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3 years ago

Help fast with the ones that are circled

Mathematics

1 answer:

Dmitry [639]3 years ago

3 0

Answer:

6. B) 0.23

7. B) 14

Step-by-step explanation:

6. To solve, change the fractions into decimals. To do so, try to get them to = 100, and then move the decimal point to the left two place values:

1/5 = 20/100 = 0.20

1/4 = 25/100 = 0.25

You are trying to find a decimal point that is in between these two decimals. In this case, your answers can be the following:

0.21, 0.22, 0.23, 0.24

B) 0.23 is your answer.

7. To solve, combine the ones that have already chosen, and subtract from the total. Find common denominators:

(1/3) + (1/5) = (5/15) + (3/15) = 8/15 x (2/2) = 16/30

16 of the students chose either station K or L.

There are 30 students in all. Subtract 16 from 30:

30 - 16 = 14

14 students chose station M

B) 14 is your answer.

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Will give Brainliest! if right!

Semenov [28]

I think the correct answer is

C) Suki's method is best because it uses a random sample

Glad I could help, and good luck!

AnonymousGiantsFan

8 0

3 years ago

Read 2 more answers

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

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

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

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