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gogolik [260]
3 years ago
13

$44 marked up by 40%

Mathematics
2 answers:
LUCKY_DIMON [66]3 years ago
7 0
The original price is $44
Marked up means increased

The new price = 40% of 44 + 44
(40% of original price + original price)

40% of 44
40/100 * 44
17.6

17.6 + 44 = 61.6

Thus the new increased price is $61.6

The general formula to help you
Percentage of increase*original + original
Vlada [557]3 years ago
5 0
You will take 44+(40/100 * 44) = 44 + 17.6 = $61.60 ($61.60 is going to be the retail price of the lamp).
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Complete parts a through f below to find nonnegative numbers x and y that satisfy the given requirements. Give the optimum value
MArishka [77]

Answer:

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Step-by-step explanation:

<u>Given</u>

  • x plus y equals 81
  • x and y are non-negative

<u>Find</u>

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<u>Solution</u>

a. Solve x plus y equals 81 for y.

  y equals 81 minus x

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b. Substitute the result from part a into the equation P equals x squared y for the variable that is to be maximized.

  P equals x squared left parenthesis 81 minus x right parenthesis

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c. Find the domain of the function P found in part b.

  left bracket 0 comma 81 right bracket

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d. Find dP/dx. Solve the equation dP/dx = 0.

  P = 81x² -x³

  dP/dx = 162x -3x² = 3x(54 -x) = 0

The zero product rule tells us the solutions to this equation are x=0 and x=54, the values of x that make the factors be zero. x=0 is an extraneous solution for this problem so ...

  P is maximized at (x, y) = (54, 27).

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3 years ago
The scores on the verbal section of the Graduate Records Examination (GRE) are approximately normally distributed with a mean of
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"assume that s is a non-empty set of real numbers which is bounded above and lambda is the least upper bound. Prove that for all
s2008m [1.1K]

By definition, if \lambda is the least upper bound of the set S, it means two thing:

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In other words, the least upper bound of a set is greater than or equal to every single element of the set, but it is "close enough" to the elements of the set, because you guaranteed to find elements in the set between \lambda-\varepsilon and \lambda

For example, pick S = [1,10). Obvisouly, the least upper bound is \lambda = 10. In fact, every number in [1,10) is smaller than 10, but as soon as you take away something from 10, say 0.01, you get 9.99, and there are elements in S greater than 9.99, say 9.9999.

So, the claim is basically proven by definition: if \beta < \lambda, let 0 < \delta = \lambda - \beta. By definition, there exists \alpha \in S:\ \alpha > \lambda - \delta.

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3 years ago
What it is the square root of 82 in decimal form
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