All categories

3 years ago

Choose the equation below that represents the line passing through the point (−2, −3) with a slope of −6.

1 answer:

7 0

Y=mx+b where m=-6, x=-2, and y=-3. Substituting these values in would result in -3=(-6)(-2)+b, which would simplify to -3=12+b, and further to b=-15. This would mean the y intercept is equal to -15, or (0,-15). To check, (-6)*(-2)-15 is in fact equal to -3, thus allowing easy identification of a graph, which will cross the y axis at -15 and decrease in y-value by 6 with every increase in x-value.

You might be interested in

This unit, you have learned several formulas and equations. Match the formula or equation with what the answer represents.

Sedbober [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

Sabrina threw a coin from a building 160 feet high at an initial speed of 30 feet per second. How many seconds will it take the

Mkey [24]

The answer would be 5.3 Seconds.

6 0

3 years ago

7+77+7+7+7+7+7++7+7+7+7+7+7+6-6-6-6-6-6

Irina-Kira [14]

Step-by-step explanation:

It is easy

You can do by calculater

7 0

3 years ago

Read 2 more answers

A man spent 58% of his income and the amount of money left was $210. Find his income

professor190 [17]

Answer:

500 $

Step-by-step explanation:

$\frac{42\%}{210}=\frac{100\%}{x}$

$\frac{\frac{21}{50}}{210}=\frac{1}{x}$

$\frac{21}{50}x=210\\$

$\frac{21}{50}x\cdot \:50=210\cdot \:50\\$

x=500

Red

6 0

3 years ago

Please help, the question below

Licemer1 [7]

Answer:

$\sf R=\left(3, -\dfrac{5}{4}\right)$

Step-by-step explanation:

Given:

P = (1, 5)
Q = (-2, 3)
R = (a, b)
R lies on line x = 3
PR = QR

If point R lies on the line x = 3, then the x-value of point R is 3.

⇒ a = 3

Distance between two points

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

(where (x₁, y₁) and (x₂, y₂) are the two points)

Use the distance formula to derive equations for PR and QR.

Let (x₁, y₁) = P = (1, 5)

Let (x₂, y₂) = R = (3, b)

$\begin{aligned} \sf PR & =\sf \sqrt{(3-1)^2+(b-5)^2}\\ & = \sf \sqrt{4+(b-5)^2} \end{aligned}$

Let (x₁, y₁) = Q = (-2, 3)

Let (x₂, y₂) = R = (3, b)

$\begin{aligned} \sf QR & =\sf \sqrt{(3-(-2))^2+(b-3)^2}\\ & = \sf \sqrt{25+(b-3)^2} \end{aligned}$

As PR = QR, equate the derived equations and solve for b:

$\begin{aligned} \sf PR & = \sf QR \\\sf \sqrt{4+(b-5)^2} & = \sf \sqrt{25+(b-3)^2}\\\sf 4+(b-5)^2 & = \sf 25+(b-3)^2\\\sf 4+b^2-10b+25 & = \sf 25 + b^2-6b+9\\\sf b^2-10b+29 & = \sf b^2 -6b +34\\\sf -10b+29 & = \sf -6b + 34\\\sf -4b & = \sf 5\\\sf b & = -\dfrac{5}{4}\end{aligned}$

Substitute the found values of a and b to find the coordinates of R:

$\sf R=(a,b)=\left(3, -\dfrac{5}{4}\right)$

Learn more about the distance formula here:

brainly.com/question/28144723

8 0

2 years ago

Read 2 more answers