12 under the pillow and puts back 24.
They all wake up and
divide 24 chocolates. 8 each.
They divide the chocolates
4 times => 3*3*3*3 => 81
The last division must at least have 3 chocolates to divide it equaliy to all children. The rest just follows plain simple.
• nope.. the last division cant be 3, 3, 3 Commented Jun 10, 2016 at 11:25
• the answer is correct but the deduction is wrong. Commented Jun 10, 2016 at 12:04
• @WasiqShahrukh I didn't say that every child got 3 chocolates... There must be at least 3 chocolates for all of them at the last division Commented Jun 10, 2016 at 12:16
• @Marius Why is the deduction wrong? There are 4 divisions with 3 as the divisor. How much chocolates every child got, wasn't the question. Commented Jun 10, 2016 at 12:19
• @WaKai See my answer below as well. OP hasn't worded the question very well. Once the first child divides them into three, he takes one third and leaves two thirds, which is then divided by the second child, who also leaves two thirds for the third child. Commented Jun 10, 2016 at 13:44
There are two interpretations of the problem, not made quite clear by OP. This one assumes that the first child splits the chocolates up onto thirds and the next child only uses one of these thirds.
81 (i.e. $3^4$) chocolates
First division - 27 each
Second division - 9 each
Third division - 3 each
Morning - 1 each.
The second interpretation, is when the first kid wakes up and splits them into thirds, the second kid takes two thirds and so on... then:
81 chocolates to start
First division - 27/54 split
Second division - 18/36 split
Third division - 12/24 split
Morning - 8 each
(but really the first kid has 35 in total, the second kid has 26, and the third kid has 20).
We cannot go any lower since there is no common divisor with a factor of 3.
Coincidentally, they are the same number.
• Wow.. too fast for me. I was typing up my answer when this popped up. Well done! Commented Jun 10, 2016 at 11:21
• @CipherBot 27 each is right but not 9 each and 3 and 1 Commented Jun 10, 2016 at 11:22
• @WasiqShahrukh Is it the first interpretation or second interpretation in my answer? Commented Jun 10, 2016 at 11:26
• @WasiqShahrukh You need to make your questions clearer. I have answered both interpretations. Commented Jun 10, 2016 at 11:36
• @Inazuma yup that's correct :) its is understodd if the first kid takes the 1/3rd then obviously 2nd will divide the remaining 2/3rd into other 3 parts and so on.. Commented Jun 10, 2016 at 11:38
|
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### Compound Interest Shortcuts, Tricks, Tips & Results-2
In this article, we pick up from where we left in the previous article and cover Compound Interest Shortcuts that you can employ in exams. We cover an extremely useful set of compound interest shortcuts, tips, tricks, and results. You can use these Compound Interest tips and Compound Interest shortcuts in competitive exams and make sure save some vital time in the examination by employing these techniques.
Kindly keep in mind that the purpose of this article is to provide you with useful compound interest shortcuts and tips. This article does not cover any core concepts for this topic. For compound interest concepts, you should refer to the first two articles for this topic. Also, some of these compound interest shortcuts might appear obscure or too formula-based to you. Remember, you need to learn these compound interest shortcuts with a pinch of salt. All of these might not be applicable and in fact, you might not be able to learn and remember all of them. It is best to be selective with these compound interest shortcuts and use them to the best of your ability.
In each of the following results, we use the following denotations:
A = future value
P = principal amount (initial investment)
r = annual nominal interest rate
n = number of times the interest is compounded per year
t = number of years for which the money is borrowed
Let’s get started and cover some of these shortcuts.
Compound Interest Shortcuts: Tooltip-1
A sum of money placed at compound interest becomes x time in ‘a’ years and y times in ‘b’ years. These two sums can be related by the following formula:
Derivation for this result:
We use the basic formula for calculating Compound Interest:
For condition 1, a sum of money becomes x times in “a” years.
Therefore, using the formula for calculating Compound Interest:
Example-1: A sum of money placed at compound interest doubles itself in 4 years.In how many years will it amount to 16 times itself?
Compound Interest Shortcuts:Â Tooltip 2
If an amount of money grows up to Rs x in t years and up to Rs y in (t+1) years on compound interest, then
Derivation for this result:
Principal + CI for t years = x ……  (1)
Principal + CI for (t+1) years= y ……. (2)
(2) – (1) =>CI for last year = y-x
Which is basically the simple interest upon x
Example-2: An amount of money grows upto Rs 3000 in 3 years and upto Rs 4000 in 4 years on compound interest. What will be the rate percent?
Solution:
Principal + CI for 3 years = 3000 ……  (1)
Principal + CI for 4 years= 4000 ……. (2)
Hence (2) – (1) =>CI for 4th year = 4000-3000= Rs 1000
Which is basically the simple interest upon 3000
Compound Interest Shortcuts: Tooltip 3
A sum at a rate of interest compounded yearly becomes Rs. A1 in n years and Rs. A2 in (n + 1) years, then
Example-3: A sum of money invested at compound interest amounts to Rs. 100 at the end of first year and Rs. 120 at the end of second year. The sum of money is :
Solution:
Simple Interest for one year = compound interest for one year
Interest on Rs. 100 for 1 year = 120-100= Rs. 20
Compound Interest Shortcut: Tooltip 4
If a certain sum becomes x times of itself in t years, the rate of compound interest will be equal to
Derivation for this result:
Use the formula for Compound Interest Calculation:
Sum becomes x times of itself in t years so
Example 4: Â If a certain sum becomes 16 times in 2 years ,what will be the rate of compound interest?
Solution:
Using the formula derived above:
Compound Interest Shortcut: Tooltip 5If the compound interest on a certain sum for 2 years is CI and simple interest for two years is SI ,then rate of interest per annum is
If the compound interest on a certain sum for 2 years is CI and simple interest for two years is SI ,then rate of interest per annum is
Derivation for this result:
Example 5: If the compound interest on a certain sum for 2 years is 20rs and simple interest for two years is 10rs ,then what wil be the rate of interest per annum ?
Solution:
Using the formula derived above:
### Compound Interest Solved Problems using Compound Interest Shortcuts:
Let’s go through some compound interest solved problems and learn how to use and implement compound interest shortcuts in actual problem solving. Remember, till the time you actually solve questions using these tricks, you won’t be able to memorize and understand them. Go through compound interest solved problems and hone your skills for the topic.
Question 1: What sum of money at comÂpound interest will amount to Rs. 650 at the end of the first year and Rs. 676 at the end of the second year?
A. Rs. 600
B. Rs. 600.25
C. Rs. 625
D. Rs. 625.25
### Answers and Explanations
Option C
Using the formula we have derived in the article for this exercise:
If an amount of money grows up to Rs x in t years and up to Rs y in (t+1) years on compound interest, then
Question 2: If the amount is 2.25 times of the sum after 2 years at comÂpound interest (compound annuÂally) , the rate of interest per anÂnum is :
A. 25%
B. 30%
C. 45%
D. 50%
### Answers and Explanations
Option D
Using the formula we have derived in the article for this exercise:
If a certain sum becomes x times of itself in t years, the rate of compound interest will be equal to
Question 3: A sum of money doubles itself in 4 years at compound interest. It will amount to 8 times itself at the same rate of interest in:
A. 18 years
B. 12 years
C. 16 years
D. 24 years
### Answes and Explanations
Option B
Using the formula we have derived in the article for this exercise:
A sum of money placed at compound interest becomes x time in ‘a’ years and y times in ‘b’ years. These two sums can be related by the following formula:
Question 4: A sum borrowed under comÂpound interest doubles itself in 10 years. When will it become fourfold of itself at the same rate of interest?
A. 15 years
B. 20 years
C. 24 years
D. 40 years
### Answers and Explanations
Option B
Using the formula we have derived in the article for this exercise:
A sum of money placed at compound interest becomes x time in ‘a’ years and y times in ‘b’ years. These two sums can be related by the following formula:
Question 5: A sum of money becomes eight times of itself in 3 years at comÂpound interest. The rate of interest per annum is
A. 100%
B. 80%
C. 20%
D. 10%
### Answers and Explanations
Option A
Let the principal be Rs. x and the rate of compound interest be r% per annum.
Then,
#### Compound Interest Practice Tests
FREE CHEAT SHEET
Learn
How to Master VA-RC
This free (and highly detailed) cheat sheet will give you strategies to help you grow
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Ch 6.1
```Ch 6.1
• Ratio- comparison of two quantities
– Ex:
3
4
or
3:4
• Ratios are UNITLESS
– Means without units ( or units cancel out)
• A ratio in which the denominator is 1 is called
a unit ratio
Proportion
• Proportion – an equation stating that two
ratios are equal
– Equivalent fractions set equal to each other form
proportions
– Ex:
2 = 6
3
9
(equivalent fractions are fractions that reduce to the
same quantity)
Solve
The U.S. Census Bureau surveyed 8,218 schools
nationally about their girls’ soccer program.
They found that 270,273 girls participated in a
high school soccer program in the 1999-2000
school year. Find the ratio of girl soccer players
per school to the nearest tenth.
• Girls = 270,273 = 32.8879 = 32.9
schools 8,218
Solve
In a triangle, the ratio of the measures of three
sides is 4:6:9, and its perimeter is 190 inches.
Find the longest side of the triangle.
4x + 6x + 9x = 190
19x = 190
X = 10
4x = 40 6x = 60 9x = 90
Pg 285
5) A replica of The Thinker is 10 inches tall. A statue of The Thinker, located in
front of Grawemeyer Hall on the Belnap Campus of the University of Louisville
in Kentucky, is 10 ft tall. What is the ratio of the replica to the statue in
Louisville?
10 in. replica covert inches to ft with ratio 1ft
10 ft statue
12 in
* 1 foot has 12 inches
10in * 1ft = 10ft
12in 12
1:12
1/12
10ft
12 . = 10ft * 1
= 1/12
10ft
12 10 ft
The replica is 1/12 the size of the statue
Cross Products
• Every proportion has cross products
– Ex:
2 = 6
3
9
Extremes (2) (9) = Means (3) (6)
18 = 18
The product of the means equals the product of
the extremes, so the cross products are equal.
Solve
1. 3 = x
5
75
2. 3x -5 = -13
4
2
Pg 286 YOU SOLVE
38. The ratios of the
lengths of the sides of
three polygons are given
below. Make a conjecture
type of polygon.
a. 2:2:3
b. 3:3:3:3
c. 4:5:4:5
40. In a golden rectangle, the
ratio of the length of the
rectangle to its width is
approximately 1.618:1.
Suppose a golden rectangle
has a length of 12
centimeters. What is its width
to the nearest tenth?
Pg 286 YOU SOLVE
38. The ratios of the
lengths of the sides of
three polygons are given
below. Make a conjecture
type of polygon.
a. 2:2:3
b. 3:3:3:3
c. 4:5:4:5
40. In a golden rectangle, the
ratio of the length of the
rectangle to its width is
approximately 1.618:1.
Suppose a golden rectangle
has a length of 12
centimeters. What is its width
to the nearest tenth?
Figures that are similar (~) have the same shape but not necessarily the same size.
Two polygons are similar
polygons if and only if
their corresponding
angles are congruent and
their corresponding side
lengths are proportional.
Example 1: Describing Similar Polygons
Identify the pairs of congruent angles and
corresponding sides.
#1
N Q and P R.
By the Third Angles Theorem, M T.
