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3 years ago

Which of the following is a point-slope equation of a line that passes through the points (-1,4) and (8,-2)

Mathematics

1 answer:

dexar [7]3 years ago

4 0

Answer:

y = (2/3)x + 4 2/3

Step-by-step explanation:

For reference, a linear function has an equation of y = mx + b

The first thing we're going to do is use the slope formula, $\frac{y2-y1}{x2-x1}$ or rise over run. In this case that looks like:

(-2-4) - (8 - -1)

-2 -4 comes out to - 6, and the denominator comes out to 9.

6/9 is equal to 2/3, which is our m or slope value.

Next we want to find the b value, or our y-intercept. To do that we'll plug one of the points into the equation with our new-found slope:

4 = (2/3)-1 +b

4 = -2/3 + b

4 2/3 = b

That means our complete equation is:

y = (2/3)x + 4 2/3

y = (2/3)x + 14/3 (equivalent but an improper fraction rather than a mixed fraction).

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marishachu [46]

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3 years ago

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Find the value of the discriminant for <img src="https://tex.z-dn.net/?f=7x%5E%7B2%7D%20%2B5x%2B1%3D0" id="TexFormula1" title="7

kondor19780726 [428]

Answer:

No real roots

Step-by-step explanation:

Given

7x² + 5x + 1 = 0 ← in standard form

with a = 7, b = 5, c = 1

To determine the nature of the roots use the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then roots are real and distinct

• If b² - 4ac = 0 then roots are real and equal

• If b² - 4ac < 0 then the roots are not real

Here

b² - 4ac = 5² - (4 × 7 × 1) = 25 - 28 = - 3

Thus the 2 roots are not real

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3 years ago

A. Mr. Gordon works with Ms. Patel from the Example. Mr. Gordon writes the expression 3x + 4x to represent the cost of ordering

Leviafan [203]

Answer:

$162

Step-by-step explanation:

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100+42+20=162

5r+10p+14t=$162

3 0

2 years ago

Which is the factored form of 3y − 18?

Lesechka [4]

3y-18
Pull out 3 (common factor):
3(y-6)

A) 3(y-6) is the correct answer

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3 years ago

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A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

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3 years ago