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3 years ago

Simon has a rectangular garden that is 10 feet wide. The perimeter of the garden is 50 feet. What is the area of Simon’s garden?

2 answers:

5 0

The area of the garden is 150 ft squared

Explanation:
The formula for perimeter is Length+length+width+width = 50 ft (in this case the perimeter is 50 ft, as stated in the question). We also know that the width of the garden is 10, so length + length + 10 + 10 = 50. Next we can turn it into an algebraic equation to find the length : x + x + 10 + 10 = 50

Solve the equation:
1) combine like terms
2x + 20 = 50

2) We want to isolate x by moving all of the other numbers to the other side of the equation. Since 20 is the only other number on the left side, we subtract 20 from both sides
2x + 20 - 20 = 50 - 20
2x = 30

3) Now, to find x, we want to just have one x on the left side instead of two, so we divide the equation by two to find that
2x/2 = 30/2
x = 15

Now we know that the length is 15 ft, and the width is 10 ft. Since the formula for finding area is length x width, we do 15 x 10 to get 150. Therefore, the area is 150 ft squared.

Have a nice day

enot [183]3 years ago

4 0

The correct answer is 150 sq.ft.
Hope this helps:)

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a pyramid whose altitude is 5ft weighs 800lbs. at what distance from its vertex must it be cut by a plane parallel to its base s

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A structure which has a square base and four triangular sides meeting at a point is called pyramid.

At distance of 3.97 feet from its vertex , pyramid is cut by plane so that the two solids of equal weight will be formed.

<u>It is assumed that weight of pyramid is proportional to its volume.</u>

So, $w = k V$ , where V is volume of original pyramid and v is volume of small pyramid and k is constant.

Let us consider that at h distance, pyramid is cut from its vertex. So a small pyramid is also formed.

Assume that base area of original pyramid is A and base of small pyramid is a.

Volume of original pyramid is, $V= \frac{1}{3} *A* 5$

So, weight of original pyramid, $W = k *(\frac{1}{3} *A* 5) = 800$

Volume of small pyramid is, $v = \frac{1}{3}* a* h$

So, weight of small pyramid, $w = k*(\frac{1}{3}* a* h)=400$

<u>Since, base and height of small pyramid and original pyramid are in proportion.</u>

So, $\frac{a}{A} = (\frac{h}{5}) ^{2}$

$a = (\frac{h}{5} )^{2}A$

Substituting value of a in equation $k*(\frac{1}{3}* a* h)=400$

So, $k*(\frac{1}{3}* \frac{h^{2} }{25}A * h)=400$

$(k*\frac{1}{3}* A*5)*\frac{1}{5} *\frac{h^{2} }{25} * h)=400$

Since, $(k *\frac{1}{3} *A* 5) = 800$ , substitute in above equation.

So, $800*\frac{h^{3} }{125}=400\\\\h^{3}=\frac{125}{2}\\\\h=\sqrt[\frac{1}{3} ]{62.5}\\\\h=3.97 feet$

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Step-by-step explanation:

Let x represent the number of children that swam in the public pool that day.

Let y represent the number of adult that swam in the public pool that day.

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