All Answers 356389 Answers

500000+30000000-66000000/-500 = 30632000

Hope this helps 

Which one(s) have the following property (pythagorean theorem)   a^2 + b^2 = c ^2     Where 'a' and 'b' are the shorter legs?

First one :  10^2 + 8^2 = 12^2   ?     100+64 = 144 ?   Nope, not the first one.....  YOU try the rest to see if they work..... 

I got no clue what is is.

Anyway, this is an order of operation question.

You are probably taught the order as PEDMAS

1) Parenthesis,

2) Exponents (the little 3 in your question is an exponent)

3) Division and Multiplcation are euqally important - you do them from left to right.

4) Addition and Subtraction are also equal important - do them from left to right

\(5^3\div(13-8)\times 2\)

Do it one step at a time in this order :

Parentheses

Exponent

Division    

Multiplication

125

Is the answer 

Good so what is 5*5*5? 

It means 5*5*5

Thanks Melody 

What does 5³ mean Nickolas?       Do you know?      

By the way,   5³  can also be written as  5^3

Yes that is correct Nickolas :)

I thinking it is answer 3???

The asymptotes of a hyperbola are y=x+1 and y=3-x. Also, the hyperbola passes through (3,3).
Find the distance between the foci of the hyperbola.

Formula hyperbola:

\(\begin{array}{|l|rcll|} \hline 1. \text{ asymptote} & x+1 &=& y \\ & x+1-y &=& 0 \\\\ \hline 2. \text{ asymptote} & 3-x &=& y \\ & 3-x-y &=& 0 \\\\ \hline \text{hyperbola: } & (x+1-y)(3-x-y) + c &=& 0 \\\\ P(x=3,y=3) & (3+1-3)(3-3-3)+c &=& 0 \\ & (1)(-3)+c &=& 0 \\ & c &=& 3 \\ \text{hyperbola: } &\mathbf{ (x+1-y)(3-x-y) + 3} &\mathbf{=}& \mathbf{0} \\\\ & (x+1-y)(3-x-y) + 3 & = & 0 \\ & \Big((x-y)+1\Big)\Big(3-(x+y)\Big) + 3 & = & 0 \\ & 3(x-y)-(x-y)(x+y)+3-(x+y) + 3 & = & 0 \\ & 3(x-y)-(x-y)(x+y)+3-(x+y) + 3 & = & 0 \\ & 3x-3y-(x^2-y^2)+3-x-y + 3 & = & 0 \\ & 2x-4y-x^2+y^2+6 &=& 0 \quad | \quad \cdot (-1) \\ & x^2-2x -y^2+4y -6 &=& 0 \\ & x^2-2x -y^2+4y &=& 6 \\ & (x-1)^2-1 -(y-2)^2+4 &=& 6 \\ & (x-1)^2 -(y-2)^2 &=& 3 \quad | \quad :3 \\ & \dfrac{(x-1)^2}{3} - \dfrac{(y-2)^2}{3} &=& 1 \\ & \boxed{\text{Formula hyperbola:} \dfrac{ (x-h)^2 }{a^2}- \dfrac{ (x-k)^2 }{b^2} = 1 } \\ & a^2 = 3 \\ & a= \sqrt{3} \\\\ & b^2 = 3 \\ & b= \sqrt{3} \\ \hline \end{array}\)

\(\begin{array}{|l|rcll|} \hline \text{Focus}_1: & F_1(h+ae,k) \\ \text{Focus}_2: & F_2(h-ae,k) \\ \hline \text{Foci distance}: & && h+ae-(h-ae) \\ & &=&h+ae-h+ae \\ & &=& 2ae \quad | \quad e=\dfrac{\sqrt{a^2+b^2}}{a} \\ & &=& 2a\dfrac{\sqrt{a^2+b^2}}{a} \\ & &=& 2 \sqrt{a^2+b^2} \quad | \quad a^2=b^2= 3 \\ & &=& 2 \sqrt{3+3} \\ & &\mathbf{=}& 2 \mathbf{ \sqrt{6} } \\ \hline \end{array}\)

The distance between the foci of the hyperbola is \(\mathbf{2\sqrt{6} }\)

The asymptotes of a hyperbola are y=x+1 and y=3-x. Also, the hyperbola passes through (3,3). Find the distance between the foci of the hyperbola.

