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Reil [10]
3 years ago
7

Name at least 3 key "things" to remember when you are Setting up and Solving Similar Figures Proportions, ANSWER NOW! PLEASE

Mathematics
2 answers:
Drupady [299]3 years ago
5 0
Vertical Horizontal and Diagonal
vladimir2022 [97]3 years ago
4 0

Answer:

Vertical.

Horizontal.

Diagonal

Step-by-step explanation:

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What is 12.165 in word from
Kamila [148]
Twelve and one hundred sixty-five thousandths
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3 years ago
AB ← → is tangent to circle O at point A. If m∠AOB = 55°, what is m∠ABO ?
AlladinOne [14]
A tangent forms a right angle with the centre of circle at the point it touches the circumference. i.e. OAB = 90 degrees
If AOB = 55 degrees, then ABO = 90 - 55 = 35 degrees
3 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
The circulation of a newsletter decreased from 6000 to 3710. Find the percent of decrease in circulation to the nearest percent.
lbvjy [14]
Percent change = ((new value) - (old value))/(old value) × 100%
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The decrease in circulation is about 38%.
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The answer is ∠S …..
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