hectictar

$501+503+\dots+697+699\\~\\ =\quad (501+2\cdot0) + (501+2\cdot1) +\dots+(501+2\cdot98) +(501+2\cdot99) \\~\\ =\quad (499+2\cdot1) + (499+2\cdot2) +\dots+(499+2\cdot99) +(499+2\cdot100) \\~\\ =\quad \sum\limits_{n=1}^{100}(499+2n) \\~\\ =\quad \sum\limits_{n=1}^{100}499+ \sum\limits_{n=1}^{100}2n \\~\\ =\quad 100\cdot499+ 2\cdot\sum\limits_{n=1}^{100}n \\~\\ =\quad 49900+ 2\cdot\frac{100(100+1)}{2} \\~\\ =\quad 49900+ 10100\\~\\ =\quad60000$.

Let's find what value(s) of  x  make  f(x) = 8

f(x)  =  8

x² - 6x + 4  =  8

x² - 6x - 4  =  0

By the Quadratic formula,

$x\ =\ \frac{6\pm\sqrt{(-6)^2-4(1)(-4)}}{2(1)}\ =\ \frac{6\pm\sqrt{52}}{2}\ =\ \frac{6\pm2\sqrt{13}}{2}\ =\ 3\pm\sqrt{13}$

And so....

$f(3+\sqrt{13})=8$       and       $f(3-\sqrt{13})=8$

First let's plug in  3 + √13  for  x  into the function  g( f(x) )  =  2x + 3

$g(\ 8\ )\ =\ g(\ f(3+\sqrt{13})\ )\ =\ 2(3+\sqrt{13})+3\ =\ 9+2\sqrt{13}$

Next let's plug in  3 - √13

$g(\ 8\ )\ =\ g(\ f(3-\sqrt{13})\ )\ =\ 2(3-\sqrt{13})+3\ =\ 9-2\sqrt{13}$

And so the sum of all the possible values of  g(8) is:

$(9+2\sqrt{13})\ +\ (9-2\sqrt{13})\ =\ 18$ .

y² + 28y + 16 + k   will be the square of a binomial when:

16 + k  =  (1/2 * 28)²

16 + k  =  (14)²

16 + k  =  196

k  =  180

Check:

y² + 28y + 16 + 180   =   y² + 28y + 196   =   (y - 14)(y - 14)   =   (y - 14)²

7x²  + 8x  =  h       Let's subtract  h  from both sides to get

7x² + 8x - h  =  0

This will have only one root/solution when the discriminate is 0,  that is, when:

8² - 4(7)(-h)  =  0

64  +  28h  =  0

Can you finish solving the above equation for  h  ?

Whatever value of  h  we get from that equation is the value of  h  which gives us exactly one solution for  x .

b)  qualitative

c)  discrete

d)  continuous

First let's look at  △ABC.  By the Pythagorean Theorem,

AC²   =   AB² + BC²   =   ( 3√2 )² + ( 3√2 )²   =   18 + 18   =   36    and so    AC  =  6

△ABC  is a right isosceles triangle which means the base angles are both  45°

And so   m∠CAD  =  m∠BAD - m∠BAC   =   75° - 45°   =   30°

Now we can see that  △ACD  is a  30-60-90 triangle,

and in all 30-60-90 triangles, the side across from the 30° angle is half the hypotenuse.

In this case, that means   CD  =  1/2 * AC   =   1/2 * 6   =   3

E  is the midpoint of  AC,  so  AE  is half the length of  AC,  or   AE  =  1/2 * AC

One way to find the scale factor from  △ABC  to  △AFE  is to find  AE / AC

AE / AC   =   (1/2 * AC) / AC   =   1/2

So the scale factor is 1/2

(Remember the scale factor is the number which each side of the starting triangle must be multiplied by in order to make it the same size as the "goal" triangle. In this case, each side of △ABC must be multiplied by 1/2 to make it the same size as △AFE.)

IDK if this is the best way...but we can first simplify it like this:

z⁴⁹ + z⁵⁰ + z⁵¹ + z⁵² + z⁵³

                                                        Factor  z⁴⁹  out of the first three terms and  z⁵¹ out of the last two terms

=  z⁴⁹(1 + z + z²)  +  z⁵¹(z + z²)

                                                        Since  z² + z + 1 = 0,   1 + z + z² =  0   and   z + z²  =  -1

=  z⁴⁹( 0 )  +  z⁵¹( -1 )

=  -z⁵¹

Now by the quadratic formula,  $z\ =\ \frac{-1\pm\sqrt{1^2-4(1)(1)}}{2(1)}\ =\ \frac{-1\pm\sqrt{-3}}{2}\ =\ -\frac12\pm \frac{\sqrt3}{2}i$

Let's pick  $z\ =\ -\frac12 + \frac{\sqrt3}{2}i$     (If we picked  $z\ =\ -\frac12 - \frac{\sqrt3}{2}i$   we would get the same answer)

Now let's re-express  z  to be in the form  $r(\cos\theta+i\sin\theta)$  so that we can use DeMoivre's Theorem.

By the Pythagorean Theorem,

$r^2\ =\ (-\frac12)^2+(\frac{\sqrt3}{2})^2\ =\ 1$   so taking the positive sqrt, we get     $r \ =\ 1$

An angle which has a cos of  $-\frac12$  and a sin of  $\frac{\sqrt{3}}{2}$  is  $\frac{2\pi}{3}$,  so let  $\theta\ =\ \frac{2\pi}{3}$

And so...

$z\ =\ 1(\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3}))$          (we can check in a calculator that this does equal $-\frac12 + \frac{\sqrt3}{2}i$ )

Then by DeMoivre's Theorem,

$z^{51}\ =\ (1)^{51}(\cos(51\cdot\frac{2\pi}{3})+i\sin(51\cdot\frac{2\pi}{3}))\\~\\ z^{51}\ =\ (1)^{51}((1)+i(0))\\~\\ z^{51}\ =\ (1)^{51}\\~\\ z^{51}\ =\ 1$

And so...

$-z^{51}\ =\ -1$

The question does not ask for the value of y. It is asking for the first equation solved for  y .

5x + y  =  31

Subtract  5x  from both sides of the equation to get

y  =  31 - 5x

This is the first equation solved for  y

From the sum formula for cosine:

cos(a + B) = cos a cos B  -  sin a sin B

Since  csc a  =  3 :

sin a  =  1/3

Since  sin a  =  1/3  and  cos²  a  =  1 - sin² a   :

cos²a  =  8/9       And since  a  is in quadrant II,

cos a  =  -√8 / 3

We can again use the Pythagorean identity and the fact that B is in quadrant II to get:

sin B  =  1/ √10

cos B  =  -3 / √10

And so...

cos(a + B) = (-√8 / 3)(-3 / √10)  -  (1/3)(1/ √10)

Plug this into a calculator or simpify it to get a final answer.....I don't have time to write a better explanation right now, but if you have any questions about this please ask!