If - is an acute angle and sin- p -6-8- p- then p must satisfy A- 6 - p-8B- 4 - p-7C- 6 - p-7D- 3 - p - 4

The correct option is C (b) θ is an acute angle so 0^∘≤θ lt; 90^∘ ∴ 0 ≤p 6/8 p lt; 1 = gt; 0 ≤ (p 6) lt; (8 p) = gt; 6 ≤ p lt; 7. ...

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Functions

If θ is an ac...

Question

If $θ$ is an acute angle and $s i n θ = \frac{p - 6}{8 - p},$ then p must satisfy

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θ

s i n θ = \frac{p - 6}{8 - p},

∠ 15 = 2^{P_{1}} 3^{P_{2}} 5^{P_{3}} 7^{P_{4}} 11^{P_{5}} 13^{P_{6}} t h e n \sum_{r = 1}^{6} P_{r}

P

2 \times 2

[\begin{matrix} 1 & 0 \end{matrix}] P = - \frac{1}{\sqrt{2}} [\begin{matrix} 1 & 1 \end{matrix}]

[\begin{matrix} 0 & 1 \end{matrix}] P = \frac{1}{\sqrt{2}} [\begin{matrix} - 1 & 1 \end{matrix}]

0

I

2

p (x)

3

p (0) = 4

p^{^{'}} (0) = 3, p^{^{''}} (0) = 6, p^{^{'''}} (x) = 6

p^{^{''}} (- 1) =

p_{1}, p_{2}, p_{3}

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