All Answers 352725 Answers

did you graph it?  The answers are pretty obvious if you do.

Here's another way...

Let's extend

WZ to point  A,

XY  to point  B,

ZY  to point  C,

WX to point  D,  and

YX  to point  E

Like this:

All sequences must end with {3rd 0, 1,1,1,1}

Clearly we must start with 0.

We can form

0001111

010101111

00101111

001101111

and that's it.

The probabilities of flipping these sequences is

\(\dfrac{1}{128}+\dfrac{1}{256}+\dfrac{2}{512}=\dfrac{1}{64}\)

\(\text{To start let's assume $a=1,~b=e^{i\theta}$}\\ a+b = 1+\cos(\theta)+i \sin(\theta)\\ |a+b| = (1+\cos(\theta))^2 + \sin^2(\theta) = 2 + 2 \cos(\theta) = 1\\ \cos(\theta) = -\dfrac 1 2\\ \theta = \pm \dfrac{2\pi}{3}\)

\(\text{So now we can let $a=Me^{i\phi},~b = Me^{i (\phi\pm 2\pi/3)}$}\\ \text{and from the previous result we know $|a+b|=M$} \)

\(\dfrac a b = \dfrac{Me^{i\phi}}{Me^{i(\phi \pm 2\pi/3)}} = e^{i\pm 2\pi/3}\\ e^{i2\pi/3}+e^{-i 2\pi/3} = 2\cos\left(\dfrac{2\pi}{3}\right) = -1\)

thank you!!!

Add the two equations together to get

2a  =  8b

a  =  4b

Substitute this value in for  a  in the first given equation.

4b + (4b)b²  =  40b

4b + 4b³  =  40b

                            Subtract  40b  from both sides of the equation.

4b³ - 36b  =  0

                            Factor  b  out of the terms on the left side.

b(4b² - 36)  =  0

                                     Factor  4b² - 36  as a difference of squares.

b(2b + 6)(2b - 6)  =  0

                                     Set each factor equal to zero and solve for  b

If     b  =  0     then     a  =  4b  =  4(0)  =  0          so     (0, 0)  is a solution.

If     b  =  -3     then     a  =  4b  =  4(-3)  =  -12          so     (-12, -3)  is a solution.

If     b  =  3     then     a  =  4b  =  4(3)  =  12          so     (12, 3)  is a solution.

\(a^5b^5\ =\ 7776\\~\\ (ab)^5\ =\ 7776\\~\\ ab\ =\ \sqrt[5]{7776}\\~\\ ab\ =\ 6\\~\\ b\ =\ \frac6a\)

We can substitute this value in for  b  in the other given equation.

\(ab^4\ =\ 12\\~\\ a(\frac6a)^4\ =\ 12\\~\\ \frac{6^4}{a^3}\ =\ \frac{12}{1}\\~\\ 12a^3\ =\ 6^4\\~\\ a^3\ =\ \frac{6^4}{12}\\~\\ a^3\ =\ 108\\~\\ a\ =\ \sqrt[3]{108}\)_

nvm I got the answer

Notice that     m∠PBQ  =  90° - 60°  =  30°     so   △PBQ  is a  30-60-90 triangle. So

BQ  =  x√3

And notice     m∠PCQ  =  45°     so   △PCQ is a  45-45-90  triangle. So

QC  =  x

And

BQ + QC  =  4

                                  Substitute   x√3   in for  BQ   and   x   in for  QC

x√3 + x  =  4

                                  Factor  x  out of the terms on the left side.

x(√3 + 1)  =  4

                                  Divide both sides of the equation by   (√3 + 1)

x  =  \(\frac{4}{\sqrt3+1}\)

                                  Multiply the numerator and denominator by  √3 - 1  and simplify

x  =  \(\frac{4}{\sqrt3+1}\cdot\frac{\sqrt3-1}{\sqrt3-1}\)

x  =  \(\frac{4\sqrt3-4}{3-1}\)

x  =  \(\frac{4\sqrt3-4}{2}\)

x  =  \(2\sqrt3-2\)

Its SuerBoranJacobs

I'll refer to the diagram below:

We know that WZ || XY and WX || ZY becuase opposite sides are congruent (Def. of a parallelogram).

Draw ZX and WY as shown (Ruler Postulate).

Label Angles 1, 2, 3, 4, 5, 6, 7, 8 as shown (Ruler Postulate).

Angle W = Angle 3 + Angle 4 (By Construction).

Angle Z = Angle 1 + Angle 2 (By Construction).

Angle Y = Angle 7 + Angle 8 (By Construction).

Angle X = Angle 5 + Angle 6 (By Construction).

Angle 1 = Angle 6 (Alternate Interior Angles Are Congruent).

Angle 2 = Angle 5 (Alternate Interior Angles Are Congruent).

Therefore, Angle Z = Angle X (Parts Make Up A Whole).

Using the same reasoning Angle Y = Angle W.

Q.E.D

Let's denote x as the value of the score for the first 4 tests. Also, the 5th test score can be denoted as y. The total score would be 5 * 82 = 410.

We have a Diophantine equation that is:

\[4x + y = 410 \]

One possible value of y is 98 which means x is 78. We can decrease y by 4 and increase x by 1 until x is equal to or larger than y to find other possible values since the overall change is 0.

\[(4 * 78) + 98 = 410 \]

\[(4 * 79) + 94 = 410 \]

\[(4 * 80) + 90 = 410 \]

\[(4 * 81) + 86 = 410 \]

This results in 4 possible values for the last test.

I attempted this question many times but every single time I got different answers. My method was really long, so it would have been very hard to write all that here, including the diagrams. Sorry for the inconvenience.

Yeah, I was offline for a while because I have a lot of activities during the summer. I'll try to be online more often.

Just considering flips the answer would 5!/2 = 120/2 = 60

Is the function  even, odd, or neither?

Can you define rotations?

I am SuerBoranJacobs and have forgotten my password. Now I have to be a guest...

We can simplify $ f(x) = 5^x - 5^(-x) $ down to:

\[ f(x) = 5^x - 1/(5^x) \]

(This will be easier to deal with).

Also, we have to substitute in -x for x in f(x) to find out if the function is even, odd, or neither:

\[ f(-x) = 5^(-x) - 5^x \]

\[ f(-x) = 1/(5^x) - 5^x \]

\[ f(-x) = - (5^x - 1/(5^x) \]

When we compare the 2 functions we see that the new function differs by a multiple of -1 meaning that its odd:

\[ f(x) = 5^x - 1/(5^x) \]

\[ f(-x) = - (5^x - 1/(5^x) \]

In conclusion, the answer is $ f(x) = 5^x - 5^(-x) $ is ODD.

Good thinking :)

If f(x) is a polynomial of degree 7, and g(x) is a polynomial of degree 7, then what is the product of the minimum and the maximum possible degrees of f(x) + g(x)?

if f(x)  = - g(x)    then  f(x)+g(x)=0    degree zero.

Most times though the f(x)+g(x) would have degree 7

7*0=0 

WHY! the chinese squad has just gotten a new member.....

*COughs* yeah, I don't think we in the need of highschool diplom'a's here posted in the forum of Chinense symbols...