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3 years ago

8

7 x 3/4

1 answer:

Mkey [24]3 years ago

7 0

7 3/4

Step 1: Multiply 7 and 4.

7 x 4=28

Step 2:Add 3 to their product.

28+3=31

Step 3:The number so obtained is the numerator and the denominator remains the same.

Answer: 31/4

I think that is what you are looking for!

Hope that helps!!!

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Last year, a beauty magazine had a total of 2,650 subscribers. This year, after its parent company made some changes, it had 3,2

lilavasa [31]

Answer:

22%

Step-by-step explanation:

Divide 2,650 by 100:

2,650/100=26.5

Then divide 3233 by the dividend:

122=122%

Now you can see that there is a 22% percent increase over 100%, theefore making our answer 22%.

Have a great day, and mark me brainliest if I am most helpful!

5 0

3 years ago

What is the relationship between .04 and .004

IRINA_888 [86]

They are both decimals.

But 0.04 > 0.004.

4 0

4 years ago

Read 2 more answers

Is this the right answer!! :)

beks73 [17]

Answer:

Option D). Quadrant V is the correct answer.

Step-by-step explanation:

Your answer is correct.

Hope this helps!

4 0

2 years ago

Read 2 more answers

How can I use two unit multipliers to convert 42 centimeters to feet?

eimsori [14]

Answer:

1.38 ft

Step-by-step explanation:

42 cm x <u> 3.28 ft </u>

100 cm

= 1.38 ft

there are plenty of printable unit conversion on google

6 0

4 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

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

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