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BabaBlast [244]
3 years ago
9

Select the correct graph of the function y = –1/8(x – 1)2 – 4 below.

Mathematics
1 answer:
Kryger [21]3 years ago
6 0

Step-by-step explanation:

The given function is :

y=\dfrac{-1}{8}(x-1)^2-4 ...(1)

It is an equation of parabola. The vertex form for equation of parabola is given by :

y=a(x-h)^2+k ...(2)

Here,

Vertex = (h,k), axis = h and rate = a

From equation (1) and (2) :

a = -1/8, h = 1 and k = -4

The atached graph shows the graph for the given equation.

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Answer:

26:  x=34

28: x=17

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Step-by-step explanation:

<u>26:</u> 2x+22=90

subtract 22

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divide by 2

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add 18 and 21

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m=30

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Step-by-step explanation:

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4 years ago
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ahrayia [7]
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8 0
3 years ago
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Find the equation of the line normal to the curve of y=3cos1/3x, Where x=\pi
sertanlavr [38]

Answer:

y = \frac{\sqrt{2}x}{3} - \frac{\sqrt{2}\pi}{3} + 1.5

Step-by-step explanation:

The equation to the line normal to the curve has the following format:

y - y(x_{0}) = m(x - x_{0})

In whicm m is the derivative of y at the point x_{0}

In this problem, we have that:

x_{0} = \pi

y(x) = 3\cos{\frac{x}{3}}

y(\pi) = 3\cos{\frac{\pi}{3}} = \frac{3}{2}

The derivative of \cos{ax} is a\sin{ax}

So

y(x) = 3\cos{\frac{x}{3}}

y'(x) = 3*\frac{1}{3}\sin{\frac{x}{3}} = \sin{\frac{x}{3}}

m = \sin{\frac{\pi}{3}} = \frac{\sqrt{2}}{3}

The equation of the line normal to the curve of y=3cos1/3x is:

y - y(x_{0}) = m(x - x_{0})

y - \frac{3}{2} = \frac{\sqrt{2}}{3}(x - \pi)

y = \frac{\sqrt{2}}{3}(x - \pi) +  \frac{3}{2}

y = \frac{\sqrt{2}x}{3} - \frac{\sqrt{2}\pi}{3} + 1.5

8 0
4 years ago
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