Learn Intercept Formula

Learn Intercept formula is nothing but slope intercept formula. Before going to learn intercept formula we need to know why it is called so.

It is called slope intercept form because the equation includes slope and the y-intercept. So now we know the reason why it is called slope intercept form. Now coming to actual concept.....

The general form of slope intercept form is:

y= mx^+b

Where m--> slope of the line.

b --> y-intercept.

y --> y-coordinate.

x -->x-coordinate.

Formula - learn intercept formula


Slope intercept form is the simplest of all forms as we just need to plug in the values of slope(m) and y-intercept(b).

Now how are we going to get the slope??. In problems the slope might be mentioned directly or two points through which the line passes might be given. When the second case occurs slope can be find out using formula
slope(m)= (change in x)/(change in y)

Now let us consider that two points are (x1,y1) and (x2,y2)  then slope is given by

m=(y2-y1)/(x2-x1)

We have learnt how to find slope in the slope intercept form. now the next one to be calculated is y-intercept(b).
For this we need to plug the point through which the line passes.

This point will be mentioned in the question. If slope is m and the point is (p,q) then

plugging these values in slope intercept form we get

q= m*p+b

==> b= q-mp.

Now we have slope and y-intercept substituting these values we get slope intercept form.

I think to learn intercept form is very easy  and... cool

Examples on learn intercept formula


Ex1:  What is slope intercept form of line having slope 2 and y-intercept 3?
Sol1:

Given slope (m)=2, and y-intercept (b)=3

Plug these values in slope intercept form y=mx+b

Then        y= 2*x+3

y=2x+3.

So slope intercept form of a line having slope 2 and y-intercept 3 is

y=2x+3

Ex2:  What is slope intercept form of line passing through the point (2, 3) and having a slope of 4.
Sol 2:

Here we have slope = 4 and y-intercept is not mentioned but the point through which line passes is given.

Plug m=4 and (x,y)= (2,3) in the slope intercept form y=mx+b.

Then we get

3=4*2+b.

==>b+8 =3

==>  b=3-8=-5

So y-intercept is b=-5 and we have slope as m=4.

Plugging these in slope intercept form we get

y= 4x -5.

Ex3:  Express the equation 3x+4y+5=0 in slope intercept form.
Sol 3:

The given equation is 3x+4y+5=0.

In order to convert it in to slope intercept form bring y terms on one side and the remaining terms to other side.

For this subtract with 3x and 5 on both sides

3x+4y+5-3x-5=-3x -5

By doing this 3x and 5 get cancelled and the equation becomes

4y= -3x-5

Divide both sides by 4

4y /4 = (-3x-5)/4

y=-(3/4)x  - 5/4

Thus the slope intercept form of 3x+4y+5=0 is y=-(3/4)x  - 5/4

Mean Value Theorem Derivatives

In calculus, the mean value theorem states, roughly, that given an arc of a smooth continuous (derivatives) curve, there is at least one point on that arc at which the derivative (slope) of the curve is equal (parallel) to the "average" derivative of the arc. It is used to prove theorems that make global conclusions about a function on an interval starting from local hypotheses about derivatives at points of the interval. I like to share this Anti derivative with you all through my article.

Formal Statement of Mean Value Theorem Derivatives:

Let f: [x, y] → R be a continuous function on the closed interval [x, y] , and differentiable on the open interval (x, y), where x < y. Then there exists some z in (x, y) such that,

f'(z) = `(f(y) - f(x))/(y - x)`

The mean value theorem is a generalization of Rolle's Theorem, which assumes f(x) = f (y), so that the right-hand side above is zero.

Only one needs to assume that f: [x, y] → R is continuous on [x, y] , and that for every m in (x, y) the limit,

`lim_(h ->0) (f(m + h) - f(m))/(h)`

exists as a finite number or equals + ∞ or − ∞. If finite, that limit equals f′(m). An example where this version of the theorem applies is given by the real-valued cube root function mapping m to m1/3, whose derivative tends to infinity at the origin.

Note that the theorem is false if a differentiable function is complex-valued instead of real-valued.

For example, define f(m) = eim for all real m. Then

f (2π) − f(0) = 0 = 0(2π − 0)

while, |f′(m)| = 1.

- Source Wikipedia


Proof of Mean Value Theorem Derivatives:


Let g(m) = f(m) − rm, where r is a constant, because f is continuous on the closed interval [x, y] and differentiable on the open interval (x, y), now we want to select r, so as g satisfies the conditions of the Theorem,

g (x) = g (y) `hArr` f (x) − rx =  f (y) − ry

`hArr` ry − rx = f (y) − f(x)

`hArr` r(y − x) = f (y) − f(x)

`hArr` r = `(f(y) - f(x))/(y - x)`

By Rolle's theorem, g is continuous on the closed interval [a, b] and g(a) = g(b), there is some c in (a, b) for which g′(c) = 0, and it follows from the equality g(x) = f(x) − rx that,

f' (c) = g' (c) + r = 0 + r =  `(f(b) - f(a))/(b - a)`

Algebra 2 Help and Answers

Algebra 2 help us find the unknown quantities with the help of known quantities. In algebra, we frequently use letters to represent numbers. Algebra 2 help includes real numbers, complex numbers, vectors, matrices etc. Algebra 2 is the study of the rules of relations and operations, and the constructions arising from them. An algebraic expression represents a scale that gets added or subtracted or multiplied or divided on both sides. In algebra 2 help and answers numbers are considered as constants.


Algebra 2 help and answers example Questions:


The following problems gives different answers to algebra 2 problems.

Ex 1:  Determine all real solutions to the equation

Sqrt (2 x + 13) = x - 5

Sol :   Given
sqrt (2 x + 13) = x - 5

We raise both sides to power 2.
[Sqrt (2 x + 13)] 2 = (x - 5) 2

And simplify.
2x +13 = x 2 - 10 x +25

Write the equation with right side equation to 0.
X 2 - 8 x + 12 = 0

It is a quadratic equation with 2 solutions
x = 6 and x = 2

Ex 2 :  Determine all real solutions to the equation

Sqrt (x + 8) = 12

Sol :  Given
sqrt (x + 8) = 12

We raise both sides to power two(2) in order to clear the square root.
[Sqrt (x + 8)] 2 = 12 2

And simplify
x + 8 = 144

Solve for x.
x = 136

Ex 3:   Determine all real solutions to the equation

Sqrt (x 2 – 13x+72) = 6

Sol :   Given
Sqrt (x 2 – 13x+72) = 6

We raise both sides to power 2.
[Sqrt (x 2 – 13x+72)] 2 = (6) 2

And simplify.
X 2 – 13x+72= 36

Write the equation with right side equation to 0.
X 2 - 13 x + 36 = 0

It is a quadratic equation with 2 solutions
x = 9 and  x = 4


Algebra 2 help and answers practice problems:


1) Determine all real solutions to the equation

Sqrt (x + 25) = 13

Ans : x=12

2) Determine all real solutions to the equation

Sqrt (x 2 – 11x+55) = 5

Ans : x=5 and x=6

3) Determine all real solutions to the equation

Sqrt (2 x + 24) = x

Ans : x=6 and x= - 4

Complex Polygon

Complex polygon is a polygon whose sides cross over eachother one or more times. If the number of cross over increases, the complexity of the polygon also increases. Below are some of the complex polygons

Area of a Complex Polygon


There are three steps to find the area of the complex polygon. They are as follows:

Step 1: Break the polygon into simple rectangle, square or triangle.

Step 2: Find the area of all rectangle, square or triangle.

Step 3: Add all the area and get the area of complex polygon.

With the help of the above steps, we can easily find the area of any complex polygons.


Examples

Given below are some of the examples to find the area of a complex polygon.

Example 1:

Find the area of following polygon


DE= 5cm, EF =8cm, FD = 5 cm, AB = 14cm, CD = 10cm and the distance between these two parallel side AB and CD is 5cm

Solution:

Step 1: Break the polygon into triangle and trapezoid.


Here, the given complex polygon is braked into two simple polygon. It is separated at the point D

Step 2: Find the area of triangle (D1EF) and trapezoid (ABCD2).

Area of triangle (D1EF)

Value of s = (D1E + EF + FD1) / 2

Value of s = (5+8+5) / 2

Value of s = 9 cm

Area of triangle (D1EF) = SquareRoot (s(s - D1E)(s - EF)(s - FD1))

= SquareRoot (9(9 - 5)(9 - 8)(9 - 5))

= SquareRoot (9 x 4 x1 x 4)

=  SquareRoot (144)

= 12 square cm

Area of Trapezoid (ABCD2) =  ½ ( AB +CD2 ) x ( Perpendicular distance between AB  and  CD2 )

= ½ (10 +14) x 5

= ½ (24) x 5

= 12 x 5

= 60 square cm

Step 3: Add all the area of triangle and trapezoid to get area of complex polygon.

Area of Complex polygon = Area of triangle (D1EF) + Area of Trapezoid ABCD2)

= 12 square cm  +  60 square cm

= 72 square cm

Binomial probability formula

Binomial is deals with the polynomial that has two terms is known as a binomial e.g.5x + 3y, 2x^2 – 5xy.Probability is the arithmetical quantify of the possibility of an event to occur. If in an examine there are n feasible ways completely and mutually exclusive and out of them in m ways in the event A occur, it is given by (m / n) if in a random sequence of n trails of an events, M are favor to the event, then ratio is (M / n).by studying in both combined as binomial probability as P(A) = m / n.

