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Feliz [49]
3 years ago
13

Please help!

Mathematics
2 answers:
Hatshy [7]3 years ago
5 0

9514 1404 393

Answer:

  (a) x = 22

  (b) 46°

  (c) 44°

  (d) 180°

Step-by-step explanation:

Because the sum of the interior angles of a triangle is 180°, we know that the sum of the acute angles in a right triangle is 90°.

<h3>(a) </h3>

The sum of the marked acute angles is 90°.

  2x +(3x -20) = 90

  5x = 110 . . . . . . . . . . add 20, collect terms

  x = 22 . . . . . . . . . divide by 5

__

<h3>(b) </h3>

∠CBA = (3x -20)° = (3·22 -20)° = 46°

__

<h3>(c)</h3>

∠CAB = 2x° = 2(22)° = 44°

__

<h3>(d)</h3>

The sum of a triangle's interior angles is 180°.

Nimfa-mama [501]3 years ago
5 0

i dont know but you will get it one day

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Three boxes have a total weight of 640 pounds. Box A weights twice as much as Box B. Box C weight 30 pounds more than Box A. How
Zarrin [17]

Answer:

  A: 244 pounds

  B: 122 pounds

  C: 274 pounds

Step-by-step explanation:

We have A+B+C = 640; A=2B; C=A+30. Substituting the last into the first gives ...

  A + B + (A +30) = 640

  2A +B = 610 . . . . . . . . . . . . subtract 30

Substituting the second into this equation gives ...

  2(2B) +B = 610

  B = 610/5 = 122 . . . . . divide by 5

  A = 2B = 244 . . . . . . . .find A from B

  C = A+30 = 274 . . . . . find C from A

Box A weighs 244 pounds; box B weighs 122 pounds; box C weighs 274 pounds.

4 0
3 years ago
Simplify: cos2x-cos4 all over sin2x + sin 4x
GrogVix [38]

Answer:

\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)

Step-by-step explanation:

\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}

Apply formula:

\cos\left(A\right)-\cos\left(B\right)=-2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right) and

\sin\left(A\right)+\sin\left(B\right)=2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)

We get:

=\frac{-2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\sin\left(\frac{2x-4x}{2}\right)}{2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\cos\left(\frac{2x-4x}{2}\right)}

=\frac{-\sin\left(\frac{2x-4x}{2}\right)}{\cos\left(\frac{2x-4x}{2}\right)}

=\frac{-\sin\left(\frac{-2x}{2}\right)}{\cos\left(\frac{-2x}{2}\right)}

=\frac{-\sin\left(-x\right)}{\cos\left(-x\right)}

=\frac{-\cdot-\sin\left(x\right)}{\cos\left(x\right)}

=\frac{\sin\left(x\right)}{\cos\left(x\right)}

=\tan\left(x\right)

Hence final answer is

\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)

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It'll take him 6 days to run 3/5 of a mile.... 

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Which term best describes a figure formed by three segments connecting
elixir [45]

Answer:

The term that best describes a figure formed by three segments connecting three non-collinear points is:

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Step-by-step explanation:

Step-by-step explanation:

We know that with the help of just one point we can't form any figure.

With the help of two points a line segment can be formed.

And with the help of three points if the three points are collinear a triangle can be formed.

Hence, the term that best describes a figure formed by three segments connecting three non-collinear points is:

Triangle.

8 0
3 years ago
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