Answer:
The rocket's maximum height will be 1424 feet.
It will reach that height in 4 seconds.
It will reach the ground in 8 seconds.
Step-by-step explanation:
One easy way to find the peak is to find the derivative and solve for zero:
h(t) = -16t² + 128t + 1680
To find the derivative of each term in that format, you simply multiply the coefficient by the exponent, and reduce the exponent by one. In the case of terms with no variable, they disappear. That gives us:
h'(t) = -32t + 128
This derivative describes the rate of change of h with respect to t, so to find the peak, we simply need to solve it for zero:
0 = -32t + 128
32t = 128
t = 4
So the maximum height occurs when t = 4. Let's plug it in to the original equation:
h(t) = -16t² + 128t + 1680
h(4) = -256 + 512 + 1680
h(4) = 1424
So the rocket's maxiumum height is 1424 feet.
And we've already answered the second question, it happens at 4 seconds.
Also, the amount of time going up will be the same as the amount of time going down, which means that it will reach the ground in twice that time, or 8 seconds.
Answer:
A [and B]
Step-by-step explanation:
If x < 5.3 then anything higher than 5.3 makes the statement false.
16 is too high and therefore is the most obvious fallacy.
5.3 would actually make x=5.3 and therefore x wouldn't be less than 5.3 making it also false but this really depends on what your teacher is looking for.
0 is true.
-8.95 is true.
If there is any clarification in the problem you can provide I can give a more definite answer but now it is really a coinflip!
Answer:
Step-by-step explanation:
Dan bought 8 shirts and 3 pants. This means that the ratio representing the number of shirts to the number of pants that Dan bought is 8/3 = 2.67
Devonte bought 12 shirts and 5 pants. This means that the ratio representing the number of shirts to the number of pants that Devonte bought is 12/5 = 2.4
Therefore, they are not equivalent ratios because the relationship is not exactly the same.
ANSWER:
Let t = logtan[x/2]
⇒dt = 1/ tan[x/2] * sec² x/2 × ½ dx
⇒dt = 1/2 cos² x/2 × cot x/2dx
⇒dt = 1/2 * 1/ cos² x/2 × cosx/2 / sin x/2 dx
⇒dt = 1/2 cosx/2 / sin x/2 dx
⇒dt = 1/sinxdx
⇒dt = cosecxdx
Putting it in the integration we get,
∫cosecx / log tan(x/2)dx
= ∫dt/t
= log∣t∣+c
= log∣logtan x/2∣+c where t = logtan x/2
