All Answers 358087 Answers

Subtract 5 and 8y from both sides of the equation. now you will have -2y = 6. divide both sides by -2 to get the answer. y=-3

b*1/b = 1.

On the left hand side of the equation bar, click where it says "scientific" and scroll down till you see "fractions"

1 foot = 12 inches

50 feet * 12 inches/1foot = 600 inches

50 FEET = 600 INCHES

5/x=x+13/6

→5=x^2+13x/6

subtract 5 to both sides

x^2+13x/6-5=0

x=$${\mathtt{\,-\,}}{\frac{{\mathtt{13}}}{{\mathtt{12}}}}{\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{889}}}}}{{\mathtt{12}}}}$$ or $${\frac{{\sqrt{{\mathtt{889}}}}}{{\mathtt{12}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{13}}}{{\mathtt{12}}}}$$

0.3020833333=94400961/312499733

I assumed any real solution.

I think mine is right

Max 33.     Min 1

x^x=2.  How to solve this equation?

$$\small{\text{$
\begin{array}{rcl}
x^x &=&2 \qquad | \qquad \ln{()}\\
x\ln(x) &=& \ln{(2)} \qquad | \qquad x = e^{ln{(x)}}\\
e^{ln{(x)}}\cdot \ln{(x)} & =& \ln{(2)} \qquad | \qquad z = \ln{(x)} \\
e^z\cdot z &=& \ln{(2)}\\
\end{array}
$}}\\\\
\small{\text{
If we have $\boxed{~a=z\cdot e^z~}$,
the solution is $\boxed{~z=W(a)~}$. and $\boxed{~x=e^z~} $}}\\
\small{\text{
In mathematics, the Lambert W function, is also called the omega function or product logarithm
}}$$

$$\small{\text{$
\begin{array}{lrcl}
\text{We have }& z &=& \ln{(x)}\\
\text{and } & e^z\cdot z &=& \ln{(2)}\\
\text{so } & z &=& W(\ln{(2)})\\
\text{and } & \ln{(x)}&=& W(\ln{(2)})\\\\
& e^{\ln{(x)}} &=& e^{W(\ln{(2)})}\\
\text{finally }& x &=& e^{W(\ln{(2)})}\\
\text{also } & \mathbf{ x }& \mathbf{=} & \mathbf{\dfrac{\ln{(2)}}{W(\ln{(2)})} }
\end{array}
$}}\\\\$$

$$\small{\text{ Numerical Evaluation:}}\\ \small{\text{ The $W$ function may be approximated using Newton's method, with successive approximations to}}\\ \small{\text{ $z=W(a)$ (so $a=ze^z$) being $z_{j+1}=z_j-\dfrac{z_j e^{z_j}-a}{e^{z_j}+z_j e^{z_j}}$. }}\\
\small{\text{We start the Iteration with $z_0 = 1$}}\\$$

$$\small{\text{$
\begin{array}{rclcrcl}
z_1 &=& 1 - \dfrac{ 1\cdot e^1 - \ln{(2)} }{e^1+ 1\cdot e^1 } = 0.62749729872 &\quad & x_1 &=& \dfrac{ \ln{(2)} } {z_1} = 1.10462177603\\\\
z_2 &=& z_1-\dfrac{z_1 e^{z_1}- \ln{(2)} }{e^{z_1}+z_1 e^{z_1}} = 0.24045491796 &\quad & x_2 &=& \dfrac{ \ln{(2)} } {z_2} = 2.88264921519\\\\
z_3 &=& z_2-\dfrac{z_2 e^{z_2}- \ln{(2)} }{e^{z_2}+z_2 e^{z_2}} = 0.48596644327 &\quad & x_3 &=& \dfrac{ \ln{(2)} } {z_3} = 1.42632724987\\\\
z_4 &=& z_3-\dfrac{z_3 e^{z_3}- \ln{(2)} }{e^{z_3}+z_3 e^{z_3}} = 0.44585119296 &\quad & x_4 &=& \dfrac{ \ln{(2)} } {z_4} = 1.55466036989\\\\
z_5 &=& z_4-\dfrac{z_4 e^{z_4}- \ln{(2)} }{e^{z_4}+z_4 e^{z_4}} = 0.44443778365 &\quad & x_5 &=& \dfrac{ \ln{(2)} } {z_5} = 1.55960452971\\\\
z_6 &=& z_5-\dfrac{z_5 e^{z_5}- \ln{(2)} }{e^{z_5}+z_5 e^{z_5}} = 0.44443609102 &\quad & x_6 &=& \dfrac{ \ln{(2)} } {z_6} = 1.55961046945\\\\
z_7 &=& z_6-\dfrac{z_6 e^{z_6}- \ln{(2)} }{e^{z_6}+z_6 e^{z_6}} = 0.44443609102 &\quad & x_7 &=& \dfrac{ \ln{(2)} } {z_7} = 1.55961046946\\\\
\end{array}
$}}$$

x = 1.55961046946

2x + 3y + 4z = 100            ...(1)

x + y + z = 35                   ...(2)

Multiply (2) by 2 and subtract from (1) to get  y = 30 - 2z    ...(3)

Multiply (2) by 3 and subtract (1) from the result to get x = 5 + z   ...(4)

I'll assume you want integer solutions, and that "at least over 1" means 2 or more (you can modify the following appropriately if you mean something else).

Then, using (3), z = 2 means y = 26 (which is greater than 1, so ok),  and y = 2 means z = 14, so z varies from 2 to 14.

From (4), because x is a linear function of z, we have minimum(maximum) of x corresponds to minimum(maximum) of z, so

minimum x = 5+2 = 7

maximum x = 5+14 = 19

. 

Max=33

Min=1

Assuming that you want to simplify 3c-2cd-8c+7d, I believe the answer is -5c-2cd+7d

No problem.  We all make mistakes!  At least yours led to an interesting few analyses!

.

Not as it stands, but it looks like it's a rounded up version of the fraction 9/17, which is a repeating decimal:

9/17 = 0.5294117647058823529411764705882352941176470588235294117647058823....

.

I've made a mistake.

Sorry, and thanks for correcting my error!!!

if you add 2 to 200, you get 20009.

Multiply 4, and you will get 80036

That's the answer to a different question!  

In your original question you had (a - 1)x.  In your latest post you have (a - i)x.

.

The answer was actually 6 or -6

(2+i)x^2+(a-i)x+4-2i=0

→(2x^2+ax+4)+(x^2-x-2)i=0

x=2 or -1

substitute 2 in 2x^2+ax+4=0

8+2a+4=0, a=-6

substitue -1 in 2x^2+ax+4=0

2-a+4=0, a=6

So a is 6 or -6

a = 3428,  b = 1   (assuming you want integer values)

.

200+2(4)=208

If you mean ( X + 4 )^2, is a notable product

802 ???

Maybe??

If he has n friends then and they each eat a fraction, f, of the pizza then n*f + (5/4)*f = 1 so

f = 1/(n + 5/4)

You need to specify the number of friends to get any further!

.

.

Thanks Alan :)