Ask question

Ask question

All categories

3 years ago

8

Plot the following equation using the x- and y-intercepts . Include at least one additional point on the line. Prove that the -i

ntercept, y- intercept and additional point are all solutions to the equation . 2x - 3y = 6 Complete your work in the space provided or upload a file that can display math symbols if your work requires it. Include the graph , and y- intercepts , additional point and proof in your answer .

1 answer:

sergejj [24]3 years ago

4 0

Answer:

2x-3y=6

X will = 0 Y will = 0

Therefore the x intercept would be 2 and the y intercept would be 6.

This is what i put lol im not sure, i have an A plus in math so this wont hurt it.

You might be interested in

Select all the values for that indicate r a positive slope for the line of best fit.

Whitepunk [10]

Answer:

u didn't show anything so I can't help u. can u show a picture or something

8 0

4 years ago

Determine if the given lines are parallel, perpendicular, or neither.

Leni [432]

Answer:

The lines are parallel.

Step-by-step explanation:

Given the lines

$-2x-3y=9$

$4x+6y=24$

Writing the equations in the slope-intercept form

$y=mx+b$

where m is the slope and b is the y-intercept

$-2x-3y=9$

$y=-\frac{2}{3}x-3$

Thus, the slope = m₁ = -2/3

also

$4x+6y=24$

$y=-\frac{2}{3}x+4$

Thus, the slope = m₂ = -2/3

We know that parallel lines have equal slopes

As

m₁ = m₂

Thus, the lines are parallel.

4 0

3 years ago

Parallelogram ABCD is a rectangle.

Ugo [173]

The correct answer is y = 11

because:
2y + 7 = 3y − 4
−y + 7 =−4

Subtract 7 from both sides.
−y + 7 − 7 = − 4 − 7
−y = −11

Divide both sides by -1.
y = 11

8 0

3 years ago

Hggghggggggggggggggggg

lidiya [134]

Answer:

(3+19)-12 = 3 + (19-12)

Step-by-step explanation:

6 0

3 years ago

Identify the scale factor used to graph the image below

Vlad [161]

Answer:

$Scale = \frac{2}{3}$

Step-by-step explanation:

To solve this question, I'll use the coordinates of SV and S'V' as points of reference.

From the attachment:

$S = (3,6)$

$V = (15,3)$

$S' = (2,4)$

$V' = (10,2)$

First, calculate distance SV

Distance formula is:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

For SV:

$d_1 = \sqrt{(15 - 3)^2 + (3 - 6)^2}$

$d_1 = \sqrt{(12)^2 + (-3)^2}$

$d_1 = \sqrt{144 + 9}$

$d_1 = \sqrt{153}$

For S'V'

$d_2 = \sqrt{(10 - 2)^2 + (2 - 4)^2}$

$d_2 = \sqrt{64 + 4}$

$d_2 = \sqrt{68}$

The scale factor is then calculated using:

$Scale = \frac{S'V'}{SV}$

$Scale = \frac{d_2}{d_1}$

Substitute values of d1 and d2

$Scale = \frac{\sqrt{68}}{\sqrt{153}}$

$Scale = \sqrt{\frac{68}{153}}$

$Scale = \sqrt{\frac{1}{2.25}}$

$Scale = \frac{1}{1.5}$

Express as a proper fraction

$Scale = \frac{1*2}{1.5*2}$

$Scale = \frac{2}{3}$

7 0

3 years ago