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diamong [38]
2 years ago
3

A map of a town has no scale. The actual distance between the school and the mall in the town is 5 mi., which is shown on the ma

p as 2 in.
The distance from the school to the library on the map is 1 1/2 in. What is the actual distance in miles between the school and the library?

A. 3 3/4 mi.
B. 4 mi.
C. 6 2/3 mi.
D. 15 mi.
Mathematics
1 answer:
vodomira [7]2 years ago
3 0

Answer:

Me personaly I think its B but it could be D

Step-by-step explanation:

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It's $6.48 divided by 4. So the answer is $1.62
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Lines s and t are perpendicular. If the slope of line s is -5, what is the slope of line ?
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Answer:

B: 1/5

Step-by-step explanation:

If the lines are perpendicular, they have negative reciprocal slopes

s has a slope of -5

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1000

1000(344%) = 34400
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You usually buy a 5.45.4 ounce bottle of lotion. There is a new bottle that says it gives you 20%20% more free.
valentina_108 [34]

Answer: The required equation is,

x = \frac{120\times 5.4}{100}

Step-by-step explanation:

Let x be the size of the larger bottle,

Since, the size of smaller bottle of lotion = 5.4 ounces,

According to the question,

The larger bottle gives 20% more free lotion,

The size of the larger bottle = The size of smaller bottle of lotion  + 20 % of the size of smaller bottle of lotion

= 120 % of the size of smaller bottle of lotion

= 120 % of 5.4

=\frac{120\times 5.4}{100}

\implies x = \frac{120\times 5.4}{100}

Which is the required equation that could be used to find the size of the larger bottle.

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2 years ago
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A waste management company is designing a rectangular construction dumpster that will be twice as long as it is wide and must ho
photoshop1234 [79]

Answer:

The dimensions that minimize the surface are:

Wide: 1.65 yd

Long: 3.30 yd

Height: 2.20 yd

Step-by-step explanation:

We have a rectangular base, that its twice as long as it is wide.

It must hold 12 yd^3 of debris.

We have to minimize the surface area, subjet to the restriction of volume (12 yd^3).

The surface is equal to:

S=2(w*h+w*2w+2wh)=2(3wh+2w^2)

The volume restriction is:

V=w*2w*h=2w^2h=12\\\\h=\frac{6}{w^2}

If we replace h in the surface equation, we have:

S=2(3wh+2w^2)=6w(\frac{6}{w^2})+4w^2=36w^{-1}+4w^2

To optimize, we derive and equal to zero:

dS/dw=36(-1)w^{-2} + 8w=0\\\\36w^{-2}=8w\\\\w^3=36/8=4.5\\\\w=\sqrt[3]{4.5} =1.65

Then, the height h is:

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The dimensions that minimize the surface are:

Wide: 1.65 yd

Long: 3.30 yd

Height: 2.20 yd

8 0
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