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2 years ago

12

Find the lateral surface area of the cylinder to two decimal places. will mark as brainiest

1 answer:

lidiya [134]2 years ago

5 0

Answer:

Step-by-step explanation:

lateral surface area=2πrh

=2π×2×9

=36π

≈36×3.14

≈130.04 sq.ft

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What is the value of 3 to the power 2 over 3 to the power 4? A. 1 over 81 B. 1 over 27 C. 1 over 9 D. 1 over 3

rodikova [14]

One way you could solve this is to just multiply the top and bottom out so that you get 9/81, reducing it by 9/9 to get 1/9 or option C.
Another way would be to do $3^{(2-4)}$ since dividing numbers with exponents would be subtracting the bottom exponent from the top exponent, provided that the base number (in this case 3) is the same for both. For this method, you would get $3^{-2}$ , which is equal to 1/9 or .1 repeating, the same answer that you'd get with the first method.

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3 years ago

Read 2 more answers

Daria earns $17.50 per customer when she gives haircuts if she gave X haircuts on Monday and Y haircuts on Tuesday which express

fomenos

Answer:

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3 years ago

In the pulley system shown in this figure, MQ = 30 mm, NP = 10 mm, and QP = 21 mm. Find MN.

ankoles [38]

<h2>Answer:</h2>

$\boxed{\overline{MN}=37.96}$

<h2>Step-by-step explanation:</h2>

For a better understanding of this problem, see the figure below. Our goal is to find $\overline{MN}$ . Since:

$\angle MRS=\angle MQP=90^{\circ} \\ \\ \overline{MQ}=\overline{MR}=30mm$

and $\overline{MN}$ is a common side both for ΔMRN and ΔMQN, then by SAS postulate, these two triangles are congruent and:

$\overline{RN}=\overline{QN}$

By Pythagorean theorem, for triangle NQP:

$\overline{QN}=\sqrt{\overline{NP}^2+\overline{QP}^2} \\ \\ \overline{QN}=\sqrt{10^2+21^2} \\ \\ \overline{QN}=\sqrt{541}$

Applying Pythagorean theorem again, but for triangle MQN:

$\overline{MN}=\sqrt{\overline{MQ}^2+\overline{QN}^2} \\ \\ \overline{MN}=\sqrt{30^2+(\sqrt{541})^2} \\ \\ \boxed{\overline{MN}=37.96}$

3 0

3 years ago

Which is a solution of the system of equations shown?

taurus [48]

Answer:

D. (2, 0)

Step-by-step explanation:

The solutions are the two points of intersection of the graphs:

(-2, -4) and (2, 0)

The latter of these corresponds to choice D, the one you have marked.

5 0

3 years ago

On a coordinate plane, the segment with endpoints (10, 40) and (70, 120) is

OlgaM077 [116]

Answer:

The length of the resulting segment is 500.

Step-by-step explanation:

Vectorially speaking, the dilation is defined by following operation:

$P'(x,y) = O(x,y) + k\cdot [P(x,y)-O(x,y)]$ (1)

Where:

$O(x,y)$ - Center of dilation.

$P(x,y)$ - Original point.

$k$ - Scale factor.

$P'(x,y)$ - Dilated point.

First, we proceed to determine the coordinates of the dilated segment:

( $P(x,y) = (10, 40)$ , $Q(x,y) = (70, 120)$ , $O(x,y) = (0,0)$ , $k = 5$ )

$P'(x,y) = O(x,y) + k\cdot [P(x,y)-O(x,y)]$

$P(x,y) = (0,0) +5\cdot [(10,40)-(0,0)]$

$P'(x,y) = (50,200)$

$Q'(x,y) = O(x,y) + k\cdot [Q(x,y)-O(x,y)]$

$Q' (x,y) = (0,0) +5\cdot [(70,120)-(0,0)]$

$Q'(x,y) = (350, 600)$

Then, the length of the resulting segment is determined by following Pythagorean identity:

$l_{P'Q'} = \sqrt{(350-50)^{2}+(600-200)^{2}}$

$l_{P'Q'} = 500$

The length of the resulting segment is 500.

3 0

2 years ago