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Rus_ich [418]
3 years ago
11

Guys please help this is a math question

Mathematics
1 answer:
Juliette [100K]3 years ago
6 0

Answer:

81.5

Step-by-step explanation:

<G and <H are congruent

<F and <I are congruent

so the last two angles in both triangles must also be congruent

this means that <K and <J are congruent

so we can create this equation: <K = <J

substitute the angles with what we know: 4y^{2} = 6y^{2} - 40

add 40 to both sides and subtract 4y^{2} from both sides: 40 = 2y^{2}

you get: 20 = y^{2}

root square on both sides: \sqrt{20} = \sqrt{y^{2} }

you get: y ≈ 4.5

substitute this into one of the equations for a missing variable:

6y^{2} -----> 6 (4.5 x 4.5) - 40

6 ( 20.25) - 40

121.5 - 40

81.5

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Solve the system of equations: <br> X+y+z=5<br> -3x-4y+4z=15<br> 2x-y-4z=-8
atroni [7]

Answer:

  (x, y, z) = (3, -2, 4)

Step-by-step explanation:

The attachment shows a solution using a graphing calculator where the first equation is solved for z, and that expression is substituted into each of the other two equations. The solution to that system is (x, y) = (3, -2). Using these values in the expression for z, we find ...

  z = 5 -(3 +(-2)) = 4

The solution is (x, y, z) = (3, -2, 4).

__

Your graphing calculator and/or online tools can give you the reduced row echelon form of the augmented matrix representing the system:

  \left[\begin{array}{ccc|c}1&1&1&5\\-3&-4&4&15\\2&-1&-4&-8\end{array}\right]

__

If you're solving by hand, you can do the substitution described above to eliminate z from one equation. You can eliminate z from another equation by adding the last two:

  -3x -4y +4(5 -x -y) = 15   ⇒   -7x -8y = -5

  (-3x -4y +4z) +(2x -y -4z) = (15) +(-8)   ⇒   -x -5y = 7

This pair of equations can be solved for x and y in any of the usual ways. Using the second equation to substitute for x, for example, gives you ...

  -7(-5y -7) -8y = -5   ⇒   27y = -54   ⇒   y = -2

  x = -5(-2) -7 = 3

  z = 5 -(3) -(-2) = 4

_____

<em>Additional comment</em>

If all that is needed is a solution, we prefer a graphical or matrix approach. These take about the same amount of time. Both require a little additional effort to determine exact values when the solutions are not integers.

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