I ran it through a calculator and the answer it gave was "Yes"
Step-by-step explanation:
This question belongs to simple algebric equation, where we need to add or subtract fractions by finding a common denominator.
Given total hours = 120
Number of hours tutored for 1st week =2 1/3 hours = 7/3 hours
Number of hours in the 2nd week = 3/2 = 1.5 hours
Let the number of more hours she needs to teach be x
![\frac{7}{3} + \frac{3}{2} + x = 20](https://tex.z-dn.net/?f=%5Cfrac%7B7%7D%7B3%7D%20%2B%20%5Cfrac%7B3%7D%7B2%7D%20%2B%20x%20%3D%2020)
= ![\frac{14+9+6x}{6}= 20}](https://tex.z-dn.net/?f=%5Cfrac%7B14%2B9%2B6x%7D%7B6%7D%3D%2020%7D)
23 + 6x = 120
6x = 97
x = 16.167
Hence Kiara needs 16.167 more hours.
And this belongs to a part-whole problem, because the time she still needs to tutor is unknown.
Put the value where the variable is and do the arithmetic.
f(-4) = -5(-4) +(-4)/(-2)
= 20 + 2
= 22
The appropriate choice is
C. 22
Answer:
YES, they are equal
Step-by-step explanation:
Given the expressions
h(x) = 2x – 3;j(x) = - 4x + 6; k(x) = – 2x + 3
h(x) + j(x) = 2x – 3 + (-4x + 6)
h(x) + j(x) = 2x - 3 -4x + 6
h(x) + j(x) = 2x - 4x -3 + 6
h(x) + j(x) = -2x + 3 = k(x)
This shows that h(x) + j(x) = k(x)