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3 years ago

Write the equation 3X plus 2Y equals eight in the slope intercept form. Show your work

Mathematics

1 answer:

Svet_ta [14]3 years ago

7 0

Answer:

Step-by-step explanation:

3x + 2y = 8

Making y as subject

2y = -3x + 8

Dividing by 2

y = -3x/2 + 8/2

y = -3x/2 + 4

Comparing with

y = mx + c

Our slope is m = -3/2

And y intercept c = 4

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A cylinder has a radius of 8 cm and a height of 14 cm. What is the volume of the cylinder?

sdas [7]

The volume of a cylinder is represented by the following equation...

V=πr^2 x h

Volume = π(64)(14)

Your volume is 896π

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3 years ago

A cell phone weighs about 2.8x 10pounds. Which value of n is most reasonable?

Damm [24]

Answer:0

Step-by-step explanation: took the quiz

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3 years ago

Cora gave her a $4.40 tip. The tip was 20% of the cost of the haircut. Write an equation to find b, the cost of the haircut .

Damm [24]

Answer:

3.52

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3 years ago

One month before the election, a poll of 630 randomly selected voters showed 54% planning to vote for a certain candidate. a wee

VikaD [51]

54% of 630 = 0.54 * 630 or approximately 340 of the 630 were planning to vote for the candidate.

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This means that the voter support for this candidate increased.

7 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.

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

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,

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

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

using the same approach as above.

5 0

2 years ago