two lines pass through the origin. The lines have slopes that are opposites. Compare and contrast the lines.

2 answers:

8 0

If the lines have different signs for their slopes, that means that in one equation, as x increases, y decreases. And in the second equation, as increases, y decreases. They will both have the same y intercept however, because they both pass through the origin.
Hope this helps.

jeka942 years ago

4 0

Well, since one has a negative rate of change, and one positive, they would form an X through the graph. That could or could not mean they have equal slopes. We also know that their y intercepts would be 0, since they pass through that.

You might be interested in

Show work to above problem

Len [333]

$\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}
\\\\\\
|a|\implies 
\begin{cases}
+a\\
-a
\end{cases}\implies \pm a\\\\
-------------------------------\\\\
(16x^2)^{\frac{1}{2}}\implies \pm\sqrt[2]{(16x^2)^1}\implies \pm\sqrt{4^2x^2}\implies \pm\sqrt{(4x)^2}\implies \pm 4x
\\\\\\
\begin{cases}
+4\\
-4
\end{cases}\implies |4x|$

5 0

2 years ago

In the figure below, F,G,H and J are collinear. FG, GK, and HK are lone segments of equivalent length, and the measure of JHK is

ANEK [815]

The answer would be 30 degrees because points F, G, H, and J are collinear, all you would have to do is figure out /_GHK by subtracting 180 degrees from it.

Like this :

JHK = 180 degrees - 120 degrees = 60 degrees

8 0

2 years ago

Solve the quadratic equation:<br> 3x^2+x-5=0

adell [148]

3x²+x-5=0

a = 3, b = 1, c= -5

-> ∆ ( delta ) = b²-4ac = 61 > 0

-> x1 =( -b+√∆ )÷ 2a =...

x2 = (-b-√∆)÷2a =...

p/s: do your teachers teach you how to use ∆ ( delta ) in maths calculation ? i live in europe and our teachers teach us that way. however, it is a rịght and fast way. you should learn it.