All categories

4 years ago

Find the perimeter of the polygon

Mathematics

1 answer:

slega [8]4 years ago

7 0

Answer:

Multiply the side length by the number of sides to get the perimeter. The formula for finding the perimeter of a regular polygon is just the number of sides x the length of any side. Once you've multiplied those 2 numbers together, you've found the perimeter of the polygon

Step-by-step explanation:

You might be interested in

Please help me!!!<br> Show work!!!<br> ____ % of 60 is 30

sergey [27]

Answer:

50%

Step-by-step explanation:

50% is equal to 1/2.

1/2 of 60 is 30.

5 0

3 years ago

Read 2 more answers

This Is challenging I hope you can figure it out

max2010maxim [7]

Answer:

980 hers your proof so 170 * 7 =180

Step-by-step explanation:

3 0

3 years ago

Which of the following descriptions represent the transformation shown in the image? Part 1b

sveticcg [70]

Answer: b) 2 units left, 4 units up, reflect across y-axis

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

The easiest way to do this is to graph the black triangle using the the given options to see which one lands on the red triangle.

The second option is the one that works.

4 0

3 years ago

The base of a triangle is 4 cm greater than the

Anna35 [415]

Answer:

Step-by-step explanation:

Formula for area of a triangle:

Height x Base /2

Base (b) = h +4

Height = h

h + 4 x h /2 = 30cm

=> h +4 x h = 60

=> h+4h =60

=> 5h = 60

=> h = 12

Height = 12

Base = 12 +4 = 16

4 0

3 years ago

How to find variables of these #2 & #3

Lubov Fominskaja [6]

So... hmmm if you check the first picture below, for 2)

we could use the proportions of those small, medium and large similar triangles like $\bf \cfrac{small}{large}\qquad \cfrac{x}{12}=\cfrac{6}{x}\impliedby \textit{solve for "x"}
\\\\\\
\cfrac{small}{large}\qquad \cfrac{z}{18}=\cfrac{6}{z}\impliedby \textit{solve for "z"}
\\\\\\
\cfrac{large}{medium}\qquad \cfrac{y}{12}=\cfrac{18}{y}\impliedby \textit{solve for "y"}$

now.. for 3) will be the second picture below

$\bf \cfrac{large}{medium}\qquad \cfrac{x+10}{2\sqrt{30}}=\cfrac{2\sqrt{30}}{10}\impliedby \textit{solve for "x"}
\\\\\\
\textit{now, because you already know what "x" is, we can use it below}
\\\\\\
\cfrac{large}{small}\qquad \cfrac{z}{x}=\cfrac{x+10}{z}\impliedby \textit{solve for "z"}
\\\\\\
\textit{and let us use "x" again below}
\\\\\\
\cfrac{small}{medium}\qquad \cfrac{y}{10}=\cfrac{x}{y}\impliedby \textit{solve for "y"}$

8 0

3 years ago