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Ivahew [28]
3 years ago
8

A manufacturer produces dice in many different sizes. If the side length of the dice is double, then the volume of the larger di

ce is how many time the volume of the smaller dice?
Mathematics
1 answer:
joja [24]3 years ago
7 0

Answer:

Here you are...

Step-by-step explanation:

If you double one of the dimensions, say change one side from 2 to 4 it doubles the volume. if you were to do this to any side, say double it, it would double the volume. ... It will make the volume 4 times as much also. the volume would go from 30 to 120.

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Solve for x: −1 < x + 3 < 5
MA_775_DIABLO [31]

Answer:

B

Step-by-step explanation:

Subtract 3 from all sides.

-4 < x < 2

6 0
3 years ago
Erin invested 15000 some at 5 percent and some at 8 percent annual interest if she recieves an annual return of 930 dollars how
eduard
.08x+.05(15000-x)=930
.08x+750-.05=930
.08x-.05x=930-750
.03x=180
X=180/.03=6,000 at 8%
(15000-6000)=9000 at 5%
8 0
3 years ago
A carpool service has 2,000 daily riders. A one-way ticket costs $5.00. The service estimates that for each $1.00 increase to th
solmaris [256]

Answer:

Total number of riders that ride on carpool daily = 2000

Total Cost of one way ticket = $ 5.00

Total Amount earned if 2000 passengers rides daily on carpool = 2000 × 5

                                                                                                            = $10,000

If fare increases by $ 1.00

New fare = $5 + $1    

               = $6

Number of passengers riding on carpool = 2,000 - 100 = 1,900

If 1,900 passengers rides on carpool daily , total amount earned ,if cost of each ticket is $ 6 = 1900 × $6 = $11400

As we have to find the inequality which represents the values of x that would allow the carpool service to have revenue of at least $12,000.

For $ 1 increase in fare = (2,000 - 1 × 100) passengers

For $ x increase in fare, number of passengers = 2,000 - 100·x

                                                         = (2,000 - 100·x) passengers

New fare = 5 + x

New Fare × Final Number of passengers ≥ 12,000

(5+x)·(2,000 - 100 x) ≥ 12,000

5 (2,000 - 100 x) + x(2,000 - 100 x) ≥ 12,000

10,000 - 500 x + 2,000 x - 100 x² ≥ 12,000

100 - 5 x + 20 x - x² ≥ 120

- x² + 15 x +100 - 120 ≥ 0

-x² + 15 x -20 ≥ 0

x² - 15 x + 20 ≤ 0

⇒ x = 1.495

x ≥ $ 1.495, that is if we increase the fare by this amount or more than this the revenue will be at least 12,000 or more .

Also, f'(x) = 0 gives x = 7.5

⇒ The price of a one-way ticket that will maximize revenue is $7.50

7 0
2 years ago
A car dealership sold 480 vehicles last year. Of these vehicles, 15% were minivans.
natulia [17]
You can use my photo as reference but its going to be 15/100 of 480

7 0
2 years ago
Solve for x please<br> 4.5x + 11.5 = 8.5x + 3.5
leonid [27]

Answer: have you tried to use photomath?


6 0
3 years ago
Read 2 more answers
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