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Dovator [93]
3 years ago
7

A Square was altered so that one side is increased by 9 inches in the other side is decreased by 2 inches.The area of the result

ing rectangle is 60 in.² what was the area of the original Square?
Mathematics
1 answer:
LekaFEV [45]3 years ago
4 0

Let s represent the length of any one side of the original square.  The longer side of the resulting rectangle is s + 9 and the shorter side s - 2.

The area of this rectangle is (s+9)(s-2) = 60 in^2.

This is a quadratic equation and can be solved using various methods.  Let's rewrite this equation in standard form:  s^2 + 7s - 18 = 60, or:

s^2 + 7s - 78 = 0.  This factors as follows:  (s+13)(s-6)=0, so that s = -13 and s= 6.  Discard s = -13, since the side length cannot be negative.  Then s = 6, and the area of the original square was 36 in^2.

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So,

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the radius is 11.5 so

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the circumference is 72.2566

Step-by-step explanation:


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Pani-rosa [81]
20000*0.45 = 9000 in the bond
20000*0.15 = 3000 in the CD
20000*0.20 = 4000 in stocks
20000*0.029 = 580 in savings

A=9000(1 + 4.35%)^3 = 10,226.33
A=3000(1 + 2.90%)^3 = 3,268.64
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Suppose the total area is represented by the equation x^2+14x+24. If the area 1 is x^2 square units and area IV is 24 units, wha
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Answer:

See Explanation

Step-by-step explanation:

Given

Total = x^2 + 14x + 24

Area\ I = x^2

Area\ IV = 24

Required

True about area II and III

The question is incomplete; so, I will answer using general knowledge.

If the total area is:

Total = x^2 + 14x + 24

And there are 4 areas, then;

Total = Area\ I + Area\ II + Area\ III + Area\ IV

Rewrite as:

Total = Area\ I + Area\ IV+ Area\ II + Area\ III

This gives:

x^2 + 14x + 24 = x^2 + 24+ Area\ II + Area\ III

Collect like terms

x^2 -x^2 + 14x + 24 -24=  Area\ II + Area\ III

14x =  Area\ II + Area\ III

Rewrite as:

Area\ II + Area\ III  = 14x

<em>So, the sum of areas III and IV must be 14x</em>

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Step-by-step explanation:

80/150=160/300

160/300=53.33/100

The decrease in calories is 53.3%

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