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# Geometric series/Ratio test/R/Section
The series ${\displaystyle {}\sum _{k=0}^{\infty }x^{k}}$ is called geometric series for ${\displaystyle {}x\in \mathbb {R} }$, so this is the sum
${\displaystyle 1+x+x^{2}+x^{3}+\ldots .}$
The convergence depends heavily on the modulus of ${\displaystyle {}x}$.
## Theorem
For all real numbers ${\displaystyle {}x}$ with ${\displaystyle {}\vert {x}\vert <1}$, the geometric series ${\displaystyle {}\sum _{k=0}^{\infty }x^{k}}$ converges absolutely, and the sum equals
${\displaystyle {}\sum _{k=0}^{\infty }x^{k}={\frac {1}{1-x}}\,.}$
### Proof
For every ${\displaystyle {}x}$ and every ${\displaystyle {}n\in \mathbb {N} }$ we have the relation
${\displaystyle {}(x-1){\left(\sum _{k=0}^{n}x^{k}\right)}=x^{n+1}-1\,}$
and hence for the partial sums the relation (for ${\displaystyle {}x\neq 1}$ )
${\displaystyle {}s_{n}=\sum _{k=0}^{n}x^{k}={\frac {x^{n+1}-1}{x-1}}\,}$
holds. For ${\displaystyle {}n\rightarrow \infty }$ and ${\displaystyle {}\vert {x}\vert <1}$ this converges to ${\displaystyle {}{\frac {-1}{x-1}}={\frac {1}{1-x}}}$ because of fact and exercise.
${\displaystyle \Box }$
The following statement is called ratio test.
## Theorem
Let
${\displaystyle \sum _{k=0}^{\infty }a_{k}}$
be a series of real numbers. Suppose there exists a real number ${\displaystyle {}q}$ with ${\displaystyle {}0\leq q<1}$, and a ${\displaystyle {}k_{0}}$ with
${\displaystyle {}\vert {\frac {a_{k+1}}{a_{k}}}\vert \leq q\,}$
for all ${\displaystyle {}k\geq k_{0}}$ (in particular ${\displaystyle {}a_{k}\neq 0}$ for ${\displaystyle {}k\geq k_{0}}$). Then the series ${\displaystyle {}\sum _{k=0}^{\infty }a_{k}}$ converges absolutely.
### Proof
The convergence does not change (though the sum) when we change finitely many members of the series. Therefore, we can assume ${\displaystyle {}k_{0}=0}$. Moreover, we can assume that all ${\displaystyle {}a_{k}}$ are positive real numbers. Then
${\displaystyle {}a_{k}={\frac {a_{k}}{a_{k-1}}}\cdot {\frac {a_{k-1}}{a_{k-2}}}\cdots {\frac {a_{1}}{a_{0}}}\cdot a_{0}\leq a_{0}\cdot q^{k}\,.}$
Hence, the convergence follows from the comparison test and the convergence of the geometric series.
${\displaystyle \Box }$
## Example
The Koch snowflakes are given by the sequence of plane geometric shapes ${\displaystyle {}K_{n}}$, which are defined recursively in the following way: The starting object ${\displaystyle {}K_{0}}$ is an equilateral triangle. The object ${\displaystyle {}K_{n+1}}$ is obtained from ${\displaystyle {}K_{n}}$ by replacing in each edge of ${\displaystyle {}K_{n}}$ the third in the middle by the corresponding equilateral triangle showing outside.
Let ${\displaystyle {}A_{n}}$ denote the area and ${\displaystyle {}L_{n}}$ the length of the boundary of the ${\displaystyle {}n}$-th Koch snowflake. We want to show that the sequence ${\displaystyle {}A_{n}}$ converges and that the sequence ${\displaystyle {}L_{n}}$ diverges to ${\displaystyle {}\infty }$.
The number of edges of ${\displaystyle {}K_{n}}$ is ${\displaystyle {}3\cdot 4^{n}}$, since in each division step, one edge is replaced by four edges. Their length is ${\displaystyle {}1/3}$ of the length of a previous edge. Let ${\displaystyle {}r}$ denote the base length of the starting equilateral triangle. Then ${\displaystyle {}K_{n}}$ consists of ${\displaystyle {}3\cdot 4^{n}}$ edges of length ${\displaystyle {}r{\left({\frac {1}{3}}\right)}^{n}}$ and the length of all edges of ${\displaystyle {}K_{n}}$ together is
${\displaystyle {}L_{n}=3\cdot 4^{n}r{\left({\frac {1}{3}}\right)}^{n}=3r{\left({\frac {4}{3}}\right)}^{n}\,.}$
Because of ${\displaystyle {}{\frac {4}{3}}>1}$, this diverges to ${\displaystyle {}\infty }$.
When we turn from ${\displaystyle {}K_{n}}$ to ${\displaystyle {}K_{n+1}}$, there will be for every edge a new triangle whose side length is a third of the edge length. The area of an equilateral triangle with side length ${\displaystyle {}s}$ is ${\displaystyle {}{\frac {\sqrt {3}}{4}}s^{2}}$. So in the step from ${\displaystyle {}K_{n}}$ to ${\displaystyle {}K_{n+1}}$ there are ${\displaystyle {}3\cdot 4^{n}}$ triangles added with area ${\displaystyle {}{\frac {\sqrt {3}}{4}}{\left({\frac {1}{3}}\right)}^{2(n+1)}r^{2}={\frac {\sqrt {3}}{4}}r^{2}{\left({\frac {1}{9}}\right)}^{n+1}}$. The total area of ${\displaystyle {}K_{n}}$ is therefore
{\displaystyle {}{\begin{aligned}&\,{\frac {\sqrt {3}}{4}}r^{2}{\left(1+3{\frac {1}{9}}+12{\left({\frac {1}{9}}\right)}^{2}+48{\left({\frac {1}{9}}\right)}^{3}+\cdots +3\cdot 4^{n-1}{\left({\frac {1}{9}}\right)}^{n}\right)}\\&={\frac {\sqrt {3}}{4}}r^{2}{\left(1+{\frac {3}{4}}{\left({\frac {4}{9}}\right)}^{1}+{\frac {3}{4}}{\left({\frac {4}{9}}\right)}^{2}+{\frac {3}{4}}{\left({\frac {4}{9}}\right)}^{3}+\cdots +{\frac {3}{4}}{\left({\frac {4}{9}}\right)}^{n}\right)}.\end{aligned}}}
If we forget the ${\displaystyle {}1}$ and the factor ${\displaystyle {}{\frac {3}{4}}}$, which does not change the convergence property, we get in the bracket a partial sum of the geometric series for ${\displaystyle {}{\frac {4}{9}}}$, and this converges.
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Home > Technology > Computer Science > Computer Architecture > Binary-Decimal conversion with floating point
# Binary-Decimal conversion with floating point
Chapter I – Numeral systems
Binary-Decimal conversion with floating point
We have learned in the previous post how to convert a number between numeral systems but for integer values. However, if we want to convert a floating point number between binary and decimal, we need more things to take into account.
In this post I’m going to explain these cases just for binary and decimal. Let’s start with another basic example.
Let’s consider the following number:
0.145
So we have one tenth (weight of 0.1), 4 hundredths (weight of 0.01) e 5 thousandths (weight of 0.001).
Representing this in powers of 10, we will have:
## Converting from binary to decimal with floating point
Let’s now consider the following binary number:
0.01
If we used positive powers of 2 for the integer part, for the decimal places we will use negative powers of 2. In this case, we will have:
This is: 0.01 (2) = 0.25 (10)
Let’s now take a look on this more complete binary number and let’s use again our frindly table:
101.11
22 = 421 = 220 = 12-1 = 0,502-2 = 0,25
101,11
We will then have:
This is: 101.11 (2) = 5.75 (10)
But if we now consider the fact that we have a limit given by the precision we can have with a determined number of bits, things could became a little harder. Let’s consider a limit given by 7 bits of precision and the following binary number (notice that we will write all the 7 bits even we have zeros on the right):
0.1100100
Performing the calculations:
Is this result final? The answer relies on an issue that exists even in the digital to analogue converters. It’s the resolution or precision! A binary number is limited to a determined number of possible combinations and this issue also applies to floating point numbers.
When we worked with integer numbers, we learned that with 7 bits we can have 27 = 128 combinations. The same thing applies to the decimal places! We can only have 128 combinations after the floating point and they are not that linear as for the integer part, as we can check:
0.10 = 0.5
0.01 = 0.25
0.11 = 0.75
This means we have some gaps but this does not mean we can’t represent 0.1 or 0.7 since these numbers might appear with more precision bits as we will see next. Now for the previous number, the problem resides on knowing if we can represent the entire number of 0.78125 with a precision of 7 bits.
In order to ensure the decimal result is in accordance with the resolution, or precision, we have with 7 bits, we have the following general condition:
10p ≤ 2n, having on “p” the number of decimal places (precision) and “n” with the number of bits. In this case, we know we have 7 bits and the condition is given by: 10p ≤ 27.
To get “p” we have to make a more complex calculation since we will need to perform a calculation with the following common logarithm:
Where ⌊x⌋ represents the “floor” function, that returns the highest integer number less or equal to x.
For our case, as we have 7 bits:
This is, we just have two precision digits!
So, we can conclude that, for a 7 bit resolution, our result will be given by:
0.1100100 (2) = 0.78 (10)
To get a better understanding on this subject, let’s see if we had precisions of 10 and 16 bits…
With 10 bits, our binary number would be written like this:
0.110010000
In this case we would have:
And the solution given by:
0.1100100000 (2) = 0.781 (10)
Finally, with 16 bits, our binary number would be written like this:
0.110010000000000
In this case we would have:
Rounding the number, the solution is:
0.1100100000000000 (2) = 0.7813 (10)
## Converting from decimal to binary with floating point
Now, let’s see how we can convert a floating point number from decimal to binary.
To convert an integer number we used successive divisions by 2. Now, for decimal places, we will successively multiply our number by 2, with a specific rule, until we reach an exact integer result of 1.
Let’s get one of the previous results in decimal:
0.25
And let’s use this table for now.
0,25 x 2 = 0 ,50 0,50 x 2 = 1 ,00
We multiply 0.25 by 2 and we get 0.50. The final result is not 1 yet, so we multiply 0.50 by 2 and we finally get 1 (1.00).
Now, from top to bottom, we join all integer results and we get:
0.25 (10) = 0.01 (2)
This was a simple example, let’s now consider the following number:
0.75
It seems to be easy too… Let’s now carefully make the calculations step-by-step:
0.75 x 2 = 1.50
Well, we got a result greater than 1 and not 1.00! If we multiply it by 2 we will get 3 and 3 does not exist in binary! So, what could we do?
It’s easy! Let’s get the decimal places and continue with the integer part with zero!
0.50 x 2= 1.00
Finally we reached 1.00! Let’s join again all the integers from top to bottom and we get the following result:
0.75 (10) = 0.11 (2)
And this is the rule for this calculations.
Let’s see another already known decimal number:
0.78125
Let’s perform the same operations:
0.78125 x 2 = 1.5625
0.5625 x 2 = 1.125
0.125 x 2 = 0.25
0.25 x 2 = 0.5
0.5 x 2 = 1
And so we get again our binary number 0.11001.
In this case, we did not mention the precision, so that’s how we will represent it. But if we assume we have 8 precision bits for the decimal places, we will have:
0.78125 (10) = 0.11001000 (2)
And if we have less precision bits? For example, 4? Then we will have to truncate the result to 0.1100.
The resolution or precision issue can be verified in more complex numbers. The following decimal number might seem to be easy to convert, but let’s see:
0.32
Applying the rules:
0.32 x 2 = 0.64
0.64 x 2 = 1.28
0.28 x 2 = 0.56
0.56 x 2 = 1.12
0.12 x 2 = 0.24
0.24 x 2 = 0.48
0.48 x 2 = 0.96
0.96 x 2 = 1.92
When do we stop? That depends on the precision we want for the decimal places!
Let’s now see a number with both integer and decimal places! For instance:
58.125
In this case, we have to handle each one of the parts!
Let’s start with the integer 58, applying the successive divisions by 2:
We get that 58 is 111010 in bae-2.
And now the decimal places:
0.125 x 2 = 0.25
0.25 x 2 = 0.50
0.50 x 2 = 1.00
We get that 0.125 is 0,001 in binary.
Now let’s just gather everything to finally get:
58.125 (10) = 111010.001 (2)
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Warm-Up - Artists Vito Acconci is a popular American sculptor and artist. One of his furniture creations called the Name Calling Chair uses a variety of geometric shapes, such as squares and triangles. The back of the chair is shown in the diagram. • What do you know about angles FCB and CFE? Are triangles ABC and DEF similar? • Are triangles ABC and DEF congruent?
# Congruent Triangles – Day 2
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Congruent Triangles – Day 2. Congruence & Triangles. Warm-Up - Artists. - PowerPoint PPT Presentation
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Warm-Up - Artists• Vito Acconci is a
popular American sculptor and artist. One of his furniture creations called the Name Calling Chair uses a variety of geometric shapes, such as squares and triangles. The back of the chair is shown in the diagram.
• What do you know about angles FCB and CFE?• Are triangles ABC and DEF similar?• Are triangles ABC and DEF congruent?
Congruent Triangles – Day 2Congruence & Triangles
Congruent Figures
•Exact same shape•Exact same size
Congruent Not Congruent
Properties of Congruent Figures
•Corresponding Angles are Congruent•Corresponding Sides are congruent
PQRABC
A
B
C
P
QR
Example 1: Write a congruence statement and identify all corresponding parts.
X
Y
Z
H
K
J
Example 2: ABCD KJHL
a.Find the value of x
b. Find the value of y
A
B
C
D
K
J
H
L
91 8
6
9 cm
6 cm(4x – 3) cm
(2y – 4) o
Third Angles Theorem
•If 2 angles in a triangle are congruent to 2 angles in a 2nd triangle, then the 3rd angles are also congruent.
A
B
C
D
E
F
Example 3: Find the value of x
A
B
C
D
E
F
22o
87o
(4x+15)o
Find the value of the variables.
Find the value of the variables.
Decide whether the triangles are congruent. Justify reasoning.
E
F
G
H
J
58o
58o
Are there any congruent figures?If so, what is the congruence statement?
Are there any congruent figures?If so, what is the congruence statement?
Are there any congruent figures?If so, what is the congruence statement?
Homework – Day 2
4.1/4.2 Practice
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# Relations and Functions – Class XI – Exercise 2.2
1. Let A = {1, 2, 3… 14}. Define a relation R from A to A by R = {(x, y): 3x – y = 0, where x, y ∈ A}. Write down its domain, co-domain and range.
Solution:
The relation R from A to A is given as R = {(x, y): 3x – y = 0, where x, y ∈ A}
i.e., R = {(x, y): 3x = y, where x, y ∈ A}
∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}
The domain of R is the set of all first elements of the ordered pairs in the relation.
∴ Domain of R = {1, 2, 3, 4}
The whole set A is the codomain of the relation R.
∴ Codomain of R = A = {1, 2, 3… 14}
The range of R is the set of all second elements of the ordered pairs in the relation.
∴Range of R = {3, 6, 9, 12}
2: Define a relation R on the set N of natural numbers by R = {(x, y): y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.
Solution:
R = {(x, y): y = x + 5, x is a natural number less than 4, x, y ∈ N}
The natural numbers less than 4 are 1, 2, and 3.
∴ R = {(1, 6), (2, 7), (3, 8)}
The domain of R is the set of all first elements of the ordered pairs in the relation.
∴ Domain of R = {1, 2, 3}
The range of R is the set of all second elements of the ordered pairs in the relation.
∴ Range of R = {6, 7, 8}
1. A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.
Solution:
A = {1, 2, 3, 5} and B = {4, 6, 9}
R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}
∴ R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}
4: The given figure shows a relationship between the sets P and Q. write this relation
(i) in set-builder form
(ii) in roster form. What is its domain and range?
Solution:
According to the given figure, P = {5, 6, 7}, Q = {3, 4, 5}
(i)R = {(x, y): y = x – 2; x ∈ P}
or
R = {(x, y): y = x – 2 for x = 5, 6, 7}
(ii) R = {(5, 3), (6, 4), (7, 5)}
Domain of R = {5, 6, 7}
Range of R = {3, 4, 5}
1. Let A = {1, 2, 3, 4, 6}.
Let R be the relation on A defined by {(a, b): a, b ∈ A, b is exactly divisible by a}.
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R.
Solution:
A = {1, 2, 3, 4, 6}, R = {(a, b): a, b ∈ A, b is exactly divisible by a}
(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}
(ii) Domain of R = {1, 2, 3, 4, 6}
(iii) Range of R = {1, 2, 3, 4, 6}
6: Determine the domain and range of the relation R defined by R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.
Solution:
R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}
∴ R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
∴ Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {5, 6, 7, 8, 9, 10}
1. Write the relation R = {(x, x3): x is a prime number less than 10} in roster form.
Solution:
R = {(x, x3): x is a prime number less than 10}
The prime numbers less than 10 are 2, 3, 5, and 7.
∴ R = {(2, 8), (3, 27), (5, 125), (7, 343)}
1. Let A = {x, y, z} and B = {1, 2}.
Find the number of relations from A to B.
Solution:
It is given that A = {x, y, z} and B = {1, 2}.
∴ A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}
Since n(A × B) = 6, the number of subsets of A × B is 26.
Therefore, the number of relations from A to B is 26.
9: Let R be the relation on Z defined by R = {(a, b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R.
Solution:
R = {(a, b): a, b ∈ Z, a – b is an integer} It is known that the difference between any two integers is always an integer.
∴ Domain of R = Z
Range of R = Z
|
# Find the intervals in which the following functions are increasing or decreasing.
Question:
Find the intervals in which the following functions are increasing or decreasing.
$f(x)=\log (2+x)-\frac{2 x}{2+x}$
Solution:
Given:- Function $f(x)=\log (2+x)-\frac{2 x}{2+x}$
Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$.
(i) If $f^{\prime}(x)>0$ for $a l l x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$
(ii) If $f^{\prime}(x)<0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$
Algorithm:-
(i) Obtain the function and put it equal to $f(x)$
(ii) Find $f^{\prime}(x)$
(iii) Put $f^{\prime}(x)>0$ and solve this inequation.
For the value of $x$ obtained in (ii) $f(x)$ is increasing and for remaining points in its domain it is decreasing.
Here we have,
$f(x)=\log (2+x)-\frac{2 x}{2+x}$
$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\log (2+\mathrm{x})-\frac{2 \mathrm{x}}{2+\mathrm{x}}\right)$
$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{1}{2+\mathrm{x}}-\frac{(2+\mathrm{x}) 2-2 \mathrm{x} \times 1}{(2+\mathrm{x})^{2}}$
$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{1}{2+\mathrm{x}}-\frac{4+2 \mathrm{x}-2 \mathrm{x}}{(2+\mathrm{x})^{2}}$
$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{1}{2+\mathrm{x}}-\frac{4}{(2+\mathrm{x})^{2}}$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{2+\mathrm{x}-4}{(2+\mathrm{x})^{2}}$
$\Rightarrow \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-2}{(2+\mathrm{x})^{2}}$
For $f(x)$ to be increasing, we must have
$\Rightarrow f^{\prime}(x)>0$
$\Rightarrow \frac{x-2}{(2+x)^{2}}>0$
$\Rightarrow(x-2)>0$
$\Rightarrow 2$\Rightarrow x \in(2, \infty)$Thus$f(x)$is increasing on interval$(2, \infty)$Again, For$f(x)$to be decreasing, we must have$f^{\prime}(x)<0\Rightarrow \frac{x-2}{(2+x)^{2}}<0\Rightarrow(x-2)<0\Rightarrow-\infty
$\Rightarrow x \in(-\infty, 2)$
Thus $f(x)$ is decreasing on interval $(-\infty, 2)$
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# NCERT Notes for Class 10 Maths Chapter 15 Probability
#### Class 10 Maths Chapter 15 Probability Notes
Chapter Name Probability Notes Class CBSE Class 10 Textbook Name Mathematics Class 10 Related Readings Notes for Class 10Notes for Class 10 MathsRevision Notes for Probability
### Probability
Probability is the study of mathematics which calculates the degree of uncertainty. There are two types of approaches to study probability:
#### 1. Experimental or Empirical Probability
The result of probability based on the actual experiment is called experimental probability. In this case, the results could be different if we do the same experiment again.
#### 2. Probability: A Theoretical Approach
In the theoretical approach, we predict the results without performing the experiment actually. The other name of theoretical probability is classical probability.
where, the outcomes are equally likely.
#### Equally Likely Outcomes
If we have the same possibility of getting each outcome then it is called equally likely outcomes.
Example: A dice have the same possibility of getting 1, 2, 3, 4, 5 and 6.
Not Equally Likely
If we don't have the same possibility of getting each outcome then it is said to be the not equally likely outcome.
Example: 3 green balls and 2 pink balls are not equally likely as the possibility of the green ball is 3 and the possibility of the pink ball is 2.
### Elementary Event
If an event has only one possible outcome then it is called an elementary event.
Remark: The sum of the probabilities of all the elementary events of an experiment is 1.
#### The General form
P (Heads) + P (Tails) = 1
P(H) + P = 1 where, is ‘not H’.
P(H) – 1 = P
P(H) and P are the complementary events.
#### Impossible Events
If there is no possibility of an event to occur then its probability is zero. This is known as an impossible event.
Example: It is not possible to draw a green ball from a group of blue balls.
#### Sure or Certain Event
If the possibility of an event to occur is sure then it is said to be the sure probability. Here the probability is one.
This shows that the probability of an event could be
0 ≤ P (E) ≤ 1
#### Some Solved Examples
Example: What is the probability of drawing a heart from a deck of cards?
Solution
We know that there are total 52 cards in a deck out of which 13 cards are of heart.
So the favourable outcomes are 13 and the total no. of events is 52.
= 13/52 = 1/4
Example: If we toss two coins together, then what is the probability of getting at least one tail?
Solution
If we toss two coins together then the total outcomes could be
The favorable outcomes for at least one head will be
{HH}, {HT}, {TH} = 3
P (for at least one head) = 3/4
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INSTRUCTORS Carleen Eaton Grant Fraser
Start learning today, and be successful in your academic & professional career. Start Today!
• ## Related Books
### Properties of Logarithms
• Use the properties either to convert the log of a complex expression into a combination of logs of simple expressions, or vice versa.
• To solve a logarithmic equation, first use the properties to combine logs on each side of the equation to get an equation of the form log x = log y. Then equate x and y.
### Properties of Logarithms
Simplify
log2( [(x3)/((x + 2)(x − 3)(y − 6)4)] )
• Recall that there are three properties of logs
• 1)log(m*n) = log(m) + log(n)
• 2)log(mn) = nlog(m)
• 3)log( [m/n] ) = log(m) − log(n)
• Begin by applying property (3)
• log2( [(x3)/((x + 2)(x − 3)(y − 6)4)] ) = log2x3 − log2(x + 2)(x − 3)(y − 6)4
• On the first term, apply property (2)
• 3log2x − log2(x + 2)(x − 3)(y − 6)4
• First half is done, now apply property (1) to the second term
• 3log2x − log2(x + 2)(x − 3)(y − 6)4 = 3log2x − (log2(x + 2) + log2(x − 3) + log2(y − 6)4)
• Apply property (2) to the very last term
• = 3log2x − (log2(x + 2) + log2(x − 3) + 4log2(y − 6))
• Distribute the negative
log2( [(x3)/((x + 2)(x − 3)(y − 6)4)] ) = 3log2x − log2(x + 2) − log2(x − 3) − 4log2(y − 6)
Simplify
log2( [(x3z4)/((x + 2)4(x − 3)2(y − 6)4)] )
• Recall that there are three properties of logs
• 1)log(m*n) = log(m) + log(n)
• 2)log(mn) = nlog(m)
• 3)log( [m/n] ) = log(m) − log(n)
• Begin by applying property (3)
• log2( [(x3z4)/((x + 2)4(x − 3)2(y − 6)4)] ) = log2x3z4 − log2(x + 2)4(x − 3)2(y − 6)4
• On the first term, apply property (1) followed by property (2)
• 3log2x + 4log2z − log2(x + 2)4(x − 3)2(y − 6)4
• First half is done, now apply property (1) to the third term
• 3log2x + 4log2z − log2(x + 2)4(x − 3)2(y − 6)4 = 3log2x + 4log2z − (log2(x + 2)4 + log2(x − 3)2 + log2(y − 6)4)
• Apply property (2) to the very last three terms
• = 3log2x + 4log2z − (4log2(x + 2) + 2log2(x − 3) + 4log2(y − 6))
• Distribute the negative
log2( [(x3z4)/((x + 2)4(x − 3)2(y − 6)4)] ) = 3log2x + 4log2z − 4log2(x + 2) − 2log2(x − 3) − 4log2(y − 6)
Write as a single logarithm
2log37 + log311 − [1/3]log32
• Recall that there are three properties of logs
• 1)log(m*n) = log(m) + log(n)
• 2)log(mn) = nlog(m)
• 3)log( [m/n] ) = log(m) − log(n)
• Condence using rule (2)
• 2log37 + log311 − [1/3]log32 = log372 + log311 − log32[1/3]
• Condense uring rule (1)
• log372 + log311 − log32[1/3] = log3(72*11) − log32[1/3]
• Condense using rule (3)
• log3(72*11) − log32[1/3] = log3[((72*11))/(2[1/3])]
• Simplify
log3[((72*11))/(2[1/3])] = log3[(49*11)/(3√{2})] = log3( [539/(3√{2})] )
Write as a single logarithm
2log3(x − 2) + log3(x + 6) + 3log3(x − 4) − [1/3]log3(y + 2)
• Recall that there are three properties of logs
• 1)log(m*n) = log(m) + log(n)
• 2)log(mn) = nlog(m)
• 3)log( [m/n] ) = log(m) − log(n)
• Condence using rule (2)
• 2log3(x − 2) + log3(x + 6) + 3log3(x − 4) − [1/3]log3(y + 2) = log3(x − 2)2 + log3(x + 6) + log3(x − 4)3 − log3(y + 2)[1/3]
• Condense uring rule (1)
• log3(x − 2)2 + log3(x + 6) + log3(x − 4)3 − log3(y + 2)[1/3] = log3(x − 2)2(x + 6)(x − 4)3 − log3(y + 2)[1/3]
• Condense using rule (3)
• log3(x − 2)2(x + 6)(x − 4)3 − log3(y + 2)[1/3] = log3[((x − 2)2(x + 6)(x − 4)3)/((y + 2)[1/3])]
• Simplify
log3[((x − 2)2(x + 6)(x − 4)3)/((y + 2)[1/3])] = log3[((x − 2)2(x + 6)(x − 4)3)/(3√{(y + 2)})]
Solve
log75 + log75x2 = 2
• Recall that there are three properties of logs
• 1)log(m*n) = log(m) + log(n)
• 2)log(mn) = nlog(m)
• 3)log( [m/n] ) = log(m) − log(n)
• Condence the left side using property (1)
• log75 + log75x2 = 2 = > log75*5x2 = 2
• log725x2 = 2
• You now have two options: re - write equation in exponential form or find the logyx that equals 2.
• I'll solving by writing into exponential form
• 72 = 25x2
• 49 = 25x2
• x2 = [49/25]
x = ±√{[49/25]} = ±[7/5]
Solve
log4(x + 5) − log4x = 2
• Recall that there are three properties of logs
• 1)log(m*n) = log(m) + log(n)
• 2)log(mn) = nlog(m)
• 3)log( [m/n] ) = log(m) − log(n)
• Condence the left side using property (3)
• log4(x + 5) − log4x = 2 = > log4( [(x + 5)/x] ) = 2
• log4( [(x + 5)/x] ) = 2
• You now have two options: re - write equation in exponential form or find the log4x that equals 2.
• I'll solve by writing into exponential form
• 42 = [(x + 5)/x]
• 16 = [(x + 5)/x]
• 16x = [(x + 5)/] = 16x = x + 5
• 15x = 5
• x = [5/15] = [1/3]
This answer is a valid answer because by pluging into the original equation does not lead to a negative number.
Solve
log2x + log2(x + 36) = log276
• Recall that there are three properties of logs
• 1)log(m*n) = log(m) + log(n)
• 2)log(mn) = nlog(m)
• 3)log( [m/n] ) = log(m) − log(n)
• Condence the left side using property (1)
• log2x + log2(x + 36) = log276 = > log2x*(x + 36) = log276
• log2x*(x + 36) = log276
• Since both logs on the left and right side of the equation have the same base, we can continue without the logs.
• x(x + 36) = 76
• x2 + 36x = 76
• x2 + 36x − 76 = 0
• Factor x2 + 36x − 76 = 0
• (x + 38)(x − 2) = 0
• Solve using the Zero Product Property
• x = − 38
x = 2
By inspection, you can see that x = − 38 is an erroneous solutions to log2x + log2(x + 36) = log276
since ther will be a negative inside the first term. Since that can't happen, the only solution is x = 2
Solve
log7x − log7(x + 1) = log711
• Recall that there are three properties of logs
• 1)log(m*n) = log(m) + log(n)
• 2)log(mn) = nlog(m)
• 3)log( [m/n] ) = log(m) − log(n)
• Condence the left side using property (3)
• log7x − log7(x + 1) = log711 = > log7[x/((x + 1))] = log711
• log7[x/((x + 1))] = log711
• Since both logs on the left and right side of the equation have the same base, we can continue without the logs.
• [x/(x + 1)] = 11
• [x/] = 11(x + 1)
• x = 11x + 1
• − 1 = 10x
• x = − [1/10]
• By inspection, you can see that x = − [1/10] is an erroneous solutions to log7x − log7(x + 1) = log711
• since ther will be a negative inside the first term.
No Solution
Solve
log92 − log9( − 5x) = log939
• Recall that there are three properties of logs
• 1)log(m*n) = log(m) + log(n)
• 2)log(mn) = nlog(m)
• 3)log( [m/n] ) = log(m) − log(n)
• Condence the left side using property (3)
• log92 − log9( − 5x) = log939 = > log9[2/(( − 5x))] = log939
• log9[2/(( − 5x))] = log939
• Since both logs on the left and right side of the equation have the same base, we can continue without the logs.
• [2/( − 5x)] = 39
• [2/] = 39( − 5x)
• 2 = − 195x
• x = − [2/195]
• By inspection, you can see that x = − [2/195] is not an erroneous solutions to log92 − log9( − 5x) = log939
• since there will not be a negative inside the second erm.
x = − [2/195]
Solve
log46 − log4( − 4x) = log467
• Recall that there are three properties of logs
• 1)log(m*n) = log(m) + log(n)
• 2)log(mn) = nlog(m)
• 3)log( [m/n] ) = log(m) − log(n)
• Condence the left side using property (3)
• log46 − log4( − 4x) = log467 = > log4[6/(( − 4x))] = log467
• log4[6/(( − 4x))] = log467
• Since both logs on the left and right side of the equation have the same base, we can continue without the logs.
• [6/( − 4x)] = 67
• [6/] = 67( − 4x)
• 6 = − 268x
• x = − [6/268] = − [3/134]
• By inspection, you can see that x = − [3/134] is not an erroneous solutions to log46 − log4( − 4x) = log467
• since there will not be a negative inside the second erm.
x = − [3/134]
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
### Properties of Logarithms
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Product Property 0:08
• Example: Product
• Quotient Property 2:40
• Example: Quotient
• Power Property 3:51
• Moved Exponent
• Example: Power
• Equations 5:15
• Example: Use Properties
• Example 1: Simplify Log 11:17
• Example 2: Single Log 15:54
• Example 3: Solve Log Equation 18:48
• Example 4: Solve Log Equation 22:13
### Transcription: Properties of Logarithms
Welcome to Educator.com.0000
We are going to continue our discussion of logarithms by exploring some properties of logarithms.0002
The first one is the product property: if you have a logb(mn), that is equal to the logb(m) + logb(n).0008
And this might look familiar to you, or this general concept, from working with exponents.0022
Recall that, when you are multiplying two exponents with the same base, you add the exponents.0026
Since logs are a type of exponent, it stands to reason that there would be similarities.0032
So, you are going to see some similarities between the properties we are going to cover today and properties from working with exponents.0040
For example, log9(10) = log9...and I could break this out into factors, 5 times 2...0046
equals the sum of the log of this factor, plus this factor...so the sum of log9(5) and log9(2).0059
Or I could move, instead of from left to right, from right to left.0075
And instead of breaking it apart into 2 separate logs, I may have to combine it into a single log.0079
And combining into a single log is especially helpful when we move on to solving more complex logarithmic equations0084
than the ones we have seen in previous lessons.0091
For example, given log6(5) + log6(7), I may want to combine those.0095
And this would be equivalent to log6...since these have the same base;0105
notice that they have to have the same base for this to work...of 5 times 7, which equals log6(35).0110
This, of course, applies when we are working with variables, as well--with logs involving variables, which is a lot of what we are going to be doing.0120
So, log8(x2 - 36)...I could factor this into log8(x - 6) (x + 6),0128
and then write it as the sum of log8(x - 6) + log8(x + 6).0141
And again, I can move from right to left: if I were given this, I could multiply these two and then combine this into a single logarithm.0149
The quotient property: again, you are going to notice similarities from your work with exponents.0160
If I have logb of the quotient m/n, this is equal to the difference between the log of the dividend and the log of the divisor.0165
Given log2(14/5), I could rewrite this as log2(14) - log2(5).0180
Again, this works when we are talking about logarithms with variables (algebraic logarithms).0197
log3 of something such as (x2 - 3)/y would be equal to log3(x2 - 3) - log3y.0203
So, the log of the dividend minus the log of the divisor--that is the quotient property.0221
The third property we are going to cover is the power property.0232
This one is a bit different that what you have probably seen before.0236
But it states that logb of m to the power p equals p times logb(m).0238
So, if you see what happened: we took this exponent up here and moved it to the front to become a coefficient.0245
Again, you are going to find this property helpful when you are trying to simplify logarithmic equations,0262
or when you need to work with equations to solve them, to be able to move back and forth between these two forms.0270
For example, log4(53): I am going to take this 3 and move it to the front.0278
It is going to become the coefficient: 3log4(5).0290
Or, another example using a variable: log3(x6) =...I pull this out in front...6log3(x).0297
Again, these properties can be used to solve logarithmic equations.0316
Earlier on, we talked about solving logarithmic equations where there was just a log on one side,0320
and situations where we had one log on each side of the equation.0326
Sometimes, however, you are given logarithmic equations where you have multiple logs on each side, sums and differences of logs...0331
And if they have the same base, then you can actually combine those, so that you end up with one log on each side.0339
And then, you can move on to use the techniques previously learned in order to solve those.0345
The product, quotient, and power properties are some extra steps that you might have to take before applying previously-learned techniques.0350
For example, given log7(4x + 50) - 2log7(3) = log7(x + 1) + log7(3),0357
I see that these all have the same bases, so I want to use the property where, if I get a log on one side0373
equal to the log of the other with the same base, then I can just say x = y.0388
But I need to get this into that form; so let's see what we can do to use these properties to combine them.0392
First, I see that I have a coefficient here; so I am going to use the power property.0400
And with the power property, we talked about how logb(mp) = plogb(m).0404
So, I was moving that power to the front to yield this form.0419
I can do the opposite, though: I can take this coefficient and move it up here and make it a power again.0426
And I am going to do that: so, this first log I am just going to keep the same for now: log7(4x + 50).0434
Here, I am going to turn this into log7(3), and I am going to put this back over here and raise the 3 to the second power.0444
That is the only power I have; so I am just going to work with that for now.0458
And I know that 32 is just going to be 9, so I can change that to 9 in this step: it is log7(9).0463
And then, I am going to apply the quotient property on the left, because I have the difference of two logs that have the same base.0482
Using the quotient property, I can rewrite this as the log7(4x + 50) divided by 9.0491
Recall that the quotient property (this was the power property up here) told me that, if I have logb(m/n),0502
that is going to equal logb(m) - logb(n).0519
So here, I am moving from right to left: I am taking these and combining them--that is what I am doing right here.0525
On the right, I am going to use the product property: the product property says that, if I have a logb of a product,0533
that is equal to the sum of the logs of those factors (logb(m) + logb(n)).0543
So, I am going to go ahead and combine these two into a single log, log7(x + 1)(3).0551
Now, I have gotten this in the form that I can actually solve, because I have a log on the left to base 7 and a single log in the right to base 7.0563
And if these are equal, then this expression must equal this expression.0571
Now, I just need to go ahead and solve: 4x + 50, divided by 9, equals...I am going to pull that 3 out in front: 3(x + 1).0575
So, I am going to multiply both sides by 9 to get 4x + 50 = 27(x + 1); 4x + 50 = 27x + 27.0587
I am going to go ahead and subtract a 4x from both sides to get 50 = 23x + 27.0603
Subtract a 27 from both sides: I will get 23 = 23x, so x = 1.0611
Now, I always need to check back and make sure that the solution is valid, because I want to make sure I don't end up taking the log of a negative number.0622
Checking for validity right here: I have log7(4x + 50).0628
Let x equal 1, and check that: log7(4(1) + 50); that is log7(4 + 50), or 54.0636
So, that is OK; let's look right here: the other one I have to check is log7(x + 1), which is log7(1 + 1), and that is just log7(2).0647
So, this one is valid; this is a valid solution.0661
We used this technique, but we had to take a bunch of steps prior to that in order to combine these into a single log on each side of the equation.0667
Before we go on to work some logarithmic equations, let's just talk about simplifying using the properties that we have already covered.0680
And when I am asked to simplify a logarithmic expression, that means that I want to get rid of quotients.0686
I don't want to be taking the log of any quotients, of any products, or of any factors raised to powers.0692
The first thing I am going to do is work with getting rid of this fraction bar.0700
And I am going to use the quotient property for that, because log6(x4y5),0706
divided by this polynomial expression down here, is going to be equal to...let's just say0712
that this whole thing is going to be equal to the log base 6 of the dividend (the numerator), minus log base 6 of the divisor.0723
So, logb(quotient) equals the log of the dividend, minus the log of the divisor.0752
All right, so I got rid of that fraction bar; I don't have any more quotients; but what I do have is a product.0768
I am going to apply the product property to just get rid of this right here.0774
Instead, I am going to say, "OK, I know that this equals the sum of log base 6 of this factor, plus log base 6 of this factor."0781
So, I have to remember that I have my negative sign out here; that is going to apply to each of these terms, once I split them apart.0799
That is log6(x - 3)3 + log6(y + 2)6.0806
All right, I still have one product (I don't have any quotients), so I am going to take care of this one next.0819
This is going to be equal to log6(x4) + log6(y5), minus0827
log6(x - 3)3, plus log6(y + 2)6.0841
Next, I am going to apply the power property.0849
And recall that the power property says that, if I have logb(mp),0853
that is going to be equal to p...this is going to turn into a coefficient...times logb(m).0863
So, I am going to take these powers and move them out front; that is going to give me 4log6(x4), plus0873
5 (that is going to go to the front, also) times log6(y), minus (3 is going to go out in front)0883
3log6(x - 3), plus...the 6 comes out in front...6log6(y + 2).0894
The last thing I want to do is move this negative sign and apply it to each term in here.0906
And that is as far as I can go with simplifying: 4log6...actually, this is gone now,0911
because we have pushed that out in front...plus 5log6y, minus 3log6(x - 3).0918
And then, the negative applies to this term, as well: minus 6log6(y + 2).0929
And this is simplified as far as I can simplify it.0938
I look here, and I no longer have the logs of any quotients; I don't have the logs of any products;0942
and I don't have the logs of any terms or expressions raised to powers.0948
Now, instead of expanding out the expression, I am going to do the opposite.0956
I am asked to write this as a single logarithm: so I am going to instead compress this into one logarithm--the opposite of what I did in the last question.0960
I can rewrite this as log4(x - 3), and I am going to turn this into an exponent and write it that way.0975
Plus log4(x + 4): that is going to be raised to the fourth power.0985
Minus log4(x + 7): there is no coefficient there to turn into an exponent.0995
There is here, though: log4(x - 8); and this is going to be raised to the third power.1003
Now, I have the difference here, which means that I can combine this by using the quotient property.1010
I also, within these parentheses, have some addition; so I can use the product property to combine those.1019
And I want to start inside the parentheses; so let's go ahead and do that and use the product property1027
to combine this into log4[(x - 3)3 (x + 4)4].1033
So, it's log base 4 of this times this, minus...here, also, I have a sum; so I can combine that1044
by taking the product of these two factors: log4[(x + 7)(x - 8)3].1056
So, this is this minus this; I can actually leave these brackets off at this point, because I have a single log.1068
Now, we have to apply the quotient property, because we have a difference.1079
This is log4 and log4; and this is going to become the numerator--it is going to become the dividend.1089
(x - 3)3(x + 4)4, divided by (x + 7)(x - 8)3...this is all together.1098
I have written this as a single logarithm, first by applying the power property,1111
then by combining these two logs and these two logs, using the product property,1117
and then finally combining this log and this log, using the quotient property.1122
Now, we are going to apply what we have learned to actually solving logarithmic equations.1129
And we have a technique for solving an equation, as long as we have a single log on each side.1134
I need to combine this side, and I need to also get rid of this 1/2 out in front,1139
so that I can have something of the form logb(x) = logb(y),1145
and then use the property that y must equal x.1151
First, I am going to apply the power property to get rid of these coefficients.1156
log9(23) - log9(2x - 1) = log9(491/2).1161
So, I can do some simplifying: I know that log9(23)...that 2 cubed is 8, so that is log9(8).1176
Now, 491/2 = √49, which equals 7; so I can write this as log9(7)--it is already looking more manageable.1186
On the left, I need to combine these; and I can do that using the quotient property.1199
log9...this is going to go in the numerator, and this will be the denominator.1203
So, this is log9(x) divided by 2x - 1 equals log9(7).1211
Once I am at this point, I have this situation, where I have the same base, and I only have one logarithmic expression on each side.1221
So, I can just say, "OK, x equals y, so 8 divided by 2x - 1 equals 7."1229
I am going to multiply both sides by 2x - 1, which is going to give me 8 = 14x - 7.1236
I am going to add 7 to both sides to get 15 = 14x, and then I am going to divide both sides by 14: so 15/14 = x, or x = 1 and 1/14.1250
It is important that I check the solution; so I want to make sure that I don't end up taking the log of a negative number.1265
And the only log here that has variables as part of it is this one, so I am checking log9(2x - 1), and I am letting x equal 15/14.1271
log9(2) times 15 times 14, minus 1...1284
Now, instead of figuring out this whole thing and subtracting all of that, all I have to do is say,1291
"All right, this is slightly more than 1; so 2 times slightly-more-than-one is going to be slightly more than 2."1297
If I take a value of slightly more than 2 and subtract 1 from it, I am going to be fine; I will have a positive number.1305
I will not be taking the log of a negative number, so this solution is valid.1314
You don't have to get the exact value; you just have to check it far enough to be sure that what you have in here,1320
what you are taking the log of, is not negative; it is greater than 0.1324
OK, Example 4: we are asked to solve a logarithmic equation, and we almost have this form, but not quite.1335
I can't use x = y, because I have this 1/2 here.1346
In order to combine these two, they have to have the same base.1352
In order for us to use the product property to combine these two logs into one, I have to somehow make 1/2 into log16.1355
So, my goal is log16--that I am going to turn 1/2 into that.1366
And what I want is log16 of some number (but I don't know what number) is going to equal 1/2,1389
because then, if I have this, instead of writing 1/2 here, I will just write this.1395
Thinking of my definition of logarithms and how I can use that to solve an equation like this,1401
an equation with a log in one side: now I have a separate equation that I need to solve in order to solve this one,1407
just to substitute there: well, recall that logb(x) = y if by = x.1414
So, I am going to solve this by converting it into its exponential form,1424
which is going to give me 161/2 = x.1431
This is the same as √16, which is 4; therefore, log16(4) = 1/2.1438
These are equivalent; since these are equivalent, I can write that up there--I can substitute.1450
Now, my next step is to combine these two logs on the right, using the product property.1469
log16(12x - 21) = log16[4(x2 - 3)].1476
Now, I have it in this form, and I can say, "OK, 12x - 21 = 4(x2 - 3)."1491
And then, I just solve, as usual: this is going to give me 12x - 21 = 4x2 - 12.1500
I have a quadratic equation; I need to first simplify; so let's subtract 12x from both sides.1512
I am going to set the whole equation equal to 0; that is -21 = 4x2 - 12x - 12.1520
I am going to add 21 to both sides to get 0 = 4x2 - 12x + 9.1527
I am going to rewrite this in a more standard form with the variables on the left.1540
Now, this is just a matter of solving the quadratic equation; and this is actually a perfect square: it is (2x - 3) (2x - 3) = 0.1546
And you can check this out to see that 2x times 2x is 4x2; 2x times -3 is -6x, plus -6x, is -12x; -3 times -3 is 9.1558
According to the product property, if 2x - 3 equals 0 (and these are the same, so I only have to look at one factor), then I will have a solution.1576
So, 2x = 3; if I solve for x, x equals 3/2.1585
x = 3/2; now, my last step--this is a potential solution--is that I need to go ahead and check it right here.1590
Let's let x equal 3/2; and we are going to insert that into log16(12x - 21).1599
So, x equals log16(12(3/2) - 21)...not x equals...log16...this cancels;1608
this becomes 1, and this is 6; this is 6 times 3, minus 21; so this is log16(18 - 21).1627
And that gives me log16(-18); therefore, this solution is not valid.1639
And since that is the only solution I have, there is no solution to this equation.1654
The only solution I came up with was an extraneous solution.1664
I solved this by first converting it to this form; and I did that by setting up an equation1668
where I figured out what x had to be for me to write 1/2 in the form log16.1676
And I used the technique of converting this to the exponential form 161/2 = x and solving for x.1682
That told me that log16(4) = 1/2; so instead of writing 1/2 here, I wrote log16(4).1690
Then I used the product property to combine the two logarithmic expressions on the right.1699
Then I used this property, and I set 12x - 21 equal to this expression, and solved, using factoring, a quadratic equation.1705
But then, I went and checked my solution and found that it was not a valid solution.1714
Thanks for visiting Educator.com; that concludes today's lesson.1720
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# Equations Reducible to Quadratic by Substitution
Equations Reducible to Quadratic by Substitution to Learn Math are made easy to solve by substitution to simplify the equations.
## Question 1
Solve $(x+2)^2-3(x+2)-4 = 0$.
\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} \text{Let } A &= x+2 \\ A^2-3A-4 &= 0 &\color{red} \text{replace } x+2 \text{ by } A \\ (A-4)(A+1) &= 0 \\ A = 4 &\text{ or } A = -1 \\ x+2 = 4 &\text{ or } x+2 = -1 &\color{red} \text{replace } A \text{ by } x+2 \\ \therefore x = 2 &\text{ or } x = -3 \end{aligned}
## Question 2
Solve $(x-1)^4-11(x-1)^2 + 18 = 0$.
\begin{aligned} \displaystyle \require{AMSsymbols} \text{Let } A &= x-1 \\ A^4-11A + 18 &= 0 &\color{red} \text{replace } x-1 \text{ by } A \\ (A^2-9)(A^2-2) &= 0 \\ A^2 = 9 &\text{ or } A^2 = 2 \\ A = 3, -3 &\text{ or } A = \sqrt{2}, -\sqrt{2} \\ x-1 = 3, -3 &\text{ or } x-1 = \sqrt{2}, -\sqrt{2} &\color{red} \text{replace } A \text{ by } x-1 \\ \therefore x = 4, -2 &\text{ or } x = 1+\sqrt{2}, 1-\sqrt{2} \end{aligned}
## Question 3
Solve $(x^2-x)^2-18(x^2-x) + 72 = 0$.
\begin{aligned} \displaystyle \require{AMSsymbols} \text{Let } A &= x^2-x \\ A^2-18A + 72 &= 0 &\color{red} \text{replace } x^2-x \text{ by } A \\ (A-12)(A-6) &= 0 \\ A = 12 &\text{ or } A = 6 \\ x^2-x = 12 &\text{ or } x^2-x = 6 &\color{red} \text{replace } A \text{ by } x^2-x \\ x^2-x-12 = 0 &\text{ or } x^2-x-6 = 0 \\ (x-4)(x+3) = 0 &\text{ or } (x-3)(x+2) = 0 \\ \therefore x &= 4, -3, 3, -2 \\ \end{aligned}
## Question 4
Solve $\displaystyle \Big(x+\frac{1}{x}\Big)^2 + \Big(x+\frac{1}{x}\Big)-2 = 0$.
\begin{aligned} \displaystyle \require{AMSsymbols} \text{Let } A &= x+ \frac{1}{x} \\ A^2 + A-2 &= 0 \\ (A+2)(A-1) &= 0 \\ A = -2 &\text{ or } A = 1 \\ x + \frac{1}{x} = -2 &\text{ or } x + \frac{1}{x} = 1 \\ x^2 + 1 = -2x &\text{ or } x^2 + 1 = -x \\ x^2 + 2x + 1 = 0 &\text{ or } x^2 +x + 1 = 0 \\ (x+1)^2 = 0 &\text{ or } x = \frac{1 \pm \sqrt{-3}}{2} &\color{red} \text{undefined}\\ \therefore x = -1 \end{aligned}
## High School Math for Life: Making Sense of Earnings
Salary Salary refers to the fixed amount of money that an employer pays an employee at regular intervals, typically on a monthly or biweekly basis,…
## Probability Pro: Mastering Two-Way Tables with Ease
Welcome to a comprehensive guide on mastering probability through the lens of two-way tables. If you’ve ever found probability challenging, fear not. We’ll break it…
## Mastering Integration by Parts: The Ultimate Guide
Welcome to the ultimate guide on mastering integration by parts. If you’re a student of calculus, you’ve likely encountered integration problems that seem insurmountable. That’s…
The sum $a+b$ is called a binomial as it contains two terms.Any expression of the form $(a+b)^n$ is called a power of a binomial. All…
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## Thinking Mathematically (6th Edition)
In order to arrive at the answer of 2 cups of water, we first need to find the amount of water needed for one serving of instant potatoes. To do this, we use the information given in the problem: 2$\frac{2}{3}$ cups of water is needed to make 8 servings of instant potatoes. We can now divide the amount of water needed by the number of servings it makes. NOTE: Before we divide, we need to convert the 2$\frac{2}{3}$ to an improper fraction. The improper fraction we need is $\frac{8}{3}$. We also need to note that 8 (servings) can be written as a fraction by writing it as $\frac{8}{1}$. Now, we can divide $\frac{8}{3}$ by $\frac{8}{1}$ Remember the procedure for dividing fractions is to keep the first one as is (the $\frac{8}{3}$ stays as is). Then, we flip the second fraction. This is called finding the reciprocal. Then we find the reciprocal of $\frac{8}{1}$. In other words, flip it over and get $\frac{1}{8}$. Now change the division to multiplication. We have: $\frac{8}{3}$ $\times$ $\frac{1}{8}$ = $\frac{8}{24}$ = $\frac{1}{3}$ (reduced form) This means we need $\frac{1}{3}$ cup of water for each serving of the instant potatoes. Therefore, to make 6 servings, we need to multiply $\frac{1}{3}$ by 6. $\frac{1}{3}$ $\times$ $\frac{6}{1}$ = $\frac{6}{3}$ = 2 This gives us the solution of needing 2 cups of water to make six servings of instant potatoes.
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2001 USAMO Problems/Problem 3
Problem
Let $a, b, c \geq 0$ and satisfy
$a^2 + b^2 + c^2 + abc = 4.$
Show that
$0 \le ab + bc + ca - abc \leq 2.$
Solution
Solution 1
First we prove the lower bound.
Note that we cannot have $a, b, c$ all greater than 1. Therefore, suppose $a \le 1$. Then $$ab + bc + ca - abc = a(b + c) + bc(1-a) \ge 0.$$ Note that, by the Pigeonhole Principle, at least two of $a,b,c$ are either both greater than or less than $1$. Without loss of generality, let them be $b$ and $c$. Therefore, $(b-1)(c-1)\ge 0$. From the given equation, we can express $a$ in terms of $b$ and $c$ as $$a=\frac{\sqrt{(4-b^2)(4-c^2)}-bc}{2}$$ Thus, $$ab + bc + ca - abc = -a (b-1)(c-1)+a+bc \le a+bc = \frac{\sqrt{(4-b^2)(4-c^2)} + bc}{2}$$
From the Cauchy-Schwarz Inequality, $$\frac{\sqrt{(4-b^2)(4-c^2)} + bc}{2} \le \frac{\sqrt{(4-b^2+b^2)(4-c^2+c^2)} }{2} = 2.$$
This completes the proof.
Solution 2
The proof for the lower bound is the same as in the first solution.
Now we prove the upper bound. Let us note that at least two of the three numbers $a$, $b$, and $c$ are both greater than or equal to 1 or less than or equal to 1. Without loss of generality, we assume that the numbers with this property are $b$ and $c$. Then we have $$(1 - b)(1 - c)\geq 0.$$ The given equality $a^2 + b^2 + c^2 + abc = 4$ and the inequality $b^2 + c^2\geq 2bc$ imply that $$a^2 + 2bc + abc\leq 4,$$ or $$bc(2 + a)\leq 4 - a^2.$$ Dividing both sides of the last inequality by $2 + a$ yields $$bc\leq 2 - a.$$ Thus, $$ab + bc + ca - abc\leq ab + 2 - a + ac(1 - b) = 2 - a(1 + bc - b - c) = 2 - a(1 - b)(1 - c)\leq 2,$$ as desired.
The last equality holds if and only if $b = c$ and $a(1 - b)(1 - c) = 0$. Hence equality for the upper bound holds if and only if $(a,b,c)$ is one of the triples $(1,1,1)$, $(0,\sqrt{2},\sqrt{2})$, $(\sqrt{2},0,\sqrt{2})$, and $(\sqrt{2},\sqrt{2},0)$. Equality for the lower bound holds if and only if $(a,b,c)$ is one of the triples $(2,0,0)$, $(0,2,0)$ and $(0,0,2)$.
Solution 3
The proof for the lower bound is the same as in the first solution.
Now we prove the upper bound. It is clear that $a,b,c\leq 2$. If $abc = 0$, then the result is trivial. Suppose that $a,b,c > 0$. Solving for $a$ yields $$a = \frac{-bc + \sqrt{b^2c^2 - 4(b^2 + c^2 - 4)}}{2} = \frac{-bc + \sqrt{(4 - b^2)(4 - c^2)}}{2}.$$ This asks for the trigonometric substitution $b = 2\sin u$ and $c = 2\sin v$, where $0^\circ < u,v < 90^\circ$. Then $$a = 2(-\sin u\sin v + \cos u\cos v) = 2\cos (u + v),$$ and if we set $u = B/2$ and $v = C/2$, then $a = 2\sin (A/2)$, $b = 2\sin (B/2)$, and $c = \sin (C/2)$, where $A$, $B$, and $C$ are the angles of a triangle. We have \begin{align*} ab &= 4\sin\frac{A}{2}\sin\frac{B}{2} \\ &= 2\sqrt{\sin A\tan\frac{A}{2}\sin B\tan\frac{B}{2}} = 2\sqrt{\sin A\tan\frac{B}{2}\sin B\tan\frac{A}{2}} \\ &\leq \sin A\tan\frac{B}{2} + \sin B\tan\frac{A}{2} \\ &= \sin A\cot\frac{A + C}{2} + \sin B\cot\frac{B + C}{2}, \end{align*} where the inequality step follows from AM-GM. Likewise, \begin{align*} bc &\leq \sin B\cot\frac{B + A}{2} + \sin C\cot\frac{C + A}{2}, \\ ca &\leq \sin A\cot\frac{A + B}{2} + \sin C\cot\frac{C + B}{2}. \end{align*} Therefore \begin{align*} ab + bc + ca &\leq (\sin A + \sin B)\cot\frac{A + B}{2} + (\sin B + \sin C)\cot\frac{B + C}{2} + (\sin C + \sin A)\cot\frac{C + A}{2} \\ &= 2\left(\cos\frac{A - B}{2}\cos\frac{A + B}{2} + \cos\frac{B - C}{2}\cos\frac{B + C}{2} + \cos\frac{C - A}{2}\cos\frac{C + A}{2} \right)\\ &= 2(\cos A + \cos B + \cos C) \\ &= 6 - 4\left(\sin^2\frac{A}{2} + \sin^2\frac{B}{2} + \sin^2\frac{C}{2}\right) \\ &= 6 - (a^2 + b^2 + c^2) = 2 + abc, \end{align*} as desired.
