Answer:
3/8
Step-by-step explanation:
If 1/4 is equaled to 1. This is 1 dived by 4 to get the 1/4.
Then 1.5 would be divided by 4 to get 3/8
Answer:
2x
Step-by-step explanation:
12 + 2x - 12
2x + 12 - 12
12 - 12 = 0
Answer:
It is 6/15, or 0.4.
Step-by-step explanation:
There are 6 English speaking people in the class. There are a total of 15 children. That makes a fraction of 6/15, and a decimal of 0.4.
Answer: see proof below
<u>Step-by-step explanation:</u>
Use the Sum & Difference Identity: tan (A - B) = (tanA - tanB)/(1 + tanA tanB)
Use the Half-Angle Identity: tan (A/2) = (1 - cosA)/(sinA)
Use the Unit Circle to evaluate tan (π/4) = 1
Use Pythagorean Identity: cos²A + sin²A = 1
<u>Proof LHS → RHS</u>
![\text{Given:}\qquad \qquad \qquad\dfrac{2\tan\bigg(\dfrac{\pi}{4}-\dfrac{A}{2}\bigg)}{1+\tan^2\bigg(\dfrac{\pi}{4}-\dfrac{A}{2}\bigg)}](https://tex.z-dn.net/?f=%5Ctext%7BGiven%3A%7D%5Cqquad%20%5Cqquad%20%5Cqquad%5Cdfrac%7B2%5Ctan%5Cbigg%28%5Cdfrac%7B%5Cpi%7D%7B4%7D-%5Cdfrac%7BA%7D%7B2%7D%5Cbigg%29%7D%7B1%2B%5Ctan%5E2%5Cbigg%28%5Cdfrac%7B%5Cpi%7D%7B4%7D-%5Cdfrac%7BA%7D%7B2%7D%5Cbigg%29%7D)
![\text{Difference Identity:}\qquad \dfrac{2 \bigg( \frac{\tan\frac{\pi}{4}-\tan\frac{A}{2}}{1+\tan\frac{\pi}{4}\cdot \tan\frac{A}{2}}\bigg)}{1+ \bigg( \frac{\tan\frac{\pi}{4}-\tan\frac{A}{2}}{1+\tan\frac{\pi}{4}\cdot \tan\frac{A}{2}}\bigg)^2}](https://tex.z-dn.net/?f=%5Ctext%7BDifference%20Identity%3A%7D%5Cqquad%20%5Cdfrac%7B2%20%5Cbigg%28%20%5Cfrac%7B%5Ctan%5Cfrac%7B%5Cpi%7D%7B4%7D-%5Ctan%5Cfrac%7BA%7D%7B2%7D%7D%7B1%2B%5Ctan%5Cfrac%7B%5Cpi%7D%7B4%7D%5Ccdot%20%5Ctan%5Cfrac%7BA%7D%7B2%7D%7D%5Cbigg%29%7D%7B1%2B%20%5Cbigg%28%20%5Cfrac%7B%5Ctan%5Cfrac%7B%5Cpi%7D%7B4%7D-%5Ctan%5Cfrac%7BA%7D%7B2%7D%7D%7B1%2B%5Ctan%5Cfrac%7B%5Cpi%7D%7B4%7D%5Ccdot%20%5Ctan%5Cfrac%7BA%7D%7B2%7D%7D%5Cbigg%29%5E2%7D)
![\text{Substitute:}\qquad \qquad \dfrac{2 \bigg( \frac{1-\tan\frac{A}{2}}{1+\tan\frac{A}{2}}\bigg)}{1+ \bigg( \frac{1-\tan\frac{A}{2}}{1+\tan\frac{A}{2}}\bigg)^2}](https://tex.z-dn.net/?f=%5Ctext%7BSubstitute%3A%7D%5Cqquad%20%5Cqquad%20%5Cdfrac%7B2%20%5Cbigg%28%20%5Cfrac%7B1-%5Ctan%5Cfrac%7BA%7D%7B2%7D%7D%7B1%2B%5Ctan%5Cfrac%7BA%7D%7B2%7D%7D%5Cbigg%29%7D%7B1%2B%20%5Cbigg%28%20%5Cfrac%7B1-%5Ctan%5Cfrac%7BA%7D%7B2%7D%7D%7B1%2B%5Ctan%5Cfrac%7BA%7D%7B2%7D%7D%5Cbigg%29%5E2%7D)
![\text{Simplify:}\qquad \qquad \qquad \dfrac{1-\tan^2\frac{A}{2}}{1+\tan^2\frac{A}{2}}](https://tex.z-dn.net/?f=%5Ctext%7BSimplify%3A%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cdfrac%7B1-%5Ctan%5E2%5Cfrac%7BA%7D%7B2%7D%7D%7B1%2B%5Ctan%5E2%5Cfrac%7BA%7D%7B2%7D%7D)
![\text{Half-Angle Identity:}\qquad \quad \dfrac{1-(\frac{1-\cos A}{\sin A})^2}{1+(\frac{1-\cos A}{\sin A})^2}](https://tex.z-dn.net/?f=%5Ctext%7BHalf-Angle%20Identity%3A%7D%5Cqquad%20%5Cquad%20%5Cdfrac%7B1-%28%5Cfrac%7B1-%5Ccos%20A%7D%7B%5Csin%20A%7D%29%5E2%7D%7B1%2B%28%5Cfrac%7B1-%5Ccos%20A%7D%7B%5Csin%20A%7D%29%5E2%7D)
![\text{Simplify:}\qquad \qquad \dfrac{\sin^2 A-1+2\cos A-\cos^2 A}{\sin^2 A+1-2\cos A+\cos^2 A}](https://tex.z-dn.net/?f=%5Ctext%7BSimplify%3A%7D%5Cqquad%20%5Cqquad%20%5Cdfrac%7B%5Csin%5E2%20A-1%2B2%5Ccos%20A-%5Ccos%5E2%20A%7D%7B%5Csin%5E2%20A%2B1-2%5Ccos%20A%2B%5Ccos%5E2%20A%7D)
![\text{Pythagorean Identity:}\qquad \qquad \dfrac{1-\cos^2 A-1+2\cos A}{2-2\cos A}](https://tex.z-dn.net/?f=%5Ctext%7BPythagorean%20Identity%3A%7D%5Cqquad%20%5Cqquad%20%5Cdfrac%7B1-%5Ccos%5E2%20A-1%2B2%5Ccos%20A%7D%7B2-2%5Ccos%20A%7D)
![\text{Simplify:}\qquad \qquad \qquad \dfrac{2\cos A-2\cos^2 A}{2(1-\cos A)}\\\\.\qquad \qquad \qquad \qquad =\dfrac{2\cos A(1-\cos A)}{2(1-\cos A)}](https://tex.z-dn.net/?f=%5Ctext%7BSimplify%3A%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cdfrac%7B2%5Ccos%20A-2%5Ccos%5E2%20A%7D%7B2%281-%5Ccos%20A%29%7D%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D%5Cdfrac%7B2%5Ccos%20A%281-%5Ccos%20A%29%7D%7B2%281-%5Ccos%20A%29%7D)
= cos A
LHS = RHS: cos A = cos A ![\checkmark](https://tex.z-dn.net/?f=%5Ccheckmark)
Answer:
what are we trying to solve?
Step-by-step explanation: