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# The mass of $3{m^3}$ of cement of density 3000 $kg/{m^3}$ isA) 1000kgB) 9000kgC) 10000kgD) 90000kg
Last updated date: 27th Feb 2024
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Hint: Here, we will proceed by defining the density of any substance. Then, we will express the density mathematically. Finally, we calculate its mass by using the formula of density. Density is mass per unit volume.
Density of any material is simply the ratio of the mass of the material to the volume of the material. Mathematically, density of any substance can be expressed as
${\rm{Density}}\left( \rho \right){\rm{ = }}\dfrac{{{\rm{Mass}}\left( {\rm{m}} \right)}}{{{\rm{Volume}}\left( {\rm{V}} \right)}}$
Density is denoted by rho, mass is (m) and volume is (V)
According to given data,
Volume of cement is $= 3\;{{\rm{m}}^3}$
Density of cement is $= \;3000\;{\rm{kg/}}{{\rm{m}}^3}$
By using above formula we get,
$3000{\rm{kg/}}{{\rm{m}}^3}\; = \;\dfrac{{{\rm{mass (m)}}}}{{3\;{{\rm{m}}^3}}}$
$\Rightarrow \;{\rm{mass}}\; = \,3000{\rm{kg/}}{{\rm{m}}^3} \times \;3{{\rm{m}}^3}$ $\therefore \;{\rm{mass = }}\;{\rm{9000 kg}}$
Hence the correct option is (B).
Additional information: Density is a quantitative physical attribute of a material or of a more or less stable mixture. When we take a piece of material, it has some mass and volume. The mass divided by volume is called density, and it depends solely on substance (remains the same for different parts of the same material).
Note: Compared to solid-liquid and gas, the solid has more density and liquid have less density; liquid density exists between solids and gases. For example, wood always floats on water because the wood is less dense than water. The unit of density is ${\rm{kg/}}{{\rm{m}}^3}$.
The dimensional formula of density is derived as, ${\rm{Density}}\left( \rho \right){\rm{ = }}\dfrac{{{\rm{Mass}}\left( {\rm{m}} \right)}}{{{\rm{Volume}}\left( {\rm{V}} \right)}}$
The dimension of mass (m) = [M]
Also, the dimension of volume (V) = $[{L^3}]$
$\Rightarrow \;\rho \; = \;\dfrac{{{\rm{[M]}}}}{{{\rm{[}}{{\rm{L}}^3}]}} = \left[ {{\rm{M}}{{\rm{L}}^{ - 3}}} \right]$
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# How do you factor 3x^3 - 3x^2 - 6x?
##### 1 Answer
May 12, 2016
$3 {x}^{3} - 3 {x}^{2} - 6 x = 3 x \left(x - 2\right) \left(x + 1\right)$
#### Explanation:
First, observe that the greatest shared constant factor of each term is $3$, and the highest power of $x$ shared by each term is ${x}^{1}$. Thus, we start by factoring that out from each term.
$3 {x}^{3} - 3 {x}^{2} - 6 x = 3 x \left({x}^{2} - x - 2\right)$
Next, to factor the remaining quadratic expression, there are several techniques. In our case, we will look for two values whose product is equal to the product of the coefficient of ${x}^{2}$ and the constant term, that is, $1 \cdot - 2 = - 2$, and whose sum is equal to the coefficient of $x$, that is, $- 1$. Doing so, we find that $- 2$ and $1$ fulfill these conditions, and so we can finish factoring the expression as
$3 x \left({x}^{2} - x - 2\right) = 3 x \left(x - 2\right) \left(x + 1\right)$
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# Section 1.5: Venn Diagrams - Shading
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1 Section 1.5: Venn Diagrams - Shading A Venn diagram is a way to graphically represent sets and set operations. Each Venn diagram begins with a rectangle representing the universal set. Then each set is represented by a circle inside the rectangle. We will only be studying with Venn diagrams that have 2 or 3 overlapping circles (sets). There are Venn diagrams that contain more than 3 circles (sets). In this section we will learn how to graphically represent union, intersection and complement of sets using Venn diagrams. We will start with Venn diagrams that have two sets as they are easier. Venn diagrams for two sets: Here is a Venn diagram for two sets A and B. The universal set is depicted with a rectangle. The two sets A and B are depicted with overlapping circles inside the rectangle. This is a Venn diagram associated with the set A. Notice we shade the set that the Venn diagram depicts.
2 Now for some set operations. Example: Create a Venn diagram for A. That is shade the area depicted by the set A. Recall that the complement of A (denoted A ) is everything in the universal set that is not in set A. We shade everything except what's in the A circle. Here is the Venn diagram that depicts set A. Example: Create a Venn diagram for A B. To do this: first create a shaded Venn diagram for each set described in the problem. I will create a Venn diagram for set A and another for set B. Now put them together. Notice that I darkened the area that is shaded in both diagrams. This is a union problem so anything that gets shaded at least once is in the union, so here's the Venn diagram that shows the answer. A B
3 Example: Create a Venn diagram for A B. First create a shaded Venn diagram for each set described in the problem. Now put them together. This is an intersection problem. The final Venn diagram should be shaded only where the two sets cross. Here is the answer: A B
4 There are the basics, now let's try a couple that are more involved. Example: Create a Venn diagram for A B First create a Venn diagram for each set described in the problem. A B Next, put them together. The intersection is the common region that is shaded. Here is my final answer. A B
5 Example: Find a Venn diagram for (A B). First shade what's in the parenthesis, that is A B. (I took this shading from an earlier example.) A B Now take the complement of that set. The complement is the region that is not already shaded. This is my answer. (A B). Homework #1-9: Sketch the region. 1) A B 2) A B 3) A B 4) A B 5) (A B) 6) (A B ) 7) (A B) 8) (A B ) 9) A B
6 Venn Diagrams for 3 Sets: Here's the beginning Venn diagram associated with 3 sets. Complements, unions and intersections are handled in the same way as they were with 2 set Venn diagrams. Here is the Venn diagram associated with the set A. Notice we shade the inside of the A circle. Example: Create a Venn diagram for A. Recall that the complement of A is everything that is in the universal set but not in the set A, so we shade everything except what's in the circle A.
7 Example: Create a Venn diagram that represents A B C. To do a Venn diagram for a union of three sets, first shade each set. A B C Now put them together Anything that gets shaded at least once is in the union, so here's the answer. A B C
8 Example: Create a Venn diagram for A B C. We need to work from left to right. First we will find A B First shade each set. A B Now put them together A B is the region where the two graphs overlap. A B
9 Next put the graphs of A B and of C next to each other. A B C Put the graphs together The union consists of any region that is shaded. This is my answer. A B C
10 Example: Create a Venn diagram for A B C. Let s work from left to right. Find A B. Create a diagram for each set A B Put the sets together. A B is the region where the two graphs overlap. A B
11 Next we put the graphs for A B and C next to each other. A B C Now put them together. We are finding the union of the two sets. Anything that gets shaded is part of my answer. Here is the answer. A B C
12 Example: Sketch (A B) C We need to work on the inside of the parenthesis of (A B) first. We will first graph A B First shade each set A B Now put them together. We are finding the union of the two sets. Anything that gets shaded is part of my answer. This is a diagram for A B Now we can find (A B), which is just the complement of the set A B, which is anything not shaded in our last drawing. (A B)
13 Now we can get back to the problem: (A B) C Put the sets (A B) and C next to each other. (A B) C Now put the graphs together Remember the intersection only includes the region where the shadings overlap. Here is the answer. (A B) C
14 Homework #10-23: Sketch the shaded region. 10) A B C 11) A B C 12) A B C 13) A B C 14) A B C 15) A B C 16) (B C) A 17) (A B) C 18) B (A C) 19) A (B C) 20) B (A C) 21) A (B C) 22) B (A C) 23) A (B C) Answers: 1) 3) 5)
15 7) 9) 11) 13) 15)
16 17) 19) 21) 23)
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### 5-4 Solving Compound Inequalities. Solve each compound inequality. Then graph the solution set. 6. f 6 < 5 and f 4 2 SOLUTION: and
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# Difference between revisions of "2007 iTest Problems/Problem 56"
The following problem is from the Ultimate Question of the 2007 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted.
## Problem
In the binary expansion of $\dfrac{2^{2007}-1}{2^{225}-1}$, how many of the first $10,000$ digits to the right of the radix point are $0$'s?
## Solution
We can approach this problem by using long division in base 2 because long division takes advantage of regrouping. The number $2^{2007} - 1$ has $2007$ ones while the number $2^{225} - 1$ has $225$ ones.
[225 ones])[2007 ones]
Notice that $2^{2007} - 1$ only has ones, and notice that $\frac{2^n \cdot (2^{225} - 1)}{2^{225} - 1} = 2^n.$ That means the remainder when $2^{2007} - 1$ is divided by $2^{225} - 1$ is the same when $2^{207} - 1$ is divided by $2^{225} - 1.$
[225 1’s])[207 1’s]
There are not enough digits for $2^{207} - 1$ to divide evenly into $2^{225} - 1,$ so we need to bring down more zeroes. Since $2^{225} - 2^{18} < 2^{225} - 1 < 2^{226} - 2^{19}$, we need to bring down 19 zeroes, resulting in 18 zeroes to the right of the radix point.
0.[18 0’s]
[225 1’s])[207 1’s][18 0’s]0
Now we can subtract $2^{225} - 1$ in the long division.
0.[18 0’s]1
[225 1’s])[207 1’s][18 0’s]0
-[206 1’s][18 1’s]1
[206 1’s][18 0’s]1
We can bring down more zeroes and repeat the iteration.
0.[18 0’s]111
[225 1’s])[207 1’s][18 0’s]0
-[206 1’s][18 1’s]1
[206 1’s][18 0’s]10
-[205 1’s][18 1’s]11
[205 1’s][18 0’s]110
-[204 1’s][18 1’s]111
[204 1’s][18 0’s]111
Notice that no zeroes are placed to the right of the last values after each iteration. Also, there is a pattern after doing the subtraction in each iteration, where there are $n$ ones followed by $18$ zeroes and $207-n$ ones.
To confirm this, we note that in each iteration, we multiply by $2$ and subtract $2^{2007} - 1.$ Doing this results in
[n 1’s][18 0’s][207-n 1’s]0
-[n-1 1’s][18 1’s][207-n 1’s]1
[n-1 1’s][18 0’s][207-(n-1) 1’s]
Thus, we see that the pattern holds. In fact, tthe pattern repeats $206$ times, and after that, the number from the subtraction just has $207$ zeroes. That means the digits repeats every $225$ digits, and $18$ of these digits are zeroes.
That means of the $9900$ digits past the radix point, $792$ of them are zeroes. After $9918$ digits, there are $810$ zeroes, and since the $82$ digits after are ones, we confirm that there are $\boxed{810}$ zeroes of the first $10,000$ digits past the radix point.
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# 60 Lesson 3 Homework Practice Equations In Y Mx Form
## Introduction
In lesson 3 of our homework practice, we will delve into equations in the y mx form. This form, also known as the slope-intercept form, is a powerful tool in algebra for understanding and solving linear equations. By the end of this lesson, you will have a solid grasp of how to work with equations in the y mx form, enabling you to confidently tackle a wide range of problems.
### Understanding the y mx Form
Before we dive into solving equations in the y mx form, let's take a moment to understand what this form represents. In this form, y represents the dependent variable (usually representing the vertical axis), m represents the slope of the line, and x represents the independent variable (usually representing the horizontal axis). By expressing an equation in this form, we can easily interpret and manipulate its components to gain valuable insights.
### Solving Equations in the y mx Form
Now that we have a clear understanding of the y mx form, let's explore how to solve equations expressed in this form. The process involves isolating y on one side of the equation and determining the values of m and x. By following a few simple steps, we can quickly find the solution to the equation.
### Step 1: Identify the Slope (m)
The first step in solving an equation in the y mx form is to identify the slope (m). The slope represents the rate of change of the dependent variable (y) with respect to the independent variable (x). It determines the steepness of the line on a graph and provides valuable information about the relationship between the variables. To find the slope, we can look at the coefficient of x in the equation.
### Step 2: Determine the y-Intercept (b)
Once we have identified the slope, the next step is to determine the y-intercept (b). The y-intercept represents the point at which the line intersects the y-axis. It gives us the value of y when x equals zero. To find the y-intercept, we can look at the constant term in the equation.
### Step 3: Graph the Line
After identifying the slope and y-intercept, we can graph the line represented by the equation. By plotting the y-intercept as the starting point and using the slope to determine additional points, we can create a line that accurately represents the relationship between the variables. This graph visually illustrates the solution to the equation and provides a useful tool for further analysis.
### Step 4: Solve for y
Now that we have a graph of the line, we can easily solve for y by substituting different values of x into the equation. By plugging in the x-values and using the equation, we can calculate the corresponding y-values. This allows us to determine the specific points that lie on the line and gain a deeper understanding of the relationship between the variables.
### Step 5: Interpret the Solution
Finally, after solving for y, we can interpret the solution in the context of the problem. By analyzing the values of y and x, we can draw conclusions about the relationship between the variables and make predictions or draw insights based on the data. This step is crucial for understanding the implications of the equation and its relevance to the real world.
## Examples of Equations in the y mx Form
Now that we have covered the steps for solving equations in the y mx form, let's explore a few examples to solidify our understanding.
### Example 1: y = 2x + 3
Let's start with the equation y = 2x + 3. In this equation, the slope (m) is 2, and the y-intercept (b) is 3. To graph this line, we can start by plotting the y-intercept at the point (0, 3). Then, using the slope of 2, we can find additional points by moving 2 units up and 1 unit to the right. By connecting these points, we get a line with a slope of 2 that intersects the y-axis at 3. Solving for y by substituting different values of x, we can find the corresponding y-values and gain insights into the relationship between the variables.
### Example 2: y = -0.5x + 1
In the equation y = -0.5x + 1, the slope (m) is -0.5 and the y-intercept (b) is 1. By following the same steps as in the previous example, we can graph this line and solve for y. The slope of -0.5 indicates a negative relationship between the variables, with y decreasing as x increases. By analyzing the values of y and x, we can draw conclusions about the specific problem at hand and make predictions based on the data.
## Conclusion
Equations in the y mx form are a powerful tool in algebra for understanding and solving linear equations. By following the steps outlined in this lesson, you can confidently solve equations in this form, graph the corresponding lines, and gain valuable insights into the relationship between the variables. Whether you're working on homework problems or real-world applications, mastering the y mx form will enhance your algebra skills and enable you to tackle a wide range of mathematical challenges.
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# The base of a triangular pyramid is a triangle with corners at (2 ,5 ), (6 ,5 ), and (3 ,8 ). If the pyramid has a height of 15 , what is the pyramid's volume?
Jun 22, 2016
Volume of pyramid is $30.015$ cubic units.
#### Explanation:
Volume of such a pyramid is one third of base of its area multiplied by its height. While height has been given, we have to find area of triangular base, which can bee found using Heron's formula, which gives are $\Delta = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$, where $s = \frac{1}{2} \left(a + b + c\right)$ and $a$, $b$ and $c$ are the three sides of the base triangle.
For this find the sides of triangle formed by $\left(2 , 5\right)$, $\left(6 , 5\right)$ and $\left(3 , 8\right)$ by using distance formula $\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
The distance between pair of points will be
$a = \sqrt{{\left(6 - 2\right)}^{2} + {\left(5 - 5\right)}^{2}} = \sqrt{16 + 0} = \sqrt{16} = 4$
$b = \sqrt{{\left(3 - 6\right)}^{2} + {\left(8 - 5\right)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 4.2426$ and
$c = \sqrt{{\left(3 - 2\right)}^{2} + {\left(8 - 5\right)}^{2}} = \sqrt{1 + 9} = \sqrt{10} = 3.1623$
Hence, $s = \frac{1}{2} \times \left(4 + 4.2426 + 3.1623\right) = \frac{11.4049}{2} = 5.7025$
And area of triangle $\Delta = \sqrt{5.7025 \left(5.7025 - 4\right) \left(5.7025 - 4.2426\right) \left(5.7025 - 3.1623\right)}$
= $\sqrt{5.7025 \times 1.7025 \times 1.4599 \times 2.5402} = \sqrt{36.0034} = 6.0003$
Hence volume of pyramid is $\frac{1}{3} \times 6.0003 \times 15 = 30.015$
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# Trigonometric Functions
Let's look at everything to do with trigonometric functions – sine, cosine and tangent functions and their respective graphsThen let's explore the secant, cosecant, cotangent, arcsine, arccosine and arctangent functions.
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## What are trigonometric functions?
