Answer:
The correct answer is option C. 3
Step-by-step explanation:
It is given an expression in variable x
(x² + 3x²)/(x + 1) ÷ (x² - 9)/(3x + 3) * (x - 3)/x
<u>To find the simplified answer</u>
x² + 3x² = x(x + 3)
x² - 9 = (x + 3)(x - 3)
The given expression can be written as,
(x² + 3x²)/(x + 1) ÷ (x² - 9)/(3x + 3) * (x - 3)/x
= x(x + 3)/(x + 1) * 3(x + 1)/(x + 1)(x - 1) * (x -3)/x
=[ x(x +3) * 3(x +1) * (x -3)]/[(x + 1) * (x + 3)(x - 3) *x]
= 3
The correct answer is option C. 3
<u>Given</u>:
The length of DE is 8 cm and the measure of ∠ADE is 60°.
We need to determine the surface area of the pyramid.
<u>Length of AD:</u>
The length of AD is given by
![cos 60^{\circ}=\frac{FD}{8}](https://tex.z-dn.net/?f=cos%2060%5E%7B%5Ccirc%7D%3D%5Cfrac%7BFD%7D%7B8%7D)
![4=FD](https://tex.z-dn.net/?f=4%3DFD)
Length of AD = 8 cm
<u>Slant height:</u>
The slant height EF can be determined using the trigonometric ratio.
Thus, we have;
![sin \ 60^{\circ}=\frac{EF}{8}](https://tex.z-dn.net/?f=sin%20%5C%2060%5E%7B%5Ccirc%7D%3D%5Cfrac%7BEF%7D%7B8%7D)
![sin \ 60^{\circ} \times 8=EF](https://tex.z-dn.net/?f=sin%20%5C%2060%5E%7B%5Ccirc%7D%20%5Ctimes%208%3DEF)
![\frac{\sqrt{3}}{2} \times 8=EF](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%20%5Ctimes%208%3DEF)
![4\sqrt{3}=EF](https://tex.z-dn.net/?f=4%5Csqrt%7B3%7D%3DEF)
Thus, the slant height EF is 4√3
<u>Surface area of the square pyramid:</u>
The surface area of the square pyramid can be determined using the formula,
![SA=Area \ of \ square + \frac{1}{2} (Perimeter \ of \ base ) (slant \ height)](https://tex.z-dn.net/?f=SA%3DArea%20%5C%20of%20%5C%20square%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%28Perimeter%20%5C%20of%20%5C%20base%20%29%20%28slant%20%5C%20height%29)
Substituting the values, we have;
![SA=8^2+\frac{1}{2}(8+8+8+8)(4 \sqrt{3})](https://tex.z-dn.net/?f=SA%3D8%5E2%2B%5Cfrac%7B1%7D%7B2%7D%288%2B8%2B8%2B8%29%284%20%5Csqrt%7B3%7D%29)
![SA=64+\frac{1}{2}(32)(4 \sqrt{3})](https://tex.z-dn.net/?f=SA%3D64%2B%5Cfrac%7B1%7D%7B2%7D%2832%29%284%20%5Csqrt%7B3%7D%29)
![SA=64+(16)(4 \sqrt{3})](https://tex.z-dn.net/?f=SA%3D64%2B%2816%29%284%20%5Csqrt%7B3%7D%29)
![SA=64+64 \sqrt{3}](https://tex.z-dn.net/?f=SA%3D64%2B64%20%5Csqrt%7B3%7D)
The exact form of the area of the square pyramid is ![64+64 \sqrt{3}](https://tex.z-dn.net/?f=64%2B64%20%5Csqrt%7B3%7D)
Substituting √3 = 1.732 in the above expression, we have;
![SA = 64 + 110.848](https://tex.z-dn.net/?f=SA%20%3D%2064%20%2B%20110.848)
![SA = 174.848](https://tex.z-dn.net/?f=SA%20%3D%20174.848)
Rounding off to one decimal place, we get;
![SA = 174.8](https://tex.z-dn.net/?f=SA%20%3D%20174.8)
Thus, the area of the square pyramid is 174.8 cm²
Answer:
First we need to calculate the are of each wall, since we alredy knew the length (l) and the width (w) which is the height of the wall in this case:
A = wl = 9 . 12 = 108 (ft²)
We also know that he painted 3 walls, we need to multiply our first result by 3, in other words, the area of wall that Brett painted is the sum of the area of three walls: 108 . 3 = 324 (ft²)
Answer:
Isolate the variable by dividing each side by factors that don't contain the variable.
x = w/3 - 2/3