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3 years ago

9

Simplify.

2 answers:

geniusboy [140]3 years ago

5 0

Answer:

4x + 4

Step-by-step explanation:

di ako sure pero i tried

dusya [7]3 years ago

3 0

Answer:

1/4 - 4x

Step-by-step explanation:

hope this helped

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Determine the point estimate of the population proportion, the margin of error for the following confidence interval, and the

PilotLPTM [1.2K]

Answer:

point of estimate p = x/n

Step-by-step explanation:

p is the point estimate for the population , x is the number of success, n is the sample size.

p = Upper bound + Lower bond/ 2 for population proportion

x = Upper bond + Lower bond /2 for population mean

E = Upper bond - Lower bond /2 E is margin of error

4 0

3 years ago

AB is tangent to the circle at B. <A=14. Arc BC = 112.

goldenfox [79]

So hmm check the picture below

a circle has a total of 360°, thus arcBC + x + arcCD is 360, so arcCD is just 360 - 112 - x, whatever "x" was that you found

3 0

3 years ago

∠A and ∠B are adjacent. The sum of their measures is 92∘. ∠A measures (2x+5)∘. ∠B is three times the size of ∠A.

Ilia_Sergeevich [38]

Answer:

Equation: 2x + 5 + 6x + 15 = 92

Solution: x = 9

m∠A =23° m∠B = 69°

Step-by-step explanation:

measure of ∠B = 3(2x + 5) = 6x + 15

The sum of the angles = 92 = 2x + 5 + 6x + 15

92 = 8x + 20

72 = 8x

x = 9

m∠A = 2(9) + 5 = 18 + 5 = 23

m∠B = 6(9) + 15 = 54 + 15 = 69

Check: 23 + 69 = 92 and 23(3) = 69

5 0

3 years ago

If KJ=4 and KM= 2, find ML

adelina 88 [10]

Answer:

6

Step-by-step explanation:

(whole secant) * (external part) = (tangent)^2

(KL) * (KM) = KJ^2

(KM+ ML) * KM = KJ^2

(2+ ML) * 2 = 4^2

(2+ ML) * 2 =16

Divide each side by 2

(2+ ML) = 8

Subtract 2 from each side

ML = 6

7 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