I could use help please

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Brilliant_brown [7]

Answer:

$f(x)=\sqrt{x-1}$ -----> Graph C

$f(x)=\sqrt{x}$ -----> Graph A

$f(x)=\sqrt{x}-1$ -----> Graph B

$f(x)=-\sqrt{x}$ -----> Graph D

$f(x)=-\sqrt{x-1}$ -----> Graph E

Step-by-step explanation:

we have

case a) $f(x)=\sqrt{x-1}$

Find the domain

we know that

The radicand must be greater than or equal to zero

so

$x-1\geq 0$

$x\geq 1$

The domain is the interval -----> [1,∞)

All real numbers greater than or equal to 1

The range is the interval -----> [0,∞)

All real numbers greater than or equal to 0

case b) $f(x)=\sqrt{x}$

Find the domain

we know that

The radicand must be greater than or equal to zero

so

$x\geq 0$

The domain is the interval -----> [0,∞)

All real numbers greater than or equal to 0

The range is the interval -----> [0,∞)

All real numbers greater than or equal to 0

case c) $f(x)=\sqrt{x}-1$

Find the domain

we know that

The radicand must be greater than or equal to zero

so

$x\geq 0$

The domain is the interval -----> [0,∞)

All real numbers greater than or equal to 0

The range is the interval -----> [-1,∞)

All real numbers greater than or equal to -1

case d) $f(x)=-\sqrt{x}$

Find the domain

we know that

The radicand must be greater than or equal to zero

so

$x\geq 0$

The domain is the interval -----> [0,∞)

All real numbers greater than or equal to 0

The range is the interval -----> (-∞,0]

All real numbers less than or equal to 0

case e) $f(x)=-\sqrt{x-1}$

Find the domain

we know that

The radicand must be greater than or equal to zero

so

$x-1\geq 0$

$x\geq 1$

The domain is the interval -----> [1,∞)

All real numbers greater than or equal to 1

The range is the interval -----> (-∞,0]

All real numbers less than or equal to 0

5 0

3 years ago

If x>7, then |x|>7. |y|>7, so y=7<br><br><br> Valid or invalid?

Alex

$\text{if }x>7\text{, then }|x|>7$ is a valid argument

$|y|>7\text{, so }y=7$ is not a valid argument

For the first argument: $\text{if }x>7\text{, then }|x|>7$

From the definition of absolute value function

$|x|=x$ if $x\ge0$

That is every positive number is its own absolute value. Since

$x>7\implies x\ge0$ ,

we can argue that

$x>7\implies |x|>7$

so the first argument is valid

For the second argument: $|y|>7\text{, so }y=7$

From the definition of absolute value function

$|y|:=\left \{ {y\text{ if }y\ge0}\atop{-y\text{ if }y$

This means that

$|y|>7:=\left \{ {y>7\text{ if }y\ge0}\atop{-y>7\text{ if }y$

or

$|y|>7:=\left \{ {y>7\text{ if }y\ge0}\atop{y$

no part of the definition allow for the option $y=7$ . So the second argument is not valid.

Learn more here: brainly.com/question/11897796

7 0

3 years ago

Find the derivative of the function y = ln(3t+4)

solniwko [45]

Answer:

3 / (3t+4) is the answer

5 0

3 years ago

If 20 is 40% of a value, what is that value?

podryga [215]

50 is the value cause 20 is 40 of the percent, 20+20 is 40 now you made it into 80%.20% is remaining so add 10.

3 0

4 years ago

A committee is to consist of two members. If there are seven people available to serve on the committee, how many different comm

ankoles [38]

3
Though there are seven people, it would only be 3 full committees. They would either find an eighth person or have a group of 3 with one of the committees.
Hope this helps!

7 0

2 years ago