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Coordinates plane H(-5,-6) & J(5,-1).
Find the Point Q that is 4/5 the distance from J to H

Formula:

$\begin{array}{|rcll|} \hline \mathbf{\vec{Q}} &\mathbf{=}& \mathbf{(1-\lambda)\vec{J} + \lambda\vec{H}} \quad & | \quad\lambda = \dfrac45 \\\\ \vec{Q} &=& \left(1-\dfrac45 \right)\vec{J} + \dfrac45\cdot \vec{H} \\\\ &=& \dfrac15 \vec{J} + \dfrac45 \vec{H} \quad & | \quad \vec{J} = \dbinom{5}{-1} \quad \vec{H} = \dbinom{-5}{-6} \\\\ &=& \dfrac15 \dbinom{5}{-1} + \dfrac45 \dbinom{-5}{-6} \\\\ &=& \dbinom{1}{-\dfrac15} + \dbinom{-4}{-\dfrac{24}{5}} \\\\ &=& \dbinom{1}{-\dfrac15} - \dbinom{4}{\dfrac{24}{5}} \\\\ &=& \dbinom{1-4}{-\dfrac15-\dfrac{24}{5 }} \\\\ &=& \dbinom{-3}{-\dfrac{25}{5 }} \\\\ \mathbf{\vec{Q}} & \mathbf{=}& \mathbf{\dbinom{-3}{-5}} \\ \hline \end{array}$

point A is located at (1,4). Point P ar (3,5) is 1/3 the distance from A to point B.

What are the coordinates of point b (SHOW WORK)

Formula:

$\begin{array}{|l|rcll|} \hline & \mathbf{(1-\lambda)\vec{A} + \lambda\vec{B}} &\mathbf{=}& \mathbf{\vec{P}} \\ \hline \lambda = 0 & (1-0)\vec{A} + 0\cdot \vec{B} &=& \vec{A} \ \checkmark \\\\ \lambda = 1 & (1-1)\vec{A} + 1\cdot \vec{B} &=& \vec{B} \ \checkmark \\\\ \lambda = \dfrac13 & \left(1-\dfrac13 \right)\vec{A} + \dfrac13\cdot \vec{B} &=& \vec{P} \quad & | \quad \vec{B} =\ ? \\ \hline \end{array}$

$\vec{B} =\ ?$

$\begin{array}{|rcll|} \hline \left(1-\dfrac13 \right)\vec{A} + \dfrac13\cdot \vec{B} &=& \vec{P} \quad & | \quad \vec{B} =\ ? \\\\ \dfrac23 \vec{A} + \dfrac13 \vec{B} &=& \vec{P} \quad & | \quad \cdot 3 \\ \\ 2 \vec{A} + \vec{B} &=& 3 \vec{P} \quad & | \quad - 2 \vec{A} \\\\ \mathbf{\vec{B}} &\mathbf{=}& \mathbf{3 \vec{P} - 2 \vec{A}}\quad & | \quad \vec{A} = \binom{1}{4} \quad \vec{P} = \binom{3}{5} \\\\ \vec{B} &=& 3\binom{3}{5} - 2 \binom{1}{4} \\\\ \vec{B} &=& \binom{9}{15} - \binom{2}{8} \\\\ \vec{B} &=& \binom{9-2}{15-8} \\\\ \mathbf{\vec{B}} &\mathbf{=}& \mathbf{\binom{7}{7}} \\ \hline \end{array}$

Simple correction. heureka made a simple typo:

80 x 111 =8,880. The smallest integer.

Mike draws five cards from a standard 52-card deck. What is the probability that he draws a card from at least three of the four suits? Express your answer as a simplified fraction.

I'll give it a go :)

There are 52*51*50*49*48 =  311875200  permutationss of 5 cards altogether (with no restrictions)

I'm going to try and work out how many of these do not include at least 1 card from 3 different suits.

