Question

6. Compare the graph from question 2 and the equation from question 5. How are the graphs similar and different? Determine which has the greater y-value when x is 160 and explain how you determined it.

Answer 1

Sorry, was on different question.

Answer 2

No no.. didn't see you posted the equation after. Still reading it =]

Answer 3

Ok thank you for helping. ^_^

Answer 4

Let's find the equation for the graph you posted. Do you an idea how to do it?

Answer 5

Ok, so what is the problem then? you can find which has greater value easily.
Besides that, they both are straight line functions. The slope of the two is different (and the other one has negative slope too..) and they intersect with X and Y axis at different points, that you could find if you wish.

Answer 6

I guess you should find at what X they both have same value..

Answer 7

How do I do that?

Answer 8

Would you say they are perpendicular?

Answer 9

you have
\[
y_1 = -1\frac{1}{2} \cdot x + 7.5 \\\
y_2 = \frac{1}{20} \cdot x + 20
\]
The point that they have the same value is where \(y_1 = y_2 \), right?
So to find when that is true:
\[
y_1 = y_2 \\
-1\frac{1}{2} \cdot x + 7.5 = \frac{1}{20} \cdot x + 20
\]
Could you proceed from here?

Answer 10

They are not perpendicular, as for them to be perpendicular there is a small 'test'. you multiply the slopes and see if you get -1
\[
y = mx + h\\
m_1 = - \frac{3}{2} \\
m_2 = \frac{1}{20} \\
m_1 \cdot m_2 = - \frac{3}{2} \cdot \frac{1}{20} = - \frac{3}{40}
\]
If you want to find the slope for the perpendicular lines to let's say \(y = \frac{1}{20}x + 20 \) then:
\[
m_p = \text{Perpendicular slope} \\
m = \frac{1}{20} \\
m_p \cdot m = -1 \\
m_p = \frac{-1}{m} = \frac{-1}{ \frac{1}{20} } = -20
\]
Means... only a line in form of \( y = -20x + C \) ,could be perpendicular to \(y = \frac{1}{20}x + 20 \)

Answer 11

oop der it iz :-P

Answer 12

there is what? o.o

Answer 13

the steps in solving