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# AP 10th Class Maths 7th Chapter Coordinate Geometry Exercise 7.1 Solutions
Well-designed AP 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Exercise 7.1 offers step-by-step explanations to help students understand problem-solving strategies.
## Coordinate Geometry Class 10 Exercise 7.1 Solutions – 10th Class Maths 7.1 Exercise Solutions
Question 1.
Find the distance between the following pairs of points:
i) (2, 3), (4, 1)
ii) (- 5, 7), (- 1, 3)
iii) (a, b), (- a, – b)
Solution:
i) Let A(2, 3) and B(4, 1)
(x1 y1) (x2 y2)
are the points on line AB.
We know that distance between two
points is $$\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}$$
AB distance = $$\sqrt{(4-2)^2+(1-3)^2}$$
Therefore, AB = $$\sqrt{2^2+(-2)^2}$$
= $$\sqrt{4+4}$$
= $$\sqrt{8}$$ or 2$$\sqrt{2}$$ units.
ii) Let P(-5, 7) and Q(-1, 3)
(x1 y1) (x2 y2)
are the points on line PQ.
We know that distance between two
points is $$\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$$
PQ distance = $$\sqrt{(-1-(-5))^2+(3-7)^2}$$
= $$\sqrt{(-1+5)^2+(-4)^2}$$
= $$\sqrt{4^2+16}$$
Therefore PQ = $$\sqrt{16+16}$$
= $$\sqrt{32}$$ = 4$$\sqrt{2}$$ units
iii) Let K(a, b) and L(-a, -b)
(x1 y1) (x2 y2)
are the points on line KL.
We know that distance between two
points is $$\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}$$
distance = $$\sqrt{(-a-a)^2+(-b-b)^2}$$
= $$\sqrt{(-2 a)^2+(-2 b)^2}$$
Therefore, KL = $$\sqrt{4 a^2+4 b^2}$$
= $$\sqrt{4\left(a^2+b^2\right)}$$
= $$2 \sqrt{a^2+b^2}$$ units
Question 2.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.
Solution:
Given A(0, 0) and B(36, 15) are two towns.
Distance between them is AB.
In ∆ABC, ∠C = 90°
By Pythagoras theorem,
AB2 = AC2 + BC2 = 362 + 152
AB = $$\sqrt{1296+225}$$ = 39 km.
Therefore, distance between two towns AB = 39 km.
(Or)
Distance between A(0, 0) and B(36, 15) is AB.
x1 = 0, y1 = 0; x2 = 36, y2 = 15
We know that distance between two
points = $$\sqrt{\left(x_2-x_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}$$
= $$\sqrt{(36-0)^2+(15-0)^2}$$
= $$\sqrt{36^2+15^2}$$
= $$\sqrt{1296+225}$$
AB = $$\sqrt{1521}$$ = 39 km.
∴ Distance between two towns A and B is 39 km.
Question 3.
Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Solution:
Let the points are A(1, 5), B(2, 3) and C(-2, -11).
Distance between two points
AB + BC ≠ AC
So, A, B, C are not collinear points.
Question 4.
Check whether (5, – 2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Solution:
Let A(5, -2), B(6, 4) and C(7, -2) are the vertices of ∆ABC.
Distance between two points
Clearly we got AB = BC
So, ∆ABC is an isosceles triangle.
Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in given figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
Solution:
From the given figure, points are A(3, 4), B(6, 7), C(9, 4) and D(6, 1).
Distance between two points
Therefore, AB = BC = CD = AD = 3$$\sqrt{2}$$ units
Diagonals AC = BD = 6 units
So, all sides and two diagonals are equal.
Therefore, given vertices can form a square.
So, ABCD is a square.
Clearly, Champa said ABCD is a square is correct.
Question 6.
Name the type of quadrilateral formed, if any, by the following points and give reasons for your answer:
i) (-1, -2), (1, 0), (-1, 2), (-3,0)
ii) (-3, 5), (3, 1), (0, 3), (-1,-4)
iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:
i) Let A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0) are vertices of a quadrilateral.
In this quadrilateral AB = BC = CD = AD = 2$$\sqrt{2}$$ units and diagonals AC = BD = 4 units.
So, given points can form a square,
ii) Let P(-3, 5), Q(3, 1), R(0, 3), S(-1, -4) are vertices of quadrilateral PQRS.
Distance between two points
So, all sides of the quadrilateral are of different lengths.
Hence, given points cannot form a special quadrilateral.
iii) Let A(4, 5), D(7, 6), I(4, 3), N(1, 2) are the vertices of the quadrilateral ADIN.
Distance between two points
In the quadrilateral opposite sides are equal and diagonals are not equal. Hence given point can form a parallelogram.
Question 7.
Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Solution:
Let the P(x, 0) is the point on the X-axis which is equidistant from the points A(2, -5) and B(-2, 9), that is AP = PB.
Distance between two points
Question 8.
Find the values of y for which the dis-tance between the points P(2, – 3) and Q(10, y) is 10 units.
Solution:
Given points are P(2, -3) and Q(10, y) and PQ = 10 units.
Distance between two points
⇒ 64 + y2 + 6y + 9 = 100
⇒ y2 + 6y + 73 – 100 = 0
⇒ y2 + 6y – 27 = 0
⇒ y2 + 9y – 3y – 27 = 0
⇒ y(y + 9) – 3(y + 9) = 0
⇒ (y – 3)(y + 9) = o
Therefore y = 3 (or) -9
Question 9.
If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Solution:
Q(0, 1) is equidistant from P(5, -3) and R(x, 6) that is QR = PQ
Distance between two points
Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).
Solution:
Let P(x, y) is equidistant from the points A(3, 6) and B(-3, 4) that is PB = PA
Distance between two points
-6x – 6x – 12y + 8y = 25 – 45
-12x – 4y = -20
12x + 4y = 20
⇒ 3x + y = 5
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# What Is The Slope When The Line Is Vertical?
## Is a straight vertical line a function?
For a relation to be a function, use the Vertical Line Test: Draw a vertical line anywhere on the graph, and if it never hits the graph more than once, it is a function.
If your vertical line hits twice or more, it’s not a function..
## How is horizontal line drawn?
Understanding a Horizontal Line The horizontal line is drawn by connecting similar swing lows in price to create a horizontal support line. For a horizontal resistance line, similar swing highs are connected. … In more simple terms, a horizontal line on any chart is where the y-axis values are equal.
## Is Y 4 horizontal or vertical?
The equation x=4 represents a vertical line which crosses the x -axis at the point (4,0) . Every ordered pair with 4 as its first coordinate is a solution . (The equation means “ x is equal to 4 , and y can be whatever it wants.”) is a horizontal line which crosses the y -axis at (0,−3) .
## Is vertical up and down?
Vertical describes something that rises straight up from a horizontal line or plane. … The terms vertical and horizontal often describe directions: a vertical line goes up and down, and a horizontal line goes across. You can remember which direction is vertical by the letter, “v,” which points down.
## How do you use the vertical line test to identify a function?
To use the vertical line test, take a ruler or other straight edge and draw a line parallel to the y-axis for any chosen value of x. If the vertical line you drew intersects the graph more than once for any value of x then the graph is not the graph of a function.
## Is a horizontal line positive or negative?
The slope of a line can be positive, negative, zero, or undefined. A horizontal line has slope zero since it does not rise vertically (i.e. y1 − y2 = 0), while a vertical line has undefined slope since it does not run horizontally (i.e. x1 − x2 = 0).
## What is the equation of the horizontal line (- 4 6?
In a horizontal line the y value remains the same. In the point ( -4. -6 ) the y value is -6. the point notation is always written as ( x, y).
## What does it mean if a line is vertical?
A vertical line is one the goes straight up and down, parallel to the y-axis of the coordinate plane. All points on the line will have the same x-coordinate. … A vertical line has no slope. Or put another way, for a vertical line the slope is undefined.
## How do you know if a line is vertical or horizontal?
Note that when a line has a positive slope it rises up left to right. Note that when a line has a negative slope it falls left to right. Note that when a line is horizontal the slope is 0. Note that when the line is vertical the slope is undefined.
## What is the Y intercept of a vertical line?
Facts about vertical lines The x -intercept of a vertical line x=c is the point (c,0) . Except for the line x=0 , vertical lines do not have a y -intercept.
## Is Y 2 horizontal or vertical?
Since it’s always hard to remember when these guys are horizontal and when they are vertical, I’ve got a sentence that will always save you… When you see y = -2, say this: y is always -2 and x can be anything! This tells you what the graph should look like!
## Is Y horizontal or vertical?
A coordinate grid has two perpendicular lines, or axes, labeled like number lines. The horizontal axis is called the x-axis. The vertical axis is called the y-axis. The point where the x-axis and y-axis intersect is called the origin.
## What is the slope and y intercept of a vertical line?
The equation of a vertical line does not have a y-intercept since a vertical line never crosses the y-axis. (link). The slope of a vertical line is undefined because the denominator of the slope (the change in X) is zero. Vertical lines help determine if a relation is a function in math.
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Lesson Explainer: Combining the Product, Quotient, and Chain Rules | Nagwa Lesson Explainer: Combining the Product, Quotient, and Chain Rules | Nagwa
# Lesson Explainer: Combining the Product, Quotient, and Chain Rules Mathematics
In this explainer, we will learn how to find the first derivative of a function using combinations of the product, quotient, and chain rules.
Many functions can be constructed from simpler functions by combining them in one or more of the following ways:
• addition and subtraction: and ,
• multiplication and division: and ,
• composition: .
Fortunately, we recall that there are rules for differentiating functions that are formed in these ways. For addition and subtraction, we can use the linearity of the derivative; for multiplication and division, we have the product rule and quotient rule; for composition, we can apply the chain rule. Let us review these rules.
### Rule: Rules of Differentiation
For differentiable functions and and constants , we have the following rules:
• Linearity: .
• Product rule: .
• Quotient rule: for , .
• Chain rule: .
In addition to using these rules separately, it is also possible to use them in conjunction with each other, allowing us to differentiate any combination of elementary functions. However, we should be aware that this is often not a trivial exercise and it can be challenging to identify the correct rules to apply, the best order to apply them, and whether there are algebraic simplifications that will make the process easier. In this explainer, we will look at a number of examples that will highlight the skills we need to navigate this landscape.
Let us consider an example of differentiating a complicated function combining many different operations together, and how we can tackle the differentiation by splitting it into separate parts. Suppose we have
At first, this may seem impossible to deal with, but we can break it into parts. Generally, the best way to do this is to begin by considering the outermost layer first and working inward. If we do this, we can see that it is the sum of two functions:
Using the linearity of differentiation, this means we can differentiate and separately and add them together afterwards. If we consider separately in this way, we can see that
In other words, is a composition of functions, so we can apply the chain rule to help us differentiate it. This of course means that we need to find the derivative of , but this too can be broken down into smaller parts. Continuing this process, we can continue to remove layers of complexity from the function until we arrive at elementary expressions that we know how to differentiate. We can represent this visually as follows.
Notice that all the functions at the bottom of the tree are functions that we can differentiate easily. Hence, we can see that by applying the appropriate rules at each stage, we can find the derivatives of even the most complicated functions.
We will now look at some examples where we apply this approach, albeit to slightly simpler cases. To start with, let us consider a high-degree polynomial function that has been factored.
### Example 1: Finding the First Derivative of Polynomial Functions at a Point Using the Product and Chain Rules
Find the first derivative of at .
Let us first analyze the given function and see what rules we can apply to it. Since it is a polynomial function, it would be possible for us to expand all the parentheses via multiplication and take the derivatives of the separate components, but this would require a large number of calculations. Instead, it would be more efficient to split it into parts by combining a couple of rules of differentiation together. We first of all note that this is a product of functions:
Recall that the product rule tells us that the derivative of the product of two differentiable functions is given by
To use this formula, we need to know the derivatives and . The first function, , can be differentiated by using the power rule to get . For , we note that it is a composition of functions:
That is, if and , then . We recall that the derivative of a composition of functions can be found using the chain rule, which is
Using the power rule, we find that the derivative of is and the derivative of is . Substituting these into the formula above (along with ), we get
Now that we have both and , we can substitute , , , and into the product rule formula to get
Finally, we need to find the first derivative at the point . Note that is the coordinate of the point on the graph representing the given function, and at this point, the -coordinate is equal to 1. Thus, we substitute in to find
As we can see, it is often useful to be able to break down functions into components that we can deal with individually by using the product and chain rules. Let us consider another example where it is beneficial for us to combine different rules together.
### Example 2: Differentiating Combinations of Polynomial and Root Functions Using Product and Chain Rules
Find at .
We have been asked to differentiate a function that is a combination of polynomial and square root functions, so let us consider how best to deal with it. One possibility is to take the term into the square root and use the chain rule to differentiate the resulting square root function. The other way is to recognize that it is a product of functions, as follows:
Thus, since and are differentiable functions, we can use the product rule, which says that
To use this, we must calculate and . First of all, can be differentiated using the power rule to get . For , we cannot use the power rule directly, but we can break it down into parts by recognizing that it is a composition of functions, as follows:
That is, , where and . Thus, we can use the chain rule, which is
Now, we can find both and using the power rule, which gives us
Substituting these into the chain rule equation, we get
Taking this function for , along with , , and , we can put them into the product rule equation to get
Now, recall that we need to evaluate this function at . Doing this, we get the following solution:
In the first couple of examples, we have considered functions that require both the product rule and the chain rule to be differentiated. In the next example, we will consider a function defined in terms of polynomials and square root functions where we will need to use the quotient rule.
### Example 3: Differentiating a Function Involving Rational Functions at a Point Using the Rules of Differentiation
Evaluate at if .
In this example, we want to determine the first derivative of the given rational function by applying the rules of differentiation and evaluating this at the point .
Since we will most likely have to apply more than one rule of differentiation, it will be helpful to consider what order we want to apply them in. The best way to do this is to consider the outermost part of the function and see how it can be broken down; if we do this, we can see that the given function is a quotient of functions as follows:
In order to determine the derivative of this, we will make use of the quotient rule which states that for a function , we have, for ,
Thus, we can determine the derivative of the given function by setting and in the numerator and denominator respectively. In order to apply this rule, we need to evaluate the derivatives of both and .
The derivative of is straightforward and can be found from the power rule as . Since is the composition of two functions, and , to find the derivative of , we will make use of the chain rule which states that for a composite function , we have
In order to apply this rule, we need to evaluate the derivatives of and with respect to , both of which can be found by an application of the power rule as follows:
Substituting these expressions back into the chain rule with , we find the derivative of with respect to as follows:
Substituting , , , and as shown above, we can now apply the quotient rule as follows:
By multiplying both the numerator and denominator by , we can simplify the fraction, giving us
Substituting the given point in this expression gives
So far, we have only seen instances of examples where it is optimal to apply the product or quotient rule followed by the chain rule, but the opposite order may be more natural depending on the given function. In each case, we have to make sure that the order in which we are applying the derivative rules makes sense. Let us consider a situation where we may have to change the order of applying the rules.
### Example 4: Differentiating a Composition of Rational and Root Functions Using the Chain and Quotient Rules
If , determine .
Here, we have been asked to calculate the derivative of a function that appears to be a combination of different functions. Since we will probably have to use more than one rule of differentiation to do this, we should begin by considering which one to apply first. We do this by looking at the outermost part of the function and working inward. Since the outer part is a square root, we can see that we have
That is, , where and . We note that since this is a composition of differentiable functions, we can use the chain rule, which says that
To use this formula, we need and , the former of which can just be obtained by using the power rule to get . For the latter, we can see that it is a quotient of differentiable functions:
Therefore, we can apply the quotient rule. Recall that for , this is
Each of and is a polynomial function, so we can differentiate them using the power rule. This gives us
Then, substituting these functions into the quotient rule formula, we get
Now that we have , and , let us recall that our original derivative could be found using the chain rule as follows:
Recall that , which means that . If we substitute this into the above equation, we get
We can simplify this a bit by making use of the fact that to get
We note that there is an alternate way to solve the above problem. If we had used the fact that from the start, we would have
If we took the derivative of this, we could use the quotient rule first instead, followed by the chain rule on the functions in the numerator and denominator. However, this would result in a similar amount of work to get to the same answer.
Recall that there exists another form of differentiating problem that we might encounter. Suppose we have to find , where
If is not given explicitly in terms of , we cannot differentiate it directly. Instead, we can use the following form of the chain rule:
Since we can differentiate with respect to and with respect to , this allows us to compute the derivative. Alternatively, we could substitute into the equation for to get
Recall that, written this way, we can return to the original form of the chain rule that we have been using so far:
We note that this is just another way of writing as shown above (in particular, and ). It is also important to note that substituting into the equation for in this way can lead to the expression simplifying, meaning there is a possibility we could differentiate it without using the chain rule (but this is not always the case).
In any case, let us consider an example in this style where we have to use the chain rule, with the additional complication of having to use another rule of differentiation.
### Example 5: Differentiating a Composition of Rational Functions Using the Chain and Quotient Rules
Evaluate at if and .
In this example, we need to differentiate a function with respect to that is not explicitly given in terms of .
One possibility of differentiating this would be to substitute the equation for into the function for , so that we can differentiate directly with respect to . However, this would result in fractions inside of fractions, which might be messy.
Instead, let us use the chain rule to differentiate this. Recall that the chain rule can be written in terms of ’s and ’s (i.e., Leibniz’s notation) as follows:
Thus, let us compute and . Both functions are quotients of polynomials (which are differentiable functions), which means we can use the quotient rule to differentiate them. Recall that for , the quotient rule is given by
For , we can switch out the variable for , and we have and . The derivative of these functions with the power rule is straightforward; we get and . Combining this all into the quotient rule equation, we get
We can perform a very similar calculation for . Let , so that and , with derivatives and . Then, using the quotient rule, we get
Returning to our formula for the chain rule, we multiply these two derivatives together to get
We could express this purely in terms of , but this is unnecessary since we only need to evaluate the derivative at the given point, . Note that at this point, . Thus, letting and , we have
Let us recap a few important points that we have learned in this explainer.
### Key Points
• Using the rules of differentiation, namely, the product, quotient, and chain rules, we can calculate the derivatives of any combination of elementary functions.
• It is important to consider the order in which we use the rules as this will help ensure we choose the most efficient method.
• Generally, we want to consider the outermost parts of the function and work inward. Doing this, we can decompose a complicated function into simpler parts that can be evaluated directly.
• It is worth keeping in mind that some problems can be simplified to the point where we do not need to use multiple rules of differentiation, making the process significantly easier (even though such problems have not been featured in this explainer).
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When you do things like going to a birthday party or going to school, you’re usually there for a long time, which is a lot of minutes. It can be difficult to describe events that last a long time only with minutes. Because of that, we have another unit for time. We call this unit an hour. $1$ hour is made up of $60$ minutes.
If you have Lego or some other building blocks, you can go get them. Pick out $60$ bricks, and imagine that one brick is $1$ minute. Build one or more towers out of the $60$ bricks you picked out.
You can imagine that the construction you’ve made is one hour, and each brick in the construction is one minute.
To see one minute in relation to one hour, you can pick out another single brick and put it next to the construction you made. To do this, you need $61$ bricks in total: $60$ bricks in the construction and $1$ brick on its own next to it.
Often, we use hours and minutes together to describe an event more precisely. How long are you at school? Some days school finishes early, and you might have been there for $3$ hours. Other days school finishes late, in which case you might have been at school for $4$ hours and $30$ minutes.
We often use minutes to describe how long something takes when it takes less than an hour. An example can be recess: Your teacher might say “You can have recess for $15$ minutes.” Or it could describe how long you have to wait for the bus: “The bus will be here in $3$ minutes.”
Rule
DoingCalculationswithTime
You can switch from minutes to hours like this:
You can switch from hours to minutes like this:
Example 1
You go to the park to play soccer with some friends. You’re there for 5 hours and 40 minutes. How many minutes did you spend playing soccer?
You have to calculate how many minutes are in the $5$ hours first. After that you can add the other minutes that were given in the problem.
$\begin{array}{llll}\hfill 5\phantom{\rule{0.17em}{0ex}}\text{h}=\left(5\cdot 60\right)\phantom{\rule{0.33em}{0ex}}\text{min}& =300\phantom{\rule{0.17em}{0ex}}\text{min}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 300\phantom{\rule{0.17em}{0ex}}\text{min}+40\phantom{\rule{0.17em}{0ex}}\text{min}& =340\phantom{\rule{0.17em}{0ex}}\text{min}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
You’ve been playing football for $340$ minutes.
Example 2
You’re going to the movies and the screen says that one of the movies lasts for 177 minutes. How many hours and minutes does that movie last?
You switch from minutes to hours first, and then you can find how many minutes are in the remainder.
$177\phantom{\rule{0.17em}{0ex}}\text{min}=\left(177÷60\right)\phantom{\rule{0.33em}{0ex}}\text{h}=2.95\phantom{\rule{0.17em}{0ex}}\text{h}$
$2.95$ h is the same as $2\phantom{\rule{0.17em}{0ex}}\text{h}+0.95\phantom{\rule{0.17em}{0ex}}\text{h}$. That gives you
$0.95\phantom{\rule{0.17em}{0ex}}\text{h}=\left(0.95\cdot 60\right)\phantom{\rule{0.33em}{0ex}}\text{min}=57\phantom{\rule{0.17em}{0ex}}\text{min},$
which means that the movie lasts for $2$ hours and $57$ minutes.
Math Vault
Would you like to solve exercises about seconds, minutes and hours? Try Math Vault!
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# Completing the Square
Completing the Square is a method used to solve a quadratic equation by changing the form of the equation so that the left side is a perfect square trinomial .
To solve $a{x}^{2}+bx+c=0$ by completing the square:
1. Transform the equation so that the constant term, $c$ , is alone on the right side.
2. If $a$ , the leading coefficient (the coefficient of the ${x}^{2}$ term), is not equal to $1$ , divide both sides by $a$ .
3. Add the square of half the coefficient of the $x$ -term, ${\left(\frac{b}{2a}\right)}^{2}$ to both sides of the equation.
4. Factor the left side as the square of a binomial.
5. Take the square root of both sides. (Remember: ${\left(x+q\right)}^{2}=r$ is equivalent to $x+q=±\sqrt{r}$ .)
6. Solve for $x$ .
Example 1:
Solve ${x}^{2}-6x-3=0$ by completing the square.
$\begin{array}{l}{x}^{2}-6x=3\\ {x}^{2}-6x+{\left(-3\right)}^{2}=3+9\\ {\left(x-3\right)}^{2}=12\\ x-3=±\sqrt{12}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=±2\sqrt{3}\\ x=3±2\sqrt{3}\end{array}$
Example 2:
Solve: $7{x}^{2}-8x+3=0$
$\begin{array}{l}7{x}^{2}-8x=-3\\ {x}^{2}-\frac{8}{7}x=-\frac{3}{7}\\ {x}^{2}-\frac{8}{7}x+{\left(-\frac{4}{7}\right)}^{2}=-\frac{3}{7}+\frac{16}{49}\\ {\left(x-\frac{4}{7}\right)}^{2}=-\frac{5}{49}\\ x-\frac{4}{7}=±\frac{\sqrt{5}}{7}i\\ x=\frac{4}{7}±\frac{\sqrt{5}}{7}i\\ {\left(x-3\right)}^{2}=12\\ x-3=±\sqrt{12}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=±2\sqrt{3}\\ x=3±2\sqrt{3}\end{array}$
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## Skills Review for Area Between Two Curves, Determining Volumes by Slicing, and Volumes of Revolution: Cylindrical Shells
### Learning Outcomes
• Graph horizontal and vertical lines
• Graph linear equations in different forms using ordered pairs
• Graph a linear equation using x and y-intercepts
• Determine if and where two equations intersect
In the first three sections of Module 6, it is very important that we graph the equations we are given so there is a visual to assist us with setting up the needed integral. First, we will review the special equation forms for horizontal and vertical lines. Then, we will review how to graph equations using points and intercepts and determine where two functions intersect.
## Graph Vertical and Horizontal Lines
The equation of a vertical line is given as
$x=c$
where c is a constant. The slope of a vertical line is undefined, and regardless of the y-value of any point on the line, the x-coordinate of the point will be c.
Suppose that we want to find the equation of a line containing the following points: $\left(-3,-5\right),\left(-3,1\right),\left(-3,3\right)$, and $\left(-3,5\right)$. First, we will find the slope.
$m=\frac{5 - 3}{-3-\left(-3\right)}=\frac{2}{0}$
Zero in the denominator means that the slope is undefined and, therefore, we cannot use point-slope form. However, we can plot the points. Notice that all of the x-coordinates are the same and we find a vertical line through $x=-3$.
The equation of a horizontal line is given as
$y=c$
where c is a constant. The slope of a horizontal line is zero, and for any x-value of a point on the line, the y-coordinate will be c.
Suppose we want to find the equation of a line that contains the following set of points: $\left(-2,-2\right),\left(0,-2\right),\left(3,-2\right)$, and $\left(5,-2\right)$. We can use point-slope form. First, we find the slope using any two points on the line.
$\begin{array}{l}m=\frac{-2-\left(-2\right)}{0-\left(-2\right)}\hfill \\ =\frac{0}{2}\hfill \\ =0\hfill \end{array}$
Use any point for $\left({x}_{1},{y}_{1}\right)$ in the formula, or use the y-intercept.
$\begin{array}{l}y-\left(-2\right)=0\left(x - 3\right)\hfill \\ y+2=0\hfill \\ y=-2\hfill \end{array}$
The graph is a horizontal line through $y=-2$. Notice that all of the y-coordinates are the same.
The line x = −3 is a vertical line. The line y = −2 is a horizontal line.
### Example: Graphing Horizontal and Vertical Lines
Graph the line $x=2$.
## Use Points to Graph Equations
We can plot a set of points to represent an equation. Suppose we want to graph the equation $y=2x - 1$. We can begin by substituting a value for x into the equation and determining the resulting value of y. Each pair of and y-values is an ordered pair that can be plotted. The table below lists values of x from –3 to 3 and the resulting values for y.
$x$ $y=2x - 1$ $\left(x,y\right)$ $-3$ $y=2\left(-3\right)-1=-7$ $\left(-3,-7\right)$ $-2$ $y=2\left(-2\right)-1=-5$ $\left(-2,-5\right)$ $-1$ $y=2\left(-1\right)-1=-3$ $\left(-1,-3\right)$ $0$ $y=2\left(0\right)-1=-1$ $\left(0,-1\right)$ $1$ $y=2\left(1\right)-1=1$ $\left(1,1\right)$ $2$ $y=2\left(2\right)-1=3$ $\left(2,3\right)$ $3$ $y=2\left(3\right)-1=5$ $\left(3,5\right)$
We can plot these points from the table. The points for this particular equation form a line, so we can connect them.
Note that the x-values chosen are arbitrary regardless of the type of equation we are graphing. Of course, some situations may require particular values of x to be plotted in order to see a particular result. Otherwise, it is logical to choose values that can be calculated easily, and it is always a good idea to choose values that are both negative and positive. There is no rule dictating how many points to plot, although we need at least two to graph a line. Keep in mind, however, that the more points we plot, the more accurately we can sketch the graph.
### How To: Given an equation, graph by plotting points
1. Make a table with one column labeled x, a second column labeled with the equation, and a third column listing the resulting ordered pairs.
2. Enter x-values down the first column using positive and negative values. Selecting the x-values in numerical order will make graphing easier.
3. Select x-values that will yield y-values with little effort, preferably ones that can be calculated mentally.
4. Plot the ordered pairs.
5. Connect the points if they form a line.
### Example: Graphing an Equation in Two Variables by Plotting Points
Graph the equation $y=-x+2$ by plotting points.
### Try It
Construct a table and graph the equation by plotting points: $y=\frac{1}{2}x+2$.
## Use Intercepts to Graph Equations
The intercepts of a graph are points where the graph crosses the axes. The x-intercept is the point where the graph crosses the x-axis. At this point, the y-coordinate is zero. The y-intercept is the point where the graph crosses the y-axis. At this point, the x-coordinate is zero.
To determine the x-intercept, we set y equal to zero and solve for x. Similarly, to determine the y-intercept, we set x equal to zero and solve for y. For example, lets find the intercepts of the equation $y=3x - 1$.
To find the x-intercept, set $y=0$.
$\begin{array}{llllll}y=3x - 1\hfill & \hfill \\ 0=3x - 1\hfill & \hfill \\ 1=3x\hfill & \hfill \\ \frac{1}{3}=x\hfill & \hfill \\ \left(\frac{1}{3},0\right)\hfill & x\text{-intercept}\hfill \end{array}$
To find the y-intercept, set $x=0$.
$\begin{array}{lllll}y=3x - 1\hfill & \hfill \\ y=3\left(0\right)-1\hfill & \hfill \\ y=-1\hfill & \hfill \\ \left(0,-1\right)\hfill & y\text{-intercept}\hfill \end{array}$
We can confirm that our results make sense by observing a graph of the equation. Notice that the graph crosses the axes where we predicted it would.
### How To: Given an equation, find the intercepts
1. Find the x-intercept by setting $y=0$ and solving for $x$.
2. Find the y-intercept by setting $x=0$ and solving for $y$.
### Example: Finding the Intercepts of the Given Equation
Find the intercepts of the equation $y=-3x - 4$. Then sketch the graph using only the intercepts.
### Try It
Find the intercepts of the equation and sketch the graph: $y=-\frac{3}{4}x+3$.
### Try It
Determine Where Two Functions Intersect
To determine where two functions intersect, set them equal to each other and solve for $x$.
### Example: Determining Where Two Functions Intersect
Find the points of intersection of the functions $y=x+2$ and $y=x^2+3x+2$.
### Try It
Find the points of intersection of the functions $y=x-2$ and $y=2x^2-4x+1$.
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# Using Divide and Conquer in Linear Search
Before we go to the topic let’s do some basic revision.
Linear Search or sequential search is a method for finding an element within a list. It sequentially checks each element of the list until a match is found or the whole list has been searched. https://www.wikiwand.com/en/Linear_search
and,
Divide and conquer is an algorithm design paradigm based on multi-branched recursion. A divide-and-conquer algorithm works by recursively breaking down a problem into two or more sub-problems of the same or related type until these become simple enough to be solved directly. https://www.wikiwand.com/en/Divide-and-conquer_algorithm
According to me, Linear Search is the most basic algorithm. It is pretty straight forward and easiest to implement. So naturally, every programmer knows this.
It doesn’t matter whether you are self-taught or learned to program in school or college, 90% chances are that you know how to implement linear search by simple iteration, And frankly, it is the best and only practical way!
So what’s the point of this post then?
Let’s come to it.
Think about the divide and conquer technique. As per the definition you have to recursively divide the bigger problem into two or more smaller problems. So basically we are diving and ruling. Now it doesn’t matter if you are dividing the problem into exactly two equal halves or many different smaller parts or some kind of uneven distribution.
Now we use this fact in the Linear search algorithm.
Let's look at the Iterative approach
Let’s see the Recursive approach
What’s going on here?
We are calling a function and giving it an array, size of the array, element to search, and index as argument.
Inside the function, we are doing some basic checks and finally, we are calling the function again and this time we are incrementing the index by 1 (cause initially it was 0).
If you are confused that how this can be a divide and conquer, then please take a look at the definition again. It says to divide the problem into two or more parts and here we are dividing our array into an array of size one less than the original one.
But,
Here’s a catch. To apply it recursively can be good if you want to trick some beginner but for practical uses, this is as useless as it can be.
We are taking comparatively more resources and then also we have to complexity of O(n).
The main takeaway is to use the iterative method for linear search but if you want to have some fun then give someone this program.
For me, it was a different way to look at problems.
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# How to Find Acceleration with Mass and Force
## 1. Step-by-Step Guide: How to Find Acceleration with Mass and Force
Acceleration is the rate at which an object changes its velocity over time. It depends on two key factors: mass and force. The formula for finding acceleration with mass and force is:
a = F/m
To find acceleration using mass and force, follow these steps:
1. Identify Mass and Force: Determine the mass (m) of the object and the force (F) acting on it.
2. Apply Newton’s Second Law: Use Newton’s second law, (F = m ⋅ a), where (F) is the force, (m) is the mass, and (a) is the acceleration.
3. Isolate Acceleration: Rearrange the equation to solve for acceleration:
[ a = F/m ]
4. Plug in Values: Substitute the known values of force and mass into the formula.
5. Calculate Acceleration: Perform the division to find the acceleration.
By following these steps, you can easily calculate the acceleration of an object based on the force applied to it and its mass. This application of Newton’s second law helps us understand how the motion of an object is influenced by the forces acting upon it.
### Tabular Explanation
Here is a simple table to help solve problems concerning acceleration with mass and force:
Example Table:
This table guides you through the process of solving acceleration problems using mass and force, providing a structured approach for easy calculations.
## 2. Concept of Acceleration
Acceleration is the measure of how quickly the velocity of an object changes over time. It is defined as the rate of change of velocity (a = v / t). When an object speeds up, slows down, or changes direction, it undergoes acceleration. The SI unit for acceleration is meters per second squared (m/s²).
Acceleration is influenced by two main factors: mass and force. Mass refers to the quantity of matter in an object, while force is a push or pull acting upon an object. According to Newton’s Second Law of Motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
## 3. Newton’s Second Law of Motion
Newton’s Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, this can be expressed as the equation:
F = ma
Where:
• F represents the net force acting on the object,
• m denotes the mass of the object, and
• a symbolizes the acceleration produced.
This equation indicates that the greater the force applied to an object or the smaller its mass, the greater the resulting acceleration.
## 4. How to Calculate Acceleration with Mass and Force
To calculate acceleration, you can rearrange Newton’s Second Law equation to solve for acceleration:
a = F / m
Where:
• a is the acceleration,
• F represents the force acting on the object, and
• m denotes the mass of the object.
By knowing the values of force and mass, you can substitute them into the equation and calculate the resulting acceleration. The unit of measurement for force is Newtons (N), while mass is typically measured in kilograms (kg).
## 5. Practice Problems
### Problem 1: Calculating acceleration with known mass and force
Let us consider a problem where a car with a mass of 1000 kg experiences a net force of 500 N. To find the acceleration, we can use the formula:
a = F / m
Substituting the values into the equation, we have:
a = 500 N / 1000 kg = 0.5 m/s²
Therefore, the car’s acceleration is 0.5 m/s².
### Problem 2: Determining force given acceleration and mass
Suppose an object with a mass of 2 kg undergoes an acceleration of 4 m/s². To find the force acting on the object, we can rearrange the formula:
F = ma
Plugging in the values, we get:
F = 2 kg x 4 m/s² = 8 N
Hence, the force acting on the object is 8 N.
## 6. Real-Life Applications
The knowledge of how to find acceleration using mass and force has numerous practical applications across various fields. Here are a few examples:
1. Automotive industry: Calculating acceleration helps engineers design and optimize vehicles for improved performance and safety.
2. Sports and athletics: Athletes and coaches utilize acceleration calculations to enhance training programs and maximize performance.
3. Space exploration: Determining the acceleration of spacecraft is important for planning trajectories and ensuring successful missions.
## 7. Common Misconceptions
When it comes to finding acceleration, there are a couple of common misconceptions to address:
1. Misunderstanding the relationship between force and acceleration: Some may believe that a greater force always results in greater acceleration, but this is not the case. The acceleration also depends on the mass of the object.
2. Neglecting the effect of friction: In real-world scenarios, friction can significantly affect the acceleration of objects. Therefore, it is important to consider frictional forces when calculating acceleration.
## 8. Troubleshooting
While calculating acceleration, you may encounter certain challenges. Here are a few troubleshooting tips to ensure accurate measurements:
1. Precise measurements: Use accurate instruments to measure force and mass, such as calibrated scales and force gauges.
2. Account for external factors: Consider any external forces or influences that might affect the object’s acceleration, such as air resistance or friction.
3. Repeat experiments: Conduct multiple trials to minimize errors and obtain more reliable results.
## 9. Conclusion
With the knowledge of Newton’s Second Law of Motion and the relationship between force, mass, and acceleration, you can calculate acceleration accurately. Remember to consider practical examples, common misconceptions, and troubleshooting tips to deepen your understanding of this concept.
## 10. Frequently Asked Questions
1. What is the role of mass in determining acceleration?
Mass plays a crucial role in determining acceleration. According to Newton’s Second Law of Motion, acceleration is inversely proportional to mass. A larger mass requires more force to achieve the same acceleration as a smaller mass.
2. How does the force affect an object’s acceleration?
Force directly affects an object’s acceleration. The greater the force applied to an object, the greater its resulting acceleration. This relationship is described by Newton’s Second Law of Motion.
3. Are there any limitations to Newton’s Second Law of Motion?
Newton’s Second Law of Motion holds true under normal circumstances. However, it may not be applicable in extreme conditions such as near the speed of light or at the atomic level, where relativistic or quantum mechanical effects come into play.
4. Can acceleration be negative?
Yes, acceleration can be negative. Negative acceleration, also known as deceleration, occurs when an object slows down. It indicates a change in velocity in the opposite direction of the initial motion.
5. What are some practical applications of calculating acceleration?
Calculating acceleration has practical applications in various fields, including engineering, sports, and space exploration. It helps design vehicles, optimize training programs, and plan trajectories for spacecraft.
You may also like to read:
How to Calculate Density Altitude
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# 6.1: Percents
Difficulty Level: At Grade Created by: CK-12
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Randy looked at his smartphone and saw that he had only 15% of his battery power remaining. That got him wondering, ‘What does 15% mean?’ How much of the phone’s total battery power does Randy have left?
In this concept, you will learn to recognize percent as a ratio whose denominator is 100.
### Guidance
Percent means the number of parts per 100. You know that 100% of anything is the entire or whole amount. Therefore, percents can tell you what part of the whole you are dealing with. To understand percents, you will set up ratios.
A ratio is a comparison of two numbers and that a ratio can be written in three ways. For example: 1 to 2, 1:2, or 12\begin{align*}\frac{1}{2}\end{align*}. Percents are ratios in which the second number (or the denominator) is 100.
Let’s look at an example.
If there are 13 red jelly beans and 15 yellow jelly beans in a jar, the ratio of red jelly beans to yellow jelly beans can be written as 13 to 15, 13:15, or 1315\begin{align*}\frac{13}{15}\end{align*}. Each of these ratios is read as “thirteen to fifteen.”
A percent is a type of ratio. You are now going to apply percents directly to ratios.
A percent is a ratio that compares a number to 100. Percent means “per hundred” and the symbol for percent is %.
100% represents the ratio 100 to 100 or 100100\begin{align*}\frac{100}{100}\end{align*}. Therefore, the value of 100% is 1.
Let’s look at another example.
If there are 100 jelly beans in a jar and 19 are black, you can say that 19100\begin{align*}\frac{19}{100}\end{align*}or 19% of the jelly beans in the jar are black.
Let’s look at one more situation that uses a real-life scenario.
There were 100 questions on a test and Amanda answered 92 of them correctly. What percent did she answer correctly? What percent did she answer incorrectly?
Amanda answered 92 out of 100 questions correctly. First, you can write this as the ratio 92100\begin{align*}\frac{92}{100}\end{align*}.
Next, since the denominator is 100, you can write this ratio in percent form as 92%.
Then, since there were 100 questions on the test and Amanda answered 92 correctly, you can determine that she answered 10092\begin{align*}100 - 92\end{align*}, or 8 out of 100 incorrectly.
Next, you can write this as the ratio 8100\begin{align*}\frac{8}{100}\end{align*}and as the percent 8%.
The answer is Amanda answered 92% of the questions correctly and 8% incorrectly.
### Guided Practice
Write a ratio and a percent that describes the shaded part of the square.
First, count the number of shaded squares and the number of total squares. The figure shows the ratio of 37 shaded squares to 100 squares.
Next, write this ratio as 37 to 100, 37:100, or 37100\begin{align*}\frac{37}{100}\end{align*}.
Then, when the denominator of a ratio in fraction form is 100, you can express the ratio as a percent.
37100=37%\begin{align*}\frac{37}{100}=37 \%\end{align*}
The answer is: the ratio that describes the shaded part is 37 to 100, 37:100, or 37100\begin{align*}\frac{37}{100}\end{align*}, and the percent is 37%.
This is a visual way of showing a percent as a ratio. You can use the picture to write the ratio and the percent.
### Examples
#### Example 1
Solve the following problem. Karen ate 12 out of 100 blueberries. What percent of the blueberries did she eat?
First, you can write this as the ratio 12100\begin{align*}\frac{12}{100}\end{align*}.
Then, since the denominator is 100, you can write this ratio in percent form as 12%.
The answer is she ate 12% of the blueberries.
#### Example 2
Solve the following problem. Joey answered 93 questions correctly out of 100 questions on his test. What percent of the questions did he answer correctly and what percent did he answer incorrectly?
First, you can write this as the ratio 93100\begin{align*}\frac{93}{100}\end{align*}.
Next, since the denominator is 100, you can write this ratio in percent form as 93%.
Then, since there were 100 questions on the test and Joey answered 93 correctly, you can determine that he answered 10093\begin{align*}100 - 93\end{align*}, or 7 out of 100 incorrectly.
Next, you can write this as the ratio 7100\begin{align*}\frac{7}{100}\end{align*}and as the percent 7%.
The answer is Joey answered 93% of the questions correctly and 7% incorrectly.
#### Example 3
Solve the following problem. Sarah gathered 25 roses out of 100 flowers. What percent of the flowers were roses?
First, you can write this as the ratio 25100\begin{align*}\frac{25}{100}\end{align*}.
Then, since the denominator is 100, you can write this ratio in percent form as 25%.
The answer is 25% of the flowers were roses.
Remember Randy and his smartphone battery indicator? It showed that he had 15% of his battery power remaining. How could you write this percent as a fraction?
First, you can write this as the ratio 15:100, which is equivalent to 15100\begin{align*}\frac{15}{100}\end{align*}.
Then, since the denominator is 100, you can write this ratio in percent form as 15%.
The answer is 15% is equivalent to the fraction 15100\begin{align*}\frac{15}{100}\end{align*}.
### Explore More
Write each percent as a ratio with a denominator of 100.
1. 10%
2. 6%
3. 22%
4. 41%
5. 33%
6. 70%
7. 77%
8. 19%
9. 25%
10. 15%
11. 7%
12. 29%
13. 88%
14. 92%
15. 90%
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Decimal
In common use, a decimal refers to part of a whole number. The numbers to the left of a decimal point represent whole numbers, and each number to the right of a decimal point represents a fractional part of a power of one-tenth. For instance: The decimal value 1.24 indicates 1 whole unit, 2 tenths, and 4 hundredths (commonly described as 24 hundredths).
fraction
A fraction is a part of a whole. A fraction is written mathematically as one value on top of another, separated by a fraction bar. It is also called a rational number.
Proportion
A proportion is an equation that shows two equivalent ratios.
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# Logarithmic Functions Section 8.4. WHAT YOU WILL LEARN: 1.How to evaluate logarithmic functions.
## Presentation on theme: "Logarithmic Functions Section 8.4. WHAT YOU WILL LEARN: 1.How to evaluate logarithmic functions."— Presentation transcript:
Logarithmic Functions Section 8.4
WHAT YOU WILL LEARN: 1.How to evaluate logarithmic functions.
Question? What is 2 2 ? What is 2 3 ? For what value of x is 2 x = 6? To solve problems like this, mathematicians developed the concept of logarithms.
Definition of Logarithm Let b and y be positive numbers, with b not equal 1. The logarithm of y with base b is denoted by log b y = x and is defined as follows: log b y = x if and only if b x = y We read the expression as “ log base by of y ”.
Examples Rewrite the following in exponential form: b x = y 1. log 2 32 = 5 2. log 5 1 = 0 3. log 10 10 = 1 4. log 10 0.1 = -1
Special Logarithms Logarithm of 1log b 1 = 0 because b 0 = 1 Log of base blog b b = 1 because b 1 = b
Evaluating Logarithmic Expressions Evaluate the expression. A. Log 3 81 B. Log 5 0.04 C. Log ½ 8 D. Log 9 3
Evaluate the expression. A. Log 4 64 B. Log 2 0.125 C. Log ¼ 256 D. Log 32 2 You Try
Types of Logarithms A logarithm with base 10 is called the common logarithm (your calculator assumes base 10). log 10 x = log x (normally written without the base) The logarithm with base e is called the natural logarithm. We usually write this is “ ln ”. log e x = ln x
Time for Calculators Evaluate. 1. log 5 2. ln 0.1
Using Inverse Properties The expressions: g(x) = log b xf(x) = b x are inverses of one another. We can use this to simplify expressions. This means (get ready for this) g(f(x)) = log b b x = xand f(g(x)) = b log b x = x
Examples Simplify the expression. 1. 10 log 10 2 2. log 3 9 x
You Try Simplify the expression. 1. 10 log 10 5x 2. log 10,000 x
Finding Inverses of Logarithmic Functions Find the inverse of the functions: 1. y = log 3 x 2. y = ln(x + 1)
Graphing Logarithmic Functions The graph of y = log b (x – h) + k has the following characteristics: 1. The line x = h is a vertical asymptote 2. The domain is x>h, and the range is all real numbers. 3. If b > 1, the graph moves up to the right. If 0<b<1, the graph moves down and to the right.
Example Graph
Another Example Objective: 8.4 Logarithmic Functions17 Graph the function. State the domain and range. y = log 5 (x + 2)
Homework page 490, 16-34 even 36, 37, 48-54 even, 56, 61, 65, 66, 69
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Cardinal Vs Ordinal Numbers
With ordinal numbers, it is possible to count unlimited sets. They are also able to generalize ordinal numbers.But before you can utilize them, you must comprehend what they are and how they operate.
1st
One of the basic concepts of math is the ordinal number. It is a number that indicates the place of an item in the list. The ordinal number is represented by a number between zero and twenty. Ordinal numbers can be used for a variety of purposes however they are most often used to indicate the number of items included in the list.
It is possible to present ordinal numbers with numbers as well as words and charts. These are also useful to show the order in which a collection of pieces are arranged.
Most ordinal numbers are classified into one of the two groups. Ordinal numbers that are infinite are represented by lowercase Greek letters, whereas the finite ones are represented by Arabic numbers.
According to the Axiom of Choice, each set that is well-organized should have at least one ordinal. For instance, the highest possible grade will be given to the class’s initial member. The student who received the highest grade was declared the winner of the contest.
Combinational ordinal numbers
Compound ordinal numbers are multi-digit number. These numbers are created by multiplying the ordinal’s last digit. The most popular uses for these numbers are for ranking or dating reasons. They don’t use a unique ending for each number, as with cardinal numbers.
Ordinal numbers are used to denote the order of elements within a collection. They are used to identify the elements in the collection. Regular numbers are available in both suppletive and regular formats.
Regular ordinals can only be made by adding the suffix -u. The number is then typed into words. A hyphen is then added. There are also other suffixes. For instance, “-nd” is for numbers that end in 2, and “-th” for numbers that end with 4 or 9.
The addition of words that have the -u, -e or–ie suffix results in suffixtive ordinals. This suffix, which can be used for counting, is a bit longer than the normal.
Limits to the significance of ordinal numbers
The limit on ordinal quantities that do not have zeros is an ordinal quantity that is not zero. Limit ordinal numbers come with the drawback of not having a maximum element. They can be constructed by joining sets that are not empty and do not have any maximum elements.
Infinite transfinite-recursion definitions use limited ordinal numbers. Based on the von Neumann model, every infinite cardinal number also functions as an ordinal limit.
A limit ordinal number is the sum of the ordinals that are below it. Limit ordinal numbers can be enumerated using arithmetic, but they are also able to be represented as a series of natural numbers.
The ordinal numbers that are used for arranging the data are utilized. These numbers give an explanation of the object’s numerical position. These numbers are commonly utilized in set theory and arithmetic. While they share the same basic structure, they are not part of the same category as natural number.
The von Neumann model uses a well-ordered set. Take a look at fy, which can be described as a subfunction of the function g’. If g’ is able to meet the requirements, g’ is an ordinal limit if it is the only subfunction (i or II).
The Church-Kleene oral is a limit order similarly. The Church-Kleene oral defines an ordinal that is a limit as a properly arranged collection of smaller ordinals, and it has a non-zero ordinal.
Numerological examples of common numbers in stories
Ordinal numbers are used to show the order of things between entities or objects. They are vital for organizing, counting or ranking among other reasons. They can be used to explain the location of the object as well as the order that they are placed.
The ordinal number can be indicated by the letter ‘th’. Sometimes, however, the letter “nd”, can be substituted. Books’ titles are often associated with ordinal numbers.
Ordinal numbers may be expressed as words even though they are often employed in list format. They can also be written as numbers and acronyms. Comparatively, these numbers are much simpler to comprehend than the cardinal ones.
There are three types of ordinal number. They can be learned more by playing games, practice, or other activities. Making yourself aware of the subject is an important aspect of developing your arithmetic skills. Coloring is a fun and easy method to increase your proficiency. You can track your progress using a marker sheet.
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# Chapter 8 Introduction to Trigonometry and its Equation NCERT Exemplar Solutions Exercise 8.4 Class 10 Maths
Chapter Name NCERT Maths Exemplar Solutions for Chapter 8 Introduction to Trigonometry and its Equation Exercise 8.4 Book Name NCERT Exemplar for Class 10 Maths Other Exercises Exercise 8.1Exercise 8.2Exercise 8.3 Related Study NCERT Solutions for Class 10 Maths
### Exercise 8.4 Solutions
1. If cosecθ + cotθ = p, then prove that cosθ = (p2 – 1)/(p2 + 1) .
Solution
As,
cosecθ + cot θ = p
2. Prove that √(sec2 θ + cosec2 θ) = tan θ + cot θ.
Solution
Taking,
L.H.S = √(sec2 θ + cosec2 θ)
Since,
3. The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.
Solution
Let us take PR = h meter, the height of the tower.
The distance between the observer and tower is QR = (20 +x) m,
where
QR = QS + SR
= 20 + x
∠PQR = 30°
∠ PSR = θ
In ∆PQR,
tan 30° = h/(20 +x )
⇒ 1/√3 = h/(20 + x)
On cross multiplication,
We get,
20 + x = √3h
⇒ x = √3h - 20 ...eq. (i)
In ∆PSR,
tan θ = h/x
As, angle of elevation increases by 15° when the observer moves 20 m towards the tower.
So,
θ = 30° + 15° = 45°
Therefore,
tan 45° = h/x
⇒ 1 = h/x
⇒ h = x
Putting x = h in eq. (i),
h = √3h - 20
⇒ √3h - h = 20
⇒ h(√3 - 1) = 20
On rationalizing we get,
h = 10(√3 + 1)
The required height of the tower is 10(√3 + 1) meter.
4. If 1 + sin2 θ = 3sin θ cosθ, then prove that tan θ = 1 or 1/2.
Solution
We have,
1 + sin2 θ = 3sin θ cosθ
on dividing L.H.S and R.H.S equations with sin2 θ,
We get,
cosec2 θ + 1 = 3 cot θ
Also,
cosec2 θ - cot2 θ = 1
⇒ cosec2 θ = cot2 θ +1
⇒ cot2 θ +1+1 = 3 cot θ
⇒ cot2 θ +2 = 3 cot θ
⇒ cot2 θ –3 cot θ +2 = 0
We split the middle term and then solve the equation,
cot2 θ – cot θ –2 cot θ +2 = 0
⇒ cot θ(cot θ -1)–2(cot θ +1) = 0
⇒ (cot θ – 1)(cot θ – 2) = 0
⇒ cot θ = 1, 2
We have,
tan θ = 1/cotθ
tan θ = 1, 1/2
Proved !!
5. Given that sinθ + 2cosθ = 1, then prove that 2sinθ - cosθ = 2.
Solution
We have,
sinθ + 2 cosθ = 1
Squaring on both sides,
(sinθ + 2cosθ)2 = 1
⇒ sin2 θ + 4cos2 θ + 4 sinθ cosθ = 1
Also,
sin2 θ = 1 - cos2 θ
and
cos2 θ = 1 - sin2 θ
⇒ (1 - cos2 θ) + 4(1 - sin2 θ) + 4sin θ cosθ = 1
⇒ 1 – cos2 θ + 4 – 4 sin2 θ + 4sin θcos θ = 1
⇒ –4 sin2 θ – cos2 θ + 4sin θcos θ = –4
⇒ 4 sin2 θ + cos2 θ – 4sin θcos θ = 4
Also,
a2 + b2 - 2ab = (a - b)2
So,
(2sin θ – cos θ)2 = 4
⇒ 2sin θ – cos θ = 2
Proved
6. The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is √st.
Solution
Let BC = s;
PC = t
Let height of the tower be AB = h.
∠ABC = θ and
∠APC = 90° – θ
(∵ the angle of elevation of the top of the tower from two points P and B are complementary)
In ABC,
tan θ = AC/BC
= h/s ...(i)
In APC,
tan(90 - θ) = AC/PC
= h/t
cotθ = h/t ...(ii)
So, multiplying (i) and (ii),
tanθ × cot θ = h/t × h/s
⇒ 1 = h2 /st
⇒ h2 = st
⇒ h = √st
Therefore, the height of the tower is √st.
7. The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower.
Solution
Let SQ = h be the tower.
∠SPQ = 30° and
∠SRQ = 60°
As given in the question, the length of shadow is 50 m long hen angle of elevation of the sun
is 30° than when it was 60°. So,
PR = 50 m and
RQ = x m
In SRQ,
tan 60° = h/x
⇒ √3 = h/x
⇒ x = h/√3
In SPQ,
tan 30° = h/(50 + x)
1/√3 = h(50 + x)
⇒ 50 + x = √3h
Putting value of x,
50 + h/√3 = √3h
⇒ 50√3 + h = 3h
⇒ 50√3 = 3h - h
⇒ 3h - h = 50√3
⇒ 2h = 50√3
⇒ h = 25√3
So, the required height is 25√3 m.
8. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are α and β, respectively. Prove that the height of the tower is (h tanα)/(tanβ - tanα).
Solution
We have, a vertical flag staff of height h is surmounted on a vertical tower of height let (H),
so that,
FP = h and
FO = H.
The angle of elevation of the bottom and top of the flag staff on the plane is ∠PRO = α and
∠FRO = β respectively.
In ∆PRO, we have,
tanα = PO/RO
= h/x
x = h/tan α ...(i)
In FRO,
tanβ = FO/RO
⇒ tanβ = (FP + PO)/RO
⇒ tanβ = (H + h)/x
⇒ x = (H+h)/tanβ ...(ii)
From (i) and (ii),
(H + h)/tanβ = h/tanα
On solving,
H = (h tanα)/(tanα + tanβ)
Hence, proved
9. If tanθ + secθ = l, then prove that secθ = (l2 + 1)/2l .
Solution
Given:
tanθ + secθ = l ...eq.(i)
Multiplying and dividing by (sec θ – tan θ) on numerator and denominator of L.H.S,
So,
secθ - tanθ = 1/l ...eq. (ii)
Adding eq. (i) and eq.(ii), we get
(tanθ + secθ) + (secθ - tanθ) = l + 1/
Adding eq. (i) and eq. (ii), we get
(tanθ + secθ) + (secθ - tanθ) = l + 1/
So,
secθ = (l2 + 1)/2l .
Hence, proved.
10. If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q(p2 - 1) = 2p.
Solution
Given that,
sinθ + cosθ = p ...(i)
and
secθ + cosecθ = q
Since, sinθ + cosθ = p
On squaring both sides, we get
(sinθ + cosθ)2 = p2
(sin2 θ + cos2 θ) + 2sinθ cosθ = p2
[as, (a + b)2 = a2 + 2ab + b2 ]
1 + 2sinθ cosθ = p2
⇒ 1 + 2(p/q) = p2
⇒ q + 2p = p2q
⇒ 2p = p2q - q
⇒ q(p2 - 1) = 2p
11. If a sinθ + bcosθ = c, the prove that a cosθ - b sinθ = √(a2 + b2 - c2)
Solution
Given that,
a sinθ + b cosθ = c
On squaring both sides,
(a + sinθ + cosθ b)2 = c2
⇒ a2 sin2 θ + b2 cos2 θ + 2ab sinθ cosθ = c2
⇒ a2 (1 - cos2 θ) + b2 ( 1 - sin2 θ) + 2ab sinθ . cosθ = c2
⇒ a2 - a2 cos2 θ + b2 - b2 sin2 θ + 2absinθ . cosθ = c2
⇒ a2 + b2 - c2 = a2 cos2 θ + b2 sin2 θ - 2absinθ.cosθ
⇒ a2 + b2 - c2 = a2 cos2 θ + b2 sin2 θ - 2absinθ.cosθ
⇒ a2 + b2 - c2 = (acosθ - bsinθ)2
[as, a2 + b2 - 1ab = (a- b)2]
⇒ a cosθ - bsinθ = √(a2 + b2 - c2 )
12. Prove that (1 + secθ - tanθ)/(1 + secθ + tanθ) = (1 - sinθ)/cosθ.
Solution
Taking LHS,
On multiplying by (sec θ - tanθ) on numerator and denominator,
13. The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the other tower.
Solution
Let us take the distance between the two towers = AB
= x m
and height of the other tower = PA
= h m
Given,
height of the tower = QB
= 30 m,
∠QAB = 60° and
∠PBA = 30°
Now, in ΔQAB,
tan 60 = QB/AB = 30/x
⇒ √3 = 30/x
⇒ x = 30/√3
⇒ x = 10√3
In PBA,
tan 30 = PA/AB
1/√3 = h/x
⇒ 1/√3 = h/10√3
⇒ h = 10
14. From the top of a tower h m high, the angles of depression of two objects, which are in line with the foot of the tower are α and β (β > α). Find the distance between the two objects.
Solution
Let the distance between the two objects is x m and CD = y m
Given,
∠BAX = α = ∠ABD
and
∠CAX = β = ∠ACD [Alternate angle]
and
height of tower, AD = h m
In ACD,
⇒ tanβ = h/y
⇒ y = h/tanβ ...(i)
In ABD,
⇒ tanα = AD/(BC + CD)
⇒ tanα = h/(x + y)
⇒ y = (h/tanα) - x ...(ii)
From (i) and (ii),
(h/tanα) - x = h/tanβ
⇒ x = (h/tanα) - (h/tanβ)
⇒ x = h(cotα - cotβ)
Which is the required distance between the two objects.
15. A ladder rests against a vertical wall at an inclination α to the horizontal. Its foot is pulled away from the wall through a distance p so that its upper end slides a distance q down the wall and then the ladder makes an angle β to the horizontal. Show that p/q = (cos β - cos α)/(sinα - sinβ)
Solution
Let,
OQ = x and
OA = y
Given,
BQ = q,
SA = P
and
AB = SQ = Length of ladder
Also,
∠BAO = α and ∠QSO = β
cos α = OA/AB
⇒ cos α = y/AB
⇒ y = AB cos α = OA ...(i)
also,
sinα = OB/AB
⇒ OB = AB sinα
In QSO,
cos β = OS/SQ
⇒ OS = SQ cos β = AB cos β
and,
sin β = OQ/SQ
⇒ OQ = SQ sin β = AB sinβ
Now,
SA = OS - AO
P = Ab(cos β - cos α )
Also,
BQ = BO - QO
⇒ q = AB(sin α - sin β)
So,
p/q = (cosβ - cos α)/(sin α - sinβ)
16. The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.
Solution
Let the height the vertical tower be,
OT = H m
and
OP = AB = xm
Given that,
AP = 10 m
and
∠TPO = 60°,
∠TAB = 45°
Now, in ΔTPO,
tan 60° = OT/OP
⇒ tan 60° = H/x
⇒ x = h/√3
In TAB,
tan 45° = TB/AB
⇒ 1 = (H - 10)/x
⇒ x = H - 10
Putting value of x,
h/√3 = H - 10
On solving,
H = 5(√3 + 3)m
17. A window of a house is h metres above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be α and β, respectively. Prove that the height of the other house is h(1 + tan α cot β) metres.
Solution
Let the height of the other house,
OQ = H m and
OB = MW = x m
Given,
height of the first house = WB = h = MO
and ∠QWM = α,
∠OWM = β = ∠WOB [Alternate angle]
Now, In ∆WOB,
tan β = WB/OB
⇒ tan β = h/x
⇒ x = h/tanβ ...(i)
In QWM,
tan α = QM/WM
⇒ tan α = (OQ - MO)/WM
⇒ tan α = (H - h)/x
⇒ x = (H - h)/tanα ...(ii)
From (i) and (ii),
(H - h)/tanα = h/tan β
On solving,
H = h(1 + tanα × 1/tanβ)
18. The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60° and 30°, respectively. Find the height of the balloon above the ground.
Solution
Let the height of the balloon from above the ground is H m
and
OP = W2 R
= W2 Q
= x m
Given, height of lower window from the ground = W2 P
= 2m = OR
Height of upper window from the lower window = W1 W2
= 4m = QR
BQ = OB - (QR + RO)
= H - (4 + 2)
= H - 6
∠BW1Q = 30° and
∠BW2R = 60°
In BW2 R,
tan 60° = BR/W2 R
= (BQ + QR)/x
⇒ √3 = [(H - 6) + 4]/x
⇒ x = (H - 2)/√3 ...(i)
In BW1 R,
tan 30° = BR/W1 R
⇒ 1/√3 = (H - 6)/x
⇒ x = √3(H - 6) ...(ii)
From (i) and (ii),
√3(H - 6) = (H - 2)/√3
On solving,
H = 8m
The required height is 8m.
Therefore, the required height of the balloon from the ground is 8m.
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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C
These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C.
Other Exercises
Question 1.
In the given circle with diameter AB, find the value of x. (2003)
Solution:
∠ABD = ∠ACD = 30° (Angle in the same segment)
But ∠ADB = 90° (Angle in a semi-circle)
∴ x + 90° + 30° = 180°
⇒ x + 120° = 180°
∴ x = 180°- 120° = 60°
Question 2.
In the given figure, O is the centre of the circle with radius 5 cm, OP and OQ are perpendiculars to AB and CD respectively. AB = 8cm and CD = 6cm. Determine the length of PQ.
Solution:
Radius of the circle whose centre is O = 5 cm
OP ⊥ AB and OQ ⊥ CD, AB = 8 cm and CD = 6 cm.
Join OA and OC, then OA = OC=5cm.
∴ OP ⊥ AB
∴ P is the mid-point of AB.
Similarly, Q is the mid-point of CD.
In right ∆OAP,
OA² = OP² + AP² (Pythagorous theoram)
⇒ (5)² =OP² +(4)² ( ∵ AP = AB = $$\frac { 1 }{ 2 }$$ x 8 = 4cm)
⇒ 25 = OP² + 16
⇒ OP3 = 25 – 16 = 9 = (3)²
∵ OP = 3 cm
Similarly, in right ∆ OCQ,
OC2 = OQ2 + CQ2
⇒ (5)2 =OQ2+(3)2 (∵ CQ = CD = $$\frac { 1 }{ 2 }$$ x 6 = 3cm)
⇒ 25 = OQ2 + 9
⇒ OQ3 = 25 – 9 = 16 = (4)2
∴ OQ = 4 cm
Hence, PQ = OP + OQ = 3-4 = 7 cm.
