All Answers 358512 Answers

Let 1A   =    10  +  A

Let  A0   =    10A

So.....if we  have  integer  solutions, the  discriminant  must  be a  perfect square   

So

( 10+A)^2   -  4 (10A)       =

100 + 20A  + A^2  -  40A   = 

A^2  - 20A  +  100     =

(A - 10) ( A  -10)   = 

(A   - 10)^2

This  will   be a perfect square for any integer    A  >   0

But since A  must  be a single digit....then....

   A   can  be any positive  integer from   1  -  9   inclusive

In th first, we have that the sum of an arithemetic series  is given  by

(first term +  last term)  (n /2)

So

(first term +  last term)  (21 /2  = 693

first term +  last term  =   693 ( 2)  / 21 =    66

And  in the  second  summation.....we  are summing  these terms  of the  first series  :

Term #   +      Term #

    1        +        21            =   66

    3        +        19            =  66

    5        +        17            =  66

    7        +        15            =  66

    9         +       13            =  66

   11                                 =  66/2   = 33

So   the sum  =   66(5) +  33    =   363                       

Arc   ACB would be 1/2 of the circle

          the WHOLE circle would be   pi x diameter = ABpi

             so 1/2 of ABpi = 6pi

                        1/2 ABpi = 6pi

                              AB =12         this is the diameter..... radius would be  6 units

We have  (2 + 3 + 4)  =  9  equal parts of  AD

Let A  = (0,0)    and  let D  = (36,0)

Then    AB  =  (2/9) (36)  =   8

So    B  =(8, 0)

And the   midpoint  of AB   =   (4,0)

And AC  is   (2 + 3)  /  9    =   (5/9)  of AD =  (5/9) (36)   =  20

So  C  =  (20,0)

And the midpoint  of CD  =   [ (20 +36)  / 2  , 0  ]  =   [  56 / 2  ,  0  ]   =  (28,0)  

So  the distance between  the midpoint  of AB  and the  midpoint  of CD = 28  - 4  =  24

Triangle  DGA ≈    Triangle  BGE

GE/ GA =  EB /AD

4/6  =  EB / AD

2/3  = EB /AD              {   AD   =  BC  }

2/3  = EB / BC

(2/3)BC = EB

So  EC  =(1/3) BC

So  EB/ EC   = 2/1

So  EC /EB  =  1/2

And

Triangle   AEB  ≈  FEC

EF / AE =   EC /EB

EF /10  =  1/2

EF   = 5

x^2 -wx - 10w = 0       (I chnged a to w for clarity)

Discriminant must be > 0

b^2 - 4ac > 0

w^2 -4(1)(-10w) >0

w^2 + 40w >0

w(w + 40) > 0       both w and (w+40)  must be positive for this to be true

w >0   and    w >-40      is restricted to    w > 0        1 2 3 4 5 6 7 8 9    (since it must be positive single digit per Question)

80 % is  0.8

0.8 ( 220 - 26) = ..........BPM  target heart rate

Should look like these:

Much neater Tiggsy.

Just copy and paste this expression into the web2.0calc calculator on the home page here, then press the = sign.

240? is that my answer and how did you get that?

Very nice, Tiggsy   !!!!

Drop  perpendicular  BE   from B to  DC

Then EC  =  (2y  -2x)  /2   =  y  - x

And  by a property of  tangents drawn to a circle,  BC  = x +  y 

So

BE^2    =   BC^2  -  EC^2

BE^2  =  ( x^2 + 2xy + y^2)  - ( x^2  - 2xy + y^2)

BE^2   =  4xy

BE   = 2sqrt (xy)

But the radius is  (1/2)  of BE  =  sqrt (xy)

xy^2     = 240y

xy =  240

x^2y^2    = 240^2

So

x^2y^2 =  -250y

240^2  =  -250y

y = 240^2 / -250   = -1152/5

So

xy  = 240

x (-1152/5)    = 240

x  = 240  (-5/1152)   =  -25/24

Here's a slightly different parts routine

Let

\(\displaystyle I_{n}=\int1.\ln(x)^{n}dx\\ \text{and let } u=\ln(x)^{n}, \quad dv = 1.dx \)

Then

\(\displaystyle \frac{du}{dx}=n.\ln(x)^{n-1}\frac{1}{x},\quad v = x\)

so

\(\displaystyle I_{n}=x\ln(x)^{n}-\int xn\ln(x)^{n-1}\frac{1}{x}dx \\ I_{n}=x\ln(x)^n-nI_{n-1}.\)

The red and blue triangles are similar, so since their areas are in the ratio 27/12 = 9/4, their side lengths will be in the ratio 3/2.

