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AleksandrR [38]

4 years ago

15

The figure shows two semicircles with centers K & M. The semicircles are tangent to each other at point J, and (QN) ⃗ is tan

gent to both circles at N & O. If KL=JP=12, what is OQ?

1 answer:

kodGreya [7K]4 years ago

8 0

Answer: OQ = 34

Step-by-step explanation:

Please find the attached files for the solution

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ANSWER ASAP!! Find the area of a triangle with legs that are: 12 m, 15 m, and 9 m.

Setler79 [48]

Answer:

<h2>54m²</h2>

Step-by-step explanation:

<h3>METHOD 1:</h3>

You can use the Heron's formula:

$A=\sqrt{p(p-a)(p-b)(p-c)}$

where

<em>p</em><em> - half of perimeter</em>

<em>a, b, c</em><em> - lengths of sides</em>

We have

$a=12m;\ b=15m;\ c=9m$

Calculate:

$p=\dfrac{12+15+9}{2}=\dfrac{36}{2}=18\ (m)\\\\A=\sqrt{18(18-12)+(18-15)(18-9)}\\\\A=\sqrt{(18)(6)(3)(9)}\\\\A=\sqrt{2916}\\\\A=54\ (m^2)$

<h3>METHOD 2:</h3>

Let's check that it is not a right triangle.

If the sum of the squares of the two shorter sides is equal to the square of the longest side, then this triangle is rectangular.

We have

$9m < 12m$

Check:

$9^2+12^2=81+144=225\\15^2=225$

This is a right trianglr wherew 9m and 12m are legs and 15m is a hypotenuse.

The formula of an area of a right triangle is:

$A=\dfrac{ab}{2}$

<em>a, b</em><em> - legs</em>

Substitute:

$A=\dfrac{(9)(12)}{2}=\dfrac{108}{2}=54\ (m^2)$

6 0

3 years ago

During optimal conditions, the rate of change of the population of a certain organism is proportional to the population at time

Lana71 [14]

Answer:

The population is of 500 after 10.22 hours.

Step-by-step explanation:

The rate of change of the population of a certain organism is proportional to the population at time t, in hours.

This means that the population can be modeled by the following differential equation:

$\frac{dP}{dt} = Pr$

In which r is the growth rate.

Solving by separation of variables, then integrating both sides, we have that:

$\frac{dP}{P} = r dt$

$\int \frac{dP}{P} = \int r dt$

$\ln{P} = rt + K$

Applying the exponential to both sides:

$P(t) = Ke^{rt}$

In which K is the initial population.

At time t = 0 hours, the population is 300.

This means that K = 300. So

$P(t) = 300e^{rt}$

At time t = 24 hours, the population is 1000.

This means that P(24) = 1000. We use this to find the growth rate. So

$P(t) = 300e^{rt}$

$1000 = 300e^{24r}$

$e^{24r} = \frac{1000}{300}$

$e^{24r} = \frac{10}{3}$

$\ln{e^{24r}} = \ln{\frac{10}{3}}$

$24r = \ln{\frac{10}{3}}$

$r = \frac{\ln{\frac{10}{3}}}{24}$

$r = 0.05$

So

$P(t) = 300e^{0.05t}$

At what time t is the population 500?

This is t for which P(t) = 500. So

$P(t) = 300e^{0.05t}$

$500 = 300e^{0.05t}$

$e^{0.05t} = \frac{500}{300}$

$e^{0.05t} = \frac{5}{3}$

$\ln{e^{0.05t}} = \ln{\frac{5}{3}}$

$0.05t = \ln{\frac{5}{3}}$

$t = \frac{\ln{\frac{5}{3}}}{0.05}$

$t = 10.22$

The population is of 500 after 10.22 hours.

7 0

3 years ago

Why is graphing not always the best option for solving systems?

RideAnS [48]

1, Might not be precise
2, Can be hard to plot the graphs
3, If there are many solutions that will be really hard to plot sometimes.

This is just from the top of my head.

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4 years ago

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Help i’m doing my exam

jolli1 [7]

Dude I don’t think using this for an exam is it but ok

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3 years ago

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The graph below shows the solution to which system of inequalities?

solong [7]

Answer:

C

Step-by-step explanation:

The graph shows it clearly.

6 0

3 years ago