Pastor Arrested in Hotel While Soliciting a Prostitute(video)

Wonders shall never end...
Read this statement of a pastor who was caught with a prostitute in a hotel room
I am a pastor of a very large congregation he says’ and ‘I don’t want to go to jail. Please I don’t want my wife to find out about this’ Sounds like a scene from a movie but this was a real life occurrence when this pastor was among dozens arrested during a massive prostitution sweep last week in Florida.
So a group of detectives set up a sting operation to track prostitutes
and their clients and rein them in. The pastor accompanied a prostitute into a hotel where cameras are already installed. He walks into the trap, pays the ‘prostitute’ for sex and is ready to rumble when the detectives burst in and swooped on him. Then he says the unthinkable. He identifies himself as a pastor. They say those who live in glass houses should not throw stones. This guy has been suspected of cheating on his wife before now but he has kept it under wraps until he got busted.
Watch the video by click the below
HERE

function of a function

FUNCTION OF A FUNCTION

Sin x is a function of x since the value of sin x depends on the value of the angle x . similarly, sin(2x + 5) is a function of the angle (2x + 5) since the value of of sine depends on the value of this angle ..

i.e. (2x + 5) is a function of (2x + 5)

but (2x + 5) is itself a function of x, since its value depends on x

i.e. (2x + 5) is a function of x

sinx(2x + 5) is therefore a function of a function of x and such a expressions are referred to generally as functions of a function

formula dy/dx = dy/du . du/dx

 

examples

1. differentiate with respect to x, y = cos (5x – 4)

 

 

Solution

We shall say let u = 5x – 4 (that’s assigning x to a variable U)

If U = 5x – 4

Then y = cos U (That’s if we substitute 5x -4 as U)

If we differentiate U with respect to X we shall have

.du/dx = 5

and if we differentiate y with respect to U we shall have

.dy/du = -sin U (from trig. Functions, hope you understand)

dy/du = -sin U is same as

dy/du = -sin (5x – 4) since U = 5x – 4 ( hope you understand, if no,use the comment section)

now substituting back to our formula

 

dy/dx = dy/du . du/dx

 

dy/dx =-sin (5x – 4). 5 (. is same as multiplication)

dy/dx = -5sin (5x – 4)

 

Please follow up this example, try to get it right, its very simple

 

2. differentiate with respect to x : f(x) = sin (1 + x² )

solution

y = sin (1 + x² )

let U = 1+ x²

du/dx = 2x (if we differentiate a constant, its zero)

y = sin U

dy/du = cos U

dy/du = cos (1 + x² )

therefore

dy/dx = dy/du . du/dx

 

dy/dx = cos(1 + x² ) . 2x

dy/dx = 2xcos (1 + x² )

 

3. y = (4x – 3) ⁵

solution

y = (4x – 3) ^{5 (}basic standard form is y = x⁵ , dy/dx = 5x⁴

here (4x – 3) replaces the single x

Therefore dy/dx = 5(4x – 3) ⁴ multiply by the derivative of the function of (4x – 3), which is 4

= 5(4x – 3) ⁴ X 4

=20(4x-3) ⁴

lets try another procedure for this same question

y = (4x – 3) ⁵

let u = 4x – 3

y = u⁵

dy/du = 5u⁴

du/dx = 4

therefore dy/dx = 4.5u⁴

since u = 4x – 3

dy/dx = 4. 5(4x – 3) ⁴

dy/dx = 20(4x – 3) ⁴

(this is function of function, hope you understand)

we shall continue with more examples

now lets move to EXPONENTIAL

Remember that among our standard derivatives we have something like this

Y = e^x dy/dx = e^x

That is, when you differentiate an exponential it doesn’t change, it remains the same

Now we shall employ this along as we solve related questions

EXAMPLES

1.    y = e^sin2x

solution

let u = sin 2x

therefore y = e^u

dy/du = e^u (it wont change,)

du/dx = 2os 2x (trig. Function)

dy/dx = 2cos 2x.e^sin2x

 

2.    if y = e^5x determine dy/dx

solution

let u = 5x

y = e^u

dy/du = e^u

du/dx = 5

therefore dy/dx = 5.e^u

dy/dx = 5e^u

but u = 5x

dy/dx = 5e^5x

 

 

 

 

 

 

standard derivatives & formula method

STANDARD DERIVATIVES & formula method

Here is a revision of the standard derivatives which you have no doubt used many times before..

