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2 years ago

I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

Mathematics

1 answer:

WINSTONCH [101]2 years ago

6 0

c = 39

hope it helps...!!!

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I need major help!!!!

Mama L [17]

1) Area of a circle = πr^2
Radius = 13.2 ÷ 2
= 6.6
Area = (3.14)(6.6)^2
= 136.78cm^2

I hope this helps!

6 0

3 years ago

Yari says that the product of 230x10 to the 4th power has 4 zeros. Is she correct?

Kamila [148]

Answer:

no she is not correct because 230×10^4=23,000,000

6 0

1 year ago

A square is constructed with an are of 18 square feet what is the approximate length of one side

SCORPION-xisa [38]

Answer:

4.5ft per side

Step-by-step explanation:

square has four equal sides and is L times w times h= A but we have A= 18 already so you reverse and divide so

18/4=4.5

4 0

3 years ago

Solve for X <br><br>can you show me how to do this problem.

sergey [27]

Answer is 4.
Right angle= 90
50+ 10(4)= 90

7 0

3 years ago

Read 2 more answers

In the graph below, Point A represents Owen's house, Point B represents David's house and Point C represents the school. Who liv

ycow [4]

Answer:

• David

• 4 miles

Explanation:

In the graph:

The given locations are:

• Owen's House, A(11,3)

• David's House, B(15,13)

• School, C(3,18)

We determine both Owen's and David's distance from the school using the distance formula.

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Owen's distance from school (AC)

$\begin{gathered} AC=\sqrt[]{(3-11)^2+(18-3)^2} \\ =\sqrt[]{(-8)^2+(15)^2} \\ =\sqrt[]{64+225} \\ =\sqrt[]{289} \\ AC=17\text{ miles} \end{gathered}$

David's distance from school (BC)

$\begin{gathered} BC=\sqrt[]{(3-15)^2+(18-13)^2} \\ =\sqrt[]{(-12)^2+(5)^2} \\ =\sqrt[]{144+25} \\ =\sqrt[]{169} \\ BC=13\text{ miles} \end{gathered}$

We see from the calculations that David lives closer to the school, and by 4 miles.

The graph below is attached for further understanding:

5 0

1 year ago