Answer:
In that one week, his commision was $896, over all 52 weeks, he made $46592
Step-by-step explanation:
Need help calculating that? Let me help you. Grab a calculator!
So, he makes 14% commision on any stock he sells. This week, he sold $6,400 worth of stocks, and, hes going to make 14% on it. So, take 6400 and divide that by 14%, and what do you get? $896! So, he made $896 that week. Now, say he made that amount EVERY WEEK. so, all you have to do is know how many weeks are in a year (52) and multiply that by 896, which gives you 46592.
Hope this helped, and brainliest would be amazing ❤️ please let me know if you need anymore help. im happy to help.
<span>Kedar earns $3887.50. His earnings are given by y=2200+0.0375*x where x is the amount of sales, in dollars, for the month.</span>
Answer:
befor i go im trynna love you baby hopefully e
Step-by-step explanation:
Answer:
The scale used on his map is <u>150 miles : 2 inches</u>.
Step-by-step explanation:
Given:
Ted knows the actual distance between two cities is 150 miles. His map shows a distance of 2 inches between these cities.
Now, to find the scale Ted used on his map.
The actual distance Ted know between two cities = 150 miles.
The distance on map between these cities = 2 inches.
So, to get the scale used on his map:
Therefore, the scale used on his map is 150 miles : 2 inches.
Part A. You have the correct first and second derivative.
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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.
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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.
