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blsea [12.9K]
3 years ago
8

Which values of PP and QQ result in an equation with no solutions? -60x+32=Qx+P−60x+32=Qx+P Choose all answers that apply: Choos

e all answers that apply: (Choice A) A Q=-60Q=−60 and P=60P=60 (Choice B) B Q=32Q=32 and P=60P=60 (Choice C) C Q=-60Q=−60 and P=-32P=−32 (Choice D) D Q=32Q=32 and P=-60P=−60
Mathematics
1 answer:
Inga [223]3 years ago
8 0
<h3>Answer:</h3>

  A and C

<h3>Step-by-step explanation:</h3>

Given:

  -60x+32 = Qx+P

Find:

Which values of P and Q result in an equation with no solutions? Choose all answers that apply:

(Choice A) A Q=-60 P=60

(Choice B) B Q=32 P=60

(Choice C) C Q=-60 P=−32

(Choice D) D Q=32 P=−60

Solution:

The equation will have no solution if it reduces to ...

  0 = (non-zero constant)

If we add 60x-32 to both sides, we get

  0 = 60x +Qx + P-32

  0 = (Q+60)x +(P-32)

The x-term must be zero, so Q+60 = 0, or Q = -60.

The constant term must be non-zero, so P-32 ≠0, or P ≠ 32.

The appropriate answer choices are those with Q=-60 and P≠32, A and C.

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\large\underline{\sf{Solution-}}

<h2 /><h2><u>Consider</u></h2>

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<h2><u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) = sinx

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\boxed{\tt{ \cos \bigg( \frac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \frac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \} = 1}}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

<h2>ADDITIONAL INFORMATION :-</h2>

Sign of Trigonometric ratios in Quadrants

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