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kicyunya [14]
3 years ago
5

Determine the intervals on which the function is increasing, decreasing, and constant. (3 points)

Mathematics
2 answers:
aksik [14]3 years ago
8 0

the answer is A

if it is increasing the  value must be more than -1 not less than but if it were decreasing then x would have to equal less then -1

Cerrena [4.2K]3 years ago
7 0

Answer:

A)  increasing x > -1 ; decreasing x <-1

Step-by-step explanation:

We use the given graph to find the intervals where the function is increasing , decreasing and constant

We analyze the graph from the left

From the left the graph starts decreasing and it is decreasing till -1

From the point (-1,0) the graph changes direction and starts increasing

So we can say that the function is decreasing from -infinity to -1

and increasing from -1 to infinity

decreasing from -infinity to -1 means decreasing x <-1

increasing from -1 to infinity means increasing x > -1

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Find solution to the system of linear equations. 5x1 + x2 = 0 , 25x1 + 5x2 = 0
Zigmanuir [339]

Answer:

the system of equation has infinite solution

Step-by-step explanation:

5x_1 + x_2 = 0 , 25x_1 + 5x_2 = 0

Solve the first equation for x_2

Subtract 5x1 on both sides

5x_1 + x_2 = 0

x_2 =-5x_1

Now substitute -5x1 on second equation

25x_1 + 5x_2 = 0

25x_1 + 5(-5x_1)= 0

25x_1-25x_1= 0

0=0

So the system of equation has infinite solution

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3 years ago
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What is the sum of 2 1/2, 3 1/4, and 7?
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51/4 is the correct answer
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When is the constant of proportionality the same as the unit rate when comparing two quantities?
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3 years ago
A store has been selling 300 Blu-ray disc players a week at $600 each. A market survey indicates that for each $40 rebate offere
timurjin [86]

Answer:

75 $

Step-by-step explanation:

According to problem statement p(300) = 600

And we know that with a rebate of 40 $, numbers of units sold will increase by 80 then if x is number of units sold, the increase in units is

(  x  - 300 )  , and the price decrease

(1/80)*40  =  0,5

Then the demand function is:

D(x)  =  600  - 0,5* ( x - 300 )  (1)

And revenue function is:

R(x) =  x * (D(x)   ⇒   R(x) =  x* [  600  - 0,5* ( x - 300 )]

R(x) = 600*x  - 0,5*x * ( x - 300 )

R(x) = 600*x - 0,5*x² - 150*x

R(x) = 450*x  - (1/2)*x²

Now taking derivatives on both sides of the equation we get

R´(x) =  450  - x

R´(x) =  0       ⇒   450  - x = 0

x = 450 units

We can observe that for   0 < x  < 450  R(x) > 0 then R(x) has a maximum for x = 450

Plugging this value in demand equation, we get the rebate for maximize revenue

D(450)  =  600  - 0,5* ( x - 300 )

D(450)  =  600 - 225 + 150

D(450)  =

D(450)  =  600 - 0,5*( 150)

D(450)  =  600 - 75

D(450)  = 525

And the rebate must be

600 - 525  = 75 $

5 0
3 years ago
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