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3 years ago

The area of a soccer field is 56 square yards. The length of one side of the field is 8 yards. What is the width of the field?

Mathematics

2 answers:

Dmitrij [34]3 years ago

7 0

Answer:

7 yards

Step-by-step explanation:

goldfiish [28.3K]3 years ago

4 0

Answer:

Step-by-step explanation:

you divide 56 by 8 and you get 7

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Write a polynomial that represents the area of the square.

Tema [17]

Answer:

The answer to your question is Area = 49n² - 70n + 25

Step-by-step explanation:

Data

Length of the side = 7n - 5

Formula

Area of a square = side x side

Process

1.- Substitute values

Area = (7n - 5)(7n - 5)

2.- Expand

Area = 49n² - 35n - 35n + 25

3.- Simplify

Area = 49n² - 70n + 25

4.- The polynomial is 49n² - 70n + 25

6 0

3 years ago

Read 2 more answers

( please help, 20 points!! )

deff fn [24]

Measure of anfle 5=150

7 0

3 years ago

What is the height of the cylinder in the diagram? Round your answer to the nearest whole number.

ryzh [129]

Answer:

$\mathrm{78.96 \: cm}$

Step-by-step explanation:

Cylinder volume formula is $\pi r^2h$

$\pi \times 6^2 \times h=904.78$

$36\pi \times h=904.78$

$h=904.78 \div 36\pi$

$h=78.95695$

5 0

4 years ago

Read 2 more answers

The average value of a function f over the interval [−2,3] is −6 , and the average value of f over the interval [3,5] is 20. Wha

Xelga [282]

Answer:

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

Step-by-step explanation:

Let suppose that function $f$ is continuous and integrable in the given intervals, by integral definition of average we have that:

$\frac{1}{3-(-2)} \int\limits^{3}_{-2} {f(x)} \, dx = -6$ (1)

$\frac{1}{5-3} \int\limits^{5}_{3} {f(x)} \, dx = 20$ (2)

By Fundamental Theorems of Calculus we expand both expressions:

$\frac{F(3)-F(-2)}{3-(-2)} = -6$

$F(3) - F(-2) = -30$ (1b)

$\frac{F(5)-F(3)}{5-3} = 20$

$F(5) - F(3) = 40$ (2b)

We obtain the average value of $f$ over the interval $[-2, 5]$ by algebraic handling:

$F(5) - F(3) +[F(3)-F(-2)] = 40 + (-30)$

$F(5) - F(-2) = 10$

$\frac{F(5)-F(-2)}{5-(-2)} = \frac{10}{5-(-2)}$

$\bar f = \frac{10}{7}$

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

4 0

3 years ago

HELPPPP THIS IS 7TH GRADE MATH AND IM CONFUSEDDDD

irga5000 [103]

Answer:

n-p for the first, p-n for the second

Step-by-step explanation:

6 0

2 years ago