Grade 10 maths...
1
Grenouille88 wrote...
Unsigned wrote...
Why does it say "Find the actual speed and direction of the car" if it's already given?It's given relative to the observer on the trail rather than in relation to a standard axis.
k so here's a diagram then. I kind of wish I knew how to do this
0
Lughost
the Lugoat
Unsigned wrote...
Grenouille88 wrote...
Unsigned wrote...
Why does it say "Find the actual speed and direction of the car" if it's already given?It's given relative to the observer on the trail rather than in relation to a standard axis.
k so here's a diagram then. I kind of wish I knew how to do this
Spoiler:
It's not too bad, just trig and whatnot. If it wasn't 5 in the morning I might have worked it out for shits.
0
Kouki wrote...
I don't see how that "problem" can be applied to real life situations.... which is also the same with all of them maths out there.
0
Unsigned wrote...
Grenouille88 wrote...
Unsigned wrote...
Why does it say "Find the actual speed and direction of the car" if it's already given?It's given relative to the observer on the trail rather than in relation to a standard axis.
k so here's a diagram then. I kind of wish I knew how to do this
That's what i thought but here's the problem.
The car going north isn't the direction in which it is actually going, but the direction that's relative to the observer on the train. I tried that diagram but then it wouldn't work....My math professor in university would kill me if he knew I couldn't solve something like this.
0
Kadushy
Douchebag
Here's an easier problem...
Graph and shade... they're all in one bracket
[size=29]{[/h]
Graph and shade... they're all in one bracket
[size=29]{[/h]
- x^2 + y^2 ≥ 9
x^2 + y^2 < 25
y ≤ 0
0
Heizan-Tr5 wrote...
Unsigned wrote...
Grenouille88 wrote...
Unsigned wrote...
Why does it say "Find the actual speed and direction of the car" if it's already given?It's given relative to the observer on the trail rather than in relation to a standard axis.
k so here's a diagram then. I kind of wish I knew how to do this
Spoiler:
That's what i thought but here's the problem.
The car going north isn't the direction in which it is actually going, but the direction that's relative to the observer on the train. I tried that diagram but then it wouldn't work....My math professor in university would kill me if he knew I couldn't solve something like this.
I don't know any math postulates or theories or whatever that factors in human error or human assumptions. I'm so confuzlled. Is that what it is?
0
Unsigned wrote...
Heizan-Tr5 wrote...
Unsigned wrote...
Grenouille88 wrote...
Unsigned wrote...
Why does it say "Find the actual speed and direction of the car" if it's already given?It's given relative to the observer on the trail rather than in relation to a standard axis.
k so here's a diagram then. I kind of wish I knew how to do this
Spoiler:
That's what i thought but here's the problem.
The car going north isn't the direction in which it is actually going, but the direction that's relative to the observer on the train. I tried that diagram but then it wouldn't work....My math professor in university would kill me if he knew I couldn't solve something like this.
I don't know any math postulates or theories or whatever that factors in human error or human assumptions. I'm so confuzlled. Is that what it is?
It's like three people, a, b and c.
a is standing there, b and c are walking towards each other in front of him.
In the eyes of a, b and c are in motion, while he is still
In the eyes of b, he is still, and c is moving backwards.
In the eyes of c, he is still, and b is walking towards him with a higher speed than he really is.
Anyways my brother solved it already....Case closed
Thx for the help <3
0
Heizan-Tr5 wrote...
Unsigned wrote...
Heizan-Tr5 wrote...
Unsigned wrote...
Grenouille88 wrote...
Unsigned wrote...
Why does it say "Find the actual speed and direction of the car" if it's already given?It's given relative to the observer on the trail rather than in relation to a standard axis.
k so here's a diagram then. I kind of wish I knew how to do this
Spoiler:
That's what i thought but here's the problem.
The car going north isn't the direction in which it is actually going, but the direction that's relative to the observer on the train. I tried that diagram but then it wouldn't work....My math professor in university would kill me if he knew I couldn't solve something like this.
I don't know any math postulates or theories or whatever that factors in human error or human assumptions. I'm so confuzlled. Is that what it is?
It's like three people, a, b and c.
a is standing there, b and c are walking towards each other in front of him.
In the eyes of a, b and c are in motion, while he is still
In the eyes of b, he is still, and c is moving backwards.
In the eyes of c, he is still, and b is walking towards him with a higher speed than he really is.
Anyways my brother solved it already....Case closed
Thx for the help <3
So relativity. I guess that's cool... you should like totally steal your bro's HW and like upload it here. I need to see the work for it.
0
Gravity cat
the adequately amused
You have OCD if you really want to find out how fast a car's going and in what direction.
0
He's travelling at 40.62 kmph at 34.something degrees.
To calculate this, consider that the car is moving at the same horizontal velocity as the train(in the opposite direction(50/sqrt(2))). The vertical velocity is what the observer views - 20kmph.
