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grin007 [14]
3 years ago
8

Anna is solving this equation:

Mathematics
2 answers:
nordsb [41]3 years ago
3 0

Answer:

7-6x+15=-3x-7

29=3x

x=29/3

the asnwer is D

Mashcka [7]3 years ago
3 0
The answer is D. 7-6x+15=-3x-7
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SOMEONE HELP ME plzz
Leokris [45]

Answer:

B? I believe? sorry if it's wrong

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3 years ago
Use what you've learned in this unit to model the population of Western
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The population function of the Western Lowland Gorillas can either represent population growth or population decay

<h3>How to model the population</h3>

The question is incomplete, as the resources to model the population of the Western Lowland Gorillas are not given.

However, the following is a general guide to solve the question.

An exponential function is represented as:

y = a(1 \pm r)^x

Where:

  • (a) represent the initial value i.e. the initial population of the Western Lowland Gorillas
  • (r) represents the rate at which the population increases or decreases.
  • (x) represents the number of years since 2022
  • (y) represents the population in x years

Given that the population of the Western Lowland Gorillas decreases, then the rate of the function would be 1 -r (i.e. an exponential decay)

Take for instance:

y = 2000 * 0.98^x

By comparison.

a = 2000 and 1 - r = 0.98

The above function can be used to model the population of the Western Lowland Gorillas

Read more about exponential functions at:

brainly.com/question/26829092

5 0
2 years ago
Which on is it? need asap Thank you !!!
skad [1K]

Irrational number has unsolvable root overs.

Last two are now our options

\\ \bull\tt\dashrightarrow \sqrt{25}=5

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6 0
3 years ago
What is 7% of 300000000 and how do you work it out?
ArbitrLikvidat [17]

Answer:

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Step-by-step explanation:

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If you could subscribe to my friend penny_pg3d on yt that would be great

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\displaystyle\int\arctan x\,\mathrm dx=uv-\int v\,\mathrm du
\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx

For the remaining integral, setting y=1+x^2 gives \dfrac{\mathrm dy}2=x\,\mathrm dx, so

\displaystyle\int\frac x{1+x^2}\,\mathrm dx=\frac12\int\frac{\mathrm dy}y=\frac12\ln|y|+C=\frac12\ln(1+x^2)+C

Putting everything together, you end up with

\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C
6 0
3 years ago
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