All Answers 356018 Answers

The probability is 23/216.

The perimeter is 15 + 5*sqrt(3).

The probability is 1/4.

a = -1/4.

The value of m is 25/3.

The area is 17*pi.

There is another way to solve this problem. Notice that the left side is the square of a binomial.

\((x + 5)^2 = 64\)

\(x + 5 = 8 \qquad x + 5 = -8\)

x = 3 or -13.

Type them

Find the smallest value of x in (x^2 + 10x + 25 = 64).

Hello Guest!

\(x^2 + 10x + 25 = 64\\ x^2+10x-39=0\)

          p          q

\(x=-\frac{p}{2}\pm \sqrt{(\frac{p}{2})^2-q}\\ x=-\frac{10}{2}\pm \sqrt{(\frac{10}{2})^2+39}\\ x=-5\pm 8\\ x\in\{-13,3\}\)

The smallest value of x in (x^2 + 10x + 25 = 64) is - 13.

  !
 

I was curious about the same thing: 

Hi Melody,

1)   The first one, start by adding m to BOTH sides  then come back with whtat you have if you need more hgelp

2)   Multiply both sides out and simpleify the two sides seperately first.

remember that   

7-5(3-2x)

= 7+     -5(3-2x)

the minus sign belongs to the 5.

After you do the first bit come back with what you have done.

You have to think of a logical method that might work.

Suggestion

Hoe many ways if only one letter is used

How many ways if 2 letters are used

How many ways if 3 letters are used

How many ways if 4 letters are used.

Thinks about that and come back with some of your own input.

\(\frac{(1 + i\sqrt3)^6}{(1 + i\sqrt3)^4}\\ =(1 + i\sqrt3)^2\)

You can finish it  :)

Let z be a complex number such that |z| = 1. Find the largest possible value of |z^2 - 1|

\(z=1e^{i\theta}\\ z^2=e^{2\theta i}\\ z^2-1=e^{2\theta i}-1\\ z^2-1=cos(2\theta)+isin(2\theta)-1\\ z^2-1=[cos(2\theta)-1]+isin(2\theta)\\ |z^2-1|^2=[cos(2\theta)-1]^2+sin^2(2\theta)\\ |z^2-1|^2=[cos^2(2\theta)+1-2cos(2\theta)]^2+sin^2(2\theta)\\ |z^2-1|^2=[2-2cos(2\theta)]^2\\ |z^2-1|^2=4[1-cos(2\theta)]^2\\ |z^2-1|=2[1-cos(2\theta)]\\\)

\(-1\le cos(2\theta)\le1\\ -1\le -cos(2\theta)\le1\\ 0\le 1-cos(2\theta)\le 2\\ 0\le 2[1-cos(2\theta)]\le 4\\ 0\le |z^2-1|\le 4\\ \)

So the largest possible value is 4 

You need to check what i have done.

LaTex

z=1e^{i\theta}\\
z^2=e^{2\theta i}\\
z^2-1=e^{2\theta i}-1\\
z^2-1=cos(2\theta)+isin(2\theta)-1\\
z^2-1=[cos(2\theta)-1]+isin(2\theta)\\
|z^2-1|^2=[cos(2\theta)-1]^2+sin^2(2\theta)\\
|z^2-1|^2=[cos^2(2\theta)+1-2cos(2\theta)]^2+sin^2(2\theta)\\
|z^2-1|^2=[2-2cos(2\theta)]^2\\
|z^2-1|^2=4[1-cos(2\theta)]^2\\
|z^2-1|=2[1-cos(2\theta)]\\

-1\le cos(2\theta)\le1\\
-1\le -cos(2\theta)\le1\\
0\le 1-cos(2\theta)\le 2\\
0\le 2[1-cos(2\theta)]\le 4\\
0\le |z^2-1|\le 4\\

\(\frac{\sqrt2+\sqrt3}{\sqrt{10}+\sqrt{14}+\sqrt{15}+\sqrt{21}}\)

Looking at the denominator, notice that \(\sqrt{10}\) and \(\sqrt{14}\) are both divisible by \(\sqrt{2}\) and\( \sqrt{15}\), \(\sqrt{21}\) are divisible by \(\sqrt{2}\). 

