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3 years ago

I hope its bigger so u can seeeeeeeeeee!!!

Mathematics

1 answer:

Nata [24]3 years ago

3 0

Answers:

Gabe needs 30 liters of the 50% solution

Gabe needs 30 liters of the 70% solution

Both answers are 30

==================================

Explanation:

We have two beakers, each of which we don't know how much is inside. Let's call this x and y

x = amount of liquid in the first beaker (water+acid)

y = amount of liquid in the second beaker (water+acid)

We're told that the beakers have a 50% solution and a 70% solution of acid. This means that the first beaker has 0.5*x liters of pure acid, and the second beaker has 0.7*y liters of pure acid. In total, there is 0.5x+0.7y liters of pure acid when we combine the two beakers. This is out of x+y = 60 liters total, which we can solve for y to get y = 60-x. We will use y = 60-x later on when it comes to the substitution step

We can divide the total amount of pure acid (0.5x+0.7y liters) over the total amount of solution (x+y = 60) to get the following

(0.5x+0.7y)/(x+y) = (0.5x+0.7y)/60

We want this to be equal to 0.6 because Gabe wants a 60% solution when everything is said and done, so

(0.5x+0.7y)/60 = 0.60

0.5x+0.7y = 0.60*60 .... multiply both sides by 60

0.5x+0.7y = 36

0.5x+0.7( y ) = 36

0.5x+0.7(60-x) = 36 ........ replace y with 60-x (substitution step)

0.5x+0.7(60)+0.7(-x) = 36 .... distribute

0.5x+42-0.7x = 36

-0.2x+42 = 36

-0.2x = 36-42 .... subtract 42 from both sides

-0.2x = -6

x = -6/(-0.2) .... divide both sides by -0.2

x = 30

He needs 30 liters of the 50% solution

Use this x value to find y

y = 60-x

y = 60-30

y = 30

So he needs 30 liters of the 70% solution

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$\bf :\longmapsto\:\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } = \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg]$

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$\purple{\rm :\longmapsto\:\displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty }\rm \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg]$

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Hence, 

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]} = \frac{1}{2}}}$

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Adding or subtracting rational expressions is a four-step process:

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To see this process, we'll look at some examples.

Example 1

Step 1: To find a common denominator, start by factoring the denominators of both fractions.

The least common denominator of these two fractions is 2(x+6)(x-6).

Looking at the first fraction, we can see that the factor missing from its denominator is (x-6). In the next step, we'll multiply the top and bottom of the first fraction by (x-6).

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Step 2: Now that the fractions have the same denominators, we'll create one combined fraction by adding the two fractions together:

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3 years ago

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