Therefore: Tri NPM ~ Tri QRT
0.5
Identify the pairs of congruent angles and
corresponding sides.
#2
B G and C H.
By the Third Angles Theorem, A J.
A similarity ratio (scale factor) is the ratio of the lengths of
the corresponding sides of two similar polygons.
The similarity ratio of ∆ABC to ∆DEF is
, or
The similarity ratio of ∆DEF to ∆ABC is
, or 2.
.
Example 2A: Identifying Similar Polygons
Determine whether the polygons are similar. If so, write the
similarity ratio and a similarity statement.
rectangles ABCD and EFGH
Example 2A Continued
Step 1 Identify pairs of congruent angles.
All s of a rect. are rt. s and are .
A E, B F,
C G, and D H.
Step 2 Compare corresponding sides.
Thus the similarity ratio is
,
and rect. ABCD ~ rect. EFGH.
Example 2B: Identifying Similar Polygons
Determine whether the
polygons are similar. If so,
write the similarity ratio and a
similarity statement.
Example 2B Continued
Step 1 Identify pairs of congruent angles.
P R and S W
isos. ∆
Step 2 Compare corresponding angles.
mW = mS = 62°
mT = 180° – 2(62°) = 56°
Since no pairs of angles are congruent, the triangles are not similar.
Check It Out! Example 2
Determine if ∆JLM ~ ∆SPN. If so, write the
similarity ratio and a similarity statement.
Step 1 Identify pairs of congruent angles.
N M, L P, S J
Check It Out! Example 2 Continued
Step 2 Compare corresponding sides.
Thus the similarity ratio is
, and ∆LMJ ~ ∆PNS.
Example 3: Hobby Application
Find the length of the model to the nearest tenth of
a centimeter.
Let x be the length of the model in centimeters. The
rectangular model of the racing car is similar to the
rectangular racing car, so the corresponding lengths
are proportional.
Example 3 Continued
5(6.3) = x(1.8)
Cross Products Prop.
31.5 = 1.8x
Simplify.
17.5 = x
Divide both sides by 1.8.
The length of the model is 17.5 centimeters.
Check It Out! Example 3
A boxcar has the dimensions shown.
A model of the boxcar is 1.25 in. wide. Find the length of the model to the nearest
inch.
Check It Out! Example 3 Continued
1.25(36.25) = x(9)
Cross Products Prop.
45.3 = 9x
5x
Simplify.
Divide both sides by 9.
The length of the model is approximately 5 inches.
CW pg 285 prob 6 – 11 all; prob #12- 34 even
Due at end of class
Lesson Quiz: Part I
1. Determine whether the polygons are similar. If so, write the
similarity ratio and a similarity statement.
no
2. The ratio of a model sailboat’s dimensions to the actual boat’s
dimensions is
.
If the length of the model is 10 inches, what is the length of the actual
sailboat in feet?
25 ft
Lesson Quiz: Part II
3. Tell whether the following statement is sometimes, always, or never
true. Two equilateral triangles are similar.
Always
```
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# Mathematics All India Set 1 2015-2016 CBSE (English Medium) Class 10 Question Paper Solution
Mathematics [All India Set 1]
Date & Time: 19th March 2016, 10:30 am
Duration: 3h
[1] 1
In Fig. 1, PQ is a tangent at a point C to a circle with centre O. if AB is a diameter and ∠CAB = 30°, find ∠PCA.
Concept: Tangent to a Circle
Chapter: [3.01] Circles
[1] 2
For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an A.P?
Concept: Arithmetic Progression
Chapter: [2.02] Arithmetic Progressions
[1] 3
A ladder leaning against a wall makes an angle of 60° with the horizontal. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder
Concept: Right-angled Triangles and Pythagoras Property
Chapter: [3.02] Triangles
[1] 4
A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen.
Concept: Basic Ideas of Probability
Chapter: [5.01] Probability [5.01] Probability
[2] 5
If -5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x)k = 0 has equal roots, find the value of k.
Concept: Nature of Roots
[2] 6
Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q.
Concept: Section Formula
Chapter: [6.01] Lines (In Two-dimensions)
[2] 7
In the following Fig, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that AB + CD = BC + DA.
Concept: Number of Tangents from a Point on a Circle
Chapter: [3.01] Circles
[2] 8
Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right-angled isosceles triangle.
Concept: Coordinate Geometry
Chapter: [6.01] Lines (In Two-dimensions) [6.01] Lines (In Two-dimensions)
[2] 9
The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term
Concept: Arithmetic Progression
Chapter: [2.02] Arithmetic Progressions
[2] 10
In Fig.3, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP = 2r, show that ∠ OTS = ∠ OST = 30°.
Concept: Number of Tangents from a Point on a Circle
Chapter: [3.01] Circles
[3] 11
In Fig. 4, O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region. (Take π = 3.14)
Concept: Circumference of a Circle
Chapter: [7.01] Areas Related to Circles [7.01] Areas Related to Circles
[3] 12
In Fig. 5, a tent is in the shape of a cylinder surmounted by a conical top of same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of Rs. 500/sq. metre ( "Use "pi=22/7)
Concept: Concept of Surface Area, Volume, and Capacity
Chapter: [7.02] Surface Areas and Volumes [7.02] Surface Areas and Volumes
[3] 13
If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b). Prove that bx = ay.
Concept: Distance Formula
Chapter: [6.01] Lines (In Two-dimensions)
[3] 14
In Fig. 6, find the area of the shaded region, enclosed between two concentric circles of radii 7 cm and 14 cm where ∠AOC = 40°. (use pi = 22/7)
Concept: Circumference of a Circle
Chapter: [7.01] Areas Related to Circles [7.01] Areas Related to Circles
[3] 15
If the ratio of the sum of first n terms of two A.P’s is (7n +1): (4n + 27), find the ratio of their mth terms.
Concept: Sum of First n Terms of an AP
Chapter: [2.02] Arithmetic Progressions
[3] 16
Solve for x
:1/((x-1)(x-2))+1/((x-2)(x-3))=2/3 , x ≠ 1,2,3
Concept: Solutions of Quadratic Equations by Factorization
[3] 17
A conical vessel, with base radius 5 cm and height 24 cm, is full of water. This water is emptied into a cylindrical vessel of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel. (use pi=22/7)
Concept: Concept of Surface Area, Volume, and Capacity
Chapter: [7.02] Surface Areas and Volumes [7.02] Surface Areas and Volumes
[3] 18
A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 3 5/9 cm. Find the diameter of the cylindrical vessel.
Concept: Concept of Surface Area, Volume, and Capacity
Chapter: [7.02] Surface Areas and Volumes [7.02] Surface Areas and Volumes
[3] 19
A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of a hill as 30°. Find the distance of the hill from the ship and the height of the hill
Concept: Heights and Distances
Chapter: [4.01] Heights and Distances
[3] 20
Three different coins are tossed together. Find the probability of getting exactly two heads.
Concept: Basic Ideas of Probability
Chapter: [5.01] Probability [5.01] Probability
Three different coins are tossed together. Find the probability of getting at least two heads.
Concept: Basic Ideas of Probability
Chapter: [5.01] Probability [5.01] Probability
Three different coins are tossed together. Find the probability of getting at least two tails.
Concept: Basic Ideas of Probability
Chapter: [5.01] Probability [5.01] Probability
[4] 21
Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively offered to the state government to provide place and the canvas for 1500 tents to be fixed by the governments and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 cm and height 3.5 m, with conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs Rs. 120 per sq. m, find the amount shared by each school to set up the tents. What value is generated by the above problem? (use pi =22/7)
Concept: Concept of Surface Area, Volume, and Capacity
Chapter: [7.02] Surface Areas and Volumes [7.02] Surface Areas and Volumes
[4] 22
Prove that the lengths of the tangents drawn from an external point to a circle are equal.
Concept: Number of Tangents from a Point on a Circle
Chapter: [3.01] Circles
[4] 23
Draw a circle of radius 4 cm. Draw two tangents to the circle inclined at an angle of 60° to each other.
Concept: Construction of Tangents to a Circle
Chapter: [3.03] Constructions
[4] 24
In Fig. 7, two equal circles, with centres O and O’, touch each other at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. Find the value of (DO')/(CO')
Concept: Tangent to a Circle
Chapter: [3.01] Circles
[4] 25
Solve for x: 1/(x+1)+2/(x+2)=4/(x+4), x ≠ -1, -2, -3
Concept: Nature of Roots
[4] 26
The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is 45. Find the height of the tower PQ and the distance PX. (Use sqrt3=1.73)
Concept: Heights and Distances
Chapter: [4.01] Heights and Distances
[4] 27
The houses in a row numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X.
Concept: Sum of First n Terms of an AP
Chapter: [2.02] Arithmetic Progressions
[4] 28
In Fig. 8, the vertices of ΔABC are A(4, 6), B(1, 5) and C(7, 2). A line-segment DE is drawn to intersect the sides AB and AC at D and E respectively such that (AD)/(AB)=(AE)/(AC)=1/3 Calculate th area of ADE and compare it with area of ΔABCe.
Concept: Area of a Triangle
Chapter: [6.01] Lines (In Two-dimensions)
[4] 29
A number x is selected at random from the numbers 1, 2, 3, and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is less than 16.
Concept: Basic Ideas of Probability
Chapter: [5.01] Probability [5.01] Probability
[4] 30
In Fig. 9, is shown a sector OAP of a circle with centre O, containing ∠θ. AB is perpendicular to the radius OQ and meets OP produced at B. Prove that the perimeter of shaded region is
r[tantheta+sectheta+(pitheta)/180-1]
Concept: Circumference of a Circle
Chapter: [7.01] Areas Related to Circles [7.01] Areas Related to Circles
[4] 31
A motor boat whose speed is 24 km/h in still water takes 1 hour more to go 32 km upstream than to return downstream to the same spot. Find the speed of the stream.
Concept: Pair of Linear Equations in Two Variables
Chapter: [2.01] Pair of Linear Equations in Two Variables
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## CBSE previous year question papers Class 10 Mathematics with solutions 2015 - 2016
CBSE Class 10 Maths question paper solution is key to score more marks in final exams. Students who have used our past year paper solution have significantly improved in speed and boosted their confidence to solve any question in the examination. Our CBSE Class 10 Maths question paper 2016 serve as a catalyst to prepare for your Mathematics board examination.
Previous year Question paper for CBSE Class 10 Maths-2016 is solved by experts. Solved question papers gives you the chance to check yourself after your mock test.
By referring the question paper Solutions for Mathematics, you can scale your preparation level and work on your weak areas. It will also help the candidates in developing the time-management skills. Practice makes perfect, and there is no better way to practice than to attempt previous year question paper solutions of CBSE Class 10.
How CBSE Class 10 Question Paper solutions Help Students ?
• Question paper solutions for Mathematics will helps students to prepare for exam.
• Question paper with answer will boost students confidence in exam time and also give you an idea About the important questions and topics to be prepared for the board exam.
• For finding solution of question papers no need to refer so multiple sources like textbook or guides.
|
# Given that there exists a triangle whose sides are a,b,c. Then prove that there exists a triangle whose sqrta,sqrtb,sqrtc?
Jul 23, 2018
Given that there exists a triangle whose sides are a,b,c.
So we have following 3 inequalities satisfied.
