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3 years ago

Easy question!

Mathematics

1 answer:

Zielflug [23.3K]3 years ago

4 0

Answer: There were 38 cans in each box.

(You probably already knew the answer, but I'm bored so I'll explain it anyway).

To find the answer, you'd have to add up the total amount of cans brought into Ms. Walker's class. Since the class is composed of both girls AND boys, you'd have to add the two together.

57+57=114

Since there are three boxes, and the cans are all collectively being divided into the boxes evenly, you'd divide 114 by 3.

114/3= 38.

So, there would be 38 cans per box.

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Express tan B as a fraction in simplest form.

liberstina [14]

Answer:

4/3

Step-by-step explanation:

Tan B is defined as the opposite side divided by the adjacent side. 5 is the hypotenuse, 4 is the opposite side of angle B, and 3 is the adjacent side of angle B

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3 years ago

Explain how to get the monthly rate like I have no idea. Calculate: Monthly Rate

LekaFEV [45]

Step by step explanation:

1. First subtract the asset's salvage from the cost value.

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3 years ago

Round the nine to the nearest hundred, (947.362)

DochEvi [55]

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2 years ago

A ladder 15 feet from the base of a building reaches a window that is 35 feet high. What is the length of the ladder to the near

nikdorinn [45]

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15+b=35
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7 0

2 years ago

Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0

3 years ago