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ratelena [41]
3 years ago
13

The area of rectangle R is 48. If its sides are whole numbers, then its perimeter cannot be__?

Mathematics
1 answer:
aev [14]3 years ago
6 0

An odd number. That's because the formula for the perimeter is

P = 2* (L + W)

Multiplying by 2 is never going to produce an odd number.  The question is much trickier than it looks. What happens when you have a square with sides of 7.

The area = s^2

Area = 7 * 7

Area = 49 and that's odd. But can you get an odd perimeter? The answer is no. The perimeter of a square is 4s

P = 4s. If s = 7

P = 4*7

P = 28 which is even. If there is another answer, I'm unaware of it.

A: P = 2*(6 + 8) = 2*14 = 28 A is a correct Perimeter.

B: P = 2*(3 + 16) = 2*19 = 38 B is a correct Perimeter.

C:P = 2*? Can't find anything. C is the answer.

D:P = 2*(1 + 48) = 2*49 = 98. D is correct.

Answer: C

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Answer:

#4: (r,t) = (1,6) #5: (x, y) = (-\frac{1}{2},4) #6: (x, y) = (3, 5) #7: (p, r) = (10,\frac{14}{3} )

Step-by-step explanation:

#4:

Solve for t.

t = 5 + r

Substitute the given value of t into equation

-2r + 5 + r = 4

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r = 1

substitute the given value of r into equation

t = 5 + 1

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t = 6

solution is ordered pair (r, t) = (1,6)

#5:

Solve for y.

y = 2 - 4x

Substitute the given value of y into equation

2x - (2- 4x) = -5

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substitute the given value of x into equation

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solution is ordered pair (x, y) = (-\frac{1}{2},4)

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Solve for x

x = -12 + 3y

Substitute the given value of x into equation

2 (-12 + 3y) + y = 11

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substitute the given value of y into equation

x = -12 + 3 x 5

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x = 3

solution is ordered pair (x, y) = (3, 5)

#7

Solve for r

r = \frac{14}{3}

Substitute the given value of  into equation

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Solve for p

p = 10

solution is ordered pair (p, r) = (10,\frac{14}{3} )

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2 years ago
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