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Question 1
Present age of Vinod and Ashok are in ratio of 3:4 respectively. After 5 years, the ratio of their ages becomes 7:9 respectively. What is Ashok’s present age is ?
A 40 years B 28 years C 32 years D 36 years
Arithmetic Aptitude 3 Ratio and Proportion Age
Discuss it
Question 1 Explanation:
Let the present age of Vinod and Ashok be 3x years and 4x years respectively.
Then (3x+5) / (4x+5) = 7 / 9
∴ 9(3x + 5) = 7(4x + 5)
∴ 27x + 45 = 28x + 35
∴ x = 10
∴ Ashok’s present age = 4x = 40 years
Question 2
At present, the ratio between ages of Ram and Shyam is 6:5 respectively. After 7 years, Shyam’s age will be 32 years. What is the present age of Ram?
A 32 B 40 C 30 D 36
Arithmetic Aptitude 3 Ratio and Proportion Age
Discuss it
Question 2 Explanation:
Let the present age of Ram and Shyam be 6x years and 5x years respectively.
Then 5x + 7 = 32
∴ 5x = 25
∴ x = 5
∴ Present age of Ram = 6x = 30 years
Question 3
The present ages of A, B and C are in proportions 4:5:9. Nine years ago, sum of their ages was 45 years. Find their present ages in years
A 15,20,35 B 20,24,36 C 20,25,45 D 16,20,36
Arithmetic Aptitude 3 Ratio and Proportion Age
Discuss it
Question 3 Explanation:
Let the current ages of A, B and C be ax years, 5x years and 9x respectively.
Then (4x-9) + (5x-9) + (9x-9) =45
∴ 18x – 27 = 45
∴ 18x = 72
∴ x = 4
Present ages of A, B and C are 4x = 16, 5x = 20, 9x = 36 respectively.
Question 4
Two numbers are in the ratio of 2:9. If their H. C. F. is 19, numbers are:
A 6, 27 B 8, 36 C 38, 171 D 20, 90
Arithmetic Aptitude 4 HCF Ratio and Proportion
Discuss it
Question 4 Explanation:
Let the numbers be 2X and 9X
Then their H.C.F. is X, so X = 19
∴ Numbers are (2x19 and 9x19) i.e. 38 and 171
Question 5
In a box, there are 10p, 25p and 50p coins in the ratio 4:9:5 with the total sum of Rs 206. How many coins of each kind does the box have?
A 200, 360, 160 B 135, 250, 150 C 90, 60, 110 D Cannot be determined
Ratio and Proportion
Discuss it
Question 5 Explanation:
Let the number of 10p, 25p, 50p coins be 4x, 9x, 5x respectively. Then, 4x/10 + 9x/4 + 5x/2 = 206 (Since, 10p = Rs 0.1, 25p = Rs 0.25, 50p = Rs 0.5) => 8x + 45x + 50x = 4120 (Multiplying both sides by 20 which is the LCM of 10, 4, 2) => 103x = 4120 => x = 40. Therefore, No. of 10p coins = 4 x 40 = 160 (= Rs 16) No. of 25p coins = 9 x 40 = 360 (= Rs 90) No. of 50p coins = 5 x 40 = 200 (= Rs 100)
Question 6
Mark, Steve and Bill get their salaries in the ratio of 2:3:5. If their salaries are incremented by 15%, 10%, and 20% respectively, the new ratio of their salaries becomes:
A 8:16:15 B 23:33:60 C 33:30:20 D 21:25:32
Ratio and Proportion
Discuss it
Question 6 Explanation:
Let their old salaries be 2a, 3a, 5a respectively. Then, their new salaries become: 115% of 2a = 2a x 1.15 = 2.3a 110% of 3a = 3a x 1.10 = 3.3a 120% of 5a = 5a x 1.20 = 6a So, the new ratio becomes 2.3a:3.3a:6a Upon simplification, this becomes 23:33:60
Question 7
In a library, the ratio of the books on Computer, Physics and Mathematics is 5:7:8. If the collection of books is increased respectively by 40%, 50% and 75%, find out the new ratio:
A 3:9:5 B 7:5:3 C 2:3:4 D 2:5:4
Ratio and Proportion
Discuss it
Question 7 Explanation:
40% increase will lead to a factor of 140 and similiarly 150 and 175 so new ratio is (5*140):(7*150):(8*175) on solving we get 2:3:4
Question 8
The ratio 5:3 represents 16 liters of a mixture containing milk and water. If 4 liters of water is added and 4 liters of milk is extracted from the mixture, then the ratio of the mixture will be:
A 7:3 B 5:6 C 2:3 D None of these
Ratio and Proportion
Discuss it
Question 8 Explanation:
Amount of Milk in 16 litres of mixture: (5/8) x 16 = 10 litres Amount of Water in 16 litres of mixture: 16-10 = 6 litres If we add 4 litres of water and extract 4 litres of milk, the total volume remains the same. Amount of Milk in 16 litres of new mixture: = 10 - 4 = 6 litres Amount of Water in 16 litres of new mixture: = 6 + 4 = 10 litres So, the new ratio becomes 3:5.
Question 9
If the ages of Jacob, Max and Samuel are in the proportion 3:5:7 and the average of their ages is 25 years, then find the age of the youngest person.
A 15 years B 10 years C 7 years D 18 years
Ratio and Proportion
Discuss it
Question 9 Explanation:
Let their ages be 3a, 5a and 7a. Then, (3a + 5a + 7a) / 3 = 25 => 15a/3 = 25 => 5a = 25 => a = 5 Therefore, age of the youngest person = 3a = 15 years
Question 10
The ratio of the speed of two trains is 7:8. If the second train covers 400 km in 4 h, find out the speed of the first train.
A 69.4 km/h B 78.6 km/h C 87.5 km/h D 40.5 km/h
Ratio and Proportion Time Speed Distance
Discuss it
Question 10 Explanation:
Let the speed of the two trains be 7x and 8x. Then, 8x = 400 / 4 ⇒ 8x = 100 ⇒ x = 12.5 km/h. Hence, speed of the first train = 7x = 7 × 12.5 = 87.5 km/h.
Question 11
A certain amount of money is distributed between Alfred, Adam, Harry and Leo in the proportion of 5:2:4:3. Harry's share is Rs 1000 more than Leo's share. How much money does Adam get?
A Rs 800 B Rs 1000 C Rs 1050 D Rs 2000
Ratio and Proportion
Discuss it
Question 11 Explanation:
Let the shares of Alfred, Adam, Harry and Leo be Rs 5x, Rs 2x, Rs 4x and Rs 3x respectively. Then, 4x - 3x = 1000 ⇒ x = 1000. Therefore, Adam gets = Rs 2x = Rs 2000.
Question 12
If (x:y) = 2:1, then (x²-y²):(x²+y²) = __ ?
A 1:2 B 3:5 C 2:1 D 5:4
Ratio and Proportion
Discuss it
Question 12 Explanation:
Given, x:y = 2:1 ⇒ x²:y² = 4:1. Then, (x²+y²):(x²-y²) = (4+1)/(4-1) [by applying Componendo & Dividendo]. ⇒ (x²-y²):(x²+y²) = 3/5.
Question 13
Syrup and Water are mixed in 2:1 ratio to form 60 litres of a mixture. How much water needs to be added to make the ratio 1:2?
A 60 litres B 85 litres C 55 litres D Cannot be determined
Ratio and Proportion
Discuss it
Question 13 Explanation:
Volume of Syrup in the mixture = 60 × 2/3 = 40 litres. Volume of Water in the mixture = 60 × 1/3 = 20 litres Let the required volume of water be x litres. Then, 40:(20+x) = 1:2 ⇒ 20+x = 80 ⇒ x = 60 litres.
Question 14
Felix and Adam win Rs 1210 together. 4/15 of Felix's share is same as 2/5 of Adam's share. How much did Adam win?
A Rs 484 B Rs 330 C Rs 360 D Data inadequate
Ratio and Proportion
Discuss it
Question 14 Explanation:
Let Felix's and Adam's share be F and A respectively. Then, 4/15 × F = 2/5 × A ⇒ 20 F = 30 A ⇒ F/A = 3/2 ⇒ F:A = 3:2. Therefore, Adam won = 2/5 of Rs 1210 = Rs 484.
There are 14 questions to complete.
## GATE CS Corner
See Placement Course for placement preparation, GATE Corner for GATE CS Preparation and Quiz Corner for all Quizzes on GeeksQuiz.
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## Engage NY Eureka Math Grade 6 Module 5 Lesson 2 Answer Key
### Eureka Math Grade 6 Module 5 Lesson 2 Exploratory Challenge Answer Key
Exploratory Challenge:
Question a.
Use the shapes labeled with an X to predict the formula needed to calculate the area of a right triangle. Explain your prediction.
Formula for the area of right triangles:
Area of the given triangle: ________
Formula for the area of right triangles:
A = $$\frac{1}{2}$$ × base × height or A = $$\frac{\text { base } \times \text { height }}{2}$$
Area of the given triangle:
A = $$\frac{1}{2}$$ × 3 in. × 2 in. = 3 in2
Question b.
Use the shapes labeled with a Y to determine if the formula you discovered in part (a) is correct.
Does your area formula for triangle Y match the formula you got for triangle X?
Answers will vary; however, the area formulas should be the same if students discovered the correct area
formula.
If so, do you believe you have the correct formula needed to calculate the area of a right triangle? Why or why not?
If not, which formula do you think is correct? Why?
Area of the given triangle:
A = $$\frac{1}{2}$$ × 3 in. × 3 in. = 4.5 im <sup>2</sup>
### Eureka Math Grade 6 Module 5 Lesson 2 Exercise Answer Key
Exercises:
Calculate the area of each right triangle below. Each figure is not drawn to scale.
Question 1.
A = $$\frac{1}{2}$$ bh
= $$\frac{1}{2}$$ (8 ft.) (15 ft.)
= 60 ft2
Question 2.
A = $$\frac{1}{2}$$ bh
= $$\frac{1}{2}$$ (11.4 cm) (17.7 cm)
= 100.89 cm2
Question 3.
A = $$\frac{1}{2}$$ bh
= $$\frac{1}{2}$$ (6 in.) (8 in.)
= 24 in2
Question 4.
Question 5.
A = $$\frac{1}{2}$$ bh
= $$\frac{1}{2}$$ (32.7 km) (21.4 km)
= 349.89 km2
Question 6.
Mr. Jones told his students they each need half of a piece of paper. Calvin cut his piece of paper horizontally, and Matthew cut his piece of paper diagonally. Which student has the larger area on his half piece of paper? Explain.
After cutting the paper, both Calvin and Matthew have the same area. Calvin cut his into two rectangles that are each half the area of the original piece of paper. Matthew cut his paper into two equivalent right triangles that are also half the area of the original piece of paper.
Question 7.
Ben requested that the rectangular stage be split into two equal sections for the upcoming school play. The only instruction he gave was that he needed the area of each section to be half of the original size. If Ben wants the stage to be split into two right triangles, did he provide enough information? Why or why not?
Ben did not provide enough information because the stage may be split horizontally or vertically through the middle of the rectangle. This would result in two equal pieces, but they would not be right triangles.
Question 8.
If the area of a right triangle is 6. 22 sq. in. and its base is 3. 11 in., write an equation that relates the area to the height, h, and the base. Solve the equation to determine the height.
6.22 = $$\frac{1}{2}$$ (3. 11)h
6.22 = (1.555)h
6.22 ÷ 1.555 = (1. 555)h ÷ 1.555
4 = h
Therefore, the height of the right triangle is 4 in.
### Eureka Math Grade 6 Module 5 Lesson 2 Problem Set Answer Key
Calculate the area of each right triangle below. Note that the figures are not drawn to scale.
Question 1.
A = $$\frac{1}{2}$$ bh
= $$\frac{1}{2}$$ (31.2 cm) (9.1 cm) = 141.96 cm2
Question 2.
Question 3.
A = $$\frac{1}{2}$$ bh
= $$\frac{1}{2}$$ (2.4 in.) (3.2 in.)
= 3.84 in2
Question 4.
A = $$\frac{1}{2}$$ bh
= $$\frac{1}{2}$$ (11 mm) (60 mm)
= 330 mm2
Question 5.
Question 6.
Elania has two congruent rugs at her house. She cut one vertically down the middle, and she cut diagonally through the other one.
After making the cuts, which rug (labeled A, B, C, or D) has the larger area? Explain.
All of the rugs are the same size after making the cuts. The vertical line goes down the center of the rectangle, making two congruent parts. The diagonal line also splits the rectangle into two congruent parts because the area of a right triangle is exactly half the area of the rectangle.
Question 7.
Give the dimensions of a right triangle and a parallelogram with the same area. Explain how you know.
Question 8.
If the area of a right triangle is $$\frac{9}{16}$$ sq. ft. and the height is $$\frac{3}{4}$$ ft., write an equation that relates the area to the base, b, and the height. Solve the equation to determine the base.
Therefore, the base of the right triangle is 1$$\frac{1}{2}$$ ft.
### Eureka Math Grade 6 Module 5 Lesson 2 Exit Ticket Answer Key
Question 1.
Calculate the area of the right triangle. Each figure is not drawn to scale.
A = $$\frac{1}{2}$$ bh = $$\frac{1}{2}$$ (8 in.)(6 in.) = 24 in2
A = $$\frac{1}{2}$$ × base × height.
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# 7.1 Kinetic energy
Page 1 / 7
• Explain work as a transfer of energy and net work as the work done by the net force.
• Explain and apply the work-energy theorem.
## Work transfers energy
What happens to the work done on a system? Energy is transferred into the system, but in what form? Does it remain in the system or move on? The answers depend on the situation. For example, if the lawn mower in [link] (a) is pushed just hard enough to keep it going at a constant speed, then energy put into the mower by the person is removed continuously by friction, and eventually leaves the system in the form of heat transfer. In contrast, work done on the briefcase by the person carrying it up stairs in [link] (d) is stored in the briefcase-Earth system and can be recovered at any time, as shown in [link] (e). In fact, the building of the pyramids in ancient Egypt is an example of storing energy in a system by doing work on the system. Some of the energy imparted to the stone blocks in lifting them during construction of the pyramids remains in the stone-Earth system and has the potential to do work.
In this section we begin the study of various types of work and forms of energy. We will find that some types of work leave the energy of a system constant, for example, whereas others change the system in some way, such as making it move. We will also develop definitions of important forms of energy, such as the energy of motion.
## Net work and the work-energy theorem
We know from the study of Newton’s laws in Dynamics: Force and Newton's Laws of Motion that net force causes acceleration. We will see in this section that work done by the net force gives a system energy of motion, and in the process we will also find an expression for the energy of motion.
Let us start by considering the total, or net, work done on a system. Net work is defined to be the sum of work done by all external forces—that is, net work is the work done by the net external force ${\mathbf{F}}_{\text{net}}$ . In equation form, this is ${W}_{\text{net}}={F}_{\text{net}}d\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta$ where $\theta$ is the angle between the force vector and the displacement vector.
[link] (a) shows a graph of force versus displacement for the component of the force in the direction of the displacement—that is, an $F\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta$ vs. $d$ graph. In this case, $F\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta$ is constant. You can see that the area under the graph is $Fd\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta$ , or the work done. [link] (b) shows a more general process where the force varies. The area under the curve is divided into strips, each having an average force $\left(F\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta {\right)}_{i\left(\text{ave}\right)}$ . The work done is $\left(F\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta {\right)}_{i\left(\text{ave}\right)}{d}_{i}$ for each strip, and the total work done is the sum of the ${W}_{i}$ . Thus the total work done is the total area under the curve, a useful property to which we shall refer later.
Net work will be simpler to examine if we consider a one-dimensional situation where a force is used to accelerate an object in a direction parallel to its initial velocity. Such a situation occurs for the package on the roller belt conveyor system shown in [link] .
what is math number
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years? Kala Reply lim x to infinity e^1-e^-1/log(1+x) given eccentricity and a point find the equiation Moses Reply 12, 17, 22.... 25th term Alexandra Reply 12, 17, 22.... 25th term Akash College algebra is really hard? Shirleen Reply Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table. Carole I'm 13 and I understand it great AJ I am 1 year old but I can do it! 1+1=2 proof very hard for me though. Atone hi Adu Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily. Vedant find the 15th term of the geometric sequince whose first is 18 and last term of 387 Jerwin Reply I know this work salma The given of f(x=x-2. then what is the value of this f(3) 5f(x+1) virgelyn Reply hmm well what is the answer Abhi If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10 Augustine how do they get the third part x = (32)5/4 kinnecy Reply make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be AJ how Sheref can someone help me with some logarithmic and exponential equations. Jeffrey Reply sure. what is your question? ninjadapaul 20/(×-6^2) Salomon okay, so you have 6 raised to the power of 2. what is that part of your answer ninjadapaul I don't understand what the A with approx sign and the boxed x mean ninjadapaul it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared Salomon I'm not sure why it wrote it the other way Salomon I got X =-6 Salomon ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6 ninjadapaul oops. ignore that. ninjadapaul so you not have an equal sign anywhere in the original equation? ninjadapaul hmm Abhi is it a question of log Abhi 🤔. Abhi I rally confuse this number And equations too I need exactly help salma But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends salma Commplementary angles Idrissa Reply hello Sherica im all ears I need to learn Sherica right! what he said ⤴⤴⤴ Tamia hii Uday hi salma hi Ayuba Hello opoku hi Ali greetings from Iran Ali salut. from Algeria Bach hi Nharnhar A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place. Kimberly Reply Jeannette has$5 and \$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Home » Blog » Find the extreme values for the perimeter of an isosceles triangle inscribed in a circle
# Find the extreme values for the perimeter of an isosceles triangle inscribed in a circle
1. Consider a circle of radius with an inscribed isosceles triangle. If the angle at the apex of the isosceles triangle is , and , find the extreme values (maximum and minimum) of the perimeter of the triangle.
2. Find the minimum value of the radius such that the circle of radius will cover every isosceles triangle with fixed perimeter .
1. In the picture we draw in six radial lines, which divide the triangle into six congruent triangles, each with hypotenuse of length and legs of lengths and . As shown in the diagram, two of the edges of the inscribed isosceles triangle have lengths , and the third leg as length . Therefore, the perimeter is,
Taking the derivative,
Now, we must note that the extreme values of the function occur when the derivative is 0 or at an end point of the interval we are looking at; in this case, is the interval stipulated in the problem.
First, let’s set the derivative equal to zero and find the values of in the interval at which the derivative is 0.
Then, the only value of in the interval such that or is
Plugging this in to our expression for the perimeter of the isosceles triangle, we find,
Examining the sign of the derivative of the perimeter function as changes, we see that this is a maximum. Furthermore, we see that the minimum occurs at the end point, . The minimum perimeter is then
2. First, we identify the value of at which the radius necessary to cover the isosceles triangle is at a maximum (the worst case scenario for covering the triangle, if you like). From part (a) we know the perimeter must satisfy the equation
Since is now a fixed constant we can write as a function of ,
Now, the value of is maximal when the denominator is minimal, so we are looking for a minimum of the function
Taking the derivative,
We set this equal to zero to find the critical points
From part (a) we know this implies or . Again, from part (a) we know this is minimal when . Since we are trying to minimize this we want the value .
Finally, there is a bit of a problem with since that corresponds to just a line, not really a triangle. We can either ignore that issue, and just plug in or (in what will amount to the same thing) we can find the minimal value of as the limit as :
where we can compute the limit by just evaluating at since the function is continuous.
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Polynomial and Rational Functions
# Power Functions and Polynomial Functions
### Learning Objectives
In this section, you will:
• Identify power functions.
• Identify end behavior of power functions.
• Identify polynomial functions.
• Identify the degree and leading coefficient of polynomial functions.
Suppose a certain species of bird thrives on a small island. Its population over the last few years is shown in (Figure).
Year Bird Population
The population can be estimated using the functionwhererepresents the bird population on the islandyears after 2009. We can use this model to estimate the maximum bird population and when it will occur. We can also use this model to predict when the bird population will disappear from the island. In this section, we will examine functions that we can use to estimate and predict these types of changes.
### Identifying Power Functions
Before we can understand the bird problem, it will be helpful to understand a different type of function. A power function is a function with a single term that is the product of a real number, a coefficient, and a variable raised to a fixed real number.
As an example, consider functions for area or volume. The function for the area of a circle with radius
is
and the function for the volume of a sphere with radius
is
Both of these are examples of power functions because they consist of a coefficient,ormultiplied by a variableraised to a power.
### Power Function
A power function is a function that can be represented in the form
where
andare real numbers, and
is known as the coefficient.
Isa power function?
No. A power function contains a variable base raised to a fixed power. This function has a constant base raised to a variable power. This is called an exponential function, not a power function.
### Identifying Power Functions
Which of the following functions are power functions?
All of the listed functions are power functions.
The constant and identity functions are power functions because they can be written asandrespectively.
The quadratic and cubic functions are power functions with whole number powersand
The reciprocal and reciprocal squared functions are power functions with negative whole number powers because they can be written asand
The square and cube root functions are power functions with fractional powers because they can be written asor
### Try It
Which functions are power functions?
is a power function because it can be written as
The other functions are not power functions.
### Identifying End Behavior of Power Functions
(Figure) shows the graphs ofandwhich are all power functions with even, whole-number powers. Notice that these graphs have similar shapes, very much like that of the quadratic function in the toolkit. However, as the power increases, the graphs flatten somewhat near the origin and become steeper away from the origin.
To describe the behavior as numbers become larger and larger, we use the idea of infinity. We use the symbolfor positive infinity andfor negative infinity. When we say that “approaches infinity,” which can be symbolically written aswe are describing a behavior; we are saying thatis increasing without bound.
With the positive even-power function, as the input increases or decreases without bound, the output values become very large, positive numbers. Equivalently, we could describe this behavior by saying that asapproaches positive or negative infinity, thevalues increase without bound. In symbolic form, we could write
(Figure) shows the graphs ofandwhich are all power functions with odd, whole-number powers. Notice that these graphs look similar to the cubic function in the toolkit. Again, as the power increases, the graphs flatten near the origin and become steeper away from the origin.
These examples illustrate that functions of the formreveal symmetry of one kind or another. First, in (Figure) we see that even functions of the formeven, are symmetric about theaxis. In (Figure) we see that odd functions of the form odd, are symmetric about the origin.
For these odd power functions, as approaches negative infinity, decreases without bound. As approaches positive infinity, increases without bound. In symbolic form we write
The behavior of the graph of a function as the input values get very small () and get very large () is referred to as the end behavior of the function. We can use words or symbols to describe end behavior.
(Figure) shows the end behavior of power functions in the formwhereis a non-negative integer depending on the power and the constant.
### How To
Given a power functionwhereis a non-negative integer, identify the end behavior.
1. Determine whether the power is even or odd.
2. Determine whether the constant is positive or negative.
3. Use (Figure) to identify the end behavior.
### Identifying the End Behavior of a Power Function
Describe the end behavior of the graph of
The coefficient is 1 (positive) and the exponent of the power function is 8 (an even number). Asapproaches infinity, the output (value of) increases without bound. We write asAsapproaches negative infinity, the output increases without bound. In symbolic form, as We can graphically represent the function as shown in (Figure).
### Identifying the End Behavior of a Power Function.
Describe the end behavior of the graph of
The exponent of the power function is 9 (an odd number). Because the coefficient is(negative), the graph is the reflection about theaxis of the graph of(Figure) shows that asapproaches infinity, the output decreases without bound. Asapproaches negative infinity, the output increases without bound. In symbolic form, we would write
#### Analysis
We can check our work by using the table feature on a graphing utility.
–10 1,000,000,000
–5 1,953,125
0 0
5 –1,953,125
10 –1,000,000,000
We can see from (Figure) that, when we substitute very small values forthe output is very large, and when we substitute very large values forthe output is very small (meaning that it is a very large negative value).
### Try It
Describe in words and symbols the end behavior of
Asapproaches positive or negative infinity,decreases without bound: asbecause of the negative coefficient.
### Identifying Polynomial Functions
An oil pipeline bursts in the Gulf of Mexico, causing an oil slick in a roughly circular shape. The slick is currently 24 miles in radius, but that radius is increasing by 8 miles each week. We want to write a formula for the area covered by the oil slick by combining two functions. The radius
of the spill depends on the number of weeks
that have passed. This relationship is linear.
We can combine this with the formula for the area
of a circle.
Composing these functions gives a formula for the area in terms of weeks.
Multiplying gives the formula.
This formula is an example of a polynomial function. A polynomial function consists of either zero or the sum of a finite number of non-zero terms, each of which is a product of a number, called the coefficient of the term, and a variable raised to a non-negative integer power.
### Polynomial Functions
Let
be a non-negative integer. A polynomial function is a function that can be written in the form
This is called the general form of a polynomial function. Each
is a coefficient and can be any real number, but
cannot = 0. Each expression
is a term of a polynomial function.
### Identifying Polynomial Functions
Which of the following are polynomial functions?
The first two functions are examples of polynomial functions because they can be written in the form
where the powers are non-negative integers and the coefficients are real numbers.
• can be written as
• can be written as
• cannot be written in this form and is therefore not a polynomial function.
### Identifying the Degree and Leading Coefficient of a Polynomial Function
Because of the form of a polynomial function, we can see an infinite variety in the number of terms and the power of the variable. Although the order of the terms in the polynomial function is not important for performing operations, we typically arrange the terms in descending order of power, or in general form. The degree of the polynomial is the highest power of the variable that occurs in the polynomial; it is the power of the first variable if the function is in general form. The leading term is the term containing the highest power of the variable, or the term with the highest degree. The leading coefficient is the coefficient of the leading term.
### Terminology of Polynomial Functions
We often rearrange polynomials so that the powers are descending.
When a polynomial is written in this way, we say that it is in general form.
### How To
Given a polynomial function, identify the degree and leading coefficient.
1. Find the highest power of
to determine the degree function.
2. Identify the term containing the highest power of
to find the leading term.
3. Identify the coefficient of the leading term.
### Identifying the Degree and Leading Coefficient of a Polynomial Function
Identify the degree, leading term, and leading coefficient of the following polynomial functions.
For the functionthe highest power ofis 3, so the degree is 3. The leading term is the term containing that degree,The leading coefficient is the coefficient of that term,
For the functionthe highest power ofisso the degree isThe leading term is the term containing that degree,The leading coefficient is the coefficient of that term,
For the functionthe highest power ofisso the degree isThe leading term is the term containing that degree,The leading coefficient is the coefficient of that term,
Identify the degree, leading term, and leading coefficient of the polynomial
The degree is 6. The leading term isThe leading coefficient is
#### Identifying End Behavior of Polynomial Functions
Knowing the degree of a polynomial function is useful in helping us predict its end behavior. To determine its end behavior, look at the leading term of the polynomial function. Because the power of the leading term is the highest, that term will grow significantly faster than the other terms as gets very large or very small, so its behavior will dominate the graph. For any polynomial, the end behavior of the polynomial will match the end behavior of the power function consisting of the leading term. See (Figure).
Polynomial Function Leading Term Graph of Polynomial Function
### Identifying End Behavior and Degree of a Polynomial Function
Describe the end behavior and determine a possible degree of the polynomial function in (Figure).
As the input values
get very large, the output valuesincrease without bound. As the input values
get very small, the output valuesdecrease without bound. We can describe the end behavior symbolically by writing
In words, we could say that asvalues approach infinity, the function values approach infinity, and asvalues approach negative infinity, the function values approach negative infinity.
We can tell this graph has the shape of an odd degree power function that has not been reflected, so the degree of the polynomial creating this graph must be odd and the leading coefficient must be positive.
### Try It
Describe the end behavior, and determine a possible degree of the polynomial function in (Figure).
AsIt has the shape of an even degree power function with a negative coefficient.
### Identifying End Behavior and Degree of a Polynomial Function
Given the functionexpress the function as a polynomial in general form, and determine the leading term, degree, and end behavior of the function.
Obtain the general form by expanding the given expression for
The general form is
The leading term is
therefore, the degree of the polynomial is 4. The degree is even (4) and the leading coefficient is negative (–3), so the end behavior is
### Try It
Given the functionexpress the function as a polynomial in general form and determine the leading term, degree, and end behavior of the function.
The leading term isso it is a degree 3 polynomial. Asapproaches positive infinity,increases without bound; asapproaches negative infinity,decreases without bound.
#### Identifying Local Behavior of Polynomial Functions
In addition to the end behavior of polynomial functions, we are also interested in what happens in the “middle” of the function. In particular, we are interested in locations where graph behavior changes. A turning point is a point at which the function values change from increasing to decreasing or decreasing to increasing.
We are also interested in the intercepts. As with all functions, the y-intercept is the point at which the graph intersects the vertical axis. The point corresponds to the coordinate pair in which the input value is zero. Because a polynomial is a function, only one output value corresponds to each input value so there can be only one y-interceptThe x-intercepts occur at the input values that correspond to an output value of zero. It is possible to have more than one x-intercept. See (Figure).
### Intercepts and Turning Points of Polynomial Functions
A turning point of a graph is a point at which the graph changes direction from increasing to decreasing or decreasing to increasing. The y-intercept is the point at which the function has an input value of zero. The x-intercepts are the points at which the output value is zero.
### How To
Given a polynomial function, determine the intercepts.
1. Determine the y-intercept by setting and finding the corresponding output value.
2. Determine the x-intercepts by solving for the input values that yield an output value of zero.
### Determining the Intercepts of a Polynomial Function
Given the polynomial functionwritten in factored form for your convenience, determine the y– and x-intercepts.
The y-intercept occurs when the input is zero so substitute 0 for
The y-intercept is (0, 8).
The x-intercepts occur when the output is zero.
The x-intercepts areand
We can see these intercepts on the graph of the function shown in (Figure).
### Determining the Intercepts of a Polynomial Function with Factoring
Given the polynomial functiondetermine the y– and x-intercepts.
The y-intercept occurs when the input is zero.
The y-intercept is
The x-intercepts occur when the output is zero. To determine when the output is zero, we will need to factor the polynomial.
The x-intercepts areand
We can see these intercepts on the graph of the function shown in (Figure). We can see that the function is even because
### Try It
Given the polynomial functiondetermine the y– and x-intercepts.
y-interceptx-interceptsand
#### Comparing Smooth and Continuous Graphs
The degree of a polynomial function helps us to determine the number of x-intercepts and the number of turning points. A polynomial function ofdegree is the product offactors, so it will have at mostroots or zeros, or x-intercepts. The graph of the polynomial function of degreemust have at mostturning points. This means the graph has at most one fewer turning point than the degree of the polynomial or one fewer than the number of factors.
A continuous function has no breaks in its graph: the graph can be drawn without lifting the pen from the paper. A smooth curve is a graph that has no sharp corners. The turning points of a smooth graph must always occur at rounded curves. The graphs of polynomial functions are both continuous and smooth.
### Intercepts and Turning Points of Polynomials
A polynomial of degreewill have, at most,x-intercepts andturning points.
### Determining the Number of Intercepts and Turning Points of a Polynomial
Without graphing the function, determine the local behavior of the function by finding the maximum number of x-intercepts and turning points for
The polynomial has a degree ofso there are at most 10 x-intercepts and at most 9 turning points.
### Try It
Without graphing the function, determine the maximum number of x-intercepts and turning points for
There are at most 12intercepts and at most 11 turning points.
### Drawing Conclusions about a Polynomial Function from the Graph
What can we conclude about the polynomial represented by the graph shown in (Figure) based on its intercepts and turning points?
The end behavior of the graph tells us this is the graph of an even-degree polynomial. See (Figure).
The graph has 2 x-intercepts, suggesting a degree of 2 or greater, and 3 turning points, suggesting a degree of 4 or greater. Based on this, it would be reasonable to conclude that the degree is even and at least 4.[/hidden-answer]
### Try It
What can we conclude about the polynomial represented by the graph shown in (Figure) based on its intercepts and turning points?
The end behavior indicates an odd-degree polynomial function; there are 3intercepts and 2 turning points, so the degree is odd and at least 3. Because of the end behavior, we know that the lead coefficient must be negative.
### Drawing Conclusions about a Polynomial Function from the Factors
Given the function
determine the local behavior.
The y-intercept is found by evaluating
The y-intercept is
The x-intercepts are found by determining the zeros of the function.
The x-intercepts areand
The degree is 3 so the graph has at most 2 turning points.
### Try It
Given the functiondetermine the local behavior.
Theintercepts areandthe y-intercept isand the graph has at most 2 turning points.
Access these online resources for additional instruction and practice with power and polinomial functions.
### Key Equations
general form of a polynomial function
### Key Concepts
• A power function is a variable base raised to a number power. See (Figure).
• The behavior of a graph as the input decreases beyond bound and increases beyond bound is called the end behavior.
• The end behavior depends on whether the power is even or odd. See (Figure) and (Figure).
• A polynomial function is the sum of terms, each of which consists of a transformed power function with positive whole number power. See (Figure).
• The degree of a polynomial function is the highest power of the variable that occurs in a polynomial. The term containing the highest power of the variable is called the leading term. The coefficient of the leading term is called the leading coefficient. See (Figure).
• The end behavior of a polynomial function is the same as the end behavior of the power function represented by the leading term of the function. See (Figure) and (Figure).
• A polynomial of degree
will have at most
x-intercepts and at most
turning points. See (Figure), (Figure), (Figure), (Figure), and (Figure).
### Section Exercises
#### Verbal
Explain the difference between the coefficient of a power function and its degree.
The coefficient of the power function is the real number that is multiplied by the variable raised to a power. The degree is the highest power appearing in the function.
If a polynomial function is in factored form, what would be a good first step in order to determine the degree of the function?
In general, explain the end behavior of a power function with odd degree if the leading coefficient is positive.
As
decreases without bound, so does
As
increases without bound, so does
What is the relationship between the degree of a polynomial function and the maximum number of turning points in its graph?
What can we conclude if, in general, the graph of a polynomial function exhibits the following end behavior? As
and as
The polynomial function is of even degree and leading coefficient is negative.
#### Algebraic
For the following exercises, identify the function as a power function, a polynomial function, or neither.
Power function
Neither
Neither
For the following exercises, find the degree and leading coefficient for the given polynomial.
Degree = 2, Coefficient = –2
Degree =4, Coefficient = –2
For the following exercises, determine the end behavior of the functions.
For the following exercises, find the intercepts of the functions.
y-intercept ist-intercepts are
y-intercept isx-intercepts areand
y-intercept isx-intercepts areand
#### Graphical
For the following exercises, determine the least possible degree of the polynomial function shown.
3
5
3
5
For the following exercises, 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.
Yes. Number of turning points is 2. Least possible degree is 3.
Yes. Number of turning points is 1. Least possible degree is 2.
Yes. Number of turning points is 0. Least possible degree is 1.
No.
Yes. Number of turning points is 0. Least possible degree is 1.
#### Numeric
For the following exercises, make a table to confirm the end behavior of the function.
10 9,500
100 99,950,000
–10 9,500
–100 99,950,000
10 –504
100 –941,094
–10 1,716
–100 1,061,106
#### Technology
For the following exercises, graph the polynomial functions using a calculator. Based on the graph, determine the intercepts and the end behavior.
Theintercept is
Theintercepts are
Theintercept is
. Theintercepts are
Theintercept is Theintercept is
Theintercept is
Theintercept are
Theintercept is Theintercepts are
#### Extensions
For the following exercises, 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.
Theintercept isTheintercepts areDegree is 2.
End behavior:
Theintercept isTheintercepts areDegree is 2.
End behavior:
Theintercept isTheintercepts areDegree is 3.
End behavior:
Theintercept isTheintercept isDegree is 3.
End behavior:
Theintercept isThere is nointercept. Degree is 4.
End behavior:
#### Real-World Applications
For the following exercises, use the written statements to construct a polynomial function that represents the required information.
An oil slick is expanding as a circle. The radius of the circle is increasing at the rate of 20 meters per day. Express the area of the circle as a function ofthe number of days elapsed.
A cube has an edge of 3 feet. The edge is increasing at the rate of 2 feet per minute. Express the volume of the cube as a function ofthe number of minutes elapsed.
A rectangle has a length of 10 inches and a width of 6 inches. If the length is increased byinches and the width increased by twice that amount, express the area of the rectangle as a function of
An open box is to be constructed by cutting out square corners of inch sides from a piece of cardboard 8 inches by 8 inches and then folding up the sides. Express the volume of the box as a function of
A rectangle is twice as long as it is wide. Squares of side 2 feet are cut out from each corner. Then the sides are folded up to make an open box. Express the volume of the box as a function of the width ().
### Glossary
coefficient
a nonzero real number multiplied by a variable raised to an exponent
continuous function
a function whose graph can be drawn without lifting the pen from the paper because there are no breaks in the graph
degree
the highest power of the variable that occurs in a polynomial
end behavior
the behavior of the graph of a function as the input decreases without bound and increases without bound
the coefficient of the leading term
the term containing the highest power of the variable
polynomial function
a function that consists of either zero or the sum of a finite number of non-zero terms, each of which is a product of a number, called the coefficient of the term, and a variable raised to a non-negative integer power.
power function
a function that can be represented in the formwhereis a constant, the base is a variable, and the exponent,, is a constant
smooth curve
a graph with no sharp corners
term of a polynomial function
anyof a polynomial function in the form
turning point
the location at which the graph of a function changes direction
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Applications Using Rational Equations
Word problems including fractions with variables in the denominator
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Practice Applications Using Rational Equations
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Applications Using Rational Equations
What if you took an airplane trip? The current of the jet stream is 100 miles per hour. It took you the same amount of time to travel 3000 miles without the jet stream as it did to travel 2000 miles with the jet stream. How could you determine the speed of the plane in calm air? After completing this Concept, you'll be able to solve real-world applications like this one using rational equations.
Guidance
A motion problem with no acceleration is described by the formula . These problems can involve the addition and subtraction of rational expressions.
Example A
Last weekend Nadia went canoeing on the Snake River. The current of the river is three miles per hour. It took Nadia the same amount of time to travel 12 miles downstream as it did to travel 3 miles upstream. Determine how fast Nadia’s canoe would travel in still water.
Solution
Define variables:
Let speed of the canoe in still water
Then, the speed of the canoe traveling downstream
the speed of the canoe traveling upstream
Construct a table:
Direction Distance (miles) Rate Time
Downstream 12
Upstream 3
Write an equation:
Since , we can say that .
Solve the equation:
Check: Upstream: ; downstream: . The answer checks out.
Example B
Peter rides his bicycle. When he pedals uphill he averages a speed of eight miles per hour, when he pedals downhill he averages 14 miles per hour. If the total distance he travels is 40 miles and the total time he rides is four hours, how long did he ride at each speed?
Solution
Define variables:
Let distance Peter bikes uphill at 8 miles per hour.
Construct a table:
Direction Distance (miles) Rate (mph) Time (hours)
Uphill 8
Downhill 14
Write an equation:
We know that .
Solve the equation:
Check: Uphill: ; downhill: . The answer checks out.
Example C
A group of friends decided to pool together and buy a birthday gift that cost $200. Later 12 of the friends decided not to participate any more. This meant that each person paid$15 more than their original share. How many people were in the group to begin with?
Solution
Define variables:
Let the number of friends in the original group.
Make a table:
Number of people Gift price Share amount
Original group 200
Later group 200
Write an equation:
Since each person’s share went up by $15 after 12 people refused to pay, we write the equation Solve the equation: The answer that makes sense is people. Check: Originally$200 shared among 20 people is $10 each. After 12 people leave,$200 shared among 8 people is $25 each. So each person pays$15 more. The answer checks out.
Watch this video for help with the Examples above.
Vocabulary
• Since , we can say that .
Guided Practice
Carrie is a runner. When she runs uphill she averages a speed of 2 miles per hour, when she runs downhill she averages 5 miles per hour. If she is running through the hilly streets of San Francisco and the total distance she travels is 13.5 miles and the total time she run 4.5 hours, how long did she run uphill and how long did she run downhill?
Solution
Define variables:
Let time Carrie runs uphill.
Let time Carrie runs downhill.
Let the distance Carrie runs uphill.
Construct a table:
Direction Distance (miles) Rate (mph) Time (hours)
Uphill 2
Downhill 5
Write an equation:
We know that .
Solve the equation:
Since is the distance Carrie to ran uphill, then is the distance Carrie ran downhill.
Check: Uphill: ; downhill: . The answer checks out.
Practice
For 1-4, solve for .
For 5-10, solve the following applications.
1. Juan jogs a certain distance and then walks a certain distance. When he jogs he averages 7 miles/hour and when he walks he averages 3.5 miles per hour. If he walks and jogs a total of 6 miles in a total of 1.2 hours, how far does he jog and how far does he walk?
2. A boat travels 60 miles downstream in the same time as it takes it to travel 40 miles upstream. The boat’s speed in still water is 20 miles per hour. Find the speed of the current.
3. Paul leaves San Diego driving at 50 miles per hour. Two hours later, his mother realizes that he forgot something and drives in the same direction at 70 miles per hour. How long does it take her to catch up to Paul?
4. On a trip, an airplane flies at a steady speed against the wind and on the return trip the airplane flies with the wind. The airplane takes the same amount of time to fly 300 miles against the wind as it takes to fly 420 miles with the wind. The wind is blowing at 30 miles per hour. What is the speed of the airplane when there is no wind?
5. A debt of $420 is shared equally by a group of friends. When five of the friends decide not to pay, the share of the other friends goes up by$25. How many friends were in the group originally?
6. A non-profit organization collected $2250 in equal donations from their members to share the cost of improving a park. If there were thirty more members, then each member could contribute$20 less. How many members does this organization have?
Vocabulary Language: English
Rational Expression
Rational Expression
A rational expression is a fraction with polynomials in the numerator and the denominator.
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## What are the 9 types of factoring?
Types of Factoring polynomials
• Greatest Common Factor (GCF)
• Grouping Method.
• Sum or difference in two cubes.
• Difference in two squares method.
• General trinomials.
• Trinomial method.
## How do you factor polynomials step by step?
1. Step 1: Identify the GCF of the polynomial.
2. Step 2: Divide the GCF out of every term of the polynomial.
3. Step 1: Identify the GCF of the polynomial.
4. Step 2: Divide the GCF out of every term of the polynomial.
5. Step 1: Identify the GCF of the polynomial.
6. Step 2: Divide the GCF out of every term of the polynomial.
What are the 5 rules of factoring?
Factoring Rules
• x2 – (r + s)x + rs = (x – r)(x – s)
• x2 + 2ax + a2 = (x + a)2 and x2 – 2ax + a2 = (x – a)2
• Difference of squares: a2 – b2 = (a – b)(a + b)
• Difference of cubes: a3 – b3 = (a – b)(a2 + ab + b2)
• a4 – b4 = (a – b)(a3 + a2b + ab2 + b3) = (a – b) [ a2(a + b) + b2(a + b) ] = (a – b)(a + b)(a2 + b2)
What are the 6 types of factoring?
The lesson will include the following six types of factoring:
• Group #1: Greatest Common Factor.
• Group #2: Grouping.
• Group #3: Difference in Two Squares.
• Group #4: Sum or Difference in Two Cubes.
• Group #5: Trinomials.
• Group #6: General Trinomials.
### How many formulas are there in Factorisation?
Factorisation Methods. There are four methods to factorise the algebraic expressions.
### What is the formula for factorization?
The general factorization formula is expressed as N = Xa × Yb × Zc. Here, X, Y, Z represent the factors of a factorized number.
What are the rules of factorization?
Basic Factorisation Formula
• a2 – b2 = (a – b)(a + b)
• (a + b)2 = a2 + 2ab + b2
• (a – b)2 = a2 – 2ab + b2
• a3 – b3 = (a – b)(a2 + ab + b2)
• a3 + b3 = (a + b)(a2 – ab + b2)
• (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca.
• (a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca.
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# How do you solve x+3y=6, 4x-5y=7 by graphing?
##### 1 Answer
Jun 14, 2018
$x = 3 , y = 1$
#### Explanation:
We seek a graphical solution of:
$\setminus \setminus x + 3 y = 6 \implies y = \frac{6 - x}{3}$
$4 x - 5 y = 7 \implies y = \frac{4 x - 7}{5}$
Thus we graph each of the functions on the same axis:
graph{(x+3y-6)(4x-5y-7)=0 [-10, 10, -5, 5]}
And we look for the point of intersection an we conclude the intersection is $\left(3 , 1\right)$ which we can readly validate via substitution
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# Search by Topic
#### Resources tagged with Limits similar to Fractional Calculus III:
Filter by: Content type:
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### There are 19 results
Broad Topics > Pre-Calculus and Calculus > Limits
### Fractional Calculus III
##### Stage: 5
Fractional calculus is a generalisation of ordinary calculus where you can differentiate n times when n is not a whole number.
### Fractional Calculus I
##### Stage: 5
You can differentiate and integrate n times but what if n is not a whole number? This generalisation of calculus was introduced and discussed on askNRICH by some school students.
### Fractional Calculus II
##### Stage: 5
Here explore some ideas of how the definitions and methods of calculus change if you integrate or differentiate n times when n is not a whole number.
### There's a Limit
##### Stage: 4 and 5 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
### Golden Eggs
##### Stage: 5 Challenge Level:
Find a connection between the shape of a special ellipse and an infinite string of nested square roots.
### Triangle Incircle Iteration
##### Stage: 4 Challenge Level:
Keep constructing triangles in the incircle of the previous triangle. What happens?
### Discrete Trends
##### Stage: 5 Challenge Level:
Find the maximum value of n to the power 1/n and prove that it is a maximum.
### Reciprocal Triangles
##### Stage: 5 Challenge Level:
Prove that the sum of the reciprocals of the first n triangular numbers gets closer and closer to 2 as n grows.
### Squaring the Circle and Circling the Square
##### Stage: 4 Challenge Level:
If you continue the pattern, can you predict what each of the following areas will be? Try to explain your prediction.
### Squareflake
##### Stage: 5 Challenge Level:
A finite area inside and infinite skin! You can paint the interior of this fractal with a small tin of paint but you could never get enough paint to paint the edge.
### Production Equation
##### Stage: 5 Challenge Level:
Each week a company produces X units and sells p per cent of its stock. How should the company plan its warehouse space?
### Over the Pole
##### Stage: 5 Challenge Level:
Two places are diametrically opposite each other on the same line of latitude. Compare the distances between them travelling along the line of latitude and travelling over the nearest pole.
### Resistance
##### Stage: 5 Challenge Level:
Find the equation from which to calculate the resistance of an infinite network of resistances.
### Witch of Agnesi
##### Stage: 5 Challenge Level:
Sketch the members of the family of graphs given by y = a^3/(x^2+a^2) for a=1, 2 and 3.
### Golden Fractions
##### Stage: 5 Challenge Level:
Find the link between a sequence of continued fractions and the ratio of succesive Fibonacci numbers.
### Converging Product
##### Stage: 5 Challenge Level:
In the limit you get the sum of an infinite geometric series. What about an infinite product (1+x)(1+x^2)(1+x^4)... ?
### Rain or Shine
##### Stage: 5 Challenge Level:
Predict future weather using the probability that tomorrow is wet given today is wet and the probability that tomorrow is wet given that today is dry.
### Exponential Trend
##### Stage: 5 Challenge Level:
Find all the turning points of y=x^{1/x} for x>0 and decide whether each is a maximum or minimum. Give a sketch of the graph.
### Spokes
##### Stage: 5 Challenge Level:
Draw three equal line segments in a unit circle to divide the circle into four parts of equal area.
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Tag: Fractions
# Get On My Fraction Level!
Many students have trouble with fractions. When I taught at a high school, my 10th and 11th grade students regularly had difficulty performing operations with fractions. As an 8th grade teacher, I’ve tried to help my students develop, refine, and maintain strong fraction skills. Don’t get me wrong: I don’t consider fluency in performing operations with fractions to be the most important skill for my students to have, but it’s certainly one that will contribute to their success at the high school level. With that in mind, here’s an approach I used this year to work on fractions.
After the warmup and overview of the day’s class some time in October, I pulled up a slide with four fraction multiplication problems. These first ones were relatively simple like $4\cdot \frac{1}{2}$. Students had little trouble performing the multiplication (yay!), and thankfully, students presented a number of different methods. The most common early responses were $\frac{4}{1}\cdot \frac{1}{2}=\frac{4}{2}=2$ and $4\div 2=2$. As I continued to present problems over the next few weeks, I added complications. Students noticed that simplifying often made the multiplication easier (e.g. $\frac{10}{5}\cdot 44$). The big breakthrough came when I presented a particularly annoying pair of fractions to multiply like $\frac{27}{7}\cdot \frac{14}{9}$. To this point, I had not pushed students to use a particular method; any simplifying they did came from them not me. Whoever offered the response of $\frac{378}{63}$ did not respond kindly to the question of whether that fraction could be simplified. By this point, students had been doing so much simplifying that it was no surprise to anyone that their lives would be easier if they found a way to simplify before multiplying. A brief discussion of the commutative property allowed a student to rewrite the multiplication as $\frac{27}{9}\cdot \frac{14}{7}$, which everyone in the classroom felt comfortable multiplying. It was a great moment of mathematical discovery.
As the weeks progressed, I continued to throw more and more challenging multiplication problems at them, and I also started to incorporate some addition, subtraction, and division. Students began feeling much more comfortable with fractions than they ever had before, even if they still weren’t the biggest fraction fans around. This fraction work paid off when we wrote equations of lines, and in general, I think it gave students some confidence in an area where they had so little before.
I definitely plan to continue using “Get On My Fraction Level” in my classes this coming school year. I’d like to find a way to incorporate more active participation. I might give students a weekly template to use each day when we do our fractions. I did that two years ago with scientific notation, and it worked pretty well. One big concern is time: with so many topics to cover, it’s difficult to carve out time to work on something that isn’t really an 8th grade standard. Having seen how working with fractions helped so many of my students grow, however, I will definitely find a way to incorporate regular fraction work into my lessons.
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# What is the dot product of 5i and 8j?
Dec 13, 2016
$0$
#### Explanation:
A property of the scaler (or dot) product is that it is zero if the vectors are perpendicular which $5 \underline{\hat{i}}$ and $8 \underline{\hat{j}}$ clearly are, so without any calculation we know the answer is zero.
We can show this if required:
Let $\vec{u} = 5 \underline{\hat{i}} + 0 \underline{\hat{j}} = \left(\begin{matrix}5 \\ 0\end{matrix}\right)$
and $\vec{v} = 0 \underline{\hat{i}} + 8 \underline{\hat{j}} = \left(\begin{matrix}0 \\ 8\end{matrix}\right)$
The scaler product is:
$\vec{u} \cdot \vec{v} = \left(\begin{matrix}5 \\ 0\end{matrix}\right) \cdot \left(\begin{matrix}0 \\ 8\end{matrix}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(5\right) \left(0\right) + \left(0\right) \left(8\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 0 + 0$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 0$
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# Rounding the Mean – Definition, Applications, and Examples
In this article, we will delve into the reasons for rounding the mean, the different rounding methods available, and the potential implications of rounding on data analysis and decision-making processes
## Definition
Rounding the mean refers to the process of adjusting the calculated average of a set of numbers to a specified degree of precision. The mean, or average, is obtained by summing all the values in a dataset and dividing by the number of data points.
However, the mean often results in a decimal value that may not be practical or necessary, depending on the context. Rounding the mean involves reducing the number of decimal places or digits after the decimal point to a more suitable level, typically to simplify the value or align it with the desired level of precision.
This adjustment facilitates easier interpretation, communication, and comparison of the mean within a given context or application. Rounding the mean is a common practice in various fields, including statistics, finance, research, and everyday data analysis.
## Exercise
### Example 1
Consider the following data set of exam scores: 78, 83, 89, 92, 95. Round the mean to the nearest whole number.
### Solution
To find the mean, sum all the scores and divide by the number of scores:
Mean = (78 + 83 + 89 + 92 + 95) / 5
Mean = 437 / 5
Mean = 87.4
Rounding the mean to the nearest whole number, we get: Rounded Mean = 87
### Example 2
Suppose we have the following dataset of rainfall measurements (in millimeters) for five cities: 12.5, 13.2, 11.8, 10.6, 14.1. Round the mean to one decimal place.
### Solution
To find the mean, sum all the measurements and divide by the number of measurements:
Mean = (12.5 + 13.2 + 11.8 + 10.6 + 14.1) / 5
Mean = 62.2 / 5
Mean = 12.44
Rounding the mean to one decimal place, we get: Rounded Mean = 12.4
### Example 3
Consider a dataset of temperatures (in degrees Celsius) recorded throughout a week: 22.3, 21.6, 24.1, 20.9, 23.5, 24.9. Round the mean to the nearest whole number.
### Solution
To find the mean, sum all the temperatures and divide by the number of temperatures:
Mean = (22.3 + 21.6 + 24.1 + 20.9 + 23.5 + 24.9) / 6
Mean = 137.3 / 6
Mean = 22.88
Rounding the mean to the nearest whole number, we get: Rounded Mean = 23
### Example 4
Suppose we have a dataset representing the ages of a group of individuals: 32, 28, 35, 40, 37, 30. Round the mean to the nearest multiple of 5.
### Solution
To find the mean, sum all the ages and divide by the number of ages:
Mean = (32 + 28 + 35 + 40 + 37 + 30) / 6
Mean = 202 / 6
Mean = 33.67
Rounding the mean to the nearest multiple of 5, we get: Rounded Mean = 35
Consider a dataset of product prices (in dollars): $12.99,$9.99, $14.99,$8.49, $11.99. Round the mean to two decimal places. ### Solution To find the mean, sum all the prices and divide by the number of prices: Mean = ($12.99 + $9.99 +$14.99 + $8.49 +$11.99) / 5
Mean = $58.45 / 5 Mean =$11.69
Rounding the mean to two decimal places, we get: Rounded Mean = \$11.69
## Applications
Rounding the mean finds applications in various fields where data analysis and statistical calculations are performed. Here are some examples of its applications in different domains:
### Finance and Economics
Rounding the mean is commonly employed in financial calculations, such as computing average returns on investments or determining mean values of financial indicators. It helps simplify monetary values and aligns them with standard currency denominations.
Rounding the mean in financial contexts aids in budgeting, forecasting, and making informed decisions based on rounded average values.
### Quality Control and Manufacturing
In industries that rely on quality control and manufacturing processes, rounding the mean can be useful for assessing product quality. By rounding the mean of measured parameters or quality metrics, such as dimensions or weights, it becomes easier to evaluate whether the product meets specific standards or falls within acceptable tolerances.
