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How many pairs of perfect squares between 1 and 100 inclusive differ by a prime number?

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1       1 * 1

                        3

4       2 * 2

                        5

9       3 * 3

                        7

16     4 * 4

                        9

25     5 * 5

                        11

36     6 * 6

                        13

49     7 * 7

                        15

64     8 * 8

                        17

81     9 * 9

                        19

100 10 * 10

7  pairs of perfect squares between 1 and 100 inclusive differ by a prime number.

  !

Let the bottom  let  vertex of the  blue triangle  =A

Let the bottom right vertex of the  blue triangle  = B

Let the remaining vertex of the blue triangle  = C

Let the bottom right vertex of the figure  =  D

Let  DE  be  the edge  that is perpendicular to AD

And traingles 

ABC  and ADE   are similar triangles

So

BC / AB   =  DE / AD

BC / 12  =  12 / 16

16* BC  = 12 * 12

16* BC  = 144

BC  = 144/16  =  9

So  the dark  area is the area of  triangle ABC  =

(1/2) (AB) ( BC)  = 

(1/2)(12) (9)  =  

(1/2) 108  =

54 units^2

Answer:Well for starters the student must make sure it is the correct temperature for each different plant even if it is different. She should then make observations and pay more attention to the specifics like which one grows faster.There should be three Ivy plants one would be warm, another would be room temp. , and the last should be cold . All of the Ivy plants must have the same amount of water for each plant. Also we make sure everything is the same what I mean is all three plants have the same pot, same feeding and fertilization if that is what you are doing, although it good to see how it would grow in different soil you should put it in the same type of soil but maybe have another set of pots and compare which one helps growth rate.I think it also might help depending on lighting like sunshine and without sunshine. Just make sure to write down your findings.Also you need to make a hypothesis

I put that does it make sense and is it correct??

 tan(x)^2 - 4*tan(x) + 1   =  0      let tan (x)  = a

a^2  - 4a  +  1  =  0        complete the square on  a

a^2  - 4a  + 4  =   - 1  + 4

(a - 2)^2   =  3           take both roots

(a - 2)  =  ±√3

a  =  2  + √3                a  =  2 - √3

So   either

tan (x)  = 2 + √3            

arctan ( 2 + √3)  =  x  = 75°

So  sin (2 * 75°)   =  sin (150°)   =  1/2     

Or

tan (x)   =  2 - √3

arctan ( 2 - √3)  =  x  =  15°

So    sin (2 * 15°)   =    sin (30°)   =   1/2

The ratio is 1:4. If you want an explanation on how I got it let me know.

a^2   =    sin^2(t) + 2sin(t )tan(t)  + tan^2 (t)

b^2    =   sin^2( t)  - 2sin (t)tan (t)  + tan^2( t)

a^2 - b^2 =   4sin(t)tan(t)   ⇒ (a^2 - b^2)^2  =  4sin^2(t) tan^2(t)

ab =  tan^2(t) -  sin^2 (t)

(a^2  - b^2)^2

_____________    = 

     ab 

4sin^2 (t) tan^2 (t)

________________       =

tan^2(t) - sin^2(t)

4 sin^2(t) sin^2(t) / cos^2(t)

_______________________________      =

[sin^2(t) - sin^2(t) cos^2(t)]  / cos^2(t)

4 sin^2(t) (sin^2(t)

____________________   =

sin^2 (t) ( 1 - cos^2(t) )

4 sIn^2(t) sin^2(t)

________________   =

sin^2(t) sin^2(t)

4

Solve for x:
log(2)/log(N) + log(4)/log(N) + log(6)/log(N) + log(8)/log(N) + log(10)/log(N) = log(x)/log(N)

log(2)/log(N) + log(4)/log(N) + log(6)/log(N) + log(8)/log(N) + log(10)/log(N) = log(x)/log(N) is equivalent to log(x)/log(N) = log(2)/log(N) + log(4)/log(N) + log(6)/log(N) + log(8)/log(N) + log(10)/log(N):
log(x)/log(N) = log(2)/log(N) + log(4)/log(N) + log(6)/log(N) + log(8)/log(N) + log(10)/log(N)

Multiply both sides by log(N):
log(x) = log(2) + log(4) + log(6) + log(8) + log(10)

log(2) + log(4) + log(6) + log(8) + log(10) = log(2 4 6 8 10) = log(3840):
log(x) = log(3840)

Cancel logarithms by taking exp of both sides:

 x = 3840

x  +  x^-1   = √2      multiply through by   x

x^2  + 1  =  √2x

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

x^2 - √2/x  + 1/2  =  -1 + 1/2

[ x - (1/√2)]^2   =  -1/2          take both roots

x - (1/√2 )  =   ±i/  √2

x  =   ( 1 + i)               x   =   ( 1 - i )

       ______     or               ________

             √2                                √2

x ^20  =  [  ( 1 + i)) ] ^20

             ____________

                     ( √2)^20

Note that  ( 1 + i)^2  =  2i  .....so we have

                         [ (1 + i)^2 ] ^10                            [ 2i ] ^10

x^20  =             _______________         =           ________   =    i^10  =   -1  

                              [( √2)^2]^10                                2^10

The same result  is obtained  for  the other solution for x raised to the 20th power

I'm not posting the answer, but here:

Factor the quadratic, and test possible values for it's factors.

This question has already been answered. Here is the link: https://web2.0calc.com/questions/math_13723

The maximum possible value for N is 405, in which case AC would have to be 45.

The way I reasoned went like this:

When ABC is divided by AC the quotient is 9 and the remainder 0, so

ABC = 9(AC).

