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andre [41]
3 years ago
6

The length of a rectangle is 8 feet more than its width. If the width is increased by 4 feet and the length is decreased by 5 fe

et, the area will remain the same. Find the dimensions of the original rectangle.
Mathematics
2 answers:
alexgriva [62]3 years ago
5 0
Let the width be x.

Length is 8 feet more than width. Length = x + 8

Area = x(x + 8)

width increased by 4, that is,  (x + 4)
Length decreased by 5,    (x + 8 - 5) = (x + 3)

Area  = (x + 4)(x +3)

Area remains the same

x(x + 8) = (x+4)(x +3)

x² + 8x =   x(x +3) + 4(x +3)

 x² + 8x =   x² +3x + 4x +12

x² + 8x =   x² +7x +12        Eliminate x² from both sides

8x = 7x + 12

8x - 7x = 12

x = 12

Dimensions of original rectangle :  x,  x + 8

12, 12 +8 =   12, 20

Original rectangle is   20 feet   by 12 feet 
kodGreya [7K]3 years ago
4 0

Answer:

12 feet by 20 feet

Step-by-step explanation:

Area = (x+4)(x+3)

x=12

x+8 = 12+8 = 20

Original triangle is 12feet by 20 feet

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a. Answer:  D: (∞, ∞)

                   R: (-∞, ∞)

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

Theoretical domain is the domain of the equation (without an understanding of what the x-variable represents).

Theoretical range is the range of the equation given the domain.

c(p) = 25p

There are no restrictions on the p so the theoretical domain is All Real Numbers.

Multiplying 25 by All Real Numbers results in the range being All Real Numbers.

a) D: (∞, ∞)

   R: (-∞, ∞)

*********************************************************************************

b. Answer:  D: (0, 200)

                    R: (0, 5000)

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

Practical domain is the domain of the equation WITH an understanding of what the x-variable represents.

Practical range is the range of the equation given the practical values of the domain.

The problem states that p represents the number of cups.  Since we can't have a negative amount of cups, p ≥ 0.  The problem also states that Bonnie will purchase a maximum of 200 cups. So, 0 ≤ p ≤ 200

The range is 25p →  (25)0 ≤ (25)p ≤ (25)200

                             →      0   ≤  25p  ≤  5000

b) D: (0, 200)

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5 0
3 years ago
at a small school, one male and one female are selected to represent the student body at the PTSCA meetings. if there are 41 mal
I am Lyosha [343]

It is given that there are 41 males and 48 females in the small school.

So, the number of ways a male student can be chosen from 41 males is ^{41}C_1

Likewise, the number of ways a female student can be chosen from 48 females is ^{48}C_1.

Thus, the total number of ways in which 2-person combinations are possible to represent the student body at the PTSAC meetings will be given by:

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7 0
3 years ago
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frutty [35]

Answer:

Step-by-step explanation:

Change the 2/3 to 4/6

Now the ratio becomes 3:4:6

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x = 40

Red = 3*40 =      120

Yellow = 4*40 = 160

Blue = 6 * 40  = 240

Total =               520

7 0
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6

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3 years ago
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