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3 years ago

12

six friends all use $2-off coupons to buy themselves movie tickets. they spend a total of $42. what is the price of one movie ti

cket without the coupon.

2 answers:

MaRussiya [10]3 years ago

6 0

So if six friends each use $2-off coupons, that would be 12 off of the 42, so 30. Then divide that by the 6 and you get $5 each. At least, I think so. Please don't hate me if that's wrong.

Nikitich [7]3 years ago

3 0

Answer: The price of one movie ticket would be $5.

Step-by-step explanation:

Since we have given that

Total amount = $42

Number of friends = 6

Amount of coupon of each friend = $2

Total amount of coupon would be

$2\times 6=\$12$

So, Price of total tickets without the coupon would be

$\$42-\$12\\\\=\$30$

Price of one ticket would be

$\dfrac{30}{6}\\\\=\$5$

Hence, the price of one movie ticket would be $5.

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What is the difference between hyperbola and parabola??

g100num [7]

Answer:

1)A parabola is defined as a set of points in a plane which are equidistant from a straight line or directrix and focus.

The hyperbola can be defined as the difference of distances between a set of points, which are present in a plane to two fixed points is a positive constant.

2)A parabola has single focus and directrix

A hyperbola has two foci and two directrices

3)Eccentricity, e = 1(parabola)

Eccentricity, e>1(hyperbola)

4)The two arms present in a parabola should be parallel to each other

The arms present in hyperbola are not parallel to each other

5)It has no asymptotes(parabola)

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3 years ago

Which of the following are solutions to the equation below?

sladkih [1.3K]

Answer:

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=3,\:x=-\frac{1}{2}$

Step-by-step explanation:

considering the equation

$2x^2\:-\:4x\:-\:3\:=\:x$

solving

$2x^2\:-\:4x\:-\:3\:=\:x$

$2x^2-4x-3-x=x-x$

$2x^2-5x-3=0$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=-5,\:c=-3:\quad x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

solving

$x=\frac{-\left(-5\right)+\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$x=\frac{5+\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$x=\frac{5+\sqrt{49}}{2\cdot \:2}$

$x=\frac{5+7}{4}$

$x=3$

also solving

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$x=\frac{5-\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$x=\frac{5-\sqrt{49}}{4}$

$x=-\frac{2}{4}$

$x=-\frac{1}{2}$

Therefore,

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=3,\:x=-\frac{1}{2}$

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3 years ago

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77julia77 [94]

Answer:

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Therefore the relationship is not linear.

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3 years ago