Tow ratios that have the same value when simplified.
Hope this helps!!;)
Answer:
700.4 cm
Step-by-step explanation:
This involves two similar triangles.
Both triangles are right triangles.
One has legs measuring 1 cm and 30 cm. We can find the hypotenuse by using the Pythagorean theorem.
(1 cm)^2 + (30 cm)^2 = c^2
c^2 = 901 cm^2
c = sqrt(901) cm
The second triangle has one leg with length 700 cm. This leg corresponds to the 30-cm leg in the other triangle. Since the triangles are similar, we can use a proportion to find the hypotenuse of the second triangle.
(30 cm)/(700 cm) = [sqrt(901) cm]/x
3/70 = sqrt(901) cm/x
3x = 70 * sqrt(901) cm
x = 70 * sqrt(901) cm/3
x = 700.4 cm
Answer: 700.4 cm
Answer:
lets talk first
Step-by-step explanation:
If you are asking for the highest number possible to be created form 1 through 9 while using each number once then the highest number is 98
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