• $a + b > c$
• $b + c > a$
• $c + a > b$
Considering the first one
$a + b > c$
$\implies {\left(\sqrt{a} + \sqrt{b}\right)}^{2} - 2 \sqrt{a b} > c$
$\implies {\left(\sqrt{a} + \sqrt{b}\right)}^{2} > c + 2 \sqrt{a b}$
$\implies \sqrt{a} + \sqrt{b} > \sqrt{c}$
Similarly from 2nd inequality we get
$\sqrt{b} + \sqrt{c} > \sqrt{a}$
And from 3rd inequality we get
$\sqrt{c} + \sqrt{a} > \sqrt{b}$
So we can say if there exists a triangle having sides $a , b \mathmr{and} c$ then there exists a triangle having sides $\sqrt{a} , \sqrt{b} \mathmr{and} \sqrt{c}$
Jul 24, 2018
My interest in the problem:
#### Explanation:
Choosing $a = 2 , b = 4 \mathmr{and} c = 3$, and using scale and compass
only, four conjoined $\triangle s D E F$,
with sides $\sqrt{a} = \sqrt{2} , \sqrt{b} = 2 \mathmr{and} \sqrt{c} = \sqrt{3}$ are constructed.
The graph shows one pair over the base $E F = \sqrt{2}$,
with vertices.
$D \left(\frac{1}{\sqrt{8}} , \pm \sqrt{\frac{23}{8}}\right) , E \left(0 , 0\right) \mathmr{and} F \left(\sqrt{2} , 0\right)$
There are three such pairs, and all have the central
common $\triangle D E F$
graph{(x^2+y^2-3)((x-sqrt2)^2+y^2-4)(x^2+y^2-0.01)((x-sqrt2)^2+y^2-0.01)((x-sqrt(1/8))^2+(y-sqrt((23)/8))^2-0.01)((x-sqrt(1/8))^2+(y+sqrt((23)/8))^2-0.01)=0[-4 4 -2 2]}
|
# EXPLORING AREAS OF COMPOSITE FIGURES
## About "Exploring areas of composite figures"
Exploring areas of composite figures :
In this section, we are going to explore the area of composite figures that are composed of smaller shapes.
The diagram given below can be a good example of composite figure. This composite figure is made up of two triangles and one rectangle.
## Exploring areas of composite figures
Aaron was plotting the shape of his garden on grid paper. While it was an irregular shape, it was perfect for his yard. Each square on the grid represents 1 square meter.
A. Describe one way you can find the area of this garden.
I can divide it into rectangles and triangles, use a formula to find the area of each, and then add the areas together.
B. Find the area of the garden.
The area of the garden is 46 square meters.
C. Compare your results with other students. What other methods were used to find the area?
Other students counted squares; rearranged the triangle to be a rectangle and found the area, and then found the area of the other rectangles in the garden.
D. How does the area you found compare with the area found using different methods ?
It is the same as the other students.
E. Use dotted lines to show two different ways Aaron’s garden could be divided up into simple geometric figures.
## Finding the area of composite figure
Find the area of the figure given below.
Solution :
By drawing a horizontal line (BD), we can divide the given composite figure into two parts as given below.
In the above figure,
(1) BECF is a rectangle
(2) ABD is a triangle
Area of the given composite figure
= Area of rectangle BECF + Area of triangle ABD
Area of rectangle BECF :
length CF = 16 cm and width BC = 7 cm
= length x width
= 16 x 7
= 112 cm² ----(1)
Area of triangle ABD :
Base BD = BE - DE => 16 - 8 => 8 cm
Height AB = AC - BC => 13 - 7 => 6
Area of triangle ABD = (1/2) x b x h
= (1/2) x 8 x 6 ==> 24 cm²----(2)
(1) + (2) :
Area of the given composite figure is
= 112 + 24
= 136 cm²
After having gone through the stuff given above, we hope that the students would have understood "Area of composite figures".
Apart from the stuff given above, if you want to know more about "Exploring the area of composite shapes ", please click here
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6
|
# Averages Formulas
## Averages Formulas Used In Aptitude
Averages formulas is widely used in statistics also to reduce the calculation in finding the average where the data is huge. Here We will demonstrate the application of the assumed mean to solve some aptitude questions based on averages and weighted averages.
## In this Page Averages Formulas are given which is useful to solve many problems
• An average is defined as the sum of n different units divided by n numbers of the units.
Example:- What will be the average weight of three boy, respective weight are 46,54,53?
Solution: $\frac{\text{sum of numbers}}{\text{total numbers}}$
=$\frac{(46+54+53 )}{3}$
=$\frac{153}{3}$
=51
Example:- A school trip by St Marry’s Academy, Meerut was organized to Appu Ghar Delhi. The total number of Girls on the trip were 160, total number of boys on the trip were 40 and total number of teachers present on the trip were 100. If they want to ride a roller coaster and all of them can not board the ride at one time. Find the average no. people boarding the ride.
Solution: $\frac{\text{sum of numbers}}{\text{total numbers}}$
= $\frac{(160+40+100)}{3}$
= 100
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### Basic Averages Formulas:
• Mathematically, it is defined as the ratio of summation of all the numbers to the number of units present in the list.
Average = $\mathbf{\frac{X_{1}+X_{2}+X_{3}+X_{4}+…..X_{n}}{n}}$
OR
Average = $\frac{\text{Sum of Observations}}{\text{Total Number of Observations}}$
### Average Speed and Velocity Formula:
• Average Speed : It can be defined as total distance Travelled by a body in definite interval of time. Average speed is calculated using the below formula
Average Speed = $\mathbf{\frac{\text{Total Distance}}{\text{Total Time}}}$
CASE 1:
When one travels at speed ‘a’ for half the time and speed ‘b’ for other half of the time. Then, average speed is the arithmetic mean of the two speeds.
Average Speed= $\mathbf{\frac{a+b}{2}}$
CASE 2 :
When one travels at speed ‘a’ for half of the distance and speed ‘b’ for other half of the distance.Then, average speed is the harmonic mean of the two speeds.
Average Speed= $\mathbf {\frac{2ab}{a+b} }$
CASE 3:
When one travels at speed a for one-third of the distance, at speed b for another one-third of the distance and speed c for rest of the one-third of the distance
or:
Average Speed = $\mathbf{\frac{3abc}{ab+bc+ca}}$
Average Velocity : It can be defined as total displacement divided by total time. We calculate Average Velocity using the below formula
Average Velocity = $\mathbf{\frac{Displacement}{Total Time}}$
### Formula of Averages Related to Numbers:
• Average of ‘n’ consecutive Natural Numbers =
$\mathbf{\frac{n+1}{2}}$
• Average of the square of consecutive n natural numbers =
$\mathbf{\frac{(n+1)(2n+1)}{6}}$
• Average of cubes of consecutive n natural numbers =
$\mathbf{\frac{n\times (n+1)^{2}}{4}}$
• Average of n consecutive even numbers = (n+1)
• Average of consecutive even numbers till n =
$\mathbf{\frac{n}{2}+1}$
• Average of n consecutive odd numbers = n
• Average of consecutive odd numbers till n =
$\mathbf{\frac{n+1}{2}}$
• Sum of 1st n even consecutive natural numbers is = n(n + 1)
• Sum of 1st n odd consecutive natural numbers is = $\mathbf{n^{2}}$
### Question 1 :
The average of 10 numbers is 23. If each number is increased by 4, what will the new average be?
### Solution :
Average of 10 numbers = 23
Sum/Total numbers = 23
Sum/10 = 23
Sum of the 10 numbers = 230
If each number is increased by 4, the total increase = 4 x 10 = 40
New sum = 230 + 40 = 270
Therefore, the new average = 270/10 = 27
### Question 2 :
The average weight of a group of seven boys is 56 kg. The individual weights (in kg) of six of them are 52, 57, 55, 60, 59 and 55. Find the weight of the seventh boy.
### Solution :
Average weight of 7 boys = 56 kg.
Total weight of 7 boys = (56 × 7) kg = 392 kg.
Total weight of 6 boys = (52 + 57 + 55 + 60 + 59 + 55) kg
= 338 kg.
Weight of the 7th boy = (total weight of 7 boys) – (total weight of 6 boys)
= (392 – 338) kg
= 54 kg.
Therefore, the weight of the seventh boy is 54 kg.
### Question 3:
The mean of 25 numbers is 36. If the mean of the first numbers is 32 and that of the last 13 numbers is 39, find the 13th number.
### Solution :
Mean of the first 13 numbers = 32
Sum of the first 13 numbers = (32 × 13) = 416
Mean of the last 13 numbers = 39
Sum of the last 13 numbers = (39 × 13) = 507
Mean of 25 numbers = 36
Sum of all the 25 numbers = (36 × 25) = 900
Therefore, the 13th observation = (416 + 507 – 900) = 23
Hence, the 13th observation is 23
### Question 4 :
The average of 7 consecutive numbers is 20. What is the largest of these numbers?
### Solution :
Let the 7 consecutive numbers be x, x + 1, x + 2, x + 3, x + 4, x + 5 and x + 6,
As per the given condition;
[x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6)] / 7 = 20
⇒ 7x + 21 = 140
⇒ 7x = 119
⇒ x =17
The largest number = x + 6 = 23.
### Question 5 :
A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after 17th inning?
### Solution :
Let the average after 7th inning = x
Then average after 16th inning = x – 3
16(x-3)+87 = 17x
x = 87 – 48 = 39
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# AP Calculus AB : Tangent line to a curve at a point and local linear approximation
## Example Questions
### Example Question #1 : Derivatives
Differentiate,
Possible Answers:
Correct answer:
Explanation:
Differentiate,
Strategy
This one at first glance appears difficult even if we recognize that the chain rule is needed; we have a function within a function within a function within a function. To avoid making mistakes, it's best to start by defining variables to make the calculation easier to follow.
Let's start with the outermost function, we will write as a function of by setting,
______________________________________________________
_______________________________________________________
Similarly, define to write as a function of
_______________________________________________________
Write as a function of
_______________________________________________________
Finally, define the inner-most function, , as the function of
________________________________________________________
Since we will just substitute that in and move to the front.
That was easy enough, now just write everything in terms of by going back to the definitions of and
### Example Question #2 : Derivatives
Find the tangent line. Given the point (1,2)
Possible Answers:
Correct answer:
Explanation:
To find the tangent line at the given point, we need to first take the derivative of the given function.
Power Rule:
Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent.
Therefore, becomes
From there we plug in the "1" from the point to get our m value of the equation . When we plug in "1" to y' we get m=-1. Then from there, we will plug our point into now that we have found m to find our b value. So,
Therefore, the tangent line is equal to
### Example Question #3 : Derivatives
Find the line tangent at the point (0,1)
Possible Answers:
Correct answer:
Explanation:
To find the tangent line at the given point, we need to first take the derivative of the given function. The rule for functions with "e" in it says that the derivative of However with this function there is also a 3 in the exponent so we will also use chain rule. Chain Rules states that we work from the outside to the inside. Meaning we will take the derivative of the outside of the equation and multiply it by the derivative of the inside of the equation.
To put this into equation it will look like
From there we plug in the "0" from the point to get our m value of the equation . When we plug in "0" to y' we get m=3. Then from there, we will plug our point into now that we have found m to find our m value. So,
then plug this all back into the equation once more and we are left with
### Example Question #1 : Derivatives
Find the tangent line given the point (2,4) and the equation
Possible Answers:
Correct answer:
Explanation:
To find the tangent line at the given point, we need to first take the derivative of the given function using Power Rule
Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent.
