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3 years ago

14

What is the domain,range ETC

2 answers:

Goshia [24]3 years ago

5 0

Domain is the x axis and range is the y axis. The minimum is
The domain is the x axis and the range is the y axis. C) (4,1) and maximum is (2,2) D) f(3) is 2 and E) is 1

Alex787 [66]3 years ago

3 0

Answer:

Domain: - infinity<X<infinity

Range: 2.5>Y> - infinity

Step-by-step explanation:

Because we don't know the strarting point, so we can just assume that the X values are staring from negative infinity and ending at possitive infinity.

Then, we can see the highest Y value is at 2.5 and then decresing infinitly.

That's why we can write the domain and range this way.

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And also I don’t understand this cuz it’s so confusing when my teacher does is it like she shows us examples but I forgot how to

ziro4ka [17]

In order to solve the equation solve for the area of the square and then the area of each of the triangles. For the first one the squares area is 256 while the triangles area is 136. Because there is four triangles multiply 136 by four which is 544. Add 256 and 544 to get the final answer of 800

8 0

4 years ago

PLEASE HELP, I WILL MARK BRAINLIEST(no links)

Bas_tet [7]

Answer:

-70 degrees

Step-by-step explanation:

What you need to do is subtract 10 from -58 to find the difference;

-58-12 = -70 degrees

7 0

3 years ago

Read 2 more answers

The Environmental Protection Agency records data on the fuel economy of many different makes of cars. Data on the mileage of 20

artcher [175]

Answer:

A) 8.5 miles per gallon

Step-by-step explanation:

Given

$x = \{12\ 13\ 15\ 16\ 16\ 17\ 18\ 18\ 19\ 19\ 20\ 20\ 22\ 23\ 24\ 26\ 26\ 27\ 27\ 29\}$

$n = 20$

Required

The IQR

This is calculated as:

$IQR = Q_3 - Q_1$

Calculate $Q_1$

$Q_1 = \frac{1}{4}(n + 1)th$

$Q_1 = \frac{1}{4}(20 + 1)th$

$Q_1 = \frac{1}{4}(21)th$

Remove bracket

$Q_1 = 5.25th$

This means that:

$Q_1$ is the mean of 5th and 6th item

From the given data:

$5th \to 16$

$6th \to 17$

So, we have:

$Q_1 = \frac{1}{2}(16 + 17)$

$Q_1 = \frac{1}{2}(33)$

$Q_1 = 16.5$

$Q_1 = 16.25$

Calculate $Q_3$

$Q_3 = \frac{3}{4}(n + 1)th$

$Q_3 = \frac{3}{4}(20 + 1)th$

$Q_3 = \frac{3}{4}(21)th$

Remove bracket

$Q_3 = 15.75th$

This means that:

$Q_3$ is the mean of 15th and 16th item

From the data:

$15th \to 24$

$16th \to 26$

So:

$Q_3 = \frac{1}{2}(24 + 26)$

$Q_3 = \frac{1}{2}(50)$

$Q_3 = 25$

Recall that:

$IQR = Q_3 - Q_1$

$IQR = 25 -16.5$

$IQR = 8.5$

7 0

3 years ago

PLEASE HELP ASAP 25 PTS + BRAINLIEST FOR BEST/RIGHT ANSWER

Airida [17]

Answer:

(-2,-1,-3)

Step-by-step explanation:

5x-4y-3z=3

x+y-z=0

x=3y+1

Multiply the second equation by -3

-3(x+y-z)=0*3

-3x-3y+3z =0

Add this to the first equation to eliminate z

5x-4y-3z=3

-3x-3y+3z =0

--------------------

2x-7y =3

Substitute x =3y+1 into the above equation

2(3y+1) -7y=3

Distribute the 2

6y +2 -7y = 3

Combine like terms

-y+2 =3

Subtract 2 from each side

-y+2-2 = 3-2

-y = 1

Multiply by -1

-y*-1 = 1*-1

y =-1

We need to find x

x=3y+1

x =3(-1) +1

x =-3+1

x= -2

Now we need to find z

x+y-z=0

-2+-1-z=0

-3-z=0

Add z to each side

-3-z+z=0+z

-3=z

x=-2, y=-1 z=-3

(-2,-1,-3)

7 0

3 years ago

Twenty-three girls tried out for the cheerleading squad. Only 12 could be chosen for the team. How many girls were not selected?

allsm [11]

Answer: 11 girls were not selected.

Step-by-step explanation:

First let's find out how many girls their were in total. In this case their are 23.

Let's subtract 12 from 23. That ends up to be 11. (<em>subtract 23 from 11 / 11 from 23)</em>

-23

-12

-------

11

3 0

4 years ago

Read 2 more answers