Question 3.
The given figure shows two circles with centres A and B; and radii 5cm and 3cm respectively, touching each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.
Solution:
Join AP and produce AB to meet the bigger circle at C.
AB = AC – BC = 5 cm – 3 cm = 2 cm.
But, M is the mid-point of AB
∴ AM = $$\frac { 2 }{ 2 }$$ = 1cm.
Now in right ∆APM,
AP2 = MP2 + AM2 (Pythagorous theorem)
⇒ (5)2 = MP2 -1- (1 )2
⇒ 25 = MP: + 1
⇒ MP: = 25 – 1 = 24
Question 4.
In the given figure, ABC is a triangle in which ∠BAC = 30°. Show that BC is equal to the radius of the circumcircle of the triangle ABC, whose centre is O.
Solution:
Given: In the figure ABC is a triangle in winch ∠A = 30°
To Prove: BC is the radius of circumcircle of ∆ABC whose O is the centre.
Const: Join OB and OC.
Proof: ∠BOC is at the centre and ∠BAC is at the remaining part of the circle
∴ ∠BOC = 2 ∠BAC = 2 x 30° = 60°
Now in ∆OBC,
OB = OC (Radii of the same circle)
∴ ∠OBC = ∠OCB
But ∠OBC + ∠OCB + ∠BOC – 180°
∠OBC + ∠OBC + 60°- 180°
⇒ 2 ∠OBC = 180°- 60° = 120°
⇒ ∠OBC = $$\frac { { 120 }^{ circ } }{ 2 }$$ = 60°
∴ ∆OBC is an equilateral triangle.
∴ BC = OB = OC
But OB and OC are the radii of the circumcircle
∴ BC is also the radius of the circumcircle.
Question 5.
Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Solution:
Given: In ∆ABC, AB-AC and a circle with AB as diameter is drawn which intersects the side BC and D.
To Prove: D is the mid point of BC.
Proof: ∠ 1 = 90° (Angle in a semi-circle)
But ∠ 1 + ∠ 2 – 180° (Linear pair)
∴ ∠ 2 = 90°
Now, in right ∆ ABD and ∆ ACD,
Hyp. AB – Hyp. AC (Given)
∴ ∆ABD = ∆ACD (RHS criterion of congruency)
∴ BD = DC (c.p.c.t.)
Hence D is he mid point of BC. Q.E.D.
Question 6.
In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.
Solution:
Join OE,
Arc EC subtends ∠EOC at the centre and ∠EBC at the remaining part of the circle.
∴ ∠EOC = 2 ∠EBC = 2 x 65° = 130°
Now, in ∆OEC, OE = OC (radii of the same circle)
∴ ∠OEC = ∠OCE
But ∠ OEC + ∠ OCE + ∠ EOC = 180°
⇒ ∠ OCE + ∠ OCE + ∠ EOC = 180°
Question 7.
Chords AB and CD of a circle intersect each other at point P, such that AP = CP. S.how that AB = CD.
Solution:
Given: Two chords AB and CD intersect each other at P inside the circle with centre O and AP = CP.
To Prove: AB = CD.
Proof: ∵ Two chords AB and CD intersect each other inside the circle at P.
∴ AP x PB = CP x PD ⇒ $$\frac { AP}{ CP }$$ = $$\frac { AD }{ PB }$$
But AP = CP ….(i) (given)
∴ PD = PB or PB = PD …,(ii)
AP + PB = CP + PD
⇒ AB = CD Q.E.D.
Question 8.
The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.
Solution:
Given: ABCD is a cyclic quadrilateral and PRQS is a quadrilateral formed by the angle bisectors of angle ∠A, ∠B, ∠C and ∠D.
To Prove: PRQS is a cyclic quadrilateral.
Proof: In ∆APD,
∠1 +∠2 + ∠P= 180° ,…(i)
Similarly, in ∆BQC.
∠3 + ∠4 + ∠Q= 180° ….(ii)
Adding (i) and (ii), we get:
∠1 + ∠2 + ∠P + ∠3 + ∠4 ∠Q = 180° + 180° = 360°
⇒ ∠1 + ∠2 + ∠3 + ∠4 + ∠P + ∠Q = 360° ..(iii)
But ∠1+∠2+∠3+∠4 = $$\frac { 1 }{ 2 }$$ (∠A+∠B+∠C+∠D)
= $$\frac { 1 }{ 2 }$$ x 360°= 180°
∴ ∠P + ∠Q = 360° – 180° = 180° [From (iii)]
But these are the sum of opposite angles of quadrilateral PRQS
Question 9.
In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:
(I) ∠BDC
(ii) ∠BEC
(iii) ∠BAC (2014)
Solution:
∠DBC = 58°
BD is diameter
∴ ∠DCB=90° (Angle in semi circle)
(i) In ∆BDC
∠BDC + ∠DCB + ∠CBD = 180°
∠BDC = 180°- 90° – 58° = 32°
(ii) ∠BEC =180°-32°
= 148°
(iii) ∠BAC = ∠BDC = 32°
(Angles in same segment)
Question 10.
D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.
Solution:
Given: In ∆ABC, AB = AC and D and E are points on AB and AC such that AD = AE, DE is joined.
To Prove: B,C,E,D. are concyclic.
Proof: In ∆ABC, AB = AC
∴ ∠B = ∠C (Angles opposite to equal sides)
In ∆ABC,
∵ $$\frac { AP }{ AB }$$ = $$\frac { AE }{ AC }$$
∴ DE || BC.
∴ ∠ADE = ∠B (Corresponding angles)
But ∠B = ∠C (Proved)
∴ Ext. ∠ADE = its interior opposite ∠C.
∴ BCED is a cyclic quadrilateral.
Hence B,C, E and D are concyclic.
Question 11.
In the given figure, ABCD is a cyclic quadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ADC = 92°, ∠FAE’= 20°; determine ∠BCD. Give reason in support of your answer.
Solution:
AF || CB and DA is produced to E such that
∠ADC = 92° and ∠FAE – 20°
Now, we have to find the measure of ∠BCD In cyclic quad. ABCD,
∠B – ∠D = 180° ⇒ ∠B + 92° = 180″
⇒∠B = 180°-92° = 88°
∵ AF || CB.
∴∠FAB = ∠B = 88°
But ∠FAE – 20° (Given)
Ext. ∠BAE – ∠BAF + ∠FAE
= 88° + 20° = 108°
But Ext. ∠BAE – ∠BCD
∴ ∠BCD = 108°
Question 12.
If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66° and ∠ABC – 80°, calculate
(i) ∠DBC,
(ii) ∠IBC,
(iii) ∠BIC.
Solution:
Join DB and DC, IB and IC.
∠BAC = 66°, ∠ABC = 80°. I is the incentre of the ∆ABC.
(i) ∵ ∠DBC and ∠DAC are in the same segment
∴ ∠DBC – ∠DAC.
But ∠DAC = $$\frac { 1 }{ 2 }$$∠BAC = $$\frac { 1 }{ 2 }$$ x 66° = 33°
∴ ∠DBC = 33°.
(ii) ∵ I is the incentre of ∆ABC.
∴ IB bisect ∠ABC
∴ ∠ IBC = $$\frac { 1 }{ 2 }$$ ∠ABC = $$\frac { 1 }{ 2 }$$ x 80° = 40°.
(iii) ∴∠BAC = 66° ∠ABC = 80°
∴ In ∆ABC,
∠ACB = 180° – (∠ABC + ∠CAB)
= 180°-(80°+ 66°)= 180°- 156° = 34°
∵ IC bisects the ∠C
∴ ∠ ICB = $$\frac { 1 }{ 2 }$$ ∠C = $$\frac { 1 }{ 2 }$$ x 34° = 17°.
Now in ∆IBC,
∠ IBC + ∠ ICB + ∠ BIC = 180°
⇒ 40° + 17° + ∠BIC = 180°
⇒ ∠ BIC = 180° – (40° + 17°) = 180° – 57°
= 123°
Question 13.
In the given figure, AB = AD = DC= PB and ∠DBC = x°. Determine in terms of x :
(i) ∠ABD
(ii) ∠APB
Hence or otherwise prove that AP is parallel to DB.
Solution:
Given: In figure, AB = AD = DC = PB.
∠DBC = x. Join AC and BD.
To Find : the measure of ∠ABD and ∠APB.
Proof: ∠DAC=∠DBC= x(angles in the same segment)
But ∠DCA = ∠DAC (∵ AD = DC)
= x
But ∠ABD = ∠DAC (Angles in the same segment)
In ∆ABP, ext. ∠ABC = ∠BAP + ∠APB
But ∠ BAP = ∠APB (∵ AB = BP)
2 x x = ∠APB + ∠APB = 2∠APB
∴ 2∠APB = 2x
⇒ ∠APB = x
∵ ∠APB = ∠DBC = x
But these are corresponding angles
∴ AP || DB. Q.E.D.
Question 14.
In the given figure, ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠CQE are supplementary.
Solution:
Given: In the figure, ABC, AEQ and CEP are straight lines.
To prove: ∠APE + ∠CQE = 180°.
Const: Join EB.
∠APE+ ∠ABE= 180° ….(i)
∠CQE +∠CBE = 180° ….(ii)
∠APE +∠ABE + ∠CQE +∠CBE = 180° + 180° = 360°
⇒ ∠APE + ∠CQE + ∠ABE + ∠CBE = 360°
But ∠ABE + ∠CBE = 180° (Linear pair)
∴ ∠APE + ∠CQE + 180° = 360°
⇒ ∠APE + ∠CQE = 360° – 180° = 180°
Hence ∠APE and ∠CQE are supplementary. Q.E.D.
Question 15.
In the given figure, AB is the diameter of the circle with centre O.
If ∠ADC = 32°, find angle BOC.
Solution:
Arc AC subtends ∠AOC at the centre and ∠ ADC at the remaining part of the circle
= 2 x 32° = 64°
∵ ∠AOC + ∠ BOC = 180° (Linear pair)
⇒ 64° + ∠ BOC = 180°
⇒ ∠ BOC=180° – 64° =116° .
Question 16.
In a cyclic-quadrilateral PQRS, angle PQR = 135°. Sides SP and RQ produced meet at point A : whereas sides PQ and SR produced meet at point B.
If ∠ A : ∠ B = 2 : 1 ; find angles A and B.
Solution:
PQRS is a cyclic-quadrilateral in which ∠ PQR =135°
Sides SP and RQ are produced to meet at A and sides PQ and SR are produced to meet at B.
Question 17.
In the following figure, AB is the diameter of a circle with centre O and CD is the chord with length equal to radius OA. If AC produced and BD produced meet at point P ; show that ∠ APB = 60°.
Solution:
Given : In the figure, AB is the diameter of the circle with centre O.
CD is the chord with length equal to the radius OA.
AC and BD are produced to meet at P.
To prove : ∠ APB = 60°
Const : Join OC and OD
Proof : ∵ CD = OC = OD (Given)
∴ ∆OCD is an equilateral triangle
∴ ∠ OCD = ∠ ODC = ∠ COD = 60°
In ∆ AOC, OA = OC (Radii of the same circle)
∴ ∠ A = ∠ 1
Similarly, in ∆ BOD,
OB = OD
∴∠2= ∠B
∠ A CD + ∠B = 180°
∠60°+ ∠ 1 + ∠B = 180°
= ∠ 1 + ∠B = 180° – 60°
⇒∠ 1 + ∠B = 120°
But ∠ 1 = ∠ A
∴ ∠ A + ∠B = 120° …(i)
Now, in ∆ APB,
∠ P + ∠ A + ∠ B = 180° (Sum of angles of a triangle)
⇒ ∠P+120°=180° [From (i)]
⇒ ∠P = 180°- 120°= 60°
Hence ∠ P = 60° or ∠ APB = 60° Hence proved.
Question 18.
In the following figure,
ABCD is a cyclic quadrilateral in which AD is parallel to BC.
If the bisector of angle A meets BC at point E and the given circle at point F, prove that :
(i) EF = FC
(ii) BF = DF
Solution:
Given : ABCD is a cyclic quadrilateral in which AD || BC.
Bisector of ∠ A meets BC at E and the given circle at F. DF and BF are joined.
To prove :
(i) EF = FC
(ii) BF = DF
Proof : ∵ ABCD is a cyclic -quadrilateral and AD || BC
∵ AF is the bisector of ∠ A
∴ ∠ BAF = ∠ DAF
∴ Arc BF = Arc DF (equal arcs subtends equal angles)
⇒ BF = DF(equal arcs have equal chords)
Hence proved
Question 19.
ABCD is a cyclic quadrilateral. Sides AB and DC produced meet at point E ; whereas sides BC and AD produced meet at point F. If ∠ DCF : ∠ F : ∠ E = 3 : 5 : 4, find the angles of the cyclic quadrilateral ABCD.
Solution:
Given : In a circle, ABCD is a cyclic quadrilateral AB and DC are produce to meet at E and BC and AD are produced to meet at F.
Question 20.
The following figure shows a circle with PR as its diameter.
If PQ = 7 cm, QR = 3 cm, RS = 6 cm. find the perimeter of the cyclic quadrilateral PQRS. (1992)
Solution:
In the figure, PQRS is a cyclic quadrilateral in which PR is a diameter
PQ = 7 cm,
QR = 3 RS = 6cm
∴ 3 RS = 6cm
and RS = $$\frac { 6 }{ 3 }$$ = 2cm
Now in ∆ PQR,
∠ Q = 90° (Angle in a semi-circle)
∴ PR2 = PQ2 + QR2 (Pythagoras theorem)
= (7)2 + (6)2 = 49 + 36 = 85
Again, in right ∆ PSQ, PR2 = PS2 + RS2
⇒ 85 = PS2 + (2)2
⇒ 85 = PS2 + 4
⇒ PS2 = 85 – 4 = 81 = (9)2
∴ PS = 9cm
Now, perimeter of quad. PQRS = PQ + QR + RS + SP = (7 + 9 + 2 + 6) cm = 24cm
Question 21.
In the following figure, AB is the diameter of a circle with centre O.
If chord AC = chord AD, prove that :
(i)arc BC = arc DB
(ii) AB is bisector of ∠ CAD. Further, if the length of arc AC is twice the length of arc BC, find :
(a) ∠ BAC
(b) ∠ ABC
Solution:
Given : In a circle with centre O, AB is the diameter and AC and AD are two chords such that AC = AD.
To prove : (i) arc BC = arc DB
(ii) AB is the bisector of ∠CAD
(iii) If arc AC = 2 arc BC, then find
(a) ∠BAC (b) ∠ABC
Construction: Join BC and BD.
Proof : In right angled A ABC and A ABD
Hyp. AB = AB (Common)
∴ ∆ ABC ≅ ∆ ABD (R.H.S. axiom)
(i) ∴ BC = BD (C.P.C.T)
∴ Arc BC = Arc BD (equal chords have equal arcs)
(ii) ∠ BAC = ∠ BAD (C.P.C.T)
∴ AB is the bisector of ∠CAD.
(iii) If arc AC = 2 arc B
Then ∠ABC = 2 ∠BAC
But ∠ABC – 2 ∠BAC = 90°
∴ 2 ∠BAC + ∠BAC = 90°
⇒ 3 ∠BAC = 90° ⇒ ∠BAC = 30°
and ∠ABC = 2 ∠BAC = 2 * 30° = 60°
Question 22.
∠ BAC = 30° and ∠ CBD = 70°, find:
(i) ∠ BCD
(ii) ∠ BCA
(iii) ∠ ABC
Solution:
∠ BAC = 30°, ∠ CBD = 70°
∠ DAC = ∠ CBD , (Angles in the same segment)
But ∠ CBD = 70°
∴ ∠ DAC = 70°
⇒ ∠ BAD = ∠ BAC + ∠ DAC = 30° + 70° = 100°
But ∠ BAD + ∠ BCD = 180°
(Sum of opposite angles of a cyclic quad.)
⇒100°+ ∠ BCD=180° ⇒ ∠ BCD=180° – 100° = 80°
∴ ∠ BCD = 80°
∴ ∠ ACD = ∠ BDC
(Equal chords subtends equal angles)
But ∠ ACB = ∠ ADB
(Angles in the same segment)
∴ ∠ ACD + ∠ ACB = ∠ BDC + ∠ ADB
⇒ ∠ BCD = ∠ ADC = 80° (∵ ∠ BCD = 80°)
But in ∆ BCD,
∠ CBD + ∠ BCD + ∠ BDC = 180° (Angles of a triangle)
⇒ 70° + 80° + ∠ BDC = ∠ 180°
⇒ 150°+ ∠ BDC = 180°
∴ ∠ BDC = 180° – 150° = 30°
⇒ ∠ ACD = 30° (∵ ∠ ACD = ∠ BDC)
∴ ∠ BCA = ∠ BCD – ∠ ACD = 80° – 30° = 50°
∠ ADC + ∠ABC = 180°
(Sum of opp. angles of a cyclic quadrilateral)
⇒ 80°+ABC = 180°
⇒ ∠ ABC=180°-80° = 100°
Question 23.
In the given figure, if ∠ ACE = 43° and ∠CAF = 62°. Find the values of a, b and c.
Solution:
Now, ∠ ACE = 43° and ∠ CAF = 62° (given)
In ∆ AEC,
∠ ACE + ∠ CAE + ∠ AEC = 180°
∴ 43° + 62° + ∠ AEC = 180°
105° + ∠ AEC = 180°
⇒ ∠ AEC = 180°- 105° = 75°
Now, ∠ ABD + ∠AED=180°
(Opposite ∠ s of a cyclic quad, and ∠ AED = ∠ AEC)
⇒ 0 + 75°= 180° a = 180° – 75° = 105°
∠ EDF = ∠ BAE (Angles in the alternate segments)
∴ c = 62°
In ∆BAF, ∠a + 62° + ∠b = 180°
⇒ 105°+ 62°+ ∠b= 180°
⇒ 167° + ∠6 = 180°
⇒ ∠b= 180°-167°= 13°
Hence, a= 105°, 6=13° and c = 62°.
Question 24.
In the given figure, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25° .
Find:
(ii)∠CBD
Solution:
In the given figure,
ABCD is a cyclic quad, in which AB || DC
∴ ABCD is an isosceles trapezium AD = BC
(i) Join BD
{ Ext. angle of a cyclic quad, is equal to interior opposite angle}
∴ ∠BAD = 80° (∵ ∠BCE = 80°)
But ∠BAC = 25°
∴ ∠CAD = ∠BAD – ∠BAC = 80° – 25° = 55°
(ii) ∠CBD = ∠CAD (Angles in the same segment)
= 55°
(iii) ∠ADC = ∠BCD (Angles of the isosceles trapezium)
= 180°- ∠BCE =180°- 80° = 100°
Question 25.
ABCD is a cyclic quadrilateral of a circle with centre O such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle. If AJD and BC produced meet at P, show that ∠APB = 60°.
Solution:
Given : In a circle, ABCD is a cyclic quadrilateral in which AB is the diameter and chord CD is equal to the radius of the circle
To prove: ∠APB = 60°
Construction : Join OC and OD
Proof: ∵ chord CD = CO = DO (radii of the circle)
∴ ∆DOC is an equilateral triangle
∠DOC = ∠ODC = ∠OCD – 60°
Let ∠A = x and ∠B = y
∵ OA = OD = OC = OB (radii of the same circle)
∴ ∠ODA = ∠OAD = x and ∠OCB = ∠OBC =y
∴∠AOD = 180° – 2x and ∠BOC = 180° – 2y
But AOB is a straight line
∴ ∠AOD + ∠BOC + ∠COD = 180°
180°- 2x + 180° -2y + 60° = 180°
⇒ 2x + 2y = 240°
⇒ x + y = 120°
But ∠A + ∠B + ∠P = 180° (Angles of a triangle)
⇒ 120° + ∠P = 180°
⇒ ∠P = 180° – 120° = 60°
Hence ∠APB = 60°
Question 26.
In the figure, given alongside, CP bisects angle ACB. Show that DP bisects angle ADB.
Solution:
Given : In the figure,
CP is the bisector of ∠ACB
To prove : DP is the bisector of ∠ADB
Proof: ∵ CP is the bisector of
∴ ∠ACB ∠ACP = ∠BCP
But ∠ACP = ∠ADP {Angles in the same segment of the circle}
and ∠BCP = ∠BDP
But ∠ACP = ∠BCP
∴ DP is the bisector of ∠ADB
Question 27.
In the figure, given below, AD = BC, ∠BAC = 30° and ∠CBD = 70°. Find :
(i) ∠BCD
(ii) ∠BCA
(iii) ∠ABC
Solution:
In the figure,
AC and BD are its diagonals
∠BAC = 30° and ∠CBD = 70°
Now we have to find the measures of ∠BCD, ∠BCA, ∠ABC and ∠ADB
(Angles in the same segment)
Similarly ∠BAC = ∠BDC = 30°
∴ ∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°
(i) Now ∠BCD + ∠BAD = 180° (opposite angles of cyclic quad.)
⇒ ∠BCD + 100°= 180°
⇒ ∠BCD = 180°-100° = 80°
(ii) ∵ AD = BC (given)
∴ ∠ABCD is an isosceles trapezium
and AB || DC
∴ ∠BAC = ∠DCA (alternate angles)
⇒ ∠DCA = 30°
∠ABD = ∠D AC = 30° (Angles in the same segment)
∴ ∠BCA = ∠BCD – ∠DAC = 80° – 30° = 50°
(iii) ∠ABC = ∠ABD + ∠CBD = 30° + 70° = 100°
(iv) ∠ADB = ∠BCA = 50° (Angles in the same segment)
P.Q.
In the given below figure AB and CD are parallel chords and O is the centre.
If the radius of the circle is 15 cm, find the distance MN between the two chords of length 24 cm and 18 cm respectively.
Solution:
Given : AB = 24 cm, CD = 18 cm
⇒AM = 12 cm, CN = 9 cm
Also, OA = OC = 15 cm
Let MO = y cm, and ON = x cm
In right angled ∆AMO
(OA)2 = (AM)2 + (OM)2
(15)2 = (12)2 + (y2
⇒ (15)2-(12)2
⇒ y2 = 225-144
⇒ y2 = 81 = 9 cm
In right angled ∆CON
(OC)2 = (ON)2 + (CN)2
⇒(15 )2= x2 + (9)2
⇒ x2 = 225-81
⇒x2= 144
⇒ x = 12 cm
Now, MN = MO + ON =y + x = 9 cm + 12 cm = 21 cm
Question 28.
In the figure given below, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°. Find :
(i) ∠OBD
(ii) ∠AOB
(iii) ∠BED (2016)
Solution:
(i) AD is parallel to BC, that is, OD is parallel to BC and BD is transversal.
∴ ∠ODB = ∠CBD = 32° (Alternate angles)
In ∆OBD,
OD = OB (Radii of the same circle)
⇒ ∠ODB = ∠OBD = 32°
(ii) AD is parallel to BC, that is, AO is parallel to BC and OB is transversal.
∴∠AOB = ∠OBC (Alternate angles)
∠OBC = ∠OBD + ∠DBC
⇒ ∠OBC = 32° + 32°
⇒ ∠OBC = 64°
∴ ∠AOB = 64°
(iii) In AOAB,
OA = OB(Radii of the same circle)
∴ ∠OAB = ∠OBA = x (say)
∠OAB + ∠OBA + ∠AOB = 180°
⇒ x + x + 64° = 180°
⇒ 2x = 180° – 64°
⇒ 2x= 116°
⇒ x = 58°
∴ ∠OAB = 58°
That is ∠DAB = 58°
∴ ∠DAB = ∠BED = 58°
(Angles inscribed in the same arc are equal)
Question 29.
In the given figure PQRS is a cyclic quadrilateral PQ and SR produced meet at T.
(i) Prove ∆TPS ~ ∆TRQ
(ii) Find SP if TP = 18 cm, RQ = 4 cm and TR = 6 cm.
(iii) Find area of quadrilateral PQRS if area of ∆PTS = 27 cm2.(2016)
Solution:
(i) Since PQRS is a cyclic quadrilateral ∠RSP + ∠RQP = 180°
(Since sum of the opposite angles of a cyclic quadrilateral is 180°)
⇒ ∠RQP = 180° – ∠RSP …(i)
∠RQT + ∠RQP = 180°
(Since angles from a linear pair)
⇒ ∠RQP = 180° – ∠RQT …(ii)
From (i) and (ii),
180° – ∠RSP = 180° – ∠RQT
⇒ ∠RSP = ∠RQT …(iii)
In ∆TPS and ∆TRQ,
∠PTS = ∠RTQ (common angle)
∠RSP = ∠RQT [From (iii)]
∴ ATPS ~ ATRQ (AA similarity criterion)
(ii) Since ∆TPS ~ ∆TRQ implies that corresponding sides are proportional that
Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
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# What is Proportion in Math?
## Ratio and Proportion
A question that is often asked in more basic Statistics is what is proportion in Math?
When we have two ratios, they are classed as being in proportion if they are both equal to each other in measure.
The ratios illustrated above, 2 : 3 and 4 : 6 are ratios that are in proportion.
3 is to 4” as “4 is to 6
or
3 out of 4” is the same ratios as “4 out of 6“.
Ratios can also be written as fractions too, so we can represent things this way.
\bf{\frac{2}{3}} = \bf{\frac{4}{6}}.
### Notation for Proportion
Say we have two different ratios that are in proportion, a:b = c:d.
When this is the case we can write the ratios as, a:b :: c:d.
This means that a is to b, as c is to d.
The figures that are in the outer places, a and d, are referred to as “extremes”.
The figures in the inner places, b and c, are referred to as “means”.
## What is Proportion in Math?Looking for Proportion
In order to check if 2 ratios are in proportion, we can multiply the means, and the extremes together.
Because if 2 ratios form a proportion, then these multiplications will be equal in value.
a : b = c : d if b × c = a × d
Example
The ratios demonstrated earlier in the page were 2 : 3 and 4 : 6.
Multiplying the means and the extremes together gives:
3 × 4 = 12 , 2 × 6 = 12
2 : 3 and 4 : 6 do form a proportion.
2 : 3 :: 4 : 6
But what if we looked at the ratios.
3 : 4 and 5 : 7.
Multiplication of the means and the extremes together results in:
4 × 5 = 20 , 3 × 7 = 21
3 : 4 and 5 : 7 do NOT form a proportion.
We could also write the ratios in fraction form, and this approach is the same as comparing the cross products.
\bf{\frac{3}{4}} , \bf{\frac{6}{8}} => 3 × 8 = 24 4 × 6 = 24
\bf{\frac{3}{4}} , \bf{\frac{5}{7}} => 3 × 7 = 21 4 × 5 = 20
### Proportion in Math Examples
(1.1)
If 3 : 7 and b : 14 form a proportion, what is the value of b?
Solution
For the two ratios to form a proportion, the ‘means’ and ‘extremes’ being multiplied together will produce an equal result.
3 × 14 = 7 × b => 42 = 7b
\boldsymbol{\frac{42}{7}} = b => b = 6
(1.2)
If \boldsymbol{\frac{6}{a}} and \bf{\frac{2}{19}} form a proportion, what is the value of a?
Solution
For the two ratios to form a proportion, the ‘means’ and ‘extremes’ again being multiplied together will produce an equal result.
6 × 19 = a × 2 => 114 = 2a
\boldsymbol{\frac{114}{2}} = a => a = 57
(1.3)
A lesser scale model is to created of a life size 4×4 car.
Based on the information seen in the image below, what height in cm should the smaller 4×4 model car be?
Solution
As the smaller size model is to be the same scale as the full size car, the width and height of each car will be in proportion with the other.
Thus we can solve for the height of the model car by drawing up the ratios of width to height.
\boldsymbol{\frac{WIDTH}{HEIGHT}} = \bf{\frac{450}{180}} = \boldsymbol{\frac{80}{h}}
450 × h = 180 × 80
450h = 14’400
h = \bf{\frac{14’400}{450}} = 32
The lesser scale 4×4 car model should be 32cm in height for the cars to be in proportion.
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# 2 - Solving Linear Equations
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1 2 - Solving Linear Equations Linear algebra developed from studying methods for solving systems of linear equations. We say an equation in the variables x 1, x 2, x 3,..., x n is linear if it consists of a linear combination of the x i equaling a constant. The following equations are linear: 4x - 2y + z = 25 ½ a b + π c - d = 0 17 x x 2 + (5.2-1/π) x 3 = 31 Here are some nonlinear equations: x 2 + y = a + 3 b + 1/c = 47 (3x + 9y)(z - 2w) = 73 A system of linear equations is simply a list or set of linear equations. A solution of the system consists of values for the variables which is a solution of each of the equations. The system -2x + 4y = 18 x + 3y = 1 has as its unique solution x = -5 and y = 2 or, more succinctly, the ordered pair (-5, 2). Geometrically, of course, the graph of each equation is a line and the solution of the system, the common solution of each, is the point where the lines intersect as the accompanying graph shows.
2 Problem What other possibilities (geometrically) are there for the solution of a linear system with two variables and two equations? What if there are three equations? Or just one? The following system with three variables 10x - 3y + 5z = 36 7x + 11y - 20z = -11 has the point (3, 8, 6) as one solution. Each equation has a plane as its graph as we see below. Problem What other solutions does this system have? What other possibilities (geometrically) are there for the solution of a linear system with three variables and two equations? What if there are three equations? It is convenient to write the general system of linear equations using double subscripts for the coefficients of the variables. a 11 x 1 + a 12 x a 1n x n = b 1 a 21 x 1 + a 22 x a 2n x n = b 2 a m1 x 1 + a m2 x a mn x n = b m
3 The first subscript of each constant indicates the equation in which it is found, while the second tells which variable it multiplies. All of the information of the system is contained in these constants so we have the following definitions. Definitions 1. A matrix is a rectangular array of numbers. The size of a matrix with m rows and n columns is m n (read m by n.) 2. The coefficient matrix of the system above is the matrix, A, whose i,j th entry (i.e., entry in the i th row and j th column) is a ij. As a shorthand we shall write A = (a ij ). 3. The augmented matrix of the system consists of its coefficient matrix along with an additional column, its (n+1) st, whose entries are the constants on the right-hand side of the b1 b2 equations, so the last column is the vector b =. b m The coefficient and augmented matrices of the system in the previous example with two variables are 2 4 and respectively. Note that, for the general system that was given above, the first column of its coefficient a 11 a1 j matrix, c 1 = a21 a, and, in general, its j th 2 j column, c j =, are vectors. For each i, the i th row amj a m 1
4 of the coefficient matrix, r i = (a i1 a i2... a in ), is also a vector and we can write the matrix in several ways A = (a ij ) = (c 1 c 2... c n ) = r1 r2 r m. Using the vector operations introduced in the last chapter, we can form the linear combination of the columns of A and observe that x 1, x 2,..., x n is a solution of the system if and only if x 1 c 1 + x 2 c x n c n = b, i.e. b is a linear combination of the columns of A. Returning to the 2 2 example (2 equations with 2 unknowns) that we introduced earlier, -2x + 4y = 18 x + 3y = 1, we have as its corresponding equation with column vectors x v + y w = b where v =, w =, and b =. Then the solution (-5, 2) can be pictured in the vector diagram that follows.
5 Thus we have two distinct pictures for systems, both of which have their value and to which we ll return. They are, as Strang refers to them in his text, the row picture (the intersecting lines shown earlier) and the column picture (the linear combination of vectors in the graph above) of the system. The latter picture was a result of viewing the system as a vector equation with the right hand side of the system, b, written as a linear combination of the column vectors of the coefficient matrix. With the dot product and the row expression of A, we can write the system r1 x r2 x r x m x1 x2 = b where x =. x m Let s exploit this view point in considering the possibilities for the system s solutions. We have seen from the row pictures with intersecting (and parallel) lines and planes that a linear system can have no solution (in which case it is called inconsistent) or, if it is consistent, can have a single, unique solution or infinitely many solutions. We now show that these are, in fact, the only possibilities. Indeed, suppose that the general system has as distinct solutions the vectors x = (x 1, x 2,..., x n ) and y = (y 1, y 2,..., y n ). Let z be the linear combination of x and y given by z = t x + (1 - t) y where t is an arbitrary real number. Consider for any i = 1, 2,.., m, the i th equation of the system. Since x and y are solutions of the system (and, hence, each equation) r i x = b i = r i y. Then r i z = r i (tx + (1 - t)y ) = t(r i x) + (1 - t)(r i y) = t b i + (1 - t) b i = b i. Thus z is a solution of the i th equation for each i and so is a solution of the system. (Note: the second step in the string of equalities uses the result of the problem below.) Consequently any system with two distinct solutions has infinitely many solutions. (Writing z = t x + (1 - t) y = y + t (x - y) it is not hard to recognize that the terminal points of z drawn in standard position (with the origin as its initial point) trace the line through the heads of the vectors x and y (in standard position) as the parameter t is varied throughout the real numbers. This is the parametric form of the equation of this line.)
6 Problem Show that the dot product is linear: for any vectors u, v, and w in R n and any scalar c 1. u (v + w) = u v + u w 2. v (cw) = c(v w)
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|
Lesson Objectives
• Demonstrate an understanding of place value
• Demonstrate an understanding of subtraction
• Demonstrate an understanding of multiplication
• Demonstrate an understanding of single-digit division
• Demonstrate an understanding of division with remainders
• Learn how to divide multi-digit whole numbers using long division
## How to Divide Multi-Digit Whole Numbers using Long Division
In our last lesson, we learned the basic properties of the division operation. At this point, you should be fairly comfortable with dividing single-digit whole numbers. What happens when we want to divide 2-digit, 3-digit, or large multi-digit whole numbers? To divide multi-digit whole numbers, we normally use a process known as long division. Long division is a way to simplify our multi-digit division problem into a series of manageable division, multiplication, and subtraction operations. First and foremost, we will draw the long division symbol. This can be drawn as a right parenthesis with an attached horizontal line (the horizontal line is known as a vinculum). Additionally, we may see a straight line instead of a right parenthesis: To set up our long division problem, we place the dividend under the long division symbol. The divisor will be placed to the left of the long division symbol. As an example, suppose we want to divide 216 by 9 using long division.
216 ÷ 9:
216 - dividend
9 - divisor We place our dividend (216) under the long division symbol. The divisor (9) goes to the left of the long division symbol. Now that we have set up the problem, we are ready to start the long division operation. Many times in math we will use an acronym to remember a sequence of events or steps. When we encounter long division we will usually run into:
DMSBR
On its own, DMSBR doesn't really do much for the memory. We can relate this acronym to members of the family unit. This may help us to remember the division steps:
• M - Mom
• S - Sister
• B - Brother
• R - Rover (dog)
Now that we have a little memory trick to recall DMSBR, we can associate the letters with their true purpose. When we encounter long division, we will work in the following order:
• D - Divide
• M - Multiply
• S - Subtract
• B - Bring Down
• Repeat the steps or Remainder
• Not every problem ends with a remainder. We are finished with the problem when there are no more digits in the dividend to bring down.
Let's return to the problem 216 ÷ 9 and work through this example step by step.
DMSBR:
Following the DMSBR long division acronym, our first step is to divide. Instead of trying to determine ? x 9 = 216 from the start, we will attack the dividend one digit at a time. We will try to divide the leftmost digit of our dividend by our divisor. We would hear this asked as:
How many times does 9 go into 2? In other words, what is 2 ÷ 9 or how many groups of 9 can be made from 2: 9 doesn't go into 2, because 2 isn't large enough to create even 1 group of 9. When this happens we can expand our selection to include the next digit of the dividend. Now we would ask:
How many times does 9 go into 21?
In other words, what is 21 ÷ 9? We will think about this question using a related multiplication statement:
9 x ? = 21
9 x 1 = 9
9 x 2 = 18
9 x 3 = 27
We can see that there isn't a whole number that multiplies 9 and produces 21 as a result. We will select the number that produces a product that is as close to 21 without going over. In this case, we see that 9 x 2 = 18. Here 18 is as close as we can get to 21 without going over. This means we will use 2 as our answer. This also means we will have a remainder for our division problem. Since 21 - 18 = 3, this will be our remainder or leftover amount from the division operation:
21 ÷ 9 = 2 R3
We want to only think about the quotient part and not the remainder or leftover amount. We would say 9 goes into 21 twice (2 times) since we can at most make two groups of 9 (9 x 2 = 18) out of 21. We write our answer directly over the dividend above the horizontal line. We want to line up our answer from each step according to the place value of the current part of our problem. Here we worked with 21 ÷ 9, so we place our answer with respect to the number 21. We will write the answer (2) above the ones' place for the number 21 (directly over the 1 in the number 216):
correct: incorrect: DMSBR:
Once we have completed the division, we are ready to move onto the next step, which is to multiply. We take our answer from the division problem (2) and multiply by the divisor 9. The result (18) will be written under the dividend from the last problem (21).
2 x 9 = 18
Write the result under 21: DMSBR:
Once we have completed the multiplication and lined up the result correctly, we are ready to move on to our next step, which is to subtract. We subtract the dividend we were working with (21) minus the result from the multiplication (18), we write the result (3) below, and maintain place value.
21 - 18 = 3 DMSBR:
Once we have completed our subtraction and lined up the result correctly, we are ready to move on to our next step, which is to bring down. The term "bring down" tells us to bring down the next digit of the dividend (6): DMSBR:
Once we have completed our "bring down" step, we repeat the process of: division, multiplication, subtraction, and bring down. This will continue until there are no more digits in the dividend to bring down. At that point, the long division process is over. We may or may not have a remainder, depending on the problem. Let's start our process over with division. Our new dividend is 36, the number formed as a result of the subtraction and the bring down steps combined. The divisor (9) remains the same throughout the problem.
DMSBR:
We want to divide 36 by 9:
36 ÷ 9 = 4
Write the result (4) next to the 2 in the answer. Each digit flows one to the right after the initial. DMSBR:
Once we have completed the division, we are ready to multiply. We take our answer (4) and multiply by the divisor (9). The result (36) will be written under the dividend from the last problem (36).
4 x 9 = 36 DMSBR:
We subtract the dividend we were working with (36) minus the result from the multiplication (36), we write the result (0) below, and maintain place value.
36 - 36 = 0 DMSBR:
Once we have completed the subtraction, we move into the "bring down step". We now have no more digits in the dividend (216) to bring down. When this occurs, we are finished with our long division operation. Since the result of the last subtraction operation (36 - 36 = 0) was zero, we don't have a remainder:
We can report our answer as 24, the number directly over the dividend.
Let's take a look at a few more examples:
Example 1: Find each quotient:
465 ÷ 15
Let's set up our long division problem:
465 - dividend
15 - divisor DMSBR:
Divide:
Our first step is to divide. We will first ask:
How many times will 15 go into 4? 15 doesn't go into 4, so let's expand our selection. We will now ask:
How many times will 15 go into 46?
In other words what is 46 ÷ 15? 15 x ? = 46
15 x 1 = 15
15 x 2 = 30
15 x 3 = 45
15 x 4 = 60
We can see that 15 x 3 = 45, this result (45) is as close to 46 as we can get without going over. We will use 3 as our answer and have a remainder of 1 (46 - 45 = 1)
46 ÷ 15 = 3 R1
We can ignore any remainders at this point. We write the answer 3 above the 6 in the number 465. DMSBR:
Multiply:
Multiply 3 x 15, the result (45) is placed below 46.
3 x 15 = 45 DMSBR:
Subtract:
Subtract 46 - 45, the result (1) is placed directly below.
46 - 45 = 1 DMSBR:
Bring Down:
Bring down the next digit of the dividend (5) DMSBR:
Repeat or Remainder:
We will repeat the process. We will be working with 15 as the dividend.
DMSBR:
How many times will 15 go into 15?
15 ÷ 15 = 1
We write the answer (1) next to the 3 in our answer. DMSBR:
Multiply:
Multiply 1 x 15, the result (15) is placed below 15.
1 x 15 = 15 DMSBR:
Subtract:
Subtract 15 - 15, the result (0) is placed directly below.
15 - 15 = 0 DMSBR:
Bring Down:
There are no more digits to bring down, and the result of our last subtraction (15 - 15 = 0) was 0:
This tells us our answer is 31 with no remainder.
465 ÷ 15 = 31
Let's take a look at an example with a remainder.
Example 2: Find each quotient:
971 ÷ 29
Let's set up our long division problem:
971 - dividend
29 - divisor DMSBR:
Divide:
Our first step is to divide. We will first ask:
How many times will 29 go into 9? 29 doesn't go into 9, so let's expand our selection. We will now ask:
How many times will 29 go into 97?
In other words what is 97 ÷ 29? 29 x ? = 97
29 x 1 = 29
29 x 2 = 58
29 x 3 = 87
29 x 4 = 116
We can see that 29 x 3 = 87, this result (87) is as close as to 97 as we can get without going over. We will use 3 as our answer and have a remainder of 10 (97 - 87 = 10)
97 ÷ 29 = 3 R10
We can ignore any remainders at this point. We write the answer 3 above the 7 in the number 971. DMSBR:
Multiply:
Multiply 3 x 29, the result (87) is placed below 97.
3 x 29 = 87 DMSBR:
Subtract:
Subtract 97 - 87, the result (10) is placed directly below.
97 - 87 = 10 DMSBR:
Bring Down:
Bring down the next digit of the dividend (1) DMSBR:
Repeat or Remainder:
We will repeat the process. We will be working with 101 as the dividend.
DMSBR:
How many times will 29 go into 101?
We know from earlier that:
29 x 3 = 87 and 29 x 4 = 116
116 is too large, so we will use 3 for the answer.
101 ÷ 29 = 3 R14
Remember to ignore any remainders at this point.
We write the answer (3) next to the 3 in our answer. DMSBR:
Multiply:
Multiply 3 x 29, the result (87) is placed below 101.
3 x 29 = 87 DMSBR:
Subtract:
Subtract 101 - 87, the result (14) is placed directly below.
101 - 87 = 14 DMSBR:
Bring Down:
There are no more digits to bring down, and the result of our last subtraction (101 - 87 = 14) was 14:
This tells us our answer is 33 with a remainder of 14. We can write "R" followed by the remainder (14) to the right of our answer. 971 ÷ 29 = 33 R14
#### Skills Check:
Example #1
Find each quotient.
2,916 ÷ 83
A
33 R12
B
35 R11
C
40 R13
D
34 R15
E
37 R55
Example #2
Find each quotient.
2,730 ÷ 65
A
52 R11
B
39 R12
C
34 R14
D
42
E
38
Example #3
Find each quotient.
2432 ÷ 54
A
61
B
38 R15
C
42 R19
D
47 R13
E
45 R2
|
# 2.5 - Solving Inequalities Algebraically and Graphically
## Linear Inequalities
When solving a linear inequality, treat it just like you were solving an equation with a few exceptions.
1. When you multiply or divide both sides of an inequality by a negative constant, change the sense (direction) of the inequality.
2. When both sides of an inequality are the same sign, change the sense of the inequality when you take the reciprocal of both sides.
3. You may chain like inequalities together:
• If a<b and b<c, then a<c (transitive property)
• If a<b and c<d, then a+c<b+d. In English, that means that if you take two things that are smaller and put them together, it will be smaller than the two larger things put together.
4. You can not combine mixed inequalities.
• If a<b and b>c, then you can't say for sure that a<c or that c<a.
• If a<b and c>d, then you can't say for sure that a+c<b+d or that a+c>b+d
5. If you rewrite the entire problem, just switching sides, make sure you change the sense of the inequality so that it still points to the same quantity.
6. The following operations do not change the sense of the inequality
• Adding a constant to both sides of an inequality
• Subtracting a constant from both sides of an inequality
• Multiplying or dividing both sides of an inequality by a positive constant
## Double Inequalities
Sometimes, two inequalities are combined into one. However, you need to be careful:
If x>3 and x<6, then you can write 3<x<6. But, if x<3 or x>6, you can not write 3>x>6 because that would imply that 3>6 and that would be a false statement and the set would be empty.
To solve a double inequality, just apply the operations to all three portions: If 3<x+2<6, then subtract 2 from all three parts to get 1<x<4.
## Absolute Value Inequalities
These are going to give you trouble. They don't have to, but they will. People just don't get absolute values.
I could tell you that the major reason people don't get it is because they don't understand restrictions. That would be the truth, but just telling you that isn't going to help you. The book is going to give a short cut and people are going to ask "why are there so many rules?". There aren't! There is one definition of absolute value, and if you know it, and apply the restrictions like you're supposed to, then the problems work out every time without a need for additional rules. The additional rules from the book are used to make the problems go quicker, just like when we omit the absolute value when taking the square root of both sides and just stick in the plus/minus instead. It sounds like an extra rule, but it's just an application of absolute value that we try to get around because we don't like them (and we wonder why we have difficulty with them).
## Absolute Values - the right way
Whether the original inequality was a > or < doesn't affect how the problem is done when you're doing it the correct way. I could show an example of both, but the technique is similar.
Consider the absolute value inequality | (x - 3) / 4 | ≤ 5
Begin by splitting the absolute value up into its two cases. Remember that the absolute value can be found by dropping the absolute value signs when the argument is non-negative (case 1) and by taking the opposite of what was in absolute values when the argument is negative (case 2).
1. (x-3)/4 ≤ 5 if (x-3)/4 ≥ 0
2. - (x-3)/4 ≤ 5 if (x-3)/4 < 0
### Case 1
(x-3)/4 ≤ 5 if (x-3)/4 ≥ 0
First the function portion
(x-3) / 4 ≤ 5
x-3 ≤ 20
x ≤ 23
Now the restriction
(x-3)/4 ≥ 0
x-3 ≥ 0
x ≥ 3
Be sure to pay special attention to the restrictions about what's in absolute value being non-negative and negative. Inequalities that occur at the same time like the function and its restriction must both be satisfied. That's why the two portions from solving the first case can be combined so that x is between 3 and 23 inclusive. It must satisfy both x≥3 and x≤23.
3 ≤ x ≤ 23
### Case 2
- (x-3)/4 ≤ 5 if (x-3)/4 < 0
First the function. We'll start off by multiplying by -1 to get rid of the negative sign, but then we have to change the sense (direction) of the inequality.
- (x-3) / 4 ≤ 5
(x-3) / 4 ≥ -5
x-3 ≥ -20
x ≥ -17
Now the restriction.
(x-3)/4 < 0
x-3 < 0
x < 3
Likewise, for the second case, the two portions must both be satisfied at the same time. Also notice that when I multiplied both sides of the inequality by -4, I changed the sense of the inequality so that it was no longer ≤, but now ≥. You could have simplified this in other ways, but I usually find it easier to get rid of the negative right away.
-17 ≤ x < 3
### Putting it back together
The two inequalities within a particular case must both be satisfied at the same time. However, the inequalities that occur between the two sides may be combined with an "or". Just like if (x-2)(x+4)=0, you would say the answer is x=2 or x=-4, you wouldn't insist that it be both 2 and -4 and the same time. The same principle holds with the two parts of the absolute value. I told you it all fits together. I know you're going to be tired of hearing me say it, but it is much easier to understand if you see the big picture.
So, put the two answers from Case 1 and Case 2 together
-17 ≤ x < 3 or 3 ≤ x ≤ 23
These two intervals can be unioned together to get the final answer.
-17 ≤ x ≤ 23
In interval notation, that would be [ -17, 23 ]
## Absolute Values the shortcut way
Less Than
When the absolute value is less than the right hand side, then the answer will be between the opposite of the right hand side and the right hand side.
The absolute value inequality |x-2|<3 can be written as -3<x-2<3. Applying techniques discussed in the double inequality section above, this can be solved by adding 2 to each part to arrive at -1<x<5.
Greater Than
When the absolute value is greater than the right hand side, then the answer will be in two parts, with an or separating them. It will be greater than the right hand side or less than the opposite of the right hand side.
The absolute value inequality |x-2|>3 can be written as x-2<-3 or x-2>3. Solving this arrives at x<-1 or x>5.
Note that this cannot be combined to be -1>x>5 because that implies that 1>5 which is just plain false and would be the empty set.
## Absolute Values the Geometric Way
I have no problem with using the geometric approach to solving absolute value inequalities. Not the geometric approach where you put it into the calculator, but the geometric approach where you use the geometric definition of absolute value. The geometric definition of absolute value is the distance from 0 on the number line. If you modify it slightly, then |x-a| is the distance from x=a on the number line.
However, this technique requires that you know some of the properties of absolute values. These are good things to know anyway (isn't it all?), so it would be good if you learned them.
Let's consider the same inequality that we solved before: | (x - 3) / 4 | ≤ 5
Multiply both sides by 4 to get |x-3| ≤ 20.
This says that the distance from 3 on the number line is less than or equal to 20. So, start at 3 and go 20 to the left (-17) and 20 to the right (23). You need the distance to be less than (or equal to) 20 units away. This would include values between -17 and 23. Since the equal to is included, you include the endpoints to get the answer (interval notation) of [-17,23].
Wow! That was easy, you say. Yep.
A little bit harder problem: | (3-5x) / 2| ≥ 3
Multiply both sides by 2 to get | 3-5x | ≥ 6.
Here's where those properties of absolute values come into play. The absolute value of the opposite of a number is the same as the absolute value of the number. That means that we can say | 3-5x | = | 5x-3 | and |5x-3| ≥ 6.
Now divide both sides by 5 to get | x- 3/5 | ≥ 6/5.
This says that the distance from 3/5 is at least 6/5. So, start at 3/5 and go 6/5 to the left (-3/5) and 6/5 to the right (9/5). Since you need the distance to be more than 6/5 units away from 3/5, you need the values to the left of -3/5 and to the right of 9/5. The answer is therefore x≤-3/5 or x≥9/5. In interval notation, you would have to use the union of two intervals (-∞,-3/5] U [9/5, +∞).
## Polynomial Inequalities
Polynomials are continuous. That means that you can draw them without picking up your pencil (there's a more rigorous definition in calculus, but that definition will work for us, now). If you're going to change from being less than zero to being greater than zero and you can't pick up your pencil, then at some point, you must cross the x-axis. That means that the only place the inequality can change is at an x-intercept, a zero, a root, a solution.
The key then, to finding the solution set for a polynomial inequality, is to find the zeros of the inequality (pretend it was an equation), putting them on the number line, and picking a test point in each region.
1. Write the polynomial inequality in standard form so that the right side is zero.
2. Find the real solutions (ignore complex solutions involving i) to the inequality any way that you want to. Factoring is preferred, but you can use the quadratic formula if you can get it down to a quadratic factor. These values are called "critical numbers" (not to be confused with the similar but slightly different "critical values" from calculus).
3. Put the zeros of the polynomial (critical numbers) on the number line. Be sure to put them in order from smallest to largest. It is not important to label any other value on the number line. Some people were taught that you always put zero on the number line. That is not necessary here.
4. Pick a test point in each interval. You will have one more interval than the number of test points. Plug that test point back into either the factored inequality or the original inequality. If the test point gives you a true statement, then any point in that interval will work, and you want to include that interval in the answer. If the test point gives you a false statement, then all points in that interval will not work and you do not want to include that interval.
5. Include the endpoints if the inequality includes the equal to and do not include the endpoints if the inequality does not include the equal to.
## Rational Inequalities
Rational inequalities are similar to polynomials, but there is an extra temptation and an extra place where critical numbers could occur.
Temptation: Yield not to temptation, for yielding is sin.
I know you don't like fractions. You try to get rid of them at every chance you get. But, fractions are your friends and I really hope you don't treat all of your friends that way.
Here's the problem. If you multiply by a constant, it's pretty obvious whether it is positive or negative so you know whether or not to change the inequality. However, with rational expressions, there's going to be a variable in the denominator. Variables can take on different values - that's why they're called variable (and you thought math made no sense). Sometimes an expression is positive, but for other values of x, the expression is negative. So, if you multiply by the least common denominator, you don't know whether your multiplying by a positive number or a negative number unless you keep track of the restrictions! So, you have your choice - fractions that you hate or restrictions that you hate? In this case, fractions are the lesser of the two evils (in your mind - neither one is really evil in reality)
## Continuity
Polynomial functions are continuous everywhere. However, rational functions are not. They are continuous everywhere except where undefined and that occurs when the denominator is zero.
If you're moving along and you can't pick up your pencil and you change from negative to positive, then it has to happen at a zero of the function, same as with a polynomial. But, if the function is undefined, then you have to pick up your pencil. While the pencil is up, there is nothing that says you can't move to a completely different location, perhaps on the other side of the x-axis when you put it back down.
We now have two places that critical numbers can occur. One is at a zero of the function, the other is where the function is undefined.
1. Write the rational inequality in standard form so that the right side is zero.
2. Get a common denominator by multiplying top and bottom of terms. Remember you can not multiply through and get rid of the fractions because you don't know if what you're multiplying by is negative or positive (unless you want to mess with restrictions - and you don't in this case)
3. Find the critical numbers. In simple terms a critical number is anything that makes the numerator or denominator zero.
• Find the places where the function is undefined because of division by zero. These will be critical points that cannot be included in the final answer even if the equal to is included because it would cause division by zero.
• Find the real solutions (ignore complex solutions involving i) to the function any way that you want to. To find the zeros, all you need to worry about is the numerator.
4. Put the critical numbers on the number line.
5. Pick a test point in each interval and determine if the interval works or doesn't work.
6. Include the endpoints if the inequality includes the equal to and do not include the endpoints if the inequality does not include the equal to. Be sure you do not include any endpoint that would cause division by zero if included.
Actually, converting the whole thing to an equation and solving to find the critical numbers isn't that bad of a route. The book, and most mathematics teachers assume that you're going to keep the inequality in the problem in all the way to the end. If you're willing to plug the values back into the original problem, you can change over to an equal sign, get rid of the fractions, and find the critical numbers. You still have to watch the restrictions, but now the restrictions are of the form x≠2 instead of x<2, and much easier to deal with.
However, learn it this way, because when we get to chapter 3, there is going to be a fundamental concept which will greatly speed finding the solutions to these problems, and it requires (sort of) that you look at positives and negatives.
## Graphing Utilities
Another way to solve inequalities is to graph the left hand side of the inequality as y1 and the right hand side as y2 and then find the intersection point. The intersection point will be your critical number. By looking at the graph, you can tell when the inequality is satisfied and record that interval. Remember that when you're giving intervals, only the x-coordinate is necessary. There is no y in the original problem, it was something that was added to make the graphing convenient.
|
```Do Now
Factor completely and solve.
1. x2 - 15x + 50 = 0
2. x2 + 10x – 24 = 0
5.2 Polynomials, Linear Factors,
and Zeros
Learning Target: I can analyze the
factored form of a polynomial and
write function from its zeros
Polynomials and Real Roots
Relative Maximum
Relative
Minimum
ROOTS !
•
1.
2.
3.
4.
5.
6.
POLYNIOMIAL
EQUIVALENTS
Roots
Zeros
Solutions
X-Intercepts
Relative Maximum
Relative Minimum
Linear Factors
Just as you can write a number into its prime
factors you can write a polynomial into its linear
factors.
Ex. 6 into 2 & 3
x2 + 4x – 12 into (x+6)(x-2)
We can also take a polynomial in factored form
and rewrite it into standard form.
Ex. (x+1)(x+2)(x+3) =
foil
distribute
(x2+5x+6)(x+1)=x (x2+5x+6)+1 (x2+5x+6)
= x3+6x2+11x+6
Standard form
We can also use the GCF (greatest common
factor) to factor a poly in standard form into its
linear factors.
Ex. 2x3+10x2+12x GCF is 2x so factor it out.
We get 2x(x2+5x+6) now factor once more to get
2x(x+2)(x+3) Linear Factors
The greatest y value of the points in a region is
called the local maximum.
The least y value among nearby points is called
the local minimum.
Theorem
The expression (x - a) is a linear factor of a
polynomial if and only if the value a is a zero
(root) of the related polynomial function.
If and only if = the theorem goes both ways
If (x – a) is a factor of a polynomial, then a is a
zero (solution) of the function.
and
Ifa is a zero (solution) of the function then (x – a)
is a factor of a polynomial,
Zeros
• A zero is a (solution or x-intercept) to a
polynomial function.
• If (x – a) is a factor of a polynomial, then a is a
zero (solution) of the function.
• If a polynomial has a repeated solution, it has
a multiple zero.
• The number of repeats of a zero is called its
multiplicity.
A repeated zero is called a multiple zero.
A multiple zero has a multiplicity equal to the
number of times the zero occurs.
On a graph, a double zero “bounces” off the x
axis. A triple zero “flattens out” as it crosses
the x axis.
Write a polynomial given the roots
0, -3, 3
•
•
•
•
•
Put in factored form
y = (x – 0)(x + 3)(x – 3)
y = (x)(x + 3)(x – 3)
y = x(x² – 9)
y = x³ – 9x
Write a polynomial given the roots
2, -4, ½
Note that the ½ term
•
•
•
•
•
•
Put in factored form
y = (x – 2)(x + 4)(2x – 1)
y = (x² + 4x – 2x – 8)(2x – 1)
y = (x² + 2x – 8)(2x – 1)
y = 2x³ – x² + 4x² – 2x – 16x + 8
y = 2x³ + 3x² – 18x + 8
becomes (x-1/2). We
don’t like fractions,
so multiply both
terms by 2 to get
(2x-1)
Write the polynomial in factored form.
Then find the roots. Y = 3x³ – 27x² + 24x
•
•
•
•
•
•
Y = 3x³ – 27x² + 24x
Y = 3x(x² – 9x + 8)
Y = 3x(x – 8)(x – 1)
ROOTS?
3x(x – 8)(x – 1) = 0
Roots = 0, 8, 1
FACTORED FORM
What is Multiplicity?
Multiplicity is when you have multiple roots
that are exactly the same. We say that the
multiplicity is how many duplicate roots that
exist.
Ex: (x-2)(x-2)(x+3)
Note: two
therefore the
multiplicity is 2
Ex: (x-1)4 (x+3)
Ex: y =x(x-1)(x+3)
Note: four
therefore the
multiplicity is 4
Note: there are no
repeat roots, so
we say that there
is no multiplicity
Let’s Try One
• Find any multiple zeros of f(x)=x4+6x3+8x2 and
state the multiplicity
Let’s Try One
• Find any multiple zeros of f(x)=x4+6x3+8x2 and
state the multiplicity
Polynomials
-4 is a solution of x2+3x-4=0
-4 is an x-intercept of the graph of y=x2+3x-4
-4 is a zero of y=x2+3x-4
(x+4) is a factor of x2+3x-4
These all say the same thing
Example 1
We can rewrite a polynomial from its zeros.
Write a poly with zeros -2, 3, and 3
f(x)= (x+2)(x-3)(x-3)
foil
= (x+2)(x2 - 6x + 9)
now distribute to get
= x3 - 4x2 - 3x + 18
this function has zeros at -2,3 and 3
Polynomials and Linear Factors
Write a polynomial in standard form with
zeros at 2, –3, and 0.
2
–3
0
ƒ(x) = (x – 2)(x + 3)(x)
Zeros
Write a linear factor for each zero.
= (x – 2)(x2 + 3x)
Multiply (x + 3)(x).
= x(x2 + 3x) – 2(x2 + 3x)
Distributive Property
= x3 + 3x2 – 2x2 – 6x
Multiply.
= x3 + x2 – 6x
Simplify.
The function ƒ(x) = x3 + x2 – 6x has zeros at 2, –3, and 0.
Polynomials and Linear Factors
Find any multiple zeros of ƒ(x) = x5 – 6x4 + 9x3 and state
the multiplicity.
ƒ(x) = x5 – 6x4 + 9x3
ƒ(x) = x3(x2 – 6x + 9)
Factor out the GCF, x3.
ƒ(x) = x3(x – 3)(x – 3)
Factor x2 – 6x + 9.
Since you can rewrite x3 as (x – 0)(x – 0)(x – 0), or
(x – 0)3, the number 0 is a multiple zero of the function, with multiplicity 3.
Since you can rewrite (x – 3)(x – 3) as (x – 3)2, the number 3 is a multiple
zero of the function with multiplicity 2.
Assignment #7
pg 293 7-37 odds
Finding local Maximums and
Minimum
• Find the local maximum and minimum of x3 +
3x2 – 24x
• Enter equation into calculator
• Hit 2nd Trace
• Choose max or min
• Choose a left and right bound and tell
calculator to guess
```
|
3.4 Proving Lines are Parallel
1 / 18
# 3.4 Proving Lines are Parallel - PowerPoint PPT Presentation
3.4 Proving Lines are Parallel. Mrs. Spitz Fall 2005. Standard/Objectives:. Standard 3: Students will learn and apply geometric concepts Objectives: Prove that two lines are parallel.
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### 3.4 Proving Lines are Parallel
Mrs. Spitz
Fall 2005
Standard/Objectives:
Standard 3: Students will learn and apply geometric concepts
Objectives:
• Prove that two lines are parallel.
• Use properties of parallel lines to solve real-life problems, such as proving that prehistoric mounds are parallel.
HW ASSIGNMENT:
• 3.4--pp. 153-154 #1-28
Quiz after section 3.5
Postulate 16: Corresponding Angles Converse
• If two lines are cut by a transversal so that corresponding angles are congruent, then the lines are parallel.
Theorem 3.8: Alternate Interior Angles Converse
• If two lines are cut by a transversal so that alternate interior angles are congruent, then the lines are parallel.
Theorem 3.9: Consecutive Interior Angles Converse
• If two lines are cut by a transversal so that consecutive interior angles are supplementary, then the lines are parallel.
Theorem 3.10: Alternate Exterior Angles Converse
• If two lines are cut by a transversal so that alternate exterior angles are congruent, then the lines are parallel.
Prove the Alternate Interior Angles Converse
Given: 1 2
Prove: m ║ n
3
m
2
1
n
Statements:
1 2
2 3
1 3
m ║ n
Reasons:
Given
Vertical Angles
Transitive prop.
Corresponding angles converse
Example 1: Proof of Alternate Interior Converse
Proof of the Consecutive Interior Angles Converse
Given: 4 and 5 are supplementary
Prove: g ║ h
g
6
5
4
h
Paragraph Proof
You are given that 4 and 5 are supplementary. By the Linear Pair Postulate, 5 and 6 are also supplementary because they form a linear pair. By the Congruent Supplements Theorem, it follows that 4 6. Therefore, by the Alternate Interior Angles Converse, g and h are parallel.
Solution:
Lines j and k will be parallel if the marked angles are supplementary.
x + 4x = 180
5x = 180
X = 36
4x = 144
So, if x = 36, then j ║ k.
Find the value of x that makes j ║ k.
4x
x
Using Parallel Converses:Using Corresponding Angles Converse
SAILING. If two boats sail at a 45 angle to the wind as shown, and the wind is constant, will their paths ever cross? Explain
Solution:
Because corresponding angles are congruent, the boats’ paths are parallel. Parallel lines do not intersect, so the boats’ paths will not cross.
Example 5: Identifying parallel lines
Decide which rays are parallel.
H
E
G
58
61
62
59
C
A
B
D
A. Is EB parallel to HD?
B. Is EA parallel to HC?
Example 5: Identifying parallel lines
Decide which rays are parallel.
H
E
G
58
61
B
D
• Is EB parallel to HD?
mBEH = 58
m DHG = 61 The angles are corresponding, but not congruent, so EB and HD are not parallel.
Example 5: Identifying parallel lines
Decide which rays are parallel.
H
E
G
120
120
C
A
• B. Is EA parallel to HC?
m AEH = 62 + 58
m CHG = 59 + 61
AEH and CHG are congruent corresponding angles, so EA ║HC.
Conclusion:
Two lines are cut by a transversal. How can you prove the lines are parallel?
Show that either a pair of alternate interior angles, or a pair of corresponding angles, or a pair of alternate exterior angles is congruent, or show that a pair of consecutive interior angles is supplementary.
|
# Solve this
Question:
If $A=\left[\begin{array}{rrr}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]$ and $B=\left[\begin{array}{rrr}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]$ are two square matrices, find $A B$ and hence solve the system of linear equations:
$x-y=3,2 x+3 y+4 z=17, y+2 z=7$
Solution:
Here,
$A=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]$ and $B=\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]$
Now,
$A B=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right]\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]$
$\Rightarrow A B=\left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{array}\right]$
$\Rightarrow A B=6\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\Rightarrow A B=6 I_{3}$
$\Rightarrow \frac{1}{6} A B=I_{3}$
$\Rightarrow\left(\frac{1}{6} B\right) A=I_{3} \quad(\because A B=B A)$
$\Rightarrow A^{-1}=\frac{1}{6} B$
$\Rightarrow A^{-1}=\frac{1}{6}\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]$
$X=A^{-1} B$
$X=\frac{1}{6}\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right]\left[\begin{array}{c}3 \\ 17 \\ 7\end{array}\right]$
$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{6}\left[\begin{array}{c}6+34-28 \\ -12+34-28 \\ 6-17+35\end{array}\right]$
$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{6}\left[\begin{array}{c}12 \\ -6 \\ 24\end{array}\right]$
$\therefore x=2, y=-1$ and $z=4$
|
# Square Root of 29 + Solution With Free Steps
The square root of 29, expressed asthe √29, is 5.385. is the symbol of square root. The square root of a value is the number that when multiplied by itself, produces the required number. An example, in this case, can be that the square root of 49 is 7 x 7 = 49
In this article, we will analyze and find the square root of 29 using various mathematical techniques, such as the approximation method and the long division method.
## What Is the Square Root Of 29?
The square root of the number 29 is 5.385.
The square root can be defined as the quantity that can be doubled to produce the square of that similar quantity. In simple words, it can be explained as:
√29 = √(5.385 x 5.385)
√29 = √(5.385)$^2$
√29 = ±5.385
The square can be canceled with the square root as it is equivalent to 1/2; therefore, obtaining y. Hence y is 29’s square root. The square root generates both positive and negative integers.
## How To Calculate the Square Root of 29?
You can calculate the square root of 29 using any of two vastly used techniques in mathematics; one is the Approximation technique, and the other is the Long Division method.
The symbol √ is interpreted as 29 raised to the power 1/2. So any number, when multiplied by itself, produces its square, and when the square root of any squared number is taken, it produces the actual number.
Let us discuss each of them to understand the concepts better.
### Square Root by Long Division Method
The process of long division is one of the most common methods used to find the square roots of a given number. It is easy to comprehend and provides more reliable and accurate answers. The long division method reduces a multi-digit number to its equal parts.
Learning how to find the square root of a number is easy with the long division method. All you need are five primary operations- divide, multiply, subtract, bring down or raise, then repeat.
Following are the simple steps that must be followed to find the square root of 29 using the long division method:
### Step 1
First, write the given number 29 in the division symbol, as shown in figure 1.
### Step 2
Starting from the right side of the number, make the number 29 into a pair that is 29.
### Step 3
Now divide the digit 29 by a number, giving a number either 29 or less than 29. Therefore, in this case, the remainder is 4, whereas the quotient is 5.If the number is not a perfect square, add pair of zeros to the right of the number before starting division.
### Step 4
After this, bring down the next pair 00. Now the dividend is 400. To find the next divisor, we need to double our quotient obtained before. Doubling 5 gives 10; hence consider it as the next divisor.
### Step 5
Now pair 10 with another number to make a new divisor that results in $\leq$ 400 when multiplied with the divisor.
### Step 6
Adding 3 to the divisor and multiplying 103 with 3 results in 309 $\leq$ 400. The remainder obtained is 91. Move the next pair of zeros down and repeat the same process mentioned above.
### Step 7
Keep on repeating the same steps till the zero remainder is obtained or if the division process continues infinitely, solve to two decimal places.
### Step 8
The resulting quotient y is the square root of 29. Figure 1 given below shows the long division process in detail:
### Square Root by Approximation Method
The approximation method involves guessing the square root of the non-perfect square number by dividing it by the perfect square lesser or greater than that number and taking the average.
The given detailed steps must be followed to find the square root of 29 using the approximation technique.
### Step 1
Consider a perfect square number 25 less than 29.
### Step 2
Now divide 29 by √25 .
29 ÷ 5 = 5.8
### Step 3
Now take the average of 5 and 5.8. The resulting number is approximately equivalent to the square root of 29.
(5 + 5.8) ÷ 2 = 5.4
### Important points
• The number 29 is not a perfect square.
• The number 29 is an irrational number.
• The number 29 can be split into its prime factorization.
## Is Square Root of 29 a Perfect Square?
The number 29 is not a perfect square. A number is a perfect square if it splits into two equal parts or identical whole numbers. If a number is a perfect square, it is also rational.
A number expressed in p/q form is called a rational number. All the natural numbers are rational. A square root of a perfect square is a whole number; therefore, a perfect square is a rational number.
A number that is not a perfect square is irrational as it is a decimal number. As far as 29 is concerned, it is a perfect square / not a perfect square. It can be proved as below:
Factorization of 29 results in 1 x 29.
Taking the square root of the above expression gives:
= √(1 x 29)
= (1 x 29)$^{1/2}$
= 5.385
This shows that 29 is not a perfect square as it has decimal places; hence it is an irrational number.
Therefore the above discussion proves that the square root of 29 is equivalent to 5.385.
Images/mathematical drawings are created with GeoGebra.
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# Find the second derivative at $x= 0$ i.e., ${{\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{x=0}}$ of the following equation: ${{e}^{y}}+xy=e$
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Hint: Differentiate the given function twice using sum and product rule of differentiation and then evaluate the value of the second derivative at the given point.
We have the function of the form ${{e}^{y}}+xy=e$.
Differentiating the given function with respect to $x$ on both sides, we get $\dfrac{d}{dx}\left({{e}^{y}}+xy \right)=\dfrac{d}{dx}(e)$.
We will use sum rule of differentiation of two functions which states that if $y=f\left( x \right)+g\left( x \right)$ then we have $\dfrac{dy}{dx}=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)$.
Thus, we have $\dfrac{d}{dx}\left( {{e}^{y}}+xy \right)=\dfrac{d}{dx}\left( {{e}^{y}} \right)+\dfrac{d}{dx}\left( xy \right)=\dfrac{d}{dx}\left( e \right).....\left( 1 \right)$ .
We know that the derivative of a constant with respect to any variable is zero.
Thus, we have $\dfrac{d}{dx}(e)=0.....\left( 2 \right)$.
To evaluate the value of $\dfrac{d}{dx}\left( {{e}^{y}} \right)$, multiply and divide the equation by $dy$.
Thus, we have $\dfrac{d}{dx}\left( {{e}^{y}} \right)=\dfrac{d}{dy}\left( {{e}^{y}} \right)\times \dfrac{dy}{dx}$.
We know that derivative of function of the form $y={{e}^{x}}$ is $\dfrac{dy}{dx}={{e}^{x}}$.
So, we have $\dfrac{d}{dx}\left( {{e}^{y}} \right)=\dfrac{d}{dy}\left( {{e}^{y}} \right)\times \dfrac{dy}{dx}={{e}^{y}}\dfrac{dy}{dx}.....\left( 3 \right)$.
We know that the product rule of differentiation says that if $y=f\left( x \right)g\left( x \right)$ then
we have $\dfrac{dy}{dx}=f\left( x \right)\dfrac{d}{dx}g\left( x \right)+g\left( x \right)\dfrac{d}{dx}f\left( x \right)$.
Thus, we have $\dfrac{d}{dx}\left( xy \right)=x\dfrac{dy}{dx}+y\dfrac{d}{dx}\left( x \right).....\left( 4\right)$.
We know that the derivative of function of the form $y=a{{x}^{n}}$ is $\dfrac{dy}{dx}=an{{x}^{n-1}}$.
So, we have $\dfrac{d}{dx}\left( x \right)=1.....\left( 5 \right)$.
Substituting equation $\left( 5 \right)$ in equation $\left( 4 \right)$, we get $\dfrac{d}{dx}\left( xy\right)=x\dfrac{dy}{dx}+y\dfrac{d}{dx}\left( x \right)=x\dfrac{dy}{dx}+y.....\left( 6 \right)$.
Substituting equation $\left( 2 \right),\left( 3 \right),\left( 6 \right)$ in equation $\left( 1 \right)$,
we get ${{e}^{y}}\cdot \dfrac{dy}{dx}+y+x\cdot \dfrac{dy}{dx}=0$.
Simplifying the above equation, we get $\dfrac{dy}{dx}=-\dfrac{y}{{{e}^{y}}+x}.....\left( 7 \right)$.
We will now differentiate the above equation to find the second derivative of the given function.
As we know $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)$.
Thus, we have $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( -\dfrac{y}{{{e}^{y}}+x} \right)$.
To find the above derivative, we will use quotient rule which says that if $y=\dfrac{f\left( x \right)}{g\left( x \right)}$ then $\dfrac{dy}{dx}=\dfrac{g\left( x \right)f'\left( x \right)-f\left( x\right)g'\left( x \right)}{{{g}^{2}}\left( x \right)}$.
Thus, we have $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( -\dfrac{y}{{{e}^{y}}+x} \right)=\dfrac{\left( {{e}^{y}}+x \right)\dfrac{d}{dx}\left( -y \right)-\left( -y \right)\dfrac{d}{dx}\left({{e}^{y}}+x \right)}{{{\left( {{e}^{y}}+x \right)}^{2}}}.....\left( 8 \right)$.
To evaluate the value of $\dfrac{d}{dx}\left( -y \right)$, we will multiply and divide the equation by $dy$.
Thus, we have $\dfrac{d}{dx}\left( -y \right)=\dfrac{d}{dy}\left( -y \right)\times \dfrac{dy}{dx}$.
We have $\dfrac{d}{dy}\left( -y \right)=-1$ as we know that the derivative of function of the form $y=a{{x}^{n}}$ is $\dfrac{dy}{dx}=an{{x}^{n-1}}$.
Thus, we have $\dfrac{d}{dx}\left( -y \right)=\dfrac{d}{dy}\left( -y \right)\times \dfrac{dy}{dx}=- \dfrac{dy}{dx}.....\left( 9 \right)$.
To evaluate the value of $\dfrac{d}{dx}\left( {{e}^{y}}+x \right)$, we will use sum rule of
differentiation.
Thus, we have $\dfrac{d}{dx}\left( {{e}^{y}}+x \right)=\dfrac{d}{dx}\left( {{e}^{y}} \right)+\dfrac{d}{dx}\left( x \right)$
Substituting equation $\left( 3 \right),\left( 5 \right)$ in the above equation, we have
$\dfrac{d}{dx}\left( {{e}^{y}}+x \right)=\dfrac{d}{dx}\left( {{e}^{y}} \right)+\dfrac{d}{dx}\left( x \right)={{e}^{y}}\dfrac{dy}{dx}+1.....\left( 10 \right)$.
Substituting equation $\left( 9 \right),\left( 10 \right)$ in equation $\left( 8 \right)$, we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{e}^{y}}+x \right)\dfrac{d}{dx}\left( -y \right)-\left( -y \right)\dfrac{d}{dx}\left( {{e}^{y}}+x \right)}{{{\left( {{e}^{y}}+x \right)}^{2}}}=\dfrac{\left( {{e}^{y}}+x \right)\left( -\dfrac{dy}{dx} \right)+y\left( 1+{{e}^{y}}\dfrac{dy}{dx} \right)}{{{\left( {{e}^{y}}+x \right)}^{2}}}.....\left( 11 \right)$ .
Now, we will evaluate the values of functions at $x=0$.
Substituting the value $x=0$ in the equation ${{e}^{y}}+xy=e$, we have ${{e}^{y}}=e$.
$\Rightarrow y=1$
Substituting the value $x=0$ in the equation $\left( 7 \right)$ ,i.e., $\dfrac{dy}{dx}=- \dfrac{y}{{{e}^{y}}+x}$, we have $\dfrac{dy}{dx}=\dfrac{-1}{e}$.
Substituting the value $x=0$ and $\dfrac{dy}{dx}=\dfrac{-1}{e}$ in the equation $\left( 11 \right)$,i.e.,$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{e}^{y}}+x \right)\left( -\dfrac{dy}{dx} \right)+y\left( 1+{{e}^{y}}\dfrac{dy}{dx} \right)}{{{\left( {{e}^{y}}+x \right)}^{2}}}$, we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{e\left( \dfrac{1}{e} \right)+\left( 1+e\left( \dfrac{-1}{e} \right) \right)}{{{\left( e \right)}^{2}}}$
Thus, we have $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{{{e}^{2}}}$ at $x=0$.
Note: We need to observe that we can’t substitute the values and then find the derivative. We need to find the derivative first and then substitute the values. This is an implicit function which is expressed in terms of both dependent and independent variables. To differentiate an implicit function, we differentiate both sides of an equation by treating one of the variables as a function of the other one.
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# Prove that every integer greater than 27 can be written as 5a+8b, where a,b in ZZ^+
Prove that every integer greater than 27 can be written as 5a+8b, where $a,b\in {\mathbb{Z}}^{+}$
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1.Statement: P(n)For $n\in \mathbb{N}$ there exist a,b > 0 such that
n+27 = a * 5 + b * 8
We use mathematical induction to show that every integer greater than 27 can be written as
a*5+b*8, for a,b>0
Base Case: Let n = 1. Note that
28 = 4 * 5 + 1 * 8
Thus base case is true.
Induction Hypothesis: Let P(n) is true for all $n\le k$. That is there exist a,b > 0 such that
k+27=a*5+b*8
We have to find a'b' > 0 such that
(k+1)+27=a*5+b*8
2.Inductive Steps: First note that if n = a* 5 + b*8 for n > 27 then either $a\ge 3\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}b\ge 3$. Otherwise note that if a < 2 and b < 2 then
Case 1:First assume $b\ge 3$ Now by using induction hypothesis we get
(k+1)+27=(k+27)+1
=(k+27)+(5*5-3*8)
=(a*5+b*8)+(5*5-3*8)
=(a+5)*5+(b-3)*8
In this case we get a' = a+5>0 and b' = b - 3>0.
Case 2: Now assume $a\ge 3$. Then we consider following derivation
(k+1)+27=(k=27)+1
=(k+27)+(-3*5=2*8)
=(a*5+b*8)+(-3*5+2*8)
=(a-3)*5+(b+2)*8
In this case we consider a' = a-3>0 and b' = b + 2>0.
Result: For $n\in \mathbb{N}$ there exist a,b > 0 such that
n+27 = a*5+b*8
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Sunday, May 24, 2015
Find x and y if $\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\cdots+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$
If $x,\,y$ are positive integers such that $\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\dfrac{1}{7!15!}+\dfrac{1}{9!13!}+\dfrac{1}{11!11!}+\dfrac{1}{13!9!}+\dfrac{1}{15!7!}+\dfrac{1}{17!5!}+\dfrac{1}{19!3!}+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$.
Find $x,\,y$.
This math problem has exactly the same form of the IMO contest problem.
Method 1:
You might argue that this is not at all a hard problem, because one can approach it by straightforward counting, but that depends on if we recognize the need to multiple both sides of the equation by $20!$.
You wonder, why that specific number, $22!$? If you're observant, you will see that the two figures on each denominator on the LHS all add up to 22, therefore, if we multiply each fraction by $22!$, each fraction will turn into an integer instead. That's why we want to multiple both sides of the equation by $22!$.
We can group the pairs of $\dfrac{1}{1!21!},\,\dfrac{1}{21!1!}$, $\dfrac{1}{3!19!},\,\dfrac{1}{19!3!}$ and so on as well to simplify the LHS of the given equation.
$\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\dfrac{1}{7!15!}+\dfrac{1}{9!13!}+\dfrac{1}{11!11!}+\dfrac{1}{13!9!}+\dfrac{1}{15!7!}+\dfrac{1}{17!5!}+\dfrac{1}{19!3!}+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$
$2\left(\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\dfrac{1}{7!15!}+\dfrac{1}{9!13!}\right)+\dfrac{1}{11!11!}=\dfrac{2^x}{y!}$
$2(22!)\left(\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\dfrac{1}{7!15!}+\dfrac{1}{9!13!}\right)+22!\left(\dfrac{1}{11!11!}\right)=22!\left(\dfrac{2^x}{y!}\right)$
$2\left(\dfrac{22!}{1!21!}+\dfrac{22!}{3!19!}+\dfrac{22!}{5!17!}+\dfrac{22!}{7!15!}+\dfrac{22!}{9!13!}\right)+\dfrac{22!}{11!11!}=\dfrac{2^x22!}{y!}$
This is really the combinatorial formula $n$ choose $r$, where we have:
[MATH] 2\left({22 \choose 1}+{22 \choose 3}+{22 \choose 5}+{22 \choose 7}+{22 \choose 9}\right)+{22 \choose 11}=\dfrac{2^x22!}{y!}[/MATH]
[MATH] 2(695,860)+705,432=\dfrac{2^x22!}{y!}[/MATH]
[MATH] 2,097,152=\dfrac{2^x22!}{y!}[/MATH]
[MATH] 2^{21}=\dfrac{2^x22!}{y!}[/MATH]
Since both $x$ and $y$ are positive integers, we can conclude by now that:
$x=21$; $y=22$
And that's all there is to it.
But, math educators can make optimal use of this problem so to show their students that there is another way to approach this problem as well.
Attention to detail and the ability to look at a problem from all angles are two important traits students must have that are waiting to be cultivated by the teachers and to flourish.
Having said that, this problem can be approached by making smarter choice to avoid using calculator to figure out the values for $x$ and $y$.
Hey math educators out there, the ball is in your court now whether you want to generate easily satisfied students or students who always want to improve and go out of their comfort zone in solving hard challenge math problems.
I will write in my next blog about how to approach this problem without the help of calculator, see ya!
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# Methods/steps in solving multiplication by two digit number word problems
###### Get more contents on this skill...
Hey kids, learn these appealing methods in solving multiplication by two-digit number word problems in this step-by-step guide carefully arranged for you. This fantastic step-by-step guide will cover all the major problems that kids need to know.
We often expect kids to solve standard text-based word problems instead of real-life or real-world text-based challenging tasks. Hence we encourage all 5th graders to engage in this resource and learn how to use real-world text-based problem-solving tasks.
This approach will greatly help young learners combat any math situation they encounter in a word problem.
## Steps to solving multiplication by two digit numbers word problems
Before we dive into providing steps to solving multiplication by two-digit numbers word problems, let’s find out why most kids struggle to solve multiplication by two-digit word problems.
We understand that one of the things 5th graders find difficulties in solving word problems is interpreting the problem. Hence we will offer your kids a carefully-designed step-by-step guide that will facilitate the interpretation and solving of grade 5 multiplication by 2-digit word problems.
However, we will offer some real-life examples below to witness how amazing this method works.
#### Step 1: IDENTIFY THE PROBLEM
Also, try to identify important numbers and keywords in the word problem.
For instance, If you come across one of the following keywords in a word problem: - times, multiplied by, product, product of, factor, of, multiply, times, multiple, double, triple, groups, by, twice, area, equal groups, every, in all, total, increased by, as much, each, lots of, groups of, per, etc., then it implies that you have to perform a Multiplication Operation.
To identify the problem, you must read the word problem very carefully to know what the problem wants you to solve.
Note: One key Element for learners to understand is that they should not always rely on keywords alone. That is to say; the same keyword can have different meanings in different word problems.
For this reason, we reiterate on the importance of reading the question very carefully to understand the situation that the word problem is describing, then figure out exactly which operation to use
#### Step 2: STRATEGIZE OR GATHER RELEVANT INFORMATION
How will you solve or tackle the problem?
One key thing you should remember is that each word problem may require a different format. So, these key points below will help kids to solve any format however it comes accurately.
• As mentioned in step 1 above, from the keyword(s) in the word problem, you can determine if you need to perform a Multiplication Operation.
• On the other hand, know that you must not rely solely on keywords. Also, try to understand the situation that the problem is describing.
• Now, after knowing the operation you will perform, construct short sentences to represent the given word problem.
#### Step 3: CREATE THE EQUATION
Here, we will write down a numerical equation representing the information given in the word problem.
#### Step 4: PROVIDE A SOLUTION
From step 3 above, multiply the numbers using regrouping or other multiplication methods like long multiplication, grid multiplication, or lattice to determine the result. Always recall adding the unit of measurement to your final result.
#### Step 5: CHECK YOUR WORK
Finally, check if your answer makes sense. For instance, estimate the answer and see if it is close to what you expected. However, if the answer is not what you expected, go back to step one and start all over again.
### Examples on how to multiply by two-digit numbers word problems
#### Example one
Mr. Davis is a blacksmith. He received a contract to make 526 exotic swords for the military yesterday. How much is the contract worth if he charges \$85 per sword?
Step 1:The important numbers here are 526 and \$85. The most important keyword(s) found in the word problem is “per.”
Step 2: Now, how are you going to tackle the problem? You see that the situation the problem is describing and the keyword found in the word problem call for you to perform a multiplication operation.
Next, after knowing which operation you will perform, form short sentences to represent the given word problem.
• Number of exotic swords Mr. Davis is supposed to make = 526
• Amount of money he charges per sword = \$85
• Therefore, the amount of money the contract is worth = the number of exotic swords Mr. Davis is supposed to make × the amount he charges per sword.
Step 3:Now, write down a numerical equation to represent the bolded sentence in step 2 above to solve this word problem:
526 × \$85 = ?
Step 4: From step 3 above, multiply the numbers using regrouping or other multiplication methods like long multiplication, grid multiplication, and lattice to determine the result. Always remember to add the unit of measurement to the final answer.
Using the long multiplication method, we have
So, the contract is worth \$44,710.
Step 5: Finally, check if your answer makes sense. For instance, estimate the answer and see if it is close to what you expected. However, if the answer is not what you expected, go back to step one and start all over again.
#### Example Two
Maya has 523 marbles. Her sister Beverly has 15 times what Maya has. How many marbles does Beverly have?
Step 1: The important numbers here are 569 and 15. The most important keyword(s) found in the word problem is “times.”
Step 2: Now, how are you going to tackle the problem? You see that the keyword(s) found in the word problem calls for you to perform a multiplication operation.
But you have to note what is being asked in the question.
Next, after knowing the operation you are going to perform, form short sentences to represent the given word problem.
• Number of marbles Maya has = 523
• Number of marbles her sister Beverly has compared to Maya = 15 times what Maya has
• Therefore, the number of marbles Beverly has = the number of marbles Maya has × the number of marbles her sister Beverly has compared to Maya.
Step 3:Now, write down a numerical equation to represent the step 2 above. Use “/” to separate the numerator from the denominator.
523 × 15 = ?
Step 4: From step 3 above, multiply the numbers using regrouping or any other multiplication methods like long multiplication, grid multiplication, and lattice method to determine the result. Remember to add the unit of measurement to your final result.
Using the long multiplication method, we have
So, Beverly has 7,845 marbles.
Step 5: Finally, check if your answer makes sense. For instance, estimate the answer and see if it is close to what you expected. However, if the answer is not what you expected, go back to step one and start all over again.
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# Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution
## Selina Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution (Including Use of Brackets as Grouping Symbols)
Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 20 Substitution (Including Use of Brackets as Grouping Symbols)
APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Mathematics for Class 6 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.
IMPORTANT POINTS
1. Substitution : The value of an expression depends on the value of its variable (s).
2. Use of Brackets :
The Symbols —, ( ), { }, [ ] are called brackets.
If an expression is enclosed within a bracket, it is considered a single quantity, even if it is made up of many terms.
Keep in Mind :
• While simplifying an expression containing a bracket, first of all, the terms inside the bracket are operated (combined).
• ( ) is called a small bracket or Parenthesis.
• { } is called a middle bracket or Curly bracket.
• [ ] is called big or square bracket.
• If one more bracket is needed, then we use the bar bracket.
i.e. a line ———— is drawn over a group of terms.
Thus, in $$3x+\bar { 4y-5z }$$, the line over 4y – 5z serves as the bar bracket and is called Vinculum.
### Substitution Exercise 20A – Selina Concise Mathematics Class 6 ICSE Solutions
Question 1.
Fill in the following blanks, when :
x = 3,y = 6, z = 18, a = 2, b = 8, c = 32 and d = 0.
Solution:
Question 2.
Find the value of :
Solution:
Question 3.
Find the value of :
Solution:
Question 4.
If a = 3, b = 0, c = 2 and d = 1, find the value of :
Solution:
Question 5.
Find the value of 5x2 – 3x + 2, when x = 2.
Solution:
Question 6.
Find the value of 3x3 – 4x2 + 5x – 6, when x = -1.
Solution:
Question 7.
Show that the value of x3 – 8x2 + 12x – 5 is zero, when x = 1.
Solution:
Question 8.
State true and false :
(i) The value of x + 5 = 6, when x = 1
(ii) The value of 2x – 3 = 1, when x = 0
(iii) $$\frac { 2x-4 }{ x+1 }$$ = -1,when x = 1
Solution:
Question 9.
If x = 2, y = 5 and z = 4, find the value of each of the following :
Solution:
Question 10.
If a = 3, find the values of a2 and 2a.
Solution:
a2 = (3)2 = 3 x 3 = 9
2a = (2)3 = 2 x 2 x 2 = 8
Question 11.
If m = 2, find the difference between the values of 4m3 and 3m4.
Solution:
4m3 = 4 (2)3 = 4 x 2 x 2 x 2 = 32
3m4 = 3 (2)4 = 3 x 2 x 2 x 2 x 2 = 48
Now, a difference 3m4 – 4m3 = 48 – 32 = 16
### Substitution Exercise 20B – Selina Concise Mathematics Class 6 ICSE Solutions
Question 1.
Evaluate :
(i) (23 – 15) + 4
(ii) 5x + (3x + 7x)
(iii) 6m – (4m – m)
(iv) (9a – 3a) + 4a
(v) 35b – (16b + 9b)
(vi) (3y + 8y) – 5y
Solution:
(i) (23 – 15) + 4 = 8 + 4 = 12
(ii) 5x + (3x + 7x) = 5x + 10x = 15x
(iii) 6m – (4m – m) = 6m – 3m = 3m
(iv) (9a – 3a) + 4a = 6a + 4a = 10a
(v) 35b – (16b + 9b)= 35b – 25b = 10b
(vi) (3y + 8y) – 5y = 11y – 5y = 6y
Question 2.
Simplify :
Solution:
Question 3.
Simplify :
Solution:
### Substitution Exercise 20C – Selina Concise Mathematics Class 6 ICSE Solutions
Question 1.
Fill in the blanks :
Solution:
(viii) 2t + r – p – q + s = 2t + r – (p + q – s)
Question 2.
Insert the bracket as indicated :
Solution:
### Substitution Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions
Question 1.
Find the value of 3ab + 10bc – 2abc when a = 2, b = 5 and c = 8.
Solution:
Question 2.
If x = 2, = 3 and z = 4, find the value of 3x2 – 4y2 + 2z2.
Solution:
Question 3.
If x = 3, y = 2 and z = 1; find the value of:
(i) xy
(ii) yx
(iii) 3x2 – 5y2
(iv) 2x – 3y + 4z + 5
(v) y2 – x2 + 6z2
(vi) xy + y2z – 4zx
Solution:
Question 4.
If P = -12x2 – 10xy + 5y2, Q = 7x2 + 6xy + 2y2, and R = 5x2 + 2xy + 4y2 ; find :
(i) P – Q
(ii) Q + P
(iii) P – Q + R
(iv) P + Q + R
Solution:
Question 5.
If x = a2 – bc, y = b2 – ca and z = c2 – ab ; find the value of :
(i) ax + by + cz
(ii) ay – bx + cz
Solution:
Question 6.
Multiply and then evaluate :
(i) (4x + y) and (x – 2y); when x = 2 and y = 1.
(ii) (x2 – y) and (xy – y2); when x = 1 and y = 2.
(iii) (x – 2y + z) and (x – 3z); when x = -2, y = -1 and z = 1.
Solution:
Question 7.
Simplify :
(i) 5 (x + 3y) – 2 (3x – 4y)
(ii) 3x – 8 (5x – 10)
(iii) 6 {3x – 8 (5x – 10)}
(iv) 3x – 6 {3x – 8 (5x – 10)}
(v) 2 (3x2 – 4x – 8) – (3 – 5x – 2x2)
(vi) 8x – (3x – $$\bar { 2x-3 }$$)
(vii) 12x2 – (7x – $$\bar { 3x^{ 2 }+15 }$$)
Solution:
Question 8.
If x = -3, find the value of : 2x3 + 8x2 – 15.
Solution:
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#### Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 47 Maths Textbook Solution.
Answer: $\frac{dy}{dx}=\frac{-1}{\sqrt{1-x^{2}}}$
Hint:
\begin{aligned} &\frac{d}{\mathrm{dx}}(\text { constsant })=0 \\ &\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}}-1 \end{aligned}
Given:
$y=\cos ^{-1}\left\{\frac{2 x-3 \sqrt{1-x^{2}}}{\sqrt{13}}\right)$
Solution:
Let,$x=\cos \theta$
\begin{aligned} &\mathrm{y}=\cos ^{-1}\left\{\frac{2 \cos \theta-3 \sqrt{1-\cos ^{2} \theta}}{\sqrt{4}}\right\} \\ &\text { Using } \cos ^{2} \theta+\sin ^{2} \theta=1 \\ &\mathrm{y}=\cos ^{-1}\left\{\frac{2}{\sqrt{13}} \cos \theta-\frac{3}{\sqrt{13}} \sqrt{\sin ^{2} \theta}\right] \\ &\mathrm{y}=\cos ^{-1}\left\{\frac{2}{\sqrt{13}} \cos \theta-\frac{3}{\sqrt{13}} \sin \theta\right] \end{aligned}
$\cos \phi=\frac{2}{\sqrt{13}}$
Let,$\sin \phi=\sqrt{1-\cos ^{2}\phi}$
\begin{aligned} &=\sqrt{1-\left(\frac{2}{\sqrt{13}}\right)^{2}}+\sqrt{1-\frac{4}{13}} \\ &=\sqrt{\frac{13-4}{13}}=\sqrt{\frac{9}{13}} \end{aligned}
$\sin \phi=\frac{3}{\sqrt{13}}$
So,
\begin{aligned} &\begin{aligned} \mathrm{y} &=\cos ^{-1}\{\cos \phi \cos \theta-\sin \phi \sin \theta] \\ &=\cos ^{-1}[\cos (\theta+\phi)] \end{aligned} \\ &\text { Using, } \cos (A+B)=\cos A \cos B-\sin A \sin B \end{aligned}
$\mathrm{y}=\phi+\theta \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\cos ^{-1}(\cos \theta)=\theta \text { if } 0 \varepsilon[0, \pi]\right]$
\begin{aligned} &\mathrm{y}=\cos ^{-1}\left(\frac{2}{\sqrt{13}}\right)+\cos ^{-1} \mathrm{x} \\ &\text { since, } \mathrm{x}=\cos \theta \\ &\cos \phi=\frac{2}{\sqrt{13}} \end{aligned}
Differentiating its with Respect to x,
\begin{aligned} &\frac{d y}{d x}=\frac{-1}{\sqrt{1-\left(\frac{2}{\sqrt{13}}\right)^{2}}} \frac{d}{d x}\left(\frac{2}{\sqrt{13}}\right)+\left(\frac{-1}{\sqrt{1-x^{2}}}\right) \\ &\frac{d y}{d x}=\frac{-1}{\sqrt{1-\left(\frac{2}{\sqrt{13}}\right)^{2}} \times 0+\left(\frac{-1}{\sqrt{1-x^{2}}}\right)} \\ &\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^{2}}} \\ &\frac{d y}{d x}=0-\frac{1}{\sqrt{1-x^{2}}} \end{aligned}
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# Puerto Rico - Grade 1 - Math - Numbers and Operations in Base Ten - Adding Within 100 - 1.NBT.4
### Description
Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten.
• State - Puerto Rico
• Standard ID - 1.NBT.4
• Subjects - Math Common Core
### Keywords
• Math
• Numbers and Operations in Base Ten
## More Puerto Rico Topics
1.OA.4 Understand subtraction as an unknown-addend problem. For example, subtract 10 – 8 by finding the number that makes 10 when added to 8. Add and subtract within 20.
1.OA.5 Relate counting to addition and subtraction (e.g., by counting on 2 to add 2).
1.OA.6 Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
1.OA.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2.
1.OA.8 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = _ – 3, 6 + 6 = _.
Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes.
Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: A. 10 can be thought of as a bundle of ten ones — called a “ten.” B. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones).
Solve word problems that call for addition of three whole numbers whose sum is less than or equal to 20, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
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# Lesson Explainer: Intersections of Circles and Lines Mathematics
In this explainer, we will learn how to find points of intersection of circles and lines.
Suppose we have a circle and a line in the plane. There are three possibilities.
Either the line cuts through the circle, intersecting it in two points; or the line misses the circle, not intersecting it at all; or the line is tangent to the circle, intersecting it in one point.
Recall that the equation of a straight line may be given in the form where is the lineβs slope and its -intercept, and that the equation of a circle may be given in standard form where is the center of the circle and its radius, or in general form with constants , , and .
If the line and the circle intersect at a point , then the coordinates of simultaneously solve the equations of both the line and the circle. This means that we can substitute ββ for in the circle equation to calculate the coordinates of the point of intersection . The result of this substitution is to eliminate from the circle equation, leaving us with a quadratic equation in :
The roots of this quadratic equation are precisely the -coordinates of the intersection points of the line with the circle. The number of roots a quadratic equation has over the real numbers is controlled by its discriminant
Thus, the number of times a given straight line intersects a circle can be read off from the discriminant of the associated quadratic. When , the line intersects the circle twice; when , the line intersects the circle once (i.e., it is tangent to the circle); and when , the line and the circle are disjoint.
### Example 1: Determining How Many Times a Line Intersects a Circle
Consider the circle and the lines and . How, if at all, do and intersect the circle?
To work out how a line intersects a circle , we proceed as follows:
1. Substitute into , eliminating .
2. This results in a quadratic equation in , . Calculate the discriminant of this quadratic, .
3. If , then the line intersects the circle in two points. If , then the line is tangent to the circle. If , then the line and the circle are disjoint.
We will deal with the two lines one at a time. First, we substitute into :
We now compute the discriminant of the resulting quadratic: which tells us that the line meets the circle in one point, or, in other words, is tangent to .
Now, for , we follow exactly the same procedure. First, we substitute:
Then, we check the discriminant:
The positive discriminant tells us that the quadratic has two real roots and that the line intersects the circle in two distinct points.
So, to answer the question, is tangent to the circle, while intersects it in two points.
Now that we know how to work out if and how many times a line intersects a circle, let us turn our attention to calculating the coordinates of the intersection points.
Given the equation of a circle and the equation of a line , we proceed as before by substituting in for (or for ; the result will be the same) in the equation of the circle to obtain a quadratic equation in . This time, however, instead of computing the discriminant of the quadratic, we are going to solve it in order to get the -coordinates of the intersection points of the line and the circle. We can then substitute these -coordinates back into the equation of the line to get the -coordinates of the intersection points.
### Example 2: Finding the Coordinates of the Points Where a Line Intersects a Circle given in Standard Form
Find the coordinates of the points where the circle meets the line .
The first step is to substitute into to obtain a quadratic equation in :
We now solve this quadratic to find the -coordinates of the intersection points of the line and the circle. In this example, the quadratic expression factors immediately, as
Thus, gives us -coordinates of and . We can now substitute these values back into the linear equation to find the corresponding -coordinates and
Therefore, the coordinates of the intersection points of the circle and the line are and .
The final step of the previous example was to substitute the -coordinates of the intersection points back into the linear equation to find the corresponding -coordinates. You might like to think about what happens if we instead substitute into the equation of the circle and why.
The previous example dealt with finding the intersection points of a line and a circle given in standard form. Let us now look at an example when the circle is given in general form.
### Example 3: Finding the Coordinates of the Points Where a Line Intersects a Circle given in General Form
Find the coordinates of the points where the circle meets the line .
The method for finding the intersection points of a line and a circle given in general form is as follows:
1. Substitute for in the equation of the circle, here .
2. This results in a quadratic equation in , . The solutions to this quadratic equation are the -coordinates of the points where the line and the circle intersect.
3. Having calculated the -coordinates, we substitute these values back into the equation of the line to find the corresponding -coordinates.
So, we begin by substituting into the circle equation :
Now, we expand and simplify:
We can solve the resulting quadratic equation to find the -coordinates of the intersection points of the line and the circle. In this example, the quadratic factorizes as , giving us the -coordinates of and . Finally, we substitute these values back into the linear equation to find the -coordinates of the intersection points: and
Therefore, the coordinates of the points of intersection of the circle and the line are and .
We will now use what we have learned about intersections of lines and circles to solve problems involving unknown constants in circle and line equations.
### Example 4: Finding the Possible Unknowns in the Equation of a Circle given a Line that Touches it at a Given Point
The line is tangent to the circle , where is a constant. Work out the two possible values of .
Since we are given that the line is tangent to the circle , we know that there exists a point at which the line and the circle meet; in particular, at this point, the value of in the equation of the line will be equal to the value of in the equation of the circle. That is to say, it is valid to substitute an expression for derived from the equation of the line into the equation of the circle. In order to do this, we first make the subject of :
Now, we can substitute and simplify:
The roots of the resulting quadratic are the -coordinates of the intersection points of the line and the circle . However, we are told that this line is tangent to the circle; that is to say, it meets the circle in exactly one point. Therefore, the quadratic must have exactly one real root. This tells us that the discriminant must be zero. Let us write this down:
The roots of this quadratic will give us the values of that we are looking for. Factorizing, we have , and so or . These two values are the -coordinates of the centers of the two circles of radius whose centers lie on the line and that are tangent to the line .
We will finish with another problem-solving example involving circles, lines, and their intersections.
### Example 5: Finding the Range of Unknowns That Lead to a Line Not Intersecting a Circle
Consider the circle and the line , where is a constant. Find the range of values of for which the circle and the line are disjoint.
If we substitute into the equation of the circle , we will get a quadratic equation whose solutions are the -coordinates of the points where the line and the circle intersect. Since we are trying to find a range of values of for which the line and the circle do not intersect, we are aiming for a quadratic equation with no real solutions.
So, we start by substituting:
Expanding and simplifying, we obtain the quadratic . We are interested in the values of for which the discriminant of this quadratic is negative. That is,
The ranges of values for which are when is positive and when is negative.
These ranges of values give ranges of slopes of the line with -intercept 3 for which it does not intersect the circle .
The pink region contains all the lines with the negative slope , while the blue region contains all the lines with positive slope .
Let us finish by recapping a few important concepts from this explainer.
### Key Points
• A line and a circle in the plane can interact in one of three ways: they may intersect in two points, they may intersect in one point, or they may not intersect at all. If the line and the circle do intersect, then the coordinates of their point (or points) of intersection simultaneously solve the equation of the line and the equation of the circle. This allows us to use algebra to calculate whether a line and a circle intersect and, if they do, the coordinates of their points of intersection.
• Given the equation of a line and the equation of a circle in either standard form or general form , we can substitute the expression for in the equation of the circle to obtain a quadratic equation in , .
• The discriminant of the quadratic tells us about the intersections of the line and the circle. If , then the line and the circle intersect in two points. If , then the line is tangent to the circle. If , then the line and the circle are disjoint.
• We can solve the quadratic to find the -coordinates of the points of intersection of the line and the circle. We can then substitute these values back into the equation of the line to find the -coordinates of the points of intersection.
• We can use conditions of the discriminant (i.e., whether it is positive, negative, or zero) to find unknown constants in equations of lines and circles when we know how they intersect.
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# What is the probability of getting two 5’s in rolling two dice?
Contents
Total Number of combinations Probability
3 2 5.56%
4 3 8.33%
5 4 11.11%
6 5 13.89%
## What is the probability of rolling a 5 twice?
Assuming a fair die: The probability of getting a 5 in the first roll is 1/6. The probability of getting a 5 in the second roll is 1/6. These are independent events.
## What is the probability of rolling a 5 and then rolling a second 5 on a pair of dice?
There is a 16 probability of getting a 2 on the first roll and a 16 probability of getting a 5 on the second roll.
## What is the probability of rolling a 5 with a pair of dice?
If you don’t get a 5 on the first die (that’s a 5/6 chance) then the chances of getting a 5 on the second die are 1 in 6. So again, we have 5/6*1/6=5/36 chance.
## What is the probability of rolling a 3 then a 5?
6 Sided Dice probability (worked example for two dice). Two (6-sided) dice roll probability table. Single die roll probability tables.
Two (6-sided) dice roll probability table.
Roll a… Probability
2 1/36 (2.778%)
3 3/36 (8.333%)
4 6/36 (16.667%)
5 10/36 (27.778%)
## What is the probability of rolling a 1 and then a 2?
because the probability of rolling a 1 is 1/6 and the probability of rolling a 2 is also 1/6. 1/6 + 1/6 = 2/6 = 1/3. Each elementary outcome (in terms of single face) has equal probability of 1/6, so two faces like 1 or 2 have a combined probability of 1/6 + 1/6 = 1/3 or 33.3% if you like that notation better.
IT IS INTERESTING: What is the probability of obtaining a sum of 7 or 5 if two dice are thrown?
## How many combinations of 7 are there in craps?
The Point Numbers
The odds of a 7 coming up before a 4 is easy to calculate. You have six possible combinations totaling 7 versus three possible combinations totaling 4.
## When rolling a die 100 times what is the probability of rolling a 6 exactly 20 times?
Assuming that each roll is independent, you have a 1/6 chance of rolling a 6. You want to find the probability of it rolling a 6 twenty times out of 100, so we have (1/6)^20. Here the order doesn’t matter so multiply this by (100/20). For eighty times, you will have (5/6)^80 chance in rolling a die that is not a 6.
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# Verbal Arithmetic - Page 2
In column 4, the addition of S and M has produced a carry-over into column 5. The carry-over can only be a 1, since the letters represent single digit numbers. So for instance, if you add 8 and 9, you get 17, so you carry 1 into the next column. But you can never carry more than 1. So M = 1.
In order for there to be a carry from column 4 to column 5, S + M has to come to at least 9, so S is either 8 or 9. Therefore S + M is 9 or 10, and so O is 0 or 1. But we have established that M = 1, so therefore O = 0.
If there were a carry from column 3 to column 4, then E would be 9 and N would be 0. But O = 0, so there is no carry, and S = 9.
If there were no carry from column 2 to column 3, then given that E + O = N and also that we know that O = 0, this would mean that E = N. But that is impossible. So therefore there is a carry from column 2 to column 3, and N = E + 1.
If there were no carry from column 1 to column 2, then (N + R) mod 10 = E, and given that N = E + 1, this would mean that E + 1 + R = E mod 10, so therefore R would be 9. But we know that S = 9, so therefore there must be a carry from column 1 to 2, and R = 8.
To produce a carry from column 1 to column 2, D + E = 10 + Y. Since Y cannot be 0 or 1, D + E must be at least 12. As D is at most 7, then E is at least 5. Also, N is at most 7, and N = E + 1. So therefore E is 5 or 6.
If E were 6, then to make D + E at least 12, D would have to be 7. But N = E + 1, so N would also be 7, which is impossible. Therefore E = 5 and N = 6.
To make D + E at least 12, we must have D = 7, and so Y = 2.
A word of explanation about the use of mod 10. Modular arithmetic is sometimes referred to as clock arithmetic, although a clockface actually works in mod 12; so for example if it is 8 o'clock and you add 6 hours, you end up at 2 o'clock, so (8 + 6) mod 12 = 2. In the same way, if you add 7 and 8 you get 15. But if you are performing an addition, as we are here, you would carry the 1, and leave the 5. So therefore (7 + 8) mod 10 = 5.
This puzzle was originally set in the Strand magazine in 1924, by the well known English puzzler Henry Dudeney (1857 - 1930).
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Master the 7 pillars of school success
Perfect Squares
√4 = 2
√9 = 3
√16 = 4
√25 =5
√36 =6
√49 = 7
√64 =8
√81 =9
√100 =10
Your radical is in the simplest form when the radicand cannot be divided evenly by a perfect square.
Radicals ( or roots ) are the opposite of exponents. For instance, 3 squared equals 9, but if you take the square root of nine it is 3.
For example
4^2=16 and √16 =4
2^3=8 and ∛8 = 2
3^3= 27 and ∛27=3
The check mark √ is called the radical symbol and gives you three pieces of important information, the degree, the radical symbol, and the radicand.
The Degree is the number of times the number ( called the radicand) has been multiplied by itself.
A degree of 3 is a cube root, a 4 is the fourth root. Most times for a square root a degree will be missing, and it is assumed that it is a square root.
Method 1. Find the largest perfect square that will divide into the radical
Simplify √32
Step 1 Find the largest perfect square that will divide evenly into 32
16 * 2 =32
Step 2 Write these numbers under the radical √(16*2)
Step 3 Give each number it’s own radical sign √16*√2
Step 4 “Take out” or reduce the perfect square 4√2 ( perfect square of 16=4)
Simplify √75
25*3 =75 Step 1. Find largest perfect square
√(25*3) Step 2. Write the numbers under the radical.
√25*√3 Step 3. Give the numbers a radical sign.
5√3 Step. 4. Take out any perfect squares (square of 25=5)
Method 2 Simplify a radical " Jailbreak Method"
I have taught this method to my students for many years it it helps many remember the steps of simplifying
Simplify √75
Step 1 Create a factor tree
Step 2 When the radical is a square root any like pair of numbers escape from under the radical. In this example the pair of 5’s escape and the 3 remains under the radical.
√(5 5 3) the 5’s jailbreak and escape in a pair and the three remains under the radical
5√3
Simplify √96
Step 1. Draw a factor tree
Step 2 Pairs of like numbers escape In this example you have 2 pairs of 2
Step 3 Multiply the pairs together 2*2 = 4√(2 3)
Step 4 Multiply the numbers remaining under the radical 2*3 = 6
Step 5 therefore √(96 )=4√6
## Simplifying Radicals “ Square Roots”
In order to simplify a square root you take out anything that is a perfect square.
Being familiar with the following list of perfect squares will help when simplifying radicals.
Radicand The value that you are taking the root of.
Radical symbol. The symbol that looks like a check mark. √ The bar tells you how much of the value to use. You can think of the bar as parenthesis. For example,
The bar is over 30 -3 so subtract 30 -3 =27,then take the cube root.
The bar is only over the 27 so take the cube root of 27 which equals 3 then subtract 2
Perfect Squares
4 = 2x2
9= 3x3
16= 4x4
25= 5 x5
36= 6x6
49= 7x7
64= 8x8
81= 9x9
100=10 x 10
Memorizing perfect cubes will also help when simplifying radicals
8 = 2 x 2 x 2 125 = 5 x 5 x 5 512 8 x 8 x 8
27= 3 x 3 x3 216 = 6 x 6 x 6 729 = 9 x 9 x 9
64 = 4 x 4 x 4 343 = 7 x 7 x 7
The Product Rule of Radicals states that the product of two
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# GCF of 3 and 12
The gcf of 3 and 12 is the largest positive integer that divides the numbers 3 and 12 without a remainder. Spelled out, it is the greatest common factor of 3 and 12. Here you can find the gcf of 3 and 12, along with a total of three methods for computing it. In addition, we have a calculator you should check out. Not only can it determine the gcf of 3 and 12, but also that of three or more integers including three and twelve for example. Keep reading to learn everything about the gcf (3,12) and the terms related to it.
## What is the GCF of 3 and 12
If you just want to know what is the greatest common factor of 3 and 12, it is 3. Usually, this is written as
gcf(3,12) = 3
The gcf of 3 and 12 can be obtained like this:
• The factors of 3 are 3, 1.
• The factors of 12 are 12, 6, 4, 3, 2, 1.
• The common factors of 3 and 12 are 3, 1, intersecting the two sets above.
• In the intersection factors of 3 ∩ factors of 12 the greatest element is 3.
• Therefore, the greatest common factor of 3 and 12 is 3.
Taking the above into account you also know how to find all the common factors of 3 and 12, not just the greatest. In the next section we show you how to calculate the gcf of three and twelve by means of two more methods.
## How to find the GCF of 3 and 12
The greatest common factor of 3 and 12 can be computed by using the least common multiple aka lcm of 3 and 12. This is the easiest approach:
gcf (3,12) = $\frac{3 \times 12}{lcm(3,12)} = \frac{36}{12}$ = 3
Alternatively, the gcf of 3 and 12 can be found using the prime factorization of 3 and 12:
• The prime factorization of 3 is: 3
• The prime factorization of 12 is: 2 x 2 x 3
• The prime factors and multiplicities 3 and 12 have in common are: 3
• 3 is the gcf of 3 and 12
• gcf(3,12) = 3
In any case, the easiest way to compute the gcf of two numbers like 3 and 12 is by using our calculator below. Note that it can also compute the gcf of more than two numbers, separated by a comma. For example, enter 3,12. The calculation is conducted automatically.
The gcf is...
Frequently searched terms on our site also include:
## Use of GCF of 3 and 12
What is the greatest common factor of 3 and 12 used for? Answer: It is helpful for reducing fractions like 3 / 12. Just divide the nominator as well as the denominator by the gcf (3,12) to reduce the fraction to lowest terms.
$\frac{3}{12} = \frac{\frac{3}{3}}{\frac{12}{3}} = \frac{1}{4}$.
## Properties of GCF of 3 and 12
The most important properties of the gcf(3,12) are:
• Commutative property: gcf(3,12) = gcf(12,3)
• Associative property: gcf(3,12,n) = gcf(gcf(12,3),n) $\hspace{10px}n\hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$
The associativity is particularly useful to get the gcf of three or more numbers; our calculator makes use of it.
To sum up, the gcf of 3 and 12 is 3. In common notation: gcf (3,12) = 3.
If you have been searching for gcf 3 and 12 or gcf 3 12 then you have come to the correct page, too. The same is the true if you typed gcf for 3 and 12 in your favorite search engine.
Note that you can find the greatest common factor of many integer pairs including three / twelve by using the search form in the sidebar of this page.
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### Learning Outcomes
Identify turning points the a polynomial duty from that graph.Identify the number of turning points and intercepts of a polynomial function from that is degree.Determine x and also y-intercepts the a polynomial role given that equation in factored form.
You are watching: Give an example and explain why a polynomial can have fewer x-intercepts than its number of roots.
## Identifying Local actions of Polynomial Functions
In enhancement to the end behavior of polynomial functions, we are additionally interested in what happens in the “middle” the the function. In particular, we room interested in locations where graph behavior changes. A turning allude is a allude at i m sorry the duty values adjust from raising to to decrease or decreasing to increasing.
We are likewise interested in the intercepts. Just like all functions, the y-intercept is the allude at i m sorry the graph intersects the upright axis. The suggest corresponds come the name: coordinates pair in i beg your pardon the input value is zero. Because a polynomial is a function, just one output value coincides to every input value so there deserve to be only one y-intercept left(0,a_0 ight). The x-intercepts take place at the input values that correspond to one output value of zero. The is possible to have an ext than one x-intercept.
### A basic Note: Intercepts and transforming Points the Polynomial Functions
A turning point of a graph is a allude where the graph transforms from increasing to decreasing or decreasing come increasing.The y-intercept is the suggest where the duty has an input value of zero.The x-intercepts are the points whereby the output value is zero.A polynomial of degree n will have, at most, n x-intercepts and n – 1 turning points.
## Determining the number of Turning Points and also Intercepts from the degree of the Polynomial
A continuous function has no division in the graph: the graph have the right to be drawn without lifting the pen native the paper. A smooth curve is a graph that has actually no sharp corners. The transforming points that a smooth graph must always occur in ~ rounded curves. The graphs of polynomial features are both consistent and smooth.
The level of a polynomial function helps us to identify the number of x-intercepts and also the number of turning points. A polynomial function of nth level is the product the n factors, so the will have at most n roots or zeros, or x-intercepts. The graph that the polynomial function of degree n must have actually at many n – 1 turning points. This means the graph contends most one fewer turning point than the level of the polynomial or one fewer 보다 the number of factors.
### Example: identify the variety of Intercepts and transforming Points that a Polynomial
Without graphing the function, recognize the local actions of the role by detect the maximum number of x-intercepts and transforming points because that fleft(x ight)=-3x^10+4x^7-x^4+2x^3.
Show Solution
The polynomial has a level of 10, so there space at many 10 x-intercepts and at most 10 – 1 = 9 turning points.
The following video clip gives a 5 minute class on just how to recognize the number of intercepts and transforming points the a polynomial role given the degree.
### Try It
Without graphing the function, identify the maximum variety of x-intercepts and turning points because that fleft(x ight)=108 - 13x^9-8x^4+14x^12+2x^3
Show Solution
There space at many 12 x-intercepts and also at most 11 turning points.
### How To: offered a polynomial function, determine the intercepts
Determine the y-intercept by setup x=0 and also finding the corresponding output value.Determine the x-intercepts by setup the duty equal come zero and solving because that the intake values.
## Using the principle of Zero products to uncover the roots of a Polynomial in Factored Form
The rule of Zero products states the if the product of n numbers is 0, climate at the very least one that the factors is 0. If ab=0, climate either a=0 or b=0, or both a and b space 0. Us will usage this idea to uncover the zeros that a polynomial that is either in factored form or have the right to be created in factored form. For example, the polynomial
P(x)=(x-4)^2(x+1)(x-7)
is in factored form. In the complying with examples, us will present the process of factoring a polynomial and also calculating that is x and y-intercepts.
### Example: determining the Intercepts of a Polynomial Function
Given the polynomial duty fleft(x ight)=left(x - 2 ight)left(x+1 ight)left(x - 4 ight), written in factored type for her convenience, recognize the y and x-intercepts.
Show Solution
The y-intercept occurs when the entry is zero, so instead of 0 for x.
eginarraylfleft(0 ight)=left(0 - 2 ight)left(0+1 ight)left(0 - 4 ight)hfill \ extfleft(0 ight)=left(-2 ight)left(1 ight)left(-4 ight)hfill \ extfleft(0 ight)=8hfill endarray
The y-intercept is (0, 8).
The x-intercepts take place when the output f(x) is zero.
0=left(x - 2 ight)left(x+1 ight)left(x - 4 ight)
eginarrayllllllllllllx - 2=0hfill & hfill & extorhfill & hfill & x+1=0hfill & hfill & extorhfill & hfill & x - 4=0hfill \ extx=2hfill & hfill & extorhfill & hfill & ext x=-1hfill & hfill & extorhfill & hfill & x=4 endarray
The x-intercepts room left(2,0 ight),left(-1,0 ight), and left(4,0 ight).
We can see these intercepts ~ above the graph that the role shown below.
### Example: determining the Intercepts that a Polynomial role BY Factoring
Given the polynomial duty fleft(x ight)=x^4-4x^2-45, identify the y and x-intercepts.
Show Solution
The y-intercept occurs once the input is zero.
eginarrayl \ fleft(0 ight)=left(0 ight)^4-4left(0 ight)^2-45hfill hfill \ extfleft(0 ight)=-45hfill endarray
The y-intercept is left(0,-45 ight).
The x-intercepts occur when the output is zero. To recognize when the calculation is zero, we will require to variable the polynomial.
eginarraylfleft(x ight)=x^4-4x^2-45hfill \ fleft(x ight)=left(x^2-9 ight)left(x^2+5 ight)hfill \ fleft(x ight)=left(x - 3 ight)left(x+3 ight)left(x^2+5 ight)hfill endarray
Then collection the polynomial duty equal to 0.
0=left(x - 3 ight)left(x+3 ight)left(x^2+5 ight)
eginarraylllllllllx - 3=0hfill & extorhfill & x+3=0hfill & extorhfill & x^2+5=0hfill \ extx=3hfill & extorhfill & extx=-3hfill & extorhfill & ext(no genuine solution)hfill endarray
The x-intercepts are left(3,0 ight) and left(-3,0 ight).
We can see these intercepts top top the graph of the role shown below. We can see that the function has y-axis the opposite or is even because fleft(x ight)=fleft(-x ight).
## The totality Picture
Now we can carry the two ideas of transforming points and also intercepts together to obtain a general snapshot of the behavior of polynomial functions. These types of analyses top top polynomials occurred before the arrival of mass computer as a means to quickly understand the general habits of a polynomial function. Us now have a fast way, through computers, come graph and calculate important attributes of polynomials that when took a many algebra.
In the an initial example, us will determine the least degree of a polynomial based upon the variety of turning points and intercepts.
### Example: illustration Conclusions around a Polynomial role from that is Graph
Given the graph that the polynomial duty below, identify the least feasible degree of the polynomial and whether it is even or odd. Use end behavior, the variety of intercepts, and the number of turning points to aid you.
Show Solution
The end habits of the graph tells united state this is the graph of an even-degree polynomial.
The graph has 2 x-intercepts, arguing a degree of 2 or greater, and 3 turning points, saying a level of 4 or greater. Based upon this, it would certainly be reasonable come conclude that the level is even and at least 4.
Now you try to identify the least feasible degree the a polynomial offered its graph.
### Try It
Given the graph of the polynomial duty below, determine the least feasible degree of the polynomial and whether that is also or odd. Use finish behavior, the number of intercepts, and the number of turning point out to assist you.
Show Solution
The end behavior indicates one odd-degree polynomial function; there space 3 x-intercepts and 2 transforming points, for this reason the level is odd and also at least 3. Since of the end behavior, we understand that the leading coefficient should be negative.
Now we will show that friend can additionally determine the least feasible degree and variety of turning points of a polynomial role given in factored form.
### Example: illustration Conclusions around a Polynomial role from the Factors
Given the function fleft(x ight)=-4xleft(x+3 ight)left(x - 4 ight), determine the regional behavior.
Show Solution
The y-intercept is discovered by examining fleft(0 ight).
eginarraycfleft(0 ight)=-4left(0 ight)left(0+3 ight)left(0 - 4 ight)hfill hfill \ extfleft(0 ight)=0hfill endarray
The y-intercept is left(0,0 ight).
The x-intercepts are discovered by setup the duty equal come 0.
0=-4xleft(x+3 ight)left(x - 4 ight)
eginarraylllllllllllllll-4x=0hfill & hfill & extorhfill & hfill & x+3=0hfill & hfill & extorhfill & hfill & x - 4=0hfill \ x=0hfill & hfill & extorhfill & hfill & extx=-3hfill & hfill & extorhfill & hfill & extx=4endarray
The x-intercepts space left(0,0 ight),left(-3,0 ight), and left(4,0 ight).
The degree is 3 so the graph contends most 2 turning points.
Now that is your revolve to recognize the local behavior of a polynomial provided in factored form.
### Try It
Given the role fleft(x ight)=0.2left(x - 2 ight)left(x+1 ight)left(x - 5 ight), identify the regional behavior.
Show Solution
The x-intercepts are left(2,0 ight),left(-1,0 ight), and left(5,0 ight), the y-intercept is left(0, ext2 ight), and the graph contends most 2 transforming points.
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# chapter1 - Chapter 1 Functions Limits and Continuity 1.1...
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Chapter 1. Functions: Limits and Continuity 1.1 Functions It is common that the values of one variable depend on the values of another. E.g. the area A of a re- gion on the plane enclosed by a circle depends on the radius r of the circle ( A = πr 2 , r > 0 . ) Many years ago, the Swiss mathematician Euler invented the symbol y = f ( x ) to denote the statement that y is a function of x ”. A function represents a rule that assigns a unique value y to each value x. We refer to x as the independent variable and y the dependent variable.
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2 MA1505 Chapter 1. Functions: Limits and Continuity One can also think of a function as an input-output system/process: input the value x and output the value y = f ( x ) . (This becomes particularly useful when we combine or composite functions together.) Unless otherwise stated, in this course, we are only concerned with real values (or real numbers). The whole collection of real numbers is denoted by R . So for our functions, the values of x and y belong to R . 1.1.1 Domain and Range There may be constraints on the possible values of x , e.g. y = 1 /x , where we require that x 6 = 0 . The collection D in which x takes values is called the domain of the function f . 2
3 MA1505 Chapter 1. Functions: Limits and Continuity Symbolically, we write f : D -→ R x 7-→ y = f ( x ) . The R appearing on the right side of the arrow in the above notation is called the codomain of f . It indicates that the output values of this function are real numbers. On the other hand, the actual collection of y values, where y = f ( x ) and x is allowed to take the values in D , is known as the range of f (denoted by R .) E.g. the range of the function f = 1 /x is given by R = R - { 0 } . 3
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4 MA1505 Chapter 1. Functions: Limits and Continuity 1.1.2 Example The function A : R 0 R ; r 7→ πr 2 (also written as A ( r ) = πr 2 ) gives the area of a circle as a function of its radius; e.g., A (2) = 4 π . It is clear that the range of this function is the set of all nonnegative numbers R 0 . What about A ( - 2) =?. This is not defined since - 2 lies outside the domain of A . 1.1.3 Interval Notation For the next few examples, we shall use the interval notation . Let a and b be two real numbers with a < b . Then the interval notation refers to the following: 4
5 MA1505 Chapter 1. Functions: Limits and Continuity
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# What is a system of equations? A system of equations is when you have two or more equations using the same variables. The solution to the system.
## Presentation on theme: "What is a system of equations? A system of equations is when you have two or more equations using the same variables. The solution to the system."— Presentation transcript:
What is a system of equations? A system of equations is when you have two or more equations using the same variables. The solution to the system is the point that satisfies ALL of the equations. This point will be an ordered pair. When graphing, you will encounter three possibilities.
N UMBER OF S OLUTIONS OF A L INEAR S YSTEM I DENTIFYING T HE N UMBER OF S OLUTIONS y x y x Lines intersect one solution Lines are parallel no solution y x Lines coincide infinitely many solutions
Intersecting Lines The point where the lines intersect is your solution. The solution of this system is (1, 2) (1,2)
Parallel Lines These lines never intersect! Since the lines never cross, there is NO SOLUTION! Parallel lines have the same slope with different y-intercepts.
