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3 years ago

Please help I'm stuck

Mathematics

2 answers:

nalin [4]3 years ago

7 0

All you dois times it three times

blagie [28]3 years ago

4 0

1>>3d
2>>5k^3
3>>20k^5
4>>35h^4
5>>2w
6>>15x^5
7>>t^2

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A population of insects increases at the rate of 200 10t 13t2 what is the change in the population of insects between day 0 and

RUDIKE [14]

Answer:

762 days

Step-by-step explanation:

Given

$Rate = 200 + 10t + 13t^2$

Let the rate be R.

So the rate change with time is represented as:

$\frac{dR}{dt} = 200 +10t + 13t^2$

So:

$dR = (200 + 10t + 13t^2)\ dt$

To get the number of insects between day 0 and day 3, we need to integrate dR and set the bounds to 0 and 3

i.e.

$dR = (200 + 10t + 13t^2)\ dt$ becomes

$\int\limits^3_0 {dR} \, =\int\limits^3_0 (200 + 10t + 13t^2)\ dt$

$R =\int\limits^3_0 (200 + 10t + 13t^2)\ dt$

Integrate

$R = 200t + \frac{10t^2}{2} + \frac{13t^3}{3} [3,0]$

Solve for R by substituting 0 and 3 for t

$R = (200*3 + \frac{10*3^2}{2} + \frac{13*3^3}{3}) - ( 200*0 + \frac{10*0^2}{2} + \frac{13*0^3}{3})$

$R = (200*3 + \frac{10*9}{2} + \frac{13*27}{3}) - (0 + \frac{10}{2} + \frac{0}{3})$

$R = (200*3 + 5*9 + 13*9) - (0)$

$R = 600 + 45 + 117 - 0$

$R =762$

<em>The population of insect between the required interval is 762</em>

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Dovator [93]

To get the answer, you need to solve the equation for x = 7.

f(7) = 3/7 + 2 - sqrt(7) - 3.

= 3/7 - 1 - sqrt(7).

If you need to simplify it further:

= -4/7 - sqrt(7).

If you need it in decimal form: = -3.2171798824

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