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12.25

C==Chocolates,    S==Stawberry

C + S ==T.................................(1)
[C - 10] / [T - 10]==2/7.............(2)
C / [T - 18] ==1/3.....................(3), solve for C, S, T

Use substitution to get:
C==86 - original number of chocolate cupcakes.
S ==190 - original number of strawberry cupcakes.

When x = 0     you will find the   y - axis intercept

 a (-3)(2) = -6 a   = y

when y = 0    you will find the x-axis intercept

0 = a (x-3)(x+2)        x = 3    and -2       ( or  a = 0   , but we will ignore that...we know this is a parabola) )

Try:     y = -6a    x = 3

1/2  ( -6a)(3) = 12

   -6a = 8

        a = - 6/8      but this would result in downward opening parabola    ( there would be a negative x^2 coefficient)

Try:   y = - 6a    x = - 2

1/2 (-6a)(-2) = 12

a = 2           

Here is a graph......note we are looking at the LEFT side of the parabola.....

     https://www.desmos.com/calculator/kot9e2fc8v                 

sumfor(n, 1, infinity, 6 / ((2*n + 1)^2 - 1))==converges to 3/2 ==1.5

The steps to solve an infinite sum are as follows:

1. Generalize the sum with a variable.  example: 1/(n(n+1))

2. Partial fraction decomposition

3. Plug in values, of n=1, 2, 3

4. Notice the pattern and cancel out the terms

Answer deleted; someone else answered the question.

Use this formula to find the PV of 360 payments of $800 each at 8% compounded monthly:

PV=?; P=800; R=0.08/12; N=30*12;    PV=P*(((1 + R)^N - 1)*((1 + R)^-N)* R^-1)

PV==$109,026.80 - amount of mortgage you can afford.

$800  x  360 months ==$288,000 principal + interest paid to the loan company.

$288,000  -  $109,026.80 ==$178,973.20 - interest paid over 30 years.

"scale factor" is one answer.

For n = 1:  a₁² - a₀·a₂  =  (-1)¹   --->   (5)² - 1·a₂  =  -1

                                                   --->        25 - a₂  =  -1

                                                                     - a₂  =  -26

                                                                       a₂  =  26

For n = 2:  a₂² - a₁·a₃  =  (-1)²   --->   26² - 5·a₃  =  1

You'll need to finish ...

What you have entered    sqrt ( 11,0)      is    this :   \(\sqrt[0]{11}\)

   which would be     11^1/0       and we all know you cannot divide by 0 

I don't get what you want your argument to be.

If you are taking the square root of a coordinate point, you have do individually root it for the x argument and the y argument.

If you mean 11.0, just type 11. 

In general, your system is interpreting your answer as the sqaure root of a coordinate point, which is obviously invalid.

Go to x = 2     and draw a vertical line.....   f(2) is the 'y' value where this line crosses the graph

atan(60)+atan(45)  = 177.772128726071   degrees

y = f(x+3) is the same as the graph of y = f(x), just shifted to the left 3 units. So, the vertex of y = f(x+3) is (-1, -1).

geometric sequence is \(a_n=a_1*q^{n-1}\), and so fore \(a_1=a_1*q^{1-1}, a_2=a_1*q^{2-1}, a_3=a_1*q^{3-1}....\). 

now we know that if there is a geometric sequence  \(6*q^1=a, 6*q^2=b, 6*q^3=1296\), and a sequence of \(a_1*q^1=a, a_1*q^2=b, a_1*q^3=1296 \), we can group them together and form a one sequnce \(6*q^0=6, 6*q^1=a, 6*q^2=b, 6*q^3=1296\),. Lets solve then q out of the final form \(6*q^3=1296, 1296/6=q^3, 216^{1/3}=q, q=6\). Now that we know that a is \(a_1*q^1=a, a_1=6, q=6, 6*6^1=36\)

Like so:

As follows:

You are welcome :)

Here is the graph of them

The original function is green and the inverse is orange.

Doesn't look to me like the inverse is a function either way ......

3/4^-2= (1/(3/4))^2

1/(3/4)= 4/3

4/3^2=16/9

16/9 is your final answer.

439 mod 14 and mod 62 ==5 - the remainder  

873 mod 14 and mod 62 ==5 - the remainder.  

Wow! Thanks melody..

no wonder why i couldnt guess and check, they were all the same!

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