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2 years ago

In simplest radical form, what are the solutions to the quadratic equation 6 = x2 – 10x?

Mathematics

1 answer:

bezimeni [28]2 years ago

5 0

Answer:

Step-by-step explanation:

quadratic equation: ax² + bx + c =0

x' = [-b+√(b²-4ac)]/2a and x" = [-b-√(b²-4ac)]/2a

6 = x² – 10x ; x² - 10x -6 =0

(a=1, b= - 10 and c = - 6

x' = [10+√(10²+4(1)(-6)]/2(1) and x" = [10-√(10²+4(1)(-6)]/2(1)

x' =5+√31 and x' = 5-√31

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Which of the following have the property that a(x)=a−1(x)? I. y=x II. y=1/x III.y=x^2 IV. y=x^3 A. I and II, only B. IV, only C.

valentina_108 [34]

Answer:

Correct answer:

A. I and II

Step-by-step explanation:

First of all, let us have a look at the steps of finding inverse of a function.

1. Replace y with x and x with y.

2. Solve for y.

3. Replace y with $f^{-1}(x)$

Given that:

$I.\ y=x \\II.\ y=\dfrac{1}x \\III.\ y=x^2 \\IV.\ y=x^3$

Now, let us find inverse of each option one by one.

I. y = x, a(x) = x

Replacing y with and x with y:

x = y

x = $a^{-1}(x)$ = $a(x)$ Hence, I is true.

II. $y =\dfrac{1}{x}$

Replacing y with and x with y:

$x =\dfrac{1}{y}$

$x=\dfrac{1}{a^{-1}(x)}$

$\Rightarrow a^{-1}(x) = \dfrac{1}{x}$

$a^{-1}(x)$ = $a(x)$ Hence, II is true.

III. $y =x^{2}$

Replacing y with and x with y:

$x =y^{2}\\\Rightarrow y = \sqrt x\\\Rightarrow a^{-1}(x) = \sqrt{x} \ne a(x)$

Hence, III is not true.

IV. $y =x^{3}$

Replacing y with and x with y:

$x =y^{3}\\\Rightarrow y = \sqrt[3] x\\\Rightarrow a^{-1}(x) = \sqrt[3]{x} \ne a(x)$

Hence, IV is not true.

Correct answer:

A. I and II

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3 years ago

What do I put in blanks explain your answer giving brainiest

KengaRu [80]

Answer:

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Step-by-step explanation:

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