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2 years ago

Pls I need help with this

Mathematics

1 answer:

tester [92]2 years ago

6 0

Answer:

third side = 6

Step-by-step explanation:

third side is hypotenuse as it is opposite to 90 degree

using pythagoras theorem

(perpendicular)^2 + (base)^2 = (hypotense)^2

5^2 + $(\sqrt{11})^2$ = hypotenuse^2

25 + 11 = hypotenuse^2

36 = hypotenuse^2

$\sqrt{36}$ = hypotenuse

6 = hypotenuse

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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

4 0

3 years ago

I need help with this

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3 years ago

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Oksanka [162]

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3 years ago

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Tcecarenko [31]

Answer:

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Step-by-step explanation:

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