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sasho [114]
3 years ago
15

How many minutes are there in 4 hours,30 minutes please help I will give you 12 points

Mathematics
2 answers:
Sergeeva-Olga [200]3 years ago
5 0
There are 240 minutes in 4 hours if u want to explain then i will have a nice day
Inessa [10]3 years ago
3 0
There are 60 minutes in one hour. If we are trying to find how many minutes are in 4 hours and 30 minutes, we first need to find out how many minutes are in 4 hours.
To do that, we just need to multiply the amount of minutes in an hour by the amount of hours we have.

60 * 4 = 240

Then, we just add the remaining 30 minutes.

240 + 30 = 270

There are 270 minutes in 4 hours and 30 minutes.
Hope that helped! =)
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Albert's hourly pay is $8.10. He has an hourly pay of $12.15 when he works more than 40 hours. He worked 45 hours last week. Wha
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Answer:

$384.75

Step-by-step explanation:

Albert's hourly pay is $8.10

When he works for more than 40 hours, his hourly pay will increase to $12.15.

Last week, Albert worked for 45 hours, his hourly pay would be:

First 40 hours = $8.10 * 40 = $324

Remaining 5 hours =

$12.15 * 5 = $60.75

Albert gross pay for this period will be calculated as:

$324 + $60.75 = $384.75

Therefore, Albert gross pay for this period = $384.75

3 0
3 years ago
Express 8 7/40 as a decimal
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Answer: 8.175
Explanation: 8 is the whole number so set that aside look at 7/40, using simple division solve for a decimal
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2 years ago
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You ask 40 students which of three items from the cafeteria they like the best. You record the results on the piece of paper sho
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Answer:

i believe 45%

Step-by-step explanation:

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3 years ago
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Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com
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f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}

f has critical points where the derivative is 0:

2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1

The second derivative is

f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}

and f''(-1)=\frac2e>0, which indicates a local minimum at x=-1 with a value of f(-1)=\frac1e.

At the endpoints of [-2, 2], we have f(-2)=1 and f(2)=e^8, so that f has an absolute minimum of \frac1e and an absolute maximum of e^8 on [-2, 2].

So we have

\dfrac1e\le f(x)\le e^8

\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx

\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}

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3 years ago
Joe completed 7 pages of homework over 3 hours. Nina completed 5 pages over two hours. Who was completing the work at a faster p
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Answer:

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Step-by-step explanation:

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