All Answers 357301 Answers

70%  =  .70

So

.70 * 200  =

140

This would be true if we were setting the equation to 0  and solving

But we aren't solving anything here......we are just completing the square

Using

sin^2 ( theta) + cos^2 (theta)  = 1    we have

sin^2(theta) +  ( -12 / 13)^2  = 1

sin^(theta)  =  1 - (-12/13)^2

sin^2(theta)  =  1 - 144/169

sin^2(theta)  =    169/169  -  144/169

sin^2 (theta)  =  25/169       take the square root of both sides

sin (theta)  = ± √ [25/169]     and the sine is negative in the 3rd quadrant

sin (theta)  =  -5/13

tan (theta)  =   sin (theta) / cos (theta)  =  ( -5/13) / (-12/13)  =  5/12

CPhill: Aren't you actually dividing by 4 in the first step and it should disappear altogether in the 2nd step (outside the bracket)?

x^2-3x+9 = x+41         subtract the right side from both sides

x^2 - 4x - 32  = 0        factor

(x - 8) (x + 4)  = 0

Setting each factor to 0  and solving for  x, we have that

x= 8   and x = -4

And the positive difference between these is

8 - (-4)  =

12

4s^2 + 28s + 45    factor out  the 4

4 [ s^2  + 7s  +  45/4]

Take (1/2) of 7  = 7/2......square this = 49/4  ....add it and subtract it

4 [ s^2 + 7s + 49/4 + 45/4 - 49/4 ] 

Factor the first three terms and simplify

4 [  ( s + 7/2)^2  - 1 ]

4(s + 7/2)^2  - 4

So q =  -4

Thank you so much I was struggling with completing the square now I get it

3x^2 + x - 4      factor out the 3

3 [x^2 +  (1/3)x - (4/3) ]

Take (1/2) of (1/3)  = 1/6.......square this = 1/36......add it and subtract it

3 [ x^2 + (1/3)x  + 1/36 - (4/3) - 1/36 ]   factor the first 3 terms

3 [ ( x + 1/6)^2 - 1/36 - 48/36 ]

3 [ (x + 1/6)^2 - 49/36 ]

3(x + 1/6)^2 - 49/12

So k =  -49/12

Yessssss I was wondering the whole day about the equation since I've seen it before and it was more like the second one. 

I am sry for my mistake but thanks that you're so mindfully!

Well I am learning Math and Physik in german and my teacher alway say to Subscripts index so I was assuming it was the same in english haha :c 

But thanks. 

Learning by mistakes ! :D

the answer is 14

Express a4(b - c) + b4 (c - a) + c4 (a - b) as the product of four factors.

(b-c),(c-a) and (a-b) are products of this expression

$\begin{array}{|rcll|} \hline a^4(b-c)+b^4(c-a)+c^4(a-b) &=& (b-c)(c-a)(a-b)\cdot x \\\\ x &=& \dfrac{a^4(b-c)+b^4(c-a)+c^4(a-b)}{(b-c)(c-a)(a-b)} \\\\ &=& \dfrac{a^4b-a^4c+b^4c-b^4a+c^4a-c^4b}{-a^2b+a^2c-b^2c+b^2a-c^2a+c^2b} \\\\ &=& \dfrac{ a^4·b - a^4·c - a·b^4 + a·c^4 + b^4·c - b·c^4 }{-a^2·b + a^2·c + a·b^2 - a·c^2 - b^2·c + b·c^2} \\\\ \hline \end{array}$

Polynom division after variable c:

