Complete Question
In ΔUVW, w = 9 cm, v = 22 cm and ∠V=136°. Find all possible values of ∠W, to the nearest 10th of a degree.
Answer:
16.5°
Step-by-step explanation:
In ΔUVW, w = 9 cm, v = 22 cm and ∠V=136°. Find all possible values of ∠W, to the nearest 10th of a degree.
We solve using Sine rule formula
a/sin A = b/sin B
We are solving for angle W
∠V=136°
Hence:
22 /sin 136 = 9 /sin W
Cross Multiply
22 × sin W = sin 136 × 9
sin W = sin 136 × 9/22
W = arc sin [sin 136 × 9/2.2]
W = 16.50975°
W = 16.5°
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
I believe the answer is sometimes
Answer:
7 1/2
Step-by-step explanation:
try putting (1.6 x 7.5)+48 in a calculator i just put random numbers in the x spot until i got 60 lol
Answer:
4(5+4)
Step-by-step explanation:
4(5+4) equals 36, and 20 + 16 equals 36 as well, so therefore 20 + 16 is equivalent to 4(5+4)