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antiseptic1488 [7]
3 years ago
11

Through:(5,0), parallel to y=4/5x-3

Mathematics
2 answers:
lutik1710 [3]3 years ago
7 0
If you are looking for a line: remember point-slope form.
In this question you have a point (5,0) and you can get the slope from the equation. Since parallel lines have the same slope you can just use 4/5.
Just plug the numbers into the equation: y-y1=m(x-x1)
y-0=4/5(x-5)
Therefore y=4/5x-4
OverLord2011 [107]3 years ago
7 0
To be parallel lines both must have the same slope.  Our reference line is:

y=0.8x-3 so the slope is 0.8, given the point (5,0) we can solve for the y-intercept, or "b" of y=mx+b for our parallel line.

0=0.8(5)+b

0=4+b

b=-4 so

y=0.8x-4  is the parallel line to y=0.8x-3 that passes through (5,0)
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Solve the following system of equations for all three variables.
GrogVix [38]

Answer:

(x,y,z) -> (2,4,1)

Step-by-step explanation:

-7x + y + z = -9

-7x + 5y - 9z = -3

7x - 6y + 4z = -6

Pick two pairs:

-7x + 5y - 9z = -3

7x - 6y + 4z = -6

and

-7x + y + z = -9

-7x + 5y - 9z = -3

Eliminate the same variable from each system:

-7x + 5y - 9z = -3

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+ 5y - 9z = -3

- 6y + 4z = -6

<u><em>-1y - 5z = - 9</em></u>

-7x + y + z = -9

-7x + 5y - 9z = -3

-7x + y + z = -9

7x - 5y + 9z = 3

<u><em>-4y - 10z = -6</em></u>

Solve the system of the two new equations:

-1y - 5z = - 9           ->      -4 ( -1y - 5z = - 9)     ->   4y + 20z = 36  

-4y - 10z = -6          ->           -4y - 10z = -6      ->   -4y - 10z = -6

10z = 30

Thus, z = 3

-4y - 10z = -6

-4y - 10(3) = -6

-4y - 30 = -6

-4y = 24

Thus, y = -6

Substitute into one of the original equations:

-7x + y + z = -9

-7x + (-6) + (3) = -9

7x + -3  = -9

7x  = -6

x =

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Read 2 more answers
H(x) = x2 1 k(x) = x – 2 (h k)(2) = (h – k)(3) = Evaluate 3h(2) 2k(3) =.
natima [27]

Quadratic equation is the equation in which only one variable is unknown. The highest power of the variable is 2.The value of the given functions are,

(h+k)(x)=5

(h-k)(x)=9

3h(2)+2k(3)=17

<h3>Given information-</h3>

The given function is,

h(x)=x^2+1

k(x)=x-2

<h3>Quadratic equation</h3>

Quadratic equation is the equation in which only one variable is unknown. The highest power of the variable is 2.

1) The value of the function (h+k)(2),

(h+k)(x)=h(x)+k(x)

(h+k)(x)=x^2+1+x-2

(h+k)(2)=2^2+1+2-2

(h+k)(x)=5

2)The value of the function (h-k)(3),

(h-k)(x)=h(x)-k(x)

(h-k)(x)=x^2+1-x+2

(h-k)(3)=3^2+1-3+2

(h-k)(x)=9

3) The value of the function 3h(2)+2k(3)

3h(x)+2k(x)=3x^2+3+2x-2\times 2

3h(2)+2k(3)=3\times2^2+3+2\times2-2\times 2

3h(2)+2k(3)=17

Hence the value of the given functions are,

(h+k)(x)=5

(h-k)(x)=9

3h(2)+2k(3)=17

Learn more about the quadratic equation here;

brainly.com/question/2263981

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