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svet-max [94.6K]

10 months ago

13

Solve the following expression when k = 12 and v = 6 4 + 5 - 4v + 12

1 answer:

mrs_skeptik [129]10 months ago

8 0

The answer would be -3

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Please tell me answer of 6 7 and 8

zloy xaker [14]

Answer:

6.angle m=180-60=120

7.angle y=180-(60+65)=55

8 0

2 years ago

How can finding the slopes of MN and AB help in proving that MN is the midsegment of triangle ABC?

Mademuasel [1]

Mid-segment and bottom side of the triangle is always parallel to each other, and parallel lines must have same slopes, so if their slopes are equal, then MN would be mid-segment of side AB in Triangle ABC

Hope this helps!

4 0

3 years ago

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The worlds two highest dams are both in Tajikistan. The rogue dam is 35 meters taller than the Nurek dam. together they are 635

galina1969 [7]

This is a sum and difference problem, with the equations
R-N=35
R+N=635
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Larger number = (sum + difference)/2
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5 0

3 years ago

f What is the simplified expression for 5 to the power of negative 3 multiplied by 2 to the power of 2 multiplied by 5 to the po

Artyom0805 [142]

//5^-3 = 1/(5^3)
(2^2*5^6)/(5^3*2^3*5^2)=
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8 0

3 years ago

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Consider a rabbit population P(t) satisfying the logistic equation StartFraction dP Over dt EndFraction equals aP minus bP squa

maria [59]

Solution:

Given :

$\frac{dP}{dt}= aP-bP^2$ .............(1)

where, B = aP = birth rate

D = $bP^2$ = death rate

Now initial population at t = 0, we have

$P_0$ = 220 , $B_0$ = 9 , $D_0$ = 15

Now equation (1) can be written as :

$\frac{dP}{dt}=P(a-bP)$

$\frac{dP}{dt}=bP(\frac{a}{b}-P)$ .................(2)

Now this equation is similar to the logistic differential equation which is ,

$\frac{dP}{dt}=kP(M-P)$

where M = limiting population / carrying capacity

This gives us M = a/b

Now we can find the value of a and b at t=0 and substitute for M

$a_0=\frac{B_0}{P_0}$ and $b_0=\frac{D_0}{P_0^2}$

So, $M=\frac{B_0P_0}{D_0}$

= $\frac{9 \times 220}{15}$

= 132

Now from equation (2), we get the constants

k = b = $\frac{D_0}{P_0^2} = \frac{15}{220^2}$

= $\frac{3}{9680}$

The population P(t) from logistic equation is calculated by :

$P(t)= \frac{MP_0}{P_0+(M-P_0)e^{-kMt}}$

$P(t)= \frac{132 \times 220}{220+(132-220)e^{-\frac{3}{9680} \times132t}}$

$P(t)= \frac{29040}{220-88e^{-\frac{396}{9680} t}}$

As per question, P(t) = 110% of M

$\frac{110}{100} \times 132= \frac{29040}{220-88e^{\frac{-396}{9680} t}}$

$220-88e^{\frac{-99}{2420} t}=200$

$e^{\frac{-99}{2420} t}=\frac{5}{22}$

Now taking natural logs on both the sides we get

t = 36.216

Number of months = 36.216

8 0

3 years ago