Trigonometric functions are functions that relate to angles and lengths in a triangle. The most common trigonometric functions are sine, cosine and tangent. However, there are reciprocal trigonometric functions, such as cosecant, secant, cotangent and inverse trigonometric functions such as arcsine, arccosine and arctangent, which we will also explore in this article.
### SOH CAH TOA
An easy way to remember the sine, cosine and tangent functions and what sides they correspond to in a right angle triangle is by using SOH CAH TOA. If we have a right angle triangle as below, and we label one angle 𝞱, we must label the three sides of the triangle opposite (for the only side that is opposite the angle 𝞱 and is not in contact with that angle), hypotenuse (for the longest side, which is always the one opposite the 90° angle) and adjacent (for the last side).
Labelling the sides of a right-angled triangle
The sine, cosine and tangent functions relate the ratio of two sides in a right-angled triangle to one of its angles. To remember which functions involve which sides of the triangle, we use the acronym SOH CAH TOA. The S, C and T stand for Sine, Cosine and Tangent respectively and the O, A and H for Opposite, Adjacent and Hypotenuse. So the Sine function involves the Opposite and the Hypotenuse, and so on.
SOH CAH TOA triangles for remembering trigonometric functions
All of the functions sine, cosine and tangent are equal to the sides they involve divided by each other.
$\sin\theta = \frac{opposite}{hypotenuse}; \space \cos \theta = \frac {adjacent}{hypothenuse}; \space \tan \theta = \frac {opposite}{adjacent}$
## What is the sine function?
As seen above, you can work out the sine of an angle in a right-angled triangle by dividing the opposite by the hypotenuse. The graph for a sine function looks like this (the red curve):
A graphical illustration of the sine function
From this graph, we can observe the key features of the sine function:
• The graph repeats every 2𝞹 or 360°
• The minimum value for sine is -1
• The maximum value for sine is 1
• This means that the amplitude of the graph is 1 and its period is 2𝞹 (or 360°)
• The graph crosses the y axis at 0, and every 𝞹 radians before and after that.
• The sine function reaches its maximum value at $$\frac{\pi}{2}$$ and every 2𝞹 before and after that.
• The sine function reaches its minimum value at $$\frac{3\pi}{2}$$ and every 2𝞹 before and after that.
### Memorising the values of sine
You will need to remember the values of sine for commonly used angles by heart, and although this might sound tricky, there is a way to make it easier to memorise. You will need to know the sine values for the angles 0, $$\frac{\pi}{6}$$ (30°), $$\frac{\pi}{4}$$ (45°), $$\frac{\pi}{3}$$ (60°) and $$\frac{\pi}{2}$$ (90°). For this, the easiest way is to start constructing a table for the angle, 𝞱 and sin𝞱:
θ 0 $$\frac{\pi}{6}$$ $$\frac{\pi}{4}$$ $$\frac{\pi}{3}$$ $$\frac{\pi}{2}$$ sinθ
Now we have to fill out the sine values. For this, we will start by putting the numbers 0 to 4 from left to right:
θ 0 $$\frac{\pi}{6}$$ $$\frac{\pi}{4}$$ $$\frac{\pi}{3}$$ $$\frac{\pi}{2}$$ sin θ 0 1 2 3 4
The next step is to add a square root to all these numbers and divide them by 2:
θ 0 $$\frac{\pi}{6}$$ $$\frac{\pi}{4}$$ $$\frac{\pi}{3}$$ $$\frac{\pi}{2}$$ sin θ $$\frac{\sqrt{0}}{2}$$ $$\frac{\sqrt{1}}{2}$$ $$\frac{\sqrt{2}}{2}$$ $$\frac{\sqrt{3}}{2}$$ $$\frac{\sqrt{4}}{2}$$
Now, all we have left to do is simplify what we can:
θ 0 $$\frac{\pi}{6}$$ $$\frac{\pi}{4}$$ $$\frac{\pi}{3}$$ $$\frac{\pi}{2}$$ sin θ 0 $$\frac{1}{2}$$ $$\frac{\sqrt{2}}{2}$$ $$\frac{\sqrt{3}}{2}$$ 1
And that's it!
## What is the cosine function?
You can find the cosine value for an angle in a right-angled triangle by dividing the adjacent by the hypotenuse. The graph for the cosine value looks exactly like the sin graph, except that it is shifted to the left by $$\frac{\pi}{2}$$ radians (the blue curve):
A graphical illustration of the cosine function
By observing this graph, we can determine the key features of the cosine function:
• The graph repeats every 2𝞹 or 360 °
• The minimum value for cosine is -1
• The maximum value for cosine is 1
• This means that the amplitude of the graph is 1 and its period is 2𝞹 (or 360°)
• The graph crosses the y-axis at $$\frac{\pi}{2}$$, and every 𝞹 radians before and after that.
• The cosine function reaches its maximum value at 0 and every 2𝞹 before and after that.
• The cosine function reaches its minimum value at 𝞹 and every 2𝞹 before and after that.
### Memorising the values of cosine
You will also need to remember the values of cosine for commonly used angles by heart, and although this might sound tricky, there is a way to make it easier to memorise. You will need to know the sine values for the angles 0, $$\frac{\pi}{6}$$ (30°), $$\frac{\pi}{4}$$ (45°), $$\frac{\pi}{3}$$ (60°) and $$\frac{\pi}{2}$$ (90°). For this, we will use the same method as for sin and start constructing a table for the angle, 𝞱 and cos𝞱:
θ 0 $$\frac{\pi}{6}$$ $$\frac{\pi}{4}$$ $$\frac{\pi}{3}$$ $$\frac{\pi}{2}$$ cos θ
Now we will fill in the numbers 0 to 4, but this time, we will do this from right to left instead:
θ 0 $$\frac{\pi}{6}$$ $$\frac{\pi}{4}$$ $$\frac{\pi}{3}$$ $$\frac{\pi}{2}$$ cos θ 4 3 2 1 0
The final two steps are the same as before, so we will take the square root of each number and divide it by 2, and we simplify:
θ 0 $$\frac{\pi}{6}$$ $$\frac{\pi}{4}$$ $$\frac{\pi}{3}$$ $$\frac{\pi}{2}$$ cos θ 1 $$\frac{\sqrt{3}}{2}$$ $$\frac{\sqrt{2}}{2}$$ $$\frac{1}{2}$$ 0
As you can see, sine and cosine values for common angles are the same, simply the other way around.
## What is the tangent function?
You can work out the tangent of an angle by dividing the opposite by the adjacent in a right-angled triangle. However, the tangent function looks a bit different from the cosine and sine functions. It is not a wave but rather a non-continuous function, with asymptotes:
A graphical illustration of the tangent function
By observing this graph, we can determine the key features of the tangent function:
• The graph repeats every 𝞹 or 180°
• The minimum value for tangent is $$-\infty$$
• The maximum value for tangent is $$\infty$$
• This means that the tangent function has no amplitude and its period is 𝞹 (or 180°)
• The graph crosses the y-axis at 0 and every 𝞹 radians before and after that
• The tangent graph has asymptotes, which are values that the function will get closer to infinity.
• These asymptotes are at $$\frac{\pi}{2}$$ and every 𝞹 before and after that.
The tangent of an angle can also be found with this formula:
$\tan\theta = \sin \theta / \cos \theta$
### Memorising the values of tangent
Similar to before, you will need to remember the tan values for the angles 0, $$\frac{\pi}{6}$$ (30°), $$\frac{\pi}{4}$$ (45°), $$\frac{\pi}{3}$$ (60°) and $$\frac{\pi}{2}$$ (90°). For this, we will use the formula above and the tables that we already constructed for sine and cosine and use the fact that $$\tan = \sin /\cos$$ to work out the tan𝞱 values:
θ 0 $$\frac{\pi}{6}$$ $$\frac{\pi}{4}$$ $$\frac{\pi}{3}$$ $$\frac{\pi}{2}$$ sin θ 0 $$\frac{1}{2}$$ $$\frac{\sqrt{2}}{2}$$ $$\frac{\sqrt{3}}{2}$$ 1 cos θ 1 $$\frac{\sqrt{3}}{2}$$ $$\frac{\sqrt{2}}{2}$$ $$\frac{1}{2}$$ 0 tan θ 0 $$\frac{1}{\sqrt{3}}$$ 1 $$\sqrt{3}$$ Undefined
Note that the value for tan ($$\frac{\pi}{2}$$) cannot be determined as it is equal to 1/0, which cannot be worked out. This will result in an asymptote at $$\frac{\pi}{2}$$.
## Inverse trigonometric functions
The inverse trigonometric functions refer to the arcsin, arccos and arctan functions, which can also be written as $$\sin^{-1}(x)$$, $$\cos^{-1}(x)$$ and $$\tan^{-1}(x)$$. These functions do the opposite of the sine, cosine and tangent functions, which means that they give back an angle when we plug a sin, cos or tan value into them.
An illustration of the relationship between trigonometric functions and their respective inverse functions
The graphs for these functions look very different to the sin, cos and tan graphs:
An illustration of arcsin, arccos and arctan on the x and y axis
## What are the reciprocal trigonometric functions?
The reciprocal trigonometric functions refer to the cosecant, secant and cotangent functions, abbreviated as csc, sec and cot, respectively. We need to look back at our right-angled triangle to understand what these functions represent.
Labelling the sides of a right-angled triangle
We earlier defined sin, cos and tan based on the ratios of the sides of this triangle. The cosecant, secant and cotangent are simply the reciprocals of the sin, cos and tan ratios respectively. This means that to find the equation for cosecant 𝞱, we would flip the equation of sin 𝞱 and so on.
$\sin\theta = \frac {opposite}{hypothenuse}; \space \cos\theta = \frac {adjacent}{hypothenuse}; \space \tan\theta = \frac {opposite}{adjacent}$
$\csc\theta = \frac {hypothenuse}{opposite}; \space \sec\theta = \frac{hypothenuse}{adjacent}; \space \cot \theta = \frac {adjacent}{opposite}$
## Trigonometric Functions - Key takeaways
• SOH CAH TOA can help us remember the sin, cos, and tan functions.
• The sine and cosine functions are waves with a period of 2𝝿 and an amplitude of 1.
• The sine and cos functions are the same except shifted by 𝝿 / 2.
• The tan function has asymptotes every 𝝿 radians.
• The inverse trigonometric functions refer to arcsin, arccos, and arctan, and these functions give us the angle with a specific sin, cos, or tan value.
• The reciprocal trigonometric functions refer to cosecant, secant, and cotangent, and these functions have the reciprocated equation of the sin, cos, and tan functions in a right-angled triangle.
#### Flashcards in Trigonometric Functions 35
###### Learn with 35 Trigonometric Functions flashcards in the free StudySmarter app
We have 14,000 flashcards about Dynamic Landscapes.
What are the nine trigonometric functions?
Sin, cos, tan, arcsin, arccos, arctan, csc, sec and cot.
What is the range of trigonometric functions?
The range for sine and cosine is -1≤y≤1 and for tan y ∈ R.
How do you draw the graph of trigonometric functions?
The easiest way is to remember the overall shape of the graph, and then work out the function for a few different numbers and add them to your graph.
## Test your knowledge with multiple choice flashcards
What are the reciprocal functions?
What is the amplitude of a tangent graph?
What is the sine value of $$x=\frac{\pi}{6}$$?
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# How can we multiply÷ the expression by its conjugate in the $\infty-\infty$ indeterminate case? Is it not $\frac{\infty}{\infty}$?
Question
$$\lim _{x \rightarrow \infty} \left(x-\sqrt{x^2+5 x}\right)$$
To evaluate the limit, we multiply and divide the expression by its conjugate.
First Question
But since $$x \rightarrow \infty$$, we multiply it by $$\frac{\infty}{\infty}$$ ? Could you explain the reason clearly and in detail?
\begin{aligned} & \lim _{x \rightarrow \infty} \frac{\left(x-\sqrt{x^2+5 x}\right)\left(x+\sqrt{x^2+5 x}\right)}{x+\sqrt{x^2+5 x}} \\ & =\lim _{x \rightarrow \infty} \frac{x^2-\left(x^2+5 x\right)}{x+|x| \sqrt{1+\frac{5}{x}}} \\ & =\lim _{x \rightarrow \infty} \frac{-5 x}{x\left(1+\sqrt{1+\frac{5}{x}}\right)}=\frac{-5}{1+\sqrt{1}} \\ & =\lim _{x \rightarrow \infty} \frac{-5}{1+1}=-\frac{5}{2} \text { dir. } \end{aligned}
Second Question
In the second step of the solution, how can we cancel the terms such as $$\lim _{x \rightarrow \infty}(x^2- x^2)...$$
Is it not $$\infty-\infty$$ indeterminate case?
• You are using the conjugate before taking the limit. In each specific case where the conjugate is used, $x$ is finite. Commented Nov 20, 2023 at 8:44
• when we multiply and divide a function by same thing except for zero, we dont change the function, but we get rid of the undetermine form Commented Nov 20, 2023 at 8:44
• In the second step of the solution, how can we cancel the terms such as $\lim _{x \rightarrow \infty}(x^2- x^2)...$ Is it not $\infty-\infty$ indeterminate case? Commented Nov 20, 2023 at 8:50
• If you try to evaluate $\lim_{x\to\infty} (x^2-x^2)$ by writing it as $\lim_{x\to\infty} x^2 - \lim_{x\to\infty} x^2$, that doesn't work because you've created an $\infty-\infty$ indeterminate form. But the function $x^2-x^2$ is always equal to the function $0$ (that's just algebra, nothing to do with calculus); therefore $\lim_{x\to\infty} (x^2-x^2)$ is equal to $\lim_{x\to\infty} 0$. This is the same lesson as for your original question: we are always allowed to do valid algebraic things to functions, even if there is a limit nearby. Commented Nov 20, 2023 at 8:59
• X^2 - x^2 is not a limit. It is a very simple arithmetic expression that always has the result 0. Commented Nov 20, 2023 at 9:00
for $$x > 0,$$ $$x^2 + 5x < \left( x + \frac{5}{2} \right)^2$$ for$$x > \frac{5}{8},$$
$$\left( x + \frac{5}{2} - \frac{25}{8x} \right)^2 < x^2 + 5x < \left( x + \frac{5}{2} \right)^2$$ so then $$x + \frac{5}{2} - \frac{25}{8x} \; \; < \; \; \sqrt{x^2 + 5x} \; \; < \; \; x + \frac{5}{2}$$
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# Episode 116-1: The algebra of power (Word, 22 KB)
```TAP 116-1: The algebra of power
These questions lead you through to an understanding of the algebraic relationships between
power, current, resistance and potential difference. Look at the example and then answer the
questions thoughtfully.
Example
The power P dissipated in a resistance R carrying current I with a pd V across it can be
calculated in several ways.
Suppose:
P = 24 W, I = 2 A, V = 12 V and R = 6 Ω
Using P = I V
1.
P 12 V 2 A 12 J C1 2 C s1 24 J s1 24 W
Since V = I R
2.
P = I V = I (I R) = I2 R
This should give:
P = (2 A)2 x 6 Ω = 24 W
3.
Since I = V / R
P = I V = ( V / R ) V = V2 / R
In our example:
P = ( 12 V)2 / 6 Ω = 24W
How can the power be proportional to resistance R in one relationship and be inversely
proportional to R in another? What does this tell you about different lamps?
2
P = V / R says that lamps operating at the same potential difference (e.g. all 12 V) will need
a lower resistance to provide greater power. Why? Because at a fixed potential difference you
need more current for more power. Lower resistance gives more current.
2
P = I R says that lamps to operate at the same current will need a larger resistance to
provide more power. Why? The charge flowing per second is the same, so each unit of
charge must deliver more energy. That means a larger potential difference. For the same
current and a larger potential difference, the resistance must be larger.
Questions
1.
Calculate the current through and the resistance of a 500 W stage light lamp with 250
V across it.
2.
Calculate the current through and the resistance of a 24 W, 12 V car headlamp.
Compare these values with those for the stage light.
Extension: More of the power of algebra
Finally if you are getting confident with algebra.
3.
2
2
Rewrite the equations P = I R and P = V /R replacing resistance R by conductance
G.
You may think it more appropriate to talk this through with your students. This will depend on
their level of confidence.
Social and human context
The effective distribution of power is a primary concern to much electrical engineering. These
equations give insight into some of the controlling concerns.
Answers and worked solutions
1.
2 A, 125
2.
2 A, 6
3.
P = I 2/G; P = V 2G
External references
This activity is taken from Advancing Physics Chapter 2, 160S
```
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# What is Decimal(Definition, Example, Conversion)
In mathematics calculation may be you heard about lots of number system like whole number, natural number, even number, odd number, rotational numbered etc.so in mathematics on of number which is called decimal number and here we will discuss about more details on it.
