Mathematics Education Standards

Introduction to mathematics education standards:

The mathematics education include a different branches of unit conversions, algebra, subtraction, measurement, number sense, multiplication, functions, adding and subtraction of decimals, fractions & mixed numbers, division, algebra, geometry, median problems, algebra function, probability and statistics number using words decimals. This mathematics education supports all type of standards up to higher standards.


Example problems for Mathematics education standards:


Example 1:

Solve the quadratic equation `x^2 +5x + 6 =0`

Solution:

`X^2 +5x +6 =0`

`X^2 +2x +3x + 6 =0`

` x(x +2) +3 (x +2) = 0`

`(x +2)(x + 3) =0`

`x + 2 = 0 `                   `x` ` + 3 =0`

`X = -2 `                      `X =-3`

Example 2:

Solve the quadratic equation `x^2 +4x + 4 =0`

Solution:

` X^2 +4x +4 =0`

`X^2 +2x +2x + 4 =0`

` x(x +2) + 2(x +2) = 0`

`(x +2)(x + 2) =0`

`x + 2 = 0 `              ` x + 2 =0`

`X = -2 `                      `X =-2`

Example of polynomial exponent problems- Mathematics education standards:

Addition of polynomial exponent:

Two or more polynomials, adding the terms,

Suitable example adding polynomial exponent,

Example1:

` (2x^2+3x^3)+(x^2+7x^3)`

`=2x^2+x^2+3x^3+7x^3`

The variable and exponent must be same then we add the polynomial exponent,

`=3x^2+10x^3`

So the result is `=3x^2+10x^3`

Subtraction of polynomial exponent

Example2:

`(3x^2+3x^3)-(x^2+7x^3)`

`=3x^2-x^2+3x^3-7x^3`

The variable and exponent must be same then we subtract the polynomial exponent,

`=2x^2-4x^3 `

So the result is` =2x^2-4x^3 `

Adding polynomials- Mathematics education standards:

Example 1: Find the sum of `6x^2 + 7x + 16 and 1x - 3x^2 -4.`

Solution: By means properties of real numbers, we realize

`(6x^2 + 7x + 16) + (-3x^2 + 1x - 4) = 3x^2 + 7x + 1x + 16 - 4`

`= 3x^2 + 7x + 1x + 16 - 4`

`= 3x^2 + 8x + 12`

So the final result is `= 3x^2 + 8x + 12`

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Examples for finite difference problem- Mathematics education standards:


Example 1: calculate the values of Δ`y` and dy if `y = f(x) = x^3 + x^2 - 2x + 1`

Where x changes (i) from `1 to 1.05` and (ii) from` 1 to 1.01`

Solution:

(i) We have `f(1) = 1^3 + 1^2 - 2(1) + 1 = 1`

`f(1.05) = (1.05)^3 + (1.05)^2 - 2(1.05) + 1 = 1.15.`

and Δ`y = f(1.05)- f(1) = 0.15.`

in general `dy = f ^ 1(x) dx = (3x^2 + 2x - 2)dx`

When `x` ` = 1` , `dx = ` Δ`x =1 and dy = [(3(1)^2+2(1)-2] 1= 3`

(ii) `f(1.01) = (1.01)^3 - (1.01)^2 - 2(1.01) + 1 = -1.01`

∴ Δ`y = f(1.01) - f(1) = 1.99`