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3 years ago

Sam has 100 feet of fencing. Does he have enough fencing to enclose a circular region whose area is 630 square feet? Show work!

1 answer:

6 0

A=pi(radius^2)
630/3.14=200.64
radius=14.16 feet
C=2(pi)(radius)
=3.14(2)(14.16)
=6.28(14.16)
C=88.92
88.92<100
Yes! He has enough fencing to enclose the circular area.

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In a triangle , the measure of the first angle is twice the measure of the second angle. The measure of the third angle is 100 m

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Answer:

The three unknown angles X, Y , and Z are:

X = 40, Y = 20, and Z = 120

Step-by-step explanation:

Let's name X the measure of the first angle, Y the measure of the second one, and Z that of the third one.

Then we can create the following equations:

X = 2 Y

Z = 100 + Y

and

X + Y + Z = 180

So we use the first equation and the second one to substitute for the variable X and Z in the thrid equation:

2 Y + Y + (100 + Y) = 180

4 Y + 100 = 180

4 Y = 80

Y = 80/4 = 20

Then X = 40, Y = 20, and Z = 120

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Suppose a city with population 500,000 has been growing at a rate of 7% per year. If this rate continues, find the population

krok68 [10]

Answer:

2,070,281

Step-by-step explanation:

let x = number of years

let y = population

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2 years ago

Find the perimeter of quadrilateral PQRS with the vertices P(2,4), Q(2,3), R(-2,-2), and S(-2,3).

storchak [24]

Answer:

$P=16.53\ units$

Step-by-step explanation:

we know that

The perimeter of quadrilateral PQRS is equal to the sum of its four length sides

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

the vertices P(2,4), Q(2,3), R(-2,-2), and S(-2,3)

step 1

Find the distance PQ

P(2,4), Q(2,3)

substitute in the formula

$d=\sqrt{(3-4)^{2}+(2-2)^{2}}$

$d=\sqrt{(-1)^{2}+(0)^{2}}$

$d=\sqrt{1}$

$dPQ=1\ unit$

step 2

Find the distance QR

Q(2,3), R(-2,-2)

substitute in the formula

$d=\sqrt{(-2-3)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-5)^{2}+(-4)^{2}}$

$dQR=\sqrt{41}\ units$

step 3

Find the distance RS

R(-2,-2), and S(-2,3)

substitute in the formula

$d=\sqrt{(3+2)^{2}+(-2+2)^{2}}$

$d=\sqrt{(5)^{2}+(0)^{2}}$

$dRS=5\ units$

step 4

Find the distance PS

P(2,4), S(-2,3)

substitute in the formula

$d=\sqrt{(3-4)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-1)^{2}+(-4)^{2}}$

$dPS=\sqrt{17}\ units$

step 5

Find the perimeter

$P=PQ+QR+RS+PS$

substitute the values

$P=1+\sqrt{41}+5+\sqrt{17}$

$P=6+\sqrt{41}+\sqrt{17}$

$P=16.53\ units$

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3 years ago