We can define decimal according to its nature that’s containing whole number part and fraction part is separated by decimal point.in decimal number that dot is consider as decimal point.in decimal point all digit have unique place number.
## Learn What is Decimal With Example
let’s assume we have number which is 75.5 So here as you can 75 is whole Number and after decimal point 5 is called fraction part. that (.) called is decimal point in between of 75.5
Ok here is Another one example with figure as you can see value is 12.5 in which 5 is place of tenths place.
So here 1 is tens place, 2 is on ones place and after decimal point 5 is on tenths places
### Decimal Representations
in decimal number all number have unique place and representations of that number so here we take example of 12.5
12.5=125/10
So if we are looking of place of that number so here we can found that
1 representations the power 101 that is tens position.
2 representations the power 100 that is tens position.
5representations the power 10-1 that is tens position.
So as you can see how digit is work According to power of 10 in decimal number
#### What is place value in decimal number
place value in decimal number is very crucial because converting any decimal number to fraction form its required to know that after decimal point which digit on place .So our taken value is here 0.123 as you can see we are moving towards left side so all digit after decimal point has unique place value.
so as you can see that 1 digit is on tenth places means that it is on 1/10. so as you can see that 2 digit is on hundred place means that is on 2/100. so as you can see that 3 digit is on thousand place means that is on 3/1000.
so if we want to convert that number in decimal to fraction form so we can write like 0.123=123/1000.
## Decimal to fraction conversion
so if you are looking to convert decimal to fraction of any number you must aware about decimal place value.so here we have some example for you
0.8 is decimal number and we are looking to convert in fraction form so how we can do?
0.8=8/10 (8 is place of tenths) 4/5(this is final Answer )
0.6 is decimal number and we are looking to convert in fraction form so how we can do?
0.6=6/10 (6 is place of tenths) 3/5(this is final Answer )
here as you can see how we are done with simple step within some times.
Frequently Asked Questions on Decimals
### What is meant by decimal ?
decimal is one type of number who separated by decimal point with whole number and fraction part.
### How to convert decimal to fraction ?
converting decimal to fraction you have to multiply and divide by its place value
### how to expand value like 12.5
here you can do 12+5/10
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# Lesson video
In progress...
Hi, I'm Miss Davies.
In this lesson, we're going to be calculating the mean of certain numbers that are displayed in a group frequency table.
Here are some times it took 20 students to walk to school.
The times are given in minutes.
Why can't we work out the mean exactly? It is because we do not know the exact amount of time that it took each of the 20 students to walk to school.
This means that we're going to find an estimate for the mean.
To do this, We're going to write down the midpoint of each of the classes.
This means the middle of the values in the first column.
The middle of one and five is three, Midpoint of six and 10 is eight, 11 and 15 gives a midpoint of 13 and 16 to 25 gives a midpoint if 20.
5.
Next we're going to multiply the frequency by the midpoints.
Here I've written frequency multiplied by time.
This gives us these times.
The total of each of these times is 202.
5 minutes.
To calculate the mean, we're going to divide the total by the number of items. In this example, we're going to divide 202.
5 by 20.
The estimate for the mean is 10.
1.
This has been rounded to three significant figures.
Here is some lengths of some goldfish.
Inequality symbols have been used to group the data.
The first group of data is that any goldfish that have got a length that is greater than 10 millimetres, but less than or equal to 30 millimetres.
Let's start by finding the midpoint of the length.
This is called midpoint of each of the classes.
The classes are the groups of lengths.
The first midpoint is 20 millimetres then 35, 45 and 70.
Next we're going to multiply together the frequency and the midpoint.
I've called this the length.
we're then going to add together these lengths.
So the total length of all of the goldfish is 1295 millimetres.
This is 30 goldfish.
The mean is found by dividing the total length by the number of items. In this example, this is 1295 divided by 30.
The estimate for the mean length of this 30 goldfish is 43.
2.
This has been rounded to three significant figures.
What do you think the modal class is? This is the grouping that has got the highest frequency.
In this example, it is 40 to 50 millimetres.
Here is a question for you to try.
Pause the video to complete your task I will resume once you're finished.
The sum of the midpoints multiplied by frequency is 97.
5.
The sum of the frequencies is 15.
The calculation you need to do to find the estimate for the mean is 97.
5 divided by 15, which gives an answer of 6.
5 minutes.
Here's a question for you to try.
Pause the video to complete your task I will resume once you're finished.
Here are the answers, the sum of the midpoint multiplied by the frequency is 260.
The total of the frequency is 20.
The calculation that you need to do to find the estimate for the mean is 260 divided by 20 which gives you an answer of 13 grammes.
What mistake has been made here? They have divided by the wrong number.
Need to find the total of the frequencies and divide by this.
Here are some questions for you to try.
Pause the video to complete your task I will resume once you're finished.
Here are the answers Rob has divided by the sum of the midpoint rather than the sum of the frequencies.
5 centimetres.
Here are some questions for you to try.
Pause the video to complete your task I will resume once you're finished.
|
# Video: AQA GCSE Mathematics Higher Tier Pack 3 • Paper 3 • Question 23
The cooking club conducted a survey about how many hot meals each member cooked in July. One of the frequencies is missing. The cooking club uses midpoints to calculate an estimate for the mean number of meals each member cooked in July. They find that the estimate of the mean is 15. Work out the missing frequency.
05:01
### Video Transcript
The cooking club conducted a survey about how many hot meals each member cooked in July. One of the frequencies is missing. The cooking club uses midpoints to calculate an estimate for the mean number of meals each member cooked in July. They find that the estimate of the mean is 15. Work out the missing frequency.
To find the mean of a set of data, we find the sum of all of the data values and divide it by how many data values there are or the total frequency. However, in this question, we haven’t been given the original data values. Instead, they’ve been presented in a grouped frequency table. We just know that, for example, there were 15 members of the cooking club who cooked between zero and six meals in July.
For this reason, we can’t work out the exact sum of the data values. We can only estimate it. And this is why the cooking club were only able to work out an estimate for their mean. We’re told that the cooking club uses midpoints to calculate their estimate. And what we’ll do is go through the process that they would have used.
We can use the letter 𝑥 to represent the missing frequency. The cooking club have already calculated the midpoints for us. These are found by averaging the endpoints of each interval. These midpoints give a best estimate for each data value within an interval with the smallest error on average.
The next step in estimating a mean from a grouped frequency table is to multiply the midpoint of each interval by its frequency. Remember the midpoint gives the best estimate for each data value. So the midpoint multiplied by the frequency gives the best estimate for the total of the data values in each interval.
For the first interval, we have three multiplied by 15 which is 45. For the next interval, 10 multiplied by 29 which is 290. The values for the third and fifth intervals can be found in the same way. For the interval with the missing frequency, we multiply the midpoint 24 by the frequency which we’ve now called 𝑥. We can, therefore, express this as 24𝑥.
Now, we have our best estimate of the sum of the data values in each interval. So our next step is to find an estimate of the sum of all the data values which we do by adding the five values in the final column. We have 45 plus 290 plus 680 plus 24𝑥 plus 248. We can simplify this expression by combining all the numbers to give 24𝑥 plus 1263.
Next, we need an expression for the total frequency which we find by summing the five values in the frequency column: 15 plus 29 plus 40 plus 𝑥 plus eight. We can again simplify this by grouping the numbers to give 𝑥 plus 92.
Remember that we were told what the cooking club’s estimate of the mean was. It’s 15. So we can set this expression equal to 15. And it gives an equation that we now need to solve in order to find the value of 𝑥, the missing frequency.
Here is that equation we just worked out: 24𝑥 plus 1263 over 𝑥 plus 92 equals 15, where 𝑥 represents the missing frequency. The first step in solving this equation is to eliminate the denominator on the left-hand side which we can do by multiplying both sides by 𝑥 plus 92. On the left, this will cancel out the factor of 𝑥 plus 92 in the denominator. And on the right, we have a bracket which we need to expand.
15 multiplied by 𝑥 is 15𝑥 and 15 multiplied by 92 is 1380. We have 𝑥s on both sides of this equation. So we want to group them all on one side. And we want to group them on the side which has the larger number of 𝑥 to start off with. That’s the left-hand side. So we’ll subtract 15𝑥 from each side to eliminate the 15𝑥 on the right of the equation, which gives nine 𝑥 plus 1263 equals 1380.
Next, we subtract 1263 from each side to eliminate the positive 1263 on the left, giving nine 𝑥 equals 117. The final step is to divide both sides of this equation by nine, giving 𝑥 equals 13.
Remember that 𝑥 represents the missing frequency. So it needs to be an integer value. As we’ve got a whole number, an answer of 13, we can be reasonably confident that our answer is correct.
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# Given two ordered pairs (1,-2) and (3,-8), what is the equation of the line in slope-intercept form?
Nov 26, 2015
$y = - 3 x + 1$
#### Explanation:
The general equation for a line in slope-intercept form is:
$y = m x + b$
where:
y = y-coordinate
m = slope
x = x-coordinate
b = y-intercept
To find the equation, first find the slope. The formula for slope is:
$m = \left({y}_{\text{2"-y_"1")/(x_"2"-x_"1}}\right)$
where:
m = slope
$\left({x}_{\text{1", y_"1}}\right) = \left(1 , - 2\right)$
$\left({x}_{\text{2", y_"2}}\right) = \left(3 , - 8\right)$
$m = \left({y}_{\text{2"-y_"1")/(x_"2"-x_"1}}\right)$
$m = \frac{\left(- 8\right) - \left(- 2\right)}{\left(3\right) - \left(1\right)}$
$m = - \frac{6}{2}$
$m = - 3$
Rewrite the equation:
$y = - 3 x + b$
Now substitute a known point into the equation to solve for $b$:
$y = - 3 x + b$
$\left(- 2\right) = - 3 \left(1\right) + b$
$- 2 = - 3 + b$
$1 = b$
$y = - 3 x + 1$
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# Selina Concise Solutions for Chapter 23 Probability Class 8 ICSE Mathematics
### Exercise 23
1. A die is thrown, find the probability of getting:
(i) a prime number
(ii) a number greater than 4
(iii) a number not greater than 4.
Solution
A die has six numbers : 1, 2, 3, 4, 5, 6
∴ Number of possible outcomes = 6
(i) A prime number
Number of favourable outcomes = a prime number = 1, 3, 5 which are in numbers
P(E) = (Numbers of favourable outcome)/(Numbers of all outcome)
= 3/6 = ½
(ii) Numbers of favourable outcome = greater than four i.e two number 5 and 6
∴ P(E) = (Numbers of favourable outcome)/(Number of all possible outcome)
= 2/6 = 1/3
(iii) Number of favourable outcome = not graeter than 4 or numbers will be 1, 2, 3, 4 which are 4 in numbers
∴ P(E) = (Numbers of favourable outcome)/(Number of all outcome)
= 4/6
= 2/3
2. A coin is tossed. What is the probability of getting:
(i) a tail?
Solution
On tossing a coin once,
Number of possible outcome = 2
(i) Favourable outcome getting a tail = 1
⇒ number of favorable outcome = 2
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= ½
Similarly, favourable outcome getting a head = 1
But number of possible outcome = 2
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= ½
3. A coin is tossed twice. Find the probability of getting:
(i) exactly one head (ii) exactly one tail
(iii) two tails (iv) two heads
Solution
Possible number of favourable outcomes = 2
(i.e. TH and HT)
Total number of possible outcomes = 4
∴ P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)
= 2/4
= ½
(ii) Exactly one tail
Possible number of favourable outcomes = 2
(i.e. TH and HT)
Total number of possible outcomes = 4
P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)
= 2/4
= ½
(iii) Two tails
Possible number of favourable outcomes = 1
(i.e., TT)
Total number of possible outcome s = 4
∴ P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)
= ¼
Possible number of favourable outcomes = 1
(i.e. HH)
Total number of possible outcomes = 4
∴ P(E) = ¼
4. A letter is chosen from the word ‘PENCIL’ what is the probability that the letter chosen is a consonant?
Solution
Total no. of letters in the word ‘PENCIL = 6
Total Number of Consonant = ‘PNCL’ i.e. 4
P(E) = (Total No. of consonants)/(Total No. of letters in the word PENCIL)
= 4/6
= 2/3
5. A bag contains a black ball, a red ball and a green ball, all the balls are identical in shape and size. A ball is drawn from the bag without looking into it. What is the probability that the ball drawn is:
(i) a red ball
(ii) not a red ball
(iii) a white ball.
Solution
(i) Total number of possible outcomes = 3
∴ P(E) = 1/3
(ii) Not a red ball
Number of favourable outcomes = Green ball + Black Ball
= 1 + 1 + 2
∴ P(E) = 2/3
(iii) A white ball
Number of favourable outcome = 0
∴ P(E) = 0/3
= 0
6. In a single throw of a die, find the probability of getting a number
(i) greater than 2
(ii) less than or equal to 2
(iii) not greater than 2.
Solution
(i) A die has six numbers = 1, 2, 3, 4, 5, 6
∴ Number of possible outcomes = 6
∴ P(E) = 4/6 = 2/3
(ii) Less than or equal to 2
Number of favourable outcome = 1, 2
∴ P(E) = 2/6
= 1/3
(iii) Not greater than 2
Number of favourable outcomes = 1, 2
∴ P(E) = 2/6
= 1/3
7. A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size.
A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:
(i) a black ball.
(ii) a red ball.
(iii) a white ball.
(iv) not a red ball.
(v) not a black ball.
Solution
In a bag, 3 balls are white
2 balls are red
5 balls are black
Total number of balls = 3 + 2 + 5 = 10
(i) Numbers of possible outcome of one black ball = 10
And number of favourable outcome of one black ball = 5
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 5/10
= 1/2
(ii) Number of possible outcome of one red ball = 10
And number of favourable outcome = 2
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 2/10
= 1/5
(iii) Number of possible outcome of white ball = 10
And number of favourable outcome = 3
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 3/10
(iv) Number of possible outcome = 10
Number of favourable outcome = 3 + 5 = 8
Not a red ball
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 8/10
= 4/5
(v) Number of possible outcomes = 10
Number of favourable outcome not a black ball = 3 + 2 = 5
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome) = 5/10
= 1/2
8. In a single throw of a die, find the probability that the number:
(i) will be an even number.
(ii) will be an odd number.
(iii) will not be an even number.
Solution
A die has six numbers : 1, 2, 3, 4, 5, 6
∴ Number of possible outcome = 6
(i) Number of favourable outcome = an even number i.e. 2, 4, 6 which are 3 in numbers
∴ P(E) = (Numbers of favourable outcome)/(Number of all possible outcome)
= 3/6
= 1/2
(ii) & (iii) Number of favourable outcome = not an even number i.e. odd numbers : 1, 3, 5 which are 3 in numbers
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 3/6
= 1/2
9. In a single throw of a die, find the probability of getting :
(i) 8
(ii) a number greater than 8
(iii) a number less than 8
Solution
On a die the numbers are 1, 2, 3, 4, 5, 6 i.e. six.
∴ Number of possible outcome = 6
(i) Number of favourable outcome = 0
(∵ 8 is not possible)
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 0/6 = 0
(ii) Number greater than 8 will be 0
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 0/6
= 0
(iii) Number less than 8 wil be 1, 2, 3, 4, 5, 6
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 6/6
= 1
10. Which of the following can not be the probability of an event?
(i) 2/7
(ii) 3.8
(iii) 37%
(iv) - 0.8
(v) 0.8
(vi) -2/5
(vii) 7/8
Solution
(i) The probability of an event cannot be
(ii) 3.8 i.e. the probability of an even cannot exceed 1.
(iv) i.e. -0.8 and
(vi) -2/5, This is because probability of an even can never be less than 1.
11. A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out:
(i) a white ball,
(ii) a black ball
Solution
∵ There are 6 black balls in a bag
∴ number of possible outcome = 6
(i) A white ball
As there is no white ball in the bag
∴ Its probability is zero (0) = or P(E) = 0
(ii) a black ball
∴ Number of favourable outcome = 1
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 1/6
12. Three identical coins are tossed together. What is the probability of obtaining:
Solution
Total outcomes = 8
i.e., (H, H, H), (H, H, T), (H, T, H), (T, T, T), (T, H, H), (T, T, H), (H, T, T), (T, H, T)
(i) Favourable outcome = i.e., (H, H, H)
∴ P (of getting all heads) = 1/8
(ii) Favourable outcome = 3 (H, H, T), (H, T, H), (T, H, H)
∴ P(E) = 3/8
(iii) Favourable outcome = 3 (H, T, H), (T, T, H), (H, T, T)
∴ P(E) = 3/8
(iv) Favourable outcomes = 1 i.e., (T, T, T)
∴ P(E) = 1/8
13. A book contains 92 pages. A page is chosen at random. What is the probability that the sum of the digits in the page number is 9?
