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4 years ago

A line with a slope through the points (0,4) and (-6,v). What is the value of v

1 answer:

3 0

You should probably revise this and tell us what the slope is. If you do that, I'll come back and edit my answer. Sorry, I should have put this in a comment.

Givens

Slope = 2

y2 = 4

y1 = v

x2 = 0

x1 = -6

Formula

m = (y2 - y1) / (x2 - x1)

Solution

Substitute

2 = (4 - v) / (0 - - 6)

2 = (4 - v) / 6 Multiply both sides by 6

2*6 = 4 - v Combine the left

12 = 4 - v Subtract 4 from both sides

12-4 = 4-4 - v Combine the left and the right

8 = - v Divide by - 1

-8 = v

Answer: v = -8

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3 years ago

Q3: Identify the graph of the equation and write and equation of the translated or rotated graph in general form. (Picture Provi

natta225 [31]

Answer:

b. circle; $2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

Step-by-step explanation:

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The new center is obtained using;

$x'=x\cos(\theta)+y\sin(\theta)$ and $y'=-x\sin(\theta)+y\cos(\theta)$

We plug in the given angle with x and y values to get;

$x'=(\frac{5}{2})\cos(\frac{\pi}{3})+(0)\sin(\frac{\pi}{3})$ and $y'=--(\frac{5}{2})\sin(\frac{\pi}{3})+(0)\cos(\frac{\pi}{3})$

This gives us;

$x'=\frac{5}{4} ,y'=\frac{5\sqrt{3} }{4}$

The equation of the rotated circle is;

$(x'-\frac{5}{4})^2+(y'-\frac{5\sqrt{3} }{4})^2=\frac{37}{4}$

Expand;

$(x')^2+(y')^2-\frac{5\sqrt{3} }{2}y'-\frac{5}{2}x'+\frac{25}{4} =\frac{37}{4}$

Multiply through by 4; to get

$4(x')^2+4(y')^2-10\sqrt{3}y'-10x'+25 =37$

Write in general form;

$4(x')^2+4(y')^2-10x'-10\sqrt{3}y'-12 =0$

Divide through by 2.

$2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

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3 years ago

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Answer:

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4 years ago

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