Copy out the list into a book and memorize those which you are not familiar with.

 

Y=f(x)	.dy/dx
1. Xn	NXn-1
2. ex	ex
3. ekx	kekx
4. ax	ax . lna
5. lnx	1/x
6. logax	1/x.lna
7. sin x	Cos x
8. cos x	-sin x
9. tan x	Sec2 x
10. cot x	-cosec2 x
11. sec x	Sec x. tan x
12. cosec x	-cosec x. cot x
13. sinh x	Cosh x
14. cosh x	Sinh x

We shall proceed with differentiation using the first derivative which is

Y= Xⁿ

^dy/dx= NX^n-1

(Known as formula method)

Where

X variable

N power

 

NOTE

When ever you wants to differentiate you will minus 1 (one) from the power and also multiply the power with the constant

 

e.g if you are asked to differentiate. x²

What you will do is this

Take the power back to multiply the constant associated with x, the minus 1 from the power

MATHEMATICALLY

.y = x²

.dy/dx = 2x^2-1

Therefore dy/dx = 2x^1. Which is 2x.

Now let’s proceed with other examples

Examples

1.    find f(x) = x²⁷

 

Solution

.dy/dx = 27x^27-1

.dy/dx = 27x²⁶ans

 

2.    differentiate with respect to x:f(x) = (8x³ + x² )

Solution

We shall differentiate this one after other

.dy/dx = 3*8x^3-1 + 2x^2-1

.dy/dx = 24x² + 2x

3.    find f(x) = x³ⁿ

Solution

.dy/dx = 3nx^3n-1 (using same procedure)

 

4.    find f(x) = (3x+4)

Solution

.dy/dx = 3 (note whenever you differentiate a constant it becomes zero)

 

5.    find f(x) = x ⁿ

solution

.dy/dx = nx ^n-1

 

at this juncture you can differentiate using formula method

we shall therefore move to more explicit ones

 

Exercises

Differentiate the following with respect to

1.    6x⁵ – 3x⁴ – 2x³

2.    7x⁴ – 5x³

3.    ax³ + bx

 

please try the exercises, it will make you more perfect, also try to read other math text book, if you encounter any challenges jus drop it here so that we can help you out.

 

 

Solving a system of algebraic equation with 3 unknowns

Today I shall be dealing with how to solve for 3 unknown in linear algebraic equations using one of the simplest method (Gauss Elimination method) gauss elimination method is one of the oldest, simplest, and most frequently used method for solving systems of algebraic equations 

  In this method we shall employ the elementary Row operations which include ...

1.. Interchanging any two rows 

2.. Multiplying/dividing any row by a nonzero constant.

3.. Adding/subtracting a constant multiple of Row to another Row 

Now let's go straight to solving some equation

EXAMPLE 1

Solve the set of linear equations

 a + 2b + 3c = 5 

 3a - b + 2c = 8 

 4a - 6b + 4c = -2

(Now carefully followup the step by step solution, don't forget to comment on areas you don't understand)

Now using the elementary row operation we will make the first element in R2 (Row 2) which is 3a to be 0 

And in R3 (row3) we shall make 4a and -6b to be 0 

So that we easily solve for c , b and  a

( you will understand once we start implementing this)

Back to our question 

 a + 2b + 3c = 5 

 3a - b + 2c = 8 

 4a - 6b - 4c = -2

Since we want to eliminate 3a in Row 2 (R2), we shall multiply R1 by -3 and add to R2 to form new row 2

