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3 years ago

Can anyone explain how to find the answer? for H

Mathematics

1 answer:

lesya692 [45]3 years ago

6 0

$\sqrt[3]{ \frac{27}{125} } = \sqrt[3]{27} / \sqrt[3]{125} \\ \sqrt[3]{27} / \sqrt[3]{125} = 3/5$

Final answer: 3/5

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What is the surface area to a cylinder

Maslowich

Answer:

A. 2(Pi)rh+2(Pi)r^2

Step-by-step explanation:

The surface area of a cylinder is the sum of the lateral area and the area of the two circular ends.

The lateral area is the product of the circumference of the cylinder and its height:

lateral area = 2πrh

The area of the two ends is twice the area of each of those circles, so is ...

total end area = 2(πr²)

Then the total surface area of a cylinder is ...

SA = 2πrh +2πr²

7 0

4 years ago

FrEeEeEeeEeEeeeE BrAINlIEsTtTtTtTttT ;) ;) ;)

AURORKA [14]

Answer:

Thank you so much.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 2 cos(t) + sin(2t),[0, π/2].

castortr0y [4]

Answer:

The absolute maximum is $\frac{3\sqrt 3}2$ and the absolute minimum value is $0.$

Step-by-step explanation:

Differentiate of $f$ both sides w.r.t. $t$ ,

$f(t)=2 \cos t+\sin 2t$

$\Rightarrow f'(t)=-2\sin t+2\cos 2t$

Now take $f'(t)=0$

$\Rightarrow -2\sin t+2\cos 2t=0$

$\Rightarrow 2\cos 2t=2\sin t$

$\Rightarrow \cos 2t=\sin t$

$\Rightarrow 1-2\sin ^2t =\sin t \quad \quad [\because \cos 2t = 1-2\sin ^2t]$

$\Rightarrow 2\sin ^2t+\sin t-1=0$

$\Rightarrow 2\sin ^2t+2\sin t-\sin t-1=0$

$\Rightarrow 2\sin t(\sin t+1)-1(\sin t+1)=0$

$\Rightarrow (\sin t+1)(2\sin t-1)=0$

$\Rightarrow \sin t+1=0 \;\text{and}\; 2\sin t-1=0$

$\Rightarrow \sin t =-1 \;\text{and}\; \sin t =\frac 12$

In the interval $0\leq t\leq \frac {\pi}2$ , the answer to this problem is $\frac {\pi}6$

Now find the second derivative of $f(t)$ w.r.t. $t$ ,

$f''(t)=-2\cos t-4\sin 2t$

$\Rightarrow \left[f''(t)\right]_{t=\frac {\pi}6}=-2\times \frac {\sqrt 3}2-4\times \frac{\sqrt 3}2=-3\sqrt 3$

Thus, $f(t)$ is maximum at $t=\frac {\pi}6$ and minimum at $t=0$

$\left[f(t)\right]_{t=\frac {\pi}6}=2\times \frac {\sqrt 3}2+\frac{\sqrt 3}2=\frac{3\sqrt 3}2\;\text{and}\;\left[f(t)\right]_{t=\frac{\pi}2}= 2\times 0+0=0$

Hence, the absolute maximum is $\frac{3\sqrt 3}2$ and the absolute minimum value is $0$ .

7 0

3 years ago

18,19,20,21,23 please help me. I really need this answer.please help me.

sp2606 [1]

Answer:

18. Supplementary

19. Vertical

20. x = 7

23. x = 25

Step-by-step explanation:

Use angle relationships to make equations which you can solve for x.

Example:

18. These two angles form a line. These are called supplementary.

19. These angles are across a vertex from each other and are equal or the same.

20. 9x + 1 = 7x - 9 + 5x - 11

9x + 1 = 12x - 20

9x - 9x + 1 = 12x - 9x - 20

1 = 3x - 20

1 + 20 = 3x - 20 + 20

21 = 3x

7 = x

23. These two angles are across a vertex from each other. This means the angles are equal. Think of it as a reflection. If you folded the paper on that vertex, the lines would be in the exact same place.

So set the two expressions equal to each other. Then use inverse operations to isolate x and solve.

5x + 4 = 8x - 71

5x - 5x + 4 = 8x - 5x - 71

4 = 3x - 71

4 + 71 = 3x - 71 + 71

75 = 3x

25 = x

7 0

3 years ago

How do you do this I need help with this now

larisa86 [58]

Answer:

you have to do, x axis times y axis which should give you your answer.

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers