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2 years ago

What is an equation of the line that passes through the points (-3, -1) and (-6,3)?

Mathematics

1 answer:

Angelina_Jolie [31]2 years ago

8 0

Answer:

y=-4/3 + 11

Step-by-step explanation:

1. Find the slope

-6-(-3)

slope= -4/3

2. Point slope form

y-3=-4/3(x+6)

y-3=-4/3x + 8

y=-4/3x+11

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sukhopar [10]

Answer:

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Step-by-step explanation:

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2 years ago

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If a=2b^3 and b= -1/2c ^-2,express a in terms of c.

ElenaW [278]

Answer:

Part 1) $a=-\frac{1}{4c^6}$

Part 2) $a=-\frac{1}{4c^{-6}}$

Step-by-step explanation:

Part 1) we have

$a=2b^{3}$ ----> equation A

$b=-\frac{1}{2}c^{-2}$ ----> equation B

substitute equation B in equation A

$a=2(-\frac{1}{2}c^{-2})^{3}$

Applying property of exponents

$(x^{m})^{n}=x^{m*n}$

$x^{-m} =\frac{1}{x^{m}}$

$a=2(-\frac{1}{2}c^{-2})^{3}=2(-\frac{1}{2})^3(c^{-2})^{3}=2(-\frac{1}{8})(c^{-6})=-\frac{1}{4c^6}$

therefore

$a=-\frac{1}{4c^6}$

Part 2) we have

$a=2b^{3}$ ----> equation A

$b=-\frac{1}{2c^{-2}}=-\frac{c^{2}}{2}$ ----> equation B

substitute equation B in equation A

$a=2(-\frac{c^{2}}{2})^{3}$

Applying property of exponents

$(x^{m})^{n}=x^{m*n}$

$x^{-m} =\frac{1}{x^{m}}$

$a=2(-\frac{c^{2}}{2})^{3}=2(-\frac{c^{6}}{8})$

simplify

$a=-\frac{c^{6}}{4}$

therefore

$a=-\frac{1}{4c^{-6}}$

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3 years ago

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2 years ago

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3 years ago

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Draw clearly the graph of the linear equation. y=1/2x, where x= (-4 -2, 0, 2, 4)

aniked [119]

Answer:

(in attachment)

Step-by-step explanation:

you can find the points by inputting the x-values into the equation to solve for the y-values, then connecting the plotted points to create the line.

When x=-4

y=1/2(-4)

y=-2

(-4,-2)

Repeat for all values.

6 0

2 years ago