Answer:

The domain for x is all real numbers greater than zero and less than 5 com
Step-by-step explanation:
<em><u>The question is</u></em>
What is the volume of the open top box as a function of the side length x in cm of the square cutouts?
see the attached figure to better understand the problem
Let
x -----> the side length in cm of the square cutouts
we know that
The volume of the open top box is

we have



substitute

Find the domain for x
we know that

so
The domain is the interval (0,5)
The domain is all real numbers greater than zero and less than 5 cm
therefore
The volume of the open top box as a function of the side length x in cm of the square cutouts is

Answer:
k=8
Step-by-step explanation:
divide both sides by 11 to get nine
then subtract 1 from both sides to get
8
700 cal out of the 2000
2000*.35
Answer:
Relative frequency is 7.41% or 0.0741
Step-by-step explanation:
Given
The Attached Table
Required
Calculate the relative frequency of the class with lower limit 27
Relative Frequency is calculated by dividing individual frequency by the total frequency
Mathematically,

The total frequency of the given data is 6+8+4+2+5+2

The class with lower limit 27 has a frequency of 2;
Hence;
becomes


(Approximated)
You may also leave your answer in percentage form


Hence, the relative frequency is 7.41% or 0.0741
You normally use pie in equation that had the circumference or radius or diameter. !<span />