Answer:
t=0.64
Step-by-step explanation:
h = -16t^2 +4t +4
We want h =0 since it is hitting the ground
0 = -16t^2 +4t +4
Using the quadratic formula
a = -16 b = 4 c=4
-b ± sqrt( b^2 -4ac)
----------------------------
2a
-4 ± sqrt( 4^2 -4(-16)4)
----------------------------
2(-16)
-4 ± sqrt( 16+ 256)
----------------------------
-32
-4 ± sqrt( 272)
----------------------------
-32
-4 ± sqrt( 16*17)
----------------------------
-32
-4 ± sqrt( 16) sqrt(17)
----------------------------
-32
-4 ± 4 sqrt(17)
----------------------------
-32
Divide by -4
1 ± sqrt(17)
----------------------------
8
To the nearest hundredth
t=-0.39
t=0.64
Since time cannot be negative
t=0.64
Side JL is 4√2 recall that in a 30-60-90 right triangle the hypotenuse is 2 times the size of the short leg.
JL also serves as the hypotenuse of the 45-45-90 triangle JML. The ratio of side lengths in this triangle is 1:1:√2
So we can see that the value of x = 4
Answer: ![\sqrt[3]{6n}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B6n%7D)
Step-by-step explanation:
We have the following expression:
Which can be written as follows:
Multiplying the exponents:
Writing in radical form we finally have the result:
Answer:
x= 12/11
Step-by-step explanation:
Answer: None of the above
Step-by-step explanation: Because none of them will equal the solution which is 0.5,
1.7 + (-1.2) = 0.5
