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Marrrta [24]
3 years ago
11

Homework help, one of the few I have

Mathematics
2 answers:
Alex73 [517]3 years ago
8 0

Answer:

x=4 and the solution is not extraneous

Step-by-step explanation:

\sqrt{2x+1}=3 \\\\2x+1=9 \\\\2x=8 \\\\x=4

This solution is not extraneous, because plugging it back in you get:

\sqrt{2(4)+1}=3 \\\\\sqrt{9}=3 \\\\3=3

Hope this helps!

Bogdan [553]3 years ago
7 0

Answer:

X = 4, solution is not extraneous

Step-by-step explanation:

3 = √9

so we get the following comparison;

2X + 1 = 9 (both minus 1)

2X = 8 (divide by 2)

X = 4

if we fill in 4 for X. we get

√2•4+ 1 = √9 = 3 so the solution is not extraneous

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The expression D=23g shows the distance that Kurt can drive on a tank of gasoline.

Step-by-step explanation:

Given,

Distance covered by Kurt's car = 23 miles per gallon of gasoline

Kurt fills up tank of his car with g gallons.

Number of gallons in car's tank = g gallons

Distance covered by g gallons = D = Distance covered by one gallon * Number of gallons

D = 23*g

D = 23g

The expression D=23g shows the distance that Kurt can drive on a tank of gasoline.

Keywords: distance, multiplication

Learn more about multiplication at:

  • brainly.com/question/101683
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3 years ago
Which statement describes the graph of f(x) = -4x3 - 28x2 - 32x + 64?
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Answer:

D=The graph touches at the x-axis at x=-4 and crosses the x-axis at x=1

Step-by-step explanation:

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3 years ago
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3. Tyler and his family went on a backpacking trip. They started hiking at 3:00 P.M. It took 2 hours and 30 minutes to reach the
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I'm pretty sure the answer is 7:00 pm
4 0
3 years ago
Integrate this two questions ​
tankabanditka [31]

Simplify the integrands by polynomial division.

\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)

\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)

Now computing the integrals is trivial.

5.

\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}

where we use the power rule,

\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)

and a substitution to integrate the last term,

\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)

8.

\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}

using the same approach as above.

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2 years ago
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The scanning electron microscope

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