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Orlov [11]
3 years ago
13

How do you graph a system of equation

Mathematics
1 answer:
Greeley [361]3 years ago
3 0
So, first your going to have to put it in y=mx+b
2x-2y=14
-2x -2x
-2y=-2x+14
now divide everything by -2
y=1x-7
and then do the same to the other equation.
x-y=2
-x. -x
-y=-x+2
now divide everything by -1
y=x-2

Then you just graph it. You plot the y-intercept on the y axis. and then I this case go up one over one and your get the points for each player equation.
y=1x-7. y- intercept:-7. slope:1 over 1
y=x-2. y-intercept:-2. slope:1 over 1
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solniwko [45]

Answer:

a) (0.555, 0) and (6, 0)

b) r = -3 and r = 1.8

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Step-by-step explanation:

Set each term in the numerator and denominator equal to 0 and find r.

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r = 7/8, 5/9, or 6

In the denominator:

r = 9/5, 7/8, or -3

Zeros in the numerator that aren't in the denominator are r-intercepts.

Zeros in the denominator that aren't in the numerator are vertical asymptotes.

Zeros in both the numerator and the denominator are holes.

a) (0.555, 0) and (6, 0)

b) r = -3 and r = 1.8

c) Evaluate m(r) at r = 7/8.  To do that, first divide out the common term (-8r + 7) from the numerator and denominator.

m(r) = (-9r+5)² (r−6)² / ( (-5r+9)² (r+3)² )

m(⅞) = (-9×⅞+5)² (⅞−6)² / ( (-5×⅞+9)² (⅞+3)² )

m(⅞) = (-23/8)² (-41/8)² / ( (37/8)² (31/8)² )

m(⅞) = (-23)² (-41)² / ( (37)² (31)² )

m(⅞) = 0.676

The hole is at (0.875, 0.676).

d) Evaluate m(r) at r = 0.

m(0) = (-9×0+5)² (0−6)² / ( (-5×0+9)² (0+3)² )

m(0) = (5)² (-6)² / ( (9)² (3)² )

m(0) = 1.235

The m(r)-intercept is (0, 1.235).

5 0
3 years ago
13
Assoli18 [71]
B)-3 for the vertex, do u have to graph it?
4 0
3 years ago
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