If $p (u) = u^{2} + 3 u + 4, q (u) = u^{2} + u - 12,$ and $r (u) = u - 2$ , then find the degree of $p (u) q (u) r (u)$ .

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Solution

$d e g r e e o f p (u) = 2 d e g r e e o f q (u) = 2 d e g r e e o f r (u) = 1 ∴ d e g r e e o f p (u) q (u) r (u) = 2 + 2 + 1 = 5$

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Similar questions

p (x) = x + 1

q (x) = x^{3} + x

r (x) = \sqrt{2} + x + x^{2}

r (u) = u + u^{2} - 2

v^{2} = u^{2} + 2 a s

u =

p (x) = x + 1

q (x) = x^{3} + x

r (x) = \sqrt{2} + x + x^{2}

r (u) = u + u^{2} - 2

p = \frac{w}{2 g} (v^{2} - u^{2})

w = 40, g = 32, u = 4, v = 12

If a body is thrown up with an initial velocity, $u$ and covers a maximum height, $h$ , then $h$ is equal to $\frac{u^{2}}{2 g}$ .

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