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avanturin [10]
3 years ago
12

Solve the system by substitution math help. Please show every step.

Mathematics
1 answer:
Arlecino [84]3 years ago
5 0
The idea here is to combine the equations into one in only one variable. The last equation is in only x and z so lets sub out y in the second.

-x - y - z = -8
8 - x - z = y

Substitute y with (8 - x - z) into the next equation.
-4x + 4(8 - x - z) + 5z = 7
-4x + 32 - 4x - 4z + 5z = 7
-8x + 32 + z = 7
 The variable z is easiest to get alone; use that as substitution in the last equation.
z = 7 + 8x - 32
z = 8x - 25

Substitute z with (8x - 25)
2x + 2(8x - 25) = 4
2x + 16x - 50 = 4
18x = 4 + 50
18x = 54
x = 3

So use x = 3 in that  last equation to find z.

2(3) + 2z = 4
6 + 2z = 4
2z = 4 - 6
2z = -2
z = -1

Now, use both x = 3 and z = -1 to find the final variable y.

-3 - y - (-1) = -8
-3 - y + 1 = -8
-2 - y = -8
-y = -8 + 2
-y = -6
y = 6

Solutions x,y,z = ( 3, 6, -1)


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What is the greatest common factor of 90, 126, 180 and 990?
mylen [45]

9514 1404 393

Answer:

  18

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180 = 18·10

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The greatest common factor of these numbers is 18.

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<em>Comment on the GCF</em>

It can be useful to know Euclid's algorithm for finding the GCF:

  1. Determine the remainder from dividing the larger number by the smaller.
  2. If the remainder is zero, the smaller number is the GCF. If the remainder is non-zero, use it to replace the larger number and repeat from step 1.

For example, 126 mod 90 = 36; 90 mod 36 = 18; 36 mod 18 = 0, so 18 is the GCF of 126 and 90. (The modulo function 'mod' returns the remainder from division.)

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3 years ago
Vertex form is f(x)=a(x-p)^2 +q. How do i determine a?
slamgirl [31]
Y1 is the simplest parabola.  Its vertex is at (0,0) and it passes thru (2,4).  This is enough info to conclude that y1 = x^2.

y4, the lower red graph, is a bit more of a challenge.  We can easily identify its vertex, which is (-4,0), and several points on the grah, such as (2,-3).

Let's try this:  assume that the general equation for a parabola is 
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-3-(-4) = a(2-0)^2.  Then 1 = a(2)^2, or 1 = 4a, or a = 1/4.

The equation of parabola y4 is   y+4 = (1/4)x^2

Or you could elim. the fraction and write the eqn as 4y+16=x^2, or

4y = x^2-16, or    y = (1/4)x - 4.  Take your pick!  Hope this helps you find "a" for the other parabolas.


8 0
3 years ago
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