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NNADVOKAT [17]
3 years ago
9

Find the value of x so that the polygons have the same perimeter.

Mathematics
1 answer:
Novosadov [1.4K]3 years ago
4 0

Answer: x=4

Step-by-step explanation:

The perimeter´s formula for this kind of polygons is

The triangle perimeter would be:

(x+2)+(x+3)+(x+3)

And the rectangle:

(x+2)+(x+2)+x+x

For them to have the same value, we should equal them:

(x+2)+(x+3)+(x+3)=(x+2)+(x+2)+x+x

Working with this:

3x+8=4x+4

x=4

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Answer:

The correct answer is option C.  x² - 4

Step-by-step explanation:

Area of rectangle  = Length * Width

It is given that ,

The length of a side of a rectangle is 2 units less than a number, and the width is 4 units more than the length.

Let 'x' be the number

Length of rectangle = x - 2

Width = Length + 4 = x - 2 + 4 = x + 2

<u>To find the area of rectangle</u>

Area = Length *  Width

Area = (x - 2)*(x + 2) = x² - 4

Therefore option C.  x² - 4 is the correct answer

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a designer wants to make a circular fountain inside a square of grass as shown at the right. What is a rule for the area A of th
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11.625

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The fish tank has side lengths 20in, 10in and height 15in. The water level is two inches below the top of the tank. A glass sphe
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Answer:

Part 1) The new distance from the water to the top of the tank is 1.979\ in

Part 2) The maximum number of balls that can be put into the tank with the tank not overflowing is 95

Step-by-step explanation:

step 1

Find the total volume of the tank

V=20*10*15=3,000\ in^{3}

step 2

Find the volume of the tank if the water level is two inches below the top of the tank

V=20*10*(15-2)=2,600\ in^{3}

step 3

Find the volume of the glass sphere

The volume of the glass sphere is equal to

V=\frac{4}{3}\pi r^{3}

we have

r=1\ in

assume

\pi=3.14

substitute

V=\frac{4}{3}(3.14)(1)^{3}

V=4.2\ in^{3}

step 4

What is the new distance from the water to the top of the tank?

we know that

2 in -----> represent (3,000-2,600)=400\ in^{3}

so

using proportion

Find how many inches correspond a volume of 4.2\ in^{3}

\frac{2}{400}\frac{in}{in^{3}}=\frac{x}{4.2}\frac{in}{in^{3}}\\ \\x=4.2*2/400\\ \\x=0.021\ in

The new distance from the water to the top of the tank is

2-0.021=1.979\ in

step 5

Find how many of these balls can be put into the tank with the tank not overflowing

we know that

The volume of one ball is equal to  4.2\ in^{3}

using proportion

\frac{1}{4.2}=\frac{x}{400}\\ \\x=400/4.2\\ \\x=95.23\ balls

therefore

The maximum number of balls that can be put into the tank with the tank not overflowing is 95

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