Let t represent "total price." Then 0.20t = $1600, and t = $1600/0.20 =
$8000 (answer)
Answer:
x=122/3
Step-by-step explanation:
Step 1: Simplify both sides of the equation.
6(x+1)(3)−10=740
18x+18+−10=740(Distribute)
(18x)+(18+−10)=740(Combine Like Terms)
18x+8=740
18x+8=740
Step 2: Subtract 8 from both sides.
18x+8−8=740−8
18x=732
Step 3: Divide both sides by 18.
18x
18
=
732
18
x=
122
3
Answer:
x= 122/3
Answer:
<u>Domain:</u>
The domain of this can be any value between 0 to 565 miles per hour
<u>Range:</u>
The reasonable range can be the distance traveled which can be from 0 to 13,560 miles (no plane travel is longer than 24 hours, we assume).
Step-by-step explanation:
Domain is the input, set of x values for the function.
Range is the output, set of y values for the function.
This isn't a function essentially, but it is given that an Airplane travels at 565 miles per hour.
<em>We can say that the domain will be the speed of the airplane and the range would be the distance it travels.</em>
<em />
<u>Domain:</u>
The domain of this can be any value between 0 to 565 miles per hour
<u>Range:</u>
The reasonable range can be the distance traveled which can be from 0 to (565*24=13,560 miles) 13,560 miles (no plane travel is longer than 24 hours, we assume).
Answer:
21. y = 75000·0.935^t
22. after 74.6 days
23. y = 27.8112·1.18832^t
24. 18.8% per month
25. 1748
Step-by-step explanation:
22. It is convenient to use the graphing calculator to solve this problem. The number of days is where the exponential curve has the value 500. It is about 74.55 days. (see the first attachment)
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23. y = 27.8112·1.18832^t (see the second attachment)
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24. The rate of change is the difference between the base of the exponential and 1, often expressed as a percentage. The time period is the units of t.
(1.18832 -1) × 100% ≈ 18.8% . . . . per month
__
25. Evaluating the function for t=24 gives y ≈ 1748.30425259 ≈ 1748.
_____
<em>Comment on graphing calculator</em>
A graphing calculator can make very short work of problems like these. It is worthwhile to get to know how to use one well.
