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Rama09 [41]
3 years ago
15

Solve the equation for m. 7(m+5)=21 m = -2 m = 2 = m= 9​

Mathematics
1 answer:
Iteru [2.4K]3 years ago
3 0

Answer:

I'm thinking it's 9

Step-by-step explanation:

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The selling price of an item is ​$480. It is marked down by ​10%, but this sale price is still marked up from the cost of ​$. Fi
erma4kov [3.2K]

Answer:

Markup = [(Revenue – COGS) / COGS] X 100.

Markup = (Gross Profit / COGS) X 100.

Markup = [(Revenue – COGS) / COGS] X 100.

Markup = [($400 – $250) / $250] X 100.

Selling Price = [(Markup X COGS) + COGS] X 100.

Selling Price = (Markup X COGS) + COGS.

Selling Price = (0.50 X $100) + $100.

Step-by-step explanation:

um i think this is the answer

4 0
3 years ago
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A store gives a discount of $20 for every purchase above 100. A customer buys clothes worth $78.04, groceries worth &24.81 a
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First figure out his total bill by adding up what he buys.

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Now subtract the discount of $20 since his purchase was above $100:

\sf 120.81-20=\boxed{\sf\$100.81}
7 0
3 years ago
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Which expression is equivalent to ​
True [87]

Answer:

2^{\frac{5}{12}}

Step-by-step explanation:

  • The original expression \sqrt{2^5} ^{\frac{1}{4}} can be transformed into (2^{\frac{5}{3}})^{\frac{1}{4}} : both expressions are equivalent, the root of certain number is equivalent to that number power at a fraction whose denominator is the index of the root. The simpliest example for this statement is \sqrt{x} =x^{\frac{1}{2}} (the squared root of x equals x raised at 1/2).
  • Now, the expression(2^{\frac{5}{3}})^{\frac{1}{4}} can be simplified by using the power of a power property, which simply states that if b\neq 0 and ((b)^n)^m=b^{n\times{m}}. In this case, then  (2^{\frac{5}{3}})^{\frac{1}{4}}=2^{\frac{5}{3}\times{\frac{1}{4}}}=2^{\frac{5}{12}}, which is the final expression.
5 0
4 years ago
HELP Me please
-Dominant- [34]

Answer:

I belive its C not 100%

Step-by-step explanation:

3 0
3 years ago
What is the least common multiple? 27, 21 ​
mojhsa [17]

Answer:

189 is the LCM of 27 and 21

7 0
3 years ago
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