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Mathematics 19 Online
OpenStudy (princeevee):

help please

OpenStudy (princeevee):

@563blackghost

563blackghost (563blackghost):

Yes?

OpenStudy (princeevee):

sorry, ill get the question

OpenStudy (princeevee):

OpenStudy (princeevee):

@563blackghost

563blackghost (563blackghost):

We would need to find the base and height of the triangle. If we plot the triangle we see that (-4,2) and (1,2) lie on the same y-value so the distance between is 5. Now we need to find the height..

563blackghost (563blackghost):

One second I put down the distance between the wrong points.

563blackghost (563blackghost):

\(\huge\bf{d=\sqrt{(-4-(-5)^{2}+(2-(-4)^{2}}}\) We find the distance between `(-4,2) and (-5,-4)`

OpenStudy (princeevee):

so 15^2?

563blackghost (563blackghost):

We would simplify it. \(\huge\bf{\sqrt{(1)^{2}+(6)^{2}}}\) Simplify. \(\huge\bf{\sqrt{1+36}}\) Add. \(\huge\bf{\sqrt{37}}\) So the height is \(\large\bf{\sqrt{37}}\). We would now multiply the height and the base and then divide that by two. \(\huge\bf{A=\frac{\sqrt{37} \times 5}{2}}\)

563blackghost (563blackghost):

You would input the answer as a decimal.

563blackghost (563blackghost):

Do you mean 15 as your answer?

563blackghost (563blackghost):

If so, then yes that is correct :) Though I do not know if you would input it as `15.2` or `15` itself.

OpenStudy (princeevee):

can you help me with some more?

563blackghost (563blackghost):

Im sorry but I have to take my leave for today. Though I do believe @3mar can help :)

OpenStudy (3mar):

I will not be late for any help. 3 minutes, please! Thank you for your confidence, @563blackghost!

OpenStudy (princeevee):

563blackghost (563blackghost):

;)

OpenStudy (3mar):

Well, I am here.

OpenStudy (3mar):

Sorry for late @princeevee

OpenStudy (princeevee):

@563blackghost

OpenStudy (princeevee):

OpenStudy (princeevee):

@3mar

OpenStudy (3mar):

Do you know the concept of the parallel and perpendicular lines?

OpenStudy (princeevee):

im a little fuzzy with it

OpenStudy (3mar):

For all parallel lines, they have the same slope. For the perpendicular lines, their slopes' product is \[m_1×m_2=-1\]

OpenStudy (princeevee):

so the second one goes into the parallel?

OpenStudy (3mar):

"so the second one goes into the parallel??" ???

OpenStudy (princeevee):

THE SECOND ONE IN THE ROW GOES IN THE PARALELL AREA DOES IT?!

OpenStudy (3mar):

-2x+y=-4 Parallel...Correct!

OpenStudy (princeevee):

and the fourth one?

OpenStudy (3mar):

Don't know?

OpenStudy (princeevee):

the fourth one goes in the parallel?

OpenStudy (3mar):

Sorry the secon does not go in the parallel as the original function has a slope of -1/2 \[x+2y=6\] \[2y=-x+6\] \[y=-\frac{ 1 }{ 2 }x+6\] \[y=mx+b\] \[m=-\frac{ 1 }{ 2 }\]

OpenStudy (princeevee):

so which ones go in which?

OpenStudy (princeevee):

@3mar

OpenStudy (3mar):

Ok the slopes of the functions in the row would be as follows: 1. -1/2 2. 2 3. 1/2 4. -1/2 did the math and fund out which is the same as -1/2 and which is the negative reciprocal of -1/2.....

OpenStudy (princeevee):

i would guess 2 and 3 are the negative ones

OpenStudy (3mar):

It is not correct, because when x goes to the other side, its sign changes. Got that?

OpenStudy (princeevee):

ok

OpenStudy (3mar):

Satisfied?

OpenStudy (princeevee):

well i still dont know which ones go in which? is 2 and 3 supose to be in the niether pile?

OpenStudy (3mar):

1 and 4 go to "parallel" 2 goes to "perpendicular" 3 goes to "neither"

OpenStudy (3mar):

@princeevee

OpenStudy (princeevee):

okay thanks , can you help with some more?

OpenStudy (3mar):

Satisfied of this firstly?

OpenStudy (princeevee):

yes

OpenStudy (princeevee):

OpenStudy (3mar):

I am happy to hear that.

OpenStudy (3mar):

Can we start in new post, please?

OpenStudy (3mar):

Thank you for the medal!

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