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3 years ago

Can anyone help me with this fraction x/-4 = 9 =

Mathematics

1 answer:

olasank [31]3 years ago

5 0

Answer: x= -36

Step-by-step explanation:

-36/-4=9

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Find the linear approximating polynomial for the following function centered at the given point a.point a. b. Find the quadratic

ZanzabumX [31]

Answer:

$(a)L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})\\(b)Q(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(x+\frac{\pi}{4})^2\\(c)L(-0.23\pi)=-0.6626\\Q(-0.23\pi)=-0.6613$

Step-by-step explanation:

Given the function:

$f(x)=sin x, a=-\frac{\pi}{4}$

(a)Linear approximating polynomial

$L(x)=f(a)+f'(a)(x-a)\\f(a)=sin(-\frac{\pi}{4})=-\frac{\sqrt{2} }{2}\\f'(x)=cos x, a=-\frac{\pi}{4}\\f'(a)=cos (-\frac{\pi}{4})=\frac{\sqrt{2} }{2} \\Therefore:\\L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x-(-\frac{\pi}{4}))\\L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})$

(b)Quadratic approximating polynomial

$Q(x)=L(x)+\frac{1}{2}f''(a)(x-a)^2\\f''(x)=-sin(x), \\f''(a)=-sin(-\frac{\pi}{4})=\frac{\sqrt{2} }{2}\\Q(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(x+\frac{\pi}{4})^2$

(c)When $x=-0.23\pi$

Using Linear Approximation polynomial

$L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})\\L(-0.23\pi)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(-0.23\pi+\frac{\pi}{4})\\L(-0.23\pi)=-0.6626$

Using the Quadratic approximating polynomial

$Q(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(x+\frac{\pi}{4})^2\\Q(-0.23\pi)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(-0.23\pi+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(-0.23\pi+\frac{\pi}{4})^2=-0.6613$

4 0

3 years ago