x^3 + x^2 - 8x - 8 = 0
Factor as
x^2 ( x + 1) - 8 (x + 1) = 0
(x + 1) ( x^2 - 8) = 0
So either or
x + 1 = 0 x^2 - 8 = 0
And x = -1 ( x - √8) ( x + √8) = 0
x - √8 = 0 or x + √8 = 0
And x = √8 x = - √8
So
C) The function has three real zeros. The graph of the function intersects the x-axis at exactly three locations.
Yah I think 135 degrees is the answer
I dont understand, are you saying that the image of the strip R under sin(z) is the first quadrant+the fourth quadrant+the real axis?
This answer is wrong (If you know what the open mapping theorem is, you can see how the theorem shows that this can't be the image, and that the image has to be an open set)
Also keep in mind that sin(x+iy)=sin(x)cosh(y)+icos(x)sinh(y), not sin(x)cosh(y)+icos(x)cosh(y)
Thanks,Rom and Guest. Hey Rom,I think your answer is partially correct.
The mapping f(z)=sin(z)=sin(x+iy)=sin(x)cosh(y)+icos(x)cosh(y) map z to Real(f(z)) in positive real number and Im(f(z)) to real, which map to first and fourth quaduant and real axis of complex plane. That is my simpilfie answer and this probelm was one of test question,and I am doing test correction. Does anyone have any thoughts on this?
135 degree
I can't explain due time being.
How did you get those answers Fiora? there is no hourly rate given ....
I don't think your notation is correct. I do not know what it means anyway .....
I think it is $28,187.50
0.75*0.4=0.3 and 0.3*57000= $22500
0.75*0.6=0.45 and 0.45*15000=$6750
0.25*0.45=0.1125 and 0.1125*15000=$1687.50
0.25*0.55=0.1375 and 0.1375*-20000 = -$2750
When all of those are added together I get $28187.50
If a = 1 there are solutions x = -1 and x = 2
If a = -1 the quadratic reduces to a linear equation with solution x = -4
.
This is incorrect.
The definition of sin(z) for complex numbers is:
\(sin(z)=\frac{{e}^{iz}-{e}^{-iz}}{2i}\)
12
\(\left( \begin{array}{c} \{15\} \\ \{5,10\} \\ \{1,4,10\} \\ \{1,6,8\} \\ \{2,3,10\} \\ \{2,5,8\} \\ \{3,4,8\} \\ \{4,5,6\} \\ \{1,2,4,8\} \\ \{1,3,5,6\} \\ \{2,3,4,6\} \\ \{1,2,3,4,5\} \\ \end{array} \right)\)
This is all correct, well done.
It looks like it maps to the strip
\(\{z: 0 \leq Re[z] \leq 1,~Im[z] \in \mathbb{R}\}\)
If you separate the 20 switches into groups of 4 and fully connect within a group you can pull this off in 30 connections.
Hello.
Since I do not know if by x2 you meant \(2x\) or \(x^2\), I'll solve for both.
Okay. here is the solution for when it is \(2x\):
If the length is 5ft, then \(x = 5\).
Using this to solve for \(5x^3 + 4 - 2x\), we get \(625 + 4 - 10 = 619\).
The perimeter is \(5 + 5 + 619 + 619 = \boxed{1248}\) ft.
Now, here is the solution for when it is \(x^2\).
Using this to solve for \(5x^3 + 4 - x^2\), we get \(625 + 4 - 25 = 604\).
The perimeter is \(5 + 5 + 604 + 604 = \boxed{1218}\) ft.
(2) looks good, except for the fact that the inequality sign must be a "less than or equal to".
(If you meant to type that)
You can do this next time, using:
\leq
in LaTeX. (\(\leq\))
For (1), the slope is \(\frac{3}{2}\).
Therefore, our answer would be something in the form of: \(y = \frac{3}{2}x + b\).
Converting this to standard form, we get \(-\frac{3}{2}x + y - b > 0\), which is \(-3x + 2y - 2b >0\).
Now the \(y\) intercept is found by plugging in a coordinate into the equation.
Solving this, we get the y-intercept to be 7.
So your answer for (1) would be correct.
I'm seeing 28
\(\dbinom{6+3-1}{3-1} = \dbinom{8}{2} = 28\)
\(\left( \begin{array}{ccc} 6 & 0 & 0 \\ 5 & 1 & 0 \\ 5 & 0 & 1 \\ 4 & 2 & 0 \\ 4 & 1 & 1 \\ 4 & 0 & 2 \\ 3 & 3 & 0 \\ 3 & 2 & 1 \\ 3 & 1 & 2 \\ 3 & 0 & 3 \\ 2 & 4 & 0 \\ 2 & 3 & 1 \\ 2 & 2 & 2 \\ 2 & 1 & 3 \\ 2 & 0 & 4 \\ 1 & 5 & 0 \\ 1 & 4 & 1 \\ 1 & 3 & 2 \\ 1 & 2 & 3 \\ 1 & 1 & 4 \\ 1 & 0 & 5 \\ 0 & 6 & 0 \\ 0 & 5 & 1 \\ 0 & 4 & 2 \\ 0 & 3 & 3 \\ 0 & 2 & 4 \\ 0 & 1 & 5 \\ 0 & 0 & 6 \\ \end{array} \right)\)
Off Topic/Spam. Thanks!
