First we need to find the rate of logging for employee A per year. To do this, the number of hours can be divided by the number of years to find out how many hours of travel they log per year.
120 ÷ 4 = 30. Employee A logs roughly 30 hours of travel per year.
To find a percentage increase of 20%, you need to multiply by 1.2.
30 x 1.2 = 36.
Therefore, Employee B logs roughly 36 hours of travel per year.
To find out how many hours of travel Employee B logs after 1.5 years, we simply need to multiply 36 by 1.5, which equals 54.
To find out how many hours of travel Employee A logs after 1.5 years, we simply need to multiply 30 by 1.5, which equals 45.
Finally, we need to find out how many MORE hours Employee B logged after 1.5 years compared to Employee A. To do this, we simply need to do 54 - 45 = 9 hours.
Working = (((120/4) x 1.2) x 1.5) - (120/4 x 1.5)
Answer = 9 hours.
Answer:
2nd answer choice
Step-by-step explanation:
One way of doing this problem is to switch over to fractional exponents first.
The given product can be thought of as a square:
the square of (fifth root of 4x^2)
= [ fifth root of 4x^2 ]^2, or
= [(4x^2)^(1/5) ]^2, or
= [ (4x^(2/5) ]^2, or
= 16x^(4/5) This is equal to the 2nd answer choice.
Answer: 9/14
Step-by-step explanation:
As Eric ran 5/7 of the distance ran on Friday, you have to times 5/7 by the distance ran on Friday.
So we have:
5/7 x 9/10 = 9/14
This is calculated by multiplying the numerators (numbers on top) and the denominators (numbers on the bottom)
Numerators:
5x9 = 45
Denominators:
7x10 = 70
So we now have the new fraction: 45/70, if we simplify this, we get:
45/70 = 9/14
(dividing the top and bottom both by 5)
Hope that makes sense! If you'd like any more help with maths, I'd be happy to offer online tuition. You can find me at: www.birchwoodtutors.com
Answer:
Option D 
Step-by-step explanation:
Verify each quadratic equation
case A) we have
This quadratic equation has a leading coefficient of 1
Substitute the value of x=3 and x=-2 in the equation
For x=3
----> is not true
therefore
x=3 is not a solution of the quadratic equation
case B) we have
This quadratic equation has a leading coefficient of 1
Substitute the value of x=3 and x=-2 in the equation
For x=3
----> is not true
therefore
x=3 is not a solution of the quadratic equation
case C) we have
This quadratic equation has a leading coefficient of 1
Substitute the value of x=3 and x=-2 in the equation
For x=3
----> is not true
therefore
x=3 is not a solution of the quadratic equation
case D) we have
This quadratic equation has a leading coefficient of 1
Substitute the value of x=3 and x=-2 in the equation
For x=3
----> is true
therefore
x=3 is a solution of the quadratic equation
For x=-2
----> is true
therefore
x=-2 is a solution of the quadratic equation
