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2 years ago

What is the equation in slope intercept form of the line that contains the point (8, -2) and has a y-intercept of -4?

1 answer:

4 0

Y=mx+b,slop intercept form
we need to find the slope.we nned to point
(-8,2) is given you can get the other by understanding the defintion of y-intercept (0,-4)
$m = slope = \frac{ 2 - ( - 4)}{ - 8 - 0} = \frac{6}{ - 8}$
m=-6/8,b=y-intercept=-4
$y = \frac{ - 6}{8} x + ( - 4)$

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2 years ago

Lets see who is smart lol

Simora [160]

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Answer:

$\square\quad{y=\dfrac{1}{2}x+4}\\\\\square\quad{\dfrac{y-5}{x-2}=\dfrac{1}{2}}$

Step-by-step explanation:

The slope of a line is the same everywhere, so an equation can be written making use of that fact.

$\dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1}\\\\\dfrac{y-5}{x-2}=\dfrac{7-5}{6-2}=\dfrac{2}{4}\\\\\boxed{\dfrac{y-5}{x-2}=\dfrac{1}{2}}$

Cross-multiplying gives ...

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$\boxed{y=\dfrac{1}{2}x+4}$

These equations match choices B and D.

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2 years ago

−3 1/3 ÷ 9= What I need it now!

Contact [7]

Step 1. Convert 3 1/3 to improper fraction.

- 3 * 3 + 1/3 ÷ 9

Step 2. Simplify 3 * 3 to 9

-9 + 1/3 ÷ 9

Step 3. Simplify 9 + 1 to 10

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Step 4. Use this rule: a ÷ b/c = a * c/b

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Step 5. Use this rule: a/b * c/d = ac/bd

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Step 6. Simplify 10 * 1 to 10

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Step 7. Simplify 3 * 9 to 27

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3 years ago

James Works in a flower shop. He will put 36 tulips in vases for a wedding. He must use the same number of tulips in each vase.H

soldi70 [24.7K]

Answer:

1,2,3,4,6,9,12,18,36

Step-by-step explanation:

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2 years ago

Which is the approximate solution to the system y = 0.5x + 3.5 and y = − 2/3 x + 1/3 shown on the graph? (–2.7, 2.1) (–2.1, 2.7)

AlekseyPX

Answer:

The approximate solution to the system is $(-2.7, 2.1)$ .

Step-by-step explanation:

To solve the system of equations $\begin{bmatrix}y=0.5x+3.5\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix}$ you must:

$\mathrm{Rationalize\:equations}\\\\\begin{bmatrix}y=\left(\frac{1}{2}\right)x+\left(\frac{7}{2}\right)\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix}$

$\mathrm{Subsititute\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\begin{bmatrix}-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\end{bmatrix}$

$\mathrm{Isolate}\:x\:\mathrm{for}\:-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\\\\-\frac{2}{3}x=\frac{19}{6}+\frac{1}{2}x\\\\-\frac{7}{6}x=\frac{19}{6}\\\\6\left(-\frac{7}{6}x\right)=\frac{19\cdot \:6}{6}\\\\-7x=19\\\\x=-\frac{19}{7}\approx-2.7$

$\mathrm{For\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\mathrm{Subsititute\:}x=-\frac{19}{7}\\\\y=-\frac{2}{3}\left(-\frac{19}{7}\right)+\frac{1}{3}\\\\y=\frac{15}{7}\approx 2.1$

The approximate solutions to the system of equations are:

$x=-2.7 ,\:y=2.1$

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3 years ago

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