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i thought they wanted me to set it euqal to each other like 5 = 5(x - 1/2)^2  (x-1)^2 + 5

then multiply it out the right side but i cant manage to do that. when plugging it into a calculator i get different multiplication than waht i did and they want me to pay to show the steps on how they were able to figure it out. The answers were 1 and .5 but i am unable to get these results.

"Your battery is not draining evenly. In the first hour it drained 19%. But, in the next 3 hours it drained 81% - 24% =57% ." !!!!

57% in 3 hours is 19% per hour.

35^1723-16^1723

\(35^{1723}-16^{1723}\;\;(mod \;6)\\ =(-1)^{1723}-(-2)^{1723}\;\;(mod \;6)\\ =-1--(2)^{1723}\;\;(mod \;6)\\ =-1+2\;\;(mod \;6)\qquad \text{(2 to the power of any odd interger, mod 6, is 2)}\\ =1 \;\;(mod\; 6) \)

For f(x) to equal 5 the term 5(x - 1/2)^2  (x-1)^2 must be zero.  Look carefully at this term.  What two values of x will make it zero?

To solve this problem, we can use a Triangle similarity theorem or trigonometry. 

4m/12m = 2m/x      x = 6

Nick's shadow is 6m long

Lamppost's shadow is 18m long   

Rationalize the denominator

Hello Guest!

\(\frac{5\sqrt{3}-4 \sqrt{2}}{4\sqrt{3}+3\sqrt{2}}\)

\(\sqrt{2}=a,\ \sqrt{3}=b\)

\(=\frac{5b-4a}{4b+3a}{\color{blue}\times\frac{4b-3a}{4b-3a}}=\frac{20b^2-15ab-16ab+12a^2}{16b^2-9a^2}\)

\(=\frac{20b^2-31ab+12a^2}{16b^2-9a^2}\)

\(a=\sqrt{2},\ b=\sqrt{3}\)

\(=\frac{20\cdot 3-31\sqrt{2\cdot 3}+12\cdot 2}{16\cdot 3-9\cdot 2}\)

\(=\frac{84-31\sqrt{6}}{30}=0.26886\)

  !

area of ΔPEF + area of ΔPGH = area of ΔPFG

Area of trapezoid EFGH = (8 + 21) * 2 = 58 u²  

this was my post but i wasnt signed in so thank john

40g/cm^3

The sum of pentagon's interior angles is 540º

540º / 5 = 108º

∠ x = ( 180º - 108º ) / 2  

That will be the highest power of x

Hint:

x^2* x^4 = x^6

Why have you answered the question if the answer is known?

btw, your answer is correct!!!

The sum of angles in a pentagon is 540 degrees. Therefore 540/5= 108 degrees in each regular angle. We know there is only one way to draw a line from one point to the other if it is straight, so angle x is valid. We can split this into 2 congruent triangles, one not so congruent. Draw line EB, and get triangle EAB... and I will name angle D measure y. Therefore x+y+108=180, and x+y=72. We want to know what y is. Angle EBA=x, so angle AEB=y and angle A is 108. Also, AED= 108-y, and

OK so first of all, WHY are you asking web2.0calc community to solve a problem you already KNOW???

The answer is:

Call a side a. Then 12a=a^3. Therefore a= 2 sqrt 3 or sqrt 12. So the answer is 2 sqrt 3...! (I thinnkkkkk)

That figure is a parallelogram+trapezoid. We shall calculate the trapezoid's area first! For the area of a trapezoid, 12+13=25, and 25+31=56/2=28, then times 11= 308! 

Figure 1: Trapezoid: A=308

                                              +

Figure 2: Parallelogram: A= We dont know yet

For the area of a parallelogram, base times height=12*14=168. YAY!!!

308+168= 476!

Your answer is 476!!!

I think

\(z^3=-2(1+i) \)

just thinking about 1+i and where it lays on the complex plane.

I can see that the distance from the origin is sqrt2   and the angle with the positive x axis is pi/4

so this can be written as 

\(z^3=-2*\sqrt2 * e^{\frac{\pi}{4}i}\\ z^3=-2^{\frac{3}{2}} * e^{\frac{\pi}{4}i}\\ z=-2^{\frac{3}{2}\cdot \frac{1}{3}} * e^{\frac{\pi}{4}i\cdot\frac{1}{3}}\\ z=-\sqrt{2} e^{\frac{\pi}{12}i}\\ \)

This is only one of the 3 cube roots, It is in the 3rd quadrant 

the other two have angles of  

\(\frac{\pi}{12}+\frac{2\pi}{3}=\frac{\pi}{12}+\frac{8\pi}{12}=\frac{9\pi}{12} =\frac{3\pi}{4} \\ and\\ \frac{\pi}{12}+\frac{4\pi}{3}=\frac{\pi}{12}+\frac{16\pi}{12}=\frac{17\pi}{12} \)

4th quad solution:

\(z=-\sqrt{2} e^{\frac{3\pi}{4}i}\\ z=-\sqrt{2} [cos(\frac{3\pi}{4})+isin(\frac{3\pi}{4})]\\ z=-\sqrt{2} [-\frac{1}{\sqrt2}+\frac{i}{\sqrt2}]\\ z=1-i \)

LaTex:

z^3=-2*\sqrt2 * e^{\frac{\pi}{4}i}\\
z^3=-2^{\frac{3}{2}} * e^{\frac{\pi}{4}i}\\
z=-2^{\frac{3}{2}\cdot \frac{1}{3}} * e^{\frac{\pi}{4}i\cdot\frac{1}{3}}\\
z=-\sqrt{2} e^{\frac{\pi}{12}i}\\

Thank you so much ilorty!

I got 14 for the first part, is that correct?

Yes,

CPhill has not been active for over a month and hence he is not automatically included on the Top User board.

However, if you click on the arrow at the bottom of the board he will reappear.

Since you might not be able to see asinus's answer, (as it is written in raw LaTeX, probably another glitch), let me help write it out in pure letter and number form. We see that both 12 and 5 share a 6. So divide both by 6, and we see that the coefficient is 2/5. The only two that satisfies that is A and B. We see that 1 of the a's cancel out, which leaves us with a squared. Therefore, the only possible answer is A. 
 

if you want a more 'mathematical' answer, check out asinus's post (which is also great!). Maybe you are smart enough to crack Tex code!

:)

Sorry, I didn't see that has only "10" coins! Forget about the above list. I can only see 3 combinations:

1  = 1   9   0  
2  = 4   5   1  
3  = 7   1   2  
 

Find the greatest common factor (GCF)

\(9x^3y\ and\ 15x^2y^2\)

Hello Guest!

\(9x^3y=3^2\cdot x^3\cdot y\\ 15x^2y^2=3\cdot5\cdot x^2\cdot y^2\)     \(GCF = 3\cdot x^2\cdot y\)        (D)

\(\frac{9x^3y}{15x^2y^2}=\frac{9x^3y\color{blue}/3x^2y}{15x^2y^2\color{blue}/3x^2y}=\color{blue}\frac{3x}{5y}\)

  !

       n   d    q

1  = 9   5   0  
2  = 7   6   0  
3  = 5   7   0  
4  = 3   8   0  
5  = 1   9   0  
6  = 10   2   1  
7  = 8   3   1  
8  = 6   4   1  
9  = 4   5   1  
10  = 2   6   1  
11  = 0   7   1  
12  = 9   0   2  
13  = 7   1   2  
14  = 5   2   2  
15  = 3   3   2  
16  = 1   4   2  
17  = 4   0   3  
18  = 2   1   3  
19  = 0   2   3