All Answers 358187 Answers

6x6=36

         6.5

  --------

4| 2   6. 0

-   2  4. 0

-----------

         2 0

-        2 0

------------

             0

The answer is 6.5. Also the correct spelling is divided* :)

Solve for x:
1 = -0.3278 x^2 + 1.223 x + 0.0032

-0.3278 x^2 + 1.223 x + 0.0032 = -(1639 x^2)/5000 + (1223 x)/1000 + 2/625:
1 = -(1639 x^2)/5000 + (1223 x)/1000 + 2/625

1 = -(1639 x^2)/5000 + (1223 x)/1000 + 2/625 is equivalent to -(1639 x^2)/5000 + (1223 x)/1000 + 2/625 = 1:
-(1639 x^2)/5000 + (1223 x)/1000 + 2/625 = 1

Multiply both sides by -5000/1639:
x^2 - (6115 x)/1639 - 16/1639 = -5000/1639

Add 16/1639 to both sides:
x^2 - (6115 x)/1639 = -4984/1639

Add 37393225/10745284 to both sides:
x^2 - (6115 x)/1639 + 37393225/10745284 = 4718121/10745284

Write the left hand side as a square:
(x - 6115/3278)^2 = 4718121/10745284

Take the square root of both sides:
x - 6115/3278 = sqrt(4718121)/3278 or x - 6115/3278 = -sqrt(4718121)/3278

Add 6115/3278 to both sides:
x = 6115/3278 + sqrt(4718121)/3278 or x - 6115/3278 = -sqrt(4718121)/3278

Add 6115/3278 to both sides:
Answer: | x = 6115/3278 + sqrt(4718121)/3278    or    x = 6115/3278 - sqrt(4718121)/3278

Solve for x:
(14 x + 22)/(x + 2) = 1 - 3 x

Multiply both sides by x + 2:
14 x + 22 = (1 - 3 x) (x + 2)

Expand out terms of the right hand side:
14 x + 22 = -3 x^2 - 5 x + 2

Subtract -3 x^2 - 5 x + 2 from both sides:
3 x^2 + 19 x + 20 = 0

The left hand side factors into a product with two terms:
(x + 5) (3 x + 4) = 0

Split into two equations:
x + 5 = 0 or 3 x + 4 = 0

Subtract 5 from both sides:
x = -5 or 3 x + 4 = 0

Subtract 4 from both sides:
x = -5 or 3 x = -4

Divide both sides by 3:
Answer: | x = -5         or          x = -4/3     

sum_(n=1)^∞ 1/2 d (1/2)^(n - 1) = d

By the geometric series test, the series converges to d.

972/16 = 60.75

0.75*16 = 12

So the answer is 12          that is because 972/16 =  60 Remainder 12

972 modulo 16

\(\begin{array}{|rcll|} \hline && 972 \text{ modulo } 16 \\ &\equiv & 972 - 60\cdot 16 \text{ modulo } 16 \\ &\equiv & 972 - 960 \text{ modulo } 16 \\ &\equiv & 12 \text{ modulo } 16 \\ \hline \end{array}\)