Solution
Number of pages of the book = 92
Which are from 1 to 92
Number of possible outcomes = 92
∴ Number of pages whose sum of its page is 9 = 10
i.e. 9, 18, 27, 36, 45, 54, 63, 72, 81, 90
∴ P(E) = 10/92
= 5/46
14. Two coins are tossed together. What is the probability of getting:
(ii) both heads or both tails.
Solution
∵ A coins has two faces Head and Tailor H, T
∴ Two coins are tossed
∴ Number of coins = 2 x 2 = 4
which are HH, HT, TH, TT
(i) At least one head, then
Number of outcomes = 3
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 3/4
(ii) When both head or both tails, then
Number of outcomes = 2
∴ P(E) = (Numberof favourable outcome)/(Number of all possible outcome)
= 2/4
= 1/2
15. From 10 identical cards, numbered 1, 2, 3, …… , 10, one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:
(i) 2 (ii) 3
(iii) 2 and 3 (iv) 2 or 3
Solution
Total outcomes = 10
i.e. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
(i) Favourable outcomes = 5 i.e., 2, 4, 6, 8, 10
P(E) = 5/10 = 1/2
(ii) Favourable outcomes = 3 i.e., 3, 6, 9
P(E) = 3/10
(iii) Favourable outcomes = 1 i.e., 6
P(E) = 1/10
(iv) Favourable outcome = 7
i.e., 2, 3, 4, 6, 8, 9, 10
P(E) = 7/10
16. Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 0
(ii) 12
(iii) less than 12
(iv) less than or equal to 12
Solution
Total outcomes = 36 i.e.
(1, 1) (1, 2) (1, 3) (1, 4) (1,5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3 ,4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
(i) Favourable outcomes = 0
P(E) = 0/36 = 0
(ii) Favourable outcomes = i.e., (6,6)
P(E) = 1/36
(iii) Favourable outcomes = 35 [Except (6, 6)]
P(E) = 35/36
(iv) Favourable outcome = 36
P(E) = 36/36
= 1
17. A die is thrown once. Find the probability of getting:
(i) a prime number
(ii) a number greater than 3
(iii) a number other than 3 and 5
(iv) a number less than 6
(v) a number greater than 6.
Solution
Total outcomes = 6
i.e., 1, 2, 3, 4, 5 and 6
(i) Favourable outcomes = 3 i.e., 2, 3, 5
P(E) = 3/6
= 1/2
(ii) Favourable outcomes = 3 i.e., 4, 5, 6
P(E) = 3/6
= 1/2
(iii) Favourable outcomes = 4 i.e., 1, 2, 4, 6
P(E) = 4/6
= 2/3
(iv) Favourable outcomes = 5
i.e., 1, 2, 3, 4 and 5
P(E) = 5/6
(v) Favourable outcomes = 0
P(E) = 0/6 = 0
18. Two coins are tossed together. Find the probability of getting:
(i) exactly one tail
Solution
Total outcomes = 4
i.e., HH, HT, TT, TH
(i) Favourable outcomes = 2 i.e., HT and TH
P(E) = 2/4
= 1/2
(ii) Favourable outcomes = 3
i.e., HH, HT and TH
P(E) = 3/4
(iii) Favourable outcomes = 1
i.e., TT
P(E) = 1/4
(iv) Favourable outcomes = 3
i.e., HH, HT and TH
P(E) = 3/4
|
# How to find x and y intercepts of an equation
## How to find x and y intercepts of an equation
How do you calculate x intercept? The x-intercept is written (x, 0) because the y-coordinate with the x-intercept is always zero. If you know the slope and y-intercept of a function, you can calculate the x-intercept with the formula (yb) / m = x, where m is the slope, y is zero and b is the intersection of the axis and .
## How do you find the x- intercept?
Find intersections X. Press the 2nd key and then the Calc key. This gives access to the trace menu. Scroll to zero and press Enter. Use the ARROW KEYS to scroll left from X-Intercept and press Enter.
## How do you identify x - intercept?
The points where lines representing a set of values intersect these axes are called intersections. The y-intercept is where the line intersects the y-axis, and the x-intercept is where the line intersects the x-axis. For simple tasks, it's easy to find the x-intercept by looking at the graph.
## How do you calculate slope and y intercept?
In general, the shape of the slope portion has the formula: y = mx + b. b is section y (lesson on section y) mnemonic: b means where the line begins. Examples y = 5x + 3 is an example of a slope intersection and is the equation of a straight line with a slope of 5 and a y-intercept of 3.
## What is the formula for X and y intercept?
X-intersection. The linear equation is y = mx + b, where M and B are constants. The x-intercept is the point where the line intersects the x-axis. By definition, the y value of a linear equation when it intersects the x axis is always 0 because the x axis is at the y = point on the graph.
## How do you determine x - intercept?
In this way, you can easily determine the values needed for the square formula method to calculate the x-axis cross-section. Remember that the formula for the square is: x = / 2a. This can be read as x equal to minus b plus or minus the square root of b squared minus four times ac times two a.
## How do you find the x - intercept of a function?
If you know the slope and y-intercept of a function, you can calculate the x-intercept with the formula (yb) / m = x, where m is the slope, y is zero and b is the intersection of the axis and .
## What is the formula for y - intercept?
The Y intersection is a feature of graphs of linear equations, which are always straight lines. Linear equations have the form: y = ax + b.
## How do you calculate x intercept in excel
Excel's intersection function calculates the intersection (value on the y-axis) of a linear regression line through a given set of x and y values.
## What is the function of a formula in Excel?
A formula in Excel is an equation that calculates the numbers in different cells in a workbook. At its simplest level, Excel can perform addition, subtraction, multiplication, and division formulas.
## What are the functions of Excel formulas?
An Excel function is a predefined formula that performs actions such as calculating values in a specific order. Common Excel functions are ADD, AVERAGE, COUNT, INDEX, ROUND, SUM, etc.
## How do you solve for the x intercept?
Using a Linear Equation Determine that the linear equation has a standard form. The standard form of a linear equation is Ax + By = C {\displaystyle Ax+By=C}. Log in with y {\displaystyle y}. The x-intercept is the point on the line where the line intersects the x-axis. Solve for x {\displaystyle x}.
## How do you calculate intercept?
Find the intersection point and the point of the slope y. Enter the formula y = mx + b {\displaystyle y=mx+b}. Connect the slope and the coordinates of the point on the line. Complete the equation by solving for b. Check your work.
## How can you go from slope intercept to standard form?
• Subtract b from both sides (optional). If m is an integer then B is 1.
• Multiply both sides of the equation by the denominator B. Which equation is the standard form?
• Subtract 1/2 x from both sides of the equation
• Multiply both sides by the denominator of the fraction. What is the equation of a straight line in standard form?
## How do you find slope intercept form from a table?
To find the y intersection, replace the slope with m in the formula y = mx + b and replace the ordered pair shown in the table with x and y in the formula, then solve for b. Finally, insert the m and b values into the formula y = mx + b to write the equation for the string.
## What is the y intercept for f (x)?
The Y intersection is the point where the line intersects the Y axis, which is the vertical line described by the equation x = 0. So the point with x = 0 and y = f(0) is the intersection at y , but normally you just give it importance because everyone knows that x = 0 is the value of x.
## Where is the x intercept located?
The first is called the x-intercept because it is a point on a line on the horizontal axis (x-axis). The second is the y-intercept, the point on the line on the vertical axis (y-axis).
## How do you find the slope
Method 2 of 2: Determine the slope Find the line whose slope you want to know. Make sure the line is straight. Enter two coordinates through which the line will intersect. The coordinates are the points x and y (x, y). Choose the coordinates of the points that dominate your equation. Compare the y coordinates at the top and the x coordinates at the bottom.
## What is the formula for finding the slope?
The slope of the straight line Point 1 is now Bertha and Point 2 is now Ernie. Look at the graph and write down their X and Y values: (X Bert, Y Bert) and (X Ernie, Y Ernie) The slope formula is now: M = (Y Ernie Y Bert) / (X Ernie X Bert).
## What are three ways to find the slope of a line?
There are three ways to find the slope of a line: it can have two points # (x_1, y_1) # and # (x_2, y_2) # (often one or both points can be # x # y / or #y # segments are off the axis). The slope is obtained from the equation. # m = (y_2y_1) / (x_2x_1) #. They can have a linear equation that is on the form or can be manipulated on the form. # y = mx + b#.
## What is an easy way to find slope?
• Use a graph and two points to find the slope without an equation.
• Find two points and place them in a simple shape (x, y). Use a graph (or test question) to find the x and y coordinates of two points on a graph.
• Label your points x1, y1, x2, y2 and store each point with its pair.
## How do you Find X in an equation?
You can find x or solve an equation for x by marking x on one side of the algebraic equation. To find x, you need to understand the basic rules of algebraic operations. Mark x on one side of the algebraic equation by subtracting the amount that appears as x on the same side of the equation.
## How do find the y intercept?
Locate the Y-junction Press the Trace key. Press the 0 key. This will move the cursor to the Y intersection where X = 0. Look at the bottom of the screen for the Y intersection.
## How do you find the x - intercept of a line?
The x-intercept is the point where the line intersects the x-axis. By definition, the y value of a linear equation when it intersects the x axis is always 0 because the x axis is at the y = point on the graph. To find the intersection y, just plug in for y and solve for x. This will give you the x value per x intersection.
## How do you find the x intercept of a function
The Y-intersections are the points where the graph of a function or equation intersects or "touches" the Y-axis of the Cartesian plane. You can think of it as a point with zero x. To find the intersection of an equation, set x = y and solve for y.
## Can a function have more then one x intercept?
Find and understand the x interceptions. For a graph of a function, an x interceptor is simply the point or points where the graph intersects the x-axis. There can be only one of these points, none of these points, or more than one, meaning the function can have multiple intersections along the x-axis.
## How do you find the x intercept from point slope form
The XIntercept formula for a straight line as a slope segment: y = mx + c, where m is the slope of the straight line. intercepts the line. The XIntercept formula for a straight line in the form of a slope y - b = m(x - a), where m is the slope of the straight line. (a, b) is a point in a straight line.
## How do you calculate slope intercept form?
Slope segment shape is the simplest way to represent linear equations. Immediately displays the slope of the line and the y-intercept. The formula for a line as a slope segment is y = mx + b, where x and y are the coordinates on the graph, m is the slope, and b is the y-axis segment.
## How do you find the slope intercept of a linear equation?
The oblique segment shape of the equation for a straight line (linear equation): y = m * x + b. In the equation of the line y = (1/4) * x +, the intersection of y is equal to the value of y when x = 0.
## What would you use slope intercept form for?
The shape of the slope segment is probably the most common way of expressing the equation for a straight line. To use the slope segment shape, you just need to 1) find the slope of the line and 2) find the y-intercept of the line.
## How do you find x intercepts in a quadratic function?
To find the x coordinates of a quadratic equation, let y = 0. Write a new equation ax squared + bx + c = and a quadratic formula that gives a solution of the form x = b plus or minus the square root of (b squared 4ac), everything is divisible by 2a. A quadratic formula can give zero, one or two solutions.
## How do you find the vertex of a parabola in standard form?
Explanation: The standard form of a parabola is y = ax2 + + bx + c, where a equals 0. The vertex is the point of the minimum or maximum of the parabola. If a > 0 the upper part is the minimum and the parabola opens. When< 0, the vertex is the maximum point and the parabola opens downward.
## What are X and y intercepts?
The intersections of X and Y are part of the basis for solving and displaying linear equations. The X-intercept is the point where the line in the equation intersects the X-axis, and the Y-intercept is the point where the line intersects the Y-axis. When you find these two points, you can find any point on the line.
## How do you identify x - intercept form
X coordinates are the points where the graph of a function or equation intersects or "touches" the x-axis of the Cartesian plane. You can think of it as a point with a value of zero and. To find the x coordinates of an equation, set y = and then solve for x. It is written as points to the left ({x, 0} to the right).
## How do you identify x - intercept f 5
As for intercepting x, you can easily read it from the equation. If x is 5, it intersects the x-axis (at the point where y = 0). You can also see in the diagram that the intersection point is (5.0). Note, however, that x = 5 is not a function.
## How do you find the intercepts for x = 5?
In terms of intersection, this means that x = 5 never crosses the y-axis. x = 5 - a vertical line parallel to the vertical y-axis. In terms of x-brackets, you can just put this into the equation. If x is 5, it intersects the x-axis (at the point where y = 0).
## When is the x intercept of a curve not defined?
The x-intercept is the point on the graph where y is zero. If the line is parallel to the y-axis, the origin is undefined. For every curve holds: the origin of the curve = the roots of the equation = the solution of the equation = the zeros of the curve.
## Where are the X and y intercepts on the Cartesian plane?
The abscissas are the points where the graph of a function or equation intersects or "touches" the abscissa of the Cartesian plane. You can think of it as a point with a value of zero and. x x. \\ left ({x, 0} right) (x, 0). The Y-intersections are the points where the graph of a function or equation intersects or "touches" the Y-axis of the Cartesian plane.
## How to calculate the x intercept of a parabola?
The X-intersection of each curve can be obtained using a similar technique. they just need to specify the value of y as in the equation of the curve. The general equation of the parabola is: y = ax2 + bx + c y = a x 2 + b x + c. You can find the x-intercept by setting y in Find the x-intercept of the parabola y = x2−3x +2 y = x 2 - 3 x + 2.
## How do you identify x - intercept worksheet
The origin of the function f(x) is found by finding the values of x that make f(x) = 0. Write f(x) = 0 and solve for x to find the origin of the function. The way to solve for x depends on the type of function (linear, quadratic, or trigonometric, etc.). Why it works.
## How to find x intercepts and y intercepts?
The y-intercept is the point where the graph intersects the y-axis. At this point, the x coordinate is zero. To find the x-intercept, set y to zero and solve for x. To find the intersection point, set x to zero and solve for y. For example, find the intersections of the equation y = 3x 1 .
## Which is the intercept of a graph on the x axis?
The intersections of the graph are the points where the graph intersects the axes. The x-intercept is the point where the graph intersects the x-axis.
## How do you identify x - intercept two
A parabola with two axes X. With your finger, draw a green parabola in the image in the next section. Note that your finger touches the x-axis at points (3.0) and (4.0). Therefore, the x-intercepts are (3.0) and (4.0).
## How do you identify x - intercept function
Finding the Intersections of a Function The intersections of a function f(x) are found by finding the values of x that make f(x) = 0. Write f(x) = 0 and solve for x to find the intersection points d' of the function. The way to solve for x depends on the type of function (linear, quadratic, or trigonometric, etc.).
## What is the x intercept?
Definition of xintercept. : The x coordinate of the point where a line, curve, or surface intersects the x-axis.
## How do you identify x - intercept formula
To find the x-intercept, set y to zero and solve for x. To find the y-intercept, set x to zero and solve for y. For example, find the intersections of the rendering style equation y = 3x 1 y = 3x - 1.
## How do you identify x - intercept points
To find the x-intercept, set y to zero and solve for x. To find the y-intercept, set x to zero and solve for y. For example, find the intercepts of the equation displaystyle y = 3x 1 y = 3x - 1. displaystyle y = 0 y = 0. displaystyle x = 0 x = 0.
## How do you find y intercept using slope?
The equation of a line, called a linear equation, can be written: y = mx + b, where m is the slope of the line and b is the y intersection. The y-intercept of this line is the y-value at the point where the line intersects the y-axis.
## How do you find slope when given an equation?
Slope Formula / Equation. The formula to determine the slope for a given radius is: Slope = (y2y1) / (x2x1), where x1, y1 and x2, y2 are the two specified points.
## How do you write an equation in slope intercept form?
A cross shape is a way to write a linear equation (linear equation). The shape of the slope is written as y = mx + b, where m is the slope and b is the y-intercept (the point where the line intersects the y-axis). In general, it is easy to draw a line with y = mx + b.
## How do you convert slope intercept form to standard form?
It takes more than simple arithmetic to convert the shape of a ramp segment to a standard shape. To convert from the intersection of the slope y = mx + b to the standard form Ax + By + C = 0, let m = A / B, collect all the terms on the left side of the equation and multiply by the denominator B to remove the break.
## How do you calculate slope and y intercept calculator
They first calculate the slope using the formula m = r (σ y / σ x). When you're done, you can calculate the intersection point. They do this by multiplying the slope by x. Then they subtract that value from y.
