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4 years ago

Simplify 1/6+1/4-1/3

Mathematics

1 answer:

frez [133]4 years ago

6 0

Answer: 1/12 or 0.083 the 3 is repeating

Step-by-step explanation:

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Can anyone plzzzz help me with this question of linear equation in two varaiable plzzzzzz... Its very important... Plz solve it

kotegsom [21]

That's not a linear system, but you have an awesome school system for giving you this problem.

$\dfrac 5 y - \dfrac 2 x = \dfrac{13}6$

Multiply by 6xy to clear the fractions.

$30x - 12y = 13 xy$

That's a second degree equation, also known as a conic. That one happens to be a hyperbola.

$30x = y(13 x +12)$

$y = \dfrac{30x}{13 x + 12}$

Let's clear the fractions from the second equation, multiplying out common denominator xy:

$\dfrac {36} x - \dfrac {24} y = 1$

$36y - 24 x = xy$

We are being asked to find the meet of two hyperbolas, so we expect two answers, a quadratic equation.

Substituting,

$36 \left( \dfrac{30x}{13 x + 12} \right) - 24 x = x \left( \dfrac{30x}{13 x + 12} \right)$

$36(30x) -24 x(13x + 12) = 30x^2$

$1080 x - 312 x^2- 288 x = 30x^2$

$x(792 - 342 x)= 0$

$x = 0 \textrm{ or } x=792/342 = \dfrac{44}{19}$

We have to rule out x=0 because it's in the denominator.

$y = \dfrac{30x}{13 x + 12} = \dfrac{30(44/19)}{13(44/19)+12}$

$y = \dfrac{33}{20}$

Answer: (44/19, 33/20)

8 0

3 years ago

Write and solve an equation.

erik [133]

49.57(total amount paid) -2.5 (del fee) = 47.07
47.07/3= 15.69 is the price of each

3 0

3 years ago

Read 2 more answers

How is this answer D?

kompoz [17]

It's a equilateral triangle with each side being equal to the radius.

So the area is $\dfrac{8^2\sqrt 3}{4}=\dfrac{64\sqrt3}{4}=16\sqrt3$

7 0

3 years ago

AleksAgata [21]

$\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\\\ \begin{array}{ccccllll} &\stackrel{\stackrel{ratio}{of~the}}{Sides}&\stackrel{\stackrel{ratio}{of~the}}{Areas}&\stackrel{\stackrel{ratio}{of~the}}{Volumes}\\ \cline{2-4}&\\ \cfrac{\stackrel{similar}{shape}}{\stackrel{similar}{shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}~\hspace{6em} \cfrac{s}{s}=\cfrac{\sqrt{Area}}{\sqrt{Area}}=\cfrac{\sqrt[3]{Volume}}{\sqrt[3]{Volume}} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \cfrac{\textit{sphere A}}{\textit{sphere B}}\qquad \stackrel{\stackrel{sides'}{ratio}}{\cfrac{1}{4}}\qquad \qquad \stackrel{\stackrel{sides'}{ratio}}{\cfrac{1}{4}}=\stackrel{\stackrel{volumes'}{ratio}}{\cfrac{\sqrt[3]{288}}{\sqrt[3]{v}}}\implies \cfrac{1}{4}=\sqrt[3]{\cfrac{288}{v}}\implies \left( \cfrac{1}{4} \right)^3=\cfrac{288}{v} \\\\\\ \cfrac{1^3}{4^3}=\cfrac{288}{v}\implies \cfrac{1}{64}=\cfrac{288}{v}\implies v=18432$

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4 years ago

Answers for question 17 and 18?

Tom [10]

Answer: think the answers near tenth

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4 years ago