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Rudik [331]
3 years ago
11

You work 35 hours/week for 52 weeks and are given the option to be paid hourly or to go on salary. In which situation will you e

arn the most?
A. $18.50/hour
B. $32,000 a year
C. $17.50/hour & a $1,500 bonus at end of year
D. $30,000 a year w/ 5% bonus
Mathematics
2 answers:
rjkz [21]3 years ago
4 0

Answer:

Hourly

Step-by-step explanation:

if you work for that long and for that many hours you would get alot of money in about a month or two.

Sauron [17]3 years ago
4 0
18.5 dollars an hour
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I hope you have a good day and please give me brainest if that is correct!

3 0
2 years ago
I think I know this much so far for A (but I could be wrong):
Eduardwww [97]
Part A. You have the correct first and second derivative.

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

Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.

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

Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out

To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h  ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0. 
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well. 
6 0
3 years ago
If cosθ=2/3 find cscθ using identities. This is in quad. 1.
laiz [17]

Step-by-step explanation:

  • cos θ = 2/3

\quad \twoheadrightarrow\sf {cos \; \theta = \dfrac{Base}{Hypotenuse} } \\

Hence, base = 2 units and hypotenuse = 3 units.

\quad \twoheadrightarrow\sf { H^2 = B^2 + P^2} \\

\quad \twoheadrightarrow\sf { P^2 = H^2 - B^2} \\

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\quad \twoheadrightarrow\bf { P = \sqrt{5}} \\

Now, we know that :

\quad \twoheadrightarrow\sf {cosec \; \theta = \dfrac{Hypotenuse}{Perpendicular} } \\

\quad \twoheadrightarrow\bf{cosec \; \theta = \dfrac{3}{\sqrt{5}} } \\

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Answer:

Step-by-step explanation:

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7 0
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2 years ago
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