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Alenkasestr [34]

3 years ago

11

chelsea and shawna are selling cheesecakes for a school fundraiser. customers can buy french silk cheesecakes and apple cheeseca

kes. chelsea sold 2 french silk cheescakes and 3 apple cheesecakes for a total of $20. shawn sold 5 french silk cheescakes and 3 apple cheesecakes for a total of $32. find the cost of one french silk cheesecake and one apple cheesecake.

1 answer:

Juliette [100K]3 years ago

6 0

Answer:

Shawna price went up by 12 dollars because of 2 extra french silks. I divides 12 by 2 and got 6 dollars.I subtracted 12 dollars from 20 and divides the rest by 3 and got 2.66$

FRENCH SILK CHEESECAKE: 6$
APPLE CHEESECAKE: 2.66$

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

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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.

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

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

Squaring both sides to get rid of the fraction exponent:

$x^2=x+1$

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

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

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

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

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You may use either equation.

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

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

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