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.
Step-by-step explanation:
4x 3.5 + 4x 2.3
14 + 9.2
= 23.2
- Midpoint Formula:
So firstly, let's start with the x-coordinates. Since we know the midpoint's x-coordinate and point A's x-coordinate, we can solve for point B's x-coordinate as such:
Next, do the same thing except solve for the y-coordinate and using point A's y-coordinate and the midpoint's y-coordinate:
<u>Putting it together, point B's coordinates are (2,4).</u>
A. 15
0 +0=0
0 +1 = 1
1+2=3
3+3=6
6+4=10
10+5 = 15
The distance between these points is (4,7) and (3,-7)
