Ok here are the problems I need help with
1)A revenue of $1500 is obtained from the sales of item A at $50 each and item B at $25 each. Write an equation that shows the relationship between the numbers of items sold.
2)An airplane trip is traveling away from its home airport at a constant speed. After 3 hours, it is 540 miles away Assuming it departed at t=0, write an equation representing the distance, “D” (in miles), at any time “t” (in hours).
3)A real estate sales agent receives a salary of $250 per week plus a commission of 2% sales. Write a linear model for the weekly income “y” in terms of sales “x”.
4)Machinery is bought new for $36,000 and is considered worth $2,000 as junk at the end of 10 years. Assuming a linear model for depreciation, write the equation of its value, “V” (in dollars), when “t” years have elapsed after the purchase.
Please label your answers with a number to represent the question you are answering. Please explain the answer too.
? sorry
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No, I will not do your homework for you! If you wanted to put what you think the answers are, I’ll tell you if I agree or not, but I will not do your work for you.
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Dude its summer stop the math god damn
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1) A total of $1500 is earned, comprising of $50 per item A and $25 per item B, thus:
50A + 25B = 1500
2) It travels at 540 miles per 3 hours, that is 180 miles per hour, thus:
D = 180t
3) He gets $250 no matter what the sales, plus 0.02 (2%) times the sales, called x, thus:
y = 250 + 0.02x
4) $36000 is the starting value, which decreases to $2000 in 10 years, which is 34000/10=3400 per year, thus:
V = 36000 – 3400t
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1) 1500 = 50(x)+25(y-x) where y is the no. of items sold and x is the number of item A was sold
2) d=180t
3) y= 250+.02x
4) v=36000-3400t
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1) The revenue from item A is 50A where A represents the number of item A sold. Likewise the revenue from item B is 25B. Thus the relationship between them is 50A + 25B = 1500 because 1500 is the total revenue.
2) distance = rate x time. In this problem d = 540 and t = 3, so the rate must be 540/3 = 180 mph. Thus, in terms of time, t, the distance the plane has traveled is d = 180t
3) What the agent receives each week is 250 + 2% of the sales, so that is y = 250 + .02x.
4) If it went from 36000 to 2000 it decreased in value 34000. Since it did this in 10 years, that’s 3400/yr. If t represents the number of years that have elapsed since the purchase, the values of the machine is given by V = 36000 – 3400t.
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1) 50x + 25y =1500, where x + no of item A sold, and y =no. of item B sold
2)D(in miles)=180t, where t is the number of hours travelled
If the plane , at constant speed travels 540 miles in 3 hrs, then it is travelling at 180 miles per hour. So distance will be 180x hours travelled.
3) y =250 + 2x/100 , where y is total salary, and x is total amount taken from sales.2% is “2 hundredths of” or–2/100, times (x)–or 2x/100.
4)V=36, 000 – 3,400t, where t =number of years elapsed since purchase, and v= value in dollars. If $34, 000 is lost over 10 years, and depracation is “linear”, then each year, 3.4 thousand dollars is lost. Therefore value = original value, minus 3, 400 x time passed, in years.
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OK
1) Let no of itemA = a
Let no of itemB = b
Then 50a + 50b = 1500
2) D = s x t
So 540 = s x 3
3) y = 250 + 2% of x
4) DEP for 1 year = (36000-2000)/10
= 3400
So V =36000 – (3400 x t )
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1. 1500 = (x . 50) + (x . 25)
1500 = 2x . 75
1500 / 75 = 2x
20 = 2x
10 = x
1500 = 10 . 50 + 10 . 25
1500 = 20 . 75
sold 20 of each
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