Given the cost function, C(x), and the revenue function, R(x), find the number of units x that must be sold to break even

1 answer:

8 0

P(x) = R(x) - C(x) Breaking even is the point where revenue starts to overtake cost. It is, therefore, the point where P(x) = 0. Since these are all linear functions, you know that after that point the business is profitable. P(x) = 0R(x) - C(x) = 015000x - (9000x+72000) = 015000x - 9000x - 72000 = 06000x - 72000 = 06000x = 72000x = 72000/6000x = 12 At 12 units the business breaks even.

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1+4=5 2+5=12 3+6=21 8+11=?

lawyer [7]

If you would like to solve this expression, you can calculate this using the following steps:

1 + 4 = 5
2 + 5 = 12
3 + 6 = 21
4 + 7 = 32
5 + 8 = 45
6 + 9 = 60
7 + 10 = 77
8 + 11 = 96

The correct result would be 8 + 11 = 96.

6 0

3 years ago

Triangle A″B″C″ is formed using the translation (x + 0, y + 2) and the dilation by a scale factor of 2 from the origin. Which eq

garik1379 [7]

Answer:

segment AC is equal to A''B'' over 2

Step-by-step explanation:

Given that,

Scale factor = 2

The coordinates of A is (-3,3)

The coordinates of B is (1,-3)

The coordinates of C is (-3,3)

We know that, the ratio of length of new segment to the length of segment of original figure is called scale factor.

So.

$\dfrac{A''B''}{AC}=2\\\\\text{On cross multiplying,}\\\\AC=\dfrac{A''B''}{2}$

It means, segment AC is equal to A''B'' over 2 is the correct answer.

3 0

3 years ago

Given that (6 3) is on the graph of f(x) find the corresponding point for the function f(-1/2x)

Vlada [557]

$\bf \begin{array}{rllll}
(6&,&3)&f(x)\\
x&&y\\\\
(-\frac{1}{2}x&,&3)&f(-\frac{1}{2}x)\\\\
(-\frac{1}{2}\cdot 6&,&3)\\\\
(-3&,&3)
\end{array}$

keep in mind that, a negative coefficient to "x", will make the graph reflect over the y-axis.

6 0

4 years ago

Read 2 more answers

The table below shows the future value of a certificate of deposit (CD) after x years Eric states the applied interest is simple

Mama L [17]

Answer:

Option B.

Step-by-step explanation:

Eric states the applied interest is simple interest. We need to check whether the this statement is true or false.

In simple interest, the amount of deposit increased at a constant rate. It means the difference between future values of any two consecutive years will same.

Difference between future values of first two years.

$d_1=210.13-205.00=5.13$