Year Net Profit
1 <span>$14,250.00
2 $15,390.00
3 $16,621.20
4 $17,950.90</span>2

We need to get the increase of the net profit of the current year from the previous year.

Percentage increase = (Current year - Previous Year)/ Previous Year * 100%

Year 2: (15,390 - 14, 250) / 14,250 * 100% = 0.08 * 100% = 8%
Year 3: (16,621.20 - 15,390) / 15,390 * 100% = 0.08 * 100% = 8%
Year 4: (17,950.90 - 16,621.20) / 16,621.20 * 100% = 0.08 * 100% = 8%

Every year the net income increases by 8%. So, the net income in Year 5 will be:

17,950.90 x 1.08 = 19,386.97 Choice D.

8 0

3 years ago

Read 2 more answers

Simplify -8x^2-3y+4x-2y^3+x^2-7x^3

rewona [7]

Answer:-7x(to the power of 2)-3y+4x-2y(to the 3rd power)-7x( to the third power)

Step-by-step explanation:

Collect like terms

5 0

2 years ago

Which of these is true? A) 357 > 571 B) 432 > 571 C) 571 > 432 D) 571 < 357

elena55 [62]

Answer is C) 571>432

4 0

2 years ago

Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest

s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and $\varepsilon>0$ , we want to find a $\delta$ such that $|f(x)-14|$

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that $|x-5|$ . Then

$|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta$ .

Then, if $|x-5|$ , then $|f(x)-14|$ . So, in this case, if $3\delta \leq \varepsilon$ we get that $|f(x)-14|$ . The maximum value of delta is $\frac{\varepsilon}{3}$ .

By definition, this procedure proves that $\lim_{x\to 5}f(x) = 14$ . Note that f(5)=14, so this proves that f is continous at x=5.

3 0

2 years ago