I don’t know how to do this. Ignore my work it’s probably wrong. Please help.
1 answer:
Check the picture below.
A)
well, we start off by making a table of values for the function, with a few "t" values, as you see in the picture, then graph those points to get the graph.
B)
at t = 0, h(0) = 98, and at t = 1, h(1) = 82, so it went from 98 to 82, for a difference of 16.
C)
at t = 1, h(1) = 82, and at t = 2 h(2) = 34, so it went from 82 to 34, for a difference of 48.
does it fall the same distance on both intervals? well, they're different, so nope.
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Answer:
2 square cm
Step-by-step explanation:
Given :
A square is inscribed in a circle whose radius is r = 1 cm
Therefore, the diameter of the circle is 2 r = 2 x 1
= 2 cm.
So the diagonal of the square is 2r.
Using the Pythagoras theorem, we find each of the side of the triangle is .
Therefore, the area of the square is given by
=
Hence the area of the largest square that is contained by a circle of radius 1 cm is 2 cm square.
Multiply 2/1 or 2 on both sides to get "y" by itself
y ≤ -14