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Nata [24]
3 years ago
13

What is 52,437 rounded to the nearest thousand

Mathematics
2 answers:
enyata [817]3 years ago
8 0
52,000.


abcdefghijklmnop
Tom [10]3 years ago
5 0

Answer:

52,000

Step-by-step explanation:

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Evaluate 6 C 3 .<br><br> 18<br> 20<br> 60<br> 120
Novosadov [1.4K]
The answer will be 6 x 5 x 4 / 3 x 2 x 1,    5 x 4 = 20. Hope this helps
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3 years ago
(50 points and Brainiest
Scorpion4ik [409]

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7 0
3 years ago
4x+3=-5 helppp please
Goryan [66]

Answer: x = -2

Step-by-step explanation: To solve for <em>x</em>, in the equation you see here, our goal is to get <em>x</em> by itself.

Our first step will be to isolate the term containing <em>x</em> which in this case is 4x. To isolate 4x, we have to get rid of the +3 by subtracting 3 from both sides of the equation. On the left, +3 and -3 cancel out and we're left with 4x. On the right, -5 - 3 simplifies to -8.

Now we have the equation 4x = -8.

To get <em>x</em> by itself, we divide both sides of the equation by 4. On the left, the 4's cancel and on the right, -8 divided by 3 is -2 so <em>x = -2</em>

5 0
3 years ago
Read 2 more answers
Help me please I’ll mark as brainliest!!
Vladimir79 [104]

Answer:

i wish i could help so sorry

Step-by-step explanation:

7 0
3 years ago
Suppose a certain population satisfies the logistic equation given by dP
Ksenya-84 [330]

Answer:

The population when t = 3 is 10.

Step-by-step explanation:

Suppose a certain population satisfies the logistic equation given by

\frac{dP}{dt}=10P-P^2

with P(0)=1. We need to find the population when t=3.

Using variable separable method we get

\frac{dP}{10P-P^2}=dt

Integrate both sides.

\int \frac{dP}{10P-P^2}=\int dt             .... (1)

Using partial fraction

\frac{1}{P(10-P)}=\frac{A}{P}+\frac{B}{(10-P)}

A=\frac{1}{10},B=\frac{1}{10}

Using these values the equation (1) can be written as

\int (\frac{1}{10P}+\frac{1}{10(10-P)})dP=\int dt

\int \frac{dP}{10P}+\int \frac{dP}{10(10-P)}=\int dt

On simplification we get

\frac{1}{10}\ln P-\frac{1}{10}\ln (10-P)=t+C

\frac{1}{10}(\ln \frac{P}{10-P})=t+C

We have P(0)=1

Substitute t=0 and P=1 in above equation.

\frac{1}{10}(\ln \frac{1}{10-1})=0+C

\frac{1}{10}(\ln \frac{1}{9})=C

The required equation is

\frac{1}{10}(\ln \frac{P}{10-P})=t+\frac{1}{10}(\ln \frac{1}{9})

Multiply both sides by 10.

\ln \frac{P}{10-P}=10t+\ln \frac{1}{9}

e^{\ln \frac{P}{10-P}}=e^{10t+\ln \frac{1}{9}}

\frac{P}{10-P}=\frac{1}{9}e^{10t}

Reciprocal it

\dfrac{10-P}{P}=9e^{-10t}

P(t)=\dfrac{10}{1+9e^{-10t}}

The population when t = 3 is

P(3)=\dfrac{10}{1+9e^{-10\cdot 3}}

Using calculator,

P=9.999\approx 10

Therefore, the population when t = 3 is 10.

8 0
3 years ago
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