One approach that might be a great help would be if you would look
at the choices ... those things at the bottom of the question that you
didn't give us.
-- One leg is 3 times the length of the other leg.
-- Any right triangle that has one leg three times as long as the other
leg is similar to this triangle.
I'll bet that if you look through the choices, you'll find one there.
Answer:
, 
Step-by-step explanation:
<h3>
I will use the elimination method.</h3>
We want to make the X's the same:
Because the signs of the X's are the same we subtract the 2 equations to make:
(I put the second one on top of the 1st)
So:
Substitute y into either equation 1 or 2:
(I chose equation 1)
So:
Answer:
Step-by-step explanation:
679
Answer:
its c
ududhsihsisbishsusbusvsusbusbduusususuuusueuieieiejejdhebteybybeybsbysyneybdyndybbydbyeynydnnydndyjdyj8ggnxr
Step-by-step explanation:
fhjfjjgcmhmhcxmgkhfkhffhkhfkgfoyfkyftfutdutxjtxoyf
Answer:
t≈8.0927
Step-by-step explanation:
h(t) = -16t^2 + 128t +12
We want to find when h(t) is zero ( or when it hits the ground)
0 = -16t^2 + 128t +12
Completing the square
Subtract 12 from each side
-12 = -16t^2 + 128t
Divide each side by -16
-12/-16 = -16/-16t^2 + 128/-16t
3/4 = t^2 -8t
Take the coefficient of t and divide it by 8
-8/2 = -4
Then square it
(-4) ^2 = 16
Add 16 to each side
16+3/4 = t^2 -8t+16
64/4 + 3/4= (t-4)^2
67/4 = (t-4)^2
Take the square root of each side
±sqrt(67/4) =sqrt( (t-4)^2)
±1/2sqrt(67) = (t-4)
Add 4 to each side
4 ±1/2sqrt(67) = t
The approximate values for t are
t≈-0.092676
t≈8.0927
The first is before the rocket is launched so the only valid answer is the second one
