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andrey2020 [161]
3 years ago
11

Eileen picks up 5 movies at a movie store and calculates that it will cost her $32.50 before tax. A customer in front of Eileen

pays $81.90 after tax. Sales tax is 5%. How many movies did the customer in front of Eileen buy?
Mathematics
2 answers:
Svetradugi [14.3K]3 years ago
6 0

Answer:

$34.125

Step-by-step explanation:

5% of 32.50 is 1.625

1.625+32.50=34.125

$34.125

yarga [219]3 years ago
3 0

Answer:

15 movies

Step-by-step explanation:

<h3>32.50/5 = 6.50 = price of one movie</h3><h3>Find 5% of 81.90 = 4.095</h3><h3>81.95 - 4.095 = 77.855</h3><h3>77.855/5 = 15.571</h3><h3 />

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sladkih [1.3K]

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A century year such as 1900, 2000, or 2100 is a leap year only if it is a multiple of 400
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If this is a true or false question, then the answer is true. This is because we technically have 365.2425 days in a year and if you add an extra day every 4 years, you get an average of 365.25. It is A LOT of math, but even though there was an extra day in 1896 and 1904, there was no leap year in 1900 because of the 400 multiple rule.
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3 years ago
What is the result of substituting for y in the bottom equation<br> y=x+3<br> y=x^2+2x-4
8_murik_8 [283]

The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

<em><u>Solution:</u></em>

Given that,

y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4  ----- eqn 2

<em><u>We have to substitute eqn 1 in eqn 2</u></em>

x + 3 = x^2 + 2x - 4

\mathrm{Switch\:sides}\\\\x^2+2x-4=x+3\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x^2+2x-4-3=x+3-3\\\\\mathrm{Simplify}\\\\x^2+2x-7=x\\\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\\\x^2+2x-7-x=x-x\\\\\mathrm{Simplify}\\\\x^2+x-7=0

\mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}

x = \frac{-1 \pm \sqrt{ 1 + 28}}{2}\\\\x = \frac{ -1 \pm \sqrt{29}}{2}

x = \frac{ -1 \pm 5.385 }{2}\\\\We\ have\ two\ solutions\\\\x = \frac{ -1 + 5.385 }{2}\\\\x = 2.1925

Also\\\\x = \frac{ -1 - 5.385 }{2}\\\\x = -3.1925

Substitute x = 2.1925 in eqn 1

y = 2.1925 + 3

y = 5.1925

Substitute x = -3.1925 in eqn 1

y = -3.1925 + 3

y = -0.1925

Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

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