Kenya Mathematics Dominion

31/08/2022

(KMD):

26/08/2022

.KE
________________________________

Solve for (?)

9=90
8=72
7=56
6=42
3= ?

04/05/2021

:(K.M.D)..........................................................

Pythagoras Theorem is the base of trigonometric identities.
It's the best time to bring it back to mind !!
Happy Learning!! :)

04/05/2021

Pythagoras Theorem is the base of trigonometric identities.
It's the best time to bring it back to mind !!
Happy Learning!! :)

20/02/2020

(K.M.D):......................................
QUESTIONSPINE:.................................
You've Log3=0.4771 and Log5=0.6990, evaluate
Log 1125 without using logarithm tables or a calculator.

REMEMBER:
1.log(A*B)=logA+logB
2.logA^m=mlogA
3.log(A^m*B^n)=mlog A+nlog B

16/02/2020

(K.M.D):..............................................

QUESTIONSPINE..................................
The sum of the interior angle of two regular polygon of sides (n-1) and "n" are in the ratio 2:3. Solve and name the polygon.

15/02/2020

(k.M.D):.....................................
Problems leading to quadratic equations:

-There are many problems that require you to form and solve quadratic equations.
-At times you may be given roots and be required to form the quadratic equation.
-At other times, you may be given word problems from which we form and solve quadratic equations.

Example:
Find the quadratic equation whose roots are 3 and -2.

Solution:
-Let the variable be X.
-The root 3 means that;
X=3=>X-3=0
-The root -2 means that;
X=-2=>X+2=0
-Thus,the two factors of the quadratic equation are;
(X-3) & (X+2).
-Therefor the equation is;
(X-3)(X+2)=0

X(X+2)-3(X+)=0
X^2+2X-3X-6=0
X^2-X-6=0

11/02/2020

(k.M.D):.................................................The equation of a circle:
Example:
-Now, you've been told to determine the centre and the radius of a circle, whose equation is given as
X^2+Y^2-6X+10Y=2
Solution:
-To find the centre and the radius of the circle,
X^2+Y^2-6X+10Y=2
should be expressed in the form
(X-a)^2+(Y-b)^2=r^2
where a,b and c are constants, (a,b) is the centre of the circle, and r is the radius of the circle.

X^2+Y^2-6X+10Y=2
>X^2-6X+Y^2+10Y=2
>X^2-6X+(-3)^2+Y^2+10Y+(5)^2
=2+(-3)^2+(5)^2
(completing squares on X and Y)
>(X-3)^2+(Y+5)^2=2+9+25=36
Therefor;
X^2-6X+Y^2+10Y=2
(X-3)^2+(Y+5)^2=6^2
-Hence, the centre of the circle is (3,-5) and the radius of the circle is 6 units.
-If the end points of a diameter of a circle are known,the equation of the circle can be found.

06/02/2020

(K.M.D):.................................................
Find the limits of error in the volume of a block 9.7cm by 8.4cm by 7.6cm (2s.f).

11/01/2020

(K.M.D):