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
Answer:
-3x²-5xeˣ-eˣ
-3eˣx²-11eˣx-6eˣ
Step-by-step explanation:
I'm going to go by the picture and not what you wrote in your title.
To find the derivative of this we have to apply the product rule
(a*b)'=
a'*b+a*b'
We plug in our numbers and get
(-3x²+x-2)'*eˣ+(-3x²+x-2)*eˣ'
Now we can evaluate the derivatives and simplify
(-3x²+x-2)'= -6x+1
eˣ'=eˣ
which means we have
(-6x+1)*eˣ+(-3x²+x-2)*eˣ
Simplify
-6xeˣ+eˣ-3x²eˣ+xeˣ-2eˣ
Combine like terms
-3x²eˣ-5xeˣ-eˣ
Now we just need to find the derivative of this
We can apply the same product rule as we did before
(-3x²eˣ)'
Let's start by factoring out the -3 to get
-3(x²eˣ)'
which is equal to
-3(x²eˣ'+x²'eˣ)
Compute this and get
-3(x²eˣ+2xeˣ)= -3x²eˣ-6xeˣ
Now let's find the derivative of the second part
(-5xeˣ)'
-5(x'eˣ+xeˣ')
-5(eˣ+xeˣ)
-5eˣ-5xeˣ
Which means we have
(-3x²eˣ-6xeˣ)+(-5eˣ-5xeˣ)-eˣ
Combine like terms and get
-3eˣx²-11eˣx-6eˣ
Answer:
Step-by-step explanation:
Answer:
E
Step-by-step explanation:
it's the correct answer I'm sure of
