<p>At time = 0 a ball was thrown upward from an initial height of 6ft. Until the ball hit the ground, its height in feet, after t secdonds was given by the function h above, in which c and d are positive constants. If the ball reached its max height of 106 ft at time = 2.5, what was the heigh, in feet of theball at t= 1.</p>
<p>How do you do that efficiently for the SAT?</p>
<p>Q2.</p>
<p>This is a blue book question. Page 674 number 20.</p>
<p>Q3.</p>
<p>The cube shown above has edges of length 2, and A an dB are midpoints of the two of the edges. What is the length of seg AB.</p>
<p>The answer is sq rt 6. But idk how they got that</p>
<p>for the first one
at time = 0 h(t) = 6
so you have
6 = c - d^2
So c = D^2 + 6
Now plug in for the second equation
106 = D^2 + 6 - (D - 10)^2
106 = D^2 + 6 - D^2 + 20D - 100
D^2 goes away
100 = 20d - 100
D = 10
Now plug in for t = 1
h(t) = 10^2 + 6 - (10 - 4)^2
so you get 106 - 36
H(T) = 70
This is not really a hard problem.. just gotta think hard.
Q 2.
You really must understand what the + h and K does
first of all the point (-1,3) ends up becoming (2,1)
So it moved across 3 and down 2.
So you have G(X) = F(x -3) - 2 (this moves it across 3 down 2)
So -3 * -2 = 6
Last one.
Draw a triangle
The distance of B to the bottom right vertex is 1
the distance of A to the bottom right vertex is simply 1^2 + 2^2 and then the square root of it (its a right triangle) you get square root of 5
Now use pythag. theorum again and do square root of 5 squared + 1 squared = 6. Now take square root and you get square root of 6.</p>
<p>OOH i feel so dumb on the second one. I figured out that it was moved and whatnot....but for some reason I didn't reconnect that to h, k. AAH stupid errors.</p>
<p>how did you get "the bottom right vertex is simply 1^2 + 2^2 "</p>
<p>If you create a right triangle and you're missing a side or an angle, you can use either the pythagorean thm or trig. Google those two things up if you don't know how to use them. They'll probably produce better explanations.</p>
<p>Its just 2 separate calculations
The distance from A to the bottom right vertex is the base
the distance from the bottom right vertex to b is the height
to find the distance from A to the bottom right vertex you have to use the pythag theorem with 2 being the base and 1 being the height (DRAW the diagram for yourself). You get a to the vertex square root of 5 then. And I already said you have height of 1 from bottom right vertex to B. So use pythag theorem again and you get square root of 6.</p>