### Education and Grading
Rounding the mean is often employed in educational settings to calculate and report grades. It allows for simplifying complex grade calculations, making them more understandable for students and parents. Rounding the mean in grading ensures consistency and fairness in evaluating student performance across different assessments or assignments.
### Opinion Surveys and Polling
Rounding the mean is applied in analyzing survey data and polling results to present summary statistics. When reporting average ratings or responses, rounding the mean can help convey the overall sentiment or preference in a more concise and easily interpretable manner. It simplifies the representation of survey data while still providing meaningful insights.
### Public Opinion and Polling Analysis
In political or social research, rounding the mean can be useful for analyzing public opinion polls. It facilitates the reporting of average responses or aggregated sentiment, allowing for clear communication of trends or patterns in a way that is understandable to policymakers, analysts, and the general public.
### Market Research and Consumer Studies
Rounding the mean is often employed in market research to analyze and present consumer feedback or rating data. By rounding the mean of customer satisfaction scores or product ratings, market researchers can summarize and communicate the overall perception of a product or service in a more digestible format.
### Statistical Reporting and Data Visualization
Rounding the mean is essential in statistical reporting and data visualization. When presenting summary statistics or creating charts and graphs, rounding the mean allows for cleaner and less cluttered visual representations. It ensures that the reported or visualized mean values are more easily understood and provide a clearer overall picture of the data.
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# Lines in Geometry(Definition, Types & Examples)
This post is also available in: हिन्दी (Hindi)
In geometry, points and lines are the two fundamental concepts that you need to learn before learning about different shapes and sizes. A line is a one-dimensional figure, which has length but no width. It is made of a set of infinite points. A line can be extended in opposite directions infinitely.
Let’s understand what is line and its properties and also the different types of lines.
## What is Line?
A line is a straight one-dimensional figure that does not have a thickness, and it extends endlessly in both directions.
The above figure shows two lines. Both lines do not have any endpoint. The two arrows at each end signify that the line extends endlessly and is unending in both directions. The length of a line cannot be measured.
A line can also be defined as a set of collinear points connected in a one-dimensional plane.
Note:
• Three points are called colinear points when they lie on the same line.
• Two points are always collinear.
In the above figure,
• Points A, B, and C are collinear points.
• Points P, Q, and R are non-collinear points. (Points P, Q, and R are not collinear).
### Representing a Line in Geometry
Lines are generally denoted by the arc over the points. The line is drawn between points A and B, extending it both sides infinitely, and it is denoted by $\overleftrightarrow{\text{AB}}$.
### Properties of a Line
Following are the properties of a line.
• A line can be defined as a straight set of points that extend in opposite directions
• It has no ends in both directions(infinite)
• A line has an infinite length
• It has no thickness
• A line is a one-dimensional geometrical figure
• A line has only length, and it does not have any thickness
• A line is made up of an infinite number of points.
• If three or more points are placed on the same line, they are called collinear points.
• Intersecting lines will cross at only one point
• The lines used in the graphs are used to locate the points and have many more applications, the $x$−axis is the horizontal line, and the $y$−axis is the vertical line
## What is a Line Segment?
A part of a line is called a line segment. In other words, a line segment is a straight line with two endpoints. A line segment has a fixed length and can be measured using a ruler.
The above figure shows two line segments. Both line segments have endpoints in both directions. The length of a line segment can be measured.
### Representing a Line Segment in Geometry
Line segments are generally denoted by the arc without an arrow over the points. The line is drawn between points A and B and it is denoted by $\overline{\text{AB}}$.
In the above figure, $\overline{\text{AB}}$ is a line segment with two endpoints, A and B. A line segment can be measured, whereas a line cannot. The length of the line segment is the distance between the two endpoints, A and B.
Similarly, a $\overline{\text{CD}}$ is also a line segment.
## What is a Ray?
A part of a line having only one endpoint (starting point) but not having an endpoint is called a ray. Since a ray extends infinitely in one direction it has an infinite length and cannot be measured.
The above figure shows two rays. Both rays have one starting point but no endpoints. The length of a ray cannot be measured.
### Representing a Ray in Geometry
Rays are generally denoted by the arc with an arrow on one side over the points. The line is drawn between points A and B and it is denoted by $\overrightarrow{\text{AB}}$.
In the above figure, $\overrightarrow{\text{AB}}$ is a ray with one endpoint, A. Similar to a line, a ray cannot be measured.
Similarly, a $\overrightarrow{\text{CD}}$ is also a ray.
## Line, Line Segment, Ray – A Comparison
Following a comparison between a line, a line segment, and a ray.
## Types of Lines
There are various types of lines based on the property that holds. The different types of lines are intersecting and non-intersecting lines, parallel lines, perpendicular lines, etc.
### Horizontal Lines
When a line moves from left to right in a straight direction, it is a horizontal line.
### Vertical Lines
When a line runs from top to bottom in a straight direction, it is a vertical line.
### Intersecting Lines
Intersecting lines are the lines that formed when they cross each other at one point. The point of intersection is the point where the lines cross each other.
### Non-Intersecting Lines
The lines that do not touch or intersect each other are known as non-intersecting lines. The non-intersecting lines are called parallel lines, and the distance between them is constant.
### Parallel Lines
The lines are said to be parallel if they do not intersect each other and have the same distance between them from any point of the lines. Thus, parallel lines are non-intersecting lines having a constant distance.
The railway tracks are the best example of parallel lines.
### Perpendicular Lines
The lines are said to be perpendicular if they intersect each other at an angle of right angle $\left(90^{\circ} \right)$. The sides of the square and the rectangle are perpendicular lines. Perpendicular lines are intersecting lines with an angle of $90^{\circ}$.
### Transversal Line
The line that intersects two or more parallel lines is called the transversal. In the image given below, the dotted line intersecting the two lines is called a transversal.
## Practice Problems
1. State True or False
• The horizontal line moves from left to right
• The vertical line moves from left to right
• The horizontal line moves from down to up
• The vertical line moves from down to up
• Angle between two perpendicular lines is $0^{\circ}$
• Angle between two perpendicular lines is $90^{\circ}$
2. Define the following
• Line
• Line Segment
• Ray
• Collinear Points
3. What are intersecting lines?
4. What are parallel lines?
5. What is a transversal line?
## FAQs
### What is a line in math geometry?
A line is a straight one-dimensional figure that does not have a thickness, and it extends endlessly in both directions.
### What are the 4 characteristics of lines?
Following are the main four characteristics of lines.
a) It has no ends in both directions(infinite)
b) A line has an infinite length
c) It has no thickness
d) A line is a one-dimensional geometrical figure
### Why is the line the most important element?
The line is the most important element in geometry as it forms the basis of all the angles and figures (polygons).
### What is the angle between two perpendicular lines?
The angle between two perpendicular lines is $90^{\circ}$.
### What are the parallel lines?
Two lines are said to be parallel lines if they lie in the same plane and never meet.
## Conclusion
A line is a straight one-dimensional figure that does not have a thickness, and it extends endlessly in both directions. In geometry, the line is one of the fundamental concepts. It forms the basis of all the angles and figures (polygons).
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chain rule with square root December 24, 2020 – Posted in: Uncategorized
The Chain Rule. Watch the video for a couple of chain rule examples, or read on below: The formal definition of the chain rule: = 2(3x + 1) (3). Let's introduce a new derivative if f(x) = sin (x) then f '(x) = cos(x) It’s more traditional to rewrite it as: This function has many simpler components, like 625 and $\ds x^2$, and then there is that square root symbol, so the square root function $\ds \sqrt{x}=x^{1/2}$ is involved. Step 4: Simplify your work, if possible. The outer function is the square root $$y = \sqrt u ,$$ the inner function is the natural logarithm $$u = \ln x.$$ Hence, by the chain rule, Problem 4. = (sec2√x) ((½) X – ½). In this example, the negative sign is inside the second set of parentheses. Just ignore it, for now. Calculate the derivative of sin (1 + 2). dy/dx = d/dx (x2 + 1) = 2x, Step 4: Multiply the results of Step 2 and Step 3 according to the chain rule, and substitute for y in terms of x. In algebra, you found the slope of a line using the slope formula (slope = rise/run). Recognise u (always choose the inner-most expression, usually the part inside brackets, or under the square root sign). Step 1: Differentiate the outer function. Then differentiate (3 x +1). ) Step 3. And inside that is sin x. To make sure you ignore the inside, temporarily replace the inside function with the word stuff. This is the most important rule that allows to compute the derivative of the composition of two or more functions. This function has many simpler components, like 625 and $\ds x^2$, and then there is that square root symbol, so the square root function $\ds \sqrt{x}=x^{1/2}$ is involved. When you apply one function to the results of another function, you create a composition of functions. Multiply the result from Step 1 … Thread starter Chaim; Start date Dec 9, 2012; Tags chain function root rule square; Home. The outside function is the square root. Answer to: Find df / dt using the chain rule and direct substitution. To prove the chain rule let us go back to basics. – your inventory costs still increase. The outside function will always be the last operation you would perform if you were going to evaluate the function. More than two functions. The outer function in this example is 2x. To cover the answer again, click "Refresh" ("Reload").Do the problem yourself first! Find the Derivative Using Chain Rule - d/dx y = square root of sec(x^3) Rewrite as . Tip: No matter how complicated the function inside the square root is, you can differentiate it using repeated applications of the chain rule. n2 = number of future facilities. Assume that y is a function of x, and apply the chain rule to express each derivative with respect to x. SQRL is a single product rule when EOQ order batching with identical batch sizes wll be used across a set of invenrory facilities. D(4x) = 4, Step 3. √ X + 1 Differentiate using the chain rule, which states that is where and . Note: In (x 2 + 1) 5, x 2 + 1 is "inside" the 5th power, which is "outside." In fact, to differentiate multiplied constants you can ignore the constant while you are differentiating. Joined Jul 20, 2013 Messages 20. How do you find the derivative of this function using the Chain Rule: F(t)= 3rd square root of 1 + tan t I'm assuming that I might have to use the quotient rule along side of the Chain Rule. Step 2 Differentiate the inner function, using the table of derivatives. The general power rule is a special case of the chain rule, used to work power functions of the form y=[u(x)]n. The general power rule states that if y=[u(x)]n], then dy/dx = n[u(x)]n – 1u'(x). The chain-rule says that the derivative is: f' (g (x))*g' (x) We already know f (x) and g (x); so we just need to figure out f' (x) and g' (x) f" (x) = 1/sqrt (x) ; and ; g' (x) = 6x-1. . Find the Derivative Using Chain Rule - d/dx y = square root of sec(x^3) Rewrite as . ). The next step is to find dudx\displaystyle\frac{{{… Dec 9, 2012 #1 An example that my teacher did was: … Remember that a function raised to an exponent of -1 is equivalent to 1 over the function, and that an exponent of ½ is the same as a square root function. This indicates that the function f(x), the inner function, must be calculated before the value of g(x), the outer function, can be found. Tap for more steps... To apply the Chain Rule, set as . How would you work this out? Forums. = (2cot x (ln 2) (-csc2)x). This exponent behaves the same way as an integer exponent under differentiation – it is reduced by 1 to -½ and the term is multiplied by ½. Thus we compute as follows. However, the reality is the definition is sometimes long and cumbersome to work through (not to mention it’s easy to make errors). d/dx (sqrt (3x^2-x)) can be seen as d/dx (f (g (x)) where f (x) = sqrt (x) and g (x) = 3x^2-x. Step 2: Differentiate y(1/2) with respect to y. The derivative of 2x is 2x ln 2, so: In order to use the chain rule you have to identify an outer function and an inner function. Differentiate using the product rule. The question says find the derivative of square root x, for x>0 and use the formal definition of derivatives. D(cot 2)= (-csc2). Just ignore it, for now. This means that if g -- or any variable -- is the argument of f, the same form applies: In other words, we can really take the derivative of a function of an argument only with respect to that argument. Note: In (x2+ 1)5, x2+ 1 is "inside" the 5th power, which is "outside." For example, what is the derivative of the square root of (X 3 + 2X + 6) OR (X 3 + 2X + 6) ½? Thread starter sarahjohnson; Start date Jul 20, 2013; S. sarahjohnson New member. The 5th power therefore is outside. The inner function is the one inside the parentheses: x4 -37. Example problem: Differentiate the square root function sqrt(x2 + 1). Note: keep cotx in the equation, but just ignore the inner function for now. Note: keep 5x2 + 7x – 19 in the equation. What is called the chain rule states the following: "If f is a function of g and g is a function of x, then the derivative of f with respect to xis equal to the derivative of f(g) with respect to gtimes the derivative of g(x) with respect to x. The outer function in this example is “tan.” (Note: Leave the inner function in the equation (√x) but ignore that too for the moment) The derivative of tan x is sec2x, so: Here’s how to differentiate it with the chain rule: You start with the outside function (the square root), and differentiate that, IGNORING what’s inside. The Derivative tells us the slope of a function at any point.. Jul 20, 2013 #1 Find the derivative of the function. Step 1 Differentiate the outer function. In this case, the outer function is the sine function. Step 4 Rewrite the equation and simplify, if possible. We take the derivative from outside to inside. Chain Rule in Derivatives: The Chain rule is a rule in calculus for differentiating the compositions of two or more functions. Here’s how to differentiate it with the chain rule: You start with the outside function (the square root), and differentiate that, IGNORING what’s inside. There are rules we can follow to find many derivatives.. For example: The slope of a constant value (like 3) is always 0; The slope of a line like 2x is 2, or 3x is 3 etc; and so on. $$\root \of{ v + \root \of u}$$ I know that in order to derive a square root function we apply this : $$(\root \of u) ' = \frac{u '}{2\root \of u}$$ But I really can't find a way on how to do the first two function derivatives, I've heard about the chain rule, but we didn't use it yet . Notice that this function will require both the product rule and the chain rule. Therefore, since the limit of a product is equal to the product of the limits (Lesson 2), and by definition of the derivative: Please make a donation to keep TheMathPage online.Even $1 will help. Tap for more steps... To apply the Chain Rule, set as . To differentiate the composition of functions, the chain rule breaks down the calculation of the derivative into a series of simple steps. f’ = ½ (x2 – 4x + 2)½ – 1(2x – 4) Here, you’ll be studying the slope of a curve.The slope of a curve isn’t as easy to calculate as the slope of a line, because the slope is different at every point of the curve (and there are technically an infinite amount of points on the curve! Although the memoir it was first found in contained various mistakes, it is apparent that he used chain rule in order to differentiate a polynomial inside of a square root. Problem 1. Therefore, since the derivative of x4 − 2 is 4x3. Let’s take a look at some examples of the Chain Rule. Click HERE to return to the list of problems. = f’ = ½ (x2-4x + 2) – ½(2x – 4), Step 4: (Optional)Rewrite using algebra: Chain Rule in Derivatives: The Chain rule is a rule in calculus for differentiating the compositions of two or more functions. The outside function is sin x. Step 3: Differentiate the inner function. The chain rule is one of the toughest topics in Calculus and so don't feel bad if you're having trouble with it. Learn how to find the derivative of a function using the chain rule. We take the derivative from outside to inside. Letting z = arccos(x) (so that we're looking for dz/dx, the derivative of arccosine), we get (d/dx)(cos(z))) = 1, so ... Where did the square root come from? Square Root Law was shown in 1976 by David Maister (then at Harvard Business School) to apply to a set of inventory facilities facing identical demand rates. Inside that is (1 + a 2nd power). To differentiate a more complicated square root function in calculus, use the chain rule. Sample problem: Differentiate y = 7 tan √x using the chain rule. Step by step process would be much appreciated so that I can learn and understand how to do these kinds of problems. D(sin(4x)) = cos(4x). This only tells part of the story. sin x is inside the 3rd power, which is outside. The outer function is √, which is also the same as the rational exponent ½. You would first evaluate sin x, and then take its 3rd power. d/dy y(½) = (½) y(-½), Step 3: Differentiate y with respect to x. The derivative of x4 – 37 is 4x(4-1) – 0, which is also 4x3. When differentiating functions with the chain rule, it helps to think of our function as "layered," remembering that we must differentiate one layer at a time, from the outermost layer to the innermost layer, and multiply these results.. More commonly, you’ll see e raised to a polynomial or other more complicated function. The online Chain rule derivatives calculator computes a derivative of a given function with respect to a variable x using analytical differentiation. The chain rule can be used to differentiate many functions that have a number raised to a power. We’re using a special case of the chain rule that I call the general power rule. When we take the outside derivative, we do not change what is inside. When we write f(g(x)), f is outside g. We take the derivative of f with respect to g first. Here’s a problem that we can use it on. However, the technique can be applied to a wide variety of functions with any outer exponential function (like x32 or x99. ", Therefore according to the chain rule, the derivative of. 7 (sec2√x) ((1/2) X – ½). The Chain Rule is thought to have first originated from the German mathematician Gottfried W. Leibniz. i absent from chain rule class and hope someone will help me with these question. In this example, cos(4x)(4) can’t really be simplified, but a more traditional way of writing cos(4x)(4) is 4cos(4x). Example 1 Find the derivative f '(x), if f is given by f(x) = 4 cos (5x - 2) Solution to Example 1 Let u = 5x - 2 and f(u) = 4 cos u, hence du / dx = 5 and df / du = - 4 sin u We now use the chain rule The derivative of sin is cos, so: This section explains how to differentiate the function y = sin(4x) using the chain rule. We will have the ratio, But the change in x affects f because it depends on g. We will have. To see the answer, pass your mouse over the colored area. Find dy/dr y=r/( square root of r^2+8) Use to rewrite as . Here, our outer layer would be the square root, while the inner layer would be the quotient of a polynomial. (This is the sine of x5.) Problem 3. Step 1. Tap for more steps... To apply the Chain Rule, set as . Because it's so tough I've divided up the chain rule to a bunch of sort of sub-topics and I want to deal with a bunch of special cases of the chain rule, and this one is going to be called the general power rule. If we now let g(x) be the argument of f, then f will be a function of g. That is: The derivative of f with respect to its argument (which in this case is x) is equal to 5 times the 4th power of the argument. Problem 5. The derivative of cot x is -csc2, so: √ (x4 – 37) equals (x4 – 37) 1/2, which when differentiated (outer function only!) A simpler form of the rule states if y – un, then y = nun – 1*u’. Calculate the derivative of (x4 − 3x2+ 4)2/3. It will be the product of those ratios. Need help with a homework or test question? The more times you apply the chain rule to different problems, the easier it becomes to recognize how to apply the rule. In this example, the inner function is 4x. Step 1 7 (sec2√x) ((½) X – ½) = In this problem we have to use the Power Rule and the Chain Rule.. We begin by converting the radical(square root) to it exponential form. That’s why mathematicians developed a series of shortcuts, or rules for derivatives, like the general power rule. The derivative of a function of a function, The derivative of a function of a function. The Square Root Law states that total safety stock can be approximated by multiplying the total inventory by the square root of the number of future warehouse locations divided by the current number. Step 3: Combine your results from Step 1 2(3x+1) and Step 2 (3). Finding Slopes. It might seem overwhelming that there’s a multitude of rules for differentiation, but you can think of it like this; there’s really only one rule for differentiation, and that’s using the definition of a limit. This rule-of-thumb only covers safety stock and not cycle stock. 7 (sec2√x) ((½) 1/X½) = Problem 9. $$\root \of{ v + \root \of u}$$ I know that in order to derive a square root function we apply this : $$(\root \of u) ' = \frac{u '}{2\root \of u}$$ But I really can't find a way on how to do the first two function derivatives, I've heard about the chain rule, but we didn't use it yet . Whenever I’m differentiating a function that involves the square root I usually rewrite it as rising to the ½ power. we can really take the derivative of a function of an argument only with respect to that argument. Guillaume de l'Hôpital, a French mathematician, also has traces of the Step 4 ANSWER: ½ • (X 3 + 2X + 6)-½ • (3X 2 + 2) Another example will illustrate the versatility of the chain rule. Derivative Rules. Add the constant you dropped back into the equation. The chain rule provides that the D x (sqrt(m(x))) is the product of the derivative of the outer (square root) function evaluated at m(x) times the derivative of the inner function m at x. Use the chain rule and substitute f ' (x) = (df / du) (du / dx) = (1 / u) (2x + 1) = (2x + 1) / (x2 + x) Exercises On Chain Rule Use the chain rule to find the first derivative to each of the functions. Oct 2011 155 0. At first glance, differentiating the function y = sin(4x) may look confusing. = cos(4x)(4). Get an answer for 'Using the chain rule, differentiate the function f(x)=square root(5+16x-(4x)squared). The derivative of ex is ex, so: The online Chain rule derivatives calculator computes a derivative of a given function with respect to a variable x using analytical differentiation. For example, what is the derivative of the square root of (X 3 + 2X + 6) OR (X 3 + 2X + 6) ½? where y is just a label you use to represent part of the function, such as that inside the square root. For an example, let the composite function be y = √(x4 – 37). The Chain rule of derivatives is a direct consequence of differentiation. Thus, = 2 (3 x +1) (3) = 6 (3 x +1) . Calculus. Therefore, the derivative is. To differentiate a more complicated square root function in calculus, use the chain rule. d/dx sqrt(x) = d/dx x(1/2) = (1/2) x(-½). = e5x2 + 7x – 13(10x + 7), Step 4 Rewrite the equation and simplify, if possible. It provides exact volatilities if the volatilities are based on lognormal returns. Solution. Thus we compute as follows. Tap for more steps... To apply the Chain Rule, set as . Differentiate using the Power Rule which states that is where . The inside function is x5 -- you would evaluate that last. Step 1: Rewrite the square root to the power of ½: In this example, the inner function is 3x + 1. ( The outer layer is the square'' and the inner layer is (3 x +1) . In this example, the outer function is ex. The results are then combined to give the final result as follows: dF/dx = dF/dy * dy/dx Solution. As for the derivative of. Differentiation Using the Chain Rule. Think about the triangle shown to the right. In algebra, you found the slope of a line using the slope formula (slope = rise/run). Differentiate both sides of the equation. Chain Rule Calculator is a free online tool that displays the derivative value for the given function. ... Differentiate using the chain rule, which states that is where and . Find dy/dr y=r/( square root of r^2+8) Use to rewrite as . Note that I’m using D here to indicate taking the derivative. BYJU’S online chain rule calculator tool makes the calculation faster, and it displays the derivatives and the indefinite integral in a fraction of seconds. Problem 2. (2x – 4) / 2√(x2 – 4x + 2). cos x = cot x. Differentiate y equals x² times the square root of x² minus 9. University Math Help. derivative of square root x without using chain rule? According to this rule, if the fluctuations in a stochastic process are independent of each other, then the volatility will increase by square root of time. Step 2:Differentiate the outer function first. The square root law of inventory management is often presented as a formula, but little explanation is ever given about why your inventory costs go up when you increase the number of warehouse locations. D(3x + 1) = 3. X2 = (X1) * √ (n2/n1) n1 = number of existing facilities. Multiplying 4x3 by ½(x4 – 37)(-½) results in 2x3(x4 – 37)(-½), which when worked out is 2x3/(x4 – 37)(-½) or 2x3/√(x4 – 37). D(2cot x) = 2cot x (ln 2), Step 2 Differentiate the inner function, which is Calculate the derivative of sin5x. 2. We started off by saying cos(z) = x. The derivative of with respect to is . The obvious question is: can we compute the derivative using the derivatives of the constituents$\ds 625-x^2$and$\ds \sqrt{x}$? Functions that contain multiplied constants (such as y= 9 cos √x where “9” is the multiplied constant) don’t need to be differentiated using the product rule. The obvious question is: can we compute the derivative using the derivatives of the constituents$\ds 625-x^2$and$\ds \sqrt{x}$? Example problem: Differentiate y = 2cot x using the chain rule. Tip You can also use this rule to differentiate natural and common base 10 logarithms (D(ln x) = (1/x) and D(log x) = (1/x) log e. Multiplied constants add another layer of complexity to differentiating with the chain rule. Then we need to re-express y\displaystyle{y}yin terms of u\displaystyle{u}u. This rule states that the system-wide total safety stock is directly related to the square root of the number of warehouses. Square Root Law was shown in 1976 by David Maister (then at Harvard Business School) to apply to a set of inventory facilities facing identical demand rates. Step 1: Write the function as (x2+1)(½). Your first 30 minutes with a Chegg tutor is free! $$f(x) = \blue{e^{-x^2}}\red{\sin(x^3)}$$ Step 2. The derivative of y2with respect to y is 2y. 2x * (½) y(-½) = x(x2 + 1)(-½), Step 5: Simplify your answer by writing it in terms of square roots. Thank's for your time . -2cot x(ln 2) (csc2 x), Another way of writing a square root is as an exponent of ½. Step 4: Multiply Step 3 by the outer function’s derivative. Chain Rule Calculator is a free online tool that displays the derivative value for the given function. Therefore sqrt(x) differentiates as follows: The chain rule can also help us find other derivatives. The chain rule provides that the D x (sqrt(m(x))) is the product of the derivative of the outer (square root) function evaluated at m(x) times the derivative of the inner function m at x. The chain rule can be extended to more than two functions. you would first have to evaluate x2+ 1. D(5x2 + 7x – 19) = (10x + 7), Step 3. The Practically Cheating Calculus Handbook, The Practically Cheating Statistics Handbook, Chain rule examples: Exponential Functions, https://www.calculushowto.com/derivatives/chain-rule-examples/. What is the derivative of y = sin3x ? Note: keep 3x + 1 in the equation. what is the derivative of the square root?' y = (x2 – 4x + 2)½, Step 2: Figure out the derivative for the “inside” part of the function, which is (x2 – 4x + 2). cot x. Note: keep 4x in the equation but ignore it, for now. dF/dx = dF/dy * dy/dx With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Maybe you mean you've already done what I'm about to suggest: it's a lot easier to avoid the chain rule entirely and write$\sqrt{3x}$as$\sqrt{3}*\sqrt{x}=\sqrt{3}*x^{1/2}\$, unless someone tells you you have to use the chain rule… Get an answer for 'Using the chain rule, differentiate the function f(x)=square root(5+16x-(4x)squared). Step 3 (Optional) Factor the derivative. Then you would take its 5th power. Now, the derivative of the 3rd power -- of g3 -- is 3g2. In this example, no simplification is necessary, but it’s more traditional to write the equation like this: Differentiate using the Power Rule which states that is where . Here are useful rules to help you work out the derivatives of many functions (with examples below). x(x2 + 1)(-½) = x/sqrt(x2 + 1). Step 3. And, this rule-of-thumb is only meant for the safety stock you hold because of demand variability. 7 (sec2√x) / 2√x. There are rules we can follow to find many derivatives.. For example: The slope of a constant value (like 3) is always 0; The slope of a line like 2x is 2, or 3x is 3 etc; and so on. Tip: The hardest part of using the general power rule is recognizing when you’re essentially skipping the middle steps of working the definition of the limit and going straight to the solution. Step 2: Differentiate the inner function. The chain rule is a method for finding the derivative of composite functions, or functions that are made by combining one or more functions.An example of one of these types of functions is $$f(x) = (1 + x)^2$$ which is formed by taking the function $$1+x$$ and plugging it into the function $$x^2$$. To make sure you ignore the inside, temporarily replace the inside function with the word stuff. For example, to differentiate 3. Differentiating using the chain rule usually involves a little intuition. D(3x + 1)2 = 2(3x + 1)2-1 = 2(3x + 1). D(e5x2 + 7x – 19) = e5x2 + 7x – 19. 5x2 + 7x – 19. 22.3 Derivatives of inverse sine and inverse cosine func-tions The formula for the derivative of an inverse function can be used to obtain the following derivative formulas for sin-1 … y = 7 x + 7 x + 7 x \(\displaystyle \displaystyle y \ … thanks! Here’s a problem that we can use it on. If you’ve studied algebra. SQRL is a single product rule when EOQ order batching with identical batch sizes wll be used across a set of invenrory facilities. ... Differentiate using the chain rule, which states that is where and . Example 2. Label the function inside the square root as y, i.e., y = x2+1. What function is f, that is, what is outside, and what is g, which is inside? Got asked what would happen to inventory when the number of stocking locations change. Let’s take a look at some examples of the Chain Rule. Combine the results from Step 1 (e5x2 + 7x – 19) and Step 2 (10x + 7). Differentiate the square'' first, leaving (3 x +1) unchanged. For example, let’s say you had the functions: The composition g (f (x)), which is also written as (g ∘ f) (x), would be (x2-3)2. ANSWER: ½ • (X 3 + 2X + 6)-½ • (3X 2 + 2) Another example will illustrate the versatility of the chain rule. Assume that y is a function of x. y = y(x). Step 5 Rewrite the equation and simplify, if possible. Let us now take the limit as Δx approaches 0. Therefore, accepting for the moment that the derivative of sin x is cos x (Lesson 12), the derivative of sin3x -- from outside to inside -- is. Derivative Rules. If you’ve studied algebra. Even if you subtract the obvious suspects that would make your costs rise – extra rent, extra staffing, upkeep of multiple locations, etc. The derivative of ex is ex, but you’ll rarely see that simple form of e in calculus. f'(x2 – 4x + 2)= 2x – 4), Step 3: Rewrite the equation to the form of the general power rule (in other words, write the general power rule out, substituting in your function in the right places). We will have the ratio, Again, since g is a function of x, then when x changes by an amount Δx, g will change by an amount Δg. We’re using a special case of the chain rule that I call the general power rule. √x. To find the derivative of the left-hand side we need the chain rule. Differentiate using the chain rule, which states that is where and . Here, you’ll be studying the slope of a curve.The slope of a curve isn’t as easy to calculate as the slope of a line, because the slope is different at every point of the curve (and there are technically an infinite amount of points on the curve! To apply the chain rule to the square root of a function, you will first need to find the derivative of the general square root function: f ( g ) = g = g 1 2 {\displaystyle f(g)={\sqrt {g}}=g^{\frac {1}{2}}} Multiply the result from Step 1 … 2x. Finding Slopes. Example 5. Combine the results from Step 1 (2cot x) (ln 2) and Step 2 ((-csc2)). In this example, 2(3x +1) (3) can be simplified to 6(3x + 1). BYJU’S online chain rule calculator tool makes the calculation faster, and it displays the derivatives and the indefinite integral in a fraction of seconds. Differentiating functions that contain e — like e5x2 + 7x-19 — is possible with the chain rule. To decide which function is outside, how would you evaluate that? Differentiate algebraic and trigonometric equations, rate of change, stationary points, nature, curve sketching, and equation of tangent in Higher Maths. Thank's for your time . The derivative of with respect to is . SOLUTION 1 : Differentiate . Step 2 Differentiate the inner function, which is X1 = existing inventory. #y=sqrt(x-1)=(x-1)^(1/2)# To find the derivative of a function of a function, we need to use the Chain Rule: (dy)/(dx) = (dy)/(du) (du)/(dx) This means we need to. Then we differentiate y\displaystyle{y}y (with respect to u\displaystyle{u}u), then we re-express everything in terms of x\displaystyle{x}x. This has the form f (g(x)). Then when the value of g changes by an amount Δg, the value of f will change by an amount Δf. I'm not sure what you mean by "done by power rule". what is the derivative of the square root?' We haven't learned chain rule yet so I can not possibly use that. Recognise u\displaystyle{u}u(always choose the inner-most expression, usually the part inside brackets, or under the square root sign). f(x) = (sqrtx + x)^1/2 can anyone help me? Then we need to re-express y in terms of u. I thought for a minute and remembered a quick estimate. whose derivative is −x−2 ; (Problem 4, Lesson 4). For any argument g of the square root function. The outside function will always be the last operation you would perform if you were going to evaluate the function. Here, our outer layer would be the square root, while the inner layer would be the quotient of a polynomial. For example, let. Once you’ve performed a few of these differentiations, you’ll get to recognize those functions that use this particular rule. What’s needed is a simpler, more intuitive approach! g is x4 − 2 because that is inside the square root function, which is f. The derivative of the square root is given in the Example of Lesson 6. Identify the factors in the function. This is the 3rd power of sin x. To decide which function is outside, decide which you would have to evaluate last. Here, g is x4 − 2. Let f be a function of g, which in turn is a function of x, so that we have f(g(x)). In this case, the outer function is x2. Combine your results from Step 1 (cos(4x)) and Step 2 (4). D(tan √x) = sec2 √x, Step 2 Differentiate the inner function, which is Chain Rule Problem with multiple square roots. : (x + 1)½ is the outer function and x + 1 is the inner function. D(√x) = (1/2) X-½. We then multiply by … Using chain rule on a square root function. Then the change in g(x) -- Δg -- will also approach 0. Calculate the derivative of sin x5. C. Chaim. This section shows how to differentiate the function y = 3x + 12 using the chain rule. The number e (Euler’s number), equivalent to about 2.71828 is a mathematical constant and the base of many natural logarithms. Include the derivative you figured out in Step 1: We then multiply by the derivative of what is inside. This is a way of breaking down a complicated function into simpler parts to differentiate it piece by piece. Volatility and VaR can be scaled using the square root of time rule. However, the technique can be applied to any similar function with a sine, cosine or tangent. Combine the results from Step 1 (sec2 √x) and Step 2 ((½) X – ½). The chain rule can also help us find other derivatives. The results are then combined to give the final result as follows: The Square Root Law states that total safety stock can be approximated by multiplying the total inventory by the square root of the number of future warehouse locations divided by the current number. Let's introduce a new derivative if f(x) = sin (x) then f '(x) = cos(x) Step 1 Differentiate the outer function, using the table of derivatives. (10x + 7) e5x2 + 7x – 19. Whenever I’m differentiating a function that involves the square root I usually rewrite it as rising to the ½ power. Here are useful rules to help you work out the derivatives of many functions (with examples below). M using D here to return to the results of another function, chain rule with square root found slope. See the answer, pass your mouse over the colored area ( cos 4x! Chain rule yet so I can learn and understand how to do these kinds of problems 1 Write! 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Function to the square root function sqrt ( x2 + 1 ) applied! ) = ( 2cot x ( x2 + 1 you 're having trouble with.... Study, you found the slope formula ( slope = rise/run ) I ’ m D! 1 in the equation and simplify, if possible analytical differentiation of shortcuts or! S why mathematicians developed a series of simple steps , Therefore according to the results are then combined give! = 7 tan √x using the power rule rule problem with multiple square roots evaluate sin x is,. Any argument g of the chain rule, the inner layer is inside... You found the slope of a given function with the word stuff of differentiations... When you apply the chain rule breaks down the calculation of the chain rule let us take... = √ ( x4 – 37 ) calculus and so do n't feel bad if you going... Of functions un, then y = 3x + 1 ) 2-1 2! Express each derivative with respect to y is a way chain rule with square root breaking down a complicated function into simpler parts differentiate! Your questions from an expert in the evaluation and this is the one inside the square root I rewrite! The part inside brackets, or rules for derivatives, like the general power rule of ). Complicated square root function in calculus for differentiating the compositions of two more! However, the negative sign is inside to a variable x using analytical differentiation example question: what inside... ( x-1 ) ^ ( 1/2 ) # Finding Slopes − 2 is 4x3 and is. Of time rule that y is a single product rule when EOQ order batching with batch... Then we need to re-express y\displaystyle { y } yin terms of u always... Different problems, the outer function is outside, how would you evaluate that last Add! Example problem: differentiate y equals x² times the square root, the! New member quick estimate require both the product rule when EOQ order batching with identical sizes! Just ignore the inside, temporarily replace the inside, temporarily replace the,. 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Only meant for the safety stock is directly related to the ½.!, 2012 ; Tags chain function root rule square ; Home of many functions that are square roots use chain.: the chain rule, set as the 3rd power, which when differentiated ( outer function is,... X² times the square root is the derivative of a function of a function at any point ( the function! Says find the derivative of a function of a function at any point Chegg Study, you found the of. – 4x + 2 ) ( 3 x +1 ) the square root time! The ½ power ) X-½ step 3 by the outer function and an function... X using analytical differentiation by step process would be the square root? = *. Invenrory facilities because it depends on g. we will have compositions of two or more functions g the... ( 3x+1 ) and step 2 ( 10x + 7 ) 're having trouble with.... 6 ( 3x + 1 ) 2 = 2 ( 10x + )! Negative sign is inside the second set of parentheses use it on also 4x3 take the limit as approaches. Whose derivative is −x−2 ; ( problem 4, step 3 amount Δg, the technique can also us! You would first evaluate sin x is -csc2, so: D ( 5x2 + 7x – 19 the! Only! explains how to do these kinds of problems to re-express y in terms of u\displaystyle u. 5Th power, which states that the system-wide total safety stock and not cycle stock the! ( ( ½ ) or ½ ( x4 – 37 ) 1/2, which ! Once you ’ ve performed a few of these differentiations, you found the slope of a line using chain. How to find the derivative tells us the slope of a line using the chain rule usually involves a intuition... Shows how to do these kinds of problems 9, 2012 ; Tags chain function root rule square Home... To any similar function with respect to that argument re-express y in terms u\displaystyle. The volatilities are based on lognormal chain rule with square root calculus, use the chain rule class and someone! ( square root is the derivative of a given function with a sine, or... ( 1/2 ) X-½ example that my teacher did was: … chain rule wll be across! ( x2+ 1 ), 2013 ; S. sarahjohnson New member is g, which is 4x3... From an expert in the equation but ignore it, for now 30 minutes with sine! You work out the derivatives of many functions that use this particular rule ) and step differentiate!
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# How do you solve the following linear system 3x-2y=8, 2x-3y=-6 ?
Jun 9, 2016
(x,y)=color(green)(""(7.2,6.8))
#### Explanation:
Given
[1]$\textcolor{w h i t e}{\text{XXX}} 3 x - 2 y = 8$
[2]$\textcolor{w h i t e}{\text{XXX}} 2 x - 3 y = - 6$
Multiplying [1] by $3$ and [2] by $2$
[3]$=$[1]$\times 3 \textcolor{w h i t e}{\text{XXX")9x-6y=color(white)("XX}} 24$
[4]$=$[2]$\times 2 \textcolor{w h i t e}{\text{XXX}} \underline{4 x - 6 y = - 12}$
Subtracting [4] from [5]
[6]$=$[5]$-$[4]$\textcolor{w h i t e}{\text{XXX}} 5 x = 36$
Dividing both sides by $5$
[7]$=$[6]$\div 5 \textcolor{w h i t e}{\text{XXX}} x = 7.2$
Substituting $7.2$ for $x$ in [1]
[8]$\textcolor{w h i t e}{\text{XXXXXX}} 3 \left(7.2\right) - 2 y = 8$
Simplifying
[9]$\textcolor{w h i t e}{\text{XXXXXX}} 21.6 - 2 y = 8$
[10]$\textcolor{w h i t e}{\text{XXXXXX}} 2 y = 13.6$
[11]$\textcolor{w h i t e}{\text{XXXXXX}} y = 6.8$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
For verification purposes substitute $\left(x , y\right) = \left(7.2 , 6.8\right)$ into [2]
$\textcolor{w h i t e}{\text{XXXXXXX}} 2 x - 3 y$
$\textcolor{w h i t e}{\text{XXXXXXXXXX}} = 2 \left(7.2\right) - 3 \left(6.8\right)$
$\textcolor{w h i t e}{\text{XXXXXXXXXX}} = 14.2 - 20.4$
$\textcolor{w h i t e}{\text{XXXXXXXXXX")=-6color(white)("XXX}}$as given.
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Rules of inequalities I
Solving Inequalities Rules
A mathematical expression which involves symbols >, <, ≥ and ≤ is an inequality. To solve such we need go find a range of values that an unknown variable can take and then satisfy the in-equality. The rules for solving such are that, if the same number is added to both sides of an in-equal form, the form remains true.
If the same number is subtracted from both sides of the inequality, the in-equality remains true. If multiplication is done by the positive number, the in-equality remains true. But if multiplication or division of such an form is done by a negative number, then it is no longer true. In fact, the relation becomes reversed. These are explained briefly with examples below.
Rules for Inequalities
While solving an such problem there are inequality rules that needs to be followed. If the form involves addition or subtraction then we add or subtract by the same number both sides in order to have the variable to one side.
For example take x – 5 > 12. Here we have -5 so we add +5 both sides and get x > 17. Take another example x +6 > 15. Here we subtract 6 both sides. We get, x > 9.
If it involves multiplying then we have to divide both sides by the same number. For example take 4x >12.
So divide by 4 both sides we get x > 3. If it involves division then we multiply both sides.
x/4 > 12. Here multiply by 4 both sides. We get x > 48.
When Multiplication is done by a negative number the sign of the in-equality should be flipped x/-3 > 12. Here we times both by -3. So x < -36. The sign of the in-equality is flipped.
The > sign is now changed to <. Similarly when we divide by a negative number we flip the sign. Take -5x > 35. Divide by -5 both sides and flip the relation. X < -7. Here also the > sign is flipped t the < sign.
While graphing inequalities if the relation has a < or > sign when we have to draw a dotted line. This indicates that the line is not a part of the solution. If the in-equality is less than, or equal to and or greater than / equal to then we use a solid line for graphing. This indicates that the line contains solutions for the inequality. These are the rules of Solving Inequalities
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Question
# A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate Rs. 5 per meter sheet, sheet being 2 m wide.
Hint – In this question first find out, total surface area of the closed iron tank which is given as $S.A = 2\left( {lb + bh + hl} \right)$ where symbols have their usual meaning, later on divide this area to the width of the sheet to get the required length of the sheet, so use this concept to reach the solution of the question.
Given data
Length (l) of iron tank = 12m
Breath (b) of iron tank = 9m
Height (h) of iron tank = 4m
So, it is given that the iron tank is closed (see figure) therefore it has total six walls (see figure), and so the total surface area (S.A) of these walls is given as
$S.A = 2\left( {lb + bh + hl} \right)$
So, substitute the given values in above equation we have,
$\Rightarrow S.A = 2\left( {12 \times 9 + 9 \times 4 + 4 \times 12} \right) \\ \Rightarrow S.A = 2\left( {108 + 36 + 48} \right) = 2\left( {192} \right) = 384{m^2} \\$
Now it is given that the sheet is 2 m wide.
So, length of the sheet required to make a closed iron tank $= \dfrac{{384{\text{ }}{{\text{m}}^2}}}{{2{\text{ m}}}} = 192{\text{ m}}$
Now it is also given that the cost of iron sheet is Rs. 5 per meter sheet.
So, the total cost of iron sheet to make a closed iron tank $= 192 \times 5 = 960{\text{ Rs}}{\text{.}}$
So, this is the required answer.
Note – in such types of questions the key concept we have to remember is that always recall the formula of total surface area which is stated above, then use this formula to calculate the surface area of the iron tank, then calculate the length of sheet required to make a closed iron tank by dividing the width of sheet to the surface area of the iron tank, then multiply this value to the cost of per sheet which is given, we will get the required cost of iron sheet to make a closed iron tank.
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# Barbara Palazzo 9/24/09 ED 637 Title: Multiplication Commutative operations in math Two well-known examples of commutative The addition
of real numbers, is commutative; For example 3 + 2 = 2+ 3, since both expressions equal 6. y+z = z+y The multiplication of real numbers, is commutative For example, 3 × 5 = 5 × 3, since both expressions equal 15. yz = zy 2 +2+2=6 3 sets of 2 =6 3 Times 2 =6 3+3=6 2 sets of 3 =6 2 Times 3 =6 OO +OO+OO = 6
OO OO = 6 OO
OOOOOO circle 6 in the picture Lets work with the manipulatives: Place 12 chips on your board. Line them up like this: 1 OOOO 2 OOOO 3 OOOO We have 3 sets of 4 Or 4+4+4 Or 3x4
Rotate them: 1 OO 2 OO 3 OO 4 OO 5 OO 6 OO Now what do we have? We have 6 sets of 2 Or 6x2 Or 2 +2 +2 +2 +2 . 1 OOOOOO 2 OOOOOO Or 6 +6 Now lets turn our chips so they are facing down.Now let’s rotate this set 1 OOO 2 OOO 3 OOO 4 OOO We now have 4 sets of 3 Or 3+3+3+3 Or 4x3 It goes both ways: 3 x 4 = 12 and 4 x 3 = 12 Lets practice one more time: How would you lay out your board if I gave you this example: 2x6 What other ways can we say this? We have 2 sets of 6 Use your manipulatives to show me.
Tens / one 1 / 3 X / 7 2 / 1 7 / 0 Now we add up the columns that we have in both and we have our answer Tens / one 1 / 3 X / 7 .Using base 10 manipulative: 13 X7 or 7 x13 These problems will add to the same number. 13 sets of 7 will add to the same number as 7 sets of 13. OOO + OOO + OOO + OOO + OOO + OOO + OOO = 21 3 + 3+3+3+3+3+3 =21 7 x 3 = 21 Let’s regroup – how many tens do we have? 2 tens How many ones do we have? 1 Tens / one 1 / 3 X / 7 2 / 1 We now have to work on our tens group? What do we have in this group? We have 7 sets of one ten = 7 x 10 IIIIIII (Base 10 sticks) We have 7 tens and 0 ones. But be careful the meanings are totally different. Tens / one 1 / 3 X / 7 If we look at the ones column it says that we have 7 sets of 3 ones Using your base 10 chips let me see 7 sets of 3’s.
We have 6 tens and 0 ones. Hundreds / Tens / one 2 / 6 X 1 / 5 3 / 0 Now let’s do the tens.2 / 7 / 9 / 1 0 1 9 tens and 1 one = 91 Let’s do one more using our hundreds manipulative: Hundreds / Tens / one 2 / 6 X 1 / 5 What do we do first? We have use the bottom number first. Hundreds / Tens / one 2 / 6 X 1 / 5 3 / 0 1/ 0/ 0 . We have 5 sets of 2 tens or 5 x 20 = 100 OOOOOOOOOO OOOOOOOOOO 10+10+10+10+10+10+10+10+10+10 =100 OOOOOOOOOO OOOOOOOOOO OOOOOOOOOO OOOOOOOOOO OOOOOOOOOO OOOOOOOOOO OOOOOOOOOO OOOOOOOOOO Lets regroup: We have 1 hundreds. OOOOOO ALSO 6+6+6+6+6 = 30 or 5 x 6 = 30 OOOOOO OOOOOO OOOOOO OOOOOO Let’s regroup. We have 5 sets of 6 ones. 0 tens and 0 ones.
Hundreds / Tens / one 2 / 6 X 1 / 5 3 / 0 1/ 0/ 0 6/ 0 2/ 0/ 0 Let’s add all of the columns and see what we have? Hundreds / Tens / one 2 / 6 X 1 / 5 3 / 0 1/ 0/ 0 6/ 0 2/ 0/ 0 . Hundreds / Tens / one 2 / 6 X 1 / 5 3 / 0 1/ 0/ 0 6/ 0 Lastly we have 10 sets of 20 tens: OOOOOOOOOOOOOOOOOOOO or 10+10+10+10+10+10+10+10+10+10+ OOOOOOOOOOOOOOOOOOOO 10+10+10+10+10+10+10+10+10+10 = 200 OOOOOOOOOOOOOOOOOOOO OOOOOOOOOOOOOOOOOOOO OOOOOOOOOOOOOOOOOOOO OOOOOOOOOOOOOOOOOOOO OOOOOOOOOOOOOOOOOOOO OOOOOOOOOOOOOOOOOOOO OOOOOOOOOOOOOOOOOOOO OOOOOOOOOOOOOOOOOOOO 10 X 20 = 200. 0 ones. We regroup to 2 hundreds. 0 tens.Now we move to the tens: What do we have? We have 10 sets of 6 ones OOOOOO OOOOOO 6+6+6+6+6+6+6+6+6+6=60 OOOOOO OOOOOO OOOOOO OOOOOO OOOOOO OOOOOO OOOOOO OOOOOO 10x6 = 60 we regroup and have 6 tens and 0 ones.
Second part: 3 x 10 = 30 Sarah does not have enough carrots to make 3 cakes. It takes 10 carrotsto make her carrot cake. Does Sarah have enough to make 3 carrot cakes? Bags = 3 carrots = 9 Or 3 sets of 9 or 3 x 9. Sarah can only make 2 cakes.3/ 9/ 0 We have 3 hundreds or 300 We have 9 Tens = 90 We have 0 ones = 0 Our answer to 26 x 15 = 390 Word Problems: Sarah has 3 bags of carrots There are 9 carrots in each bag. Take out your manipulatives OOOOOOOOO 9+9+9 = 27 OOOOOOOOO OOOOOOOOO Let’s regroup: We have 2 tens and 7 ones Tens / ones 2/ 7 If we have 27 carrots and it takes 10 carrots to make a cake. How many marbles does he have in all? Lets use our manipultative: Blue Marbles = OOOO Red Marbles = OOOO Yellow Marbles = OOOO . 4 red Marbles and 4 yellow marbles. Word Problem 2: John has 4 Blue marbles.
Lets regroup: 12 ones equals 1 ten and 2 ones Tens / ones 1 2 3 sets of 4 = 4+4+4 = 12 3 x 4 = 12 .
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# Precalculus : Products and Quotients of Complex Numbers in Polar Form
## Example Questions
### Example Question #1 : Find The Product Of Complex Numbers
Find the value of ,where the complex number is given by .
Explanation:
We note that by FOILing.
We also know that:
We have by using the above rule: n=2 , m=50
Since we know that,
We have then:
Since we know that:
, we use a=2 ,b=i
We have then:
### Example Question #1 : Polar Coordinates And Complex Numbers
Compute the following sum:
. Remember is the complex number satisfying .
Explanation:
Note that this is a geometric series.
Therefore we have:
Note that,
= and since we have .
this shows that the sum is 0.
### Example Question #3 : Find The Product Of Complex Numbers
Find the following product.
Explanation:
Note that by FOILing the two binomials we get the following:
Therefore,
### Example Question #4 : Find The Product Of Complex Numbers
Compute the magnitude of .
Explanation:
We have
We know that
Thus this gives us,
.
### Example Question #5 : Find The Product Of Complex Numbers
Evaluate:
Explanation:
To evaluate this problem we need to FOIL the binomials.
Now recall that
Thus,
### Example Question #1 : Polar Coordinates And Complex Numbers
Find the product , if
.
Explanation:
To find the product , FOIL the complex numbers. FOIL stands for the multiplication of the Firsts, Outers, Inners, and Lasts.
Using this method we get the following,
and because
.
### Example Question #1 : Find The Product Of Complex Numbers
Simplify:
Explanation:
The expression can be rewritten as:
Since , the value of .
### Example Question #1 : Find The Product Of Complex Numbers
Find the product of the two complex numbers
and
Explanation:
The product is
### Example Question #1 : Find The Quotient Of Complex Numbers
Let . Find a simple form of .
Explanation:
We remark first that:
and we know that :
.
This means that:
What is ?
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1. ## inequality confucion
Solve this inequality .
$\displaystyle \frac{6}{|x|+1}<|x|$
squarring both sides , i got
$\displaystyle \frac{36}{x^2+2|x|+1}<x^2$ , rearranging them
$\displaystyle \frac{36}{x^2+2|x|+1}-\frac{x^2(x^2+2|x|+1)}{x^2+2|x|+1}<0$
$\displaystyle \frac{36-x^4-2x^2|x|-x^2}{x^2+2|x|+1}<0$ , till here am i correct , and i don see a way to solve it from here .
Solve this inequality .
$\displaystyle \frac{6}{|x|+1}<|x|$
squarring both sides , i got
$\displaystyle \frac{36}{x^2+2|x|+1}<x^2$ , rearranging them
Squaring unnecessarily complicates this simple problem. Instead take the following route:
$\displaystyle \frac{6}{|x|+1}<|x|$
$\displaystyle \frac{6}{|x|+1} - |x| < 0$
$\displaystyle \frac{6 - |x|^2 - |x|}{|x|+1} < 0$
But $\displaystyle |x| + 1 > 0$, thus we can multiply it on both sides without affecting the inequality. Thus
$\displaystyle 6 - |x|^2 - |x| < 0$
And so on....
3. ## Re :
Originally Posted by Isomorphism
Squaring unnecessarily complicates this simple problem. Instead take the following route:
$\displaystyle \frac{6}{|x|+1}<|x|$
$\displaystyle \frac{6}{|x|+1} - |x| < 0$
$\displaystyle \frac{6 - |x|^2 - |x|}{|x|+1} < 0$
But $\displaystyle |x| + 1 > 0$, thus we can multiply it on both sides without affecting the inequality. Thus
$\displaystyle 6 - |x|^2 - |x| < 0$
And so on....
Yeah , i make things complicated for myself .
From there , i got x>3 and x<-3 .... is it correct .
Yeah , i make things complicated for myself .
From there , i got x>3 and x<-3 .... is it correct .
Unfortunately no.
From $\displaystyle 6-|x|^2-|x|<0$ ......you'll get:
$\displaystyle |x|^2+|x|-6>0~\implies~(|x|+3)(|x|-2)>0$
A product of 2 factors is positive if both factors are positive or both factors are negative. Since $\displaystyle |x|+3>0$ is true for all x, the second case is impossible. Therefore (keep in mind that $\displaystyle |x|+3>0$ is true for all x):
$\displaystyle |x|-2>0~\implies~|x|>2$
Now use the definition of the absolute value:
$\displaystyle |x|>2~\iff~x>2~\vee~-x>2~\iff~\boxed{x<-2~\vee~x>2}$
By the way:
From there , i got x>3 and x<-3
You'll never find a number which is greater than 3 and simultaneously is smaller than -3.
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×
# CHEFST - Editorial
Author: Tapas Jain
Tester: Sergey Kulik
Editorialist: Kevin Atienza
# PROBLEM:
Consider a single-player game on two piles of stones. One has $n_1$ stones and the other has $n_2$ stones. Before the start of the game, we choose an integer $m$.
In a move, we choose a number between $1$ and $m$, and remove that number of stones from both piles. (this is only possible when both the piles have at least that many stones). Also, the number of stones to remove at each step must be unique. The game ends when there are no more moves.
What is the minimum number of stones that could remain, among all possible games?
# QUICK EXPLANATION:
The answer is $n_1 + n_2 - 2x$, where $x$ is the maximum number of stones we can remove from both piles.
We can only remove up to $1 + 2 + \cdots + m = \frac{m(m+1)}{2}$ stones from both piles. If $n_1$ and $n_2$ are both at least $\frac{m(m+1)}{2}$, then this is the maximum number we can remove. Otherwise, we can remove up to $\min(n_1,n_2)$ from both piles. Therefore, $x = \min(\frac{m(m+1)}{2}, n_1, n_2)$.
The answer can also be expressed as a single expression: $$\max(n_1 + n_2 - m(m+1), n_1 - n_2, n_2 - n_1)$$
# EXPLANATION:
At any point in the game, the number of stones we have removed from both piles are always equal. Suppose we remove $x$ stones from both piles. Then there are $n_1 - x$ stones remaining in the first pile and $n_2 - x$ in the second. Therefore, the number of stones remaining is: $$(n_1 - x) + (n_2 - x) = n_1 + n_2 - 2x$$ The answer is the minimum value of this expression. $n_1$ and $n_2$ are fixed throughout the game, but we can control $x$ depending on how we play. But a higher $x$ corresponds to a lower $n_1 + n_2 - 2x$, so we actually want to maximize $x$, the number of stones we remove from both piles!
