10 points

Mathematics

1 answer:

MrRissso [65]3 years ago

5 0

Can you explain better

You might be interested in

What price should a retailer place on an item that costs him $6.95, if he wants a markup of 33%? $

zubka84 [21]

So he wants to mark the product up by 33%, now, if we take 6.95 to be the 100%, what is 33% of that?

$\bf \begin{array}{ccll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
6.95&100\\
x&33
\end{array}\implies \cfrac{6.95}{x}=\cfrac{100}{33}\implies \cfrac{6.95\cdot 33}{100}=x$

so the marked up price will be 6.95 + x.

6 0

3 years ago

Read 2 more answers

Solve question 3 for <br> 41 points

Sever21 [200]

Answer:

(2)

Step-by-step explanation:

Our logarithmic expression is: $ln(\frac{\sqrt{e} }{y^3} )$ .

Remember the logarithmic property that ln(a/b) = lna - lnb. So, we can write this as:

$ln(\frac{\sqrt{e} }{y^3} )=ln(\sqrt{e} )-ln(y^3)$

Also, we can write square roots as powers of one-half, so √e = $e^{1/2}$ . There's another log property that: $ln(a^b)=b*ln(a)$ . We can apply that here for both the √e and the y³: