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vitfil [10]
3 years ago
13

The coordinate grid shows the plot of four equations. A coordinate plane is shown with four lines graphed. Line A crosses the x

axis at negative 3.2 and has a slope of 4. Line B crosses the x axis at 3 and the y axis at 6. Line C crosses the x axis at 1.5 and the y axis at negative 6. Line D crosses the x axis at negative 3 and the y axis at 3. Which set of equations has (2, 2) as its solution? A and D B and D A and C B and C
Mathematics
2 answers:
astraxan [27]3 years ago
4 0
Line B and C is your answer

Vitek1552 [10]3 years ago
3 0
Let m be the slope and b the y-intercept:

m=(y-y)/(x-x) and y=mx+b

Line A 

y=4x-3.2

Line B 

(3,0) and (0,6)

m= 6/-3= -2

y= -2x+b

0=-2(3)+b

b=6

y= -2x+6

Line C

(1.5,0) and (0,-6)

m=-6/-1.5=4

y=4x+b

0=4(1.5)+b

b = -6

y=4x-6

Line D

(-3,0) and (0,3)

m=3/3=1

y=x+b

0=-3+b

b=3

y=x+3



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Alegebra liner equations.
luda_lava [24]
1. The answer would be (D). This is because you will plug is the (x) of the ordered pair into the equation(y=-2x), so first off would be (y=-2(-2)). since your multiplying a negative time a negative, you will get your (y) which is 4. Now you do this with the rest. -2(1)=-2 and -2(3)=-6.
2. This would be (A). This answer is simple. First you will take one of the equations (y=-x-1) and plug x into it's place so, y=-3-1, getting your answer(y) as being -4. Now you do the same with the other pair. y=-5-1, getting -6, showing (A) as your answer.
3.The answer will be (A)y is 2 times x. this is because you wou would make out the equation from the answer (y=2x). Now knowing your equation, all thats left is plugging in your (x) factors. so y=2(1), getting your (y) value of 2 and you would do this with the rest of your x factors. 
4. y=5+x
    y=5+(2)= 7
    y=5+(3)= 8
    y=5+(4)= 9
5. (B)y is 3 times x
    y=3(x)
    y=3(3)=9
    y=3(4)=12
    y=3(5)=15
6. (C)y is 6 less than x
    y=6-(x)
    y=6-(8)=2
    y=6-(9)=3
    y=6-(10)=4
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Step-by-step explanation:

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PLEASE HELP ME!!!!!!!!!
vitfil [10]
<h2>1)</h2>

(x - 4) {}^{2}  - 28 = 8 \\ (x - 4) {}^{2}  = 8 + 28 \\ (x - 4) {}^{2}  = 36

This must be true for some value of x, since we have a quantity squared yielding a positive number, and since the equation is of second degree,there must exist 2 real roots.

\sqrt{(x - 4) {}^{2} }  =  ± \sqrt{36}  \\x - 4 = ±6 \\ x _{1}- 4 = 6 \:  \:  \:  \:  \:  \:  \: \:  \:  \:  x _{2}- 4 =  - 6 \\ x_{1} = 6 + 4 \:  \:  \:  \:  \:  \:  \:  \:  \:  \: x_{2} =  - 6 + 4 \\ x_{1} = 10 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \: \:  \: x_{2} =  - 2

<h2>2)</h2>

Well he started off correct to the point of completing the square.

(x - 3)  {}^{2}  = 16 \\ x - 3 = ±4 \\  \: x_{1}  - 3 = 4 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: x_{2}  - 3 =  - 4 \\ x_{1}  = 7 \:  \:  \:  \:  \:  \:  \:  \:  \:  \: x_{2}  =  - 1

8 0
1 year ago
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