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*sorry, i meant the first function. The second seems to intersect a few more times, but I'm not sure whether to count the intersections of each of the pieces with the third individually, or as a whole. i.e. individually, there would be around 2007*2 intersections, yet as a whole it would be around 2007.

GREAT SOLUTION

thank you so much :D

sin 60°  =  altitude   / side length

sqrt (3) / 2   =  27 sqrt (3)   / side length

1/2   =  27  / side length

side length  =   27 / (1/2)  =    54

a^2  - b^2   +  3a   +  b

(a + b) (a  - b)  +  3a  + b

(-1)(a - b)   + 3a  + b  =

-a + b  +  3a  +  b  =

2a + 2b  = 

2 ( a + b)   =

2 (-1)    =  

-2.......just  as HSD  found  !!!!!!

Note the entire expression is 2(a + b), so substitution gives a^2 - b^2 + 3a + b = -2.

3x^2  -  6x  - 2  +   8x  +  14   =

3x^2  +  2x  + 12

3 ( x^2  + (2/3)x  + 4)    complete  the  square on  x inside the parentheses

3 ( x^2   +  (2/3)x  +  1/9  +  4  -  1/9)

3  [ ( x + 1/3)^2   +  35/9 ]  

3 ( x + 1/3)^2   +  35/3

a  +  h  +   k  = 

3   +  (1/3)   +  35/3   =

45/3   =

15

Find all pairs (x,y) of real numbers such that
\(x + y = 10 \\ x^2 + y^2 = 56 + xy.\)

\(\begin{array}{|rcll|} \hline a+b &=& 10 \\ \mathbf{b} &=& \mathbf{10-a} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (x+y)^2 &=& x^2+2xy+y^2 \\ (x+y)^2 &=& x^2+y^2 +2xy \quad | \quad x^2 + y^2 = 56 + xy \\ 10^2 &=& 56 + xy +2xy\\ 10^2 &=& 56+3xy \\ 3xy &=& 100-56 \\ 3xy &=& 44 \quad | \quad : 3 \\ xy &=& \dfrac{44}{3} \quad | \quad y=10-x \\ x(10-x) &=& \dfrac{44}{3} \\ 10x-x^2 &=& \dfrac{44}{3} \\ \mathbf{x^2-10x+ \dfrac{44}{3}} &=& \mathbf{0} \\ \hline x &=& \dfrac{ 10\pm \sqrt{10^2 -4*\dfrac{44}{3}} }{2} \\ \\ x &=& \dfrac{ 10\pm \sqrt{4*25 -4*\dfrac{44}{3}} }{2} \\ \\ x &=& \dfrac{ 10\pm 2\sqrt{25 -\dfrac{44}{3}} }{2} \\ \\ x &=& 5\pm \sqrt{25 -\dfrac{44}{3}} \\ \\ x &=& 5\pm \sqrt{\dfrac{75-44}{3}} \\ \\ \mathbf{x} &=& \mathbf{5\pm \sqrt{\dfrac{31}{3}}} \\ \hline \end{array}\)

1.

\(\begin{array}{|rclrcl|} \hline \mathbf{x} &=& \mathbf{5+ \sqrt{\dfrac{31}{3}}} \\ &&&y &=& 10-a \\ &&&&=& 10-\left(5+ \sqrt{\dfrac{31}{3}}\right) \\ &&&\mathbf{y} &=& \mathbf{5- \sqrt{\dfrac{31}{3}}} \\ \hline \end{array}\)

2.

\(\begin{array}{|rclrcl|} \hline \mathbf{x} &=& \mathbf{5- \sqrt{\dfrac{31}{3}}} \\ &&&y &=& 10-a \\ &&&&=& 10-\left(5- \sqrt{\dfrac{31}{3}}\right) \\ &&&\mathbf{y} &=& \mathbf{5+ \sqrt{\dfrac{31}{3}}} \\ \hline \end{array}\)

Find all solutions to the system
\(a + b = 14\\ a^3 + b^3 = 812 + 3ab\)

\(\begin{array}{|rcll|} \hline a+b &=& 14 \\ \mathbf{b} &=& \mathbf{14-a} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (a+b)^3 &=& a^3+3a^2b+3ab^2+b^3 \\ (a+b)^3 &=& a^3+b^3 +3ab(a+b) \quad | \quad a+b = 14 \\ 14^3 &=& a^3+b^3 +3*14ab \quad | \quad a^3+b^3 = 812+3ab \\ 14^3 &=& 812+3ab +3*14ab \\ 14^3 &=& 812+3*15ab \\ 3*15ab &=& 14^3- 812 \\ 3*15ab &=& 1932 \quad | \quad : 3 \\ 15ab &=& 644 \quad | \quad b=14-a \\ 15a(14-a) &=& 644 \\ a(14-a) &=& \dfrac{644}{15} \\ 14a-a^2 &=& \dfrac{644}{15} \\ \mathbf{a^2-14a+ \dfrac{644}{15}} &=& \mathbf{0} \\ \hline a &=& \dfrac{ 14\pm \sqrt{14^2 -4*\dfrac{644}{15}} }{2} \\ \\ a &=& \dfrac{ 14\pm \sqrt{4*49 -4*\dfrac{644}{15}} }{2} \\ \\ a &=& \dfrac{ 14\pm 2\sqrt{49 -\dfrac{644}{15}} }{2} \\ \\ a &=& 7\pm \sqrt{49 -\dfrac{644}{15}} \\ \\ a &=& 7\pm \sqrt{\dfrac{735-644}{15}} \\ \\ \mathbf{a} &=& \mathbf{7\pm \sqrt{\dfrac{91}{15}}} \\ \hline \end{array}\)

1.

