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mihalych1998 [28]
4 years ago
13

Janet is buying a $28 necklace. The store reduces the price by 20% and then applied a $2 off coupon. How much will Janet pay for

the necklace?
Mathematics
2 answers:
Viefleur [7K]4 years ago
8 0

The cost will be <u>$20.40</u>

<h3>EXPLANATION</h3>

10% of $28 = $2.80

10% x 2 = $5.60 = 20%

$5.60 + $2 = $7.60

$28 - $7.60 = <u>$20.40</u>

Hope this helps!

Umnica [9.8K]4 years ago
7 0
Janet will pay $20.40 for the necklace because:


Step 1: Subtract 28 - 20% ($5.60)

Step 2: Subtract 22.40 - 2

Step 3: Final answer - 20.4

Step 4: Restate question and add answer

I hope this helps and please give me brainliest!

Enjoy your day! <3
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Answer:

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Which of the following statements is biconditional?
bazaltina [42]

Answer with Step-by-step explanation:

We are given some statements

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2.Mary will eat pudding today if and only if is custard.

If Mary will eat pudding today then it is custard.It may or may not be true because pudding can be any soft sweet desserts.It is not necessary that it is custard only.

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3.It is raining if and only if it is cloudy.

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3 years ago
Solve<br><img src="https://tex.z-dn.net/?f=%5Csf%20%5Cdfrac%7B1%7D%7Bp%7D%20%2B%20%5Cdfrac%7B1%7D%7Bq%7D%20%2B%20%5Cdfrac%7B1%7D
Nostrana [21]

Answer:

\displaystyle   \begin{cases} \displaystyle  {x} _{1} =  - p \\   \displaystyle x _{2}   =  -  q \end{cases}

Step-by-step explanation:

we would like to solve the following equation for x:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}  +  \frac{1}{x}  =  \frac{1}{p  + q + x}

to do so isolate \frac{1}{x} to right hand side and change its sign which yields:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}    =  \frac{1}{p  + q + x}  -  \frac{1}{x}

simplify Substraction:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}    =  \frac{x - (q + p +  x)}{x(p  + q + x)}

get rid of only x:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}    =  \frac{  - (q + p )}{x(p  + q + x)}

simplify addition of the left hand side:

\displaystyle  \frac{q + p}{pq}     =  \frac{  - (q + p )}{x(p  + q + x)}

divide both sides by q+p Which yields:

\displaystyle  \frac{1}{pq}     =  \frac{  -1}{x(p  + q + x)}

cross multiplication:

\displaystyle    x(p  + q + x)  =   - pq

distribute:

\displaystyle    xp  + xq +  {x}^{2} =   - pq

isolate -pq to the left hand side and change its sign:

\displaystyle    xp  + xq +  {x}^{2} + pq =  0

rearrange it to standard form:

\displaystyle   {x}^{2} +    xp  + xq  + pq =  0

now notice we end up with a <u>quadratic</u><u> equation</u> therefore to solve so we can consider <u>factoring</u><u> </u><u>method</u><u> </u><u> </u>to use so

factor out x:

\displaystyle  x( {x}^{} +   p ) + xq  + pq =  0

factor out q:

\displaystyle  x( {x}^{} +   p ) +q (x + p)=  0

group:

\displaystyle  ( {x}^{} +   p ) (x + q)=  0

by <em>Zero</em><em> product</em><em> </em><em>property</em> we obtain:

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cancel out p from the first equation and q from the second equation which yields:

\displaystyle   \begin{cases} \displaystyle  {x}^{}   =  - p \\   \displaystyle x  =  -  q \end{cases}

and we are done!

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