Side length-80
So we know that for a Right Angle, both sides have to measure up to 90 degrees. If one side is 10, and both sides need to add up to 90, then we can take 10 away from 90. That would be 80. 80 + 10 = 90. So the missing side length is 80.
-DustinBR
Answer:
2 1/3
Step-by-step explanation:
4 + ( -1 2/3)
= 4 - 1 2/3
= 4 - 5/3
= 12/2 - 5/3
= 7/3
= 2 1/3
Answer:
A
C
D
Step-by-step explanation:
Distance between points
and
is
Distance from H to B:
![[tex]d=\sqrt{(10-(-3))^2+(1-(-9))^2}=\sqrt{169+100}=\sqrt{269}](https://tex.z-dn.net/?f=%5Btex%5Dd%3D%5Csqrt%7B%2810-%28-3%29%29%5E2%2B%281-%28-9%29%29%5E2%7D%3D%5Csqrt%7B169%2B100%7D%3D%5Csqrt%7B269%7D)
d=\sqrt{(1-(-3))^2+(10-(-9))^2}=\sqrt{16+361}=\sqrt{376}[/tex] units.
Distance from Z to B:
units.
Horse Z is closer to the barn.
(The conversion to meters is not required; the question does not ask for actual distances, so "units" is OK.)
Answer:
Step-by-step explanation:
To make the problem easier to solve, we will set it up as the equation of the length of time of each class times the number of classes equals the total amount of minutes. However, since we don't know the number of classes, we'll symbolize our two unknowns with two variables.
75x + 45y = 705
(75x + 45y)/15 = 705/15
5x + 3y = 47
y = (47-5x)/3
It looks like we can't simplify the equation any more, so now it is a matter of trial and error. The minimum number of Saturday classes means the maximum number of weekday classes. We first will test for the maximum by assuming there are no Saturday classes, then will work our way up until x is an integer.
If x = 0
(47-5(0))/3 = 47/3 = 15.6666
If x = 1
(47-5(1))/3 = 42/3 = 14
This works. Therefore, the maximum number of weekday classes is 14, or choice b.
