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2 years ago

What is the area of a circle with a radius of 1 foot?

Mathematics

2 answers:

Andreas93 [3]2 years ago

5 0

22/7 * 30.48 * 30.48

2919.81 cm square

1 foot = 30.48 cm

PLEASE MARK IT AS BRAINLIEST

SashulF [63]2 years ago

3 0

It would be Pi feet squared❤️

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I need help with these four questions please (68 points)

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Answer:

Part 1) The distance is $d=7.3\ units$

Part 2) The measure of angle 2 is 121°

Part 3) The coordinates of endpoint V are (7,-27)

Part 4) The value of x is 10

Step-by-step explanation:

Part 1) Find the distance between M(6,16) and Z(-1,14)

we know that

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

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

substitute the given values in the formula

$d=\sqrt{(14-16)^{2}+(-1-6)^{2}}$

$d=\sqrt{(-2)^{2}+(-7)^{2}}$

$d=\sqrt{53}\ units$

$d=7.3\ units$

Part 2) Find the measure m∠2

we know that

If two angles are supplementary, then their sum is equal to 180 degrees

In this problem we have

m∠1+m∠2=180°

substitute the given values

$(4y+7)\°+(9y+4)\°= 180\°$

Solve for y

$(13y+11)\°= 180\°$

$13y= 180-11$

$13y=169$

$y=13$

Find the measure of m∠2

$(9y+4)\°$

substitute the value of y

$(9(13)+4)=121\°$

Part 3) The midpoint of UV is (5,-11). The coordinates of one endpoint are U(3,5) Find the coordinates of endpoint V

we know that

The formula to calculate the midpoint between two points is

$M=(\frac{x1+x2}{2},\frac{y1+y2}{2})$

we have

$M(5,-11)$

$(x1,y1)=(3,5)$

substitute and solve for (x2,y2)

$(5,-11)=(\frac{3+x2}{2},\frac{5+y2}{2})$

Equation 1

$5=(3+x2)/2$

$10=3+x2$

$x2=7$

Equation 2

$-11=(5+y2)/2$

$-22=(5+y2)$

$y2=-27$

therefore

The coordinates of endpoint V are (7,-27)

Part 4) GI bisects ∠DGH so that ∠DGI is (x-3) and ∠IGH is (2x-13) Find the value of x

we know that

If GI bisects ∠DGH

then

∠DGI=∠IGH

Remember that bisects means, divide into two equal parts

substitute the given values

$x-3=2x-13$

solve for x

$2x-x=-3+13$

$x=10$

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3 years ago