PLEASE PLEASE HELP ME!! I AM IN TESPARATE NEED OF HELP!!! PLEASE I AM BEGGING AT THIS POINT!!
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Answer:

a)  13/10

b) [tex]y+15=\dfrac{13}{10}(x-5)[/tex]

c) [tex]y+2=\dfrac{13}{10}(x-15)[/tex]

d) Both equations are [tex]y=\dfrac{13}{10}x-\dfrac{43}{2}[/tex] so represent the same line

e) [tex]-\dfrac{43}{2}[/tex] °F  ⇒ y-intercept

Step-by-step explanation:

a)  Choose two ordered pairs from the table: (5, -15) and (15, -2)

Let [tex](x_1,y_1)[/tex] = (5, -15)

Let [tex](x_2,y_2)[/tex] = (15, -2)

Use the slope formula to find slope: [tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

[tex]\implies m=\dfrac{-2+15}{15-5}=\dfrac{13}{10}[/tex]

b)  Use point-slope with (5, -15)

[tex]\implies y-y_1=m(x-x_1)[/tex]

[tex]\implies y+15=\dfrac{13}{10}(x-5)[/tex]

c)  Use point-slope with (15, -2)

[tex]\implies y-y_1=m(x-x_1)[/tex]

[tex]\implies y+2=\dfrac{13}{10}(x-15)[/tex]

d)

[tex]\implies y+15=\dfrac{13}{10}(x-5)[/tex]

[tex]\implies y=\dfrac{13}{10}x-\dfrac{43}{2}[/tex]

[tex]\implies y+2=\dfrac{13}{10}(x-15)[/tex]

[tex]\implies y=\dfrac{13}{10}x-\dfrac{43}{2}[/tex]

Therefore, they represent the same line

e) when x = 0:

[tex]y=\dfrac{13}{10}(0)-\dfrac{43}{2}=-\dfrac{43}{2}[/tex] °F

Represents the y-intercept on the graph

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