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tankabanditka [31]

3 years ago

10

5.

1 answer:

Pepsi [2]3 years ago

7 0

Answer:

B. x = 2

Step-by-step explanation:

Answer:

x = 2

Step-by-step explanation:

check by substituting each given value of x into the function

x = 0 ⇒ y = × 0² = 0

hence (0, 0) is correct

x = 2 ⇒ y = × 4 = 3

hence (2, 4 ) is incorrect

x = 3 ⇒ y = × 9 = 6

hence (3, 6 ) is correct

x = 5 ⇒ y = × 25 = 18

hence (5, 18 ) is correct

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Will 100% give Brainiest to correct and help, please show work, needed answered asap

IgorLugansk [536]

Answer:

x^6/2

Step-by-step explanation:

simplify the expression, 4x^4x^3/8x, you make it 4x^7/8x

and it simplifys to 4x^6/8 which then becomes x^6/2

extra note:

a^x * a^y = a^ x+y

(see attachment below)

I hope this helps :)

3 0

3 years ago

kati45 [8]

Answer:

x = 3/2 ± $\sqrt{5}$ /2

Step-by-step explanation:

4x²-12x+9 = 5

Rearrange:

4x²-12x+4 = 0

Solve using the quadratic equation:

x = 3/2 ± $\sqrt{5}$ /2

8 0

2 years ago

A classroom contains an equal number of boys and girls. If 5 girls leave, twice as many boys and girls remain. What was the orig

lorasvet [3.4K]

The answer is 10 because 5 x 2 is 10

5 0

4 years ago

Which of the following are expressions? Check all that apply.

Ainat [17]

Answer:

3 + x

the difference of x and 8

Step-by-step explanation:

u need a number, a varible(letter), and a [+, -, /, x(*)]

5 0

3 years ago

The harmonic motion of a particle is given by f(t) = 2 cos(3t) + 3 sin(2t), 0 ≤ t ≤ 8. (a) When is the position function decreas

iren [92.7K]

For the last part, you have to find where $f'(t)$ attains its maximum over $0\le t\le8$ . We have

$f'(t)=-6\sin3t+6\cos2t$

so that

$f''(t)=-18\cos3t-12\sin2t$

with critical points at $t$ such that

$-18\cos3t-12\sin2t=0$

$3\cos3t+2\sin2t=0$

$3(\cos^3t-3\cos t\sin^2t)+4\sin t\cos t=0$

$\cos t(3\cos^2t-9\sin^2t+4\sin t)=0$

$\cos t(12\sin^2t-4\sin t-3)=0$

So either

$\cos t=0\implies t=\dfrac{(2n+1)\pi}2$

or

$12\sin^2t-4\sin t-3=0\implies\sin t=\dfrac{1\pm\sqrt{10}}6\implies t=\sin^{-1}\dfrac{1\pm\sqrt{10}}6+2n\pi$

where $n$ is any integer. We get 8 solutions over the given interval with $n=0,1,2$ from the first set of solutions, $n=0,1$ from the set of solutions where $\sin t=\dfrac{1+\sqrt{10}}6$ , and $n=1$ from the set of solutions where $\sin t=\dfrac{1-\sqrt{10}}6$ . They are approximately

$\dfrac\pi2\approx2$

$\dfrac{3\pi}2\approx5$

$\dfrac{5\pi}2\approx8$

$\sin^{-1}\dfrac{1+\sqrt{10}}6\approx1$

$2\pi+\sin^{-1}\dfrac{1+\sqrt{10}}6\approx7$

$2\pi+\sin^{-1}\dfrac{1-\sqrt{10}}6\approx6$

4 0

3 years ago