From there we plug in the "2" from the point to get our m value of the equation . When we plug in "2" to y' we get m=8. Then from there, we will plug our point into now that we have found m to find our b value. So,
Plug this back into
### Example Question #5 : Derivatives
Find the equation of the line tangent to the curve at the point where
Possible Answers:
Correct answer:
Explanation:
Find the equation of the line tangent to the curve at the given point
The slope of the line tangent at the given point will be equal to the derivative of at that point. Compute the derivative and find the slope for our line:
Evaluate the secant term:
Therefore slope of the tangent line is simply:
So now we know the slope of the tangent line and can write the equation then solve for
In order to solve for we need one point on the line. Use the point where the tangent line meets the curve. Use the original function to find the "y" coordinate at this point:
We now have our point:
Use the point to find
### Example Question #6 : Derivatives
Find the slope of the line tangent to the curve of d(g) when g=6.
Possible Answers:
Correct answer:
Explanation:
Find the slope of the line tangent to the curve of d(g) when g=6.
All we need here is the power rule. This states that to find the derivative of a polynomial, simply subtract 1 from each exponent and then multiply each term by their original exponent.
Constant terms will drop out when we do this, and linear terms will become constants.
From here substitute in g=6.
### Example Question #1 : Derivative At A Point
Give the equation of the line tangent to the graph of the equation
at the point .
Possible Answers:
Correct answer:
Explanation:
The tangent line to the graph of at point is the line with slope that passes through that point. Find the derivative :
Apply the sum rule:
The tangent line is therefore the line with slope 5 through .Apply the point-slope formula:
### Example Question #8 : Derivatives
Give the equation of the line tangent to the graph of the equation
at the point .
Possible Answers:
None of the other choices gives the correct response.
Correct answer:
None of the other choices gives the correct response.
Explanation:
The tangent line to the graph of at point is the line with slope that passes through that point. Find the derivative :
Apply the constant multiple and sum rules:
Set and and apply the chain rule.
Substituting back:
Evaluate using substitution:
The tangent line is therefore the line with slope through is a -intercept, so apply the slope-intercept formula to get the equation
.
This is not among the choices given.
### Example Question #1 : Ap Calculus Ab
Find the equation of the line parallel to the function at , and passes through the point
Possible Answers:
Correct answer:
Explanation:
We first start by finding the slope of the line in question, which we do by taking the derivative of and evaluate at
We then use point slope form to get the equation of the line at the point
### Example Question #10 : Derivatives
Find the equation of the line tangent to at the point .
Possible Answers:
Correct answer:
Explanation:
The first step is to find the derivative of the function given, which is . Next, find the slope at (1,4) by plugging in x=1 and solving for , which is the slope. You should get . This means the slope of the new line is also -1 because at the point where a slope and a line are tangent they have the same slope. Use the equation to express your line. Y and x are variables and m is the slope, so the only thing you need to find is b. Plug in the point and slope into to get . Now you can express the general equation of the line as .
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## Introductory Algebra for College Students (7th Edition)
$\text{$x-$intercepts: } \{ 6, 2 \}$
Setting $y=0$ and solving for $x$ using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), then the $x-$intercept/s of the given equation, $y=-x^2+8x-12 ,$ is/are \begin{array}{l}\require{cancel} -x^2+8x-12=0 \\\\ x=\dfrac{-8\pm\sqrt{8^2-4(-1)(-12)}}{2(-1)} \\\\ x=\dfrac{-8\pm\sqrt{64-48}}{-2} \\\\ x=\dfrac{-8\pm\sqrt{16}}{-2} \\\\ x=\dfrac{-8\pm\sqrt{(4)^2}}{-2} \\\\ x=\dfrac{-8\pm4}{-2} .\end{array} Hence, the $x-$intercepts are \begin{array}{l}\require{cancel} \text{$x-$intercepts: } \{ 6, 2 \} .\end{array}
|
1
Q:
# Hemavathi cut a cake into two halves and cuts one half into smaller pieces of equal size. Each of the small pieces is fifteen grams in weight. If she has nine pieces of the cake in all with her, how heavy was the original cake ?
A) 240 gms B) 280 gms C) 180 gms D) 170 gms
Explanation:
The nine pieces consist of 8 smaller equal pieces and one half cake piece.
Weight of each small piece = 15 g.
So, total weight of the cake = [2 x (15 x 8)]g = 240 g.
Q:
Negative plus a positive equals
A) negative B) positive C) zero D) Can't be determined
Explanation:
We know that a negative value plus a positive value always count backward and the sign depends on the bigger value sign.
When you are adding a negative number to a positive number, then the value counts backward and takes the sign of the biggest number.
For example::
-5 + 2 = -3 (big number is 5 and its sign is '-')
+5 + (-3) = 5 - 3 = +2(big number is 5 and it has sign '+').
0 19
Q:
How many black cards are in a deck?
A) 13 B) 26 C) 39 D) 52
Explanation:
The total cards in the deck are 52. These 52 cards are divided into 4 suits of 13 cards in each suit. Two Red suits and Two black suits.
Red suits :: Heart suit and Diamond suit = 26
Black suits :: Spade suit and Club suit = 26.
0 51
Q:
6 choose 3 =
A) 20 B) 30 C) 18 D) 24
Explanation:
6 choose 3 means number of possible unordered combinations when 3 items are selected from 6 available items i.e, nothing but 6C3.
Now 6C3 = 6 x 5 x 4/3 x 2 x 1 = 120/6 = 20.
0 26
Q:
Which of the following is a Composite number?
A) 0 B) 19 C) 29 D) 91
Explanation:
Here in the given numbers 91 is a Composite number. Since it has factors of 7 and 13 other than 1 and itself.
Composite Numbers :
A composite number is a positive integer which is not prime (i.e., which has factors other than 1 and itself).
Examples :: 4, 6, 8, 12, 14, 15, 18, 20, ...
1 97
Q:
What is the value of a two–digit number?
A. The sum of the digits is 15.
B. The difference of the squares of the digits is 45.
C. The difference of their digits is 3.
A) B and C together are sufficient B) Any one pair of A and B, B and C or C and A is sufficient C) C and A together are sufficient D) A and B together are sufficient
Answer & Explanation Answer: B) Any one pair of A and B, B and C or C and A is sufficient
Explanation:
From the given data,
Let the two gits of a number be x & y
A) x + y = 15
B)
(x+y) (x - y) = 45
C) x - y = 3
From any 2 of the given 3 statements, we can find that 2 digit number as
2x = 18 => x = 9
=> y = 6
Hence, 2 digit number is 96.
Any one pair of A and B, B and C or C and A is sufficient to find.
1 68
Q:
How can we get 24 as an answer by using 6, 11, 3 and 9?
This can be solved in two ways using only 6, 11, 9 and 3 we must get 24.
1. {[(11–6) x 3] + 9}
By solving this we get
5 x 3 + 9 = 15 + 9 = 24.
2. [(9–6) x (11–3)]
= 3 x 8 = 24.
108
Q:
Negative divided by Negative
A) -ve B) +ve C) 0 D) Can't be determined
Explanation:
We know the Mathematical rules that
7 184
Q:
Are Alternate Exterior Angles congruent?
A) TRUE B) FALSE
Explanation:
We know that,
Alternate Exterior Angles Theorem::
The Alternate Exterior Angles Theorem states that, when two parallel lines are cut by a transversal, the resulting alternate exterior angles are congruent .
Converse of the Alternate Exterior Angles Theorem ::
Converse of the Alternate Exterior Angles Theorem states that, If two lines are cut by a transversal and the alternate exterior angles are congruent, then the lines are parallel.
|
NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2
# NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2
## NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2
NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2.
Ex 4.2 Class 9 Maths Question 1.
Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution,
(ii) only two solutions,
(iii) infinitely many solutions
Solution:
Option (iii) is true. Because in the given equation, for every value of x, we get a corresponding value of y and vice-versa.
Hence, given linear equation has an infinitely many solutions.
Ex 4.2 Class 9 Maths Question 2.
Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y
Solution:
(i)
2x + y = 7
If x = 0, we have 2(0) + y = 7
y = 7
Therefore, one solution is (0, 7).
If x = 1, we have 2(1) + y = 7
y = 7 – 2 y = 5
Therefore, the second solution is (1, 5).
If x = 2, we have 2(2) + y = 7
y = 7 – 4 y = 3
Therefore, the third solution is (2, 3).
If x = 3, we have 2(3) + y = 7
y = 7 – 6 y = 1
Therefore, the fourth solution is (3, 1).
(ii) πx + y = 9
If x = 0, we have π(0) + y = 9
y = 9 – 0 y = 9
Therefore, one solution is (0, 9).
If x = 1, we have π(1) + y = 9
y = 9 – Ï€
Therefore, the second solution is (1, (9 – Ï€)).
If x = 2, we have π(2) + y = 9
y = 9 – 2Ï€
Therefore, the third solution is (2, (9 – 2Ï€)).
If x = -1, we have π(-1) + y = 9
y = 9 + π
Therefore, the fourth solution is (-1, (9 + π)).
(iii) x = 4y
If x = 0, we have 4y = 0
y = 0
Therefore, one solution is (0, 0).
If x = 1, we have 4y = 1
y = 1/4
Therefore, the second solution is (1, 1/4).
If x = 4, we have 4y = 4
y = 1
Therefore, the third solution is (4, 1).
If x = -4, we have 4y = -4
y = -1
Therefore, the fourth solution is (-4, -1).
Ex 4.2 Class 9 Maths Question 3.
Check which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0, 2)
(ii) (2, 0)
(iii) (4, 0)
(iv) (√2, 4√2)
(v) (1, 1)
Solution:
(i)
(0, 2) means x = 0 and y = 2
Putting x = 0 and y = 2 in x – 2y = 4, we get
L.H.S. = 0 – 2(2) = -4
But R.H.S. = 4
L.H.S. ≠ R.H.S.
(0, 2) is not a solution of the equation x – 2y = 4.
(ii) (2, 0) means x = 2 and y = 0
Putting x = 2 and y = 0 in x – 2y = 4, we get
L.H:S. = 2 – 2(0) = 2 – 0 = 2
But R.H.S. = 4
L.H.S. ≠ R.H.S.
(2, 0) is not a solution of the equation x – 2y = 4.
(iii) (4, 0) means x = 4 and y = 0
Putting x = 4 and y = 0 in x – 2y = 4, we get
L.H.S. = 4 – 2(0) = 4 – 0 = 4 = R.H.S.
L.H.S. = R.H.S.
(4, 0) is a solution of the equation x – 2y = 4.
(iv) (√2, 4√2) means x = √2 and y = 4√2
Putting x = √2 and y = 4√2 in x – 2y = 4, we get
L.H.S. = √2 – 2(4√2) = √2 – 8√2 = -7√2
But R.H.S. = 4
L.H.S. ≠ R.H.S.
(√2, 4√2) is not a solution of the equation x – 2y = 4.
(v) (1, 1) means x = 1 and y = 1
Putting x = 1 and y = 1 in x – 2y = 4, we get
L.H.S. = 1 – 2(1) = 1 – 2 = -1.
But R.H.S = 4
L.H.S. ≠ R.H.S.
(1, 1) is not a solution of the equation x – 2y = 4.
Ex 4.2 Class 9 Maths Question 4.
Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Solution:
We have 2x + 3y = k
x = 2, y = 1 is a solution of the given equation.
Putting x = 2 and y = 1 in 2x + 3y = k, we get
2(2) + 3(1) = k
k = 4 + 3
k = 7
Thus, the required value of k is 7.
NCERT Solutions for Maths Class 10
NCERT Solutions for Maths Class 12
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# Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Additional Questions
Students can Download Maths Chapter 4 Geometry Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 4 Geometry Additional Questions
Exercise 4.1
Question 1.
“The sum of any two angles of a triangle is always greater than the third angle”. Is this statement true. Justify your answer.
Solution:
No, the sum of any two angles of a triangle is not always greater than the third angle. In an isosceles right angled triangles, the angle will be 90°, 45°, 45°.
Here sum of two angles 45° + 45° = 90°.
Question 2.
The three angles of a triangle are in the ratio 1:2:1. Find all the angles of the triangle. Classify the triangle in two different ways.
Solution:
Let the angles of the triangle be x, 2x, x.
Using the angle sum property, we have
x + 2x + x = 180°
4x = 180°
x = $$\frac{180^{\circ}}{4}$$
x = 45°
2x = 2 × 45° = 90°
Thus the three angles of the triangle are 45°, 90°, 45°.
Its two angles are equal. It is an isoscales triangle. Its one angle is 90°.
∴ It is a right angled triangle.
Question 3.
Find the values of the unknown x and y in the following figures
Solution:
(i) Since angles y and 120° form a linear pair.
y + 120° = 180°
y = 180° – 120°
y = 60°
Now using the angle sum property of a triangle, we have
x + y + 50° = 180°
x + 60° + 50° = 180°
x + 110° = 180°
x = 180° – 110°= 70°
x = 70°
y = 60
(ii) Using the angle sum property of triangle, we have
50° + 60° + y = 180°
110° + y = 180°
y = 180° – 110°
y = 70°
Again x and y form a linear pair
∴ x + y = 180°
x + 70° = 180°
x = 180° – 70°= 110°
∴ x = 110°; y = 70°
Question 4.
Two angles of a triangle are 30° and 80°. Find the third angle.
Solution:
Let the third angle be x.
Using the angle sum property of a triangle we have,
30° + 80° + x = 180°
x + 110° = 180°
x = 180° – 110° = 70°
Third angle = 70°.
Exercise 4.2
Question 1.
In an isoscleles ∆ABC, AB = AC. Show that angles opposite to the equal sides are equal.
Solution:
Given: ∆ABC in which $$\overline{A B}$$ = $$\overline{A C}$$.
To Prove: ∠B = ∠K.
AB = AC (Hypoteneous) (given)
∴ Their corresponding parts are equal. ∠B = ∠C.
Question 2.
ABC is an isosceles triangle having side $$\overline{A B}$$ = side $$\overline{A C}$$. If AD is perpendicular to BC, prove that D is the mid-point of $$\overline{B C}$$.
Solution:
In ∆ABD and ∆ACD, we have
Side $$\overline{A D}$$ = Side $$\overline{A D}$$ [Common]
Side $$\overline{A B}$$ = Side $$\overline{A C}$$ [Common]
Using RHS congruency, we get
∆ABD ≅ ∆ACDc
Their corresponding parts are equal
∴ BD = CD
∴ O is the mid point of BC.
Question 3.
In the figure PL ⊥ OB and PM ⊥ OA such that PL = PM.
Prove that ∆PLO ≅ ∆PMO.
Solution:
In ∆PLO and ∆PMO, we have
∠PLO = ∠PMO = 90° [Given]
$$\overline{O P}$$ = $$\overline{O P}$$ [Hypotenuse]
PL = PM
Using RHS congruency, we get
∆PLO ≅ ∆PMO
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# The circle passing through (1,-2) and touching the axis of x at (3, 0) also passes through the point
## Solution :
Let the equation of circle be $$(x-3)^2 + (y-0)^2 + \lambda y$$ = 0
As it passes through (1, -2)
$$\therefore$$ $$(1-3)^2 + (-2)^2 + \lambda (-2)$$ = 0
$$\implies$$ 4 + 4 – 2$$\lambda$$ = 0
$$\implies$$ $$\lambda$$ = 4
$$\therefore$$ Equation of circle is $$(x-3)^2 + y^2 + 4y$$ = 0
By hit and trial method, we see that point (5, -2) satisfies the equation of circle.
### Similar Questions
Let C be the circle with center at (1,1) and radius 1. If T is the circle centered at (0,y) passing through origin and touching the circle C externally, then the radius of T is equal to
The equation of the circle through the points of intersection of $$x^2 + y^2 – 1$$ = 0, $$x^2 + y^2 – 2x – 4y + 1$$ = 0 and touching the line x + 2y = 0, is
Find the equation of circle having the lines $$x^2$$ + 2xy + 3x + 6y = 0 as its normal and having size just sufficient to contain the circle x(x – 4) + y(y – 3) = 0.
The equation of the circle passing through the foci of the ellipse $$x^2\over 16$$ + $$y^2\over 9$$ = 1 and having center at (0, 3) is
The length of the diameter of the circle which touches the X-axis at the point (1,0) and passes through the point (2,3) is
|
# Problems on Compound Interest
More solved problems on compound interest using formula are shown below.
### 1. The simple interest on a sum of money for 3 years at 6²/₃ % per annum is $6750. What will be the compound interest on the same sum at the same rate for the same period, compounded annually? Solution: Given, SI =$ 6750, R = ²<span style='font-size: 50%'>/₃ % p.a. and T = 3 years.
sum = 100 × SI / R × T
= $(100 × 6750 × ³/₂₀ × 1/3 ) =$ 33750.
Now, P = $33750, R = ²<span style='font-size: 50%'>/₃% p.a. and T = 3years. Therefore, amount after 3 years =$ {33750 × (1 + (20/3 × 100)}³ [using A = P (1 + R/100)ᵀ]
= $(33750 × 16/15 × 16/15 × 16/15) =$ 40960.
Thus, amount = $40960. Hence, compound interest =$ (40960 - 33750) = $7210. ### 2. The difference between the compound interest, compounded annually and the simple interest on a certain sum for 2 years at 6% per annum is$ 18. Find the sum.
Solution:
Let the sum be $100. Then, SI =$ (100 × 6 × 2/100) = $12 and compound interest =$ {100 × (1 + 6/100)² - 100}
= ${(100 × 53/50 × 53/50) - 100} =$ (2809/25 - 100) = $309/25 Therefore, (CI) - (SI) =$ (309/25 – 100) = $9/25 If the difference between the CI and SI is$ 9/25, then the sum = $100. If the difference between the CI and SI is$ 18, then the sum = $(100 × 25/9 × 18 ) =$ 5000.
Hence, the required sum is $5000. Alternative method Let the sum be$ P.
Then, SI = $(P × 6/100 × 2) =$ 3P/25
And, CI = ${P × (1 + 6/100)² - P} =$ {(P × 53/50 × 53/50) - P} = $(²⁸<span style='font-size: 50%'>⁹/₂₅₀₀ P - P) =$ (309P/2500)
(CI) - (SI) = $(309P/2500 – 3P/25) =$ (9P/2500)
Therefore, 9P/2500 = 18
⇔ P = 2500 × 18/9
⇔ P = 5000.
Hence, the required sum is $5000. ### 3. A certain sum amounts to$ 72900 in 2 years at 8% per annum compound interest, compounded annually. Find the sum.
Solution:
Let the sum be $100. Then, amount =$ {100 × (1 + 8/100)²}
= $(100 × 27/25 × 27/25) =$ (2916/25)
If the amount is $2916/25 then the sum =$ 100.
If the amount is $72900 then the sum =$ (100 × 25/2916 × 72900) = $62500. Hence, the required sum is$ 62500.
Alternative method
Let the sum be $P. Then, amount =$ {P × (1 + 8/100)²}
= ${P × 27/25 × 27/25} =$ (729P/625)
Therefore, 729P/625 = 72900
⇔ P = (72900 × 625)/729
⇔ P = 62500.
Hence, the required sum is $62500. ### In this question the formula is when the interest is compounded annually to solve this problem on compound interest. 4. At what rate per cent per annum will Ron lends a sum of$2000 to Ben. Ben returned after 2 years $2205, compounded annually? Solution: Let the required rate be R% per annum. Here, A =$ 2205, P = $2000 and n = 2 years. Using the formula A = P(1 + R/100)ⁿ, 2205 = 2000 × ( 1 + R/100)² ⇒ (1 + R/100)² = 2205/2000 = 441/400 = (21/20)² ⇒ ( 1 + R/100) = 21/20 ⇒ R/100 = (21/20 – 1) = 1/20 ⇒ R = (100 × 1/20) = 5 Hence, the required rate of interest is 5% per annum. ### 5. A man deposited$1000 in a bank. In return he got $1331. Bank gave interest 10% per annum. How long did he kept the money in the bank? Solution: Let the required time be n years. Then, amount =$ {1000 × (1 + 10/100)ⁿ}
= ${1000 × (11/10)ⁿ} Therefore, 1000 × (11/10)ⁿ = 1331 [since, amount =$ 1331 (given)]
⇒ (11/10)ⁿ = 1331/1000 = 11 × 11 × 11/ 10 × 10 × 10 = (11/10)³
⇒ (11/10)ⁿ = (11/10)³
⇒ n = 3.
Thus, n = 3.
Hence, the required time is 3 years.
Compound Interest
Compound Interest
Compound Interest with Growing Principal
Compound Interest with Periodic Deductions
Compound Interest by Using Formula
Compound Interest when Interest is Compounded Yearly
Compound Interest when Interest is Compounded Half-Yearly
Compound Interest when Interest is Compounded Quarterly
Problems on Compound Interest
Variable Rate of Compound Interest
Difference of Compound Interest and Simple Interest
Practice Test on Compound Interest
Uniform Rate of Growth
Uniform Rate of Depreciation
Uniform Rate of Growth and Depreciation
Compound Interest - Worksheet
Worksheet on Compound Interest
Worksheet on Compound Interest when Interest is Compounded Half-Yearly
Worksheet on Compound Interest with Growing Principal
Worksheet on Compound Interest with Periodic Deductions
Worksheet on Variable Rate of Compound Interest
Worksheet on Difference of Compound Interest and Simple Interest
Worksheet on Uniform Rate of Growth
Worksheet on Uniform Rate of Depreciation
Worksheet on Uniform Rate of Growth and Depreciation
|
Evaluating Algebraic Expressions
7 teachers like this lesson
Print Lesson
Objective
SWBAT evaluate algebraic expressions.
Big Idea
How can we use order of operations to evaluate expressions?
Do Now
10 minutes
Previously, students learned how to translate algebraic expressions. The Do Now Problems are a review of this concept. Students will have about 5 minutes to work on the following problems.
Do Now:
Write an algebraic expression for each.
1) The product of four and a number
2) Nine less than a number
3) One half of a number
4) The difference between eight and a number
5) Three times a number less two
I will call students to the board to present their answers. They will have to explain which operation they used and why. For problem 2, students often forget to reverse the order when "less than" is used. Problem 5 may present some difficulty for students because it involves more than one operation. I will remind students to look for and underline key words that indicate an operation.
Mini Lesson
15 minutes
The focus of this lesson will be evaluation expressions. As we work on the first example, students will develop the steps.