As "heureka" read it:
Take the integral:
 integral(x + 4)/(x^2 - 5 x + 6) dx
Rewrite the integrand (x + 4)/(x^2 - 5 x + 6) as (2 x - 5)/(2 (x^2 - 5 x + 6)) + 13/(2 (x^2 - 5 x + 6)):
 = integral((2 x - 5)/(2 (x^2 - 5 x + 6)) + 13/(2 (x^2 - 5 x + 6))) dx
Integrate the sum term by term and factor out constants:
 = 1/2 integral(2 x - 5)/(x^2 - 5 x + 6) dx + 13/2 integral1/(x^2 - 5 x + 6) dx
For the integrand (2 x - 5)/(x^2 - 5 x + 6), substitute u = x^2 - 5 x + 6 and du = (2 x - 5) dx:
 = 1/2 integral1/u du + 13/2 integral1/(x^2 - 5 x + 6) dx
The integral of 1/u is log(u):
 = (log(u))/2 + 13/2 integral1/(x^2 - 5 x + 6) dx
For the integrand 1/(x^2 - 5 x + 6), complete the square:
 = (log(u))/2 + 13/2 integral1/((x - 5/2)^2 - 1/4) dx
For the integrand 1/((x - 5/2)^2 - 1/4), substitute s = x - 5/2 and ds = dx:
 = (log(u))/2 + 13/2 integral1/(s^2 - 1/4) ds
Factor -1/4 from the denominator:
 = (log(u))/2 + 13/2 integral4/(4 s^2 - 1) ds
Factor out constants:
 = (log(u))/2 + 26 integral1/(4 s^2 - 1) ds
Factor -1 from the denominator:
 = (log(u))/2 - 26 integral1/(1 - 4 s^2) ds
For the integrand 1/(1 - 4 s^2), substitute p = 2 s and dp = 2 ds:
 = (log(u))/2 - 13 integral1/(1 - p^2) dp
The integral of 1/(1 - p^2) is tanh^(-1)(p):
 = (log(u))/2 - 13 tanh^(-1)(p) + constant
Substitute back for p = 2 s:
 = (log(u))/2 - 13 tanh^(-1)(2 s) + constant
Substitute back for s = x - 5/2:
 = (log(u))/2 + 13 tanh^(-1)(5 - 2 x) + constant
Substitute back for u = x^2 - 5 x + 6:
 = 1/2 log(x^2 - 5 x + 6) + 13 tanh^(-1)(5 - 2 x) + constant
Factor the answer a different way:
 = 1/2 (log(x^2 - 5 x + 6) + 26 tanh^(-1)(5 - 2 x)) + constant
Which is equivalent for restricted x values to:
Answer: | = 7 log(3 - x) - 6 log(2 - x) + constant

Take the integral:
 integral(4/x^2 - 4 x + 6) dx

Integrate the sum term by term and factor out constants:
 = -4 integral x dx + 4 integral1/x^2 dx + 6 integral1 dx

The integral of x is x^2/2:
 = -2 x^2 + 4 integral1/x^2 dx + 6 integral1 dx

The integral of 1/x^2 is -1/x:
 = -4/x - 2 x^2 + 6 integral1 dx

The integral of 1 is x:
Answer: | = -2 x^2 + 6 x - 4/x + constant

\(8\frac9{16}\div2\\~\\ =(8+\frac9{16})\div2\\~\\ =(\frac{128}{16}+\frac9{16})\div2 \\~\\ =\frac{137}{16}\div2\\~\\ =\frac{137}{16}\,*\,\frac12 \\~\\ =\frac{137}{32} \\~\\ = 4.28125\)

Also, you can type it into the calculator like this:      (8+9/16)/2        

integral( (x+4) / (x^2-5x+6) ) dx

\(\begin{array}{rcl} && \int { \frac{x+4} {x^2-5x+6} \ dx} \quad & | \quad x^2-5x+6 = (x-3)(x-2) \\ &=& \int { \frac{x+4} {(x-3)(x-2)} \ dx} \\ \end{array} \)

Partial fraction decomposition:

\(\begin{array}{|lrcll|} \hline & \frac{x+4} {(x-3)(x-2)} &=& \frac{A}{x-3} + \frac{B}{x-2} \\ & x+4 &=& A\cdot (x-2) + B\cdot (x-3) \\ x = 2: & 2+4 &=& A\cdot (2-2) + B\cdot (2-3) \\ & 6 &=& 0 + B\cdot (-1) \\ & 6 &=& B\cdot (-1) \\ & \mathbf{B} &\mathbf{=}& \mathbf{-6} \\ \\ x = 3: & 3+4 &=& A\cdot (3-2) + B\cdot (3-3) \\ & 7 &=& A\cdot (1) + B\cdot (0) \\ & 7 &=& A\cdot (1) \\ & \mathbf{A} &\mathbf{=}& \mathbf{7} \\\\ & \mathbf{\frac{x+4} {(x-3)(x-2)}} &\mathbf{=}& \mathbf{\frac{7}{x-3} - \frac{6}{x-2}} \\ \hline \end{array} \)

