I need help please it’s an easy question that I forgot

Mathematics

2 answers:

VARVARA [1.3K]3 years ago

3 0

not completely sure but i think b

nata0808 [166]3 years ago

3 0

Answer:

Step-by-step explanation:

add 2 to both sides then multiply 8 to both sides

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3 years ago

The selling price of a box of crackers is $1.75 You mark the crackers up to $2.54 . What is the markup percentage?

Aleks04 [339]

The markup percentage is 45.14%

Step-by-step explanation:

The given is:

The selling price of a box of crackers is $1.75
You mark the crackers up to $2.54

We need to find the markup percentage

The markup percentage = $\frac{New-Old}{old}$ × 100%

∵ The selling price of a box of crackers is $1.75

∴ Old = 1.75

∵ You mark the crackers up to $2.54

∴ New = 2.54

- Substitute these values in the rule above

∵ The markup percentage = $\frac{2.54-1.75}{1.75}$ × 100%

∴ The markup percentage = $\frac{0.79}{1.75}$ × 100%

∴ The markup percentage = 0.4514 × 100%

∴ The markup percentage = 45.14%

The markup percentage is 45.14%

Learn more:

You can learn more about percentage in brainly.com/question/1834017

#LearnwithBrainly

8 0

3 years ago

Read 2 more answers

The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of

Lilit [14]

Check the picture below.

$\bf (\stackrel{a}{1}~,~\stackrel{b}{-1})\qquad \impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c = \sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{1^2+(-1)^2}\implies c=\sqrt{2} \\\\[-0.35em] ~\dotfill$

$\bf sin(\theta ) \implies \cfrac{\stackrel{opposite}{-1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{-1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies -\cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies -\cfrac{\sqrt{2}}{2}$

$\bf cos(\theta ) \implies \cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{\sqrt{2}}}\implies \cfrac{1}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{\sqrt{2}}{(\sqrt{2})^2}\implies \cfrac{\sqrt{2}}{2} \\\\\\ tan(\theta ) = \cfrac{\stackrel{opposite}{-1}}{\stackrel{adjacent}{1}}\implies tan(\theta ) = -1$