Ex. 1 - Evaluate 9 + a2 for a = 6
What do we call 9 + a?
Students should recognize that it's an expression.
What does evaluate mean? What word do you (almost) see in evaluate?
Students may notice that evaluate contains the word value. I will share that evaluate means that we need to find the value of the expression.
How can we find the value for this expression?
Many students will recognize that we are given a value for the variable, a, that we can use.
If we substitute 6 for a, what should we do next, since there is both an exponent and addition?
Students should make the connection that the order of operations should be applied at this point.
We will work through a few more examples, so students can realize that expressions can be evaluated with fractions, decimals, and more than one variable.
Ex. 2 - Evaluate 6x + 12 for x = 1/3
It is important for students to be mindful when evaluating expressions with multiplication.
If I replace x with 1/3, should I rewrite the expressions as 61/3 + 12 ?
Most students will recognize that 61/3 is not the same as 6 times 1/3 .
What should we do to make the operation clear?
Students need to indicate the operation with a multiplication sign.
Ex. 3 - Evaluate 4a + 2b + 9 for a = 2 and b = 3
Ex. 4 - Evaluate 8.1x - 3.2 for x = 2.4
Independent Practice
10 minutes
Following the lesson, students will have the opportunity to independently practice evaluating expressions. Students should follow the steps that they've developed.
As students work on the problems, I will circulate throughout the room focusing on students who had difficulty with the order of operations lesson.
* It may be helpful to group together the students who had difficulty with the order of operations lesson.
Independent Practice
1) 4z - 5 for z = 4
2) 2a + 5 for a = 3
3) 6(n - 4) for n = 9
4) 2(3y - 2) for y = 2
5) 52 - b2 for b = 7
6) x3 - 7 for x = 5
7) 3h + 2 for h = 3
8) 2x2 for x = 3
9) 4a + a/c for a = 8 and c = 2
After 15 minutes, I will give each group a problem to present. Groups will be given a whiteboard and marker in which to show their work on. Groups will have a few minutes to agree on their work and answer and then they will present to the class. If students should disagree or have questions, we will discuss the problem as a class.
Exit Ticket
5 minutes
To assess students' understanding of the lesson and verify that there are not any misconceptions, I will give students an exit ticket to complete.
Exit Ticket
Evaluate 2x + 6y - 3 for x = 5 and y = ½
When checking the exit tickets, I will be looking for incorrect answers caused by using the order of operations incorrectly and multiplication errors.
I will use the results to clarify any misconceptions in the following lesson and also for student groupings.
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Saturday, January 18, 2014
Bertrand's Postulate Part 3: Bounding the Central Binomial Coefficient
This is week three of the four week series on Bertrand's Postulate. Last week, we restricted the range of the prime factors of "2n choose n." This week, we are going to restrict the size of it. This will use many of the techniques and identities we have discussed in the last two weeks to do.
The third step here is to determine the lower bound. The fourth step will be to find the upper bound, and we will then have an inequality to work with. To find a lower bound, we must use another property of Pascal’s Triangle.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
We will now expand a few polynomials (see definition of polynomial expansion in the first post). The Pascal’s Triangle application will become clear in a minute.
(a + b)0 = 1
(a + b)1 = 1a + 1b
(a + b)2 = 1a2 + 2ab + 1b2
(a + b)3 = 1a3 + 3a2b + 3ab2 + 1b3
(a + b)4 = 1a4 + 4a3b + 6a2b2 + 4ab3 + 1b4
Do you see a pattern? Let me make it clearer:
(a + b)0 = 1
(a + b)1 = 1a + 1b
(a + b)2 = 1a22ab + 1b2
(a + b)3 = 1a3 + 3a2b + 3ab2 + 1b3
(a + b)4 = 1a4 + 4a3b + 6a2b2 + 4ab3 + 1b4
The coefficients in each expansion are the numbers in the corresponding row of Pascal’s Triangle. In general, by using the “n choose k” notation, this theorem can be generalized as:
(a + b)n = (“n choose 1”)an + (“n choose 2”)an-1b + (“n choose 3”)an-2b2... + (“n choose n-1”)abn-1 + (“n choose n”)bn
After you let that sink in for a second, the process of creating the lower bound will be simple.
The lower bound that is used is 4n/2n. To prove that this is a lower bound, we must prove the following inequality to be true:
4n/2n < “2n choose n
This can be rearranged to be:
4n < 2n(“2n choose n”)
4n can be rewritten using the binomial theorem. A lone four is not an expandable binomial, but here is a reconfiguration of 4n that can be expanded.
4n
(22)n
22n
(1 + 1)2n
Plug these values into the binomial theorem for a, b, and n to get:
(a + b)n = (“n choose 1”)an + (“n choose 2”)an-1b + (“n choose 3”)an-2b2... + (“n choose n-1”)abn-1 + (“n choose n”)bn
(1 + 1)2n = (“2n choose 1”)12n + (“2n choose 2”)12n-11 + (“2n choose 3”)12n-212... + (“2n choose 2n-1”)1•12n-1 + (“2n choose 2n”)12n
Since one raised to any power is one, all of the ones cancel out to get:
(1 + 1)2n = (“2n choose 1”)1 + (“2n choose 2”)1 + (“2n choose 3”)1 ... + (“2n choose 2n-1”)1 + (“2n choose 2n”)1
(1 + 1)2n = (“2n choose 1”) + (“2n choose 2”) + (“2n choose 3”) ... + (“2n choose 2n-1”) + (“2n choose 2n”)
This creates just a sum of all of the values in the (2n)-th row of Pascal’s Triangle. Earlier, it was noted that the central binomial coefficient of a row is the largest number in that row. So, “2n choose n” is the largest number in the sum above.
How many numbers are in that sum? Since it is the (2n)-th row of Pascal’s Triangle, there are 2n entries in that row. This means that the product of 2n and “2n choose n” must be greater than that sum.
(“2n choose 1”) + (“2n choose 2”) + (“2n choose 3”) ... + (“2n choose 2n-1”) + (“2n choose 2n”) < 2n(“2n choose n”)
What is that sum equal to? It was defined to be the expansion of 4n, meaning that 4n can be substituted in for that sum.
4n < 2n(“2n choose n”)
Dividing both sides by 2n gives the inequality that the step was looking for:
4n/2n < “2n choose n
And that completes the step.
The fourth step is to set an upper bound. A lower bound won’t do much good without an upper bound to counter it. A fully intact upper bound for “2n choose n” cannot be found until step five is partly established, but an upper bound can be placed on a different expression which will play back into the proof later on.
This step requires something called the primorial function, which is similar to the factorial function. The number n factorial is the product of all of the positive integers less than or equal to n. Similarly, the number n primorial (written n╫) is the product of all of the prime numbers less than or equal to n. For example:
4╫ = 3 • 2 = 6
8╫ = 7 • 5 • 3 • 2 = 210
16╫ = 13 • 11 • 7 • 5 • 3 • 2 = 30030
This step will set an upper bound on the number n╫. The upper bound we will try to set is 4n. So, this is the inequality that needs to be proven:
n╫ < 4n
This proof needs to be tackled in two parts. First, it needs to be proven for all odd values of n. Then, it can be proven for all even values of n.
If n is odd, then it can be rewritten as 2m + 1 assuming that m is an integer (see definition of odd number). Throughout the proof, quantities like “2n choose n” and “2m choose m” have come up, but the row of Pascal’s Triangle it refers to is always an even numbered row (2n and 2m are even). What about odd numbered rows?
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
An odd numbered row does not have a center position, but the two positions in the middle are always equal to each other. In other words, “2m+1 choose m” is equal to “2m+1 choose m+1.”
“2m+1 choose m” = “2m+1 choose m+1”
Now, add “2m+1 choose m” to both sides of this.
“2m+1 choose m” = “2m+1 choose m+1”
“2m+1 choose m” + “2m+1 choose m” = “2m+1 choose m+1” + “2m+1 choose m
2(“2m+1 choose m”) = “2m+1 choose m+1” + “2m+1 choose m
The right hand side of that equation can be thought of as the sum of two of the elements of the (2m+1)-st row of Pascal’s Triangle. But, what if the sum of all of the elements in the (2m+1)-st row was found? That would definitely be a bigger number than what is currently there. So, an inequality can be formed using that sum as the right-hand side.
2(“2m+1 choose m”) < (“2m+1 choose 1”) + (“2m+1 choose 2”) + (“2m+1 choose 3”) ... + (“2m+1 choose 2m”) + (“2m+1 choose 2m+1”)
But in step 3, we defined this to be the polynomial expansion of (1 + 1)2m+1, or 22m+1. Substitute that in for the right-hand side. Then, use the laws of exponents (see definition of law of exponents) to get:
2(“2m+1 choose m”) < 22m+1
(“2m+1 choose m”) < 22m+1 ÷ 21
(“2m+1 choose m”) < 22m+1–1
(“2m+1 choose m”) < 22m
(“2m+1 choose m”) < (22)m
(“2m+1 choose m”) < 4m
Now, let’s look at the prime factorization of “2m+1 choose m.” Recall the formula that was used to find exponents of primes in step 1 and step 2.
vp(“2n choose n”) = vp((2n)!) - 2vp(n!)
This can be rewritten for the quantity “2m+1 choose m” fairly easily.
vp(“2m+1 choose m”) = vp((2m + 1)!) - 2vp(m!) - vp(m + 1)
Prime numbers less than m+1 are difficult to analyze, but what about between m+2 and 2m+1? By similar logic as in step 1, all of these numbers will have an exponent of zero in m! or (m+1)!, but an exponent of one in (2m+1)!. So, it can be guaranteed that each prime between m+2 and 2m+1 is a factor of “2m+1 choose m.” This also means that the product of all primes between m+2 and 2m+1 is less than or equal to “2m+1 choose m.”
The inequality we are trying to reach is n╫ < 4n. So, plugging that inequality in on a different interval (say (m+1)╫ < 4m+1), is a legal maneuver. This is called proof by induction.
(m+1)╫ < 4m+1
The left-side of the inequality above is (m+1)╫. It was just determined what the upper bound of the product of all primes between m+2 and 2m+1 was as well. If we multiply the primorial of m+1 by the product of primes between m+2 and 2m+1, we will just get the primorial of 2m+1. This number will be less than the product of the two right hand sides.
(2m+1)╫ < 4m+1 • “2m+1 choose m
Earlier, it was proven that “2m+1 choose m” has an upper bound of 4m. Since that will only make the right side bigger, we can plug that in without a problem. This gives:
(2m+1)╫ < 4m+1 • 4m
Using the law of exponents brings it to:
(2m+1)╫ < 42m+1
And substituting n back in for 2m+1 gives:
n╫ < 4n
We are almost done with the step. The hard part is complete; any odd values of n are covered. All that needs to be done is to make sure the even values are covered as well.
If n is even, then its primorial will always be equal to the odd number below it. For instance:
4╫ = 6 = 3╫
10╫ = 210 = 9╫
22╫ = 9699690 = 21╫
This is because that even number cannot be prime. If n is even, its primorial must be equal to (n-1)╫ because n-1 is the highest potentially prime number less than n.
n╫ = (n-1)╫
Earlier, it was proved that the upper bound works on an odd numbered primorial. Since n-1 is odd, we can plug that in to get:
(n-1)╫ < 4n–1
We just determined that n╫ = (n-1)╫, so substituting n╫ in for (n-1)╫ gives:
n╫ < 4n–1
4n–1 is definitely less than 4n, so that can be put on the right-hand side to get:
n╫ < 4n
Since the bound is good on all odd numbers and all even numbers, the step is complete.
