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Mashutka [201]
3 years ago
6

Mrs. Fuller is trying to organize all of her children's toys into a large toy box. Each toy

Mathematics
1 answer:
djverab [1.8K]3 years ago
5 0

Answer: 1,152 toys can be stored in the box.

Step-by-step explanation:

For an object with measures L (length) by W (width) by H (height) the volume is calculated as:

V = L*W*H

We know that the measures of each toy are: 4in x 3in x 3in

Then the volume of each toy is:

V = 4in*3in*3in = 36in^3

The toy box is 4ft x 2ft x 3 ft

First we should write the measures in the same units as before, we know that:

1ft = 12 in.

Then:

4ft = 4*12 in = 48 in

2ft = 2*12in = 24in

3ft = 3*12in = 36in

Then the measures of the box are:

48in x 24in x 36in

Then the volume of the box is:

V = 48in*24in*36in = 41,472 in^3

Now, the number of toys that can be stored in the box is the same as the number of times that  36in^3 is in 41,472 in^3

this is equal to the quotient between 41,472 in^3 and 36in^3

N = 41,472 in^3/36in^3 = 1,152

1,152 toys can be stored in the box.

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One of the vertices of an equilateral triangle is on the vertex of a square and two other vertices are on the not adjacent sides
Elina [12.6K]
<h2>Answer:</h2>

<em> The side of the triangle is either 38.63ft or 10.35ft</em>

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

This problem can be translated as an image as shown in the Figure below. We know that:

  • The side of the square is 10 ft.
  • One of the vertices of an equilateral triangle is on the vertex of a square.
  • Two other vertices are on the not adjacent sides of the same square.

Let's call:

Since the given triangle is equilateral, each side measures the same length. So:

x: The side of the equilateral triangle (Triangle 1)

y: A side of another triangle called Triangle 2.

That length is the hypotenuse of other triangle called Triangle 2. Therefore, by Pythagorean theorem:

\mathbf{(1)} \ x^2=100+y^2

We have another triangle, called Triangle 3, and given that the side of the square is 10ft, then it is true that:

y+(10-y)=10

Therefore, for Triangle 3, we have that by Pythagorean theorem:

(10-y)^2+(10-y)^2=x^2 \\ \\ 2(10-y)^2=x^2 \\ \\ \\ \mathbf{(2)} \ x^2=2(10-y)^2

Matching equations (1) and (2):

2(10-y)^2=100+y^2 \\ \\ 2(100-20y+y^2)=100+y^2 \\ \\ 200-40y+2y^2=100+y^2 \\ \\ (2y^2-y^2)-40y+(200-100)=0 \\ \\ y^2-40y+100=0

Using quadratic formula:

y_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ y_{1,2}=\frac{-(-40) \pm \sqrt{(-40)^2-4(1)(100)}}{2(1)} \\ \\ \\ y_{1}=37.32 \\ \\ y_{2}=2.68

Finding x from (1):

x^2=100+y^2 \\ \\ x_{1}=\sqrt{100+37.32^2} \\ \\ x_{1}=38.63ft \\ \\ \\ x_{2}=\sqrt{100+2.68^2} \\ \\ x_{2}=10.35ft

<em>Finally, the side of the triangle is either 38.63ft or 10.35ft</em>

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