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By arithmetic geometric series, the sum of Gamma's sequence is 145.

If we have a double root.....the discriminant  =   0

So

b^2   - 4 *2 * 18   =  0

b^2  -144   =  0

b^2   =  144

b  =  12    or   b   = -12

We could use the quad formula.....but....by Vieta....here's an easier  solution method.....

u + v =   -5/3    ⇒  (u + v)^2 = 25/9 ⇒  u^2  +  2uv  + v^2   =  25/9    ⇒   u^2 + v^2  =   25/9 - 2uv

uv = 7/3   ⇒  2uv =  2(7/3)

u/ v +  v/ u   =

(u^2 + v^2)  / uv   =

( 25/9  - 2uv)  / ( 7/3)   =

(25/9  - 2 ( 7/3)) / (7/3)  =

(25/9  - 14/3)  / (7/3)  =

(25/9 - 42/9) / (7/3)  =

(-17/9) ( 3/7)  =

(-17/7) ( 3/9)  =

(-17/7) (1/3)  =

-17/21

By Vieta's Theorem

a +  b =   -1/7   ⇒   -4a   -4b =  4/7

ab  = -5/7

So

(a - 4) ( b  - 4)   =  

ab   + (- 4a - 4b )   +  16   =

-5/7   + 4/7   + 16   =

-1/7  +  16   =

111/7     (  CORRECTED  )

If they are multiples of both 4 and 6, they will be the multiples of lcm(4,6) = 12.

The multiples of 12 that are less than 100 are

12, 24, 36, 48, 60, 72, 84, 96.

So there are 8 numbers.

What are u and v.?     Use the quadratic formula    \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

You can do that bit yourself!

this is not a hard question.

MMi of 11 mod 1000 ==91

MMI of 2 mod 1000== 0

MMI of 13 mod 1000==77

1138  1146  1164  1183  1226  1234  1243  1262  1318  1324  1342  1381  1416  1423  1432  1461  1614  1622  1641  1813  1831  2126  2134  2143  2162  2216  2223  2232  2261  2314  2322  2341  2413  2431  2612  2621  3118  3124  3142  3181  3214  3222  3241  3412  3421  3811  4116  4123  4132  4161  4213  4231  4312  4321  4611  6114  6122  6141  6212  6221  6411  8113  8131  8311  Total =  64 such numbers

CD^2  =  [ AB + (CD - AB)  / 2 ] ^2  + AB^2

CD^2  =   [ AB  + CD ]^2  /  4  + AB^2

4CD^2  =  AB^2 + 2AB *CD + CD^2  + 4AB^2

3CD^2  = 2AB * CD  + 5AB^2

3CD^2   -2AB * CD  -5 AB^2  =   0

(3CD  - 5AB)  ( CD + AB)   =   0

Either   CD  =  - AB      (reject)    or

3CD  - 5AB   = 0

3CD   = 5AB

AB / CD   =  3 /  5

\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=\frac{15}{2}\)

\(15\cdot \:2^x=\frac{15}{2} \)
\(\frac{15\cdot \:2^x}{15}=\frac{\frac{15}{2}}{15}\)

\(2^x=\frac{1}{2}\)

\(x=-1\)

i=0;j=0;m=0;t=0;a=(9, 7, 5);r= (5,5,1);c=lcm(a); d=c / a[i];n=d % a[i] ;loop1:m++; if(n*m % a[i] ==1, goto loop, goto loop1);loop:s=(c/a[i]*r[j]*m);i++;j++;t=t+s;m=0;if(i< count a, goto4,m=m);printc,"m + ",t % c;return

OUTPUT: 315 m +  131, where m=0, 1, 2, 3........etc.

When m = 0, the smallest vlaue of N == 131

just ignore the base and just look at the exponent

so the equation becomes (x)+(x+1)+(x+2)+(x+3)=15/2. 

combine like terms -> 4x+6=15/2

subtract 6 from both sides and divide both sides by 4 to get x=3/8

I am not sure why this picture is blocked permanently, is there a way you can fix it so I can help you with this problem?

It's A. -17.34/3=-5.78 (It may look like C but they use the word "descended" so yeah.)

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Complete the square, to get 2(x + 3)^2 = 2x^2 + 6x + 18 ==> b = 6.

At the beginning, Rob has 16 glue sticks, Sydney has 8 glue sticks, Koreb has 24 glue sticks.

The OTHER x -intercept will be the same distance from the vertex...only on the OTHER side...

 6, 0 ------  12, -4  ------  18,0

The largest number of integer is 7 + 7 + 7 = 21.

w+x+y = -2

w+x+z = -4

w+y+z = 19

x+y+z = 12

subtract  the second equation from the  first  ⇒   y  - z =  2    ⇒   z = y  - 2   

subtract  the  third equation  from the second  ⇒   x - y = -23  ⇒  x =  y - 23

x +  y  +  z   =  12

(y -23) + y  + ( y  -2)  =   12

3y -25  = 12

3y = 37

y  = 37/3

z =  y  -2  =   37/3  -2  =  31/3

x =  t  -23  =   37/3 - 23   =  -32/3

w +  x + z    =   -4

w - 32/3  + 31/3   =  -4

w - 1/3   =  -4

w = - 4 + 1/3  =  -11/3

wx + yz   =

(-11/3) ( -32/3)  + ( 37/3) (31/3)     = 1499 /  9

A square sheet of paper ABCD is folded such that B falls on the midpoint M of CD. In what ratio does the crease divide BC?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

BC = 4           y = 4 - x          z = 2

Using Omi's diagram we have:

x² = y² - z²              x = sqrt{ (4 - x)² - 2²}             x = 1.5

Ratio  y : x = 5 : 3

The area of QUAD is 110.

f(3)  =  f(2)  + f(1)  = 10 + 3  =   13

f(4)  =  f(3)  + f(2)  =  13 + 10  =   23

f(5)  = f(4) + f(3)  =  23 + 13  =  36

f(6)  =  f(5)  + f(4)  =  36  + 23  =    59

(a) Let w = a + bi and z = c + di.  The rest is expanding.

(b) Let w = a + bi and z = c + di.  Then

\(\dfrac{w + z}{1 + wz} = \dfrac{a + c + bi + di}{1 + (a + bi)(c + di)}\)

To express this in rectangular form, we can multiply the numerator and denominator by the conjugate:

\(\dfrac{a + c + bi + di}{1 + (a + bi)(c + di)} = \dfrac{(a + c + bi + di)((1 - (a + bi)(c + di))}{(1 + (a + bi)(c + di))(1 - (a + bi)(c + di))}\)

The denominator simplifies to (1 - (a^2 + b^2)(c^2 + d^2)), which is real.  The numerator simplifies to a^2 - b^2 + c^2 - d^2, which is also real.  Therefore, the complex number (z + w)/(zw + 1) is real.