## What does y represent in slope intercept formula?
The shape of the intersection of the slope goes directly to the point: y = mx + b. m represents the slope of a straight line, b represents the y intersection of line x, and y represents the ordered pairs along the line. Learn to solve for and in linear equations with one or more solutions.
## How to calculate a linear slope?
1) Determine the values,, and. 2) Combine these values with the slope formula to find the slope. 3) gut control. When you think of points on the coordinate plane, make sure this slope is correct.
## How do you calculate y - intercept?
The y-intercept is the point where the linear equation intersects the y-axis at the point on the graph x = 0. The equation to determine the slope and y-intercept of a straight line is y = mx + b. This calculation can be done manually or with software that can solve complex math problems.
## How do you write slope intercept equations?
Write linear equations in the form of the slope of the intersection. The equation for the slope is written as. y = mx + b. Here m is the slope of the straight line and b is the y-intercept. You can use this equation to write the equation if you know the slope and y-intercept.
## How do you calculate the slope of a linear equation?
The slope can also be calculated using the slope formula if there are two points on the line: b = intersection and. This is the point where the line crosses the y-axis. A linear equation is an equation that describes a straight line. One form of the linear equation is the slope intercept form, which is written as y = mx + b.
## What is the difference between the slope and the y-intercept?
The main difference between these two forms is and. In contrast to the standard form, the insulation here is in the form of a sloping section. Whether you want to plot a linear function on paper or with a graphing calculator, you'll quickly learn what a single one can add to your math experience without frustration.
## Does every line have a slope and a y-intercept?
The answer is yes. Every non-vertical line has a slope and a baseline. The vertical line has no slope and the origin is always 0.
## What is the equation for finding y intercept?
The y-intercept is the point where the linear equation intersects the y-axis at the point on the graph x = 0. The equation to determine the slope and y-intercept of a straight line is y = mx + b.
## What is the slope of a graph?
• The line becomes ascending as it rises from left to right. The slope is positive, m > {\displaystyle m>0}.
• The line decreases as it descends from left to right. Negative slope, m < {\displaystyle m .
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Assignment 3 Solutions
Assignment 3 Solutions - Regular session Assignment#3 Due...
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Regular session: Assignment #3 Due Tuesday, 2/3 Complete the following exercises from chapter 1.6: � 5, 6, 9, 10, 13, 14, 16, 20, 24, 26 Exercise 5: Prove that if m + n and n + p are even integers, where m, n, and p are integers, then m + p is even. What kind of proof did you use? m + n is even Hypothesis m + n = 2s by definition of even integer n + p is even Hypothesis n + p = 2t by definition of even integer m + n + n + p = 2s + 2t Algebra m + p + 2n = 2s + 2t Algebra m + p = 2s + 2t – 2n Algebra m + p = 2(s + t – n) Algebra m + p is even by definition of even integer This is a direct proof. Exercise 6: Use a direct proof to show that the product of two odd numbers is odd. Let m and n be two odd numbers. m = 2a + 1 by definition of odd integer n = 2b + 1 by definition of odd integer m × n = (2a + 1) × (2b + 1) Algebra m × n = 4ab + 2a + 2b + 1 Algebra m × n = 2(2ab + a + b) + 1 Algebra m × n is odd by definition of odd integer Exercise 8: Prove that if n is a perfect square, then n + 2 is not a perfect square.
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Let n = m 2 . If m = 0, then n + 2 =2, which is not a perfect square, so we can assume that m 1. The smallest perfect square greater than n is (m + 1) 2 , and we have (m + 1) 2 = m 2 + 2m + 1 = n + 2m +1> n+2 1+1>n +2. Therefore n +2 cannot be a perfect square. Exercise 9: Use a proof by contradiction to prove that the sum of an irrational number and a rational number is irrational. Let p be the proposition “ i is an irrational number and r is a rational number” and q be the proposition “sum s of i and r is irrational”. Prove by contradiction means we are assuming q is false and that leads to contradiction to the proposition p . Assuming q is false or “sum s of i and r is rational”. s is rational Assumption s = a / b by definition of rational with a and b being integers and b 0 r = c / d by definition of rational with c and d being integers and d 0 i r s + = Algebra i d c b a + = Algebra i d c b a = - Algebra i bd bc ad = - Algebra i
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Solving Eq & Ineq Graphs & Func. Systems of Eq. Polynomials Frac. Express. Powers & Roots Complex Numbers Quadratic Eq. Quadratic Func. Coord. Geo. Exp. & Log. Func. Probability Matrices Trigonometry Trig. Identities Equations & Tri. On this page, we hope to clear up problems that you might have with solving equations and/or inequalities. Equations are the most common thing you'll see in all of math! Click any of the links below or start scrolling down to begin understanding equations and inequalities. Addition and multiplication principles Dealing with fractions or decimals Inequalities Quiz on Solving Equations and Inequalities To solve an equation, you isolate the variable are solving for. The Addition Principle says that when a = b, a + c = b + c for any number c. Example: ```1. Solve: x + 6 = -15 Solution: Using the Addition Principle, add -6 to each side of the equation. x + 6 - 6 = -15 - 6 The variable is now isolated. x = -21 ``` Along the same lines, the Multiplication Principle says that if a = b and c is any number, a * c = b * c. This principle is also used to help isolate the variable you are asked to solve for. Example: ```2. Solve: 4x = 9 Solution: Using the Multiplication Principle, multiply each side of the equation by (1/4). (1/4)4x = (1/4)9 The variable is now isolated. x = (9/4) ``` Also, be aware of problems where you might need to use both of these principles together! Example: ```3. Solve: 3x - 4 = 13 Solution: Use the Addition Principle to add 4 to each side. 3x - 4 + 4 = 13 + 4 After simplifying, 3x = 17. Use the Multiplication Principle to multiply each side by (1/3). (1/3)3x = (1/3)17 After simplification, the variable is isolated. x = (17/3) ``` When an equation contains fractions or decimals, it usually makes it easier to solve them when the fractions or decimals aren't there. The Multiplication Principle is used to do this. Example: ```1. Solve: (3/4)x + (1/2) = (3/2) Solution: Multiply both sides by the LCM of the denominators, 4 in this case. Use the Distributive Law, which says a(b + c) = ab + ac to make the equation easier to deal with. (4 * (3/4)x) + (4 * (1/2)) = 4(3/2) After simplification, you get an equation with no fractions! 3x + 2 = 6 (It's left up to you to solve for x.) ``` If a and b are real numbers, and ab = 0, either a, b, or both equal 0. This principle, called The Principle of Zero Products, is useful when you have an equation to solve that has two instances of a variable, such as (x + 3)(x - 2) = 0. Example: ```2. Solve: 7x(4x + 2) = 0 Solution: Using the Principle of Zero Products, 7x = 0 and 4x + 2 = 0 Solve each equation for x. x = 0 and x = -(1/2) The solutions are 0 and -(1/2).``` Math problems containing <, >, <=, and >= are called inequalities. A solution to any inequality is any number that makes the inequality true. On many occasions, you will be asked to show the solution to an inequality by graphing it on a number line. This is usually covered in elementary algebra (Algebra I) courses. This custom has been followed on this site, so click here to understand graphing on a number line. As with equations, inequalities also have principles dealing with addition and multiplication. They are outlined below. 1. Addition Principle for Inequalities - If a > b then a + c > b + c. Example: ``` 1. Solve: x + 3 > 6 Solution: Using the Addition Principle, add -3 to each side of the inequality. x + 3 - 3 > 6 - 3 After simplification, x > 3.``` 2. Multiplication Principle for Inequalities - If a >b and c is positive, then ac > bc. If a > b and c is negative, then ac < bc (notice the sign was reversed). Example: ``` 2. Solve: -4x < .8 Solution: Using the Multiplication Principle, multiply both sides of the inequality by -.25. Then reverse the signs. -.25(-4x) > -.25(.8) x > -.2``` One thing in math that seems to give people trouble throughout their math careers is absolute value. The absolute value of any number is its numerical value (ignore the sign). For example, the absolute value of -6 is 6 and |+3| (the vertical lines stand for absolute value) is 3. Absolute value in inequalities is a little more complicated. For example, |x| >= 4 asks us for all numbers that have an absolute value that is greater than or equal to 4. Obviously, 4 and any number greater than 4 is a solution. The confusing part comes from the fact that -4 and any number less than -4 is a solution (|-4| = 4, |-5| = 5, etc.). Therefore, the solution is x >= 4 or x <= -4. Absolute value becomes even more complicated when dealing with equations. However, there is a theorem that tells us how to deal with equations with absolute value and complicated inequalities. 1. If X is any expression, and b any positive number, and |X| = b it is the same as |X| = b or |X| = -b. 2. If X is any expression, and b any positive number, and |X| < b it is the same as -b < X < b. 3. If X is any expression, and b any positive number, and |X| > b it is the same as X < -b, X > b. Example: ``` 3. Solve: |5x - 4| = 11 Solution: Use the theorem stated above to rewrite the equation. |X| = b X = 5x - 4 and b = 11 5x - 4 = 11, 5x - 4 = -11 Solve each equation using the Addition Principle and the Multiplication Principle. 5x = 15, 5x = -7 x = 3, x = -(7/5)``` Take the Quiz on solving equations and inequalities. (Very useful to review or to see if you've really got this topic down.) Do it!
Math for Morons Like Us - Algebra II: Solving Equations and Inequalities
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Rational integrals with different real roots. Exercises resolved.
# Rational integrals with different real roots. Exercises resolved.
Do you know how to integrate rational functions? Do you know which integration method to use?
Not only is it important to know how to integrate rational functions by performing the integration method correctly, but it is even more important to know which integration method to apply.
One of your problems may be that when you see an integral, you don’t know which integration method to use, even if you know how to apply the method.
Next I am going to explain not only how to integrate rational functions, but also how to know when to apply these methods, or rather, when not to apply them. But let’s start at the beginning:
## What is a rational function?
A rational function is one that has the form of a fraction, whose numerator and denominator are polynomials. They have this form:
These are some examples of rational functions:
The degree of polynomial of the numerator P(x), can be zero, as for example:
However, the minimum degree of the polynomial of the denominator Q(x) must be greater than or equal to 1, since if it were zero, we would not have any fraction, but a polynomial:
Having said that, I’m going to explain how to integrate rational functions in the next section.
## How to integrate rational functions step by step
To know how to integrate rational functions we must pay attention to the numerator and denominator degrees. If we have this integral:
We can distinguish two groups:
1 – That the polynomial degree of the numerator is less than the polynomial degree of the denominator
And within this group we have three cases:
1.1 – Q(x) has distinct real roots
1.2 – Q(x) has multiple real roots
1.3 – Q(x) has complex roots
The other group that we can distinguish between integrals of rational functions is:
2 – That the degree of the polynomial of the numerator is greater than or equal to the degree of the polynomial of the denominator
Depending on what type it is, one method or another will be applied.
Now I’m going to focus on explaining how to integrate rational functions that are in the first group (grade P(x) < grade Q(x)) and within them those that have different real roots.
Let’s explain it with an example, step by step.
## How to solve rational integrals with different real roots in the denominator
As I mentioned before, I am going to make an example when the polynomial of the denominator has different real roots.
For example, we have this integral:
How do I know which integration method to use? The first step is to identify what kind of integral it is.
We have to discard integration methods. It cannot be solved with any immediate integral of a simple function, since in the denominator we have a function with more than one term.
The immediate integral of a compound function that can be most similar is the logarithmic type:
But if we try to solve it with this immediate integral, we must make the following variable change:
And we have no terms to replace dt, nor can we add them.
It would have been possible to apply this immediate integral if we had to integrate this function:
That if you realize, it is also the integral of a rational function, but it is resolved with the method of the immediate integral of a compound function.
Therefore, once the other integration methods have been discarded, it is when we apply the method I am going to teach you. We therefore have the integral of a rational function:
Of the form:
Dwhere the grade of the denominator polynomial is greater than the grade of the numerator polynomial:
In order to integrate this fraction, we have to decompose it in a sum of simple fractions, in which the numerator degree will be 0, that is, we will have a number and the denominator degree will be 1, so that each one of these simple fractions can be integrated later with immediate integrals of logarithmic type:
We are going to decompose the fraction into a sum of simple fractions:
First, we have to factor the polynomial of the denominator.
In this case it is a polynomial of degree 2. To decompose it we find its roots (solutions of the second degree equation) and for this we equal it to 0 and solve the equation:
Whose solutions are:
At this point is where we know that the denominator has different real roots.
Each root of the polynomial can be written in the form of a binomial (x-a), where a is the value of each root and in addition, any one polynomial can be written as the product of all its binomials (x-a). Therefore:
If the denominator polynomial was grade 3 or higher, other methods would have to be used to decompose it.
Now, the integral fraction can be written as a sum of fractions whose denominators are each of the denominator factors:
And the numerators of each fraction are two constants that we do not know:
Now let’s calculate the value of A and B.
We have a sum of two fractions with different denominator. In order to add them we have to find a common denominator, which will be the multiplication of the denominators:
And to find each new numerator, the common denominator is divided by the denominator of each initial fraction and the result is multiplied by the numerator of the initial fraction:
And we can write this as a single fraction:
And here we have come to the following conclusion: The original fraction is equal to the fraction we have just calculated:
Therefore, as we know the denominators are the same:
The numerators will also be the same:
And we keep this last expression. We are going to give values to x to calculate A and B.
The values that must be given are the roots of the polynomial, since it will cause A or B to be annulled and we know its result. Then:
Now that we know the values of A and B, we substitute them in this expression we had at the beginning:
And we already have the fraction broken down into a sum of simple fractions. To understand what we have done so far, you need to master how to operate with polynomials and algebraic fractions.
We continue with the integral. Now, we write the integral with our new simple fractions:
And each of these integrals are immediate integrals of logarithmic type:
We solve them and add the constant:
Do you master the immediate integrals? This is another of the things you need to master and which I can also explain to you step by step.
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# How do you find the product c^2d^3(5cd^7-3c^3d^2-4d^3)?
Dec 23, 2016
$\text{ "5c^3d^10" "-" "3c^5d^5" "-" } 4 {c}^{2} {d}^{6}$
#### Explanation:
$\textcolor{b l u e}{\text{Explaining multiplication rules involving powers }}$
Using a numeric example:
Known that $2 \times 2 = 4$
$2 \times 2 \text{ is the same as } {2}^{1} \times {2}^{1} = {2}^{1 + 1} = {2}^{2} = 4$
Another example:
$3 \times 3 \times 3 \text{ is the same as } {3}^{1} \times {3}^{1} \times {3}^{1} = {3}^{1 + 1 + 1} = {3}^{3} = 27$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Answering the question}}$
Given:$\text{ } \textcolor{b l u e}{{c}^{2} {d}^{3}} \textcolor{b r o w n}{\left(5 c {d}^{7} - 3 {c}^{3} {d}^{2} - 4 {d}^{3}\right)}$
Multiply everything inside the bracket by $\textcolor{b l u e}{{c}^{2} {d}^{3}}$
color(brown)(color(blue)(c^2d^3xx)5cd^7 " "- " "color(blue)(c^2d^3xx)3c^3d^2 " "- " "color(blue)(c^2d^3xx)4d^3)
$\text{ "5c^3d^10" "-" "3c^5d^5" "-" } 4 {c}^{2} {d}^{6}$
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# A dictionary is printed consisting of 7 lettered words only that can be made with the letter of word CRICKET. If the words are printed at the alphabetical order, as in an ordinary dictionary, then the number of word before the word CRICKET is$\left( a \right)$ 530$\left( b \right)$ 480$\left( c \right)$ 531$\left( d \right)$ 481
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Hint: In this particular question use the concept that first fixed the first and second place letters and arrange the remaining letters then fixed the first and third place letters at starting and arrange the remaining letters this process continues till we get the given word so use these concepts to reach the solution of the question.
Given word
CRICKET
As we see there are 7 letters in the given word.
Now we have to arrange them in a dictionary.
So first write the place value of these letters.
C – 1, C – 2, E – 3, I – 4, K – 5, R – 6, T – 7.
Now we have to arrange them in a dictionary.
So if the first and second letters are C so the number of ways to arrange the remaining 5 letters = 5!
Now if first C and second E so the number of ways to arrange the remaining 5 letters = 5!
Now if first C and second I so the number of ways to arrange the remaining 5 letters = 5!
Now if first C and second K so the number of ways to arrange the remaining 5 letters = 5!
Now if first C, second R and third C so the number of ways to arrange the remaining 4 letters = 4!
Now if first C, second R and third E so the number of ways to arrange the remaining 4 letters = 4!
