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Rom4ik [11]
3 years ago
9

5x + y = 6 5x + 3y = -4 The y-coordinate of the solution to the system shown is _____. -5 -1 1

Mathematics
1 answer:
anzhelika [568]3 years ago
8 0
5x +y = 6
5x + 3y = -4 
multiply top equation by -1 and bottom by 1
-5x -y = -6
5x+3y = -4
solve and get 
2y = -10
y = -10/2
y = -5 
answer is -5
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tankabanditka [31]
<h3>Answer: Choice A) 3.5</h3>

Work Shown:

The point (8,28) is the furthest to the right on the graph. We have x = 8 and y = 28 pair up here. Divide the y over the x

y/x = 28/8 = 3.5

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5 0
3 years ago
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Select correct answer
Sergio [31]

Answer:

The values of p in the equation are 0 and 6

Step-by-step explanation:

First, you have to make the denominators the same. to do that, first factor 2p^2-7p-4 = \left(2p+1\right)\left(p-4\right)2p

2

−7p−4=(2p+1)(p−4)

So then the equation looks like:

\frac{p}{2p+1}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{5}{p-4}

2p+1

p

−

(2p+1)(p−4)

2p

2

+5

=−

p−4

5

To make the denominators equal, multiply 2p+1 with p-4 and p-4 with 2p+1:

\frac{p^2-4p}{(2p+1)(p-4)}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{10p+5}{(p-4)(2p+1)}

(2p+1)(p−4)

p

2

−4p

−

(2p+1)(p−4)

2p

2

+5

=−

(p−4)(2p+1)

10p+5

Since, this has an equal sign we 'get rid of' or 'forget' the denominator and only solve the numerator.

(p^2-4p)-(2p^2+5)=-(10p+5)(p

2

−4p)−(2p

2

+5)=−(10p+5)

Now, solve like a normal equation. Solve (p^2-4p)-(2p^2+5)(p

2

−4p)−(2p

2

+5) first:

(p^2-4p)-(2p^2+5)=-p^2-4p-5(p

2

−4p)−(2p

2

+5)=−p

2

−4p−5

-p^2-4p-5=-10p+5−p

2

−4p−5=−10p+5

Combine like terms:

-p^2-4p+0=-10p−p

2

−4p+0=−10p

-p^2+6p=0−p

2

+6p=0

Factor:

p=0, p=6p

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given a point A is on a coordinate plane in quadrant I. if it is rotated 90° counter clockwise then reflecred across the y axis.
den301095 [7]

Answer:

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The intersection of lines y=-14x+3 and y=-32x+3 is .
Kamila [148]

The intersection of lines y=-14x+3 and y=-32x+3 is (0, 3)

<h3>How to determine the intersection of the lines?</h3>

The lines are given as:

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Substitute y = -32x+3 in

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Evaluate the like terms

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Divide by -18

x = 0

Substitute x = 0 in y = -14x + 3

y = -14(0) + 3

Evaluate

y = 3

Hence, the intersection of lines y=-14x+3 and y=-32x+3 is (0, 3)

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brainly.com/question/14323743

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