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3 years ago

Please help due now

Mathematics

1 answer:

Elis [28]3 years ago

6 0

Answer:

Casey's Function has the greater rate of change

Step-by-step explanation:

The graph shows a ROC of 5/1

Get a common denominator (4)

5/1 * 4/4 = 20/4

20/4 > 13/4

Casey's function has a greater ROC.

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Six hundred and eight thousandths as a decimal

jolli1 [7]

600.008 (why does this need to be 20 things long)

7 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

2 years ago

Read 2 more answers

6 (x+15) evaluate the expression for x=-4

Gala2k [10]

The answer would be 66.

6(-4 +15)
=
66

4 0

3 years ago

Read 2 more answers

Which of the following statements is true about the three quadrilaterals? (5 points)

klemol [59]

Answer:

1 one os true

2 nd one is true

7 0

3 years ago

Please tell me if my answer is right NO LIES brainlest report you then

ladessa [460]

Answer:

Your answers are correct.

Step-by-step explanation:

The "<" sign means less than and the ">" sign means greater than. It is clear that you have a handle on them.

1.) If Steve has no more that 28 toys, then the less than sign would represent the fact that Steve can have a maximum of 28 toys, and no more.

2.) If the temperature is more that 28 degrees Fahrenheit, then the greater than sign is appropriate in order to imply that the variable represents a higher temperature.

3.) If tony is younger than 29 years old, then the less than sign fits to show that the variable represents an age that is smaller than 29.

4.) If the table is greater than 29 kilograms, then the greater than sign is needed. It shows that the table has a greater weight than 29 kilograms.

Hope this helps! :D

4 0

3 years ago