Now if first C, second R, third I, fourth C, fifth E so the number of ways to arrange the remaining 2 letters = 2!
Now if first C, second R, third I, fourth C, fifth K, sixth E and seventh T = 1 way
So the position of word CRICKET in the dictionary is the sum of all the above cases.
Therefore, the position of the word CRICKET in the dictionary is = $4 \times 5! + 2 \times 4! + 2! + 1$
Now simplify it we have,
Therefore, position of word CRICKET in the dictionary is
= $4 \times \left( {120} \right) + 2 \times \left( {24} \right) + 2 + 1 = 480 + 48 + 3 = 531$
So the number of words before the given word CRICKET = 531 – 1 = 530.
So this is the required answer.
So, the correct answer is “Option A”.
Note: Whenever we face such types of questions the key concept we have to remember is that the position of word CRICKET in the dictionary is the sum of all the cases as above and always recall that the number of ways to arrange n different objects are n!, so after addition just simplify and subtract the last word from the summation value we will get the required number of words before the given word CRICKET.
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Introduction & Solving Quadratic Equations
# Introduction & Solving Quadratic Equations | Mathematics (Maths) Class 10 PDF Download
Quadratic Equations occur almost everywhere in our real life. For example, even the problem of designing a playground can be formulated as a quadratic equation. When so many situations give rise to quadratic equations, it sparks a genuine interest in looking for their solutions. Let’s say Q(x) = 0 is a quadratic equation. The solutions to a quadratic equation represent the points where this equation is satisfied that is Q(x) = 0. The solutions are also called roots/zeros of the quadratic equation. Let’s look at some approaches for solving the quadratic equations.
### Quadratic Equation
A quadratic equation is a second-degree polynomial. Its general form is given by,
ax2 + bx + c = 0
a, b and c are real numbers while a ≠ 0. Its shape is a parabola that opens upwards or downwards depending upon the value of “a”.
Its solution is the point where the equation is satisfied. There are several methods of finding out a solution to the quadratic equation given as follows:
• Factoring
• Quadratic Formula
Methods to solve quadratic equations
### Factoring
We try to factor out the equation such that we get the equation in form of the product of two terms. Then on equating these two terms to zero, we get the roots.
The following steps must be used for finding the roots with factorization:
The following steps must be used for finding the roots with factorization:
• All the terms must be on one side of the equation, either LHS or RHS leaving zero on the other side.
• Factorize the equation
• Set the factors equal to zero to find the roots one by one.
Let’s look at this method in more detail using the examples below:
Question 1: Factorize the following equation and find its roots: 2x2 – x – 1 = 0
Solution:
2x2 – x – 1 = 0
⇒ 2x2 -2x + x – 1 = 0
⇒ 2x(x – 1) + 1(x – 1) = 0
⇒ (2x + 1) (x – 1) = 0
For this equation two be zero, either one of these or both of these terms should be zero.
So, we can find out roots by equating these terms with zero.
2x + 1 = 0
x = -1/2
x – 1 = 0
⇒ x = 1
So, we get two roots in the equation.
x = 1 and -1/2
Question 2: Factorize the following equation and find its roots: x+ x – 12 = 0
Solution:
x2 + x – 12 = 0
⇒ x2 + 4x – 3x – 12 = 0
⇒ x(x + 4) -3(x + 4) = 0
⇒ (x – 3) (x + 4) = 0
Equating both of these terms with zero.
x – 3 = 0 and x – 4 = 0
x = 3 and 4
### Quadratic Formula
All the quadratic equations can be solved using the quadratic formula.
For an equation of the form,
ax2 + bx + c = 0,
Where a, b and c are real numbers and a ≠ 0.
The roots of this equation are given by,
x =
Given that b2 – 4ac is greater than or equal to zero.
Question 1: Find out the roots of the equation using Quadratic Formula,
Solution:
4x2 + 10x + 3 = 0
Using Quadratic Formula to solve this,
a = 4, b = 10 and c = 3
Before plugging in the values, we need to check for the discriminator
b2 – 4ac
⇒ 102 – 4(4)(3)
⇒ 100 – 48
⇒ 52
This is greater than zero, So now we can apply the quadratic formula.
Plugging the values into quadratic equation,
Question 2: Find out the roots of the equation using Quadratic formula,
5x2 + 9x + 4 = 0
Solution:
5x2 + 9x + 4 = 0
Using Quadratic Formula,
a = 5, b = 9 and c = 4.
Before plugging in the values, we need to check for the discriminator
b2 – 4ac
⇒ 92 – 4(5)(4)
⇒ 81 – 80
⇒ 1
This is greater than zero, So the quadratic formula can be applied. Plugging in the values in the formula,
The document Introduction & Solving Quadratic Equations | Mathematics (Maths) Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
## Mathematics (Maths) Class 10
116 videos|420 docs|77 tests
## FAQs on Introduction & Solving Quadratic Equations - Mathematics (Maths) Class 10
1. What is a quadratic equation?
Ans. A quadratic equation is a polynomial equation of degree 2, which means the highest power of the variable is 2. It can be written in the form ax^2 + bx + c = 0, where a, b, and c are constants.
2. How do you solve a quadratic equation by factoring?
Ans. To solve a quadratic equation by factoring, follow these steps: 1. Write the equation in the form ax^2 + bx + c = 0. 2. Factor the quadratic expression on the left side of the equation. 3. Set each factor equal to zero and solve the resulting linear equations. 4. Check the solutions by substituting them back into the original equation.
3. Can all quadratic equations be solved by factoring?
Ans. No, not all quadratic equations can be solved by factoring. Some quadratic equations may have complex roots or cannot be easily factored. In such cases, other methods like completing the square or using the quadratic formula are used to find the solutions.
4. What is the quadratic formula?
Ans. The quadratic formula is a formula used to solve quadratic equations. It is given by: x = (-b ± √(b^2 - 4ac))/(2a) where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0. The ± sign indicates that there are two possible solutions.
5. How do you solve a quadratic equation using the quadratic formula?
Ans. To solve a quadratic equation using the quadratic formula, follow these steps: 1. Write the equation in the form ax^2 + bx + c = 0. 2. Identify the values of a, b, and c. 3. Substitute these values into the quadratic formula: x = (-b ± √(b^2 - 4ac))/(2a). 4. Simplify the equation and calculate the values of x using the plus and minus signs ±. These values represent the solutions to the quadratic equation.
## Mathematics (Maths) Class 10
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# Question #51f4c
Nov 28, 2017
29
#### Explanation:
Let’s make R stand for the Red marbles, and B for the Black.
From the info given, we can see that red has twice more than black plus 3, so $R = 2 B + 3$
Note that the $T o t a l = R + B$
You want to replace R with the equation we found earlier, so:
$T o t a l = R + B = 2 B + 3 + B$
And since we know the total amount of marbles, we can plug that in.
$42 = 2 B + 3 + B$
Move like terms together (subtract 3 on both sides, 2B+B = 3B)
$39 = 3 B$
Solve B (Divide both sides by 3)
$\frac{39}{3} = \frac{3 B}{3}$
$13 = B$
We have solved the amount of black marbles. There are 13 black marbles!
Now plug it back to the R equation we found earlier to get the amount of red marbles. (Equation is 2B+3.) Since we found the amount of black marbles (B), you can replace B with 13 to find R
$R = 2 B + 3 = 2 \left(13\right) + 3 = 29$
There are therefore 29 red marbles.
Nov 28, 2017
Pam has a total of $29$ red marbles
#### Explanation:
Method 1
Let the number of black marbles be $x$.
Then the number of red marbles is $\left(2 x + 3\right)$
By setting up an equation,
$x + \left(2 x + 3\right) = 42$
$3 x + 3 = 14 \times 3 \text{ } \left(\div 3\right)$
$x + 1 = 14$
$x = 13$
Therefore, Pam has a total of $\left[2 \left(13\right) + 3\right]$
$= 29$ red marbles
Method 2
Let the number of red marbles be $x$.
Then the number of black marbles is $\left(\frac{x - 3}{2}\right)$
By setting up an equation,
$x + \left(\frac{x - 3}{2}\right) = 42 \text{ } \left(\times 2\right)$
$2 x + x - 3 = 84$
$3 x - 3 = 3 \cdot 28 \text{ } \left(\div 3\right)$
$x - 1 = 28$
$x = 29$
Therefore, Pam has a total of $29$ red marbles
|
## Equilateral Spherical Triangles
The term ‘tessellate’ means “to fit together exactly a number of identical shapes, leaving no spaces”.
Since the area of a sphere is 4R², if n equilateral spherical triangles tessellate a sphere into n parts, the area of each triangle is:
Let us apply this to the spherical equivalent of the icosahedron on the left, and determine the spherical angle ‘a’ at each vertex of the sphere.
By central projection, the edges of the underlying icosahedron have generated a net of spherical triangles on the surface of the sphere. (I have left the edge planes in, since ‘a’ is the angle between them). The icosahedron has 20 faces, so the area A of a single spherical triangle is:
For a unit sphere, this is just:
A = 0.62831853
The area of a spherical triangle on a unit sphere is given by:
A = (a + b + c) -
Since a, b and c must all be equal, this becomes:
A = 3a -
Then
a = 1.256637061 radians. To convert radians to degrees, multiply by (180 / )
a = 72°
This is clearly what we expect since the total angle at each vertex must be 2 (360°). And an icosahedron has 5 triangles meeting at each vertex, so 360° / 5 = 72°.
At this point we might ask: “How many equilateral triangles will tessellate a sphere?”
Assume we have no knowledge of the platonic solids, but can only guess how many edges meet at a single vertex. How many equilateral spherical triangles will tessellate the sphere? If e is the number of edges meeting at any given vertex, then 2 / e will be the spherical angle for each triangle vertex.
We already know that for a unit sphere, the area A of an equilateral spherical triangle is:
A = 3a -
And we also know that for a unit sphere, tessellated by n equilateral spherical triangles, the total area is:
Therefore
4 = n(3a - )
Make n the subject…
Since (2 / e) is the spherical angle for each triangle vertex, then the theoretical number of equilateral spherical triangles will be…
Tessellation only occurs when the result, n, is a whole number. As it turns out, the only valid values for e are 3, 4 and 5. As we might expect, this gives the spherical equivalents of the tetrahedron, octahedron and icosahedron.
Tetrahedron e = 3 n = 4 a = 120° Octahedron e = 4 n = 8 a = 90° Icosahedron e = 5 n = 20 a = 72°
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# Population Mean (Known Variance)
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## Transcription
1 Confidence Intervals Solutions STAT-UB.0103 Statistics for Business Control and Regression Models Population Mean (Known Variance) 1. A random sample of n measurements was selected from a population with unknown mean µ and known standard deviation σ. Calculate a 95% confidence interval for µ for each of the following situations: (a) n = 49, x = 28, σ = 28 The formula for the 95% confidence interval is x ± 2 σ n. If we substitute the values of the problem, we get 28 ± or (20, ). 28 ± 8, (b) n = 100, x = 125, σ = ± ± 10. (c) Is the assumption that the underlying population of measurements is normally distributed necessary to ensure the validity of the confidence intervals in parts (a) (c)? No. Since the sample sizes are large (n 30), the central limit theorem guarantees that x is approximately normal, so the confidence intervals are valid.
2 2. A random sample of measurements was selected from a population with unknown mean µ and known standard deviation σ = 18. The sample mean is x = 12. Calculate a 95% confidence interval for µ. We compute a 95% confidence interval for µ via the formula x ± 2 σ n. In this case, we get 12 ± 2 18 i.e., 12 ± With respect to the previous problem, which of the following statements are true: A. There is a 95% chance that µ is between 6 and 18. B. The population mean µ will be between 6 and 18 about 95% of the time. C. In 95% of all future samples, the sample mean will be between 6 and 18. D. The population mean µ is between 6 and 18. E. None of the above. The correct answer is E. The numbers µ, 6, and 18 are all nonrandom, so it makes no sense to talk about probability. Instead, we can say that we have 95% confidence that µ is between 6 and 18. The term confidence denotes subjective belief, as opposed to probability, which is concerned with randomness. 4. Complete Problem 2, with a 99% confidence interval instead of a 95% confidence interval. σ For a 100(1 α)% confidence interval for µ, we use the formula x ± z α/2 n. For a 99% confidence interval, we have α =.01 and z α/2 = Thus, our confidence interval for µ is 12 ± i.e., 12 ± Complete Problem 2, with an 80% confidence interval instead of a 95% confidence interval. σ For a 100(1 α)% confidence interval for µ, we use the formula x ± z α/2 n. For a 80% confidence interval, we have α =.20 and z α/2 = Thus, our confidence interval for µ is 12 ± i.e., 12 ± Page 2
3 Population Mean (Unknown Variance) 6. A random sample of n measurements was selected from a population with unknown mean µ and unknown standard deviation σ. Calculate a 95% confidence interval for µ for each of the following situations: (a) n = 25, x = 28, s = 12 When we don t know the population standard deviation, we use a t-based confidence interval with n 1 degrees of freedom. The formula for the 95% confidence interval is s x ± t.025, n where t.025 is the critical value for a t random variable with n 1 degrees of freedom. Since n = 25, we need the critical value for n 1 = 24 degrees of freedom. Using the 12 t table, this value is t.025 = The 95% confidence interval is 28 ± ± (b) n = 16, x = 12, s = 18 We use n 1 = 15 degrees of freedom. Using the t table, t.025 = The 95% confidence interval is 12 ± ± (c) n = 100, x = 125, s = 50 We use n 1 = 99 degrees of freedom. The closest value in the table is 60 degrees of freedom. We use t The 95% confidence interval is ± ± 10. (d) Is the assumption that the underlying population of measurements is normally distributed necessary to ensure the validity of the confidence intervals in parts (a) (c)? We need this assumption for (a) and (b), but not for (c). In (c), the sample size is above 30, so the central limit theorem applies, and the t is a reasonable approximation. Page 3
4 7. In each of the following situations, find α and t α/2. (a) An 80% confidence interval with n = 10. α =.20, n 1 = 9 degrees of freedom, t.100 = (b) A 99% confidence interval with n = 25. α =.01, n 1 = 24 degrees of freedom, t.005 = (c) A 90% confidence interval with n = 30. α =.10, n 1 = 29 degrees of freedom, t.050 = Compute z α/2 for each of the situations in problem 7. We can look up the answer in the degrees of freedom column. The answers are: (a) z.100 = 1.282; (b) z.005 = 2.576; (c) z.050 = The white wood material used for the roof of an ancient Japanese temple is imported from Northern Europe. The wooden roof must withstand as much as 100 centimeters of snow in the winter. Architects at Tohoku University (Japan) conducted a study to estimate the mean bending strength of the white wood roof. A sample of 25 pieces of the imported wood were tested and yielded the following statistics on breaking strength: x = 74.5, s = Estimate the true mean breaking strength of the white wood with a 90% confidence interval. We don t know the population standard deviation so we use a t-based confidence interval. The formula for the interval is x±t α/2 s n. We have α =.10 and n 1 = 24 degrees of freedom, so t α/2 = t.05 = The 90% confidence interval is 74.5 ± ± Page 4
5 10. Researchers recorded expenses per full-time equivalent employee for each in a sample of 1,751 army hospitals. The sample yielded the following summary statistics: x = \$6, 563 and s = \$2, 484. Estimate the mean expenses per full-time equivalent employee of all U.S. army hospitals using a 90% confidence interval. Again, we use a t-based confidence interval. Now, we have α/2 =.10/2 =.05 and n 1 = 1750 degrees of freedom. We use the largest value of degrees of freedom in the 2484 table (120), so t α/ The 90% confidence interval is 6563 ± ± Page 5
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# Collatz Conjecture and a simple program
Author: Kazi Abu Rousan
Mathematics is not yet ready for such problems.
Paul Erdos
#### Introduction
A problem in maths which is too tempting and seems very easy but is actually a hidden demon is this Collatz Conjecture. This problems seems so easy that it you will be tempted, but remember it is infamous for eating up your time without giving you anything. This problem is so hard that most mathematicians doesn't want to even try it, but to our surprise it is actually very easy to understand. In this article our main goal is to understand the scheme of this conjecture and how can we write a code to apply this algorithm to any number again and again.
Note: If you are curious, the featured image is actually a visual representation of this collatz conjecture \ $3n+1$ conjecture. You can find the code here.
#### Statement and Some Examples
The statement of this conjecture can be given like this:
Suppose, we have any random positive integer $n$, we now use two operations:
• If the number n is even, divide it by $2$.
• If the number is odd, triple it and add $1$.
