All Answers 358797 Answers

1/x=5/x-4

If you mean

$$\\\frac{1}{x}=\frac{5}{x}-4 \qquad then\\\\
\frac{1}{x}-\frac{5}{x}=-4\\\\
\frac{1-5}{x}=-4\\\\
\frac{-4}{x}=-4\\\\
\frac{4}{x}=4\\\\
\frac{4}{x}=4\\\\
x=1$$

Where's your diagram, Rodjeni?

The one who knocked off the insured mark takes the greatest risk, he/she should take the least monetary hit.

But i've found x;exactly like melody did.is that wrong to find only x?the task is to solve the equation if its possible.so how can i know what do i need to find-x or a

Why would anyone not like maths???

$$\\\text{Do you mean }M^2\times M^5\text{ ?}\\\\\text{If so, this is }M^{2+5}\rightarrow M^7$$

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When it comes to popcorn-caramel

when it comes to soda-Dr. Pepper

list of sodas (from greatest to least favorite):

-Dr. Pepper

-7 UP

-Fanta

-Sprite

-Coca Cola

-Pepsi

-Sierra Mist

I cant join Rosala or Rome for the Battle Of Popcorn, but I can join Rosala for the Soda Wars

yea, its about time Rome loses another battle

I did it as follows:

$$\\\frac{1}{ax}-\frac{a}{x+a-1}=0\\\\1-\frac{a^2x}{x+a-1}=0\\\\x+a-1-a^2x=0\\\\\text{ or }\\\\xa^2-a+1-x=0\\\\\text{Now I just used the quadratic formula to get }\\\\a=\frac{1\pm \sqrt{1-4x(1-x)}}{2x}\\\\a=\frac{1\pm(2x-1)}{2x}\\\\a=1 \text{ and }a=\frac{1}{x}-1$$

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178/8 can be reduced to 89/4.

1.8 + 0.7 can be reduced to 2.5 which can be written as 5/2.

(89/4) / (5/2)  =  (89/4) x (2/5)                  (change from division to multiplication)

                        =  (89/2) x (1/5)                  (cancel the 2s)

                        =  89/10  or 8.9

3xy² - 48x

Factor out the common terms, 3 and x:

3x(y² - 16)

Look at what remains inside the parentheses, a difference of squares:

3x(y - 4)(y + 4)

Thanks a lot!!!

Distribute the 6y: (6y*8y)-(6y*4)+22y

48y^2-24y+22y

48y^2-2y.

178/8=22.25.

1.8+0.7=2.5

So, the answer becomes 22.25/2.5. 

$$\\y=\sqrt{4-x}-3\qquad x\le4 \qquad y\ge-3 \\\\
y+3=\sqrt{4-x}\\\\
(y+3)^2=4-x\\\\
x=4-(y+3)^2\\\\
x=4-(y^2+6y+9)\\\\
x=4-y^2-6y-9\\\\
x=-y^2-6y-5\\\\
Inverse\;\;function\\\\
y=-x^2-6x-5 \qquad where \;\;\; x\ge-3\\\\$$

Here is the graph

$${\frac{\left({\frac{{\mathtt{178}}}{{\mathtt{8}}}}\right)}{\left({\mathtt{1.8}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.7}}\right)}} = {\frac{{\mathtt{89}}}{{\mathtt{10}}}} = {\mathtt{8.9}}$$

$$\\(\frac{1}{4})^{3/2}\\\\
=\frac{1}{4^{3/2}}\\\\
=\frac{1}{(4^{1/2})^3}\\\\
=\frac{1}{(\sqrt{4})^3}\\\\
=\frac{1}{(2)^3}\\\\
=\frac{1}{8}\\\\$$

Yep....negative numbers follow the convention of positive ones. That is....  -1 is "odd"....-2 is "even"....etc.

Beginning Balance + Deposits - Withdrawals (checks)  = Final Balance......so.....

$425  + $203 - $234 = $394

(4x -1)^2 = 11    expand the left side....

16x^2 - 8x + 1   = 11      subtract 11 from both sides

16x^2 - 8x - 10   = 0      divide through by 2

8x^2 - 4x - 5   =  0      this does not factor...using the on-site solver, we get.....

$${\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{5}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{11}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{11}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{4}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.579\: \!156\: \!197\: \!588\: \!85}}\\
{\mathtt{x}} = {\mathtt{1.079\: \!156\: \!197\: \!588\: \!85}}\\
\end{array} \right\}$$

Note......we could have also used the square root property to solve this, if we wished.....!!!!!

The distance run in one "rep"  =  2* (30 / sin 35) = about 104.606807737274288 m

So....25 "reps"  =  25 * 104.606807737274288  =  about 2615.2 m

Of course your special! A special virus that can affect my whole Kingdom! You know you villains are really special to us queens and kings and heroes as becoz of you guys  we are what we are! A big salute,......continue being bad!!!!!

Oh yeah! So that means I can count another vote on my pepper and butter popcorns.....you definitely have to give it now tin!!!

Thank you, Alan!

M is the midpoint of AB. The cordinates of A Are (-2,3) and the cordinates of M are (1,0). Find the cordinates of B

$$\small{\text{$
\begin{array}{rcl}
\vec{M}&=&\dfrac{1}{2}(\vec{A}+\vec{B})\\\\
2\cdot \vec{M}&=&\vec{A}+\vec{B}\\\\
2\cdot \vec{M}-\vec{A}&=&\vec{B}\\\\
\vec{\mathbf{B}}& \mathbf{=} &\mathbf{ 2\cdot} \vec{\mathbf{M}} \mathbf{ -} \vec{\mathbf{A}} \\\\
\vec{B} &=& 2\cdot \binom10 - \binom{-2}{3}\\\\
\vec{B} &=& \binom20 - \binom{-2}{3}\\\\
\vec{B} &=& \binom{2+2}{0-3}\\\\
\vec{\mathbf{B}} &\mathbf{=}& \mathbf{\binom{4}{-3}}\\\\
\end{array}
$}}$$

 B  is (4,-3)