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2 years ago

Given that triangle QRS is similar to triangle DEF, what is x?

Mathematics

2 answers:

marta [7]2 years ago

4 0

So the ratio is 40/26 = 1.54

24 ÷ 1.54 = 15.58

2x-4 = 15.58

2x = 19.58

x = 9.8

klasskru [66]2 years ago

3 0

Answer:

x = 9.8

Step-by-step explanation:

Given that triangle QRS is similar to triangle DEF.

Hence, the ratio of corresponding sides of these triangles are equal.

Thus, we have

$\frac{QR}{DE}=\frac{RT}{EB}$

Substituting the known values from the given figure

$\frac{26}{40}=\frac{2x-4}{24}$

Cross multiplying, we get

$40(2x-4)=26\cdot24\\\\80x-160=624\\\\80x=784\\\\x=\frac{784}{80}\\\\x=9.8$

The value of x is 9.8

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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>

4 0

3 years ago

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2 years ago

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slavikrds [6]

Answer:

Step-by-step explanation:

n - integer

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2 years ago

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2 years ago

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