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kkurt [141]
3 years ago
6

Use elimination to solve this system of equations. X-4y=-11 and 5x-7y=-10

Mathematics
1 answer:
timurjin [86]3 years ago
4 0
X-4y=-11<=> -5x + 20y = 55
                     <span> 5x - 7y    = -10</span>
                =>0*x + 13*y = 45 => y =45/13 => x = 4*(45/13) -11 = 180/13 -11 =>
x = 37/13;
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To graph the equation 2x + 5y = 10, Zeplyn draws a line through the points (5, 0) and (0, 2). What is the slope of the line repr
BARSIC [14]
Hi there! The slope of the line is -2/5.

We can determine the slope of a line by dividing the change of y-coordinates by the change of x-coordinates.

In this situation we have a...
change of x-coordinates of (5 - 0) = 5
change of y-coordinates of (0 - 2) = -2

Therefore, the slope of this line is -2/5
6 0
3 years ago
Quadrilateral missing angles
alukav5142 [94]

Answer:

idk

Step-by-step explanation:

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3 0
3 years ago
Plz help!
jek_recluse [69]

Step-by-step explanation:

let's look at the last line :

x³ + 8x - 3 = Ax³ +5Ax + Bx² + 5B + Cx + D

since we find A, B, C, and D by simply comparing the factors of the various terms in x (or constants) in both sides of the equation, we need to combine a few terms on the right hand side (so that we have one term per x exponent grade).

x³ + 8x - 3 = Ax³ + (5A + C)x + Bx² + (5B + D)

by comparing now both sides, to make both sides truly equal, the factors have to be equal.

A = 1 (the same as for x³ on the left hand side).

B = 0 (a we have no x² on the left side).

5A + C = 8 (a 8 is the factor of x in the left side).

5×1 + C = 8

5 + C = 8

C = 3

5B + D = -3 (as the constant term is -3 on the left side).

5×0 + D = -3

D = -3

so, the 4th answer option is correct.

7 0
2 years ago
Write expanded form for six hundred and four hundred thirteen thousandths
lakkis [162]
64,013,000 =
<span>six hundred and four hundred thirteen thousandths =
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4 0
3 years ago
Find all integers $n$ for which $\frac{n^2+n+1}{n-1}$ is an integer.
Grace [21]
\dfrac{n^2+n+1}{n-1}=\dfrac{(n-1)^2+3(n-1)+3}{n-1}=n-2+\dfrac3{n-1}

The remainder term \dfrac3{n-1} will never vanish, and will always be rational as long as the denominator is not 1, 3, or -3.

n-1=1\implies n=2
n-1=3\implies n=4
n-1=-3\implies n=-2
7 0
3 years ago
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