Mathematically, we can write it like this:
$f(n) = \frac{n}{2}$ if $n \equiv 0$ (mod 2)
$f(n) = (3n+1)$ if $n \equiv 1$ (mod 2)
Now, this conjecture says that no-matter what value of n you choose, you will always get 1 at the end of this operation if you perform it again and again.
Let's take an example, suppose we take $n=12$. As $12$ is an even number, we divide it by $2$. $\frac{12}{2} = 6$, which is again even. So, we again divide it by $2$. This time we get, $\frac{6}{2} = 3$, which is odd, hence, we multiply it by 3 and add $1$, i.e., $3\times 3 + 1 = 10$, which is again even. Hence, repeating this process we get the series $5 \to 16 \to 8 \to 4 \to 2 \to 1$. Seems easy right?,
Let's take again, another number $n=19$. This one gives us $19 \to 58 \to 29 \to 88 \to 44 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26 \to 13 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1$. Again at the end we get $1$.
This time, we get much bigger numbers. There is a nice trick to visualize the high values, the plot which we use to do this is called Hailstone Plot.
Let's take the number $n=27$. For this number, the list is $27 \to 82 \to 41 \to 124 \to 62 \to 31 \to 94 \to 47 \cdots \to 9232 \cdots$. As you can see, for this number it goes almost as high as 9232 but again it falls down to $1$. The hailstone plot for this is given below.
So, what do you think is the conjecture holds for all number? , no one actually knows. Although, using powerful computers, we have verified this upto $2^{68}$. So, if you are looking for a counterexample, start from 300 quintillion!!
###### Generating the series for Collatz Conjecture for any number n using python
import numpy as np
def collatz(n):
result = np.array([n])
while n >1:
# print(n)
if n%2 == 0:
n = n/2
else:
n = 3*n + 1
result = np.append(result,n)
return result
The function above takes any number $n$ as it's argument. Then, it makes an array (think it as a box of numbers). It creates a box and then put $n$ inside it.
1. Then for $n>1$, it check if $n$ is even or not.
2. After this, if it is even then $n$ is divided by 2 and the previous value of n is replaced by $\frac{n}{2}$.
3. And if $n$ is odd, the it's value is replaced by $3n+1$.
4. For each case, we add the value of new $n$ inside our box of number. This whole process run until we have the value of $n=1$.
5. Finally, we have the box of all number / series of all number inside the array result.
Let's see the result for $n = 12$.
n = 12
print(collatz(n))
#The output is: [12. 6. 3. 10. 5. 16. 8. 4. 2. 1.]
Similarly, for $n = 19$, the code gives us,
n = 19
print(collatz(n))
#The output is: [19. 58. 29. 88. 44. 22. 11. 34. 17. 52. 26. 13. 40. 20. 10. 5. 16. 8. 4. 2. 1.]
You can try out the code, it will give you the result for any number $n$.
Test this function.
#### Hailstone Plot
Now let's see how to plot Hailstone plot. Suppose, we have $n = 12$. In the image below, I have shown the series we get from $12$ using the algorithm.
Now, as shown in the above image, we can assign natural numbers on each one of them as shown. So, Now we have two arraies (boxes), one with the series of numbers generated from $n$ and other one from the step number of the applied algorithm, i.e., series of natural numbers. Now, we can take elements from each series one by one and can generate pairs, i.e., two dimensional coordinates.
As shown in the above image, we can generate the coordinates using the steps as x coordinate and series a y coordinate. Now, if we plot the points, then that will give me Hailstone plot. For $n = 12$, we will get something like the image given below. Here I have simply added each dots with a red line.
The code to generate it is quite easy. We will be using matplotlib. We will just simply plot it and will mark the highest value along with it's corresponding step.
import numpy as np
import matplotlib.pyplot as plt
def collatz(n):
result = np.array([n])
while n >1:
# print(n)
if n%2 == 0:
n = n/2
else:
n = 3*n + 1
result = np.append(result,n)
return result
n = 63728127
y_vals = collatz(n); x_vals = np.arange(1,len(y_vals)+1)
plt.plot(x_vals,y_vals)
x_val_max = np.where(y_vals == max(y_vals))[0][0]+1
plt.text(x_val_max, max(y_vals), '({}, {})'.format(x_val_max, int(max(y_vals))))
plt.grid(color='purple',linestyle='--')
plt.ylabel("Sequence Generated")
plt.xlabel("Number of Iteration")
plt.show()
The output image is given below.
Crazy right?, you can try it out. It's almost feels like some sort of physics thing right!, for some it seems like stock market.
This is all for today. In it's part-2, we will go into much more detail. I hope you have learn something new.
Maybe in the part-2, we will see how to create pattern similar to this using collatz conjecture.
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Inverse functions and relations
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Inverse functions and relationsPresentation Transcript
• Inverse Functions and Relations
• Inverse Functions and Relations
• Given any function, f, the inverse of the function, f -1 , is a relation that is formed by interchanging each (x,y) of f to a (y,x) of f -1 .
•
Introduction
• Let f be defined as the set of values given by Let f -1 be defined as the set of values given by 10 -5 4 0 y-values 7 4 0 -2 x-values 7 4 0 -2 y-values 10 -5 4 0 x-values
• For a function to have an inverse that is a function, then the original function must be a 1:1 function. A 1:1 function is defined as a function in which each x is paired with exactly one y and y is paired with exactly one x. To be a function we can have no domain value paired with two range values .
• That is, the function could not have (-3,5) and (-3,-3) as two points .
• To be a function the graph of the function must pass the “vertical line” test. To be a 1:1 function the graph of the function must also pass a “horizontal line” test.
•
• To find the inverse of a function:
• Replace f(x) with y.
• 2. Interchange x and y.
• 3. Solve for y.
• 4. Replace y with f -1 (x).
• Example: Inverse Relation Algebraically Example1 : Find the inverse relation algebraically for the function f ( x ) = 3 x + 2. y = 3 x + 2 Original equation defining f x = 3 y + 2 Switch x and y . 3 y + 2 = x Reverse sides of the equation. To calculate a value for the inverse of f , subtract 2, then divide by 3 . To find the inverse of a relation algebraically , interchange x and y and solve for y . y -1 = Solve for y.
• y = x The graphs of a relation and its inverse are reflections in the line y = x . The ordered pairs of f a re given by the equation . Example 1a : Find the graph of the inverse relation geometrically from the graph of f ( x ) = x y 2 -2 -2 2 The ordered pairs of the inverse are given by .
• Example2:
• Let f (x)= y = 3x , find the inverse. This is, f -1 : x = 3y .
• 5x - 1 5y - 1
• Solve for y: x (5y – 1) = 3y.
• 5xy - x = 3y
• 5xy – 3y = x
• y(5x – 3) = x
• y = x
• 5x - 3
• Example 3: Find the inverse of f(x) = f -1 (x) = -2x +2 (-2) (-2) Replace f(x) with y. Interchange x and y. Solve for y. Replace y with f-1(x).
• Example 4: Two functions f and g are inverse functions if and only if both of their compositions are the identity function; f(x) = x. Determine whether and are inverse functions. You must do [f ◦ g](x) and [g ◦ f ](x), if they both equal x, they are inverses!
• [f ◦ g](x) = x + 6 – 6 = x [g ◦ f ](x) = x – 8 + 8 = x So, they ARE inverses of each other!
• Example: Composition of Functions It follows that g = f -1 . Example 5 :Verify that the function g ( x ) = is the inverse of f ( x ) = 2 x – 1. f( g ( x ) ) = 2 g ( x ) – 1 = 2( ) – 1 = ( x + 1) – 1 = x g ( f ( x ) ) = = = = x
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# Maths: the three basic percentages
HideShow resource information
## Type 1 find x% of y
1) find 10% of y by dividing it by ten. (You can easily do this by moving the decimal one place to the left.)
2) now you have 10% you can times it to find 20, 30 etc or half it to find 5.
3) Another way to do this is dividing the % you're given to find by 100 and times it by the numerical value you were given to find the % of.
Example : find 15% of £46
0.15 X 46 = 690 in pounds = £6.90
1 of 3
## Type 2 Find the new amount after a % increase/decr
1) Find the numerical % value - use the methods shown in type 1.
2) Whatever was the % given out of step 1 if the question is the new amount after a % increase add the answer in step 1 to the original value however, if it's a decrease take the answer in step 1 away from the original value.
Put into practise: A toaster is reduced in price by 40% in the sales. It originally cost £68. what is the new price of the toaster.
1) 40% of £68 = £27.2 method 1 = 10% of 68 = 6.8 to find 40% X that by 4 = £27.2.
Method 2: 0.4 X 68 = £27.2
2) Because the % of £68 = £27.2 and the question is a reduction question take it away from the original value £68 - £27.2 = £40.2
2 of 3
## Type 3 - "Express x as a % of y"
1) Make sure both of the numbers are in the same unit of measurement.
2) Divide the number that will be given as a percentage by the number it will give a percentage of and then times it by 100. Divide X by Y and then TMES by 100.
Put into practise: Give 40p as a percentage of £3.34
1) covert £3.34 into pence divide by 100.
2) 40/334 X 100 =12% (1.d.p)
3 of 3
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Question
# Let P(k) be a statement that \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{k\cd
Series
Let P(k) be a statement that $$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{k\cdot(k+1)}=$$
for: The basis step to prove $$P(k)$$ is that at $$k = 1, ?$$ is true.
for:Show that $$P(1)$$ is true by completing the basis step proof. Left side of $$P(k)$$ and Right side of $$P(k)$$
for: Identify the inductive hypothesis used to prove $$P(k)$$.
for: Identify the inductive step used to prove $$P(k+1).$$
2021-05-19
Let the property $$P(k)$$ be $$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{k(k+1)}=\frac{k}{k+1}$$
Show that $$P(k)$$ is true for all integers $$k\geq1$$ using mathematical induction
Basis Step: $$P(k)$$ is true:
That is to show that $$\frac{1}{1\cdot2}=\frac{1}{1+1}$$
The left hand side of the equation is $$\frac{1}{1\cdot2}=\frac{1}{2}$$ and right-hand side is
$$\frac{1}{1+1}=\frac{1}{2}$$
It follows that $$\frac{1}{2}=\frac{1}{2}$$
Hence $$P(1)$$ is true.
Show that for all integers $$n\geq1$$, $$P(k)$$ is true then $$P(k+1)$$ is also true:
Suppose P(k) is true.
Then the inductive hypothesis is
$$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{k(k+1)}$$
Now show that $$P(k+1)$$ is true.
That is to show that
$$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{(k+1)+1}$$
Or, equivalently that
$$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{(k+1)(k+2)}=\frac{k+1}{k+2}$$
The left-hand side of $$P(k+1)$$ is
$$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}$$
$$=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}$$
$$=\frac{1}{k+1}(k+\frac{1}{k+2})$$
$$=\frac{1}{k+1}(\frac{k^2+2k+1}{k+2})$$
$$=\frac{1}{k+1}(\frac{(k+1)^2}{k+2})$$
$$=\frac{1}{k+1}\frac{(k+1)^2}{k+2}$$
$$=\frac{k+1}{k+2}$$
which is right hand side of $$P(k+1)$$
Hence from the principle of mathematical induction,
$$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{n(n+1)}=\frac{n}{n+1}$$ is true, for all integers $$n\geq1$$
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Factors of 217
Factors of 217 are 1, 7, 31, and 217
How to find factors of a number
1. Find factors of 217 using Division Method 2. Find factors of 217 using Prime Factorization 3. Find factors of 217 in Pairs 4. How can factors be defined? 5. Frequently asked questions 6. Examples of factors
Example: Find factors of 217
• Divide 217 by 1: 217 ÷ 1 : Remainder = 0
• Divide 217 by 7: 217 ÷ 7 : Remainder = 0
• Divide 217 by 31: 217 ÷ 31 : Remainder = 0
• Divide 217 by 217: 217 ÷ 217 : Remainder = 0
Hence, Factors of 217 are 1, 7, 31, and 217
2. Steps to find factors of 217 using Prime Factorization
A prime number is a number that has exactly two factors, 1 and the number itself. Prime factorization of a number means breaking down of the number into the form of products of its prime factors.
There are two different methods that can be used for the prime factorization.
Method 1: Division Method
To find the primefactors of 217 using the division method, follow these steps:
• Step 1. Start dividing 217 by the smallest prime number, i.e., 2, 3, 5, and so on. Find the smallest prime factor of the number.
• Step 2. After finding the smallest prime factor of the number 217, which is 7. Divide 217 by 7 to obtain the quotient (31).
217 ÷ 7 = 31
• Step 3. Repeat step 1 with the obtained quotient (31).
31 ÷ 31 = 1
So, the prime factorization of 217 is, 217 = 7 x 31.
Method 2: Factor Tree Method
We can follow the same procedure using the factor tree of 217 as shown below:
So, the prime factorization of 217 is, 217 = 7 x 31.
3. Find factors of 217 in Pairs
Pair factors of a number are any two numbers which, which on multiplying together, give that number as a result. The pair factors of 217 would be the two numbers which, when multiplied, give 217 as the result.
The following table represents the calculation of factors of 217 in pairs:
Factor Pair Pair Factorization
1 and 217 1 x 217 = 217
7 and 31 7 x 31 = 217
Since the product of two negative numbers gives a positive number, the product of the negative values of both the numbers in a pair factor will also give 217. They are called negative pair factors.
Hence, the negative pairs of 217 would be ( -1 , -217 ) , ( -7 , -31 ) .
How do you explain factors?
In mathematics, a factor is a number or also it can be an algebraic expression that divides another number or any expression completely and that too without leaving any remainder. A factor of a number can be positive or negative.
Properties of factors
• Every factor of a number is an exact divisor of that number, example 1, 7, 31, 217 are exact divisors of 217.
• Every number other than 1 has at least two factors, namely the number itself and 1.
• Each number is a factor of itself. Eg. 217 is a factor of itself.
• 1 is a factor of every number. Eg. 1 is a factor of 217.
• What are the prime factors of 217?
The factors of 217 are 1, 7, 31, 217.
Prime factors of 217 are 7, 31.
• What is the sum of all factors of 217?
The sum of all factors of 217 is 256.
• What are the pair factors of 217?
Pair factors of 217 are (1,217), (7,31).
• What two numbers make 217?
Two numbers that make 217 are 7 and 31.
• What are multiples of 217?
First five multiples of 217 are 434, 651, 868, 1085.
• Is 217 a perfect square?
No 217 is not a perfect square.
• Which is greatest factor of 217?
The greatest factor of 217 is 31.
• How do you factors of 217?
Factors of 217 are 1, 7, 31, 217.
• What are five multiples of 217?
First five multiples of 217 are 434, 651, 868, 1085, 1302.
Examples of Factors
Can you help Rubel to find out the product of the even factors of 217?
Factors of 217 are 1, 7, 31, 217.
Even factors of 217 are 0.
Hence, product of even factors of 217 is; 0 = 0.
Annie's mathematics teacher has asked her to find out all the positive and negative factors of 217? Help her in writing all the factors.
Positive factors are 1, 7, 31, 217.
Negative factors are -1, -7, -31, -217.
How many factors are there for 217?
Factors of 217 are 1, 7, 31, 217.
So there are in total 4 factors.
Sammy is puzzled while calculating the prime factors of 217. Can you help him find them?
Factors of 217 are 1, 7, 31, 217.
Prime factors of 217 are 7, 31
Joey wants to write all the prime factors of 217 in exponential form, but he doesn't know how to do so can you assist him in this task?
Prime factors of 217 are 7, 31.
So in exponential form it can be written as 7 x 31.
What is prime factorization of 217?
Prime factorization of 217 is 7 x 31 = 7 x 31.
Can you help Rubel to find out the product of the even factors of 217?
Factors of 217 are 1, 7, 31, 217.
Even factors of 217 are 0.
Hence, product of even factors of 217 is; 0 = 0.
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# Maxima and Minima Problem No 13 - Application of Derivatives - Diploma Maths - II | Summary and Q&A
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April 11, 2022
by
Ekeeda
Maxima and Minima Problem No 13 - Application of Derivatives - Diploma Maths - II
## TL;DR
Find the dimensions of an open box with a square base to maximize its volume.
## Install to Summarize YouTube Videos and Get Transcripts
### Q: What is the objective of the problem?
The objective is to find the dimensions of an open box's square base to maximize its volume.
### Q: How is the area of the box calculated?
The area of the box consists of the five faces: the base (X^2) and four sides (X*Y), where Y is the height. The total area should be equal to 192 square centimeters.
### Q: How is the volume of the box calculated?
The volume is calculated by multiplying the length, breadth, and height of the box together.
### Q: How is the equation for volume manipulated to find a maximum?
The equation for volume is differentiated with respect to X, and then the derivative is set equal to zero to find critical points.