We have most of the groundwork done. Next week, we will officially prove Bertrand's Postulate as well as finalize our upper bound.
|
### Prompt Cards
These two group activities use mathematical reasoning - one is numerical, one geometric.
### Consecutive Numbers
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Exploring Wild & Wonderful Number Patterns
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
# Two Number Lines
## Two Number Lines
You could try Number Lines before this problem.
Max and Mandy both had number lines. Max's number line went along from left to right like this:
Mandy's number line went up and down like this:
Max and Mandy both started at zero on their number lines. Max made a secret jump along his number line and then moved on seven and landed on $10$. How long was his secret jump?
Mandy made a jump of three and another secret jump. She landed on $6$. How long was her secret jump?
Max and Mandy decided to put their number lines together. Their teacher gave them some squared paper. They had made a graph. It looked like a bit like this:
Max moved four along and Mandy moved six up. They put a counter on the place they landed. Now their graph looked like this:
How far had each of them moved along and up from $0$ to get the counter to the place marked on the grid below?
If Max and Mandy both moved the same distance along and up, where could the counter be?
### Why do this problem?
This problem follows on from Number Lines and could be used to introduce coordinates. In addition to focusing on ideas associated with inverse operations and algebra, the final part of this activity asks pupils to find all possibilities, giving them an opportunity to identify and explain the pattern produced on the graph.
### Possible approach
It would be good for learners to work on this task practically, with number lines, squared paper and counters. These sheets of number lines could be printed off and cut out for this purpose.
### Key questions
How far up Mandy's number line has the counter been put?
How far along Max's number line has the counter been put?
Where are you going to put your counter now?
If they both moved the same distance along or up their number line, can you think of a number of jumps they could have made?
What about a different number of jumps?
Can you think of a word to describe that line?
### Possible extension
Learners could follow on with Fred the Class Robot.
|
# solve the given system of differential equations. (dx_1)/(dt)=2x_1-3x_2 (dx_2)/(dt)=x_1-2x_2
solve the given system of differential equations.
$\frac{{dx}_{1}}{dt}=2{x}_{1}-3{x}_{2}$
$\frac{{dx}_{2}}{dt}={x}_{1}-2{x}_{2}$
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Step 1
Consider the following system of differential equations:
$\frac{{dx}_{1}}{dt}=2{x}_{1}-3{x}_{2}$...(1)
$\frac{{dx}_{2}}{dt}={x}_{1}-2{x}_{2}$
Substitute $x1=\frac{{dx}_{2}}{dt}+2{x}_{2}$ in the equation (1):
$\frac{d}{dt}\left(\frac{{dx}_{2}}{dt}+2{x}_{2}\right)=2\left(\frac{{dx}_{2}}{dt}+2{x}_{2}\right)-3{x}_{2}$
$\frac{{d}^{2}{x}_{2}}{{dt}^{2}}+2\frac{{dx}_{2}}{dt}=2\frac{{dx}_{2}}{dt}+4{x}_{2}-3{x}_{2}$
$\frac{{d}^{2}{x}_{2}}{{dt}^{2}}-{x}_{2}=0$
Step 2
The auxiliary equation is ${m}^{2}-1=0$.
The roots of auxiliary equation are 1,−1.
Hence, the solution of homogeneous equation is ${x}_{2}=A{e}^{t}+B{e}^{-t}$.
A and B are constants.
Substitute value of ${x}_{2}$ in the equation ${x}_{1}=\frac{{dx}_{2}}{dt}+2{x}_{2}$:
${x}_{1}=\frac{d}{dt}\left(A{e}^{t}+B{e}^{-t}+2\left(A{e}^{t}+B{e}^{-t}\right)$
$=A{e}^{t}-B{e}^{-t}+2A{e}^{t}+2B{e}^{-t}$
$=3A{e}^{t}+B{e}^{t}$
Step 3
Hence, the solution is
${x}_{1}=3A{e}^{t}+B{e}^{t}$
${x}_{2}=A{e}^{t}+B{e}^{-t}$.
Jeffrey Jordon
|
## RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS
Other Exercises
Mark the correct alternative in each of the following:
Question 1.
The number of surfaces of a cone has, is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
Number of surfaces of a cone are 2 (b)
Question 2.
The area of the curved surface of a cone of radius 2r and slant height $$\frac { 1 }{ 2 }$$, is
Solution:
Radius of a cone = 2r
Question 3.
The total surface area of a cone of radius $$\frac { r }{ 2 }$$ and length 2l, is
Solution:
Question 4.
A solid cylinder is melted and cast into a cone of same radius. The heights of the cone and cylinder are in the ratio
(a) 9 : 1
(b) 1 : 9
(c) 3 : 1
(d) 1 : 3
Solution:
Let r be the radius and h be the height of cylinder, then volume = πr2h
Now volume of cone = πr2h
Question 5.
If the radius of the base of a right circular cone is 3r and its height is equal to the radius of the base, then its volume is
Solution:
Radius of the base of a cone (R) = 3r
and height (H) = 3r
Question 6.
If the volumes of two cones are in the ratio 1 : 4 and their diameters are in the ratio 4 : 5, then the ratio of their heights, is
(a) 1 : 5
(b) 5 : 4
(c) 5 : 16
(d) 25 : 64
Solution:
Ratio in the volumes of two cones =1:4
and ratio in their diameter = 4:5
Let h1, h2 be their heights
Question 7.
The curved surface area of one cone is twice that of the other while the slant height of the latter is twice that of the former. The ratio of their radii is
(a) 2 : 1
(b) 4 : 1
(c) 8 : 1
(d) 1 : 1
Solution:
Let r be the radius and l be the slant height
∴ Curved surface area of first cone = πr1l1
and let curved surface area of second cone = πr2l2
Question 8.
If the height and radius of a cone of volume V are doubled, then the volume of the cone, is
(a) 3V
(b) 4V
(c) 6V
(d) 8V
Solution:
Let r and h be the radius and height of a cone, then
Question 9.
The ratio of the volume of a right circular cylinder and a right circular cone of the same base and height, is
(a) 1 : 3
(b) 3 : 1
(c) 4 : 3
(d) 3 : 4
Solution:
Let r be the radius and h be the height of a right circular cylinder and a right circular cone, and V1 and V2 are their volumes, the V1 = πr2h and
Question 10.
A right cylinder and a right circular cone have the same radius and same volumes. The ratio of the height of the cylinder to that of the cone is
(a) 3 : 5
(b) 2 : 5
(c) 3 : 1
(d) 1 : 3
Solution:
Let r be the radius of cylinder and cone and volumes are equal
and h1, and h2 be their have h2 is respectively
Question 11.
The diameters of two cones are equal. If their slant heights are in the ratio 5 : 4, the ratio of their curved surface areas, is
(a) 4 : 5
(b) 25 : 16
(c) 16 : 25
(d) 5 : 4
Solution:
∵ Diameters of two cones are equal
∴ Their radii are also be equal
Let r be their radius of each cone,
and ratio in their slant heights = 5:4
Let slant height of first cone (h1) = 5x
Then height of second cone (h2) = 4x
Question 12.
If the heights of two cones are in the ratio of 1 : 4 and the radii of their bases are in the ratio 4 : 1, then the ratio of their volumes is
(a) 1 : 2
(b) 2 : 3
(c) 3 : 4
(d) 4 : 1
Solution:
Ratio in the heights of two cones =1 : 4
and ratio in their radii of their bases = 4 : 1
Let height of the first cone = x
and height of the second cone = 4x
Radius of the first cone = 4y
and radius of the second cone = y
Question 13.
The slant height of a cone is increased by 10%. If the radius remains the same, the curved surface area is increased by
(a) 10%
(b) 12.1%
(c) 20%
(d) 21%
Solution:
Let r be radius and l be the slant height of a cone, then curved surface area = πrl
If slant height is increased by 10%, then
Question 14.
The height of a solid cone is 12 cm and the area of the circular base is 6471 cm2. A plane parallel to the base of the cone cuts through the cone 9 cm above the vertex of the cone, the area of the base of the new cone so formed is
(a) 9π cm2
(b) 16π cm2
(c) 25π cm2
(d) 36π cm2
Solution:
Height of a solid cone (h) = 12 cm
Area of circular base = 64π cm2
Question 15.
If the base radius and the height of a right circular cone are increased by 20%, then the percentage increase in volume is approximately
(a) 60
(b) 68
(c) 73
(d) 78
Solution:
In first case,
Let r be radius and h be height, in volume
Question 16.
If h, S and V denote respectively the height, curved surface area and volume of a right circular cone, then 3πVh3 – S2h2 + 9V2 is equal to
(a) 8
(b) 0
(c) 4π
(d) 32π2
Solution:
h = height, S = curved surface area
V = volume of a cone
Let r be the radius of the cone, then
Question 17.
If a cone is cut into two parts by a horizontal plane passing through the mid¬point of its axis, the ratio of the volumes of upper and lower part is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 7
(d) 1 : 8
Solution:
∴ ∆PDC ~ ∆PBA (AA axiom)
and O’ is mid point of PO
Hope given RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
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Main content
## SAT
### Course: SAT>Unit 6
Lesson 6: Additional Topics in Math: lessons by skill
# Right triangle word problems | Lesson
## What are right triangle word problems, and how frequently do they appear on the test?
Right triangle word problems on the SAT ask us to apply the properties of right triangles to calculate side lengths and angle measures.
In this lesson, we'll learn to:
1. Use the Pythagorean theorem and recognize Pythagorean triples
2. Use trigonometric ratios to calculate side lengths
3. Recognize special right triangles and use them to find side lengths and angle measures
On your official SAT, you'll likely see 1 question that is a right triangle word problem.
This lesson builds upon the Right triangle trigonometry skill.
You can learn anything. Let's do this!
## How do I calculate side lengths using the Pythagorean theorem?
### Intro to the Pythagorean theorem
Khan Academy video wrapper
Intro to the Pythagorean theoremSee video transcript
### The Pythagorean theorem
In a right triangle, the square of the hypotenuse length is equal to the sum of the squares of the leg lengths.
At the beginning of each SAT math section, you'll find this diagram provided as reference:
c, squared, equals, a, squared, plus, b, squared
#### Calculating missing side lengths in right triangles
With the Pythagorean theorem, we can calculate any side length in a right triangle when given the other two.
Let's look at some examples!
What is the length of start overline, B, C, end overline in the figure above?
What is the length of start overline, E, F, end overline in the figure above?
#### Recognizing Pythagorean triples
Pythagorean triples are integers a, b, and c that satisfy the Pythagorean theorem. For example, the side lengths of the right triangle shown below form a Pythagorean triple:
Each side of the triangle has an integer length, and 5, squared, equals, 3, squared, plus, 4, squared. 3-4-5 is the most commonly used Pythagorean triple on the SAT. All triangles similar to it also have side lengths that are multiples of the 3-4-5 Pythagorean triple, like 6-8-10, 9-12-15 or 30-40-50.
Being able to recognize Pythagorean triples can save you valuable time on test day. For example, if you see a right triangle with a hypotenuse length of 15 and a leg length of 12, recognizing it's a 9-12-15 triangle will give you the missing side length, 9, without having to calculate it using the Pythagorean theorem.
