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3 years ago

What’s the equation of the line in fully simplified slope-intercept form

Mathematics

1 answer:

Vinil7 [7]3 years ago

6 0

Answer:

y= -x - 8

Step-by-step explanation:

To write an equation in slope-intercept form, we naturally need both the slope and the y-intercept. Here, our y-intercept is -8, and our slope is -1. The general form of a line in slope intercept form is y = mx + b, where m is the slope and b is the y intercept. Using our values of m = -1 and b = -8 then gives us the equation y = -x - 8.

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Simplify: -10 - 90x + 6x - 60

antoniya [11.8K]

<u><em>Answer:</em></u>

- 84x - 70

<em><u>Step-by-step explanation:</u></em>

We want to combine like terms:

-10 - 90x + 6x - 60

We can see (-10) and (-60) are both regular constants so:

- 10 - 60 = - 70

Then (-90x) and (+ 6x):

6x - 90x = - 84x

- 84x - 70

4 0

3 years ago

What is the definition of A net (of a shape)

Hatshy [7]

Answer: A pure mathematics internet could be a 2-dimensional form that may be rolled-up to make a three-d form or a solid. Or a internet could be a pattern created once the surface of a solid figure is ordered out flat showing every face of the figure. A solid could have totally different nets.

Step-by-step explanation: A pure mathematics web could be a 2-dimensional form which will be closed to make a third-dimensional form or a solid. Or a web could be a pattern created once the surface of a solid figure is arranged out flat showing every face of the figure. A solid could have completely different nets.

6 0

3 years ago

Read 2 more answers

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vodomira [7]

It's A.

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3 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$