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2 years ago

7

Is (5,0) , (4,2), (3,4), (2,6), (1,8) a linear function?

1 answer:

Wittaler [7]2 years ago

4 0

Answer:

Yes

Step-by-step explanation:

Is (5,0) , (4,2), (3,4), (2,6), (1,8) a linear function?

Yes, it is a linear function.

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Find the length of RS with endpoints R(4,5) and S(-1, – 4).

Artist 52 [7]

Answer:

$d=\sqrt{106}$

Step-by-step explanation:

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

Simply plug in your 2 coordinates into the distance formula to find distance <em>d</em>:

$d=\sqrt{(-1-4)^2+(-4-5)^2}$

$d=\sqrt{(-5)^2+(-9)^2}$

$d=\sqrt{25+81}$

$d=\sqrt{106}$

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3 years ago

Point B is collinear with A and C and partitions AC in a 3:4 ratio. B is located at (4,1) and C is located at (12,5). Find the c

GrogVix [38]

Answer:

$A(-\frac{20}{3} ,-\frac{13}{3})$

Step-by-step explanation:

Point B partitions AC in a 3:4 ratio.

B is located at (4,1) and C is located at (12,5).

Let $(x_1,y_1)$ be the coordinates of A.

Then $(\frac{mx_1+nx_2}{m+n}, \frac{my_1+ny_2}{m+n})=(4,1)$

But m:n=3:4 implies m=3 and n=4 and $(x_2=12,y_2=5)$

$(\frac{3x_1+4*12}{3+4}, \frac{3y_1+4*5}{3+4})=(4,1)$

$(\frac{3x_1+48}{7}, \frac{3y_1+20}{7})=(4,1)$

$\frac{3x_1+48}{7}=4, \frac{3y_1+20}{7}=1$

$3x_1+48=28, 3y_1+20=7$

$3x_1=-20, 3y_1=-13$

$x_1=-\frac{20}{3} , y_1=-\frac{13}{3}$

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