All Answers 358156 Answers

Set the equations  =   to find the x coordinates (in terms of a) of the intersections of these two functions

3x + a =   x^2 +  x          rearrange as

x^2 - 2x   =   a              complete the  square on x

x^2  -2x + 1 =  a + 1

(x  - 1)^2  = a + 1           take both roots

x -  1 =  ± √ (a + 1)

x = ± √ (a + 1)  + 1

Plugging these two x values into  the  equation of the  line we  have that

y =  3 ( sqrt (a + 1) + 1) + a        or      y =  3 ( -sqrt (a + 1) + 1) + a   

y =  3sqrt (a + 1) + a + 3                      y =  -3sqrt (a + 1) + a + 3

Using  the square of the distance formula we have that

( (sqrt (a + 1) + 1)  - ( -sqrt (a + 1) + 1) )^2  +  ( (3sqrt (a + 1) + a + 3) - (-3sqrt (a + 1) + a + 3))^2 = 30

Simplify

( 2 sqrt ( a + 1) )^2   +  ( 6 sqrt (a + 1 )^2     =  30

4 (a + 1)  + 36 (a + 1)  = 30

40a  + 40  =  30

40a =  -10

a =  -1/4

Here's a graph :

Points  A and B  are the intersections of  the functions.......the midpoint of the segment connecting A and B  = (1, 2.75)

A circle with this center and a diameter of sqrt (30) = radius of sqrt (30) /2   will pass through both points

This hyperbola opens sideways...so.....let the center  be    at ( x , 2)

"c"  =  x -  3       →   c^2  =  x^2 - 6x + 9

"a"  =  x  - 4       →   a^2 =   x^2  - 8x + 16

"b"  = sqrt  [ c^2 - a^2 ]  =   sqrt [  (x^2 - 6x + 9) - (x^2 - 8x + 16)  ]  =  sqrt [ 2x - 7]  

The slope is given as

b / a   =  sqrt 2

sqrt [ 2x  -7 ]  / [x - 4 ]   =  sqrt 2

sqrt [ 2x - 7 ]  =  sqrt 2 *  [ x -4]           square both sides

2x - 7  = 2  [ x^2 - 8x + 16 ]

2x^2 - 18x + 39  =  0

          18  +√ [ 18^2 - 4*2 * 39 ]               18 +√ [ 12 ]               9 + √ [ 3 ]

x =    ______________________  =    ___________  =       ___________  ≈   5.366

                      2 * 2                                         4                               2

Tanzi uses    1200 ml      when  Maeve uses  (1/3) 1200 ml =   400 ml

Therefore.....she uses  3 times as much in a specified amout of time

So

Maeve should  pour   (3/4) (400 ml)  =   300 ml  into  Tanzi's bottle

Tanzi will have 300 ml  and Maeve will have 100 ml    and.....3 times what Maeve uses = whatTanzi uses in the  same amount of time

We have this relationship :

AX^2  + CX^2  =  BX^2 + DX^2

11^2  + 10^2   = 11^2  + DX^2

100  = DX^2

10   = DX

3a + b + c =    -3       (1)

a + 3b + c =    9        (2)

a +   b  + 3c  =  -6     (3)        add all three equations

4a + 4b + 4c =   0

a + b +c =   0

a + b =  -c                sub this into  (3)

-c + 3c  = -6

2c =  -6

 c  = - 3

Also   a + c  = -b      sub this into (2)

-b + 3b =  9

2b = 9

b =  9/2

And   a + b + c  =0

a + 9/2  - 3  = 0

a + 3/2  =0

a =  -3/2

abc   =  (-3/2)(9/2) (-3)  =   81 / 4

Expected value  = 

Value of win * probability of winning +   Cost of losing * probability of losing =   

2 ( 1/2)  +  -2(1/2)  =  

0

We have a  low over high situation  in terms of  polynomials.....when this happens  there is a horizontal asymptote at y  = 0

So.....0  is not in the  range  of the  function

x(3x - 7)  = -3 + 2x

3x^2  - 7x  = -3 + 2x           rearrange as

3x^2 - 9x  =  -3             divide through by 3

x^2  - 3x  =  - 1        complete the square on x

x^2  -3x  + 9/4 =  -1  + 9/4

(x - 3/2)^2  =  5/4       take  both roots

x - 3/2  =     ± √5/2

x =   3/2   ± √5/2

m= 3    n  = 5     p  =  2

m + n +  p  =   10

3 sides   angle sum 180 degrees

4 sides   angle sum =  2*180 degrees

n sides   angle sum =  (n-2)*180 degrees

so what is the angle sum of a decagon

Once you know that you just have a simple equation to solve.

Continued here

I have a solution.

