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Oksi-84 [34.3K]
2 years ago
14

A discount store had monthly sales of $86,600 and spent 14% of it on promotions. how much was spent on promotion?

Mathematics
1 answer:
Vladimir79 [104]2 years ago
8 0

The amount spent on promotion by the store on monthly sales of $86,600, when they are spending 14 percent on promotion is $12,124.

The percent signifies the hundredth part of the whole.

When informed that the promotion is 14 percent of the monthly sales of $86,600.

Therefore, to calculate the total amount spent on promotion, we calculate 14 percent of 86600.

Therefore, the total amount on promotion = 14% of $86,600 = $ (14/100 * 86,000) = $ (14*866) = $12,124.

Therefore, the amount spent on promotion by the store on monthly sales of $86,600, when they are spending 14 percent on promotion is $12,124.

Learn more about percentages at

brainly.com/question/843074

#SPJ4

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Step-by-step explanation:

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Suppose that w and t vary inversely and that t = 5/12
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Answer:

Option B. t=5/3w ;1/3

Step-by-step explanation:

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w=\frac{a}{t}        Eqn. (1)

Now since we are given the values of w=4 and t=\frac{5}{12}, we can plug them in Eqn. (1) and find our constant of proportionality a as follow:

w=\frac{a}{t}\\ \\a=wt\\\\a=(4)(\frac{5}{12} )\\\\a=\frac{5}{3}

Now that we have our constant we can find the new t value for the second value of w=5 as follow:

w=\frac{a}{t} \\\\t=\frac{a}{w}\\ \\t=\frac{\frac{5}{3} }{5}\\\\t=\frac{5}{15}\\ \\t=\frac{1}{3}\\

Therefore based on the options give, we can see that Option B. is correct since

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8 0
3 years ago
Help! If you know this can you tell me how to do it?
aleksandr82 [10.1K]

Answer:

c

Step-by-step explanation:

Here's how this works:

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In the first term, the cos(x)'s cancel out, and in the second term the sin(x)'s cancel out, leaving:

\frac{sin^2(x)}{sin(x)cos(x)}+\frac{cos^2(x)}{sin(x)cos(x)}=?

Put everything over the common denominator now:

\frac{sin^2(x)+cos^2(x)}{sin(x)cos(x)}=?

Since sin^2(x)+cos^2(x)=1, we will make that substitution:

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We could separate that fraction into 2:

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