Coinciding Lines These lines are the same! Since the lines are on top of each other, there are INFINITELY MANY SOLUTIONS! Coinciding lines have the same slope and y-intercepts.
Solving a system of equations by graphing. Let's summarize! There are 3 steps to solving a system using a graph. Step 1: Graph both equations. Step 2: Do the graphs intersect? Step 3: Check your solution. Graph using slope and y – intercept or x- and y-intercepts. Be sure to use a ruler and graph paper! This is the solution! LABEL the solution! Substitute the x and y values into both equations to verify the point is a solution to both equations.
1) Solve the system of equations: 2x + y = 4 x - y = 2 Graph both equations. I will graph using x- and y-intercepts (plug in zeros). Graph the ordered pairs. 2x + y = 4 (0, 4) and (2, 0) x – y = 2 (0, -2) and (2, 0)
Graph the equations. 2x + y = 4 (0, 4) and (2, 0) x - y = 2 (0, -2) and (2, 0) Where do the lines intersect? (2, 0) 2x + y = 4 x – y = 2
Check your answer! To check your answer, plug the point back into both equations. 2x + y = 4 2(2) + (0) = 4 x - y = 2 (2) – (0) = 2 Nice job…let’s try another!
2) Solve the system of equations: y = 2x – 3 -2x + y = 1 Graph both equations. Put both equations in slope-intercept or standard form. I’ll do slope-intercept form on this one! y = 2x – 3 y = 2x + 1 Graph using slope and y-intercept
Graph the equations. y = 2x – 3 m = 2 and b = -3 y = 2x + 1 m = 2 and b = 1 Where do the lines intersect? No solution! Notice that the slopes are the same with different y-intercepts. If you recognize this early, you don’t have to graph them!
y = 2x + 0 & y = -1x + 3 Slope = 2/1 y-intercept= 0 Up 2 and right 1 y-intercept= +3 Slope = -1/1 Down 1 and right 1 The solution is the point they cross at (1,2) (1,2)
y = x - 3 & y = -3x + 1 Slope = 1/1 y-intercept= - 3 y-intercept= +1 Slope = -3/1 The solution is the point they cross at (1,-2)
The solution is the point they cross at (1,2) y =-2x + 4 & y = 2x + 0 Slope = -2/1 y-intercept= 4 y-intercept= 0 Slope = 2/1
What is the solution of this system? 3x – y = 8 2y = 6x -16 1. (3, 1) 2. (4, 4) 3. No solution 4. Infinitely many solutions
Graph the system of equations. Determine whether the system has one solution, no solution, or infinitely many solutions. If the system has one solution, determine the solution.
x y The two equations in slope- intercept form are: Plot points for each line. Draw in the lines. These two equations represent the same line. Therefore, this system of equations has infinitely many solutions.
The two equations in slope- intercept form are: x y Plot points for each line. Draw in the lines. This system of equations represents two parallel lines. This system of equations has no solution because these two lines have no points in common.
BACK
Solve each system using your Nspire: 1.) x + y = -2 2x – 3y = -9 2.) x + y = 4 2x + y = 5 3.) x – y = 5 2x + 3y = 0 (-3, 1) (1, 3) (3,-2) No Solution (-2, 5) 4.) y = -x + 2 y = -x – 4 5.) x = -2 y = 5
Objective The student will be able to: solve systems of equations using elimination with addition and subtraction.
Solving Systems of Equations So far, we have solved systems using graphing and the Nspire. These notes show how to solve the system algebraically using ELIMINATION with addition and subtraction. Elimination is easiest when the equations are in standard form.
Solving a system of equations by elimination using addition and subtraction. Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. Step 3: Add or subtract the equations. Step 4: Plug back in to find the other variable. Step 5: Check your solution. Standard Form: Ax + By = C Look for variables that have the same coefficient. Solve for the variable. Substitute the value of the variable into the equation. Substitute your ordered pair into BOTH equations.
1) Solve the system using elimination. x + y = 5 3x – y = 7 Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. They already are! The y’s have the same coefficient. Step 3: Add or subtract the equations. Add to eliminate y. x + y = 5 (+) 3x – y = 7 4x = 12 x = 3
1) Solve the system using elimination. Step 4: Plug back in to find the other variable. x + y = 5 (3) + y = 5 y = 2 Step 5: Check your solution. (3, 2) (3) + (2) = 5 3(3) - (2) = 7 The solution is (3, 2). What do you think the answer would be if you solved the system by graphing? x + y = 5 3x – y = 7
1) Solve the system using elimination. The solution is (3, 2). What do you think the answer would be if you solved the system by graphing? x + y = 5 3x – y = 7
2) Solve the system using elimination. 4x + y = 7 4x – 2y = -2 Step 1: Put the equations in Standard Form. They already are! Step 2: Determine which variable to eliminate. The x’s have the same coefficient. Step 3: Add or subtract the equations. Subtract to eliminate x. 4x + y = 7 (-) 4x – 2y = -2 3y = 9 y = 3
2) Solve the system using elimination. Step 4: Plug back in to find the other variable. 4x + y = 7 4x + (3) = 7 4x = 4 x = 1 Step 5: Check your solution. (1, 3) 4(1) + (3) = 7 4(1) - 2(3) = -2 4x + y = 7 4x – 2y = -2
Which step would eliminate a variable? 3x + y = 4 3x + 4y = 6 1. Isolate y in the first equation 2. Add the equations 3. Subtract the equations 4. Multiply the first equation by -4
Solve using elimination. 2x – 3y = -2 x + 3y = 17 1. (2, 2) 2. (9, 3) 3. (4, 5) 4. (5, 4)
3) Solve the system using elimination. y = 7 – 2x 4x + y = 5 Step 1: Put the equations in Standard Form. 2x + y = 7 4x + y = 5 Step 2: Determine which variable to eliminate. The y’s have the same coefficient. Step 3: Add or subtract the equations. Subtract to eliminate y. 2x + y = 7 (-) 4x + y = 5 -2x = 2 x = -1
2) Solve the system using elimination. Step 4: Plug back in to find the other variable. y = 7 – 2x y = 7 – 2(-1) y = 9 Step 5: Check your solution. (-1, 9) (9) = 7 – 2(-1) 4(-1) + (9) = 5 y = 7 – 2x 4x + y = 5
What is the first step when solving using the elimination method? 1. Add or subtract the equations. 2. Plug numbers into the equation. 3. Solve for a variable. 4. Check your answer. 5. Determine which variable to eliminate. 6. Put the equations in standard form.
Find two numbers whose sum is 18 and whose difference 22. 20 and -2
Solving Systems of Equations So far, we have solved systems using graphing, and elimination. These notes go one step further and show how to use ELIMINATION with multiplication. What happens when the coefficients are not the same? We multiply the equations to make them the same! You’ll see…
Solving a system of equations by elimination using multiplication. Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. Step 3: Multiply the equations and solve. Step 4: Plug back in to find the other variable. Step 5: Check your solution. Standard Form: Ax + By = C Look for variables that have the same coefficient. Solve for the variable. Substitute the value of the variable into the equation. Substitute your ordered pair into BOTH equations.
1) Solve the system using elimination. 2x + 2y = 6 3x – y = 5 Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. They already are! None of the coefficients are the same! Find the least common multiple of each variable. LCM = 6x, LCM = 2y Which is easier to obtain? 2y (you only have to multiply the bottom equation by 2)
1) Solve the system using elimination. Step 4: Plug back in to find the other variable. 2(2) + 2y = 6 4 + 2y = 6 2y = 2 y = 1 2x + 2y = 6 3x – y = 5 Step 3: Multiply the equations and solve. Multiply the bottom equation by 2 2x + 2y = 6 (2)(3x – y = 5) 8x = 16 x = 2 2x + 2y = 6 (+) 6x – 2y = 10
1) Solve the system using elimination. Step 5: Check your solution. (2, 1) 2(2) + 2(1) = 6 3(2) - (1) = 5 2x + 2y = 6 3x – y = 5 Solving with multiplication adds one more step to the elimination process.
2) Solve the system using elimination. x + 4y = 7 4x – 3y = 9 Step 1: Put the equations in Standard Form. They already are! Step 2: Determine which variable to eliminate. Find the least common multiple of each variable. LCM = 4x, LCM = 12y Which is easier to obtain? 4x (you only have to multiply the top equation by -4 to make them inverses)
2) Solve the system using elimination. x + 4y = 7 4x – 3y = 9 Step 4: Plug back in to find the other variable. x + 4(1) = 7 x + 4 = 7 x = 3 Step 3: Multiply the equations and solve. Multiply the top equation by -4 (-4)(x + 4y = 7) 4x – 3y = 9) y = 1 -4x – 16y = -28 (+) 4x – 3y = 9 -19y = -19
2) Solve the system using elimination. Step 5: Check your solution. (3, 1) (3) + 4(1) = 7 4(3) - 3(1) = 9 x + 4y = 7 4x – 3y = 9
What is the first step when solving with elimination? 1. Add or subtract the equations. 2. Multiply the equations. 3. Plug numbers into the equation. 4. Solve for a variable. 5. Check your answer. 6. Determine which variable to eliminate. 7. Put the equations in standard form.
Which variable is easier to eliminate? 3x + y = 4 4x + 4y = 6 1. x 2. y 3. 6 4. 4
3) Solve the system using elimination. 3x + 4y = -1 4x – 3y = 7 Step 1: Put the equations in Standard Form. They already are! Step 2: Determine which variable to eliminate. Find the least common multiple of each variable. LCM = 12x, LCM = 12y Which is easier to obtain? Either! I’ll pick y because the signs are already opposite.
3) Solve the system using elimination. 3x + 4y = -1 4x – 3y = 7 Step 4: Plug back in to find the other variable. 3(1) + 4y = -1 3 + 4y = -1 4y = -4 y = -1 Step 3: Multiply the equations and solve. Multiply both equations (3)(3x + 4y = -1) (4)(4x – 3y = 7) x = 1 9x + 12y = -3 (+) 16x – 12y = 28 25x = 25
3) Solve the system using elimination. Step 5: Check your solution. (1, -1) 3(1) + 4(-1) = -1 4(1) - 3(-1) = 7 3x + 4y = -1 4x – 3y = 7
What is the best number to multiply the top equation by to eliminate the x’s? 3x + y = 4 6x + 4y = 6 1. -4 2. -2 3. 2 4. 4
Solve using elimination. 2x – 3y = 1 x + 2y = -3 1. (2, 1) 2. (1, -2) 3. (5, 3) 4. (-1, -1)
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## How do you find the y-intercept quickly?
To find y-intercept: set x = 0 and solve for y. The point will be (0, y). To find x-intercept: set y = 0 and solve for x. The point will be (x, 0).
### How do you find the y-intercept of a calculator?
To find the y-intercept of a line given by ax + by + c = 0 , follow these simple steps:
1. Substitute the value x = 0 into the line equation to get by + c = 0 .
2. Rearrange this equation to find the y-intercept y꜀ , as y꜀ = −c/b .
3. Verify your results using our y-intercept calculator.
#### How do you find the y-intercept manually?
Using the “slope-intercept” form of the line’s equation (y = mx + b), you solve for b (which is the y-intercept you’re looking for). Substitute the known slope for m, and substitute the known point’s coordinates for x and y, respectively, in the slope-intercept equation.
How do you find the y-intercept of a straight line?
The y-intercept ‘c’
1. The y -intercept is the point of intersection between the straight line y = mx + c , and the y -axis. The y -intercept is the value of y when x = 0 .
2. E.g.
3. To find the y intercept we substitute x = 0 into the equation.
4. So when x = 0, y = 7, giving us the coordinate (0,7) .
5. This is the y -intercept.
What is the y-intercept of a quadratic function calculator?
Y-Intercept: Every parabola has y-intercept, it is said to be the point at which the function crosses the y-axis. It is figure out by setting the x-variable in the equation to 0.
## How do you find y-intercept without B?
To find the x-intercept of a given linear equation, plug in 0 for ‘y’ and solve for ‘x’. To find the y-intercept, plug 0 in for ‘x’ and solve for ‘y’.
### How do you find the y-intercept with two points without graphing?
Solutions
1. Step 1: Find the slope using the formula:
2. Step 2: Use the slope and one of the points to find the y-intercept b:
3. Step 3: Plug the slope (m= − 3 ), and the y-intercept (b= − 5 ), into y=mx+b:
#### How do you find y-intercept given two points?
Convert to slope-intercept form y=mx+b , where m is the slope and b is the y-intercept, by solving for y .
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## Work Done by a Constant Force Along a Line
Suppose $$F$$ is a constant force. If $$F$$ acts on an object that moves in a straight line in the direction of the force is the product as in Figure 1, the work $$W$$ done by this force is the product of the force and the distance through which the force acts; that is $\text{work}=\text{Force}\times\text{displacement}$ $W=Fd$
If $$F$$ is measured in newtons (N = kg$$\cdot$$m/s$$^{2}$$) and displacement in meters, then the unit of $$W$$ is a newton-meter (N$$\cdot$$m), called a joule (J).
## Work Done by a Variable Force Along a Line
Suppose $$F$$ is a variable force, and it acts in a given direction on an object moving in this direction, which we take to be the $$x$$-axis (Figure 2). The work done by $$F(x)$$ as its point of application moves from $$x=a$$ to $$x=b$$ can be computed by integration.
If the object moves a little bit $$dx$$, then we can assume that $$F(x)$$ is a constant force in this interval, and the small work $$dW$$ done by this force is $dW=F(x)\ dx$ This is the element of work. The total work is $\bbox[#F2F2F2,5px,border:2px solid black]{W=\int_{a}^{b}dW=\int_{a}^{b}F(x)dx.}$
Example 1
Newton’s law of gravitation states that every two particles of masses $$m_{1}$$ and $$m_{2}$$ attract each other with a force $$F$$ that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers $$r$$, $F=G\frac{m_{1}m_{2}}{r^{2}},$ where $$G$$ is a universal constant called gravitational constant ($$G=6.674\times10^{-11}\text{ m}^{3}\text{kg}^{-1}\text{s}^{-2}$$). How much work is required to move $$m_{2}$$ from $$r=R_{0}$$ to $$r=R_{1}$$ if $$m_{1}$$ is fixed?
Solution 1
The element of work is \begin{aligned} dW & =Fdr\\ & =G\frac{m_{1}m_{2}}{r^{2}}dr\end{aligned} Therefore, the total work is \begin{aligned} \int_{R_{0}}^{R_{1}}dW & =\int_{R_{0}}^{R_{1}}G\frac{m_{1}m_{2}}{r^{2}}dr\\ & =Gm_{1}m_{2}\left[-\frac{1}{r}\right]_{R_{0}}^{R_{1}}\\ & =Gm_{1}m_{2}\left(\frac{1}{R_{0}}-\frac{1}{R_{1}}\right).\end{aligned}
• Gravitational Potential Energy. The work required to completely separate two particles is the opposite of the gravitational potential energy $$U$$. So if in the above example, we let $$R_{1}\to\infty$$, then $U=-W=-G\frac{m_{1}m_{2}}{R_{0}}.$
• Hooke’s law. The force required to stretch a spring $$x$$ units beyond its natural length is proportional to $$x$$ (Figure 3). That is, $f(x)=kx$ where $$k$$ is a constant called the spring constant. This is called Hooke’s law and holds as long as $$x$$ is not too large.
Example 2
The natural length of a spring is 10 cm. If we apply a force of 5 N, the length of the spring increases to 12 cm. How much work is done in stretching the spring from 12 cm to 18 cm?
Solution 2
First we need to find the spring constant. Hooke’s law states $F=kx$ Here $$F=5$$ N and $$x=2$$ m $$=0.02$$ cm. Therefore, $k=\frac{5}{0.02}=250\ \text{N/m}$ Thus $F=250x$ The work done in stretching from $$x=0.02$$ m to $$x=0.08$$ m is \begin{aligned} W & =\int_{0.02}^{0.08}250x\ dx\\ & =\left.125x^{2}\right|_{0.02}^{0.08}\\ & =125(0.08^{2}-0.02^{2})\\ & =0.75\text{ J}.\end{aligned}
Example 3
A cylindrical tank of radius $$R$$ and height $$H$$ is filled with water to a height $$D$$ (Figure 4). Find the work required to pump all of the water to the rim of the tank. Denote the weight density of water by $$\gamma$$.
Figure 4
Solution 3
Consider a layer of thickness $$dx$$ at the height $$x$$ above the bottom of the tank as shown in Figure 5. The volume of this layer is $$\pi R^{2}dx$$, so its weight is $$\gamma \pi R^{2}dx$$. The force required to lift this layer is equal to its weight. The distance through which this force must act is $H-x$, which is the distance of this layer to the rim of the tank. Therefore, the work done in pumping this later to the top of the tank is $dW=\underbrace{\gamma\pi R^{2}dx}_{\rm force}\ \underbrace{(H-x)}_{\rm distance}.$ As $$x$$ varies between $$0$$ and $$D$$, the total work done in pumping out all of the water is \begin{aligned} W & =\int_{0}^{D}dW\\ & =\int_{0}^{D}\gamma\pi R^{2}(H-x)dx\\ & =\gamma\pi R^{2}\left[Hx-\frac{x^{2}}{2}\right]_{0}^{D}\\ & =\gamma\pi R^{2}\left(HD-\frac{D^{2}}{2}\right).\end{aligned}
Figure 5
Example 4
An inverted circular cone-shaped tank with height $$H$$ and base radius $$R$$ is filled with water to a height of $$D$$ (Figure 6). Find the work required to pump all of the water to the rim of the tank. Denote the weight density of water by $$\gamma$$.
Figure 6
Solution 4
Consider a layer of thickness $$dx$$ at a height $$x$$ above the bottom of the tank as shown in Figure 7(a). This thin layer has the shape of a circular cylinder with radius $$r$$. From similar triangles (Figure 7(b)), we have $\frac{r}{R}=\frac{x}{H}\Rightarrow r=\frac{R}{H}x.$
(a) (b)
Figure 8
The volume of this layer is $$\pi r^{2}dx=\pi\left(\frac{R}{H}x\right)^{2}dx$$, so its weight is $$\gamma\pi(Rx/H)^{2}dx$$. The force required to lift this layer is equal to its weight. The distance through which this force must act is $H-x$, which is the distance of this layer to the top of the tank. The work done in pumping this thin layer to the rim of the tank is thus $dW=\underbrace{\gamma \pi\left(\frac{R}{H}x\right)^{2}dx}_{\rm force}\ \underbrace{(H-x)}_{\rm distance}.$ Because $$x$$ increases from $$0$$ to $$D$$, the total work is \begin{aligned} W & =\int_{0}^{D}dW\\ & =\gamma\pi\frac{R^{2}}{H^{2}}\int_{0}^{D}x^{2}(H-x)dx\\ & =\gamma \pi\frac{R^{2}}{H^{2}}\left[\frac{H}{3}x^{3}-\frac{1}{4}x^{4}\right]_{0}^{D}\\ & =\gamma\pi\frac{R^{2}D^{3}}{H^{2}}\left(\frac{H}{3}-\frac{D}{4}\right).\end{aligned}
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Difference between revisions of "2019 AMC 8 Problems/Problem 9"
Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are $6$ cm in diameter and $12$ cm high. Felicia buys cat food in cylindrical cans that are $12$ cm in diameter and $6$ cm high. What is the ratio of the volume one of Alex's cans to the volume one of Felicia's cans?
$\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1$
Solution 1
Using the formula for the volume of a cylinder, we get that the volume of Alex's can is $3^2\cdot12\cdot\pi$, and that the volume of Felicia's can is $6^2\cdot6\cdot\pi$. Now we divide the volume of Alex's can by the volume of Felicia's can, so we get $\frac{1}{2}$, which is $\boxed{\textbf{(B)}\ 1:2}$ ~~SmileKat32
Solution 2
The ratio of them is $1/2$, meaning that the ratio of them will be $1(2)/2(2)=\boxed{\textbf{(B)}\ 1:2}$
-Lcz
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26 February, 04:57
# You are choosing between two health clubs. Club A offers membership for a fee of \$ 28\$28 plus a monthly fee of \$ 20.\$20. Club B offers membership for a fee of \$ 19\$19 plus a monthly fee of \$ 23.\$23. After how many months will the total cost of each health club be the same? What will be the total cost for each club?
+2
1. 26 February, 06:31
0
That is to say that at 3 months, the values of the club are equal.
Club A = 268 dollars in total
Club B = 295 dollars in total
Step-by-step explanation:
The first thing is to propose an equation for each club. Let m be the number of months.
Club A = 28 + 20 * m
Club B = 19 + 23 * m
Now, to know when it is the same, we must equal the value of Club A with Club B, it would look like this
28 + 20 * m = 19 + 23 * m
By rearranging the equation we have:
23 * m - 20 * m = 28-19
3 * m = 9
m = 3
That is to say that at 3 months, the values of the club are equal.
Now to know the total value, we will do it with an annual membership, which would be a total of 12 months, therefore it is to replace in the equation of each one:
Club A = 28 + 20 * 12 = 268 dollars in total
Club B = 19 + 23 * 12 = 295 dollars in total
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# Introduction To Limits
Go back to 'LCD'
The concept of limits forms the basis of calculus and is a very powerful one. Both differential and integral calculus are based on this concept and as such, limits need to be studied in good detail.
This section contains a general, intuitive introduction to limits.
Consider a circle of radius r.
We know that the area of this circle is $$\pi {r^2}$$ How?
The ancient Greeks derived this result using the concept of limits.
To see how, recall the definition of $$\pi$$ .
\begin{align}&\pi = \frac{{{\rm{length}}\,{\rm{of}}\,{\rm{circumference}}}}{{{\rm{length}}\,{\rm{of}}\,{\rm{diameter}}}}\\&\pi = \frac{c}{d} = \frac{c}{{2r}} \\&c = 2\pi r\end{align}
With this definition in hand, the Greeks divided the circle as follows (like cutting a cake or a pie):
Now they took the different pieces of this ‘pie’ and placed them as follows:
See what happens if the number of cuts are increased
The figure on the right side starts resembling a rectangle as we increase the number of cuts to the circle. The sequence of curves that joins x to y starts becoming more and more of a straight line with the same total length $$\pi r$$ .
What happens as we increase the number of cuts indefinitely, or equivalently, we decrease $$\theta$$ indefinitely? The figure ‘almost’ becomes a rectangle, though never becoming a rectangle exactly. The area ‘almost’ becomes $$\pi r \times r = \pi {r^2}.$$
In the language of limits, we say that the figure tends to a rectangle or the area A tends to $$\pi {r^2},$$ or the limiting value of area is $$\pi {r^2}.$$
In standard terminology.
$\mathop {{\rm{lim}}}\limits_{\theta \to 0} {\rm{A}} = \pi {r^2}$
Hence, we see that a limit describes the behaviour of some quantity that depends on an independent variable, as that independent variable ‘approaches’ or ‘comes close to’ a particular value.
For example, how does \begin{align}\frac{1}{x}\end{align} behave when x becomes larger and larger? \begin{align}\frac{1}{x}\end{align} becomes smaller and smaller and ‘tends’ to 0.
We write this as
$\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0$
How does \begin{align}\frac{1}{x}\end{align} behave when x becomes smaller and smaller and approaches 0? \begin{align}\frac{1}{x}\end{align} obviously becomes larger and larger and ‘tends’ to infinity.
We write this as:
$\mathop {\lim }\limits_{x \to 0} \frac{1}{x} = \infty$
The picture is not yet complete. In the example above, x can ‘approach’ 0 in two ways, either from the left hand side or from the right hand side:
$$x \to {0^ - }$$ : approach is from left side of 0
$$x \to {0^ + }$$ : approach is from right side of 0
How do we differentiate between the two possible approaches? Consider the graph of \begin{align}f\left( x \right) = \frac{1}{x}\end{align} carefully.
As we can see in the graph above, as x increase in value or as $$x \to \infty ,\,\,f\left( x \right)$$ decreases in value and approaches 0 (but it remains positive, or in other words, it approaches 0 from the positive side)
This can be written
$\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = {0^ + }$
Similarly,
$\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = + \,\infty$
What if x approaches 0, but from the left hand side $$\left( {x \to {0^ - }} \right)?$$ From the graph, we see that as \begin{align}x \to {0^ - },\,\,\frac{1}{x}\end{align} increases in magnitude but it also has a negative sign, that is \begin{align}\frac{1}{x} \to - \,\infty .\end{align}
What if \begin{align}x \to - \,\infty ?\,\,\,\frac{1}{x}\end{align} decreases in magnitude (approaches 0) but it still remains negative, that is, \begin{align}\frac{1}{x}\end{align} approaches 0 from the negative side or \begin{align}\frac{1}{x} \to {0^ - }\end{align}
These concepts and results are summarized below:
Download SOLVED Practice Questions of Introduction To Limits for FREE
Limits
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# How to Convert a Decimal to a Fraction
Converting a decimal to a fraction isn't as hard as it looks. If you want to know how to do it, just follow these steps.
### Method 1 If the Decimal Terminates
1. 1
Write down the decimal. If the decimal terminates, then it should end after a one or several points after the decimal. Let's say you're working with the terminating decimal .325. Write it down.
2. 2
Convert the decimal to a fraction. To do this, count how many numbers there are after the decimal point. With the number .325, there are three numbers after the decimal point. So, put the number "325" over the number 1000, which is really the number 1 with three 0's after it. If you were working with the number .3, which is one number after the decimal points, then you could represent it as 3/10.
• You can also say the decimal aloud. In this case .325 = "325 thousandths." That sounds like a fraction! Write .325 = 325/1000.
3. 3
Find the greatest common factor (GCF) of the numerator and denominator of the new fraction. This is how you can simplify the fraction. Find the biggest number that divides evenly into both 325 and 1000. In this case, the GCF of both numbers is 25, because that's the largest number that goes evenly into both numbers.
• You don't have to look for the GCF right away. You can also use trial and error to simplify the fractions. For example, if you're working with two even numbers, keep dividing them by 2 until one of them becomes odd or you can't simplify further. If you're working with an even and odd number, try dividing them by 3.
• If you're working with numbers that end in a 0 or 5, divide them by 5.
4. 4
Divide both numbers by the GCF to simplify the fraction. Divide 325 by 25 to get 13 and divide 1000 by 25 to get 40. The simplified fraction is 13/40. So, .325 = 13/40.
### Method 2 If the Decimal is Periodic
1. 1
Write it down. A periodic decimal is a decimal with a repeating pattern that never ends. For example, 2.345454545 is a periodic decimal. This time, we'll be solving for x. Write x = 2.345454545.
2. 2
Multiply the number by a power of ten that would move any non-repeating part of the decimal to the left of the decimal point. In this example a single power of 10 will suffice, so write "10x = 23.45454545...." You have to do this because if you multiply the right side of the equation by 10, you have to multiply the left side of the equation by 10 too.
3. 3
Multiply the equation by another power of 10 to move more numbers to the left of the decimal point. In this example, let's multiply the decimal by 1000. Write,"1000x = 2345.45454545...." You have to do this because if you multiply the right side of the equation by 1000, you have to multiply the left side of the equation by 1000 too.
4. 4
Place the variable and constant terms over each other. This will set them up to be subtracted. Now, place the second equation over the first, so that 1000x = 2345.45454545 is lined up over 10x = 23.45454545 just as it would be in a regular subtraction problem.
5. 5
Subtract. Subtract 10x from 1000x to get 990x and subtract 23.45454545 from 2345.45454545 to get 2322. Now you have 990x = 2322.
6. 6
Solve for x. Now that you have 990x = 2322, you can find "x" by dividing both sides by 990. So, x = 2322/990.
7. 7
Simplify the fraction. Divide the numerator and denominator by any common factors. Compute the GCD of the numerator and denominator to ensure that you have fully simplified. In this example the GCD of 2322 and 990 is 18, so you can divide both 990 and 2322 by 18 to simplify the numerator and denominator of the fraction. 990/18 = 129 and 2322/18 = 129/55. Therefore, 2322/990 = 129/55. You're done.
## Tips
• Practice makes perfect.
• When first using this method, a piece of scrap paper and a good eraser is recommended.
• Once you get better, these questions should take you ten seconds or so unless you need to simplify.
• Always check your answer at the end. 2 5/8 = 2.375 seems correct. But if you get 32/1000 = .50, something went wrong.
## Warnings
• Don't simplify with the wrong numbers.
## Things You'll Need
• Pencil
• Paper
• Good eraser
• Someone to check your work
• If no one to check work, a calculator
• Scrap paper
• A good work area
## Article Info
Categories: Conversion Aids
In other languages:
Español: convertir un decimal a fracción, Italiano: Trasformare un Numero Decimale in Frazione, Português: Converter um Decimal em Fração, Deutsch: Eine Dezimalzahl in einen Bruch umwandeln, Nederlands: Een decimaal omzetten naar een breuk, Français: convertir un chiffre décimal en fraction, Русский: переводить десятичные дроби в обыкновенные, Bahasa Indonesia: Mengubah Desimal Menjadi Bentuk Pecahan, 中文: 把小数转换成分数
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How do you find the derivative of y= e^sqrt(x) ?
Sep 24, 2014
In this problem we have to use the chain rule.
$y = {e}^{\sqrt{x}} = {e}^{{x}^{\frac{1}{2}}}$, convert the square root to its rational power
Apply the chain rule and begin to simplify
$y ' = {e}^{\sqrt{x}} \cdot \left(\frac{1}{2}\right) {x}^{\frac{1}{2} - 1}$
$y ' = {e}^{\sqrt{x}} \cdot \left(\frac{1}{2}\right) {x}^{\frac{1}{2} - \frac{2}{2}}$
$y ' = {e}^{\sqrt{x}} \cdot \left(\frac{1}{2}\right) {x}^{\frac{- 1}{2}}$
$y ' = \left({e}^{\sqrt{x}} / 2\right) {x}^{\frac{- 1}{2}}$
Convert the exponents to positive numbers
$y ' = {e}^{\sqrt{x}} / \left(2 {x}^{\frac{1}{2}}\right)$
$y ' = {e}^{\sqrt{x}} / \left(2 \sqrt{x}\right)$
Rationalize
$y ' = {e}^{\sqrt{x}} / \left(2 \sqrt{x}\right) \cdot \left(\frac{\sqrt{x}}{\sqrt{x}}\right)$
$y ' = \frac{{e}^{\sqrt{x}} \sqrt{x}}{2 x}$
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# Understanding Continuous Random Variables and Normal Distribution Dive into the world of continuous random variables and normal distribution. Master essential statistical concepts, from probability density functions to real-world applications in various fields.
Now Playing:Normal distribution and continuous random variable– Example 0
Intros
1. $\cdot$ Why are z-scores needed for random continuous variables?
$\cdot$ Relationship between area under the curve and probability
$\cdot$ Leads to Standard normal distribution
Examples
Use the Z table to find:
1. the z-score when the area under the curve to the left of z is 0.3015.
2. the area from the mean to a z-score of 1.45.
3. the z-score when the area under the curve to the left of z is 0.7774.
Introduction to normal distribution
Notes
## Normal distribution and continuous random variables
Our lesson for today focuses on the topic of the normal probability distribution, and for that, let us have a review of what a probability distribution is, and the usage of continuous random variables on it, and then, jump into the normal distribution.
#### Probability distribution of a continuous random variable
A probability distribution is that which produces a graph where the vertical axis depicts the probability of a certain outcome during and experiment and not the frequency. In other words, a probability distribution is a depiction of a statistical experiment in which we are interested to know the probability for each possible outcome to occur, but not the precise outcomes from the experiment (depicted in frequency distributions).
Remember that probability refers to the chances of an outcome to occur in a random process.
When performing a random process, there are a variety of possible results (outcomes) from such experiment, but there is no way for the experimenter to know what each outcome from each trial will be; therefore, probability allows us make a guess of each possible outcome value based on the proportion of occurrence of each possible outcome in the set of all possible outcomes. Thus, probability provides the likelihood of a particular value to be the final outcome in a particular trial from the experiment.
The probability of a specific value of a random variable x to occur is mathematically defined as:
$probability \, = p(x) =$ $\large \frac{Number\;of\;successful\;outcomes}{Total\;number\;of\;possible\;outcomes}$
Therefore, we can see that a probability distribution is a tool that allows us to understand the values that a random variable may produce in a statistical experiment. But what is a random variable? And furthermore, how can it be continuous?
When recording the possible outcomes of a random experiment, each of the values produced as results throughout the experiment are what we call random variables. A random variable is formed by data points, each assuming a unique numerical value that has been produced through a trial, and which is a random result of our particular experiment; therefore, there is no way to know for sure what the result values of a random variable will be, but we can use its recorded outcomes and the probability of each to study the behaviour of a population.
For that, random variables are classified into two categories depending on the type of values they can contain: Discrete random variables and continuous random variables.
A discrete random variable is that which contains countable values: Whole numbers, integers. Therefore, discrete random variables refer to variables that deal with items that can be counted as complete units, not fractions or any infinitesimally small parts of a unit interval.
On the other hand, a continuous random variable may assume any value as long as it belongs to a particular defined interval that is being studied. Simply said, a continuous random variable can be any number within a specified interval, that includes decimal expressions or fractions; they are said to be continuous because they will contain every single value within the particular interval being studied, and so, no matter how small scale can you go within an interval, this variable is taking account of every single point within it.
For example, let us say a continuous random variable being studied by us may have values within 2 and 6, that means the variable can have an outcome equal to 3.2 or 5.23456 or 4.123 or even 2.0000000000000000001 because is still in between the values of 2 to 6!
If this same range was to be used for a discrete random variable, the possible values of such variable would only be the complete numbers: 2, 3, 4, 5, 6.
With the definition of the two types of random variables existen in statistics we can automatically understand that there will be two types of probability distributions: discrete and continuous, each of them making use of the type of random variable that gives them their name.
And so, a continuous probability distribution is the probability distribution of a continuous random variable; it describes a continuous interval of infinite possible outcomes and the probability of such to occur from all the possibilities within the continuous interval being studied. Since you can create tables and graphs with a probability distribution (just as can be done with the frequency distribution), a practical approach is to define a probability distribution as a list or graphic representation of all the possible values of a random variable and the probability of each of them occurring; thus, a probability distribution table lists the possible outcomes of the statistical experiment as classes and provides a value for the probability of each of these on the adjunct column; while a probability distribution graph can be done in the form of a histogram, or a curve.
A continuous probability distribution can also be though as a probability density function, where the area under the smooth curve of its graph is equal to 1 and the frequency of occurrence of values between any two points equals the total area under the curve between the two points on the x-axis. This is because there is an infinite amount of possible value outcomes in a continuous interval, which makes that the probability of each individual point on the interval to be an outcome goes to zero, therefore, the probability must be measured over interval pieces, and is equal to the proportion of area under the curve for that interval piece.
The most common types of continuous probability distributions are:
• The normal distribution
• The uniform probability distribution
• The exponential probability distribution
On this lesson we will focus on the first one.
#### What is a normal distribution?
A normal distribution, also called a Gaussian distribution, is the most common (and probably most important) type of continuous probability distribution that exists. Because of that, many academic texts and study materials may provide a normal distribution definition where they simply call them a continuous probability distribution.
The normal distribution allows an statistician to work with the best approximation for a random variables behavior on real life scenarios as established in the central limit theorem : as long as the sample is sufficiently large, the shape of a random variables distribution will be nearly normal. The normal distribution curve looks like:
The main characteristics of a normal probability distribution are:
• It has a bell-shaped curve (reason why many times is simply called a bell curve).
• The normal curve is symmetric with the mean of the distribution as its symmetry axis and this mean has a value that is equal to the median and mode of the distribution (so, median = mode = mean in a normal distribution!).
• The total area under the bell curve (also called a Gaussian curve) is equal to 1, then half of it is on one side of the mean value (the axis of symmetry) and half is on the other side.
• The left and right tails on the normal distribution graph never touch the horizontal axis, they extend indefinitely because the distribution is asymptotic.
• The shape of the normal distribution and its position on the horizontal axis are determined by the standard deviation and the mean. The mean sets the center point, while the bigger the standard deviation, the wider the bell curve will be.
• About 68% of the population are within 1 standard deviation of the mean.
• About 95% of the population are within 2 standard deviations of the mean.
• About 99.7% of the population are within 3 standard deviations of the mean.
#### The standard normal curve
So far, we have found that the normal distribution is out best ally when studying real life large populations, but what happens if you study different variables from a same population? For that, we have the standard normal distribution.
A standard normal distribution (also called a z-distribution) is a normal distribution which has a value of zero as its mean and a value of one as its standard deviation. The values of any normal distribution can be translated into a standard normal distribution, meaning that any distribution can be re-scaled into a curve centered in the value of zero. This process is called standardization, and the resulting values corresponding to each point in the z-distribution are named z-scores. The formula to obtain such z-scores is:
$\large z = \frac{x\;-\;\mu}{\sigma}$
$\quad$Where:
$\quad z = z -$ score or standard score
$\quad x =$ original value from the normal distribution
$\quad\mu =$ mean of the original distribution
$\quad\sigma =$ standard deviation of original distribution
How is this useful? as we said before, if researchers are studying different characteristics from a population, all of these data distributions will have different means and standard deviations, making the comparison of the values throughout the different normal probability distributions very difficult.
Why? because you would need to use calculus (integration) to calculate the probability of each piece interval in each distribution for each set of data, where their scales are different. Even when data distributions come from the same population, the difference in their mean and standard deviation complicates such comparisons. Therefore, what we can do is to re-scale each distribution into a standard normal one and then compare them.
In simple words, statisticians made the practical decision to only calculate the probabilities of different values for one normal distribution: the z-distribution. With that, they created a standard normal distribution table (or z-table) with the standard scores and the probability of each. Then, you can use this as a standard to obtain the probabilities of intervals from any other normal distribution out there (you just need to translate the original normal distribution into the standard one and then check the z-table for probabilities).
As you can see, the standard normal distribution serves to estimate the probability of events from a normal distribution and to compare more than one of these distributions with one another; thus, allowing researchers of large populations to obtain a more complete picture of the population, and making the process faster and efficient.
Next we present the z-table for you to use it in our example problems in the next section:
#### Example 1
Reading and using the Z Table, find the following:
1. the z-score when the area under the curve to the left of z is 0.3015.
2. The area from the mean to a z-score of 1.45.
3. The z-score when the area under the curve to the left of z is 0.7774.
#### Example 2
Finding Probabilities from Z-Scores
Answer the following questions based on the properties of standard normal distribution.
1. What is the probability of having a z-score that is less than 0.75?
2. What is the probability of having a z-score that is greater than -1.83?
3. What is the probability of having a z-score that is between -1.27 to 1.06?
#### Example 3
Finding Z-Scores from Areas
Answer the following questions based on the properties of standard normal distribution.
1. by Z table
2. by calculator
***Notice that we will only include the method by the Z table, but you can watch our videos for this topic where you will find the calculation using the calculator explained.
1. Find the z-score that represents the bottom 70%.
2. Find the z-score that represent the top 70%.
3. Find the z-scores that represent the top 4% and the bottom 4%.
We have arrived to the end of our lesson.
Before you go, we recommend you to take a look at these materials on the normal (Gaussian) distribution. On our next lesson, we will finally talk about the relationship of the z-scores and random continuous variables, so stay tune and enjoy!
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# Optimization
Optimization refers to the maximums or minimums of a function in calculus.
Common types of optimization problems include:
Optimization Area & Perimeter
Optimization Volume & Surface Area
Optimization of the Distance Between a Point on a Curve
First step is to create two or more equations to solve the problem.
Second step is to find the global maximum or minimum through the first derivative test.
Third step is to substitute the variable back into one of the equations to find the other variables.
## Example 1
A rectangle has its base on the x-axis and its upper two vertices on the parabola (y = 3 − x^2), as shown in the figure below. What is the largest area the rectangle can have?
### Step 1
(A = 2xy) (y = 3 – x^2) Substitute (y) into the area equation: (A = 2x(3-x^2)=6x-2x^3)
### Step 2
Maximize the area by taking the first derivative and set it equal to (0).
(A’ = 6 – 6x^2 = 0 rightarrow x = 1)
begin{array}{|c|c|c|} hline +& 0 & – \ hline & x = 1 end{array}
(A’(0) = 6 – 6(0)^2 = 6 rightarrow text{positive value})
(A’(2) = 6 – 6(2)^2 = -18 rightarrow text{negative value})
### Step 3
Substitute (x = 1) into (y)
(y = 3 – 1^2 = 2)
(text{Maximum Area} = 2xy = 2(1)(2) = 4)
## Example 2
Find the dimensions of a right circular cylindrical can (with bottom and top closed) that has a volume of 1 liter and that minimizes the amount of material used. (Note: One liter corresponds to (1000 mathrm{cm}^3).
### Step 1
(V = pi r^2h = 1000 mathrm{cm}^3 rightarrow h = frac {1000} {pi r^2} )
(text{Surface Area} = S_text{top} + S_text{bottom} + S_text{side})
(text{Surface Area} = pi r^2 + pi r^2 + 2pi r h)
### Step 2
Minimize the surface area
(text{Surface Area} = pi r^2 + pi r^2 + 2pi r h)
(= pi r^2 + pi r^2 + 2pi r (frac {1000} {pi r^2}))
(= 2 pi r^2 + frac {2000} {r} )
(S’ = 4pi r – frac {2000} {r^2})
(= frac {4pi r^3 – 2000} {r^2} = 0 )
(4pi r^3 – 2000 = 0 rightarrow r^3 = frac {2000} {4pi})
(r = 5.42)
begin{array}{|c|c|c|} hline -& 0 & + \ hline & r = 5.42 end{array}
(S’(-1) rightarrow text{negative value})
(S’(6) rightarrow text{positive value})
### Step 3
Substitute back in (r = 5.42) to Volume equation to get the value of (h)
(h = frac {1000} {pi (5.42)^2} = 10.84 mathrm{cm} )
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# LINES
LINES. The gradient or gradient of a line is a number that tells us how “steep” the line is and which direction it goes. This one has the greatest gradient. If you move along the line from left to right and are climbing, it is a positive gradient. This one has the smallest gradient. gradient.
## LINES
E N D
### Presentation Transcript
1. LINES
2. The gradient or gradient of a line is a number that tells us how “steep” the line is and which direction it goes. This one has the greatest gradient If you move along the line from left to right and are climbing, it is a positive gradient. This one has the smallest gradient gradient These are all positive gradients. The “steeper” the line, the larger the gradient value.
3. We compute the gradient by taking the ratio of how much the line rises (goes up) and how much the line runs (goes over) We could compute the run by looking at the difference between the x values. run (x2, y2) y2 - y1 If we took two points on the line, we could compute the rise by looking at the difference between the y values. rise (x1, y1) x2 - x1 So the gradient or gradient is the rise over the run This is the gradient formula gradient is designated with an m
4. This one has the greatest absolute value gradient If you move along the line from left to right and are descending, it is a negative gradient. This one has the smallest absolute value gradient These are all negative gradients. The “steeper” the line, the larger the absolute value of the gradient. (basically this means if you ignore the negative, the larger the number, the steeper the line---but in the negative gradient direction). Gradient
5. Let’s figure out the gradient of this line. We know it should be a positive number. (2, 4) Choose two points on the line. 1 (1, 2) The rise over the run is 2 over 1 which is 2. Let’s compute it with the gradient formula. 2 (0, 0) (-2, -4) What if we'd chosen two different points on the line? It doesn't matter which two points we pick, we'll always get a constant ratio of 2 for this line.
6. If we look at any points on this line we see that they all have a y coordinate of 3 and the x coordinate varies. Let's choose the points (-4, 3) and (2, 3) and compute the gradient. (-4, 3) (2, 3) (-1, 3) This makes sense because as you go from left to right on the line, you are not rising or falling (so zero gradient). The equation of this line is y = 3 since y is 3 everywhere along the line. In general, the equation of a horizontal line is y = b, where b is the y coordinate of any point on the line. In general, the equation of a horizontal line is y = b, where b is the y coordinate of any point on the line.
7. If we look at any points on this line we see that they all have a x coordinate of - 2 and the y coordinate varies. Let's choose the points (-2, 3) and (-2, - 2) and compute the gradient. (-2, 3) (-2, 0) (-2, -2) Dividing by 0 is undefined so we say the gradient is undefined. You can't go from left to right on the line since there isn't a left and right. The equation of this line is x = - 2 since x is - 2 everywhere along the line. In general, the equation of a vertical line is x = a, where a is the x coordinate of any point on the line. In general, the equation of a vertical line is x = a, where a is the x coordinate of any point on the line.
8. positive gradient It is easy to remember undefined gradient because you can’t move along from left to right (it is vertical) negative gradient undefined gradient zero gradient (or no gradient) It is easy to remember 0 gradient because the line does not slope at all (it is horizontal)
9. We often have points on a line but want to find an equation of the line. We'll see how to do this by looking at an example. Find the equation of a line the contains the points (- 2, 4) and (2, - 2). First let's plot the points and graph the line. Now let's compute the gradient---we know it will be negative by looking at the line. (x, y) Pick a general point on the line, (x, y). (2, -2) Use the point (2, -2) and this general point in the gradient formula subbing in the gradient we found. This is an equation for the line. Let's get it in a neater form. If we get rid of brackets and fractions and get the x and y on one side (with positive x term) and constants on the other side we'll have standard form.
10. If we get rid of brackets and fractions and get the x and y on one side (with positive x term) and constants on the other side we'll have standard form. get rid of brackets get rid of fractions by multiplying by - 2 get the x and y terms on one side (with positive x term) general or standard form constants on the other side Choose any x and sub it in this equation and solve for y and you will get a point (x, y) that lies on the line. x = 0 (0, 1) is on the line
11. Let's generalize what we did to get a formula for finding the equation of a line. Let's call the specific point we know on the line (x1, y2). (x - x1) Multiply both sides by x - x1 (x - x1) rearranging a bit we have: This is called the point-gradient formula because it will find the equation of a line when you have a specific point (x1, y2)on the line and the gradient. We can also use it when we know two points on the line because we could find the gradient first and then use one of the points for the specific point.
12. Remember that gradient is the change in y over the change in x. The gradient is 2 which can be made into the fraction -1 -1 0 0 Example when you have a point and the gradient A point on a line and the gradient of the line are given. Find two additional points on the line. +1 +2 To find another point on the line, repeat this process with your new point (0,3) (0,3) +1 +2 So this point is on the line also. You can see that this point is found by changing (adding) 2 to the y value of the given point and changing (adding) 1 to the x value. (1,1) (1,5)
13. -1 -1 0 0 A way to do the last problem using the equation of the line A point on a line and the gradient of the line are given. Find two additional points on the line. x = 0 If we find the equation of the line using the point-gradient formula, we can easily find additional points on the line by subbing in various x values and finding the y values. (1,5)
14. -1 -1 0 0 Let's take this equation and solve for y. This form of the equation is called gradient-intercept form because it contains the gradient and the y intercept of the line. gradient-intercept form
15. y intercept gradient Example of given an equation, find the gradient and y intercept Find the gradient and y intercept of the given equation and graph it. First let's get this in gradient-intercept form by solving for y. -3x +4 -3x +4 Now plot the y intercept -4 -4 Change in y Change in x From the y intercept, count the gradient Now that you have 2 points you can draw the line
16. Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. www.slcc.edu Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar www.ststephens.wa.edu.au
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Example SEQ.T.2. Suppose $10\tan(4x)=5$. Find $x$.
Solution: Since $A \ne 0$ and $B=4, C=0$, the equation can be solved with $x =\frac 1B \arctan \frac D A$ where $A=10,B=4,$ and $D=5$. So the key solution is $x =\frac 14\arctan( \frac 5{10}) =\frac 14 \arctan (\frac 12)$
Comment: We can consider the expression on the left hand side of the equation as a function of $x$ giving $f(x) = 10\sin(4x)$ . Now the problem can be restated: to find a $x$ where $f(x) = 5$. This problem and its solution can be visualized both on the graph and the mapping diagram for the function $f$.
For the graph of $f$: Find $y=5$ on the Y axis , then find the point(s) on the graph of $f$ with second coordinate $5$, determine it's first coordinate, and that is the key desired value for $x$. For the mapping diagram of $f$: Find $y=5$ on the target axis , then find $x$ on the source axis with the function arrow pointing to $5$. To do this, look for the point where the graph of $f$ intersects the line $y=5$. To do this,look for the arrow that joins the point on the source axis to the point $y=5$ on the target axis.
Comments: You can move the (red) point labelled x on the left axis of the mapping diagram to a position where the arrow head points to $f(x) = 5$ and the corresponding point on the graph of $f$ will move to the position where the graph of $f$ crosses the line $y = 5$.
Check the box and the diagram will show the solution on both the mapping diagram and the graph.
You can use the sliders to investigate other examples by $A$ and $B,$ as well as the value of $D$.
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# Grade 3, Unit Seven: Three-Dimensional Geometry, Multiplication & Data Analysis
In this unit your child will: recognize, describe, and compare 3-dimensional shapes find the surface area and volume of rectangular prisms (blocks) multiply 2-digit numbers by 1-digit numbers find the mode, range, and mean (average) of a set of a data Your child will learn and practice these skills by solving problems like those shown below. Keep this sheet for reference when youre helping with homework. Problem
Which block has both faces shown below? Use pictures and/or words to show how you can tell.
Students identify, describe, and classify different 3dimensional shapes in terms of their faces, edges, and vertices (corners). As they work with blocks, they are better able to visualize the shapes and analyze them mentally. Later in the unit, students calculate the surface area and volume of rectangular prisms. (The first three shapes at left are rectangular prisms.)
Face 1
Face 2
All o f the block s ha ve F ace 1, bu t the bl ock I ci rcled is the onl y o ne with a la rg e squ are face like F ace 2 . Solve 26 x 8 in two ways. 26 x 8 = 25 x 8 + 1 x 8 = 200 + 8 = 208 or 26 x 8 = 20 x 8 + 6 x 8 = 160 + 48 = 160 + 40 + 8 = 208
8
Both strategies shown here involve breaking the number 26 into more manageable parts to multiply by 8. In the first example, the student relied on a familiarity with multiplying by 25, probably based on experiences with money. In the second example, the student broke 26 into component parts based on place value. The pictures students use in this unit help them understand why strategies like these work.
26 20 6 6 x8 48
20 x 8 = 160
Bridges in Mathematics
(Parents and teachers may reproduce for classroom and home use.)
## The Math Learning Center
At the grocery store, the oranges are sold in 5-pound bags. The table below shows how many oranges were in five different bags.
bag number of oranges bag a 8 bag b 7 bag c 7 bag d 6 bag e 7
The numbers in this example are simple, but the student demonstrates a good understanding of range, mode, and mean (average). Students review these data analysis concepts and skills in this unit to prepare for Unit Eight when they will build their own model bridges, collect extensive data on those bridges, and then analyze that data.
What is the range in the number of oranges per bag? Show your work. 8 6 = 2 What is the mode for the number of oranges per bag? Explain how you know. 7, b eca use it is the m os t c omm on num be r of or a nge s i n a ba g. What is the mean (average) number of oranges per bag? 7 + 7 + 7 + 6 + 8 = 2 1 + 14 = 35 35 5 = 7 7 is als o th e m ea n n umb er of or a nge s pe r b ag .
Q: Why does this unit cover three such different topics: geometry, multiplication, and data analysis? A: The work students do finding the volume of rectangular prisms leads naturally to multiplying larger numbers, because they need to calculate, for example, how many cubic centimeters are in 8 layers of 24 cubic centimeters altogether. (See the diagram at right.) The last two lessons in the unit review data analysis skills and concepts to prepare students for the exciting work they will do building their own model bridges and collecting and analyzing extensive data on those bridges in Unit Eight.
4 cubes deep 6 cubes wide
8 cubes tall
Q: When will students learn the algorithm for multiplying larger numbers? A: The lessons on multiplying larger numbers in this unit are a preview of whats to come in Grades 4 and 5. In Grade 4 Bridges, students will continue to use pictures and a variety of strategies to multiply larger numbers. Fourth grade teachers also have the option to use supplemental materials provided with the curriculum to teach the standard algorithm if your state or district expects fourth graders to master it. The standard algorithm is explicitly taught and practiced in Grade 5 Bridges.
Bridges in Mathematics
(Parents and teachers may reproduce for classroom and home use.)
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# Maths Year 5 Summer Fractions and Percentages
Each unit has everything you need to teach a set of related skills and concepts. 'Teaching for Understanding' provides whole-class teaching and fully differentiated adult-led group activities. ‘Problem-solving and Reasoning’ develops these skills, and includes questions to enable you to assess mastery. Practice sheets ensure procedural fluency. Extra support activities enable targeted work with children who are well below ARE.
‘UNIT PLAN’ gives you a text version of all parts of the unit to use in your school planning documentation. ‘DOWNLOAD ALL FILES’ gives you that unit plan plus all of associated documents. These bulk downloads are added value for Hamilton Friends and School Subscribers.
## Unit 1 Begin to understand percentages (suggested as 3 days)
### Objectives
Begin to understand percentages
Unit 3: Id# 5779
National Curriculum
Fr (xi) (xii)
Hamilton Objectives
33. Solve problems involving fractions, decimals and percentages using known equivalences to help.
28. Identify simple fraction and percentage equivalents: 1/2 ≡ 50%, 25% ≡ 1/4 and 75% ≡ 3/4, 40% ≡ 4/10 etc.
### Teaching and Group Activities for Understanding
Day 1 Teaching
Write on the board the symbol %. Ask children where they’ve seen this. Per cent means per or out of 100. Show a blank 100-square. Each square is 1% of the whole square. Write 1%. What fraction is this? Write 1/100. Shade the top row. What fraction is shaded? In tenths? In hundredths? Continue like this.
Group Activities
-- Identify equivalent percentages, fractions and decimals shaded on 100-squares.
-- Colour some 5 × 2 rectangles according to given percentages.
Day 2 Teaching
Write on the board two headings: ‘swimming’ and ‘cycling’. Ask ten children to stand next to their favourite. What fraction prefer swimming? Show a blank 100-square, shade this number of tenths. What is this as fraction out of 100? As a percentage? Repeat with different children and different sports.
Group Activities
Use the ‘Percentage puzzles’ in-depth problem-solving investigation below as today’s group activity.
Or, use these activities:
-- Apply understanding of equivalent fractions, decimals and percentages to create a display of ‘mobiles’.
Day 3 Teaching
Display three word problems, all of which require that children recognise percentage and fraction equivalences, e.g. that 50% is equivalent to 1/2 and 25% is equivalent to 1/4. Solve the problems using equivalences.
Group Activities
-- Use blank 100-squares to help find percentages of 200 and 300
-- Find percentages of amounts. Use conversion to percentages to compare fractions with different denominators.
-- Find any percentage of an amount by using ‘key’ percentages as a starting point.
### You Will Need
• Whiteboards and pens
• ‘Blank 100 grid’ (see resources)
• ‘Percentages’ sheet (see resources)
• Flipchart and pens
• Squared paper and coloured pencils
• ‘Mobile circles’ (see resources) printed on coloured card
• String, pens, glue and scissors
• Additional activity sheets (see resources)
### Mental/Oral Maths Starters
Day 1
Divide by 10, 100 and 1000 (pre-requisite skills)
Day 2
Percentages (pre-requisite skills)
Suggested for Day 3
Multiply three numbers together (simmering skills)
### Procedural Fluency
Day 1
Write the percentage of each 100-square that is shaded, then an equivalent vulgar and decimal fraction.
Day 2
Convert fractions to percentages.
Day 3
Find percentages of amounts: multiples of 100.
Find percentages of amounts: multiples of 10 and 100.
### Mastery: Reasoning and Problem-Solving
• Explain what 10% means. Is 10% larger or smaller than 1/5?
• Write the missing percentage in each of the bar models:
Diagram 1:
60 Children Walk 20% Cycle 50% Drive ?
Diagram 2:
24 Children Crisps 75% Fruit ?
• Now write the numbers of children in each category.
• Write 10% of each price:
£3.40 £5.10 £12 £9.90
Now work out 40% of each price.
• Aayush scored 13/20 on a test. Samar scored 70% on the same test. Which of them did better?
In depth Investigation: Percentage Puzzles
Children work systematically to find equivalent percentages of different amounts.
### Extra Support
Go for 100%
Understanding what a percentage is and knowing equivalent percentages for half and tenths
## Unit 2 Add/subt fractions with related denominators (suggested as 3 days)
### Objectives
Add and subtract fractions with related denominators
Unit 5: ID# 5801
National Curriculum
Fr (ii) (iv) (iii)
Hamilton Objectives
25. Recognise mixed numbers and improper fractions and convert from one to the other, writing mathematical statements.
26. Add and subtract fractions where the denominators are multiples of the same number.
23. Identify, name and write equivalent fractions.
### Teaching and Group Activities for Understanding
Day 1 Teaching
In a school, 1/4 of children voted for cycling as their favourite sport, 3/8 voted for swimming, the rest voted for football. Which was the most popular? Model how to use fraction equivalences (fraction families) to solve this. Repeat for comparing other pairs of fractions, e.g. 7/10 and 3/5, 5/9 and 2/3.
Group Activities
-- Use a fraction wall to help compare two fractions with related denominators. Convert improper fractions to mixed numbers.
-- Play a game comparing fractions that have related denominators.
Day 2 Teaching
Display ‘Adding and subtracting fractions’ (see resources). Children work in pairs to add two fractions with the same denominator. Then find two fractions with different, but related, denominators to add by converting one so they have the same denominator. Model a strategy for 4/5 + 7/10 (total >1), converting the resulting improper fraction to a mixed number. Children try 3/4 + 5/8. Perform subtractions using similar strategies.
Group Activities
Use the in-depth problem-solving investigation ‘A4 Fraction Addition’ and/or ‘A4 Fraction Subtraction’ from NRICH as today’s group activity.
Or, use this activity:
-- Whole class activity: Find the total of, and difference between, two fractions with related denominators.
Day 3 Teaching
Write 3¹/2 + 2¹/4. We add the 3 and 2, then add the fractions, then combine the answers. Ask children to do this and agree the answer 5³/4. Repeat for 2³/4 + 1¹/2. This time we’ve got another whole to add from adding the fractions. Repeat to subtract mixed numbers: 3⁷/8 – 2¹/4.
Group Activities
-- Add and subtract mixed numbers (fractions in some additions total >1).
-- Subtracting mixed numbers which need converting to improper fractions first, e.g. 3¹/2 – 2³/4.
### You Will Need
• Mini-whiteboards and pens
• Fraction wall (see resources)
• flipchart and pens
• Number cards 1–9
• Adding and subtracting fractions (see resources)
• Adding and subtracting mixed numbers (see resources)
### Mental/Oral Maths Starters
Day 1
Equivalent fractions (pre-requisite skills)
Day 2
Count in steps of eighths (pre-requisite skills)
Day 3
Fractions with a total of 1 (pre-requisite skills)
### Procedural Fluency
Day 1
Practise comparing, then ordering groups of proper fractions; compare mixed numbers and improper fractions by converting them all to mixed numbers.
Day 2
Practise adding and subtracting fractions with related denominators, some totals > 1.
Day 3
Further practise adding and subtracting fractions with related denominators.
Practise adding and subtracting mixed numbers.
### Mastery: Reasoning and Problem-Solving
• Sunil and Zoe have picked 36 cherries in total. Sunil has 1/3 of the cherries and Zoe has 1/2. How many cherries are left for Sunil’s little brother?
• Which of these additions will have a total less than 1 whole?
2/3 + 5/6
3/4 + 3/8
1/5 + 3/10
1/3 + 5/12
• The total of each pair of fractions is 11/12. Write the missing fraction in each pair.
1/4 and ☐
☐ and 1/6
1/2 and ☐
Write each fraction in its simplest form.
In-depth Investigation: A4 Fraction Addition and/or A4 Fraction Subtraction
Explore calculating with fractions using A4 paper to model wholes and related parts. A4 Fraction Addition and/or A4 Fraction Subtraction from nrich.maths.org.
### Extra Support
Pete’s Pizza Passion
Adding and subtracting fractions with the same denominator
## Unit 3 Multiply fractions by whole numbers (suggested as 2 days)
### Objectives
Multiply fractions and mixed numbers by whole numbers
Unit 6: ID# 5807
National Curriculum
Fr (iii) (v)
Hamilton Objectives
21. Solve problems involving multiplication and division.
25. Recognise mixed numbers and improper fractions and convert from one to the other, writing mathematical statements.
33. Solve problems involving fractions.
### Teaching and Group Activities for Understanding
Day 1 Teaching
Sketch two cakes, dividing each into fifths. Two children eat 2/5 of a cake each. How much do they eat altogether? Record: 2 × 2/5 = 4/5. Two other children eat 4/5 of a cake. How much is this? Shade 4/5 of the first cake in one colour, then last 1/5 and 3/5 of the other cake in a different colour. Record 2 × 4/5 = 8/5 = 1³/5. Repeat for 2 × 2/3. Challenge children to work in pairs to double other fractions less than 1, finding answers less than one and more than 1.
Group Activities
-- Use circles divided into thirds to help write the 2/3 times table.
-- Use three 2 to 6 or 2 to 12 number cards to create whole number × fraction calculations.
Day 2 Teaching
Write on the board 3 × 2³/4. Say that three hungry children ate 2³/4 rounds of sandwiches each! Children discuss how to work out how much has been eaten all together. Take feedback. Use brackets to show how we can work out the multiplication in stages.
Group Activities
Use the ‘Fraction fireworks’ in-depth problem-solving investigation below as today’s group activity.
Or, use these activities:
-- Play a Bingo game, multiplying numbers 2 to 10 by 3/4.
-- Investigate using the digits 2, 3, 4 and 5 to make as many different multiplications of mixed numbers and whole numbers as possible.
-- Investigate a sequence of mixed number by whole number multiplications.
### You Will Need
• Mini-whiteboards and pens
• ‘2/3 times table’ activity sheet (see resources)
• Number cards 2–12
• 1–10 dice
• Fraction multiplication Bingo: 3/4 (cut sheet into four separate cards - see resources)
### Mental/Oral Maths Starters
Day 1
Convert improper fraction to mixed numbers (pre-requisite skills)
Suggested for Day 2
Equivalent fractions, decimals and percentages (simmering skills)
### Procedural Fluency
Day 1
Write the 2/5 times table using a supporting image.
Practise multiplying fractions by whole numbers.
Day 2
Further practise multiplying fractions by whole numbers.
Write the 1³/4 times table.
Practise multiplying mixed numbers by whole numbers.
### Mastery: Reasoning and Problem-Solving
• Count in steps of 4/5, starting from 4/5. What is the first number you say which is not a mixed number? What is the second number you say which is not a mixed number?
Stop counting here!
• Six children each say they want 2/3 of a pizza for a party tea. How many pizzas will Dad need to buy?
Four dogs each eat 3/5 of a tin of dog food every day. How many tins must Mum buy for a week?
• True or false?
6 × 1³/4 = 10¹/2
12 × 2¹/5 = 22²/5
9 × 3²/3 = 33¹/3
7 × 4¹/4 < 30
In-depth Investigation: Fraction Fireworks
Children find the answers to a sequence of fraction multiplications and look for patterns in the answers.
### Extra Support
Mini multiplications
Count in unit fraction steps on a number line. Write mixed numbers.
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# Math for Quantitative Finance
Take a guided tour through the powerful mathematics and statistics used to model the chaos of the financial markets.
# Matrix Operations (QF)
Let $A=\left[\begin{array}{cc} 1&-2\\ 1&3\end{array}\right]\quad\text{and}\quad B=\left[\begin{array}{cc} 2&4\\ -1&1\end{array}\right].$ $$A+B$$ contains four numbers; what is their product?
Matrix addition is almost completely analogous to addition of real numbers, but matrix multiplication is not. Matrix multiplication does not retain all the properties of scalar multiplication, and is not even always defined.
Let $A=\left[\begin{array}{cc} 1&-2\\ 1&3\end{array}\right]\quad\text{and}\quad B=\left[\begin{array}{cc} 2&4\\ -1&1\end{array}\right].$ $$AB-BA$$ contains four numbers; what is their product?
Suppose that $$A$$ is an $$a\times b$$ matrix, $$B$$ is a $$c\times d$$ matrix, and $$C$$ is an $$e\times f$$ matrix. If the product $$AB$$ is a $$2\times 4$$ matrix, and the product $$BC$$ is $$1\times 3$$ matrix, what is $$a+b+c+d+e+f$$?
An operation unique to matrices is the determinant. The determinant is especially important because matrices with zero determinant have very different properties from matrices with non-zero determinant, and many standard algorithms for working with matrices break down when the determinant is zero.
Let $A=\left[\begin{array}{cc} 1&-2\\ 1&3\end{array}\right]\quad\text{and}\quad B=\left[\begin{array}{cc} 2&4\\ -1&1\end{array}\right].$ What is the determinant of $$AB$$?
Another matrix operation is the transpose - this operation reflects the entries of a matrix across its main diagonal. This amounts to switching the row and column for each entry.
This can be useful for detecting symmetry in data. If we have some data in a matrix, and we take the transpose and compare it to the original matrix, this tells us about how close our original data was to being symmetric.
Let $A=\left[\begin{array}{cc} 1&-2\\ 1&3\end{array}\right].$ What is the transpose of $$A$$?
Let $A=\left[\begin{array}{cc} 1&-2\\ 1&3\end{array}\right].$ What is the trace of $$A$$?
These matrix operations don't always go together in the way you'd expect - many matrix calculations require successively applying these operations and simplifying the result. For example, linear regression can require looking at matrices of the form $$(AA^T)^{-1}$$, so understanding how they interact is essential.
For square matrices $$A$$ and $$B$$, which of the matrices below are equal to $$AB$$? $\text{I. } \left(B^TA^T\right)^T\qquad\text{II. } \left(A^TB^T\right)^T$
Which of the quantities below are equal to $$\text{trace}(AB)$$? $\text{I. } \text{trace}(A)\text{trace}(B)\qquad\text{II. } \text{trace}(BA)$
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# How do you differentiate ycosx^2-y^2=xy-x^2?
Jun 8, 2018
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{2 x y \setminus \sin \left({x}^{2}\right) + y - 2 x}{\setminus \cos \left({x}^{2}\right) - 2 y - x}$
#### Explanation:
Treat $y$ as a function of $x$ and use the Chain Rule whenever $y$ appears. See here for more details on implicit differentiation.
Here, differentiating both sides with respect to $x$ and taking the Product Rule and Chain Rule into account yields
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} \setminus \cos \left({x}^{2}\right) - 2 x y \setminus \sin \left({x}^{2}\right) - 2 y \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = y + x \setminus \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x$
Put all terms that contain $\setminus \frac{\mathrm{dy}}{\mathrm{dx}}$ on one side, all the other terms on the other sides:
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} \setminus \cos \left({x}^{2}\right) - 2 y \setminus \frac{\mathrm{dy}}{\mathrm{dx}} - x \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x y \setminus \sin \left({x}^{2}\right) + y - 2 x$
Factoring $\setminus \frac{\mathrm{dy}}{\mathrm{dx}}$ and dividing, we obtain
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{2 x y \setminus \sin \left({x}^{2}\right) + y - 2 x}{\setminus \cos \left({x}^{2}\right) - 2 y - x}$
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 9.14: Relation of Polygon Sides to Angles and Diagonals
Difficulty Level: At Grade Created by: CK-12
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Have you ever tried to name a figure based on angle measures? Take a look at this problem.
Travis is drawing a design for the skatepark. He has labeled all of the angles in his figure.
The sum of the angle measures is \begin{align*}900^\circ\end{align*}.
What figure is Travis drawing? Do you know how to figure this out?
This Concept will teach you how to relate the sides of a polygons to angle measures. By the end of the Concept, you will know how to help Travis figure this out.
### Guidance
We can divide polygons into triangles using diagonals. This becomes very helpful when we try to figure out the sum of the interior angles of a polygon other than a triangle or a quadrilateral.
Look at the second piece of information in this box. The sum of the interior angles of a quadrilateral is \begin{align*}360^\circ\end{align*}. Why is this important? You can divide a quadrilateral into two triangles using diagonals. Each triangle is \begin{align*}180^\circ\end{align*}, so the sum of the interior angles of a quadrilateral is \begin{align*}360^\circ\end{align*}.
Here is one diagonal in the quadrilateral. We can only draw one because otherwise the lines would cross.
A diagonal is a line segment in a polygon that joins two nonconsecutive vertices.
A consecutive vertex is one that is next to another one, so a nonconsecutive vertex is a vertex that is not next to another one.
How do we use this with other polygons?
We can divide up other polygons using diagonals and figure out the sum of the interior angles.
Here is a hexagon that has been divided into triangles by the diagonals. You can see here that there are four triangles formed. If sum of the interior angles of each triangle is equal to \begin{align*}180^\circ\end{align*}, and we have four triangles, then the sum of the interior angles of a hexagon is:
\begin{align*}4(180) = 720^\circ\end{align*}
We can follow this same procedure with any other polygon.
What if we don’t have the picture of the polygon? Is there another way to figure out the number of triangles without drawing in all of the diagonals? The next section will show you how using a formula with the number of sides in a polygon can help you in figuring out the sum of the interior angles.
To better understand how this works, let’s look at a table that shows us the number of triangles related to the number of sides in a polygon.
Do you see any patterns?
The biggest pattern to notice is that the number of triangles is 2 less than the number of sides. Why is this important? Well, if you know that the sum of the interior angles of one triangle is equal to 180 degrees and if you know that there are three triangles in a polygon, then you can multiply the number of triangles by 180 and that will give you the sum of the interior angles.
Here is the formula.
\begin{align*}x =\end{align*} number of sides
\begin{align*}(x - 2)180 =\end{align*} sum of the interior angles
You can take the number of sides and use that as \begin{align*}x\end{align*}.
Then solve for the sum of the interior angles.
Let’s try this out.
What is the sum of the interior angles of a decagon?
A decagon has ten sides. That is our \begin{align*}x\end{align*} measurement. Now let’s use the formula.
\begin{align*}(x - 2)180 & = (10 - 2)180 \\ 8(180) & = 1440^\circ\end{align*}
Our answer is that there are \begin{align*}1440^\circ\end{align*} in a decagon.
Try a few of these on your own.
#### Example A
The sum of the interior angles of a pentagon
Solution: \begin{align*}540^\circ\end{align*}
#### Example B
The sum of the interior angles of a triangle
Solution: \begin{align*}180^\circ\end{align*}
#### Example C
The sum of the interior angles of an octagon
Solution: \begin{align*}1080^\circ\end{align*}
Here is the original problem once again.
Travis is drawing a design for the skatepark. He has labeled all of the angles in his figure.
The sum of the angle measures is \begin{align*}900^\circ\end{align*}.
What figure is Travis drawing? Do you know how to figure this out?
We can figure this out by using the formula for angle measures and sides of a polygon.
\begin{align*}(x - 2)180 = 900^\circ\end{align*}
Now we can solve this just as we would an equation. Begin by dividing both sides by 180 degrees.
\begin{align*}x - 2 = 900 \div 180\end{align*}
\begin{align*}x - 2 = 5\end{align*}
Next add 2 to both sides of the equation.
\begin{align*}x - 2 + 2 = 5 + 2\end{align*}
\begin{align*}x = 7\end{align*}
Travis' figure is a heptagon with seven sides.
### Vocabulary
Here are the vocabulary words in this Concept.
Polygon
A simple closed figure formed by three or more line segments.
Pentagon
five sided polygon
Hexagon
six sided polygon
Heptagon
seven sided polygon
Octagon
eight sided polygon
Nonagon
nine sided polygon
Decagon
ten sided polygon
Regular Polygon
polygon with all sides congruent
Irregular Polygon
a polygon where all of the side lengths are not congruent
Congruent
exactly the same or equal
Diagonal
a line segment in a polygon that connects nonconsecutive vertices
Nonconsecutive
not next to each other
### Guided Practice
Here is one for you to try on your own.
What is the sum of the interior angles of a regular nonagon?
To figure this out, we can use the formula presented in the Concept.
\begin{align*}(x - 2)180\end{align*}
In this formula, the value of \begin{align*}x\end{align*} is the number of sides of the polygon.
In this case, a nonagon has 9 sides.
\begin{align*}(9 - 2)180\end{align*}
\begin{align*}(7)(180)\end{align*}
The answer is \begin{align*}1260^\circ\end{align*}.
### Video Review
Here is a video for review.
### Practice
Directions: Look at each image and name the type of polygon pictured.
1.
2.
3.
4.
5.
6.
Directions: Name the number of diagonals in each polygon.
7.
8.
9.
Directions: Use the formula to name the sum of the interior angles of each polygon.
10. Hexagon
11. Pentagon
12. Decagon
13. Pentagon
14. Octagon
15. Square
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Congruent Congruent figures are identical in size, shape and measure.
Decagon A decagon is a ten sided polygon.
Diagonal A diagonal is a line segment in a polygon that connects nonconsecutive vertices
Heptagon A heptagon is a seven sided polygon.
Hexagon A hexagon is a six sided polygon.
Irregular Polygon An irregular polygon is a polygon with non-congruent side lengths.
Nonagon A nonagon is a nine sided polygon.
Nonconsecutive Nonconsecutive means not next to each other or not in order.
Octagon An octagon is an eight sided polygon.
Pentagon A pentagon is a five sided polygon.
Polygon A polygon is a simple closed figure with at least three straight sides.
Regular Polygon A regular polygon is a polygon with all sides the same length and all angles the same measure.
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Summary
# Basic angle rules
## In a nutshell
There are five rules involving angles on lines and shapes. These rules refer to angles in a triangle, angles on a straight line, angles in a quadrilateral, angles around a point and angles in an isosceles triangle. The rules help to find missing angles on lines and in shapes.
## Angles in a triangle
The angles in a triangle add up to $180\degree$.
##### Example 1
Find the size of the missing angle in the triangle:
Angles in a triangle add up to $180\degree$
\begin{aligned}70\degree+30\degree & =100\degree \\180\degree-100\degree &=80\degree\end{aligned}
The missing angle is $\underline{80\degree}$.
## Angles on a straight line
Angles on a straight line add up to $180\degree$.
##### Example 2
Find the size of the missing angle on the straight line:
Angles on a straight line add up to $180\degree$
$180\degree - 100\degree = 80\degree$
The missing angle is $\underline{80\degree}$.
Angles in a quadrilateral add up to $360\degree$.
##### Example 3
Find the size of the missing angle in the quadrilateral shown:
Angles in a quadrilateral add up to $360\degree$. Subtract the sum of the known angles from $360\degree$
\begin{aligned}80\degree+120\degree+40\degree &= 240\degree \\ 360\degree- 240\degree &= 120\degree\end{aligned}
The missing angle is $\underline{120\degree}$.
## Angles around a point
Angles around a point add up to $360\degree$.
##### Example 4
Find the size of the missing angle:
Angles around a point add up to $360\degree$, so take $140\degree$ away from $360\degree$.
$360\degree- 140\degree= 220\degree$
The missing angle is $\underline{220\degree}$.
## Angles in an isosceles triangle
Isosceles triangles have two sides equal in length and two equal angles. The two equal angles are sometimes referred to as the 'base angles' in an isosceles triangle. If you know one of the angles in an isosceles triangle, the other two angles can be found easily.
##### Example 5
The isosceles triangle has equal sides $AC$ and $BC$. If $\angle BAC$ is $50\degree$, find the size of $\angle ACB$.
$\angle BAC$ and $\angle ABC$ are both $50\degree$, as they are the base angles of the triangle. As angles in a triangle add up to $180\degree$, you can find $\angle ACB$ by subtracting the sum of $\angle ABC$ and $\angle BAC$ from $180\degree$.
\begin{aligned}50\degree+50\degree &= 100\degree \\180\degree - 100\degree &= 80\degree\end{aligned}
$\underline{\angle\ ACB\ = 80\degree}$
## Want to find out more? Check out these other lessons!
Angle rules
Angles: types, notation and measuring
FAQs
• Question: What do angles on a straight line add up to?
• Question: What do the angles in a triangle add up to?
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# How can I prove a b c
## ABC formula or midnight formula derivation
The derivation or the proof for the midnight formula (ABC formula) can be found here. We look at this content:
• A Explanationhow to derive the ABC formula or midnight formula.
• A example how to do the derivation.
• Tasks / exercises so that you can practice using the formula.
• A Video to derive the midnight formula / ABC formula.
• A Question and answer area to this topic.
Note: If you are interested in calculating tasks with the formula, please have a look at the midnight formula or the ABC formula.
### Derivation of the midnight formula / ABC formula
How do you actually come up with formulas like the midnight formula? Somebody must have come up with this at least once. How do you get from the quadratic equation / function to the solution equation?
A few pointers for a better understanding:
1. There are several ways to derive the ABC formula (midnight formula). For this reason, other sources will have different derivations.
2. We have decided on the path that can be carried out with what we believe to be the least amount of previous knowledge.
3. However, it is not possible without prior knowledge. You should definitely be fit with the application of the binomial formulas, especially the 1st binomial formula. If you don't know these, please learn further under binomial formulas.
All right? Let's look at the derivation for the midnight formula bit by bit.
Display:
### Midnight Formula Proof
We first subtract c. This will give us -c on the right.
We multiply both sides of the equation by 4a. We thereby get:
We add b on both sides of equation2.
Does anyone see anything special in the last equation? Look closely! On the left side is the 1st binomial formula. If you don't remember this any more, you will find it in the next line:
You still have no idea about it? In this case, please first take a look at the content under binomial formulas. We apply the 1st binomial formula to the left side.
As a reminder, on the right side we still have -4ac + b2. We converted the left side:
We now apply the first binomial formula to the left side and get:
We take the roots on both sides of the equation. Note that there is a plus and a minus before the root:
We subtract b to the other side:
They still divide by 2a.
Since there are two solutions, it is usually written with x1/2.
Finished!
Show:
### Proof video ABC formula
This video provides a derivation or a proof for the midnight formula or the ABC formula. For a better understanding, an equation is also solved in order to find the corresponding zeros. The next video is from Youtube.com.
Next video ยป
### Questions with answers midnight formula
In this section, we'll look at typical questions with answers about the Midnight Formula.
Q: What additional content should I watch?
A: Take a look at these topics:
• PQ formula: The PQ formula is an alternative to the ABC formula or midnight formula. But we have an extra article on this. You can find this under the link PQ formula.
• Polynomial division: The polynomial division is also used to find zeros. We also discuss the polynomial division in detail in our article Polynomial division.
• Calculate zeros: The ABC formula or midnight formula is used to find zeros. You can find a complete article on this topic under Calculating Zeros.
Otherwise, of course, the use of the formula derived here under the midnight formula or the ABC formula.
Q: When is the Midnight Formula covered in school?
A: The midnight formula is mostly on the curriculum from 8th grade onwards.
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# 11.1 - Geometric Sequences
Document Sample
``` 11.1 – Geometric Sequence
Objectives: 1. Recognize and extend geometric sequences. 2. Find the nth term of a geometric sequence.
The table shows the heights of a bungee jumper’s bounces.
The height of the bounces shown in the table above form a
geometric sequence. In a geometric sequence, the ratio of
successive terms is the same number r, called the _______________________________________.
Example 1: Extending Geometric Sequences
Find the next three terms in the geometric sequence.
1, 4, 16, 64,…
When the terms in a geometric sequence alternate between positive and negative, the value of r is ________________.
Check It Out! Example 1a
Find the next three terms in the geometric sequence.
5, –10, 20,–40,… 512, 384, 288,…
The variable a is often used to represent terms in a sequence. The variable a4 (read “_________________”)is the fourth
term in a sequence.
Geometric sequences can be thought of as functions. The term
number, or position in the sequence, is the input of the function,
and the term itself is the ________________________of the
function. To find the output an of a geometric sequence when n is a
large number, you need an equation, or function rule. Look for a
____________________ to find a function rule for the sequence above.
The pattern in the table shows that to get the nth term, multiply the first term by the common ratio raised to the power
n – 1.
If the first term of a geometric sequence is a1, the nth term is an , and the common ratio is r, then an = a1rn–1
Example 2A: Finding the nth Term of a Geometric Sequence
The first term of a geometric sequence is 500, and the common ratio is 0.2. What is the 7th term of the sequence?
For a geometric sequence, a1 = 5, and r = 2. Find the 6th term of the sequence?
What is the 9th term of the geometric sequence 2, –6, 18, –54, …?
Check It Out! Example 2
What is the 8th term of the sequence 1000, 500, 250, 125, …?
Bounce Height
(cm)
Example 3: Application
A ball is dropped from a tower. The table shows the heights of the balls bounces, which form a 1 300
geometric sequence. What is the height of the 6th bounce?
2 150
3 75
```
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#### Functions and graphs primer
October 9, 2016
A function is a rule that maps each of the elements in a domain to one and only one element in a range. The set of all possible elements in the input is the domain D of a function; the set of all possible elements in the output is the range R of a function.
According to this trivial definition, a function takes a set of elements and applies its rule to map to another set of elements, so that each and every domain element is mapped once and only. There are, of course, cases in which an element in the range is the mapping of multiple elements in the domain (like in a constant function).
Typically, although the domain is not mentioned one should always keep in mind that the domain is the set of elements for which the function may be evaluated. There are cases in which the domain has to be carefully considered, like in fractions (remember division by zero) and in even roots (remember the elements in even roots have to be greater or equal to zero). Take a look,
$\displaystyle f(x) = \frac{2x+3}{x-1}, \; D=R-\{1\} \\ f(x) = \sqrt{x-1}, \; D=\{x \geq 1 \}$
In some cases, the domain is not restricted by the definition of the function but it might be restricted by the application (i.e. in which set of elements it might “make sense” to apply the function, in real world applications).
It is possible to combine and to construct composite functions in various forms, like,
$\displaystyle f(x) + g(x), \; f(x) \cdot g(x), \; f(g(x)) \; [\text{or } ((f \circ g)(x)]$
For various obvious (and non-obvious) reasons it s possible to visualize a function by creating graphs. Typically a graph can be considered as a visualization of the set of points created by the rule of a function in a coordinate system created by the domain and range of the function.
Simple functions of one independent variable, which do not involve powers, etc, like
$\displaystyle f(x) = y = -\frac{3}{2}x+2 \quad [\text{or in general } f(x) = y = ax + b]$
are visualized as straight lines and are called linear. In addition, in these functions it is easy to pinpoint the role of their ingredients a, b as being the slope of the line (the rate of change in the function in respect to the independent variable x) and the y-intercept (where the line crosses y-axis) respectively. It is easy to also pinpoint the x-intercept by equating the function with zero, so putting all together we have
$\displaystyle f(x) = ax + b \\ \text{the slope is } a \left[= \frac{\Delta f(x)}{\Delta x} \right] \\ \text{the y-intercept is } b \left[ = f(x=0) \right] \\ \text{the x-intercept is } f(x) = ax + b = 0 \Rightarrow x_i = -\frac{b}{a}$
Thus, for example
$\displaystyle f(x) = y = -\frac{3}{2}x+2 \quad \\ \text{has slope } a=-\frac{3}{2} \\ \text{has y-intercept } b=2 \\ \text{has x-intercept } x_i = -\frac{2}{-\frac{3}{2}} = \frac{4}{3}$
Wanna see how this function’s graph looks like? Take a look,
One may easily deduce that
$\displaystyle f(x)=c, \text{ c being a constant, is a horizontal line} \\ x=c, \text{ c being a constant, is a vertical line}$
What comes naturally after the study of linear functions is the study of quadratic or second degree functions, that is functions in which the independent variable is raised to the second power.
The standard form of the quadratic functions is
$\displaystyle f(x) = ax^2 + bx + c, \quad a,b,c \in R, \; a \neq 0$
How do quadratic functions look like in graphs? Well, they are parabolas.
If a is positive then the parabola opens up (has a lowest point). If a is negative then the parabola opens down (has a highest point). This lowest or highest point is usually called the vertex. The y-intercept is the constant c,
$\displaystyle f(x=0) = ax_{x=0}^2 + bx_{x=0} + c \Rightarrow f(x=0) = y_i = c$
The x-intercepts can be calculated,
$\displaystyle f(x) = 0 \Rightarrow ax^2 + bx + c = 0 \Rightarrow x_i = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
which gives the obvious hint that one quadratic function might have none, one, or at most two x-intercepts!
In addition, the parabola that represents a quadratic function is symmetrically horizontally mirrored around a vertical line that crosses the vertex of the parabola! But where is this vertex? Here it is
$\displaystyle x_v = -\frac{b}{2a}, \; y_v = f\left(\frac{-b}{2a}\right) = \frac{4ac-b^2}{4a}$
*** The computation of the vertex is easy if one considers that this is an extremal of the function and the function is differentiable, thus
$\displaystyle \dot{f} = 2ax+b \xrightarrow[extremals]{for} \dot{f} = 0 \Rightarrow 2ax+b=0 \Rightarrow x = -\frac{b}{2a}$
…but this is a matter of another post regarding differentiation and derivatives.
How do these functions look like in graphs? Here, take a look
Increasing the power of the independent variable leads to third degree or cubic functions.
The standard form of the cubic functions is
$\displaystyle f(x) = ax^3 + bx^2 + cx + d, \quad a,b,c,d \in R, \; a \neq 0$
How do cubic functions look like in graphs? Well, they look like recumbent sigmoids S-shapes lying down…
If a is positive then the curve’s right-hand tail opens up, whereas if a is negative then the curve’s right-hand tail opens down. The y-intercept is the constant d,
$\displaystyle f(x=0) = ax_{x=0}^3 + bx_{x=0}^2 + cx + d \Rightarrow f(x=0) = y_i = d$
The x-intercepts can be calculated,
$\displaystyle f(x) = 0 \Rightarrow ax^3 + bx^2 + cx + d = 0 \Rightarrow x_k = -\frac{1}{3a} \left ( b + \zeta^k C + \frac{\Delta_0}{\zeta^k C} \right ), \\ k \in \{0,1,2\}, \; \zeta = -\frac{1}{2} + \frac{1}{2} \sqrt{3} i, \; i \text{ being the complex numbers symbol} \\ \Delta_0 = b^2 -3ac, \; \Delta_1 = 2 b^3 -9abc + 27 a^2 d, \; C = \sqrt[3]{\frac{\Delta_1 \pm \sqrt{\Delta_1^2 -4 \Delta_0^3}}{2}}$
which gives the obvious hint that one cubic function might have none, one, two, or at most three x-intercepts! Since this is extremely tedious, we usually employ factorization (if possible to be easily employed). We would even be really happy if the function was initially given in a factorized form like say,
$\displaystyle f(x) = (x+2)(x-3)(x+4)$
which otherwise would have been given as,
$\displaystyle f(x) = x^3 + 3x^2 -10x -24$
and it would be really messy…
Following the latter representation we readily know that the curve of this function opens up at the right-hand side (a>0) and that the y-intercept is
$\displaystyle f(x=0) = y_i = -24$
Following the factorized representation we readily know the x-intercept points, which are, in this case, three,
$\displaystyle x_k = \{ -4, -2, 3 \}$
And here is how this thing shows up in a graph
Apparently, the third-degree functions are the last of those functions for which to conclude about their appearance in a graph just by looking at their definition (and doing some simple calculations)
The whole concept of such functions generalized to what is called the polynomial functions.
Polynomial functions are the generalized form of n-th degree functions.
The standard form of the n-th degree polynomial functions is
$\displaystyle f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \\ n \in N, \; a_0,\cdots ,a_n \in R, \; a_n \neq 0$
When moving to degrees higher than 4, although there are some clues on how some of them might look there is no formal or informal way to picture them. The things we know is that polynomial functions represent smooth curves without sharp corner edges and breaks.
For example, here is a graph (x-axis is scaled for better illustration) of a fifth degree polynomial function that is defined as
$\displaystyle f(x) = (x+5)(x-1)(x-2)(x+3)(x-4)$
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# 2 Z Tables and how to use them correctly
The Standard Normal Table is also commonly know as the Z table. It is used in the calculation of probability on the LEFT side of the normal distribution. To use the Table Calculate the Z value using the formula:
## Understanding the table
When looking at a Z table we should first focus on the LEFT COLUMN and TOP ROW. The left column has the Z value up to the first decimal and the Top row has the second decimal of the Z value. For example, If you have a Z value of 2.34, then 2.3 will be in the left column and 0.04 will be in top row. The probability associated with 2.34 will be at the intersection of these values. In this case probability of the Z value 2.34 will be 0.9904 from the Z table. This is the area under the curve of the Normal Distribution on the LEFT side of the given Z value.
## Negative Z table
The Standard Normal Distribution has a mean of 0 and a standard deviation of 1. This makes the range of the Z values from -3.99 to +3.99. The table is thus divided in two parts. from -3.99 to 0 and 0 to 3.99. The part of the table from -3.99 to 0 is called as the Negative Z table and should be used when the calculated Z value is negative. The Negative Z table is given below.
## Positive Z Table
Use the Positive table when the calculated Z value is between 0 and +3.99. The table is given below.
## Practice with Z table
To become familiar with using a Z table find the probability values of the following Z values.
1. 2.68
2. 1.94
3. – 0.08
4. – 2.71
5. – 1.64
6. 2.30
7. 1.23
8. -2.22
9. 0.67
10. 2.00
The probability values that we get from the Z table for the above examples are as follows
1. 0.9963 by using the positive table
2. 0.9738 by using the positive table
3. 0.4681 by using the negative table
4. 0.0034 by using the negative table
5. 0.0505 by using the negative table
6. 0.9893 by using the positive table
7. 0.8907 by using the positive table
8. 0.0132 by using the negative table
9. 0.7486 by using the positive table
10. 0.9772 by using the positive table
The Z table is used for solving problems on normal distribution. Learn more about the Normal Distribution Here.
You can find some solved examples of normal distribution problems with Z table here.
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# Subtraction: 3-digit
Contributor: Erika Wargo. Lesson ID: 12215
Is there a "difference" between addition and subtraction? If you can subtract 1-digit numbers can you subtract 3-digit numbers? Learn to use subtraction in addition to addition on big number problems!
categories
## Arithmetic, Whole Numbers and Operations
subject
Math
learning style
Auditory, Visual
personality style
Otter
Primary (K-2), Intermediate (3-5)
Lesson Type
Skill Sharpener
## Lesson Plan - Get It!
Audio:
The new bike you want costs \$250 and you have \$116 in your bank account. How much more money do you need before you can buy the bike?
Did you know that subtraction is the inverse, or opposite of, addition?
The answer to a subtraction problem is called the "difference." When we subtract whole numbers, we align the digits by place value. Subtract the bottom number from the top number and regroup when necessary. But it is important to remember that order matters when we subtract. The greater number needs to be the starting number. If you are subtracting money, remember to line up the decimals and place values.
Watch Subtraction with Regrouping Song by NUMBEROCK to review how to regroup in subtraction:
After the video, discuss with a parent or teacher:
• What does regrouping mean?
• How do you know if you need to regroup?
To subtract whole numbers:
1. Rewrite the problem vertically and line up the place values and decimal points. Be sure the greater number is the top number.
2. Begin in the ones place and subtract. Regroup if needed.
3. Move to the left and subtract the tens and hundreds.
4. Check your work by "Adding up." Add the two bottom numbers to be sure the sum equals the top number.
\$250 - \$116 =
Rewrite the problem, with the greater number on top:
#### 6
Subtract and regroup as needed:
#### 4
"Add up" and check your work. \$250 is the number you started with, so you are correct!
#### 0
Discuss with a parent or teacher how you know if you need to regroup. Explain how you can use a model or different strategy to help with subtraction. What happens if you don’t regroup?
In the Got It? section, you will practice three-digit subtraction with games!
## Elephango's Philosophy
We help prepare learners for a future that cannot yet be defined. They must be ready for change, willing to learn and able to think critically. Elephango is designed to create lifelong learners who are ready for that rapidly changing future.
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Home » Math Theory » Fractions » Improper Fractions
Improper Fractions
Introduction
Numbers that represent a portion of a whole are called fractions. Let us use the square below as an example.
Since it has been cut into 4 equal parts, each piece represents one out of four. Mathematically, each piece is described as $\frac{1}{4}$. The number at the top is referred to as the numerator, and its bottom number is the denominator. $\frac{1}{4}$ is an example of a proper fraction because the denominator is greater than the numerator.
Otherwise, we will have an improper fraction when the numerator exceeds the denominator. Let us suppose that we instead have two squares and that each square is divided into four equal parts, as illustrated below.
There are 7 yellow parts out of 8 equal parts, which can be expressed as $\frac{7}{8}$. This number is an example of an improper fraction.
The topic of improper fractions will be explored in this article, along with methods for converting them to mixed fractions, simplifying them, and finding solutions to problems involving them.
What is an Improper Fraction?
Definition
When the numerator exceeds the denominator, the fraction is called improper. Suppose $\frac{a}{b}$ is an improper fraction; thus, a must be greater than b. Examples of improper fractions are $\frac{4}{3}$, $\frac{7}{4}$, $\frac{8}{5}$, and $\frac{11}{2}$.
A fraction has two main types, proper fractions and improper fractions. We have an improper fraction when the numerator is higher than the denominator; otherwise, it is a proper fraction.
Simplifying Improper Fractions
Simplifying improper fractions is the same thing as reducing the fractions. We do this by dividing the denominator and the numerator by their greatest common factor
Let us say, for example; we have the improper fraction $\frac{18}{12}$. The numerator in the fraction is 18, while the denominator is 12. These are the factors of 18 and 12.
Factors of 18: { 2, 3, 6, 9 }
Factors of 12: { 2, 3, 4, 6 }
From the list above, the common factors of 18 and 12 are 2, 3, and 6. To simplify the $\frac{18}{12}$, we will use 6 as the divisor since it is the greatest common factor. Hence, we have,
$\frac{18÷6}{12÷6}$ = $\frac{3}{2}$
Therefore, $\frac{18÷6}{12÷6}$ is $\frac{3}{2}$ when simplified.
Improper Fraction vs Mixed Number
When the numerator exceeds the denominator, the fraction is called an improper fraction. A mixed number is a way of writing fractions that are greater than one. A whole number and a proper fraction make up a mixed number.
Examples of improper fractions are:
Examples of mixed numbers are:
Improper Fraction Conversions
We can easily convert improper fractions to mixed numbers or decimals by following simple steps.
Converting Improper Fractions to Mixed Numbers
We shall divide the numerator by the denominator to make an improper fraction into a mixed number. To complete the mixed number, the quotient is the whole number part; the numerator is the remainder, while the divisor is the denominator.
For example, let us say that we must convert the improper fraction $\frac{23}{7}$ to a mixed number. When we divide 23 by 7, we will get 3 with the remainder of 2. We can form 3 equal groups of 7 with the remainder of 2. The image below illustrates the division process.
To form the equivalent mixed number, we will use the quotient ( 3 ) as the whole number part, and the fractional part is $\frac{2}{7}$. Notice that we use the same denominator.
$\frac{23}{7}$ =3$\frac{2}{7}$
To check the conversion, we must multiply the denominator by the whole number part and then add the answer to the numerator. The denominator stays the same. Hence, we have,
( 7 × 3 ) + 2 = 21 + 2 = 23
Therefore, the improper fraction $\frac{23}{7}$ is equal to 3$\frac{2}{7}$ in mixed number.
For another example, let us convert $\frac{9}{2}$ to a mixed number. When we divide 9 by 2, we will have 4 with the remainder of 1. The figure below shows the division process.
To check the conversion, we must multiply the denominator by the whole number part and then add the answer to the numerator. The denominator stays the same. Hence, we have,
( 2 × 4 ) + 1 = 8 + 1 = 9
Therefore, the improper fraction $\frac{9}{2}$ is equal to 4$\frac{1}{2}$ in mixed number.
Converting Improper Fractions to Decimals
To convert improper fractions to decimals, we must divide the numerator by the denominator. The process works the same way in converting improper fractions to mixed numbers.
For example, let us say that we must convert the improper fraction $\frac{11}{2}$ to decimals.
Therefore, $\frac{11}{2}$ is equal to 5.5 in decimals.
Addition and Subtraction of Improper Fractions
When adding or subtracting improper fractions, it is important to identify if you are adding or subtracting like or unlike terms. Like fractions have the same denominators; otherwise, they are unlike fractions. For example, $\frac{4}{3}$ and $\frac{5}{3}$ are like improper fractions, while $\frac{5}{4}$ and $\frac{7}{6}$ are unlike improper fractions.
To add like improper fractions, we must add the numerators of the given fractions and then copy the same denominator. Simplify the answer when necessary.
Let us say, for example, let us add the like improper fractions $\frac{7}{4}$ and $\frac{9}{4}$. Hence we have,
$\frac{7}{4}$+$\frac{9}{4}$=$\frac{16}{4}$
Since $\frac{16}{4}$ can still be simplified, the sum of $\frac{7}{4}$ and $\frac{9}{4}$ is equal to 16.
For another example, let us say we have $\frac{11}{3}$+$\frac{14}{3}$. Since they have the same denominators, it is easy to get the sum of these fractions by adding the numerators and copying the same denominator.
$\frac{11}{3}$+$\frac{14}{3}$=$\frac{25}{3}$
Thus, the sum of $\frac{11}{3}$ and $\frac{14}{3}$ is equal to $\frac{25}{3}$.
To add unlike improper fractions, we must first make the given improper fractions into proper fractions, then proceed to the process of adding like fractions. Identifying the least common multiple ( LCM ) of the denominators in converting the improper fraction to proper fractions is important. Then, simplify the fraction when needed.
Let us say, for example, we have the $\frac{5}{2}$+$\frac{13}{6}$.
Let us first make the given fractions to like improper fractions by identifying the LCM of the denominators.
Multiples of 2 are: { 2, 4, 6, 10, 8, 12, 14, 16, 18, 20, … }
Multiples of 6 are: { 6, 12, 24, 18, 36 30, 40, … }
From the list, 6, 12, and 18 are the common multiples, but we have to get the least common multiple ( LCM ), which is 6.
Thus, to make the given improper fractions into proper fractions, we must identify the numbers to multiply to these fractions, making the denominators equal to 6.
Hence, we have,
$\frac{5}{2}$ × $\frac{3}{3}$ = $\frac{15}{6}$
$\frac{13}{6}$ × $\frac{1}{1}$ = $\frac{13}{6}$
We shall then add the fractions $\frac{15}{6}$ and $\frac{13}{6}$ and add like fractions.
$\frac{15}{6}$+$\frac{13}{6}$=$\frac{28}{6}$
Since $\frac{28}{6}$ can still be simplified since 28 and 6 are divisible by 2, we have $\frac{28÷2}{6÷2}$=$\frac{14}{3}$.
Therefore $\frac{5}{2}$+$\frac{13}{6}$=$\frac{14}{3}$.
Subtracting Like Improper Fractions
To subtract like improper fractions, we must subtract the numerators of the given fractions and then copy the same denominator. Simplify the answer when necessary.
Let us say, for example, we subtract the like improper fractions $\frac{12}{5}$ and $\frac{9}{5}$. Hence we have,
$\frac{12}{5}$-$\frac{9}{5}$=$\frac{3}{5}$
Therefore, the difference between $\frac{12}{5}$ and $\frac{9}{5}$ is equal to $\frac{3}{5}$.
For another example, let us say we have $\frac{10}{7}-\frac{8}{7}$. Since they have the same denominators, it is easy to get the sum of these fractions by subtracting the numerators and copying the same denominator.
$\frac{10}{7}$-$\frac{8}{7}$=$\frac{2}{7}$
Thus, the difference between $\frac{10}{7}$ and $\frac{8}{7}$ is equal to $\frac{2}{7}$.
Subtracting Unlike Improper Fractions
To subtract, unlike improper fractions, we must first make the given improper fractions into proper fractions, then proceed to the process of subtracting like fractions. Finding the denominators’ least common multiple (LCM) is crucial for converting an improper fraction to a proper fraction. Then, simplify the fraction when needed.
Let us say, for example, we have the $\frac{15}{4}-\frac{9}{5}$.
Let us first make the given fractions to like improper fractions by identifying the LCM of the denominators.
Multiples of 4 are: { 4, 8, 16, 12, 20, 24, 28, 32, 36, 40, … }
Multiples of 5 are: { 5, 15, 10, 20, 25, 30, 35, 40, … }
From the list, 20 and 40 are the common multiples of 4 and 5. but we have to get the least common multiple ( LCM ), which is 20.
Thus, to make the given improper fractions into proper fractions, we must identify the numbers to multiply to these fractions, which will make the denominators equal to 20.
Hence, we have,
$\frac{15}{4}$ × $\frac{5}{5}$ = $\frac{75}{20}$
$\frac{9}{5}$ × $\frac{4}{4}$ = $\frac{36}{20}$
We shall then subtract the fractions $\frac{45}{20}$ and $\frac{36}{20}$ and proceed to add like fractions.
$\frac{75}{20}$ – $\frac{36}{20}$ = $\frac{39}{20}$
Therefore $\frac{15}{4}$ – $\frac{9}{5}$ = $\frac{39}{20}$.
Multiplying Improper Fractions
When multiplying improper fractions, we must obtain the product of the numerators of the provided fractions over the product of the denominators. And then simply the answer when necessary.
Let us say, for example; we have $\frac{7}{5}$ x $\frac{9}{8}$.
To get the product, we must multiply the numerators 7 and 9 and then multiply the denominators 5 and 8. Thus,
$\frac{7}{5}$ x $\frac{9}{8}$ =6340
Since 63 and 40 do not have common factors, $\frac{63}{40}$ is simplified already.
Thus, $\frac{7}{5}$ x $\frac{9}{8}$ is equal to $\frac{63}{40}$.
More Examples
Example 1
Which of the following are improper fractions?
Solution
When the denominator exceeds the numerator, we have a proper fraction; otherwise, it is improper. Hence, the improper fractions above are as follows:
In each of the improper fractions, we have,
7 > 2 ; 11 > 7; 9 > 4
12 > 5 ; 5 > 3 ; 21 > 4
Example 2
Convert the following improper fractions to mixed numbers.
( a ) $\frac{7}{3}$ ( b ) $\frac{16}{5}$ ( c ) $\frac{27}{2}$ ( d ) $\frac{39}{4}$ ( e ) $\frac{50}{11}$
Solution
The improper fractions can be converted into mixed numbers by simply dividing the numerator by the denominator. To complete the mixed number, the quotient is the whole number part, while the numerator is the remainder. The denominator is the divisor or the same denominator of the given improper fraction.
( a ) $\frac{7}{3}$
When we divide 7 by 3, we will have 2 as the quotient and 1 as the remainder. Therefore, the improper fraction $\frac{7}{3}$ is equal to 2$\frac{1}{3}$ in mixed number.
( b ) $\frac{16}{5}$
When we divide 16 by 5, we will have 3 as the quotient and 1 as the remainder. Therefore, the improper fraction $\frac{16}{5}$ is equal to 3$\frac{1}{5}$ in mixed number.
( c ) $\frac{27}{2}$
When we divide 27 by 2, we will have 13 as the quotient and 1 as the remainder. Therefore, the improper fraction $\frac{27}{2}$ is equal to 13$\frac{1}{2}$ in mixed numbers.
( d ) $\frac{39}{4}$
When we divide 39 by 4, we will have 9 as the quotient and 3 as the remainder. Therefore, the improper fraction $\frac{39}{4}$ is equal to 9$\frac{3}{4}$ in mixed numbers.
( e ) $\frac{50}{11}$
When we divide 50 by 11, we will have 4 as the quotient and 6 as the remainder. Therefore, the improper fraction $\frac{50}{11}$ is equal to 4$\frac{6}{11}$ in mixed numbers.
Example 3
Write the following mixed numbers to improper fractions.
( a ) 2$\frac{3}{5}$ ( b ) 3$\frac{9}{11}$ ( c ) 7$\frac{4}{7}$ ( d ) 5$\frac{5}{9}$ ( e ) 12$\frac{1}{4}$
Solution
When we convert a mixed number into an improper fraction, we must multiply the denominator by the whole number part and then add the result to the numerator. The improper fraction and the mixed number have the same denominator.
( a ) 2$\frac{3}{5}$
We must multiply the denominator, 5, by the whole number part, 2. The product of 5 and 2 is10, then add the numerator, 3, which results in 13. Hence, we have ( 5 × 2 ) + 3 = 13. The answer gives the numerator of the improper fraction and then uses the same denominator.
2$\frac{3}{5}$= $\frac{(2×5)+3}{5}$=$\frac{13}{5}$
Therefore, the mixed number 2$\frac{3}{5}$ is equal to $\frac{13}{5}$ in improper fraction.
( b ) 3$\frac{9}{11}$
We must multiply the denominator, 11, by the whole number part, 3. The product of 3 and 11 is 33, then add the numerator 9, which results in 42. Hence, we have ( 11 × 3 ) + 9 = 42. The answer forms the numerator of the improper fraction and then copies the same denominator.
3$\frac{9}{11}$= $\frac{(11×3) +9}{11}$=$\frac{42}{11}$
Therefore, the mixed number 3$\frac{9}{11}$ is equal to $\frac{42}{11}$ in improper fraction.
( c ) 7$\frac{4}{7}$
We must multiply the denominator, 7, by the whole number part, 7. The product of 7 and 7 is 49, then add the numerator 4, which results in 53. Hence, we have ( 7 × 7 ) + 4 = 53. The answer forms the numerator of the improper fraction and then copies the same denominator.
7$\frac{4}{7}$= $\frac{(7×7) +4}{7}$=$\frac{53}{7}$
Therefore, the mixed number 7$\frac{4}{7}$ is equal to $\frac{53}{7}$ in improper fraction.
( d ) 5$\frac{5}{9}$
We must multiply the denominator, 9, by the whole number part, 5. The product of 5 and 9 is 45, then add the numerator, 5, which results in 50. Hence, we have ( 9 × 5 ) + 5 = 50. The answer forms the numerator of the improper fraction and then copies the same denominator.
5$\frac{5}{9}$= $\frac{(9×5) +5}{9}$=$\frac{50}{9}$
Therefore, the mixed number 5$\frac{5}{9}$ is equal to $\frac{50}{9}$ in improper fraction.
( e ) 12$\frac{1}{4}$
We must multiply the denominator, 4, by the whole number part, 12. The product of 4 and 12 is 48, then add the numerator 1, which results in 49. Hence, we have ( 4 × 12 ) + 1 = 49. The answer forms the numerator of the improper fraction and then copies the same denominator.
12$\frac{1}{4}$= $\frac{(4×12) +1}{4}$=$\frac{49}{4}$
Therefore, the mixed number 12$\frac{1}{4}$ is equal to $\frac{49}{4}$ in improper fraction.
Example 4
Perform the indicated operations and simplify.
( a ) $\frac{5}{3}$+$\frac{8}{3}$
( b ) $\frac{16}{5}$-$\frac{9}{5}$
( c ) $\frac{7}{4}$ x $\frac{5}{3}$
( c ) $\frac{5}{2}$ x $\frac{8}{3}$
Solution
( a ) $\frac{4}{3}$+$\frac{8}{3}$=$\frac{12}{3}$=4
( b ) $\frac{16}{5}$ – $\frac{9}{5}$=$\frac{7}{5}$
( c ) $\frac{7}{4}$ x $\frac{5}{3}$ = $\frac{35}{12}$
( c ) $\frac{5}{2}$ x $\frac{8}{3}$ = $\frac{40}{6}$ = $\frac{20}{3}$
Summary
Numbers that represent a portion of a whole are called fractions.
When the numerator exceeds the denominator, the fraction is called an improper fraction. Suppose $\frac{a}{b}$ is an improper fraction; thus, a must be greater than b.
Examples of improper fractions are:
A mixed number is a way of writing fractions that are greater than one. A whole number and a proper fraction make up a mixed number.
Examples of mixed numbers are:
To create a mixed number from an improper fraction, divide the numerator by the denominator. To form the mixed equivalent mixed number, the quotient is the whole number part; the numerator is the remainder, while the divisor is the denominator.
Simplifying Improper Fractions
Simplifying improper fractions is the same thing as reducing the fractions. We do this by dividing the denominator and the numerator by their greatest common factor.
Frequently Asked Questions on Improper Fractions ( FAQs )
How do proper and improper fractions differ from one another?
Numbers that represent a portion of a whole are called fractions. Fractions are written with a numerator, a fraction bar, and a denominator. When the denominator is greater than the numerator, we have a proper fraction; otherwise, it is improper.
Examples of proper fractions ( Numerator < Denominator )
Examples of improper fractions ( Numerator > Denominator )
How do you convert mixed numbers to improper fractions?
A mixed number has two parts: the whole number and the proper fraction. To make a mixed number into an improper fraction, we must multiply the denominator by the whole number part, then add the answer to the numerator. The denominator stays the same.
For example, let us say that we must convert the mixed number 5$\frac{2}{3}$ to an improper fraction. We will have,
( 3× 5 ) + 2 = 15 + 2 = 17 , this is the numerator of the improper fraction
Therefore, the mixed number 5$\frac{2}{3}$ is equal to $\frac{17}{3}$.
How do like and unlike fractions
differ from one another?
If two fractions have the same denominator, they are said to be like; otherwise, they are unlike.
Examples of like fractions ( same denominators )
$\frac{5}{2}$ and $\frac{3}{2}$ are like fractions.
$\frac{9}{7}$ and $\frac{10}{7}$ are like fractions.
Examples of unlike fractions ( different denominators )
$\frac{11}{3}$ and $\frac{11}{2}$ are unlike fractions.
$\frac{12}{5}$ and $\frac{16}{9}$ are unlike fractions.
How do we simplify improper fractions?
When simplifying improper fractions is the same thing as reducing the fractions. We do this by dividing the denominator and the numerator by their greatest common factor.
Let us say, for example; we have the improper fraction $\frac{50}{10}$. Since the greatest common factor of 10 and 50 is 10, we have,
$\frac{50÷10}{10÷10}$ = $\frac{5}{1}$ = 5
Therefore, the simplified form of $\frac{50}{10}$ is 5.
How do we add and subtract improper fractions?
When adding or subtracting improper fractions, it is important to identify if you are adding or subtracting like or unlike terms. Like fractions have the same denominators; otherwise, they are unlike fractions. For example, $\frac{4}{3}$ and $\frac{5}{3}$ are like improper fractions, while $\frac{5}{4}$ and $\frac{7}{6}$ are unlike improper fractions.
To add like improper fractions, we must add the numerators of the given fractions and then copy the same denominator. Simplify the answer when necessary.
When adding unlike improper fractions, it is important to make the fractions proper using the least common multiple ( LCM ) of the denominators. Then, proceed to the process of adding like fractions.
How to multiply improper fractions?
When multiplying improper fractions, we must obtain the product of the numerators of the provided fractions over the product of the denominators. And then simply the answer when necessary.
Let us say, for example; we have $\frac{10}{2}$ x $\frac{5}{3}$.
To get the product, we must multiply the numerators 10 and 5, then multiply the denominators 2 and 3. Thus,
$\frac{10}{2}$ × $\frac{5}{3}$ = $\frac{50}{6}$
Since 2 is the greatest common factor of 50 and 6, $\frac{50}{6}$ can be simplified as $\frac{25}{3}$.Thus, $\frac{10}{2}$ × $\frac{5}{3}$ is equal to $\frac{25}{3}$.
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# Explained: Any 7 Bet In Detail
Remember playing dice and predicting the number that would come up next time the dice got rolled? We have all done this in childhood and the one whose prediction was true rejoiced as if some superpower has been granted to him or her. It used to extreme fun. If you miss this fun, try the game of craps. The game of craps takes this simple game to a new level.
In the game of craps, there is a shooter who rolls the dice. This person can be a dedicated person who is responsible for rolling dice or can be one among the players. If the shooter has is one of the players, each player gets a turn to roll the dice. In this game, a player can bet in several ways-
• The player can bet on how the series of outcomes can be. What could be the pattern of the outcome after each roll.
• The players can bet on a single outcome – this bet can change with every roll.
• Players can also bet on the sequence of series of the outcomes. Like a player can bet on two sixes, two twos, and so on.
• The player can also bet on a number. For example – A player can be on a number say 10 and predict that the summation of the two rolls will be 10. In this kind of bets, the Any 7 bet is very popular.
## Let’s know any 7 bet better:
Any 7 bet is a bet made in craps when a single dice is rolled. The bet is placed by a player that says that the next number rolled will be 7. This means that there are be 3 possibilities that the player wins that is- 1-6, 2-5 and 3-4.
In a game of craps, the probability of winning any 7 bet is pretty high and thus seven is one of the easy combinations to get on the table of the craps game. When the player places the ‘any seven bet’, whichever number might come of the dice, (any number from one to six) there is always a hope of getting a number on the next roll that could sum up to 7. Which is not true for any number other than 7.
Explanation: If the number rolled first is 6, the next roll’s outcome cannot be 1. This rules out one rolls and reduces the probability to 1 out of 6 to 1 out of 5.
The usual payout of any 7 bet is 4 to 1. The bet is also known as a ‘Big red’ bet. This bet might also be indicated using a red stripe on the grid on the craps table. The any-7 bet makes the craps game easier and even more interesting.
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### Summary
Name: Explained: Any 7 Bet In Detail
Posted On: 19/05/2019
Author: Darren Henley
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# Difference between revisions of "2015 AMC 12B Problems/Problem 4"
## Problem
Lian, Marzuq, Rafsan, Arabi, Nabeel, and Rahul were in a 12-person race with 6 other people. Nabeel finished 6 places ahead of Marzuq. Arabi finished 1 place behind Rafsan. Lian finished 2 places behind Marzuq. Rafsan finished 2 places behind Rahul. Rahul finished 1 place behind Nabeel. Arabi finished in 6th place. Who finished in 8th place?
$\textbf{(A)}\; \text{Lian} \qquad\textbf{(B)}\; \text{Marzuq} \qquad\textbf{(C)}\; \text{Rafsan} \qquad\textbf{(D)}\; \text{Nabeel} \qquad\textbf{(E)}\; \text{Rahul}$
## Solution 1
Let $-$ denote any of the 6 racers not named. Then the correct order following all the logic looks like:
$$-, \text{Nabeel}, \text{Rahul}, -, \text{Rafsan}, \text{Arabi}, -, \text{Marzuq}, -, \text{Lian}, -, -$$
Clearly the 8th place runner is $\fbox{\textbf{(B)}\; \text{Marzuq}}$.
## Solution 2
We can list these out vertically to ensure clarity, starting with Marta and working from there.
$$1 -$$ $$2 Nabeel$$ $$3 Rahul$$ $$4 -$$ $$5 Rafsan$$ $$6 Arabi$$ $$7 -$$ $$8 Marzuq$$
Thus our answer is $\fbox{\textbf{(B)}\; \text{Marzuq}}$.
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# NCERT solution class 11 chapter 2 Sets exercise 2.1 mathematics
## EXERCISE 2.1
#### Question 1:
If, find the values of x and y.
It is given that.
Since the ordered pairs are equal, the corresponding elements will also be equal.
Therefore,
and.
∴ x = 2 and y = 1
#### Question 2:
If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A × B)?
It is given that set A has 3 elements and the elements of set B are 3, 4, and 5.
⇒ Number of elements in set B = 3
Number of elements in (A × B)
= (Number of elements in A) × (Number of elements in B)
= 3 × 3 = 9
Thus, the number of elements in (A × B) is 9.
#### Question 3:
If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.
G = {7, 8} and H = {5, 4, 2}
We know that the Cartesian product P × Q of two non-empty sets P and Q is defined as
P × Q = {(pq): p∈ P, q ∈ Q}
∴G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}
#### Question 4:
State whether each of the following statement are true or false. If the statement is false, rewrite the given statement correctly.
(i) If P = {mn} and Q = {nm}, then P × Q = {(mn), (nm)}.
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (xy) such that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ Φ) = Φ.
(i) False
If P = {mn} and Q = {nm}, then
P × Q = {(mm), (mn), (n, m), (nn)}
(ii) True
(iii) True
#### Question 5:
If A = {–1, 1}, find A × A × A.
It is known that for any non-empty set A, A × A × A is defined as
A × A × A = {(abc): ab∈ A}
It is given that A = {–1, 1}
∴ A × A × A = {(–1, –1, –1), (–1, –1, 1), (–1, 1, –1), (–1, 1, 1),
(1, –1, –1), (1, –1, 1), (1, 1, –1), (1, 1, 1)}
#### Question 6:
If A × B = {(ax), (ay), (bx), (by)}. Find A and B.
It is given that A × B = {(ax), (a, y), (bx), (by)}
We know that the Cartesian product of two non-empty sets P and Q is defined as P × Q = {(pq): p ∈ P, q ∈ Q}
∴ A is the set of all first elements and B is the set of all second elements.
Thus, A = {ab} and B = {xy}
#### Question 7:
Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that
(i) A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) A × C is a subset of B × D
(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)
We have B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ
∴L.H.S. = A × (B ∩ C) = A × Φ = Φ
A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
∴ R.H.S. = (A × B) ∩ (A × C) = Φ
∴L.H.S. = R.H.S
Hence, A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) To verify: A × C is a subset of B × D
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
We can observe that all the elements of set A × C are the elements of set B × D.
Therefore, A × C is a subset of B × D.
#### Question 8:
Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.
A = {1, 2} and B = {3, 4}
∴A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
⇒ n(A × B) = 4
We know that if C is a set with n(C) = m, then n[P(C)] = 2m.
Therefore, the set A × B has 24 = 16 subsets. These are
Φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)},
{(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)},
{(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)},
{(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}
#### Question 9:
Let A and B be two sets such that n(A) = 3 and n (B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where xy and z are distinct elements.
It is given that n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B.
We know that A = Set of first elements of the ordered pair elements of A × B
B = Set of second elements of the ordered pair elements of A × B.
∴ xy, and z are the elements of A; and 1 and 2 are the elements of B.
Since n(A) = 3 and n(B) = 2, it is clear that A = {xyz} and B = {1, 2}.
#### Question 10:
The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0, 1). Find the set A and the remaining elements of A × A.
We know that if n(A) = and n(B) = q, then n(A × B) = pq.
∴ n(A × A) = n(A) × n(A)
It is given that n(A × A) = 9
∴ n(A) × n(A) = 9
⇒ n(A) = 3
The ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A × A.
We know that A × A = {(a, a): a ∈ A}. Therefore, –1, 0, and 1 are elements of A.
Since n(A) = 3, it is clear that A = {–1, 0, 1}.
The remaining elements of set A × A are (–1, –1), (–1, 1), (0, –1), (0, 0),
(1, –1), (1, 0), and (1, 1)
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04.24.2019: One Lesson of Math – Trigonometric Equations and Identities, 3/6: Trigonometric Identities, Part 1
Today’s soundtrack is Dream Theater: Falling Into Infinity, probably their most radio-friendly album.
You may have noticed that this series is titled “Trigonometric Equations and Identities.” So far, we’ve been working with equations; today, we’ll be learning about trigonometric identities. Let’s start by differentiating the two.
An equation is only true for certain values of its variables.
An identity is true for any values of its variables.
The following are the identities that we should be familiar with:
• Reciprocal Identities
1. secθ =1 / cosθ
2. cscθ =1 / sinθ
3. cotθ =1 / tanθ
• Quotient Identities
1. tanθ= sinθ / cosθ
2. cotθ= cosθ/ sinθ
• Pythagorean Identities
1. sin²θ + cos²θ = 1
2. 1 + cot²θ = csc²θ
3. 1 + tan²θ = sec²θ
All of these identities will be true for all values of θ.
We can prove identities by breaking down complex terms into simple terms through simplifying. We start with a two-sided equation. We split the equation with a bar down the middle, then simplify the more complicated side. After that, we simplify the easier side, and show that the two values are equal.
While we prove identities, we will be asked to simplify rational expressions using this method:
• Identify the denominator of the top term
• Identify the denominator of the bottom term
• Combine them as a common denominator
• Multiply the top term by the common denominator
• Multiply the bottom term by the common denominator
• Simplify
If we are asked to verify an equation for a value, we simply substitute the value in for the variable and solve by breaking down the identities as discussed above. We can also solve graphically by using technology and checking for intersects. If the graphs overlap at all points, then the value is verified.
That’s it for today; next time, we’ll tackle part 2 of this lesson.
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# Thin Lenses Problem: Solving for $q=d-p$
• Archimedess
In summary, the thin lens equation is a mathematical equation that relates the focal length of a lens to the object and image distances. It can be used to solve for q by rearranging the equation. There are two types of lenses, converging and diverging, which cause light rays to converge and diverge, respectively. The object and image distances can be determined using a ruler or the thin lens equation. The equation can be used for both types of lenses, but for diverging lenses, the focal length and image distance are considered negative. The position of the object relative to the lens affects the type and location of the image formed.
Archimedess
Homework Statement
We try with a thin convergent lens, by moving it on its optical axis, to project on a fixed screen the image of a small object located in the optical axis. This is only possible with two position of the lens ##I_1=20cm## and ##I_2=80cm##. Determine the ##d## distance from the object to the screen and focal length ##f## of the lens.
Relevant Equations
##\frac{1}{p}+\frac{1}{q}=\frac{1}{f}##
##d=q+p##
Where ##p## is the distance of the object from the lens and ##q## is the distance between the image and the lens.
Since ##d=q+p \implies q=d-p##
\begin{cases}
\frac{1}{I_1}+\frac{1}{d-I_1}=\frac{1}{f}\\
\frac{1}{I_2}+\frac{1}{d-I_2}=\frac{1}{f}
\end{cases}
Is this correct?
Yes. Now manipulate until you have an expression for ##d##.
## 1. What is a thin lens?
A thin lens is a transparent optical device with two curved surfaces that are designed to refract light in a specific way. It is typically made of glass or plastic and is used to focus or diverge light rays.
## 2. What is the "q=d-p" equation used for?
The "q=d-p" equation is used to solve for the image distance (q) of a thin lens, given the object distance (d) and the focal length (p) of the lens. This equation is derived from the thin lens equation, 1/q = 1/p + 1/d, which relates the object distance, image distance, and focal length of a thin lens.
## 3. How do you solve for q in the "q=d-p" equation?
To solve for q, you simply need to substitute the known values for d and p into the equation and solve for q using basic algebraic principles. Make sure to pay attention to the units of measurement and use consistent units throughout the calculation.
## 4. What are the units for q, d, and p in the "q=d-p" equation?
The units for q, d, and p in the "q=d-p" equation are all in distance units, typically meters (m) or centimeters (cm). It is important to use consistent units throughout the equation to ensure accurate calculations.
## 5. Are there any assumptions or limitations when using the "q=d-p" equation?
Yes, the "q=d-p" equation assumes that the thin lens is being used in air and that the light rays are parallel to the optical axis of the lens. It also assumes that the lens is thin, meaning that its thickness is negligible compared to its radius of curvature. Additionally, this equation is only applicable to thin lenses and may not accurately predict the behavior of thick lenses or other optical devices.
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# Using Area Models for Multiplication, Perimeter, and Area
### Written by Donna Boucher
Donna has been a teacher, math instructional coach, interventionist, and curriculum coordinator. A frequent speaker at state and national conferences, she shares her love for math with a worldwide audience through her website, Math Coach’s Corner. Donna is also the co-author of Guided Math Workshop.
Yesterday, I blogged about the Number of the Day posters that one of the presenters at CAMT displayed. The main topic of the presentation was using area models for multiplication, area, perimeter, and prime and composite numbers. The presenter, Jana Davison from Sheldon ISD here in Houston, packed a lot of information in a short amount of time–kind of like we want to do in our classrooms!
Area models are fantastic for helping kiddos develop a deep understanding of multiplication. Of course, we use arrays for facts, and area models are a perfect tool for multi-digit multiplication. Let’s look at the example below. In keeping with best practices, Jana framed it in a real world problem, but we’ll cut to the chase and simply show how to use the area model to do the multiplication. To solve the problem, we are multiplying 12 x 13.
1. Students write the 12 and 13 along the side and top of the grid paper. Starting in the top left hand corner, they outline a 12 x 13 rectangle.
2. Using base-10 blocks, students use the largest piece possible–a hundred–placing it in the upper left hand corner. They continue covering the rectangle with base-10 pieces, tens and ones, until it is covered. To determine the product, they count the base-10 pieces. Students should have lots of concrete practice with the blocks before moving to the representational, shown below.
3. Remember, my Texas friends, the students get grid paper to use on STAAR.
The problem Jana used for perimeter and area was a great one:
Danny has a rectangular rose garden that measures 8 meters by 4 meters. How many meters of fencing does he need to secure the garden?
Notice that the problem doesn’t mention perimeter. Students have to know that perimeter is the distance around the outside, so it’s critical that students get lots of real world problems dealing with perimeter and area.
This time we used inch grid paper, because we were going to be using 1-inch tiles for the area portion of the problem. To find the perimeter, we drew an 8 x 4 rectangle and numbered around it (you can see the numbers in the pictures below).
The next part of the problem:
If one bag of fertilizer can cover 16 sq meters, how many bags will he need to cover the entire garden?
Wowza. Awesome problem. So here we used the color tiles. The problem told us one bag covered 12 sq meters, so we used 16 tiles to cover part of the rectangle and labeled it 1 bag. We still had more to cover, so we used another 16 tiles to cover the rest of the rectangle.
Who doesn’t love using food as a math manipulative, right? For the activity on prime and composite numbers, we used Cheerios. Notice that the activity hits all the CRA bases. The concrete part is the Cheerios, the representational is the model drawing, and finally the abstract is writing the factors.
1. The math program in our school district, Math Expressions, uses area models. Most of my students love to use it because they are able to break the problem down into smaller chunks. Though our area models look a little different. Ours use expanded notation to break the numbers and the rectangles down.
-Jenn B. NJ
• I’ll have to look those up! I think it’s great for the kids to see a variety of models.
