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Exactly. Although, I need general hypothesis because I don't understand at all what the queastion is.

60.

You can either use the minus button or there is a +/- button, as well.

2*(158*839)+672 = 265796

g = 4π√x2+t2

\(g = 4π\sqrt{x^2}+t^2\\ g = 4π*|x|+t^2\\\)

Thast is it there is nothing else you can do with it ://

Hi guest

i is not an imaginary number it stands for interest.

This equaton is for determining the rate of compond interest.

(1+i)^ 15 = 3 is FALSE:

Simplify the following:
(1+i)^15

(1+i)^15 = (1+i) (1+i)^14 = (1+i) ((1+i)^7)^2 = (1+i) ((1+i) (1+i)^6)^2 = (1+i) ((1+i) ((1+i)^3)^2)^2 = (1+i) ((1+i) ((1+i) (1+i)^2)^2)^2:
(1+i) ((1+i) ((1+i) (1+i)^2)^2)^2

(1+i)^2 = 1+0+1 i+0+1 i-1 = 2 i:
(1+i) ((1+i) ((1+i)×2 i)^2)^2

i (1+i) = -1+i:
(1+i) ((1+i) (i-1 2)^2)^2

((i-1)×2)^2 = 2^2 (i-1)^2:
(1+i) ((1+i)×2^2 (i-1)^2)^2

2^2 = 4:
(1+i) ((1+i)×4 (i-1)^2)^2

(i-1)^2 = 1+0-i+0-i-1 = -2 i:
(1+i) ((1+i)×4×-2 i)^2

4 (-2) = -8:
(1+i) ((1+i)×-8 i)^2

i (1+i) = -1+i:
(1+i) (-8i-1)^2

((i-1) (-8))^2 = (-8)^2 (i-1)^2:
(1+i)×(-8)^2 (i-1)^2

(-8)^2 = 64:
(1+i)×64 (i-1)^2

(i-1)^2 = 1+0-i+0-i-1 = -2 i:
(1+i)×64×-2 i

64 (-2) = -128:
(1+i)×-128 i

i (1+i) = -1+i:
Answer: | 128 - 128i
 |

3=(1+i)^15

3^(1/15)-1=i

(〖(-9)〗^3×〖(-12)〗^4×〖(10)〗^(-2))/(〖(15)〗^(-2)×〖18〗^2 )

\(\begin{array}{|rcll|} \hline && \displaystyle \frac{ (-9)^3 \cdot (-12)^4 \cdot 10^{-2} } { 15^{-2} \cdot 18^2 } \\\\ &=& \displaystyle \frac{ (-9)^3 \cdot (-12)^4 \cdot 10^{-2} } { 15^{-2} \cdot 18^2 } \\\\ &=& \displaystyle \frac{ (-9)^3 \cdot (-12)^4 \cdot 15^{2} } { 10^{2} \cdot 18^2 } \\\\ &=& \displaystyle \frac{ (-9)^3 \cdot 12^4 \cdot 15^{2} } { 10^{2} \cdot 18^2 } \\\\ &=& \displaystyle -\frac{ 9^3 \cdot 12^4 \cdot 15^{2} } { 10^{2} \cdot 18^2 } \\\\ &=& \displaystyle -\frac{ 729\cdot 20736 \cdot 225 } { 100 \cdot 324 } \\\\ &=& \displaystyle -\frac{ 729\cdot 20736 \cdot 9\cdot 25 } { 4\cdot 25 \cdot 324 } \\\\ &=& \displaystyle -\frac{ 729\cdot 20736 \cdot 9 } { 4 \cdot 324 } \\\\ &=& \displaystyle -\frac{ 729\cdot 81\cdot 256 \cdot 9 } { 4 \cdot 81\cdot 4 } \\\\ &=& \displaystyle -\frac{ 729\cdot 256 \cdot 9 } { 4 \cdot 4 } \\\\ &=& \displaystyle -\frac{ 729\cdot 256 \cdot 9 } { 16 } \\\\ &=& \displaystyle -\frac{ 729\cdot 16\cdot 16 \cdot 9 } { 16 } \\\\ &=& \displaystyle - 729\cdot 16 \cdot 9 \\\\ &=& \displaystyle - 104976 \\ \hline \end{array} \)

I think the parentheses is just to make sure your  not stupid :D

 6x-5y=9x+5y = x=-(((10*y)/3))

fx. It could just be -(10*y/3)

But just to show that, start by 10*y, then /3 and den just use it in the formel

I don'tknow why there are so many brackets (there aren't any in the solution displayed above the calculator), but, since you have only one equation for two unknowns there are an infinite number of possible values of x and y that satisfy the equation.  You need two independent equations if you expect a unique solution.

Multiplying the width by the height gives you the area of the Rectangular wall 12 ft * 20ft = 240 ft^2

Is this cordinates?

If so? 

https://gyazo.com/f33a5c668ee09bfffdf181bf410a17dd

A=(3,-5)

B=(-5,5)

Yes, sorry im a middleschooler :D Dont know what x₂ is yet

hi Kasper,

Maybe you are talking about 

\(x^2\qquad not \qquad x_2\)??      

The "X2" Means there is a unknown. X

So if we put it in a formel like this one X2=X3-4

We now need to find out what is x.

Anybody can help me explain? Melody!!!!

What is 1.032^ -10

1.032^ -10 = 0.7297986019863001

Its a square, so all you gotta do is multiply to two measures in foot= 12*20= 240

-7/-6 = 7/6

It is usually just the x value for a second point.

So the first point might be

\((x_1,y_1)\)

and the second point would be

\((x_2,y_2)\)

for every 45 points there is 30 points how much is one point

\(\begin{array}{|r|r|} \hline \text{points} & \text{points} \\ \hline 45 & 30 \\ 90 & 60 \\ 135 & 90 \\ \cdots & \cdots \\ x & x\cdot \frac{30}{45} \\ 1 & 1\cdot \frac{30}{45} \\ \hline \end{array} \)

would the square root of -12 be equal to 12i?

\(\begin{array}{|rcll|} \hline && \sqrt{-12} \\ &=& \sqrt{(-1)\cdot 12 } \\ &=& \sqrt{12\cdot (-1) } \\ &=& \sqrt{12}\cdot \sqrt{ -1 } \quad & | \quad \sqrt{ -1 } = i \\ &=& \sqrt{12}\cdot i \\ \hline \end{array} \)

Is it slope intercept form you want?