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2 years ago

If you had $9 what would you buy

Mathematics

2 answers:

Finger [1]2 years ago

8 0

Answer:

Good question, probably another month of this VPN service or something

Step-by-step explanation:

As a wise green ogre from a dreamworks movie once said...

"I like my privacy."

Karolina [17]2 years ago

4 0

Answer:

Food

Step-by-step explanation:

You need food to survive so find some good food and eat it :)

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Jhvjkelhndbkjehlhdbhlfoei;hljkdv joer

alexandr1967 [171]

Answer:

2.5

Step-by-step explanation:

5 0

3 years ago

A store marks up all of their merchandise by 70%. The floor price of a scarf is $117.25.

KiRa [710]

Answer:

$68.97

Step-by-step explanation:

Markup is the difference between the cost of an item and the price at which it is sold:

cost price + markup = selling price

Here, the markup is said to be 70%. The base for that percentage can be either the cost price or the selling price. The way this question is worded suggests the markup is 70% of the cost price.

So, we have ...

cost price + 0.70 × cost price = selling price

cost price × (1 + 0.70) = selling price

Then, to find the cost price, we can divide by its coefficient in this equation. Doing that gives ...

cost price = (selling price)/1.70

cost price = $117.25/1.70 = $68.97

The store's purchase price before the markup for this item was $68.97.

4 0

3 years ago

How does the graph of y = sqr x + 2 compare to the graph of the parent square root function?

posledela

Is the 2 in the sqr? If not, the graph just moves up by 2 units

8 0

3 years ago

Multiply (2x–3)(2x+3

MakcuM [25]

Multiply (2x–3)(2x+3)

Answer:

7 0

3 years ago

Find dy/dx <br>2^(xy) = x^2

Dovator [93]

$2^{xy}=x^2$
$e^{\ln2^{xy}}=x^2$
$e^{xy\ln2}=x^2$

Differentiate, using the chain rule on the left side:

$\dfrac{\mathrm d}{\mathrm dx}[e^{xy\ln2}]=\dfrac{\mathrm d}{\mathrm dx}[x^2]$
$e^{xy\ln2}\dfrac{\mathrm d}{\mathrm dx}[xy\ln2]=2x$
$e^{xy\ln2}\left(x\ln2\dfrac{\mathrm dy}{\mathrm dx}+y\ln2\right)=2x$
$x\ln2\dfrac{\mathrm dy}{\mathrm dx}+y\ln2=2xe^{-xy\ln2}$
$x\dfrac{\mathrm dy}{\mathrm dx}+y=\dfrac{2^{xy+1}}{\ln2}$
$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\left(\dfrac{2^{xy+1}}{\ln2}-y\right)$

3 0

3 years ago