Try this:
1) if ratio=100/3, then 100/3=2600/x
x=26*3=78 p.
2) if (-2) ⇒ -3*(4)=-12 ⇒ answer: (-2;-12)
if (-1) ⇒ -3*(1)=-3 ⇒ answer: (-1;-3)
if 0 ⇒ -3*0=0 ⇒ answer: (0;0)
if 1 ⇒ -3*1=-3 ⇒ answer (1;-3)
if 2 ⇒ -3*4=-12 ⇒ answer: (2;-12)
Answer:
Step-by-step explanation:
What this question is asking of you is what is the greatest common divisor of 12 and 15. Or, what is the biggest number that divides both 12 and 15.
in order to find this we have to split each number into it's prime components.
for 12 they are 2,2 and 3 (
2
⋅
2
⋅
3
=
12
)
and for 15 they are 3 and 5 (
3
⋅
5
=
15
)
Out of those two groups (2,2,3) and (3,5) the only thing in common is 3, so 3 is the greatest common divisor. That tells us that the greatest number of groups that can exist and have the same number of girls and the same number of boys for each group is 3.
Now to find out how many girls and boys there are going to be in each group we divide the totals by 3, so:
12
3
=
4
girls per group, and
15
3
=
5
boys per group.
(just as a thought exercise, if there were 16 boys, the divisors would have been (2,2,3) and (2,2,2,2), leaving us with 4 groups [
2
⋅
2
] of 3 girls [12/4] and 4 boys [16/4] )
(-6)(-0.4)(-0.5) = (-1)6⋅4⋅5⋅10⁻² = -1.2
Answer:
Step-by-step explanation:
We have:
And we want to find B’(6).
So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:
![\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20B%5E%5Cprime%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B24.6%5Csin%28%5Cfrac%7B%5Cpi%20t%7D%7B10%7D%29%288-t%29%5D)
We can move the constant outside:
![\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20B%5E%5Cprime%28t%29%3D24.6%5Cfrac%7Bd%7D%7Bdt%7D%5B%5Csin%28%5Cfrac%7B%5Cpi%20t%7D%7B10%7D%29%288-t%29%5D)
Now, we will utilize the product rule. The product rule is:
We will let:
Then:
(The derivative of u was determined using the chain rule.)
Then it follows that:
![\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20B%5E%5Cprime%28t%29%26%3D24.6%5Cfrac%7Bd%7D%7Bdt%7D%5B%5Csin%28%5Cfrac%7B%5Cpi%20t%7D%7B10%7D%29%288-t%29%5D%20%5C%5C%20%5C%5C%20%26%3D24.6%5B%28%5Cfrac%7B%5Cpi%7D%7B10%7D%5Ccos%28%5Cfrac%7B%5Cpi%20t%7D%7B10%7D%29%29%288-t%29%20-%20%5Csin%28%5Cfrac%7B%5Cpi%20t%7D%7B10%7D%29%5D%20%5Cend%7Baligned%7D)
Therefore:
![\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20B%5E%5Cprime%286%29%20%3D24.6%5B%28%5Cfrac%7B%5Cpi%7D%7B10%7D%5Ccos%28%5Cfrac%7B%5Cpi%20%286%29%7D%7B10%7D%29%29%288-%286%29%29-%20%5Csin%28%5Cfrac%7B%5Cpi%20%286%29%7D%7B10%7D%29%5D)
By simplification:
![\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17](https://tex.z-dn.net/?f=%5Cdisplaystyle%20B%5E%5Cprime%286%29%3D24.6%20%5B%5Cfrac%7B%5Cpi%7D%7B10%7D%5Ccos%28%5Cfrac%7B3%5Cpi%7D%7B5%7D%29%282%29-%5Csin%28%5Cfrac%7B3%5Cpi%7D%7B5%7D%29%5D%20%5Capprox%20-28.17)
So, the slope of the tangent line to the point (6, B(6)) is -28.17.
Answer:
A
Step-by-step explanation:
how long is the ball in the air ?
that is the same as asking : after how many seconds will the ball hit the ground (= reach the height of 0) ?
so, that means we need to find the zero solution of h(t).
at what t is h(t) = 0 ?
when at least one of the factors is 0 :
2(-2 - 4t)(2t - 5)
we have 3 factors
2 : can never be 0.
(-2 -4t) : can only be 0 for negative t, which does not make sense in our scenario (we cannot go back in time, only forward).
(2t - 5) : is 0 when 2t = 5 or t = 2.5
so, A is the right answer.
FYI : the starting height (on the hill) is given by t = 0 :
2(-2 - 0)(0 - 5) = 2×-2×-5 = 20 ft
