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Flura [38]
3 years ago
8

Janine has job offers at two companies. One company offers a starting salary of $28,000 with a raise of $3000 each year. The oth

er company offers a starting salary of $36,000 with a raise of $2000 each year. After how many years would Janine’s salary be the same with both companies?
Mathematics
1 answer:
Umnica [9.8K]3 years ago
4 0

Answer:

  8 years

Step-by-step explanation:

The first company's offer is $8000 lower, but the raises are $1000 higher each year. It will take 8000/1000 = 8 years for the salaries to be the same.

__

If you like, you can write an equation that will tell you the same thing.

  28000 +3000y = 36000 +2000y

Subtract 28000+2000y from both sides:

  3000y -2000y = 36000 -28000

  1000y = 8000

  y = 8000/1000 = 8

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5+8 to the power of -2 x2-
vagabundo [1.1K]

5+8=13
-2x-2 = 4
13 to the fourth power = 13x13x13x13 =28561
28561 is your answer
4 0
4 years ago
A skydiver falls 288.52 feet in four seconds.<br><br> The skydiver fall feet per second.
Veronika [31]

Answer:

72.13

Step-by-step explanation:

72.13

4|288.52

28

08

8

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Clarge%7B%5Cbold%20%5Cred%7B%20%5Csum%20%5Climits_%7B8%7D%5E%7B4%7D%20%7Bx%7D%5E%7B2%7D%2
kiruha [24]

Answer:

No solution

Step-by-step explanation:

We have

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

For the sum it is not correct to assume

$\sum_{x=8}^{4}x^2= 8^2 + 7^2+6^2+5^2+4^2 = 64+49+36+25+16 = 190$

Note that for

$\sum_{x=a}^b f(x)$

it is assumed a\leq x \leq b and in your case \nexists x\in\mathbb{Z}: a\leq x\leq b for a>b

In fact, considering a set S we have

$\sum_{x=a}^b (S \cup \varnothing) = \sum_{x=a}^b S + \sum_{x=a}^b \varnothing $ that satisfy S = S \cup \varnothing

This means that, by definition \sum_{x=a}^b \varnothing = 0

Therefore,

$\sum_{x=8}^{4}x^2 = 0$

because the sum is empty.

For

9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)

we have other problems. Actually, this case is really bad.

Note that \cos^2(\infty) has no value. In fact, if we consider for the case

$\lim_{x \to \infty} \cos^2(x)$, the cosine function oscillates between [-1, 1] , and therefore it is undefined. Thus, we cannot evaluate

9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)

and then

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

has no solution

7 0
3 years ago
Factor out the gcf<br><br> -240c^2+ 30c^5+ 18c^4
exis [7]

Answer:

6c^2(-40 + 5c^3 + 3c^2)

Step-by-step explanation:

These three terms have the common factor c^2.  Thus:

-240c^2+ 30c^5+ 18c^4 = c^2(-240+ 30c^3+ 18c^2).

Recognizing that 6 is common to all three terms inside the parentheses, we get:

6c^2(-40 + 5c^3 + 3c^2)  This is as far as factoring of this expression can be taken.

8 0
3 years ago
The questions are in the image above.
enyata [817]

Answer:

Hope this helps :)

you will find every answer in the photo I sent

4 0
3 years ago
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