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Katyanochek1 [597]
3 years ago
10

1) g=-y+c/x, solve for X

Mathematics
1 answer:
Sindrei [870]3 years ago
3 0

Answer:

the solution  x = \frac{c}{g-y}

Step-by-step explanation:

<u>Step 1:-</u>

Given problem g=\frac{-y+c}{x}

here l.cm is x

g = \frac{x y+c}{x}

cross multiplication we get ,

g x = x y +c

<u>Step 2</u>:-

subtracting x y on both sides , we get

g x - x y = c

taking common 'x' we get,

x(g-y) = c

<u>Final answer</u>:-

x = \frac{c}{g-y}

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Answer:

6 Apples

Step-by-step explanation:

There would be six apples and 2 oranges left over.

Hope this Helps!

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3 0
3 years ago
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Please help me! The i is an imaginary number<br><br> (9+4i)^2
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Answer: 2500

Explanation: I turned the I into a 1 and it turned into a 41 so I added 9+41 which is 50. Then I multiplied 50 by 50 because of the ^2 which gave me 2500. I’m assuming that you can put whatever number as a replacement for the I. I’m sorry if this answer isn’t right.
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2 years ago
Solve the given triangles by finding the missing angle and other side lengths.
coldgirl [10]

Answer:

  1. a ≈ 27.65; B = 28°; c ≈ 19.17
  2. a ≈ 163.30; B = 120°; c ≈ 59.77
  3. A = 13°; b ≈ 0.3874; c ≈ 1.3737

Step-by-step explanation:

As long as you're seeking outside help, the use of a suitable tool is appropriate. Many graphing calculators, apps, and web sites are available for solving triangles.

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Here, you're given two angles and a side. The third angle is the difference from 180° of the sum of the other two. For example, in the first problem, ...

  B = 180° -A -C = 28°

The missing sides are found using the law of sines. Since you are given a side, use that as the reference, and its opposite angle as the reference angle. Then you have ...

  side = sin(opposite angle)×<em>((reference side)/sin(reference angle))</em>

Note that the factor in italics remains the same for both remaining sides.

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In the first problem, this becomes ...

  a = sin(112°)·14/sin(28°) ≈ 27.65

  c = sin(40°)·14/sin(28°) ≈ 19.17

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The remaining problems are worked in similar fashion.

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<em>Comment on triangle solutions</em>

These are straightforward because you're given two angles. If you're given two sides and an angle, the solution method varies depending on which angle you're given (opposite the long side, the short side, or the unknown side). In the last case, the law of cosines is involved. In the second case, there are likely two solutions. Once you have two angles, you can proceed as above.

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Answer:

congruent SAS

Step-by-step explanation:

We know two sides of the triangles are congruent to each other

MD = MT

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We also know that <DMA = < TMU

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