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Aleksandr-060686 [28]
3 years ago
14

Write a quadratic function with zeroes 0 and 8.

Mathematics
1 answer:
Jlenok [28]3 years ago
3 0

Answer:

x^2-8x=0

Step-by-step explanation:

the quadratic function should be as follows:

x^2-8x=0

Now let's confirm that the zeros of the function are 0 and 8

x^2-8x=0=x(x-8)=0

Therefore we can see that if x = 0

0^2*8*0=0\\0=0

the equation is fulfilled

And we also have (x-8)

for this expresion to be equal to zero:

x-8=0\\x=8

thus, if x = 8

8^2-8*8=0\\64-64=0\\0=0

the equation is also fulfilled

The zeros of the quadratic function x^2-8x=0 are 0 and 8.

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Step-by-step explanation: Given that Cosine of an angle 'A' is 3 over 11,

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We have the following relations between cosine, secant and cosecant of an angle from trigonometry.

\sec A=\dfrac{1}{\cos A}~~\textup{and}~~\csc A=\dfrac{1}{\sqrt{1-\cos^2 A}}.

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\sec A=\dfrac{11}{3}

and

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Find the area of the circle.
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Answer:

See Explanation Below

Step-by-step explanation:

Your question isn't clear; However, I'll solve in two ways

1. Diameter = 14 cm

2. Radius = 14 cm

<em></em>

<em>When Diameter = 14 cm</em>

Given

Area = \pi r^2

Diameter = 14cm

Required

Calculate the area of the circle

First, the radius has to be solved;

Radius = \frac{1}{2} Diameter

Radius = \frac{1}{2} * 14cm

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<em>When Radius = 14 cm</em>

Given

Area = \pi r^2

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Required

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Area = \pi r^2

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