1. The domain of a sqrt(x) is x > 0. So, L > 0.
2. The range of a S(x) = sqrt(x) is S(x) > 0 for all valid x > 0.
The answer is -7 :) hope this helps
Answer:
A is false
Step-by-step explanation:
notice how in both diagrams, there are 4 triangles with lengths a, b, c, but in each diagram the triangles are just rotated and positioned differently. if there are the same amount of triangles in both diagrams, and the have equal lengths then the area of the remaining figure, should both match up.
basically the area of the big square in Step 2, is equal to both of the shaded squares, combined area, in Step 1.
Y = x - 9 meets the y-axis at (0,-9)
y = -2x - 3 meets the y-axis at (0,-3)
This is because for every linear equation, there is a formula corresponding to it as y=mx + c, where (0,c) is where it meets the y-axis. As we assign the value of 0 to x, only c will be left to equal y, being the intercept.
So, these are actually pretty simple once you learn the equality used to solve for "x" and when to implement this method. You can use this equality to solve for a segment "x" anytime that two secant lines cutting through a circle come from the same point outside the circle.
Secant: by geometric definition is just a straight line that cuts a curve into multiple pieces.
I did one of them for you hopefully you can use my work for "a" to help you solve for "b".
For a. I got x=7.