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kvv77 [185]
3 years ago
10

Sean is taking drivers ed. And the instructor asked him to stay within 2 miles of the posted speed limits. The current speed lim

it is 45 mph. Write an inequality that shows the acceptable speeds that Sean may drive
Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
3 0

Answer:

{x| 43 ≤ x ≤ 47} where x = Sean's speed (in mph)

Step-by-step explanation:

Sean is taking driver's ed. The instructor asked him to stay within 2 miles of the posted speed limits. The current speed limit is 45 mph.

So, the maximum speed which Sean can reach is, (45 + 2) mph

                                                                                   =47 mph

and,  the minimum speed which Sean can reach is, (45 - 2) mph

                                                                                   =43 mph

so, {x| 43 ≤ x ≤ 47} where x = Sean's speed (in mph)

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2 11/32 or 75/32

Step-by-step explanation:

2 3/8 - 1/4 x 1/8= 2 3/8 - 1/32=  19/8- 1/32= 76/32 - 1/32= 75/32= 2 11/32

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After 2 hours, Ernie had travelled 192 miles. If he travels at a constant speed, how far will he have travelled after 5 hours?
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a gift shop sells 160 wind chimes per month at $150 each. the owners estimate that for each $15 increase in price, they will sel
adell [148]

Answer:

The price per wind chime that will maximize revenue = $ 315

Step-by-step explanation:

Given - A gift shop sells 160 wind chimes per month at $150 each. the owners estimate that for each $15 increase in price, they will sell 5 fewer wind chimes per month.

To find - Find the price per wind chime that will maximize revenue.

Proof -

Given that,

Total Wind chimes selling = 160

Price of each Wind chime = $150

Now,

Given that, for each $15 increase in price, they will sell 5 fewer wind chimes per month.

So,

Let the price = 150 + 15x

So,

Number of wind Chimes sold per month = 160 - 5x

So,

Total Revenue, R = (150 + 15x)(160 - 5x)

                             = 24000 - 750x + 2400x - 75x²

                             = 24000 + 1650x - 75x²

⇒R(x) = 24000 + 1650x - 75x²

Differentiate R with respect to x , we get

R'(x) = 1650 - 150x

Now,

For Maximize Revenue, Put R'(x) = 0

⇒1650 - 150x = 0

⇒150x = 1650

⇒x = 1650/150

⇒x = 11

∴ we get

Price per Wind chime = $ 150 + 15(11)

                                    = $ 150 + 165

                                    = $ 315

So,

The price per wind chime that will maximize revenue = $ 315

3 0
2 years ago
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