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3 years ago

Solve the equation for . 2x+22=4(x+3)

Mathematics

1 answer:

neonofarm [45]3 years ago

8 0

Answer:

2x+22=4(x+3)

2x+22=4x+12

22-12=4x-2x

10=2x

5=x

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70 - q -q - 2q = 80 <br> plz help me i don't get it

Montano1993 [528]

Answer:

the answer is: q = -2.5

Step-by-step explanation:

70 - q - q - 2q = 80

70 - 4q = 80

-4q = 10

q = -2.5

3 0

3 years ago

32% of what number is 320

Vedmedyk [2.9K]

Hello,

Shall we begin?

32%..................320
100%.................x

32 * x = 320 * 100
32x = 32000
x = 32000/32
x = 1000

<span>Answers:1000</span>

6 0

3 years ago

What expressions is equivalent to −2(x−5)?

Ksenya-84 [330]

Answer:

-2x + 10

Step-by-step explanation:

Distribute -2 to x - 5 to simplify the expression. So we get: -2(x - 5) = -2x + 10.

6 0

3 years ago

Jus help im in a timed assigment

gogolik [260]

0.59
x 0.37
413
1770
2183

4 0

3 years ago

After the mill in a small town closed down in 1970, the population of that town started decreasing according to the law of expon

wlad13 [49]

Answer:

$P_o = \frac{143000}{e^{-20*0.01303024661}}=110193.69$

And we can round this to the nearest up integer and we got 110194.

Step-by-step explanation:

The natural growth and decay model is given by:

$\frac{dP}{dt}=kP$ (1)

Where P represent the population and t the time in years since 1970.

If we integrate both sides from equation (1) we got:

$\int \frac{dP}{P} =\int kdt$

$ln|P| =kt +c$

And if we apply exponentials on both sides we got:

$P= e^{kt} e^k$

And we can assume $e^k = P_o$

And we have this model:

$P(t) = P_o e^{kt}$

And for this case we want to find $P_o$

By 1990 we have t=20 years since 1970 and we have this equation:

$143000 = P_o e^{20k}$

And we can solve for $P_o$ like this:

$P_o = \frac{143000}{e^{20k}}$ (1)

By 2019 we have 49 years since 1970 the equation is given by:

$98000 = P_o e^{49k}$ (2)

And replacing $P_o$ from equation (1) we got:

$98000=\frac{143000}{e^{20k}} e^{49k} =143000 e^{29k}$

We can divide both sides by 143000 we got:

$\frac{98000}{143000} =0.685 = e^{29k}$

And if we apply ln on both sides we got:

$ln(0.685) = 29k$

And then k =-0.01303024661[/tex]

And replacing into equation (1) we got:

$P_o = \frac{143000}{e^{-20*0.01303024661}}=110193.69$

And we can round this to the nearest up integer and we got 110194.

7 0

3 years ago