The possibility of an event can be conveyed as a binomial probability if its conclusions can be wrecked down into two probability of p and q, where p and q are balancing (i.e. p + q = 1)


Binomial Probability Formula - Explained


The probability of getting exact value of k in n trials is given by using the formula,

`P(X = x) = ((n),(k)) p^k q^(n-k) `
where
` q= (1-p)`
or
`P(X = r) =nCr p^r(1-p)^(n-r)`

where
n = Number of events.
r = Number of successful events in the trials.
p = Probability of success in the single trial of the events.
nCr = `(( (n!) / (n-r)! ) / (r!))`
1-p = Probability of failure in the trial.


Examples on Binomial Probability Distribution

Example 1:

Assume that there are taking about 10 question multiple choice test. If each question has four choices and you have to guess on each question, what is the probability of getting exactly 7 questions correct using binomial Probability formula?


Solution:
General Formula for finding solution is,

`P(X = x) = ((n),(k)) p^k q^(n-k) `

n = 10
k = 7
n – k = 3
p = 0.25 = probability of guess the correct answer on a question
q = 0.75 = probability of guess the wrong answer on a question

P(7 Correct guesses out of 10 questions) = `((10),(7))` (0.25)7(0.75)3
? 0.0031 approximate value

Therefore if someone guesses 10 answers on a multiple choice test with 4 options, they have about a 5.8% chance of getting 5 and only 5 correct answers. If 5 or more correct answers are needed to pass then probability of passing can be calculated by adding the probability of getting 5 (and only 5) answers correct, 6 (and only 6) answers correct, and so on up to 10 answers correct. Total probability of  5 or more correct answer is approximate percentage is  7.8


Example 2:
Suppose a die is tossed 7 times. What is the probability of getting exactly 3 fours?

Solution:

General Formula for finding solution

`P(X = r) =nCr p^r(1-p)^(n-r)`

Step 1:
Number of trials n = 7
Number of success r =3

Probability of success in any single trial p is given as 1/6 or 0.167

Step 2:

To calculate nCr formula is used.

nCr = `(n!)/((n-r)!(r!))`

= `(7!)/((7-3)!(3!))`

= `(7!)/((4)!(3!))`

= `(5040)/((24)(6))`

= `(5040)/(144)`

= 35

Step 3:

Find pr.
pr =  0.1673
= 0.004657463

step 4:

To Find (1-p)n-r Calculate 1-p and n-r.
1-p = 1-0.167 = 0.833
n-r = 7-3 = 4

Step 5:

Find (1-p)n-r.
= 0.8334 = 0.481481944

Step 6:

Solve P(X = r) = nCr p r (1-p)n-r
= 10 × 0.004657463 × 0.481481944
=  0.0224248434

The probability of getting exactly 3 fours is 0.0224248434

Solving Online Venn Diagram of Quadrilaterals

We know that on combining three non-collinear points in pairs, the figure obtained as a triangle. Now, let us mark four points and see what we obtain on combining them in pairs in some order. Such a figure created by combining four points in an order is called quadrilateral. In a quadrilateral, a pair of opposite sides is parallel.

In online, students can learn about quadrilaterals. Through online, students can have interactive sessions with tutors. Online is one of the efficient tools for one to one learning. In this article, we are going to discuss about solving online venn diagram of quadrilaterals.

Solving Online Venn Diagram of Quadrilaterals - Definition and its Types:

Definition of Quadrilateral:

Quadrilateral is a geometrical figure that consists of four ends called verticals joined to each other by straight-line segments or sides called edges. The quadrilateral are classified depends on the length of side, angle and diagonals of object.

Solving online venn diagram of quadrilaterals - Types:

There are two types of Quadrilaterals in geometry. The following are the two types of Quadrilaterals.

Convex quadrilaterals:

Parallelograms
Square
Rectangle
Rhombus

Concave quadrilaterals:

Trapezium
Kite

Solving online venn diagram of quadrilaterals:



Solving Online Activities with Quadrilaterals - Descriptions and Example Problems:

Parallelogram:

Parallelogram is one of the classifications of quadrilateral
Parallelogram consists of parallel opposite sides
Parallelogram consists of equal length
Parallelogram consists of equal opposite angles


Formula:

Area = b × h sq. units (b – Base, h - Height)

Example:

The base of a parallelogram is 19 and the height is 12. What is the area of this parallelogram?

