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When y = - 2, the result =29

When y = - 32 the result = 61

Abs[ - 2 + - 32] = 34

4 , 13 , 22 , 31 , 40 , 103 , 112 , 121 , 130 , 202 , 211 , 220 , 301 , 310 , 400 , Total =  15 such pages.

You just missed 202 and 220.

As follows:

https://web2.0calc.com/questions/helpelp#r3

https://web2.0calc.com/questions/help_60614#r2

Current total:  t = n*90

next score = s

New average   av = (t + s)/(n + 1)   

So:  91 = (n*90 + s)/(n + 1)

Rearrange to find s.  You will need to know n if you want an actual number for s.

You must minimise the function. To do this, you must allow the derivative of the function to be equal to zero:

\(f(x) = 2x^2 -10x + 15\)

\(f'(x) = 4x - 10\)

Now let the fucntion equal zero and find the value of x at this point:

\(0 = 4x - 10\)

\(10 = 4x\)

\(10/4 = 2.5 = x\)

Sub this back into the initial function to find the lowest point of this parabola:

\(f(2.5) = 2(2.5)^2 - 10(2.5) + 15\)

This coincidentally comes to a minimum of 2.5

Hope this is right :)

I would like to know how these are done though.

Can anyone show me?

The largest value of n is 85.

The answer is (-1/2,3/7).

The vectors are parallel to k = 2.  The vectors are perpendicular for k =-5.

From the Fibonacci and Lucas formula, c = (7 + 2*sqrt(5))/2.

1. The remainder works out to 2x^2 - 4x + 3.

2. Since f(x) is divisible by x^3, f(x) is of the form ax^5 + bx^4 + cx^3.

You then want ax^5 + bx^4 + cx^3 + 2 to be divisible by (x + 1)^3.  Using long division, you get the equations

-10a  + 6b - 3c = 0

4a - 3b + 2c = 0

-a + b - c + 2 = 0

==> a = 6, b = 16, c = 12

So f(x) = 6x^5 + 16x^4 + 12x^3.

A circle has diameter 24, and arc AB subtends a central angle of 60 degrees.  Find the length of arc AB.  

All the way around a circle is 360^o so 60^o is 1/6 of the way around. 

The circumference of this circle is 24 π  so 1/6 of that is 4 π  or approximately 12.57. 

_.

x=−1+i√3, −1−i√3

I used mathway, I hate imaginary numbers, lol.

Yes it is great to see you here DS.

I did not say it was easy, I said parts  of it were easy and I gave a start point

this is exactly what i said.

Try looking at each scenario separately,

0 blue, 6 green

1 blue, 5 green

etc

Some are easy some  will be hard.

Show us what you can do on your own.

Did guest show what he could do on his own?

OR did you just race in and give him an answer?  (which may or may not be correct)

I should not sound too negative about it, if the question inspired you to practice or learn something then your effort is of value (to you)

Your answer only has cheat value to the asker however.

geno3141, to answer the original question, the total overall discount would be 37%, or $37

You can work out the radius first.

then use the equation of a circle centre (h,k) formula    (you can google it if you do not know it)

Sub in the values of h, k, and r   to get the formula for the circle.

Rearrange to find AB and C

Ask questions if you want but show evidence that you have done something for yourself first.

This is a teaching answer, do not make it valueless by posting after it.

hint:

look into Vieta's theorem.

I have answered the question.

We have the equation: \(5x^2+kx=4\) we also know that one root is 2. This means that if we plug in 2 for x the equation should be satisfied.

If we do that we get: \(5(4)+2k=4\) this is simplified to \(k=-8\).

We now plug in the value of k and get: \(5x^2-8x=4\) or \(5x^2-8x-4=0\).

Factoring the quadratic (use the fact that we know that 2 is a root to make it easier) gives us \((x-2)(5x+2)\)

The other root is: \(\boxed{-5/2}\)

Hope you can understand this. :P