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y - 3  = (1/2)x + 1      add 3 to both sides

y  = (1/2)x  +  4

See here:  http://web2.0calc.com/questions/calculate_1

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A = 80(1/2)^t

A = 1,800(0.6)^t

A = 1,900(0.999)^t

Solution:

ah, I think it's looking for the coefficients of f, not the coefficients of 47, 32, -13, and 16, is it possible that would render a different answer?

Let f(x)=-3x^2+x-4, g(x)=-5x^2+3x-8, and h(x)=5x^2+5x+1. Express f(x)+g(x)+h(x) as a single polynomial, with the terms in order by decreasing degree.

 [ -3x^2+ x - 4 ]  +  [ -5x^2+3x - 8] + [ 5x^2+5x+1 ]   =

-3x^2 + 9x - 11   

Here is the graph : https://www.desmos.com/calculator/if4wjn6elr

All the specified points appear.....is it possible that the polynomial you want is of a different degree ???

Not sure what the Fermi Process is...but  here is an approach :

We will have some  empty space around each bucket....each bucket would occupy a prism with a volume of  [10 *10 * 25 ]  = 2500 in^3

The height of each bucket is 25 in   and the height of the warehouse is 240 in

So......20*12 / 25  ≈  9 buckets could be stacked

And in one row,  50*12/10  =  60 buckets   could be accomodated

And the number of rows is 120*12/10  =  120

So....the approximate number of buckets that the warehouse could hold is

9 * 60 * 120    =  64800 buckets

3:  What is the coefficient of x^3 when 7x^4-3x^3 -3x^2-8x + 1 is multiplied by 8x^4+2x^3 - 7x^2 + 3x + 4 and the like terms are combined?

[ 7x^4 - 3x^3 -3x^2 - 8x + 1 ] *  [ 8x^4 + 2x^3 - 7x^2 + 3x + 4 ]

The coefficient of the x^3 term will come from

(-3 * 4)   +  (-3 * 3) + (-8 * -7)  + (1 * 2)  =

-12      -   9          +    56    +   2   =

37

2:  Consider the polynomials [f(x)=4x^3+3x^2+2x+1]and [g(x)=3-4x+5x^2-6x^3.] Find c such that the polynomial f(x)+cg(x) has degree 2.

[4x^3+3x^2+2x+1] + c [ 3-4x+5x^2-6x^3]

[4x^3+3x^2+2x+1]  + c [ -6x^3 + 5x^2 - 4x + 3 ]

For a resulting polynomial to have degree 2,  we must have that

4  - 6c  =  0

4  =  6c

4/6  = c   =  2/3

So  we have the resulting polynomial

[4x^3+3x^2+2x+1]  + (2/3) [ -6x^3 + 5x^2 - 4x + 3 ]  =

(1/3) (19 x^2 - 2 x + 9)

Thanks a bunch, but...  My math program says that's wrong.  I couldn't find anything wrong with your calculation though?

1:  Suppose f is a polynomial such that f(0) = 47, f(1) = 32, f(2) = -13, and f(3)=16. What is the sum of the coefficients of f?

Let us suppose that we have this form

P(x)  =   ax^3  + bx^2 +  cx  +  d       where d  =  47

So we have that

a  +    b +  c +   47  =  32    ⇒   a +  b +  c  =  -15

8a +  4b  + 2c + 47  = -13  ⇒  8a + 4b + 2c  =  -60

27a + 9b + 3c + 47  =  16  ⇒  27a + 9b + 3c  = -31

Multiply the first equation by -2 and add it to the second equation

Multiply the first equation by -3 and add it to the third equation

6a  +  2b  = -30   ⇒  3a + b =  - 15   

24a + 6b =    14  ⇒ 12a + 3b  = 7

Multiply  the first equation by -3  and add it to the second

3a   = 52   ⇒    a  = 52/3

3(52/3) + b  =  -15  ⇒  52 + b  = -15   ⇒  b =   -67

(52/3) + (-67) + c  =  -15  ⇒  c  = -15 - 52/3 + 67   ⇒  c  =  104/3

So.... the sum of the coefficients is

52/3 - 67 + 104/3   =    -15 

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I = W/d^2 =23/19,500^2 =~6 x 10^-8 - That is the closest I get !!.

Max: You are simply beyond words !! Bravo and thanks.