$\small{ \begin{array}{|rcll|} \hline && \text {in divisor term with max power of c is: } -ac^2 \\ && \text {current residue: } a^4·b - a^4·c - a·b^4 + a·c^4 + b^4·c - b·c^4 \\ && \text {in current residue term with max power of c: } ac^4 \\ && \text {quotient } \frac{ac^4}{-ac^2} = -c^2 \\ && \text {product } -c^2·(-a^2·b + a^2·c + a·b^2 - a·c^2 - b^2·c + b·c^2) = a^2·b·c^2 - a^2·c^3 - a·b^2·c^2 + a·c^4 + b^2·c^3 - b·c^4 \\ && \text {subtract product form current residue: } \\ && \text {current residue: } a^4·b - a^4·c - a^2·b·c^2 + a^2·c^3 - a·b^4 + a·b^2·c^2 + b^4·c - b^2·c^3 \\ && \text {in current residue term with next lower power of c is: } a^2·c^3 \\ && \text {quotient } \frac{a^2·c^3}{-ac^2} = -a·c \\ && \text {product } -a·c·(-a^2·b + a^2·c + a·b^2 - a·c^2 - b^2·c + b·c^2) = a^3·b·c - a^3·c^2 - a^2·b^2·c + a^2·c^3 + a·b^2·c^2 - a·b·c^3 \\ && \text {subtract product form current residue: } \\ && \text {current residue: } a^4·b - a^4·c - a^3·b·c + a^3·c^2 + a^2·b^2·c - a^2·b·c^2 - a·b^4 + a·b·c^3 + b^4·c - b^2·c^3 \\ && \text {in current residue term with next lower power of c is: } a·b·c^3 \\ && \text {quotient } \frac{a·b·c^3}{-a·c^2} = -b·c \\ && \text {product } -b·c·(-a^2·b + a^2·c + a·b^2 - a·c^2 - b^2·c + b·c^2) = a^2·b^2·c - a^2·b·c^2 - a·b^3·c + a·b·c^3 + b^3·c^2 - b^2·c^3 \\ && \text {subtract product form current residue: } \\ && \text {current residue: } a^4·b - a^4·c - a^3·b·c + a^3·c^2 - a·b^4 + a·b^3·c + b^4·c - b^3·c^2 \\ && \text {in current residue term with next lower power of c is: } a^3·c^2 \\ && \text {quotient } \frac{a^3·c^2}{-a·c^2} = -a^2 \\ && \text {product } -a^2·(-a^2·b + a^2·c + a·b^2 - a·c^2 - b^2·c + b·c^2) = a^4·b - a^4·c - a^3·b^2 + a^3·c^2 + a^2·b^2·c - a^2·b·c^2 \\ && \text {subtract product form current residue: } \\ && \text {current residue: } a^3·b^2 - a^3·b·c - a^2·b^2·c + a^2·b·c^2 - a·b^4 + a·b^3·c + b^4·c - b^3·c^2 \\ && \text {in current residue term with next lower power of c is: } a^2·b·c^2 \\ && \text {quotient } \frac{a^2·b·c^2}{-a·c^2} = -a·b \\ && \text {product } -a·b·(-a^2·b + a^2·c + a·b^2 - a·c^2 - b^2·c + b·c^2) = a^3·b^2 - a^3·b·c - a^2·b^3 + a^2·b·c^2 + a·b^3·c - a·b^2·c^2 \\ && \text {subtract product form current residue: } \\ && \text {current residue: } a^2·b^3 - a^2·b^2·c - a·b^4 + a·b^2·c^2 + b^4·c - b^3·c^2 \\ && \text {in current residue term with next lower power of c is: } a·b^2·c^2 \\ && \text {quotient } \frac{a·b^2·c^2}{-a·c^2} = -b^2 \\ && \text {product } -b^2·(-a^2·b + a^2·c + a·b^2 - a·c^2 - b^2·c + b·c^2) = a^2·b^3 - a^2·b^2·c - a·b^4 + a·b^2·c^2 + b^4·c - b^3·c^2 \\ && \text {subtract product form current residue: } \\ && \text {current residue: } 0 \\\\ x&=& \dfrac{a^4·b - a^4·c - a·b^4 + a·c^4 + b^4·c - b·c^4 } {-a^2·b + a^2·c + a·b^2 - a·c^2 - b^2·c + b·c^2 } \\\\ &=& -a^2 - a·b - a·c - b^2 - b·c - c^2 \\ \hline \end{array} }$

$a^4(b-c)+b^4(c-a)+c^4(a-b) = (b-c)(c-a)(a-b)(-a^2 - a·b - a·c - b^2 - b·c - c^2)$

My math program says that you can transform....
$1/sec\theta * tan\theta = sin\theta$
Into...
$tan^2\theta + 1 = sec^2\theta$
Could someone explain how they made this transition?