# Maximum number of removed stones
Next, let's consider the requirement that "the number of stones to remove at each step must be unique". What does this mean for us? Well, since there are only $m$ unique numbers from $1$ to $m$, namely $1, 2, \ldots, m$, this means that the game ends after at most $m$ moves, and that the maximum number of stones we can remove is simply: $$1 + 2 + 3 + 4 + \cdots + m$$ This is actually a pretty famous sum, and the formula is $\frac{m(m+1)}{2}$. (A derivation is given in the appendix.) Therefore, we can only remove between $0$ to $\frac{m(m+1)}{2}$ stones.
# Removing a given number of stones
Now we know that we can only remove between $0$ to $\frac{m(m+1)}{2}$ stones. The next question is: given some number $r$ between $0$ and $\frac{m(m+1)}{2}$, can we remove exactly $r$ stones? In other words, is it possible to express $r$ as a sum of distinct numbers from $1$ to $m$? For example, can you try to express the number $31$ as the sum of distinct numbers from $1$ to $10$?
Let's try. Since $31$ is huge, we try adding the large addends first. Say $10$. Then $31 - 10 = 21$ remains. The next largest addend is $9$, so we try that. $21 - 9 = 12$ remains. The next one is $8$, so we try that, and $12 - 8 = 4$ remains. Our number is small now, and in fact since it's already $\le 7$, we can just use $4$ as our final addend. Therefore: $$31 = 10 + 9 + 8 + 4$$ Does this greedy algorithm always work? Amazingly, yes! (as shown in the appendix) Therefore, given any $r$, we can always remove exactly $r$ stones.
# Maximizing the number of removed stones
We're almost there! We want to maximize the number of stones to remove. Clearly, regardless of the size of the piles, the absolute maximum number we can remove is $\frac{m(m+1)}{2}$ as shown above. Thus, if $n_1$ and $n_2$ are both at least $\frac{m(m+1)}{2}$, then this is the maximum.
Now, what if one of $n_1$ and $n_2$ are smaller than $\frac{m(m+1)}{2}$? In other words, $\min(n_1,n_2) < \frac{m(m+1)}{2}$. In this case, we can no longer remove $\frac{m(m+1)}{2}$ stones, because there aren't enough stones in one of the piles. In fact, we can't remove more than $\min(n_1,n_2)$, because this is the size of the smaller pile. But since $\min(n_1,n_2)$ is smaller $\frac{m(m+1)}{2}$, as shown above we can always remove exactly $\min(n_1,n_2)$ stones. Therefore, the maximum stones we can remove is actually $\min(n_1,n_2)$!
We now have the whole solution! It is: $$n_1 + n_2 - 2x$$ where $$x = \min\left(\frac{m(m+1)}{2},n_1,n_2\right)$$
As a side note, we can actually substitute $x$ to $n_1 + n_2 - 2x$ to get the following expression: $$\max(n_1 + n_2 - m(m+1), n_1 - n_2, n_2 - n_1)$$ which gives us the following one-liner Python code (excluding the part of code that takes the input from stdin):
print max(n1 + n2 - m*(m+1), n1 - n2, n2 - n1)
# Appendix: Showing that $1 + 2 + 3 + 4 + \cdots + m = \frac{m(m+1)}{2}$
We want to find a formula $S := 1 + 2 + 3 + 4 + \cdots + m$. I'll show a derivation here that is slightly different from the usual reverse and add method.
Let's begin: \begin{align*} S &= 1 + 2 + 3 + 4 + \cdots + m \\\ 2S &= 2 + 4 + 6 + 8 + \cdots + 2m \\\ 2S &= [1 + 3 + 5 + 7 + \cdots + (2m-1)] + \underbrace{1 + 1 + 1 + 1 + \cdots + 1}_{m} \\\ 2S &= [1 + 3 + 5 + 7 + \cdots + (2m-1)] + m \end{align*} Now, consider the sum in the brackets. Notice the pattern: \begin{align*} 1 &= 1 \\\ 4 &= 1 + 3 \\\ 9 &= 1 + 3 + 5 \\\ 16 &= 1 + 3 + 5 + 7 \\\ 25 &= 1 + 3 + 5 + 7 + 9 \\\ 36 &= 1 + 3 + 5 + 7 + 9 + 11 \end{align*} These are just the perfect squares! Thus, we conjecture that $1 + 3 + 5 + \cdots + (2m-1) = m^2$. In fact, this can easily be visualized as follows:
m=1 m=2 m=3 m=4 m=5
1 1 2 1 2 3 1 2 3 4 1 2 3 4 5
2 2 2 2 3 2 2 3 4 2 2 3 4 5
3 3 3 3 3 3 4 3 3 3 4 5
4 4 4 4 4 4 4 4 5
5 5 5 5 5
Note that the $m$th figure contains $m^2$ items, but at each step we're adding $1$, then $3$, then $5$, etc., items, to the previous figure!
Therefore: \begin{align*} 2S &= [1 + 3 + 5 + 7 + \cdots + (2m-1)] + m \\\ 2S &= m^2 + m \\\ S &= \frac{m(m+1)}{2} \end{align*} which is what we wanted to show.
# Appendix: Summing a number between $0$ and $\frac{m(m+1)}{2}$
Finally, we want to show why the greedy algorithm above works for expressing a number $r$ between $0$ and $\frac{m(m+1)}{2}$ as a sum of distinct numbers from the set $\{1, 2 \ldots m\}$.
The greedy algorithm first proceeds to check whether $m \le r$, and if so, adds $m$ to the list of addends. Then we proceed by expressing the number $r - m$ as a sum of numbers from the set $\{1, 2 \ldots m-1\}$. However, by assumption: $$1 + 2 + 3 + \cdots + m \ge r$$ By transposing the $m$, we get: $$1 + 2 + 3 + \cdots + (m-1) \ge r - m$$ Therefore, by doing an induction argument, we can see that the number $r - m$ can be expressed as a sum of numbers from the set $\{1, 2 \ldots m-1\}$!
This only works when $m \le r$ though. What if $m > r$? Then in that case, we can simply use $r$ as the lone addend in the sum, because $r \in \{1, 2 \ldots m\}$! Thus, we have just shown that the greedy algorithm works.
# Time Complexity:
$O(1)$
# AUTHOR'S AND TESTER'S SOLUTIONS:
This question is marked "community wiki".
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1 @durwasa_jec: U need to use llabs() for long long int and not abs(). Here's the correct code: https://www.codechef.com/viewsolution/8975993 answered 18 Dec '15, 18:41 21●2 accept rate: 0%
0 How did we arrive at the following equation? max(n1+n2−m(m+1),n1−n2,n2−n1) answered 14 Dec '15, 21:54 2★vedant12 1 accept rate: 0%
0 @author @admin Why does abs(n1-n2) only solve the first two tasks (here) , while calculating the difference and multiplying a -1 accordingly (like diff=(diff if greater than 0)?diff:diff*-1) solve the problem (here) Or there is something -Regards answered 14 Dec '15, 23:28 3●1●2 accept rate: 0% @shahbaz_23 :: thank you so much :) :) (18 Dec '15, 19:13)
0 can some please check this solution and point the error : https://www.codechef.com/viewsolution/8950590 Thanks !! answered 28 Jan '16, 00:19 1 accept rate: 0%
0 Great editorial! Checkout this brilliant wiki also: Greedy answered 29 Nov '17, 14:02 1★mr_nair 2●1 accept rate: 0%
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As I said in the previous post, there is a duality:
Points on a Curve (Geometry) $\Leftrightarrow$ Solutions of an Equation (Algebra)
This means we can answer geometric questions using algebra and answer algebraic questions using geometry.
### Problem
Consider the following two questions:
1. Find the tangents to a circle $\mathcal{C}$ of a given slope.
2. Find the tangents to a circle $\mathcal{C}$ through a given point.
Both can be answered using the duality principle.
#### Example
Find the tangents to the circle
$\mathcal{C}\equiv x^2+y^2-4x+6y-12=0$
that are
(a) parallel to the line $L\equiv 4x+3y+20=0$
(b) through the point $(-10,-5)$ [caution: the numbers here are disgusting]
##### Solution (a) i:
First of all a sketch (and the remark that a tangent is a line):
Here we see the circle $\mathcal{C}$ and the line $L$ on the bottom left. The two tangents we are looking for are as shown. They have the same slope as $L$ and have only one intersection with $\mathcal{C}$. These two pieces of information will allow us to find the equations of the tangents.
Let us first find the slope of $L$ by writing it in the form:
$y=mx+c$.
$3y=-4x-20$
$\displaystyle\Rightarrow y=-\frac43 x -\frac{20}{3}$.
Now any line parallel to $L$ also has slope $m=-\frac43$ and thus is of the form:
$\displaystyle y=-\frac{4}{3}x+c$,
for some $c$ (the ‘lower/bottom’ line will have the smaller $c$-value).
Now the second condition is that a tangent to a circle intersects it at a single point only. Now an intersection between two curves is a point on both curves and by duality the coordinates of the intersection satisfy both curve’s equations at the same time. That is we want the following simultaneous equations to have a single solution only (in general an $x^2$ has two):
$x^2+y^2-4x+6y-12=0$
$\displaystyle y=-\frac{4}{3}x+c$
$\displaystyle \Rightarrow x^2+(c-\frac43 x)^2-4x+6\left(c-\frac43x\right)-12=0$
$\displaystyle \Rightarrow x^2+c^2-\frac83 cx+\frac{16}{9}x^2-4x+6c-8x-12=0$
$\displaystyle \left[1+\frac{16}{9}\right]x^2+\left[-\frac83 x-12\right]c+(c^2+6c-12)=0$
$\displaystyle \frac{25}{9}x^2+(-12-\frac83 c)x+(c^2+6c-12)=0$
Now this is a quadratic in $x$. It is a $+x^2$ quadratic and so has $\bigcup$ geometry. For it to have a single root it must look like:
A quadratic has a single (repeated) root when $b^2-4ac=0$.
Therefore the circle and line of slope $-4/3$ has a unique solution if:
$\displaystyle \left(-12-\frac83 c\right)^2-4\left(\frac{25}{9}\right)(c^2-6c-12)=0$
Note that this is a quadratic in $c$ — and so has two solutions as expected. Multiplying out and simplifying gives:
$\displaystyle 144+\frac{192}{3}c+\frac{64}{9}c^2-\frac{100}{9}c^2-\frac{200}{3}c+\frac{400}{3}=0$
$\displaystyle \Rightarrow 4c^2-\frac{8}{3}c-\frac{832}{3}=0$.
The ‘$-b$‘ formula can be used to solve this however by multiplying both sides by three and dividing both sides before we have:
$3c^2+2c-208=0$.
Looking at the factors of $3\times 208=624$ we rewrite the middle term
$3c^2-24c+26c-208=0$
$\Rightarrow 3c(c-8)+26(c-8)=0$
$\Rightarrow (c-8)(3c+26)=0$
$\displaystyle \Rightarrow c=8\text{ or }-\frac{26}{3}$,
and therefore the two tangents are:
$\displaystyle y=-\frac43 x+8$ and $y=-\frac43 x-\frac{26}{3}$.
The two tangents, along with $L$ which actually intersects the curve twice (so that the simultaneous equations of $\mathcal{C}$ and $L$ would have two solutions and so $b^2-4ac>0$).
##### Solution (a) (ii)
The preceding solution is nice and sound. However, if we take a theorem on board we get a nicer solution:
The tangent to a circle is perpendicular to the diameter. This means that the perpendicular distance from the centre to a tangent is equal to the radius…
The perpendicular distance, $d$, from a line $ax+by+c=0$ (and this is when it is necessary to use this format rather than the much more useful $y=mx+c$) to a point $(x_1,y_1)$ is given by:
$\displaystyle d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$.
Now lines parallel to $4x+3y+20=0$ are of the form $L_{\lambda}\equiv 4x+3y+\lambda =0$… and using $(-g,-f)$, the centre of $\mathcal{C}$ is at $(2,-3)$. Therefore, for $L_{\lambda}$ to be a tangent to $\mathcal{C}$, it must be the case that the perpendicular distance to the centre $(2,-3)$
$\displaystyle d=\frac{|4(2)+3(-3)+\lambda|}{\sqrt{4^2+3^2}}$,
$r=\sqrt{g^2+f^2-c}=\sqrt{(-2)^2+3^2-(-12)}=\sqrt{25}=5$ so that
$\frac{|\lambda-1|}{5}=5$
$\Rightarrow |\lambda -1|=25$.
It is possible to square both sides but easier just to say $|a|=\pm a$ so that we have
$\lambda-1=25$ or $\lambda-1=-25$
$\Rightarrow \lambda =26$ or $-24$,
so that the two tangents have equations:
$4x+3y+26=0$ and $4x+3y-24=0$.
Exercise: Are these the same solutions as before?
##### Solution (b) (i)
First of all, a picture:
In this case we know a point on the line but not the slope, $m$.
We know that the point $(-10,-5)$ is on these lines but we don’t know the slopes, $m$. Therefore the tangents are of the form:
$y+5=m(x+10)\Rightarrow y=mx+10m-5$.
We want these lines to be such that the intersection between them and $\mathcal{C}$ is unique. Therefore we substitute $y=mx+10m-5$ into the equation of $\mathcal{C}$:
$x^2+(mx+10m-5)^2-4x+6(mx+10m-5)-12=0$.
Multiplying out and tidying up:
$(m^2+1)x^2+(20m^2-4m-4)x+(100m^2-40m-17)=0$.
Now, if this is to have a single solution, we once again require $b^2-4ac=0$:
$(20m^2-4m-4)^2-4(m^2+1)(100m^2-40m-17)=0$.
Multiplying out an rearranging we get:
$476m^2-192m-84=0$
$\Rightarrow 119m^2-48m-21=0$.
Using the ‘$-b$‘ formula we find the two values of $m$ and thus tangents:
$\displaystyle y=\left(\frac{24}{119}\pm \frac{5}{119}\sqrt{123}\right)(x+10)-5$.
##### Solution (b) (ii)
Another option is to use geometry in the first instance to find the following circle:
The red circle here cuts $\mathcal{C}$ at the contact points $A$ and $B$.
Where $c(2.-3)$ is the centre of the circle, we can find the distance $|cP|$ ($P(-10,-5)$) using the distance formula:
$|cP|=\sqrt{(2+10)^2+(-3+5)^2}=\sqrt{148}=2\sqrt{37}$.
We know the radius, the distance $|Ac|=|Bc|=5$. We know that $cA\perp cP$ so that $\Delta AcP$ is a right-angled-triangle satisfying Pythagoras:
$|cP|^2=|cA|^2+|AP|^2=\Rightarrow 148=25+|Ap|^2\Rightarrow |Ap|=\sqrt{123}$.
Therefore, the red circle has centre $(-10,-5)$ and radius $\sqrt{123}$ so equation:
$(x+10)^2+(y+5)^2=123$
$\Rightarrow x^2+y^2+20x+10y+2=0$.
|
## SOLVING QUADRATIC EQUATIONS
Note:
• A quadratic equation is a polynomial equation of degree 2.
• The ''U'' shaped graph of a quadratic is called a parabola.
• A quadratic equation has two solutions. Either two distinct real solutions, one double real solution or two imaginary solutions.
• There are several methods you can use to solve a quadratic equation:
1. Factoring
2. Completing the Square
4. Graphing
• All methods start with setting the equation equal to zero.
Solve for x in the following equation.
Problem 4.5c:
are the exact answers using the Completing the Square Method.
even though the two answers look different, they are equivalent because both yield the same approximate answers of and
Solution:
Simplify the equation .
Divide both sides by
Method 1: Factoring
The equation is not easily factored. Therefore, we will not use this method.
Method 2: Completing the square
Add to both sides of the equation
Add to both sides of the equation:
Factor the left side and simplify the right side :
Take the square root of both sides of the equation :
Subtract from both sides of the equation :
are the exact answers are approximate answers.
Method 3: Quadratic Formula
The quadratic formula is
In the equation ,a is the coefficient of the term, b is the coefficient of the x term, and c is the constant. Substitute 1 for a, for b , and for c in the quadratic formula and simplify.
are the exact answers are approximate answers.
Method 4: Graphing
Graph the equation, (the left side of the original equation). Graph (the x-axis). What you will be looking for is where the graph of crosses the x-axis. Another way of saying this is that the x-intercepts are the solutions to this equation.
You can see from the graph that there are two x-intercepts, one at 6.57797305561 and one at -10.31963044.
The answers are 6.57797305561 and -10.31963044. These answers may or may not be solutions to the original equations. You must verify that these answers are solutions.
Check these answers in the original equation.
Check the solution x = 6.57797305561 by substituting 6.57797305561 in the original equation for x. If the left side of the equation
equals the right side of the equation after the substitution, you have found the correct answer.
• Left Side:
• Right Side:
Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 6.57797305561 for x, then x = 6.57797305561 is a solution.
Check the solution x = -10.31963044 by substituting -10.31963044 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.
• Left Side:
• Right Side:
Since the left side of the original equation is equal to the right side of the original equation after we substitute the value -10.31963044 for x, then x = - 10.31963044 is a solution.
The solutions to the equation are 6.57797305561 and -10.31963044.
If you would like to review the solution to problem 4.5d, click on Problem
If you would like to go back to the beginning of the quadratic section, click on Quadratic
If you would like to go back to the equation table of contents, click on Contents
This site was built to accommodate the needs of students. The topics, explanations, examples, and problems are what students ask for. Student viewer help with the editing so that future viewers will access a cleaner site. If you feel that some of the material is this section is ambiguous or needs more clarification, please let us know by e-mail. If you find a mistake, please let us know. The constant feedback from viewers is one of the reasons that this site garners so many awards.
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|
# Exponents and Fractions Online Quiz
Following quiz provides Multiple Choice Questions (MCQs) related to Exponents and Fractions. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Q 1 - Evaluate $\frac{4^2}{8}$
### Explanation
Step 1:
$\frac{4^2}{8} = \frac{4 \times 4}{8} = \frac{16}{8}$
Step 2:
$\frac{16}{8} = \frac{2}{1} = 2$
Step 3:
So, $\frac{4^2}{8} = 2$
Q 2 - Evaluate $\left ( \frac{2}{7} \right )^3$
### Explanation
Step 1:
$\left ( \frac{2}{7} \right )^3 = \frac{2^3}{7^3}$
Step 2:
$\frac{2^3}{7^3} = \frac{2 \times 2 \times 2}{7 \times 7 \times 7} = \frac{8}{343}$
Step 3:
So, $\left ( \frac{2}{7} \right )^3 = \frac{8}{343}$
### Explanation
Step 1:
$\frac{9}{3^4} = \frac{9}{3 \times 3 \times 3 \times 3}$
Step 2:
$\frac{9}{3 \times 3 \times 3 \times 3}$
Step 3:
So, $\frac{9}{3^4} = \frac{9}{81} = \frac{1}{9}$
### Explanation
Step 1:
$\frac{3^2}{6^3} = \frac{3 \times 3}{6 \times 6 \times 6}$
Step 2:
$\frac{3 \times 3}{6 \times 6 \times 6} = \frac{9}{216} = \frac{1}{24}$
Step 3:
So, $\frac{3^2}{6^3} = \frac{1}{24}$
Q 5 - Evaluate $\left ( \frac{7}{8} \right )^2$
### Explanation
Step 1:
$\left ( \frac{7}{8} \right )^2 = \frac{7 \times 7}{8 \times 8}$
Step 2:
$\frac{7 \times 7}{8 \times 8} = \frac{49}{64}$
Step 3:
So, $\left ( \frac{7}{8} \right )^2 = \frac{49}{64}$
### Explanation
Step 1:
$\frac{7^2}{5^3} = \frac{7 \times 7}{5 \times 5 \times 5}$
Step 2:
$\frac{7 \times 7}{5 \times 5 \times 5} = \frac{49}{125}$
Step 3:
So, $\frac{7^2}{5^3} = \frac{49}{125}$
Q 7 - Evaluate $\left ( \frac{4}{11} \right )^2$
### Explanation
Step 1:
$\left ( \frac{4}{11} \right )^2 = \frac{4^2}{11^2}$
Step 2:
$\frac{4^2}{11^2} = \frac{16}{121}$
Step 3:
So, $\left ( \frac{4}{11} \right )^2 = \frac{16}{121}$
### Explanation
Step 1:
$\frac{4}{9^2} = \frac{4}{9 \times 9}$
Step 2:
$\frac{4}{9 \times 9} = \frac{4}{81}$
Step 3:
So, $\frac{4}{9^2} = \frac{4}{81}$
### Explanation
Step 1:
$\frac{2^2}{7} = \frac{2 \times 2}{7}$
Step 2:
$\frac{2 \times 2}{7} = \frac{4}{7}$
Step 3:
So, $\frac{2^2}{7} = \frac{4}{7}$
### Explanation
Step 1:
$\frac{3^2}{5^3} = \frac{3 \times 3}{5 \times 5 \times 5}$
Step 2:
$\frac{3 \times 3}{5 \times 5 \times 5} = \frac{9}{125}$
Step 3:
So, $\frac{3^2}{5^3} = \frac{9}{125}$
exponents_and_fractions.htm
|
# Conic Sections/Circle
The circle is the simplest and best known conic section. As a conic section, the circle is the intersection of a plane perpendicular to the cone's axis.
The geometric definition of a circle is the locus of all points a constant distance ${\displaystyle r}$ from a point ${\displaystyle (h,k)}$ and forming the circumference (C). The distance ${\displaystyle r}$ is the radius (R) of the circle, and the point ${\displaystyle O=(h,k)}$ is the circle's center also spelled as centre. The diameter (D) is twice the length of the radius.
## Equations
#### Standard Form
The standard equation for a circle with center ${\displaystyle (h,k)}$ and radius ${\displaystyle r}$ is
${\displaystyle (x-h)^{2}+(y-k)^{2}=r^{2}}$.
In the simplest case of a circle whose center is at the origin, the equation is simply a restatement of the Pythagorean Theorem:
${\displaystyle x^{2}+y^{2}=r^{2}}$
#### General form
The general form of a circle equation is
${\displaystyle x^{2}+y^{2}+2gx+2fy+c=0}$, where
<-g,-f> is the center of the circle.
#### Polar Coordinates
In the case of a circle centered at the origin, the polar equation of a circle is very simple because polar coordinates are essentially based on circles. For a circle with radius ${\displaystyle a}$,
${\displaystyle r=a}$.
In the more complicated case of a circle with an arbitrary location, the equation is
${\displaystyle r^{2}-2rr_{0}\cos(\theta -\varphi )+r_{0}^{2}=a^{2}}$,
where ${\displaystyle r_{0}}$ is the distance from the circle's center to the origin and ${\displaystyle \varphi }$ is the angle pointing to the circle.
There are many cases that allow the equation to be simplified. If a point on the circle is touching the origin, its polar equation may consist of a single trig function.
.....
#### Parametric Equations
When the circle's equation is parametrized with respect to ${\displaystyle t}$, the equation becomes
${\displaystyle x=h+r\cos t}$,
${\displaystyle y=k+r\sin t}$.
## Example
Find the center and the radius of the following circle: x2+y2+8x-10y+20=0 find by:
x2+y2+8x-10y+20=0
x2+y2+8x-10y= - 20
(x2+8x)+(y2-10y)= - 20
+16 +25 +16+25
(x2+8x+16)+(y2-10y+25)=21
(x+4)2+(y-5)2=21
Thus:
C(-4,5) radius=${\displaystyle radical(21)}$
|
P. 1
003 - Fractions and Their Forms (Smt I)
# 003 - Fractions and Their Forms (Smt I)
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To write a fraction in a percent form
To write a fraction in a decimal form
To compare fractions
and its inverse
number into fractions
To change a mixed
To simplify fractions
To write a fraction in a percent form
To write a fraction in a decimal form
To compare fractions
and its inverse
number into fractions
To change a mixed
To simplify fractions
See more
See less
09/23/2011
pdf
text
original
# 2.
1
What are you going to learn?
To define fractions To simplify fractions To change a mixed number into fractions and its inverse To compare fractions To write a fraction in a decimal form To write a fraction in a percent form
The Meaning of Fraction
In our everyday life, sometimes we have to divide something into some equal parts. For instances: • A slice of bread divided into three equal parts, • A piece of paper cut into two equal parts, • An orange sliced into some equal parts,
All situations above are related to fractions.
Key Terms :
• • • • • fraction numerator denumerator decimal percent
Look at the picture below.
Tools and Material :
• • paper pencil • ruler
Firstly, an orange is sliced into two equal parts. Each part is called one over two or one-half or half and it is written as
1 . 2
Then, each of the two equal parts is divided into two
Mathematics for Junior High School – Year 7/43
equal parts so that we obtain four equal parts. Each part of four equal parts is called one over four or one-fourth, written as
1 . 4
The numbers
1 1 3 , , and are examples of fraction 2 4 2
numbers, or fractions for short.
A fraction is a number that can be written in the form of
a , where a and b are integers and b ≠ 0, and b is not a b
factor of a. The number a is called the numerator, b is called the denominator. Why b cannot be zero?
For the fraction denominator. For the fraction denominator.
1 , 1 is the numerator, and 4 is the 4
3 , 3 is the numerator, and 2 is the 2
Simplifying Fractions
Look at the shadowed parts of the bars on the left. How many parts are in each bar? How many parts are
1 2 3 4 , , , and have equal values. Of these 2 4 6 8
44/Student’s Book – Fractions
fractions,
1 is the simplest form. A fraction is called the 2
simplest form (simple fraction) if the Greatest Common Factor (GCF) of the numerator and the denominator is 1. You can write the simplest form of a fraction by dividing the numerator and the denominator with the GCF of the numerator and the denominator.
EXAMPLE 1
Write
20 in the simplest form. 28
The GCF of 20 and 28 is 4.
Divide the numerator and the denominator by 4.
Therefore, the simplest form of fraction
20 28
is 5 .
7
Converting a Mixed Number into a Fraction
Two men rode horses. One rode for 1 1 km., and the
2
other covered a distance of 1 1 km.
4
The numbers 1 1 and 3
2
1 are called mixed numbers. 4
A mixed number can also be written as a fraction form. Work out the following Mini Lab.
Mathematics for Junior High School – Year 7/45
Mini-Lab
Work in pairs. Material and equipment: paper, pencil, and ruler. Draw a model for a mixed number 1 1 in the following steps.
4
Draw an identical quadrilateral beside the first one. Divide the quadrilateral on the right side into four equal parts to get one-fourth. Shade in one part to show 1 . Thus, we will get the model for 1 1 .
4
4
Divide the model of a quadrilateral into four equal parts (one-fourth).
The shadowed areas on the last picture show the mixed numbers 1 1 .
4
Discuss: b. How many shadowed one-fourth areas are there in the above picture? c. How many unshadowed one-fourth areas are there in the above drawing? d. What fraction has the same value as 1 1 ?
4
Based on the Mini-Lab, you may conclude that a mixed number can be written as a fraction that is called an improper fraction.
You can also use a different way to convert the form of a mixed number.
46/Student’s Book – Fractions
Converting an Improper Fraction into a Mixed Number
Suppose you have 28 liters of gasoline. You are asked to fill it in 8 containers. Each container will be filled with equal volume of gasoline. How many liters should be filled into each container? Solution:
28 ← Write down the division in the form of 8
fraction
3 8 28 24 4 −
Divide 28 by 8
34 = 31 8 2
Write down the remainder as a fraction, and then simplify it.
Therefore, each container is filled with 3 gasoline.
1 liters of 2
Comparing and Ordering Fractions
Suppose there was a vote for chairperson of the Student Organization at your school. The result of the vote was as follows : • 1 of the students voted for Candidate I. 3
Mathematics for Junior High School – Year 7/47
• 2 of the students voted for Candidate II 7 Based on the above result, which candidate got more votes, Candidate I or Candidate II? To answer the
question, you need to know how to compare fractions.
There are two aspects you should know to compare fractions: (1) Comparing like fractions
Examine the length of the shaded parts of the two models of fraction below.
Based on these models, you can conclude that
5 6 4 6
5 > 4 . Why? 6 6
Also, check that one-sixth can be considered as a new unit.
5 means 5 sixths, and 4 means 4 sixths. 6 6
Which one is greater, 5 sixths or 4 sixths? Based on the explanation, it is clear that 5 > 4 .
6 6
Therefore, to compare some like fractions, you just need to compare the numerator. If the numerator of one fraction is greater, then the fraction is also greater than the other. (2) Comparing unlike fractions
48/Student’s Book – Fractions
Let us begin by comparing 1 to 1 .
2 3
We know that 1 is equal to 3 and 1 is equal to 2 . 2 3 6 6 The four fractions above can be modeled as follows.
is equal to
1 2 3 6
1 3
is equal to
2 6
Which fraction is greater? It is clear that 1 > 1 and
2 3 3 > 2 , because 1 = 3 and 1 = 2 . 6 6 2 6 3 6
Note
In a measurement the sizes can be compared if they have the same units. Analogically, to compare fractions, we have to change the fractions so that they have the same denominators
Therefore, one way to compare fractions is to express the fractions to like fractions, and then compare the numerators. To get the same numerator, we use the Least Common Multiple (LCM) of two numbers. For more explanation, look at the procedures to compare fractions 1 with 2 in Example 2 below.
3 7
Use the sign <, =, or > to compare 1 to 2 .
3 7
Step I: Find LCM of 3 and 7. Multiple of 3 are 3, 6, 9, 12, 15, 18, 21 ,24 Multiple of 7 are 7, 14, 21 ,28
Mathematics for Junior High School – Year 7/49
LCM of 3 and 7 is 21, because 21 is the least number in which 3 and 7 match for the first time. Step II: Find the fraction which is equivalent to
1 3
and
2 7
using the LCM in Step I, as its
denominator.
1 3
=
... , so that 1 = 7 21 3 21
2 7
=
... , so that 2 = 6 21 7 21
Step III: Compare the like fractions in Step II. Compare the numerator of Since 7 > 6, then 7 > 6 .
21 7 21 7 and 6 . 21 21
We may conclude that 1 > 2 . Since 1 > 2 , the answer
3 3 7
to the question about the election of the chairperson of the Student Organization is that Candidate I got more votes than Candidate II.
Use the sign <, =, or > to compare 7 with 5 . 18 24
Remember
24 = 2 × 2 × 2 × 3
To write an equivalent fraction, multiply the numerator and the denominator with the same number, excluding zero.
18 = 2 × 3 × 3
Find the LCM of 18 and 24 then circle all different factors that ap pear most frequently.
Multiply all circled factors to get the LCM of 24 and 18.
50/Student’s Book – Fractions
2 × 2 × 2 × 3 × 3 = 72
7 24
=
21 72
5 18
=
20 72
Write equivalent fractions using LCM as the denominator.
21 > 20 Since
5 21 20 7 > , then > . 72 72 24 18
Compare the numerator.
Ordering Fractions
Look at the fraction bars on the left. 1. What fraction represents each model? 2. Which fraction is the greatest? The smallest? 3. Put them in sequence from the smallest to the greatest.
To put fractions in sequence is the same as to compare three or more fractions. If you want to put like fractions in sequence, then order them based on the numerator. However, if you want to put unlike fractions in sequence, you need to find the LCM of the denominator of the original fractions. The LCM will be their denominators.
EXAMPLE 4
Put the fractions
3 2 7 in sequence from the , , and 8 5 20
least to the greatest.
Mathematics for Junior High School – Year 7/51
8= 2× 2 × 2 5= 5 20 = 2 × 2 × 5
Find the LC M of 8, 5, and 20 by writing down all prime factors of each number, an d then circle all different factors th at appear most frequently.
Multiply all circled factors. 2 × 2 × 2 × 5 = 40
The LCM of 8, 5, and 20 is 40.
3 8
=
15 40
2 5
=
16 40
7 20
=
14 40
16 > 15 > 14
Compare the numerator and put them in order.
Since 16 > 15 > 14 , then 2 > 3 > 7 . 5 8 20 40 40 40
Therefore, if we put the fractions in order, from the least to the greatest, we get: 7 , 3 , 2 . 20 8 5
Decimal
A fraction or a mixed number can also be expressed as decimal. Similarly, a decimal can be expressed as a fraction or a mixed number.
Get a calculator and do the following activities.
52/Student’s Book – Fractions
Find the value of Push the bottom:
9 using a calculator. 40
9
/
4
0
=
What number do you get?
Such a number as 0.225 is called decimal and read “zero point two two five”.
Likewise, a decimal can be converted to a simple fraction. For example, convert 0.225 to a simple fraction.
Solution: 0,225 = 225 1000
Write it as a simple fraction.
Remember
You may read 1.32 as one-thirty two per one hundred”
225 1.000
=
9 40
Simplifiy by dividing the numerator and the denominator by the GCF. The GC F of 225 and 1,000 is 25.
Therefore, 0.225 = 9 . 40 When a decimal is greater than 1, it can be written in a mixed number.
Write 1.32 as a mixed number in the simplest form.
Mathematics for Junior High School – Year 7/53
Solution: 1.32 = 1 32 100
An integer is written separately from the fraction .
Simplify the fraction. The GCF of 100 and 32 is 4.
Therefore 1.32 = 1 8 . 25
In addition, to write a fraction in a decimal form, you may divide the numerator by the denominator. For instance, 3 can be calculated as 4 4
0.75 3 2.8 0.2 0.2 0
You could also use a calculator to divide the numerator by the denominator.
EXAMPLE 7
Relationwith the Real World
A carpenter wants to make a hole with a diameter of not more than 0.6 inch. Could he use
5 inch drill? 8
You may also use a calculator to divide 5 by 8.
Since 0.625 > 0.6, the carpenter could not use a drill with
54/Student’s Book – Fractions
a size of
5 inch because the hole is too big. 8
For a fraction, if you divide the numerator by the denominator and the remainder is zero, then the quotient is in the form of an irrecursive decimal number. On the other hand, if the quotient repeats a number or a set of certain numbers without ending, then such a decimal number is called a recursive decimal number. Example: 0.4444 . . . . = 0.4
The bar over the number me ans that the number 4 is recursive.
Rounding off If the decimal number is rounded off up to one decimal place, it can be written as 0.4. The number 4 cannot be changed because the number on its right (that is 4) is less than 5.
Write down the following fractions as decimal numbers.
a.
4 15
Solution:
0.266 4 3 1 0.9 0.1 0.09 0.1
15
The number 6 is recursive.
Therefore, 4 = 0.26 15
Mathematics for Junior High School – Year 7/55
b.
8 11
Solution: By using a calculator, we know that 72 is recursive.
Th erefore, 8 = 0.72 11
Rounding off • If the number 0.266 is rounded off up to one decimal place, it becomes 0.3 ( because 6 is larger than 5) • If the number 0.266 is rounded off up to two decimal places, it becomes 0.27 (because 6 is larger than 5) • If the number 0.725 is rounded off up to one decimal place, it becomes 0.7 ( because 2 is less than 5) • If the number 0.725 is rounded off up to two decimal places, it becomes 0.73.
Besides protein, meat contains fat, carbohydrate, vitamins, minerals and water. The amount of each depends on species, age and sex. Chicken, for example, has about 18% protein and 60-70% water.
Percent and Permil
Consider the copy of an article on the left. In this article, it is written 18% and 60-70%. Do you know what percentage is? If you compare a number to 100, then you will have a percentage.
Percent means “per hundred”.
56/Student’s Book – Fractions
75 12.5 You may write a ratio 15 as 15%, as 75%, as 100 100 100
12.5%, and so on.
You could make a model of percent with a scale paper of 10 x 10 as follows.
EXAMPLE 9
Modeling What is the percentage of the shadowed parts?
the number of shadowed parts 15 = the number of all parts 100
Write down the ratio as percentage.
15 = 15% 100
Therefore, the shadowed parts are 15% of all parts. What is the percentage of partitioned paper in Example 9 that is not shadowed? How do you answer it without counting the number of squares?
You may use what you already know about percent to express a percentage as a fraction.
EXAMPLE 10
Express 36% as a fraction in its simplest form.
Express the percent as a fraction having 100 as th e denominator.
36% = 36 100
Mathematics for Junior High School – Year 7/57
36 = 9 100 25
Express the fraction in its simplest form.
Sometimes, you need to express a fraction as a decimal first, before writing the equivalent
percentage. Natural Science About 7 of the earth surface is covered by water. 10 Express 7 in percent. 10
7 = 70 = 70% 10 100
EXAMPLE 11
0.7 = 70% You may use a calculator to convert a fraction to a percent as demonstrated in the following example.
EXAMPLE 12
Use a calculator to express the fraction 2 as a
3
percent.
Therefore, the fraction 2 is about 66.7%.
3
EXAMPLE 13
Express 1 as a percent. 4
1 = ..... 4 100 1 = 1 × 25 = 25 4 4 × 25 100
58/Student’s Book – Fractions
Thus, 1 is equal to 25%.
4
Permil Percent mean per hundred, while permil means ‘per thousand’ symbolized o / oo . per thousand,
23 can be read 23 1000
17.5 can be read 17.5 per thousand. 1000
EXAMPLE 14
Express
13 in per thousand. 25
Solution:
13 13 × 40 520 = = . 25 25 × 40 1000
Therefore,
13 is equal to 520 per thousand. 25
EXAMPLE 15
Express 125 per thousand as a fraction in its simplest form.
125 per thousand=
125 1000
Express per thousand as a fraction of which 1,000 is the denominator.
1 125 = 8 1000
Express th e fraction in its simples t form.
Mathematics for Junior High School – Year 7/59
1. Thirty-five percent of the members of a club play football as their hobbies. What is the percentage of the members that do not play football as their hobbies?
2. Write down the following percentages as fractions in the simplest forms. a. 15% b. 75% c. 88% d. 18%
3. Express each of the following fractions in per thousand. a.
3 20
b.
34 50
c.
18 150
d.
23 250
4. Biology.
The exhaled air consists of about 80%
nitrogen and 20% oxygen. Write down each percentage as a fraction in its simplest form.
5. Write down each fraction below as a percentage. a.
19 20
b. 7 50 g. 8 20
c. 1 4 h. 3 10
d. 1 8 i. 12 30
e.
3 200
f. 9 50
j. 2 25
6. Express each of the following decimal numbers as a fraction or a mixed number in its simplest form. a. 0.3 e. 5.500 b. 0.004 c. 2.625 d. 1.35
7. Round off the following decimal numbers up to one
60/Student’s Book – Fractions
and two decimal places. Give your reason for rounding them off. a. 0.075 b. 1.627 c. 0.155 d. 0.074 e. 10.023
8. Ali has 1-meter length of rope. This rope is cut into two parts, and one part has 0.55 meter length. Express the length of each part of the rope as a simple fraction.
9. Writin. Describe the steps to convert 0.8 to a fraction in its simplest form.
10. Express the following fractions as a decimal number. a.
3 20
b.
9 50
c.
7 32
d.
5 6
e.
11 16
11. Arrange the following numbers from the least to the greatest. a.
7 9 ; 0.8; ; 0.87 8 11 2 3 b. 1.65;1 ; 1 ; 1.7 3 5
c. 3
1 1 1 ; 3.1 ; 3 ; 3 ; 3.01 12 5 20 3 7 km, and Budi ran as far as 1 4 10
12. Ali ran as far as 1
km. Who ran farther than the other? 13. Arrange the following fractions in sequence, from the least to the greatest. a. 2 , 2 , 2 3 5 7 d. 3 , 2 , 3 5 7 8 b. 4 , 5 , 7 8 6 9 c. 1 2 ,1 3 , 1 5 3 4 6 f. 11 , 5 , 5 24 8 12
e. 2 8 , 2 17 , 2 5 9 18 6
Mathematics for Junior High School – Year 7/61
g. 7 , 1 , 7 15 3 12
h. 1 8 , 2 1 , 1 3 11 4 4
14. Critical Thinking. I am a fraction in my simplest form. My numerator and denominator are prime numbers having 2 as their differences. The sum of my numerator and denominator are the same as 12. What number am I?
15.
Writing. When you are given two different and
unequal fractions, write in your own words how you determine the greater fraction.
16.
Open Question. Write down three fractions and
arrange them in order, from the least to the greatest. Describe the steps that you apply to put the fractions in order.
17.
Describe in your own words how you find out
that a fraction is less than, the same as, or greater than 1.
18. Choose A, B, C, or D. Which mixed the
number
describes
shaded parts? A. 4 3 4 B. 3 3 4 C. 3 15 16 D. 3 1 4
62/Student’s Book – Fractions
19.
Critical Thinking. Express 100 as a fraction
using four similar numerals. Can it be expressed using six similar numerals? 20. Express each of the following fractions as a
mixed number. a. 17 5 b. 13 7 c. 27 5 d. 37 12 e. 21 f. 16 4 5
21. Writing.
Think of two different situations in your
everyday life in which you use a mixed number. 22. Physics. The formula to convert the degree in the Celsius scale into the Fahrenheit scale is Express the fraction 9 5 number. 23. Research. Measure the height of your classmate or
9 o C + 32 = o F . 5
in this formula as a mixed
family in a centimeter unit. When the height is over 100 cm, express such a measurement in a meter unit by using a mixed number.
24. Write down two fractions that are equal to each of the following fractions. a. 1 4 25. b. 10 c. 4 5 d. 15 e. 6
20
45
8
Open Question. Use the numbers 2, 3, 4, 6,
12, 18, and 24 to write 3 pairs of equal fractions.
Mathematics for Junior High School – Year 7/63
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# The Peculiar Puzzle of Geometrical Transformation-Insights into Transformation Geometry
Updated: Apr 11, 2021
Aloha! Excited about Geometric Math? Transformation? Dilation? Well, I am! Transformation is used everywhere! From operating a Mars rover to running a mile, transformation takes the lead. Geometric transformation is a marvel itself. Today, we will dive into the concepts of transformation.
## So, What is Geometrical Transformation or Transformation Geometry?
Transformation Geometry is when you move an object from its original position to a new position. The object in the new position is called an image. There are four types of geometric transformations: Dilation, translation, reflection, and rotation.
1.Dilation
Dilation is also frequently known as enlargement. When using the dilation technique we increase or decrease the size of the object. In this process, all the angles remain the same. In other words, the images are similar in shape to the original, but it is a different size. Dilation is unique from other transformations because it moves each point along a straight line. A straight line is drawn from the center of dilation. The distance from the point moved is based on the scale factor. If the scale factor is greater than 1, the size of the image is enlarged. If the scale factor is between 0 and 1 then the image will be smaller than the original.
### What is Scale Factor?
To calculate the scale factor you will need to use the formula: image length/original length=scale factor. In other words, image length is the distance of the image from the center of dilation and the original length is the distance of the object from the center of dilation. An example might be if we are facing an illustration showing the two similar triangles ABC and A'B'C'. As we can assume triangle A'B'C' is the dilated version of triangle ABC. D will remain as our center of dilation. Now we can tell that the scale factor is DA'/DA.
scale factor=image length/original length
### 2.Translation
Translation is when all the points of the original shape are moved in a straight line to draw an image. Translation may seem similar to dilation, but the difference is the shape, size, and orientation remain the same. In translation, every point is moved with the same distance and direction. We represent translation with commands such as “translate 5 units right and 2 units down.”
### 3.Reflection
When applying geometric reflection all points of the shape are reflected on a line called the axis of reflection or the line of reflection. A reflection technically mirrors the original object. The shape and size are the same as the original object. This is defined as an isometric transformation. Since the orientation is latterly inverted, the image and the original shape are facing opposite directions.
### 4.Rotation
The rotation transformation is when an object is rotated around a point. When rotation is in use, we decide to move the object clockwise or anticlockwise. The point at which the object is rotated is most commonly known as the center of rotation. The amount we rotate the object is said to be the angle of rotation. Don’t forget to always mention the center, angle, and direction when recording the data.
### What are the Geometric Transformation applications in daily life?
You may not have noticed, but everywhere geometric transformation has many applications. Architects may use the dilation transformation when building houses, roads, etc. Translation applies when flying a plane or cooking. Reflection can be related when you look into the mirror or in the footwear industry. Rotation is when you walk or pay a visit to the amusement park.
Aren’t geometric transformations interesting? Now it is thus proven that math can be fun. The geometric transformation has multiple applications in many industries, technologies, and daily life. Right now, you are using the rotation transformation! (heart is circulating blood). Dilation, translation, reflection, and rotation are all mysteries to be uncovered. Credits go to my math teacher who inspired me to write this post. Can’t wait until next time! Adios!
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# What is the distance between (-4,-11) and (13,-41)?
Mar 26, 2018
Distance$= 34.482 \ldots$
#### Explanation:
Apply Pythagorean theorem, where $d$ is the distance between the two points.
$d = \sqrt{{\left(13 - - 4\right)}^{2} + {\left(- 41 - - 11\right)}^{2}}$
$\textcolor{w h i t e}{d} = \sqrt{{\left(17\right)}^{2} + {\left(- 30\right)}^{2}}$
$\textcolor{w h i t e}{d} = \sqrt{1189}$
$\textcolor{w h i t e}{d} = 34.482 \ldots$
Mar 26, 2018
#### Explanation:
D=sqrt((y_2-y_1)^2+(x_2-x_1)^2
subbing in
$D = \sqrt{{\left(- 41 - \left(- 11\right)\right)}^{2} + {\left(13 - \left(- 4\right)\right)}^{2}}$
$D = \sqrt{900 + 289}$
$D = \sqrt{1189} u n i t s$
Mar 26, 2018
$d = \sqrt{1189}$
#### Explanation:
distance between $A \left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right)$:
$d = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}} = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
in this case : $d = \sqrt{{\left(- 4 - 13\right)}^{2} + {\left(- 11 - \left(- 41\right)\right)}^{2}}$
$d = \sqrt{289 + 900} = \sqrt{1189}$
|
# How do you solve and check your solutions to 3/2k=18?
Mar 11, 2018
$k = 12$
#### Explanation:
$\text{multiply both sides by 2 to eliminate the fraction}$
$\cancel{2} \times \frac{3}{\cancel{2}} k = 2 \times 18$
$\Rightarrow 3 k = 36$
$\text{divide both sides by 3}$
$\frac{\cancel{3} k}{\cancel{3}} = \frac{36}{3}$
$\Rightarrow k = 12$
$\textcolor{b l u e}{\text{As a check}}$
Substitute this value into the left side of the equation and if equal to the right side then it is the solution.
$\text{left "=3/cancel(2)^1xxcancel(12)^6=3xx6=18=" right}$
$\Rightarrow k = 12 \text{ is the solution}$
|
How do you graph y=-2x+7?
Jun 2, 2018
See a solution process below:
Explanation:
First, solve for two points which solve the equation and plot these points:
First Point: For $x = 2$
$y = \left(- 2 \cdot 2\right) + 7$
$y = - 4 + 7$
$y = 3$ or $\left(2 , 3\right)$
Second Point: For $x = 4$
$y = \left(- 2 \cdot 4\right) + 7$
$y = - 8 + 7$
$y = - 1$ or $\left(4 , - 1\right)$
We can next plot the two points on the coordinate plane:
graph{((x-2)^2+(y-3)^2-0.035)((x-4)^2+(y+1)^2-0.035)=0 [-10, 10, -5, 5]}
Now, we can draw a straight line through the two points to graph the line:
graph{(y + 2x - 7)((x-2)^2+(y-3)^2-0.035)((x-4)^2+(y+1)^2-0.035)=0 [-10, 10, -5, 5]}
|
### Mean, Median and Mode
#### Mean
##### Definition
The mean of a set of numbers in a data set is obtained by adding up all the numbers then dividing by the size of the data set. When people use the word 'average' in everyday conversation, they are usually referring to the mean.
#### Worked Example 1
###### Worked Example
The ages of people in the checkout queue at Aldi are as follows: $23, 54, 2, 6, 20, 25, 21, 64, 19, 19, 75, 36\text{.}$ Find the mean.
###### Solution
To find the mean add all of the observed numbers together then divide by the number of observations, which, in this case, is $12$.
$\dfrac{23+54+2+6+20+25+21+64+19+19+75+36}{12}=\dfrac{364}{12}=30.33333\ldots$
##### Population Mean
If (which is unusual) we have information for the entire population, we use the term population mean for, as you would expect, the mean of the entire population. We represent the population mean by $\mu$. If we have data for the entire population, we can calculate it in the same way: $\mu = \frac{1}{N} \sum\limits_{i=1}^N x_i$
where $N$ is the size of the population consisting of $x_1, x_2, \ldots, x_N$.
##### Sample Mean
When we have taken a sample of $n$ observations $x_1, x_2, x_3,...,x_n$ from the underlying population, we use the term sample mean for the mean of $x_1, x_2, x_3,...,x_n$. It is represented by $\bar{x} = \frac{1}{n} \sum\limits_{i=1}^n x_i$ where $n$ is the size of the sample and $x_1, x_2, \ldots, x_n$ are the $n$ observations obtained. This is exactly the same as what has been done above, it is just a more formal way of expressing it.
#### Median
##### Definition
The median is usually described as the ‘middle number’. We can obtain the median by ordering the data in terms of size, then:
• For an odd number of observations, simply take the middle number.
• For an even number of observations, take the two middle numbers, add them together and divide by $2$.
When you have a large data set it is often more useful to find the position of the median within the data set. This is given by $\frac{n+1}{2}$ where $n$ is the number of data values in the data set.
#### Worked Example 2
###### Worked Example
The ages of people in the checkout queue at Aldi are as follows: $23, 54, 2, 6, 20, 25, 21, 64, 19, 19, 75, 36\text{.}$
Find the median.
###### Worked Example
To find the median first reorder the numbers in terms of size.
$2 , 6 , 19 , 19 , 20 , 21 , 23 , 25 , 36 , 54, 64 , 75\text{.}$
The number of data entries is $12$ so the position of the median is
$\frac{n+1}{2}=\frac{12+1}{2}=\frac{13}{2}=6.5\text{.}$
This means the median is between the $6$th and $7$th values, which are $21$ and $23$ respectively. In this case we compute
$\dfrac{21+23}{2}=22\text{.}$
So $22$ is the median age of people in the checkout queue at Aldi.
#### Mode
##### Definition
The mode is the most common number that appears in your set of data. To find the mode count how often each number appears and the number that appears the most times is the mode.
#### Worked Example 3
###### Worked Example
The ages of people in the checkout queue at Aldi are as follows: $23, 54, 2, 6, 20, 25, 21, 64, 19, 19, 75, 36\text{.}$
###### Solution
The age that appears most frequently is the number $19$; so the modal age is $19$.
#### Workbook
This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.
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# What is the distance between (3, –1, 1) and (2, –3, 1) ?
Apr 3, 2018
Distance b/w the pts.=$\sqrt{5}$ units.
#### Explanation:
let the pts. be A(3,-1,1) & B(2,-3,1)
so, By distance formula
$A B = \sqrt{\left({\left({x}_{2} - {x}_{1}\right)}^{2}\right) + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$
$A B = \sqrt{{\left(2 - 3\right)}^{2} + {\left(- 3 + 1\right)}^{2} + {\left(1 - 1\right)}^{2}}$
$A B = \sqrt{1 + 4 + 0}$
$A B = \sqrt{5}$ units.
Apr 3, 2018
The distance between $\left(3 , - 1 , 1\right)$ and $\left(2 , - 3 , 1\right)$ is $\sqrt{5} \approx 2.236$.
#### Explanation:
If you have a point $\left({x}_{1} , {y}_{1} , {z}_{1}\right)$ and another point $\left({x}_{2} , {y}_{2} , {z}_{2}\right)$ and you want to know the distance, you can use the distance formula for a normal pair of $\left(x , y\right)$ points and add a $z$ component. The normal formula is $d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$, so when you add a $z$ component, it becomes $d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$. For your points, you would say $\sqrt{{\left(2 - 3\right)}^{2} + {\left(\left(- 3\right) - \left(- 1\right)\right)}^{2} + {\left(1 - 1\right)}^{2}}$ which simplifies to $\sqrt{5} \approx 2.236$
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# How to get percentage of 2 numbers: Real-Life Applications
0
903
We use percentages daily, often without even realizing it. Whether calculating tips at a restaurant or working out a sale price at the store, percentages are a part of our lives. So how to get a percentage of 2 numbers? This blog post will look at some real-life applications of calculating percentages. By the end, you’ll better understand how this math concept is used in the world around us.
## Sales and Discounts
A big part of percentages is finding the original price when the percent off is given, and vice versa. For example, say a shirt is on sale for 20% off. To find the original price of the shirt, we take 100% – 20%, which equals 80%. This means that 80% of the original price equals the sale price. Therefore, if the sale price of the shirt is \$36, we can calculate that100%=\$45.
We divide the sale price by the original price to determine an item’s percentage. Using our example from before, we would take 36/\$45=.8 or 80%. To make it a percentage, we move the decimal over two places and get 80%.
When discounts and sales are given as “X dollars off,” we can still use percentages to find either value. However, if we’re trying to find what percent off an item is being sold for, we must first convert the number of dollars into a decimal form.
## Food
Percentages also show up when talking about food. For starters, recipes almost always use percentages when giving ingredient measurements. For example, a typical recipe might call for 1/2 cup (or 50%) of sugar. When doubling or halving a recipe, it’s helpful to know that 1/4 cup (or 25%) equals two tablespoons (or 16 teaspoons).
This can be useful for cooking and baking, where specific measurements are crucial to success. Likewise, a pie crust recipe may call for 2/3 cup (or 66 2/3 %)of butter, which can then be halved to 1/3 cup (or 33 1/3 %)of butter if making only half a recipe’s worth of crust measuring food can also tell us how healthy it is.
The typical macronutrient breakdown in terms of percentages is 60% carbohydrates, 20% protein, and 20% fat. While this isn’t a hard-and-fast rule, it’s an excellent general guideline to follow to maintain a balanced diet.
## Investing
Regarding investing, percentages are key to understanding how much money you stand to make (or lose). For example, let’s say you invest \$1,000 in a stock that goes up 10% over a year. This means that your investment is now worth \$1,100. To calculate the percentage gain, we take the difference between the new value and the original value (\$1,100 – \$1,000), and then we divide it by the original value (\$1,000). This gives us a 10% return on our investment.
Conclusion:
As you can see, percentages pop up all around us in everyday life! From sales and discounts to recipes and food labels, understanding how to calculate percentages can be helpful in so many ways. I hope this blog post has given you a better understanding of where and how percentages are used in real-life situations. Thanks for reading!
|
Q&A
# what does inverse operation mean
Inverse operationsare pairs of mathematical manipulations in which one operation undoes the action of the other—for example, addition and subtraction, multiplication and division.
## What is an inverse operation example?
The set of two opposite operations is called inverse operations. For example: If we add 5 and 2 pens, we get 7 pens. Now subtract 7 pens and 2 pens and we get 5 back. Here, addition and subtraction are inverse operations.
## What does inverse operation mean for kids?
Inverse operations are opposite operations. Subtraction is the inverse of addition and division is the inverse of multiplication.
## What is the inverse operation of a function?
An inverse function essentially undoes the effects of the original function. If f(x) says to multiply by 2 and then add 1, then the inverse f(x) will say to subtract 1 and then divide by 2. If you want to think about this graphically, f(x) and its inverse function will be reflections across the line y = x.
## What is the inverse operation of 5?
Example: The multiplicative inverse of 5 is 15, because 5 × 15 = 1.
## Can you give more examples of inverse?
Examples of inverse operations are: addition and subtraction; multiplication and division; and squares and square roots.
## What is a real life example of inverse operations?
An example of an action and its inverse that you might experience in your everyday life is when you put on your shoes in the morning and then take them off at night. Untying your shoe is the inverse of tying your shoe. Another example of an action and its inverse is the wrapping and unwrapping of a present.
