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Shtirlitz [24]
3 years ago
10

You have a summer job that pays time and a half for overtime (any hours over 40 in a week). Overtime is 1.5 times your hourly ra

te of $7.00/hr.
How much money will you make if you work 50 hours in one week?

$350


$525


$385


$400
Mathematics
1 answer:
san4es73 [151]3 years ago
7 0
Each hour you work is already $7.00, without overtime (more than 40 hours a week)
So if you work 40h and earn $7.00 a week, 40h*$7.00 = $280
but since you work 50 hours, 10 hours are over time so you earn 1.5*the $7.00 which is $10.50 for those 10 hours
So $10.50*10h = $105
So now you add the 280 you already made plus the $105 in over time.
105+280
The answer is $385
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Step-by-step explanation:

The "long division method" algorithm for square root makes use of the relation described by the square of a binomial.

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<h3>Steps</h3>

The value for which the root is desired is written with digits marked off in pairs either side of the decimal point.

The initial digit of the root is the integer part of the square root of the most-significant pair. Here that is floor(√21) = 4. This is shown in the "quotient" spot above the leftmost pair. The square of this value is subtracted, and the next pair brought down for consideration. Here, that means the next "dividend" is 506.

The next "divisor" will be 2 times the "quotient" so far, with space left for a least-significant digit. Here, that means 506 will be divided by 80 + some digits. As in regular long division, determining the missing digit involves a certain amount of "guess and check." We find that the greatest value 'b' that will give b(80+b) ≤ 506 is b=5. This is the next "quotient" digit and is placed above the "dividend" pair 06. The product 5(85) = 425 is subtracted from 506, and the next "dividend" pair is appended to the result. This makes the next "dividend" equal to 8181.

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The root found here is 459.

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<em>Additional comment</em>

In practice, roots are often computed using iterative methods, with some function providing a "starter value" for the iteration. Some iterative methods can nearly double the number of good significant digits in the root at each iteration.

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