What is the sum of the following sequence: 1000^3 - 999^3 - 998^3 - 997^3 -.............-3^3 - 2^3 - 1^3.

\(\begin{array}{|rcll|} \hline && \mathbf{ 1000^3 - 999^3 - 998^3 - 997^3 -\ldots -3^3 - 2^3 - 1^3} \\ \\ &=& 1000^3 - \left(1^3+2^3+3^3+\ldots +997^3+998^3+999^3 \right) \\ &=& 1000^3 - \left(1 +2 +3 +\ldots +997 +998 +999 \right)^2 \\ &=& 1000^3 - \left(\left(\dfrac{1+999}{2}\right) \cdot 999\right)^2 \\ &=& 1000^3 - 500^2 \cdot 999^2 \\ &=& 1\ 000\ 000\ 000 - 249\ 500\ 250\ 000 \\ &\mathbf{=}& \mathbf{-248\ 500\ 250\ 000} \\ \hline \end{array}\)

\(a^2 + ab + b^2 = 0\\ (a+b)^2 = a^2+2ab+b^2= ab\\ a = \dfrac{-b\pm \sqrt{-3b^2}}{2} = b e^{\pm i 2\pi/3}\\ a^9+b^9 = b^9(e^{\pm i 6\pi}+1) = 2b^9\\ (a+b)^9 =( (a+b)^2)^{9/2} = \\ (ab)^{9/2} =\left( b^2 e^{\pm i 2\pi/3}\right)^{9/2} = b^9e^{\pm i 3\pi} = -b^9\\ \dfrac{a^9+b^9}{(a+b)^9} = \dfrac{2b^9}{-b^9} = -2\)

sec 54° 23’

Most calculators have a way of putting in minutes and seconds. 

BUT you have to change it to    1 / 54° 23’

The web 2 calc has the 'sec' function (trig) so you don't need to worry about that but it has no seconds(angle)  function.

So you have to put it in like this

sec(54+23/60) = -1.7853120273505811

What kind of calc are you using?

\(1) \text{Find vectors defining parallelogram from one vertex}\\ v_1=(1,1)-(0,2) = (1,-1)\\ v_2=(2,3)-(0,2) = (2,1)\\ 2) \text{augment these to 3D by setting the z coordinate to 0}\\ v_1= (1,-1,0)\\ v_2= (2,1,0)\\ 3) A\hat{k} = v_1 \times v_2\\ \begin{pmatrix}i&j&k\\1&-1&0\\2&1&0\end{pmatrix} = (0, 0,3) = 3\hat{k}\\ A=3\)

\(3\sqrt{18}+9\sqrt{50} = \\ 9\sqrt{2} + 45\sqrt{2} = \\ 54\sqrt{2}\)

\(T = \mu + z \sigma = 17 + (2.8)(4) = 28.2\\ P[\text{someone scores > Sarah}] = 0.00255513\\ \text{So Sarah's score is pretty exceptional}\)

\(f(x) = a x^3 + b x ^2 + c x + d\\ f(0) = 0 \Rightarrow d = 0\\ \begin{pmatrix}-1 &1 &-1 \\1 &1 &1 \\8 &4 &2\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}= \begin{pmatrix}15 \\-5 \\12\end{pmatrix}\)

\(\text{Use Gaussian elimination to obtain}\\ \begin{pmatrix}a\\b\\c\end{pmatrix} = \begin{pmatrix}2 \\5\\-12\end{pmatrix}\\ f(x) =2x^3 +5x^2 -12x\)

First, I'd multiply both sides by 5 and then multiply both sides by (13x - 2y)

Sometimes this is called cross-multiplying. 

the result of cross-multiplying is        (5)(10x - 3y) = (3)(13x -2y)

multiply it out                                       50x -15y = 39x - 6y 

put like unknowns together                  50x - 39x = 15y - 6y

combine/subtract terms                            11x = 9y

divide both sides by y                                11x/y = 9

divide both sides by 11                                x/y = 9/11     the ratio of x to y is 9 to 11