anonymous
  • anonymous
PLEASE SOMEONE HELP ITS AN SOS!!! WILL MEDAL AND FAN Find the distance between these points. C(0, 4), T(-6, -3) √(37) √(85) √(109)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
A parallelogram whose vertices have coordinates R(1, -1), S(6, 1), T(8, 5), and U(3, 3) has a shorter diagonal of ___ . 5 √(13) √(97)
anonymous
  • anonymous
Find the distance between these points. W(-6, -8), X(6, 8) 20 10 √8
anonymous
  • anonymous
Find the other endpoint of a line segment with one endpoint at (-3, 5) and whose midpoint is (5, 2). (7, 2) (1, 7/2) (13, -1)

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DanJS
  • DanJS
The distance is figured using the pythagorean theorem
anonymous
  • anonymous
ive never heard of that
DanJS
  • DanJS
(0,4) and (-6,-3) |dw:1438902675608:dw|
anonymous
  • anonymous
okay then what
DanJS
  • DanJS
you can form a right triangle to get the components of a right triangle
anonymous
  • anonymous
okay
jdoe0001
  • jdoe0001
\(\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) C&({\color{red}{ 0}}\quad ,&{\color{blue}{ 4}})\quad % (c,d) T&({\color{red}{ -6}}\quad ,&{\color{blue}{ -3}}) \end{array}\qquad % distance value d = \sqrt{({\color{red}{ x_2}}-{\color{red}{ x_1}})^2 + ({\color{blue}{ y_2}}-{\color{blue}{ y_1}})^2}\)
DanJS
  • DanJS
|dw:1438902788817:dw|
anonymous
  • anonymous
ohhh f.o.i.l right?
DanJS
  • DanJS
Right, that is just memorizing the formula
anonymous
  • anonymous
ya
anonymous
  • anonymous
is that how i do it using foil?
DanJS
  • DanJS
If you know how to get to it, more likely to remember even if you forget the formulas
anonymous
  • anonymous
this is way to complicated
DanJS
  • DanJS
That forms a right triangle, with the property d^2 = (side1)^2 + (side2)^2
anonymous
  • anonymous
its going way over my head
DanJS
  • DanJS
solving for Distance d, you get what @jdoe0001 typed above
anonymous
  • anonymous
ya
anonymous
  • anonymous
but still really confused
DanJS
  • DanJS
\[distance = \sqrt{(\Delta x)^2 + (\Delta y)^2}\] Distance is square root of the sum of the difference in x coordinates and the difference in y coordinates of the two points
anonymous
  • anonymous
um i dont get it still but thank you sooo much for your trouble
DanJS
  • DanJS
(0,4) and (-6 , -3) \[distance = \sqrt{(-6 - 0)^2 + (-3 - 4)^2} = \sqrt{36 + 49} = \sqrt{85}\]
anonymous
  • anonymous
oh so it was how i thought. I dont know why but for some reason its harder then it should be
anonymous
  • anonymous
so all i do is foil pretty much right?
DanJS
  • DanJS
i was just explaining what the distance formula comes from. ha sorry
DanJS
  • DanJS
no, no Foil. You just put the numbers in and calculate it
anonymous
  • anonymous
oh
anonymous
  • anonymous
well guess i dont get it
DanJS
  • DanJS
I guess all you have to take note of is, if you given 2 points (x , y) and (x1 , y1) The distance between them is calculated by.. \[distance = \sqrt{(x - x1)^2+(y-y1)^2}\]
DanJS
  • DanJS
\[distance = \sqrt{(-6 - 0)^2 + (-3 - 4)^2} = \sqrt{(-6)^2+(-7)^2 } = \sqrt{85}\]
anonymous
  • anonymous
oh isee what u did there
DanJS
  • DanJS
coo, try the next one

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