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3 years ago

10

The local grocery store usually sells large bottles of water for $2.69 each. This week they have a promotion, 3 large bottles fo

r $6.30. How much less is the cost of a large bottle this week with the promotion?

2 answers:

Natasha2012 [34]3 years ago

5 0

Answer:

$0.59

Step-by-step explanation:

Each water bottle was $2.69

Now it is 3 bottles for $6.30

If 3 bottles=$6.30 then each bottle must equal $2.10 ($6.30 divided by 3)

$2.10<$2.69 by how much?

well, $2.69-$2.10=$0.59

$0.59 is their difference.

for the promotion the cost is $0.59 less

zlopas [31]3 years ago

4 0

━━━━━━━☆☆━━━━━━━

▹ Answer

<em>It is $0.59 cheaper this week.</em>

▹ Step-by-Step Explanation

Generally - $2.69 EACH

This week - $6.30 for 3

$6.30 ÷ 3 = $2.10

$2.69 - $2.10 = $0.59

Hope this helps!

- CloutAnswers ❁

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Read 2 more answers

COMPUTE<br><br> 3 ( 2 1/2 - 1 ) + 3/10

Juli2301 [7.4K]

Answer:

<h3> $\boxed{ \frac{24}{5} }$ </h3>

Step-by-step explanation:

$\mathsf{3(2 \frac{1}{2} - 1) + \frac{3}{10} }$

Convert mixed number to improper fraction

$\mathrm{3( \frac{5}{2} - 1) + \frac{3}{10} }$

Calculate the difference

⇒ $\mathrm{3( \frac{5 \times 1}{2 \times 1} - \frac{1 \times 2}{1 \times 2} }) + \frac{3}{10}$

⇒ $\mathrm{ 3 \times( \frac{5}{2} - \frac{2}{2}) } + \frac{3}{10}$

⇒ $\mathrm{3 \times ( \frac{5 - 2}{2} ) + \frac{3}{10} }$

⇒ $\mathrm{3 \times \frac{3}{2} + \frac{3}{10} }$

Calculate the product

⇒ $\mathrm{ \frac{3 \times 3}{1 \times 2} + \frac{3}{10} }$

⇒ $\mathrm{ \frac{9}{2} + \frac{3}{10}}$

Add the fractions

⇒ $\mathsf{ \frac{9 \times 5}{2 \times 5} + \frac{3 \times 1}{10 \times 1} }$

⇒ $\mathrm{ \frac{45}{10} + \frac{3}{10} }$

⇒ $\mathrm{ \frac{45 + 3}{10 } }$

⇒ $\mathrm{ \frac{48}{10} }$

Reduce the numerator and denominator by 2

⇒ $\mathrm{ \frac{24}{5} }$

Further more explanation:

<u>Addition </u><u>and </u><u>Subtraction</u><u> </u><u>of </u><u>like </u><u>fractions</u>

While performing the addition and subtraction of like fractions, you just have to add or subtract the numerator respectively in which the denominator is retained same.

For example :

Add : $\mathsf{ \frac{1}{5} + \frac{3}{5} = \frac{1 + 3}{5} } = \frac{4}{5}$

Subtract : $\mathsf{ \frac{5}{7} - \frac{4}{7} = \frac{5 - 4}{7} = \frac{3}{7} }$

So, sum of like fractions : $\mathsf{ = \frac{sum \: of \: their \: number}{common \: denominator} }$

Difference of like fractions : $\mathsf{ \frac{difference \: of \: their \: numerator}{common \: denominator} }$

<u>Addition </u><u>and </u><u>subtraction</u><u> </u><u>of </u><u>unlike </u><u>fractions</u>

While performing the addition and subtraction of unlike fractions, you have to express the given fractions into equivalent fractions of common denominator and add or subtract as we do with like fractions. Thus, obtained fractions should be reduced into lowest terms if there are any common on numerator and denominator.

For example:

$\mathsf{add \: \frac{1}{2} \: and \: \frac{1}{3} }$

L.C.M of 2 and 3 = 6

So, ⇒ $\mathsf{ \frac{1 \times 3}{2 \times 3} + \frac{1 \times 2}{3 \times 2} }$

⇒ $\mathsf{ \frac{3}{6} + \frac{2}{6} }$

⇒ $\frac{5}{6}$

Multiplication of fractions

To multiply one fraction by another, multiply the numerators for the numerator and multiply the denominators for its denominator and reduce the fraction obtained after multiplication into lowest term.

When any number or fraction is divided by a fraction, we multiply the dividend by reciprocal of the divisor. Let's consider a multiplication of a whole number by a fraction:

$\mathsf{4 \times \frac{3}{2} = \frac{4 \times 3}{2} = \frac{12}{2} = 6}$

Multiplication for $\mathsf{ \frac{6}{5} \: and \: \frac{25}{3} }$ is done by the similar process

$\mathsf{ = \frac{6}{5} \times \frac{25}{3} = 2 \times 5 \times 10}$

Hope I helped!

Best regards!

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3 years ago

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uranmaximum [27]

One way to do this is to notice

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$\tan\dfrac{17\pi}{12}=\tan\left(\dfrac\pi6+\dfrac{5\pi}4\right)=\dfrac{\tan\frac\pi6+\tan\frac{5\pi}4}{1-\tan\frac\pi6\tan\frac{5\pi}4}$

We have

$\tan\dfrac\pi6=\dfrac{\sin\frac\pi6}{\cos\frac\pi6}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}$

and since $\tan x$ has a period of $\pi$ ,

$\tan\dfrac{5\pi}4=\tan\left(\pi+\dfrac\pi4\right)=\tan\dfrac\pi4=1$

and so

$\tan\dfrac{17\pi}{12}=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3$

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4 years ago