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tankabanditka [31]
3 years ago
6

How can you solve y=3x+2 and -4x+2y=8 with substitution

Mathematics
2 answers:
seraphim [82]3 years ago
7 0
Well, I can't explain how to solve but the answer is (2,8)
Xelga [282]3 years ago
7 0
We have:

y=3x+2~~(i)\\\\ -4x+2y=8~~(ii)

Substituting (i) in (ii):

-4x+2y=8\\\\ -4x+2(3x+2)=8\\\\ -4x+6x+4=8\\\\ 2x+4=8\\\\ 2x=4\\\\ x=\dfrac{4}{2}\\\\ \boxed{x=2}

Using the value obtained above in the expression (i):

y=3x+2\\\\ y=3\cdot2+2\\\\ y=6+2\\\\ \boxed{y=8}
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cestrela7 [59]

The coordinates of P' is (9, -6) and L' is (3, -1).

<h2>Given that</h2>

PL has endpoints P(4, −6) and L(−2, 1).

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<h3>We have to determine</h3>

What are the coordinates of P’ and L’?

<h3>According to the question</h3>

PL has endpoints P(4, −6) and L(−2, 1).

The segment is translated using the mapping (x, y) → (x + 5, y).

The coordinates of point P after translation using mapping is,

\rm P(x, \ y) - P'(x+5, y)\\&#10;\\&#10;P(4, -6) - P'(4+5, \ -6)\\&#10;\\&#10;P(4, -6) - P'(9, -6)\\

And the coordinates of point L after translation using mapping is,

\rm L(x, \ y) - L'(x+5, y)\\\\L(-2, 1) - L'(-2+5, \ -1)\\\\L(4, -6) - L'(3, -1)\\

Hence, The coordinates of P' is (9, -6) and L' is (3, -1).

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8 0
3 years ago
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80 * .20=16

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4 years ago
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Studentka2010 [4]
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What is the constant term of the quotient when 2x^3-3x^2 + 4x-2 is divided by x^2-x+2?
AlladinOne [14]

Answer:

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Step-by-step explanation:

See the attachment for the polynomial long division. The constant in the quotient is -1.

_____

Here, there is a remainder of -x. If there were no remainder the constant in the quotient is the ratio of the constant in the dividend to the constant in the divisor: -2/2 = -1.

That could be a first guess in a "guess and check" solution approach.

<em>Guess</em>: first term of binomial quotient is (2x^3)/x^2 = 2x; last term of binomial quotient is -2/2 = -1. So, the quotient is guessed to be (2x -1).

<em>Check</em>: (2x -1)(x^2 -x +2) = 2x^3 -3x^2 +5x -2

Subtracting this from the actual dividend gives a remainder of -x. This has a lower degree than the divisor, so no further adjustment of the quotient is required.

6 0
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