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a binomial

$a=5$   ;   $b=-11$   ;   $c=4$

pluggint hat info into    $_a x_b=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$   we get:

$_a x_b= \frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4 \times \:5 \times \:4}}{2 \times\:5}$

$    _a x_b=\frac{11 \pm \sqrt{41}}{10}  $

$ a=1.74  $    and $  b=0.46 $

now for our condition where we have to find $a^3 + b^3  \ \ \implies \ \   1.74^3 +0.46^3 \ \ \implies \ \ \   5.27 + 0.097  $

so the answer is $  \boxed { 5.367  } $

The center point (-6,2) already lies on the y-axis. Therefore, the other 2 points are just (-6,2+10) and (-6,2-10) ---> (-6,12) and (-6,-8). Taking the 2 y-coordinates: 12+(-8)=\(\boxed{4}\).

Standard form is usually Ax + By = C.

Where A, B, and C are integers and A is positive.

From your line, -2x + 3y = 3 would be 2x - 3y = -3 in standard form. :))

=^._.^=

Melody DID answer this !

Find two integer points   -3,-1    and     3,3    use them to calculate the slope = 4/6 = 2/3

  b is the y axis intercept    b = 1

y = mx + b     form

y = 2/3 x + 1

-2/3x + y = 1

-2x + 3y = 3             Not sure which you consider 'standard' form.....  You're pick ! 

Oh I see it now okay! Thanks a lot :)

The interior angles of a convex polygon are in an arithmetic progression.
If the smallest angle is 100 degrees and common difference is  10 degrees ,
then find the number of sides.

\(\text{Let the sides of the polygon $n$ }\)

Arithmetic progression:\(\begin{array}{|rcll|} \hline \text{sum} &=& 100^\circ+110^\circ+120^\circ+130^\circ+\ldots+\Big( 100^\circ+(n-1)10^\circ\Big) \\ \text{sum} &=& \left(\dfrac{100^\circ + \Big( 100^\circ+(n-1)10^\circ\Big)}{2}\right) * n \\\\ \text{sum} &=& \left( \dfrac{100^\circ+100^\circ +10^\circ n -10^\circ}{2} \right) * n \\\\ \text{sum} &=& \left( \dfrac{190^\circ +10^\circ n}{2} \right) * n \\\\ \mathbf{\text{sum}} &=& \mathbf{(95^\circ +5^\circ n)*n} \\ \hline \end{array} \)

Interior angles of a convex polygon: \(\text{sum} = (n-2)*180^\circ\)

\(\begin{array}{|rcll|} \hline (95^\circ +5^\circ n)*n &=& (n-2)*180^\circ \\ 95 n +5 n^2 &=& 180n -2*180\\ \ldots \\ 5n^2-85n + 360&=& 0 \quad | \quad : 5 \\ \mathbf{n^2-17n + 72} &=& \mathbf{0} \\ \hline n &=& \dfrac{17 \pm \sqrt{17^2- 4 *72 } }{2} \\\\ n &=& \dfrac{17 \pm \sqrt{289-288 } }{2} \\\\ n &=& \dfrac{17 \pm 1 }{2} \\ \mathbf{n}=\mathbf{9} &\text{or}&\mathbf{n}=\mathbf{8} \\ \hline \end{array}\)

I used n as a general value. You can replace it with n1 and the first two terms produce a real number; replace it with n2 and the second two terms produce a real value.  So your original expression is true for all positive integer values of n1 and n2.  

The question asks for a proof "if and only if"  (iff).  The expression is a real number for all the possibilities listed, however, it is also true for other possibilities not listed (such as n1 = n2+3, for example), which means none of a) to d) provide the correct answer (they all conform to the "if" part, but not the "only if" part).

(    (-sqrt(3)+x)*((1/8)*cos(   ((5*pi)/3)  )x(sin(((5*pi)/3))))

I know you have tried your best to put all the brackets in but I am still confused.

Is this what you want.  

\((     (-\sqrt{3}+x)   *    (\frac{1}{8}*cos   (\frac{5\pi}{3})    x(sin(\frac{5\pi}{3}))))\\~\\  =(-\sqrt{3}+x)       (\frac{1}{8}*cos   (\frac{5\pi}{3})    x(sin(\frac{5\pi}{3}))\\~\\  =(-\sqrt{3}+x)       (\frac{1}{8}*cos   (\frac{1\pi}{3})    x(-sin(\frac{1\pi}{3}))\\~\\  =(-\sqrt{3}+x)       (\frac{1}{8}*\frac{1}{2}   x(-\frac{\sqrt3}{2}))\\~\\  =(-\sqrt{3}+x)       (-\frac{\sqrt3 x}{32})\\~\\  =\frac{3x-\sqrt3 x^2}{32}\\~\\ \)

LaTex:

(     (-\sqrt{3}+x)   *    (\frac{1}{8}*cos   (\frac{5\pi}{3})    x(sin(\frac{5\pi}{3}))))\\~\\

 =(-\sqrt{3}+x)       (\frac{1}{8}*cos   (\frac{5\pi}{3})    x(sin(\frac{5\pi}{3}))\\~\\

 =(-\sqrt{3}+x)       (\frac{1}{8}*cos   (\frac{1\pi}{3})    x(-sin(\frac{1\pi}{3}))\\~\\

 =(-\sqrt{3}+x)       (\frac{1}{8}*\frac{1}{2}   x(-\frac{\sqrt3}{2}))\\~\\

 =(-\sqrt{3}+x)        (-\frac{\sqrt3 x}{32})\\~\\

 =\frac{3x-\sqrt3 x^2}{32}\\~\\

This is too easy!!! 

