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DENIUS [597]
3 years ago
10

The measure of the supplement of an angle exceeds three times the measure of the complement of the angle by 12. Find the measure

of half of the supplement.​
Mathematics
1 answer:
satela [25.4K]3 years ago
8 0

Answer:

  64.5°

Step-by-step explanation:

The supplement of angle x is 180-x. The complement of the same angle is 90-x, which is 90 less than the supplement.

We can let 'a' represent half the measure of the supplement. Then 2a is the measure of the supplement, and we are given the relation ...

  2a -(3(2a-90)) = 12

  2a -6a +270 = 12 . . . simplify

  258 = 4a . . . . . . . . . . add 4a-12

  a = 64.5

The measure of half the supplement is 64.5 degrees.

_____

<em>Additional comment</em>

Then the supplement of the angle is 129°, and the angle is 51°. Its complement is 39°. 3 times the complement is 117°, which is 12° less than the supplement, as required.

__

There are many ways to assign the variable(s) in this problem. Here, we chose one that is pretty direct: the value of the variable answers the question. If you let the variable represent the angle, its complement, or its supplement, then additional work is required to answer the question after the value of the variable is found.

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I hope I helped you.

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Answer:

The graph of the parabola is represented by the quadratic function

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Step-by-step explanation:

4 0
3 years ago
Help solve for “q”<br> —————————————
VMariaS [17]

Digram:-

\\

\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\put(5,1){\vector(1,0){4}}\put(5,1){\vector(-1,0){4}}\put(5,1){\vector(1,1){3}}\put(2,2){$\underline{\boxed{\large\sf a + b = 180^{\circ}}$}}\put(4.5,1.3){$\sf a^{\circ}$}\put(5.7,1.3){$\sf b^{\circ}$}\end{picture}

\\

STEP :-

\dashrightarrow \tt(4q - 1) {}^{ \circ}  +  {117}^{ \circ}  = 18 {0}^{ \circ}

{Linear pair}

\\  \\

\dashrightarrow \tt(4q - 1) {}^{ \circ}= 18 {0}^{ \circ}   - {117}^{ \circ}

\\

\dashrightarrow \tt(4q - 1) {}^{ \circ}=63^{ \circ}

\\

\dashrightarrow \tt4q - 1{}^{ \circ}=63^{ \circ}

\\

\dashrightarrow \tt4q =63^{ \circ} + 1{}^{ \circ}

\\

\dashrightarrow \tt4q =64{}^{ \circ}

\\

\dashrightarrow \tt \: q =  \dfrac{64}{4}^{ \circ}

\\

\dashrightarrow \tt \: q =  \dfrac{16 \times 4}{4}^{ \circ}

\\

\dashrightarrow \tt \: q =  \dfrac{16 \times \cancel4}{\cancel4}^{ \circ}

\\

\dashrightarrow \tt \: q =  \dfrac{16}{1}

\\

\dashrightarrow \bf q = 16 \degree

\\  \\

Verification:

\\

\dashrightarrow \tt(4 \times 16- 1) {}^{ \circ}  +  {117}^{ \circ}  = 18 {0}^{ \circ}

\\

\dashrightarrow \tt(64- 1) {}^{ \circ}  +  {117}^{ \circ}  = 18 {0}^{ \circ}

\\

\dashrightarrow \tt63^{ \circ}  +  {117}^{ \circ}  = 18 {0}^{ \circ}

\\

\dashrightarrow \tt180^{ \circ}  = 18 {0}^{ \circ}

\\

LHS = RHS

HENCE VERIFIED!

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