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3 years ago

How do you solve 28-2.2x=11.6x-54.8??

Mathematics

1 answer:

malfutka [58]3 years ago

6 0

28 - 2.2x = 11.6x - 54.8
28 + 54.8 = 11.6x + 2.2x
82.8 = 13.8x
82.8 / 13.8 = x
6 = x
or x = 6

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WORTH 30 PTS

kifflom [539]

Answer:

Done

Step-by-step explanation:

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3 years ago

Find X if m/_ UVH=50°, m/_HVW=25x-5, and m/_ UVW = 35x-5.

ra1l [238]

Step-by-step explanation:

first you have to solve for x

35x - 5=50

35-35x-5= 50-35

x-5=15

x- 5÷5=15÷5

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Now we have to substitute x into the equation.

35×(3)-5=

175-5=170

angle UVW= 170

to find angle HUW instead of solving the equation just subtract 170 from 50 and you will get 120.

angle UVH=50

angle HUW=120

angle UVW=170

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2 years ago

Karl's team score is 39 points. this was one point less than twice Todd's score. find Todd's score

EleoNora [17]

Answer: i'm pretty sure todd's score would be 19

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3 years ago

Read 2 more answers

Solve x^2-8x-9=0 using completing the square (show the work)

Elden [556K]

Answer: x=9,-1

Step-by-step explanation:

First, we are gonna start out with finding two numbers that multiply to equal -9 and add up to equal -8

Those numbers would be -9 & 1

-9+1=-8

-9x1=-9

Now we have

(x-9)(x+1)

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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.

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3 years ago