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kakasveta [241]

4 years ago

10

How much do Brand A (29% fruit juice) must be mixed with a 5 gallon of Brand B fruit punch (47% fruit juice) to create a mixture

containing 35% fruit juice?

2 answers:

choli [55]4 years ago

8 0

Answer:

at twice the father's rate of speed. At this rate; how many miles would the bicycle rider travel in 9 hours?

Step-by-step explanation:

at twice the father's rate of speed. At this rate; how many miles would the bicycle rider travel in 9 hours?

Citrus2011 [14]4 years ago

8 0

<h3>Answer: 10 gallons of brand A</h3>

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

Explanation:

Let x be the number of gallons for brand A.

Brand A has 29% juice, so 0.29x gallons are pure juice in this container.

Brand B has 47% juice and we have 5 gallons of this, so 0.47*5 = 2.35 gallons of pure juice are in the brand B container.

In total, we have 0.29x+2.35 gallons of juice only

------------

x is the number of gallons in the brand A container

5 is the number of gallons in the brand B container

x+5 is the total number of gallons of water+juice mix

we want 35% fruit juice, so 35% of (x+5) = 0.35(x+5) is the desired amount of fruit juice we want.

Set this equal to 0.29x+2.35 we found earlier, as this also measures the total amount of fruit juice

Solve for x

-------------

0.35(x+5) = 0.29x+2.35

0.35*x+0.35*5 = 0.29x+2.35

0.35x+1.75 = 0.29x+2.35

0.35x-0.29x = 2.35-1.75

0.06x = 0.6

x = 0.6/0.06

x = 10

The chart shown below might help keep everything organized.

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Intersection point of Y=logx and y=1/2log(x+1)

GalinKa [24]

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ ?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ , we will need to find when they have a common point; when their $x$ and $y$ are the same.

Let's start with setting the $y$ 's equal to find those $x$ 's for which the $y$ 's are the same.

$\log(x)=\frac{1}{2}\log(x+1)$

By power rule:

$\log(x)=\log((x+1)^\frac{1}{2})$

Since $\log(u)=\log(v)$ implies $u=v$ :

$x=(x+1)^\frac{1}{2}$

Squaring both sides to get rid of the fraction exponent:

$x^2=x+1$

This is a quadratic equation.

Subtract $(x+1)$ on both sides:

$x^2-(x+1)=0$

$x^2-x-1=0$

Comparing this to $ax^2+bx+c=0$ we see the following:

$a=1$

$b=-1$

$c=-1$

Let's plug them into the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$x=\frac{1 \pm \sqrt{1+4}}{2}$

$x=\frac{1 \pm \sqrt{5}}{2}$

So we have the solutions to the quadratic equation are:

$x=\frac{1+\sqrt{5}}{2}$ or $x=\frac{1-\sqrt{5}}{2}$ .

The second solution definitely gives at least one of the logarithm equation problems.

Example: $\log(x)$ has problems when $x \le 0$ and so the second solution is a problem.

So the $x$ where the equations intersect is at $x=\frac{1+\sqrt{5}}{2}$ .

Let's find the $y$ -coordinate.

You may use either equation.

I choose $y=\log(x)$ .

$y=\log(\frac{1+\sqrt{5}}{2})$

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

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boyakko [2]

Step-by-step explanation:

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3x-8(-3+5x)=24

3x+24-40x=24

x=0

y=-3+5(0)

y=-3

(x,y)=(0,3)

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