$\because e^{i\pi} = -1\\ \text{We can immediately imply that} \log(-1) = i\pi\\ \dfrac{1}{6}(-i\log(-1))^2 \\ =\dfrac{1}{6}(-i\cdot i\pi)^2\quad\boxed{-i\cdot i = 1}\\ =\dfrac{1}{6}\pi^2\\ $

Hence, proved.

a^5 + b^5 = (a+b)(a^4-a^3*b+a^2*b^2-a*b^3+b^4)

Thanks guest   

Solve the following system:
{76 x + 49 y = 474.25 | (equation 1)
54 x + 37 y = 346.25 | (equation 2)

Subtract 27/38 × (equation 1) from equation 2:
{76 x + 49 y = 474.25 | (equation 1)
0 x+(83 y)/38 = 9.28289 | (equation 2)

Multiply equation 2 by 38/83:
{76 x + 49 y = 474.25 | (equation 1)
0 x+y = 4.25 | (equation 2)

Subtract 49 × (equation 2) from equation 1:
{76 x+0 y = 266. | (equation 1)
0 x+y = 4.25 | (equation 2)

Divide equation 1 by 76:
{x+0 y = 3.5 | (equation 1)
0 x+y = 4.25 | (equation 2)

 x = 3.5
y = 4.25

Brilliant heureka, as usual. Thank you very much.

http://web2.0calc.com/questions/in-a-lottery-a-ticket-costs-3-and-the-jackpot-is_3

Can this identity be proven, and if so how?  π^2/6 = 1/6 (-i log(-1))^2

$\begin{array}{|rcll|} \hline && \frac{1}{6}\cdot \Big(-i \log(-1) \Big)^2 \\ && &\small{ \begin{array}{|rcll|} \hline \log(z) &=& \ln|z| + i\underbrace{\arg(z)}_{=\arctan(\frac{b}{a}) } \quad | \quad z = a+b\cdot i \\ \log(-1) &=& \ln|-1| + i\underbrace{\arg(-1)}_{= \underbrace{\arctan\left(\frac{0}{-1}\right)}_{=-\pi} } \quad | \quad z = -1+0\cdot i \\ \log(-1) &=& \ln|-1| + i\cdot(-\pi) \\ \log(-1) &=& \ln(1) - i\cdot \pi \quad | \quad \ln(1) = 0 \\ \log(-1) &=& 0 - i\cdot \pi \\ \log(-1) &=& - i\cdot \pi \\ \hline \end{array}} \\ &=& \frac{1}{6}\cdot \Big(-i \cdot(- i\cdot \pi) \Big)^2 \\ &=& \frac{1}{6}\cdot (i^2 \cdot \pi )^2 \quad | \quad i^2 = -1 \\ &=& \frac{1}{6}\cdot (-\pi )^2 \\ &=& \frac{1}{6}\cdot \pi ^2 \\ &=& \dfrac{\pi ^2}{6} \\ \hline \end{array}$

By definition of tangent,

tan θ  =  sin θ / cos θ

                                         Plug in  1/5  for  cos θ , and  2√6  for  tan θ .

2√6  =  sin θ / (1/5)

                                         Multiply both sides of the equation by  1/5 .

(1/5)(2√6)  =  sin θ

sin θ  =  2√6 / 5               

What is the value of x and y 

76x + 49y = 474.25

54x + 37y = 346.25 

$\begin{array}{|rcl|rcl|} \hline x&=& \displaystyle \dfrac{ \begin{vmatrix} 474.25 & 49 \\ 346.25 & 37 \end{vmatrix} } { \begin{vmatrix} 76 & 49 \\ 54 & 37 \end{vmatrix} } \quad & \quad y&=& \displaystyle \dfrac{ \begin{vmatrix} 76 & 474.25 \\ 54 & 346.25 \end{vmatrix} } { \begin{vmatrix} 76 & 49 \\ 54 & 37 \end{vmatrix} } \\\\ x&=& \displaystyle \dfrac{474.25\cdot 37-346.25\cdot 49} { 76\cdot 37-54\cdot 49 } \quad & \quad y&=& \displaystyle \dfrac{ 76\cdot 346.25- 54\cdot 474.25 } { 76\cdot 37-54\cdot 49 } \\\\ x&=& \displaystyle \dfrac{581} { 166 } \quad & \quad y&=& \displaystyle \dfrac{705.5} { 166 } \\\\ x&=& \displaystyle 3.5 \quad & \quad y&=& \displaystyle 4.25 \\ \hline \end{array}$