## What does inverse mean example?
The Inverse of Subtraction is Addition 10 – 6 = 4 can be written as 6 + 4 = 10 or 4 + 6 = 10. The number at the start of a subtraction that we are subtracting from becomes the answer of an addition sentence. The two other numbers in the subtraction are the two numbers that get added together in an addition sentence.
## What is inverse of a function with example?
The inverse function returns the original value for which a function gave the output. If you consider functions, f and g are inverse, f(g(x)) = g(f(x)) = x. A function that consists of its inverse fetches the original value. Then, g(y) = (y-5)/2 = x is the inverse of f(x).
## What is an example of inverse?
Inverse operations are opposite operations that undo each other. For example, 5 ✕ 2 = 10 and 10 ÷ 2 = 5 are inverse operations.
## What are the 4 inverse operations in math?
Examples of inverse operations are: addition and subtraction; multiplication and division; and squares and square roots.
## What are inverse operations simple definition?
Inverse operationsare pairs of mathematical manipulations in which one operation undoes the action of the other—for example, addition and subtraction, multiplication and division.
## What does inverse mean in math for kids?
In mathematics, the word inverse refers to the opposite of another operation. Let us look at some examples to understand the meaning of inverse. Example 1: The addition means to find the sum, and subtraction means taking away. So, subtraction is the opposite of addition.
## What is inverse operation 3rd grade?
Inverse operations are opposite operations. They are the operation that reverses the effect of another operation. For example, addition is the inverse operation of subtraction and multiplication is the inverse operation of division .
## What is the inverse operation of a function?
An inverse function essentially undoes the effects of the original function. If f(x) says to multiply by 2 and then add 1, then the inverse f(x) will say to subtract 1 and then divide by 2. If you want to think about this graphically, f(x) and its inverse function will be reflections across the line y = x.
## What is the inverse of a 5?
The multiplicative inverse of 5 is 1/5. The multiplicative inverse property states that any number a multiplies with its reciprocal, 1/a, to give 1. Therefore, the multiplicative inverse of a number a is 1/a. Applying this property to the number 5, we get that the multiplicative inverse of 5 is 1/5.
## What is inverse of negative 5?
Example: Let a negative integer be -5. Then, the additive inverse of -5 = -(-5) = 5.
## What is the inverse of positive 5?
This rule gives that the additive inverse of 5 is found by changing positive 5 to negative 5, or -5. To make sure that -5 is the additive inverse of 5, we simply add the two numbers together and make sure we get 0. We see that if we add -5 to 5, then we get 0, so the additive inverse of 5 is -5.
## What is inverse operation for Grade 5?
Multiplication and division are the inverse operations, which means multiplication undoes division and division undoes multiplication. We can rearrange the numbers given in the multiplication equation and then we can do two different division equations. For example: The two division equations formed are: 28 ÷ 4 = 7 and.
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# Teaching for mastery in primary maths
Year 1
Year 1
Year 2
Year 3
Year 4
Year 5
Year 6
Autumn
Autumn
Spring
Summer
All Topics
## Numbers up to 20
• Year group: Year 1
• Term: Autumn
• Topic: Number and place value
• Duration: 2 weeks
• Keystage: KS1
• Age Range: 5 to 6 yrs
• Unit can be taught out of sequence: N
• Week Range: 10 - 11
In this chapter, pupils are introduced to the numbers 11 to 20. It takes a similar approach to the previous chapter in that pupils continue to practise counting, comparing and ordering numbers.
The key to this chapter, however, is that pupils start to partition numbers above 10 into tens and ones. They use part-whole diagrams to reinforce their understanding and, as before, work with many different concrete and pictorial representations of numbers.
Pupils continue to use language, including “equal to”, “more than”, “less than”, “fewer”, “most” and “least”. They count forwards or backwards to 20, beginning at 0 or 1, or from any given number. They can also identify “one more” and “one less” than a given number.
This chapter lays the essential foundations of place value, as pupils begin to recognise the place value of each digit in a two-digit number (ie, in tens and ones).
Download all number and place value chapters as a PDF
## Resources for this topic
### White Rose Maths
#### Year1 - Numbers within 20 - More or Less
A worksheet for Y1 teachers to use with children looking at comparing numbers within 20. The sheet gets pupils to represent numbers using pictures and ten frames and encourages use of language of more or less
#### Reasoning - Problem Solving - Money Problems (KS1/2) - March 8th
Today's problems are ones that involve reasoning and problem solving with the numbers. There are two problems for KS1 and two for KS2.
In KS1 problem children have to use a number line to work out the size of the interval and then locate a particular number. Children then have to work out the highest and lowest numbers you can make given three digit cards. Version 2 of the Y2 problems has a slightly less demanding problem Q1 that might be good as a warm up.
At KS2 children also need to find intervals on a number line, but the problem is a little more involved.. The second problem should be tacked systematically.
Every day in March the White Rose Maths Hub is aiming to put out a question of the day.
Tweet us @WRMathsHub a picture of your children's answers for a chance to win a prize.
### Mathematics Mastery
#### Numbers up to 20
Mathematical discussion is a key element of the Mathematics Mastery programme.
Talk Tasks are a brilliant way of enabling pupils to develop their mathematical language, thinking, understanding and confidence.
This activity explores the concept of numbers up to 20.
• Talk tasks are best completed in mixed attainment pairs, with pupils taking turns to listen and construct arguments
• Emphasis is on discussion rather than the solution, encouraging pupils to use key language and talk in full sentences
• Ask pupils to prove their answers in a variety of ways, using the CPA approach to support mathematical understanding and number sense
Like what you see?
Mathematics Mastery is a professional development programme for teachers with a mission to transform mathematics education in the UK.
This task is just a taster of the complete classroom resources we offer. We also provide in-depth development training, online CPD, specialist support and assessment tools.
We believe all elements of our programme are vital in creating lasting change – enabling every child to enjoy and succeed in mathematics.
Want to find out more? Check out our free resources and blogs or join an information session.
#### Numbers up to 20
Enabling pupils to develop their mathematical reasoning skills through independent work is a key element of the Mathematics Mastery programme.
This activity explores the concept of numbers up to 20
• Pupils should complete these tasks using concrete manipulatives as part of the CPA approach
• Pupils should be encouraged to explain their working in full sentences
• These tasks may be adapted with pupils working in pairs to encourage mathematical discussion
Like what you see? Mathematics Mastery is a professional development programme for teachers with a mission to transform mathematics education in the UK.
This task is just a taster of the complete classroom resources we offer. We also provide in-depth development training, online CPD, specialist support and assessment tools.
We believe all elements of our programme are vital in creating lasting change – enabling every child to enjoy and succeed in mathematics.
Want to find out more? Check out our free resources and blogs or join an information session.
### Resources shared by teachers
#### Year 1 - Week 10 - Number - Place Value 2
This is pack 1 of 2 on the second block of year 1 Place Value and covers the small steps:
- Counts forwards and backwards and write numbers to 20 in numerals and words
- Numbers from 11 to 20
- Tens and Ones
- Count One More and One Less
The resources aim to help children gain a deeper understanding of the concepts covered in the topic.
The resources include pictorial and abstract representations as well as reasoning and problem solving activities and more open ended problems to support and extend your whole class.
I hope you find them useful.
#### Year 1 - Week 11 - Number - Place Value 2
This is pack 2 of 2 on the second block of year 1 Place Value and covers the small steps:
- Compare Groups of Objects
- Compare Numbers
- Order Groups of Objects
= Order Numbers
The resources aim to help children gain a deeper understanding of the concepts covered in the topic.
The resources include pictorial and abstract representations as well as reasoning and problem solving activities and more open ended problems to support and extend your whole class.
I hope you find them useful.
#### CURLY CATERPILLAR ACTIVITY, counting, numbers
A caterpillar numberline - use as you will I have used as a sequencing, ordering, missing number, follow the pattern sequence etc etc.
#### Matching numbers 11 - 20
Cards that can be used as a matching game with numbers, words and pictorial representations.
#### Self checking counting games
pupils love the fact that these 'magic' cards are self checking - good for one to one counting and numeral recognition
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# What is the probability of getting at least 35 cents?
$$3$$ nickels, $$1$$ dime and $$2$$ quarters are in a purse. In picking three coins at one time (without replacement), what is the probability of getting a total of at least $$35$$ cents?
I know that the total ways of selecting 3 coins out of 6 is $${6\choose 3}=20$$. In order to have a total at least 35 cents, there must be at least 1 quarter among the 3 chosen coins. But I don't know how to proceed.
One way to do this problem is to find the probability that 3 chosen coins are not quarters. This is $${4\choose 3}=4$$ (numbers of ways choosing three coins out of 4 non-quarters). Then I take the total 20 and subtract 4 to get 16. The probability of getting a sum at least 35 is therefore $$\frac{16}{20}$$.
Is there any other methods to solve this problem?
Another way would be to just directly calculate the probability of picking 3 non-quarters, and subtract that probability from 1, since at least one coin needs to be a quarter for 35 cents or more:
first draw, 4/6
second draw, 3/5
third draw, 2/4
Multiplying these together:
(4*3*2)/(4*5*6) = 1/5, and 1 - 1/5 = 4/5 = 16/20.
I do think that recognizing that this question can be rephrased as "what's the probability of drawing at least 1 quarter?" is the key to making this problem easier, doing either your way or mine above; there are likely other ways as well though.
• Just to state explicitly, you will always get at least 35 cents if you have at least 1 quarter, and you will never get at least 35 cents if you have less than 1 quarter. This is the reason the problem splits into only two possibilities. If there was, for example, 1 penny, then you could draw 1 quarter, 1 penny, 1 nickel, and not reach 35 cents. In that case, the problem would not simplify in this way. – AlexanderJ93 Oct 15 '18 at 19:36
• Right, I should've mentioned that in the initial answer, I didn't elaborate more at the end. Without that simplification the problem is certainly more challenging. – Paul Oct 15 '18 at 19:45
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# 4.6: Triangle Congruence using AAS
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Understand and apply the AAS Congruence Theorem.
## AAS Congruence
Another way you can prove congruence between two triangles is by using two angles and the non-included side.
Angle-Angle-Side (AAS) Congruence Theorem
If two angles and a non-included side in one triangle are congruent to two corresponding angles and a non-included side in another triangle, then the triangles are congruent.
In the AAS Theorem, you use two _______________ and a _____________________________ side to prove congruence.
This is a theorem because it can be proven. First, we will do an example to see why this theorem is true, then we will prove it formally. Like the ASA Postulate, the AAS Theorem uses two angles and a side to prove triangle congruence. However, the order of the letters (and the angles and sides they stand for) is different.
The AAS Theorem is equivalent to the ASA Postulate because when you know the measure of two angles in a triangle, you also know the measure of the third angle. The pair of congruent sides in the triangles will determine the size of the two triangles. We will explore this further in the last section of this lesson.
Notice that when you look at the congruent triangles in a clockwise direction (beginning at C\begin{align*}\angle C\end{align*} and Z\begin{align*}\angle Z\end{align*}), the congruent parts spell A-A-S, but when you look at them in a counter-clockwise direction, they spell S-A-A.
The AAS Theorem is similar to the _____________________________________________ because they both use two angles and a side to prove congruence.
Example 1
What information would you need to prove that these two triangles were congruent using the AAS Theorem?
A. the measures of sides TW¯¯¯¯¯¯¯¯¯\begin{align*}\overline{TW}\end{align*} and XZ¯¯¯¯¯¯¯¯\begin{align*}\overline{XZ}\end{align*}
B. the measures of sides VW¯¯¯¯¯¯¯¯¯\begin{align*}\overline{VW}\end{align*} and YZ¯¯¯¯¯¯¯\begin{align*}\overline{YZ}\end{align*}
C. the measures of VTW\begin{align*}\angle VTW\end{align*} and YXZ\begin{align*}\angle YXZ \end{align*}
D. the measures of angles TWV\begin{align*}\angle TWV\end{align*} and XZY\begin{align*}\angle XZY \end{align*}
If you are to use the AAS Theorem to prove congruence, you need to know that pairs of two angles are congruent and the pair of sides adjacent to one of the given angles are congruent.
You already have one side and its adjacent angle, but you still need another angle. It needs to be the angle not touching the known side, rather than adjacent to it. Therefore, you need to find the measures of TWV\begin{align*}\angle TWV \end{align*} and XZY\begin{align*}\angle XZY\end{align*} to prove congruence.
Then you would have:
TWVWVTVT¯¯¯¯¯¯¯(A)(A)(S)\begin{align*}\angle TWV & \cong \underline{\;\;\;\;\;\;\;\;\;\;\;\;} (A)\\ \angle WVT & \cong \underline{\;\;\;\;\;\;\;\;\;\;\;\;} (A)\\ \overline{VT} & \cong \underline{\;\;\;\;\;\;\;\;\;\;\;\;} (S)\end{align*}
These spell A-A-S.
The correct answer is D.
When you use AAS (or any triangle congruence postulate) to show that two triangles are congruent, you need to make sure that the corresponding pairs of angles and sides actually align.
When using triangle congruence postulates, it is important for ____________________________ angles and sides to match up.
For instance, look at the diagram below:
Even though two pairs of angles and one pair of sides are congruent in the triangles, these triangles are NOT congruent. Why?
Notice that the marked side in ΔTVW\begin{align*}\Delta TVW\end{align*} is TV¯¯¯¯¯¯¯\begin{align*}\overline{TV}\end{align*}, which is between the unmarked angle and the angle with two arcs.
However in ΔKML\begin{align*}\Delta KML\end{align*}, the marked side is between the unmarked angle and the angle with one arc.
Since the corresponding parts DO NOT match up, you CANNOT use AAS to say these triangles are congruent.
If you want to prove that two triangles are ________________________________, you must be careful to make sure that _______________________________ parts of the triangles match up!
1. In the space below, sketch two congruent triangles and mark the parts that are congruent by the AAS Theorem.
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
2. In the space below, sketch two congruent triangles and mark the parts that are congruent by the ASA Postulate.
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
## AAS and ASA Congruence
The AAS Triangle Congruence Theorem is logically the exact same as the ASA Triangle Congruence Postulate. Look at the following diagrams to see why:
You can see the following in the figure on the previous page:
• CZ\begin{align*}\angle C \cong \angle Z\end{align*} because both angles have two arcs
• BY\begin{align*}\angle B \cong \angle Y\end{align*} because both angles have one arc
The congruent parts in the figure spell A-A-S, so based on these markings you see, the triangles are congruent because of the AAS Theorem.
ALSO,
Since CZ\begin{align*}\angle C \cong \angle Z\end{align*} and BY\begin{align*}\angle B \cong \angle Y\end{align*}, we can conclude from the Third Angle Theorem that AX\begin{align*}\angle A \cong \angle X\end{align*}. This is because the sum of the measures of the three angles in each triangle is 180\begin{align*}180^\circ\end{align*} and if we know the measures of two of the angles, then the measure of the third angle is already determined. We therefore know that all three angles in both triangles are congruent.
We know that AX\begin{align*}\angle A \cong \angle X\end{align*} because all three angles in a triangle add up to _______________.
Marking AX\begin{align*}\angle A \cong \angle X\end{align*}, the diagram becomes this:
Now we can see that:
• AX (A)\begin{align*}\angle A \cong \angle X \ (A)\end{align*}
• AB¯¯¯¯¯¯¯¯XY¯¯¯¯¯¯¯¯ (S)\begin{align*}\overline{AB} \cong \overline{XY} \ (S)\end{align*}
• and BY (A)\begin{align*}\angle B \cong \angle Y \ (A)\end{align*}
which shows that ΔABCΔXYZ\begin{align*}\Delta ABC \cong \Delta XYZ\end{align*} is also true by the ASA Postulate.
1. True/False: When you use the AAS Congruence Theorem OR the ASA Congruence Postulate, you can prove that all angles in both triangles are congruent with the Third Angle Theorem.
2. True/False: The AAS Theorem and the ASA Postulate are logically completely different.
## Graphic Organizer for Lessons 3-5
Proving Triangle Congruence – Postulates and Theorems
Type of Congruency Letters stand for... Postulate or Theorem? Draw a picture Describe the corresponding congruent parts
SSS
SAS
ASA
AAS
HL
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Concept: Factorial
CONTENTS
INTRODUCTION
Factorial, in mathematics, is the product of all positive integers less than or equal to a given positive integer and is denoted by that integer and an exclamation point.
Thus, factorial four is written 4!, which means 1 × 2 × 3 × 4
Hence, n! = 1 × 2 × 3 × ... × (n-1) × n
Remember:
• n! = n × (n - 1)! = n × (n - 1) × (n - 2)!
• 0! = 1
• 1! = 1
• 2! = 2
• 3! = 6
• 4! = 24
• 5! = 120
• 6! = 720
HIGHEST POWER OF P (PRIME) IN N!
If 'p' is a prime number, then highest power of 'p' in N! is sum of all quotients when N is successively divided by p.
Example: Find the highest power of 2 in 10!.
Solution:
To find the highest power of 2 in 10!, we successively divide 10 by 2 and then add all the quotients.
$\mathrm{Q}\left[\frac{10}{2}\right]$ = 5
$\mathrm{Q}\left[\frac{5}{2}\right]$ = 2
$\mathrm{Q}\left[\frac{2}{2}\right]$ = 1
(We stop the division when quotient is less than the divisor)
∴ Highest power of 2 in 10! = 5 + 2 + 1 = 7
Note: Since, highest power of 2 in 10! is 7, it means 27 will be a factor of N!.
∴ N! can be written as 27 × X
where X is a factor of N! which is coprime to 2.
Example: Find the highest power of 5 in 100!.
Solution:
To find the highest power of 5 in 100!, we successively divide 100 by 5 and then add all the quotients.
$\mathrm{Q}\left[\frac{100}{5}\right]$ = 20
$\mathrm{Q}\left[\frac{20}{5}\right]$ = 4
∴ Highest power of 5 in 100! = 20 + 4 = 24
HIGHEST POWER OF A COMPOSITE NUMBER IN N!
To find highest power of a composite number in N!, we first prime factorise the given composite number and then find the highest powers of prime numbers.
Example: Find the highest power of 6 in 10!.
Solution:
To find the highest power of 6 in 10!, we prime factorise 6 as 2 × 3. Now we will find the highest power of 2 and 3 in 10!
Highest power of 2 in 10! is 7 (obtained in previous example).
To find the highest power of 3 in 10!, we successively divide 10 by 3 and then add all the quotients.
$\mathrm{Q}\left[\frac{10}{3}\right]$ = 3
$\mathrm{Q}\left[\frac{3}{3}\right]$ = 1
∴ Highest power of 3 in 10! = 3 + 1 = 4
∴ N! can be written as 27 × 34 × X (X is coprime to both 2 and 3)
⇒ N! = 23 × 64 × X
∴ Highest power of 6 in 10! is 4.
Example: Find the highest power of 4 in 50!.
Solution:
To find the highest power of 4 in 5!, we prime factorise 4 as 22. Now we will find the highest power of 2 in 50!
To find the highest power of 2 in 50!, we successively divide 50 by 2 and then add all the quotients.
∴ Highest power of 2 in 50! = 25 + 12 + 6 + 3 + 1 = 47
∴ 50! can be written as 247 × X
⇒ 50! = (22)23 × 2 × X
⇒ 50! = 423 × 2 × X
∴ Highest power of 4 in 50! is 23.
Example: Find the highest power of 12 in 50!.
Solution:
To find the highest power of 12 in 50!, we prime factorise 12 as 22 × 3. Now we will find the highest power of 2 and 3 in 50!
Highest power of 2 in 50! is 47 (obtained in previous example).
To find the highest power of 3 in 50!, we successively divide 50 by 3 and then add all the quotients.
∴ Highest power of 3 in 50! = 16 + 5 + 1 = 22
∴ 50! can be written as 247 × 322 × X
⇒ 50! = (22)22 × 23 × 322 × X
⇒ 50! = 422 × 322 × 23 × X
⇒ 50! = 1222 × 23 × X
∴ Highest power of 12 in 50! is 22.
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# KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.1
KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.1.
## Karnataka Board Class 9 Maths Chapter 1 Number Systems Ex 1.1
KSEEB Solutions For Class 9 Maths Question 1.
Is a zero a rational number? Can you write it in the form $$\frac{p}{q}$$, where p and q are integers and q ≠ 0 ? (p, q,ϵ Z, q ≠ 0)
Zero is a rational number.
This can be written in the form of $$\frac{p}{q}$$ because $$\frac{o}{q}$$ is a rational number.
E.g. $$\frac{0}{2}=0, \quad \frac{0}{5}=0$$. etc.
Zero belongs to set of rational number.
KSEEB Solutions For Class 9 Maths Number System Question 2.
Find six rational numbers between 3 and 4.
We can write six rational numbers between 3 and 4 as
$$3=\frac{21}{7} \text { and } 4=\frac{28}{7}$$
∴ rational numbers between $$\frac{21}{7}$$ and $$\frac{28}{7}$$.
KSEEB Solutions For Class 9 Maths Chapter 1 Question 3.
Find five rational numbers between $$\frac{3}{5}$$ and $$\frac{4}{5}$$
Rational numbers between $$\frac{3}{5}$$ and $$\frac{4}{5}$$ are
∴ Rational numbers between $$\frac{30}{50}$$ and $$\frac{40}{50}$$
a
KSEEB Solutions For Class 9 Maths Chapter 1 Number System Question 4.
State whether the following statements are true or false. Give reasons for your answers :
(i) Every natural number is a whole number.
True. Because set of natural numbers belongs to a set of whole numbers.
∴ W = {0, 1, 2, 3 …………………….}
(ii) Every integer is a whole number.
False. Because $$\frac{1}{2}$$ is a rational number but not a whole number.
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# What is the equation of the line passing through (9,2) and (9,14)?
May 8, 2016
$x = 9$
#### Explanation:
As it is a line that passes through $\left(9 , 2\right)$ and $\left(9.14\right)$, when either abscissa or ordinate is common, we can easily find the equation of the line - as it will of the form $x = a$, if abscissa is common and of the form $y = b$, if ordinates are common.
In the given case, abscissa is common and is $9$, hence equation is $x = 9$.
May 8, 2016
$x = 9$
#### Explanation:
Gradient $\to \left(\text{change in y")/("change in x}\right)$
Let point 1 be:$\text{ } {P}_{1} \to \left({x}_{1} , {y}_{1}\right) \to \left(9 , 2\right)$
Let point 2 be$\text{ } {P}_{2} \to \left({x}_{2} , {y}_{2}\right) \to \left(9 , 14\right)$
Notice that there is no change in $x$
This means that the line is parallel to the y axis (vertical)
Put another way: $x$ is always 9 and you may allocate any value you wish to $y$
The way to write this mathematically is $x = 9$
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1. ## HELPfocus of parabola!!
could someone please walk through the steps of solving this problem....?
find the focus of the conic section y=-3x^2-12x-5
2. Hello, emetty90!
Find the focus of the parabola: .$\displaystyle y\:=\:-3x^2-12x-5$
We must get the equation into the form: .$\displaystyle (x-h)^2 \:=\:4p(y-k)$
Then the vertex is at $\displaystyle (h,k)$ and $\displaystyle p$ is the directed distance from the vertex to the focus.
We have: .$\displaystyle 3x^2 + 12x \:=\:\text{-}y - 5$
Divide by 3: .$\displaystyle x^2 + 4x \:=\:\text{-}\tfrac{1}{3}y - \tfrac{5}{3}$
Complete the square: .$\displaystyle x^2 + 4x {\color{red}\;+\; 4} \:=\:\text{-}\tfrac{1}{3}y - \tfrac{5}{3} {\color{red}\;+\; 4} \quad\Rightarrow\quad (x+2)^2 \:=\:\text{-}\tfrac{1}{3}y+\tfrac{7}{3}$
. . And we have: .$\displaystyle (x+2)^2 \;=\;\text{-}\tfrac{1}{3}(y - 7)$
The vertex is: .$\displaystyle V(-2,7)$ . . . and the parabola opens downward.
Since $\displaystyle 4p = \text{-}\tfrac{1}{3} \quad\Rightarrow\quad p \:=\:\text{-}\tfrac{1}{12}$ . . . The focus is $\displaystyle \tfrac{1}{12}$ unit below the vertex.
Therefore, the focus is: .$\displaystyle \left(\text{-}2,\,7\,\text{-}\,\tfrac{1}{12}\right) \;=\;\left(\text{-}2,\frac{83}{12}\right)$
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Theoretical and Experimental Spinners
Using APPS on the TI 84 calculator for spinner simulations
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Theoretical and Experimental Spinners
If you have a spinner that is divided into equal portions of red, green, purple, and blue, what is the probability that when you spin you will land on blue? If you spin the spinner 100 times will you get this same result? What about 1,000 times?
Watch This
First watch this video to learn about theoretical and experimental spinners.
Then watch this video to see some examples.
Guidance
The 2 types of probability are theoretical probability and experimental probability. Theoretical probability is defined as the number of desired outcomes divided by the total number of outcomes.
Theoretical Probability
$P(\text{desired}) = \frac{\text{number of desired outcomes}}{\text{total number of outcomes}}$
Experimental probability is, just as the name suggests, dependent on some form of data collection. To calculate the experimental probability, divide the number of times the desired outcome has occurred by the total number of trials.
Experimental Probability
$P(\text{desired}) = \frac{\text{number of times desired outcome occurs}}{\text{total number of trials}}$
You can try a lot of examples and trials yourself using the NCTM Illuminations page found at http://illuminations.nctm.org/activitydetail.aspx?ID=79 .
What is interesting about theoretical and experimental probabilities is that, in general, the more trials you do, the closer the experimental probability gets to the theoretical probability. We'll see this as we go through the examples.
Example A
You are spinning a spinner like the one shown below 20 times. How many times does it land on blue? How does the experimental probability of landing on blue compare to the theoretical probability? Simulate the spinning of the spinner using technology.
On the TI-84 calculator, there are a number of possible simulations you can do. You can do a coin toss, spin a spinner, roll dice, pick marbles from a bag, or even draw cards from a deck.
After pressing $\boxed{\text{ENTER}}$ , you will have the following screen appear.
Since we're doing a spinner problem, choose Spin Spinner.
In Spin Spinner, a wheel with 4 possible outcomes is shown. You can adjust the number of spins, graph the frequency of each number, and use a table to see the number of spins for each number. We want to set this spinner to spin 20 times. Look at the keystrokes below and see how this is done.
In order to match our color spinner with the one found in the calculator, you will see that we have added numbers to our spinner. This is not necessary, but it may help in the beginning to remember that 1 = blue (for this example).
Now that the spinner is set up for 20 trials, choose SPIN by pressing $\boxed{\text{WINDOW}}$ .
We can see the result of each trial by choosing TABL, or pressing $\boxed{\text{GRAPH}}$ .
And we see the graph of the resulting table, or go back to the first screen, simply by choosing GRPH, or pressing $\boxed{\text{GRAPH}}$ again.
Now, the question asks how many times we landed on blue (number 1). We can actually see how many times we landed on blue for these 20 spins. If you press the right arrow $\left ( \boxed{\blacktriangleright} \right )$ , the frequency label will show you how many of the times the spinner landed on blue (number 1).
To go back to the question, how many times does the spinner land on blue if it is spun 20 times? The answer is 3. To calculate the experimental probability of landing on blue, we have to divide by the total number of spins.
$P(\text{blue}) = \frac{3}{20} = 0.15$
Therefore, for this experiment, the experimental probability of landing on blue with 20 spins is 15%.
Now let's calculate the theoretical probability. We know that the spinner has 4 equal parts (blue, purple, green, and red). In a single trial, we can assume that:
$P(\text{blue}) = \frac{1}{4} = 0.25$
Therefore, for our spinner example, the theoretical probability of landing on blue is 0.25. Finding the theoretical probability requires no collection of data.
Example B
You are spinning a spinner like the one shown below 50 times. How many times does it land on blue? How about if you spin it 100 times? Does the experimental probability get closer to the theoretical probability? Simulate the spinning of the spinner using technology.
Set the spinner to spin 50 times and choose SPIN by pressing $\boxed{\text{WINDOW}}$ .
You can see the result of each trial by choosing TABL, or pressing $\boxed{\text{GRAPH}}$ .
Again, we can see the graph of the resulting table, or go back to the first screen, simply by choosing GRPH, or pressing $\boxed{\text{GRAPH}}$ again.
The question asks how many times we landed on blue (number 1) for the 50 spins. Press the right arrow $\left ( \boxed{\blacktriangleright} \right )$ , and the frequency label will show you how many of the times the spinner landed on blue (number 1).
Now go back to the question. How many times does the spinner land on blue if it is spun 50 times? The answer is 11. To calculate the probability of landing on blue, we have to divide by the total number of spins.
$P(\text{blue}) = \frac{11}{50} = 0.22$
Therefore, for this experiment, the probability of landing on blue with 50 spins is 22%.
If we tried 100 trials, we would see something like the following:
In this case, we see that the frequency of 1 is 23.
So how many times does the spinner land on blue if it is spun 100 times? The answer is 23. To calculate the probability of landing on blue in this case, we again have to divide by the total number of spins.
$P(\text{blue}) = \frac{23}{100} = 0.23$
Therefore, for this experiment, the probability of landing on blue with 100 spins is 23%. You can see that as we perform more trials, we get closer to 25%, which is the theoretical probability.
Example C
You are spinning a spinner like the one shown below 170 times. How many times does it land on blue? Does the experimental probability get closer to the theoretical probability? How many times do you predict we would have to spin the spinner to have the experimental probability equal the theoretical probability? Simulate the spinning of the spinner using technology.
With 170 spins, we get a frequency of 42 for blue.
The experimental probability in this case can be calculated as follows:
$P(\text{blue}) = \frac{42}{170} = 0.247$
Therefore, the experimental probability is 24.7%, which is even closer to the theoretical probability of 25%. While we're getting closer to the theoretical probability, there is no number of trials that will guarantee that the experimental probability will exactly equal the theoretical probability.
Guided Practice
You are spinning a spinner like the one shown below 500 times. How many times does it land on blue? How does the experimental probability of landing on blue compare to the theoretical probability? Simulate the spinning of the spinner using technology.
In the list of applications on the TI-84 calculator, choose Prob Sim.
After pressing $\boxed{\text{ENTER}}$ , you will have the following screen appear.
Since we're doing a spinner problem, choose Spin Spinner.
In Spin Spinner, a wheel with 4 possible outcomes is shown. You can adjust the number of spins, graph the frequency of each number, and use a table to see the number of spins for each number. We want to set this spinner to spin 500 times. To do this, choose SET by pressing $\boxed{\text{ZOOM}}$ , enter 500 after Trial Set, and choose OK by pressing $\boxed{\text{GRAPH}}$ .
Remember that for this example, 1 = blue.
Now that the spinner is set up for 500 trials, choose SPIN by pressing $\boxed{\text{WINDOW}}$ . Since we've chosen a large number of spins, the spinning may take a while!
We can see the result of each trial by choosing TABL, or pressing $\boxed{\text{GRAPH}}$ .
And we see the graph of the resulting table, or go back to the first screen, simply by choosing GRPH, or pressing $\boxed{\text{GRAPH}}$ again.
Now, the question asks how many times we landed on blue (number 1). We can actually see how many times we landed on blue for these 500 spins. If you press the right arrow $\left ( \boxed{\blacktriangleright} \right )$ , the frequency label will show you how many of the times the spinner landed on blue (number 1).
To go back to the question, how many times does the spinner land on blue if it is spun 500 times? The answer is 123. To calculate the experimental probability of landing on blue, we have to divide by the total number of spins.
$P(\text{blue}) = \frac{123}{500} = 0.246$
Therefore, for this experiment, the experimental probability of landing on blue with 500 spins is 24.6%.
Do you remember how to calculate the theoretical probability from Example A? We know that the spinner has 4 equal parts (blue, purple, green, and red). In a single trial, we can assume that:
$P(\text{blue}) = \frac{1}{4} = 0.25$
Therefore, for our spinner example, the theoretical probability of landing on blue is 0.25. As we pointed out in Example A, finding the theoretical probability requires no collection of data. It's also worth mentioning that our experimental probability was slightly farther away from the theoretical probability with 500 spins that it was with 170 spins in Example C. While, in general, increasing the number of spins will produce an experimental probability that is closer to the theoretical probability, as we've just seen, this is not always the case!
Practice
1. Based on what you know about probabilities, write definitions for theoretical and experimental probability.
1. What is the difference between theoretical and experimental probability?
2. As you add more data, do your experimental probabilities get closer to or further away from your theoretical probabilities?
3. Is spinning 1 spinner 100 times the same as spinning 100 spinners 1 time? Why or why not?
A spinner was spun 750 times using Spin Spinner on the TI-84 calculator, with 1 representing blue, 2 representing purple, 3 representing green, and 4 representing red as shown:
1. According to the following screen, what was the experimental probability of landing on blue?
2. According to the following screen, what was the experimental probability of landing on purple?
3. According to the following screen, what was the experimental probability of landing on green?
4. According to the following screen, what was the experimental probability of landing on red?
A spinner was spun 900 times using Spin Spinner on the TI-84 calculator, with 1 representing blue, 2 representing purple, 3 representing green, and 4 representing red as shown:
1. According to the following screen, what was the experimental probability of landing on blue?
2. According to the following screen, what was the experimental probability of landing on purple?
3. According to the following screen, what was the experimental probability of landing on green?
4. According to the following screen, what was the experimental probability of landing on red?
Vocabulary Language: English Spanish
experimental probability
experimental probability
Experimental (empirical) probability is the actual probability of an event resulting from an experiment.
theoretical probability
theoretical probability
Theoretical probability is the probability ration of the number of favourable outcomes divided by the number of possible outcomes.
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Advice 1: How to learn to divide a two-digit number
One of the most important topics in math elementary school division of two-digit numbers. Usually this action is done by selection or in a column, if the job is written. In any case a good tool is the multiplication table.
Instruction
1
Are two-digit numbers from 10 to 99. The division of these numbers to each other is included in the program of the third grade mathematics and has the highest complexity among the so-called unestablished actions on numbers.
2
Before you learn to divide a two-digit number, you must explain to the child what the number represents amount of tens and ones. This will save it from future a pretty common error, which allowed many children. They start to divide first and second digits of dividend and divisor each other.
3
To start work with the division of two-digit numbers in single digits. Best of all this technique is fulfilled using knowledge of multiplication tables. The more of these practices, the better. The skills of this division should be brought to automatism, then the child will be easier to move on to more complex subject of two-digit divisor, which, like the dividend represents the amount of tens and ones.
4
The most common way of dividing two-digit numbers – matching method, which involves successive multiplication of the divisor by a number between 2 and 9 so that the final product was equal to delimma. Example: divide 87 by 29. The arguments lead to the following:
29 times 2 is equal 54 – a little;
29 x 3 = 87 – correctly.
5
Please note student on the second digit (units) dividend and divisor, which are easy to navigate when using the multiplication tables. For example, in the example the second digit of the divisor is 9. Think about how much you need to multiply the number 9 to the number of units of work was equal to 7? The answer in this case only one – 3. This greatly facilitates the task of two-digit division. Check your guess by multiplying the total number of 29.
6
If the job is executed in writing, it is advisable to use the method of long division. This approach is similar to the previous except that the student does not need to keep the numbers in my head and make oral settlements. Better for written work the arm with a pencil or rough paper.
Advice 2 : How to count in a column
In order to quickly solve the math problem, you must be able to consider not only using the calculator. In addition, it can at hand and not be and have to use traditional methods account.
You will need
• a piece of paper
• handle
Instruction
1
Start with addition. Then write down the numbers that need to be folded under each other, so that units come under units, tens under tens, hundreds under hundreds. Swipe under the lower number line. To fold start with the units, that is, with the latest figures. Less than ten instantly record under the units. If you add get a two-digit number, then the units record the number of units, and the tens — remember. Unable to be sure to write it down somewhere in a corner. Fold the tens. Add to the obtained sum, which you remembered after the addition of the units. Record exactly the same as in the previous step. If it is less than ten, then write all at once, and if more — the number of units and the number of tens remember. In the same way, add the hundreds and thousands.
2
When you subtract the number recorded in the same way. Subtract the units. If the number of units in umanesimo more than visitama, "take" a dozen. That is, if you subtract 8 from 5, imagine that you are reading not of 5 and 15. Below the line write 7, but the number of tens is reduced by 1. Similarly alternately subtract tens and hundreds, not forgetting where and how much you "borrowed".
3
By multiplying the number of similarly written one below the other. If you decide an example, consisting of a multiple number in single digits, you first multiply it units, then the tens and subsequent discharges. If the multiplying units, you get a two-digit number, below the line, record the number of units of that number, and the number of tens remember and add after multiplication of the multiplier in tens. Exactly the same with all the other bits. When multiplying multi-digit numbers proceed sequentially. First multiply and write the result of multiplying the second multiplier by the number of units of the first multiplier. Write below the line result. Then multiply the second factor by tens of the first multiplier. Write the second result under the first, but don't forget that you have multiplied by the number of tens, and accordingly, the last digit of the result will be under tens. Similarly, multiply the second factor by the number of hundreds, thousands, and so on, following the order of recording. Under the last result, move the line and add up all the results. This will be the desired product.
4
To divide one number by another, write the first number, put a division sign, then write the second number and the equal sign. Aside from the beginning of the dividend as many figures, how many of them in private, and see whether the dividend divided by the divisor and how many, approximately, it will turn out in the end. If the number is less than the divisor, put down another figure. Write down the first digit in the private result. Multiply this number by the divisor and the number under sample record, starting with the first digit. Between severable and the written number sign "-".Subtract the first digit of the dividend given number, move the line and write under it the result is, strictly observing the order. The number that you got below the line, add the next digit of the dividend. Divide the resulting number by the divisor, write the result response. Multiply this number by the divisor and subtract the result from the number of which is below the line. Do the same while not using the last digit of the dividend.
Advice 3 : How to learn to divide a two-digit number
One of the most important topics in math elementary school division of two-digit numbers. Usually this action is done by selection or in a column, if the job is written. In any case a good tool is the multiplication table.
Instruction
1
Are two-digit numbers from 10 to 99. The division of these numbers to each other is included in the program of the third grade mathematics and has the highest complexity among the so-called unestablished actions on numbers.
2
Before you learn to divide a two-digit number, you must explain to the child what the number represents amount of tens and ones. This will save it from future a pretty common error, which allowed many children. They start to divide first and second digits of dividend and divisor each other.
3
To start work with the division of two-digit numbers in single digits. Best of all this technique is fulfilled using knowledge of multiplication tables. The more of these practices, the better. The skills of this division should be brought to automatism, then the child will be easier to move on to more complex subject of two-digit divisor, which, like the dividend represents the amount of tens and ones.
4
The most common way of dividing two-digit numbers – matching method, which involves successive multiplication of the divisor by a number between 2 and 9 so that the final product was equal to delimma. Example: divide 87 by 29. The arguments lead to the following:
29 times 2 is equal 54 – a little;
29 x 3 = 87 – correctly.
5
Please note student on the second digit (units) dividend and divisor, which are easy to navigate when using the multiplication tables. For example, in the example the second digit of the divisor is 9. Think about how much you need to multiply the number 9 to the number of units of work was equal to 7? The answer in this case only one – 3. This greatly facilitates the task of two-digit division. Check your guess by multiplying the total number of 29.
6
If the job is executed in writing, it is advisable to use the method of long division. This approach is similar to the previous except that the student does not need to keep the numbers in my head and make oral settlements. Better for written work the arm with a pencil or rough paper.
Advice 4 : How to teach your child to count in a column
In order for your child can solve the math problems as quickly as possible, it is necessary that he not only knew the multiplication table, but was able to think fast. How to teach child to count in a column?
You will need
• - a piece of paper;
• - handle.
Instruction
1
Getting to the training, start with the basics - with additions. To do this, take a clean sheet of paper, a pen and ask the child to write the number that need to be put in the following way: units under units, tens under tens, hundreds under hundreds. Further, under the bottom number line guide.
2
Explain that folding, starting with the last digits, that is, with units. When the amount is less than ten put down under the unit. If it turned out a two-digit number, then under the units record the number of units and the number of tens remember.
3
Now add the tens and add the number which you remembered in the mind after the addition of the units. Tell us what hundreds and thousands of folding the same way.
4
Performing operations with subtraction, explain that the numbers need to record exactly how and in addition. If when you subtract the number of units in umanesimo more than visitama, you need to "take" a dozen.
5
Show that when multiplying multi-digit numbers in single digits first multiply ones, then tens and subsequent discharges. Multiply multi-digit numbers proceed sequentially. First, multiply the second factor by the number of units of the first multiplier and write the result below the line. Then multiply by the tens first multiplier and again record the result.
6
Teach the child to carry out operations with the division. For this, write down the next number of the dividend with the divisor, and divide by the area and write the result underneath.
7
Daily practice, that knowledge has evolved. But keep in mind that the lessons should not consist in the memorization, otherwise it will not give any positive results. Do not go from one operation of the account columnof ω to another. That is, until the child learns to put in a column, not move on to learning subtraction.
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# How do you find the median of a data set?
## How do you find the median of a data set?
Finding the median
1. Arrange the data points from smallest to largest.
2. If the number of data points is odd, the median is the middle data point in the list.
3. If the number of data points is even, the median is the average of the two middle data points in the list.
## What does median?
Median is the middle number in a sorted list of numbers. To determine the median value in a sequence of numbers, the numbers must first be sorted, or arranged, in value order from lowest to highest or highest to lowest.
What does median mean in statistics?
The mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. The median is the middle value when a data set is ordered from least to greatest. The mode is the number that occurs most often in a data set.
### How do you find the median of 8 numbers?
Median
1. Arrange your numbers in numerical order.
2. Count how many numbers you have.
3. If you have an odd number, divide by 2 and round up to get the position of the median number.
4. If you have an even number, divide by 2.
### Why is median used?
The median is the most informative measure of central tendency for skewed distributions or distributions with outliers. For example, the median is often used as a measure of central tendency for income distributions, which are generally highly skewed.
How do you find the median of 10 numbers?
Arrange your numbers in numerical order. Count how many numbers you have. If you have an odd number, divide by 2 and round up to get the position of the median number.
#### How do you find the median of 7 numbers?
Add up all of the numbers and divide by the number of numbers in the data set. The median is the central number of a data set.
#### Where is median used in real life?
Suppose, every month you spend INR 700 on personal care, INR 100 to pay the water bill, INR 800 on snacks, INR 500 to pay the electricity bill, and INR 6000 to pay the house rent. If you calculate the average expenditure, it comes out to be INR 1,620 by the notion of mean, and INR 700 by employing the median concept.
Why is median better than average?
The median is calculated by taking the “middle” value, the value for which half of the observations are larger and half are smaller. When there is a possibility of extreme values, the median is generally the better measure to use.
## What is the median of 1 to 10?
1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Where the number of terms is in even. Therefore, the median of the first 10 natural numbers is 5.5.
## Why do we need median?
The importance of the median value is that it provides the idea about the distribution of the data. If the mean and the median of the data set are the same, then the dataset is evenly distributed from the smallest to the highest values.
When should median be used?
skewed distributions
The median is the most informative measure of central tendency for skewed distributions or distributions with outliers. For example, the median is often used as a measure of central tendency for income distributions, which are generally highly skewed.
### What is the median of 8 numbers?
The Median of a Data Set The median of a set of numbers is the middle number in the set (after the numbers have been arranged from least to greatest) — or, if there are an even number of data, the median is the average of the middle two numbers. Example 1 : Find the median of the set {2,5,8,11,16,21,30} .
### What is the use of fsharp data?
F# allows for the direct integration of scalable data stores into F# programming. For example, a type provider for the World bank data sets is available in the FSharp.Data library. Other web data stores can be accessed through F# support for JSON, XML and CSV data formats.
What types of data does F #support?
F# supports all common data import and access techniques. In addition, the type provider feature of F# brings simplicity and flexibility to accessing data, including databases, web-scale data and structured text formats like JSON, and XML.
#### What is median in a sorted array?
Definition of median is clear if you have odd number of elements. It’s just the element with index (Count-1)/2 in sorted array. But when you even number of element (Count-1)/2 is not an integer anymore and you have two medians: Lower median Math.Floor ( (Count-1)/2) and Math.Ceiling ( (Count-1)/2).
#### Is it necessary to have a median in a data set?
This is generally not necessary unless you know your data has certain patterns so that last element won’t be random enough or if somehow your code is exposed outside for targeted exploitation. Definition of median is clear if you have odd number of elements.
• October 14, 2022
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# Log Formulas
Logarithm is a convenient way to express large numbers. The logarithm of a product is the sum of the logarithms of the numbers being multiplied. Log Formulas are available below for the convenience of the students. Here you can find Logarithms Formulas including Basic & Special Case Log with Examples Equations that can help you to be familiar with this topic easily. Apart from it, you may check Logarithm Numerical also that can help you to know about the way of applying Log Formulas, so please have a look..
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## Log Formulas
Basic idea and rules for logarithms
A logarithm is the opposite of a power. In other words, if we take a logarithm of a number, we undo an exponentiation
Let’s start with simple example. If we take the base b=2 and raise it to the power of k=3, we have the expression 23. The result is some number, we’ll call it cc, defined by 23=c, We can use the rules of exponentiation to calculate that the result is
c=23=8.
Let’s say I didn’t tell you what the exponent k was. Instead, I told that the base was b=2 and the final result of the exponentiation was c=8. To calculate the exponent k, you need to solve
2k=8.
From the above calculation, we already know that k=3. But, what if I changed my mind, and told you that the result of the exponentiation was c=4 =, so you need to solve 2k=4? Or, I could have said the result was c=16 (solve 2k=16) or c=1 (solve 2k=1).
A logarithm is a function that does all this work for you. We define one type of logarithm (called “log base 2” and denoted log2) to be the solution to the problems I just asked. Log base 2 is defined so that
log2c=k
is the solution to the problem
2k=c
for any given number cc. In other words, the logarithm gives the exponent as the output if you give it the exponentiation result as the input. To get all answers for the above problems, we just need to give the logarithm the exponentiation result cc and it will give the right exponent k of 2. The solution to the above problems are:
Just like we can change the base bb for the exponential function, we can also change the base bb for the logarithmic function. The logarithm with base bb is defined so that
logbc=k
is the solution to the problem
bk=c
For any given number c and any base b.
For example, since we can calculate that 103=1000, we know that log101000= 3 (“log base 10 of 1000 is 3”). Using base 10 is fairly common. But, since in science, we typically use exponents with base e, it’s even more natural to use e for the base of the logarithm. This natural logarithm is frequently denoted by ln(x), i.e.,
ln(x)=logex.
In other words,
k=ln(c)……………(1)
is the solution to the problem
ek=c ………….(2)
for any number c. Since using base e is so natural to mathematicians, they will sometimes just use the notation log x instead of ln x. However, others might use the notation logx for a logarithm base 10, i.e., as a shorthand notation for log10x. Because of this ambiguity, if someone uses log x without stating the base of the logarithm, you might not know what base they are implying. In that case, it’s good to ask.
Get Best Answer: How to Prepare for Maths
Basic Log Formulas
Since taking a logarithm is the opposite of exponentiation (more precisely, the logarithmic function logbx is the inverse function of the exponential function bx), we can derive the basic rules for logarithms from the basic rules for exponents.
For simplicity, we’ll write the rules in terms of the natural logarithm ln(x). The rules apply for any logarithm logbx, except that you have to replace any occurence of e with the new base bb.
The natural log was defined by equations (1) and (2). If we plug the value of k from equation (1) into equation (2), we determine that a relationship between the natural log and the exponential function is
elnc=c …………………….(3)
Or, if we plug in the value of cc from (2) into equation (1), we’ll obtain another relationship
ln(ek)=k ……………………..4
These equations simply state that ex and ln x are inverse functions. We’ll use equations (3) and (4) to derive the following rules for the logarithm..
All Log Formulas (BASIC)
Rule or special case Formula Product ln(xy)=ln(x)+ln(y) Quotient ln(x/y)=ln(x)−ln(y) Log of power ln(xy)=yln(x) Log of ee ln(e)=1 Log of one ln(1)=0 Log reciprocal ln(1/x)=−ln(x)
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Formulas Of Log – The product rule
The quotient rule
The quotient rule for logarithms follows from the quotient rule for exponentiation
ea/eb= e a−b
in the same way.
Starting with c=x/y in equation (3) and applying it again with c=x and c=y, we can calculate that
eln(x/y) = x/y
=e ln(x) / e ln(y) = e ln(x)−ln(y),
Where in the last step we used the quotient rule for exponentation with a=ln(x)a=ln(x) and b=ln(y)b=ln(y). Since eln(x/y)=e ln(x)−ln(y) we can conclude that the quotient rule for logarithms is
ln(x/y)=ln(x)−ln(y)
(This last step could follow from, for example, taking logarithms of both sides of eln(x/y)=eln(x)−ln(y) like we did in the last step for the product rule.)
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Log of a power
To obtain the rule for the log of a power, we start with the rule for power of a power,
(ea)b=eab
Starting with c=xy in equation (3) and applying it again, this time just once more with c=x, we can calculate that
e ln(xy)=xy
= (eln(x))y
=e yln(x)
Where in the last step we used the power of a power rule for a=ln(x) and b=y From eln(xy)=eyln(x), we can conclude that
ln(xy)=yln(x)
which is the rule for the log of a power.
Log of e
The formula for the log of e comes from the formula for the power of one,
e1=e
Just take the logarithm of both sides of this equation and use equation (4) to conclude that
ln(e)=1
Log of one
The formula for the log of one comes from the formula for the power of zero
e0=1
Just take the logarithm of both sides of this equation and use equation (4)(4) to conclude that
ln(1)=0.
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Log of reciprocal
The rule for the log of a reciprocal follows from the rule for the power of negative one
x−1=1/x
and the above rule for the log of a power. Just substitute y=−1y=−1 into the the log of power rule, and you have that
ln(1/x)=−ln(x).
Question and Answers Based On Logarithms Formulas
Question1. Which of the following statements is not correct?
1. log10 10 = 1
2. log (2 + 3) = log (2 x 3)
3. log10 1 = 0
4. log (1 + 2 + 3) = log 1 + log 2 + log 3
Question2. If log 2 = 0.3010 and log 3 = 0.4771, the value of log5 512 is:
1. 870
2. 967
3. 876
4. 912
Question3. log 8/ log 8 is equal to:
1. 1/8
2. 1/4
3. 1/2
4. 1/8
Question4. If log 27 = 1.431, then the value of log 9 is:
1. 934
2. 945
3. 954
4. 958
Question5. If log a/b + log b/a = log (a + b), then:
1. a + b = 1
2. a – b = 1
3. a = b
4. a2 – b2 = 1
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Logarithms Formulas
In mathematics there are different types of formulas and algorithms. One of them is Log Formulas. Logarithms Formulas are used in mathematics while solving trigonometric equations etc. logarithm is the opposite function to exponentiation, just as division is the inverse of multiplication and vice versa.
Here on this page we are going to provide Basic & Special Case Log Product/Power Rule Equations of Logarithms. Candidates willing to learn Log Formulas easily must take a look below.
Here on this page of recruitmentresult.com we are going to provide all the Logarithms Formulas – Basic & Special Case Log Product/Power Rule Equations that are used by candidates while solving mathematical equations.
Use of Log starts from 11th standards and continue till graduation/ Post-graduation in courses like BCA / MCA / B Tech / BE / M Tech / ME etc. for the ease of students here we have also given All Log Formulas PDF.
Final Words!!
We hope the information given here on this page will prove to be useful for all of you regarding the topic Integration Log Formulas.
Something That You Should Put An Eye On
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## Engage NY Eureka Math 6th Grade Module 1 Lesson 9 Answer Key
### Eureka Math Grade 6 Module 1 Lesson 9 Example Answer Key
Example 1.
To make paper mache, the art teacher mixes water and flour. For every two cups of water, she needs to mix in three cups of flour to make the paste.
Find equivalent ratios for the ratio relationship 2 cups of water to 3 cups of flour. Represent the equivalent ratios in the table below:
Example 2.
Javier has a new job designing websites. He is paid at a rate of $700 for every 3 pages of web content that he builds. Create a ratio table to show the total amount of money Javier has earned in ratio to the number of pages he has built. Answer: Javier is saving up to purchase a used car that costs$4,200. How many web pages will Javier need to build before he can pay for the car?
Javier will need to build 18 web pages in order to pay for the car.
### Eureka Math Grade 6 Module 1 Lesson 9 Exercise Answer Key
Exercise 1.
Spraying plants with cornmeal juice is a natural way to prevent fungal growth on the plants. It is made by soaking cornmeal in water, using two cups of cornmeal for every nine gallons of water. Complete the ratio table to answer the questions that follow.
Cups of Cornmeal Gallons of Water 2 9 4 18 6 27 8 36 10 45
a. How many cups of cornmeal should be added to 45 gallons of water?
10 cups of cornmeal should be added t0 45 gallons of water.
b. Paul has only 8 cups of cornmeal. How many gallons of water should he add if he wants to make as much cornmeal juice as he can?
Paul should add 36 gallons of water.
c. What can you say about the values of the ratios in the table?
The values of the ratios are equivalent.
Exercise 2.
James is setting up a fish tank. He is buying a breed of goldfish that typically grows to be 12 inches long, It is recommended that there be 1 gallon of water for every inch of fish length in the tank. What is the recommended ratio of gallons of water per fully-grown goldfish in the tank?
Complete the ratio table to help answer the following questions:
Number of Fish Gallons of Water 1 12 2 24 3 36 4 48 5 60
a. What size tank (in gallons) Is needed for James to have 5 full-grown goldfish?
James needs a tank that holds 60 gallons of water in order to have 5 full-grown goldfish.
b. How many full-grown goldfish can go in a 40-gallon tank?
3 full-grown goldfish can go in a 40-gallon tank.
c. What can you say about the values of the ratios in the table?
The values of the ratios are equivalent.
### Eureka Math Grade 6 Module 1 Lesson 9 Problem Set Answer Key
Assume each of the following represents a table of equivalent ratios. Fill in the missing values. Then choose one of the tables, and create a real-world context for the ratios shown in the table.
Question 1.
Question 2.
Question 3.
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There are some special situations that have special methods of factoring. Once such situation is the factoring the difference of two perfect squares. You need to know this formula!
Factoring the Difference of Two Squares
We already know that by multiplying: (a + b)(a - b) = a2 - b2.
If we use the commutative property on this equation we get:
a2 - b2 = (a + b)(a - b)
which shows a formula for factoring a2 - b2, the difference (subtraction) of two perfect squares.
Notice that the factors are identical except that one is addition and the other is subtraction.
In algebra, a term is a perfect square when the numerical coefficient (the number in front of the variable) is a perfect square number, and the exponents of each of the variables are even numbers. 49x2; 25x6; 81y4; 9x2y8; 121x12 are perfect squares.
This process of factoring does NOT apply to a2 + b2.
Factor: x2 - 64 Solution: Both x2 and 64 are perfect squares and this problem is subtraction. It qualifies for use of the difference of squares formula. a2 - b2 = (a + b)(a - b) x2 is the square of x; 64 is the square of 8. a = x and b = 8 [While -x and -8 are also possible, the positive values will make the work easier.] Answer: x2 - 64 = (x + 8)(x - 8) or (x - 8)(x + 8)
Factor: 9m2 - 81n6 Solution: First, there is a common factor of 9 in these terms. 9(m2 - 9n6 ) Both m2 and 9n6 are perfect squares and this problem is subtraction. It qualifies for use of the difference of squares formula. a2 - b2 = (a + b)(a - b) m2 is the square of m; 9n6 is the square of 3n3. a = m and b = 3n3 Answer: 9m2 - 81n6 = 9(m + 3n3 )(m - 3n3 )
Factor: x2 - 9 using Algebra Tiles
KEY: See more about Algebra Tiles.
