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2 years ago

7

2. The east-west streets and north-south avenues of a certain city are numbered consecutively from south to north and west to ea

st. All blocks are squares with the same side lengths. Mandy is standing on the corner of 59th Street and 15th
Avenue. Mark is standing on the comer of 35th Street and 21st Avenue. Mark calls Mandy and asks her to meet him at a spot halfway between their current
locations

a. Use the Midpoint Formula to find out where Mark wants to meet. Show your work

1 answer:

sukhopar [10]2 years ago

4 0

Answer:

They were to meet at 47th street and 18th avenue.

Step-by-step explanation:

Mid-point = (sum of two numbers) / 2

(= average formula)

Street: (59+35) / 2 = 47

Avenue: (21+15) / 2 = 18

Therefore they were to meet at 47th street and 18th avenue.

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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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A number is decreased by 40% to 525. what is the original amount?

zhenek [66]

Answer:

875

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IrinaVladis [17]

Answer:

The solution is 1.

Step-by-step explanation:

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Suppose we have a logarithmic equation in the following format:

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It means that:

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2 years ago