Lol
trouble varies directly as distance
lets say t=trouble and d=distance
t=kd
k is constant
given
when t=20, and d=400
find k
20=400k
divide by 400 both sides
1/20=k
t=(1/20)d
given, d=60
find t
t=(1/20)60
t=60/20
t=3
3 troubles
Answer: 19,461
Step-by-step explanation:
Add 19,000+400 get 19,400
Add 19,400+60 and get 19,460 add 1 and get 19,461
Answer:
Step-by-step explanation:
* = =
Answer:
34 classes
Step-by-step explanation:
First, subract 180 from 1200. You should get 1020. then do 1020/30. So 34 classes can watch the performance of 180 students.
Answer:
or 2.738
Step-by-step explanation:
Let’s just look at the triangle on the top with the on the top and x on the bottom. (Basically the top half to the equilateral triangle)
There is a small square in the bottom right corner, which indicates that this triangle is a right triangle. This means that we can use the Pythagorean Theorem:
We know that \sqrt{10} is our hypotenuse, and therefore our c in our equation. Let’s say that x=a in our equation. Therefore we are left to find b. However, b is half the length of the side of the original equilateral triangle. An equilateral triangle means that all three sides are the same length. Therefore our side would also be \sqrt{10} units long. However we know that b is half of that value, so b= or
Plugging these values into the equation:
x^2+ (\frac{\sqrt{10} }{2})^{2}=\sqrt{10} ^{2}
This approximately equals 2.738