Answer:42.55$
Step-by-step explanation:
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.
That diagram can be split into a right angled triangle of base 7 and of height 4 + 5.2 + 2 = 11.2, and a square of length of 5.2 So the are of the total shape is the area of the right angled triangle + the area of the square.
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Please mark me as brainliest.
The 4 on the left side is 10x more than the 4 on the right. This only works if the numbers are next to each other and are the same.
60 * 3 = 180 and 90 * 2 = 180
if they are supplementary angles than they equal 180 together.
angle p is three times less than twice the measure of angle q so that means 180/3 = angle p and 180/2 = angle q.
Angle q is 90 and twice that is 180 and Angle p is 60 and three times that is 180
so 60 is three times less than twice the measure of 90.
60 three times is 180 and 90 twice is 180 which means 3p = 2q and you have 180 because they are supplementary angles so you have really 180/3 = p and 180/2 = q
The answer is p = 60 and q = 90
