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Sholpan [36]
4 years ago
12

Chance knows the length of a football field (including end zones) is 120 yards. The width of the field is 53.33 yards. What is t

he area of football field rounded appropriately?
Mathematics
1 answer:
givi [52]4 years ago
4 0

Answer = about 6,400 yards squared

Step-by-step explanation:

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3 years ago
Jason designs a rectangular sandbox. He models the perimeter of the sandbox using the expression 8l+2, where l is the length of
inn [45]

Answer:

A   .The expression 2l+2(3l+1) shows the width is 1 more than 3 times the length.

Step-by-step explanation:

p = 8l + 2

Length = l

Width:

2w = p - 2l

2w = (8l + 2) - 2l

2w = 6l + 2

w = 3l + 1

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3 years ago
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Answer: a = 23, b = 21, c = 136

Step-by-step explanation:

8 0
3 years ago
2. What are the measures of
hodyreva [135]
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6 0
4 years ago
Find the limit, if it exists, or type dne if it does not exist.
Phantasy [73]
\displaystyle\lim_{(x,y)\to(0,0)}\frac{\left(x+23y)^2}{x^2+529y^2}

Suppose we choose a path along the x-axis, so that y=0:

\displaystyle\lim_{x\to0}\frac{x^2}{x^2}=\lim_{x\to0}1=1

On the other hand, let's consider an arbitrary line through the origin, y=kx:

\displaystyle\lim_{x\to0}\frac{(x+23kx)^2}{x^2+529(kx)^2}=\lim_{x\to0}\frac{(23k+1)^2x^2}{(529k^2+1)x^2}=\lim_{x\to0}\frac{(23k+1)^2}{529k^2+1}=\dfrac{(23k+1)^2}{529k^2+1}

The value of the limit then depends on k, which means the limit is not the same across all possible paths toward the origin, and so the limit does not exist.
8 0
3 years ago
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