Place the x2 tile and the -9 (red) tiles in the grid. Fill in the empty spaces with x-tiles that do not change the value of the current entries. (1 x-tile + 1 red x-tile = 0). Fill the outside sections of the grid with x-tiles and 1-tiles that complete the pattern.
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# MSBSHSE Solutions For Class 8 Maths Part 1 Chapter 3- Indices and Cube Root
MSBSHSE Solutions For Class 8 Maths Part 1 Chapter 3 Indices and Cube Root is one of the best reference material students can rely on. The MSBSHSE Solution for Class 8 module contains all the important topics that can help students score well in the examinations. Facilitators have provided the solutions accurately, depending upon the students grasping abilities to understand the concepts clearly. Every question is explained stepwise for a better understanding of the students.
The best reference guide for the students is the Maharashtra State Board Class 8 Textbooks Part 1 Maths Solutions. Students can analyse their weaker section and can improve on it by thoroughly going through the solutions, after completing every chapter. The PDF of Maharashtra Board Solutions for Class 8 Maths Chapter 3 Indices and Cube Root is provided here. Students can refer and download from the given links. This chapter is based on indices and cube root. Some of the essential topics of this chapter includes the meaning of numbers with rational indices, Cube and Cube root.
## Download the PDF of Maharashtra Board Solutions For Class 8 Maths Part 1 Chapter 3- Indices and Cube Root.
### Access answers to Maharashtra Board Solutions For Class 8 Maths Part 1 Chapter 3- Indices and Cube Root.
Practice set 3.1 PAGE NO: 15
1. Express the following numbers in index form.
(1) Fifth root of 13
(2) Sixth root of 9
(3) Square root of 256
(4) Cube root of 17
(5) Eighth root of 100
(6) Seventh root of 30
Solution:
In general, nth root of ‘a’ is expressed as a1/n. where, a is the base and 1/5 is the index.
So now,
(1) Fifth root of 13
Index form of fifth root of 13is expressed as 131/5.
(2) Sixth root of 9
Index form of sixth root of 9 is expressed as 91/6.
(3) Square root of 256
Index form of square root of 256 is expressed as 2561/2.
(4) Cube root of 17
Index form of cube root of 17 is expressed as 171/3.
(5) Eighth root of 100
Index form of eighth root of 100 is expressed as 1001/8.
(6) Seventh root of 30
Index form of seventh root of 30 is expressed as 301/7.
2. Write in the form ‘nth root of a’ in each of the following numbers.
Solution:
In general, a1/n is written as ‘nth root of a’.
So now,
(1) (81)1/4
(81)1/4 is written as ‘4th root of 81’.
(2) (49)1/2
(49)1/2 is written as ‘square root of 49’.
(3) (15)1/5
(15)1/5 is written as ‘5th root of 15’.
(4) (512)1/9
(512)1/9 is written as ‘9th root of 512’.
(5) (100)1/19
(100)1/19 is written as ‘19th root of 100’.
(6) (6)1/7
(6)1/7 is written as ‘7th root of 6’.
Practice set 3.2 PAGE NO: 16
1. Complete the following table.
Sr. No. Number Power of the root Root of the power (1) (225)3/2 Cube of square root of 225 Square root of cube of 225 (2) (45)4/5 (3) (81)6/7 (4) (100)4/10 (5) (21)3/7
Solution:
Generally we can express the number am/n as
am/n = (am)1/n means ‘nth root of mth power of a’.
am/n = (a1/n)m means ‘mth power of nth root of a’.
So by using the above rules let us fill the table:
Sr. No. Number Power of the root Root of the power (1) (225)3/2 Cube of square root of 225 Square root of cube of 225 (2) (45)4/5 Fourth power of fifth root of 45 Fifth root of fourth power of 45 (3) (81)6/7 Sixth power of seventh root of 81 Seventh root of sixth power of 81 (4) (100)4/10 Fourth power of tenth root of 100 Tenth root of fourth power of 100 (5) (21)3/7 Cube of seventh root of 21 Seventh root of cube of 21
2. Write the following number in the form of rational indices.
(1) Square root of 5th power of 121.
(2) Cube of 4th root of 324.
(3) 5th root of square of 264.
(4) Cube of cube root of 3.
Solution:
We know that ‘nth root of mth power of a’ is expressed as (am)1/n.
And ‘mth power of nth root of a’ is expressed as (a1/n) m.
So by using the above rules let us find
(1) Square root of 5th power of 121.
Square root of 5th power of 121 is expressed as (1215)1/2 or (121)5/2.
(2) Cube of 4th root of 324.
Cube of 4th root of 324 is expressed as (3241/4)3 or (324)3/4.
(3) 5th root of square of 264.
5th root of square of 264 is expressed as (2642)1/5 or (264)2/5.
(4) Cube of cube root of 3.
Cube of cube root of 3 is expressed as (31/3)3Â or (31)3/3.
Practice set 3.3 PAGE NO: 18
1. Find the cube root of the following numbers.
(1) 8000
(2) 729
(3) 343
(4) -512
(5) -2744
(6) 32768
Solution:
(1) 8000
Firstly let us find the factor of 8000
8000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5
So to find the cube root, we pair the prime factors in 3’s.
8000 = (2 × 2 × 5)3
= (2 × 10)3
= 203
Hence, cube root of 8000 = ∛(8000)
= (203)1/3
= 20
(2) 729
Firstly let us find the factor of 729
729 = 9 × 9 × 9
So to find the cube root, we pair the prime factors in 3’s.
729 = 93
Hence, cube root of 729 = ∛(729)
= (93)1/3
= 9
(3) 343
Firstly let us find the factor of 343
343 = 7 × 7 × 7
So to find the cube root, we pair the prime factors in 3’s.
343 = 73
Hence, cube root of 343 = ∛(343)
= (73)1/3
= 7
(4) -512
Firstly let us find the factor of -512
-512 = (-8) × (-8) × (-8)
So to find the cube root, we pair the prime factors in 3’s.
-512 = (-8)3
Hence, cube root of -512 = ∛(-512)
= (-83)1/3
= -8
(5) -2744
Firstly let us find the factor of -2744
-2744 = (-14) × (-14) × (-14)
So to find the cube root, we pair the prime factors in 3’s.
-2744 = (-14)3
Hence, cube root of -2744 = ∛(-2744)
= (-143)1/3
= -14
(6) 32768
Firstly let us find the factor of 32768
32768 = 32 × 32 × 32
So to find the cube root, we pair the prime factors in 3’s.
32768 = 323
Hence, cube root of 32768 = ∛(32768)
= (323)1/3
= 32
2. Simplify:
Solution:
Cube roots are used in day to day Mathematics, such as in powers and exponents or to find the side of a three-dimensional cube when its volume is given. Here, many such exercise problems are given, students can practice the solutions to secure good marks in the exams. Students can depend on these Solutions to understand all the topics under this chapter completely. Stay tuned to learn more about Indices and Cube Root, MSBSHSE Exam pattern and other information.
## Frequently Asked Questions on Maharashtra State Board Solutions for Class 8 Maths Chapter 3 Indices and Cube Root
### Where can I get Maharashtra State Board Class 8 Maths Solutions Chapter 3 Indices and Cube Root?
We have given the solutions as a scrollable PDF, and also mentioned a clickable link for the students to access the PDF. Meanwhile, the questions and the solutions are also made available online on our webpage.
### Is Maharashtra State Board Class 8 Maths Solutions Chapter 3 Indices and Cube Root important?
Yes, these solutions are significant because they provide step-by-step solution or answer for the questions from the Chapter 3 of the Maharashtra state board Class 8 Maths. These solutions cover the topics that forms the basis for the questions that are likely to be asked in the board exams. Preparing with the help of these solutions can help students to score good marks in the exams.
### How do I study Maharashtra State Board Class 8 Maths Solutions Chapter 3?
Students are urged to first solve the questions and then to refer back to the solutions in order to identify the mistakes and solve them early on, so that it can be avoided for the exams. Timing the process helps to manage time better. This is also a good tool to analyse your exam preparations.
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Select Page
If you want to set points via some algebraic equation, then you are probably doing the equation of line. So, the equation of a line is setting the points with an algebraic equation, which forms a line in a coordinate system. There are numerous points that are placed together known as the variables which are represented as x and y. These variables are then applied in the form of an algebraic equation.
So, what does the equation of line confirm? The equation of line confirms whether or not the points in consideration are set on a line or not.
Students must remember that the equation of a line is a type of linear equation with a degree as in one.
## How Can You Form the Equation of a Line?
The equation of a line can be formed by making a slope of the line and a specific point on the line. First, we are required to understand this slope of the line and the specific point on the line, in order to better understand the formation of the equation of the line.
The slope can be defined as nothing but an inclination of the line with the positive axis as an x-axis which is expressed as a numeric integer, fraction, or like the tangent of the angle which will make the positive x-axis. The point is referred to as the point in the coordinate system with the x coordinate and the y coordinate in place.
## What is the Standard Form of Equation?
The standard form of the equation of a line can be stated as ax + by + c = 0. In this case, a and b are the coefficients while a and y are the variables and c is the constant here.
This equation is degree one, x and y being its variables. The values of x and y will represent the coordinating points on the line which is represented in the coordinate plane. Check out the important points which are required in writing the standard form of the equation.
• First write the x term, then the y term, and then write the constant term.
• The constant and the coefficient terms cannot be written in the form of fractions, decimals. It should be written in the form of integers.
• The value of a and the x coefficient is to be written as the positive integer.
## What Will be the Equation of Circle?
The equation of a circle is a description of the circle in an algebraic way. This equation will describe the circle which has the length and the radius of the circle. Students must not get confused between the equation of the circle and the formula of the circumference of the circle, both are specific in their measurement. The equation of circle is very much necessary in coordinate geometry.
A circle is to be drawn on a piece of paper. We can draw the circle if only we know the center and radius of the circle. Thus, a radius can be represented in many other forms:
1. In the general form
2. In the standard form
3. In the parametric form
4. In the polar form
## Lay Out the Equation of Circle
In a Cartesian Plane, an equation of circle represents where it is located. If we know the central position of the circle and the coordinates of the circle, we can easily frame the equation of a circle. The equation of a circle is the image of the points on the circumference of the circle.
We know a circle is a representation of the locus of the points whose distance from the fixed point has a constant value. This is the center of the circle, and the constant value is actually the radius of the circle.
So, the standard equation of circle at the center placed at (x1,y1) and radius r is – (xx1)2+(yy1)2=r2.
If you want to know more about mathematical concepts then visit Cuemath.
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# November 1998 Winner
Question:
1. Find the sum of the angles formed at the tips of both stars
2. Find a formula for the sum of the sum of the angle measurements at the tips of an n-pointed star.
1. The sum of the angles formed at the tips of the five pointed star is 180; the sum of the angles formed at the tips of the six pointed star is 360.
2. The formula for the sum of the angle measurements at the tips of an n-pointed star is f(n)=180(n)-720 where n is an integer greater than 4.
Explantion:
Let's start off with the five pointed star:
If you get confused at any time, just refer back to the top diagram.
Before we jump into the problem, lets get the facts straight:
• Based on the diagram we can see there are five triangle whose bases make up a pentagon, which all together make up this figure.
• We also see that the polygons that make this figure are convex polygons since if I select two point inside the polygons, the segment between the two points do not exit the polygon.
• There is a theorem that states that the sum of the measures of the angles of any convex polygon, one such angle at the vertex, is 360.
• The exterior angle of a convex polygon is an angle that is formed when the side of a polygon is extended in one direction (like a ray), and is supplemmentary to an vertex angle of a polygon, and that only one line that contains a side can hold an exterior angle at a time.
• Finally, there are two sets of exteriors angles in every convex polygon.
With that out of the way, we can attact the problem.
Based on this diagram, we can see that we have:
• Pentagon FGHIJ
• Triangle AFG
• Triangle BGH
• Triangle CHI
• Triangle DIJ
• Triangle EJF
I labeled the diagram in the following way:
Red angles- Set 1 of the exterior angles of Pentagon FGHIJ
Blue Angles- et 2 of the exterior angles of Pentagon FGHIJ
Green Angles- Vertex angles
• After labeling the diagram, we can see that the measurements of the red exterior angles and the blue exterior angles each equal 360, show all together the sum of both sets of exterior angle is 720
• We can also see that the exterior angles of the pentagon makes up each set of the base angles of the triangle.
Lets deal with the five triangles now:
• The measurements of the angles that makes up the triangle equals 180
• The sum of all the angles of the 5 triangles equals 180(5), or 900
• We can setup an equation to find the the measurements (here the sum of the measurements) of the triangles:Ê 5[measurement of the vertex angle+measurement of one set of the base angles+measurement of the other set of the base angles=180],
• Next, we can distribute the 5 to get [(5)measurement of the vertex angle+(5)measurement of one set of the base angles+(5)measurement of the other set of the base angles=900]
• Since we know the sum of both sets of exterior angle (also the sum of the base angles), we can substitute 360 into the equation, we get (5)measurement of the vertex angle+720=900, or (5)measurement of the vertex angle=180
FINALLY, we have our answer: the sum of the vertex angles is 180!!!
If you are still interested, continue on:
For the six pointed the star, we can do ALL of that work again, or we can take a shortcut...
We see that the figure is made up of two triangle, highlighted in the figure.Ê We then can say that the sum of the vertex angles is 180(2), or 360.Ê We CAN do ALL that work to arrive at the same answer, but I think I would be too time consuming.
To find the equation, we need to find the varible:
Number of triangles and the sides of the polygon that form the n-pointed figure.
Even though the number of sides of the polygons changes, the sum of each set of exterior angles doesn't change ( 2 sets of 360)
We now can refer back to our original equation, and insert the constant and varible to get:
(n)measurement of the vertex angle+720=180(n) or sum of vertex angles [or (f)n] =180(n)-720.
If we set up the equation to set an inequality, we get n>4 (greater than because the function needs to be greater than 0) and since we can't have numbers other than postive integers, we can see that f(n)=180(n)-720 where n is an integer greater than 4.
If you were able to stay interested, I thank you for your attention and I hope you understand.
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# Math Snap
## A group of 41 people are going to a concert together in 7 vehicles. Some of the vehicles can hold 7 people each, the rest can hold 5 people each. Assuming all the vehicles are filled to capacity, exactly how many of these vehicles can hold 7 people?
#### STEP 1
Assumptions1. The total number of people is41. The total number of vehicles is73. Some vehicles can hold7 people each, the rest can hold5 people each4. All vehicles are filled to capacity
#### STEP 2
Let's denote the number of vehicles that can hold7 people as $x$ and the number of vehicles that can hold5 people as $y$. We can set up the following system of equations based on the problem$7x +5y =41$$x + y =7$
#### STEP 3
We can solve this system of equations using substitution or elimination. Let's use substitution. From the second equation, we can express $y$ in terms of $x$$y =7 - x$
#### STEP 4
Substitute $y =7 - x$ into the first equation$7x +(7 - x) =41$
#### STEP 5
implify the equation$7x +35 -5x =41$
#### STEP 6
Combine like terms$2x +35 =41$
#### STEP 7
Subtract35 from both sides of the equation$2x =41 -35$
#### STEP 8
implify the equation$2x =6$
#### STEP 9
Divide both sides of the equation by2 to solve for $x$$x =6 /2$
##### SOLUTION
implify to find the value of $x$$x =3$So, there are3 vehicles that can hold7 people each.
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# Math Tutorial 1.5 - Multiples, Factors, Prime Numbers and Prime Factorization including LCM and GCF
There are 7 lessons in this math tutorial covering Multiples, Factors, Prime Numbers and Prime Factorization including LCM and GCF. The tutorial starts with an introduction to Multiples, Factors, Prime Numbers and Prime Factorization including LCM and GCF and is then followed with a list of the separate lessons, the tutorial is designed to be read in order but you can skip to a specific lesson or return to recover a specific math lesson as required to build your math knowledge of Multiples, Factors, Prime Numbers and Prime Factorization including LCM and GCF. you can access all the lessons from this tutorial below.
In this Math tutorial, you will learn:
• What are multiples of a number?
• What are factors of a number?
• How to find the factors of a number?
• What are prime numbers? What do we call a number that is not prime?
• How to write a number as a product of prime factors?
• Why is the least common multiple of two or more numbers so important?
• Why is the greatest common factor of two or more numbers so important?
• How to calculate the least common multiple and greatest common factor of two or more numbers?
• How are these two concepts (LCM and GCF) applied in practice?
## Introduction
A store only sells forks in sets of 6 and not individually (single pieces). How many sets should you buy if 48 people are invited for dinner? What about if there were 78 people?
Can you buy exactly 33 forks in the above store? Why? What must you do in this example?
Sets of plates have 12 pieces. How many sets from each type (forks & plates) are needed for 48 people? Is there a smaller number in which every person invited has one fork and one plate?
This tutorial will focus on multiples and factors of a number as well as the common multiples and factors of two or more numbers. The thing that interests us most is the smallest of multiples and the greatest of factors as these two factors have a wide range of applications in practice.
In addition, we will explain prime numbers and how to write a number in prime factors.
Please select a specific "Multiples, Factors, Prime Numbers and Prime Factorization including LCM and GCF" lesson from the table below, review the video tutorial, print the revision notes or use the practice question to improve your knowledge of this math topic.
Arithmetic Learning Material
Tutorial IDMath Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
1.5Multiples, Factors, Prime Numbers and Prime Factorization including LCM and GCF
Lesson IDMath Lesson TitleLessonVideo
Lesson
1.5.1What are Multiples of a Number? What are Factors?
1.5.2Finding the Factors of a Number Using the Tree Method
1.5.3Prime Numbers
1.5.4Prime Factorization
1.5.5Finding the Common Multiples of Two Numbers
1.5.6Finding the LCM of Two or More Numbers
1.5.7Greatest Common Factor, GCF
## Whats next?
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3. Arithmetic Revision Notes: Multiples, Factors, Prime Numbers and Prime Factorization including LCM and GCF. Print the notes so you can revise the key points covered in the math tutorial for Multiples, Factors, Prime Numbers and Prime Factorization including LCM and GCF
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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: Divisibility Rules
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#### You may also like
What is the relationship between the arithmetic, geometric and harmonic means of two numbers, the sides of a right angled triangle and the Golden Ratio?
### Classical Means
Use the diagram to investigate the classical Pythagorean means.
# Pythagorean Golden Means
##### Stage: 5 Challenge Level:
Freddie Manners, age 11 from Packwood Haugh School, Shropshire sent in the following beautiful solution. Freddie asks Is this relationship to the Golden Ratio coincidental?'' The answer is probably not. Mathematics if full of connections which at first seem surprising. The question involves the sides of a right-angled triangle, the cube of the Golden Ratio $\varphi = {1\over 2}(1+\sqrt{5})$, and the arithmetic, geometric and harmonic means of two number (AM, GM and HM respectively). Firstly Freddie found the cube of $\varphi = {1\over 2}(1+\sqrt{5})$.
$$\begin{eqnarray} \varphi^2 &=& {1\over 4}(5+2\sqrt{5}+1) \\ \varphi^3 &=& {1\over 8}(1 + \sqrt {5})(6 + 2\sqrt{5}) \\ &=& {1\over 8}(16 + 8\sqrt{5}) \\ &=& 2 + \sqrt{5}. \end{eqnarray}$$
Take any two numbers $a$ and $b$, where $0< b< a$. Because the AM is the largest we have
$$\begin{eqnarray} \left({(a+b)\over 2}\right)^2 &=& ab + {1\over {\left({1\over 2}({1\over a}+{1\over b})\right)^2}} \\ &=& ab + {(2ab)^2\over (a+b)^2 } \\ {(2ab)^2\over (a+b)^2} &=& \left({(a+b)\over 2}\right)^2 - ab \\ &=& \left({(a-b)\over 2}\right) ^2 \\ {2ab \over (a+b)} &=& {(a- b)\over 2} \\ 4ab &=& a^2 - b^2 \\ {4a\over b} &=& \left({a\over b}\right)^2 - 1 . \end{eqnarray}$$
Let the ratio $a/b = x$ then
$$\begin{eqnarray} 4x &=& x^2 -1 \\ x^2 - 4x -1 &=& 0 \\ x &=& 2 \pm \sqrt 5 \end{eqnarray}$$
As $\sqrt 5 > 2$ the solution $2-\sqrt5$ would give a minus number.
So $a/b = 2 + \sqrt5 = \varphi^3$ and $a=b\varphi^3.$
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# Normal Form Games¶
## Motivating example: Coordination Game¶
Game theory is the study of interactive decision making. One example of this is the following situation:
Two friends must decide what movie to watch at the cinema. Alice would like to watch a sport movie and Bob would like to watch a comedy. Importantly, they would both rather spend their evening together than apart.
To quantify this mathematically, numeric values are associated to the 4 possible outcomes:
1. Alice watches a sport movie, Bob watches a comedy: Alice receives a utility of 1 and Bob a utility of 1.
2. Alice watches a comedy, Bob watches a sport movie: Alice receives a utility of 0 and Bob a utility of 0.
3. Alice and Bob both watch a sport movie: Alice receives a utility of 3 and Bob a utility of 2.
4. Alice and Bob both watch a comedy: Alice receives a utility of 2 and Bob a utility of 3.
This particular example will be represented using two matrices.
$$A$$ will represent the utilities of Alice:
$\begin{split}A = \begin{pmatrix} 3 & 1\\ 0 & 2 \end{pmatrix}\end{split}$
$$B$$ will represent the utilities of Bob
$\begin{split}B = \begin{pmatrix} 2 & 1\\ 0 & 3 \end{pmatrix}\end{split}$
Alice is referred to as the row player and Bob as the column player:
• The row player chooses which row of the matrices the player will gain their utilities.
• The column player chooses which column of the matrices the player will gain their utilities.
This representation of the strategic interaction between Alice and Bob is called a Normal Form Game
## Definition of Normal Form Game¶
An $$N$$ player normal form game consists of:
• A finite set of $$N$$ players.
• Action set for the players: $$\{\mathcal{A}_1, \mathcal{A}_2, \dots \mathcal{A}_N\}$$
• Payoff functions for the players: $$u_i : \mathcal{A}_1 \times \mathcal{A}_2 \dots \times \mathcal{A}_N \to \mathbb{R}$$
Question
For the Coordination game:
1. What is the finite set of players?
2. What are the action sets?
3. What are the payoff functions?
## Definition of a Zero Sum Game¶
A two player normal form game with payoff matrices $$A, B$$ is called zero sum if and only if:
$A = -B$
Question
Is the Coordination game zero sum?
## Examples of other Normal Form Games¶
### Prisoners Dilemma¶
Assume two thieves have been caught by the police and separated for questioning. If both thieves cooperate and do not divulge any information they will each get a short sentence (with a utility value of 3). If one defects they are offered a deal (utility value of 5) while the other thief will get a long sentence (utility value of 0). If they both defect they both get a medium length sentence (utility value of 1).
Question
For the Prisoners Dilemma
1. What is the finite set of players?
2. What are the action sets?
3. What are the payoff functions?
4. Is the game zero sum?
### Hawk Dove Game¶
Suppose two birds of prey must share a limited resource. The birds can act like a hawk or a dove. Hawks always act aggressively over the resource to the point of exterminating another hawk (both hawks get a utility value of 0) and/or take a majority of the resource from a dove (the hawk gets a utility value of 3 and the dove a utility value of 1). Two doves can share the resource (both getting a utility value of 2).
Question
For the Hawk Dove Game
1. What is the finite set of players?
2. What are the action sets?
3. What are the payoff functions?
4. Is the game zero sum?
### Pigs¶
Consider two pigs. One dominant pig and one subservient pig. These pigs share a pen. There is a lever in the pen that delivers food but if either pig pushes the lever it will take them a little while to get to the food.
• If the dominant pig pushes the lever, the subservient pig has some time to eat most of the food before being pushed out of the way. The dominant pig gets a utility value of 2 and the subservient pig gets a utility value of 3.
• If the subservient pig pushes the lever, the dominant pig will eat all the food. The dominant pig gets a utility value of 6 and the subservient pig gets a utility value of -1.
• If both pigs push the lever, the subservient pig will a small amount of the food. The dominant pig gets a utility value of 4 and the subservient pig gets a utility value of 2.
• If both pigs do not push the lever they both get a utility value of 0.
Question
For the Pigs Game
1. What is the finite set of players?
2. What are the action sets?
3. What are the payoff functions?
4. Is the game zero sum?
### Matching Pennies¶
Consider two players who can choose to display a coin either Heads facing up or Tails facing up. If both players show the same face then player 1 wins, if not then player 2 wins. Winning corresponds to a numeric value of 1 and losing a numeric value of -1.
Question
For the Matching Pennies game:
1. What is the finite set of players?
2. What are the action sets?
3. What are the payoff functions?
4. Is the game zero sum?
## Exercises¶
1. Represent the following game in normal form:
Alice, Bob and Celine are childhood friends that would like to communicate online. They have a choice between 3 social networks: facebook, twitter and G+.
Clearly state the players, strategy sets and interpretations of the utilities.
2. Obtain the full game representations $$(A, B)$$ for the zero sum games with row play payoff matrix given by:
1. $$A =\begin{pmatrix}1 & 3\\ -1 & 4\end{pmatrix}$$
2. $$A =\begin{pmatrix}1 & -2\\ -1 & 2\end{pmatrix}$$
3. $$A =\begin{pmatrix}1 & -2 & 4\\ 2 & -1 & 2\\ 7 & -7 & 6\end{pmatrix}$$
3. Consider the game described as follows:
An airline loses two suitcases belonging to two different travelers. Both suitcases have the same value. An airline manager tasked to settle the claims of both travelers explains that the airline is liable for a maximum of £5 per suitcase.
To determine an honest appraised value of the suitcases, the manager separates both travelers and asks them to write down the amount of their value at no less than £2 and no larger than £5 (to the single dollar):
• If both write down the same number, that number as the true dollar value of both suitcases and reimburse both travelers that amount.
• However, if one writes down a smaller number than the other, this smaller number will be taken as the true dollar value, and both travelers will receive that amount along with a bonus/malus: £2 extra will be paid to the traveler who wrote down the lower value and a £2 deduction will be taken from the person who wrote down the higher amount.
Represent this as a Normal Form Game.
## Using Nashpy¶
See Create a Normal Form Game for guidance of how to use Nashpy to create a Normal form game.
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# Circumcentre, Incentre, Orthocentre and Centroids
Contributed by:
This ppt includes the following topics:-
Medians
Centroid
Angle Bisector
Incentre
Altitude
Orthocentre
Perpendicular Bisector and many more.
1. Centers of Triangles or
Points of Concurrency
Prepared for Ms. Pullo’s
Geometry Classes
2.
3. Example 1
In MNP, MC and ND are medians.
M D P
C What is NC if NP = 18?
MC bisects NP…so 18/2 9
If DP = 7.5, find MP.
7.5 + 7.5 = 15
4. How many
medians does a
triangle have?
5. The medians of
a triangle are
concurrent.
The intersection of
the medians is
called the
CENTRIOD.
6. Theorem
The length of the segment
from the vertex to the
centroid is twice the length
of the segment from the
centroid to the midpoint.
7. Example 2
In ABC, AN, BP, and CM are medians.
If EM = 3, find C
EC = N
P
2(3) E
EC = 6 B
M
A
8. Example 3
In ABC, AN, BP, and CM are medians.
If EN = 12, find
AE = C
AN = AE + EN
2(12)=24 N
P E
AN = 24 +
B
12
AN = A M
9. Example 4
In ABC, AN, BP, and CM are medians.
If EM = 3x + 4
and CE = 8x, C
what is x?
N
P E
x=4 M
B
A
10. Example 5
In ABC, AN, BP, and CM are medians.
If CM = 24
what is CE? C
CE = 2/3CM N
CE = P E
2/3(24) B
CE = 16 M
A
11. Angle Bisector
12. Example 1
In WYZ, ZX bisects WZY . If m1 = 55,
find mWZY .
W mWZY 55 55
mWZY 110
X
1
2
Z Y
13. Example 2
In FHI, IG is an angle bisector. Find mHIG.
F
5( x 1) I
G (4 x 1)
5(x – 1) = 4x + 1
H 5x – 5 = 4x + 1
x=6
14. How many angle bisectors
does a triangle have?
three
The angle
bisectors of a
triangle are
concurrent
____________.
The intersection of the
angle bisectors is
called the ________.
Incenter
15. The incenter is the same distance from the
sides of the triangle.
Point P is called
B
the __________.
Incenter
D
F P
A E C
16. Example 4
The angle bisectors of triangle ABC meet at point L.
• What segments are congruent? LF, DL, EL
• Find AL and FL.
Triangle ADL is a
A right triangle, so use
FL = 6 Pythagorean thm
8
D AL2 = 82 + 62
F AL2 = 100
L AL = 10
6
C E B
17.
18. Tell whether each red segment is an altitude of the
The altitude is the
“true height” of
the triangle.
19. How many altitudes
does a triangle have?
The altitudes of
a triangle are
concurrent.
The intersection of the
altitudes is called the
ORTHOCENTER.
20. Perpendicular Bisector
21. Example 1: Tell whether each red segment is
a perpendicular bisector of the triangle.
22. Example 2: Find x
3x + 4 5x - 10
23. How many perpendicular
bisectors does a triangle
have?
The perpendicular
bisectors of a triangle
are concurrent.
The intersection of the
perpendicular bisectors is called
the CIRCUMCENTER.
24. The Circumcenter is
equidistant from the vertices
of the triangle.
B
PA = PB = PC
P
A C
25. Example 3: The perpendicular bisectors of
triangle ABC meet at point P.
• Find DA. DA = 6
• Find BA. BA = 12
• Find PC. PC = 10
• Use the Pythagorean Theorem B
to find DP.
DP2 + 62 = 102 6
D 10
DP + 36 = 100
2
DP2 = 64 P
A C
DP = 8
26. Tell if the red segment is an altitude,
perpendicular bisector, both, or neither?
NEITHER
ALTITUDE
PER.
BOTH BISECTOR
27. IN A NUT SHELL
Median – Centroid
Angle Bisector – Incenter
Altitude – Orthocenter
Perpendicular Bisector - Circumcenter
Angle Bisector: The Incentor is equidistance to
the sides
Perpendicular Bisector – the Circumcenter is
equidistance to the
vertex
28. The End
|
Draw ∆ABC, right-angled at B, such that AB = 3 cm and BC = 4 cm.
Question:
Draw $\triangle A B C$, right-angled at $B$, such that $A B=3 \mathrm{~cm}$ and $B C=4 \mathrm{~cm}$. Now, construct a triangle similar to $\triangle A B C$, each of whose sides is $\frac{7}{5}$ times the corresponding sides of
ABC.
Solution:
Step 1. Draw a line segment BC = 4 cm.
Step 2. With B as centre, draw an angle of 90o.
Step 3. With B as centre and radius equal to 3 cm, cut an arc at the right angle and name it A.
Step 4. Join AB and AC.
Thus, △ ABC is obtained .
Step 5. Extend BC to D, such that BD $=\frac{7}{5} \mathrm{BC}=\frac{7}{5}(4) \mathrm{cm}=5.6 \mathrm{~cm}$.
Step 6. Draw DE ∥ CA, cutting AB produced to E.
Thus, $\triangle \mathrm{EBD}$ is the required triangle, each of whose sides is $\frac{7}{5}$ the corresponding sides of $\triangle A B C$.
|
How To Factor Trinomials How To Factor Trinomials Today in this session we are going to discuss about Factoring Trinomials. Trinomials are the very interesting and important part of the elementary algebra. Trinomial is a polynomial which has three terms. The three terms can be any variable. For instance 2x + 4y +9z and x^3 + 4x^2 + 5x are trinomials. Factoring a trinomial is similar to the finding factors of a given integer. It completely depends on ones skills of multiplication because factoring a number is just reverse process of multiplication. In this session we will learn how to find factors of the trinomials that have the following forms: 1.
x^2 + bx + c
2.
ax^2 + bx +c
So first of all we should know the whole procedure to solve these problems. Know More About How do you Find the Circumference of a Circle Math.Tutorvista.com
Page No. :- 1/4
1. The procedure of finding factors of trinomial of the form x^2 + bx + c. a. First of all make a set up for the product of two brackets ()*() where each bracket will have two terms. b. Now we need to find the factors that will placed at the first position. For this to get back the square of x we need to set x in first position of both the brackets say (x ¹ a) * (x ¹ b). c. Now the task is to get the factors that we will place in the last positions of the brackets. The factors for the last position must be two expressions such that their multiplication become equal to c that is a constant and their sum become equals to b that is a number and coefficient of the variable x. Now in this case: If the constant c is positive then our factors both have the same sign which will depend on the sign of b. If constant c is negative then our factors both have the opposite signs which also depend on the sign of b. For example x^2 – 5x + 6 = (x -2) (x -3) here the first place of both the brackets have x to satisfy the square of x in the equation and the next two factors are 2 and 3 where (2 + 3) = 5 and (2*3) = 6 so it satisfy the conditions as above mentioned. 3. The procedure to find the factors of trinomials of the form ax^2 + bx +c:
Page No. :- 2/4
a. Similarly in this we will set up a product of two brackets ()*() where each of the bracket will have two terms. b. Now we will use trial and error method to find the required factors. In this the factors of a will be placed in the first term and in the last term we will place factors of c. For example: 3y^2 – 5y + 2 = (3y - 2) ( y - 1). Here this is the right combination. The first term is satisfying the 3x^2 of the equation if we will multiply them.
Math.Tutorvista.com
Page No. :- 4/4
Thank You
Math.TutorVista.com
How To Factor Trinomials
Trinomial is a polynomial which has three terms. The three terms can be any variable. For instance 2x + 4y +9z and x^3 + 4x^2 + 5x are trino...
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# What Is 33/60 as a Decimal + Solution With Free Steps
The fraction 33/60 as a decimal is equal to 0.55.
Proper fractions, improper fractions, and mixed fractions are the three types of Fractions. Proper fractions are those in which the numerator is less than the denominator, whereas Improper fractions are those in which the numerator is greater than the denominator. An improper fraction and a whole number combine to form a Mixed fraction.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 33/60.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 33
Divisor = 60
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 33 $\div$ 60
This is when we go through the Long Division solution to our problem. The following figure shows the long division:
Figure 1
## 33/60 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 33 and 60, we can see how 33 is Smaller than 60, and to solve this division, we require that 30 be Bigger than 60.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 33, which after getting multiplied by 10 becomes 330.
We take this 330 and divide it by 60; this can be done as follows:
330 $\div$ 60 $\approx$ 5
Where:
60 x 5 = 300
This will lead to the generation of a Remainder equal to 330 – 300 = 30. Now this means we have to repeat the process by Converting the 30 into 300 and solving for that:
300 $\div$ 60 = 5
Where:
60 x 5 = 300
This, therefore, produces another Remainder which is equal to 300 – 300 = 0.
Finally, we have a Quotient generated after combining the pieces of it as 0.55=z, with a Remainder equal to 0.
Images/mathematical drawings are created with GeoGebra.
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# Projectile word problems
A few carefully chosen projectile word problems to help you see how to solve these problems.
Problem #1:
A ball is dropped from a helicopter traveling with a speed of 75 m/s. After 2 seconds, the ball lands in the pool. Assuming no air resistance, what was the horizontal distance between the ball and the pool when it fell from the helicopter?
Solution
Key ideas
The first key idea here is that once released, the ball is a projectile launched horizontally.
Furthermore, since the ball was dropped and not shot from the helicopter, the ball will have the same velocity as the helicopter.
Finally, since the ball will travel horizontally, we can use the formula
distance = speed × time
distance = 75 m/s × 2 s = 150 meters
The helicopter was 150 meters away from the pool.
## Interesting projectile word problems
Problem #2:
From a height of 5 meters, a ball is thrown horizontally a distance of 15 meters. What is the speed of the ball?
Use g = 10 m /s2
Solution
The formula to get the speed is speed =
d / t
We already have the distance the ball will travel. It is 15 meters. All we need now is the time it takes the ball to hit the ground. Since the ball is experiencing free fall before it hits the ground, we can use the free fall equation.
d =
g × t2 / 2
d =
10 × t2 / 2
= 5t2
The height the ball fell is 5 meters, so replace d with 5.
d = 5t2
5 = 5t2
5/5 = (5/5)t2
1 = 1t2
t = 1
The ball hit the floor after 1 second.
Speed =
15 meters / 1 second
Speed = 15 m/s
Problem #3:
A projectile is launched with a speed of 40 m/s at 60 degrees above the horizontal. What are the horizontal and vertical velocities at launch?
Now we show the horizontal component in blue and the vertical component in green.
Some basic trigonometric identities will help us solve this now.
Let us call the horizontal speed vx and the vertical speed vy
cos(60°) =
vx / 40 m/s
0.5 =
vx / 40 m/s
vx = 0.5 × 40 = 20 m/s
sin(60°) =
vy / 40 m/s
0.86 =
vy / 40 m/s
vx = 0.86 × 40 = 34.4 m/s
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### Home > CCAA > Chapter 1 CC2: Unit 1 > Lesson CC2: 1.1.2 > Problem1-15
1-15.
Maria was playing a game with her brother. She said, “I’m thinking of a number. When you multiply my number by six and add seven, you get twenty-five. What is my number?”
1. Find Maria’s number.
How big do you expect her number to be? Can you think of a reasonable number to try?
Pick a reasonable number and try Maria's steps. Let's try $2$. If Maria multiplies this number ($2$) by $6$, she would get $12$. Then, adding $7$, she would get $19$.
Can you think of a number that might work better? Try it!
Since $19$ is smaller than $25$, let's choose a bigger number. Let's try $3$. If Maria multiplies $3$ by $6$, she would get $18$. Adding $7$, she would get $25$. Since Maria gets $25$ by using this number, $3$ is the number she described.
We now know that Maria's number is $3$.
2. Explain how you figured out your answer to Maria’s number puzzle.
Can you describe with words the steps that you took to find Maria's number?
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# NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5
These NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 Questions and Answers are prepared by our highly skilled subject experts.
## NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.5
Question 1.
Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Solution:
(i) 7 cm, 24 cm,-25 cm
(7)2 + (24)2 = 49 + 576 = 625 = (25)2 = 25
∴ The given sides make a right angled triangle with hypotenuse 25 cm
(ii) 3 cm, 8 cm, 6 cm(8)2 = 64
(3)2 + (6)2 = 9 + 36 = 45
64 ≠ 45
The square of larger side is not equal to the sum of squares of other two sides.
∴ The given triangle is not a right angled.
(iii) 50 cm, 80 cm, 100 cm
(100)2= 10000
(80)2 + (50)2 = 6400 + 2500
= 8900
The square of larger side is not equal to the sum of squares of other two sides.
∴The given triangle is not a right angled.
(iv) 13 cm, 12 cm, 5 cm
(13)2 = 169
(12)2 + (5)2= 144 + 25 = 169
= (13)2 = 13
Sides make a right angled triangle with hypotenuse 13 cm.
Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM . MR.
Solution:
We have PQR is a right triangle and PM ⊥ QR.
Question 3.
In the given figure, ABD is a triangle right angled at A and AC ⊥. BD. Show that
(i) AB2 = BC.BD
(ii) AC2 = BC.DC
Solution:
Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
Solution:
Given: In ∆ABC, ∠C = 90° and AC = BC
To Prove: AB2 = 2AC2
Proof: In ∆ABC,
AB2= BC2 + AC2
AB2 = AC2 + AC2 [Pythagoras theorem]
= 2AC2
Question 5.
ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2 , Prove that ABC is a right triangle.
Solution:
Given that ABC is an isosceles triangle with AC = BC and given that AB² = 2AC²
Now we have AB² = 2AC²
AB² = AC² + AC²
But AC= BC (Given)
AB² = AC² + BC²
Hence by Pythagoras theorem ∆ABC is a right triangle where AB is the hypotenuse of ∆ABC.
Question 6.
ABC is an equilateral triangle of side la. Find each of its altitudes.
Solution:
Given: In ∆ABC, AB = BC = AC = 2a
We have to find length of AD
In ∆ABC,
AB = BC = AC = 2a
BD = $$\frac { 1 }{ 2 }$$ x 2 a = a
⇒ AD2 = AB2 – BD2= (2a)2 – (a)2 = 4a2– a2= 3a2
Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution:
Given: ABCD is a rhombus. Diagonals AC and BD intersect at O.
To Prove: AB2+ BC2+ CD2+ DA2 = AC2+ BD2
Question 8.
In the given figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
Solution:
(i) Given: ∆ABC, O is any point inside it,
OD, OE and OF are perpendiculars to BC, CA and AB respectively.
To Prove:
Question 9.
A ladder 10 m long reaches a window 8 m above the ground. ind the distance of the foot of the ladder from base of the wall.
Solution:
By Pythagoras theorem
Hence the distance of the foot of the ladder from base of the wall in 6 m.
Question 10.
A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution:
Let AB i.e., the vertical pole of height 18 m and AC be the guy wire of 24 m long. BC is the distance from the vertical pole to where the wire will be staked.
By Pythagoras theorem
Question 11.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1$$\frac { 1 }{ 2 }$$ hours?
Solution:
Question 12.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Solution:
We have two poles.
We have
BC = 12 m
AB = 11 – 6
AB = 5 m
By Pythagoras theorem in right triangle ABC
AC² = AB² + BC²
AC² = (12)² +(5)²
AC² = 144 + 25
AC² = 169
AC = 13 m
Hence the distance between the tops is 13 m
Question 13.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.
Solution:
Given: ∆ABC is a right angled at C D and E are the points on the side CA and CB.
To Prove: AE² + BD² = AB² + DE²
Proof : ∆ACE is right angled at C
AE² = AC² + CE²… (i)
(Pythagoras theorem)
Question 14.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see the figure). Prove that 2AB2 = 2AC2 + BC2.
Solution:
We have
BD = 3CD
∴ BC = BD + DC
⇒ BC = 3CD + CD
BC = 4CD
⇒ CD = $$\frac { 1 }{ 4 }$$BC … (i)
And BD = 3CD
⇒ BD = $$\frac { 3 }{ 4 }$$BC …(ii)
Since ∆ABD is a right triangle, right angled at
Similarly, ∆ACD is right angled at D.
AC² = AD² + CD² …(iv)
Substracting (iv) from (iii)
We get
Question 15.
In an equilateral triangle ABC, D is a point on side BC, such that BD = $$\frac { 1 }{ 3 }$$BC. Prove that 9AD2 = 7AB2.
Solution:
Question 16.
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Solution:
Let ABC be an equailateral traingle of each side x. AD be its altitude.
So, AB = BC = CA = x
and BD = DC = $$\frac { 1 }{ 2 }$$ BC = $$\frac { x }{ 2 }$$
In right triangle ADC in which ∠D = 90°
DC = base and AC = hypotenuse.
Apply Pythagorus theorem, we get
So, we get length of each side is x an length of altitude is $$\frac{\sqrt{3} x}{2}/latex] Then, three times the square of each side = 3 x (x)² = 3x² … (i) and four times, the square of its altitudes = 4 x [latex]\frac { 3 }{ 4 }$$x² = 3x² … (ii)
It shows that equations (i) and (ii) are same. Hence times the square of one side an equilateral triangle is equal to four times the square of its altitude.
Question 17.
Tick the correct answer and justify : In ∆ABC, AB = 6$$\sqrt{3}$$ cm, AC = 12 cm and BC = 6 cm. The angle B is:
(a) 120°
(b) 60°
(c) 90°
(d) 45
Solution:
(12)= (6$$\sqrt{3}$$)² + (6)²
|
Thinking Mathematically (6th Edition)
Published by Pearson
Chapter 6 - Algebra: Equations and Inequalities - 6.2 Linear Equations in One Variable and Proportions - Exercise Set 6.2 - Page 362: 17
{-17}
Work Step by Step
4x -14 = -82 Step 1: Add 14 on both the sides 4x -14 +14 = -82 +14 Step 2: Simplify 4x = -68 Step 3: Divide both the sides by 4 $\frac{4x}{4}$ = $\frac{-68}{4}$ Step 4: Simplify x = -17 Now we check the proposed solution, -17, by replacing x with -17 in the original equation. Step 1: the original equation 4x -14 = -82 Step2: Substitute -17 for x 4.(-17) - 14 =-82 Step 3: Multiply 4.(-17) = -68 -68 - 14 = -82 Step 4: Solve -68 - 14 = -82 -82 =-82 Since the check results in true statement, we conclude that the solution set of the given equation is {-17}
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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Home MATHEMATICS TOPIC 1: EXPONENTS AND RADICALS ~ MATHEMATICS FORM TWO
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# TOPIC 1: EXPONENTS AND RADICALS ~ MATHEMATICS FORM TWO
Exponents
Exponents tell how many times to use a number itself in multiplication. There are different laws that guides in calculations involving exponents. In this chapter we are going to see how these laws are used.
Indication of power, base and exponent is done as follows:
Solution:
To write the expanded form of the following powers:
Solution
To write each of the following in power form:
Soln.
The Laws of Exponents
List the laws of exponents
First law:Multiplication of positive integral exponent
Second law: Division of positive integral exponent
Third law: Zero exponents
Fourth law: Negative integral exponents
Verification of the Laws of Exponents
Verify the laws of exponents
First law: Multiplication of positive integral exponent
Generally,
when we multiply powers having the same base, we add their exponents.
If x is any base and m and n are the exponents, therefore:
Example 1
Solution
If you are to write the expression using the single exponent, for example,(63)4.The expression can be written in expanded form as:
Generally if a and b are real numbers and n is any integer,
Example 2
Example 3
Example 4
Generally, (xm)= X(mxn)
Example 5
Rewrite the following expressions under a single exponent for those with identical exponents:
Second law: Division of positive integral exponent
Example 6
Example 7
Therefore, to divide powers of the same base we subtract their exponents (subtract
the exponent of the divisor from the exponent of the dividend). That
is,
where x is a real number and x ≠ 0, m and n are integers. m is the exponent of the dividend and n is the exponent of the divisor.
Example 8
solution
Third law: Zero exponents
Example 9
This is the same as:
If a ≠ 0, then
Which is the same as:
Therefore if x is any real number not equal to zero, then X0 = 1,Note that 00is undefined (not defined).
Fourth law: Negative integral exponents
Also;
Example 10
Exercise 1
1. Indicate base and exponent in each of the following expressions:
2. Write each of the following expressions in expanded form:
3. Write in power form each of the following numbers by choosing the smallest base:
1. 169
2. 81
3. 10,000
4. 625
a. 169 b. 81c. 10 000 d. 625
4. Write each of the following expressions using a single exponent:
5. Simplify the following expressions:
6. Solve the following equations:
7. Express 64 as a power with:
1. Base 4
2. Base 8
3. Base 2
Base 4 Base 8 Base 2
8. Simplify the following expressions and give your answers in either zero or negative integral exponents.
9. Give the product in each of the following:
10. Write the reciprocal of the following numbers:
Laws of Exponents in Computations
Apply laws of exponents in computations
Example 11
Solution
Re-arranging Letters so that One Letter is the Subject of the Formula
Re-arrange letters so that one letter is the subject of the formula
A formula is a rule which is used to calculate one quantity when other quantities are given. Examples of formulas are:
Example 14
From the following formulas, make the indicated symbol a subject of the formula:
Solution
Transposing a Formulae with Square Roots and Square
Transpose a formulae with square roots and square
Make the indicated symbol a subject of the formula:
Exercise 3
1. Change the following formulas by making the given letter as the subject of the formula.
2. Use mathematical tables to find square root of each of the
following:
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### Free Educational Resources
+
2 Tutorials that teach Basic Quadratic Factoring
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Author: Sophia Tutorial
##### Description:
In this lesson, students will learn how to factor quadratic expressions.
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Tutorial
This tutorial covers basic quadratic factoring, through the exploration of:
1. FOIL: A Review
## 1. FOIL: A Review
In review, FOIL is an acronym used to remember the steps for multiplying terms in binomial multiplication. It stands for:
First
Outside
Inside
Last
If you are multiplying the two binomials shown below, you start by multiplying your first two terms in each parentheses, followed by your outside terms, then your inside terms, and finally, your last terms. In the resulting expression, combine your two like terms to arrive at your final answer.
This resulting expression is a quadratic expression that can be generally written in this form:
Quadratic expressions in this form have an x-squared term, an x term, and a constant term. Quadratic factoring is the reverse process of FOILing. Therefore, you could factor your resulting expression above, and the result would be your original expression, multiplying the two binomials.
Consider the previous example to look at the process for factoring quadratics. You can see that the constant term 15 in expanded form is the product of the two numbers in factored form, 3 and 5.
You can also see that the coefficient of the x term in expanded form, 8, is the sum of the two numbers in factored form.
When you factor a quadratic expression in expanded form, you use these two patterns to identify the two numbers to use in the factored form of the expression.
Using the same expanded form expression from above, how would you write it into factored form?
Step 1: Identify the pairs of numbers that, when multiplied together, equal the constant term, which in this case is 15.
When considering factor pairs for a constant term, both positive and negative numbers should be considered. This will be especially important when there is a subtraction in the expanded form, as this indicates that at least one number in the factor pair will be negative.
The factor pairs of 15 are as follows:
Step 2: Add your factor pairs to find the pair of numbers that, when added together, also equal the coefficient of your x term (in this case, 8).
Once you have found the pair of numbers with the correct sum, you do not need to consider other factor pairs, because there will be only one pair of numbers that will multiply to the constant term and add to the x term coefficient. In this example, 3 and 5 are your pair of numbers that multiply to 15 and add to 8.
Step 3: Use this pair of numbers to write the expression in factored form.
Suppose you want to factor the expression:
Start by identifying the pair of numbers that, when multiplied together, equal -12.
Now, because your constant term is negative, you know that one of the numbers must be negative and the other one will be positive. The pairs of numbers that multiply to -12 are:
Next, look for which pair also has a sum of 4.
Now you can use your two numbers, -2 and 6, to write your expression in factored form.
You can verify that you correctly factored the expression by multiplying the binomials together using FOIL to ensure that you arrive back at your original expression.
This is the same as your original expression, which means your answer is the correct factored form.
Here’s another example. Suppose you want to factor the expression:
Start by identifying the pairs of numbers that, when multiplied together, equal 10. Note that because the pair of numbers must multiply to a positive number but add to a negative number, both of the numbers must be negative.
Now, look for which of these pairs of numbers also have a sum of -7.
Therefore, you use the pair (-2, -5) to write your expression in factored form.
Use what you’ve learned so far about quadratic factoring to factor this last expression:
Can you identify the pairs of numbers that, when multiplied together, equal -2? Remember, because the pair of numbers must multiply to a negative number and add to a negative number, one of the numbers will be positive and one will be negative.
Now, because there’s no written number in front of the x term, the x term has an implied coefficient of -1. Therefore, look for which of these number pairs also has the sum of -1.
Since you’ve determined that 1 and -2 add to a -1, you can use this pair of numbers to write your expression in factored form:
Today you reviewed the acronym FOIL, which is used to remember the steps for multiplying terms in binomial multiplication. You learned that quadratic factoring is used to write a quadratic expression from expanded form to factored form. In quadratic factoring, the constant term in expanded form is the product of the two numbers in factored form and the coefficient of the x term in expanded form is the sum of the two numbers in factored form. You then use these two facts to write your expression from expanded form to factored form.
Source: This work is adapted from Sophia author Colleen Atakpu.
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Guided Lessons
# Parts of a Set: Fractions
Get your students excited to learn about fractions with this hands on lesson. Working in groups will help students learn from each other and share ideas.
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Students will be able to divide sets into equal parts and identify the parts using fractional names.
(15 minutes)
• Tell students that today they are going to learn how to represent parts of a Set, or a group of things, that we call fractions.
• Tell them that a FractionRepresents a part of a set, and therefore it is not meaningful without knowing what the total of the set is.
• Show a set of four balls consisting of three soccer balls and one football.
• Ask students what fraction of this set of balls are soccer balls. Answer: three-fourths or three quarters.
• Tell students that you have to know what the whole amount, or total number of items in a set, is to know what the given part represents. Relate it back to your last example by bringing up that you have to know that there are four balls total to know that three quarters of them are soccer balls.
• Have five girls and three boys come to the front of the class. Ask the class what fraction of the group is girls. Ask the class what fraction of the group is boys. Remind them, if needed, that each is a fraction out of eight.
• Show a set of shapes containing four squares and three rectangles. Ask what fraction of these shapes are triangles. Answer: zero out of seven.
• If more examples are needed, you can collect shoes from students and determine what fraction require laces, or bring students to the front of the class and determine what fraction are blonde or brunette.
(15 minutes)
• At this point, review the notion that we have to know what the total amount of the set is before we can say what the fraction is.
• If you feel the students are ready, you can introduce the notion of a DenominatorAs being the total amount in the set. Create an example and write your denominator under the fraction line.
• Show a stack of text books and ask your class what fraction are maths books, science books, or story books.
• Tell students that a NumeratorIs the specific item within a set that the fraction is about.
(20 minutes)
• Divide your students into groups.
• Give each group a set of counters or coins.
• Have each group choose twelve 2-sided counters and set them out on the desk, breaking them into three groups.
• Ask them to flip over one group so that they have two groups of four of one colour and the third group is a different colour.
• Ask what fraction of the whole set is one colour, then ask what fraction of the set is the other colour (or heads and tails if your class is using coins.)
• Ask students how many counters are in one-third of the set.
• Have students arrange the counters so that one-fourth of them are [a colour of your choosing].
• Ask students to rearrange counters into six equal groups all with the same colored side up.
• Tell them to flip the counters in one group. Ask them what fraction of the set is [a colour of your choosing].
• Tell your students to flip another group. Ask them what fraction of the set is [a colour of your choosing].
• Have them flip another group. Ask them what fraction of the set is [a colour of your choosing].
• Ask students to arrange the counters so that one-fourth of them have [a colour of your choosing] sides showing.
• Ask students to share what they have found out from this activity.
(10 minutes)
• Pass out Colorful Plants: practising Fractions.
• Have the students write and draw about fractions of a set in their maths journals. If using the maths journal, have them draw something easy to draw and colour, such as circles or squares. You will collect these for assessing once they are finished. This will help you know what the students have learned.
• Support:For students who are struggling, spend additional time helping them grasp the notion of parts of a whole. You might also draw a couple of examples in their journal and ask them to explain it to you that way.
(10 minutes)
• Circulate as groups are working and determine whether or not students understand the lesson.
• Grade the worksheets to check understanding of individual students.
• Mark the journal entries to determine the students' levels of understanding of fractions as parts of a whole.
(5 minutes)
• Have students take out their maths journals and explain with pictures, numbers, and words what they learned about fractions today.
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Solve Quadratic equations: 4x ^ 2 – 5x – 12 = 0
4x ^ 2 – 5x – 12 = 0 is an example of a quadratic equation. This equation is a mathematical equation and it requires a formula to solve it. This is one of the mathematical equation-based questions where variables and numbers are given. In this equation, we have to simply prove that left hand side is equal to right hand side (L.H.S = R.H.S).