\(\begin{array}{rcl} && \mathbf{ \int { \frac{x+4} {x^2-5x+6} \ dx} } \\ &=& \int { \frac{x+4} {(x-3)(x-2)} \ dx} \\ &=& \int { \Big(\frac{7}{x-3} - \frac{6}{x-2} \Big) \ dx} \\ &=& 7\cdot \int { \frac{1}{x-3} \ dx} -6\int { \frac{1}{x-2} \ dx} \\ &\mathbf{=}& \mathbf{7\cdot \ln(|x-3|) -6\cdot \ln(|x-2|) +c }\\ \end{array} \)

 f(x)  =  x² + bx + c

g(x)  =  x³ + wx² + yx + z

[f(x)]³ - [g(x)]² + f(x) - 1

(x²+bx+c)³ - (x³+wx²+yx+z)²   +   x²+bx+c   -   1

(x²+bx+c)(x²​+bx+c)(x²​+bx+c)  -  (x³+wx²+yx+z)(x³+wx²+yx+z)   +   x²+bx+c   -   1

x⁴ + ...

We could continue, but by now I can see that the first term of   [f(x)]³   will be   1x⁶   and the first term of   [g(x)]²   will be   1x⁶   . Since   [g(x)]²    is being subtracted,  the   1x⁶   will cancel each other and go away. The next biggest exponent comes from  x³ * wx²  =  wx⁵ .

So  5   is the maximum degree. 

\(\int(x+\frac4{x^2}-5x+6)\,dx \\~\\ =\int(\frac4{x^2}-4x+6)\,dx \\~\\ =\int \frac4{x^2}\,dx+\int-4x\,dx+\int6\,dx \\~\\ =4\int x^{-2}\,dx-4\int x\,dx+6\int 1\,dx\)       Combine like terms.

Applying the power rule gives us...

\(=4(\frac{x^{-1}}{-1})-4(\frac{x^2}{2})+6(x) +c\\~\\ =-4x^{-1}-2x^2+6x+c\)                 , where   c   is a constant.

hmmmm... I really don't know. I'll take a guess! 2?

6 remainder 2

I tried creating a computer program for this problem, and it couldn't find an answer; it would simply go in an infinite loop! Something tells me that there is no solution to this. 

If you put in an answer that is possible (like 53/60), then the program finds the answer in very little at all. \(\frac{2}{6}+\frac{3}{4}-\frac{1}{5}=\frac{53}{60}\).

Anyone who wants to see it can; here's the link for those curious

It is possible that I am misinterpreting this question as well; I am assuming that the numbers listed (1-6) can only be used once. Is this a correct assumption?

P is 1/6 of the distance between X and Y

We can find the point as follows :

[  (1/6)(6 - - 6)  + -6 ,  (1/6)( 7 - -2) + -2 ]  

[  (1/6)(12) - 6, (9/6) - 2 ]  =

[ -4, - 1/2 ]  = P

Proof...

Distance between X and P   (-6, -2) and (-4, -1/2)  =  2.5

Distance between P and Y  (-4, -1/2)  and  ( 6,7)  = 12.5

And   XP : PY  =  2.5 : 12.5  =  1 : 5

Is a square a rectangle?

Yes! A square is a rectangle. (A square is the only shape that is both a rectangle and a rhombus!)

Can a rectangle be a square?

Yes, a rectangle can be a square, but a rectangle is not always a square.   

Question

For every 1,000 miles on the car, reduce its value by one-fifth of a percent.

So... after each 1,000 miles, it is worth   (100 - .02)%  =  99.8%  =  0.998   of its previous value.

So we can say....

car value  =  35,000 * 0.998ⁿ             , where  n  is the number of thousands of miles.

127.5 out of what number equals 15 out of 100 ?

\(\frac{127.5}{x}=\frac{15}{100} \\~\\ 127.5=\frac{15}{100}\,*\,x \\~\\ 127.5\,*\,\frac{100}{15}=x \\~\\ 850=x\)   Now we just need to solve this for x .

Let the number = N, so we have:

0.15N = 127.50  divide both sides by 0.15

N =127.50 / 0.15

N = 850

Gradient  =  [ subtract the y's in some order ] / [ subtract the x's in the same order ]  =

[ 6 - 5 [ / [ 3 - 4]  = [ 1 ] / [ -1 ]   =  -1