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This problem requires a conversion from feet to inches. The area of the poster board is 6 feet squared, which is same as 6*12*12 inches squared or 864 inches squared. One sticker is 4 inches squared. So Julian needs 864/4=216 stickers.

we can make it simple from an example like if we toss a coin we can get 2 possible outcome

if

 we toss 2 coin we can get 2 * 2 possible outcome

 if we toss 3 coin we can get 2* 2 *2  possible outcomes 

so tossing a dice 6 times is same as tossing 6 dice one time. 

for one fair dice the possible no of outcome is 6  {1,2,3,4,5,6}

for 2 dice possible outcomes 6*6 

for 3 dice possible outcomes 6*6 *6

for 4 dice possible outcomes 6*6*6*6 

for 5 dice possible outcomes 6*6*6*6*6

for 6 dice possible outcomes 6*6*6*6*6*6  = 6 ^6             [answer]

x² + 12x + k  =  18x + 3     --->     x² - 6x + k - 3

Using the quadratic formula with  a = 1,  b = -6,  c = k - 3:

x  =  [ - -6  +/-  sqrt( (-6)² -  4(1)(k - 3) ) ]  /  2(1)

x  =  [ 6  +/-  sqrt( 36 - 4k + 12 ) ]  /  2

x  =  [ 6  +/-  2 sqrt( 12 - k ) ] /  2

x  =  3  +/-  sqrt( 12 - k )

The larger root (the "+" root) is twice as large as the smaller root (the "-" root):

3 + sqrt( k - 12 )  =  2(3 - sqrt( 12 - k )

3 + sqrt( 12 - k )  =  6 - 2 sqrt( 12 - k )

   3 sqrt( 12 - k )  =  3

      sqrt( 12 - k )  =  1

 [ sqrt( 12 - k ) ]²  =  (1)²

                 12 - k  =  1

                        k  =  11

When x = - pi/12     what is sin   -3 pi/12  ?         =      - sqrt2/2

-2  *  -sqrt2/2  +1 = a

Do simialr for 'b'        2* sin 3b + 1 = 1 

                                  2 * sin 3b = 0

                                        sin 3b = 0

                                           b =    pi/3

A circular planetary ring of radius 125000 km completes one full revolution in 120000 hours.

a)  What is the angular velocity of the ring? (in radians/hour)

b)  What is the linear velocity of a rock on the ring? (in km/hr)

Hello Guest!

a)

\(\omega=\dfrac{2\pi}{12\cdot 10^4hour}\\ \color{blue}\omega=\dfrac{0.1\overline 6\cdot 10^{-4}\large \pi}{hour}\)

b)

\(v_{rock}=\omega \cdot r=\dfrac{0.1\overline 6\cdot 10^{-4}\large \pi}{h}\cdot 12.5\cdot 10^4km\)

\(v_{rock}=6.545km/h\)

  !

thank you so much! that definitely helped me! :D

300 *   1.20 =  360 miles     

3 x^2      +     x^2      = 4 x^2

Ingredients in = ingredients out

.50 x  +   1.0 (300-x)   =   .8 ( 300)   solve for 'x'    the ml of 50%     then 100% is  300 - x

Vertex form of parabola

y = a (x+4)^2 + 8        a = 1 in this case

y = x^2 + 8x + 16 + 8        you can finish this I am sure !  

One revolution is 2pi Radians in 120000 hours

2pi/120000    R/hr

circumference = pi * 2r

  velocity is then      pi * 2 * 125 000 / 120000     km/hr

If a is three times b, b is twice c, and c is four times d, what is the value of (a + c)/(b + d) ?

Hello Guest!

\(a=3b\\ b=2c\\ c=4d\\ \)

\(\dfrac{a+c}{b+d}=\dfrac{3b+4d}{2c+d}=\dfrac{6c+4d}{2c+d}=\dfrac{6c+c}{2c+\dfrac{c}{4}}=\dfrac{7c}{\dfrac{9c}{4}}=\color{blue}\dfrac{28}{9}\)

  !

How do I find the are of a convex quadrialater with vertices P1(-1,0),P2(0,1),P3(3,0) and P4(0,-5)?

Hello Guest!

\(A=\dfrac{(y_2-y_4)(x_3-x_1)}{2}\\ A=\dfrac{(1-(-5))(3-(-1))}{2}\\ A= \dfrac{6\cdot 4}{2}\)

\(A=12\)

  !

There are 52 terms in the sequence. If you divide each of the numbers in 2: 2,4,6,...,102,104 = 1,2,3,...,51,52.

2+104 = 4+102 = 6+100 which means ((2+104)x52)/2 equals the sum of the sequence, which is 2756, and if you divide that by 7 you get 393 r5.

So, the answer is 5. 

The newer train can go 350 miles.

permutations of {1, 2, 2, 3, 4, 4} ==6!/(2!)(2!)==720 / 4 ==180 sequences of rolls

\(\frac{1}{x} + \frac{1}{y} = \frac{1}{18}\\ \frac{x+y}{xy} = \frac{1}{18}\\ xy=18x+18y\\ xy-18x-18y=0\\ (x-18)(y-18)=324\)

The pair of numbers that multiplies to 324 that are closest to each other but are not equal to each other are 12 and 27, so the answer is (18+12)+(27+18)

btw a very similar question was already answered here:

Thats wrong, sorry

You could also look up "method of differences" or "Lagrange interpolation" for faster computation. I'm not really familiar with it, but they both look faster than solving 6 simultaneous equations.

Since we need to compute the last 3 digits, we can take the whole expression mod 1000:

\(1991\cdot1993\cdot1995\cdot1997\cdot1999\cdot2001\cdot2003 (\text{mod 1000})\\ =-9\cdot-7\cdot-5\cdot-3\cdot-1\cdot1\cdot3 (\text{mod 1000})\\ =-2835 (\text{mod 1000})\\ =165(\text{mod 1000})\)

The sum of the hundreds, tens, and units digit is \(1+6+5=\boxed{12}\)

-2/3t -7 <0       and      -4t+9 >= 0

t > - 21/2                           t <= 9/4

(-21/2, 9/4]

|5a+2| <9

if 5a+2 is positive            5a+2 < 9

                                           a < 7/5

if 5a+2 is negative          5a+2 > -9

                                           a >  -11/5

-11/5 < a < 7/5               or     ( -11/5 , 7/5)

45 is incorrect