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I said that the other root was -1/4 because I first factored ax^2+bx+a into (ax+1)(x+a). Since one of the roots were 4, I knew that a had to be -4. Then, substituting -4 for a in ax+1=0, I got that it was -1/4.
Let f(x)=ax^2+bx+a, where a and b are constants and a is not equal to 0.
If one of the roots of the equation f(x)=0 is x=4, what is the other root? Explain your answer.
Is it possible to do it with algebra rather than trig?
the hint is: Try proving that $\overline z = 1/z$ and $\overline w = 1/w$.
So far, I know that a number is real iff it is equal to it's conjugate. So I did (w+z)/(1+wz)=(1+wz)/(w+z). I factored it into 0=(w+1)(w-1)(z+1)(z-1). I don't know where to go to from there.
Please help!
Prove that if w, z are complex numbers such that |w|=|z|=1 and wz is not equal to -1, then (w+z)/(1+wz) is a real number.
for Q1, I put in (-inf, 1) but it was wrong
1) Similarly, by restricting the domain of the function f(x)=2x^2-4x-5 to an interval, we can make it invertible. What is the largest such interval that includes the point x=0?
2)Suppose the domain of f is (-1,1). Define the function h by h(x)=f(x)+1. What is the domain of h?
In a math class, the quadratic x^2+10x+20 is written on the board. Each student goes to the board and increases or decreases either the linear or constant coefficient by 1. After some time, x^2+20x+10 is written on the board. Did a quadratic with integer roots necessarily appear on the board at some time during this process? Why or why not?
Help?
then wouldnt we have to prove that there are perpendicular lines?
How do you know that AB=LM without using geogebra?
Also, what is M and J?
Is there a way to solve this problem by making ABP a medial triangle?
I got that
andIn triangle $ABC$, $\angle A = 36^\circ$ and $\angle B = \angle C = 72^\circ$. Let $\overline{BD}$ be the angle bisector of $\angle ABC$.
(a) Prove that $BC = BD = AD$.
(b) Let $x = BC$ and let $y = CD$. Using similar triangles $ABC$ and $BCD$, write an equation relating $x$ and $y$.
(c) Write the equation from Part b in terms of $r=\frac yx$ and find $r.$
(d) Compute $\cos 36^\circ$ and $\cos 72^\circ$ using Parts a-c. (Do not use your calculator!)
I was able to do a-c. Can I help on d?
thanks for the help! I got it
But how do I show " for any x such that x is not an integer multiple of 180"?
does the trig identity have to do with PT:
sin2(t) + cos2(t) = 1
Explain why we must have
for any such that is not an integer multiple of .I get it now. Thanks for all the help!
I still don't know what a,b, and c stand for in a^2+b^2=c^2 and how you derived it from AB=AC. Also, how go the product of the slopes equal -1 now?
I also need help on these two:
1)In right triangle ABC, we have AB = 10, BC = 24, and angle ABC = 90 degrees. If M is on AC such that BM is a median of triangle ABC, then what is cos angle ABM?
2)How many values of x with
I don't understand how you got a^2-c^2=-b^2
One more question:
as my final result, the slopes of AF and BE are
Thanks for all the help! It made the problem so much easier!
Thanks! How about the other two?