Problem
A food company distributes its tomato soup in two cans of different sizes. For the larger can, both the diameter and the height have been increased by 15%.
By what percentage does the volume increase from the smaller can to the larger can? Round your answer to the nearest percent.
Solution
the volume of a cylinder is given by:

We know that the diameter=radius and the height are increased by 15% so then the new volume can be calculated on this way:

And the new volume would increase about 52%
95.0cm is the answer of this
Answer:
The value of given expression is
.
Step-by-step explanation:
Given information: n=7, x=2, p=1/2

The given expression is

It can be written as

Substitute n=7, x=2, p=1/2 and q=1/2 in the above formula.






Therefore the value of given expression is
.
Carlos made the mistake that he did not combine like terms (3 x and 2 x) properly and did not use addition property of equality.
<u>Step-by-step explanation:</u>
Carlos did the work as 3 x + 2 x - 6 = 24
We need to find his mistake that he made in above given.
Here, he did not add the like terms (3 x and 2 x)
3 x + 2 x = 5 x
Therefore, his work should be
5 x - 6 = 24
Also, he did not use addition property of equality. It means the equation remains same even though the same number gets added on both sides. It would be
5 x - 6 = 24
+ 6 = + 6
-----------------------
5 x = 30
Dividing 30 by 5, we get answer as '6'. Hence,
= 6
So, stated the above two are the mistakes found in carlos work.
Mult 2 by x you'll get 2x then 2 by 3 you'll get 6 and add 6 to 5, after that you'll get 11 it should look like this 2x+11.