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Effectus [21]
4 years ago
13

If (x – ½) and (x-2) are the factors of px2+5x + r, find p and r . Also find p -r .

Mathematics
1 answer:
Ostrovityanka [42]4 years ago
8 0

Answer:

p = -2, r = -2 and p-r =  0.

Step-by-step explanation:

(x - 1/2)(x - 2)

= x^2 - 2x - 1/2x + 1

= x^2 - 5/2x + 1

= 2x^2 - 5x + 2

= -2x^2 + 5x - 2

Comparing this with

px^2 + 5x + r

we see that  p = -2, r = -2 and p-r =  0.

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Step-by-step explanation:

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Now find \frac{dB}{dA}

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\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}

b)

When A=\frac{\pi}{4},B=\frac{\pi}{3}, the value of \frac{dB}{dA} is,

\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}

c)

In general, the linear approximation at x= a is,

f(x)=f'(x).(x-a)+f(a)

Here the function f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

At A=\frac{\pi}{4}

f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}

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f'(A)=\frac{dB}{dA}=\sqrt{3} from part b

Therefore, the linear approximation at A=\frac{\pi}{4} is,

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d)

Use part (c), when A=46°, B is approximately,

B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°

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