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3 years ago

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Which of the following points is in the solution set of y< x2 - X-6? -2,-1 0,-5 3,0

1 answer:

blondinia [14]3 years ago

4 0

Answer:

X-6 is the answer it's so easy

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A pulley is used to pull a boat to the dock. The rope is attached to the boat 7 feet below the level of the pulley. What is the

Svetradugi [14.3K]

Answer: 24 ft

Step-by-step explanation:

By the Pythagorean theorem,

$7^{2}+x^{2}=25^{2} \\ \\ 49+x^{2}=625 \\ \\ x^{2}=576 \\ \\ x=\boxed{24 \text{ ft}}$

7 0

2 years ago

At its lowest point the Euro tunnel is 115m below sea level. At this point, the tunnel is 50m below the sea bed. How deep is the

sweet-ann [11.9K]

Euro tunnel : - 115 (115 m below sea level of 0)

-115 = S - 50
-115 + 50 = S
- 65 = S....the sea bed is -65 m (or 65 m below sea level

8 0

3 years ago

Help me with this some

ch4aika [34]

Answer:

1:408

2:279

3:216

4:345

5:152

6:360

8:558

9:546

10:567

Step-by-step explanation:

YOUR WELCOME <3

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

Adjacent angles have no common interior points

qwelly [4]

<span>They share a vertex and side, but do not overlap

Hope this helps</span>

4 0

4 years ago