Solution:

Area of parallelogram = b * h

= 19 * 12

= 228sq.units

Square:

Parallel lines are lines are exactly the same length and height that has never cross.

Square is a one of the classifications of quadrilateral
A square shape is same to both rhombus and rectangle
Square consists of equal parallel sides and also all angles are right angle (90°) only
The Square is a parallelogram having an angle, equal to right angle and adjacent sides equal.


Formula:

Area = a2    (a - area)

Example:

Find the area of square, when side length is 22cm

Solution:

Area of square = a2

= 22 * 22

= 484 cm2

Rhombus properties:

Rhombus is a parallelogram having its adjacent sides equal but more of whose angles is a right angles.

Rhombus consists of four-sided geometrical shape and also all sides have same length
Rhombus consists of parallel opposite sides
It consists of equal opposite angles
It is also one of the parallelograms


Formula:

Area = `1 / 2` (d1 × d2) sq. units

Where, d1, d2 are diagonals.

Example:

Find the area of rhombus d1 = 20 cm, d2 = 29 cm

Solution:

Area of rhombus = `1 / 2` (d1 * d2)

= `1 / 2` (20 * 29)

= (10 * 29)

= 290 cm2

Trapezium:

A quadrilateral should have at least one pair of parallel sides. The quadrilateral one side of opposite sides are parallel if non parallel of opposite sides of a trapezoid are congruent it is called as isosceles triangle.

Trapezium is also otherwise known as trapezoid
Trapezium is one of the main types of quadrilateral
Trapezium has one pair of parallel that present in opposite sides


Formula:

Area = `1 / 2` (a + b) × h sq. units.

Example:

Find the area of trapezoid height = 11 cm, a = 20 cm, b = 29 cm

Solution:

Area of trapezoid = `1 / 2 ` (a + b)* h

=` 1 / 2` (20 + 29) * 11

= `1 / 2 ` (49) * 11

= 24.5 * 11

= 269.5 square cm

Right Triangle Similarity Theorem

The right triangle similarity theorem is otherwise called as Pythagoras theorem. This right angle theorem was introduced by Philosopher and Greek Mathematician, Pythagoras. The right triangle is a part of geometrical figures. This theorem is used for finding the length of any sides of a right triangle. The right triangles are special triangles that contain only one right angle. Here the right triangle is a triangle that measuring an angle `90^o`

Statement for Right Triangle Theorem:

In terms of area, we can define this theorem. In any right triangle, the area of the square whose side is hypotenuse is equal to the sum of the areas of the other two sides. Here the hypotenuse refers to the side that opposite to the right triangle. The other two sides of a triangle meet at right angle.



Here Hypotenuse is the side opposite to the right angle. an adjacent side is the side adjacent to the given angle and the opposite side is the side that opposite to given  .

According to this theorem, the equation can be given as

(Opposite)2+ (Adjacent)2 = (Hypotenuse)2

Here, an angle  value lies between 0 and 90 degree. Here, 90 degree is one of the interior angle and the other two interior angles are complementary. Complementary means the angle value less than 90 degree. The other two interior angle sum should be equal to 90 degree.  This right triangle similarity theorem can also be used in trigonometric functions. This right triangle forms six possible ratios in trigonometry.



Other Forms of the Equation:



Here consider a is opposite side, b is an adjacent side and c is hypotenuse. Therefore, an equation can be written as a2 + b2 = c2. To find c, an equation can be written as c= `sqrt(a^(2)+b^(2))` . If c is known value, the length of one side is given; the following equations can be used;

a =`sqrt(c^(2)+b^(2))`   (or)       b = `sqrt(c^(2)-c^(2))`

Proof:

This theorem proof is based on the proportionality of two similar triangles. It depends on the ratio of any two corresponding sides of similar triangles.



Let ABC represent a right triangle, with the right angle located at B. Here, H is the altitude drawn from B and it also intersects AC. The point H divides the hypotenuse c into two parts d and e. The new triangle ABH is similar to the triangle ABC because both the triangles have a right angle. Similarly, the triangle BCH is also similar to the triangle ABC. Thus the proof of similarity of the right triangles requires the Triangle postulate i.e., the sum of the angles in a triangle is 2 right angles, and is equivalent to the parallel postulate. Similarity of the right angle triangles leads to the equality of ratios of their corresponding sides:

a/c =e/a and b/c= d/b

In this the first result is equal to cosine angle and the second result is equal to sine angle.

These ratios can be written as:

a2= c*e and b2=c*d

Summing these two equations we get:

a2 + b2 = (c*e) + (c*d) = c* (e + d) = c*c = c2

Therefore a2+ b2 = c2

Hence, right triangle similarity theorem is proved.