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thank you very much 1 million kisses for you @ icosahedron

1.Msquared-9/Msquared-m-6 multiplied by m+2m/Msquared
2.Asquared-Bsquared/ab+Asquaredmultiplied by 2Acubed/ab-Asquared
3.Asquared-ab-6Bsquared/Asquared+ab-6Bsquared multiplied by Asquared-ab-2Asquared/Asquared-2ab-3Bsquared
4.Asquared+ab-2Bsquared/Asquared-2ab-3Bsquared multiplied by Asquared-Bsquared/ab+2Bsquared divide by Asquared-2ab+Bsquared/A squared-3ab

pauljustina:
1.Msquared-9/Msquared-m-6 multiplied by m+2m/Msquared
2.Asquared-Bsquared/ab+Asquaredmultiplied by 2Acubed/ab-Asquared
3.Asquared-ab-6Bsquared/Asquared+ab-6Bsquared multiplied by Asquared-ab-2Asquared/Asquared-2ab-3Bsquared
4.Asquared+ab-2Bsquared/Asquared-2ab-3Bsquared multiplied by Asquared-Bsquared/ab+2Bsquared divide by Asquared-2ab+Bsquared/A squared-3ab

just refer us to new general mathematics for ss 2

inschool:
x=3, y=2.....or x=2, ...y=3

Can I see your workings or you just tested numbers like those guys?

CACAWA:
Can I see your workings or you just tested numbers like those guys?

my working is on the screnshot

inschool:
my working is on the screnshot

no it only shows you testing those numbers to confirm they are right.

How did you arrive at 2 and 3

pauljustina:
1.Msquared-9/Msquared-m-6 multiplied by m+2m/Msquared
2.Asquared-Bsquared/ab+Asquaredmultiplied by 2Acubed/ab-Asquared
3.Asquared-ab-6Bsquared/Asquared+ab-6Bsquared multiplied by Asquared-ab-2Asquared/Asquared-2ab-3Bsquared
4.Asquared+ab-2Bsquared/Asquared-2ab-3Bsquared multiplied by Asquared-Bsquared/ab+2Bsquared divide by Asquared-2ab+Bsquared/A squared-3ab

y not post an image instead .

CACAWA:
no it only shows you testing those numbers to confirm they are right.

How did you arrive at 2 and 3

How did you arrive at 2 and 3[/quote]X^y +Y^x =17.................(1)
X +Y =5..........................(2)

Now,
17 =2^3 +3^2
hence,
X^y +y^x =2^3 +3^2............(3)
comparing both sides of (3), we found:
x=2, y=3 or y=3, x=2(ans)

inschool:
How did you arrive at 2 and 3
X^y +Y^x =17.................(1)
X +Y =5..........................(2)

Now,
17 =2^3 +3^2
hence,
X^y +y^x =2^3 +3^2............(3)
comparing both sides of (3), we found:
x=2, y=3 or y=3, x=2(ans)

i said.....How did you get to 2 and 3. Did you use any analytical methods like logarithms or did you use trial and error as follows:

Let x and y be 2 and 3,

Hence, replacing them in both equations, we can agree that x and y are actually 2 and 3.

Which approach did you use?

CACAWA:
i said.....How did you get to 2 and 3. Did you use any analytical methods like logarithms or did you use trial and error as follows:

Let x and y be 2 and 3,

Hence, replacing them in both equations, we can agree that x and y are actually 2 and 3.

Which approach did you use?

because 17 can be expressed as the sum of two powers

CACAWA:
i said.....How did you get to 2 and 3. Did you use any analytical methods like logarithms or did you use trial and error as follows:

Let x and y be 2 and 3,

Hence, replacing them in both equations, we can agree that x and y are actually 2 and 3.

Which approach did you use?

not analytical method though

inschool:
not analytical method though

I see...So kindly send me the original analytical workings via WhatsApp. Thnx

It's not that difficult

https:///7f6ox8ebHlFCCOGsxfuKj8


Mathematical villa group link....

