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\(\text{well just trying the first two primes, 2 and 3, for p and q we get}\\ 7p+q = 14+3 = 17 \text{ is prime}\\ pq + 11 = 2\cdot 3 + 11 = 6 + 11 = 17 \text{ is also prime}\\ (2^2 + 3^2)(3^2+2^3) = (4+9)(9+8) = 13 \cdot 17 = 221\)

\(\ell = A +\lambda (B-A) = (3,-2,4) + \lambda(-2,-3,2)\)

\(\vec{u}=(3,1)\\ \vec{v} = (2,4)\\ p_{\vec{u}}(\vec{v}) = \dfrac{\vec{v}\cdot \vec{u}}{|\vec{u}|^2}\vec{u} = \\ \dfrac{(3,1)\cdot(2,4)}{|(3,1)|^2} \vec{u}= \dfrac{10}{10}(3,1) =(3,1)\)

\(55 = m^2 - n^2\\ 48 = 2 m n\\ 73 = m^2 + n^2\)

\(55 = 5 \cdot 11 = (m+n)(m-n)\\ m+n=11\\ m-n=5\\ m = 8,~n=3\)

ohh thnxx

(1+2i)(1+2i)

expand it you have

1+2i+2i+4i^2 

i^2 is equal to -1

so 1+2i+2i+4(-1)

which gives u

1-4+4i

which is 

A: -3+4i

An artifact was found to have an original amount of Carbon-14 of 32 grams. Approximately how many grams of Carbon-14 remain after 4300 years? Carbon 14 decays at a rate of -0.00012 grams per year.

R =32 x 1/2^(4300/5730)
R =32 x 1/2^(0.7504363)
R =~19.1 grams

We have to choose 1 of each color so

\(p= \dfrac{\dbinom{3}{1}\dbinom{4}{1}\dbinom{5}{1}}{\dbinom{12}{3}} = \dfrac{3\cdot 4\cdot 5}{220}=\dfrac{3}{11}\)

All of the following numbers will give the same remainder for both, 6 and 8 ranging from 0 to 5:
n =100, 101, 120, 121, 122, 123, 124, 125, 144, 145,146, 147, 148, 149, 168, 169,170, 171, 172, 173, 192, 193, 194, 195, 196, 197. 

Mmmmm...yes.... I missed that...

The distinct results are 

1 + 4 + 5 + 9 + 3   =

22 = m

m / 11  = 2

Thanks, guest...for catching that....!!!

CPhill: When he/she says "add up all the distinct results", does that mean "no duplicates"?

It will have to be an even-numbered year  since 697 * odd  will not be evenly divisible by 12

A little trial and error shows that the next time that the year  will begin on the first day of the week will be  Year 12

[and every 12 years after that ]

1^2 /11  = R 1

2^2 /11 =   R4

3^2 /11 = R9

4^2 /11 = R 5

5^2 /11  = R 3

6^2 /11 = R 3 

7^2/11 = R 5

8^2 / 11 = R9

9^2 /11 = R 4

10^2 /11 = R1

m = [ 1 + 4 + 9 + 5 + 3 + 3 + 5 + 9 + 4 + 1 ]  =  44 / 11  =  4

4) the other x-intercept would be the same distance from the axis of symetry as the x-intercept that they have given you.

If it is 3 places over than the other one would be (-3) places from the axis of symetry.

Using the Triangle Inequality we have that

8 + 15 > x

23 > x

Also

8 + x > 15

x > 7

So.....the unknown side x  can be

7 < x < 23

And there are   15  integers between 7 and 23

Got it, Android...

4.  I need to know what the first x-intercept is.....

5. The x-intercepts of the parabola y=x^2+bx+c are (-3,0) and (5,0) Find the equation of the parabola, and submit your answer in y=ax^2+bx+c form.

If (-3, 0)  and (5, 0)  are x intercepts...then they are also roots

So  we have that

(x -   - 3) ( x - 5) = 0

(x + 3) ( x - 5) = 0      expand  and simplify

x^2 - 2x  - 15   =  0

So....the equation becomes

y =  x^2   - 2x - 15

Since these triangles are similar....angle BCA = angle EFD

And ED = EF....so angle EDF = angle EFD

So

Angle DEF + Angle EDF + Angle EFD = 180

42  + Angle EFD    +  Angle EFD =  180

2 *Angle EFD =  180 - 42

2 * Angle EFD = 138     divide both sides by 2

Angle EFD =  69°   =  Angle BCA

just need help with 4 & 5 please?

If we draw  a perpendicular  from D to CB....we will have a right triangle with a hypotenuse of 13  and a height DM (leg) of 12

So....call the intersection of this perpendicular and CB, M

So  using the Pythagorean Theorem.....we can find  CM   as

CM  =  sqrt ( DC^2 - DM^2 )   =   sqrt (13^2 - 12^2 ) =  sqrt (169 - 144) =  sqrt (25)  = 5

So....M is the point  (5, 0)......and it has the same x coordinate as D

2.Find the vertex of the graph of the equation y=3x^2-6x+7

Similar to the last one

x cooordinate of the vertex is  - [ -6] /  [2 * 3 ]  =  6 / 6   =  1

And the y coordinate of the vertex is   y =  3(1)^2 - 6(1) + 7  =  3 -  6 + 7   =  4

So...the vertex is  (-1, 4)

1. Find the vertex of the graph of y=x^2+8x+23

In the form    y = ax^2 + bx + c.....the x coordinate of the vertex is given by   -b / [ 2a ]

So

The x coordinate of the vertex is   -8/ [ 2 * 1 ] =  -8/2  = -4

And the y coordinte of the vertex is    y = (-4)^2 + 8(-4) + 23  =  16 - 32 + 23 = 7

So....the vertex =   (-4, 7)

All prime numbers are odd except for 2....so the product  of each of these odds and 11 will be odd

So....the only even product is   2 * 11   =  22

Maybe easier ways to do this....but....

Let H =   (0, 0)     A = (0, 12)   C = (4, 8)    and  G = (4, 0)

The slope of the line containing  AG   =  [ -12/4] = -3

And the equation of the line on which AG lies is   y = -3x + 12   (1)

The slope of the line containing HC is  8/4 = 2

And the equation of the line on which HC lies is   y = 2x   (2)

So....setting (1) = (2) to find the x coordinate of M, we have that

-3x + 12 = 2x

12 = 5x

12/5 = x

And y = 2(12/5)  = 24/5

Connecting LC  and calling the intersection of LC and  AG   =  N

Using similar triangles

Triangle ALN is similar to triangle AHG

So LN /AL = HG / AH

LN /4 = 4 / 12

LN /4 = 1/3

LN = 4/3

So.....NC is the base of trapezoid ABCN  = LC - LN =  4 - 4/3  = 8/3.....and the area of this trapezoid is

(1/2)BC ( AB + NC) =   (1/2)(4)(4 + 8/3)  = 2 ( 20/3) = 40/3 units^2      (3)

And triangle NCM  has a base of NC =  11/3   and a height of  8 - 24/5 =  16/5

So...the area of this triangle is  (1/2)(11/3)(16/5) =  88/15  units^2    (4)

So...[ ABCM ]   =   (3) + (4)   =   [ 40/3 + 88/15 ]  =  [288/15] units^2   =  19.2 units^2

144 / .80    =    180 boxes sold

I will give this a go!

So you have 100 groups of 1,000 people each.

Let us designate the two persons "A" and "B".

"A" can be in any of the 100 groups, or 100/100.

The probability that "B" will be in a different group than "A" is 99/100.

So, the probability that "A" and "B" will be in different groups is: 100/100 x 99/100 =99%.

Therefore, the probability that they will be in the same group is: 1 - 99% =1%