2. Ok, I’m in heaven. THIS is why I LOVE MATH!!!I love finding new ways for students to see why math works the way it does. When I was talking with my third grade students about the grid paper that would be with the STAAR test last year, we practiced several different ways that they could make use of the paper. We talked about area and perimeter, but not about multiplication. Thanks for the wonderful visual. I have been to Math CAMT before, but was not able to go this year. Looks like I missed some good presenters. Thanks for sharing.
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# ML Aggarwal Decimals Exe-7.1 Class 6 ICSE Maths Solutions
ML Aggarwal Decimals Exe-7.1 Class 6 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-7.1 Questions for Decimals as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-6.
## ML Aggarwal Decimals Exe-7.1 Class 6 ICSE Maths Solutions
Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 6th Chapter-7 Decimals Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-7.1 Questions Edition 2023-2024
### Decimals Exe-7.1
ML Aggarwal Class 6 ICSE Maths Solutions
Page-136
#### Question 1. Write each of the following decimal numbers in words:
(i) 30.5
(ii) 0.03
(iii) 108.56
(iv) 47.20
(v) 5.008
(vi) 26.039
Given decimal can be written in the word form as,
(i) 30.5 = Thirty point five
(ii) 0.03 = zero point zero three
(iii) 108.56 = One hundred eight point five six
(iv) 47.20 = Forty-seven point two zero.
(v) 5.008 = Five point zero zero eight
(vi) 26.039 = Twenty-six point zero three nine
#### Question 2. Write each of the following decimal numbers in the place value table :
(i) 4.2
(ii) 0.3
(iii) 205.9
(iv) 0.29
(v) 2.08
(vi) 7200.812
(vii) 38.007
Places Thousands Hundreds Tens One Tenths Hundredths Thousandths Values 1000 100 10 1 1/10 1/100 1/1000 (i) 4.2 4 2 (ii) 0.3 0 3 (iii) 205.9 2 0 5 9 (iv) 0.29 0 2 9 (v) 2.08 2 0 8 (vi) 7200.812 7 2 0 0 8 1 2 (vii) 38.007 3 8 0 0 7
#### Question 3. Write the following decimal numbers in the expanded form :
(i) 123.7
(ii) 43.06
(iii) 509.306
(i) 123.7 = 100 + 20 + 3 + 7/10
(ii) 43.06 = 40 + 3 + 6/100
(iii) 509.306 = 500 + 9 + 3/10 + 6/1000
#### Question 4. Write each of the following as a decimal number :
(i) 200 + 60 + 5 + 3/10
(ii) 50 + 1/10 + 6/100
(iii) 70 + 6 + 7/10 + 9/1000
(iv) 600 + 7 + 3/100 + 6/1000
(i) 200 + 60 + 5 + 3/10
The above question can be written in decimal form as = 265.3
(ii) 50 + 1/10 + 6/100
The above question can be written in decimal form as = 50.16
(iii) 70 + 6 + 7/10 + 9/1000
The above question can be written in decimal form as = 76.709
(iv) 600 + 7 + 3/100 + 6/1000
The above question can be written in decimal form as = 607.036
#### Question 5. Write each of the following as decimals:
(i) Two ones and five tenths
(ii) Two tens and nine tenths
(iii) Six hundred point eight
(iv) Two hundred five anf five hundredths
(v) Seven and fifteen thousandths
(i) Two ones and five-tenths
The above questions can be written as,
= (2 × 1) + (5 × (1/10))
= 2 + 5/10
= 2 + 0.5
= 2.5
(ii) Two tens and nine-tenths
The above questions can be written as,
= (2 × 10) + (9 × (1/10))
= 20 + 9/10
= 20 + 0.9
= 20.9
(iii) Six hundred point eight
The above questions can be written as,
= 600.8
(iv) Two hundred five and five hundredths
The above questions can be written as,
= (205) + (5 × (1/100))
= 205 + 5/100
= 205 + 0.05
= 205.05
(v) Seven and fifteen thousandths
The above questions can be written as,
= (7 × 1) + (15 × (1/1000))
= 7 + 15/1000
= 7 + 0.015
= 7.015
#### Question 6. Write the number given in the following place value table in decimal form:
Thousands Hundreds Tens One Tenths Hundredths Thousandths 1000 100 10 1 1/10 1/100 1/1000 (i) 7 1 0 2 3 0 6 (ii) 2 1 1 9 0 2 (iii) 3 0 5 3 0 1 5 (iv) 7 0 0 3 (v) 5 4 0 (vi) 7 1 9 0 2 8
(i) 7102.306
(ii) 211.902
(iii) 3053.015
(iv) 70.03
(v) 5.40
(vi) 719.028
### Decimals Exe-7.1
ML Aggarwal Class 6 ICSE Maths Solutions
Page-137
(i) 0.4
(ii) 1.9
(iii) 1.1
(iv) 2.5
A = 0.8
B = 1.3
C = 2.2
D = 2.9
(i) 0.06
(ii) 0.45
(iii) 0.66
(iv) 0.92
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What The Numbers Say About Christian Bale (11/09/2019)
How will Christian Bale perform on 11/09/2019 and the days ahead? Let’s use astrology to conduct a simple analysis. Note this is for entertainment purposes only – do not take this too seriously. I will first find the destiny number for Christian Bale, and then something similar to the life path number, which we will calculate for today (11/09/2019). By comparing the difference of these two numbers, we may have an indication of how well their day will go, at least according to some astrology enthusiasts.
PATH NUMBER FOR 11/09/2019: We will analyze the month (11), the day (09) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. Here’s how it works. First, for the month, we take the current month of 11 and add the digits together: 1 + 1 = 2 (super simple). Then do the day: from 09 we do 0 + 9 = 9. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 2 + 9 + 12 = 23. This still isn’t a single-digit number, so we will add its digits together again: 2 + 3 = 5. Now we have a single-digit number: 5 is the path number for 11/09/2019.
DESTINY NUMBER FOR Christian Bale: The destiny number will take the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Christian Bale we have the letters C (3), h (8), r (9), i (9), s (1), t (2), i (9), a (1), n (5), B (2), a (1), l (3) and e (5). Adding all of that up (yes, this can get tedious) gives 58. This still isn’t a single-digit number, so we will add its digits together again: 5 + 8 = 13. This still isn’t a single-digit number, so we will add its digits together again: 1 + 3 = 4. Now we have a single-digit number: 4 is the destiny number for Christian Bale.
CONCLUSION: The difference between the path number for today (5) and destiny number for Christian Bale (4) is 1. That is lower than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But this is just a shallow analysis! As mentioned earlier, this is not scientifically verified. If you want really means something, check out your cosmic energy profile here. Go ahead and see what it says for you – you’ll be glad you did.
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
|
Q:
# How do I calculate probability?
A:
The theoretical definition of probability states that if the outcomes of an event are mutually exclusive and equally likely to happen, then the probability of the outcome "A" is: P(A) = Number of outcomes that favors A / Total number of outcomes. For example, there are two possible outcomes when a coin is tossed in the air, and the probability of the coin landing on a head or a tail is equal to 0.5.
## Keep Learning
Credit: Florian Murauer / EyeEm Eye/Em Getty Images
1. Identify the number of possible outcomes
For any event, all the possible outcomes must be identified first. These should contain all the possible outcomes in all possible combinations. For instance, if two coins are tossed in the air at the same time then the possible outcomes would be: HH, TH, HT and TT, where "H" represents a head and "T" represents a tail.
2. Identify the number of favorable outcomes
Once all the possible outcomes of an event are identified, the number of favorable outcomes that satisfy a specific condition must identified. For example, if two coins are tossed in the air at the same time, the number of outcomes that satisfy the condition of a coin landing on heads at least once is 3.
3. Calculate the probability
Once all the numbers are obtained, calculate the probability. For example, the probability of getting at least one head when both coins are tossed in the air at the same time is: P(Head) = 3/4 = 0.75.
Sources:
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# Binomial Theorem
In this article, we will learn about Binomial Theorem. So far, we have learnt how to expand the square of a binomial using algebraic identities like $\displaystyle{(a + b)^2 = a^2 + 2ab + b^2}$. However, for the expansion of $\displaystyle{(a + b)^n}$ where $n$ is a positive integer greater than $2$, how do we get the result of the expansion?
We will be touching on the following concepts:
• Pascal’s Triangle
• Binomial Coefficients
• Binomial Expansion of $\displaystyle{(a + b)^n}$
• Binomial Theorem
## Expansion of Algebraic Expressions of the Form (1 + b)n
The table below shows the expansions of algebraic expressions of the form $\small{\displaystyle{(1 + b)^n}}$, where $\small{n = 1, 2, 3, 4}$.
$\small{\displaystyle{(1 + b)^1}}$ \small{\begin{align}&=\displaystyle{1 + b}\end{align}} $\small{\displaystyle{(1 + b)^2}}$ \small{\begin{align}&=\displaystyle{ 1 + 2b + b^2}\end{align}} $\small{\displaystyle{(1 + b)^3}}$ \small{\begin{align} &= (1 + b)(1 + b)^2 \\ &= (1 + b)(1 + 2b + b^2) \\ &= 1 + 2b + b^2 + b + 2b^2+ b^3 \\ &= 1 + 3b + 3b^2 + b^3 \\ \end{align}} $\small{\displaystyle{(1 + b)^4}}$ \small{\begin{align} &= (1 + b)(1 + 3b + 3b^2 + b^3) \\ &= 1 + 3b + 3b^2 + b^3 + b + 3b^2 + 3b^3 + b^4 \\ &= 1 + 4b + 6b^2 + 4b^3 + b^4 \end{align}}
## Investigation: Introduction to Pascal’s Triangle
Pascal was a French mathematician and scientist who formulated what became known as Pascal’s Principle of Pressure. The SI unit for pressure is pascal ($Pa$). Pascal is also famously known for Pascal’s Triangle which is used to derive binomial coefficients in binomial expansions.
$\small{\begin{matrix} & (1+b)^0 = 1 &\\[2ex] & (1+b)^1 = 1+1b &\\[2ex] & (1+b)^2 = 1+2b+1b^2 &\\[2ex] & (1+b)^3 = 1+3b+3b^2+1b^3 &\\[2ex] & (1+b)^4 = 1+4b+6b^2+4b^3+1b^4 &\\[2ex] & (1+b)^5 = 1+5b+10b^2+10b^3+5b^4+1b^5 &\\[2ex] & (1+b)^6 = 1+6b+15b^2+20b^3+15b^4+6b^5+1b^6 & \end{matrix}}$
The triangle above shows the expansions of algebraic expressions of the form $\small{(1 + b)^n}$, where $\small{n = 1, 2, 3, 4, 5, 6}$. If we reduce the triangle to just the coefficients of each term, we will get Pascal’s Triangle, as seen below:
$\small{\begin{matrix} 1\\[2ex] 1 + 1\\[2ex] 1 + 2 + 1\\[2ex] 1 + 3 + 3 + 1\\[2ex] 1 + 4 + 6 + 4 + 1\\[2ex] 1 + 5 + 10 + 10 + 5 + 1\\[2ex] 1 + 6 + 15 + 20 + 15 + 6 + 1 \end{matrix}}$
Notice that each new entry, $\large{x}$, of each subsequent row forms a triangle with the number above and to the left, $\large{a_{_L}}$, and the number above and to the right, $\large{a_{_R}}$. We obtain $\large{x}$ by taking the sum of $\large{a_{_L}}$ and $\large{a_{_R}}$.
For example, in the fifth row, $6$ is obtained by taking the sum of $3$ and $3$ from the fourth row.
• From this investigation, we observe that Pascal’s Triangle can be used to expand $(1 + b)^n$ if $\large{n}$ is small. However, can Pascal’s Triangle be used for the expansion of $(1 + b)^{20}$?
• No. Pascal's Triangle is not possible for the expansion of $(1 + b)^{20}$. This is because the expansion will be extremely tedious and we would need to start from $(1 + b)^{1}$.
Hence, we need to explore a different method to obtain binomial expansions with greater powers.
## Introduction to Binomial Coefficient
• The notation $\large{\binom {n}{r}}$ read as “$n \;\text{choose} \;r$”, represents a binomial coefficient in the expansion of $(1 + b)^n$.
$n$ represents the power and the $r$ represents the position, starting from $0$.
• The first coefficient is $\large{\binom {n}{0}}$ not $\large{\binom {n}{1}}$.
• In general the binomial expansion of $(1 + b)^n$ is
\begin{align*} (1+b)^n=\binom{n}{0}+\binom{n}{1}b+\binom{n}{2}b^2+...+\binom{n}{n}b^n \end{align*}
As you can see in the above equation, the first position of the binomial formula is $0$, followed by $1, 2, \cdots n$. Hence, the binomial coefficients are \begin{align*} \binom{n}{0},\;\binom{n}{1}, \;\cdots \;,\binom{n}{n} \end{align*}.
Question 1:
Find, in ascending powers of $x$, the first four terms in the expansion of $(1+x)^{20}$.
Solution:
\small{\begin{align*} (1+x)^{20} &=\binom{20}{0}+\binom{20}{1}x^1+\binom{20}{2}x^2+\binom{20}{3}x^3 + \cdots \\[2ex] &= 1+20x+190x^2+1140x^3+\cdots \end{align*}}
Question 2:
Find, in ascending powers of $y$, the first four terms in the expansion of $(1+y)^{13}$.
Solution:
\small{\begin{align} (1+y)^{13} &=\binom{13}{0}+\binom{13}{1}y^1+\binom{13}{2}y^2+\binom{13}{3}y^3 + \cdots \\[2ex] &= 1+13y+78y^2+286y^3+\cdots \end{align}}
### Binomial Expansion of (a + b)n
We have learnt how to expand $(1 + b)^n$.
What about $(a + b)^n$?
$\small{(a+b)^1}$ \small{\begin{align} &= 1a+1b \end{align}} \small{\begin{align} &= 1a^2+2ab+1b^2 \end{align}} \small{\begin{align} &=1a^3+3a^{2}b+3ab^2+1b^3 \end{align}}
From the table, \small{\begin{align} (a+b)^3 &=a^3+3a^{2}b+3ab^2+b^3 \end{align}}.
Here, $\small{a^3+3a^{2}b+3ab^2+b^3 }$ has the binomial coefficients $1, \;3, \;3, \;1,$ which follows the fourth row of Pascal’s Triangle. Notice that the power of $a$ decreases and power of $b$ increases in $\small{a^3+3a^{2}b+3ab^2+b^3 }$.
Hence, we can deduce that the binomial expansion of $(a+b)^4$ is
$\small{(a+b)^4= 1a^4+4a^{3}b+6a^{2}b^{2}+4ab^3+b^4}$.
## Binomial Theorem
From the above investigation, the result of the expansion of $(a+b)^n$ is as follows,
\small{\begin{align} (a+b)^n &= \binom{n}{0}a+\binom{n}{1}a^{(n-1)}b^1+\binom{n}{2}a^{(n-2)}b^2+\cdots+\binom{n}{r}a^{(n-r)}b^r+\cdots+\binom{n}{n}b^n \\ \end{align}}
Question 3:
Find, in ascending powers of $x$, the first four terms in the expansion of $(3+2x)^7$.
Solution:
\small{\begin{align} (3+2x)^7&=\binom{7}{0}(3)^7(2x)^0 +\binom{7}{1}(3)^{6}(2x)^1+\binom{7}{2}(3)^{5}(2x)^2+\binom{7}{3}(3)^{4}(2x)^3+\cdots \\[2ex] &= 2187+10206x+20412x^2+22680x^3+... \end{align}}
Question 4:
Find, in ascending powers of $y$, the first four terms in the expansion of $(2+3y)^5$.
Solution:
\small{\begin{align} (2+3y)^5 &= \binom{5}{0}(2)^{5}(3y)^0+\binom{5}{1}(2)^{4}(3y)^1+\binom{5}{2}(2)^{3}(3y)^2+\binom{5}{3}(2)^{2}(3y)^3+\cdots \\[2ex] &=32+240y+720y^2+1080y^3+\cdots \end{align}}
## Conclusion
In this article, we have learnt about Pascal’s Triangle and how it relates to the binomial coefficients, $\binom {n}{r}$, of a binomial expansion $(a + b)^n$
An important reminder when performing binomial expansion is that the first binomial coefficient is always $\binom {n}{0}$, while the last binomial coefficient is always $\binom {n}{n}$.
Also make sure that you are careful in your manipulation when simplifying the expression!
Keep learning! Keep improving!
Continue Learning
Quadratic Functions in Real-World Context Equations and Inequalities
Logarithmic Functions Surds
Polynomials & Cubic Equations Partial Fraction
Exponential Functions Coordinate Geometry (Circles)
Linear Law Binomial Theorem
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# Divide Fractions
In this worksheet, students will divide two simple fractions.
Key stage: KS 3
Curriculum topic: Number
Curriculum subtopic: Use Four Operations for All Numbers
Difficulty level:
#### Worksheet Overview
In this activity, we will practise dividing fractions.
The method we're going to use is KFC
No, not the takeaway chicken - something much better!
KFC stands for Keep Flip Change
First, we keep the first number or fraction unchanged.
Next, we flip the second fraction upside down.
Finally, we change the sign between the two fractions or numbers from ÷ to x.
So, 3/4 ÷ 2/5 becomes 3/4 x 5/2
Having done that, we can multiply the top numbers together and then the bottom numbers together, remembering to reduce the final answer. Sometimes, we can cancel down within the calculation before doing any multiplying, as shown below.
Example
Work out
3 ÷ 9 10 15
First, we keep the first fraction unchanged.
Next, we flip the second fraction.
Finally, we change the ÷ sign to x.
3 x 15 10 9
We could just multiply the top numbers together: 3 x 15 = 45 and the bottom numbers 10 x 9 = 90
This would give us 45/90 which cancels down to 1/2.
However, it is also possible to cancel down before doing the calculation!
Look for any top number that has a common factor with any bottom number.
If we find such a reducible pair of numbers, we divide them by their common factor.
3 x 15 10 9
= 3 ÷ 3 x 15 10 9 ÷ 3
= 1 x 15 10 3
= 1 x 15 ÷ 5 10 ÷ 5 3
= 1 x 3 2 3
= 1 x 3 ÷ 3 2 3 ÷ 3
= 1 x 1 2 1
= 1 2
Let's get started!
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# Quantitative Comparison and Manipulation
Many quantitative questions have variables in both columns. While your first instinct may be to work algebraically, this strategy is not always best. Often the fastest way to a solution is by plugging in different values to see which column is greater.
Developing a sense of when to plug in and when to solve algebraically takes practice. Here are a few helpful guidelines when trying to determine which approach to use.
## Work algebraically if the question is a polynomial
If you are dealing with a polynomial, simplify. Plugging in may require too much calculation. Instead, work with the familiar algebraic forms shown below:
Now let’s take a look at a question.
Column AColumn B
1. The quantity in Column A is greater
2. The quantity in Column B is greater
3. The two quantities are equal
4. The relationship cannot be determined from the information given
In Quantitative Comparison we can make it so each side is equal. Then we can balance the equation, adding and subtracting, multiplying and dividing, where necessary.
First, note that can be factored into . Now we can set both columns equal to each other:
At this point we have to be careful. While algebra tells us to divide both sides by (x – 2), we need to be aware of the following: if x is between 0 and 2, (x-2) yields a negative. However, (x + 2) yields a positive, meaning (x – 2)(x + 2) gives us a negative. In this case (A) would be bigger.
However, if we divide each side by (x – 2) and solve we get the following:
DIVIDE EACH SIDE BY (x – 2)
SUBTRACT ‘X’ FROM BOTH SIDES
This hardly looks like a solution (in fact it looks like I forgot to go to grade school!). However, what this yields is the important insight: Column B is now 4 greater than Column A. Therefore, the answer is (D). Alternatively you could pick numbers, such as ‘0’ and ‘4’. Each gives us different answers, leading to the same conclusion: Answer (D).
## If the question has variables, but there no polynomials, plug in values.
This advice pertains to variables that are not in polynomial form, as seen above. Here come up with easy numbers to plug in to see which values the columns yield.
0 > x > y > z > -1
Column AColumn B
1. The quantity in Column A is greater
2. The quantity in Column B is greater
3. The two quantities are equal
4. The relationship cannot be determined from the information given
.
It would be very nice if we could just stop here and choose answer (B). However, things are not so simple. When plugging in one set of numbers we will inevitably come up with one outcome. Once we’ve plugged in and come up with one answer, whether it is (A), (B), or (C), our job is to disprove that answer.
Can we make Column A larger than Column B? Well what if we plug in a values for x and y that are very close to one another. . Now, for z we can plug in a value close to -1, say -8/9. This gives us . Now you can see (B) is much smaller. Therefore the answer is (D).
## Takeaway
You should be adept at both algebra and plugging-in to efficiently—and accurately—answer a quantitative comparison question that contains variables. Typically it is best to simply check for polynomials before plugging in. To test out your skills, try these GRE math questions for practice!
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Quadratic equations are problems with squares, or “quadratus” in Latin. We derive a method to solve quadratic equations. In 2000 BC, the Babylonians developed an approach to solve problems which, in current notation would be a quadratic equation [wiki]. A arbitrary example of this is:
A field has a perimeter of 40 and an area of 96. What are the dimensions of this field?”
Around 300 BC, the Greek mathematician Euclid produced a general method to solve quadratic equations.
## Introduction
Before we introduce quadratic polynomials, let’s start by multiplying two linear functions (lines).
### Multiplying linear functions
Let $$g(x)$$ and $$h(x)$$ be linear functions
$$\left\{ \begin{array}{c} g(x)=x-s\\[5pt] h(x)=x-t \end{array} \right.$$
where $$x$$ is a variable and $$s$$ and $$t$$ are constants.
The function value equals $$0$$ at the $$x$$-intercepts.
$$\left\{ \begin{array}{c} g(x_0)\equiv 0 \Rightarrow & x_0-s\equiv 0 \Rightarrow & x_0=s\\[5pt] h(x_0)\equiv 0 \Rightarrow & x_0-t\equiv 0 \Rightarrow & x_0=t \end{array} \right.$$
First, we define function $$f(x)$$ as a scalar $$a$$ multiplied with the functions $$g(x)$$ and $$h(x)$$
$$f(x)\equiv a\,g(x)\,h(x)=a(x-s)(x-t) \label{eq:GxHxFactorized}$$
This equation results in the parabola $$f(x)$$ that shares $$x$$-intercepts with the two lines.
The interactive graph above visualizes the zero product property:
If the product of two quantities is equal to zero, then at least one of the quantities must be equal to zero. zero-product property
### Expand
The expression $$(x-s)(x-t)$$ from function $$\eqref{eq:GxHxFactorized}$$ can be expanded geometrically by representing the quantities $$x, -s, -t$$ as line segments, and representing the products of two quantities by the area of a rectangle.
Now we multiply this expanded form with scalar $$a$$:
$$ax^2-a(s+t)x+a\,s\,t\label{eq:exanded}$$
The equation now fits in the standard form for a single variable quadratic polynomial expression.
$$\begin{array}{c} ax^2+bx+c \\ \text{where}\ \ t=-a(s+t),\ \ c=a\,s\,t \end{array} \label{eq:quadraticExp}$$
This is called a quadratic expression, or a second order polynomial since the greatest power in the equation is two.
In the previous section we showed that multiplying two linear functions creates a quadratic function. Here we will do the opposite and bring the standard form quadratic $$\eqref{eq:quadraticExp}$$ back to its factorized form.
$$ax^2 + bx + c \equiv a(x-r_1)(x-r_2) \label{eq:quadratic}$$
We replaced the constants $$s$$ and $$t$$ with $$r_1$$ and $$r_2$$ to clearly denote them as the roots.
According to $$\eqref{eq:quadraticExp}$$ these roots should add up to $$-\frac{b}{a}$$ and while their product equals $$\frac{c}{a}$$
$$r_1+r_2=-\frac{b}{a}\quad\land\quad r_1\,r_2=\frac{c}{a} \label{eq:factorize}$$
In the following section, we will derive a general formula for the roots using a method called “completing the square”.
In the factorized form, $$r_1$$ and $$r_2$$ are the values of $$x$$ for which the expression equals $$0$$.
$$a(x-r_1)(x-r_2)=0\ \Rightarrow\ \left| \begin{array}{l} x=r_1 \\ x=r_2 \end{array} \right.$$
This implies that the expanded form $$\eqref{eq:quadratic}$$ must equal $$0$$ for the same values of $$x$$.
$$ax^2+bx+c\equiv 0\label{eq:derive0}$$
Now we solve the equation $$\eqref{eq:derive0}$$ by isolating $$x$$ on the left
$$x^2+\frac{b}{a}x = -\frac{c}{a} \label{eq:derive1}$$
The variable $$x$$ occurs twice, which makes the equation hard to solve. We can find the solutions by working towards the identity:
$$\color{green}{p}^2+2\color{green}{p}\color{purple}{q}+\color{purple}{q}^2=(\color{green}{p}+\color{purple}q)^2\nonumber$$
Then we multiply equation $$\eqref{eq:derive1}$$ by $$4a^2$$and add $$\color{purple}b^2$$ to both sides.
$$$$\begin{split} x^2+\frac{b}{a}x &=-\frac{c}{a} & \times 4a^2 \nonumber\\ \Leftrightarrow\quad(\color{green}{2ax})^2+4abx &=-4ac & +b^2 \nonumber\\ \Leftrightarrow\quad(\color{green}{\underline{2ax}})^2+2(\color{green}{\underline{2ax}})\color{purple}{\underline{b}}+\color{purple}{\underline{b}}^2&=-4ac+\color{purple}{b}^2 \end{split} \label{eq:derive2}$$$$
The left side now fits the format of the identify, where $$\color{green}p=\color{green}{2ax}$$ and $$\color{purple}q=\color{purple}b$$. Now we apply the identity to equation $$\eqref{eq:derive2}$$
$$$$\begin{split} (\color{green}{2ax}+\color{purple}{b})^2&=b^2-4ac & \text{take the }\sqrt{\color{white}{1}}\nonumber\\ 2ax+b &= \pm\sqrt{b^2-4ac} & \text{solve for }x\nonumber\\ 2ax &= -b\pm\sqrt{b^2-4ac}\nonumber\\ x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{split}\label{eq:roots}$$$$
The roots found by this equation are the values of $$x$$ that are solutions to $$\eqref{eq:roots}$$. This implies that the expression $$ax^2+bx+c$$ can be factorized as:
$$\shaded{ \begin{split} a^2+bx+c &=a(x-r_1)(x-r_2) \nonumber \\ \text{where}\ \ r_{1,2} &=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \nonumber \end{split} } \label{eq:quadroots}$$
### Discriminant
A polynomial of the second power always has two roots, though they may or may not be distinct or real.
The expression under the square root sign in $$\eqref{eq:quadroots}$$ is called the discriminant, or $$D$$.
$$D = b^2-4ac$$
The discriminant determines the nature of the roots.
• when $$D>0$$, there are two distinct real roots.
• when $$D=0$$, there is a double real root.
• when $$D\lt0$$, there are two complex roots.
The unit Complex Numbers defined the imaginary unit $$i$$ as $$i^2=-1$$ and showed that each complex number $$z$$ consists of a real part $$x$$ and an imaginary part $$iy$$, where $$x$$ and $$y$$ are real numbers. Using $$i$$ as the imaginary unit, we can denote any complex number $$z$$ as:
$$z = x+iy\label{eq:zdef}$$
These complex numbers extend the number line to a two-dimensional $$\mathbb{C}$$-plane as shown below.
### Real roots
The distinct or double real roots can be easily visualized by plotting the function and finding the $$x$$-intercepts. These intercepts are the roots.
In the figures above the function depicted in the graph on the left has real roots at $$1$$ and $$3$$, and the function on the right has a double root at $$2$$.
### Where are the imaginary roots?
When the roots of a function are complex, the quadratic function $$\eqref{eq:quadratic}$$ has a graph that doesn’t intersect the $$x$$-axis as shown below. So where are those roots?
The quadratic formula $$\eqref{eq:quadroots}$$ tells us that the roots for the function depicted in the graph above are $$2+2j$$ and $$2-2j$$.
$$x^2-4x+8=(x-2-2j)(x-2+2j)\label{eq:zquadratic}$$
To find these complex roots visually, we need to broaden our perspective and allow the independent variable $$x$$ to have complex values. After all, these roots are complex values. From here on we will name the variable $$z$$ instead of $$x$$ to emphasize that $$z\in\mathbb{C}$$. We will reuse $$x$$ for the real part of $$z$$.
Evaluate quadratic expression $$\eqref{eq:quadratic}$$ with variable $$z$$ follows
$$$$\begin{split} f(z) &= az^2+bz+c &\forall_{z\in\mathbb{C}}\\ \text{where}\quad z &\equiv x+jy &\forall_{x,y\in\mathbb{R}}\\ \text{and}\ \ f(z) &\equiv u+vj \end{split} \label{eq:fzdef}$$$$
With a complex function argument $$x+yj$$ $$\eqref{eq:zdef}$$ and a complex function value $$u+iv$$ $$\eqref{eq:fzdef}$$, we need four mutually perpendicular axes $$x,y,u,v$$ to graph the function. The catch: we can’t graph a 4-dimensional function.
To reduced the graph to a 3-dimensions, either
1. Only consider variables $$z$$ for which the function value is a real number, and graph the function value on z-axis perpendicular to the $$\mathbb{C}$$-plane. The roots will be where the graph intersects the $$\mathbb{C}$$-plane.
2. Consider all variables $$z\in\mathbb{C}$$, but take only the modulus of the function value $$|f(z)|$$.
The following sections describe each of two visualization techniques. [1]
#### Plotting real function values
Now we graph the function $$\eqref{eq:zquadratic}$$, considering only values of $$z$$ for which the function value $$f(z)=u+jv$$ is real (v=0).
We can let the variable be $$z=x+jy$$ and split the function value $$\eqref{eq:fzdef}$$ into real and imaginary parts
$$$$\begin{split} f(z)&=az^2+bz+c & z\equiv x+iy\nonumber\\ &=a(x+jy)^2+b(x+jy)+c &\text{expand}\nonumber\\ &=ax^2+2axjy-ay^2+bx+jby+c &\text{split Re/Im}\nonumber\\ &=\underbrace{(ax^2-ay^2+bx+c)}_{\text{real part}} + \underbrace{y(2ax+b)j}_{\text{imaginary part}} \end{split}$$$$
The imaginary part of the function value must be $$0$$.
$$y(2ax+b)=0 \ \Rightarrow\ \left| \begin{array}{l} y=0 \\ x=-\frac{b}{2a} \end{array} \right.$$
Substitute the value $$y=0$$ in \eqref{eq:zdef}
$$$$\begin{split} z_r&=x+jy &\text{subst }y=0 \nonumber\\ &=x+0j &\forall_{x\in\mathbb{R}} \end{split}\label{eq:zr}$$$$
Do the same for $$x=-\tfrac{b}{2a}$$
$$$$\begin{split} z_c &=x+yj &\text{subst }x=-\tfrac{b}{2a} \nonumber \\ &=-\tfrac{b}{2a}+yj &\forall_{y\in\mathbb{R}} \end{split}\label{eq:zc}$$$$
This implies that the function $$\eqref{eq:zquadratic}$$ has a real-value
1. when evaluated for $$z=x$$ where $$x\in\mathbb{R}$$, or
2. when evaluated for $$z=-\frac{b}{2a}+yj$$ where $$y\in\mathbb{R}$$
In the first case, evaluating for $$z=x$$ where $$x\in\mathbb{R}$$, means evaluating along the familiar $$x$$-axis. This is how we visualized the real roots.
$$$$\begin{split} f(z_r) &= az^2+bx+c &\text{subst }z_r=x\text{ from }\eqref{eq:zr}\nonumber\\ \Rightarrow\quad f(x) &= ax^2+bx+c \end{split}\label{eq:fx}$$$$
In the second case, evaluating for $$z=-\tfrac{b}{2a}+yj$$ where $$y\in\mathbb{R}$$, means evaluating along a line that intersects the point $$\left(-\frac{b}{2a}+0j\right)$$ and runs parallel to the imaginary $$y$$-axis as shown in the figure below.
Substituting $$-\frac{b}{2a}+yj$$ for $$z$$ in $$\eqref{eq:zquadratic}$$.
$$\newcommand\ccancel[2][black]{\color{#1}{\cancel{\color{black}{#2}}}} \newcommand\cbcancel[2][black]{\color{#1}{\bcancel{\color{black}{#2}}}} \newcommand\ccancelto[3][black]{\color{#1}{\cancelto{#2}{\color{black}{#3}}}} $$\begin{split} f(z) &= az^2+bz+c \quad\quad\quad\text{subst }z=-\frac{b}{2a}+yj\text{ from }\eqref{eq:zc}\\ f\left(-\frac{b}{2a}+yj\right) &= a\left(-\frac{b}{2a}+yj\right)^2+b\left(-\frac{b}{2a}+yj\right)+c\nonumber\\ &= {\ccancel[red]{a}}\frac{b^2}{4a^{\ccancel[red]{2}}}-{\ccancel[red]{2a}}\frac{\cbcancel[blue]{b}}{\ccancel[red]{2a}}{\cbcancel[blue]{yj}}-ay^2-\frac{b^2}{2a}+{\cbcancel[blue]{byj}}+c \end{split}\label{eq:fy0}$$$$
This makes \eqref{eq:fy0} a function of $$y$$ because $$a,b$$ and $$j$$ are constants.
$$\shaded{ f(y)=-ay^2-\frac{b^2}{4a}+c\quad\forall_{y\in\mathbb{R}} } \label{eq:fy}$$
Similar to how the graph for $$f(x)$$ intersects the $$\mathbb{C}$$-plane at the real roots, the graph for \eqref{eq:fy} intersects the $$\mathbb{C}$$-plane at the complex roots of the function.
The interactive graph shown below visualizes this concept. Click Interact and wait for the model to load.
#### Plotting the modulus of the function values
Here we graph the function $$\eqref{eq:zquadratic}$$ by considering all values of $$z$$, but only plotting the modules of the function value $$f(z)$$. The modulus $$|f(x+jy)|$$ is defined as:
$$|f(x+jy)|\triangleq\sqrt{x^2+y^2}\nonumber$$
To find the modulus of the function value, we can apply the definition $$\eqref{eq:zdef}$$ to the quadratic equation $$\eqref{eq:zquadratic}$$.
$$$$\begin{split} f(z) &= az^2+bz+c\nonumber\\\ f(x+jy) &= (ax^2-ay^2+bx+c) +y(2ax+b)j\nonumber\\ \Rightarrow\quad |f(x+jy)| &= \sqrt{(ax^2-ay^2+bx+c)^2 +y^2(2ax+b)^2}\\ \text{where}\quad x &\in\mathbb{R}\ \land\ y\in\mathbb{R}\nonumber \end{split}$$$$
This equation implies that the function arguments are two independent variables. In the graph we depict them as a horizontal complex plane with points $$z=x+jy$$. The $$z$$-axis of the graph is used for the modulus of the function value. In this so called modulus surface, color is used to show the angle of the function value.
While this surface is well suited to depict real and complex values, it has some drawbacks. The most obvious one is that it shows the modulus of the function where $$|f(z)|\geq0$$. As a consequence, the parabolic shape is harder to recognize.
In the graph below, we added the real function values $$f(z_c)$$ from the previous visualization technique that showed negative values. In the surface plot, these same function values are represented on the positive $$z$$-axis but shown in red to signify an angle of $$\varphi=\pi$$, where $$\mathrm{e}^{j\pi}=-1$$.
To draw a modulus surface, you can either use the model below, or you can using the code shown in Appendix A. Click Interact, wait for the model to load and click on $$|f(z)|$$.
## Appendix A
clear
[x,y] = meshgrid(-10: 0.1: 10);
z = x + i*y;
fz = z.^2 - 4.*z + 8;
surf(x,y,abs(fz), angle(fz));
zlim([0,20]);
xlabel("x"); ylabel("y"); zlabel ("|f(x+jy)|");
## References
[1] The Complex Roots of a Quadratic Equation: A Visualization Carmen Q Artino, Professor in Mathematics at The College of Saint Rose, Albany, NY. Parabola Volume 45, Issue 3 (2009)
## Functions of complex numbers
This introduces the functions with complex arguments. The article Complex Numbers introduced a 2-dimensional number space called the complex-plane ($$\mathbb{C}$$-plane). The arithmetic functions, that we studied since first grade, gracefully extend from the one-dimensional number line onto this new $$\mathbb{C}$$-plane. Here we will introduce functions that operate on these complex numbers.
$$j$$ We refer to the imaginary unit as “$$j$$”, to avoid confusion with electrical engineering, where the variable $$i$$ is already used for current.
An overview of the functions is given for reference. We will proof the some of these functions in subsequent paragraphs.
## Overview
Consider a complex number $$z$$ expressed in either notation style
$$z = x+jy=r\,(\cos\varphi+j\sin\varphi)=r\,\mathrm{e}^{j\varphi}\nonumber$$
As you wish
\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align} z_1+z_2&=(x _1+x _2)+j\,(y _1+y _2) \\[6mu] z_1-z_2&=(x _1-x _2)+j\,(y _1-y _2) \\[6mu] z_1\,z_2&=r_1r_2\ \mathrm{e}^{j \cdot (\varphi_1+\varphi_2)} \\[6mu] \tfrac{1}{z} &= \tfrac{1}{r}\,\mathrm{e}^{-j\varphi}\\[6mu] \frac{z_2}{z_1} &= \frac{r_1}{r_2}\,\mathrm{e}^{j(\varphi _1-\varphi_2)}\\[6mu] z_1\parallelsum z_2 &= \frac{z_1\, z_2}{z_1+z_2}\\[6mu] \mathrm{e}^z &=\mathrm{e}^x\sin y + j\,\mathrm{e}^x\cos y\\[6mu] \ln z&= \ln r+j\,\varphi\\[6mu] {z_2}^{z_1} &= {r_1}^{x _2}\,\mathrm{e}^{-y_2\,\varphi_1}\,\mathrm{e}^{j \cdot (x _2\,\varphi_1+\,y_2\ln r_1)} \\[6mu] \sqrt[n]{z} &= \sqrt[n]{r}\,\mathrm{e}^{j\varphi/n} \end{align}
Circular based trigonometry
\begin{align} \sin z &= \sin x\cosh y + j\,\cos x\sinh y \\[6mu] \cos z &= \cos x\cosh y + j\,\sin x\sinh y \\[6mu] \tan z &= \frac{\sin(2 x)}{\cosh(2 y) + \cos(2 x)} + j\,\frac{\sinh(2 y)}{\cosh(2 y) + \cos (2 x)} \\[6mu] \csc z &= {(\sin z)}^{-1} \\[6mu] \sec z &= {(\cos z)}^{-1} \\[6mu] \cot z &= {(\tan z)}^{-1} \\[6mu] \end{align}
Inverse circular based trigonometry
\DeclareMathOperator{\asin}{asin} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\acos}{acos} \DeclareMathOperator{\atan}{atan} \DeclareMathOperator{\acsc}{acsc} \DeclareMathOperator{\asec}{asec} \DeclareMathOperator{\acot}{acot} \begin{align} \asin z &= \asin b +j\,\sgn(y)\ln\left(a + \sqrt{a^{\mathrm{e}}}-1\right), \quad a\geq1 \land b \text{ in } [\mathrm{rad}]\\[6mu] \acos z &= \acos b +j \sgn(y) \ln\left(a + \sqrt{a^{\mathrm{e}}}-1\right),\quad a\geq1 \land b \text{ in } [\mathrm{rad}]\\[6mu] \text{where}\quad a &= \tfrac{1}{2} \left( \sqrt{(x +1)^{2} + y ^{2} } + \sqrt{ (x -1)^{2} + y^{2}} \right),\nonumber \\[6mu] b &= \tfrac{1}{2} \left( \sqrt{(x +1)^{2} + y ^{2} } – \sqrt{ (x -1)^{2} + y^{2}} \right),\nonumber \\[6mu] \sgn(a) &= \begin{cases}-1 & a \lt 0\\[6mu]1 & a \geq 0\end{cases} \nonumber \\[6mu] \atan z &= \tfrac{1}{2}\left(\pi – \atan\left(\frac{1+ y}{x}\right) -\atan\left(\frac{1-y}{x}\right)\right) \\ &\quad +j\,\tfrac{1}{4}\,\ln\left( \frac{\left(\frac{1+y}{x}\right)^2 +1}{\left(\frac{1-y}{x}\right)^2 +1} \right) \\[6mu] \acsc z &= \asin(z^{-1}) \\[6mu] \asec z &= \acos(z^{-1}) \\[6mu] \acot z &= \atan(z^{-1}) \\[6mu] \end{align}
Hyperbolic based trigonometry
\DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\sech}{sech} \begin{align} \sinh z &= \cos y \sinh x + j\,\sin y\cosh x \\[6mu] \cosh z &= \cos y \cosh x + j\,\sin y\sinh x \\[6mu] \tanh z &= \frac{\sinh(2y)}{\cosh(2x)} +j\,\frac{\sin(2 y)}{\cosh(2 x) + \cos(2y)}\\[6mu] \csch z &= {(\sinh z)}^{-1} \\[6mu] \sech z &= {(\cosh z)}^{-1} \\[6mu] \coth z &= {(\tanh z)}^{-1} \\[6mu] \end{align}
Inverse hyperbolic based trigonometry
\DeclareMathOperator{\asin}{asin} \DeclareMathOperator{\acos}{acos} \DeclareMathOperator{\atan}{atan} \DeclareMathOperator{\acsc}{acsc} \DeclareMathOperator{\asec}{asec} \DeclareMathOperator{\acot}{acot} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\asinh}{asinh} \DeclareMathOperator{\acosh}{acosh} \DeclareMathOperator{\atanh}{atanh} \DeclareMathOperator{\acsch}{acsch} \DeclareMathOperator{\asech}{asech} \DeclareMathOperator{\acoth}{acoth} \begin{align} \asinh z &= -j \asin(jz) \\[6mu] \acosh z &= j \acos z \\[6mu] \atanh z &= -j \atan(jz) \\[6mu] \acsch z &= j \acsc(jz) \\[6mu] \asech z &= -j \asec z \\[6mu] \acoth z &= j \acot(jz) \\[6mu] \end{align}
This table formed the basis of software like Complex Arithmetic for HP-41cv/cx.
## Proofs
Without further ado, we introduce the proofs for some common complex functions
Consider adding the numbers $$z_1$$ and $$z_2$$ in cartesian form
$$z_1+z_2 = (x_1+i\,y_1)+(x_2+i\,y_2)$$
Thus
$$\shaded{z_1+z_2=(x_1+x_2)+i(y_1+y_2)}$$
This can be visualized similar to adding real numbers by putting the vectors head to tail
### Subtraction
Consider subtracting the numbers $$z_1$$ and $$z_2$$ in cartesian form
$$z_1-z_2 = (x_1+i\,y_1) – (x_2+i\,y_2)$$
So that
$$\shaded{ z_1-z_2=(x_1-x_2)+i(y_1-y_2) }$$
This can be visualized similar to the subtraction of real numbers by rotating the subtrahend by $$\pi$$ and the putting them head to tail
### Multiplication
Consider the product $$z_1\,z_2$$ in polar form, using the trig identities
\begin{align} \cos\alpha\cos\beta-\sin\alpha\sin\beta&=\sin(\alpha+\beta) \nonumber \\ \sin\alpha\cos\beta+\cos\alpha\sin\beta&=\sin(\alpha+\beta) \nonumber \end{align}\nonumber
\require{enclose} \begin{align} z_1\,z_2&=r_1\enclose{phasorangle}{\small\varphi_1}\ r_2\enclose{phasorangle}{\small\varphi_2}\\ &=r_1 (\cos\varphi_1+i\sin\varphi_1)\ r_2 (\cos\varphi_2+i\sin\varphi_2)\nonumber\\ &=r_1r_2\,((\cos\varphi_1\cos\varphi_2-\sin\varphi_1\sin\varphi_2)+i\,(\cos\varphi_1\sin\varphi_2+\sin\varphi_1\cos\varphi_2)) \end{align}
From what follows that
$$\shaded{ z_1\,z_2=r_1r_2\,(\cos(\varphi_1+\varphi_2)+i\,\sin(\varphi_1+\varphi_2) }$$
where
\begin{align} |z_1\,z_2|&=|z_1|\,|z_2|\\[6mu] \angle(z_1\,z_2)&=\angle z_1+\angle z_2 \end{align}
This can be visualized as adding the angles and multiplying the lengths of the vectors
### Division
Consider the quotient $$\frac{z_1}{z_2}$$ in polar form
\require{enclose} \begin{align} \frac{z_1}{z_2}&=\frac{r_1\enclose{phasorangle}{\small\varphi_1}}{r_2\enclose{phasorangle}{\small\varphi_2}} =\frac{r_1 \left({\cos \varphi_1 + i \sin \varphi_1}\right)} {r_2 \left({\cos \varphi_2 + i \sin \varphi_2}\right)}\nonumber\\[6mu] &=\frac{r_1}{r_2}\,\frac{\cos\varphi_1+i\sin\varphi_1}{\cos\varphi_2+i\sin\varphi_2}\ \frac{\cos \varphi_2 – i \sin \varphi_2}{\cos \varphi_2 – i \sin \varphi_2},&\text{product rule}\nonumber\\[6mu] &=\frac{r_1}{r_2}\,\frac{\cos \left({\varphi_1 – \varphi_2}\right) + i \sin \left({\varphi_1 – \varphi_2}\right)}{\cos \left({\varphi_2 – \varphi_2}\right) + i \sin \left({\varphi_2 – \varphi_2}\right)}\nonumber\\[6mu] &=\frac{r_1}{r_2}\,\frac{\cos \left({\varphi_1 – \varphi_2}\right) + i \sin \left({\varphi_1 – \varphi_2}\right)} {\cos 0 + i \sin 0}\\[6mu] \end{align}
So that
$$\shaded{ \frac{z_1}{z_2}=\frac{r_1}{r_2}{\Large(}{\cos \left({\varphi_1 – \varphi_2}\right) + i \sin \left({\varphi_1 – \varphi_2}\right)}{\Large)} }$$
where
\begin{align} \left|\frac{z_1}{z_2}\right|&=\frac{|z_1|}{|z_2|}\\[6mu] \angle\frac{z_1}{z_2}&=\angle z_1-\angle z_2 \end{align}
This can be visualized as subtracting the angles and dividing the lengths of the vectors
### $$n$$th power
Consider the power $$z^n$$ in polar form, where $$n\in\mathbb{Z}^+\}$$
$$z^n=(r(\cos\varphi+i\sin\varphi))^n\nonumber\\$$
Using Euler’s formula
$$\cos\phi+i\sin\phi=\mathrm{e}^{i\phi}\nonumber$$
\require{enclose} \begin{align} z^n&=\left(r\enclose{phasorangle}{\small\varphi}\right)^n\\ &=\left(r\,\mathrm{e}^{i\varphi}\right)^n\nonumber\\ &=r^n\,\mathrm{e}^{in\varphi},&\text{Euler’s formula} \end{align}
So that
$$\shaded{ z^n=r^n\,(\cos(n\varphi)+i\sin(n\varphi)) }$$
where
\begin{align} |z^n| &= {|z|}^n\\[6mu] \angle(z^n) &= n\angle z \end{align}
This can be visualized multiplying the angles with $$n$$ and taking the $$n$$th power of the length of the vector
### $$n$$th root
Consider the $$n$$th root $$\sqrt[n]{z}$$ in polar form, where $$n\in\mathbb{Z}^+\}$$
\require{enclose} \begin{align} \sqrt[n]{z}&=\sqrt[n]{r\enclose{phasorangle}{\small\varphi}} =\left(r\,\mathrm{e}^{i\varphi}\right)^{\frac{1}{n}},&\text{Euler’s formula}\nonumber\\ &=r^{\frac{1}{n}}\,\mathrm{e}^{i(\varphi+2k\pi)/n},\quad k\in\mathbb{Z},&\text{Euler’s formula} \end{align}
Therefore
$$\shaded{ \sqrt[n]{z}=\sqrt[n]{r}\,\left(\cos\frac{\varphi+2k\pi}{n}+i\sin\frac{\varphi+2k\pi}{n}\right),\quad k\in\mathbb{Z} } \label{eq:root}$$
where
\begin{align} |\sqrt[n]{z}|&=\sqrt[n]{|z|},\\[6mu] \angle\,\sqrt[n]{z}&=\frac{\angle z+2k\pi}{n},\quad k\in\mathbb{Z} \end{align}
This can be visualized dividing the angles by $$n$$ and taking the $$n$$th root of the length of the vector. The other vectors will be separated by $$\frac{2\pi}{n}$$ radians.
#### Wait a minute
Depending on how we measure the angle $$\varphi$$, we get different answers? Correct, because adding $$2k\pi$$ to $$\varphi$$ still maps to the same complex number, but may give a different function value.
In comparison, the functions that we saw described do not produce different results when adding extra rotations to the angle. Other multivalued functions are $$\log{z}$$, $$\mathrm{arcsin}z$$ and $$\mathrm{arccos}z$$.
In general:
the $$n$$th root has $$n$$ values,
because when we add $$2k\pi$$ to the angle $$\varphi$$, for $$k\in\mathbb{Z}$$, we may get different results.
The big question becomes: how do we define the angle $$\varphi$$?
#### Multi-valued
Even real valued functions can have multiple values. Remember $$\sqrt{1}=\{-1,1\}$$? Using equation $$\eqref{eq:root}$$, we find the function values that we are familiar with.
\begin{align} \sqrt{1}&=\sqrt{\cos\varphi+i\sin\varphi},&\text{polar notation}\nonumber\\ &=\cos\frac{\varphi+2k\pi}{2}+i\sin\frac{\varphi+2k\pi}{2},&\text{equation }\eqref{eq:root}\nonumber\\ &=\cos\frac{\varphi+2k\pi}{2},&k\in\mathbb{Z}\nonumber\\ &=\left\{1,-1\right\} \end{align}
all roots have magnitude $$1$$, but their angles $$\varphi$$ are $$\pi$$ apart.
Similarly, the cube root $$\sqrt[3]{1}$$ has three roots, two of which are complex. All roots have magnitude $$1$$, but their angles $$\varphi$$ are $$\frac{2\pi}{3}$$ apart.
\require{enclose} \begin{align} y_1&=1\,\enclose{phasorangle}{0}=\cos0+i\sin0=1\nonumber\\[8mu] y_2&=1\,\enclose{phasorangle}{\small\tfrac{2\pi}{3}}=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}=-\tfrac{1}{2}+\tfrac{1}{2}\sqrt{3}i\nonumber\\ y_3&=1\,\enclose{phasorangle}{\small\tfrac{4\pi}{3}}=\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}=-\tfrac{1}{2}-\tfrac{1}{2}\sqrt{3}i\nonumber \end{align}\nonumber
#### Making it single-valued
For real valued arguments, we conventionally choose $$\varphi$$ in the range $$[0,2\pi)$$ where the function is single-valued and where we find a positive function value. This default single-value is called the principal value.
Besides that the function $$\sqrt[n]{z}$$ is not differentiable at $$0$$, it has no discontinuities. To make the function single-valued, we can limit the range of $$\varphi$$ similar to what we usually do for real valued arguments. The technical term for this is branch cut. We then only express $$\varphi$$ so that it doesn’t cross the branch cut. Some common branch cuts are shown in the table below. In the table $$\mathbb{R}^-$$ stands for the negative real axis.
Example branch cuts and their effect on the function value
Branch cut Range for $$\varphi$$ Effect Consistent with
just under $$\mathbb{R}^-$$ $$(-\pi,\pi]$$ $$\Re(z)\geq0$$ Sqrt of real numbers
just under $$\mathbb{R}^+$$ $$[0,2\pi)$$ $$\Im(z)\geq0$$ Phase shift in waves
No matter where you define the branch cut, when $$z$$ approaches a point on the branch cut from opposite sides, either the real or imaginary part of the function value abruptly changes signs. In practice, the best place for the branch cut depends on the application. For instance, it there is already a discontinuity at the point $$-1$$, you may as well put the branch cut just under $$\mathbb{R}^-$$.
We will use the $$\mathbb{C}$$-plane extensively as we explore the physic fields of electronics and domain transforms.
## Complex numbers
Instead of projecting the future merits of complex numbers, we will introduce them in an intuitive way. We draw a parallel to negative numbers that have been universally accepted around the same time.
We start this writing with a review of concepts that should be evident. Nevertheless, I encourage you to read through them, as we build on these concepts while introducing complex numbers.
## Review
Arithmetic gives us tools to manipulate numbers. It allows us to transform one number into another using transformations such as negation, addition, subtraction, multiplication and division.
### Positive numbers
In first grade, we learned the concept of the number line and how numbers can be represented by vectors starting at $$0$$. We visualized addition by putting these vectors head to tail, where the net length and direction is the answer.
Soon thereafter, we learned how to subtract numbers by rotating the subtrahend (the value that you subtract) vector and then putting the head to tail.
We will expand on this as we discuss negative numbers and imaginary numbers. Before we introduce such numbers, let’s also refresh on the concept of equations with squares and square roots.
### Square and square root
When we solve the equation $$2x^2=8$$, we look for a transformation ($$\times x$$) that, when applied twice, turns the number $$2$$ into $$8$$.
$$2 \times x \times x = 8$$
As shown in the animation below, the two solutions $$x=2$$ and $$x=-2$$, both satisfy the equation $$2x^2=8$$.
## Negative numbers
Negative numbers have lingered around since 200 BC, but with mathematics based on geometrical ideas such as length and count, there was little place for negative numbers. After all, how can a pillar be less than nothing in height? How could you own something less than nothing?
Even a hundred years after the invention of algebra in 1637, the answer to $$4=4x+20$$ would be thought of as absurd as illustrated by the quotes:
Negative numbers darken the very whole doctrines of the equations and make dark of the things which are in their nature excessively obvious and simple. Francis Maseres, British mathematician (1757)
To really obtain an isolated negative quantity, it would be necessary to cut off an effective quantity from zero, to remove something of nothing: impossible operation. Lazare Carnot, French mathematician (1803)
### Acceptance
Some mathematicians in the 17th century discovered that negative numbers did have their use in solving cubic and quadratic equations, provided they didn’t worry about the meaning of these negative numbers. While the intermediate steps of their calculations may involve negative numbers, the solutions were typically real positive numbers.
Only in the 19th century were negative numbers truly accepted when mathematicians started to approach mathematics in terms of logical definitions.
Physical meaning has given way to algebraic use.
The English mathematician, John Wallis (1616 – 1703) is credited with giving meaning to negative numbers by inventing the number line. Even today, the number line helps students as they actively construct mathematical meaning, number sense and an understanding of number relationships. We learned to use negative numbers without a thinking about the thousands of years it took to develop the principle.
You were probably introduced to the concepts of absolute value and direction through the geometric representation on this number line. With negative numbers so embedded in our mathematics, we accept the solution to $$3-5$$ without a second thought.
In general, we learned that the negative symbol represents the “opposite” of a number. To change the sign of a number, we rotate its vector 180 degrees ($$\pi$$) around the point $$0$$. We will extend on this important concept as we introduce imaginary numbers.
## Imaginary numbers
First off, imaginary numbers are called such because for a long time they were poorly understood and regarded by some as fictitious. Even Euler, who used them extensively, once wrote:
Because all conceivable numbers are either greater than zero or less than 0 or equal to 0, then it is clear that the square roots of negative numbers cannot be included among the possible numbers [real numbers]. Consequently we must say that these are impossible numbers. And this circumstance leads us to the concept of such number, which by their nature are impossible, and ordinarily are called imaginary or fancied numbers, because they exist only in imagination. Leonhard Euler, Swiss mathematician and physicist, Introduction to Algebra pg.594
### History
Around the same time that mathematicians were struggling with the concept of negative numbers, they also came across square roots of negative numbers.
Girolamo Cardano, an Italian mathematician who published the solution to cubic and quartic equations, studied a problem that in modern algebra would be expressed as $$x^2-10x+40=0\,\land\,y=10-x$$
Find two numbers whose sum is equal to 10 and whose product is equal to 40. Translation of Ars Magna chapter 37, pg 219 (1545) by Girolamo Cardano
Cardano states that it is impossible to find the two numbers. Nonetheless, he says, we will proceed. He goes on to provide two objects that satisfy the given condition. Cardano found a sensible answer by working through his algorithm, but he called these numbers “ficticious” because not only did they disappear during the calculation, but they did not seem to have any real meaning.
This subtlety results from arithmetic of which this final point is as I have said as subtle as it is useless. Translation of Ars Magna chapter 37, pg 219 (1545) by Girolamo Cardano
### Acceptance
In 1849, Carl Friedrich Gauß, produced a rigorous proof for complex numbers what gave a big boost to the acceptance of these numbers in the mathematical community.
With this historic perspective, let’s see how these imaginary numbers fit into what we know about number theory.
We learned that the negative symbol represents the “opposite” of a number. On the number line we can represent this by rotating the vector that represents the number 180° ($$\pi$$) around the origin $$0$$.
Let’s dive straight in and consider the equation
\begin{align} x^2&=-1\\ \Rightarrow\quad 1\times x\times x&=-1\label{eq:1tomin1} \end{align}
What multiplication with $$x$$, when applied twice, turn $$1$$ into $$-1$$? Multiplying twice by a positive number gives a positive result. Same for a negative number.
Time to take a step back: we said that a negation represents a rotation of $$\pi$$ around the origin. What if we rotate the vector $$1$$ by $$\frac{\pi}{2}$$ twice, and worry about its meaning later.
Indeed, twice rotating the vector $$1$$ around the origin by $$\frac{\pi}{2}$$ gives us $$-1$$. All that we have left to do, is to find a name for the vector where $$1$$ is rotated by $$\frac{\pi}{2}$$. To credit Euler’s “imaginary or fancied numbers”, we call it $$i$$. The first multiplication turns $$1$$ into $$i$$, and the second multiplication turns $$i$$ into $$-1$$.
More over, rotating in the opposite direction works as well. The first multiplication turns $$1$$ into $$-i$$, and the second multiplication turns $$-i$$ into $$-1$$. So there are two square roots of $$-1$$: $$i$$ and $$-i$$. We have two solutions to $$x^2=-1$$!
### Interpretation of “$$i$$”
What is the meaning of this mysterious value $$i$$? The first rotation turned $$1$$ into $$i$$, so the rotation is a visualization of multiplying with $$i$$. Rotating $$i$$ once more turns $$i$$ into $$-1$$.
Substituting $$x=i$$ in $$\eqref{eq:1tomin1}$$ implies that $$i\times i=-1$$ or written as
$$\shaded{ i^2=-1 }$$
By introducing an axis perpendicular to the number line, we have extended or number space to a two dimensional plane called $$\mathbb{C}$$. This $$\mathbb{C}$$-plane includes the real numbers from the real number line, along with imaginary numbers on the $$i$$-axis and every combination thereof. We call this new number set “Complex Numbers“.
### Notation
By introducing $$i$$, we have added a dimension to the number line. With that come three new notation forms that each have their own use case. We will express the complex number $$z$$ as shown below in these forms.
#### Cartesian form
Each complex number has a real part $$x$$ and an imaginary part $$y$$, where $$x$$ and $$y$$ are real numbers. Using $$i$$ as the imaginary unit, we can denote a complex number $$z$$ as
$$\shaded{z=x+iy}$$
The point $$z$$ can be specified by its rectangular coordinates $$(x,y)$$, where $$x$$ and $$y$$ are the signed distances to the imaginary $$y$$ and real $$x$$-axis. This $$xy$$-plane is commonly called the complex plane $$\mathbb{C}$$.
This cartesian form is a logical extension of the number line and will prove useful when adding or subtracting complex numbers.
#### Polar form
The same point $$z$$ can be specified by its polar coordinates $$(r,\varphi)$$, where $$r$$ is the distance to the origin and $$\varphi$$ is the angle of the vector, in radians, with the positive $$x$$-axis. With $$r\in\mathbb{R}^+$$ and $$\varphi\in\mathbb{R}$$, we can describe point $$z$$ as
$$\require{enclose} \shaded{ r\enclose{phasorangle}{\small\varphi}=r\,(\cos\varphi+i\sin\varphi) }$$
here $$r$$ corresponds to modulus $$|z|$$, and $$\varphi$$ is called the argument. The value $$z=0$$ was excluded because the angle $$\varphi$$ is not defined that that point.
The polar form simplifies the arithmetic when used in multiplication or powers of complex numbers.
From the illustration, it is clear how to convert from cartesian to polar form
\shaded{ \begin{align} x&=r\cos\varphi\nonumber\\ y&=r\sin\varphi\nonumber \end{align}} and back \shaded{\begin{align} r&=\sqrt{x^2+y^2}\nonumber\\[6mu] \varphi&=\mathrm{atan2}(y,x)\nonumber\\[10mu] \end{align} }
Here $$\mathrm{atan2(y,x)}$$ prevents negative signs in $$\arctan\frac{y}{x}$$ from canceling each other out. Otherwise, we would not be able to distinguish $$\varphi$$ in the 1st from that in the 3rd quadrant, or $$\varphi$$ in the 2nd from that in the 4th.
\begin{align} \mathrm{atan2}(y,x) &= \begin{cases} \arctan\left(\frac{y}{x}\right) & x\gt0\nonumber\\ \arctan\left(\frac{y}{x}\right)+\pi & x\lt0 \land y\geq0\nonumber\\ \arctan\left(\frac{y}{x}\right)-\pi & x\lt0 \land y\lt0\nonumber\\ \frac{\pi}{2} & x= 0 \land y\gt0\nonumber\\ -\frac{\pi}{2} & x= 0 \land y\lt0\nonumber\\ \text{undefined} & x= 0 \land y = 0\nonumber \end{cases} \end{align}
Consider complex number $$z$$ with angle $$\varphi_0$$. If you make any integer number of rotation rotate around the origin, you will be back the your initial starting point. Since a full rotation corresponds to an angle of $$2\pi$$, the same point $$z$$ can be described as
$$\require{enclose} r\enclose{phasorangle}{\small\varphi+2k\pi}=r\,(\cos(\varphi+2k\pi)+i\sin(\varphi+2k\pi)),\quad k\in\mathbb{Z}$$
We will the effects of this further when discussing multi-valued functions such as square root.
#### Exponential form
Euler’s formula was introduced in a separate write-up as
$$\mathrm{e}^{i\varphi} = \cos\varphi+j\sin\varphi \nonumber$$
Using Euler’s formula we can rewrite the polar form of a complex number into its exponential form
$$\shaded{ z=r\,\mathrm{e}^{i\varphi} }$$
Similar to the polar form, the angle can be expressed in infinite different ways
\begin{align} z &= r\,\mathrm{e}^{i(\varphi+2k\pi)}, & k\in\mathbb{Z} \end{align}
This exponential form is often preferred over the polar form, because it reduces the need for trigonometry.
## What next?
My follow-up article Complex Functions introduces functions that operate on complex numbers. Such functions include addition, subtraction, multiplication to the most obscure trigonometry functions.
