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Never mind, after experimenting a bit, I figured out the answer.  

Essentially, I started by assuming \(F_c=F_{b+1}.\)  Thus, \( F_c-F_b = F_b+F_{b-1}-F_{b} = F_{b-1}\)

Knowing that Fa, Fb, and Fc form an arithmetic sequence, we can use properties of Fibonacci numbers to write Fa, Fb, and Fc in terms of

Fb, and use the fact that a+b+c=2000 to find what one of them is.  The rest is simple.  

Fa=F665, a=b-2, c=b+1

I know I used LaTeX at the start, and it looks clear, but it was too hard to write quickly.  

do u mean what is j?

After typing in the answer, the website said it was incorrect.  If you gave the method, it would be greatly appreciated though.  

I used a computer program, and found a = 624.

can you give more differeneces?

Well, a fraction is just a number over another number (which is the numerator divided by the denominator). And percents is basically a fraction with the denominator being 100. 21/100 = 21%

The ratio of the number of magnets Aspilla and Barry had was 2 : 3. When Aspilla bought 4 more magnets and Barry bought 13 more magnets, the ratio became 1 : 2. Find the number of magnets Barry had in the end.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

A : B = 2 : 3

(2x + 4) : (3x + 13) = 1 : 2

4x + 8 = 3x + 13

x = 5

Barry had 28 magnets.

By same side interior angles, 180 - 120 = 60. 

=^._.^=

2(-8) + 1 = -15

2(4) + 1 = 9

So our answer is [-15, 9]

=^._.^=

This is a extremely similar problem as what someone else had already posted ---> https://tinyurl.com/ce7hdrts

There are 7 different products possible

16A + 4A + A = 3b + 3

If A = 1, b = 6. 

If A = 2, b = 13. 

If A = 3, b = 20

3 + 20 = 23

=^._.^=

1/7 of 35 = 5

So, the concentration of krill would drop to 30/m³

Thanks! I totally overlooked that.

Pyhatgorean Therom,

13^2 = 12^2 + x^2

169 = 144+x^2

x^2 = 25

x = 5

Dieter had 3 times as many flags in her collection as Andrew. When Dieter gave 8 flags to Andrew, each of them had exactly the same amount. How many flags did they have altogether?

Hello Guest!

Andrew has x flags at the beginning. Then the following applies:

\(3x-8=x+8\ |\ -x+8\\ 2x=16\ _\ | :2\\ \color{blue} x=8\\ \color{blue} 3x=24\\\color{blue} x+3x=32\)

Dieter and Andrew together have 32 flags.

  !

Professor Crump maintains...

Hello Guest!

\(\dfrac{35}{m^3}-\dfrac{1\cdot 35}{7\cdot m^3}=\dfrac{6\cdot 35}{7\cdot m^3}\color{blue}\approx \dfrac{30}{m^3}\)

After the whales have passed through, the concentration is 30 / m³.

  !

The real numbers x and y satisfy \((x - 3)^2 + (y - 4)^2 = 18\).

Find the minimum value of \(x^2 + y^2\).

\(\text{Let the origin $O=(0,0)$ } \\ \text{Let the center of the circle $C=(3,4)$ } \\ \text{Let the radius of the circle $r=\sqrt{18}=3\sqrt{2}$ } \\ \text{Let $x^2+y^2=r_{\text{min}}^2$ }\)

1. Distance between origin and center of the circle:

\(\begin{array}{|rcll|} \hline \overline{OC} &=& \sqrt{(0 - 3)^2 + (0 - 4)^2} \\ \overline{OC} &=& \sqrt{9+16} \\ \overline{OC} &=& \sqrt{25} \\ \mathbf{\overline{OC}} &=& \mathbf{5} \\ \hline \end{array}\)

2. \(x^2+y^2 = r^2_{\text{min}}\)

\(\begin{array}{|rcll|} \hline r_{\text{min}} &=& \overline{OC} - r \\ r_{\text{min}} &=& 5 -3\sqrt{2} \\ \hline x^2+y^2 &=& r_{\text{min}}^2 \\ x^2+y^2 &=& \left(5 -3\sqrt{2}\right)^2 \\ \mathbf{ x^2+y^2 } &=& \mathbf{ 43 -30\sqrt{2} } \\ \text{The minimum value of }~\mathbf{ x^2+y^2 } &=& \mathbf{ 0.5735931288 } \\ \hline \end{array}\)

How did you find the height(5)?

https://web2.0calc.com/questions/math_84283

Given

t = 1/3 f     

t = 2 dol bills    f = 5 dol bills

value:

    2 t +   5f   = 136        sub in the top eqution

     2(1/3f) + 5f = 136

         f = 24   five dollar bills       ...... you should be able to finsih from here !

4/5 is blue    then   1/5 is white

1/5 * 7.9   =    7.9 / 5   =   79/50   as a fraction 79/50 = 1.58 pints of white

3)  x^2 -2x    + y^2  + 6y   =  6     Arrange into standard circle equation form

     (x-1)^2      +      ( y+3)^2 = 6 +1 + 9           

                                                6+ 1 + 9 = 16  = r^2

     Area of a circle = pi r^2 = pi (16)  = 16 pi  units2

The ratio of the number of magnets Aspilla and Barry had was 2 : 3.
When Aspilla bought 4 more magnets and Barry bought 13 more magnets,
the ratio became 1 : 2.
Find the number of magnets Barry had in the end.

\(\begin{array}{|rcll|} \hline \dfrac{A}{B} &=& \dfrac{2}{3} \\\\ \mathbf{A} &=& \mathbf{ \dfrac{2}{3}B } \\ \hline \dfrac{A+4}{B+13} &=& \dfrac{1}{2} \quad &|\quad \mathbf{A= \dfrac{2}{3}B } \\\\ \dfrac{\dfrac{2}{3}B+4}{B+13} &=& \dfrac{1}{2} \\\\ 2*\left(\dfrac{2}{3}B+4 \right) &=& B+13 \\\\ \dfrac{4}{3}B+8 &=& B+13 \\\\ \dfrac{4}{3}B-B &=& 5 \\\\ \dfrac{1}{3}B &=& 5 \quad &|\quad *3 \\\\ \mathbf{B} &=& \mathbf{15} \\ \hline A &=& \dfrac{2}{3}B \quad & | \quad \mathbf{B=15} \\\\ A &=& \dfrac{2}{3} * 15 \\\\ \mathbf{A} &=& \mathbf{10} \\ \hline \end{array}\)

The number of magnets Barry had in the end is \(B + 13 = 15+13 = \mathbf{28}\)