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Paladin
Vector problem that i can't seem to crack?
Hi, i can't seem to solve this problem no matter what i do!
The vector (imagine this is in component form),
(a - b, b) is parallel to the x axis and the vector ( a + b, 2a - b) is parallel to (2, 1). Find the values of "a" and "b".
Thanks in advance :)
1 AnswerMathematics1 decade agoA simple trigonometric equation - need help solving it! ?
I'm really not sure on how to solve this one?
Find all the values of x in the interval [0, pi] which satisfy the equation cos(2x) = sin^2(x)
So far i've tried doing this:
cos(2x) = sin^2(x)
cos(2x) = 1 - cos^2(x)
cos^2(x) + cos(2x) - 1 = 0
Now that we have a quadratic equation we apply the discriminant formula
D = b^2 - 4ac
D = 1 - 4(1)(-1)
D = 1+ 4
D = 5
so
x1 = (1 + sqrt(5)) / (2)
x2 = (1 - sqrt(5)) / (2)
Is the process correct so far? And what do i do next?
1 AnswerMathematics1 decade agoThe polynomial x^2 - 4x + 3 is a factor of x^3+(a-4)x^2 + (3-4a)x + 3. Calculate what a is. ?
The polynomial x^2 - 4x + 3 is a factor of x^3+(a-4)x^2 + (3-4a)x + 3. Calculate what a is.
I just can't seem to solve it, i want to see the whole process of solving it.
Thank you in advance. :)
1 AnswerMathematics1 decade agoFind the equation of a circle with the centre on the x-axis. The circles passes through points (2,2) & (-4,-4)?
I just can't seem to solve this problem, so far i've done this:
Find the equation of a circle with the centre on the x-axis. The circles passes through points (2,2) & (-4,-4)
Since it's centre is on the x axis, the y coordinate of the center is 0.
Let's name the points A and B:
A(2, 2)
B(-4, -4)
So for A:
q=0 therefore,
(2-p)^2+(2-0)^2 = r^2
4 - 4p +p^2 + 4 = r^2
p^2 - 4p + 8 = r^2
And for B:
(-4-p)^2+(-4-0)^2 = r^2
16 + 8p + p^2 +16 = r^2
p^2 + 8p + 32 = r^2
Now since both of these points can be found on the circle, we know that the radii are the same, so we can equal them.
p^2 - 4p + 8 = p^2 + 8p + 32
p^2 + 8p + 32 - p^2 + 4p - 8 = 0
12p = -24
p = -2
So
p = -2
q = 0
Therefore the equation of the circle is
(x + 2)^2 + y^2 = r^2
Now the thing i don't get is: how do you get the radius?
2 AnswersMathematics1 decade agoSimple quadratic function problem - that i just can't solve... help?!?
Consider the equation (1+2k)x^2 - 10x + k - 2 = 0
k € R
Find the set of values of k for which the equation has real roots.
1 AnswerMathematics1 decade agoNeed help with an algebra problem?
For what values of k is the straight line y = kx + 1 a tangent to the circle with the origin at (5, 1) and a radius of 3 (r = 3)?
I'm just not sure how to approach this problem.
1 AnswerMathematics1 decade agoSimple absolute value problem?
I'm not really sure how to answer
|x+1| + |x-1| = 3
thing is, it's easy when you have one absolute value, but i get confused when I see two. I'd just like the answer and a brief process.
thanks!
3 AnswersMathematics1 decade ago(quite) Simple Trig Problem?
A surveying team are trying to find the height of a hill. They take a sight on top of the hill and find that the angle of elevation is 23º27'. They move a distance of 250m on level ground directly away from the base of the hill and take a second sight. From this point, the angle of elevation is 19º46'. Draw a diagram to represent the situation and find the height of the hill to the nearest metre.
Spare the diagram, but how would I solve this?
2 AnswersMathematics1 decade ago