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3 years ago

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Karla rode her bicycle 135 miles in 9 weeks, riding the same distance each week. Brad rode his bicycle 102 miles in 6 weeks, rid

ing the same distance each week. Which statement correctly compares the number of miles per week they rode?

1 answer:

blsea [12.9K]3 years ago

8 0

Answer:

Karla's 15 miles per week to Brad's 17 miles per week.

Step-by-step explanation:

First, we would need find the total number of miles each individual rides per week. We do this by dividing the total miles ridden by the number of weeks it took for each individual like so...

Karla: 135 / 9 = 15 miles per week

Brad: 102 / 6 = 17 miles per week

Finally, the comparison would be

Karla's 15 miles per week to Brad's 17 miles per week.

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Roberto will save 1/6 of his allowance each day. If he gets $2.00 a day, about how much money will he save each day? Round your

Rus_ich [418]

Amount saved= fraction saved x amount of money received
A= 1/6 x 2
A=1/3
A=.33

33 cents or $.33

Hope this helps :)

6 0

3 years ago

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

6 0

3 years ago

If -3 1/4 is subtracted from 1 1/3. What will be the result?

Alenkinab [10]

1_1/3 = 4/3

-3_1/4 = -13/4

(4/3) - (-13/4)

(4/3) + (13/4)

We can see that (4/3) + (13/4) leads to a positive answer.

Answer: choice D

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3 years ago

Help plsssss????!??,????

Strike441 [17]

Answer:

y=4x+5

Step-by-step explanation:

so like the b in y= mx +b is the starting point well in that graph it is also is where the line touches the y-axis and the slope is rise/run so start at 5 move over 1 then go up 4 that 4/1 = 4 so m=4 so if y= 4x+b and b being 5 then y=4x+5

3 0

3 years ago

Read 2 more answers

What is the equation of the line that passes through the points (-1,3) and (1,11)?

marta [7]

Answer:

y=4x+7

Step-by-step explanation:

You can find the slope of the line by using the formula of change and y over change in x. ∆y/∆x OR (y2-y1)/(x2-x1)

11-3=8

1--1=2

8/2=4

Let's plug in a y value 3.

3=4*-1+b

b=7

Once you found b rewrite your equation in slope intercept form.

y=4x+7

You can check this equation by plugging in the unused coordinate (1,11) and find it balances.

7 0

3 years ago