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Pic...? I never posted a pic??

Ah, yeah, my friend just helped me figure it out right now! It's a quadratic equation/parabola and I just watched a video to finish it. The equation is y = 3x^2 - 3x + 12 if you were wondering. ^^

So I should plug in the options? And see which is closet to 0?

Each of the following possibilites has 6 different arrangements. We don't need to consider this to get the final answer.

If the largest number is 3, there is 1 possibility: (1,2,3).

If the largest number is 4, there are 3 possibilities: (1,2,4), (1,3,4), (2,3,4).

If the largest number is 5, there are 6 possibilities: (1,2,5), (1,3,5), (1,4,5), (2,3,5), (2,4,5), (3,4,5).

If the largest number is 6, there are 10 possibilities.

If the largest number is 7, there are 15 possibilities.

If the largest number is 8, there are 21 possibilities.

If the largest number is 9, there are 28 possibilities.

If the largest number is 10, thre are 36 possibilities.

To find the total number of possibilities, add: 1 + 3 + 6 + ... + 36  =  sum1

Now, we need to find the sum of the value times the number of possibilities:

    1 x 3  +  3 x 4  +  6 x 5  +  10 x 6  + ... +  36 x 10   =  sum2

To find the expected value, divide  sum2  by  sum1.

I like geometry homework too!

Find all rational zeros of f. Then​ (if necessary) use the depressed equation to find all roots of the equation

​f(x)equals=0.

f(x)=6x^3-7x^2-6x-1=0

Hello mharrigan920!

\(6x^3-7x^2-6x-1=0\)

\(x^3-\frac{7}{6}x^2-x-\frac{1}{6}=0\) 

\(x=-\frac{1}{3}\ ?\)      supposed (graph).

\((x^3-\frac{7}{6}x^2-x-\frac{1}{6}):(x+\frac{1}{3})\) = \(x^2-\frac{9}{6}x-\frac{1}{2}\)

\(\ \underline{x^3+\frac{1}{3}x^2}\)

       \(-\frac{9}{6}x^2\ -x\)

       \(\underline{-\frac{9}{6}x^2-\frac{1}{2}x}\)

                   \(-\frac{1}{2}x-\frac{1}{6}\)

                   \(\underline{-\frac{1}{2}x-\frac{1}{6}}\)

                    \(- - - -\)

\(x_1=-\frac{1}{3}\)

\(x^2-\frac{3}{2}x-\frac{1}{2}=0\\ x=-\frac{p}{2}\pm\sqrt{(\frac{p}{2})^2-q}\\ x=\frac{3}{4}\pm \sqrt{\frac{9}{16}+\frac{1}{2}}\)

\(x=\frac{3}{4}\pm \sqrt{\frac{17}{16}}\)

\(x_2=\frac{3-\sqrt{17}}{4}\approx -0,280776\\ x_2=\frac{3+\sqrt{17}}{4}\approx 1,780776\)

  !

I looked at desmos and I don't think that there is a standard format linear equation modeled by these points. Maybe they want an exponential function?

How many seconds?

First, find the area of the circle using:  Area  =  pi · r²

Since there are 360^o in a circle but you only want a sector of size 48^o,

multiply the answer that you got for the area of the circle by the fraction (48/360).

I'm assuming that you mistyped the problem:

4^x₁  =  5     --->     log(4^x₁​)  =  log(5)     --->     x₁ · log(4)  =  log(5)     --->     x₁  =  log(5) / log(4)

5^x₁  =  6     --->     log(5^x₁​)  =  log(6)     --->     x₁ · log(5)  =  log(6)     --->     x₁  =  log(6) / log(5)

6^x₁  =  7     --->     log(6^x₁​)  =  log(7)     --->     x₁ · log(6)  =  log(7)     --->     x₁  =  log(7) / log(6)

127^x₁  =  128     --->     log(127^x₁​)  =  log(128)     --->     x₁ · log(127)  =  log(128)     

                          --->     x₁  =  log(128) / log(127)

[ log(5) / log(4) ] · [ log(6) / log(5) ] · [ log(7) / log(6) ] · ... · [ log(128) / log(127) ]  =  log(128) / log(4) 

          =  log(2⁷) / log(2²)     =      [ 7 · log(2) ] / [ 2 · log(2) ]  =  7/2

thank you!!

There will be an infinite number of solutions if the two equations are identical.

So multiple the second one by 2 the equate coefficients.

I'm sorry, I can't find one... maybe someone else can?

a₀  =  6          a_n  =  a_n-1 / ( 1 + a_n-1 )

a₁  =  6 / ( 1 + 6 )  =  6/7

a₂  =  (6/7) / ( 1 + 6/7 )  =  (6/7) / (13/7)  =  6/13

a₃  =  (6/13) / ( 1 + 6/13 )  =  (6/13) / (19/13)  =  6/19

Notice that the numerator is always 6 and the denominator is 6 times the term's subscript + 1.

So for the fourth term, the numerator will be 6 and the denominator will be 6 x 4 + 1  =  25.  a₄  =  6/25.

Use this to find a₁₀₀.

Answered a question like this the other day....plot all of the given points and come back with your graph for more help....

Yeah, I saw that too! What I don't understand is how to write an equation in standard format for it..? All I could come up with for it was an equation based off of previous answers and it wasn't a y = mx + b equation, it was an equation for a sequence(?)...

The choices are  π/18,π/9,π/6,2π/9,7π/18,4π/9,5π/9,10π/9,11π/9,16π/9,17π/9

Normal speed = x

xt = 200      t = 200/x

faster speed    (x+10 )    new time = t-1   = 200/x -1

(x+10)(200/x -1) = 200 miles

200 -x + 2000/x -10 = 200

-x  + 2000/x -10 = 0

-x^2 + 2000 - 10 x = 0       via quadratic formula   x = 40 m/hr

Hey! 

I'll first convert hese to co-ordinates: (1,12), (2,18), (3,30)

If you look at the pattern carefully you can see that you add 6 more to each y co-ordinate each time... the first time you add 6, the second time you add 12... and I imagine it continues like this!

See original answer above...I edited it .....   

https://web2.0calc.com/questions/please-help-must-get-dinner

a = k/b^3      sub in a = 9/2    and b = 2   to find k

9/2 = k / 8

k = 36

a = 36/b^3   

   = 36/3³ = 36/27 = 1  1/3

Thank you!