5 hearts

1*(13*12*11*10*9) 

4 hearts and one other card

5*(13*12*11*10*39)

3 hearts and 2 from one other suits

5C2* (13*12*11*13*12)*3  

And this can be multiplied by 4 beacuse there are 4 suits 

so

 Number of combinations without 3 different suits represented

= [1*(13*12*11*10*9)   +  5*(13*12*11*10*39)   +  5C2* (13*12*11*13*12)*3 ]    *4

=(13*12*11*4)[ 90  +  5*(390)   +  10* (156)*3  ]

= [(13*12*11*4)[ 90  +  1950   +  4680  ]

= [(13*12*11*10*4)[ 9  +  195  +  468  ]

= 13*12*11*10*4* 672 

So the number of combinations with at least 3 different suites represented 

=  52*51*50*49*48 - 13*12*11*10*4* 672 

= 12*4   [ 52*51*50*49   -   13*11*10* 672 ]

= 12*4*10*13   [ 4*51*5*49   -   11* 672 ]

= 12*4*10*13*4   [ 51*5*49   -   11* 168 ]

= 12*4*10*13*4 *7  [ 51*5*7   -   11* 24 ]

= 12*4*10*13*4 *7 *3 [ 17*5*7   -   11* 8 ]

= 3*4*4*7 *10*12*13 [ 595   -   88 ]

= 3*4*4*7 *10*12*13 *507

So the prob of not getting at least 3 suites 

$=  \dfrac{3*4*4*7 *10*12*13 *507}{52*51*50*49*48}\\ =  \dfrac{507}{17*5*7}\\ =  \dfrac{507}{595}\\$

What is the smallest positive integer N  for which

$(12{,}500{,}000)\cdot n$ 

leaves a remainder of 111 when divided by 999,999,999?

1.

$\begin{array}{|rcll|} \hline (12~500~000)\cdot n & \equiv & 111 \pmod{999~999~999} \quad & | \quad :(12~500~000) \\ n & \equiv & 111\cdot \dfrac{1}{12~500~000} \pmod{999~999~999} \\ \hline \end{array}$

$\begin{array}{|lcll|} \hline \text{According to Euler's theorem,} \\ \text{if $a$ is coprime to $m$, that is, $gcd(a, m) = 1$, then }\\ {\displaystyle a^{\phi (m)}\equiv 1{\pmod {m}},}\\ \text{where ${\displaystyle \phi } $ is Euler's totient function.}\\ \text{Therefore, a modular multiplicative inverse can be found directly:} \\ {\displaystyle a^{\phi (m)-1}\equiv a^{-1}{\pmod {m}}.} \\ \hline \end{array}$

2.

$\text{greatest common divisor $gcd(12~500~000,999~999~999) = 1$ }$

$\begin{array}{|rcll|} \hline && \dfrac{1}{12~500~000} \pmod{999~999~999} \\\\ & \mathbf{\equiv} & \mathbf{12~500~000^{-1} \pmod{999~999~999}} \\\\ & \equiv & 12~500~000^{\phi (999~999~999)-1} \pmod{999~999~999} \\\\ & \equiv & 12~500~000^{648~646~704-1} \pmod{999~999~999} \\\\ & \equiv & 12~500~000^{648~646~703} \pmod{999~999~999} \\\\ & \mathbf{\equiv} & \mathbf{80 \pmod{999~999~999}} \\ \hline \end{array}$

3. n = ?

$\begin{array}{|rcll|} \hline n & \equiv & 111\cdot \dfrac{1}{12~500~000} \pmod{999~999~999} \\\\ & \equiv & 111\cdot \left( 12~500~000^{-1} \pmod{999~999~999} \right) \\\\ & \equiv & 111\cdot 80 \pmod{999~999~999} \\\\ & \mathbf{\equiv} & \mathbf{ 8880 \pmod{999~999~999} } \\ \hline \end{array}$

The smallest positive integer n is 8880.

$\begin{array}{rcll} 12 ~500~000 \cdot 8880 &\equiv & 111 \pmod{999~999~999} \\ 110~000~000~000 &\equiv & 111 \pmod{999~999~999}\ \checkmark\\ \end{array}$

A ball travels on a parabolic path in which the height (in feet) is given by the expression $-16t^2+80t+21$,
where $t$ is the time after launch.
What is the maximum height of the ball?

