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3 years ago

13

do yall ever just get into a terrible mood and yoy dont know how you got into it but you are and now you yell at averyone and th

en they get mad at you and yell back and then ur in a worse mood and life is just a never ending cycle of madness? just me?

1 answer:

Sonja [21]3 years ago

7 0

Answer:

I think it's just about everyone

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If a material has a high index of refraction, it has a 1 low optical density, and light travels more slowly through it. . high o

Archy [21]

Answer:

B. High optical density, and light travels more slowly through it.

Step-by-step explanation:

Refractive index is the property of a material that describes the change in the initial path of light and the rate at which light travels through it. And optical density of a material describes its absorption ability to rays or beam of light passing through it.

As the index of refraction of a material increases, therefore its optical density would increase. This would cause light to travel through it slowly. Unless the light has a high frequency.

A material with a high index of refraction has a high optical density, and light travels more slowly through it.

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3 years ago

Which graph represents a function with a growth factor of 5?

seropon [69]

Answer:

what graph

Step-by-step explanation:

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3 years ago

Read 2 more answers

PLEASE HELP BRANNIEST

Goshia [24]

i believe its graph D

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3 years ago

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In 2013, the moose population in a park was measured to be 5,100. By 2018, the population was measured again to be 5,200. If the

natka813 [3]

Answer:

$P(t) = 5100e^{0.0039t}$

Step-by-step explanation:

The exponential model for the population in t years after 2013 is given by:

$P(t) = P(0)e^{rt}$

In which P(0) is the population in 2013 and r is the growth rate.

In 2013, the moose population in a park was measured to be 5,100

This means that $P(0) = 5100$

So

$P(t) = 5100e^{rt}$

By 2018, the population was measured again to be 5,200.

2018 is 2018-2013 = 5 years after 2013.

So this means that $P(5) = 5200$ .

We use this to find r.

$P(t) = 5100e^{rt}$

$5200 = 5100e^{5r}$

$e^{5r} = \frac{52}{51}$

$\ln{e^{5r}} = \ln{\frac{52}{51}}$

$5r = \ln{\frac{52}{51}}$

$r = \frac{\ln{\frac{52}{51}}}{5}$

$r = 0.0039$

So the equation for the moose population is:

$P(t) = 5100e^{0.0039t}$

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3 years ago

Edward is playing a game where he draws cards with integers on them from a deck of cards. If the integer is positive he moves fo

Murljashka [212]

B) 4 steps in front of where he started.

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4 years ago