Answer:
9.28
Step-by-step explanation:
hope this help!!!
Ok so
sale is 15% off
and add 6% tax
100-15=85
100+6=106
therefor
1250 with 15% discount plus 6% tax means
1250 times 85% times 106% =
1250 times 0.85 times 1.06=1126.25
final cost=$1126.25
If the density of the man is 1.0321 grams per cubic cm. Then the body fat percentage will be 28.59%.
<h3>What is a function?</h3>
The function is an expression, rule, or law that defines the relationship between one variable to another variable. Functions are ubiquitous in mathematics and are essential for formulating physical relationships.
An older method of estimating body fat percentage was to measure a person's average density (total mass divided by total volume) and apply a formula to convert it to a body fat percentage.
where BF = body fat and p = density in g/cm3 A man weighs 175 pounds.
A man weighs 175 pounds.
His total body volume is modeled by a cylinder with a height of 170 cm and a radius of 12 cm.
Then the density will be

Then the Body fat percentage will be

More about the function link is given below.
brainly.com/question/5245372
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Answer:
0.375 feet-lb
Step-by-step explanation:
We have been given that the work required to stretch a spring 2 ft beyond its natural length is 6 ft-lb. We are asked to find the work needed to stretch the spring 6 in. beyond its natural length.
We can represent our given information as:

We will use Hooke's Law to solve our given problem.

Substituting this value in our integral, we will get:

Using power rule, we will get:
![6=\left[ \frac{kx^2}{2} \right ]^2_0](https://tex.z-dn.net/?f=6%3D%5Cleft%5B%20%5Cfrac%7Bkx%5E2%7D%7B2%7D%20%5Cright%20%5D%5E2_0)


We know that 6 inches is equal to 0.5 feet.
Work needed to stretch it beyond 6 inches beyond its natural length would be 
Using power rule, we will get:
![\int\limits^{0.5}_0 {3x} \, dx = \left [\frac{3x^2}{2}\right]^{0.5}_0](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B0.5%7D_0%20%7B3x%7D%20%5C%2C%20dx%20%3D%20%5Cleft%20%5B%5Cfrac%7B3x%5E2%7D%7B2%7D%5Cright%5D%5E%7B0.5%7D_0)

Therefore, 0.375 feet-lb work is needed to stretch it 6 in. beyond its natural length.