Find the selling price.<br> Cost to store: $75<br> Markup: 35%

iris [78.8K]

Here is your question: 7/6 divided by 2.

$= 7/6$ ÷ $2 = 7/6 * 1/2 =$ .

Multiply and you get $= 7/12.$

6 0

3 years ago

Suzi bought 5 packages of M&M's, and each package contained 18 candy pieces. She opened the first package and counted 4 gree

Leni [432]

Answer: The correct answer is option B: There are between 15 and 20 green pieces in all 5 packages

Step-by-step explanation: The most important factor has been given which is, "Which statement about the candy pieces in the remaining packages is best supported by this information."

The information given is such that, the first package she opened had 4 green pieces and on this basis we can safely assume that all other packages have 4 green pieces as well. The second package had 3 green pieces and this based on this too we can safely assume that all other packages had 3 green pieces. Hence, all 5 packages can either have a total of 4 x 5 green candies which equals a total of 20 green pieces or, all 5 packages can have a total of 3 x 5 green candies which equals a total of 15 green pieces.

So according to Suzi's experiment, there are between 15 and 20 green pieces in all 5 packages.

7 0

3 years ago

How do I solve: 2 sin (2x) - 2 sin x + 2√3 cos x - √3 = 0

ziro4ka [17]

Answer:

$\displaystyle x = \frac{\pi}{3} +k\, \pi$ or $\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ , where $k$ is an integer.

There are three such angles between $0$ and $2\pi$ : $\displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{2\, \pi}{3}$ , and $\displaystyle \frac{4\,\pi}{3}$ .

Step-by-step explanation:

By the double angle identity of sines:

$\sin(2\, x) = 2\, \sin x \cdot \cos x$ .

Rewrite the original equation with this identity:

$2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0$ .

Note, that $2\, (2\, \sin x \cdot \cos x)$ and $(-2\, \sin x)$ share the common factor $(2\, \sin x)$ . On the other hand, $2\sqrt{3}\, \cos x$ and $(-\sqrt{3})$ share the common factor $\sqrt[3}$ . Combine these terms pairwise using the two common factors:

$(2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0$ .

Note the new common factor $(2\, \cos x - 1)$ . Therefore:

$\left(2\, \sin x + \sqrt{3}\right) \cdot (2\, \cos x - 1) = 0$ .

This equation holds as long as either $\left(2\, \sin x + \sqrt{3}\right)$ or $(2\, \cos x - 1)$ is zero. Let $k$ be an integer. Accordingly:

$\displaystyle \sin x = -\frac{\sqrt{3}}{2}$ , which corresponds to $\displaystyle x = -\frac{\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = -\frac{2\, \pi}{3} + 2\, k\, \pi$ .
$\displaystyle \cos x = \frac{1}{2}$ , which corresponds to $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ and $\displaystyle x = -\frac{\pi}{3} + 2\, k \, \pi$ .

Any $x$ that fits into at least one of these patterns will satisfy the equation. These pattern can be further combined:

$\displaystyle x = \frac{\pi}{3} + k \, \pi$ (from $\displaystyle x = -\frac{2\,\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ , combined,) as well as
$\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ .