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gulaghasi [49]
2 years ago
5

Suppose the length and the width of the sandbox are doubled.

Mathematics
1 answer:
zalisa [80]2 years ago
3 0

The percentage change in perimeter is 100%.

The percentage change in area is 300%

<h3>How to find perimeter and area?</h3>

The perimeter of the sandbox = 2l + 2w

where

  • l = length
  • w = width

Therefore,

perimeter = 2(10) + 2(6)

perimeter = 20 + 12 = 32 ft

The area of the sand box = lw

area = 10 × 6 = 60 ft²

If the length and width are doubled,

length = 2(10) = 20 ft

width = 2(6) = 12 ft

hence,

perimeter = 2(20) + 2(12) = 40 + 24 = 64 ft

area = 12 × 20 = 240 ft²

Therefore,

percentage change in perimeter = 64 - 32 / 32 × 100 = 100%

percentage change in area = 240 - 60 / 60 × 100 = 18000 / 60 = 300%

learn more on perimeter and area here: brainly.com/question/15322533

#SPJ1

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1.666667 headphones per hour (simplify this)

Step-by-step explanation:

divide the top and bottom by 3 and you should get your answer

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2 years ago
Is this answer correct
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Yes, that looks right.
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I have attached an image of the process I used

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3 years ago
Read 2 more answers
Point Coordinate
telo118 [61]

Step 1:

First, let's order the number from least to greatest.

A. -0.31

B. -0.33

C. -0.38

D. -0.29

We could see that A, B, and C's numbers are already from least to greatest, so all we have to do is to move -0.29 (D) up to the top, since -0.29 is the number with the lowest value.

C. -0.38

B. -0.33

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D. -0.29

Step 2:

Let's look at the four number lines. Remember that we already organized the numbers from least to greatest, so the order they go in on the number line is: C, B, A, and D. Since Line C and D Don't follow that rule, we know those are not our answers. Also on line B, C is on the left of -0.4, but C's value is less than -0.4, so B is wrong. We are left with line A.

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3 years ago
A 46 gram sample of a substance that is used to sterilize surgical instruments has a k-value of 0.1374. Find the substance's hal
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Answer:

  t=5.0days

Step-by-step explanation:

Using the formula for the exponential decay that is N=N_{0}e^{-kt}, we have N=\frac{1}{2}{\times}46=23, N_{0}=46 and k=0.1374.

Thus, N=N_{0}e^{-kt} becomes

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\frac{23}{46}=e^{-0.1374t}

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ln(\frac{1}{2})={-0.1374t}

t=\frac{ln\frac{1}{2}}{-0.1}

t=\frac{-0.6931}{-0.1374}

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3 years ago
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