We will use the $$\mathbb{C}$$-plane extensively as we explore electronics and domain transforms. From here on
we will refer to the imaginary unit as “$$j$$”, to avoid confusion with electronics where the variable $$i$$ is already used for electrical current.
The same Leonard Euler that once said these numbers “only exist in our imagination” also used imaginary numbers to unite trigonometry and analysis in his most beautiful formula.
## Partial fraction expansion
Describes particle fraction expansion using the Heaviside method. Decompose rational functions of polynomials in Laplace or Z-transforms.
Oliver Heaviside (1850-1925), was an English electrical engineer, mathematician and physicist who among many things adapted complex numbers to the study of electrical circuits. He introduced a method to decompose rational function of polynomials as they occur when using the Laplace transform to solve differential equations.
Whenever the denominator of a rational function can be factored into distinct linear factors, the fraction can be expressed as the sum of partial fractions.
\begin{align} f(x)=\frac{b_0+b_1x+b_2x^2+b_3x^3+\ldots+b_Mx^{M}}{1+a_1x+a_2x^2+a_3x^3+\ldots+a_Nx^N}\nonumber\\[10mu] \triangleq\frac{b_0+b_1x+b_2x^2+b_3x^3+\ldots+b_Mx^{M}}{(x-r_1)(x-r_2)(x-r_3)\ldots(x-r_N)}\label{eq:factors} \end{align}
A rational function $$\eqref{eq:factors}$$ with $$N$$ distinct poles $$r_k$$, can be expressed as a summation of simple fractions
$$\shaded{ \frac{b_0+b_1x+b_2x^2+b_3x^3+\ldots+b_Mx^{M}}{(x-r_1)(x-r_2)(x-r_3)\ldots(x-r_N)}\triangleq\sum_{k=1}^N \frac{c_k}{x-r_k} }$$
Written out as
$$\frac{b_0+b_1x+b_2x^2+b_3x^3+\ldots+b_Mx^{M}}{(x-r_1)(x-r_2)(x-r_3)\ldots(x-r_N)}=\frac{c_1}{x-r_1}+\frac{c_2}{x-r_2}+\cdots+\frac{c_{\small N}}{x-r_{\small N}}\label{eq:distinctpoles}$$
The constants $$c_k$$ can be obtained by dividing out the $$(x-r_k)$$ factor in equation $$\eqref{eq:distinctpoles}$$ and evaluating at $$x=r_k$$
$$c_k = \left.\frac{b_0+b_1x+b_2x^2+b_3x^3+\ldots+b_Mx^{M}}{(x-r_1)(x-r_2)(x-r_3)\ldots(x-r_N)}\right|_{x=r_k}$$
## Non-distinct poles
If a pole $$r_k$$ is not distinct, the partial fraction expansion for that multiple pole becomes a bit more involved. When the pole occurs $$q$$ times as in $$(x-r_k)^q$$, that pole expands to a summation of decreasing powers of $$(x-r_k)$$
$$\sum_{i=1}^q\frac{d_i}{(x-r_k)^i}\label{eq:nondistict}$$
where the constants $$d_i$$ follow to satisfy
$$d_i=\left.\frac{1}{(q-i)!}\ \left( \frac{\mathrm{d}^{q-i}}{\mathrm{d}x^{q-i}}\ (x-r_k)^q\ f(x) \right)\right|_{x=r_k}$$
In practice, it is easier to first find $$d_q$$ by dividing out the highest power $$(x-r_k)^q$$ in equation $$\eqref{eq:factors}$$ and evaluating at $$x=r_k$$
$$d_q=\left.\frac{b_0+b_1x+b_2x^2+b_3x^3+\ldots+b_Mx^{M}}{(x-r_1)(x-r_2)(x-r_3)\ldots(x-r_N)}\right|_{x=r_k}$$
The remaining $$d_k$$ constants are found by substituting known constants and matching like powers.
## For example
his is one of those cases where examples might shed some light on the topic.
### Example 1
Consider the rational polynomial
$$f(x)=\frac{-8+24x}{1-2x+x^2}\label{eq:example}$$
The denominator is a second-order polynomial. The roots of any quadratic equation $$ax^2+bx+c=0$$ are
$$r_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \nonumber$$
The two roots of the denominator polynomial in equation $$\eqref{eq:example}$$ are
$$r_{1,2 } = \frac{2\pm\sqrt{4-4}}{2}\quad\Rightarrow\quad\shaded{r_{1,2}=r=1}$$
This gives equation $$\eqref{eq:example}$$ with a factorized denominator
$$f(x) = \frac{-8+24x}{\left(1-x\right)^2}$$
According to $$\eqref{eq:nondistict}$$, this can be expressed as partial fractions
$$f(x)=\frac{-8+24x}{(1-x)^2}\triangleq\frac{c_1}{(1-x)^2}+\frac{c_2}{1-x} \label{eq:heavyside2}$$
Find the constant $$c_1$$ using using Heaviside’s cover up method: multiply all terms by $$(1-x)^2$$ and then evaluate for $$1-x=0$$ so that only the term $$c_1$$ is left on the right
$$c_1=\left.\frac{-8+24x}{((1-x)^2}\right|_{x=1}=-8+24\Rightarrow \shaded{c_1=16}$$
Given the value of $$c_1$$, constant $$c_2$$ can be found by substituting any numerical value (other than $$0$$ in equation $$\eqref{eq:heavyside2}$$
\begin{align} \frac{-8+24x}{(1-x)^2} &= \frac{c_1}{(1-x)^2}+\frac{c_2}{1-x} \Big|_{x=2,\,c_1=16} \implies \nonumber \\ \frac{-8+48}{(1-2)^2} &= \frac{16}{(1-2)^2}+\frac{c_2}{1-2} \implies \nonumber \\ 40 &= 16-c_2\Rightarrow\shaded{c_2=-24} \end{align}
The sum of partial fractions follows as
$$\shaded{ f(x)=\frac{-24}{1-x}+\frac{16}{\left(1-x\right)^{2}} }$$
### Example 2
Consider the rational polynomial where the order of the numerator is the same as that of the denominator
$$f(x) = \frac{2x^3+x^2-x+4}{(x-2)^3}$$
Make the degree of the numerator $$1$$ less than that of the denominator by dividing both sides by $$x$$. This makes $$\frac{f(x)}{z}$$ is a strictly proper rational function. According to Heaviside, this can be expressed as partial fractions
$$\frac{f(x)}{x} = \frac{2x^3+x^2-x+4}{x(x-2)^3}\triangleq\frac{c_1}{x}+\frac{c_2}{(x-2)^3}+\frac{c_3}{(x-2)^2}+\frac{c_4}{x-2}\label{eq:example2a}$$
Multiply both sides of equation $$\eqref{eq:example2a}$$ by $$x$$ so that only $$c_0$$ is left on the right; then evaluate for $$x=0$$
$$c_1 = \left.\frac{2x^3+x^2-x+4}{(x-2)^3}\right|_{x=0}=\frac{4}{(-2)^3}\Rightarrow \shaded{c_1=-\frac{1}{2}}$$
Use Heaviside’s cover up method to find the constant $$c_3$$. (multiply both sides with $$(x-2)^3$$ so that only $$c_1$$ is left on the right; then evaluate for $$x=2$$
$$c_2 = \left.\frac{2x^3+x^2-x+4}{x}\right|_{x=2}=\frac{16+4-2+4}{2}\Rightarrow \shaded{c_2=11}$$
Bring all the terms into a common denominator
\begin{align} \frac{2x^3+x^2-x+4}{x(x-2)^3}&=\frac{c_1}{x}+\frac{c_2}{(x-2)^3}+\frac{c_3}{(x-2)^2}+\frac{c_4}{x-2}\Rightarrow\nonumber\\[10mu] \frac{2x^3+x^2-x+4}{x(x-2)^3}&=\frac{c_1(x-2)^3}{x(x-2)^3}+\frac{c_2x}{x(x-2)^3}+\frac{c_3x(x-2)}{x(x-2)(x-2)^2}+\frac{c_4x(x-2)^2}{x(x-2)^2(x-2)}\Rightarrow\nonumber\\[10mu] 2x^3+x^2-x+4&=c_1(x-2)^3+c_2x+c_3x(x-2)+c_4x(x-2)^2 \end{align}
Substitute the known values $$c_1$$ and $$c_2$$, expand and regroup
\begin{align} 2x^3+x^2-x+4&=-\frac{1}{2}(x^3-6x^2+12x-8)+11x+c_3x(x-2)+c_4x(x-2)^2\quad\Rightarrow\nonumber\\ 2x^3+x^2-x\cancel{+4}&=-\frac{1}{2}x^3+3x^2-6x\cancel{+4}+11x+c_3x^2-2c_3x+c_4x^3-4c_4x^2+4c_4x\ \Rightarrow\nonumber\\ 2x^3+x^2-x&=(c_4-\frac{1}{2})x^3+(3-4c_4+c_3)x^2+(-6+11+4c_4-2c_3)x \end{align}
Match corresponding powers of $$x$$
\left. \begin{align} 2\cancel{x^3}&=(c_4-\frac{1}{2})\cancel{x^3}\Rightarrow 2=c_4-\frac{1}{2}\Rightarrow \shaded{c_4=\frac{5}{2}}\nonumber\\ \cancelto{1}{x^2}&=(3-4c_4+c_3)\cancel{x^2}\Rightarrow 1=3-4c_4+c_3\nonumber\\ -\cancelto{1}{x}&=(-6+11+4c_4-2c_3)\cancel{x}\Rightarrow -1=-6+11+4c_4-2c_3\nonumber\\ \end{align}\right\}\Rightarrow\nonumber
Given $$c_4$$, we only need one equation to solve for $$c_3$$
$$c_3 = 4c_4-2=10-2\Rightarrow \shaded{c_3=8}$$
Substitute the values of $$c_{1\ldots4}$$ in $$\eqref{eq:example2a}$$
$$\frac{f(x)}{x}=-\frac{\frac{1}{2}}{x}+\frac{11}{(x-2)^3}+\frac{8}{(x-2)^2}+\frac{\frac{5}{2}}{x-2}$$
Multiplying both sides by $$x$$ gives the sum of partial fractions
$$\shaded{f(x)=-\frac{1}{2}+11\frac{x}{(x-2)^3}+8\frac{x}{(x-2)^2}+\frac{5}{2}\frac{x}{x-2}}$$
Reference: MIT-cu
## Euler’s formula
Proofs Euler’s formula using the MacLaurin series for sine and cosine. Introduces Euler’s identify and Cartesian and Polar coordinates.
Around 1740, the Swiss mathematician, physicist and engineer Leonhard Euler obtained the formula later named after him.
## Euler’s Formula
Use the MacLaurin series for cosine and sine, which are known to converge for all real $$x$$, and the MacLaurin series for $$\mathrm{e}^x$$.
\begin{align} \cos x &= \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots \\ \sin x &= \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots \\ \mathrm{e}^z &=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots \label{eq:ez} \end{align}
Substitute $$z=j\varphi$$ in $$\eqref{eq:ez}$$
\begin{align} \mathrm{e}^{j\varphi}&=\sum_{n=0}^{\infty}\frac{(j\varphi)^n}{n!}=1+j\varphi+\frac{(j\varphi)^2}{2!}+\frac{(j\varphi)^3}{3!}+\cdots \nonumber\\ &=1+j\varphi-\frac{\varphi^2}{2!}-j\frac{\varphi^3}{3!}+\frac{\varphi^4}{4!}+j\frac{\varphi^5}{5!}-\cdots \nonumber\\ &=\underbrace{\left(1-\frac{\varphi^2}{2!}+\frac{\varphi^4}{4!}+\cdots\right)}_{\cos\varphi} +j\underbrace{\left(\varphi-\frac{\varphi^3}{3!}+\frac{\varphi^5}{5!}-\cdots\right)}_{\sin\varphi} \end{align}
This leads us to Euler’s formula
$$\shaded{ \mathrm{e}^{j\varphi}=\cos\varphi+j\sin\varphi }$$
## Euler’s Identify
For the special case where $$\varphi=\pi$$:
$$\mathrm{e}^{j\pi} = \cos\pi+j\sin\pi = -1$$
Rewritten as
$$\shaded{ \mathrm{e}^{j\pi}+1 = 0 }$$
This combines many of the fundamental numbers with mathematical beauty
• The number $$0$$, the additive identify
• The number $$1$$, the multiplicative identity
• The number $$\pi$$, the ratio between a circle’s circumference and its diameter
• The number $$j$$, used to find the roots of polynomials defined as $$j=\sqrt{-1}$$
• The number $$e$$, from continuous compounding interest and from $$\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^x=\mathrm{e}^x$$
## Cartesian and Polar coordinates
Euler’s formula traces out a unit circle in the complex plane as a function of $$\varphi$$. Here, $$\varphi$$ is the angle that a line connecting the origin with a point on the unit circle makes with the positive real axis, measured in radians.
A point in the complex plane can be represented by a complex number written in cartesian coordinates. Euler’s formula lets you convert between cartesian and polar coordinates. The polar form simplifies the mathematics when used in multiplication or powers of complex numbers. [wiki]
Any complex number $$z=x+jy$$ can be written as
$$\shaded{ z=x+jy=r(\cos\varphi+j\sin\varphi) = r\,\mathrm{e}^{j\varphi} }$$
where
\begin{aligned} x&=\Re\{z\} & \mathrm{real\ part\ of\ }z\\ y&=\Im\{z\} & \mathrm{imaginary\ part\ of\ }z\\ r&=|z|=\sqrt{x^2+y^2} & \mathrm{magnitude\ of\ }z\\ \varphi&=\angle z=\mathrm{atan2}(y,x) & \mathrm{angle\ of\ }z \end{aligned}
Another notation method
$$\shaded{r\,\mathrm{e}^{j\varphi} = \mathrm{e}^{\ln r}\mathrm{e}^{j\varphi} = \mathrm{e}^{\ln r+j\varphi}}$$
## Relation to trigonometry
Euler’s formula connects analysis with trigonometry. The connections most easy follow from adding or subtracting Euler’s formulas. [wiki]
\left\{ \begin{aligned} \mathrm{e}^{j\varphi}&=\cos\varphi+j\sin\varphi \\ \mathrm{e}^{-j\varphi}&=\cos(-\varphi)+j\sin(-\varphi)=\cos\varphi-j\sin\varphi \end{aligned} \right.
Adding gives the $$\cos\varphi$$
\left. \begin{aligned} \mathrm{e}^{j\varphi}&=\cos\varphi+j\sin\varphi \\ \mathrm{e}^{-j\varphi}&=\cos(-\varphi)+j\sin(-\varphi)=\cos\varphi-j\sin\varphi \end{aligned} \right\} \overset{add}{\Rightarrow}
\begin{align} \mathrm{e}^{j\varphi}+\mathrm{e}^{-j\varphi} &= (\cos\varphi+\cancel{j\sin\varphi}) + (\cos\varphi-\cancel{j\sin\varphi}) \nonumber \\ &= 2\cos\varphi \end{align}
So
$$\shaded{ \cos\varphi = \frac{\mathrm{e}^{j\varphi} + \mathrm{e}^{-j\varphi}}{2} } \label{eq:Eulers_sine}$$
Subtracting gives the $$\sin\varphi$$
\left. \begin{align} \mathrm{e}^{j\varphi}&=\cos\varphi+j\sin\varphi \nonumber\\ \mathrm{e}^{-j\varphi}&=\cos(-\varphi)+j\sin(-\varphi)=\cos\varphi-j\sin\varphi\nonumber \end{align} \right\} \overset{subtract}{\Rightarrow}
Subtraction
\begin{align} \mathrm{e}^{j\varphi}-\mathrm{e}^{-j\varphi} &= (\cancel{\cos\varphi}+j\sin\varphi) – (\cancel{\cos\varphi}-j\sin\varphi) \nonumber \\ &=2j\sin\varphi \end{align}
Thus
$$\shaded{ \sin\varphi = \frac{\mathrm{e}^{j\varphi} – \mathrm{e}^{-j\varphi}}{2j} } \label{eq:Eulers_cosine}$$
This concludes “Euler’s formula and identify”.
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Writing and Manipulating Inequalities
Learning Outcomes
• Use interval notation to express inequalities.
• Use properties of inequalities.
Indicating the solution to an inequality such as $x\ge 4$ can be achieved in several ways.
We can use a number line as shown below. The blue ray begins at $x=4$ and, as indicated by the arrowhead, continues to infinity, which illustrates that the solution set includes all real numbers greater than or equal to 4.
We can use set-builder notation: $\{x|x\ge 4\}$, which translates to “all real numbers x such that x is greater than or equal to 4.” Notice that braces are used to indicate a set.
The third method is interval notation, where solution sets are indicated with parentheses or brackets. The solutions to $x\ge 4$ are represented as $\left[4,\infty \right)$. This is perhaps the most useful method as it applies to concepts studied later in this course and to other higher-level math courses.
The main concept to remember is that parentheses represent solutions greater or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number. Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be “equaled.” A few examples of an interval, or a set of numbers in which a solution falls, are $\left[-2,6\right)$, or all numbers between $-2$ and $6$, including $-2$, but not including $6$; $\left(-1,0\right)$, all real numbers between, but not including $-1$ and $0$; and $\left(-\infty ,1\right]$, all real numbers less than and including $1$. The table below outlines the possibilities.
Set Indicated Set-Builder Notation Interval Notation
All real numbers between a and b, but not including a or b $\{x|a<x<b\}$ $\left(a,b\right)$
All real numbers greater than a, but not including a $\{x|x>a\}$ $\left(a,\infty \right)$
All real numbers less than b, but not including b $\{x|x<b\}$ $\left(-\infty ,b\right)$
All real numbers greater than a, including a $\{x|x\ge a\}$ $\left[a,\infty \right)$
All real numbers less than b, including b $\{x|x\le b\}$ $\left(-\infty ,b\right]$
All real numbers between a and b, including a $\{x|a\le x<b\}$ $\left[a,b\right)$
All real numbers between a and b, including b $\{x|a<x\le b\}$ $\left(a,b\right]$
All real numbers between a and b, including a and b $\{x|a\le x\le b\}$ $\left[a,b\right]$
All real numbers less than a or greater than b $\{x|x<a\text{ and }x>b\}$ $\left(-\infty ,a\right)\cup \left(b,\infty \right)$
All real numbers $\{x|x\text{ is all real numbers}\}$ $\left(-\infty ,\infty \right)$
Example: Using Interval Notation to Express All Real Numbers Greater Than or Equal to a
Use interval notation to indicate all real numbers greater than or equal to $-2$.
Try It
Use interval notation to indicate all real numbers between and including $-3$ and $5$.
Example: Using Interval Notation to Express All Real Numbers Less Than or Equal to a or Greater Than or Equal to b
Write the interval expressing all real numbers less than or equal to $-1$ or greater than or equal to $1$.
Try It
Express all real numbers less than $-2$ or greater than or equal to 3 in interval notation.
Using the Properties of Inequalities
When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equations. We can use the addition property and the multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number, we must reverse the inequality symbol.
A General Note: Properties of Inequalities
$\begin{array}{ll}\text{Addition Property}\hfill& \text{If }a< b,\text{ then }a+c< b+c.\hfill \\ \hfill & \hfill \\ \text{Multiplication Property}\hfill & \text{If }a< b\text{ and }c> 0,\text{ then }ac< bc.\hfill \\ \hfill & \text{If }a< b\text{ and }c< 0,\text{ then }ac> bc.\hfill \end{array}$
These properties also apply to $a\le b$, $a>b$, and $a\ge b$.
Illustrate the addition property for inequalities by solving each of the following:
1. $x - 15<4$
2. $6\ge x - 1$
3. $x+7>9$
Try It
Solve $3x - 2<1$.
Example: Demonstrating the Multiplication Property
Illustrate the multiplication property for inequalities by solving each of the following:
1. $3x<6$
2. $-2x - 1\ge 5$
3. $5-x>10$
Try It
Solve $4x+7\ge 2x - 3$.
Solving Inequalities in One Variable Algebraically
As the examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable.
Example: Solving an Inequality Algebraically
Solve the inequality: $13 - 7x\ge 10x - 4$.
Try It
Solve the inequality and write the answer using interval notation: $-x+4<\frac{1}{2}x+1$.
Example: Solving an Inequality with Fractions
Solve the following inequality and write the answer in interval notation: $-\frac{3}{4}x\ge -\frac{5}{8}+\frac{2}{3}x$.
Try It
Solve the inequality and write the answer in interval notation: $-\frac{5}{6}x\le \frac{3}{4}+\frac{8}{3}x$.
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# Percentages Revision: Step-by-Step Guidance
Whether you’re studying for your end-of-year/term exams or preparing for GCSE Maths, a spot of percentages revision never goes amiss!
To help you on your way to becoming 100% confident in the topic (or close to!), here’s a slice of magic taken from our expert-made resources at Beyond Maths.
The tips, methods and worked examples given below should help you boost your percentages revision and increase the effectiveness of your self-study for KS3 Maths.
Included in these notes are the subtopics ‘Percentage Increase and Decrease’, ‘Finding the Percentage of an Amount’, and ‘Original Value Problems’. You can also find a huge number of complementary worksheets and resources in our dedicated Percentages section.
### Percentages Revision Part 1 – Percentage Increase and Decrease
#### Percentage Multipliers
Percentage multipliers can be used to find percentages of an amount.
For example: Find 7% of 30.
• 7% as a decimal is 7 ÷ 100 = 0.07
This is the multiplier we use to find 7% of 30.
• 0.07 × 30 = 2.1
#### Percentage Increase and Decrease
To increase or decrease by a percentage, begin by either adding or subtracting the relevant percentage from 100%, then find the multiplier.
Example 1
Increase 50 by 20%.
Since we are increasing, we want to add 20% to 100%.
• 100 + 20 = 120%
We need to find 120% of 50.
• 120 ÷ 100 = 1.2
This is the multiplier we use to find 120% of 50.
• 1.2 × 50 = 60.
Example 2
A video game usually costs £40. Its price is reduced by 6% in the sale. Work out the sale price of the game.
This time, we are decreasing the amount so we want to subtract 6% from 100%.
• 100 – 6 = 94%
We need to find 94% of £40.
• 94 ÷ 100 = 0.94
This is the multiplier we use to find 94% of 40.
• 0.94 × 40 = £37.60
Note that, when we are working with money, we give our answers correct to 2 decimal places.
If you’d like a deeper dive into Percentage Increase and Decrease – perhaps as part of your percentages revision programme, then the following resource provides the full walkthrough complete with plentiful examples to work through!
Percentage Increase and Decrease
### Part 2 – Finding the Percentage of an Amount
It is useful to remember that percent comes from the words per- (meaning out of) and -cent (meaning 100). A percentage is measured out of 100.
You should know how to find the percentage of an amount, without using a calculator. You can calculate any percentage by finding either 10% or 1%.
Example 1
Work out 10% of 128.
To find 10% of 128, simply divide 128 by 10.
• 10%: 128 ÷ 10 = 12.8
• 10% of 128 = 12.8
Example 2
Work out 30% of 155.
If we know how to find 10% of a number, then we can easily find 30% of it too. First, find 10%:
• 10%: 155 ÷ 10 = 15.5.
We can add three lots of 10% together to find 30%:
• 30%: 15.5 + 15.5 + 15.5 = 46.5
Or, we could multiply the 10% by 3 to find 30%.
• 30%: 15.5 × 3 = 46.5
• 30% of 155 = 46.5
Example 3
Calculate 14% of 200.
Start by finding 10%:
• 10%: 200 ÷ 10 = 20
Next, find 1% of 200. To do this, simply divide 200 by 100:
• 1%: 200 ÷ 100 = 2
To make 14%, we need 10% and 4%.
We have 10% already and if we know 1%, we can easily find 4%:
• 4%: 2 + 2 + 2 + 2 = 8
• Or 4%: 2 × 4 = 8
Finally, add 10% and 4% to find 14%:
• 14%: 20 + 8 = 28
• 14% of 200 is 28.
Example 4
Calculate 7.5% of 50.
Start by finding 10%:
• 10%: 50 ÷ 10 = 5
We can use this to find 5%. 5% is half of 10% so, to find 5%, we halve 5:
• 5%: 5 ÷ 2 = 2.5
We could also find this value in one calculation:
• 5%: 50 ÷ 10 ÷ 2 = 2.5
2.5% is half of 5% so that means we can also halve the 5% value.
• 2.5%: 2.5 ÷ 2 = 1.25
We have a couple of choices on how to make 7.5%. We could take 2.5% from 10%:
• 7.5%: 5 – 1.25 = 3.75
• 7.5%: 2.5 + 1.25 = 3.75
We could also have added together seven lots of 1% and a 0.5%. There are lots of ways of making a percentage. If you do them correctly, they will all give the same answer.
If you’ve found your rhythm as well as your percentages and would like to carry on, this link below will provide you with the unfiltered 100% experience!
Finding the Percentage of an Amount
### Part 3 – Original Value Problems
We can use multipliers to help us find the original amount after a percentage increase or decrease.
Example 1:
A number is increased by 12%. The new number is 44.8. What was the original number?
To increase by 12%, begin by adding 12% to 100%.
• 100 + 12 = 112%
The percentage multiplier is 112 ÷ 100 = 1.12
To find the original number, *we divide 44.8 by 1.12.
• 44.8 ÷ 1.12 = 40
*How about a recap? – when we know the original number and are finding the new number, we multiply. So in reverse, to find the original number, we must divide the new number by the ‘multiplier’.
Example 2:
A number is decreased by 35%. The new number is 39. What was the original number?
To decrease by 35%, begin by subtracting 35% from 100%.
• 100 – 35= 65%
The percentage multiplier is 65 ÷ 100 = 0.65
To find the original number, we divide 39 by 0.65.
• 39 ÷ 0.65 = 60
If you’ve found value in these percentages revision tips, then you can find the full resource in the link below.
Percentages Original Value
Don’t forget to read even more of our blogs here! You can also subscribe to Beyond for access to thousands of secondary teaching resources. You can sign up for a free account here and take a look around at our free resources before you subscribe too.
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# Harnessing Properties: How to Add Three Fractions or Mixed Numbers with Ease
Adding fractions or mixed numbers can sometimes be a challenge, especially when dealing with three or more.
However, by utilizing the properties of addition, this process can be simplified and made more intuitive. In this guide, we’ll explore how to add three fractions or mixed numbers by leveraging these properties.
## Step-by-step Guide to Add Three Fractions or Mixed Numbers with Ease:
– Commutative Property: The order in which numbers are added doesn’t affect the sum. For example, $$a + b = b + a$$.
– Associative Property: The way numbers are grouped in addition doesn’t affect the sum. For example, $$a + (b + c) = (a + b) + c$$.
2. Converting Mixed Numbers to Improper Fractions:
If you’re dealing with mixed numbers, convert them to improper fractions. This makes the addition process more straightforward.
3. Finding the Least Common Denominator (LCD):
Determine the smallest number into which all the denominators can divide. This will ensure that the fractions are of comparable sizes.
4. Adjusting Each Fraction to the LCD:
Modify each fraction so that they all have the LCD as their denominator.
With the same denominator in place, simply add up all the numerators to get the final answer.
6. Converting Back to Mixed Numbers (if necessary):
If the result is an improper fraction, and you need a mixed number, convert it back.
### Example 1:
Add $$\frac{1}{4}$$, $$\frac{2}{8}$$, and $$1 \frac{1}{2}$$.
Solution:
First, convert the mixed number: $$1 \frac{1}{2}$$ becomes $$\frac{3}{2}$$.
The LCD for 4, 8, and 2 is 8. Adjusting the fractions:
– $$\frac{1}{4}$$ becomes $$\frac{2}{8}$$.
– $$\frac{2}{8}$$ remains the same.
– $$\frac{3}{2}$$ becomes $$\frac{12}{8}$$.
Adding them up, the result is $$\frac{16}{8}$$, which is equal to 2.
The Absolute Best Book for 5th Grade Students
### Example 2:
Add $$\frac{1}{3}$$, $$\frac{2}{6}$$, and $$2 \frac{1}{2}$$.
Solution:
First, convert the mixed number: $$2 \frac{1}{2}$$ becomes $$\frac{5}{2}$$.
The LCD for 3, 6, and 2 is 6. Adjusting the fractions:
– $$\frac{1}{3}$$ becomes $$\frac{2}{6}$$.
– $$\frac{2}{6}$$ remains the same.
– $$\frac{5}{2}$$ becomes $$\frac{15}{6}$$.
Adding them up, the result is $$\frac{19}{6}$$, which can be expressed as $$3 \frac{1}{6}$$.
### Practice Questions:
1. Add $$\frac{1}{5}$$, $$\frac{2}{10}$$, and $$1 \frac{3}{4}$$.
2. Add $$\frac{3}{7}$$, $$\frac{2}{14}$$, and $$2 \frac{1}{2}$$.
3. Add $$\frac{4}{9}$$, $$\frac{2}{3}$$, and $$3 \frac{1}{3}$$.
A Perfect Book for Grade 5 Math Word Problems!
1. $$2 \frac{9}{20}$$
2. $$4 \frac{1}{14}$$
3. $$6 \frac{5}{9}$$
The Best Math Books for Elementary Students
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# What is the Remainder Theorem? (With Examples)
We are always looking for tools that can help make our math journey just a little bit easier. The remainder theorem is a tool that most people forget about or don’t consider how useful this theorem is.
The remainder theorem states that if we take a polynomial, f(x), and divide it by a binomial, (x-c), the remainder will result in f(c).
Let’s look into how this works and how it can make your life just a bit easier, whether you are taking algebra or calculus or any math in between.
## Let’s break down the remainder theorem.
In basic math, a remainder is what you get when you divide a number into another number, but it doesn’t divide evenly. The bit that is left over is a remainder. For instance, 9 / 2 = 4 with a remainder of 1. If we reverse the process, we can multiply 4 * 2 and add in 1 to get back to 9.
When it comes to polynomials, the remainder theorem is almost the same, but it provides more information to us. We divide our polynomial by a binomial, but the remainder that we get tells us what our polynomial is equal to when we place that value back into the polynomial.
Evaluating a simple polynomial at a value of x=c isn’t that bad. But when dealing with complex polynomials, the calculations can be intense. The remainder theorem is a way to evaluate the polynomial without all the hassle.
Dividing a polynomial by a binomial can either be done through long division or synthetic division. Both methods will provide the same result.
There is something small that you should remember. At x = c, the binomial form changes, depending on whether the value is positive or negative.
Also, remember that in long division, we use the binomial form of x=c, which is either x-c or x+c. In synthetic division, we use the value of c.
If we have the polynomial, f(x) = x4+8x3+17x2-2x-24, and we are trying to calculate the value at x=-2, we could plug in -2 and calculate it out. If this polynomial were simpler, that would be an easy task. And with a calculator, the calculation would not be too bad.
But let’s show you the magic of using the remainder theorem.
I am going to use long division, synthetic division, and plugging in the value to show that all three ways will get us to the goal of finding f(c)
I know that all of this sounds confusing but stay with me.
## How to use the remainder theorem using long division.
We take the binomial and divide it into our polynomial with long division to find the value f(c).
The value that we are plugging into the polynomial is x=-2. That means that our binomial is going to be x+2.
Long division looks like regular long division, but we are using polynomials. If this concept is confusing, I have another post breaking down this concept so that you can better understand it. At the end of this section, a short video will show the steps I took using long division.
The value that we are testing is x=-2. Our binomial will be represented by x+2, as shown below. It will be the divisor in our long division problem.
We then start our calculations. We find out how many times x+2 divides into x4 and then put the result just below our polynomial. Once we subtract our result, we can bring the next value down, 17x2. We continue the process until no values remain.
The value at the end is the remainder. This value represents what our polynomial would equal if we plugged in the value of x=-2. So, in this case, f(-2)=0.
I have included a short video breaking down the process to make things clearer. It takes the process I used above and breaks it down step by step.
## How to use the remainder theorem using synthetic division.
This next section looks at how the remainder theorem and synthetic division make solving certain polynomials a bit easier.
The first step is to set up the polynomial for synthetic division. You will take each coefficient of your polynomial, starting from the highest and going down to the lowest, and you will list them out. The divisor will be the value of x that you are testing.
If there are any missing degrees in your polynomial, we will list them as a zero in the list.
Once we get everything listed, we add and multiply until the final digit. That final digit is our remainder. If the value is zero, that value of x is a root factor or a zero of that polynomial. It zeros out the polynomial when placed back in and evaluate our polynomial at x=c.
If there is an actual value in the last place, this value represents the remainder. The remainder, in this case, is the value of f(c).
Here is another short video that combines the remainder theorem with synthetic division.
## Plugging in the value c into our polynomial.
I wanted to show that you should get the same result regardless of your chosen method. This final section shows what happens when we plug -2 back into the polynomial. Our results end up being the same as the ones that we got before.
## Example problems.
I wanted to include a few example problems to see how this works out. I broke each problem down, step by step.
Please feel free to play and pause when you need to.
Great job! Keep practicing.
Danielle
Just a nerd who loves math. Trying to help the world one problem at a time.
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# Solving Multi-Step Algebra Equations Practice Questions
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Updated on Sep 23, 2011
## Tips for Solving Multi-Step Equations
• There are at least two ways to show multiplication. You may be used to seeing multiplication shown with an × like this: 5 × 3 = 15. In equations, this becomes confusing. In algebra, the convention is to show multiplication with either a · like this: 5 · 3 = 15, or with parentheses like this: 5(3) = 15. Both conventions will be used in the answers, so you should get used to either one.
• Similarly, division can be shown using the standard division symbol ÷, as in 10 ÷ 2 = 5. Or it can be shown using a fraction bar like this: 10 ÷ 2 == 5. Use and get used to both.
• Check your answers before looking at the answer solutions. Just substitute the value you find for the variable and work each side of the equation as if it were a numerical expression. If the quantities you find are equal, your solution is correct.
• Dividing by a fraction is the same as multiplying by its reciprocal. For example, .
• To write an equation for a word problem, let the unknown quantity be equal to the variable. Then write the equation based on the information stated in the problem.
## Practice Questions
1. 4x + 7 = 10
2. 13x + 21 = 60
3. 3x – 8 = 16
4. 39 = 3a – 9
5. 4 = 4a + 20
6. 10a + 5 = 7
7. 0.3a + 0.25 = 1
8. 41 – 2m = 65
9. 4m – 14 = 50
10. 10s – 6 = 0
11. 8s – 7 = 41
Solve the following word problems by letting a variable equal the unknown quantity, making an equation from the information given, and then solving the equation.
1. A farmer is raising a hog that weighed 20 lbs. when he bought it. He expects it to gain 12 pounds per month. He will sell it when it weighs 200 lbs. How many months will it be before he will sell the animal?
2. Mary earns \$1.50 less than twice Bill's hourly wage. Mary earns \$12.50 per hour. What is Bill's hourly wage?
3. At year's end, a share of stock in Axon Corporation was worth \$37. This was \$8 less than three times its value at the beginning of the year. What was the price of a share of Axon stock at the beginning of the year?
4. Jennifer earned \$4,000 more than 1.5 times her former salary by changing jobs. She earned \$64,000 at her new job. What was her salary at her previous employment?
5. Twenty-five more girls than the number of boys participate in interscholastic sports at a local high school. If the number of girls participating is 105, how many boys participate?
### Answers
Numerical expressions in parentheses like this [ ] are operations performed on only part of the original expression. The operations performed within these symbols are intended to show how to evaluate the various terms that make up the entire expression.
Expressions with parentheses that look like this ( ) contain either numerical substitutions or expressions that are part of a numerical expression. Once a single number appears within these parentheses, the parentheses are no longer needed and need not be used the next time the entire expression is written.
When two pair of parentheses appear side by side like this ( )( ), it means that the expressions within are to be multiplied.
Sometimes parentheses appear within other parentheses in numerical or algebraic expressions. Regardless of what symbol is used, ( ), { }, or [ ], perform operations in the innermost parentheses first and work outward.
Underlined equations show the simplified result.
1. Subtract 7 from both sides of the equation. 4x + 7 – 7 = 11 – 7 Associate like terms. 4x + (7 – 7) = (11 – 7) Perform numerical operations. 4x + (0) = (4) Zero is the identity element for addition. 4x = 4 Divide both sides of the equation by 4. 4x ÷ 4 = 4 ÷ 4 x = 1 2.Subtract 21 from both sides of the equation. 13x + 21 – 21 = 60 – 21 Associate like terms. 13x + (21 – 21) = (60 – 21) Perform numerical operations. 13x + (0) = (39) Zero is the identity element for addition. 13x = 39 Divide both sides of the equation by 13. 13x ÷ 13 = 39 ÷ 13 x = 3 3.Add 8 to each side of the equation. 3x – 8 + 8 = 16 + 8 Change subtraction to addition and change the sign of the term that follows. 3x+ –8 + 8 = 16 + 8 Associate like terms. 3x + (–8 + 8) = 16 + 8 Perform numerical operations. 3x + (0) = 24 Zero is the identity element for addition. 3x = 24 Divide both sides of the equation by 3. 3x ÷ 3 = 24 ÷ 3 x = 8 4. Add 6 to each side of the equation. 5x – 6 + 6 = –26 + 6 Change subtraction to addition and change the sign of the term that follows. 5x + –6 + 6 = –26 + 6 Associate like terms. 5x + (–6 + 6) = –26 + 6 Perform numerical operations. 5x + (0) = –20 Zero is the identity element for addition. 5x = –20 Divide both sides of the equation by 5. 5x ÷ 5 = –20 ÷ 5 5. Subtract 4 from both sides of the equation. + 4 – 4 = 10 – 4 Associate like terms. + (4 – 4) = 10 – 4 Perform numerical operations. + (0) = 6 Zero is the identity element for addition. = 6 Multiply both sides of the equation by 3. 3() = 3(6) x = 18 6. Add 5 to each side of the equation. – 5 + 5 = 1 + 5 Change subtraction to addition and change the sign of the term that follows. + –5 + 5 = 1 + 5 Associate like terms. + (–5 + 5) = 1 + 5 Perform numerical operations. + (0) = 6 Zero is the identity element for addition. = 6 Multiply both sides of the equation by 7. 7() = 7(6) x = 42 7. Add 9 to each side of the equation. 39 + 9 = 3a – 9 + 9 Change subtraction to addition and change the sign of the term that follows. 39 + 9 = 3a + (–9 + 9) Associate like terms. 39 + 9 = 3a + (–9 + 9) Perform numerical operations. 48 = 3a + (0) Zero is the identity element for addition. 48 = 3a Divide both sides of the equation by 3. 48 ÷ 3 = 3a ÷ 3 16 = a 8. Subtract 20 from both sides of the equation. 4 – 20 = 4a + 20 – 20 Associate like terms. 4 – 20 = 4a + (20 – 20) Perform numerical operations –16 = 4a + (0) Zero is the identity element for addition. –16 = 4a Divide both sides of the equation by 4. 9. Subtract 5 from both sides of the equation. 10a + 5 – 5 = 7 – 5 Associate like terms. 10a + (5 – 5) = 7 – 5 Perform numerical operations. 10a + (0) = 2 Zero is the identity element for addition. 10a = 2 Divide both sides of the equation by 10. 10. Subtract 0.25 from both sides of the equation. 0.3a + 0.25 – 0.25 = 1 – 0.25 Associate like terms. 0.3a + (0.25 – 0.25) = 1 – 0.25 Perform numerical operations. 0.3a + (0) = 0.75 Zero is the identity element for addition. 0.3a = 0.75 Divide both sides of the equation by 0.3. Simplify the result. a = 2.5 11. Subtract 8 from both sides of the equation. m + 8 – 8 = 20 – 8 Associate like terms. m + (8 – 8) = 20 – 8 Perform numerical operations. m + (0) = 12 Zero is the identity element for addition. m = 12 Multiply both sides of the equation by the reciprocal of m = 18 12. Add 3 to both sides of the equation. 9 + 3 = m – 3 + 3 Change subtraction to addition and change the sign of the term that follows. 9 + 3 = m + –3 + 3 Associate like terms. 9 + 3 = m + (–3 + 3) Perform numerical operations. 12 = m + (0) Zero is the identity element for addition. 12 = Multiply both sides of the equation by the reciprocal of 16 = m 13. Subtract 41 from both sides of the equation. 41 – 41 – 2m = 65 – 41 Associate like terms. (41 – 41) – 2m = 65 – 41 Perform numerical operations. (0) – 2m = 24 Zero is the identity element for addition. –2m = 24 You can change the subtraction to addition and the sign of the term following to its opposite, which in this case is –2m. –2m = 24 Divide both sides of the equation by –2. Use the rules for operating with signed numbers. 14.Add 14 to each side of the equation. 4m – 14 + 14 = 50 + 14 Change subtraction to addition and change the sign of the term that follows. 4m + –14 + 14 = 50 + 14 Associate like terms. 4m + (–14 + 14) = 50 + 14 Perform numerical operations. 4m + (0) = 64 Zero is the identity element for addition. 4m = 64 Divide both sides of the equation by 4. m = 16 15. This equation presents a slightly different look. The variable in the numerator has a coefficient. There are two methods for solving. Subtract 16 from both sides of the equation. + 16 – 16 = 24 – 16 Associate like terms. + (16 – 16) = 24 – 16 Perform numerical operations. + (0) = 8 Zero is the identity element for addition. = 8 Multiply both sides of the equation by 5. Use rules for multiplying whole numbers and fractions. 2m = 40 Divide both sides by 2. m = 20 Or you can recognize that Then you would multiply by the reciprocal of the coefficient. m = 20 16. Add 6 to each side of the equation. 7m – 6 + 6 = –2.5 + 6 Change subtraction to addition and change the sign of the term that follows. 7m + –6 + 6 = –2.5 + 6 Associate like terms. 7m + (–6 + 6) = –2.5 + 6 Perform numerical operations. 7m + (0) = 3.5 Divide both sides of the equation by 7. m = 0.5 17. Add 6 to each side of the equation. 10s – 6 + 6 = 0 + 6 Change subtraction to addition and change the sign of the term that follows. 10s + –6 + 6 = 0 + 6 Associate like terms. 10s + (–6 + 6) = 0 + 6 Perform numerical operations. 10s + (0) = 6 Divide both sides of the equation by 10. Express the answer in the simplest form. 18. Subtract 2.7 from both sides of the equation. + 2.7 – 2.7 = 3 – 2.7 Associate like terms. + (2.7 – 2.7) = 3 – 2.7 Perform numerical operations. + (0) = 0.3 Multiply both sides of the equation by 4. 4() = 4(0.3) s = 1.2 19. Add 7 to each side of the equation. 8s – 7 + 7 = 41 + 7 Change subtraction to addition and change the sign of the term that follows. 8s + –7 + 7 = 41 + 7 Associate like terms. 8s + (–7 + 7) = 41 + 7 Perform numerical operations. 8s + (0) = 48 Divide both sides of the equation by 8. Express the answer in simplest form. s = 6 20. Subtract 25 from both sides of the equation. –55 – 25 = 25 – 25 – s Change subtraction to addition and change the sign of the term that follows. –55 + –25 = 25 + –25 + –s Associate like terms. (–55 + –25) = (25 + –25) + –s Perform numerical operations. –80 = (0) + –s Zero is the identity element for addition. –80 = –s You are to solve for s, but the term remaining is –s. If you multiply both sides by –1, you will be left with a +s or just s. –1(–80) = –1(–s) Use the rules for operating with signed numbers. 80 = s 21. Let x = the number of months. The number of months (x), times 12 (pounds per month), plus the starting weight (20), will be equal to 200 pounds. An equation that represents these words would be 12x + 20 = 200. Subtract 20 from both sides of the equation. 12x + 20 – 20 = 200 – 20 Associate like terms. 12x + (20 – 20) = 200 – 20 Perform numerical operations. 12x + (0) = 180 Divide both sides of the equation by 12. x = 15 The farmer would have to wait 15 months before selling his hog. 22. Let x = Bill's hourly wage. Then 2x less \$1.50 is equal to Mary's hourly wage. The equation representing the last statement would be 2x – 1.50 = 12.50. Add 1.50 to both sides of the equation. 2x – 1.50 + 1.50 = 12.50 + 1.50 Perform numerical operations. 2x = 14.00 Divide both sides of the equation by 2. x = 7.00 Bill's hourly wage is \$7.00 per hour. 23. Let x = the share price at the beginning of the year. The statements tell us that if we multiply the share price at the beginning of the year by 3 and then subtract \$8, it will equal \$37. An equation that represents this amount is 3x – 8 = 37. Add 8 to both sides of the equation. 3x – 8 + 8 = 37 + 8 Perform numerical operations. 3x = 45 Divide both sides of the equation by 3. x = 15 One share of Axon costs \$15 at the beginning of the year. 24. Let x = her previous salary. The statements tell us that \$64,000 is equal to 1.5 times x plus \$4,000. An algebraic equation to represent this statement is 64,000 = 1.5x + 4,000. Subtract 4,000 from both sides of the equation. 64,000 – 4,000 = 1.5x + 4,000 – 4,000 Perform numerical operations. 60,000 = 1.5x Divide both sides of the equation by 1.5. 40,000 = x Jennifer's former salary was \$40,000 per year. 25. Let x = the number of boys who participate in interscholastic sports. The question tells us that the number of boys plus 25 is equal to the number of girls who participate. An equation that represents this statement is x + 25 = 105. Subtract 25 from both sides of the equation. x + 25 – 25 = 105 – 25 Perform numerical operations. x = 80 Multiply by the reciprocal of x = 120 The number of boys who participate is 120.
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Reasoning Distance And DirectionPage 2
11.
Ram and Shyam start walking towards North and cover 20 m. Ram turns to his left and Shyam to his right. After sometime, Ram walks 10 m in the same direction in which he turned. On the other hand, Shyam walks only 7 m. Later, Ram turns towards his left and Shyam to his right. Both walk 25 m forward. How far is Ram from Shyam now? (a)10 m (b)20 m (c)15 m (d)17 m
Answer is: DData is arranged as
So, Distance between Ram and Shyam = (10 + 7) = 17m
12.
After starting from a point, A walks 3 km towards East, that turning to his left he and moves 3 km. After this he again turns left and moves 3 km. In which direction is ‘A’ from his stating point? (a)North (b)East (c)West (d)South
Thus, A is in North direction from his starting point.
13.
Rajnikant left his home for office in car. He drove 15 km straight towards North and then turned eastwards and covered 8 km. He then turned to left and covered 1 km. He again turned left and drove for 20 km and reached office. How far and in what direction is his office from the home? (a)21 km West (b)15 km North - East (c)20 km North - West (d)26 km North - West
In triangle AFE,
AE = √(AF2 + EF2)
AE = √[(AB + BF)2 + (ED - FD)2]
AE = √(162 + 122) = √(256+144) = √400 = 20 km
So, his office is 20 km North - West
14.
Ram goes 15 m North, then turns right and walks 20 m, then again turns right and walks 10 m, then again turns right and walks 20 m. How far is he from his original position? (a)5 m (b)10 m (c)15 m (d)20 m
Thus, Ram is at a distance of 5 m from his starting point.
15.
Ram and Ravi start walking in opposite directions. Ram covers 6 km and Ravi 8 km. Then, Ram turns right and walks 8 km and Ravi turns left and walks 6 km. How far everyone in from their starting point? (a)11 km (b)8 km (c)9 km (d)10 km
From above figure, we can see that both are standing equal distance from starting point.
So, required distance = √(82 + 62)
= √(64 + 36) = √100 = 10 km
Hence, everyone is 10 km far from their starting point.
16.
Pooja starting from a fixed point goes 15 m towards North and then after turning to his right, he goes 15 m. Then, he goes 10, 15 and 15 after turning to his left each time. How far is he from his starting point? (a)15 m (b)5 m (c)10 m (d)20 m
From figure, A is the starting point and F is the ending point of Pooja.
So, AE = AB + BE
AE = 15 + 10 = 25 m
Now, AF = AE – EF = 25 – 15 = 10 m
Hence, Pooja is 10 m far from his starting point.
17.
Alok walked 30 m towards East and took a right turn and walked 40 m. He again took a right turn and walked 50 m. towards which direction is he from his starting point? (a)South (b)West (c)South - West (d)South - East
Answer is: CThe movement of Alok is shown below
From figure, it is clear that he is in South-West direction from his starting point
18.
A postman was returning to the post office which was in front of him to the North. When the post office was 100 m away from him, he turned to the left and moved 50 m to deliver the last letter at Shanti Villa. He, then moved in the same direction for 40 m, turned to his right and moved 100 m. How many meters was he away from the post office? (a)50 m (b)100 m (c)90 m (d)120 m
It is clear from above figure that Postman was 90 m away from post office.
19.
Rohit walked 25 m towards South. Then, he turned to his left and walked 20 m. He then turned to his left and walked 25 m. He again turned to his right and walked 15 m. At what distance is he from the starting point and in which direction? (a)35 m, East (b)35 m, North (c)40 m, East (d)60 m, East
Answer is: AThe movements of Rohit are shown in figure below
So, Rohit’s distance from starting point,
AE = BC + DE = 20 + 15 = 35 m East.
20.
The door of Aditya's house faces the East. From the back side of his house, he walks straight 50 m, then turns to the right and walks 50 m, then turns to right and walks 50 m again. Finally, he turns towards left towards left and stop after walking 25 m. Now, Aditya is in which direction from the starting point? (a)South - East (b)North - East (c)South - West (d)North - West
Answer is: DSince, Aditya’s house face towards East and he walks from backside of his house. It means that he starts walking towards West which is shown as below
Clearly, Aditya's final position is D which is to the North-West of the starting point A.
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Study Notes on Time, Speed & Distance
By Gaurav Mohanty|Updated : December 2nd, 2021
Today, we will discuss the basic concepts and short tricks of an important topic i.e. Speed, Distance, and Time To help you score well, we are here providing you with the basic concepts & tricks of how to solve Speed, Distance & Time questions.
Today, we will discuss the basic concepts and short tricks of an important topic i.e. Speed, Distance, and Time To help you score well, we are here providing you with the basic concepts & tricks of how to solve Speed, Distance & Time questions.
What are Speed, Distance, and Time?
Speed:
Speed is simply defined as the distance covered per unit time. Mathematically, it is defined as :
• Speed = Distance Travelled / Time Taken
• The formula of Time in terms of distance and Speed: Time = Distance/ Speed
• The formula of Distance in terms of Speed and Time: Distance = (Speed X Time)
• It is very important to know the importance of the unit while solving these kinds of problems.
SI Unit for Speed: Meter per second
SI Unit for Distance: Meter
SI Unit for Time: Second
All the formulas of speed, time, and distance are interrelated. One needs to connect the information given in the problems wisely and this can be done easily with good practice. Let us now discuss the various types of Speed, Distance & Time questions.
Type 1 - Basic
Now, we will discuss the most frequently used concept in this chapter i.e. questions based on the unit conversion. So, in order to learn this concept, we need to know how these questions are framed in the question paper.
Question: A train runs at X km/hr and it takes 18 seconds to pass a pole. The length of the train is 180m. Find the speed of the train in km/hr.
Solution:
Speed = distance /Time
Therefore,
S=180m/18s= 10m/s
Now we need the answer in km/hr and for that, we will multiply the given speed with (18/5) in order to get the answer in km/h
Therefore, S= 10*(18/5)=36km/h.
So here the trick for the same is :
• Convert metre per second (m/sec) to km per hr (km/h)
For converting (meter per second) to (kilometer per hour) we use the following formula
s m/sec= S *(18/5) km/h
• Convert km per hr (km/h) to metre per second (m/sec)
For converting kph(kilometer per hour) to MPs(meter per second) we use the following formula
S km/hr=(s∗5/18) m/sec
Question: A boy covers a distance of 600m in 2min 30sec. What will be the speed in km/hr?
Solution:
Speed =Distance / Time =Distance covered = 600m, Time taken = 2min 30sec = 150sec
Therefore, Speed= 600 / 150 = 4 m/sec = 4m/sec = (4*18/5) km/hr = 14.4 km/ hr
Type 2 - Relative Speed
Considering 2 objects A and B having the speed x, y.
• If the ratio of the speeds of A and B is x:y, then the ratio of the times taken by them to cover the same distance is: 1/x: 1/y or y: x
Question: The ratio of the speed of a bike and a motor is 4:5 then what will be the ratio for the time taken by both the vehicles for the same destination?
Solution:
As the destination is the same so distance will be the same for both Car and bike.
Let the Distance be d
And the speeds for both the vehicles be 4s and 5s
Now, t1 = d/4s----(1)
t2 = d/5s----(2)
so,t1/t2 = 5/4= 5:4
Type 3 - Average Speed
• Average Speed is another very important concept.It is defined as:
Average Speed = Total Distance Travelled /Total Time Taken
Question: Dewansh travels 320 km at 64 km/hr and returns at 80 km/hr. Calculate the average speed of Dewansh?
Solution:
We know that speed = Distance/ time taken
⇒ ∴Total time taken = 320/64 + 320/80 = 9
⇒ Average Speed = (320 + 320)/9
⇒ Average speed = 71.11 km/hr
Question: A car moving with a uniform speed of 50km/h covers half the distance with this speed. Half of the time of the remaining distance is covered with speed 35km/h and the other half time at 10km/h. If the total distance traveled is 90 km then What was the car’s average speed (approximately) during his entire journey?
Solution:
Half of total distance = 45km, Speed = 50km/h
Time taken = 45km/50 = 0.9 hr = 0.9 x 60 = 54 minutes
Let time taken for remaining 45 km be T
And, distance via speed 10km/hr by D
Then, ATQ, T/2 = (45 - D) / 35 …(1)
And, T/2 = (D) / 10 …(2)
Equating equations 1 & 2,
(45 - D) / 35 = (D) / 10
10(45 - D) = 35D
45D = 450 or,
D = 10 km, T= 2 hour
Total time = 2+.9= 2.9 hour
Average speed= 90/2.9=31km/hr
Type 4
• Suppose a Person covers a certain distance at x km/hr and an equal distance at y km/hr.
Then, the average speed for the complete Journey: 2xy/(x + y)
Question: A train goes from Ballygunge to Sealdah at an average speed of 20 km/hour and comes back at an average speed of 30 km/hour. The average speed of the train for the whole journey is?
Solution:
Let x and y be the average speed for the same distance in two different times.
Then, average speed = (2xy)/(x + y)
A train goes from Ballygunge to Sealdah at an average speed of 20 km/hour and comes back at an average speed of 30 km/hour.
The average speed of train = (2 × 20 × 30)/(20 + 30) = 24 km/hr
Question: A boy goes to school at a speed of 3 km per hr and returns to the village at a speed of 2 km per hr. If it takes 5 hrs in all, what is the distance between the village and the school?
Solution:
Let the required distance be x km.
Then time is taken during the first journey = x/3 hr.
and time is taken during the second journey = x/2 hr.
x/3 + x/2 = 5 => (2x + 3x) / 6 = 5
=> 5x = 30.
=> x = 6
Required distance = 6 km.
Type 5
• In this type, we will discuss the formulas related to Train problems which are most common in competitive exams.
Let the length of the train be X m and the Speed of the train is S m/s then the Time Taken to cross a pole or man standing or a signal post will be: t=X/S
Question: A train takes 10 seconds to cross a pole and 20 seconds to cross a platform of length 200m. What is the length of the train?
Solution:
Let the length of the train be x meter and speed be y m/sec
Therefore, time taken to cross the pole = x/y sec
and time taken to cross the platform = ( Length of train + Length of platform) / speed of the train
Therefore x/y = 10
And, (x+200) / y = 20
⇒ 10y + 200 = 20 y
∴ speed of the train, y = 20 m/sec
And length of rain = x = 10y = 200m
Question: Train B and C while traveling in opposite direction crosses each other in 10 seconds while the time taken by train C to cross a standing pole is 7 seconds and the length of train B and train C are 180m and 140m respectively then find the time taken by train B to cover the distance of 1728 km?
Solution:
Speed of train C= 140/7= 20m/s
Let the speed of train B be x m/s
So,
Time taken to cross each other= (Length of train B + length of train C)/Speed of train B relative to train C
10= (180+140)/(x+20 )
10x+200= 320
10x= 120
x= 12m/s
Speed of train B in kmph= 12*18/5= 43.2 kmph
Time is taken by train B to cover the distance= 1728/43.2= 40 hours
Question: A 180 m long train crosses another 270m long train running in the opposite direction in 10.8 seconds. If the shorter train crosses a pole in 12 seconds, what is the speed of the longer train?
Solution:
Speed of shorter train = 180/12 = 15m/sec
(15 + x) = (180+270)/10.8 where
X = speed of longer train = 15 + x = 4500/108
X = (4500/108) – 15
X = (2880/108) x (18/5) = 96 kmph
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Dorian Mode Explained – Theory, CAGED Positions and Diagrams
The Dorian mode is the 2nd mode of a major scale. If you have read guitar modes explained, you should already have a pretty good idea of how modes work. In this post we are going to go into the specifics of the dorian mode.
Remember, to understand how modes work, you first need to understand major scales. Period.
The dorian mode can be summed up very simply like this:
1. It has a flat 3 and a flat 7.
2. It is also the 2nd mode of a major scale.
Parallel and Derivative Dorian Mode
What this basically means is that there are 2 ways of figuring out a mode, parallel and derivative. Using the parallel approach requires altering notes in a major scale to produce a mode (in this case 3rd note and the 7th note).
Using the derivative approach requires playing a major scale but starting on a different note (in this case the 2nd note). It is important to be familiar with both approaches and to realize that they can be used to produce the same results. Again, as you can see, a knowledge of major scales is essential in both approaches.
The dorian mode has a flat 3 and a flat 7. What that really means is that compared to a major scale, the 3rd note is lowered by a semitone and the 7th note is lowered by a semitone.
Let’s look at some examples.
Using the parallel approach, all we have to do is change the notes of a major scale to produce the desired mode. If we wanted to play a D dorian scale, we would need to play a D major scale, and change the 3rd note by lowering it a semitone and also change the 7th note by lowering it a semitone.
D major has the following notes
D – E – F# – G – A – B – C#
Root = D, 2 = E, 3 = F#, 4 = G, 5 = A, 6 = B, 7 = C#
According to the parallel approach, we need to lower the 3rd note (F#) to F and the 7th note (C#) to C. That would then give us following:
D – E – F – G – A – B – C
It’s important to be aware that the above scale has a flat 3rd and a flat 7th. Even though there are no ‘b’s or ‘#’s, the 3rd and 7th notes are still said to be ‘flattened’ because they have gone from sharps to naturals.
Let’s look at D Dorian from a derivative approach:
The derivative approach says that we must play the desired mode by playing a major scale and starting on a different note. In the case of the dorian mode, we need to start on the 2nd note of a major scale. In this case, you need to know which major scale has D as the 2nd note. It is of course C major.
The key of C major looks like this:
C- D – E – F – G – A – B
Root = C, 2 = D, 3 = E, 4 = F, 5 = G, 6 = A, 7 = B
So if we were to play the dorian mode by starting from the 2nd note of the C major scale (D) we would end up with D dorian, which would look like this:
D – E – F – G – A – B – C
Of course, you can see that the scale that has just been produced is the same as the scale that we got to before by altering the 3rd and 7th notes of a D major scale. We have in fact produced a D dorian scale through 2 different methods – parallel and derivative.
Let’s do another example…
Let’s say we want to play an E flat dorian scale. The parallel approach requires us to lower the 3rd and 7th notes of a major scale, in this case Eb major. The notes in Eb major are:
Eb – F – G – Ab – Bb – C – D
Therefor, if we ‘flatten’ the 3rd (G) and the 7th (D) we end up with the following scale:
Eb – F – Gb – Ab – Bb – C – Db
Let’s make sure this is correct by figuring out an Eb dorian using the derivative approach as well. Of course, to do this we need to know which major scale has Eb as its 2nd note. This is Db major:
Db – Eb – F – Gb – Ab – Bb – C
If we start this scale from the 2nd note (Eb), we get the following:
Eb – F – Gb – Ab – Bb – C – Db
We have just played an Eb dorian mode (or scale) using the derivative approach.
If you understand this concept you should be able to figure out the dorian mode in every key.
I have written out each dorian mode in every key and the 5 positions along the neck (plus the open position) for each of these keys. Remember, because modes are derived from major scales, the shapes and positions should seem familiar to you if you have studied and played the major scales along the fretboard.
List of Individual Dorian Mode Keys
Here is a list of all the dorian modes in every key:
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# 5.8 – Binomial Theorem
Key Terms
• Binomial Coefficient – The coefficient of a given term in a binomial expansion.
• The coefficient of the k-term in the expansion of $(x+a)^n$ is denoted $\binom{n}{k}$.
• Binomial Expansion – An expression that is equivalent to a binomial raised to a positive integer power, and that results from carrying out the multiplication indicated by the power.
• Combinations – Unique selections of items, no repeats.
• Ex. If you have 3 different fruits: apple, pear, and banana, and must choose only 2, you wouldn’t choose [apple and pear], then [pear and apple] because those would be repeats. The combination of apple and pear will only happen once, no matter the order.
• Choosing 2, you can also have a pear with banana or an apple with banana.
• Binomial Theorem – A polynomial identity which states: $(x+y)^n=\binom{n}{0}x^n+ \binom{n}{1}x^{n-1}y^1+ \binom{n}{2}x^{n-2}y^2+ \binom{n}{3}x^{n-3}y^3+...+ \binom{n}{n}y^n$
• Factorial – The product of n and all positive whole numbers less than n.
• The notation n! means “n factorial.”
• Ex. 4! = 4 3 2 1 = 24
• Ex. 8! = 8 4 3 2 1 = 40,320
• There is a button on your calculator “n!” you can use!
• Pascal’s Triangle – A triangular arrangement of numbers in which each row gives you the coefficients of a binomial expansion.
• Any number in the triangle can be computed by taking the sum of the two adjacent numbers in the previous row.
• Permutation – A selection of objects in which the order of the objects matters.
• Use this to show how many ways can you arrange a certain number of items.
• $\frac{n!}{k!(n-k)!}$
• 0! = 1
Notes
• Pascal’s Triangle
• Do you see a diagonal pattern?
• 1 + 1 = 2
• 1 + 2 = 3
• 1 + 3 = 4, so the coefficient of the 2nd term in the 4th line down will be 4.
• What will the 2nd term look like including the variables? Look for the pattern. Try writing it out on paper! Scroll ALL the way down to see the answer!
• Do you see a horizontal pattern (for the exponents)?
• From left to right, powers of x get smaller while powers of y get bigger
• The row will always start and end with the same power (which matches the row that it is in)
• The pattern found with Pascal’s triangle works for any binomial
• Using Pascal’s triangle may be faster than using the FOIL method when binomials have coefficients.
• Watch and copy the video below
• Ex. $(2+y)^3$
• Look at the 3rd row of Pascal’s Triangle and replace ALL x’s with 2’s.
