anonymous
  • anonymous
algebra
Linear Algebra
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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abb0t
  • abb0t
This is \(\sf \color{red}{not}\) linear algebra.
anonymous
  • anonymous
i thought it is algebra
abb0t
  • abb0t
To find the magnitude, speed, \(|\sqrt{a^2+b^2}|\), where a = 3, b = 4. Now, to finf direction, tan(\(\theta\)) = \(\frac{b}{a}\)

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anonymous
  • anonymous
You don't need those \(|\) dude. \(\sqrt{\ldots }\iff |\sqrt{\ldots}|\)
anonymous
  • anonymous
magnitude = 5 m/s and direction theta = 53 degree
abb0t
  • abb0t
Hmm...that's how I remember it. But I guess I should check b4 hand next tym. Thanks @wio
anonymous
  • anonymous
and 53 degree would be northeast right????
anonymous
  • anonymous
Remember that \(|x| \iff \sqrt{x^2}\) and \(\sqrt{x} \iff |\pm \sqrt{x}|\)
anonymous
  • anonymous
Algebra is one of the broad parts of mathematics, together with number theory, geometry and analysis. For historical reasons, the word "algebra" has several related meanings in mathematics, as a single word or with qualifiers.
anonymous
  • anonymous
Algebra can essentially be considered as doing computations similar to that of arithmetic with non-numerical mathematical objects.[1] Initially, these objects represented either numbers that were not yet known (unknowns) or unspecified numbers (indeterminate or parameter), allowing one to state and prove properties that are true no matter which numbers are involved. For example, in the quadratic equation ax^2+bx+c=0,
anonymous
  • anonymous
IN this case \(x=-3\), \(y=-4\)\[ \tan \theta = \frac y x \implies \theta = \tan ^{-1} \left(\frac {4}{3}\right) \]
abb0t
  • abb0t
\(\huge o.O\)
anonymous
  • anonymous
Does algebra have a wee wee or a coochie?
anonymous
  • anonymous
We can start with a graph for this one, with arrows representing the two velocities... 3m/s west and 4m/s south|dw:1378455204639:dw|
anonymous
  • anonymous
Then we can draw an arrow for the change in velocity, which will go from the tip of the original (3m/s line) arrow to the tip of the final (4m/2) arrow
anonymous
  • anonymous
@Archie thats what i started with but i didnt write the axises
anonymous
  • anonymous
Do you have any ideas about how to find its magnitude?
anonymous
  • anonymous
ya x^2 + y^2 = 5
anonymous
  • anonymous
ohh|dw:1378455437063:dw|
anonymous
  • anonymous
Good idea! It looks like you're using the Pythagorean theorem. What does that give us for the length of the black arrow?
anonymous
  • anonymous
so you agree with abbots process right archie
anonymous
  • anonymous
its 5
anonymous
  • anonymous
right, would you like to go through how to get that answer together?
anonymous
  • anonymous
or are you fine?
anonymous
  • anonymous
sure go ahead
anonymous
  • anonymous
|dw:1378455660607:dw|
anonymous
  • anonymous
shoudnt 3 and 4 be negative bcz they are in opposite to x and y direction
anonymous
  • anonymous
|dw:1378455719936:dw|
anonymous
  • anonymous
Basically we can plug in the two shorter sides of the triangle for a and b in the Pythagorean theorem, then solve for the hypotenuse, c like it is written on the board. Also Good question, generally we just use the positive lengths of the sides... in the Pythagorean theorem even if they're going in the negative direction.
anonymous
  • anonymous
so magnitude of object's velocity is 5 m/s right?
anonymous
  • anonymous
But since, a,b and c are each squared, a^2, b^2 and c^2 will turn out to be positive either way.
anonymous
  • anonymous
right!
anonymous
  • anonymous
So, do you have any questions on that part or does it make sense?
anonymous
  • anonymous
arc tan -4/-3 = 53 degree that means it is in northeast direction
anonymous
  • anonymous
then we also need to state the direction of this change in velocity, so say which direction the black arrow is pointing. Which direction does that look like?
anonymous
  • anonymous
northheast
anonymous
  • anonymous
Good idea about using arctan... that gives the value of this angle here: But not the direction of the black arrow, since there are other triangles that could have the same arctan to see the direction of the black arrow, we need to look at the graph. And state the direction we have to go to get from the end of the 3/ms arrow to the end of the 4m/s one.
anonymous
  • anonymous
Does it make sense why that is, or would you like me to explain it differently? xD
anonymous
  • anonymous
yep thx hey u know how to do transfer momentum
anonymous
  • anonymous
Yes! :)
anonymous
  • anonymous
Do you have a specific question on that?
anonymous
  • anonymous
ya just a sec lemme post it
anonymous
  • anonymous
already did ;)
anonymous
  • anonymous
ok, lol.
anonymous
  • anonymous
hey sry again but the direction is southwest or northeast
anonymous
  • anonymous
Southeast, because the red arrow points toward the southeast.
anonymous
  • anonymous
|dw:1378457820720:dw| there is no arrow pointing se...sorry lol confused again

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