I was in grade 10 6 years ago and I might be wrong.
If this is right, you should make your brother calculate the distance himself.
To calculate this, consider that the car is moving at the same horizontal velocity as the train(in the opposite direction(50/sqrt(2))). The vertical velocity is what the observer views - 20kmph.
I was in grade 10 6 years ago and I might be wrong.
If this is right, you should make your brother calculate the distance himself.
0
artcellrox
The Grey Knight :y
Kadushy wrote...
Here's an easier problem...Graph and shade... they're all in one bracket
[size=29]{[/h]
- x^2 + y^2 ≥ 9
x^2 + y^2 < 25
y ≤ 0
This is basically an upside-down arch. The inner semicircle has a radius of 3, and the outer has a radius of 5. Basically:
Spoiler:
Ignore the equation being written upside down. I actually noticed after I started drawing that it was y being less than zero, and I had started drawing for the opposite. Also, it's kinda blurry/transparent because it's other side of the pad. For some reason, taking pics with Cameroid flips the image, so i had to flip the pad itself.
0
Kadushy
Douchebag
artcellrox wrote...
Kadushy wrote...
Here's an easier problem...Graph and shade... they're all in one bracket
[size=29]{[/h]
- x^2 + y^2 ≥ 9
x^2 + y^2 < 25
y ≤ 0
This is basically an upside-down arch. The inner semicircle has a radius of 3, and the outer has a radius of 5. Basically:
Spoiler:
Ignore the equation being written upside down. I actually noticed after I started drawing that it was y being less than zero, and I had started drawing for the opposite. Also, it's kinda blurry/transparent because it's other side of the pad. For some reason, taking pics with Cameroid flips the image, so i had to flip the pad itself.
(y) You got it :D
0
animefreak_usa
Child of Samael
((x/7)^2*√((|(|x|-3)|)/(|x|-3))+
(y/3)^2*√((|(y+((3*√(33))/7))|)/
(y+((3*√(33))/7)))-1)*
(|(x/2)|-(((3*√(33))-7)/112)*
x^2-3+√(1-((|(|x|-2)|)-1)^2)-y)*
(9*√((|((|x|-1)*(|x|-.75))|)/
((1-|x|)*(|x|-.75))
Plot each component separately:
f1(x,y) = ((x/7)^2) * sqrt( sign(abs(x)-3) ) + ((y/3)^2) * sqrt( sign(y+3*sqrt(33)/7) ) - 1
f2(x,y) = abs(x/2) - ((3*sqrt(33)-7)/112)*(x^2) - 3 + sqrt(1 - (abs(abs(x)-2)-1)^2) - y
f3(x,y) = 9*sqrt( sign((1-abs(x))*(abs(x)-0.75)) ) - 8*abs(x) - y
f4(x,y) = 3*abs(x) + 0.75*sqrt( sign((0.75-abs(x))*(abs(x)-0.5)) ) - y
f5(x,y) = 2.25*sqrt( sign((0.5-abs(x))*(abs(x)+0.5)) ) - y
f6(x,y) = 6*sqrt(10)/7 + (1.5-0.5*abs(x))*sqrt(sign(abs(x)-1)) - 6*(sqrt(10)/14)*sqrt(4-(abs(x)-1)^2) - y
F(x)=(1 sin(x))(1 .9cos(8x))(1 .1cos(24x))
(y/3)^2*√((|(y+((3*√(33))/7))|)/
(y+((3*√(33))/7)))-1)*
(|(x/2)|-(((3*√(33))-7)/112)*
x^2-3+√(1-((|(|x|-2)|)-1)^2)-y)*
(9*√((|((|x|-1)*(|x|-.75))|)/
((1-|x|)*(|x|-.75))
Plot each component separately:
f1(x,y) = ((x/7)^2) * sqrt( sign(abs(x)-3) ) + ((y/3)^2) * sqrt( sign(y+3*sqrt(33)/7) ) - 1
f2(x,y) = abs(x/2) - ((3*sqrt(33)-7)/112)*(x^2) - 3 + sqrt(1 - (abs(abs(x)-2)-1)^2) - y
f3(x,y) = 9*sqrt( sign((1-abs(x))*(abs(x)-0.75)) ) - 8*abs(x) - y
f4(x,y) = 3*abs(x) + 0.75*sqrt( sign((0.75-abs(x))*(abs(x)-0.5)) ) - y
f5(x,y) = 2.25*sqrt( sign((0.5-abs(x))*(abs(x)+0.5)) ) - y
f6(x,y) = 6*sqrt(10)/7 + (1.5-0.5*abs(x))*sqrt(sign(abs(x)-1)) - 6*(sqrt(10)/14)*sqrt(4-(abs(x)-1)^2) - y
F(x)=(1 sin(x))(1 .9cos(8x))(1 .1cos(24x))