This makes ABC divisible evenly by both 9 and AC. Also, expanding the numbers we get

C+10B+100A =9(C+10A)=9C+90A.

Simplifying, we have,

10(A+B)= 8C, or 5(A+B)= 4C.

Now this last equation implies that A+B is a multiple of 4.

Moreover, A+B cannot be a bigger multiple than 4, since A+B=8 would result in C=10, which it cannot be.

Now since we are looking for the largest N, I set A=4 and B=0, forcing C=5, and low and behold it worked. QED.

Not as hard as it seems

(1 + i)^2011

___________

( 1 - i)^2009

Note that   ( 1 + i)^2  =  2i

And ( 1 - i)^2  = -2i

So  we have

[ (1 + i)^2 ] ^1005  *  ( 1 + i)

_______________________  =

[  ( 1 - i)^2]^1004  *  ( 1 - i)

[2i ] ^1005  *  (1 + i)

________________  =

[ -2i]^ 1004  * ( 1 - i)

[ 2i] * [ 2i]^1004 * ( 1 + i)

_____________________

    [ 2i]^1004  * ( 1 - i)

2i ( 1 + i)

________

   ( 1 - i)

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

_____________

 (1 - i) ( 1 + i)

2i ( 2i)

________

1 - i^2

4i^2

_______

1  - (-1)

-4

____

  2

-2

The equation for circumference is πd and the so the circumference of the circle formed by the second hand is 12π. In 30 minutes, the second hand will travel around the clock 30 times, 1 per minute. This gives us the equation 12π * 30 = 360π cm.

The number of ways to order Shakespeare's and Dickens' book is 6!. There will be 7 spaces (in front of all the books, in between the first and second book, in between the second and third book...all the way until at the end of all the books) that Conrad's books can go with them since they can't be next to each other. The number of ways of doing this is binom{7}{3}⋅3! since once they are in their spots they can be ordered 3! ways.

Thus, the number of ways is 6!⋅binom{7}{3}⋅3!=151200.

x + y  + sqrt (x +y)  = 12

x - y  + sqrt (x - y )  = 6        

Let   ( x +y)  =  a

Let   (x - y)  = b

So we have

a + sqrt (a)  =12          

a + sqrt (a) - 12  = 0       Factor as

(sqrt (a) + 4) ( sqrt (a) - 3)  =  0

sqrt (a) + 4  = 0                     sqrt (a) - 3  = 0

No real solutions exist           sqrt (a)  = 3

                                                 a  = 9

And

b + sqrt (b)  = 6

b +sqrt (b) - 6    = 0     Factor  as

(sqrt (b) + 3) (sqrt (b) - 2)  = 0

sqrt (b) + 3  = 0                               sqrt (b) - 2  = 0

No real solutions exist                     sqrt (b)  =  2

                                                               b   = 4 

So

x + y  = 9

x - y  = 4       add these

2x  = 13

x = 13/2

And

13/2 + y  =9

13/2 + y  = 18/2

y = 5/2

(x, y)   =  ( 13/2, 5/2)

Yep that's good.

Since x > y > z   (meaningx is the largest , y is the second largest, and z is the smallest of the three quantities), then

max(x, y) = x          (since x is the largest),

max(x, z) = x          (since x is the largest),

max(y, z) = y           (since y is the second largest) , and

max(x, y, z) = x      (since x is the largest).

so x + y +z - max(x.y) - max(x, z) - max(y, z) +max(x, y, z) =  x + y + z - x - x - y + x = 2x - 2x +y - y +z =  z . So z remains because he is the smallest.

Let  z  =  a + bi

Conjugate z  =  a - bi

So we have

2 (a + bi)  + 3 (a - bi)  =  -25  - 2i       simplify

5a - bi  =  -25  - 2i

Equate  real  and non-real   terms

5a  = -25            -bi  = -2i

a  = -5                  b  = 2

So

z =   -5 + 2i

CORRECT   !!!

We have an isosceles quadrilateral; the angles B and C together make up exactly half of 360. OB and OC are bisectors of the angles B and C, and so two of the angles of the triangle OBC add up to 90. So the third should also be 90.

I am also impressed by the amount of work you do and how you stay pleasantly appreciative of other people's work. I could never keep up with you even in the impossible case of having to devote all my time to the forum. And have a happy birthday.

My pleasure, thank you.

Let

\(\alpha= arctan(x)\), and

\(\beta=arctan(x^2)\). Then,

\(tan(\alpha)=x\), and \(tan(\beta)=x^2\), and

\(tan(arctan(x)+arctan(x^2))=Tan(\alpha+\beta)\)

\(=\frac{tan(\alpha)+tan(\beta)}{1-tan(\alpha)tan(\beta)} =\frac{x+x^2}{1-x\cdot x^2}=\frac{x+x^2}{1-x^3}=x\);

Cross-multiplying , we get

\(x+x^2=x-x^4\), which implies \(x^2=-x^4\). This equation is only true for \(x=0\) (unless we are in the complex field). So there is only one value satisfying the original equation.

(a)  Assuming  that  we have an isosceles trapezoid

AB  = DC               sides of an isoceles trapezoid are equal

Angle ABC  = Angle DCB      base angles of an isoceles trapezoid are equal

Angle ABC + Angle DAB  = 180          opposite angles in an isoceles trapezoid are  supplementary

Angle  DCB  + Angle ADC  = 180

Angle DAB  = Ange  ADC                subtraction property of equality

AD  = AD                                       reflexive property       

Triangle  ABD  is congruent to Triangle DCA         By   SAS

(b)   

AD  is parallel to BC            bases of anisoscees trapezoid are  parallel

Angle ADB  = Angle DBC         a transversal  cutting parallel lines   makes alternate interior angles equal