We can factor those out to get \(\frac{\sqrt2+\sqrt3}{\sqrt2(\sqrt5+\sqrt7)+\sqrt{3}(\sqrt5+\sqrt7)} = \frac{\sqrt2+\sqrt3}{(\sqrt2+\sqrt{3})(\sqrt5+\sqrt7)}\). Divide top and bottom by \(\sqrt2+\sqrt3\): \(\frac{1}{\sqrt5+\sqrt7}\).

Moving the radicals to the numerator: \(\frac{1}{\sqrt5+\sqrt7} = \frac{1\cdot(\sqrt5-\sqrt7)}{(\sqrt5+\sqrt7)(\sqrt5-\sqrt7)} = \frac{\sqrt{7}-\sqrt{5}}{2}\).

So the final result is: \(\boxed{\frac{\sqrt7 - \sqrt5}{2}}\). Hope this helped 

Let  be the amount of soda contained by Bottle A and  be the amount of soda contained by Bottle B. We can track the changes in amounts of soda in each bottle. First, Bottle A contains  ounces, and Bottle B contains . At the next step, A is reduced by  ounces, and B gains  ounces, so A has  and B has  ounces. Then, A is increased by  ounces and  is reduced by the same, so A has  and B has  ounces. Finally, B is increased by  ounces and A is decreased by the same, so A has  ounces and B has  ounces. We are given that , and we are looking for . Adding  to  gives . Adding  to  gives , or . So, , or . Thus,  ounces.

p*u = 22/3.

|x|/|y| = 4/9.

First I simplified the first equation to:

x + y = 5xy. I got this by multiplying both sides of the equation by 'xy'.

Then we can substitute that into the second equation, getting:

3xy + 5xy = 4 + 5xy.

This simplifies to 3xy = 4.

And xy = 4/3.

Since the question is looking for x^2y + xy^2, that is equal to: (x+y)(xy).

Substituting the value back in, we have (5xy)(xy), and finally:

\(x^2y + xy^2 = {80\over9}\)

This analytical geometry question has been answered correctly here: https://web2.0calc.com/questions/analytic-graphing-geometry-help

Also, the perimeter wouldn't change whether it would be cut by a different line because they are two CONGRUENT quadrilaterals, and maybe there was a typo for y = x/3, not y = x/2. Either way, go check out the URL :D

good luck with your alcumus.

To make the equation clear, it looks like this: \({y\over3}+1={y+3\over y}+7\).

Then multiply every term by '3y'.

It should now look like this:\(y^2 + 3y = 3y + 9 + 21y\).

Combining like terms and moving the equation to one side, we get: \(y^2 - 21y - 9 = 0\).

Then you can use the quadratic formula \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) where a is the coefficient of x^2. b is the coefficient of x, and c is the constant.

Plugging in the values, we have:

\(y = {21 + 3\sqrt{53}\over2}\)

\(y = {21-3\sqrt{53}\over2}\)

Those are the two possible values of y :)

So first we make the coefficient of x^2 = 1.

The new quadratic is: \(x^2 + {5\over3}x + {7\over3} = 0\)

Then since we know by Vieta, the two solutions of the quadratic multiply to the c term, and they add to the b term * -1. The solutions are u and v.

Thus, \(u + v = -{5\over3}\), and \(uv = {7\over3}\). 

Since \(u^2 + v^2 = (u + v)^2 - 2uv\), we can substitute in our previous values. 

Then \(u^2 + v^2 = {25\over9} - {14\over3}\).

Thus, \(u^2 + v^2 = -{17\over9}\).

n = -80^o