So, let AB = 3k, then EF will equal 2k, (and notice in passing that DE + FC = 3k - 2k = k).

Run a perpendicular from AB to CD, passing through G and meeting the two sides at P and Q.

Let PG = 3h, then GQ = 2h.

Area of the red triangle = 3k*3h/2 = 27, so hk = 6.

(Similarly, area of the blue triangle = 2k*2h/2 = 12, so hk = 6).

Area of the trapezium ABCF = 5h*(AB + CF)/2  = 27 +Y1.

Area of the trapezium ABDE = 5h*(AB + DE)/2 = 27 + Y2.

Adding, 5h*(2AB + CF + DE)/2 = 54 + Y1 + Y2,    (DE + CF = k from earlier.)

so, 54 + Y1 + Y2 = 5h*(6k + k)/2 = 35hk/2 = 105,

so Y1 + Y2 = 51.

Drop a vertical from A meeting DC at E.

DE = (2y - 2x)/2 = y-x.

AD = x + y.

AE = 2R.

Now Pythagoras, AD^2 = AE^2 + DE^2, etc..

"One root of x^2 + 12x + k = 0 is twice the other root. Find k."

[ABD] = 1/2(BD * AO) = 1.5

[BCD] = 1/2(BD * CO) = 4.5

[ABCD] = 1.5 + 4.5 = 6

We first find the intersection point of x^2 - 12x + y^2 = 28 and y = 8 - x.

\(\begin{cases} x^2 - 12x + y^2 = 28\\ y = 8-x \end{cases}\)

After some work, we get (x, y) = \((7-\sqrt{31}, 1+\sqrt{31})\). We don't need the other intersection point.

We also get that the region is represented by the compound inequality \(\max(0, 8- x)\leq y \leq \sqrt{28 + 12x - x^2}\).

We use the straight line x = 8 to divide the region into 2 parts.

Using integration, the area is 

\(\displaystyle \int^{8}_{7-\sqrt{31}}\left(\sqrt{28 +12x - x^2}-(8-x)\right)\,dx+\int^{14}_8 \sqrt{28 + 12x - x^2}\,dx\)

which would be \(16\pi + 32\arcsin\dfrac{\sqrt{31} - 1}{8} - \sqrt{31} - 1 \approx 63.14\text{ sq.units}\)

The computation is too long so I didn't include it. 

Here's a verification of my answer by Wolfram Alpha:

Notice a pattern on the partial sums:

1 = 1 = 1^2

1 + 3  = 4 = 2^2

1 + 3 + 5 = 9 = 3^2

1 + 3 + 5 + 7 = 16 = 4^2

The sum is equal to n^2, where n is the number of terms. (This can be proven by mathematical induction.)

Now, 144 = 12^2. This means n^2 = 12^2, n = 12.

f(x) = f(2x + 1)

5x - 12 = 5(2x + 1) - 12

5x = 5(2x + 1)

x = 2x + 1

-x = 1

x = -1

Each number is a 3-digit number.

This means a + b + c + d = 410 where a,b,c,d >= 100.

Let a' = a - 100, b' = b - 100, etc.

Then a' + b' + c' + d' = 10, where a,b,c,d >= 0.

Then the required number of ways is \(H^4_{10} =\displaystyle \binom{10 + 4-1}{10} = \binom{13}{10}=286\)

The coefficients are rational => Sum of roots and product of roots are also rational.

=> conjugates of 2 + sqrt 2 and 1 - sqrt 5 must also be roots of the polynomial.

=> The roots are 2 + sqrt 2, 2 - sqrt 2, 1 - sqrt 5, 1 + sqrt 5.

The required monic quartic polynomial is 

\(\quad(x - (2 + \sqrt 2))(x - (2 - \sqrt 2))(x - (1 - \sqrt 5))(x - (1 + \sqrt 5))\\ =(x^2 - 4x + 2)(x^2 - 2x - 4)\\ =x^4 - 6x^3 +6x^2 +12x-8\)

The dependent variable is v because v is dependent on b and h.