Richiez:
nice try bro, although you made some some mistakes especially at the point where you took log of both sides. e.g it ought to be log(x+y)= log5 and not logx + logy = log5. but i must admit, you did a very good job.

here's another approach;

x + y = 5.......(1)
x^x + y^y =31.......(2)

from eqn(1), x=5-y, hence we substitute this value for x in eqn(2).
(5-y)^(5-y) + y^y = 31.......(3)
now, we have reduced the problem to what value of y that must be inserted in eqn (3) to obtain 31.

a thorough look at eqns (1) & (2) shows that the values of x & y must be positive integers within the range 0<x<5 and 0<y<5
clearly, y=3
i.e (5-3)^(5-3) + 3^3
2^2 + 27
4+27=31
now that we have established that y=3, we shall substitute the value of y in eqn(1) to get x
recall that x+y=5
therefore; x+3=5 or x=5-3=2
hence, x=2 when y=3

If you test y=2, (5 - y) ^(5 - y) + y^y = 3^3 + 2²= 27 +4 = 31.
So y = 2 or 3.
If y = 2, x + y =5 ➡ x + 2 = 5 ➡ x = 3.
So we have 2 answers:
x=2 when y=3, and x =3 when y = 2.

Hmm

Point P lies inside triangle ABC.Lines AP,BP,CP meet the circumcircle of ABC again at points K, L,M,respectively.The tangent to the circumcircle at C meets line AB at S. Prove that SC = SP if and only if MK = ML

Please I need urgent solution to this math problem.
Using first principle derive e^x

agentofchange1:
https:///7f6ox8ebHlFCCOGsxfuKj8


Mathematical villa group link....

Is there any whatsapp group?

NNAMDIII:
Is there any whatsapp group?

Yes

agentofchange1:
Yes

.

NNAMDIII:
please add me

done

agentofchange1:
done

Thanks! please modify your post......i mean remove my number from your post

Given that
sinA + sinB = x and
cosA + cosB = y

To solve this systems of equations. Sqaure both systems to obtain

sin²A + 2sinAsinB sin²B = x² (*)
cos²A + 2cosAcosB + cos²B = y² (**)

adding we obtain

2 + 2(sinAsinB + cosAcosB) = x² + y²
or
2 +2(cos(A-B)) = x² + y²

*** cos(A-B) = (x² + y² - 2)/2

Complete the solution thankyoujesus and you shall obtain similar argument for tan(A-B) and use the half-angle identity to obtain the value of tan(A-B)/2 and then proceed to solve the original problem.

Prove that

(cosA + cosB)² + (sinA - sinB)² = 4cos²(A+B)/2

Prove also that (1−cot23°)(1−cot22°) = 2


Come on, where are the trigonometry warlords?

Please I need help for this Quantitative Reasoning questions... Thank you all...

Icosahedron:
Given that
sinA + sinB = x and
cosA + cosB = y

find the values of cos(A+B)/2 and sin(A+B)/2

and hence or otherwise, evaluate tan(A+B)/2

use

sin A + sin B = 2sin((A+B)/2) cos((A-B)/2)) = x

cosa + cosb = 2cos((a+b)/2) cos((a-b)/2) = y

divide
x/y = tan((a+b)/2)

use knowledge of right angled triangle and Pythagora's to complete it.

DrinkWater10:
Please I need urgent solution to this math problem.
Using first principle derive e^x

y = e^x
dy = e(x+dx) - ex
dy = e^x (e^dx - 1)
dy/dx = e^x ( e^dx - 1)/ dx

take limit as dx approaches zero
dy/dx = e^x

please pay no attention to wrong use of letter.

Icosahedron:
Prove that

(cosA + cosB)² + (sinA - sinB)² = 4cos²(A+B)/2

I stated the one for cos above.

sina - sinb = 2cos((a+b)/2) sin((a-b)/2)

continue from here

Chop this simple math
4^x=8x......
Show workings....am nt interested in ur guess...