\(\begin{array}{|rclrcl|} \hline \mathbf{a} &=& \mathbf{7+\sqrt{\dfrac{91}{15}}} \\ &&&b &=& 14-a \\ &&&&=& 14-\left(7+\sqrt{\dfrac{91}{15}}\right) \\ &&&\mathbf{b} &=& \mathbf{7-\sqrt{\dfrac{91}{15}}} \\ \hline \end{array} \)

2.

\(\begin{array}{|rclrcl|} \hline \mathbf{a} &=& \mathbf{7-\sqrt{\dfrac{91}{15}}} \\ &&&b &=& 14-a \\ &&&&=& 14-\left(7-\sqrt{\dfrac{91}{15}}\right) \\ &&&\mathbf{b} &=& \mathbf{7+\sqrt{\dfrac{91}{15}}} \\ \hline \end{array} \)

Write the scale of the drawing as a fraction. Put the shorter length in the numerator.
 

7 mm/1 m

First, convert the denominator to the same units as the numerator. Write an equivalent fraction with millimetres in the denominator. Since 1 metre is equal to 1000 millimetres, multiply the denominator by 1000 to get millimetres in the denominator.
 

7 mm/1 m = 7 mm/1000 mm

Next, drop the units. The units will cancel out because the numerator and denominator are both measured in millimetres.
 

7 mm/1000 mm = 7/1000

The scale factor is 

7/1000.

Using Law of Sines

sin 60 / 12 = sin b / 10

b = 46.194 degrees

angle A = 180 - 46.194 - 60 = 73.8059 degrees

12 / sin 60 = a / sin 73.8059

a = 13.31 units

Find \(14^{-1} \pmod{35}\), as a residue modulo 35.

(Give an answer between 0 and 34, inclusive.)

Because \(\gcd(14,35) = 7 \ne 1\), so

14 is not invertible modulo 35

2020

2024

    28

    32

    36

    40

    44

2048  

I believe that would be the   5/6                     4/5 3/4 2/3  1/2   are all outside the interval

If \(3x+7 \equiv 11 \pmod{16}\),
then 2x+11 is congruent \(\pmod{16}\) to what integer between 0 and 15, inclusive?

\(\begin{array}{|rcll|} \hline 3x+7 &\equiv& 11 \pmod{16} \quad | \quad -7 \\ 3x &\equiv& 11-7 \pmod{16} \\ 3x &\equiv& 4 \pmod{16}\quad | \quad :3 \\ x &\equiv& \dfrac{4}{3} \pmod{16} \\ x \pmod{16} &\equiv& \dfrac{4}{3} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline 2x+11\pmod{16} &\equiv& 2*\dfrac{4}{3}+11 \pmod{16} \\ &\equiv& \dfrac{8}{3}+11 \pmod{16} \\ &\equiv& \dfrac{8}{3}+\dfrac{33}{3} \pmod{16} \\ &\equiv& \dfrac{41}{3}\pmod{16} \quad | \quad 41\equiv 9 \pmod{16} \\ &\equiv& \dfrac{9}{3}\pmod{16} \\ \mathbf{2x+11\pmod{16}} &\equiv& \mathbf{3 \pmod{16}} \\ \hline \end{array}\)

x ( y + 4)  = -7 - 2y

x = (-7-2y) / (y+ 4)               y cannot equal -4

The cosine rule here can be written as:

12^2 = a^2 + 10^2 - 2*a*10*cos(60deg)

or  a^2 - 20*cos(60deg)*a - 44 = 0

solve the quadratic equation to find a.

x   and   x+1   are the numbers

x^2  +   4 (x+1)  = 64

x^2 + 4x -60 = 0

(x+10)(x-6) = 0         x = -10 or 6      take 6 since we are looking for positive integers    then x+1 = 7

There will be TWO for each 360 degrees 

    3600 / 360 = 10      sets of 360 degrees

    10 * 2 = 20  values

Find two integer points on the line   like   -5, -3       and   0,1   to compute the slope = m = 2

y = 2x + b    b is the y axis intercept = 1

so the line equation is y = 2x + 1

The shaded aea is everything ABOVE the line....but does not include the dashed line:

     y > 2x+1      now re-arrange into the requaested form

   -2x + y -1 > 0       Done !

To find the value of a that makes the curve continuous at the upper intersection, set x = 2 in the upper two expressions and equate them:

i.e.   a*(2) + 3 = (2) + 5

or   2a+3 = 7

      a = 2

Similarly to find the value of b that makes the curve continuous at -2, set

 (-2) + 5 = 8*(-2) + b

I'll let you finish.

Answer: \((-\frac5 3,\frac8 3) \)

Solution:

The right side of the equation can be turned into \(A(x+1) + B(x-2)\over x^2 -x-2\). After that, you can multiply both sides of the equation to get \(x-7 = A(x+1) + B(x-2)\).

The coefficient of the x on the left, which is one, is equal to A + B, because the x term on the left side is made up of Ax + Bx. The constant term on the left side, -7, is equal to A-2B. This is because when expanding out the terms on the right, you would get Ax +\(A\)+Bx \(-2B\). From this information, you get two equations:

\(A+B=1\)

and

\(A-2B=-7\)

Subtracting the second equation from the first gives 3B = 8, which means that B = 8/3. Substituting this value into the first equation gives A + 8/3 = 1. Solving this gives A = 1-8/3 = -5/3.

Putting this into an ordered pair gives \((-\frac5 3,\frac8 3) \).

Hello Guest!

The graph is continuous when a = 0 and b = - 9.

 !