There is a given point (10,3) this shows that when the width (X) is 10, the height (Y) is 3.
On the left side of the graph, they show an equation for the height Hw as being the constant over w ( width).
Using the given point solve for the constant.
Replace Hw with 3 and w with 10:
3 = Constant/10
Solve for the constant by multiplying both sides by 10:
Constant = 3 x 10
Constant = 30
The answer is B.
Answer:
B ------- DE ≅ RS
it's B because DE and RS both contain the equal angles.
Answer:

Step-by-step explanation:



1,400 it will be 20 qrtly making 80 a year time 5 years
You solve for the domain by setting the radicand less than or equal to 0 and solving for x. Dividing by a -x, we switch the sign so we have that the domain is less than or equal to 0, or all negative numbers. We know that it breaks every law in math to have a negative radicand with an even index, so if the domain is all negative values of x, taking a negative of a negative gives us a positive. The negative sign OUTSIDE the radical means you are flipping the graph upside down. So instead of having a range of y is greater than or equal to 0 as does the parent graph, you have flipped it upside down so it heads more negative in regards to the range. Therefore, the domain and the range both have the same sign, thee last choice from above.