CHEMISTRY FOR 10TH CLASS (UNIT # 10)

PHYSICS FOR 10TH CLASS (UNIT # 11)

PHYSICS CLASS 10 (UNIT # 10)

Objective Simultaneous Equations

1. By the eliminations of one or more than one variables from the given simultaneous equations, we get such a relation which is __________ of that variable.
2.  At least ________ equations are required for elimination of one variable.
3. In elimination, both equations should have the ________ that has to eliminate.
4. Eliminant or relation shows that the solution set of both equations is not _________.
5. The relation free from x for x-b = 0 and x-d = 0 is ___________.
6. The relation free from x for xt = s and x = t is ___________.

7. The relation free from t for at = x and 2at = y is ______________.
8. The relation independent of ‘x’ for equations x + 1/x = a  and x² + 1/x² = b² is ________.
9. Te relation independent of ‘x’ for equations x + 1/x = m and x³ + 1/x³= n is __________.
10. The relation free from ‘t’ for equations x + t = 3p and x – t = 4q is ____________.
11. The eliminant by eliminating ‘m’ for equations m + bc = x and m – ad = y is _______.
12. The eliminant by elimination y for equations y² = s and y³ = r is ____________.
13. The relation free from y for equation y = 1/2m and y = 4n is ____________.
14. The equation y + 4 = 9 is y –5 =6 are not true for a __________ value of y.
15. A relation independent of ‘t’ from equations t⁵ = d and t³ = b is ___________.
16. A relation independent of ‘x’ from equations x³ – b = 0 and x² + d = 0 is ________.
17. The relation free from ‘x for equations x² + 1 = 3m² and x⁴ + 1 = n⁴ is __________.                                                                                         x²                         x⁴

18. The relation free from ‘y’ for equations √y – 1    = √a and y + 1/y  = b is ___________.                                                                                    √y

19. The relation from ‘y’ for equation x = √2 t and y = √7 t is ____________.
20. The relation free from ‘x’ for equations x + a = 0 and x² + y² = b² is _________.
21. The eliminant by elimination ‘u’ for equations v = u –t and u² = 2vt.
22. The eliminant by eliminating ‘y’ for equations y³ + 1/y³  = m and y³ – 1/y³  = n is _______.
23. The relation free from ‘x’ for equations x – 1 = m and x³ – 1 = 4n³ is _________                                                                                            x                       x³

24. The relation free from ‘x’ for equations x = 3p and x = 1 is __________                                                                                                                               7t

25. The relation free from ‘y’ for equation y² – 1/y² = a and y⁴ + 1/y⁴ = b⁴.

Important points and formulae related to Percentage

Increase = new value – original value

    Decrease = original value – new value

    Profit = selling price – cost price

     Loss = cost price – selling price

     Discount = marked price – sale price

        % 

Important Formulas related to a Circle

      

    Area of circle = πr²

    

    Perimeter of circle = circumference = 2 πr

      Area of ring = πR² – πr²

      Length of arc = x⁰/ 360 × 2 πr

      Area of sector = x⁰/ 360 × πr²

      Perimeter of sector = length of arc + 2 radius

      Area of segment = Area of sector – area of triangle

      1 revolution (in terms of angle) = 360⁰

      1 revaluation( as a distance) = circumference= 2 πr  

 

Standard Form or Scientific Notation

Many measurements involve very large numbers

 Example:  speed of light is 300000000 m/s.

This figures can easily b written as:

3.0 x 10⁸  (Correct to 2 significant figures)

This way of writing a numbers is called standard form or scientific notation.

Another example can be the wavelength of violet light which is 0.000038cm.

It can be written as:

3.8 x 10^-5 (Correct to 2 significant figure)

So a general rule for all figure in written in standard form or scientific notation is:

     A x 10ⁿ , Where 1≤ A ≤ 10 and n is an integer

                                                              By: Sir Baqir

Determining Minimum & Maximum Values

Determining Minimum & Maximum Values

One of the most important uses of calculus is determining minimum and maximum values. This has its applications in manufacturing, finance, engineering, and a most of other industries. Before we examine a real-world example, we should learn how to calculate such values.

Let's consider example, the equation 2X² -5X -7 = 0

This is a quadratic equation in one variable. 

ax² + bx + c = 0

With equations of this type, we know that when the "a" term is positive, the graph of the curve will be "concave up" (U-Shaped) and therefore the equation will have a minimum value but no maximum value (okay - technically, the maximum value is infinity). Looking at the graph we see that the minimum point is roughly X = 1.5 and Y = -10. Is there a way to determine the minimum point without graphing the equation and getting an exact value? Yes there is !

Look at the graph. If slope values were calculated for points on the left side of the curve, you could see that the slope would always be negative but it becomes "less negative" the closer the curve approaches the minimum (the bottom). If the slope were calculated along the right side of the curve, the value would always be positive and the slope values would get larger the further away from the "bottom" the points were.

So, it is logical to think that the slope is zero at that "bottom" point and therefore the derivative is zero at that point too.
So, let's take the derivative of 2X² -5X -7 = 0 which is:

4X - 5

When 4X -5 equals zero, X =1.25 which means that at this point, a minimum value exists. As for the 'Y' value, we go to the original equation and enter the value of X as 1.25.

Y = 2X² -5X -7

Y = 2*(1.25)² -5*1.25 -7

Y = -10.125

So, at point X=1.25, Y= -10.125 there exists a minimum value.

