Answer:
This problem is incomplete, we do not know the fraction of the students that have a dog and also have a cat. Suppose we write the problem as:
"In Mrs.Hu's classroom, 4/5 of the students have a dog as a pet. X of the students who have a dog as a pet also have cat as a pet. If there are 45 students in her class, how many have both a dog and a cat as pets?"
Where X must be a positive number smaller than one, now we can solve it:
we know that in the class we have 45 students, and 4/5 of those students have dogs, so the number of students that have a dog as a pet is:
N = 45*(4/5) = 36
And we know that X of those 36 students also have a cat, so the number of students that have a dog and a cat is:
M = 36*X
now, we do not have, suppose that the value of X is 1/2 ("1/2 of the students who have a dog also have a cat")
M = 36*(1/2) = 18
So you can replace the value of X in the equation and find the number of students that have a dog and a cat as pets.
130 degrees.
ADB = 90 degrees
BDT=40 Degrees
Therefore, TDS = ADB(90)+BDT(40)=130 degrees
He will need 527.52 cubic centimeters of wax. You divide the diameter by 2 ten multiply that number by itself to get the circumference. From there you multiply it by 12 then by 4 and there's your answer
Answer: x = 6V/bh
Explanation:
V = 1/6 bhx
V = (bh/6)x
V • 6/bh = x
6V/bh = x
Answer:
x = 3
Step-by-step explanation:
1.8 - 3.7x = -4.2x + 0.3
add -4.2x to both sides
1.8 - 0.5x = 0.3
subtract 1.8 from both sides
-0.5x = 0.3 - 1.8
-0.5x = -1.5
divide by -0.5 on both sides
x = 3
GREAT JOB!!!! YOU GOT IT!!! =)
