Using derivatives, it is found that:
i) 
ii) 9 m/s.
iii) 
iv) 6 m/s².
v) 1 second.
<h3>What is the role of derivatives in the relation between acceleration, velocity and position?</h3>
- The velocity is the derivative of the position.
- The acceleration is the derivative of the velocity.
In this problem, the position is:
item i:
Velocity is the <u>derivative of the position</u>, hence:
Item ii:
The speed is of 9 m/s.
Item iii:
Derivative of the velocity, hence:
Item iv:
The acceleration is of 6 m/s².
Item v:
t for which a(t) = 0, hence:
Hence 1 second.
You can learn more about derivatives at brainly.com/question/14800626
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It would be 18 degrees.
5% of 360(the total number of degrees in a circle) = 18 degrees
we'll start off by grouping some
so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?
well, let's recall that a perfect square trinomial is
so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.
so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²
![\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%28%20x%5E2%20-x%20%2B%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2-%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%20%5Cright%29%3D6%5Cimplies%20%5Cleft%28%20x%5E2%20-x%20%2B%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%20%5Cright%29-%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%3D6%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%3D6%2B%5Ccfrac%7B1%7D%7B4%7D%5Cimplies%20%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%3D%5Ccfrac%7B25%7D%7B4%7D%5Cimplies%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Csqrt%7B%5Ccfrac%7B25%7D%7B4%7D%7D%20%5C%5C%5C%5C%5C%5C%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Ccfrac%7B%5Csqrt%7B25%7D%7D%7B%5Csqrt%7B4%7D%7D%5Cimplies%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Ccfrac%7B5%7D%7B2%7D%5Cimplies%20x%3D%5Ccfrac%7B5%7D%7B2%7D%2B%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20x%3D%5Ccfrac%7B6%7D%7B2%7D%5Cimplies%20%5Cboxed%7Bx%3D3%7D)
Step-by-step explanation:
since the ratio of children to adult is 2:7,
2+7 = 9
number of children would be
(2 × 2700)/9 = 600 children in the crowd
number of adults would be
(7 × 2700)/9 = 2100 adults in the crowd
So, if a ticket costs £8 per adults, total cash made from adult
= 8 × 2100 = £16800
if a ticket costs £(8 - 3) per children, total cash made from children
= 5 × 600 = £3000
total money made = £(16800 + 3000)
= £19800
