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2 years ago

Yooo someone

Mathematics

1 answer:

natita [175]2 years ago

8 0

Answer:

39 sq cm

Step-by-step explanation:

Ok. Since ΔAED is isosceles, then the height of the trapezoid is 3 cm

One base is 16 and the other base is 16 - 3 - 3 = 10.

A = 1/2 h( $b_{1}$ + $b_{2}$ ) where $b_{1}$ and $b_{2}$ are lengths of the bases

A = 1/2(3)(10 + 16)

= 1/2(3)(26)

= 39 sq cm

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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

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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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