ON = 1/3(QN)                (Median theorem)

NR = 1/2(PR)

OR = sqrt[(ON)² + (PR)²] 

x^2 + 8x + 16   =     (x+4)^2                  ( take 1/2 of the 'x' coefficient ...then square it....then add it to the equation)

i asked one person i know about this one, as for me myself am not sure, and he said his first impression was to do $ y(\lambda x +z) \le \sqrt{\frac{5}{2}} $

$  y \le\frac{\sqrt{\frac{5}{2}} }{(\lambda x +z)} $

any idea what to do next?

Slope = rise/run =   4/2 = 2

b = y-axis intercept = 3

y = 2x+3

y-2x = 3

-2x+y = 3

Solve -16t^2 - 24t + 160 = 0

Solutions are t = -3 and t = 7/2

We want t to be positive, so t = 3.5.

But as per the solution, at one point you've seem to take n instead of n₁ and n₂ what about that? That's why I presumed the solution might be holding for  n₁ = n₂

19x-8=12x+13       Subtract 12x from both sides of the equation to get

7x-8 = 13               add 8 to both sides to get

7x = 21                    divide both sides by 7

x = 3

Midpoint is   1,6          slope, m, between points is 4/-8 = -1/2    ( you should know how to do this)  

                                      perpindicular slope =  -1/m  =    2   ( you should know this too)

point slope form of your perpindicular line is then:

y-6 = 2 (x-1)    re-arrange

y-6 = 2x - 2

y = 2 x + 4          you can finish .....

Area of   ADE =  1/2 5 x 5 = 25/2

area of  ABC = 1/2  3 * AC  = 25/2

                             AC = 25/3

    AC - AE = EC = 25/3 - 5 = 10/3

the complex number -4 has a degree of 180 degrees. that means that the degrees of the solutions are going to be 180/4=45, 180/4+360/4=135, 180/4+360/4+360/4=225, and 180/4+360/4+360/4+360/4=315.

The magnitude of -4 is obviously 4, so the magnitude of the solutions are going to be 4^(1/4) = sqrt(2)

Therefore, the solutions in polar form are: \(\sqrt{2}(\cos(45)+i\sin(45)), \sqrt{2}(\cos(135)+i\sin(135)), \sqrt{2}(\cos(225)+i\sin(225)), \sqrt{2}(\cos(315)+i\sin(315))\)

Simplifying gives \(\boxed{-1-i, -1+i, 1-i, 1+i}\)

(notice that I gave a general way to solve this problem, but obviously, for this problem, you could factor it and solve it in a quicker way.)

x^2 = 152AB1

sqrt(152001) = 389.873056263

So the number x is slightly greater than 389.873056263. 

Since 152001 ends with a 1, the number x has to end with a 1 or 9. 

391^2 = 152881

=^._.^=

x =  1 + sqrt 2      and   2 - sqrt 7     are also  roots

We  have

(x - ( 1 + sqrt 2) ) ( x - (1 - sqrt 2)  )  ( x - ( 2 + sqrt 7))  ( x - (2 - sqrt 7))

Expanding  this we  get

x^4 - 6 x^3 + 4 x^2 + 10 x + 3

See here  :  https://web2.0calc.com/questions/andrew-chooses-a-number-from-1-to-100-and-mary-also

( If the product  is a multiple of 3, it is also  divisible  by 3  )

p/q = 5/6, q/r = 7/10, and r/s = 9/4

q = (6/5)p

q/r  =  7/10

(6/5)p  / r   =  7/10

r =  (6/5)p (10/7) =  (60/35)p  =  ( 12/7)p

r/s  =  9/4

(12/7)p / s   =  9/4

s   =  (12/7)p ( 4/9)  =   (48/63)p  =  ( 16/21)p

So

(p - q)  / (q - r)  =    (p - 6/5)p)  / ( 6/5 p  -  12/7 p)   =  (1/5 p )  / (-18/37 p )  =  (1/5) ( -37/18)   = -37/90

And

(2q - r)  / (r + s)  = [ (12/5)p - (12/7) p]  / (12/7 p  +  16/21 p)  =  (24/35 p)   /  (52/21 p) = 

(24/35) (21/52)  =  (24/52) ( 21/35)  =   (6/13) (3/5)   =   18/65