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Brilliant_brown [7]

2 years ago

10

Find the point on a circle that has a radius of 8 with an angle of rotation of 130 degrees centered at (-3,5)?

1 answer:

nikklg [1K]2 years ago

4 0

-2,8
Is it
Jusmfjfkddnfnfnfnjdf just to let me post :)

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I need help with this :( will mark brainliest

JulsSmile [24]

a. 20 because y is the amount of $$ in account now

b. 12 a $$

c. 260, bc 12*20= 240 240+20=260

6 0

2 years ago

Please help me!!!!!

Bas_tet [7]

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>

$\text{Given:}\qquad \qquad \qquad\dfrac{2\tan\bigg(\dfrac{\pi}{4}-\dfrac{A}{2}\bigg)}{1+\tan^2\bigg(\dfrac{\pi}{4}-\dfrac{A}{2}\bigg)}$

$\text{Difference Identity:}\qquad \dfrac{2 \bigg( \frac{\tan\frac{\pi}{4}-\tan\frac{A}{2}}{1+\tan\frac{\pi}{4}\cdot \tan\frac{A}{2}}\bigg)}{1+ \bigg( \frac{\tan\frac{\pi}{4}-\tan\frac{A}{2}}{1+\tan\frac{\pi}{4}\cdot \tan\frac{A}{2}}\bigg)^2}$

$\text{Substitute:}\qquad \qquad \dfrac{2 \bigg( \frac{1-\tan\frac{A}{2}}{1+\tan\frac{A}{2}}\bigg)}{1+ \bigg( \frac{1-\tan\frac{A}{2}}{1+\tan\frac{A}{2}}\bigg)^2}$

$\text{Simplify:}\qquad \qquad \qquad \dfrac{1-\tan^2\frac{A}{2}}{1+\tan^2\frac{A}{2}}$

$\text{Half-Angle Identity:}\qquad \quad \dfrac{1-(\frac{1-\cos A}{\sin A})^2}{1+(\frac{1-\cos A}{\sin A})^2}$

$\text{Simplify:}\qquad \qquad \dfrac{\sin^2 A-1+2\cos A-\cos^2 A}{\sin^2 A+1-2\cos A+\cos^2 A}$

$\text{Pythagorean Identity:}\qquad \qquad \dfrac{1-\cos^2 A-1+2\cos A}{2-2\cos A}$

$\text{Simplify:}\qquad \qquad \qquad \dfrac{2\cos A-2\cos^2 A}{2(1-\cos A)}\\\\.\qquad \qquad \qquad \qquad =\dfrac{2\cos A(1-\cos A)}{2(1-\cos A)}$

= cos A

LHS = RHS: cos A = cos A $\checkmark$

4 0

3 years ago

Will the square of a negative number always be positive? Explain why.

qwelly [4]

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...!!!

6 0

2 years ago

Read 2 more answers

Hey, Its the same person. This is Algebra 2. Please Help!

suter [353]

Answer:

first option

Step-by-step explanation:

Given

f(x) = $\frac{2x^2+5x-12}{x+4}$ ← factorise the numerator

= $\frac{(x + 4)(2x-3)}{x+4}$ ← 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 = $\frac{3}{2}$

There is a zero at ( $\frac{3}{2}$ , 0 )

Thus

discontinuity at (- 4, - 11 ), zero at ( $\frac{3}{2}$ , 0 )

7 0

2 years ago

What is 1.04 as a percent

zavuch27 [327]

If you convert the number on a percent,
then multiply the number by 100 with sign %

1.04 = 1.04 × 100% = <u>104%</u>

4 0

2 years ago

Read 2 more answers