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vovangra [49]
3 years ago
7

SOLVE THE EQUATION FOR Y

Mathematics
1 answer:
max2010maxim [7]3 years ago
6 0
(4)\ -6x + y = 11\ \ \ \Rightarrow\ \ \ y=6x+11\\\\(5)\ \ 8x + 2y = 10\ \ \ \Rightarrow\ \ \ 2y=-8x+10\ \ \ \Rightarrow\ \ \ y=-4x+5\\\\(6)\ \  6x - 3y = -9\ \ \ \Rightarrow\ \ \ -3y=-6x-9\ \ \ \Rightarrow\ \ \ y=2x+3\\\\(7)\ \  -4x + 2y = 16\ \ \ \Rightarrow\ \ \ 2y=4x+16\ \ \ \Rightarrow\ \ \ y=2x+8\\\\
(8)\ \  10x - 5y = 25\ \ \ \Rightarrow\ \ \ -5y=-10x+25\ \ \ \Rightarrow\ \ \ y=2x-5\\\\(9)\ \  3x + 2y = -8\ \ \ \Rightarrow\ \ \ 2y=-3x-8\ \ \ \Rightarrow\ \ \ y=-1.5x-4

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What is the value of x? (SOMEONE PLEASE HELP ME OUT BRO PLEASEEEEE IGNORE THE 9 smh)
max2010maxim [7]

Answer:

x=28

Step-by-step explanation:

Sum of all the angles in the triangle add up to equal 180

90 + x + 6 + 2x = 180

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96 + 3x = 180

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180-96=84

96-96 cancels out

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divide each side by 3

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<em>The1AndOnlyMarkus</em>

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Can someone help me
erik [133]

Answer:

   6 < x < 23.206

Step-by-step explanation:

To properly answer this question, we need to make the assumption that angle DAC is non-negative and that angle BCA is acute.

The maximum value of the angle DAC can be shown to occur when points B, C, and D are on a circle centered at A*. When that is the case, the sine of half of angle DAC is equal to 16/22 times the sine of half of angle BAC. That is, ...

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Of course, the minimum value of angle DAC is 0°, so the minimum value of x is ...

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Then the range of values of x will be ...

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* One way to do this is to make use of the law of cosines:

  22² = AB² + AC² -2·AB·AC·cos(48°)

  16² = AD² + AC² -2·AD·AC·cos(2x-12)

The trick is to maximize x while satisfying the constraints that all of the lengths are positive. This will happen when AB=AC=AD, in which case the equations be come ...

  22² = 2·AB²·(1-cos(48°))

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The value of AB drops out of the ratio of these equations, and the result for x is as above.

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