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Natali5045456 [20]

2 years ago

10

Consider U = {x|x is a real number}. A = {x|x ∈ U and x + 2 > 10} B = {x|x ∈ U and 2x > 10} Which statements are true? 5 ∉

A; 5 ∈ B 6 ∈ A; 6 ∉ B 8 ∉ A; 8 ∈ B 9 ∈ A; 9 ∉ B

2 answers:

-Dominant- [34]2 years ago

5 0

x+2 > 10 solves to x > 8 after we subtract 2 from both sides

So set A is the set of real numbers that are larger than 8. The value 8 itself is not in set A. The same can be said about 5 as well.

Set B is the set of values that are larger than 5 since 2x > 10 turns into x > 5 after dividing both sides by 2. The value x = 5 is not in set B since x > 5 would turn into 5 > 5 which is false. The values x = 6, x = 8, and x = 9 are in set B.

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Summarizing everything, we can say...

5 is not in set A. True

5 is in set B. False

6 is in set A. False

6 is not in set B. False

8 is not in set A. True

8 is in set B. True

9 is in set A. True

9 is not in set B. False

yanalaym [24]2 years ago

4 0

Answer is C. I just took the test.

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Which pair of funtions is not a pair of inverse functions? please help!!

antiseptic1488 [7]

Answer:

$f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}$

Step-by-step explanation:

we know that

To find the inverse of a function, exchange variables x for y and y for x. Then clear the y-variable to get the inverse function.

we will proceed to verify each case to determine the solution of the problem

<u>case A)</u> $f(x)=\frac{x+1}{6} , g(x)=6x-1$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y+1}{6}$

Isolate the variable y

$6x=y+1$

$y=6x-1$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=6x-1$

therefore

f(x) and g(x) are inverse functions

<u>case B)</u> $f(x)=\frac{x-4}{19} , g(x)=19x+4$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y-4}{19}$

Isolate the variable y

$19x=y-4$

$y=19x+4$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=19x+4$

therefore

f(x) and g(x) are inverse functions

<u>case C)</u> $f(x)=x^{5}, g(x)=\sqrt[5]{x}$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=y^{5}$

Isolate the variable y

fifth root both members

$y=\sqrt[5]{x}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\sqrt[5]{x}$

therefore

f(x) and g(x) are inverse functions

<u>case D)</u> $f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y}{y+20}$

Isolate the variable y

$x(y+20)=y$

$xy+20x=y$

$y-xy=20x$

$y(1-x)=20x$

$y=20x/(1-x)$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=20x/(1-x)$

$\frac{20x}{1-x}\neq \frac{20x}{x-1}$

therefore

f(x) and g(x) is not a pair of inverse functions

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Angelina_Jolie [31]

Answer:

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Dafna1 [17]

Answer:

option C is correct answer ..

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so option C is the correct answer of this question ...

plz mark my answer as brainlist plzzzz vote me also as I have done this question by taking a lot of time Hope it will be helpful for you ....

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2 years ago