Less frequently used Pythagorean triples include 5-12-13 and 7-24-25.
### Try it!
try: use pythagorean triples and similarity to find side lengths
In the figure above, start overline, B, E, end overline is parallel to start overline, C, D, end overline.
What is the length of start overline, B, E, end overline ?
If C, D, equals, 6, what is the length of start overline, A, C, end overline ?
## How do I use trigonometric ratios and the properties of special right triangles to solve for unknown values?
### Recognizing side length ratios
#### Using trigonometric ratios to find side lengths
Sine, cosine, and tangent represent ratios of right triangle side lengths. This means if we have the value of the sine, cosine, or tangent of an angle and one side length, we can find the other side lengths.
Let's look at an example!
In the figure above, tangent, left parenthesis, C, right parenthesis, equals, start fraction, 4, divided by, 7, end fraction. What is the length of start overline, A, C, end overline ?
#### Using special right triangles to determine side lengths and angle measures
Special right triangles are right triangles with specific angle measure and side length relationships. At the beginning of each SAT math section, the following two special right triangles are provided as reference:
This means when we see a special right triangle with unknown side lengths, we know how the side lengths are related to each other. For example, if we have a 30, degrees-60, degrees-90, degrees triangle and the length of the shorter leg is 3, we know that the length of the hypotenuse is 2, left parenthesis, 3, right parenthesis, equals, 6 and the length of the longer leg is 3, square root of, 3, end square root.
We can also identify the angle measures of special right triangles when we spot specific side length relationships. For example, if we're given a right triangle with identical leg lengths, we know it's a 45, degrees-45, degrees-90, degrees special right triangle.
### Try it!
try: recognize trigonometric ratios and special right triangles
Right triangle A, B, C is shown in the figure above. The value of sine, left parenthesis, A, right parenthesis is start fraction, 1, divided by, 2, end fraction.
What is the length of start overline, A, C, end overline ?
What is the measure of angle C ?
degrees
## Your turn!
Practice: find segment length
In the figure above, start overline, B, D, end overline is parallel to start overline, A, E, end overline. What is the length of start overline, D, E, end overline ?
Practice: find side length
In the figure above, sine, left parenthesis, C, right parenthesis, equals, start fraction, 3, divided by, 5, end fraction. What is the length of start overline, B, C, end overline ?
Practice: find angle measure
In quadrilateral A, B, C, D above, start overline, A, D, end overline is parallel to start overline, B, C, end overline and C, D, equals, square root of, 2, end square root, A, B. What is the measure of angle C ?
Choose 1 answer:
## Want to join the conversation?
• In the last example, What do you mean by 'SAT provides us'? Does the question provide it is a 45-45-90 triangle. If it does not provide, it could also be 30-60-90 triangle. How do we know?
(3 votes)
• The explanation is talking about the formula sheet that you get to reference during the math sections. It contains things like area and volume formulas, as well as the side-length ratios for 45-45-90 and 30-60-90 right triangles. So, if you didn't feel like memorizing the ratios, or forgot in the heat of the moment, you can always check the formula sheet at the front of the section.
From the information that the question gives you, you can conclude that the triangle is 45-45-90. The hypotenuse of the right triangle is sqrt(2) * AB, which is the same length as one of the legs of the triangle. Combine this with the information that the triangle is a right triangle, and you have enough information to be able to state that the triangle is 45-45-90, and thus know all of its angle measures.
(10 votes)
• For the SAT, would the right triangle word problems be in the calculator or non-calculator section?
(4 votes)
• I need help on the topics of
Recognizing Pythagorean Triples
Recognizing side length ratios
Using trigonometric ratios to find side lengths
Using special right triangles to determine side lengths and angle measures
(1 vote)
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# 3.4 Solving Multi-Step Inequalities - PowerPoint PPT Presentation
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3.4 Solving Multi-Step Inequalities. SWBAT solve inequalities with variables on one side. SWBAT solve inequalities with variables on both sides. Do Now. Solve each equation. 3(c + 4) = 6 3t + 6 = 3(t – 2) 5p + 9 = 2p – 1 7n + 4 – 5n = 2(n + 2).
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
3.4 Solving Multi-Step Inequalities
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## 3.4 Solving Multi-Step Inequalities
SWBAT solve inequalities with variables on one side.
SWBAT solve inequalities with variables on both sides.
### Do Now
Solve each equation.
• 3(c + 4) = 6
• 3t + 6 = 3(t – 2)
• 5p + 9 = 2p – 1
• 7n + 4 – 5n = 2(n + 2)
### Solving Inequalities with Variables on One Side
• Sometimes you need more than one step to solve an equation, and the same is true with inequalities.
• Just as with equations, you undo addition and subtraction first and then multiplication and division.
### Example 1
• Solve 7 + 6a > 19 and graph.
• Check the solution.
### Example 2
• Solve 2(t + 2) – 3t > -1 and graph.
### More Examples
• -3x – 4 < 14
• 5 < 7 – 2t
• 4p + 2(p + 7) < 8
• 15 < 5 – 2(4m + 7)
• The school band needs a banner to carry in a parade. The banner committee decides that the length of the banner should be 18 feet. What are the possible widths of the banner if they can use no more than 48 feet of trim?
• To make a second banner, the committee decided to make the length 12 feet. They have 40 feet of a second type of trim. Write and solve an inequality to find the possible widths of the second banner.
### Solving Inequalities with Variables on Both Sides
• Step 1: Find like terms.
• Step 2: Combine like terms on each side of the equal sign (if necessary).
• Step 3: Move variable terms to one side of the equal sign by addition or subtraction.
• Step 4: Isolate the variable by multiplication or division (if necessary).
### Example 1
• Solve 6z – 15 < 4z + 11
### Example 2
• Solve -3(4 – m) > 4(2m + 1)
### Example 3
• Solve -6(x – 4) > 7(2x – 3)
### Independent Practice
• p. 155 # 13 – 27 odd
### Homework
• p. 155 #29 – 29 odd, 52
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# Divisibility Rules
In addition to the revision notes for Divisibility Rules on this page, you can also access the following Arithmetic learning resources for Divisibility Rules
Arithmetic Learning Material
Tutorial IDTitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
1.6Divisibility Rules
In these revision notes for Divisibility Rules, we cover the following key points:
• What does divisibility mean?
• How do we write in symbols when a number is divisible by another number?
• What are the divisibility rules for the numbers from 1 to 20?
• What are relatively prime numbers?
• How do relatively prime numbers determine some of divisibility rules?
## Divisibility Rules Revision Notes
An integer x is divisible by another integer y if the result of x ÷ y is another integer, i.e. it is a number without remainder (r = 0). We write the symbol (⁝) to represent the divisibility of two numbers.
The divisibility by 1 rule states that all numbers are divisible by 1.
The divisibility by 2 rule states that the number must be even in order to be divisible by 2.
The divisibility by 3 rule states that the sum of digits of the original number must be divisible by 3.
The divisibility by 4 rule states that the last two digits of the original number must form a number divisible by 4.
The divisibility by 5 rule states that a number must end with 0 or 5.
The divisibility by 6 rule states that a number must be divisible by 2 and 3 at the same time in order to be divisible by 6.
The divisibility by 7 rule states that a number is divisible by 7 if the difference between twice the value of the digit in the ones place and the number formed by the rest of the digits is either 0 or a multiple of 7.
The divisibility by 8 rule states that the last three digits of the original number must form a number that is divisible by 8.
The divisibility by 9 rule states that the sum of digits of the original number must be divisible by 9.
The divisibility by 10 rule states that the number must end with zero to be divisible by 10.
The divisibility by 11 rule states that a number is divisible by 11 if the difference between the sum of the digits in the odd place value and even place value is a multiple of 11 (including 0).
The divisibility by 12 rule states that a number must be divisible by both 3 and 4 to be divisible by 12.
The divisibility by 13 rule states that a number is divisible by 13 if after grouping the digits in groups of three starting from the rightmost place value and applying the subtraction and addition of the numbers obtained by these groups alternatively from right to left, we obtain a number divisible by 13, including 0.
The divisibility by 14 rule states that a number is divisible by 14 if it is divisible by both 2 and 7.
The divisibility by 15 rule states that a number is divisible by 15 if it is divisible by both 3 and 5.
The divisibility by 16 rule states that a number is divisible by 16 if the last three digits form a number that is divisible by 16 while the fourth last digit is even.
The divisibility by 17 rule states that a number is divisible by 17 if after multiplying the last digit by 5 and subtract it from the rest, the result is divisible by 17.
The divisibility by 18 rule states that a number is divisible by 18 if it is divisible by both 2 and 9.
The divisibility by 19 rule states that a number is divisible by 19 if twice the last digit plus the rest of number give a number divisible by 19.
The divisibility by 20 rule states that a number is divisible by 20 if the last two digits of the original number are a multiple of 20.
Two numbers are relatively prime when they don't have any other common factor besides 1. In general, if a number is divisible by each of two relatively prime numbers, it is also divisible by their product.
## Whats next?
Enjoy the "Divisibility Rules" revision notes? People who liked the "Divisibility Rules" revision notes found the following resources useful:
1. Revision Notes Feedback. Helps other - Leave a rating for this revision notes (see below)
2. Arithmetic Math tutorial: Divisibility Rules. Read the Divisibility Rules math tutorial and build your math knowledge of Arithmetic
3. Arithmetic Video tutorial: Divisibility Rules. Watch or listen to the Divisibility Rules video tutorial, a useful way to help you revise when travelling to and from school/college
4. Arithmetic Practice Questions: Divisibility Rules. Test and improve your knowledge of Divisibility Rules with example questins and answers
5. Check your calculations for Arithmetic questions with our excellent Arithmetic calculators which contain full equations and calculations clearly displayed line by line. See the Arithmetic Calculators by iCalculator™ below.
6. Continuing learning arithmetic - read our next math tutorial: Decimal Number System and Operations with Decimals
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# Geometry » Calculating area of compound figures
Calculating the area of a compound figure works as follows:
1 Divide the figure into known figures like rectangle and (semi-)circles. 2 Calculate the area of the separate figures. 3 Calculate the area of the whole figure by adding the area of the separate figures.
Example
Calculate the area of the following figure in m2.
Divide the figure into separate figures. In this case three rectangles.
To calculate all the areas, we need the length of CT (or DS).
CT = DS = 135 – 65 = 70 cm
Now we can calculate the areas.
Area rectangle left = 120 × 135 = 16200 cm 2
Area rectangle middle = 110 × 70 = 7700 cm2
Area rectangle right = 85 × 115 = 9775 cm2
Area whole figure = 16200 + 7700 + 9775 = 33675 cm2 = 3,3675 m2
Other method
The figure can also be divided into other rectangles.
Below, you can see one of these ways.
However, this method gives us more unknown sides that we need. We now need HT (or GS), GH (or ST) and DE (or FS) before we can calculate the areas.
HT = GS = 135 – 65 = 70 cm
GH = ST = 120 + 110 + 85 = 315 cm
DE = FS = 65 - (135 - 115) = 45 cm
Now we can calculate the areas.
Area rectangle ABCT = 120 × 65 = 7800 cm 2
Area rectangle TSGH = 315 × 70 = 22050 cm2
Area rectangle EFSD = 85 × 45 = 3825 cm2
Area whole figure = 7800 + 22050 + 3825 = 33675 cm2 = 3,3675 m2
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