All categories

3 years ago

The area of each circle

Mathematics

1 answer:

Katyanochek1 [597]3 years ago

4 0

Answer:

8. $25\pi$ $in^2$

11. $9\pi$ $m^2$

Step-by-step explanation:

Area = $\pi r^2$

r = d/2

$\pi$ = 3.14

r = 10/2

r = 5 in

Area = $\pi 5^2$

Area = $25\pi$ $in^2$

10.

r = 6/2

r = 3 m

Area = $\pi 3^2$

Area = $9\pi$ $m^2$

Hope it helps! (:

You might be interested in

Help with this problem plz!

Tomtit [17]

Use photos math it helps alot

5 0

3 years ago

Convert the length of 7 centimeters to meters. Compare the numerical values when both numbers are written in scientific notation

Pie

I think you did correct.

8 0

3 years ago

HELPPPPPPPPPPPPPPP RATE 5 STARSS

Stels [109]

Answer:

13. c,

14. a.

Step-by-step explanation:

13. The length of a side PQ with coordinates of

P as (x,y) and Q as (w,z)

is: $\sqrt{(x-w)^{2} +(y-z)^{2} }$

so, the length of the side AB =

$\sqrt{(5-5)^{2}+(-2+7)^{2} } =\sqrt{0^{2}+5^{2} } =5$

the length of side BC =

$\sqrt{(5-1)^{2}+(-7+2)^{2} } =\sqrt{(4)^{2}+(-5)^{2} } =\sqrt{4^{2}+5^{2} }$

the length of side AC=

$\sqrt{(5-1)^{2} +(-2+2)^{2} } =\sqrt{4^{2} } =4$

now, the easiest one will be when the vertices are on the coordinate axes.

so, option c will be the most appropriate one.

14. If u see the figure clearly, the lines l and FH are parallel.

the parallel postulate, i.e, the alternate interior angles will always be congruent.

here, the alternate interior angles are angle1 and angle4.

therefore to prove this step, he used parallel postulate as a reason.

so, the correct option is "a"

7 0

4 years ago

Write a rule for linear function in the table

Vesnalui [34]

d. f(x)=2x+1

f(5)=2(2)+1

f(5)=4+1

f(5)=5

f(11)=2(5)+1

f(11)=10+1

f(11)=11

f(8)=2(8)+1

f(8)=16+1

f(8)=17

f(23)=2(11)+1

f(23)=22+1

f(23)=23

3 0

3 years ago

Use the Counting Principle to find the probability. Choosing the 7 winning lottery numbers when the numbers are chosen at random

frozen [14]

Answer: 120

<u>Step-by-step explanation:</u>

Since the order of the numbers doesn't matter we can use the formula:

$\dfrac{n!}{r!(n-r)!}\quad \text{where n is the quantity of items and r is the quantity chosen}\\\\\text{In this problem, there are 10 numbers (n = 10) and 7 to be chosen (r = 7)}\\\\_{10}C_{7}=\dfrac{10!}{7!(10-7)!}\\\\.\qquad=\dfrac{10!}{7!3!}\\\\.\qquad=\dfrac{10\times 9\times 8\times 7!}{7!\times 3\times 2\times 1}\\\\.\qquad=10\times 3\times 4\\\\.\qquad=120$

7 0

3 years ago

Read 2 more answers