Let's solve this problem step-by-step.
STEP-BY-STEP EXPLANATION:
First, we will establish that the shape of the window is a semi-circle. This means we must use the formula for the perimeter of a semi-circle to obtain the perimeter of the window.
The formula for the perimeter of a semi-circle is as follows:
Let perimeter of window or semi-circle = P
P = [ 2( Pi )r / 2 ] + 2r
Where r = radius of circle or semi-circle
From this, we will simply use the value of the radius given from the diagram in the problem and substitute it into the formula to obtain the perimeter of the window.
P = [ 2( Pi )r / 2 ] + 2r
r = 20
THEREFORE:
P = [ 2( Pi )( 20 ) / 2 ] + 2( 20 )
P = 20( Pi ) + 40
P = 102.83...cm^2
P = 102.8cm^2 ( to the nearest tenth )
FINAL ANSWER:
Therefore, the perimeter of the window is 102.8cm^2 ( to the nearest tenth ).
Hope this helps! :)
Have a lovely day! <3
Answer: D
<u>Step-by-step explanation:</u>
In order to increase each ticket by $2, you are ADDING 2 to each value.
So you create a matrix of all 2's and add that to the given matrix.
![\left[\begin{array}{cc}2&2\\2&2\\2&2\end{array}\right] +\left[\begin{array}{cc}8&10\\12&16\\6&8\end{array}\right]\quad =\quad \large \left[\begin{array}{cc}10&12\\14&18\\8&10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%262%5C%5C2%262%5C%5C2%262%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D8%2610%5C%5C12%2616%5C%5C6%268%5Cend%7Barray%7D%5Cright%5D%5Cquad%20%3D%5Cquad%20%5Clarge%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D10%2612%5C%5C14%2618%5C%5C8%2610%5Cend%7Barray%7D%5Cright%5D)
Answer:
Bbbbbbbbbbbb
Step-by-step explanation:
Answer:
|x - 4| ≤ 0.3
Step-by-step explanation:
The actual width = 4
Maximum variation, x = 0.3
Hence, the lowest width position based on actual width and possible variation is :
4 - 0.3 ≤ x ≤ 4 + 0.3
3.7 ≤ x ≤ 4.3
Answer: 17 and 18.
Step-by-step explanation: <em>The simple way is to divide 35 by 2 i.e, 35/2 = 17.5 and take the preceding and succeeding whole numbers. In this case, 17 and 18.</em>
<em />
