Answer:
First find the LCD between the denominator. 2, and 3 both share 6, so the answers 6
Answer:
a) h(3) = 9
b) h(0) = 6
c) h(6) = 4
Step-by-step explanation:
Piecewise functions have <u>multiple pieces</u> of curves/lines where each piece corresponds to its definition over an <u>interval</u>.
Given piecewise function:
![\sf h(x)=\begin{cases}\sf \dfrac{x^2-36}{x-6} &\textsf{if }x\neq 6\\\\\sf 4 & \textsf{if }x=6\end{cases}](https://tex.z-dn.net/?f=%5Csf%20h%28x%29%3D%5Cbegin%7Bcases%7D%5Csf%20%5Cdfrac%7Bx%5E2-36%7D%7Bx-6%7D%20%26%5Ctextsf%7Bif%20%7Dx%5Cneq%20%206%5C%5C%5C%5C%5Csf%204%20%26%20%5Ctextsf%7Bif%20%7Dx%3D6%5Cend%7Bcases%7D)
Therefore, the function has two definitions:
![\sf h(x)=\dfrac{x^2-36}{x-6} \quad \textsf{when x does not equal 6}](https://tex.z-dn.net/?f=%5Csf%20h%28x%29%3D%5Cdfrac%7Bx%5E2-36%7D%7Bx-6%7D%20%5Cquad%20%5Ctextsf%7Bwhen%20x%20does%20not%20equal%206%7D)
![\sf h(x)=4 \quad \textsf{when x equals 6}](https://tex.z-dn.net/?f=%5Csf%20h%28x%29%3D4%20%5Cquad%20%5Ctextsf%7Bwhen%20x%20equals%206%7D)
<u>Part (a)</u>
h(3) means to substitute x = 3 into the function.
As <u>x does not equal 6</u>, substitute x = 3 into the <u>first piece of the function</u>:
![\begin{aligned}\sf \implies h(3) & =\sf \dfrac{(3)^2-36}{(3)-6}\\\\ & = \sf \dfrac{9-36}{-3}\\\\ & = \sf \dfrac{-27}{-3}\\\\ & = \sf 9\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Csf%20%5Cimplies%20h%283%29%20%26%20%3D%5Csf%20%5Cdfrac%7B%283%29%5E2-36%7D%7B%283%29-6%7D%5C%5C%5C%5C%20%26%20%3D%20%5Csf%20%5Cdfrac%7B9-36%7D%7B-3%7D%5C%5C%5C%5C%20%26%20%3D%20%5Csf%20%5Cdfrac%7B-27%7D%7B-3%7D%5C%5C%5C%5C%20%26%20%3D%20%5Csf%209%5Cend%7Baligned%7D)
<u>Part (b)</u>
h(0) means to substitute x = 0 into the function.
As <u>x does not equal 6</u>, substitute x = 0 into the <u>first piece of the function</u>:
![\begin{aligned}\sf \implies h(0) & =\sf \dfrac{(0)^2-36}{(0)-6}\\\\ & = \sf \dfrac{0-36}{-6}\\\\ & = \sf \dfrac{-36}{-6}\\\\ & = \sf 6\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Csf%20%5Cimplies%20h%280%29%20%26%20%3D%5Csf%20%5Cdfrac%7B%280%29%5E2-36%7D%7B%280%29-6%7D%5C%5C%5C%5C%20%26%20%3D%20%5Csf%20%5Cdfrac%7B0-36%7D%7B-6%7D%5C%5C%5C%5C%20%26%20%3D%20%5Csf%20%5Cdfrac%7B-36%7D%7B-6%7D%5C%5C%5C%5C%20%26%20%3D%20%5Csf%206%5Cend%7Baligned%7D)
<u>Part (c)</u>
h(6) means to substitute x = 6 into the function.
As <u>x equals 6</u>, then use the <u>second piece of the function</u>:
![\sf \implies h(6)=4](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20h%286%29%3D4)
Learn more about piecewise functions here:
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Answer: d. 50
Steps:
1. Find value of x.
5x + 4x = 90
9x = 90
x = 10
2. Use x value to find ∠EDH.
∠EDH = 5x
∠EDH = 5(10)
∠EDH = 50°
Answer:
Circumference is about 185.26 cm.
Area is about 2732.585 ![cm^{2}](https://tex.z-dn.net/?f=cm%5E%7B2%7D)
Step-by-step explanation:
Circumference = 2πr
2 x 3.14 x 29.5 = 185.26
Area of a circle = π![r^{2}](https://tex.z-dn.net/?f=r%5E%7B2%7D)
3.14 x 29.5 x 29.5 = 2732.585
The diagonals of a parallelogram are congruent
The length WY is 32
<h3>How to determine the length WY</h3>
Given that:
and
![AY = x^2 - 6x](https://tex.z-dn.net/?f=AY%20%3D%20x%5E2%20-%206x)
Then, we have:
diagonal
Subtract x^2 from both sides
![- 48 =- 6x](https://tex.z-dn.net/?f=-%2048%20%3D-%206x)
Divide both sides by -6
![x = 8](https://tex.z-dn.net/?f=x%20%3D%208)
The length WY is calculated as:
![WY=2 * WA](https://tex.z-dn.net/?f=WY%3D2%20%2A%20WA)
So, we have:
![WY=2 * [x^2 - 48]](https://tex.z-dn.net/?f=WY%3D2%20%2A%20%5Bx%5E2%20-%2048%5D)
Substitute 8 for x
![WY=2 * [8^2 - 48]](https://tex.z-dn.net/?f=WY%3D2%20%2A%20%5B8%5E2%20-%2048%5D)
Simplify
![WY=32](https://tex.z-dn.net/?f=WY%3D32)
Hence, the length WY is 32
Read more about parallelograms at:
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