a. 20 because y is the amount of $$ in account now
b. 12 a $$
c. 260, bc 12*20= 240 240+20=260
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)
the square of a negative number always be positive
eg: -2² = -2×-2 = (-×- = +) +4
the cube of a negative number always be positive
eg: -2³ = -2×-2×-2 = (-×-×- = +) -8
hope it helps...!!!
Answer:
first option
Step-by-step explanation:
Given
f(x) =
← factorise the numerator
=
← cancel (x + 4) on numerator/ denominator
= 2x - 3
Cancelling (x + 4) creates a discontinuity ( a hole ) at x + 4 = 0, that is
x = - 4
Substitute x = - 4 into the simplified f(x) for y- coordinate
f(- 4) = 2(- 4) - 3 = - 8 - 3 = - 11
The discontinuity occurs at (- 4, - 11 )
To obtain the zero let f(x) = 0, that is
2x - 3 = 0 ⇒ 2x = 3 ⇒ x = ![\frac{3}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B2%7D)
There is a zero at (
, 0 )
Thus
discontinuity at (- 4, - 11 ), zero at (
, 0 )
If you convert the number on a percent,
then multiply the number by 100 with sign %
1.04 = 1.04 × 100% = <u>104%</u>