The graph:

$\begin{array}{|lrcll|} \hline & h(t) &=& -16t^2+80t+21 \\ h = 0 \ ? \\ & -16t^2+80t+21 &=& 0 \\\\ & t_{1,2} &=& \dfrac{-80 \pm\sqrt{80^2-4\cdot(-16)\cdot21}}{2\cdot(-16)} \\\\ & &=& \dfrac{-80 \pm\sqrt{6400+1344} }{-32} \\\\ & t_{1,2} &=& \dfrac{-80 \pm88}{-32} \\\\ & t_1 &=& \dfrac{-80 + 88}{-32} \\\\ & &=& \dfrac{8}{-32} \\\\ & \mathbf{t_1} &\mathbf{=}& \mathbf{-0.25} \\\\ & t_2 &=& \dfrac{-80 - 88}{-32} \\\\ & &=& \dfrac{168}{32} \\\\ & \mathbf{t_2} &\mathbf{=}& \mathbf{5.25} \\\\ &t_{h_\text{max}} &=& \dfrac{t_1+t_2}{2} \\\\ & &=& \dfrac{-0.25+5.25}{2} \\\\ & &=& \dfrac{5}{2} \\\\ &\mathbf{t_{h_\text{max}}} &\mathbf{=}& \mathbf{2.5} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline h_\text{max} &=& -16\mathbf{t_{h_\text{max}}}^2+80\mathbf{t_{h_\text{max}}}+21 \quad & | \quad \mathbf{t_{h_\text{max}} = 2.5} \\ &=& -16\cdot 2.5^2+80\cdot 2.5+21 \\ &=& -100+200+21 \\ &\mathbf{=}& \mathbf{121~ \text{feet}}\\ \hline \end{array}$

The maximum height of the ball is $\mathbf{121~ \text{feet}}$

[12500000*n] mod 999999999 =111, solve for n

One obvious N =999,999,999 + 111 =1,000,000,110 because:

1,000,000,110 mod 999,999,999 =111      !!!, But:

n = 1,000,000,110 / 12,500,00

n =80.0000088, which is non-integer, but when multiplied by 12,500,00 you get:

N =1,000,000,110 which is an integer !!!! I don't think your teacher will buy this !!!!.

Now, onward to another rediculous answer !!!

If we wish to have both N and n whole integers, will try this:

n =111 x [999,999,999 / 111 + 1,000,000,000 / 12,500,000]

n = 111 x [9,009,009 + 80]

n =1,000,008,879, which is a whole integer!!!!. And...........

N =1,000,008,879 x 12,500,000

N =12,500,110,987,500,000, which is a whole integer !!!!! And.......

12,500,110,987,500,000 mod 999,999,999 =111    !!!!

Melody and CPhill : Take a whack at this one !!!!

4th term of (4x-y)^9

The forth term from what end .. I never know what people want when they say this.

$9C3(4x)^3(-y)^6 = 84*64x^3y^6=5376x^3y^6$         (Edited)

or

$9C3(4x)^6(-y)^3=-84*4096x^6y^3=-344064x^6y^3$         (Edited)

As Omi pointed out, I had mistakenly omitted the minus sign in front of the y. 

I also made a mistake with the C value.

Thanks Omi :)

Omi you also made a small mistake as  28+56=84 NOT 54

I've done some homework this morning and I do not think it is quite right.

According to what I read this is only one answer. The same answer written in different ways.

Plus

I probably didn't need to make it look so complicated ...

Here is the answer that I found before, only I have done it much more simply.  

$r^2 + 4r + 4 \equiv r^2 + 2r + 1 \pmod{55}\\ 4r + 4 \equiv 2r + 1 \pmod{55}\\ 2r \equiv -3 \pmod{55}\\ 2r \equiv 52 \pmod{55}\\ r \equiv 26 \pmod{55}\\ r \equiv 26+55k \pmod{55}\\ r \equiv 26+1045 \pmod{55}\\ r \equiv 1071 \pmod{55}\\ $

Now i need to find 3 other answers.  (Assuming that if I just add multiples of 55 I will just keep getting the same answer expressed in different way.)

Mmm.....

I don't know about any other ones ....   

And another example: 2w-1=11

Thanks!