$\begin{array}{|rcll|} \hline \frac{1}{\sec(\theta)} * \tan(\theta) &=& \sin(\theta) \quad & | \quad \cdot \sec(\theta) \\ \tan(\theta) &=& \sin(\theta) \cdot \sec(\theta) \quad & | \quad \text{square both sides} \\ \tan^2(\theta) &=& \sin^2(\theta) \cdot \sec^2(\theta) \quad & | \quad \sin^2(\theta) + \cos^2(\theta) = 1 \text{ or } \sin^2(\theta) = 1-\cos^2(\theta) \\ \tan^2(\theta) &=& [~1-\cos^2(\theta)~] \cdot \sec^2(\theta) \\ \tan^2(\theta) &=& \sec^2(\theta) - \cos^2(\theta) \cdot \sec^2(\theta) \quad & | \quad \sec^2(\theta) = \frac{1}{\cos^2(\theta)} \\ \tan^2(\theta) &=& \sec^2(\theta) - \cos^2(\theta) \cdot \frac{1}{\cos^2(\theta)} \\ \tan^2(\theta) &=& \sec^2(\theta) - \frac{\cos^2(\theta)}{\cos^2(\theta)} \quad & | \quad \frac{\cos^2(\theta)}{\cos^2(\theta)} = 1 \\ \tan^2(\theta) &=& \sec^2(\theta) - 1 \quad & | \quad +1 \\ \mathbf{\tan^2(\theta)+1} &\mathbf{=}& \mathbf{\sec^2(\theta)} \\ \hline \end{array}$

What is the smallest positive integer that will satisfy the following two modular equations:

N mod 3,331 =1,851 and

N mod 1,851 =1,468?

$\begin{array}{|rcll|} \hline N &\equiv& 1851 \pmod{3331} \\ N &\equiv& 1468 \pmod{1851} \\ \hline \end{array}$

Formula:

$\begin{array}{rcll} N = 1851\cdot 1851\cdot \frac{1}{1851} \pmod{3331} + 1468\cdot 3331 \cdot \frac{1}{3331} \pmod{1851} + 3331\cdot 1851\cdot z \\ \quad z \in Z \\ \end{array}$

check:

$\begin{array}{|rcll|} \hline \pmod{3331}: & N = 1851 \cdot \frac{1851}{1851} + 0 +0 = 1851 \ \checkmark \\ \pmod{1851}: & N = 0 + 1468 \cdot \frac{3331}{3331} + 0 = 1468 \ \checkmark \\ \hline \end{array}$

$\text{with }\varphi() =\text{ Euler's totient theorem }: \\ 3331 \text{ prime number} \qquad \varphi(p) = p-1 \qquad \varphi(3331) = 3330 \\ 1851 =3\cdot 617\qquad \varphi(1851) =1851\cdot \left(1-\frac13\right)\left(1-\frac{1}{617}\right) = 1232$

Modular inverses:

$\begin{array}{|rcll|} \hline && \frac{1}{1851} \pmod{3331} \\ &\equiv& 1851^{\varphi(3331)-1} \pmod{3331} \quad & | \quad gcd(3331,1851)=1 \\ &\equiv& 1851^{3330-1} \pmod{3331} \\ &\equiv& 1851^{3329} \pmod{3331} \\ && \ldots \\ &\equiv& 835 \pmod{3331} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \frac{1}{3331} \pmod{1851} \\ &\equiv& 3331^{\varphi(1851)-1} \pmod{1851} \quad & | \quad gcd(3331,1851)=1 \\ &\equiv& 3331^{1232-1} \pmod{1851} \\ &\equiv& 3331^{1231} \pmod{1851} \\ && \ldots \\ &\equiv& 1387 \pmod{1851} \\ \hline \end{array}$

