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zepelin [54]
3 years ago
9

Jenna collected data modeling a company's company costs versus its profits. The data are shown in the table:

Mathematics
1 answer:
Zarrin [17]3 years ago
3 0

Statement 3:

The function is increasing from x = 0 to x =  1 is the true statement.

Solution:

Statement 1: The function is decreasing from x= –2 to x = 1

At x = –2, the function value is 2.

At x = –1, the function value is –3.

At x = 0, the function value is 2.

At x = 1, the function value is 17.

So, the value of the function is not constantly decreasing or increasing.

Hence it is false statement.

Statement 2: The function is decreasing from x = –1 to x = 0

At x = –1, the function value is –3.

At x = 0, the function value is 2.

From –3 to 2, the value is increasing.

Hence it is false statement.

Statement 3: The function is increasing from x = 0 to x =  1

At x = 0, the function value is 2.

At x = 1, the function value is 17.

Hence it is true statement.

Statement 4: The function is decreasing from x = 0 to x =  1

It is already proved in statement 3.

Hence it is false statement.

Therefore, the function is increasing from x = 0 to x =  1 is the true statement.

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77julia77 [94]
C and E, I believe since the other theorems don’t exist and AAS does not apply
5 0
3 years ago
In the equation c*2=22 what is the next step in the equation solving sequence?
romanna [79]
Divide 22 by 2 to get "c" isolated.
3 0
3 years ago
Evaluate v(x)=12-2x-5 when x=-2,0, and 5 .<br><br> v(-2)= <br><br> v(0)= <br><br> v(5)=
Vaselesa [24]

Step-by-step explanation:

v(-2)= 12-2x(-2)-5

=12+4-5

=16-5

=11

v(0)= 12-2x0-5

= 12-0-5

=12-5

=7

v(5)= 12-2x5-5

= 12-10-5

=12-5

=7

<em><u>Hope </u></em><em><u>it </u></em><em><u>helps </u></em>

6 0
2 years ago
One ticket to a ride of the merry-go-round ​at the Sunday Fair costs $2.
sammy [17]

Considering the definition of an inequality, the maximum number of tickets that they can buy is 10.

<h3>Definition of inequality</h3>

An inequality is the existing inequality between two algebraic expressions, connected through the signs:

  • greater than >.
  • less than <.
  • less than or equal to ≤.
  • greater than or equal to ≥.

An inequality contains one or more unknown values ​​called unknowns, in addition to certain known data.

Solving an inequality consists of finding all the values ​​of the unknown for which the inequality relation holds.

<h3>Maximum number of tickets that they can buy</h3>

In this case, you know that

  • One ticket to a ride of the merry-go-round ​at the Sunday Fair costs $2.
  • Jenny and her friends have $36 with them.
  • After buying tickets to the merry-go-round, they want to be left with no less than $15.

So, they want to spend on the purchase of tickets for the merry-go-round a value less than or equal to $36 - $15= $21.

Being "x" the maximum number of tickets that they can buy, the inequality that expresses the previous relationship is

2x≤ 21

Solving:

x≤ 21÷2

<u><em>x≤ 10.5</em></u>

Then, the maximum number of tickets that they can buy is 10.

Learn more about inequality:

brainly.com/question/17578702

brainly.com/question/25275758

brainly.com/question/14361489

brainly.com/question/1462764

#SPJ1

7 0
2 years ago
Solve the system by elimination.(show your work)
PilotLPTM [1.2K]

Answer:

x = 1 , y = 1 , z = 0

Step-by-step explanation by elimination:

Solve the following system:

{-2 x + 2 y + 3 z = 0 | (equation 1)

-2 x - y + z = -3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x - 3 y - 2 z = -3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+3 y + 2 z = 3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+3 y + 2 z = 3 | (equation 2)

0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y+0 z = 5 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:

{-(2 x) + 0 y+3 z = -2 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = -2 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = 1 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer: {x = 1 , y = 1 , z = 0

6 0
3 years ago
Read 2 more answers
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