• Substitute 2 for x: $1(2)^3+3(2)^2y+3(2)y^2+1y^3$
• Simplified: $8+12y+6y^2+y^3$
• Ex. $(3x-2y)^2$
• Rewrite as: $(3x+(-2y))^2$
• Substitute 3x for x and substitute -2y for y: $1(3x)^2+2(3x)(-2y)+(-2y)^2$
• Simplify: $9x^2-12xy+4y^2$
• Pascal’s Patterns Enumerated
1. Expanding a binomial to the nth power gives n + 1 terms.
2. The first term of $(x+y)^n$ will always be $x^n$, and the last will always be $y^n$.
3. The sum of the exponents of each term of the expansion of $(x+y)^n$ is n.
4. The exponents of the x-terms in the expansion of $(x+y)^n$ will start at n and decrease by 1 for every consecutive term.
5. The exponents of the y-terms in the expansion of $(x+y)^n$ will start at zero and increase by 1 for every consecutive term.
6. The coefficients follow the pattern of Pascal’s triangle for the nth row.
• Negative Values for Pascal’s Triangle Expansion
• For $(x-y)^n$, raising a negative value to a power creates a pattern of alternating signs.
• For $((-x)+(-y))^n$, if n is an even number, every sign will be positive; but, if n is an odd number, every sign will be negative.
• Ex. $(x-y)^n$
• Pascal’s Triangle by Row (n) and Column (k)
• Rows: x
• Columns: y
• Permutations Using the Combination Formula
• $\binom{n}{k}$ is read “n choose k,”
• Formula can be used to find the kth term of the nth row of Pascal’s triangle
• It is defined by the quotient (ratio) of factorials
• $\binom{n}{k}=\frac{n!}{k!(n-k)!}$
• And for some reason, 0! = 1
• Use a calculator to find the answers
• Ex. You can choose 3 out of 26 friends to go with you to the movies. How many possible combinations can you have?
• Since calculators can’t handle numbers that large without converting to scientific notation, here’s a shortcut!
Examples
• For $(x+y)^6$
• Ex 1a. Which of the following could NOT be a term in the expansion?
• $6x^5y$
• $15x^2y^4$
• $20xy$
• $15x^3y^3$
• Answer: since each term’s exponents MUST add up to 6, “20xy” is the one that could NOT be a term in the expansion.
• Ex 1b. What is the expansion?
• Ex 2. Calculate the 5th term for $(x-3y)^8$
• The first term will have 8 as the x-exponent and zero as the y-exponent.
• The fifth term will have 8−4 (so = 4) as the x-exponent and 4 as the y-exponent.
• The term will be positive because $(-3y)^4$ is positive.
• The coefficient from the triangle is 70, so the fifth term is:
• $70x^4(-3y)^4=(70)(81)x^4y^4=5670x^4y^4$
• Pascal’s Triangle: The Next Level
• Coefficients of Pascal’s Triangle
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Home | | Business Maths 12th Std | Measurements of Trends: Method of Least Squares
# Measurements of Trends: Method of Least Squares
The line of best fit is a line from which the sum of the deviations of various points is zero.
Method of Least Squares
The line of best fit is a line from which the sum of the deviations of various points is zero. This is the best method for obtaining the trend values. It gives a convenient basis for calculating the line of best fit for the time series. It is a mathematical method for measuring trend. Further the sum of the squares of these deviations would be least when compared with other fitting methods. So, this method is known as the Method of Least Squares and satisfies the following conditions:
(i) The sum of the deviations of the actual values of Y and Ŷ (estimated value of Y) is Zero. that is Σ(Y–Ŷ) = 0.
(ii) The sum of squares of the deviations of the actual values of Y and Ŷ (estimated value of Y) is least. that is Σ(YŶ)2 is least ;
Procedure:
(i) The straight line trend is represented by the equation Y = a + bX …(1)
where Y is the actual value, X is time, a, b are constants
(ii) The constants ‘a’ and ‘b’ are estimated by solving the following two normal
Equations ΣY = n a + b ΣX ...(2)
ΣXY = a ΣX + b ΣX2 ...(3)
Where ‘n’ = number of years given in the data.
(iii) By taking the mid-point of the time as the origin, we get ΣX = 0
(iv) When ΣX = 0 , the two normal equations reduces to
The constant ‘a’ gives the mean of Y and ‘b’ gives the rate of change (slope).
(v) By substituting the values of ‘a’ and ‘b’ in the trend equation (1), we get the Line of Best Fit.
Example 9.6
Given below are the data relating to the production of sugarcane in a district.
Fit a straight line trend by the method of least squares and tabulate the trend values.
Solution:
Computation of trend values by the method of least squares (ODD Years).
Therefore, the required equation of the straight line trend is given by
Y = a+bX;
Y = 45.143 + 1.036 (x-2003)
The trend values can be obtained by
When X = 2000 , Yt = 45.143 + 1.036(2000–2003) = 42.035
When X = 2001, Yt = 45.143 + 1.036(2001–2003) = 43.071,
similarly other values can be obtained.
Example 9.7
Given below are the data relating to the sales of a product in a district.
Fit a straight line trend by the method of least squares and tabulate the trend values.
Solution:
Computation of trend values by the method of least squares.
In case of EVEN number of years, let us consider
similarly other values can be obtained.
Note
(i) Future forecasts made by this method are based only on trend values.
(ii) The predicted values are more reliable in this method than the other methods.
Tags : Procedure, Example Solved Problem | Time Series Analysis , 12th Business Maths and Statistics : Chapter 9 : Applied Statistics
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
12th Business Maths and Statistics : Chapter 9 : Applied Statistics : Measurements of Trends: Method of Least Squares | Procedure, Example Solved Problem | Time Series Analysis
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AP Statistics
# AP Statistics
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## AP Statistics
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##### Presentation Transcript
1. AP Statistics Chapter 1 Exploring Data
2. WHAT IS STATISTICS? Statistics is the study of how we: -Collect Data -Organize Data -Analyze Data -Use data to make predictions Statistics is the tool we use to extract information from data!
3. Lesson Objectives Identify individuals and variables in a set of data. Classify a variable as being a quantitative or categorical variable. Identify the units of measurement for a quantitative value.
4. VARIABLES Individualsarethe objects described by a set of data. Variables are characteristics that can take different values from individual to individual. A variable can be considered either categorical or quantitative.
5. EXAMPLE Suppose we observed a bag of M&M candies and were studying the different colors of the pieces. What would be the individuals in the study? What would be the variable?
6. Categorical vs. Quantitative While ZIP codes are numeric in form, you would not use arithmetic to combine them in any form. Quantitative Categorical Categorical Categorical variables place an individual in to a group or category. Quantitative variables assign a numerical value to an individual. EXAMPLES: Which type of variable is each? A person’s height … A person’s eye color … A person’s ZIP code …
7. Distribution The type of data collected can be a determining factor of the way the values are organized. Quantitative values can be very close together or very spread out. The pattern of variation between these values in a set of data is called the distribution. Distribution – a description of the values a variable takes and how often it takes these values.
8. DISTRIBUTION Both quantitative and categorical data will have differences from individual to individual. The pattern of variation of a variable is referred to as its distribution. In order to get a grasp of a variable’s distribution, we may use a graphical display of the data.
9. AP EXAM Tip You will often be asked to “describe the distribution” of a set of data. When you are asked to do this, make sure that you have your SOCkS on! S – Shape O – Outliers C – Center S– Spread When you describe these four characteristics of the data, you will be effectively describing the distribution!
10. ACTIVITY: Sexual Discrimination????? 25 airplane pilots have applied to fill 8 positions to be pilots with an airline company. 15 of them are males and 10 are females. To be fair, the managers select the 8 pilots to be employed by a lottery. A day later, the managers announce the 8 pilots to be hired. 5 of them are female and only 3 are males. Many of the males claimed that the lottery had to have been “rigged” since there was no way that so many females were selected.
11. ACTIVITY CONTINUED To simulate the situation, select ten red cards and fifteen black cards. Use the cards within your group to conduct your own lottery by drawing 8 cards. Count the number of females, and record that number. Put the cards back, and shuffle the cards. Repeat the process four more times. Report your results to be recorded.
12. What do we see? Do you think that it is possible that the number of females hired in the problem was a coincidence???
13. HOMEWORK Complete the assignment listed in the packet. This assignment will be due at the beginning of the next class session.
14. Analyzing Categorical Data • In this section we will learn about: • Bar graphs/pie charts • Problems with graphs • Two-way Tables and Marginal Distribution • Conditional Distributions • Simpson’s Paradox
15. EXAMPLE The Radio Arbitron service places each of the contry’s 13,838 stations in categories that describe the type of music they play. Here is the distribution of the data.
16. Continued Sometimes, we may wish to use a graph instead of table to clarify relationships.
17. Be Careful!!! Because of their appeal to the eyes, graphical displays can sometimes be misleading. Always look for things like scaling and relevance. Pictographs can almost always be misleading.
18. Pictograph • What is the issue with this ad that was used by Apple Computers to show the people that were buying their new iMac Computer?
19. Activity Use the table below. A.) Make a well-labeled graph to display the data. B.) Would it be appropriate to make a pie chart here? Why?
21. Two-Way Tables A survey of 4826 randomly selected young adults (19-25 yrs old) asked, “What do you think are the chances you will have much more than a middle-class income at age 30?” This is an example of a two-way table.
22. Two-Way Table • A two-way table describes two categorical variables, organizing counts according to a row variable and a column variable. • Marginal Distribution – • The distribution of values of one of the variables among all individuals in that category of a two-way table. • To examine a marginal distribution: • Use the table data to compute percents of the row or column totals. • Make a graph to display the marginal distribution.
23. Two-Way Tables and Marginal Distributions Examine the marginal distribution of chance of getting rich.
24. Simpson’s Paradox Accident victims are often transported by hospital to a medical facility. Does this act help save lives? What are the percentage of deaths for each of the two categories? …not too positive, huh?!
25. Continued… Let’s look at the data differently. Compute the percentages now. …is that right? This phenomenon is referred to as Simpson’s Paradox. It is caused by what is referred to a lurking variable. What was the lurking variable here?
26. HOMEWORK Complete the assignment listed in the packet. This assignment will be due at the beginning of the next class session.
27. 1.2 – Displaying Quantitative Data Dotplots are a commonly used method of displaying quantitative data. To make a dotplot, DRAW a horizontal number line, labeled with the name of the variable. SCALE the number line, including the minimum and maximum values. MARK a dot above the corresponding location on the axis for each data value.
28. EXAMPLE The table here displays goals scored by the US Women’s Soccer Team in 2004. Create a dotplot to represent the data.
29. EXAMPLE 2 The table and dotplot below displays the Environmental Protection Agency’s estimates of highway gas mileage in miles per gallon (MPG) for a sample of 24 model year 2009 midsize cars. Use the dotplot to describe the distribution.
30. Describing the Shape We can describe the shape of a distribution bby using the following terms. Symmetric - if the right and left sides of the graph are approximately mirror images of each other. Skewed Right - if the right side of the graph (containing the half of the observations with larger values) is much longer than the left side. Skewed Left – just the opposite of skewed right. Bimodal – A set of data that has two peaks.
31. Identify Each Symmetric Skewed - left Bimodal Skewed - right
32. Applying the Concepts Complete the “Check Your Understanding” questions on pg. 31.
33. VIDEO #2 Decisions Through Data Stemplots
34. Stemplots Stemplots are often used as a means of representing quantitative values. The data is organized by separating each observation into a stem (all but the last digit) and a leaf (the last digit). The leaf values are then paired with their stem and ordered. Trends and patterns in the distribution can be seen here.
35. Caffeine content of an 8oz. serving of many popular soft drinks. A&W Cream 20 Diet Sun Drop 47 Barq’s Root Beer15 Diet Sunkist 28 Cherry Coke 23 Diet Cherry Pepsi 24 Cherry RC Cola 29 Dr. Nehi 28 Coke Classic 23 Dr. Pepper 28 Diet A&W Cream 15 IBC Cherry 16 Diet Cherry Coke 23 Kick 38 Diet Coke 31 KMX 36 Diet Dr. Pepper 28 Mello Yello 35 Diet Mello Yello 35 Mountain Dew 37 Diet Mtn Dew 37 Mr. Pibb 27 Diet Mr. Pibb 27 Nehi Wild Red 33 Diet Pepsi 24 Pepsi One 37 Diet Red Squirt 26 Pepsi 25
36. Arrange all of this data in to a stemplot, and observe the distribution. RC Edge 47 Red Flash 27 Royal Crown 29 Red Squirt 26 Sun Drop Cherry 43 Sun Drop 43 Sunkist 28 Surge 35 Tab 31 Cherry Pepsi 25
37. Stemplot from the Example Caffeine Content (mg) per 8oz. Serving of Various Soft Drinks Key: 3|5 means 35 mg of caffeine per 8 oz. serving
38. An Alternative (better) Plot Caffeine Content (mg) per 8oz. Serving of Various Soft Drinks Key: 3|5 means 35 mg of caffeine per 8 oz. serving
39. Tips for Stemplots When you split stems, make sure each part is assigned an equal number of possibilities. There is no set number of stems. Too few stems makes a “skyscaper” shape. Too many stems makes a “pancake” shape. As a rule, a minimum of five stems is good to follow. Always include a title and a key to show how the stems were formed.
40. HOMEWORK Complete the assignment listed in the packet. This assignment will be due at the beginning of the next class session.
41. VIDEO #3 Decisions Through Data Histograms
42. HISTOGRAMS Histograms are different than bar graphs as they are represented on a continuum of values. As with stemplots, there is no set number of classes to use. Five classes is a good minimum. Remember: area of each bar is what matters. Make sure width is constant and height varies.
43. Relative Frequency Histograms A relative frequency histogram is based on relative frequencies of each category. Relative Frequency = number of occurrences in the category/ total number of occurrences. Relative frequency is often used to find percentiles, or the portion of data that at or below a value.
44. Applying the Concepts Complete the “Check Your Understanding” questions on pg. 39.
45. HOMEWORK Complete the assignment listed in the packet. This assignment will be due at the beginning of the next class session.
46. VIDEO #4 Decisions Through Data Measures of Center
47. 1.3 - Describing Distributions with Numbers To describe a distribution, we must identify its center. One measure of the center of a set of data is the mean. The mean is the sum of all observations in a set divided by the total number of observations.
48. FORMULA for Mean The mean of a set of data is: or…
49. Median The median of a set of data is the midpoint of the data. Arrange all of the numbers from least to greatest. If there is an odd number of observations, the median is the center of the list. If there is an even number of observations, the median is the mean of the two center observations.
50. Mean vs. Median The median is a more resistant measure than the mean. This means that the mean can be more easily influenced by extreme values. Differences between mean and median can indicate skewness in the data. Skewed Left data will have a mean that is less than the median. Skewed Right data will have a mean that is greater than the data. Mean/Median Applet
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# Mathematical Induction
Subject: Mathematics Exam: MUOET Engineering
Chapter Rating: Do If You Have Time
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## Important Concepts in Mathematical Induction
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• Principles of Mathematical Induction
• Application of Induction
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## Important Points
FIRST PRINCIPLE OF MATHEMATICAL INDUCTION
Step I: Actual verification of the proposition for the starting value 'i'.
Step II: Assuming the proposition to be true for 'k', k >=i and then providing that it is true for the value (k+1) which is the next higher integer.
Step III: Combine the two steps or let P(n) be a statement involving natural number n. To prove statement P(n) is true for all natural number we use following process:
1. Prove that P(1) is true.
2. Assume P(k) is true
3. Using (1) and (2) prove that statement is true for n=k+1, i.e., P(k+1) is true.
This is first principle of Mathematical Induction.
SECOND PRINCIPLE OF MATHEMATICAL INDUCTION
Step I: Actual verification of the proposition for the starting value i and (i+1).
Step II: Assuming the proposition to be true for k-1 and then proving that it is true for the value (k+1):k >=i+1.
Step III: Combine the above two steps these are used to solve problem or in 2nd principle of Mathematical Induction following steps are used:
1. Prove that P(1) is true
2. Assume P(n) is true for all natural number such that 2<=n < k
3. Using (1) and (2) prove that P(k+1) is true.
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# If $\sin x = \cos x$ and x is acute state the value of x in degrees
Last updated date: 24th Jun 2024
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Hint: We have been given a trigonometric equation, by making certain changes to it and using various trigonometric identities, we can easily obtain the required value of angle x. This angle x is given to be acute, which means its value is less than 90°.
Trigonometric identities to be used:
${\sin ^2}x + {\cos ^2}x = 1 \\ \sin 2x = 2\sin x\cos x \;$
We have been given an equation:
$\sin x = \cos x$ and we need to find the value of angle x in degrees.
This equation can be written as:
$\sin x - \cos x = 0$
Squaring both the sides, we get:
${\left( {\sin x - \cos x} \right)^2} = 0 \\ {\sin ^2}x + {\cos ^2}x - 2\sin x\cos x = 0 \\ 1 - 2\sin x\cos x = 0\left( {\because {{\sin }^2}x + {{\cos }^2}x = 1} \right) \; 1 = 2\sin x\cos x \;$
The value of double sine angle is given as:
$\sin 2x = 2\sin x\cos x$
Substituting this value, we get:
$\sin 2x = 1$
The value of sin 90° is 1, so the above equation can be written as:
$\sin 2x = \sin {90^\circ } \\ \Rightarrow 2x = {90^\circ } \\ \therefore x = {45^\circ } \;\$
It is given that the angle is acute i.e. less than 90° which is also true for the angle obtained.
Therefore, if $\sin x = \cos x$ and x is acute then the value of x is 45 degrees
So, the correct answer is “ 45° ”.
Note: We can also find the value of angle x by the following method, using the basic formula for tanx i.e. $\dfrac{{\sin x}}{{\cos x}} = \tan x$
Given equation: $\sin x = \cos x$
Dividing both the sides by $\cos x$ , we get:
$\dfrac{{\sin x}}{{\cos x}} = \dfrac{{\cos x}}{{\cos x}} \\ \tan x = 1 \\ \left( {\because \dfrac{{\sin x}}{{\cos x}} = \tan x} \right) \\$
The value of tan is 1 when the angle is equal to 45°, x can be calculated mathematically as:
$\tan x = 1 \\ \Rightarrow x = {\tan ^{ - 1}}(1) \\ \therefore x = {45^\circ }\left( {\because \tan {{45}^\circ } = 1} \right) \;$
Thus, we get the value of x as 45° by following every method.
The angles less than 90° are called acute angles and greater than that are called obtuse.
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Question
# A sum becomes 8000 in 3 years and 10000 in 6 years at C.I. Find the sum.
Hint: We can use the formula for finding the amount $=p{{\left( 1+\dfrac{R}{100} \right)}^{n}}$ where $p$ is principal, $R$ is the rate of interest and $n$ is time.
We have been given the question that a sum becomes 8000 in 3 years and 10000 in 6 years at C.I. Here, we have to find the principal.
Before proceeding with the solution, we must know that the initial money is the principal and after adding interest to the principal, we get the final amount.
According to the question, we have to find the sum, that means we need to find the principal. Therefore, we have to use the formula for amount = $p{{\left( 1+\dfrac{R}{100} \right)}^{n}}$.
We are supposed to take the sum as the principal. Let the principal be $p$.
We have the amount$=8000$and the time period as 3 years. Therefore, we can substitute the values in the formula and we will get,
$\therefore 8000=p{{\left( 1+\dfrac{R}{100} \right)}^{3}}..........(i)$
Again, we have the amount$=10000$and the time period as 6 years. Therefore, we can substitute the values in the formula and we will get,
$\therefore 10000=p{{\left( 1+\dfrac{R}{100} \right)}^{6}}...........(ii)$
We can write $p{{\left( 1+\dfrac{R}{100} \right)}^{6}}$ as$p{{\left( 1+\dfrac{R}{100} \right)}^{3}}{{\left( 1+\dfrac{R}{100} \right)}^{3}}$.
Now, equation (ii) will be converted as:
$\Rightarrow 10000=p{{\left( 1+\dfrac{R}{100} \right)}^{3}}{{\left( 1+\dfrac{R}{100} \right)}^{3}}.....(iii)$
We can substitute the value of $p{{\left( 1+\dfrac{R}{100} \right)}^{3}}$ from equation (i) in equation (iii) and we will get,
$\Rightarrow 10000=8000{{\left( 1+\dfrac{R}{100} \right)}^{3}}$
$\Rightarrow \dfrac{10000}{8000}={{\left( 1+\dfrac{R}{100} \right)}^{3}}................(iv)$
No, we can substitute the value of ${{\left( 1+\dfrac{R}{100} \right)}^{3}}$ from equation (iv) in equation (i) and we will get,
$\Rightarrow 8000=p\left( \dfrac{10000}{8000} \right)$
By cross multiplying, we will get,
$\Rightarrow p=\dfrac{64000000}{10000}$
$\Rightarrow p=6400$
$\therefore$ Principal$=6400$
Hence, the sum is $6400$.
Note: Do not confuse compound interest with simple interest otherwise you will be messed up with your answer. Always remember that sum is taken as principal in both compound interest and simple interest, the final money will be taken as the amount. In these types of questions, if we are not given the rate of interest, then we need not find the rate. It is because the rate will get cancelled while taking the ratios of the formula for different time periods. So, do not get confused if the rate of interest is not given.
Since in formula, we have the amount $p{{\left( 1+\dfrac{R}{100} \right)}^{n}}$. So in case you are provided with time more than 3 years then, try to get the values from other equations. Do not split the powers.
Like, for example, do not do like:
${{\left( 1+\dfrac{R}{100} \right)}^{4}}=\left( 1+\dfrac{R}{100} \right)\left( 1+\dfrac{R}{100} \right)\left( 1+\dfrac{R}{100} \right)\left( 1+\dfrac{R}{100} \right)$
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# 6-1 Properties and Attributes of Polygons 6.1: Polygon
6-1 Properties and Attributes of Polygons 6.1: Polygon Angle Theorems Objectives Find and use the measures of interior and exterior angles of polygons. Handbook p. 19 Holt Geometry Polygon Angle Sum Theorem: 6-1 Properties and
Attributes of Polygons The sum of the interior angles of a convex polygon is (n 2) ) 180. Draw each diagonal from a single vertex, then complete the chart. Polygon Triangle Quadrilateral Pentagon Hexagon n-gon Holt Geometry # of Sides # of Triangles
Sum of Interior Measures 6-1 Properties and Attributes of Polygons 6.1: Polygon Angle Theorems 1. Find the sum of the interior angle measures of a convex heptagon. (n 2) )180 (7 2) )180 900 Holt Geometry 6-1 Properties and Attributes of Polygons
6.1: Polygon Angle Theorems 2) . Find the measure of each interior angle of a regular 16-gon. Step 1: Find the sum of the interior angle measure (n 2) )180 (16 2) )180 = 2) 52) 0 Step 2: Find the measure of one interior angle. Holt Geometry 6-1 Properties and Attributes of Polygons Handbook page 19
Interior Angles Sum of all angles 180( n 2) Measure of one if the polygon is REGULAR! Holt Geometry 180(n 2) n Exterior Angles 6-1 Properties and Attributes of Polygons
6.1: Polygon Angle Theorems 3. Find the measure of each interior angle of pentagon ABCDE. (5 2) )180 = 540 35c + 18c + 32) c + 32) c + 18c = 540 135c = 540 c=4 mA = 35(4) = 140 mB = mE = 18(4) = 72) mC = mD = 32) (4) = 12) 8 Holt Geometry 6-1 Properties and Attributes of Polygons
6.1: Polygon Angle Theorems 4a. What is the measure of each interior angle of a regular decagon? b. What is the measure of each exterior angle of a regular decagon? c. What is the SUM of all of the exterior angles? Holt Geometry 6-1 Properties and Attributes of Polygons 6.1: Polygon Angle Theorems
In the polygons below, an exterior angle has been measured at each vertex. Notice that in each case, the sum of the exterior angle measures is 360. Holt Geometry 6-1 Properties and Attributes of Polygons 5. Find the measure of each exterior angle of a regular 20-gon. A 2) 0-gon has 2) 0 sides and 2) 0 vertices.
Sum of ext. s = 360. Measure of one ext. = The measure of each exterior angle of a regular 2) 0gon is 18. Holt Geometry 6-1 Properties and Attributes of Polygons Angle measures of a convex polygon with n sides Interior Angles Exterior Angles Sum of all angles Measure of one if the polygon is REGULAR!
Holt Geometry 180(n 2) 360 180(n 2) n 360 n 6. Find the value of b in of Polygons 6-1 Properties
and Attributes polygon FGHJKL. 15b + 18b + 33b + 16b + 10b + 2) 8b = 360 12) 0b = 360 b=3 Holt Geometry 6-1 Properties Attributes of Polygons 7. Name theand convex polygon
whose sum of its interior angle measures is 6840. 8. The exterior angle of a regular polygon is about 27.7. Find the number of its sides and the measure of each interior angle. Holt Geometry
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# Convert Fractional Ratio into Whole Number Ratio – Definition, Facts, Examples | How to Convert Fractional Ratio to Whole Number Ratio?
Students, who are looking for learning ways on how to convert fractional ratios into whole-number ratios can find them all here? Then, look no further as we have listed the easy way to convert the fractional ratio to the whole ratio in a detailed way here. Conversion of Fractional ratio to Whole Number Ratio means you wish to present the fractional ratios as exact whole number ratios.
Before going deep into the concept, you need to understand the concept of whole number ratio and fractional ratio first. Refer to the entire article to know about the procedure on how to convert a fractional ratio to a whole number ratio along with solved examples. 6th Grade Math Students can Practice these questions on converting fractional ratio to a whole number ratio and improve your math skills.
Also, Refer:
### Fractional Ratio – Definition
The fractional ratio is defined as when we write a Ratios as fractions, first add the total parts in the ratio and then find the denominators and write each part of the ratio as the individual numerators. When we write ratios as fractions, the number of fractions that are formed is the same as the number of parts in the ratio.
### What is Whole Number Ratio?
The Whole Number Ratio is when both the numerator and denominator are natural numbers of any size but are coprime that is they share no common factors, hence the ratio is simple. In simple the whole number ratio implies that the ratio of the elements in a compound is a whole number which is the ratio of elements bearing a whole number.
For example in CH4, C and H are combined in a simple way. The whole number ratio of that is1:4. Here are some examples of a simple whole-number ratio 2:3, 23:44,1:1.
Here are some examples that are not the whole number ratio is 12:16 (this isn’t simplified fully – you can still divide both sides by 4 ), 2–√:5 (the numerator then is not a natural number), 2:1.1 (where the denominator is not a natural number).
### How to Convert Fractional Ratio to Whole Number Ratio?
The following are the steps to convert fractional ratio into whole number ratio. The steps are:
• First, find the least common factor (L.C.M) of the denominators.
• Next, multiply each term of the ratio by this least common multiple(L.C.M).
• Now, simplify it.
### Converting Fractional Ratios to Whole Number Ratios Examples
Problem 1:
What is the whole number ratio of the fractional ratio value is 1/3 : 1/5: 1/8?
Solution:
As given in the question, the fractional ratio is 1/3: 1/5: 1/8.
Now, we need to find out the given ratio into the whole number ratio.
First, we can rewrite the given ratio as 1/3 ÷1/5÷1/8
Now, multiply it. We get,
1/3 *5/1 *8/1= 40:3
Therefore, the whole number ratio is 40:3.
Problem 2:
Convert the fractional ratio 1/5:1/3 into the whole number ratio or the ratio in its simplest form?
Solution:
In the given question, the fractional ratio is 1/5: 1/3.
Now, to convert the fractional ratio into the whole number ratio, we need to follow the steps given below:
(i) First, Find the least common multiple (L. C. M.) of the denominators.
So, the Denominators are 5 and 3 and the LCM of denominators is 5×3 = 15.
(ii) Next, Multiply each term of the ratio by the least common multiple (L. C. M.) that is,
(1/5)×15: (1/3)×15
(iii) Now, simplify it. So the value is 15/5:15/3 = 3: 5.
Hence, the whole number ratio or ratio in simplest form is 3:5.
Problem 3:
What is the whole number ratio of 4/2: 8/3?
Solution:
Given in the question, the fractional value is 4/2:8/3.
Now, to convert the fractional ratio into the whole number ratio, we need to follow the steps given below:
(i) First, Find the least common multiple (L. C. M.) of the denominators.
So, the Denominators are 2 and 3 and the LCM of the denominators is 2×3 = 6.
(ii) Next, Multiply each term of the ratio by the least common multiple (L. C. M.) that is,
(4/2)×6: (8/3)×6
(iii) Now, simplify it. So the value is 24/2:48/3 = 12: 16.
Thus, the whole number ratio or the ratio in simplest form is 12:16.
Problem 4:
How to Convert the fractional ratio into the whole number ratio. The fractional ratio is 2/1: 2/5:2/8?
Solution:
As given in the question, the fractional ratio is 2/1:2/5:2/8.
Now, we need to find out the given ratio into the whole number ratio.
We can also rewrite as the given ratio is 2/1 ÷2/5÷2/8
Now, multiply it. So, we get the value is,
2/1 *2/5*2/8= 8: 40
Therefore, after converting the fractional ratio into the whole number ratio is 8:40.
### FAQ’s on Fractional Ratio to Whole Number Ratio Conversion
1. What is the difference between Fraction and Ratio?
A fraction is simply some number divided by another number. A ratio is a relationship between two numbers, although it could be one number divided by another number. Fraction has two parts one is the numerator and the second part is the denominator whereas the ratio has no parts.
2. List the steps to convert the fractional ratio into the whole number ratio?
The following are the steps to convert the fractional ratio into the whole number ratio:
Step I: First, find the least common multiple (L. C. M.) of the denominators.
Step II: Next, we have to Multiply each term of the ratio by this least common multiple (L. C. M.).
Step III: Then simplify it. We get the ratio in the simplest ratio.
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Geometry
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A piece of wire is 42 inches long. The wire is bent into a rectangle where the length is 2 times the width. What are the dimensions of the rectangle?
Rachel M.
The first step to this problem is realizing that the length of the wire will be the new perimeter of the rectangle once it's bent. What is the equation of a rectangle with respect to its length, width, and perimeter? P = 2W + 2L The perimeter is equal to two times its width plus two times its length. We know P is equal to 42 inches in this case. Equation 1 gives us: (1) 42 = 2W + 2L We have one equation and two unknowns. We need another equation to solve this problem. What else was given in the problem? We know that the length is 2 times the width for the rectangle. We can write the second equation: (2) L = 2W Now, we have two equations and two unknowns! Let's substitute Equation 2 into Equation 1, and solve. 42 = 2W + 2(2W) 42 = 2W + 4W 42 = 6W W = 7 We were able to solve for the width (W) by substituting equation 2 into equation 1, giving us only one unknown! Now that we have the width, we can find the length by substituting our known width into equation 2. L = 2(7) L=14 The dimensions of the rectangle are 7"X14"!
Basic Math
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Fred's income is 800 dollars per week. How much does Fred make in one year?
Rachel M.
This is a simple word problem. There's a few questions we have to answer before we come to the final answer. First, we were given the income per WEEK, but asked how much he makes in one YEAR. How many weeks are in one year? 52! There are 52 weeks per year. Next, we know that Fred makes 800 dollars per WEEK. So how do we figure out how much he makes in one year? We have to use multiplication! We multiply 800 dollars X 52 weeks per year. \$800 X 52 = \$41,600 Fred makes \$41,600 per year.
Algebra
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Solve the equation for X. 4X + 2 = 22 – 3(X+2)
Rachel M.
To solve the equation for X, our mission is to get X on one side of the equation, and everything else on the other side. This is done by Algebra. 1. The first step is to distribute anywhere you have parenthesis. In this case, we have one set of parenthesis on the right side of the equation (X+2). We need to distribute the negative 3 into the parenthesis (don’t forget to distribute the negative sign!!). 4X + 2 = 22 -3(X) – 3(2) 4X + 2 = 22 -3X – 6 2. Next, we have to get all “like terms” on the same side of the equation. Let’s move all the constants to one side of the equation! On the right side, we have a 22 and a -6. Subtract to simplify which gives us 16. 4X + 2 = 16 – 3X We have a +2 on the left side, so to undo this; we have to subtract 2 on BOTH sides of the equation. 4X + 2 -2 = 16 – 2 – 3X 4X = 14 – 3X 3. Next, we have to get all “X”s on the same side of the equation. Let’s move the 3X from the right side to the left side. To undo the -3X, we have to ADD 3X to both sides of the equation. 4X + 3X = 14 -3X + 3X The -3X and +3X cancel out on the right side of the equation. Next, let’s add the 3X and 4X. 7X = 14 4. Our last step is to undo the 7 on the left side of the equation to get X by itself. To undo multiplication, we have to divide 7 on both sides of the equations! 7X / 7 = 14/7 X = 2 Our final answer is X=2!
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# Logarithm concepts -- introduction
Page 1 / 1
This module introduces the concept of logarithms.
Suppose you are a biologist investigating a population that doubles every year. So if you start with 1 specimen, the population can be expressed as an exponential function: $p\left(t\right)={2}^{t}$ where $t$ is the number of years you have been watching, and $p$ is the population.
Question: How long will it take for the population to exceed 1,000 specimens?
We can rephrase this question as: “2 to what power is 1,000?” This kind of question, where you know the base and are looking for the exponent, is called a logarithm .
${\text{log}}_{2}\text{1000}$ (read, “the logarithm, base two, of a thousand”) means “2, raised to what power, is 1000?”
In other words, the logarithm always asks “ What exponent should we use ?” This unit will be an exploration of logarithms.
## A few quick examples to start things off
Problem Means The answer is because ${\text{log}}_{2}8$ 2 to what power is 8? 3 ${2}^{3}$ is 8 ${\text{log}}_{2}16$ 2 to what power is 16? 4 ${2}^{4}$ is 16 ${\text{log}}_{2}10$ 2 to what power is 10? somewhere between 3 and 4 ${2}^{3}=8$ and ${2}^{4}=16$ ${\text{log}}_{8}2$ 8 to what power is 2? $\frac{1}{3}$ ${8}^{\frac{1}{3}}=\sqrt[3]{8}=2$ ${\text{log}}_{10}10,000$ 10 to what power is 10,000? 4 ${10}^{4}=10,000$ ${\text{log}}_{10}\left(\frac{1}{100}\right)$ 10 to what power is $\frac{1}{100}$ ? –2 ${10}^{–2}=\frac{1}{{10}^{2}}=\frac{1}{100}$ ${\text{log}}_{5}0$ 5 to what power is 0? There is no answer ${5}^{\text{something}}$ will never be 0
As you can see, one of the most important parts of finding logarithms is being very familiar with how exponents work!
#### Questions & Answers
find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Question of Exercise 2 (Subjective)
# Question In figure, Ois the centre of the circle. A is any point on minor arc BC. Prove that ∠BAC – ∠OBC = 90°. .
Simplify the following expressions
(√(5) + √(2))2
Solution:
Explanation:
Which of the following rational numbers lies between 0 and - 1
A: 0
B: - 1
C: -1/4
D: 1/4
Solution:
Explanation:
0 and 1 cannot be found between 0 and 1.
In addition,0= o/4 and -1= -4/4
We can see that -1/4 is halfway between 0 and -1.
Hence, the correct option is (c) -1/4
Prove that the diagonals of a parallelogram bisect each other
Solution:
Explanation:
We must show that the diagonals of the parallelogram ABCD cross each other.
OA = OC & OB = OD, in other words.
Now AD = BC [opposite sides are equal] in ΔAOD and ΔBOC.
[alternative interior angle] ∠ADO = ∠CBO in ΔAOD and ΔBOC.
Similarly, ∠AOD = ∠BOC by ΔDAO = ΔBCO (ASA rule)
As a result, OA = OC and OB = OB [according to CPCT].
Hence, it is prove that the diagonals of a parallelogram bisect each other.
What is total surface area of sphere
Solution:
Explanation:
• The radius of the sphere affects the formula for calculating the sphere's surface area.
• If the sphere's radius is r and the sphere's surface area is S.
• The sphere's surface area is therefore stated as Surface Area of Sphere 4πr2, where ‘r’ is the sphere's radius.
• The surface area of a sphere is expressed in terms of diameter as S=4π(d/2)2, where d is the sphere's diameter.
Thus, total surface area of sphere is =4πr2.
Fill in the blanks If two adjacent angles are supplementary
they form a __________.
Solution:
Explanation:
• If the non-common sides of two angles form a straight line, they are called linear pair angles.
• The sum of the angles of two linear pairs is degrees.
• If the total of two angles is degrees, they are called supplementary angles.
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# Algebra: Finding the Equation of a Line
On this page we hope to clear up problems that you might have with finding the equation of a line.
Many times, you'll have the graph of an equation shown to you and you'll need to find the equation. This seems a very daunting task, but it's actually quite easy!
For example, take a horizontal line such as y = 2. Every point on that graph is 2 units above teh x axis. All horizontal lines have equations that are written in the same format, such as y = -4.5. Because they're all written the same way, we can come up with a general formula for horizontal lines. It is the following equation (where k represents any real number): y = k. Under the same assumptions, the general formula for vertical lines can be written as follows (where k represents any real number): x = k.
Linear equations such as y = .009x + 34 also have a general equation that can represent any linear equation. It is written as follows: y = mx + b.
The things to remember about the above formula, which is called the slope-intercept formula, are outlined below.
1. Since you know a line with an equation is that form cannot be horizontal or vertical, all you need to find are m and b to find the equation.
2. b is called the intercept. It is the point when the line crosses the y axis.
3. m is called the slope. The slope has both a sign (either + or -) and a value (the number behind the sign). For example, the equation y = -3x + 4 has a negative sign and a value of three. Therefore, the slope is -3. When looking at a graph, you can always tell if the slope is negative or positive by the direction it points. When looking for the sign of a slope, look at the left side of the graph. Then, look at the right side of the graph. If the right side is lower than the left side, the line has a negative slope, if the right side is higher, the line has a positive slope.
Example:
``` The graph: Example Graph
```
To find the value of the slope, you compute the rise over the run. To do that, pick two points on the line at random and then draw a line through each of those points that runs parallel to the coordinate axes. Count the number of units between the point on the line and where the two additional lines you drew intersect. The number of units on the horizontal line is the run and the number of units on the vertical line is the rise. Dividing the rise by the run gives you the value of the slope.
Example:
``` The graph: Example Graph
```
Example:
```1. Find the equation of the line graphed in the accompanying figure.
Solution: The desired equation is in the slope-intercept form.
You need to find m and b.
By looking at the graph, you can see that b must be 3.
Also, by inspecting the graph, you can see that the slope is
negative because the left end of the line is higher than the
right end.
Now you need to find the value of the slope. Pick any two
points on the line and draw a line through each point that
is horizontal to the coordinate axes. (For simplicity's sake,
we will use b, which is (0,3) and the point given in
the problem, (2,0).) Count the number of units on the
vertical line and horizontal line. The vertical line is the
rise and the horizontal line is the run. Compute
the slope's value by putting the rise over the run. In this
case, the rise is 3 and the run is 2.
Therefore, the value of the slope is (3/2). Combine the
sign of the slope and the value of the slope to get the
complete number for m. It is -(3/2).
Plug the values you found into the slope-intercept
general formula and you get the following:
3
y = - -x + 3
2
```
|
Bisection Method Notes
# Bisection Method Notes - Chapter 03.03 Bisection Method of...
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Chapter 03.03 Bisection Method of Solving a Nonlinear Equation After reading this chapter, you should be able to: 1. follow the algorithm of the bisection method of solving a nonlinear equation, 2. use the bisection method to solve examples of finding roots of a nonlinear equation, and 3. enumerate the advantages and disadvantages of the bisection method. What is the bisection method and what is it based on? One of the first numerical methods developed to find the root of a nonlinear equation 0 ) ( = x f was the bisection method (also called binary-search method). The method is based on the following theorem. Theorem An equation 0 ) ( = x f , where ) ( x f is a real continuous function, has at least one root between x and u x if 0 ) ( ) ( < u x f x f (See Figure 1). Note that if 0 ) ( ) ( u x f x f , there may or may not be any root between x and u x (Figures 2 and 3). If 0 ) ( ) ( < u x f x f , then there may be more than one root between x and u x (Figure 4). So the theorem only guarantees one root between x and u x . Bisection method Since the method is based on finding the root between two points, the method falls under the category of bracketing methods. Since the root is bracketed between two points, x and u x , one can find the mid- point, m x between x and u x . This gives us two new intervals 1. x and m x , and 2. m x and u x . 03.03.1
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03.03.2 Chapter 03.03 Figure 1 At least one root exists between the two points if the function is real, continuous, and changes sign. Figure 2 If the function ) ( x f does not change sign between the two points, roots of the equation 0 ) ( = x f may still exist between the two points. f ( x ) x x u x f ( x ) x x u x
Bisection Method 03.03.3 Figure 3 If the function ) ( x f does not change sign between two points, there may not be any roots for the equation 0 ) ( = x f between the two points. Figure 4 If the function ) ( x f changes sign between the two points, more than one root for the equation 0 ) ( = x f may exist between the two points. Is the root now between x and m x or between m x and u x ? Well, one can find the sign of ) ( ) ( m x f x f , and if 0 ) ( ) ( < m x f x f then the new bracket is between x and m x , otherwise, it is between m x and u x . So, you can see that you are literally halving the interval. As one repeats this process, the width of the interval [ ] u x x , becomes smaller and smaller, and you can zero in to the root of the equation
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## This note was uploaded on 05/01/2011 for the course CHBE 2120 taught by Professor Gallivan during the Spring '07 term at Georgia Institute of Technology.
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Bisection Method Notes - Chapter 03.03 Bisection Method of...
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Question Video: Finding the Volume of a Sphere Given the Area of Its Great Circle | Nagwa Question Video: Finding the Volume of a Sphere Given the Area of Its Great Circle | Nagwa
# Question Video: Finding the Volume of a Sphere Given the Area of Its Great Circle Mathematics • Second Year of Preparatory School
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Find, to the nearest tenth, the volume of a sphere given that the area of its great circle is 400π inΒ².
01:52
### Video Transcript
Find, to the nearest tenth, the volume of a sphere given that the area of its great circle is 400π square inches.
Letβs begin by sketching a sphere and recalling what we mean by its great circle. When a sphere is cut by a plane that passes through its center, the intersection of the sphere and the plane is called a great circle. A great circle cuts a sphere exactly in half, and its radius is the same as that of the sphere itself.
Weβre given that the area of the great circle of a sphere is 400π square inches. And weβre asked to find the volume of the sphere. We can use the fact that the area of a circle is ππ squared, where π is the radius of the circle, to find the radius of the sphere. And we do this by equating the area of our great circle, thatβs 400π, with ππ squared. Next, dividing through by π, we have π squared equal to 400. And then taking the positive square root on both sides, positive since π is a length and lengths are always positive, we have π equal to 20 inches.
So now that we have the radius of the great circle, which equals the radius of the sphere, recalling that the volume of a sphere is given by π equals four over three ππ cubed and making a little space for our working out, we have π equals four over three π times 20 cubed. Thatβs four over three π times 8000, which evaluates to 33,510.3216 and so on. To the nearest tenth, thatβs to one decimal place, this is 33,510.3.
Hence, given that the area of the great circle of a sphere is 400π square inches, its volume is 33,510.3 cubic inches, to the nearest tenth.
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What is theta in math? Properly, in the event you ask me, it really is a mathematical unit in which all of the digits are equal to 1 an additional.
This unit is called the tau and it is defined because the integral of any equation containing the equation with 1 or extra quantities which are denoted by a decimal point.
These units are very helpful in calculus mainly because they are able to be multiplied to calculate another function that’s a a number of of itself. These functions are called polynomials and any time you try to resolve them with theta units, you’ll get a a number of assignments help from the original function. It is actually a similar scenario after you endeavor to convert from tau to d. If you are dealing with the d element of theta and not theta then it will likely be impossible for you personally to have any solution.
So in reality the math unit in calculus is theta, and what exactly is theta in math. Simply put, what is theta in math may be the number that expresses the sum of your logarithms of each of the digits of a quantity.
But there is a larger point that you just should know. What’s theta in math is a logical function, a one-to-one relation. This signifies that theta equals one. Should you have a number n, say n = -5, then theta is equal to one, and for those who multiply it by itself, you will get -5.
So if you’d like to add numbers in this way, you have to use this equation: tan(n) + cot(n) = theta. Bear in mind that theta have to be precisely equal to one particular to be correct.
There are numerous other items which you can do with math, but what exactly is theta in math. This gives us two easy ideas about math.
Theta is a base unit in calculus, but whenever you will need to solve a a number of issue, you can not do so with theta. By way of example, let’s say that you are looking to operate out the volume of water within your tank. You will need to take the volume of water, after which divide it by the height in the water, to ensure that you are able to function out the density in the water.
The initial step is to uncover the volume of water inside the tank. For those who know the volume of water inside your tank, then you definitely https://www.buyessay.net know the density from the water. Then you take the volume of water and divide it by the height from the water, within this case, the height from the water inside your tank.
This is an vital step since the total volume of water you must solve the problem might be precisely the same because the total volume of water you may have measured inside your tank. The answer you will give will be a many of the total volume of water you’ve got measured.
You could have taken the volume of water and divided it by the height of your water, within this case, the height of your water within your tank, and got exactly the same answer as you are going to now. The answer you get will then be a various of your height on the water.
What is theta in math is a logical function, but you need to remember that when you’ve got complications solving, the tau will make it impossible for you personally to be in a position to have any answer. This signifies that you will need to use a various of your tau as a way to get a single solution.
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# Ncert Class 6 Math Playing With Numbers Exercise 3.1
Playing With Numbers Class 6 Ex. 3.1
Ncert Class 6 Math Free Solution.
Exercise 3.1
Question 1 :- Write all the factors of the following numbers:
(a) 24 (b) 15 (c) 21
(d) 27 (e) 12 (f) 20
(g) 18 (h) 23 (i) 36
Solution 1:-
(a) Factors of 24 are:
24 = 1 x 24;
24 = 2 x 12;
24 = 3 x 8;
24 = 4 x 6
So, all the factors of 24 are :- 1, 2, 3, 4, 6, 8, 12 and 24.
(b) Factors of 15 are:
15 = 1 x 15;
15 = 3 x 5
So, all the factors of 15 are :- 1, 3, 5 and 15.
(c) Factors of 21 are:
21 = 1 x 21;
21 = 3 x 7
So, all the factors of 21 are :- 1, 3, 7 and 21.
(d) Factors of 27 are:
27 = 1 x 27;
27 = 3 x 9.
So, all the factors of 27 are :- 1, 3, 9 and 27.
(e) Factors of 12 are:
12 = 1 x 12;
12 = 2 x 6;
12 = 3 x 4
So, all the factors of 12 are :- 1, 2, 3, 4, 6 and 12.
(f) Factors of 20 are:
20 = 1 x 20;
20 = 2 x 10;
20 = 4 x 5
So, all the factors of 20 are :- 1, 2, 4, 5, 10 and 20.
(g) Factors of 18 are:
18 = 1 x 18;
18 = 2 x 9;
18 = 3 x 6;
So, all the factors of 18 are :- 1, 2, 3, 6, 9 and 18.
(h) Factors of 23 are:
23 = 1 x 23;
So, all the factors of prime number 23 are :- 1 and 23.
(i) Factors of 36 are:
36 = 1 x 36;
36 = 2 x 18;
36 = 3 x 12;
36 = 4 x 9;
36 = 6 x 6;
So, all the factors of 36 are :- 1, 2, 3, 4, 6, 9, 12, 18 and 36.
Question 2 :- Write first five multiples of :-
(a) 5 (b) 8 (c) 9
Solution:
(a) First five multiples of 5 are:
5 x 1 = 5;
5 x 2 = 10;
5 x 3 = 15;
5 x 4 = 20;
5 x 5 = 25;
So, the required multiples of 5 are :- 5, 10, 15, 20 and 25.
(b) First five multiples of 8 are:
8 x 1 = 8;
8 x 2 = 16;
8 x 3 = 24;
8×4 = 32;
8 x 5 = 40;
So, the required multiples of 8 are :- 8, 16, 24, 32 and 40.
(c) First five multiples of 9 are:
9 x 1 = 9;
9 x 2 = 18;
9 x 3 = 27;
9 x 4 = 36;
9 x 5 = 45;
So, the required multiples of 9 are :- 9,18, 27, 36 and 45.
Question 3 :- Match the items in column I with the items in column II.
Column 1 Column 2
(i) 35 (a) Multiple of 8
(ii) 15 (b) Multiple of 7
(iii) 16 (c) Multiple of 70
(iv) 20 (d) Factor of 30
(v) 25 (e) Factor of 50
(f) Factor of 20
Solution 3:-
(i) → (b) {because, 7 x 5 = 35}
(ii) → (d) {because, 15 x 2 = 30}
(iii) → (a) {because, 8 x 2 = 16}
(iv) → (f) {because, 20 x 1 = 20}
(v) → (e) {because, 25 x 2 = 50}
Question 4 :- Find all the multiples of 9 upto 100.
Solution 4:-
9 x 1 = 9;
9 x 2 = 18;
9 x 3 = 27;
9 x 4 = 36;
9 x 5 = 45;
9 x 6 = 54;
9 x 7 = 63;
9 x 8 = 72;
9 x 9 = 81;
9 x 10 = 90;
9 x 11 = 99;
So, all the multiples of 9 upto 100 are :-
9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.
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## NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.1
NCERT Solutions of Chapter 10 Straight Lines Exercise 10.1 is given here that are detailed and accurate so you can easily solve difficult questions without wasting your time. NCERT Solutions for Class 11 Maths will increase an individual's problem solving skills and increase his efficiency. These NCERT Solutions will improve your marks and solve complex problems easily.
1. Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, –5) and (– 4, – 2). Also find its area.
Given points (– 4, 5), (0, 7), (5, –5) and (– 4, 2) are plotted on a graph paper. They are denoted by A, B, C and D respectively.
Now, divide the quadrilateral into two triangle viz. △ABD and △BDC.
Area of a triangle is
= 1/2 I x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)I
The verticals of △ABD are (-4, 5), (0, 7) and (-4, -2)
Area of △ABD
= 1/2 I -4(7 + 2) + 0.(-2 – 5)+ (-4)(5 – 7) I
= 1/2 I -36 + 8 I = 28/2 = 14 ….(i)
The vertices of △BCD are (0, 7), (5, -5), (-4, -2)
Area of △BCD
= 1/2 I 0(-5 + 2) + 5(-2 -7) – 4(7 + 5) I
= 1/2 I -45 – 48 I = 93/2 ……(ii)
= Area of △ABD + area of △BCD
= 14 + 93/2 [from (i) & (ii)]
= (28 + 93)/2 = 121/2 = 60.5 sq. units.
2. The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.
BC is the base of △ABC such that BO = OC = a
AB = AC = 2a
Now, AO2 = AB2 - BO2
∴ AO = √(4a2 – a2) = √3a
∴ The point A lying on x-axis is (√3a, 0)
The point B is (0, a) and the point C is (0, – a), when A lies on the left of y-axis the vertices of triangle are (–√3a, 0) a (0, a) and (0, – a).
3. Find the distance between P(x1, y 1) and Q(x2, y 2) when: (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.
(i) When PQ is parallel to y-axis, every point on this line has the same abscissa, let
2 = x 1.
The points P and Q are P(x 1, y 1 ),Q(x 2, y 2)
∴ PQ = |y 2 – y1|
(ii) When PQ is parallel to x-axis. Every point on the line has the same ordinate.
Let y 2 = y 1
∴ The coordinates of points P and Q are
P(x 1, y 1), Q(x 2, y 2)
∴ PQ = |x 2 – x 1|
4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Let the point on the x-axis be (x 1, 0). The other two points A and B are A(7, 6), B (3, 4)
we have
PA = PB
or PA2 = PB2
or (x 1 – 7)2 + 62 = (x 1 – 3)2 + 42
or x12 – 14x1 + 49 + 36 = x12 – 6x1 + 9 + 16
or –14 x+ 85 = –6x1 + 25
or 8x 1 = 60
or x1 = 60/8
∴ x1 = 15/2
∴ The point on x-axis equidistant from A and B is (15/2 , 0).
5. Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P(0, –4) and B(8, 0).
Coordinates of mid-point of two points (x 1, y 1) and (x 2, y 2) are ((x1 + x2)/2 , (y1 + y2)/2 )
The mid-point of B(8, 0) and P(0, –4) is
((8 + 0)/2 , (0 – 4)/2) i.e., (4, -2)
The slope of line joining the points (x 1, y 1) and (x 2, y 2) = (y2 – y1)/(x2 – x1)
The two points are O(0, 0) and A(4, –2)
∴ slope = (-2 – 0)/(4 – 0) = -1/2
6. Without using Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.
We have three points A(4, 4), B(3, 5) and C (–1, –1)
∴ Slope of AB = (y2 – y1)/(x2 – x1) = (5 – 4)/(3 – 4) = -1
Slope of AC = (-1 -4)/(-1 – 4) = -5/-5 = 1
Product of slopes of AB and AC = 1× -1 = -1
∴ AB ⊥ AC
Hence, △ABC is a right angled triangle.
7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
The line OP makes an angle of 30° with y-axis measured anti-clock-wise.
∴ OP makes an angle of 90 + 30 = 120 with positive direction of x-axis
∴ Slope of OP = tan 120° = tan (180° – 60°)
= –tan 60° = –√3
8. Find the value of x for which the points (x, –1), (2, 1) and (4, 5) are collinear.
We have the points A(x, – 1), B(2, 1), C(4, 5).
A, B, C are collinear if the
Slope of AB = Slope of BC.
Slope of AB = (1 + 1)/(2 – x) = 2/(2 – x);
Slope of BC = (5 – 1)/(4 – 2) = 4/2 = 2
∴ 2/(2 – x) = 2 or 2 – x = 1 or x = 1
9. Without using distance formula, show that points (–2, 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.
The given points are A(–2, –1), B(4, 0), C(3, 3), D(–3, 2)
Slope of AB = 1/(4 + 2) = 1/6;
Slope of DC = (3 – 2)/(3 + 3) = 1/6
∴ Slope of AB = Slope of DC
=> AB || DC
Slope of AD = (2 + 1)/(-3 + 2) = 3/-1;
Slope of BC = (3 – 0)/(3 – 4) = 3/-1
∴ Slope of AD = Slope of BC
=> AD parallel BC Hence, ABCD is a parallelogram.
10. Find the angle between the x-axis and the line joining the points (3, –1) and (4, –2).
Slope of the line joining the point P(3, –1) and Q (4, –2) is equal to (-2 + 1)/(4 – 3) = -1/1 = -1
tan θ = –1 ∴ θ = 135°
11. The slope of a line is double of the slope of another line. If tangent of the angle between them is 1 3, find the slopes of the lines.
Let m1 and m2 be the slopes of the two lines
Given that m1 = 2m2
Let θ be the angle between the line,
tan θ = 1/3 = (m1 – m2)/(1 +m1m2)
putting m1 = 2m2, 1/3 = (2m2 – m2)/(1 + 2m2m2) = m2/(1 + 2m22)
or 3m2 = 1 + 2m22 or 2m22 – 3m2 + 1 = 0
or (2m2 – 1)(m2 -1) = 0 m2 = 1 or 1/2
since, m1 = 2m2
∴ for m2 = 1, m1 = 2, for m2 = 1/2, m1 = 1
∴ slopes of these lines are 1, 1/2 or 2, 1
12. A line passes through (x1, y 1) and (h, k). If slope of the line is m, show that
k – y 1 = m(h – x1).
Slope of the line joining the points A(x 1, y 1) and B(h, k)
= (k – y1)/(h – x1) = m (Given)
Cross-multiplying, k – y 1 = m(h – x 1) Hence proved.
13. If three points (h, 0), (a, b) and (0, k) lies on a line, show that a/h + b/k = 1.
The given points are A(h, 0), B(a, b), C(0, k), they lie on the same plane.
∴ Slope of AB = Slope of BC
∴ Slope of AB = (b – 0)/(a – h) = b/(a – h);
Slope of BC = (k – b)/(0 – a) = (k – b)/-a
∴ b/(a – h) = (k – b)/ -a or by cross-multiplication
–ab = (a – h)(k – b)
or –ab = ak – ab – hk + hb
or 0 = ak – hk + hb
or ak + hb = hk
Dividing by hk => ak/hk + hb/hk = 1 or a/h + b/k = 1
Hence proved.
14. Consider the following population and year graph, find the slope of the line AB and using it, find what will be the population in the year 2010?
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# Sum of Squares
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
There are two types of ‘sum of squares’ formulas, where one of them is the sum of squares of two numbers and the other is the sum of squares of ‘n’ given values. The formula for sum of squares of two numbers ‘a’ and ‘b’ is given as a2 + b2 = (a + b)2 – 2ab. The formula for the sum of squares of ‘n’ values such as (12 + 22 + 32 + 42 + … + n2) = [n (n+1) (2n + 1)]/ 6. According to the given situation, either of the formulas is used for solving.
Example 1: Find the value of the sum of the squares of ‘1’ and ‘4’ written as, 12 + 42.
According to the sum of the squares formula, a2 + b2 = (a + b)2 – 2ab.
Given 12 + 42, which implies a = 1 and b = 4.
Therefore according to the sum of the squares formula, we get: 12 + 42 = (1 + 4)2 – 2*1*4
This gives: 12 + 42 = (5)2 – 8 ==> 12 + 42 = 25 – 8 = 17.
Hence the value of 12 + 42 = 17.
Example 2: Find the value of the sum of squares from ‘1’ to ‘5’.
Sum of the squares from ‘1’ to ‘5’ can be written as: 12+ 22+ 32+ 42+ 52.
Sum of squares of numbers from ‘1’ to ‘n’ written as:
(12+ 22+32+ … + n2) = [n (n+1) (2n + 1)]/ 6
Here n= 5.
So, 12+ 22+ 32+ 42+ 52 = [5 (5 + 1) (10 + 1)]/ 6.
This gives: (5* 6* 11)/ 6= 330/6 = 55.
Therefore the sum of squares from ‘1’ to ‘5’ is 55.
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# Column Addition of Decimal Numbers
In this worksheet, students practise adding decimal numbers using columns.
Key stage: KS 3
Curriculum topic: Number
Curriculum subtopic: Use Four Operations for All Numbers
Difficulty level:
### QUESTION 1 of 10
It is important to line up the decimal points.
Example
7702 + 77.02 + 770.2 = _____
Line up the decimal points and insert extra zeros to help fill columns.
596.1 + 5.961 + 59.61 = _____
3150.5 + 31.505 + 315.05 = _____
6284.7 + 62.847 + 628.47 = _____
1954.4 + 19.544 + 195.44 = _____
4956 + 49.56 + 495.6 = _____
840.3 + 8.403 + 84.03 = _____
7469.3 + 74.693 + 746.93 = _____
465.2 + 4.652 + 46.52 = _____
7561.9 + 75.619 + 756.19 = _____
1100 + 110 + 1.1 + 0.11 = _____
• Question 1
596.1 + 5.961 + 59.61 = _____
661.671
EDDIE SAYS
Line up the numbers using the decimal places.
Remember to add extra zeros on the end so each number has the same amount of decimal places.
596.1 5.961 59.61
• Question 2
3150.5 + 31.505 + 315.05 = _____
3497.055
EDDIE SAYS
Line up the numbers using the decimal places.
Remember to add extra zeros on the end so each number has the same amount of decimal places.
3150.5 31.505 315.05
• Question 3
6284.7 + 62.847 + 628.47 = _____
6976.017
EDDIE SAYS
Line up the numbers using the decimal places.
Remember to add extra zeros on the end so each number has the same amount of decimal places.
6284.7 62.847 628.47
• Question 4
1954.4 + 19.544 + 195.44 = _____
2169.384
EDDIE SAYS
Line up the numbers using the decimal places.
Remember to add extra zeros on the end so each number has the same amount of decimal places.
1954.4 19.544 195.44
• Question 5
4956 + 49.56 + 495.6 = _____
5501.16
EDDIE SAYS
Line up the numbers using the decimal places.
Remember to add extra zeros on the end so each number has the same amount of decimal places.
4956 49.56 495.6
• Question 6
840.3 + 8.403 + 84.03 = _____
932.733
EDDIE SAYS
Line up the numbers using the decimal places.
Remember to add extra zeros on the end so each number has the same amount of decimal places.
840.3 8.403 84.03
• Question 7
7469.3 + 74.693 + 746.93 = _____
8290.923
EDDIE SAYS
Line up the numbers using the decimal places.
Remember to add extra zeros on the end so each number has the same amount of decimal places.
7469.3 74.693 746.93
• Question 8
465.2 + 4.652 + 46.52 = _____
516.372
EDDIE SAYS
Line up the numbers using the decimal places.
Remember to add extra zeros on the end so each number has the same amount of decimal places.
465.2 4.652 46.52
• Question 9
7561.9 + 75.619 + 756.19 = _____
8393.709
EDDIE SAYS
Line up the numbers using the decimal places.
Remember to add extra zeros on the end so each number has the same amount of decimal places.
7561.9 75.619 756.19
• Question 10
1100 + 110 + 1.1 + 0.11 = _____
1211.21
EDDIE SAYS
Line up the numbers using the decimal places.
Remember to add extra zeros on the end so each number has the same amount of decimal places.
1100 110 1.1 0.11
---- OR ----
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## Consider this problem:
Sketch a graph of a function y=f(x).
Now sketch a graph of y=f(ax) for some constant value a>1.
## Reflect.
What kind of graph did you sketch for y=f(x)? How did you decide what to sketch? Did you choose a particular value for a? How does your graph of y=f(ax) compare to your graph of y=f(x)? What is similar? What is different?
## It is no secret that students have difficulty making sense of transformations of functions. A covariation approach can help. Here’s how.
A graph of y=f(x) represents a relationship between two “things” that can change: “y” and “x.”
## Think about a graph of y=f(x) as representing a relationship between two “things” that can change: “y” and “x“
The notation y=f(x) means that y is a function of x. Functions are special kinds of relationships between two “things” that can change. When y is a function of x, as the values of x change, the values of y change in predictable ways.
## A graph of y=f(ax) represents a relationship between two “things” that can change: “y” and “ax“
The notation y=f(ax) means that y is a function of ax. When y is a function of ax, as the values of ax change, the values of y change in predictable ways. Because a is some constant value greater than one, does not change. Therefore, the changing values of ax depend on changes in the values of x.
## Working from a relationship between “y” and “x,” students can make sense of a relationship between”y” and “ax.”
For a function y=f(x), as the values of x change, the values of y change along with them.
For a function y=f(ax), the values of x change by a factor of “a.” This means that “1/ath of the change in x will yield the same amount of change in y as for y=f(x).
## Using relationships between changing “things” can help students to make sense of graphs of transformations of functions.
For example, sketch a graph y=f(x).
Now sketch a new graph with this constraint: “1/2 the change in x will produce the same change in y as for the original function.”
## Reflect.
What kind of graph did you sketch for y=f(x)? How did you decide what to sketch? How does your new graph compare to your graph of y=f(x)? What is similar? What is different?
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