I think it is what the other answerer was trying to explain.  (Although I am not sure)

\(\binom{n + 2}{3} = (1)(n) + (2)(n - 1) + (3)(n - 2) + \dots + (k)(n - k + 1) + \dots + (n)(1)\)

Method 1:

The number of subsets of 3 elements from the set of n+2 elements is given by   \(\binom{n+2}{3}\)

Method 2:

Let us  have 2 sets,    A(n elements)  and    B(2 elements)

How many ways are there to choose 3 elements if two MUST come from set B      answer: n

Now take one of the elements from A and put it in set B  so A(n-1 elements)  and B(3 elements)

How many ways are there to choose 3 elements if two MUST come from set B but one of the 2 must be the new one    answer: (n-1) 2 = 2(n-1)

Now take one of the elements from A and put it in set B  so A(n-2 elements)  and B(4 elements)

How many ways are there to choose 3 elements if two MUST come from set B but one of the 2 must be the new one    answer: (n-2) 3 = 3(n-2)

....

and so on until all but one of the elements are in set B        n(n-(n-1) = n(1)

So  The number of subsets of 3 elements from the set of n+2 elements is also given by

 \( (1)(n) + (2)(n - 1) + (3)(n - 2) + \dots + (k)(n - k + 1) + \dots + (n)(1)\)

HENCE

\(\binom{n + 2}{3} = (1)(n) + (2)(n - 1) + (3)(n - 2) + \dots + (k)(n - k + 1) + \dots + (n)(1),\)

I have shown that they are equal by counting in 2 different ways

You have 6 possible configurations as follows:

         a    b    c

1  = (60, 45, 75)
2  = (45, 60, 75)
3  = (72, 30, 78)
4  = (30, 72, 78)
5  = (80, 18, 82)
6  = (18, 80, 82)

What is the area of a right triangle? Can you figure it out?

The smaalest positive a == 8

[5 * 8]==40 mod 11 ==7

Because we are given both the roots of the quadratic, we can narrow the equation down to \(y = a(x+1)(x-5)\)

Now, think about how we can use the remaining coordinate (0,5) to solve for a...

Now that you know the quadratic, I assume you can take it from here

Expand it, and substitute the \(i^2\) for -1. 

If Ali gives $6.20 to Ben, he will have 1/5 of Ben's amount,
Ali : 1u + 6.20
Ben : 5u - 6.20

If Ben gives $12.80 to Ali, he will have 3 times as much money as Ali,
Ali : 1p - 12.80
 

Ben : 3p + 12.80

1u + 6.20 -------> 1p - 12.80
1u -------> 1p - 12.80 - 6.20 = 1p - 19
5u - 6.20 -------> 3p + 12.80
5(1p - 19) - 6.20 = 5p - 101.20 -------> 3p + 12.80
1p -------> (12.80 + 101.20)/(5 - 3) = 57
3p + 12 .80 -------> 3 x 57 + 12 .80 = 183.80

Ans : $183.80.

P6 math.

Hypotenuse: 10

One leg: 6

x: 8 

Explanation:

Pythagorean Triple (3,4,5)

multiply that by 2 and you get (6,8,10)

and also, it's a right triangle.

Nigel have x money at first. Then Joe have (x + 475) money at first.

x - 60%x = x + 475 - (x + 475) * 85%

0.4x = x + 475 - 0.85x - 403.75

0.4x = 0.15x + 71.25

0.25x = 71.25

x = 285 

x + 475 = $760.

285 * 60% + 760 * 85% = $817

The present cost $817

A figure with 1,2,3,4 forming a sqaure

2 is 1/2 of the 50%

3 is 1/2 of the 50%

1 is 3/4 of the 50%

4 is 1/4 of the 50%

B has 5/8 of the figure

Adding the fraction of 1 and 2 or 1 and 3

(1/4 of 50%) + (3/4 of 50%)

= 25% + 37.5% = 0.25 + 0.375

                          = 0.625

                          = 625/100

                          = 5/8

So the answer is (1) 1 and 2

1/4 * 13/51 * 12/50=13/850

1/4 is for 13 hearts out of 52 cards, then you have already chosen a card so you have 51 - but still 13 choices of spades, so 13/51

finally, you have 50 cards after picking 2, and you have 13-1=12 hearts available.

You integrate by parts.

Let Nigel's money ==N

Joe's money ==475 + N

N - 0.60N ==0.40N - what Nigel has left

0.15 * [475  +  N]==71.25  +  0.15N - what Joe has left

0.40N ==71.25  +  0.15N, solve for N

0.40N - 0.15N ==71.25

0.25N ==71.25

N ==71.25 /0.25 ==$285 - what Nigel started with

$285 + $475 ==$760 - what Joe started with.

Nigel spent: 285  x  0.60 ==$171 - money spent to buy a present.

Joe spent ==760  x  0.85==$646 - money spent to buy a present.

[285 - 171] ==[760 -646]

 $ 114 == $114 - what both had left at the end.

Using Simon's Favorite Factoring Trick to all of the equations gives us:

(a + 2)(b + 2) = - 12

(a + 2)(c + 2) = 30

(b + 2)(c + 2) = -40

Multiply all those equations together gives us

(a+2)(b + 2)(c + 2) = 120

Applying these into the equations gives us: 

-12(c + 2) = 120

30(b + 2) = 120

-40(a + 2) = 120

c = - 12

b = 2

a = -5