4x ^ 2 – 5x – 12 = 0, is one of the quadratic equation forms. This type of quadratic equations problem generally solved by simplification method. This method involves breaking of equations till it equalizes the equation. Mathematical equations a just seemed to be complex but they’re actually difficult. It requires basic mathematical knowledge and formulas to solve any equations.
About Quadratic Equation 4x ^ 2 – 5x – 12 = 0
Quadratic equation is a mathematical term. Quadratic Equation a problem-based equation that is asked to solve to prove a solution. It is derived from a Latin word quadratus. 4x ^ 2 – 5x – 12 = 0, is a prime example of quadratic equation. A polynomial equation of a second degree where atleast one of the variables have a square on it. Quadratic equations are represented in the form of ax^2 + bx + c = 0.
Example:
• In the given equation, ax^2 + bx + c = 0
• a, b, c are called as a known coefficient where a is not equal 0
• x is represented as variable.
Why coefficient ‘a’ is not equal to zero in a quadratic equation?
We all have learned that Coefficient ‘a’ can never be represented with a value 0. Mathematics have rules and regulations for each mathematical concept. Quadratic equation will get converted to a linear equation if coefficient ‘a’ is represented equal to 0 which means it will alter the whole equation.
Formula to solve 4x ^ 2 – 5x – 12 = 0
Given term is an example of quadratic equation, and it requires a special formula to solve such kinds of problems in mathematics.
Quadratic Formula: x = [ -b (b2 – 4ac)] / 2a
This is an accurate and verified formula to solve quadratic equations and related problems. This formula is a formula of simplification in which we simplify the equation as much as possible. This technique is a best method to solve the equation, 4x ^ 2 – 5x – 12 = 0.
Four different Methods to Solve 4x ^ 2 – 5x – 12 = 0
Factoring is a technique to solve an equation by splitting the middle terms of the equation. Factoring technique is used to determine the factors of an equation and then it is solved.
x = [ -b (b2 – 4ac)] / 2a, this is a quadratic formula, by simply adding all the respective values in this formula we can find a right solution.
Taking the Square Root
Taking the square root is a method to find a solution for solving Quadratic Equation. In this technique we solve the roots of both the sides to get an accurate answer for this. This method applicable only when the squared variable is brought to the one side and the constant terms shifted to other side.
Completing the square
Divide the equation by the coefficient of square variable. It means divide the equation by the coefficient ‘a’ then imply the square method. Add a constant term on both the sides of the equation and calculate.
Solution of the 4x ^ 2 – 5x – 12 = 0
• We will solve the equation by quadratic formula.
• Quadratic formula: x = [ -b ± √ (b2 – 4ac)] / 2a
• In this given equation: a = 4, b = – 5 and c = – 12 where a is not equal to 0.
• X is an unknown factor or a variable.
• Put all the respective values in the quadratic formula.
Assume, X = x
• Quadratic formula = X = [ -b ± √ (b2 – 4ac)] / 2a
• X = [ – (-5) ± √ ( (-5)2– 4 (4) (-12))] / 2 (4)
• X = [ 5 ± √ ( 25 +192 )] / 8
• X = [5 ± √ ( 217 )] / 8
• X = 5 + √ 217 / 8 and X = 5 – √ 217 / 8
Thus, x = 2.466 and x = – 1.216
Two consecutive solution for the equation are: x = 2.466 and x = – 1
• x2 – 2x – 24 = 0
• 2x2 + 4x – 5 = 0
• x2 – 1x – 6 = 0
These are three examples of what quadratic equations looks like. These equations are further solved with the help of quadratic formula to find the value of ‘x’.
Conclusion
4x ^ 2 – 5x – 12 = 0, is an example of quadratic equation. This equation can be solve by four different methods. In this equation we have to find the value of the x, by breaking down the equation as per the mathematical order. Quadratic equations is a small universe in the cosmos of mathematics.
Quadratic equations are very easy to solve when we apply the right formula. If we have the formula, we have just put the right value at the right position and it’s done. 4x ^ 2 – 5x – 12 = 0 an equation and it can asked in the format of multiple choice questions also.
For any quadratic equations same quadratic formula is used to find the value of ‘x’. Mathematics just looks difficult but it’s not a difficult subject. Mathematics is a game of formula, if you know the formula you can crack all the mathematical problems.
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Contents of this Precalculus episode:
Absolute value of complex numbers, Sets on the complex plane, Equation of line, Equation of circle.
Text of slideshow
The absolute value of a complex number is its distance from zero.
We can compute this distance using the Pythagorean Theorem.
Let's see one more.
Instead of the formula, here we try factorization.
And now let’s see what else we can do with these complex numbers.
Let’s try to graph on the complex plane those complex numbers where:
We use the algebraic form,
that is
Next come some coordinate geometry horror stories.
The equation of
is the equation of a circle centered at the origin, with a radius of r.
Based on this, is also a circle centered at the origin, with a radius of r=2.
And means the circle and its inside.
Coordinate geometry horror stories:
The equation of a line:
The equation of a circle:
Let’s find on the complex plane complex numbers such that:
We use the algebraic form, so we replace z
with everywhere.
The inequality means one side of the line.
Let's see which side.
It is always a good idea to try a=0 and b=0.
This seems to fit, so we need this side of the line.
Next, let's see what is going on with this one:
The inequality means one side of the circle.
Either the inside or the outside of the circle.
Again, it is a good idea to try a=0 and b=0.
It seems we need the outside.
And because equality is not allowed,
the circle itself is not part of the region.
Finally, let’s see what this is about:
We will need to complete the square.
# Absolute value of complex numbers, sets on the complex plane
04
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Absolute value of complex numbers, Sets on the complex plane, Equation of line, Equation of circle.
Let's see this
Precalculus episode
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# Learn How to Find Prime Factors and Factor Pairs of 36 in Detail
Factors are whole numbers that are multiplied together to generate another number. If a × b = c, then a and b are c’s factors.
• A factor is an integer that divides a given whole number exactly with no remainder.
• Every whole number greater than one has at least two factors.
• When a number is divided by one of its factors, the result is another factor. A factor pair is made up of two factors.
• The products of each factor pair equal the number.
Keep in mind that the first factor pair is always 1 and the number itself.
## Key Points Regarding Factors
• A factor is a whole number that divides perfectly without a remainder. Eg, 3 is a factor of 36.
• A factor pair is a pair of factors. They produce a certain product when multiplied together. Eg, 2 and 27 have a product of 54, so 2 and 27 are factors of 54 and (2,27) is a factor pair of 54.
• A multiple is a number that appears in a specific times table. For example, 12 is a multiple of 3 because it appears in the 3 times table. 12 is also a multiple of four because it appears in the four times table
• Understanding factors and factor pairs are aided by knowing multiplication and division facts. Knowing divisibility rules can help you find multiples of a given number.
Let’s understand how to factorize a number in detail by taking the example of the number 36.
## Factors of 36
The numbers that divide 36 perfectly without leaving a residual are known as 36 factors. The factors of 36 can be both positive and negative, but they cannot be decimals or fractions. The factors of 36, for example, can be (1, 36) or (-1, -36). When we multiply a pair of negative numbers, such as -1 and -36, the outcome is the original number.
## What are the Factors of 36?
The factors of 36 are the numbers multiplied in pairs to produce the original number 36. Since the number 36 is a composite number, it contains numerous factors in addition to one and the number itself. As a result, the 36 factors are 1, 2, 3, 4, 6, 9, 12, 18, and 36.
### Pair Factors of 36
As previously stated, the pair factors of 36 can be both positive and negative. The pair factors of 36 are the two numbers multiplied together to produce the original number. The positive and negative pair factors are as follows:
Positive Pair Factors of 36:
1×36
2×18
3×12
4×9
6×6
Therefore, the positive factor pairs of 36 are (1,36), (2,18), (3,12), (4,9) and (6,6).
Negative PaFactors of 36:
-1 × -36
-2 × -18
-3 × -12
-4 × -9
-6 × -6
Therefore, the positive factor pairs of 36 are (-1,-36), (-2,-18), (-3,-12), (-4,-9) and (-6,-6).
### Prime Factorization of 36
The prime factorization of 36 refers to the practice of expressing the number as a product of prime factors of 36. Divide 36 by the least prime number, which is 2, to find the prime factors. When it can no longer be divided by two, divide it by the next prime number, 3, and repeat the process until the end product is 1. This is known as prime factorization, and a step-by-step approach for 36 is shown below.
• Divide 36 by 2
36 ÷ 2 = 18
• Again divide the quotient (18) with 2
18 ÷ 2 = 9
• Since the quotient (9) is no more divisible by 2, move to the next prime number i.e. 3
9 ÷ 3 = 3
• Lastly, divide the quotient (3) with 3 to get 1.
3 ÷ 3 = 1
From the above steps, it can be said that the prime factor of 36 will be 2 × 2 × 3 × 3 i.e. 22 × 32.
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@professionalbronco Ohh, I see! Very clever to think about units digits, since with bigger numbers, it can save you a lot of time. Still, I think that the best way to do the problem is to notice that all of $$\frac{18}{396},\frac{15}{396},\frac{9}{396}$$ aren't simplified.
BUT, let's talk about the units digit method that you used, since it's a very smart method to bring up. I'm not sure if you just compared the units digit or if you actually found the units digit of the whole fraction, because there's actually a difference.
You can't simply just look at the units digit of the numerators on both fractions. The fact that it worked out that way is a coincidence. However, if we go a bit deeper, we can see why it actually turned out to work.
Let's look at the first answer choice: $$\frac{7937}{396}=20+\frac{18}{396}$$
To compare the units digits, I'm going to turn $$20$$ into a fraction with denominator $$396$$. $$20=\frac{7920}{396}$$, so we have $$\frac{7937}{396}=\frac{7920}{396}+\frac{18}{396}$$. Clearly this doesn't work.
But wait-- it's really slow to multiply out $$396\cdot20$$. This is where we can use the units digit as a shortcut! I don't care about the rest of the digits of $$396\cdot20$$, all that matters is that it has units digit $$0$$. So all we needed is the fact that
$$\frac{7937}{396}=\frac{***0}{396}+\frac{18}{396}$$
A number with units digit $$0$$ plus a number with units digit $$8$$ definitely can't add to a number with units digit $$7$$. So, we can rule out the first option.
We can repeat this process with the other three options and it gives us another quick method to find the solution!
for the second option:
$$\frac{8053}{396}=20+\frac{133}{396}$$
again, $$20\cdot396$$ has units digit 0, so
$$\frac{8053}{396}=\frac{***0}{396}+\frac{133}{396}$$
$$0+3$$ is indeed $$3$$. However, since we didn't find exactly what $$20\cdot396$$ is, we can't be 100% sure that this is right. All we know is that it's a possibility.
for the third:
$$\frac{9521}{396}=24+\frac{15}{396}$$
for $$24\cdot396$$, it will have the same units digit as $$4\cdot6$$, which is $$4$$, so
$$\frac{9521}{396}=\frac{***4}{396}+\frac{15}{396}$$
This can't work because $$4+5=9$$.
last but not least:
$$\frac{5953}{396}=15+\frac{9}{396}$$
for $$15\cdot396$$, it will have the same units digit as $$5\cdot6$$, which is $$0$$, so
$$\frac{5953}{396}=\frac{***0}{396}+\frac{9}{396}$$
This can't work because $$0+9=9$$.
So, since the question doesn't have a "none of the above option", we can be safe with (b) since the other three are clearly wrong. Otherwise, just double check (b) since that's the only possibility and you'll have your answer!
(by the way, do you see how you can't just compare only the numerators of the two fractions? a few extra steps does the trick though :))
It might look long, but once you get the hang of it, it's a really clever shortcut! And in general, it's very smart to look out for units digit to simplify things. Great idea, well done! 😄
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2006 AMC 12A Problems/Problem 24
Problem
The expression
$$(x+y+z)^{2006}+(x-y-z)^{2006}$$
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
$\mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ } 2,015,028$
Solution 1
By the Multinomial Theorem, the summands can be written as
$$\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^ay^bz^c}$$
and
$$\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c},$$
respectively. Since the coefficients of like terms are the same in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are:
$${2006+2\choose 2} = 2015028$$
terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of $y$ and $z$ must be opposite. Now we find a pattern:
if the exponent of $y$ is $1$, the exponent of $z$ can be all even integers up to $2004$, so there are $1003$ terms.
if the exponent of $y$ is $3$, the exponent of $z$ can go up to $2002$, so there are $1002$ terms.
$\vdots$
if the exponent of $y$ is $2005$, then $z$ can only be 0, so there is $1$ term.
If we add them up, we get $\frac{1003\cdot1004}{2}$ terms. However, we can switch the exponents of $y$ and $z$ and these terms will still have a negative sign. So there are a total of $1003\cdot1004$ negative terms.
By subtracting this number from 2015028, we obtain $\boxed{\mathrm{D}}$ or $1,008,016$ as our answer.
Solution 2
Alternatively, we can use a generating function to solve this problem. The goal is to find the generating function for the number of unique terms in the simplified expression (in terms of $k$). In other words, we want to find $f(x)$ where the coefficient of $x^k$ equals the number of unique terms in $(x+y+z)^k + (x-y-z)^k$.
First, we note that all unique terms in the expression have the form, $Cx^ay^bz^c$, where $a+b+c=k$ and $C$ is some constant. Therefore, the generating function for the MAXIMUM number of unique terms possible in the simplified expression of $(x+y+z)^k + (x-y-z)^k$ is $$(1+x+x^2+x^3\cdots)^3 = \frac{1}{(1-x)^3}$$
Secondly, we note that a certain number of terms of the form, $Cx^ay^bz^c$, do not appear in the simplified version of our expression because those terms cancel. Specifically, we observe that terms cancel when $1 \equiv b+c\text{ (mod }2\text{)}$ because every unique term is of the form: $$\binom{k}{a,b,c}x^ay^bz^c+(-1)^{b+c}\binom{k}{a,b,c}x^ay^bz^c$$ for all possible $a,b,c$.
Since the generating function for the maximum number of unique terms is already known, it is logical that we want to find the generating function for the number of terms that cancel, also in terms of $k$. With some thought, we see that this desired generating function is the following: $$2(x+x^3+x^5\cdots)(1+x^2+x^4\cdots)(1+x+x^2+x^3\cdots) = \frac{2x}{(1-x)^3(1+x)^2}$$
Now, we want to subtract the latter from the former in order to get the generating function for the number of unique terms in $(x+y+z)^k + (x-y-z)^k$, our initial goal: $$\frac{1}{(1-x)^3}-\frac{2x}{(1-x)^3(1+x)^2} = \frac{x^2+1}{(1-x)^3(1+x)^2}$$ which equals $$(x^2+1)(1+x+x^2\cdots)^3(1-x+x^2-x^3\cdots)^2$$
The coefficient of $x^{2006}$ of the above expression equals $$\sum_{a=0}^{2006}\binom{2+a}{2}\binom{1+2006-a}{1}(-1)^a + \sum_{a=0}^{2004}\binom{2+a}{2}\binom{1+2004-a}{1}(-1)^a$$
Evaluating the expression, we get $1008016$, as expected.
Solution 3
Define $P$ such that $P=y+z$. Then the expression in the problem becomes: $(x+P)^{2006}+(x-P)^{2006}$.
Expanding this using binomial theorem gives $(x^n+Px^{n-1}+...+P^{n-1}x+P^n)+(x^n-Px^{n-1}+...-P^{n-1}x+P^n)$, where $n=2006$ (we may omit the coefficients, as we are seeking for the number of terms, not the terms themselves).
Simplifying gives: $2(x^n+x^{n-2}P^2+...+x^2P^{n-2}+P^n)$. Note that two terms that come out of different powers of $P$ cannot combine and simplify, as their exponent of $x$ will differ. Therefore, we simply add the number of terms produced from each addend. By the Binomial Theorem, $P^k=(y+z)^k$ will have $k+1$ terms, so the answer is $1+3+5+...+2007=1004^2=1,008,016 \implies \boxed{D}$.
Solution 4
Using stars and bars we know that $(x+y+z)^{2006}$ has ${2006+2\choose 2}$ or $2015028$ different terms. Now we need to find out how many of these terms are canceled out by $(x-y-z)^{2006}$. We know that for any term(let's say $x^{a}(-y)^{b}(-z)^{c}$) where $a+b+c=2006$ of the expansion of $(x-y-z)^{2006}$ is only going to cancel out with the corresponding term $x^{a}y^{b}z^{c}$ if only $b$ is odd and $c$ is even or $b$ is even and $c$ is odd. Now let's do some casework to see how many terms fit this criteria:
Case 1: $a$ is even
Now we know that $a$ is even and $a+b+c=2006$. Thus $b+c$ is also even or both $b$ and $c$ are odd or both $b$ and $c$ are even. This case clearly fails the above criteria and there are 0 possible solutions.
Case 2: $a$ is odd
Now we know that $a$ is odd and $a+b+c=2006$. Thus $b+c$ is odd and $b$ is odd and $c$ is even or $b$ is even and $c$ is odd. All terms that have $a$ being odd work.
We now need to figure out how many of the terms have $a$ as a odd number. We know that $a$ can be equal to any number between 0 and 2006. There are 1003 odd numbers between this range and 2007 total numbers. Thus $\frac{1003}{2007}$ of the $2015028$ terms will cancel out and $\frac{1004}{2007}$ of the terms will work. Thus there are $(\frac{1004}{2007})(2015028)$ terms. This number comes out to be $1,008,016$ $\implies \boxed{D}$ (Author: David Camacho)
Solution 5
Noticing how $y$ and $z$ are negative in the second part of the expression, let $x=a$ and $y+z = b$. Then we get $$(a+b)^{2006} + (a-b)^{2006}$$ We know that the terms that don't cancel out have even powers of $a$ and $b$. Our expansion will be in the form of $$2a^{2006} + x_1a^{2004}b^{2} + x_2a^{2002}b^{4} + \cdots + 2b^{2006}$$ Note that $b^n = (x+y)^n$ has $n+1$ terms. Furthermore, the current expression is irreducible as each term has a different $x$ power. Thus, when we write $a$ and $b$ back to their original terms, we will have $1+3+5+ \cdots + 2007 = 1004^2 = 1,008,016 \implies \boxed{D}$
-smartguy888
|
# Fractions: Computing the Total Purchase Cost
(*)(*)(*)(*)(*) ( 4.3, 379 students)
• Price per ClassroomFREE
• Grade Levels 4, 5, 6, 7
• Topics Numbers & Operations, Fractions
• Duration 15 Minutes
• Auto Scored? Yes
• Teacher Evaluation Needed? No
### Activity Description
In this activity, students will use their knowledge about fractions, multiplication of fractions with whole numbers, subtraction of fractional numbers to solve a real world problem.
Problem Statement: The student is asked to go to the supermarket to purchase some grocery items and the mother has given \$100 for the purchase. But the students is doubtful whether the \$100 will be enough or not. To make sure that the student has enough money, he/she pulls up the cost card (unit rate card) of the supermarket from internet and tries to compute the total cost of the purchase. Student will identify that he/she was right and needs some extra money to purchase all the items.
Students will also analyze a bar chart that shows the purchase cost of each item and then will compare the bars to find which items needs more money and which needs less money.
The activity is very common use case but makes sure that students thoroughly understand when and why to use fraction and multiplication of fractions.
### Learning Objective
Students will learn about using multiplication and subtraction in a real life scenario. The activity will ensure that students have mastered in the multiplication and subtraction operations and knows when to use them.
### Teacher Tips
Included with the activity, you can view the tips to clarify student's doubts or to evaluate answers (for a teacher scored worksheet).
### Common Core: MATH
Number & Operations_Fractions
Add and subtract fractions with unlike denominators (including mixed numbers) by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators. *For example, 2/3 + 5/4 = 8/12 + 15/12 = 23/12. (In general, a/b + c/d = (ad + bc)/bd.)*
Number & Operations_Fractions
Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. *For example, recognize an incorrect result 2/5 + 1/2 = 3/7, by observing that 3/7 < 1/2*.
Number & Operations_Fractions
Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction.
Number & Operations_Fractions
Solve real world problems involving multiplication of fractions and mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
Number & Operations_Fractions
Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions.
## More Activities You Might Like
\$3.00/Classroom
### Community Kitchen - Weekly Budget
Students will calculate the total cost of all ingredients needed for one Christmas meal They will then find out the total number of meals that can be prepared in a community kitchen for Christmas Day with a budget of \$5000.
• Fractions, Ratios & Proportions, Budgeting
• 6, 7, 8, 9, 10 Auto-Graded
• ### Grandma's Recipe - Unit Conversion & Unit Rate
\$2.00/Classroom
### Grandma's Recipe - Unit Conversion & Unit Rate
Students will convert the ingredients of Grandma's cookie from cooking units (eg: cups) to supermarket units (pounds, gallons) and then find the cost.
• Fractions, Ratios & Proportions, Business, FCS
• 4, 5, 6, 7 Auto-Graded
• ### Cross-country Cost Comparison: Ratio And Fraction
\$3.00/Classroom
### Cross-country Cost Comparison: Ratio And Fraction
Students act as the CEO of a company, to find the cheapest country to purchase the raw materials, students will do unit conversion between different currencies.
• Fractions, Ratios & Proportions
• 6, 7, 8
• ### Bake And Share: A Budget-friendly Christmas Cake
\$3.00/Classroom
### Bake And Share: A Budget-friendly Christmas Cake
The activity revolves around the concept of baking a Christmas cake on a tight budget compared to the cost of purchasing one from a store. Students will be guided through ingredient conversions, cost calculations, and the joy of creating a homemade treat.
• Fractions, Ratios & Proportions, Budgeting
• 6, 7, 8, 9, 10 Auto-Graded
|
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LINEAR EQUATIONS IN ONE VARIABLE
21
Linear Equations in
One Variable
2.1
Introduction
CHAPTER
2
In the earlier classes, you have come across several algebraic expressions and equations.
Some examples of expressions we have so far worked with are:
5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x 2 + 1, y + y 2
Some examples of equations are: 5x = 25, 2x – 3 = 9,
2
y +
5
37
=
2
2
, 6
z +
10
=−
2
You would remember that equations use the equality (=) sign; it is missing in expressions.
Of these given expressions, many have more than one variable. For example, 2xy + 5 has two variables. We however, restrict to expressions with only one variable when we form equations. Moreover, the expressions we use to form equations are linear.This means that the highest power of the variable appearing in the expression is 1.
These are linear expressions:
2x, 2x + 1, 3y – 7, 12 – 5z,
5
4
(
x
– 4)
+
10
These are not linear expressions:
x 2
+ 1, y + y 2 , 1 + z + z 2 + z 3
(since highest power of variable > 1)
Here we will deal with equations with linear expressions in one variable only. Such equations are known as linear equations in one variable. The simple equations which you studied in the earlier classes were all of this type.
Let us briefly revise what we know:
(a)
An algebraic equation is an equality involving variables. It has an equality sign. The expression on the left of the equality sign is the Left Hand Side (LHS). The expression on the right of the equality sign is the Right Hand Side (RHS).
2x – 3 = 7
2x – 3 = LHS 7 = RHS
22
MATHEMATICS
(b)
(c)
In an equation the values of the expressions on the LHS and RHS are equal. This happens to be true only for certain values of the variable. These values are the
solutions of the equation.
How to find the solution of an equation? We assume that the two sides of the equation are balanced. We perform the same mathematical operations on both sides of the equation, so that the balance is not disturbed. A few such steps give the solution.
x = 5 is the solution of the equation 2x – 3 = 7. For x = 5, LHS = 2 × 5 – 3 = 7 = RHS On the other hand x = 10 is not a solution of the equation. For x = 10, LHS = 2 × 10 –3 = 17. This is not equal to the RHS
2.2 Solving Equations which have Linear Expressions on one Side and Numbers on the other Side
Let us recall the technique of solving equations with some examples. Observe the solutions; they can be any rational number.
Example 1: Find the solution of 2x – 3 = 7
Solution:
Step 1 Add 3 to both sides.
2x – 3 + 3
= 2x = 10
7 + 3
or Step 2 Next divide both sides by 2.
(The balance is not disturbed)
2 x 10 = 2 2 or x = 5 (required solution)
Example 2: Solve 2y + 9 = 4
Solution: Transposing 9 to RHS
2y = 4 – 9 or 2y = – 5
Dividing both sides by 2,
To check the answer : LHS = 2
⎜ ⎛ ⎝
5 ⎞ ⎟
2
y =
+ 9
5
2
= – 5 + 9 = 4 = RHS
(solution)
(as required)
Do you notice that the solution
⎛ ⎜ ⎝
5
2
we solved did not have such solutions.
is a rational number? In Class VII, the equations
LINEAR EQUATIONS IN ONE VARIABLE
23
5 3
x +
=
3
2
2
Example 3: Solve
Solution: Transposing 5
or
Multiply both sides by 3, or
12
5
+
32
x − 35 − =− 8 = 3 22 2 3 x = – 4 = – 4 × 3 x = – 12 x (solution) −+ 8 5 − 3 = = RHS (as required)
2
2
2 to the RHS, we get
5
Check: LHS =
Do you now see that the coefficient of a variable in an equation need not be an integer?
2
=−
4
+
=
Example 4: Solve 15
– 7x = 9
4
15
Solution: We have
– 7x =9
4
15
or
– 7x = 9 –
4
21
or
– 7x =
4
21
or
x =
4 × (−7)
3
× 7
or
x =
4
× 7
3
or
x =
4
15
⎛ −
3 ⎞
15
21
36
− 7
Check: LHS =
+== 9
=
4
⎜ ⎝
4
⎠ =
44
4
(transposing 15
4
to R H S)
(dividing both sides by – 7)
(solution) RHS (as required)
EXERCISE 2.1
Solve the following equations.
1.
4.
7.
x – 2 = 7
3
17
+ x
=
7
7
2
x = 18
3
2.
y + 3 = 10
5. 6x = 12
8.
y
1.6 = 1.5
3.
6 = z + 2
t
6.
= 10
5
9. 7x – 9 = 16
24
MATHEMATICS
10. 14y – 8 = 13
11.
17 + 6p = 9
12.
x +
3
7
15
1
=
2.3 Some Applications
We begin with a simple example.
Sum of two numbers is 74. One of the numbers is 10 more than the other. What are the numbers? We have a puzzle here. We do not know either of the two numbers, and we have to find them. We are given two conditions.
One of the numbers is 10 more than the other. Their sum is 74. We already know from Class VII how to proceed. If the smaller number is taken to be x, the larger number is 10 more than x, i.e., x + 10. The other condition says that the sum of these two numbers x and x + 10 is 74. This means that x + (x + 10) = 74. 2x + 10 = 74
or Transposing 10 to RHS, or Dividing both sides by 2, The other number is
(i)
(ii)
2x = 74 – 10
2x = 64 32. This is one number. x + 10 = 32 + 10 = 42
x =
The desired numbers are 32 and 42. (Their sum is indeed 74 as given and also one number is 10 more than the other.) We shall now consider several examples to show how useful this method is.
Example 5: What should be added to twice the rational number
7
3
to get
3
7 ?
is 2 × ⎛ − 7 ⎜ ⎝ 3 ⎞ ⎟ ⎠ − 14 = 3 3 7 3 7 3 14 + 7 3 (3 3) ×+ (14 × 7)
14
3
=
=
x =
=
7
3
Solution: Twice the rational number
number gives
. Suppose x added to this
3
7
; i.e.,
x +
⎜ ⎝
− 14 ⎞
3 ⎠
x
or
or
(transposing 14
3
=
9
+ 98
107
=
to RHS)
21 21
21
.
LINEAR EQUATIONS IN ONE VARIABLE
25
Thus 107
21
2
×
7
to give
3
3
7 .
Example 6: The perimeter of a rectangle is 13 cm and its width is
length.
Solution: Assume the length of the rectangle to be x cm. The perimeter of the rectangle = 2 × (length + width)
= 2 × (x +
2
11
x +
⎟ ⎞
= 2 ⎛ ⎜ ⎝
4
3
4 )
The perimeter is given to be 13 cm. Therefore,
2
⎜ ⎝
x +
11
4
= 13
3
2 4 cm. Find its
or
or
The length of the rectangle is
x +
3
3 4 cm.
11
4
=
x =
=
13
2
13
11
(dividing both sides by 2)
2
26
4
11
15
3
−== 3
444
4
Example 7: The present age of Sahil’s mother is three times the present age of Sahil. After 5 years their ages will add to 66 years. Find their present ages.
Solution: Let Sahil’s present age be x years.
We could also choose Sahil’s age
5 years later to be x and proceed.
Why don’t you try it that way?
Sahil
Mother
Sum
Present age
x 3x
Age 5 years later
x + 5
3x + 5
4x + 10
It is given that this sum is 66 years.
Therefore,
4x + 10 = 66
This equation determines Sahil’s present age which is x years. To solve the equation,
26
MATHEMATICS
we transpose 10 to RHS,
or
4x = 66 – 10 4x = 56
56
or
Thus, Sahil’s present age is 14 years and his mother’s age is 42 years. (You may easily check that 5 years from now the sum of their ages will be 66 years.)
Example 8: Bansi has 3 times as many two-rupee coins as he has five-rupee coins. If he has in all a sum of Rs 77, how many coins of each denomination does he have?
Solution: Let the number of five-rupee coins that Bansi has be x. Then the number of two-rupee coins he has is 3 times x or 3x. The amount Bansi has:
x =
4
= 14
(solution)
Rs 2
Rs 5
(i) from 5 rupee coins, Rs 5 × x = Rs 5x (ii) from 2 rupee coins, Rs 2 × 3x = Rs 6x
Hence the total money he has = Rs 11x But this is given to be Rs 77; therefore,
11x = 77
77
or x = 11 = 7 Thus, number of five-rupee coins = x = 7 and number of two-rupee coins = 3x = 21
(solution)
(You can check that the total money with Bansi is Rs 77.)
Example 9: The sum of three consecutive multiples of 11 is 363. Find these multiples.
Solution: If x is a multiple of 11, the next multiple is x + 11. The next to this is
x + 11 + 11 or x + 22. So we can take three consecutive multiples of 11 as x, x + 11 and x + 22.
It is given that the sum of these consecutive multiples of 11 is 363. This will give the followingequation:
or x + (x + 11) + (x + 22) = 363 x + x + 11 + x + 22 = 363 or 3x + 33 = 363 or 3x = 363 – 33 or 3x = 330
Alternatively, we may think of the multiple of 11 immediately before x. This is (x – 11). Therefore, we may take three consecutive multiples of 11 as x – 11, x, x + 11.
In this case we arrive at the equation
or
(x – 11) + x + (x + 11) = 363 3x = 363
LINEAR EQUATIONS IN ONE VARIABLE
27
363
3
x = 121, x – 11 = 110, x + 11 = 132
Hence, the three consecutive multiples are 110, 121, 132.
or
x =
= 121.
Therefore,
330
3
or
= 110 Hence, the three consecutive multiples
are 110, 121, 132 (answer). We can see that we can adopt different ways to find a solution for the problem. Example 10: The difference between two whole numbers is 66. The ratio of the two numbers is 2 : 5. What are the two numbers?
Solution: Since the ratio of the two numbers is 2 : 5, we may take one number to be 2x and the other to be 5x. (Note that 2x : 5x is same as 2 : 5.) The difference between the two numbers is (5x – 2x). It is given that the difference is 66. Therefore,
x =
5x – 2x = 66 3x = 66 x = 22
or
or Since the numbers are 2x and 5x, they are 2 × 22 or 44 and 5 × 22 or 110, respectively. The difference between the two numbers is 110 – 44 = 66 as desired. Example 11: Deveshi has a total of Rs 590 as currency notes in the denominations of Rs 50, Rs 20 and Rs 10. The ratio of the number of Rs 50 notes and Rs 20 notes is 3:5. If she has a total of 25 notes, how many notes of each denomination she has?
Solution: Let the number of Rs 50 notes and Rs 20 notes be 3x and 5x, respectively. But she has 25 notes in total. Therefore, the number of Rs 10 notes = 25 – (3x + 5x) = 25 – 8x The amount she has from Rs 50 notes : 3x × 50 = Rs 150x from Rs 20 notes : 5x × 20 = Rs 100x from Rs 10 notes : (25 – 8x) × 10 = Rs (250 – 80x) Hence the total money she has =150x + 100x + (250 – 80x) = Rs (170x + 250) But she has Rs 590. Therefore, 170x + 250 = 590
or
170x = 590 – 250 = 340
x =
340
170
= 2
or
The number of Rs 50 notes she has = 3x = 3 × 2 = 6 The number of Rs 20 notes she has = 5x = 5 × 2 = 10 The number of Rs 10 notes she has = 25 – 8x = 25 – (8 × 2) = 25 – 16 = 9
28
MATHEMATICS
EXERCISE 2.2
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
If you subtract 1 2 from a number and multiply the result by 1 2 , you get
the number?
The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
8 . What is
1
The base of an isosceles triangle is 4 cm . The perimeter of the triangle is
.
What is the length of either of the remaining equal sides?
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Three consecutive integers add up to 51. What are these integers?
The sum of three consecutive multiples of 8 is 888. Find the multiples.
Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?
The number of boys and girls in a class are in the ratio 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?
Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
A rational number is such that when you multiply it by 5 2 and add 2
to the product,
4
2
15
3
cm
3
7
you get
12
. What is the number?
14. Lakshmi is a cashier in a bank. She has currency notes of denominations Rs 100, Rs 50 and Rs 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is Rs 4,00,000. How many notes of each denomination does she have?
15. I have a total of Rs 300 in coins of denomination Re 1, Rs 2 and Rs 5. The number of Rs 2 coins is 3 times the number of Rs 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
16. The organisers of an essay competition decide that a winner in the competition gets a prize of Rs 100 and a participant who does not win gets a prize of Rs 25. The total prize money distributed is Rs 3,000. Find the number of winners, if the total number of participants is 63.
LINEAR EQUATIONS IN ONE VARIABLE
29
2.4 Solving Equations having the Variable on both Sides
An equation is the equality of the values of two expressions. In the equation 2x – 3 = 7, the two expressions are 2x – 3 and 7. In most examples that we have come across so far, the RHS is just a number. But this need not always be so; both sides could have expressions with variables. For example, the equation 2x – 3 = x + 2 has expressions with a variable on both sides; the expression on the LHS is (2x – 3) and the expression on the RHS is (x + 2).
We now discuss how to solve such equations which have expressions with the variable on both sides.
Example 12: Solve 2x – 3 = x + 2
Solution: We have
2x = x + 2 + 3
or 2x = x + 5 or 2x – x = x + 5 – x (subtracting x from both sides) or x = 5 (solution)
Here we subtracted from both sides of the equation, not a number (constant), but a term involving the variable. We can do this as variables are also numbers. Also, note that subtracting x from both sides amounts to transposing x to LHS.
7
Example 13: Solve 5x +
2
Solution: Multiply both sides of the equation by 2. We get
3
=
x 14
2
2 ×
5 x
+
⎜ ⎛ ⎝
(2 × 5x) +
2 ×
⎜ ⎝
⎟ ⎞
7
2
⎟ ⎞
7
2
3
=
2 ×
x
14 ⎟ ⎠ ⎞
⎜ ⎝
2
= ⎝ ⎜
3
2
×
x
−×
(2
⎟ ⎠
2
or 10x + 7 = 3x – 28 or 10x – 3x + 7 = – 28 or 7x + 7 = – 28 or 7x = – 28 – 7 or 7x = – 35 − 35 or x = 7 or x = – 5
14)
(transposing 3x to LHS)
(solution)
30
MATHEMATICS
EXERCISE 2.3
Solve the following equations and check your results.
1.
3x = 2x + 18
2.
5t – 3 = 3t – 5
4.
4z + 3 = 6 + 2z
5.
2x – 1 = 14 – x
4
2 3 x
7
7.
x =
(x + 10)
8.
+ 1 =
x + 3
5
15
8
10.
3m = 5 m –
5
3.
6.
9.
5x + 9 = 5 + 3x
8x + 4 = 3 (x – 1) + 7
2y +
5
3
=
26
3
y
2.5 Some More Applications
Example 14: The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. What can be the original number?
Solution: Take, for example, a two-digit number, say, 56. It can be written as 56 = (10 × 5) + 6.
If the digits in 56 are interchanged, we get 65, which can be written as (10 × 6 ) + 5.
Let us take the two digit number such that the digit in the units place is b. The digit in the tens place differs from b by 3. Let us take it as b + 3. So the two-digit number is 10 (b + 3) + b = 10b + 30 + b = 11b + 30.
With interchange of digits, the resulting two-digit number will be
Could we take the tens place digit to be (b – 3)? Try it and see what solution you get.
10b + (b + 3) = 11b + 3
If we add these two two-digit numbers, their sum is
(11b + 30) + (11b + 3)
= 11b + 11b + 30 + 3 = 22b + 33
It is given that the sum is 143. Therefore, 22b + 33 = 143
or
or
22b = 143 – 33 22b = 110
110
b =
22
b =5
Remember, this is the solution
when we choose the tens digits to
be 3 more than the unit’s digits.
What happens if we take the tens
digit to be (b – 3)?
The statement of the
example is valid for both 58
and 85 and both are correct
or
or
The units digit is 5 and therefore the tens digit is 5 + 3 which is 8. The number is 85.
Check: On interchange of digits the number we get is
58. The sum of 85 and 58 is 143 as given.
LINEAR EQUATIONS IN ONE VARIABLE
31
Example 15: Arjun is twice as old as Shriya. Five years ago his age was three times Shriya’s age. Find their present ages.
Solution: Let us take Shriya’s present age to be x years. ThenArjun’s present age would be 2x years. Shriya’s age five years ago was (x – 5) years. Arjun’s age five years ago was (2x – 5) years.
It is given that Arjun’s age five years ago was three times Shriya’s age.
Thus, 2x – 5 = 3(x – 5) or 2x – 5 = 3x – 15 or 15 – 5 = 3x – 2x 10 = x or So, Shriya’s present age = x = 10 years. Therefore, Arjun’s present age = 2x = 2 × 10 = 20 years.
EXERCISE 2.4
1. Amina thinks of a number and subtracts 5
2
from it. She multiplies the result by 8. The
result now obtained is 3 times the same number she thought of. What is the number?
2. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
3. Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?
4. One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?
5. Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?
6. There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4.At the rate Rs100 per metre it will cost the village panchayat Rs 75000 to fence the plot. What are the dimensions of the plot?
7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs 50 per metre and trouser material that costs him Rs 90 per metre.
32
MATHEMATICS
8.
9.
10.
For every 2 meters of the trouser material he buys 3 metres of the shirt material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs 36,660. How much trouser material did he buy?
Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.
Agrandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.
2.6 Reducing Equations to Simpler Form
x +
Example 16: Solve 61
3
Solution: Multiplying both sides of the equation by 6,
x 3
6
+
1
=
6 (6
x +
1)
3
+
6
×
1
=
6(
x
3)
6
Why 6? Because it is the
smallest multiple (or LCM)
of the given denominators.
or or or or or or or or Check: LHS =
2 (6x + 1) + 6 = x – 3 12x + 2 + 6 = x – 3 12x + 8 = x – 3 12x x + 8 = – 3 11x + 8 = – 3 11x = –3 – 8 11x = –11 x = – 1
(opening the brackets )
(required solution)
6( − 1) + 1 6 −+ 1 1 += 3 3 RHS = ( 1) −− 3 =
6
5
−+
53
2
+
1
=
+= 3
=
33
3
3
4
2
=
63
LHS = RHS.
(as required)
Example 17: Solve 5x – 2 (2x – 7) = 2 (3x – 1) +
7
2
Solution: Let us open the brackets, LHS = 5x – 4x + 14
= x + 14
7 47 3 = 6 x −+= 6 x + 2 22 2 3 2 3 14 = 6x – x + 2 3 14 = 5x + 2 3 14 – 2 =5x 28 − 3 2 =5x 25 2 =5x 25 ×= 1 5 × 5 = 5 x = 2 5 25 × 2 5 2 .
RHS = 6x – 2 +
The equation is x + 14 = 6x +
or
or
or
or
or
or
Therefore, required solution is x =
LINEAR EQUATIONS IN ONE VARIABLE
(transposing
3
2 )
33
Did you observe how we
simplified the form of the given
multiply both sides of the
equation by the LCM of the
denominators of the terms in the
expressions of the equation.
Check: LHS = 5 5 ×− 2 2 ⎛ ⎝ ⎜ 5 2 × 2–7 ⎞ ⎟ ⎠ 25 − − = 25 25 −−= + 4 25 + 8 = 33 2 ⎛ = 5 2 2(5 7) × 3–1 ⎟ ⎞ ⎠ += 7 2 2( 2) 22 ⎛⎞ 15 2 += 7 2 × = 13 + 2 7 2 Note, in this example we brought the equation to a simpler form by opening brackets and combining like terms on both sides of the equation. RHS = ⎜ ⎝ 26 2 + 7 33 2 ⎜⎟ ⎝⎠ – 22 2 2 2 = 2 = 2 = LHS. (as required) EXERCISE 2.5 Solve the following linear equations. 1. x −=+ 1 x 1 2. nnn −+= 3 5 21 3. 7 x +− 8 x = 17 − 5 x 2534 24 6 362
34
MATHEMATICS
4.
x
5
x
3
=
3
5
5.
3
t
22
t
+
32 =− t
4
33
6.
m − 1 =− 1 m − 2 m − 2 3
Simplify and solve the following linear equations.
7 3(t – 3) = 5(2t + 1) 8. 15(y – 4) –2(y – 9) + 5(y + 6) = 0 9 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17 10 0.25(4f – 3) = 0.05(10f – 9)
2.7 Equations Reducible to the Linear Form
Example 18: Solve
x + 1
3
=
38
2
x +
Solution: Observe that the equation is not a linear equation, since the expression on its LHS is not linear. But we can put it into the form of a linear equation. We multiply both sides of the equation by (2x + 3),
x + 1
×
(2
x
+
3)
=
2
x +
3 ⎠
3
8
×
(2
x +
3)
Notice that (2x + 3) gets cancelled on the LHS We have then,
Note that
2x + 3 ≠ 0 (Why?)
x + 1 =
3 (2
x +
3)
8
We have now a linear equation which we know how to solve. Multiplying both sides by 8
8 (x + 1) = 3 (2x + 3)
or 8x + 8 = 6x + 9 8x =6x + 9 – 8 8x =6x + 1 8x – 6x =1 2x =1 or or or or 1 or x = 2
This step can be
directly obtained by
‘cross-multiplication’
The solution is x = 1 2 .
1 12
+
3
Check : Numerator of LHS =
2 + 1 =
=
2
2
1
Denominator of LHS = 2x + 3 =
2 ×
+ 3 = 1 + 3 = 4
2
LHS = numerator ÷ denominator =
3
2 ÷
4
=
313
×
=
248
LINEAR EQUATIONS IN ONE VARIABLE
35
LHS = RHS.
Example 19: Present ages of Anu and Raj are in the ratio 4:5. Eight years from now the ratio of their ages will be 5:6. Find their present ages.
Solution: Let the present ages of Anu and Raj be 4x years and 5x years respectively. After eight years. Anu’s age = (4x + 8) years; After eight years, Raj’s age = (5x + 8) years.
Therefore, the ratio of their ages after eight years = 5 4
x
+
8
x
+
8
This is given to be 5 : 6
Therefore,
Cross-multiplication gives
4
x
+
8
5
x
+
8
=
5
6
6 (4x + 8) = 5 (5x + 8)
or 24x + 48 = 25x + 40 or 24x + 48 – 40 = 25x or 24x + 8 = 25x or 8 = 25x – 24x or 8 = x Therefore, Anu’s present age = 4x = 4 × 8 = 32 years Raj’s present age = 5x = 5 × 8 = 40 years
EXERCISE 2.6
Solve the following equations.
1.
4.
8
x
3
3 x
= 2
34
y +
2
=
2–6
y 5
2.
5.
9 x
7
6 x
= 15
y +
74
4
=
y + 2
3
3.
z 4
=
z + 15
9
6.
7.
The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.
The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number
3
obtained is
2 . Find the rational number.
36
MATHEMATICS
WHAT HAVE WE DISCUSSED?
1. An algebraic equation is an equality involving variables. It says that the value of the expression on
one side of the equality sign is equal to the value of the expression on the other side.
2. The equations we study in Classes VI, VII and VIII are linear equations in one variable. In such
equations, the expressions which form the equation contain only one variable. Further, the equations
are linear, i.e., the highest power of the variable appearing in the equation is 1.
3. Alinear equation may have for its solution any rational number.
4. An equation may have linear expressions on both sides. Equations that we studied in Classes VI
and VII had just a number on one side of the equation.
5. Just as numbers, variables can, also, be transposed from one side of the equation to the other.
6. Occasionally, the expressions forming equations have to be simplified before we can solve them
by usual methods. Some equations may not even be linear to begin with, but they can be brought
to a linear form by multiplying both sides of the equation by a suitable expression.
7. The utility of linear equations is in their diverse applications; different problems on numbers, ages,
perimeters, combination of currency notes, and so on can be solved using linear equations.
|
# How to Find the Perimeter of a Shape
Perimeter is a measure of the area or distance around a two-dimensional shape. On a rectangle, for instance, the perimeter is the total length of the rectangle's outline, including the two widthwise borders and the two lengthwise borders. To determine the perimeter of a shape, therefore, you add together all the dimensions that make up the shape's outer edge. Being able to find the perimeter of a shape has many applications in the real world. Say, for example, that you wanted to build a fence around your backyard. In order to purchase materials, you need to know how much fencing you'll need, and to determine that you have to figure out the perimeter of the area you want to fence in.
Part 1
Part 1 of 2:
### Finding the Perimeter of Most Shapes
1. 1
Determine the length of each side. Since perimeter is just a measure of the outline of a two-dimensional figure, you don't usually need a specific formula to find perimeter (though there are equations for specific shapes to make it easier). However, you do need to know the length of all the sides of the shape.[1]
• For example, a pentagon has five sides, and you need to know the length of each one to determine perimeter.
• Even with an irregular polygon that has 20 sides, you can still find the perimeter as long as you know the length of each side.
2. 2
Add the length of all the sides together. To find the perimeter of non-circular objects, find the sum of all the side lengths to determine the distance around the shape.[2]
• Say the irregular pentagon has the following lengths: A = 4, B =2, C = 3, D = 3, and E = 2
• Add 4 + 2 + 3 + 3 + 2 = 14, where P (perimeter) = 14
3. 3
Deal with variables. You can still find perimeter when you are working with variables. For example, say you have a triangle with the side lengths 14a, 11b, and 7a:[3]
• Find the sum of all the sides: P = 14a + 11b + 7a
• Combine the like terms: P = (14a + 7a) + 11b
• P = 21a + 11b
4. 4
Pay attention to units of measurement. In a real world application, finding the perimeter of an object won't do you much good if you don't know what unit of measurement you're working with (such as feet, miles, or meters). With the pentagon, if each side was measured in centimeters, then you know that P = 14 cm.
Part 2
Part 2 of 2:
### Learning Perimeter Formulas
1. 1
Find the perimeter of a circle. Some regular shapes have formulas that make it faster for you to find the figure's perimeter. But there are other shapes, like circles, that require a formula to find perimeter. The perimeter of a circle is called the circumference. To find the circumference of a circle, use the equation C (circumference) = 2πr.[4]
• To start, find the radius of the circle, which is the length of a line segment running from the center of the circle to the perimeter.
• For simpler equations, use the truncated version π = 3.14
• For a circle with a radius of 4cm: C = 2 x 3.14 x 4 = 25.12cm
2. 2
Find the perimeter of a triangle. Use the equation P = a + b + c for a triangle. For instance, if a triangle has the dimensions a = 20cm, b = 11cm, and c = 9cm, then P = 20 + 11 + 9 = 40cm.
3. 3
Find the perimeter of a square. Because a square has four sides of the same length, you can use the simple equation P = 4x, where x equals the length of one side.
• On a square where x = 3cm, then P = 4 x 3 = 12cm
4. 4
Find the perimeter of a rectangle. Since the lengthwise sides are the same and the widthwise sides are the same on a rectangle, you can use the equation P = 2l + 2w, where l is the length of one side and w is the width of one side. For a rectangle where l = 8cm and w = 5cm:
• P = (2 x 8) + (2 x 5)
• P = 16 + 10
• P = 26cm
• The equation P = 2(l + h) will also give you the same result: 2(8 + 5) = 2(13) = 26cm[5]
5. 5
Find the perimeter of other quadrilaterals. A quadrilateral refers to any two-dimensional shape with four closed, straight sides. This includes rectangles, squares, trapezoids, parallelograms, kites, and rhombuses.[6] There are three equations you can use for a quadrilateral, depending on the sides:
• For a quadrilateral with no equal sides, like an irregular trapezoid, use the equation P = a + b + c + d
• For a quadrilateral with four equal sides, use the same equation as a square: P = 4x.
• For quadrilaterals where the lengthwise sides are the same and the widthwise sides are the same (like a rectangle), use the equations P = 2a + 2b or P = 2(a + b)
## Community Q&A
Search
• Question
How do I find the perimeter of an irregular shape, not regular?
Donagan
The only way to calculate the perimeter of an irregular shape is to add together the lengths of all the sides. If you're not given those lengths, you may be given other information that could help you find those lengths.
• Question
What is the difference between wide and long?
Donagan
In the case of a rectangular shape, its length is normally considered to be the longer of its two dimensions, while its width is the shorter.
• Question
Square tiles 30 cm by 30 cm are used to cover a floor. How many tiles are needed to cover a floor measuring 4.5 m long and 3.9 m wide?
Donagan
Convert the floor measurements to centimeters: 450 cm x 390 cm. Divide each of those measurements by 30. Then multiply the two quotients together. That's how many tiles you'll need. In this case you shouldn't have any partial tiles left over.
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## Video.By using this service, some information may be shared with YouTube.
Co-authored by:
Co-authors: 14
Updated: March 25, 2023
Views: 110,361
Categories: Geometry
Article SummaryX
To find the perimeter of a shape, just add the lengths of all of the sides together. If you're trying to find the perimeter of a circle, you'll need to use the formula c = 2πr, where c is the circumference and perimeter, and r is the radius of the circle. If you want to learn formulas for how to find the perimeter of specific shapes, keep reading the article!
Thanks to all authors for creating a page that has been read 110,361 times.
|
# Understanding Variables. Algebra/Standard GLE 0506.3.2 Develop and apply the concept of variable. √ 0506.3.1 Evaluate an expression by substituting non-negative.
## Presentation on theme: "Understanding Variables. Algebra/Standard GLE 0506.3.2 Develop and apply the concept of variable. √ 0506.3.1 Evaluate an expression by substituting non-negative."— Presentation transcript:
Understanding Variables
Algebra/Standard GLE 0506.3.2 Develop and apply the concept of variable. √ 0506.3.1 Evaluate an expression by substituting non-negative rational number values for letter variables in the expression. √ 0506.3.2 Use variables appropriately to represent numbers whose values are not yet known.
Understanding Variables What is so important about variables?
Understanding Variables Imagine having two cars that together cost \$2.00 and a car and a dog that together cost \$5.00. What is the price for a car? What is the price for the dog? Explain how you found your answer. \$2 \$5
Vocabulary Car + Car = \$2.00; Car + Dog = \$5.00. Instead of using numbers, students can used representations to find out the value. Give a definition of a variable. A variable is a letter or symbol that can stand for any number
Variable Machine On the three-centimeter-wide strip of lined notebook paper, write the letters of the alphabet in order down the left side of the paper. Down the right side of the five-centimeter strip of notebook paper, write the numbers from 0 to 25. Then attach the ends of the number strip together with a piece of tape; wrap the letter strip around the number wheel, matching the letters to the corresponding numbers: A to 0; B to 1; C to 2 and so on.... and tape the ends of the letter strip together, as shown below:
Explore Have the students find the value of their first names, using their variable machines; for example, the value of Amy's name is 36: A = 0, M = 12, Y = 24 0 + 12 + 24 = 36 Have the students find the value of their last names, and ask the following questions: Which name has the higher value - your first name or last name? What is the difference in the values of your first and last names? Students should explore the values of various words. For example, ask students to find words: whose values are equal to 25, 36, or 100.
Cracking the Code Use your Variable Machine to answer the questions which follow. 1. What is the value of your last name? _______ What is the value of your first name? _______ What is the difference of these two values? ______ 2. What is the value of each of these words? a. variable ______ b. machine ______ c. algebra ______ d. mathematics ______ 3. Find three different words whose values are each equal to 25. Record the words below.
Extension Assign the values on the Variable Machine as decimals: A= 1‚; B=2‚; C=4‚ If A = \$.01 and Z = \$. 26 what words will equal exactly \$1.00?
Understanding Variables Example 1:Dexter had three baseballs. After practice, he found several more baseballs. Write an expression using a variable. What the does the variable stands for? The variable stands for the number of baseballs Dexter found after practice. Solution: 3 + B
Understanding Variables Example 2: On her birthday, Kristin brought 30 cupcakes to school. She gave a cupcake to each student in her class. Write an expression using a variable. What does the variable stands for? Solution: 30- c
Understanding Variables Example 3: If p = 88, what is p – 30? Rewrite the expression substituting the value for the variable in the expression, then solve.
Understanding Variables Example 3: Evaluate x ½ ½ using the order of operations. x = 1/3 Step 1: 1/3 + ½ ½ ½ ½ = ¼ Multiplication Step 2: 1/3 1/4 = 1/12 Addition Solution: 1/3 + ½ ½ = 1/12
Your Turn!!! Let x = 12 3x x + 79 56 - x x / 6
Check your Answers x = 36 x = 91 x = 44 x = 2
Download ppt "Understanding Variables. Algebra/Standard GLE 0506.3.2 Develop and apply the concept of variable. √ 0506.3.1 Evaluate an expression by substituting non-negative."
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# Circumcenter of a Triangle
• Last Updated : 26 May, 2022
The point of intersection, or the meeting point of the perpendicular bisectors of a triangle, is the circumcenter of a triangle and is usually represented by the letter “O”. In simple words, the circumcenter of a triangle is the point of concurrency of the bisectors of the sides of a triangle. The line segment that bisects another line segment at a right angle is called a perpendicular bisector.
### What is a Circumcenter?
A circumcenter is the center of a circumcircle, whereas a circumcircle is a circle that passes through all the vertices of a polygon, i.e., a circle in which a polygon is enclosed. Every circle is cyclic, which means every triangle can circumscribe a circle. Hence, any type of triangle will have a circumcenter.
For the construction of the circumcenter of any triangle, the perpendicular bisectors of any two sides of a triangle need to be drawn.
### Properties of a circumcenter
Let us consider a triangle ABC. Now, the properties of the circumcenter of a triangle are:
1. The circumcenter of a triangle is equidistant from all the vertices, i.e., OA = OB = OC.
2. All the new triangles that are formed by joining the circumcenter of a triangle to its vertices are isosceles, i.e., ∆ AOB, ∆ BOC, and ∆ COA are isosceles triangles.
3. In a triangle, if ∠A is acute or when O and A are on the same side of BC, then ∠BOC = 2 ∠A.
4. In a triangle, if ∠A is obtuse or when O and A are on different sides of BC, then ∠BOC = 2(180° – ∠A).
5. For an acute-angled triangle, the circumcenter lies inside the triangle.
6. In the case of an obtuse-angled triangle, the circumcenter lies outside the triangle.
7. For a right-angled triangle, the circumcenter lies on the hypotenuse of the triangle.
### Construction of the circumcenter of a triangle
To construct the circumcenter of any triangle, we need to draw the perpendicular bisectors of any two sides of the triangle. The following are the steps to construct the circumcenter of a triangle:
Step 1: For the given triangle, draw the perpendicular bisectors of any two sides using a compass.
