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3 years ago

X =A.1B.3C.7Picture Provided Below

Mathematics

1 answer:

Yakvenalex [24]3 years ago

7 0

Answer:

B 3

Step-by-step explanation:

We can use the secant-secant formula:

(whole secant) x (external part) = (whole secant) x (external part)

9 * 4 = 12 *x

36 = 12x

Divide each side by 12

36/12 =12x/12

3 =x

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If 4 classrooms can seat 188 students how many classrooms would be needed to seat 517 students?

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Step-by-step explanation:

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The width and the length of a rectangle are consecutive even integers. If the width is decreased by 3 inches, then the area of t

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The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of

Alexandra [31]

Answer:

Population of mosquitoes in the area at any time t is:

$P(t) =504,943.26 -104,943.26e^{0.693t}$

Step-by-step explanation:

assume population at any time t = P(t)

population increases at a rate proportional to the current population:

⇒dP/dt ∝ P

$\implies \frac{dP}{dt} =kP$ ----(1)

where k is constant rate at which population is doubled

solving (1)

$ln|P(t)|=kt +C\\P(t)= e^{kt+C}\\P(t)=Ce^{kt}$

$t=0\\P(0) = P_{o}\\\implies C= P_{o}\\P(t) =P_{o}e^{kt}\\$ ---- (2)

initial population = 400,000

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⇒P(1)=2P(0)

Using (2)

$P_{o}e^{k(1)} = 2P_{o} e^{k(0)}\\$

$e^{k} =2\\k=ln|2|\\$

In presence of predators amount is decreased by 50,000 per day

Then amount decreased per week = 350,000

In this case (1) becomes

$\frac{dP}{dt}=kP-350,000\\\frac{dP}{dt} - kP=-350,000\\$ ---(3)

solving (3) by calculating integrating factor

$I.F=e^{\int-k dt}$

Multiplying I.F with all terms of (3)

$e^{-kt}\frac{dP}{dt} - ke^{-kt}P =-350,000 e^{-kt}\\\frac{d}{dt}(e^{-kt}P) = -350,000 e^{-kt}$

Integrating w.r.to t

$e^{-kt}P(t)= \frac{350,000e^{-kt}}{k} +C$

$P(t) =\frac{350,000}{k} +Ce^{kt}\\$

$k=ln|2| =0.693$

$P(t) =504,943.26 + Ce^{0.693t}\\$

at t=0

$P(0) =504,943.26 + Ce^{0.693(0)}$

$400,000 =504,943.26 + C$

$C = -104,943.26$

So, population of mosquitoes in the area at any time t is

$P(t) =504,943.26 -104,943.26e^{0.693t}$

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Shanti wrote the predicted values for a data set using the line of best fit y = 2.55x – 3.15. She computed two of the residual v

svet-max [94.6K]

<span>Residual value is the difference between the observed value of the dependent variable (y) and the predicted value (ŷ) in a data set.

i.e. Residual value = given value - predicted value

From the table, the residual value corresponding to a has 4.1 as the given value and 4.5 as the predicted value.
Therefore, a = 4.1 - 4.5 = -0.4

Similarly, </span><span>the residual value corresponding to b has 7.2 as the given value and 7.05 as the predicted value.
Therefore, b = 7.2 - 7.05 = 0.15
</span>
Therefore, a = -0.4 and b = 0.15

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