Plz help. I rally need it

HhhhhhhhhhhhhhhhhhhhhhheeeeeeeeeeeeeeeelpppPP!!!!!!!! pls.

soldi70 [24.7K]

N = 2
M + 5

Hope that helped :)

3 0

3 years ago

a line from the point (2,3) is perpendicular to the line y=1/3x+1. The two lines meet at the point p. Find the coordinates of P.

Helga [31]

Answer:

(2.4,1.8)

Step-by-step explanation:

the answer is in the image above

3 0

3 years ago

A gumball has a diameter that is 66 mm. The diameter of the gumball's spherical hollow core is 58 mm. What is the volume of the

grigory [225]

Answer:

Volume of gumball without including its hollow core is 48347.6 cubic mm.

Step-by-step explanation:

Given:

Diameter of Gumball = 66 mm

Since radius is half of diameter.

Radius of gumball = $\frac{diameter}{2}=\frac{66}{2} =33 \ mm$

Now We will first find the Volume of Gumball.

To find the Volume of Gumball we will use volume of sphere which is given as;

Volume of Sphere = $\frac{4}{3}\pi r^3$

Now Volume of Gumball = $\frac{4}{3}\times3.14 \times (33)^3 = 150456.24 \ mm^3$

Also Given

Diameter of gumball's spherical hollow core = 58 mm

Since radius is half of diameter.

Radius of gumball's spherical hollow core = $\frac{diameter}{2}=\frac{58}{2} =29 \ mm$

Now We will find the Volume of gumball's spherical hollow core.

Volume of Sphere = $\frac{4}{3}\pi r^3$

So Volume of gumball's spherical hollow core = $\frac{4}{3}\times3.14 \times (29)^3 = 102108.61 \ mm^3$

Now We need to find volume of the gumball without including its hollow core.

So, To find volume of the gumball without including its hollow core we would Subtract Volume of gumball spherical hollow core from Volume of Gumball.

volume of the gumball without including its hollow core = Volume of Gumball - Volume of gumball's spherical hollow core = $150456.24\ mm^3 - 102108.61\ mm^3 = 48347.63\ mm^3$

Rounding to nearest tenth we get;

volume of the gumball without including its hollow core = $48347.6\ mm^3$

Hence Volume of gumball without including its hollow core is 48347.6 cubic mm.

8 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

Plssssssssssssssssssssssssssssssssssssss

solniwko [45]

Answer:

c. quadrilateral

Step-by-step explanation:

All of the sides are different lengths, so the quadrilateral cannot be a parallelogram, rhombus, or square.

Its best descriptor is parallelogram.

_____

A parallelogram has opposite sides parallel and congruent. A rhombus also has adjacent sides congruent. A square is a special case of rhombus in which the corner angles are right angles.

5 0

2 years ago