PLEASE HELP ME I DONT UNDERSTAND!

F(x)= square root of x+2; g(x)=8x-12 find f(g(x))

Ann [662]

$f(x)=\sqrt{x+2}\\g(x)=8x-12\\f(g(x))=\sqrt{8x-12+2}=\sqrt{8x-10}$

4 0

3 years ago

Use complete sentences to describe why √-1 ≠ -√1

tekilochka [14]

Well let's say that to compare these two numbers, we have to start with the definition first.

Definition

$\displaystyle \large{ {y}^{2} = x} \\ \displaystyle \large{ y = \pm \sqrt{x} }$

Looks like we can use any x-values right? Nope.

The value of x only applies to any positive real numbers for one reason.

As we know, any numbers time itself will result in positive. No matter the negative or positive.

Definition II

$\displaystyle \large{ {a}^{2} = a \times a = |b| }$

Where b is the result from a×a. Let's see an example.

Examples

$\displaystyle \large{ {2}^{2} = 2 \times 2 = 4} \\ \displaystyle \large{ {( - 2)}^{2} = ( - 2) \times ( - 2) = | - 4| = 4}$

So basically, their counterpart or opposite still gives same value.

Then you may have a question, where does √-1 come from?

It comes from this equation:

$\displaystyle \large{ {y}^{2} = - 1}$

When we solve the quadratic equation in this like form, we square both sides to get rid of the square.

$\displaystyle \large{ \sqrt{ {y}^{2} } = \sqrt{ - 1} }$

Then where does plus-minus come from? It comes from one of Absolute Value propety.

Absolute Value Property I

$\displaystyle \large{ \sqrt{ {x}^{2} } = |x| }$

Solving absolute value always gives the plus-minus. Therefore...

$\displaystyle \large{ y = \pm \sqrt{ - 1} }$

Then we have the square root of -1 in negative and positive. But something is not right.

As I said, any numbers time itself of numbers squared will only result in positive. So how does the equation of y^2 = -1 make sense? Simple, it doesn't.

Because why would any numbers squared result in negative? Therefore, √-1 does not exist in a real number system.

Then we have another number which is -√1. This one is simple.

It is one of the solution from the equation y^2 = 1.

$\displaystyle \large{ {y}^{2} = 1} \\ \displaystyle \large{ \sqrt{ {y}^{2} } = \sqrt{1} } \\ \displaystyle \large{ y = \pm \sqrt{1} }$

We ignore the +√1 but focus on -√1 instead. Of course, we know that numbers squared itself will result in positive. Since 1 is positive then we can say that these solutions exist in real number.

Conclusion

So what is the different? The different between two numbers is that √-1 does not exist in a real number system since any squared numbers only result in positive while -√1 is one of the solution from y^2 = 1 and exists in a real number system.

5 0

3 years ago

Read 2 more answers

For the members of a neighborhood meeting group, the probability of randomly selecting a member who likes to cook is 42%, the pr

Monica [59]

Answer: Probability of randomly selecting a member who likes to cook or likes to sew = 58%

Step-by-step explanation:

Since we have given that

Probability of a randomly selecting a member who likes to cook=P(C) = 42%

Probability of a randomly selecting a member who likes to sew =P(S)= 23%

Probability of a randomly selecting a member who likes to both cook and sew

$P(C\cap S)=7\%$