Step 2: With the help of a ruler, extend the perpendicular bisectors till they intersect at a point.
Step 3: Now, mark the point of intersection, which will be the circumcenter of the given triangle.
We can get the circumcenter accurately by drawing the perpendicular bisector of the third side of the triangle.
### How to find the Circumcenter of a Triangle?
We can calculate the circumcenter of a triangle using different methods.
Method 1
Let us consider a triangle ABC whose vertices are A (x1, y1), B(x2, y2), and C (x3, y3), and O(x, y) is its circumcenter.
Step 1: Calculate the coordinates of the mid-points of the sides AB, BC, and AC using the mid-point formula.
M(xm ,ym) = [(x1 + x2)/2, (y1 + y2)/2]
Step 2: Calculate the slopes of the sides AB, BC, and AC. Let the slope of a side be m, then the slope of its perpendicular bisector is “-1/m”.
Step 3: Now, using the coordinates of the midpoint (xm, ym) and the slope of the perpendicular bisector (-1/m), find out the equation of the perpendicular bisector using the point-slope form.
(y – ym) = -1/m(x – xm)
Step-4: Similarly, find out the equations of the other bisector lines too.
Step-5: Solve any of the two equations and find their intersection point.
The obtained intersection point is the circumcenter of the given triangle.
Method 2
Let us consider a triangle ABC whose vertices are A (x1, y1), B(x2, y2), and C (x3, y3), and O (x, y) is its circumcenter.
We know that the circumcenter of a triangle is equidistant from all the vertices, i.e., OA = OB = OC = circumradius.
Let OA = D1, OB = D2, and OC = D3.
Step-1: Using the distance formula between two coordinates, find the values of D1, i.e.,
(D1)2 = (x – x1)2 + (y – y1)2
Similarly,
(D2)2 = (x – x2)2 + (y – y2)2
(D3)2 = (x – x3)2 + (y – y3)2
Step-2: Now, by equating D1 = D2 = D3 we will get linear equations. By solving these linear equations, we can get the coordinates of the circumcenter O (x, y).
Method-3
Let us consider a triangle ABC whose vertices are A (x1, y1), B (x2, y2), and C (x3, y3), and ∠A, ∠B, and ∠C are their respective angles. O (x, y) is the circumcenter of the triangle.
We can determine the circumcenter easily by using the formula below.
(
### Sample Problems
Problem 1: Determine the circumcenter of a triangle with vertices A (1,3), B (0,4), and C (-2,5).
Solution:
Given data,
The vertices of a triangle are A (x1, y1) = (1,3), B (x2 , y2) = (0,4), and C (x3, y3) = (-2,5).
Let “O” be the circumcenter of the triangle ABC and (x, y) be its coordinates.
Let D1 be the distance from the circumcenter to vertex A, i.e., OA = D1.
Let D2 be the distance from the circumcenter to vertex B, i.e., OB = D2.
Let D3 be the distance from the circumcenter to vertex C, i.e., OC = D3.
By using the formula for the distance between two points we get
(D1)2 = (x – x1)2 + (y – y1)2 = (x – 1)2 + ( y – 3)2
(D2)2 = (x – x2)2 + (y – y2)2 = ( x- 0)2 + (y – 4)2
(D3)2 = (x – x3)2 + (y – y3)2 = (x + 2)2 + (y – 5)2
We know that the distances from all the vertices to the circumcenter (O) are equal, i.e.,
OA = OB = OC = circumradius
⇒ D1 = D2 = D3
Now, take D1 = D2
(x -1)2 + (y – 3)2 = (x – 0)2 + (y -4)2
⇒ x2 – 2x + 1 + y2 – 6y + 9 = x2 + y2 – 8y + 16
⇒ 2x – 2y = -6 ……(1)
Now, take D2 = D3
( x- 0)2 + (y – 4)2 = (x + 2)2 + (y – 5)2
⇒ x2 + y2 – 8y + 16 = x2 + 4x + 4 + y2 – 10y + 25
⇒ 4x – 2y = -13 …..(2)
Now, solve both equations (1) and (2)
2x – 2y = – 6
4x – 2y = -13
(-) (+) (+)
– 2x = 7
⇒ x = -7/2 = – 3.5
Now, substitute the value of x in equation (1)
2 (-3.5) – 2y = -6
⇒ -7 – 2y = -6
⇒ 2y = -1 ⇒ y = -1/2 = -0.5
Hence, the circumcenter of the triangle ABC is (-3.5, -0.5)
Problem 2: Determine the circumcenter of the triangle with vertices A (3, -6), B (1, 4), and C (5, 2).
Solution:
Given data,
The vertices of a triangle are A (x1, y1) = (3, -6), B (x2 , y2) = (1, 4), and C (x3, y3) = (5, 2).
Let “O” be the circumcenter of the triangle ABC and (x, y) be its coordinates.
For finding the circumcenter of a triangle, we can calculate the intersection point of any two perpendicular bisectors.
Now, the mid point of the side AB = [(3 + 1)/2, (-6 + 4)/2] = (2, -1)
Slope of AB = (y2 – y1)/(x2 – x1) = (4 + 6)/(1 – 3) = -5
We know that the product of the slopes of two perpendicular lines = -1.
So, the slope of the perpendicular bisector of the side AB = 1/5
Now, the equation of the perpendicular bisector of AB with slope = 1/5 and the coordinates (2,-1) is
(y – (-1)) = (1/5) (x – 2) [From point-slope form]
⇒ 5(y +1) = x – 2
⇒ x – 5y = 7 ……(1)
The mid point of the side BC is [(1 + 5)/2, (4 + 2)/2] = (3,3)
Slope of BC = (y3 – y2)/(x3 – x2) = (2 – 4)/(5 – 1) = -1/2
Now, the slope of the perpendicular bisector of the side BC = 2
The equation of the perpendicular bisector of BC with slope = 2 and the coordinates (3,3) is
(y – 3) = 2(x – 3) [From point-slope form]
⇒ y – 3 = 2x – 6
⇒ 2x – y = 3 …….(2)
Now, multiply equation (1) with “2” on both sides and subtract equation (2) from the obtained equation.
2x – 10y = 14
2x – y = 3
(-) (+) (-)
-9y = 11
⇒ y = -11/9
Now substitute the value of y in equation (2)
2x + 11/9 = 3
⇒ x = 8/9
Hence, the circumcenter of the triangle ABC is (8/9, -11/9).
Problem 3: Find the circumcenter of the ∆ ABC with vertices A (1, 3), B (3, 7), and C (5, 9), and the measures of the respective angles are 45°, 45°, and 90°.
Solution:
Given data,
The vertices of a triangle are A (x1, y1) = (0, 3), B (x2, y2) = (3, 7), and C (x3, y3) = (5, 9).
The measures of the angles are ∠A = 45°, ∠B = 45°, and ∠C = 90°.
We know the formula for the circumcenter (O) of a triangle when its vertices and their respective angles are given, i.e.,
Circumcenter (O) = (, )
O = ()
= ()
= ()
Circumcenter (O) = (1, 3)
Hence, the circumcenter of the triangle ABC is (1,3)
Problem 4: Find the circumcenter of a triangle whose vertices are A (0, 6), B (-8, 4), and C (2, -4).
Solution:
Given data,
The vertices of a triangle are A (x1, y1) = (0, 6), B (x2 , y2) = (-8, 4), and C (x3, y3) = (2, -4).
Let “O” be the circumcenter of the triangle ABC and (x, y) be its coordinates.
For finding the circumcenter of a triangle, we can calculate the intersection point of any two perpendicular bisectors.
Now, the mid point of the side AB = [(0 – 8)/2, (6 + 4)/2] = (-4, 5)
Slope of AB = (y2 – y1)/(x2 – x1) = (4 – 6)/(-8 – 0) = 1/4
We know that the product of the slopes of two perpendicular lines = -1.
So, the slope of the perpendicular bisector of the side AB = -4
Now, the equation of the perpendicular bisector of AB with slope = -4 and the coordinates (-4, 5) is
(y – 5) = -4(x+ 4) [From point-slope form]
⇒ y – 5 = -4x – 16
⇒ 4x + y = -11 ——— (1)
The mid point of the side BC is [(-8 +2)/2, (4 – 4)/2] = (-3, 0)
Slope of BC = (y3 – y2)/(x3 – x2) = (-4 – 4)/(2 + 8) = -4/5
Now, the slope of the perpendicular bisector of the side BC = 5/4
The equation of the perpendicular bisector of BC with slope = 5/4 and the coordinates (-3, 0) is
(y – 0) = 5/4(x + 3) [From point-slope form]
⇒ 4y = 5x + 15
⇒ 5x – 4y = -15 ——— (2)
Now, multiply the equation (1) with “4” on both sides and add the result to equation (2).
16x + 4y = -44
5x – 4y = -15
——————
21x = -59 ⇒ x = -59/21
Now, substitute x = -59/21 in equation (1)
4(-59/21) + y = -11
⇒ y = -11 + (236/21) ⇒ y = 5/21
Hence, the circumcenter of the triangle ABC is (-59/21, 5/21)
Problem 5: Find the circumcenter of a triangle whose coordinates are A (3, 8), B (6, 2), and C (-4, 7).
Solution:
Given data,
The vertices of a triangle are A (x1, y1) = (3, 8), B (x2 , y2) = (6, 2), and C (x3, y3) = (-4, 7).
Let “O” be the circumcenter of the triangle ABC and (x, y) be its coordinates.
Let D1 be the distance from the circumcenter to vertex A, i.e., OA = D1.
Let D2 be the distance from the circumcenter to vertex B, i.e., OB = D2.
Let D3 be the distance from the circumcenter to vertex C, i.e., OC = D3.
By using the formula for the distance between two points we get
(D1)2 = (x – x1)2 + (y – y1)2 = (x – 3)2 + ( y – 8)2
(D2)2 = (x – x2)2 + (y – y2)2 = ( x- 6)2 + (y – 2)2
(D3)2 = (x – x3)2 + (y – y3)2 = (x + 4)2 + (y – 7)2
We know that the distances from all the vertices to the circumcenter (O) are equal, i.e.,
OA = OB = OC = circumradius
⇒ D1 = D2 = D3
Now, take D1 = D2
(x -3)2 + (y – 8)2 = (x – 6)2 + (y -2)2
⇒ x2 – 6x + 9 + y – 16y + 64 = x2 – 12x + 36 + y2 – 4x + 4
⇒ 6x – 4y = -33 …….(1)
Now, take D2 = D3
( x- 6)2 + (y – 2)2 = (x + 4)2 + (y – 7)2
⇒ x2 – 12x + 36 + y2 – 4x + 4 = x2 + 8x + 16 + y2 – 14y + 49
⇒ 20x – 10y = -25
⇒ 4x – 2y = -5 …..(2)
Now, multiply equation (2) with “2” on both sides and subtract the result from the equation (1)
6x – 4y = -33
8x – 4y = -10
(-) (+) (+)
-2x = -23 ⇒ x = 23/2 = 11.5
Now, substitute the value of x = 23/2 in equation (2)
4(11.5) – 2y = -5
⇒ 46 – 2y = -5
⇒ 2y = 51
⇒ y = 51/2 = 25.5
Hence, the circumcenter of the triangle ABC is (11.5, 23.5)
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# NCERT Solutions class 12 Maths Exercise 7.9 (Ex 7.9) Chapter 7 Integrals
## NCERT Solutions for Class 12 Maths Exercise 7.9 hapter 7 Integrals – FREE PDF Download
NCERT Solutions for integrals Class 12 exercises 7.9 are accessible in PDF format which is available at CoolGyan’s online learning portal. The solutions to this particular exercise from chapter 7 Integrals have been prepared by our experts who have years of experience in this field. Students can find, well-explained solutions to maths problems at CoolGyan’s online learning portal. All the solutions are provided in a way that will help you in understanding the concept effectively. You can download them in PDF format for free.
# NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.9) Exercise 7.9
Evaluate the definite integrals in Exercises 1 to 11.
1.
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=2 Ans.
2.
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3.
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4.
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5.
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= 0 Ans.
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7.
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8.
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9.
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10.
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11.
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Evaluate the definite integrals in Exercises 12 to 20.
12.
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13.
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14.
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=150110x5x2+1dx+3011(5x)2+12dx=15∫0110x5x2+1dx+3∫011(5x)2+12dx
=15(log|5(1)2+1|log|5(0)2+1)+35(tan15–√tan10–√)=15(log|5(1)2+1|−log|5(0)2+1)+35(tan−15−tan−10)
=15(log6log1)+35(tan15–√0)=15(log6−log1)+35(tan−15−0)
=15(log60)+35(tan15–√0)=15(log6−0)+35(tan−15−0)
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15.
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First we evaluate = ……….(i)
Putting
From eq. (i), =
=
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16.
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(On dividing)
where I =
……….(i)
Now, I = ……….(ii)
Let ……….(iii)
Comparing coefficients of A + B = –20 ……….(iv)
Comparing constants 3A + B = –15 ……….(v)
On solving eq. (iv) and (v), we get A = B =
Putting these values in eq. (iii),
I =
Putting this value of I in eq. (i),
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17.
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π41024+π2+2π41024+π2+2 Ans.
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=
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20.
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[Applying Product Rule on first definite integral]
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Choose the correct answer in Exercises 21 and 22.
21. equals:
(A)
(B)
(C)
(D)
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Therefore, option (D) is correct.
22. equals:
(A)
(B)
(C)
(D)
Ans.
Therefore, option (C) is correct.
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Logarithmic Functions
# Logarithmic Functions
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## Logarithmic Functions
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Logarithmic Functions
2. The logarithmic function to the base a, where a > 0 and a 1 is defined: y = logax if and only if x = a y logarithmic form exponential form When you convert an exponential to log form, notice that the exponent in the exponential becomes what the log is equal to. Convert to log form: Convert to exponential form:
3. LOGS = EXPONENTS With this in mind, we can answer questions about the log: This is asking for an exponent. What exponent do you put on the base of 2 to get 16? (2 to the what is 16?) What exponent do you put on the base of 3 to get 1/9? (hint: think negative) What exponent do you put on the base of 4 to get 1? When working with logs, re-write any radicals as rational exponents. What exponent do you put on the base of 3 to get 3 to the 1/2? (hint: think rational)
4. In the last section we learned about the graphs of exponentials. Logs and exponentials are inverse functions of each other so let’s see what we can tell about the graphs of logs based on what we learned about the graphs of exponentials. Recall that for functions and their inverses, x’s and y’s trade places. So anything that was true about x’s or the domain of a function, will be true about y’s or the range of the inverse function and vice versa. Let’s look at the characteristics of the graphs of exponentials then and see what this tells us about the graphs of their inverse functions which are logarithms.
5. Characteristics about the Graph of an Exponential Function a > 1 Characteristics about the Graph of a Log Function where a > 1 1. Domain is all real numbers 1. Range is all real numbers 2. Range is positive real numbers 2. Domain is positive real numbers 3. There are no x intercepts because there is no x value that you can put in the function to make it = 0 3. There are no y intercepts 4. The x intercept is always (1,0) (x’s and y’s trade places) 4. The y intercept is always (0,1) because a0 = 1 5. The graph is always increasing 5. The graph is always increasing 6. The x-axis (where y = 0) is a horizontal asymptote forx - 6. The y-axis (where x = 0) is a vertical asymptote
6. Logarithmic Graph Exponential Graph Graphs of inverse functions are reflected about the line y = x
7. Transformation of functions apply to log functions just like they apply to all other functions so let’s try a couple. up 2 Reflect about x axis left 1
8. Remember our natural base “e”? We can use that base on a log. What exponent do you put on e to get 2.7182828? ln Since the log with this base occurs in nature frequently, it is called the natural log and is abbreviatedln. Your calculator knows how to find natural logs. Locate the ln button on your calculator. Notice that it is the same key that has ex above it. The calculator lists functions and inverses using the same key but one of them needing the 2nd (or inv) button.
9. Another commonly used base is base 10.A log to this base is called a common log.Since it is common, if we don't write in the base on a log it is understood to be base 10. What exponent do you put on 10 to get 100? What exponent do you put on 10 to get 1/1000? This common log is used for things like the richter scale for earthquakes and decibles for sound. Your calculator knows how to find common logs. Locate the log button on your calculator. Notice that it is the same key that has 10x above it. Again, the calculator lists functions and inverses using the same key but one of them needing the 2nd (or inv) button.
10. The secret to solving log equations is to re-write the log equation in exponential form and then solve. Convert this to exponential form check: This is true since 23 = 8
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# Understanding Show Me Cash Chances
You may already know that the chances of winning the Show Me Cash jackpot are 1 in 575,757. But many people wonder how those chances of winning are determined.
Show Me Cash is played by selecting five numbers out of 39 possible choices (the numbers from 1 to 39). During each drawing, one by one, the five numbers are randomly selected. Drawings are carefully monitored and are open to the public for viewing.
So how does one determine what the chances of winning are for correctly matching all five numbers for any Show Me Cash drawing? We start with the math. The formula for calculating the possibilities for matching the numbers drawn out of all the numbers in the machine is based on probability theory. So how likely is it to get one particular outcome out of all the possible outcomes that could happen? In other words, how many different combinations of five numbers can be made out of 39 numbers? Let's look and see!
Let's do the math; it won't be hard. Any calculation of chances for a Show-Me-Cash-type game involves three basic elements:
1. The total number of numbers you are choosing from, which is called the field or (f).
2. The number of numbers the player selects on his\her playslip, which is called the pick or (p).
3. The number of correct matches between the player's picks and the numbers drawn, which is called the match or (m).
Here is the formula statisticians use to calculate the probability:
It looks complicated, but let us explain. Remember, f = the 39 numbers, p = the numbers you pick, and m = the numbers you correctly match.
Let's get started. The factors or elements depicted as x over y, with ( ) around them, are called binomial coefficients and, when filled in with our numbers, are what mathematicians call factorials and are depicted as x! or a number followed by an exclamation point. What's that all about? Well, simply put, a factorial is a number multiplied progressively by all other smaller numbers until you reach 1. Let's use 5 as an example; 5! would be figured as 5 x 4 x 3 x 2 x 1 = 120
In other words, we would multiply 5 times 4 times 3 times 2 times 1 to get the result. 0! is equal to 1.
So what does that have to do with Show Me Cash? Well, 5! is all the different combinations you can theoretically make out of five numbers. Starting to make sense?
To match all five numbers and win the Show Me Cash jackpot, here are the chances of winning:
(Remember, there are 39 numbers, and you must choose five of those numbers to play and match to win the jackpot.)
Show Me Cash chances of winning the jackpot:
(5 of 5 numbers matched)
We have one five-number combination (our pick) in a possible 575,757 different possible combinations or outcomes. These are your chances of hitting the jackpot with any five numbers played.
We hope this helps to better explain the chances of winning the Show Me Cash jackpot.
Back to Show Me Cash Page
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Solution of Digit Problem2
In this page solution of digit problem2 we are going to see solution for question 3 and question 4 with detailed steps.
Question 3:
The unit's digit of a two digit number is twice its ten's digit. If 18 is added to the number, the digits interchange their places. Find the number.
Solution:
Let "x y" be the required two digit number.
Let "x" be the number which is in unit's digit
Let "y" be the number which is in ten's digit
The unit's digit of a two digit number is twice its ten's digit
y = 2 x
2x - y = 0-------(1)
x y + 18 = y x
Let us write them using expanded form
10 x + y + 18 = 10 y + x
10 x - x + y - 10 y = -18
9 x - 9 y = -18
Dividing this equation by 9
x - y = - 2 --------(2)
(1) - (2) 2x - y = 0 ------(1)
x - y = - 2 ------(2)
(-) (+) (+)
_____________
x = 2
Substituting x = 2 in the first equation
2 (2) - y = 0
4 - y = 0
- y = -4
y = 4
Therefore the required number is 24
Checking:
The unit's digit of a two digit number is twice its ten's digit.
2 (2) = 4
If 18 is added to the number, the digits interchange their places
24 + 18 = 42
These are the problems in solution of digit problem2.
Question 4:
The sum of the digits of two digit number is 12. If the new number formed by reversing the digits is greater than the original number by 54, find the original number.
Solution:
Let "x y" be the required two digit number.
The sum of the digits of two digit number = 12
x + y = 12 ------(1)
If the new number formed by reversing the digits is greater than the original number by 54
y x = x y + 54
Let us write this as expanded form
10 y + x = 10 x + y + 54
x - 10 x + 10 y - y = 54
- 9 x + 9 y = 54
Dividing this equation by 9. We will get
- x + y = 6 ------(2)
(1) + (2) x + y = 12
- x + y = 6
___________
2 y = 18
y = 18/2
y = 9
Substituting y = 9 in the first equation.
x + 9 = 12
x = 12 - 9
x = 3
Therefore the required number is 39
Checking:
The sum of the digits of two digit number is 12
3 + 9 = 12
If the new number formed by reversing the digits is greater than the original number by 54
93 = 39 + 54
Solution of Digit Problem2 to Worksheet
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# How to Simplify Fractions With Decimals
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Fractions and decimals are parts of whole numbers written in two different forms. A fraction has a numerator over a denominator, which represents the number of parts you have of a whole number over the number of parts by which the whole number is divided. A decimal contains part of a whole number to the right of a decimal point. If a fraction contains a decimal in either its numerator or denominator, you can convert the decimal into a fraction so that you have two similar number formats to simplify the fraction. A fraction is simplified when the only common factor of the numerator and denominator is 1.
Determine a fraction with a decimal that you want to simplify. For the following example, use the fraction 0.2/2.
Convert the decimal to a fraction by placing the number to the right of the decimal point as a numerator over a denominator that corresponds to the place value of the decimal number. In the example, the decimal 0.2 extends to the tenths place, so place 2 as a numerator over 10 as a denominator, which equals 2/10. This leaves (2/10)/2, which consists of a fraction within a fraction.
Divide the numerator by the denominator, which is equivalent to multiplying the numerator by the reciprocal of the denominator, to convert the fraction within a fraction to a single fraction. A reciprocal is a fraction flipped upside down. In the example, divide 2/10 by 2, which is equivalent to multiplying 2/10 by 1/2. This equals 2/20.
Find the largest number that divides evenly into the numerator and denominator of the fraction. In the example, 2 is the largest number that divides evenly into 2 and 20.
Divide both the numerator and denominator by the largest number that divides evenly into both to simplify the fraction. In the example, divide 2 by 2, which equals 1, and divide 20 by 2, which equals 10. This leaves 1/10, which is the simplified form of the fraction with a decimal.
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How do you solve lnx + ln(x+1) = ln12?
Oct 23, 2015
$x = 3$
Explanation:
In general
$\textcolor{w h i t e}{\text{XXX}} \log \left(A\right) + \log \left(B\right) = \log \left(A B\right)$
(this is one of the basic logarithmic rules)
Specifically
$\textcolor{w h i t e}{\text{XXX}} \ln \left(x\right) + \ln \left(x + 1\right) = \ln \left({x}^{2} + x\right)$
and we are told, this is
$\textcolor{w h i t e}{\text{XXXXXXXXXXXXX}} = \ln \left(12\right)$
$\Rightarrow {x}^{2} + x = 12$
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + x - 12 = 0$
$\textcolor{w h i t e}{\text{XXX}} \left(x + 4\right) \left(x - 3\right) = 0$
$\textcolor{w h i t e}{\text{XXX}} x = - 4$$\textcolor{w h i t e}{\text{XXX}}$or$\textcolor{w h i t e}{\text{XXX}} x = 3$
Since $\ln \left(x\right)$ is not defined for negative values of $x$
$\Rightarrow x = 3$
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# Dividing Rational Expressions – Techniques & Examples
Rational expressions in mathematics can be defined as fractions in which either or both the numerator and the denominator are polynomials. Just like dividing fractions, rational expressions are divided by applying the same rules and procedures.
To divide two fractions, we multiply the first fraction by the inverse of the second fraction. This is done by changing from the division sign (÷) to the multiplication sign (×).
The general formula for dividing fractions and rational expressions is;
• a/b ÷ c/d = a/b × d/c = ad/bc
For example;
• 5/7 ÷ 9/49 = 5/7 × 49/9
= (5 × 49)/ (7 × 9) = 245/63
= 35/9
• 9/16 ÷ 5/8
= 9/16 × 8/5
= (9 × 8)/ (16 × 5)
= 72/80
= 9/10
## How to Divide Rational Expressions?
Dividing rational expressions follow the same rule of dividing two numerical fractions.
The steps involved in dividing two rational expressions are:
• Factor both the numerators and denominators of each fraction. You must know how to factor quadratic and cubic equations.
• Change from division to multiplication sign and flip the rational expressions after the operation sign.
• Simplify the fractions by canceling common terms in the numerators and denominators. Take care that you cancel the factors and not the terms.
• Finally, rewrite the remaining expressions.
Below are the few examples which will better explain the dividing rational expression technique.
Example 1
[(x2 + 3x – 28)/ (x2 + 4x + 4)] ÷ [(x2 – 49)/ (x2 – 5x- 14)]
Solution
= (x2 + 3x – 28)/ (x2 + 4x + 4)] ÷ [(x2 – 49)/ (x2 – 5x – 14)
Factor both the numerators and denominators of each fraction.
⟹ x2 + 3x – 28 = (x – 4) (x + 7)
⟹ x2 + 4x + 4 = (x + 2) (x + 2)
⟹ x2 – 49 = x2 – 72 = (x – 7) (x + 7)
⟹ x2 – 5x – 14 = (x – 7) (x + 2)
= [(x – 4) (x + 7)/ (x + 2) (x + 2)] ÷ [(x -7) (x + 7)/ (x – 7) (x + 2)]
Now, multiply the first fraction by the reciprocal of the second fraction.
= [(x – 4) (x + 7)/ (x + 2) (x + 2)] * [(x – 7) (x + 2)/ (x – 7) (x + 7)]
On cancelling common terms and rewrite the remaining factors to get;
= (x – 4)/ (x + 2)
Example 2
Divide [(2t2 + 5t + 3)/ (2t2 +7t +6)] ÷ [(t2 + 6t + 5)/ (-5t2 – 35t – 50)]
Solution
Factor the numerators and denominators of each fraction.
⟹ 2t+ 5t + 3 = (t + 1) (2t + 3)
⟹ 2t+ 7t + 6 = (2t + 3) (t + 2)
⟹ t+ 6t + 5 = (t + 1) (t + 5)
⟹ -5t2 – 35t -50 = -5(t2 + 7t + 10)
= -5(t + 2) (t + 5)
= [(t + 1) (2t + 3)/ (2t + 3) (t + 2)] ÷ [(t + 1) (t + 5)/-5(t + 2) (t + 5)]
Multiply by the reciprocal of the second rational expression.
= [(t + 1) (2t + 3)/ (2t + 3) (t + 2)] * [-5(t + 2) (t + 5)/ (t + 1) (t + 5)]
Cancel common terms.
= -5
Example 3
[(x + 2)/4y] ÷ [(x2 – x – 6)/12y2]
Solution
Factor the numerators of the second fraction
⟹ (x2 – x – 6) = (x – 3) (x + 2)
= [(x + 2)/4y] ÷ [(x – 3) (x + 2)/12y2]
Multiply by the reciprocal
= [(x + 2)/4y] * [12y2/ (x – 3) (x + 2)]
On cancelling common terms, we get the answer as;
= 3y/4(x – 3)
Example 4
Simplify [(12y2 – 22y + 8)/3y] ÷ [(3y2 + 2y – 8)/ (2y2 + 4y)]
Solution
Factor the expressions.
⟹ 12y2 – 22y + 8 = 2(6y2 – 11y + 4)
= 2(3y – 4) (2y – 1)
⟹ (3y2 + 2y – 8) = (y + 2) (3y – 4)
= 2y2 + 4y = 2y (y + 2)
= [(12y2 – 22y + 8)/3y] ÷ [(3y2 + 2y – 8)/ (2y2 + 4y)]
= [2(3y – 4) (y – 1)/3y] ÷ [y + 2) (3y – 4)/2y (y + 2)]
= [2(3y – 4) (2y – 1)/3y] * [y (y + 2)/ (y + 2) (3y – 4)]
= 4(2y – 1)/3
Example 5
Simplify (14x4/y) ÷ (7x/3y4).
Solution
= (14x4/y) ÷ (7x/3y4)
= (14x4/ y) * (3y4/7x)
= (14x* 3y4) / 7xy
= 6x3y3
### Practice Questions
1. If we divide the rational expression, $\dfrac{a+b }{a-b} \div \dfrac{a^3+b^3}{a^3-b^3}$, which of the following shows the correct expression?
2. If we divide the rational expression, $\dfrac{x^2-16}{x^2-3x+2} \div \dfrac{x^3+64}{x^2-4} \div \dfrac{x^2-2x-8}{x^2-4x+16}$, which of the following shows the correct expression?
3. If we divide the rational expression, $\dfrac{x^2-4x-12}{x^2-3x-18} \div \dfrac{x^2+3x+2}{x^2-2x-3}$, which of the following shows the correct expression?
4. If we divide the rational expression, $\dfrac{p^2-1}{p}\cdot \dfrac{\frac{p^2}{p-1}}{\dfrac{p+1}{1}}$, which of the following shows the correct expression?
5. If we divide the rational expression, $\dfrac{2x-1}{x^2+2x+4} \div \dfrac{2x^2+5x-3}{x^4-8x} \div \dfrac{x^2-2x}{x+3}$, which of the following shows the correct expression?
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# Question Video: Finding the Algebraic Form of a Vector That Makes Equal Angles with the Axes Given Its Norm Mathematics • 12th Grade
Find the algebraic form of a vector π if its norm is 31, given that it makes equal angles with the positive directions of the Cartesian axes.
02:52
### Video Transcript
Find the algebraic form of a vector π if its norm is 31, given that it makes equal angles with the positive directions of the Cartesian axes.
Letβs begin by recalling that the algebraic, also called the Cartesian, form of a vector π with components π΄ π₯, π΄ π¦, and π΄ π§ is given by the π₯-component π΄ π₯ multiplied by the unit vector π’ in the positive π₯-direction plus the π¦-component π΄ π¦ multiplied by the unit vector π£ in the positive π¦-direction plus the π§-component π΄ π§ multiplied by the unit vector π€ in the positive π§-direction. And so, to find the algebraic from of our vector π, we need to find the components π΄ π₯, π΄ π¦, and π΄ π§.
The information weβve been given is that the norm of the vector is 31. That is, its magnitude is 31. Weβre also told that π makes equal angles with the positive directions of the Cartesian axes. And this tells us that our direction angles π π₯, π π¦, and π π§ are all equal. And remember that these are the angles the vector makes with the positive π₯-, π¦-, and π§-axes. If π π₯, π π¦, and π π§ are all the same, letβs just call this angle π. Now, recall also that the direction cosines are the cosines of the direction angles, that is, the cos of π π₯, the cos of π π¦, and the cos of π π§.
And using right angle trigonometry, we can show that the components of the vector π, π΄ π₯, π΄ π¦, and π΄ π§, are given by the magnitude or norm of the vector π multiplied by the direction cosines. And in our case, since our direction angles are all equal, that is, π, then our three components π΄ π₯, π΄ π¦, and π΄ π§ are equal to the magnitude or norm of the vector multiplied by cos π. And since our norm is 31, thatβs 31 cos π.
Now, in order to find π, we can use the known result that the sum of the squares of our direction cosines is equal to one. And in our case, since our three angles are equal, this tells us that three cos squared π is equal to one. And now dividing both sides by three and taking the positive and negative square roots, we have the cos of our angle π is positive or negative one over root three. And rationalizing our denominator, this gives us positive or negative root three over three.
In our equation for the components of our vector then, we have π΄ π₯ is equal to π΄ π¦ is equal to π΄ π§ is equal to positive or negative 31 root three over three. And so, in algebraic or Cartesian form, the vector π with norm 31 and making equal angles with the positive directions of the Cartesian axes is positive or negative 31 times the square root of three over three multiplied by the sum of the unit vectors π’, π£, and π€.
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# 3 Forms Of A Quadratic Function
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### 3 Forms Of A Quadratic Function
1. 1. 3 Forms of a Quadratic Function!<br />
2. 2. Key Ideas<br /><ul><li>The axis of symmetry divides the parabola into mirror images and passes through the vertex.
3. 3. a tells whether the graph opens up or down and whether the graph is wide or narrow.
4. 4. H controls horizontal translations.
5. 5. K controls vertical translations
6. 6. a>0; the graph opens up.
7. 7. a<0; the graph opens down.</li></li></ul><li>Standard Form<br />f(x) = ax^2 + bx + c<br />
8. 8. Y=2x^2-8x+6<br />Step 2:<br /> Find the vertex:<br /> x= -b/2a= -(-8)/2(2)<br /> X = 2<br /> y= 2(2)^2-8(2)+6<br /> y = -2<br />Step 1: <br /> Identify coefficients:<br /> a=2 b=-8 c=6<br /> a>0 so the parabola opens up<br />Step 4:<br /> Plot 2 more points. <br />Go to the left by one and up 2. Then do it again for the right side.<br />Step 3:<br /> Draw axis of symmetry<br /> x=-b/2a<br /> x=2<br />Step 5:<br /> Draw parabola through the points.<br />Step 6: <br /> Label the points<br />(3,0)<br />(1,0)<br />(2,-2)<br />
9. 9. Vertex Form<br />f(x) = a (x-h)^2 + k<br />
10. 10. Y=2(x-3)^2-3<br />Step 3:<br /> Plot 2 more points. <br />Go to the left by one and up 2. Then do it again for the right side.<br />Step 1:<br /> Determine vertex:<br /> (h,k)= (3, -3)<br />Step 4:<br /> Connect the points to form a parabola.<br />Step 2:<br /> Plot the vertex, (3,-3)<br /> Draw the axis of symmetry.<br /> x=h so x=3<br />Step 5:<br /> Label the points<br />(4,-1)<br />(2,-1)<br />(3,-3)<br />
11. 11. Intercept Form<br />f(x) = a (x-p) (x-q)<br />
12. 12. Y= 2(x-3)(x+1)<br />(-1,0)<br />(3,0)<br />Step 2:<br /> Find the coordinates of the vertex and plot them.<br /> ((p+q)/2, f(p+q)/2)<br /> x= (3-1)/2= 2/2=1<br /> x=1<br /> Y= 2(1-3)(1+1)<br /> Y= 2(-2)(2)<br /> y=-8<br /> Vertex= (1,-8)<br />Step 3: <br /> Plot the axis of symmetry. <br /> Axis of symmetry is x=(p+q)/2<br /> x=1<br />Step 1:<br /> Plot x-intercepts<br /> x=3 and x=-1<br />Step 5:<br /> Label the points.<br />Step 4:<br /> Connect the points.<br />(1,-8)<br />
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# Left hand limit
The value of a function as its input approaches some value from left hand side is called the left-hand limit, and also called as the left sided limit.
## Introduction
Assume, $x$ is a variable, a function is defined in terms of $x$ and the function is expressed as $f(x)$ simply. Let $a$ represents a constant and $L$ represent the limit of the function $f(x)$.
The left sided limit of the function $f(x)$ is $L$ as the variable $x$ approaches $a$ from left side. In this case, $a^-$ is the closer value of $a$ from left side in Cartesian coordinate system. It can be written in the following mathematical form.
$L \,=\, \displaystyle \large \lim_{x \,\to\, a^-}{\normalsize f(x)}$
Remember, the representation $x \,\to\, a^-$ means it is neither $a$ nor $-a$ but less than $a$ and very closer to $a$.
### Example
The concept of left-sided limit can be understood from the following example.
Evaluate $\displaystyle \large \lim_{x \,\to\, 2^-}{\normalsize (x^2-3)}$
In this case, we have to calculate the value of the function $x^2-3$ as the input $x$ closer to $2$ from left-side in the Cartesian coordinate system. According to the two-dimensional space, $0$ and $1$ are near points from left-side of point $2$. So, calculate the values of the function for both values.
1. $f(0) = (0)^2-3 = -3$
2. $f(1) = (1)^2-3 = -2$
But, the point $x = 0$ is two units away from the point $x = 2$ in left-side. Similarly, the point $x = 1$ is one unit away from the point $x = 2$ in left-side. They both are near points but not closer from left-side of the point $x = 2$.
The point $x = 1.9$ is closer point to point $x = 2$ than $x = 0$ and $x = 1$. So, calculate the value of the function.
$f(1.9) = (1.9)^2-3 = 0.61$
There is no doubt about it, the point $x = 1.9$ is closer to point $x = 2$ but the point $x = 2$ is $0.1$ unit away from point $x = 1.9$ and the difference between them is not negligible. Hence, the value of the function $x^2-3$ at point $x = 1.9$ cannot be taken as the limit of the function as $x$ approaches $2$ from left side.
So, we have to consider the points, which are very closer to the point $x = 2$, for example $x = 1.99$ $,$ $1.999$ $,$ $1.9999$ and so on.
1. $f(1.99)$ $=$ $(1.99)^2-3$ $=$ $0.9601$
2. $f(1.999)$ $=$ $(1.999)^2-3$ $=$ $0.996001$
3. $f(1.9999)$ $=$ $(1.9999)^2-3$ $=$ $0.99960001$
4. $f(1.99999)$ $=$ $(1.9999)^2-3$ $=$ $0.9999600001$
For $x = 1.99$ $,$ $1.999$ $,$ $1.9999$ and $1.99999$, the values of the function are $0.9601$, $0.996001$, $0.99960001$ and $0.99996$ respectively.
In fact, $1.99$ $\ne$ $1.999$ $\ne$ $1.9999$ $\ne$ $1.99999$ but $1.99 \approx 2$ $,$ $1.999 \approx 2$ $,$ $1.9999 \approx 2$ and $1.99999 \approx 2$. Similarly, $0.9601$ $\ne$ $0.996001$ $\ne$ $0.99960001$ $\ne$ $0.99996$ but $0.9601 \approx 1$ $\,$ $0.996001 \approx 1$ $,$ $0.99960001 \approx 1$ and $0.9999600001 \approx 1$.
For better understanding, it is displayed in a tabular form.
$\Large x$ $f(x)$
$Actual$ $Approximate$ $Actual$ $Approximate$
$0$ $0$ $-3$ $-3$
$1$ $1$ $-2$ $-2$
$1.9$ $1.9$ $0.61$ $0.61$
$1.99$ $2$ $0.9601$ $1$
$1.999$ $2$ $0.996001$ $1$
$1.9999$ $2$ $0.99960001$ $1$
$1.99999$ $2$ $0.9999600001$ $1$
$\vdots$ $\vdots$ $\vdots$ $\vdots$
Mathematically, the values of the function are not equal but each value is approximately same and it is equal to $1$ because the value of $x$ is very closer to $2$. Therefore, you can take any value of $x$ as $x$ approaches $2$ from left hand side for calculating the limit.
For example, take $x = 1.99999$. It approaches to $2$ and left-side in the Cartesian coordinate system.
$\displaystyle \large \lim_{x \,\to\, 2^-}{\normalsize x^2-3}$ $\,=\,$ $(1.99999)^2-3$
$\implies$ $\displaystyle \large \lim_{x \,\to\, 2^-}{\normalsize x^2-3}$ $\,=\,$ $3.9999600001-3$
$\implies$ $\displaystyle \large \lim_{x \,\to\, 2^-}{\normalsize x^2-3}$ $\,=\,$ $0.9999600001$
$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2^-}{\normalsize x^2-3}$ $\,\approx\,$ $1$
Therefore, the limit of the function $x^2-3$ as $x$ approaches $2$ from left side is equal to one and it is called the left-hand limit or left-sided limit.
Remember, $x \,\to\, 2^-$ means $x \ne 2$ and $x \ne -2$ but $x < 2$ and $x \approx 2$ in this case.
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# How do you find the derivative of (x^4 - 1)^10 (2x^4 + 3)^7?
Dec 30, 2015
$56 {x}^{3} {\left({x}^{4} - 1\right)}^{10} {\left(2 {x}^{4} + 3\right)}^{6} + 40 {x}^{3} {\left(2 {x}^{4} + 3\right)}^{7} {\left({x}^{4} - 1\right)}^{9}$
#### Explanation:
Use the product rule :
$\frac{d}{\mathrm{dx}} \left(f \left(x\right) \cdot g \left(x\right)\right) = f \left(x\right) \cdot g ' \left(x\right) + g \left(x\right) \cdot f ' \left(x\right)$
and inside this product rule there will be need for use of the power rule :
$\frac{d}{\mathrm{dx}} {\left[u \left(x\right)\right]}^{n} = n \cdot {\left[u \left(x\right)\right]}^{n - 1} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.
Hence the derivative of the given function with respect to x is :
${\left({x}^{4} - 1\right)}^{10} \cdot 7 {\left(2 {x}^{4} + 3\right)}^{6} \cdot \left(8 {x}^{3}\right) + {\left(2 {x}^{4} + 3\right)}^{7} \cdot 10 {\left({x}^{4} - 1\right)}^{9} \cdot \left(4 {x}^{3}\right)$
$= 56 {x}^{3} {\left({x}^{4} - 1\right)}^{10} {\left(2 {x}^{4} + 3\right)}^{6} + 40 {x}^{3} {\left(2 {x}^{4} + 3\right)}^{7} {\left({x}^{4} - 1\right)}^{9}$.
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# High School Math : Quadratic Functions
## Example Questions
### Example Question #23 : Quadratic Functions
Based on the figure below, which line depicts a quadratic function?
None of them
Blue line
Red line
Purple line
Green line
Red line
Explanation:
A parabola is one example of a quadratic function, regardless of whether it points upwards or downwards.
The red line represents a quadratic function and will have a formula similar to .
The blue line represents a linear function and will have a formula similar to .
The green line represents an exponential function and will have a formula similar to .
The purple line represents an absolute value function and will have a formula similar to .
### Example Question #14 : Quadratic Functions
Which of the following functions represents a parabola?
Explanation:
A parabola is a curve that can be represented by a quadratic equation. The only quadratic here is represented by the function , while the others represent straight lines, circles, and other curves.
### Example Question #70 : Quadratic Functions
Find the radius of the circle given by the equation:
Explanation:
To find the center or the radius of a circle, first put the equation in the standard form for a circle: , where is the radius and is the center.
From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula .
, so .
and , so and .
Therefore, .
Because the constant, in this case 4, was not in the original equation, we need to add it to both sides:
Now we do the same for :
We can now find :
### Example Question #71 : Quadratic Functions
Find the center of the circle given by the equation:
Explanation:
To find the center or the radius of a circle, first put the equation in standard form: , where is the radius and is the center.
From our equation, we see that it has not yet been factored, so we must do that now. We can use the formula .
, so .
and , so and .
This gives .
Because the constant, in this case 9, was not in the original equation, we must add it to both sides:
Now we do the same for :
We can now find the center: (3, -9)
### Example Question #51 : Quadratic Functions
Find the -intercepts for the circle given by the equation:
Explanation:
To find the -intercepts (where the graph crosses the -axis), we must set . This gives us the equation:
Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:
and
We can then solve these two equations to obtain .
### Example Question #1 : Circle Functions
Find the -intercepts for the circle given by the equation:
Explanation:
To find the -intercepts (where the graph crosses the -axis), we must set . This gives us the equation:
Because the left side of the equation is squared, it will always give us a positive answer. Thus if we want to take the root of both sides, we must account for this by setting up two scenarios, one where the value inside of the parentheses is positive and one where it is negative. This gives us the equations:
and
We can then solve these two equations to obtain
### Example Question #31 : Functions And Graphs
Find the center and radius of the circle defined by the equation:
Explanation:
The equation of a circle is: where is the radius and is the center.
In this problem, the equation is already in the format required to determine center and radius. To find the -coordinate of the center, we must find the value of that makes equal to 0, which is 3. We do the same to find the y-coordinate of the center and find that . To find the radius we take the square root of the constant on the right side of the equation which is 6.
### Example Question #1 : Finding The Center And Radius
Find the center and radius of the circle defined by the equation:
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# Finding Distances on Maps
33 teachers like this lesson
Print Lesson
## Objective
SWBAT solve map scale problems using a double number line
#### Big Idea
Remember planning trips before Mapquest or Google Maps? Students find distances using rulers and map scales.
## Introduction
10 minutes
I will start by asking my students how would they or their parents plan a long distance trip? I am sure most will respond that they will use the computer. Then, how do you think people planned trips before the tools on the internet were so readily available? How did people calculate distances and routes on a map? Hopefully this pulls a lot of prior knowledge from students about how to read and interpret maps. I really don't know how much experience my current group of students will have had with "analog" map reading.
I will then walk my students through the example problem. Several students will need help using their rulers to measure distances. Some students will think they should start at the number one and measure from there. While it is possible to accurately measure from any point, I will encourage them to start from 0!
The rest of the problem is to focus on the proportional relationship represented by a map scale.
## Guided Problem Solving
10 minutes
This next problem is similar to the example problem but students now measure in inches. To keep things simple, I will have students measure to the nearest half inch. Again I will need to be on the lookout for students who are measuring on the metric side of the ruler.
Students must then explain how they know that the found measurement is equivalent to the given scale. This is MP3. It also addresses one of the essential questions of the unit: 1) How can proportional relationships be identified?
## Independent Problem Solving
20 minutes
Now the students are on their own with their partners to solve the next set of problems. There are several things to look out for here. Each problem asks students to complete their number line using a specific increment. I'll make sure students are doing this correctly. On problem 1, students are asked to identify two possible unit rates from the given scale. Some students may treat this as identical and say 5 to 1. Here I may say, "If 1 cm equals 5 miles will 1 mile be represented by a length less than or greater than 1 cm?". It may help to use the double number line as a visual reference.
The final problem requires students to actually identify the scale given the length of 2 miles. The last two problems require students to solve rate problems involving distance to time.
## Exit Ticket
5 minutes
Before going into the exit ticket, we will summarize the lesson by discussing the map scale and how we represented equivalent ratios on the double number line. Then students will take the exit ticket.
This exit ticket will be worth 4 points. A score of at least 3 correct answers will be considered as a sign of success.
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Binomials Cubed – Examples and Practice Problems
Binomials cubed exercises can be solved using two methods. The first method consists in multiplying the binomial three times and fully expanding the expression. The second method is to use a standard formula that can simplify the resolution process.
Here, we will look at a summary of these two methods to solve binomials cubed. In addition, we will explore various examples with answers to fully master this topic.
ALGEBRA
Relevant for
Exploring examples of binomials cubed.
See examples
ALGEBRA
Relevant for
Exploring examples of binomials cubed.
See examples
Summary of binomials cubed
Recall that a binomial cubed is an expression of the form $latex {{(x+y)}^3}$. This expression could contain coefficients or other variables.
To solve binomials cubed, we can use two main methods:
Method 1: We can rewrite the binomial three times as a multiplication of binomials and eliminate the exponent. For example, we can rewrite $latex {{(x+y)}^3}$, as follows:
$latex (x+y)(x+y)(x+y)$
Then, we use the distributive property to multiply all the terms and obtain a simplified expression.
Method 2: Method 1 could be very tedious since we have to multiply each term by each term. To make it easier to solve binomials cubed, we can use standard formulas for adding cubes and subtracting cubes.
Sum of cubes: The sum of a binomial cubed is equal to the first term cubed, plus three times the square of the first term times the second term, plus three times the first term times the square of the second term, plus the cube of the second term. finished:
Difference of cubes: The difference of a binomial cubed is equal to the first term cubed, minus three times the square of the first term times the second term, plus three times the first term times the square of the second term, minus the cube of the second term:
Binomials cubed – Examples with answers
The following binomial cubed examples can be used to learn how to apply the solving methods mentioned above. It is recommended that you try to solve the exercises yourself before looking at the solution.
EXAMPLE 1
Solve the binomial: $latex {{(x+1)}^3}$.
Method 1: We have to rewrite the binomial as a multiplication:
$latex {{(x+1)}^3}$
⇒ $latex (x+1)(x+1)(x+1)$
We start by multiplying the first two parentheses and then we multiply the remaining parentheses:
$latex (x+1)(x+1)(x+1)$
$latex =({{x}^2}+2x+1)(x+1)$
$latex ={{x}^3}+3{{x}^2}+3x+1$
Method 2: Using the formula for the sum of a binomial cubed $latex {{(a+b)}^3}={{a}^3}+3{{a}^2}b+3a{{b}^2}+{{b}^3}$, we have:
⇒ $latex {{x}^3}+3{{x}^2}(1)+3x{{1}^2}+{{1}^3}$
$latex ={{x}^3}+3{{x}^2}+3x+1$
We see that we got the same answer using both methods. However, the first method is usually more tedious when we have more complicated binomials, so we will only use the second method to solve the next examples.
EXAMPLE 2
Find the result of the binomial cubed: $latex {{(x+5)}^3}$.
Using the formula for the sum of a binomial cubed $latex {{(a+b)}^3}={{a}^3}+3{{a}^2}b+3a{{b}^2}+{{b}^3}$, we have:
⇒ $latex {{x}^3}+3{{x}^2}(5)+3x{{(5)}^2}+{{5}^3}$
$latex ={{x}^3}+15{{x}^2}+75x+125$
We see that with the standard formula we can more easily find the answer.
EXAMPLE 3
Solve the binomial cubed: $latex {{(2x-6)}^3}$.
In this case, we have to use the formula for subtracting a binomial from the cube $latex {{(a-b)}^3}={{a}^3}-3{{a}^2}b+3a{{b}^2}-{{b}^3}$. Therefore, we have:
⇒ $latex {{(2x)}^3}-3{{(2x)}^2}(6)+3(2x){{(6)}^2}-{{6}^3}$
$$=8{{x}^3}-3(4{{x}^2})(6)+3(2x)(36)-216$$
$latex =8{{x}^3}-72{{x}^2}+216x-216$
We could easily find the answer by using the formula.
EXAMPLE 4
Solve the binomial cubed: $latex {{(3x-2y)}^3}$.
We have to use the formula for the subtraction of a binomial cubed $latex {{(a-b)}^3}={{a}^3}-3{{a}^2}b+3a{{b}^2}-{{b}^3}$. Therefore, we have:
⇒ $${{(3x)}^3}-3{{(3x)}^2}(2y)+3(3x){{(2y)}^2}-{{(2y)}^3}$$
$$=27{{x}^3}-3(9{{x}^2})(2y)+3(3x)(4{{y}^2})-8{{y}^3}$$
$latex =27{{x}^3}-54{{x}^2}y+36x{{y}^2}-8{{y}^3}$
EXAMPLE 5
Solve the binomial cubed $latex {{(2{{x}^2}+4y)}^3}$.
Here, we have to use the formula for the sum of a binomial cubed $latex {{(a+b)}^3}={{a}^3}+3{{a}^2}b+3a{{b}^2}+{{b}^3}$. Therefore, we have:
⇒ $${{(2{{x}^2})}^3}+3{{(2{{x}^2})}^2}(4y)+3(2{{x}^2}){{(4y)}^2}+{{(4y)}^3}$$
$$=8{{x}^6}+3(4{{x}^4})(4y)+3(2{{x}^2})(16{{y}^2})+64{{x}^3}$$
$latex =8{{x}^6}+48{{x}^4}y+96{{x}^2}{{y}^2}+64{{x}^3}$
This binomial contained a squared variable, but we apply the binomial sum formula in the same way as in the previous exercises.
We simply use the power of a power rule, which tells us that when we raise an expression with a power to another power, we have to multiply the exponents.
EXAMPLE 6
Simplify this expression: $latex {{(x+2y)}^3}+{{(x-2y)}^3}$.
We can use the formulas for adding a binomial cubed and subtracting a binomial cubed separately to calculate each binomial. Therefore, we have:
⇒ $${{(x)}^3}+3{{(x)}^2}(2y)+3(x){{(2y)}^2}+{{(2y)}^3}$$
$latex ={{x}^3}+3({{x}^2})(2y)+3(x)(4{{y}^2})+8{{y}^3}$
$latex ={{x}^3}+6{{x}^2}y+12x{{y}^2}+8{{y}^3}$
⇒ $${{(x)}^3}-3{{(x)}^2}(2y)+3(x){{(2y)}^2}-{{(2y)}^3}$$
$latex ={{x}^3}-3({{x}^2})(2y)+3(x)(4{{y}^2})-8{{y}^3}$
$latex ={{x}^3}-6{{x}^2}y+12x{{y}^2}-8{{y}^3}$
Now, we can add both expressions obtained and simplify like terms:
$$={{x}^3}+6{{x}^2}y+12x{{y}^2}+8{{y}^3}+{{x}^3}-6{{x}^2}y+12x{{y}^2}-8{{y}^3}$$
$latex =2{{x}^3}+24x{{y}^2}$
Binomials squared – Practice problems
Use the formulas for adding a binomial cubed and subtracting a binomial cubed detailed above to solve the following problems. If you have trouble with these problems, you can look at the solved examples above.
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Home » News » GMAT » GMAT Tip: Let It Cancel Out
# GMAT Tip: Let It Cancel Out
When faced with Geometry problems with variables, many test takers will approach the question with fear, believing they are forgetting some obscure geometric rule that is the only path towards a correct answer.
In reality, as we’ve covered in a few past posts, the understanding required to do well on Geometry questions on the GMAT is basic – Pythagorean theorem, special right triangles, area formula, and the like that you’ve found in the first half of an introductory Geometry course. Occasionally, we see some oddball questions associated with central angles, but there are still multiple ways to get to the correct answer.
But beyond these simple rules, the majority of Geometry problems in the GMAT Quantitative section are Algebra questions in disguise, requiring you understand what the simple geometry concept is that is being tested, then using Algebra-related problem solving from there.
Let’s examine a question that fits this mold:
In an xy-coordinate plane, a line is defined by y = kx + 1. If (4, b), (a, 4), and (a, b +1) are three points on the line, where a and b are unknown, then k = ?
1. ½
2. 1
3. 1 and ½
4. 2
5. 2 and ½
Initially, we may think that this problem is impossible – if a and b are unknown, how in the world will we find a third variable? But, if we assess the answer choices, we’ll notice that these are concrete values, and the two value questions suggest that we should consider an algebraic route, perhaps with factoring?
Either way, the best way to tackle these questions is to start with what you know, then work it out from there – you’ll never be successful by trying to jump to the right answer choice.
So, what do we know here? We know that the line is y = kx + 1 and we have three points we can plug in, even if they do consist of variables. We will start with (4,b):
y = kx + 1
b = k(4) + 1
Then (a,4):
y = kx + 1
4 = k(a) + 1
Then (a, b+1):
y = kx + 1
b + 1 = k(a) + 1
That gives us three equations:
b = 4k + 1
4 = k(a) + 1
b + 1 = k(a) + 1
The trigger here is to remember the answer choices are actual values, so we need to figure out how to get the variables to cancel out. It doesn’t look like there are any squared values, so at the very least, we should be able to eliminate (C) and (E) for answer choices.
One route to take is to solve for k(a):
k(a) = 3
Then, b + 1 = 3 + 1 , making b = 3
From there,
3 = 4k + 1
2 = 4k
2/4 = k > ½
Therefore the correct answer is (A).
Many “Geometry” GMAT problems will play out this way. We are excited to have a basic understanding of lines and coordinate points, but from there, it is all about setting up equations to find a value. Whenever tackling Geometry problems with lots of values, just remember to cancel it out!
Posted in: GMAT, GMAT - Quantitative, GMAT Tips
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