All Answers 358524 Answers

C = (2/3 -1/4) Find C - 2E

C = 8/12 - 3/12 = 5/12

C-2E = 5/12 - 2E

Oops I was just about the reveal the answer for the last one :P

\(f(x+1)\\=4(x+1)^2+2(x+1)-2\\=4x^2+8x+4+2x+2-2\\=4x^2+10x+4\)

f(x)= 4*x²  +  2*x  -  2

f(3)= 4*3²  +  2*3  -  2

you can simplify it

f(x)= 4*x²  +  2*x  -  2

f(x+1)= 4*(x+1)²  +  2*(x+1)  -  2

and you can simplify it.

Can you see how I am just replacing x with whatever is in the bracket?

The last one you do the two seperately and then add them together.

\(f(-3)\\=4(-3)^2+2(-3)-2\\=28\)

I am sorry to say that your trick question is too easy for me and the answer is '? = 5' :P

Guest #5, I don't quite understand your answer. But Alan's answer did help me a lot. Thanks Alan. And Guest#5 can you give a more detailed explanation?

If 5 ^x-3 = 10, what is the value of x?

\(5^x-3=10\\ 5^x=13\\ log5^x=log13\\ xlog5=log13\\ x=\frac{log13}{log5}\\\)

(log(13)/log(5)) = 1.593692641167082

u = (-1,4) v = (3,-4)

I am sure there is a less clumsy way to do this but....

u is in the second quadrant.  The angle it makes with the positive x axis is 

\(\pi-atan(4/1) = \pi-atan(4)\)

v is in the 4th quadrant so the angle it makes with the positive x axis is

\(atan(-4/3)\)

\(angle\;\;between=|180-atan(4)-atan(-4/3)|\\\)

104.036243467926-atan(-4/3) = 157.166345822082

157.17 degrees

The answer given by Melody here is the SAME answer given to you yesterday and the day before that.....etc. This is the 3rd time you have posted the same question. The difference is that the answer given to you yesterday is in " degrees", while this answer is in "radians". Can't you convert from one to the other?. Just try converting this into "degrees" and look at the answer given to you yesreday.

Can you answer from this pic?

In general:  \(e^{i\theta}=\cos \theta+i\sin\theta\)

You already know that  \(\sqrt{i}=e^{i\pi /4}\)

Put these together to get:  \(\sqrt{i}=\cos{(\pi /4)+i\sin{(\pi /4)}}\)

Since we have: \(\cos{(\pi /4)}=\frac{1}{\sqrt{2}} \space \text{and }\sin{(\pi /4)}=\frac{1}{\sqrt{2}}\)

we can see that:  \(\sqrt{i}=\frac{1+i}{\sqrt{2}}\)

.

Here's a picture to illustrate Melody's answer:

.

HI MAX   

Your answer looks good except if we are dealing with the real number system there are restrictions on the domain.

\(e^{g\left(x\right)}=\frac{x^x}{x^2-1}\\ \mbox{e to the power of any real number is positive so the RHS must be positive}\\ g(x)= \ln(\frac{x^x}{x^2-1})\\ \)

1)     \(x^2-1\ne0\;\;\;\;so\\ x\ne\pm1\)

2)   x^x does crazy things isf x is not not greater than zero. (x certainly cannot equal 0) So I am going to say that x must be greater than 0.

      I think that this is correct but I would like another mathematician to comment

      \(x>0 \qquad x\ne1\)

3)  

       \(\frac{x^x}{x^2-1}>0\\ \mbox{Since the numerator is positive the denominator must be positive too}\\ x^2>1\\ x>1 \;\;x<-1\\ so\;\;x>1\)

\(g(x)= \ln(\frac{x^x}{x^2-1})\\ \quad \;\;\;=x\ln x-(\ln(x+1)(x-1))\\ \quad \;\;\;=x \ln x - \ln (x+1)- \ln (x-1)\\ \)

Your answer is correct so long as you state that the domain is  \(x>1 \;\;for \;\;x\in R\)

what is the full number of "2.0163e+006"?

2.0163e+006 = 2.0163 * 10⁶    = 2.016300

40000000000=(66-77)+25?

40 000 000 000 = (66 - 77) + 25 + 39 999 999 986

Hello Guest!

If you had lines x + y = 9 and x + y = 9. Are they parallel or other?

This problem confues me because parallel lines dont intersect.

The two equations are identical.
Cover yourself.
The two lines are still in parallel.

Greeting  asinus :- )

   !

Guest

Here is a direct algebraic answer:

Suppose that z=c+di, and we want to find √z=a+bi lying in the first two quadrants. So what are a and b?

Precisely we have:
a=Sqrt[c+sqrt(c^2+d^2)/2]

and

b=d/|d|Sqrt[-c+sqrt(c^2+d^2)/2]
(The factor of d/|d| is used so that b has the same sign as d) To find this, we can use brute force and the quadratic formula. Squaring, we would need to solve:
a^2−b^2+2abi=c+di.
This gives two equations and two unknowns (seperate into real and imaginary parts) which can then be solved by substitutions and the quadratic formula.

Hope that helps.

Here is a direct algebraic answer:

Suppose that z=c+di, and we want to find √z=a+bi lying in the first two quadrants. So what are a and b?

Precisely we have:
a=Sqrt[c+sqrt(c^2+d^2)/2]

and

b=d/|d|Sqrt[-c+sqrt(c^2+d^2)/2]
(The factor of d/|d| is used so that b has the same sign as d) To find this, we can use brute force and the quadratic formula. Squaring, we would need to solve:
a^2−b^2+2abi=c+di.
This gives two equations and two unknowns (seperate into real and imaginary parts) which can then be solved by substitutions and the quadratic formula.

Hope that helps,

\(\text{Why is }\sqrt{i}=\frac{1+i}{\sqrt2}? \text{I still don't understand how to obtain this answer?}\)

Just discovered how to escape from Math mode of LaTeX while the above method won't work for me(The $ one)

\text{}

\(\text{2^5}\)\text{2^5}

\(\text{The above one have completely escaped from math mode, so is this one}\)

\text{The above one have completely escaped from math mode, so is this one}

LOL

\(\left(z-\frac{2}{\sqrt{z}}\right)^9\)

\(\text{The general term is}\)

\((9Cr)(z)^{(9-r)}\left(\frac{-2}{\sqrt{z}}\right)^r\)

\(=(9Cr)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z})^r}\right)\)

\(=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(\sqrt{z})^r}\right)\)

\(=(9Cr)\left(\frac{z^{18/2}*(-2)^r}{z^{(2r/2)}*(z)^{(r/2)}}\right)\)

\(=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\)

\(=(9Cr)(z^{((18-3r)/2)}*(-2)^r)\)

\(\text{The constant term will be when}\)

\((18-3r)/2=0\)

\(18-3r=0\)

\(r=6\)

Just wanna try. This code is really crazy! Typing it twice takes me 30minutes. (I accidentally deleted when I have already finished)

I understood this but I just realized that \(\sqrt[\frac{1}{n}]{a^m}=a^{mn}\)

\(e^{g\left(x\right)}=\frac{x^x}{x^2-1}\)

\(g(x)= \ln(\frac{x^x}{x^2-1})\\ \quad \;\;\;=x\ln x-(\ln(x+1)(x-1))\\ \quad \;\;\;=x \ln x - \ln (x+1)- \ln (x-1)\\ \text{Sorry if I am wrong I am pretty bad at logarithms}\)