In this example we knew that we were obtaining a minimum value because we graphed it. Also, we stated that the "rule" for quadratic equations is such that when the 'a' term is positive, the curve will be "concave-up". There is yet a third method to determine whether a point is a maximum or minimum value.
If we take the second derivative and if that value is positive, then we are dealing with a minimum value.
In this example, taking the derivative of the derivative we have the value 4 which is positive and so we know this is a minimum.

For equations of the type aX² + bX + c =0, a handy tool to use is the Quadratic Equation Calculator. Not only does this calculate the roots of the equation, it will also show the derivative and the point at which the maximum or minimum exists.

The second example we will look at is very similar to the previous one, except that it is "concave down" instead of "concave up".

Okay, let's examine this equation:

-4X² + 4X + 13 = 0

Since this is a quadratic equation in one variable with the 'a' term being negative, we know that the graph of the curve will be shaped "concave down" (shaped like ∩) and it will have a maximum value but no minimum value (okay, if you want to be technical, its minimum value is negative infinity. You happy now?).

We learned from the first example that the way to calculate a maximum (or minimum) point is to find the point at which an equation's derivative equals zero. The derivative of this equation is:

-8X + 4

and when -8X + 4 = 0, then X= .5 and it is at that point where the maximum of the curve is located. As for the 'Y' value, we substitute .5 into the original equation and get:

Y = -4*(.5*.5)² +4*.5 + 13

Y = 14

So, at point X=.5, Y= 14 there exists a maximum value.
Taking the second derivative of -8X + 4, we get -8. Since this is negative, it means that we have found a maximum value.

Factor Theorem

Note: Before this see Remainder Theorem

if we calculate f(c) and it was 0? which implies remainder is 0,
then (x-c) must be a factor of the polynomial!

Example: x²-3x-4

f(4) = (4)²-3(4)-4 = 16-12-4 = 0

so (x-4) must be a factor of x²-3x-4


The Factor Theorem:

When f(c)=0 then x-c is a factor of the polynomial

we can also write:

When x-c is a factor of the polynomial then f(c)=0

 Why Is This Useful?

Knowing that x-c is a factor is the same as knowing that c is a root (and vice versa).

The factor "x-c" and the root "c" are the same thing

Example:

 2x³-x²-7x+2

The polynomial is degree 3, and could be difficult to solve. So let us plot it first:

The curve crosses the x-axis at three points, and one of them might be at 2. We can check easily:
f(2) = 2(2)³-(2)²-7(2)+2 = 16-4-14+2 = 0
Yes! f(2)=0, so we have found a root and a factor.

So (x-2) must be a factor of 2x³-x²-7x+2

How about where it crosses near -1.8?
f(-1.8) = 2(-1.8)³-(-1.8)²-7(-1.8)+2 = -11.664-3.24+12.6+2 = -0.304
No. (x+1.8) is not a factor.

Remainder Theorem

how to avoid Polynomial Long Division when finding factors!!

"7 divided by 2 equals 3 with a remainder of 1"

Each part of the division has special names:

We can write as a sum like:

Polynomials:

Polynomial is an expression containing more than 2 algebraic terms.

We can also divide polynomials.

f(x) ÷ g(x) = q(x) with a remainder of r(x)

But it is better to write it as a sum like this: 

Example:

                             2x²-5x-1 divided by x-3

•  f(x) is 2x²-5x-1
• g(x) is x-3

After dividing we get the answer 2x+1, but there is a remainder 2.

• q(x) is 2x+1
• r(x) is 2

we can write:

                  f(x) = g(x)·q(x) + r(x)           

          2x²-5x-1 = (x-3)(2x+1) + 2

When you divide by a polynomial of degree 1 (such as "x-3") the remainder will have degree 0.

2. The Remainder Theorem

When you divide a polynomial f(x) by x-c you get:

f(x) = (x-c)·q(x) + r(x)

But r(x) is simply the constant r (remember) when you divide by (x-

c) the remainder is a constant)

f(x) = (x-c)·q(x) + r

Now put x equal to c:

f(c) = (c-c)·q(c) + r
f(c) = (0)·q(c) + r
f(c) = r

So we get this:

The Remainder Theorem:
When you divide a polynomial f(x) by x-c the remainder r will be f(c)

So if you want to know the remainder after dividing by x-c you don't need to do any division:

Just calculate f(c).

Example: 

2x²-5x-1 divided by x-3

We don't need to divide by (x-3) ... just calculate f(3):

2(3)²-5(3)-1 = 2x9-5x3-1 = 18-15-1 = 2

And that is the remainder we got from our calculations above.
We didn't need to do Long Division at all!

Example:

 Dividing by x-4

(Continuing our examplee)

What would the remainder be if we divided by "x-4" ?
"c" is 4, so let us check f(4):

2(4)²-5(4)-1 = 2x16-5x4-1 = 32-20-1 = 11

Laws of Logs

The properties of indices can be used to show that 

the following rules for logarithms are satisfied:

·         log_ax + log_ay = log_a(xy)

·         log_ax – log_ay = log_a(x/y)

·         log_axⁿ = n.log_ax

Example

Simplify: log 3 + 3log 2 - log 4

                 = log 3 + log 2³ - log 4

                 = log 2 + log 8 - log 4

                = log (2 × 8) - log 4

                = log 16 - log 4

                = log (16/4)
            
                = log 4