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4 years ago

I really need help!!!!

Mathematics

1 answer:

dezoksy [38]4 years ago

7 0

Answer:

150 bikes, $10,500 minimum manufacturing cost

Step-by-step explanation:

5x^2 - 1500x + 123000 is represented by a parabolic graph that opens up. You could easily estimate the x value at which C(x) is at a minimum, as well as the smallest C(x) value.

Or you could do this problem algebraically by finding the vertex of the parabola. The results MUST be the same as before.

-b

The equation of the axis of symmetry of this curve is x = ---------

... which here is x = 1500

--------- = 150 units (150 bikes)

2(5)

Evaluating C(x) (see the problem statement) at x = 150 leads to finding the minimum cost. I like to use synthetic division to evaluate polynomials. Here, the divisor would be 150 and the coefficients of the quadratic would be

5 -1500 123000

Setting up synthetic division, we get:

150 / 5 -1500 123000

750 -112500

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

5 -750 10500

The remainder is $10,500. This is the minimum cost of this manufacturing operation.

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During the last ice age the sea level was ____________ it is today. A) unknown, unlike B) much lower than Eliminate C) much high

allsm [11]

During the last ice age (top) sea level was at least 120 m lower than it is today surprisingly.

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4 years ago

Read 2 more answers

On the scale drawing of a car, the length from bumper to bumper is 6 inches. The car's actual length, from bumper to bumper, is

baherus [9]

Answer:

$I=\frac{3}{8}\cdot F$ and $F = \frac{8}{3}\cdot I$ .

Step-by-step explanation:

Let I represent inches and F represent feet.

We have been given that on he scale drawing of a car, the length from bumper to bumper is 6 inches. The car's actual length, from bumper to bumper, is 16 feet.

We will use proportions to solve our given problem as:

$\frac{\text{Length on scale}}{\text{Actual length}}=\frac{6}{16}$

$\frac{I}{F}=\frac{6}{16}$

Now, let us solve for $I$ .

$I=\frac{6}{16}\cdot F$

$I=\frac{3}{8}\cdot F$

Now, let us solve our equation for F.

$I\cdot\frac{8}{3}=\frac{8}{3}\cdot\frac{3}{8}\cdot F$

$\frac{8}{3}\cdot I= F$

$F = \frac{8}{3}\cdot I$

Therefore, our required equations would be $I=\frac{3}{8}\cdot F$ and $F = \frac{8}{3}\cdot I$ .

8 0

3 years ago

A small company borrows money and remains in debt to its lenders for a period of time. The function f(x)=−6x2+8x+50 represents t

Marysya12 [62]

Given that the debt has been represented by the function:
f(x)=-6x^2+8x+50
To get the number of years, x that it would take for the company to be debt free we proceed as follows:
we solve the equation for f(x)=0
hence:
0=-6x^2+8x+50
solving for x using the quadratic formula we get:
x=[-b+/-sqrt(b^2-4ac)]/2a
x=[-8+/-sqrt(8^2-4*(-6)*50)]/(-6*2)
x=[-8+/-√1264]/(-12)
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x~28

5 0

3 years ago

I need help with this problem

gizmo_the_mogwai [7]

Answer:

what she sells each t-shirt for.

Step-by-step explanation:

if f(x) is the profit and x=# of t-shirts,

the rate of change is 8, which is what she sells each t-shirt for.

The 100 that is subtracted, is the money that she spent on materials.

For eg, if she sells 8 t-shirts she makes $64, but she spent $100 before and didn't gain any profit, having not made up $36 from what her original spending was.

8 0

3 years ago

Eight adults and two children need to cross a river. A small boat is available that can hold one adult or one or two children. E

sleet_krkn [62]

Answer:

A = 8, C = 2 => T = 33

A=100, C = 2 => T = 401

A = any, C = 2 => T = 4A + 1

A = 8, C = 4 => T = 37

As long as the boat can carry one adult or one or two children, the following equation should solve the problem.

T = 4A + 2(C-2) + 1

Step-by-step explanation:

To transfer one adult, 4 one-way trips need to be made as follows:

1. Two children cross the river

2. One of them takes the boat back to the other side

3. One adult crosses the river

4. The other child takes the boat back

These four steps are repeated until all adults are crossed. At this point no children has crossed yet.

Then for every 2 children except the last 2, 2 one-way trips need to be made if the number of children is greater than 2:

1. Two children cross

2. One takes the boat back

For the last 2,only one trip needs to be made.

A = number of adults

C = number of children >= 2

T = total trips

T = 4A + 2(C-2) + 1

4 0

3 years ago