Answer:
3 and 1/2.
Step-by-step explanation:
1/2 times 7 = (7 times 1) / 2 = 7 / 2 = (6 + 1) / 2 = 3 and 1/2.
Hope this helps!
Answer:
This expression has 2 coefficients, (-3 and -4), because any number with a variable is a coefficient. That means 25, -2, and 5 are all constants, because they don't have variables.
Variables are the x's in this case.
I hope this answers your question :) Also if you like, can you mark me brainliest, please? It would really mean a lot to me :)
Answer:
1/4 - 3 2/5 - (2 1/3 - 1/4) =
-157/30=
-5 7/30
≅ -5.2333333
Step-by-step explanation:
Answer:
$20
Step-by-step explanation:
to check
16/20
divide
= .80
times 100
80%
Answer:
3200 ft-lb
Step-by-step explanation:
To answer this question, we need to find the force applied by the rope on the bucket at time 
At 
After
seconds, the weight of the bucket is 
Since the acceleration of the bucket is the force on the bucket by the rope is equal to the weight of the bucket.
If the upward direction is positive, the displacement after
seconds is 
Since the well is 80 ft deep, the time to pull out the bucket is 
We are now ready to calculate the work done by the rope on the bucket.
Since the displacement and the force are in the same direction, we can write

Use
and 



![=\left[63 t-0.2 t^{2}\right]_{0}^{36}](https://tex.z-dn.net/?f=%3D%5Cleft%5B63%20t-0.2%20t%5E%7B2%7D%5Cright%5D_%7B0%7D%5E%7B36%7D)
