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OleMash [197]
3 years ago
13

Let g(x)=3x^3. find g(2+h)

Mathematics
1 answer:
IceJOKER [234]3 years ago
4 0

Answer:

g(2+h) = 3h^3+18h^2+36h+24

Step-by-step explanation:

g(x) = 3x^3

Putting x = 2+h

g(2+h) = 3(2+h)^3\\g(2+h) = 3[8+h^3+6h(2+h)]

g(2+h) = 3(8+h^3+12h+6h^2)\\g(2+h) = 24+3h^3+36h+18h^2\\

g(2+h) = 3h^3+18h^2+36h+24

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Rewrite the fraction with the denominator of 33 6/11=/33
Sonja [21]
You would time the denominator by 3 so 11*3 would be 33 then times the numerator by the same so 6*3 soot would be 18/33
7 0
2 years ago
Rory's battery was full at 6:00 a.M, or t = 0. Which times could Rory's phone have been plugged into the charger? Select three o
Taya2010 [7]

Complete question:

Rory records the percentage of battery life remaining on his phone throughout a day. The graph represents the percentage of battery life remaining after a certain number of hours.

Rory’s battery was full at 6:00 a.m, or t = 0. Which times could Rory's phone have been plugged into the charger? Select three options.

12:00 p.m.

3:00 p.m.

5:00 p.m.

8:00 p.m.

11:00 p.m.

The graph is attached

Answer:

3:00 p.m

5:00 p.m

11:00 p.m

Step-by-step explanation:

From the graph, there are rise times, fall times and there are times the movement is steady (no rise or fall).

The rise time represents the time the phone was plugged. The fall time represents when the charger has been unplugged so the battery level starts depreciating. When it is constant, it means the battery level is at 100 but the charger is still connected to the phone.

We are told at initial condition, time was 6 a.m (t = 0). To get the exact time, we are to add the initial condition, i.e 6

From the graph, the times the phone could have been plugged are 8:00 to 12:00 hrs and 16:00 to 20:00 hrs.

Converting from 24 hr to 12 hr time, we have:

6 + 8hrs = 14:00 = 2:00 p.m

6 + 9hrs = 15:00 = 3:00 p.m

6 + 10 hrs = 16:00 = 4:00 p.m

6 + 11 hrs = 17:00 = 5:00 p.m

6 + 12 hrs = 18:00 = 6:00 p.m

6 + 16 hrs = 22:00 = 10:00 p.m

6 + 17 hrs = 23:00 = 11:00 p.m

6 + 18 hrs = 00:00 = 12:00 am

6 + 19 = 1 : 00 = 1:00 a.m

6 + 20 = 2:00 = 2:00 a.m

From the options given in the question, we have:

3:00 p.m; 5:00 p.m; 11:00 p.m

Therefore, times could Rory's phone have been plugged into the charger are:

3:00 p.m

5:00 p.m

11:00 p.m

5 0
3 years ago
Need help on this problem
trasher [3.6K]
The answer to the question

3 0
3 years ago
If demand is D1, what is the lowest price that Stromnord can charge so that it will not run out of this model of shoe in the mon
Molodets [167]
D1 = 60 for a price of $80. Charging $80 will ensure supply exceeds demand.
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The store apparently could charge a price slightly lower than $80, but we cannot tell from the chart how much lower.
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3 years ago
You and your friend are selling tickets to a charity event. you sell 12 adult tickets and 6 student tickets for $138. your frien
mr Goodwill [35]
Let's say the cost of student tickets is x and the cost of adult tickets is y. Then:

(1) 12y + 6x = 138
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If we rearrange equation (1) we get:
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Now divide each side by 12:
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We can now substitute this into equation (2):

5(11.5 - 0.5x) + 11x = 100
57.5 - 2.5x + 11x = 100
8.5x = 42.5
x = 5, therefor the cost of a student ticket is $5.00
6 0
3 years ago
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