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2 years ago

1/3(x-1/3)-1/2(2/3x-1/2)

Mathematics

1 answer:

kenny6666 [7]2 years ago

4 0

For this case we must simplify the following expression:

$\frac {1} {3} (x- \frac {1} {3}) - \frac {1} {2} (\frac {2} {3} x- \frac {1} {2})$

So, if we apply distributive property to the terms within parentheses we have:

$\frac {1} {3} x- \frac {1} {9} - \frac {2} {6} x + \frac {1} {4} =\\\frac {1} {3} x- \frac {1} {9} - \frac {1} {3} x + \frac {1} {4} =$

We simplify taking into account that:

Equal signs are added and the same sign is placed.

Different signs are subtracted and the major sign is placed.

$- \frac {1} {9} + \frac {1} {4} =\\- \frac {4} {36} + \frac {9} {36} =\\\frac {-4 + 9} {36} =\\\frac {5} {36}$

Answer:

$\frac {5} {36}$

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I hope this helps you

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2 years ago

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Answer:

Step-by-step explanation:

We'll take this step by step. The equation is

$8-3\sqrt[5]{x^3}=-7$

Looks like a hard mess to solve but it's actually quite simple, just do one thing at a time. First thing is to subtract 8 from both sides:

$-3\sqrt[5]{x^3}=-15$

The goal is to isolate the term with the x in it, so that means that the -3 has to go. Divide it away on both sides:

$\sqrt[5]{x^3}=5$

Let's rewrite that radical into exponential form:

$x^{\frac{3}{5}}=5$

If we are going to solve for x, we need to multiply both sides by the reciprocal of the power:

$(x^{\frac{3}{5}})^{\frac{5}{3}}=5^{\frac{5}{3}}$

On the left, multiplying the rational exponent by its reciprocal gets rid of the power completely. On the right, let's rewrite that back in radical form to solve it easier:

$x=\sqrt[3]{5^5}$

Let's group that radicad into groups of 3's now to make the simplifying easier:

$x=\sqrt[3]{5^3*5^2}$ because the cubed root of 5 cubed is just 5, so we can pull it out, leaving us with:

$x=5\sqrt[3]{5^2}$ which is the same as:

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2 years ago

A. One player places 1 red, 5 green and 3 blue tiles in Bag A, and 6 red, 4 green, and 2 blue in Bag B. What is the probability

cupoosta [38]

Answer:

$\frac{8}{27}$ is the probability that a player draws out two tiles of the same color assuming they are drawing one tile from each bag.

Step-by-step explanation:

In each bag there are red, green, and blue tiles, meaning that no matter which color is pulled out first there is always some probability that the second tile will be the same color. So, we can set up three possible outcomes:

Red: The player pulls out a red tile first. This has a $\frac{1}{9}$ probability of happening. Then in order to succeed for the problem, the next tile also needs to be red which has a $\frac{6}{12}$ probability attached to it. $\frac{1}{9}$ × $\frac{6}{12}$ = $\frac{1}{18}$ probability of happening.

Green: There is a $\frac{5}{9}$ probability of the player pulling out a green tile first. In this case we want to calculate the probability of the second tile being green, which would be $\frac{4}{12}$ . $\frac{5}{9}$ × $\frac{4}{12}$ = $\frac{5}{27}$ .

Blue: There is a $\frac{3}{9}$ probability of the first tile being blue in which case we are hoping for the second tile to be blue as well. The probability of the second tile being blue is $\frac{2}{12}$ on its own, and them both being blue is $\frac{3}{9}$ × $\frac{2}{12}$ = $\frac{1}{18}$

Adding $\frac{1}{18}$ + $\frac{1}{18}$ + $\frac{5}{27}$ we get the answer $\frac{8}{27}$ .

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2 years ago