$\begin{array}{rcll} N &=& 1851\cdot 1851\cdot 835 + 1468 \cdot 3331 \cdot 1387 + 3331\cdot 1851\cdot z \quad & | \quad z \in Z \\ &=& 2860877835 + 6782302396 + 6165681\cdot z \\ &=& \underbrace{9643180231}_{\equiv 55147 \pmod{6165681}} + 6165681\cdot z \\ &=& 55147 + 6165681\cdot z \\ \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{55147} &\mathbf{\equiv}& \mathbf{1851 \pmod{3331}} \\ \mathbf{55147} &\mathbf{\equiv}& \mathbf{1468 \pmod{1851}} \\ \hline \end{array}$

Compute: 7Chose6

$\begin{array}{|rcll|} \hline && \mathbf{\binom{7}{6}} \\\\ &=& \binom{7}{7-6} \\\\ &=& \binom{7}{1} \\\\ &=& \dfrac{7}{1} \\\\ &\mathbf{=}&\mathbf{ 7 } \\ \hline \end{array}$

Compute:  85 choose 82

$\begin{array}{|rcll|} \hline && \mathbf{\binom{85}{82}} \\\\ &=& \binom{85}{85-82} \\\\ &=& \binom{85}{3} \\\\ &=& \dfrac{85}{3}\cdot \dfrac{84}{2}\cdot \dfrac{83}{1} \\\\ &=& 85\cdot \dfrac{84}{2\cdot 3}\cdot 83 \\\\ &=& 85\cdot \dfrac{84}{6}\cdot 83 \\\\ &=& 85\cdot 14 \cdot 83 \\\\ &\mathbf{=}&\mathbf{ 98770 } \\ \hline \end{array}$

Here's a ( highly questionable ) proceedure.......

1/sec * tan  =  sin

cos * sin/cos  = sin

sin  = sin                multiply both sides by the sin

sin^2  = sin^2

sin^2  =  1 - cos^2       add cos^2  to both sides

sin^2 + cos^2  =  1       divide through by cos^2

sin^2 / cos^2    + 1  =  1 / cos^1

tan^2  +  1  =  sec^2

Note :  multiplying/dividing on both sides of an identity isn't usually allowed......!!!!

We need more information than that. What does x equal?

$(x*x*x*x)$

^ That's what x to the 4th power is. But it simplifies down to:

$x^4$

Denary is a less common term for decimal, or base 10. Assuming that number is in base 10, 100000 in denary is still 100000. 

Yes, I’m going to claim my copyright. We chimps always climb down from our trees to collect any bananas we drop. But first we check to make sure that there isn’t a lion, tiger, or Bernoulli lurking in the bushes.  

It seems the monkey himself answered your question.

Good job AT! If you keep it up, I’ll give you a banana.

Yes the answers should be 2,4 and 6 as you said,

Can you chill? I made an error in my wording. Whateve. Happens to everyone. We aren't perfect. What I should have said is thank you, Hectictar. Happy now, Jacob?

Let's pick two points off of the table:  (1, 0)  and  (0, -6)

slope of  f(x)  =  $\frac{\text{change in f(x)}}{\text{change in x}}\,=\,\frac{-6 - 0}{0-1}\,=\,\frac{-6}{-1}\,=\,6$

g(x)  =  2x + 6    Notice that this is in slope-intercept form, so we can already tell that its slope is  2 .

6 > 2 ,   so the slope of f(x) > the slope of g(x) . f(x) is "steeper" than g(x).

----------

The y-intercept is the value of the function when  x = 0 .

On the table, there is the point  (0, -6) .

So the y-intercept of f(x)  is  -6 .

To find the y-intercept of g(x), plug in  0  for  x  and solve for  g(x) .

g(0)  =  2(0) + 6

g(0)  =  6                So the y-intercept of g(x)  is  6 .

6 > -6 ,   so the y-intercept of g(x) > the y-intercept of f(x).  g(x) has a greater y-intercept.