 2.4u + 1.4 = 23     subtract 1.4 from  both sides

2.4u  =  21.6          divide both sides by 2.4

u  =  21.6 / 2.4  =  9

 2x = 50      divide both sides by 2

x  = 25

e-10=80    add 10  to  both sides

e  = 90

3xy^2z - 2z^5 + (2/3)y^2z^4

Just add the exponential powers on each term

The greatest result is the degree

1st term  = 4

2nd term  = 5

3rd term = 6

So....the degree is 6

Quadratic Trinomial  =  3 terms with highest power exponent  =  2

Cubic Trinomial  =  3 terms  with highest power exponent = 3

( c - d)^10      7th term

The 7th term is as folows :

C (10,6) * c^4 * d^6  =

210c⁴d⁶

BTW.... here's the full expansion  as a check ;

c^10 - 10 c^9 d + 45 c^8 d^2 - 120 c^7 d^3 + 210 c^6 d^4 - 252 c^5 d^5 + 210 c^4 d^6 - 120 c^3 d^7 + 45 c^2 d^8 - 10 c d^9 + d^10

 6x^2 + 5x

 x^2 + 5x^(1/5)......no  fractional powers allowed

 -x^2 + 5x

 -7x^2 + (5/3)x

 y

 4x^3 + y

2 + s

- x^2 - sqrt of x + 2    .......  sqrt ( x)  = x^(1/2).....no fractional exponents

A plolynomial    has constant coefficients associated with each variable and  non-negative integer powers on these variables

So....the polynomials are in  red

Point A is located at (1,4). Point P at (3,5) is 1/3 the distance from A to point B.What are the coordinates of point b

Let m be the x coordinate of B

So

x coordinate of A   +  3 times ( x coordinate of P - x coordinate of A)  =  m

1   +  3 ( 3 - 1)  =  m

1  + 3(2)  = m

1 + 6  = m

7  = m

Similarly

Let n be the y coordinate of B

So

y coordinate of A   +  3 times ( y coordinate of P - y coordinate of A)  =  n

4  +  3 ( 5 - 4)  =  n

4  + 3(1)  = m

4 + 3  = m

7  = m

So B =  (m, n)  =  (7, 7)

Draw a circle with a radius of 15  and a center at (-3, 2)

The equation of this circle is

(x + 3)^2  + (y - 2)^2   = 225

Let point  L  be located at   ( m, 0)    where m is positive

So...we can determine m  by letting  y  = 0  in in the above equation...so we have...

(m + 3)^2  + (0 - 2)^2  =  225

(m + 3)^2  + (-2)^2  =  225

(m + 3)^2  + 4   =  225     subtract 4 from each side

(m + 3)^2  =  221            take the positive root of both sides

m  +  3   =  √ 221          subtract 3 from  both sides

m = √ 221  - 3

So  L =  ( √ 221  - 3 , 0 )

And we can find P thusly

[ -3 +  (1/3) ( √ 221  - 3 - - 3) ,  2  + (1/3)(0 - 2) ]  =

[ - 3 +  (1/3) (  √ 221 )  ,  2 + (1/3)(-2)  ]  =

[ -3 + √ 221/3   ,  2 - 2/3 ]  =

[ -3 + √ 221/3  , 4/3 ]

See the graph here : https://www.desmos.com/calculator/ggoirs81vs

We can use the Law of Cosines here

x^2  =  10^2 + 10^2  - 2(10)(10)cos (48°)

x^2  =   200 - 200cos(48°)

x = √ [  200 - 200cos(48°) ]   ≈ 8.135

35 = -2/3x + 5    Subtract 5 from both sides of the equation

30 = -2/3 x          Multiply both sides by - 3/2

(-3/2)(30) = x

-45 = x

Point C is located at the origin. Point Q at (-1,-2) partitions CD in a ratio of 1:6. What are the coordinates of point D?

Think about this

QD  must be 6 times the distance CQ

Let  m be the x coordinate of D

So

x coordinate of Q +  6 times ( the difference in the x coordinate of Q - the x coordinate of the origin)  = m

So

-1 +  6 (-1 - 0)  =  m

-1 + 6(-1)  = m

-1 - 6  =  m

-7  = m

Similarly

Let  n be the y coordinate of D

So

y coordinate of Q +  6 times ( the difference in the y coordinate of Q - the y coordinate of the origin)  =  n

So

-2  + 6(-2 - 0)  =  n

-2 + 6(-2) = n

-2 - 12  = n

-14  = n

So....D  = (m ,n)  =  (-7 , -14 )

IMHO, It's easier to figure point Q if we realize that it also must be equal to 1/5 of the distance from H to J

So....we can find point Q as follows :

Q  =   

[ -5  +  (1/5) ( 5 - - 5) ,  -6 + (1/5) ( -1 - -6)  ]  =

[ -5 + (1/5) (10)  , - 6 + (1/5) (5) ]  =

[ - 5 + 2, - 6 + 1 ]

{ -3, - 5)  =  Q