Find the range of each function for the given domain-

earnstyle [38]

Answer:

x=4.5,y=5,z=9

Step-by-step explanation:

First, we use the pythagorean theorem in the lower right triangle (with sides 4,x,6) and find x^2+4^2=6^2 and therefore x^2=6^2-4^2=20 so x=√20 which is approximately 4.5.

Also, we can see from the picture that z=y+4, and therefore the only option left is x=4.5,y=5,z=9.

5 0

2 years ago

There are 10 fruits in a basket and 5 of them are apples. Write the ratio in simplest form.

klasskru [66]

Answer:

10:5= 2:1

Step-by-step explanation:

Hope this helped!! Give brainliest..

3 0

3 years ago

Read 2 more answers

A recipe requires

Alla [95]

Answer:

22/3=8/3=2.66

if 12 servings =2.6 cup of flour

...18 servings=???????

18×2.6/12=3.9

the answer is 3.9

6 0

2 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

Which choice is the equation of a line that passes through the point (–6, –2) and is perpendicular to the line represented by th

kobusy [5.1K]

Answer:

B. y = 1/3x

Step-by-step explanation:

The slope of the given line is -3, so the slope of the perpendicular line will be the negative reciprocal of that: -1/(-3) = 1/3. This eliminates choices A and C.

We can find the y-intercept (b) if we use the given point values in the equation ...

y = 1/3x +b

we get ...

-2 = (1/3)(-6) +b

-2 = -2 +b . . . simplify

0 = b . . . . . . . add 2

So, the equation of the perpendicular line through the given point is ...

y = 1/3x . . . . matches choice B

_____

Alternate solution method

You can try the given point in the given equations. You will find that only the equation of choice B will work.

7 0

3 years ago