Answer:
i wouldsay b
Step-by-step explanation:
I hope this helps you
✔cos^2A+sin^2A=1
✔1-cos^2A=sin^2A
✔cos2A=cos^2A-sin^2A
✔sin2A=2.sinA.cosA
secA=1/cosA
tgA=sinA/cosA
sin^2A/1/cos^2A-sin^2A/cos^2A
sin^2A.cos^2A/cos2A
2.sin^2A.cos^2A/cos2A
sin2A.2.sin2A/cos2A
tg2A.2.sin2A
Ok so to solve to first one u do this:
62.4 - 31.53, which gives u 30.87. and then u add 30.87 and 31.33, and u get 62.4
for the second one u do the same thing.
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
Order the numbers from least to greatest. a –8, −10, 0, 5, 1, −1, 3, −6, −9, −19, 32
krek1111 [17]
Start with -19, -10, -9,-8,-6,-1,0,1,3,5,32
