Step-by-step explanation:
Solve for a:
Step 1: Multiply both sides by b.
b=a
Step 2: Flip the equation.
a=b
---------------------------
Solve for b:
Step 1: Multiply both sides by b.
b=a
Answer:
15 inches
Step-by-step explanation:
Volume of a squre based pyramid :
V = a²(h/3)
a = base edge ; h = height
h =1/2 a
V = 4500
4500 = a² * (1/2a) ÷ 3
4500 = a² * a/2 * 1/3
4500 = a³/2 * 1/3
4500 = a³ / 6
27000 = a³
27000 = a³
Take the cube root of both sides
(27000)⅓ = a
30 = a
Recall:
h = 1/2a
h = 1/2(30)
h = 15 inches
Answer:
<h2>it's b</h2><h2>you can go with it</h2>
...........
Answer:
g(1) = 0
There is a remainder of 10 so you can automatically cross out 3 and 4.
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π → A + B = π - C
→ B + C = π - A
→ C + A = π - B
→ C = π - (B + C)
Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]
Use the Sum/Difference Identity: cos (A - B) = cos A · cos B + sin A · sin B
Use the Double Angle Identity: sin 2A = 2 sin A · cos A
Use the Cofunction Identity: cos (π/2 - A) = sin A
<u>Proof LHS → Middle:</u>





![\text{Factor:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{4}\bigg)\bigg]](https://tex.z-dn.net/?f=%5Ctext%7BFactor%3A%7D%5Cquad%20%20%3D2%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Cbigg%5B%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B4%7D%5Cbigg%29%2B%5Csin%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Cbigg%5D)
![\text{Cofunction:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi}{2}-\dfrac{A+B}{4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{2\pi-(A+B)}{4}\bigg)\bigg]](https://tex.z-dn.net/?f=%5Ctext%7BCofunction%3A%7D%5Cquad%20%20%3D2%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Cbigg%5B%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B4%7D%5Cbigg%29%2B%5Ccos%20%5Cbigg%28%5Cdfrac%7B%5Cpi%7D%7B2%7D-%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Cbigg%5D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D2%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B4%7D%5Cbigg%29%2B%5Ccos%20%5Cbigg%28%5Cdfrac%7B2%5Cpi-%28A%2BB%29%7D%7B4%7D%5Cbigg%29%5Cbigg%5D)
![\text{Sum to Product:}\ 2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[2 \cos \bigg(\dfrac{2\pi-2B}{2\cdot 4}\bigg)\cdot \cos \bigg(\dfrac{2A-2\pi}{2\cdot 4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)](https://tex.z-dn.net/?f=%5Ctext%7BSum%20to%20Product%3A%7D%5C%202%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Cbigg%5B2%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B2%5Cpi-2B%7D%7B2%5Ccdot%204%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B2A-2%5Cpi%7D%7B2%5Ccdot%204%7D%5Cbigg%29%5Cbigg%5D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D4%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B%5Cpi-B%7D%7B4%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B%5Cpi%20-A%7D%7B4%7D%5Cbigg%29)

LHS = Middle 
<u>Proof Middle → RHS:</u>

Middle = RHS 