Answer:
18 sq ft
Step-by-step explanation: all you have to do is add the sides. 5+5=10
4+4= 8
10+8=18
This makes the most sense with the answers givin
Answer:

Step-by-step explanation:
Acceleration is second derivative of distance and are related as:

Integrating both sides w.r.to t

Using initial value

We have to calculate the distance covered in time interval [0,5], so:
![\int\limits^5_0 \frac{ds}{dt}=\int\limits^5_0 {\frac{1}{t+2}} \, dt\\\\s(t)=[ln|t+2|]^5_0\\\\s(t)=ln|5+2|+ln|0+2|\\\\s(t)=(ln|7|+ln|2|)\,ft](https://tex.z-dn.net/?f=%5Cint%5Climits%5E5_0%20%5Cfrac%7Bds%7D%7Bdt%7D%3D%5Cint%5Climits%5E5_0%20%7B%5Cfrac%7B1%7D%7Bt%2B2%7D%7D%20%5C%2C%20dt%5C%5C%5C%5Cs%28t%29%3D%5Bln%7Ct%2B2%7C%5D%5E5_0%5C%5C%5C%5Cs%28t%29%3Dln%7C5%2B2%7C%2Bln%7C0%2B2%7C%5C%5C%5C%5Cs%28t%29%3D%28ln%7C7%7C%2Bln%7C2%7C%29%5C%2Cft)
Answer:
x = 9
Step-by-step explanation:
5x+4 = 49
subtract 4 from both sides
5x = 45
divide both sides by 5
x = 9