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vfiekz [6]
3 years ago
9

If a pair of jeans cost $37.80 including 8% tax what is orginial cost before tax

Mathematics
1 answer:
Nezavi [6.7K]3 years ago
3 0
X=original cost

37.80=1.08x
x=37.80/1.08
x=35
original cost: $35
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EleoNora [17]

The proper equation for midpoints is inside the following photo

3 0
3 years ago
The new price of the plane ticket
katen-ka-za [31]

Answer:

159.25

Step-by-step explanation:

175 multiplied by 6% is 10.5

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3 0
3 years ago
Read 2 more answers
Ben spent 1/6 of his money on a burger, fries, and a drink. Then he spent half of the money he had left: $5 on a magazine, $8.25
Ket [755]

The amount of money Ben had to begin with after spending 1/6 and 1/2 of it is 57 dollars.

<h3>How to find the how much money he had with an equation?</h3>

let

x = amount he had to begin with

He  spent 1/6 of his money on a burger, fries, and a drink. Therefore,

amount spent on burger, fries, and a drink = 1  / 6 x

Hence,

amount he had left = x - 1 / 6 x =6x - x /6 = 5 / 6 x

Then he spent half of the money he had left.

1  / 2(5  /6 x) = 5 + 8.25 + 10.50

5 / 12 x = 23.75

cross multiply

5x = 23.75 × 12

5x = 285

divide both sides by 5

x = 285 / 5

x = 57

Therefore, the amount of money he have to begin with is $57.

learn more on equation here: brainly.com/question/5718696

4 0
2 years ago
Please help! Explain how to solve!
Gnesinka [82]
So this one is a little tricky because in order to solve you need to combine like terms. The only problem is that two fractions don't have the same denominator.
This is what you do.
2/3 - 1/9
= (2*9) - (1*3) / 3*9
= 18 - 3 / 27
= 15/ 27
Simplify.
5/9 + y = 7/9
Now since you need to isolate the variable you have to subtract 5/9 from both sides. Since the denominator is the same you can just subtract the numerator. 
So y = 2/9
I hope this helps love! :)
3 0
3 years ago
8. Solve for x.<br> A 30<br> C 180<br> B 60<br> D 90
xxTIMURxx [149]

Answer:

<em><u> </u></em><em><u>Given </u></em><em><u>:</u></em>

<em><u>=</u></em><em><u>></u></em><em><u> </u></em><em><u>An </u></em><em><u>equilateral</u></em><em><u> Triangle</u></em><em><u> </u></em><em><u>i.e </u></em><em><u>a </u></em><em><u>triangle</u></em><em><u> </u></em><em><u>which </u></em><em><u>have </u></em><em><u>all </u></em><em><u>it's</u></em><em><u> </u></em><em><u>side </u></em><em><u>equal.</u></em>

<em><u>To </u></em><em><u>Find </u></em><em><u>:</u></em>

<em><u>=</u></em><em><u>></u></em><em><u>The </u></em><em><u>value </u></em><em><u>of </u></em><em><u>(</u></em><em><u>2</u></em><em><u>x</u></em><em><u>)</u></em><em><u>°</u></em>

<em><u>-</u></em><em><u>-</u></em><em><u>-</u></em><em><u>-</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>-</u></em><em><u>-</u></em><em><u>-</u></em><em><u>-</u></em><em><u>❣️</u></em>

<h2><em><u>Solution</u></em><em><u> </u></em></h2>

<em><u>----_______----❣️</u></em>

<em><u>1</u></em><em><u>. </u></em><em>We </em><em>know</em><em> </em><em>that </em><em>the </em><em>sum </em><em>of </em><em>a </em><em>triangle</em><em> </em><em>is </em><em>1</em><em>8</em><em>0</em><em>°</em>

<em>2.We </em><em>also </em><em>know</em><em> </em><em>that </em><em>all </em><em>the </em><em>sides </em><em>of </em><em>a </em><em>triangle</em><em> </em><em>are </em><em>equal</em><em> </em><em>hence,</em><em> thier</em><em> </em><em>corresponding</em><em> </em><em>Angles </em><em>are </em><em>Equal</em><em> </em><em>too.</em>

<em>So.</em><em>.</em><em>.</em>

<em> </em><em><u>Each </u></em><em><u>angle </u></em><em><u>measures </u></em><em><u>6</u></em><em><u>0</u></em><em><u>°</u></em><em><u>. </u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>(</u></em><em><u>.</u></em><em><u>1</u></em><em><u>)</u></em>

<em><u>3</u></em><em><u>.</u></em><em><u> </u></em><em>using(</em><em>1</em><em>)</em><em>,</em>

<em>We </em><em>can </em><em>find</em><em> the</em><em> </em><em>value</em><em> </em><em>of </em><em>x</em>

<h2><em>=</em><em>></em><em> </em><em>2</em><em>x</em><em>°</em><em> </em><em>=</em><em>6</em><em>0</em><em>°</em></h2><h2><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>x=</em><em> </em><em>6</em><em>0</em><em>/</em><em>2</em></h2><h2><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>x=</em><em> </em><em>3</em><em>0</em></h2>

3 0
3 years ago
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