Answer:
32
Step-by-step explanation:
To make it a fraction form answer, you multiply the whole number by the denominator and make the result the new numerator. The old numerator becomes the new denominator
Thus, the answer to 12 divided by 3/8 in fraction form is:
96
/3
To make the answer to 12 divided by 3/8 in decimal form, you simply divide the numerator by the denominator from the fraction answer above:
96 / 3 = 32
The answer is rounded to the nearest two decimal points if necessary.
96/3 can be simplified to 32/1.
32/1 is an improper fraction and should be written as 32.
Hope this helps <3
Answer:
Incorrect
Step-by-step explanation:
We are given the equation:

Add both sides by 9.

When we want to tell if the absolute equation has solutions or not, we have to simplify in this form first: or isolate the absolute sign.

If c ≥ 0, the equation has solutions.
If c < 0, the equation does not have solutions.
Therefore, it does not always matter if the constant on right side is in negative because if there is a number on the left side then there is a chance that the equation has solutions.
From |3x+8| = 4 is equivalent |3x+8|-9=-5 and the right side is 4 which is positive.
Hence, the equation does have a solution!
I’m so sorry! I hope you find the answer soon! I’m stressing out as well
Answer:
you bet
Step-by-step explanation:
Let,
Lee’s class size = L
Fred’s class size = F
We are told that Lee’s fifth-grade class has 1.0 times as
many students as Fred’s
Therefore, L = 1F
L = F
Question A. If there are a total of 60 students, how many
students does Fred’s class have?
Total number of students = 60
L + F = 60
Since L = F
L + F = F + F + 60
F + F = 60
2F = 60
F = 60/2
F = 30
Therefore, if there are a total of 60 students, Fred’s class
has 30 students.
QUESTION B. If there are a total of 60 students, how many
students does Lee’s class have?
Total number of students = 60
L + F = 60
Since L = F
L + F = L + L + 60
L + L = 60
2L = 60
L = 60/2
L = 30
Therefore, if there are a total of 60 students, Lee’s class
has 30 students.