Use the graph to write a linear function that relates Y toX

Mathematics

1 answer:

stich3 [128]2 years ago

8 0

In the graph you can see the point are crossing Y -1 and x= 2,
x-y=xy
2-1=1
Y=1

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!!!I WILL MARK BRAINLIEST! PLZ ANSWER ASAP AND EXPLAIN! VERY IMPORTANT!!

Nata [24]

Answer: Check explanation

Step-by-step explanation: (3,-5), because that's the point where the two lines intersect.

Y=-2x+1

Y=-3/2x-1/2

Substitute the Y to get -2x+1=-3/2x-1/2

1/2x=3/2

x = 3

Now plug x into first equation: y=-2*3+1=-5

so the intersection point is (3,-5)

3 0

3 years ago

Find the distance of the line segment joining the two points (-4 /2 - /12) and (/32, 2/3)

astraxan [27]

Answer: $4\sqrt{3}$ .

Step-by-step explanation:

Distance formula : Distance between points (a,b) and (c,d) is given by :-

$D=\sqrt{(d-b)^2+(b-a)^2}$

Distance between points $(-4\sqrt{2},\sqrt{12}) \text{ and }(-\sqrt{32}, 2\sqrt{3})$ .

$D=\sqrt{(2\sqrt{3}-(-\sqrt{12}))^2+(-\sqrt{32}-(-4\sqrt{2}))}\\\\=\sqrt{(2\sqrt{3}+\sqrt{2\times2\times3})^2+(-\sqrt{4\times4\times2}+4\sqrt{2})^2}\\\\=\sqrt{(2\sqrt{3}-\sqrt{2^2\times3})^2+(-\sqrt{4^2\times2}+4\sqrt{2})^2}\\\\=\sqrt{(2\sqrt{3}+2\sqrt{3})^2+(-4\sqrt{2}+4\sqrt{2})^2}\\\\=\sqrt{(4\sqrt{3})^2+0}\\\\=4\sqrt{3}\text{ units}$