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Monica [59]
3 years ago
14

The function c = 2.50(n – 2) + 1.50 represents the cost c in dollars of printing n invitations. Which of the following is not tr

ue?
A For each additional invitation, it costs an extra $2.50 to print.
B Each invitation costs $1.50 to print
C One can not print just one invitation.
D The cost depends on the number of invitations printed.
Mathematics
2 answers:
o-na [289]3 years ago
8 0
B: Each invitation costs $1.50 to print
andreev551 [17]3 years ago
8 0

Answer:

Option B -  Each invitation costs $1.50 to print

Step-by-step explanation:

Given : The function c=2.50(n-2)+1.50

 Where,

 n: number of invitations

 c: cost in dollars

 The slope of the line for this case is: m=2.50

 The slope represents the cost of printing of each of the invitations.

 Therefore, it is false that the cost of printing of each invitation is $ 1.50

So, Option B is not true

i.e, B- Each invitation costs $1.50 to print


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Answer:

See explanation

Step-by-step explanation:

Car A: Started at 0 and ended at 300, thus, car A travels 300 miles.

It travels 6 hours, so car A speed is \frac{300}{6}=50 mph.

Car B: Started at 100 and ended at 300, thus, car B travels 300-100=200 miles.

It travels 5 hours, so car B speed is \frac{200}{5}=40 mph.

Since 50>40, car A traveled faster than car B.

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4 years ago
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2\sqrt{10}

Step-by-step explanation:

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3 years ago
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3 years ago
What is the slope of the line that passes through the points (5, 2) and (3,-1)?
oksano4ka [1.4K]
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to calculate the slope m use the gradient form

m = y2 -y1
————
x2 - x1


let (x1, y1) = (3,1) and (x2, y2) = (-2, 5)


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3 0
3 years ago
A rectangular package sent by a postal service can have a maximum combined length and girth (perimeter of a cross sectio) of 108
Morgarella [4.7K]

Answer:

The maximum volume of the package is obtained with a cross section of side 18 inches and a length of 36 inches.

Step-by-step explanation:

This is a optimization with restrictions problem.

The restriction is that the perimeter of the square cross section plus the length is equal to 108 inches (as we will maximize the volume, we wil use the maximum of length and cross section perimeter).

This restriction can be expressed as:

4x+L=108

being x: the side of the square of the cross section and L: length of the package.

The volume, that we want to maximize, is:

V=x^2L

If we express L in function of x using the restriction equation, we get:

4x+L=108\\\\L=108-4x

We replace L in the volume formula and we get

V=x^2L=x^2*(108-4x)=-4x^3+108x^2

To maximize the volume we derive and equal to 0

\dfrac{dV}{dx}=-4*3x^2+108*2x=0\\\\\\-12x^2+216x=0\\\\-12x+216=0\\\\12x=216\\\\x=216/12=18

We can replace x to calculate L:

L=108-4x=108-4*18=108-72=36

The maximum volume of the package is obtained with a cross section of side 18 inches and a length of 36 inches.

4 0
3 years ago
Read 2 more answers
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