( NOTE: our R1 will not be changed but our operation will then become our new row 2)

-3×R1 + R2 

-3 × R1 … -3[a+2b+3c=5] == -3a-6b-9c=-15

R2 … 3a-b+2c=8

.•. -3R1 + R2 = (-3a+3a=0), (-6b+-b=-7b), (-9c+2c=-7c),(-15+8=-7)

Now our new equation will look like this

a+2b+3c=5

0-7b-7c=-7.   ( note that our R2 changed to our operation)

4a-6b-4c=-2

Now eliminating '4a' in R3

R1×-4 + R3 (because -4 + 4 is zero)

-4×R1…… -4a-8b-12c=-20

  + R3……… 4a-6b-4c=-2

-4R1 + R3 == 0 - 14b - 16c =-22

Our new equation

a+2b+3c=5

0-7b-7c=-7

0-14b-16c=-22

Now we will use row 2 to make -14b in R3 be Zero

-2R2 + R3 

-2R2 … 0+14b+14c=14

 R3 …….0-14b-16c=-22

-2R2 + R3 = 0+0-2c=-8

Our new equation

a+2b+3c=5

0-7b-7c=-7

0+0-2c=-8

BRAVO!!!  We have successfully reduced our equation using the elementary row operation. Now we can easily solve for our c,b and a

From equ.3

0+0-2c=-8

-2c=-8

C= -8/-2

C=4

Now let's substitute C as 4 in equ. 2

0-7b-7c=-7

-7b-7(4)=-7

-7b-28=-7

-7b = -7+28

-7b = 21

b = 21/-7

b = -3

Now let's substitute c as 4 and b as -3 in equ. 1

a+2b+3c=5

a + 2(-3) + 3(4)=5

a-6+12=5

a+6=5

a= 5-6

a=-1

,a=-1, b= -3 and c= 5

So you can see that its so simple, don't  think that its too long, its jus because am explaining along.. 

Once you know how to solve with this method, u will confirm  how short it is, compared to other methods...

Pls always solve more related questions and don't forget to log unto this site for help (jus place your  questions in the comment sector)

Example 2

Solve the set of linear equation

.a+3b+2c=3

2a-b-3c=-8

5a+2b+c=9

Lets solve

To eliminate 2a from R2 we shall multiply R1 by 2 then add to R1 to form new R2

(Note that am using R as Row)

.•. (R1 × -2) + R1 shall give us a new R2 and make our equation look like thus

a+3b+2c=3

0-7b-7c=-14

5a+2b+c=9

  ( I hope you understand how I arrived at this,, if no pls comment)

Now let's use R1 to  eliminate 5a of R3

R1×-5 + R3

R1×-5……   -5[a+3b+2c=3]

=     -5a-15b-10c=-15

R3……  5a+2b+c=9

R1×-5 + R3

0-13b-9c=-6

Our new equation will then be

.a+3b+2c=3

0-7b-7c=-14

0-13b-9c=-6

Now lets eliminate -13b off equ.3 

(Note that we shall use R2 to perform this operation,)

So R2 × 13 + R3 × -7

R2×13... (0×13=0), (-7b×13=-91b), (-7c×13=91c), (-14×13=-182)

R3×-7 ..... 91b+63c=42

R2×13 + R3×-7 = 0+0-28c=-140

Our new equation

.a+3b+2c=3

0-7b-7c=-14

0+0-28c=-140

Now let's solve for our c,b and a

From equ. 3

0+0-28c=-140

-28c=-140

C = -140/-28

C=5

In  equ.2 sub. 5 as c

0-7b-7c=-14

-7b-7(5)=-14

-7b-35=-14

-7b=-14+35

-7b=21

.b=21/-7

.b=-3

In equ.1 sub. -3 as b and 5 as c

.a+3b+2c=3

.a = 3 + 9 -10 

.a=2