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Tatiana [17]
3 years ago
5

The graph represents a linear function. Two of the points the line passes through are (5, 7) and (10, 14).

Mathematics
1 answer:
Arte-miy333 [17]3 years ago
7 0

Answer:

slope=1.4

Basically m=y2-y1/x2-x1 (plug values)

m=14-7/10-5

m=7/5

m=1.4

Step-by-step explanation:

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becoz according to me when I solved this sum then answer was 5√3

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LaDonna is running for city council. She needs to get four fifths 4 5 of her votes from senior citizens and 40 comma 00040,000 v
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7 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

<u />

<u>Distance between two points</u>

\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 <u>distance formula</u> 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
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