### Q: How is the maximum volume determined?
The second-order derivative is calculated by differentiating the derivative with respect to X again. If the second derivative is greater than zero, it indicates a maximum volume.
### Q: What are the dimensions of the box that maximize the volume?
The dimensions of the box that maximize the volume are 8 centimeters for length and breadth, and 4 centimeters for height.
## Summary & Key Takeaways
• A box with a square base needs to be made with an open top, and the area of the material for making the box is given as 192 square centimeters.
• The goal is to determine the dimensions of the box that will maximize its volume.
• By considering the side length of the square base as X, the dimensions of the box are found to be 8 centimeters for length and breadth, and 4 centimeters for height.
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# What is the center and radius of the circle with equation x^2 + y^2 - 2x + 14y + 42 = 0?
Jun 18, 2018
the centre is $\left(1 , - 7\right)$ and the radius is $\sqrt{8} = 2 \sqrt{2}$
#### Explanation:
${x}^{2} + {y}^{2} - 2 x + 14 y + 42 = 0$
$\left({x}^{2} - 2 x\right) + \left({y}^{2} + 14 y\right) = - 42$
Complete the square. If you have an equation $a {x}^{2} + b x$ and you want to complete the square, all you have to do is to half the coefficient of the term $x$ ie $\frac{b}{2}$ and then square it ie ${\left(\frac{b}{2}\right)}^{2}$. Your equation will end up being $a {x}^{2} + b x + {\left(\frac{b}{2}\right)}^{2}$
$\left({x}^{2} - 2 x + 1\right) + \left({y}^{2} + 14 y + 49\right) = - 42 + 1 + 49$
${\left(x - 1\right)}^{2} + {\left(y + 7\right)}^{2} = 8$
Since the general form of the circle is given by: ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$ where the centre is $\left(h , k\right)$ and the radius is r, then looking at the above equation, the centre is $\left(1 , - 7\right)$ and the radius is $\sqrt{8} = 2 \sqrt{2}$
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# How do you calculate multiple discounts?
Contents
For example, if the original price was \$50 and we have two discounts: 20% and 10% , then we’re doing something like this: \$50 – 20% = \$50 – \$10 = \$40 . Then \$40 – 10% = \$40 – \$4 = \$36 .
## How do you calculate a triple discount?
Take the original price and subtract the original price times the 1st discount. Take the price from step 2 and subtract the price from step 2 times the 2nd discount. Calculate the final price. Take the price from step 3 and subtract the price from step 3 times the 3rd discount.
## How do you calculate a series discount?
For example, if the discount offered is 35 percent and the total value of the goods sold are \$100,000, the single discount formula would be: \$100,000 x 35 percent = \$35,000 discount. In contrast to the single discount formula, however, the discount series yields a lower discount.
## What is discount formula in Excel?
Say you want to reduce a particular amount by 25%, like when you’re trying to apply a discount. Here, the formula will be: =Price*1-Discount %.
## How do you solve successive discount questions?
If the successive discounts d1, d2, and d3 are given on an item, then the selling price of that item is calculated by, SP = (1 – d1/100) x (1 – d2/100) x (1 – d3/100) x MP, where SP is selling price and MP is marked price.
THIS IS INTERESTING: How much is Volvo NHS discount?
## How do you calculate discount in accounting?
A sales discount equals the percentage discount times the outstanding invoice amount. The discounted invoice amount equals the outstanding invoice amount minus the sales discount. For example, the sales discount on an invoice of \$1,000 that offers a 2 percent discount is \$20, since 0.02 x \$1,000 = \$20.
## How do you calculate total discount in Excel?
If you know the original price and the discounted price, you can calculate the percentage discount. If you know the original price and the percentage discount, you can calculate the discounted price, etc.
Calculate Percentage Discount
1. First, divide the discounted price by the original price. …
2. Subtract this result from 1.
## How do you take 20% off in Excel?
If you want to calculate a percentage of a number in Excel, simply multiply the percentage value by the number that you want the percentage of. For example, if you want to calculate 20% of 500, multiply 20% by 500.
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# Percent Difference – Explanation & Examples
The percent difference is the difference between two numbers expressed in percentage. To understand the concept of a percent difference, we must first understand what is meant by a percentage? A percentage is a number that is expressed as a fraction of 100.
For example, $10$ percent or $10\%$ means $\dfrac{10}{100}$. We can also use it to describe a relation between two numbers. For example, $24$ is $20\%$ of $120$. The percentage sign is denoted by “%” and is equal to $\dfrac{1}{100}$. Say we want to calculate $8\%$ of $150$, we simply do the following calculations.
$8\%\hspace{1mm} of \hspace{1mm} 150 = [\dfrac{8}{100}] \times 150 = 12$.
The percent difference is the ratio of the absolute difference of two values and their average value, multiplied by 100.
You should refresh the following concepts to understand the material discussed here.
1. Percentage.
2. Basic Arithmetic.
## What is Percent Difference
The percent difference is used to calculate the difference between two non-identical positive numbers, and it is expressed in percentage. For example, we have two numbers, $26$ and $10$; we want to calculate the percent difference between these two numbers.
The first step is to calculate the difference between them; in this case that would be $26\hspace{1mm} –\hspace{1mm} 10 = 16$ or $10\hspace{1mm} – \hspace{1mm}26 = -16$. We are not provided with the information on which number is original or which number is the new one; we are simply given two numbers and have to calculate the difference between them.
So, in this example, the difference is $16$ or $-16$. Still, as we are using the absolute value in the calculation of percent difference, so the result will always be a positive number.
Hence, the difference is 16 no matter which number we take as “a” and which number as “b.” Once we calculate the difference, now it is time to decide the reference or base value that we can use for divis. As we just mentioned, we have not been given any data regarding the context of the two numbers, so taking the average of the two numbers is a good solution.
The average value in this example is calculated as $\dfrac {(26\hspace{1mm}+\hspace{1mm}10)}{2}= 18$. We will calculate the percent difference by dividing the number $16$ by average value $18$ and then multiplying by $100$, and the result will be $88.88 \%$.
Percentage Difference = [Absolute difference of the two numbers/Average of those numbers] * 100.
## How to Calculate Percent Difference
Calculation of the percent difference is pretty simple and easy. But, first, you need to follow the steps given below.
1. Name the two given numbers as “a” and “b.”
2. Calculate the absolute difference between the given two numbers: $|a\hspace{1mm} -\hspace{1mm} b|$
3. Calculate the average of the two numbers by using the following formula: $\dfrac{(a\hspace{1mm}+\hspace{1mm} b)} { 2}$.
4. Now divide the value calculated in step 2 with the average value calculated in step 3: $\dfrac{ |a\hspace{1mm}-\hspace{1mm} b|} { ((a\hspace{1mm} +\hspace{1mm} b) / 2)}$.
5. Express the final answer in percentage by multiplying the result in step 4 by $100$
### Percent Difference Formula:
We can calculate the percent difference by using the formula given below.
$\mathbf{Percentage\hspace{1mm} Difference = [\dfrac{\left | a\hspace{1mm}-\hspace{1mm}b \right |}{(a\hspace{1mm}+\hspace{1mm}b)\hspace{1mm}/2}]\times 100}$
Here,
a and b = Two non-identical positive numbers.
$| a\hspace{1mm} -\hspace{1mm} b |$ = Absolute difference value of two numbers
$\dfrac{(a\hspace{1mm}+\hspace{1mm}b)}{2}$ = Average of two numbers
Example 1: Calculate the Percent difference between number $30$ and $15$.
Solution:
Let $a = 30$ and $b =15$
$a\hspace{1mm}-\hspace{1mm}b = 30 \hspace{1mm}-\hspace{1mm}15 = 15$
$| a\hspace{1mm} -\hspace{1mm} b |= | 15 | = 15$
$\dfrac{(a\hspace{1mm}+\hspace{1mm}b)}{2} = \frac{30\hspace{1mm} +\hspace{1mm} 15}{2} = \frac{45}{2} = 22.5$
$Percent\hspace{1mm} difference = [\dfrac{\left | a\hspace{1mm}-\hspace{1mm}b \right |}{(a\hspace{1mm}+\hspace{1mm}b)/2}]\times 100$
$Percent\hspace{1mm} difference = [\dfrac{\left | 15 \right |}{22.5}]\times 100$
$Percent \hspace{1mm}difference = 0.666\times 100 = 66.7\%$
### Percent Difference vs. Percent Change:
A related concept to percent difference is percent change, and it is very easy to confuse the two. In this section, we will clear the difference between these two concepts.
The formula for percentage difference is given as.
$\mathbf{Percentage\hspace{2mm} Difference = [\dfrac{\left | a-b \right |}{(a+b)/2}]\times 100 }$
The formula for percentage change is given as.
$\mathbf{Percentage\hspace{2mm} Change = [\dfrac{x2 -x1}{\left | x1 \right |}]\times 100 }$
Here,
x1 = Initial value.
x2 = Final value.
| x1 |= Absolute Initial Value
For example, you are given two numbers. The initial number is = 30, and the final number is = 20, and you are required to calculate the percent difference between these two numbers.
Let $a = 30$ and $b =20$
$a\hspace{1mm}-\hspace{1mm}b = 30 \hspace{1mm}-\hspace{1mm} 20 = 10$
$| a\hspace{1mm} -\hspace{1mm} b |= | 10 | = 10$
$\dfrac{(a\hspace{1mm}+\hspace{1mm}b)}{2} = \dfrac{(30\hspace{1mm} + \hspace{1mm}20)}{2} = \dfrac{50}{2} = 25$
$Percent\hspace{1mm} difference = [\dfrac{\left | 10 \right |}{25}]\times 100$
$Percent \hspace{1mm}difference = 0.4\times 100 = 40\%$
Let us now swap the values of both variables and see the result
Let $a = 20$ and $b =30$
$a\hspace{1mm}-\hspace{1mm}b = 20\hspace{1mm} – \hspace{1mm}30 = -10$
$| a\hspace{1mm} – \hspace{1mm}b |= | -10 | = 10$
$\dfrac{(a\hspace{1mm}+\hspace{1mm}b)}{2} = \dfrac{(20\hspace{1mm}+\hspace{1mm}30)}{2} = \dfrac{50}{2} = 25$
$Percent\hspace{1mm} difference = [\dfrac{\left | 10 \right |}{25}]\times 100$
$Percent\hspace{1mm} difference = 0.4\times 100 = 40\%$
So, the percentage difference between any two numbers will remain the same even if initial and final values are swapped with each other.
Let us now calculate the percentage change for the same example.
Let initial value $x1 = 30$ and final value $x2 =20$
$x2-x1 = 20 – 30 = – 10$
$| x1 |= | 30 | = 30$
$Percent\hspace{1mm} change = [\dfrac{ – 10 }{30}]\times 100$
$Percent\hspace{1mm} change = -0.333\times 100 = -33.3\%$ or $33.3 \%$ decrease in the value.
Let us now swap the values of both variables, initial value = 20 and final value = 30 and see the result
Let initial value $x1 = 20$ and final value $x2 =30$
$x2\hspace{1mm}-\hspace{1mm}x1 = 30 \hspace{1mm}-\hspace{1mm} 20 = 10$
$| x1 |= | 20 | = 20$
$Percent\hspace{1mm} change = [\dfrac{ 10 }{20}]\times 100$
$Percent\hspace{1mm} change = 0.5\times 100 = 50\%$ or $50\%$ increase in the value.
The above example should have cleared the confusion between percentage difference and percentage change and it also explains that percent difference does not tell us the direction of difference, i.e., which variable had a positive or negative percentage change as compared to the other. This directional difference is captured in percentage change.
### Percent Difference between Two Numbers
So far, we have studied how to calculate the percent difference between two numbers. But a question arises when is it feasible to use the percent difference between two numbers?
### Real-Life Examples of Percent Difference
• Let us look at some real-life examples and see where we can apply the method of a percent difference. Let us assume we have two sections of the 2nd-grade class, section “A” and section “B”; section A has a strength of $35$ students while section B has a strength of $45$ students. In this case, we are comparing the strengths of two sections of the same class so we can easily apply the percent difference method as it will tell us about the percentage difference of class strengths between the two sections. The percent difference between the two sections is $25\%$.
• Let’s take another example and assume class A had $20$ students in January, and in three months, the class strength increased to $40$. In this case, we again have two numbers, $20$ and $40$, but it’s the same section, and the use of percentage change is suitable for this kind of example. The percent change shows that there has been a $100\%$ increase in the class strength. So, for a scenario that deals with an original value and an updated new value, We should use the percentage change to calculate the percentage increase or decrease. In contrast, percentage difference should be used when comparing the same thing, for example, comparing prices of two Toyota cars.
• Similarly, there is a difference between percent error and percent difference as well. Therefore, when comparing actual and estimated values, we will use percentage error to calculate this scenario’s percent error.
### Limitation of Percent Difference
• The percent difference method has its limitation, and they are prominent when the difference between values of two numbers is very high. For example, suppose a multinational company consists of two major departments A) HR department B) Technical department. Now suppose in the year $2019$, the total number of employees working in the “HR department” were $500$ and in the “Technical department” were $900$. Thus, the percent difference between the two departments was approx—$57\%$.
• Assume that the company hires $100,000$ more technical staff in the year $2020$ while the number of staff in the” HR department” remains the same. Thus, the total number of employees in the “Technical department” would be $100,900$ and the percent difference for the year $2020$ would be $198\%$.
• Assume the company hires a further $100,000$ technical staff in 2021 while no recruitment is done for the “ HR department.” The total number of employees in the “Technical department” would be $200,900$ and the percent difference for the year $2021$ would be $199\%$. As we can see, there is not much of a difference between the percentage difference values of the year $2020$ and $2021$ even after hiring further $100,000$ persons. This indicates the limitation of a percent difference, i.e., whenever the difference of values between two numbers is huge, the percent difference may not be ideal for comparison. As the difference in the value of two numbers increases, the absolute difference also increases with it. Still, its effect is very little or negligible on percent difference because we are diving with the average of the two numbers.
Now that we have studied the percent difference and its limitations. The flow chart for the calculation of the percent difference is given below.
Example 2: Car “A” is moving at $50$ Miles Per Hour, and car “B” is moving at $70$ Miles Per Hour. Calculate the percent difference of speed between these two cars.
Solution:
$a = 50$ and $b = 70$
$a\hspace{1mm}-\hspace{1mm}b = 50 \hspace{1mm}- \hspace{1mm}70 = -20$
$| a\hspace{1mm} – \hspace{1mm}b |= | -20 | = 20$
$\dfrac{(a\hspace{1mm}+\hspace{1mm}b)}{2} = \frac{(50\hspace{1mm}+\hspace{1mm}70)}{2} = \frac{120}{2} = 60$
$Percent\hspace{1mm} difference = [\dfrac{\left | a\hspace{1mm}-\hspace{1mm}b \right |}{(a\hspace{1mm}+\hspace{1mm}b)/2}]\times 100$
$Percent\hspace{1mm} difference = [\dfrac{\left | 20 \right |}{60}]\times 100$
$Percent \hspace{1mm}difference = 0.333\times 100 = 33.3\%$
Example 3: Calculate the percent difference between the numbers in the table given below.
Solution:
• $a = 200$ and $b = 300$
$a\hspace{1mm}-\hspace{1mm}b = 200\hspace{1mm} -\hspace{1mm} 300 = -100$
$| a\hspace{1mm} -\hspace{1mm} b |= | -100 | = 100$
$\dfrac{(a\hspace{1mm}+\hspace{1mm}b)}{2} = \dfrac{(200\hspace{1mm}+\hspace{1mm}300)}{2} = \dfrac{500}{2} = 250$
$Percent\hspace{1mm} difference = [\dfrac{\left | a\hspace{1mm}-\hspace{1mm}b \right |}{(a\hspace{1mm}+\hspace{1mm}b)/2}]\times 100$
$Percent\hspace{1mm} difference = [\dfrac{\left | 100 \right |}{250}]\times 100$
$Percent \hspace{1mm}difference = 0.4\times 100 = 40\%$
• Let $a = 800$ and $b = 400$
$a\hspace{1mm}-\hspace{1mm}b = 800\hspace{1mm} – \hspace{1mm}400 = 400$
$| a\hspace{1mm} -\hspace{1mm} b |= | 400 | = 400$
$\dfrac{(a\hspace{1mm}+\hspace{1mm}b)}{2} =\dfrac{(800\hspace{1mm}+\hspace{1mm}400)}{3} = \frac{1200}{2} = 600$
$Percent\hspace{1mm} difference = [\dfrac{\left | a\hspace{1mm}-\hspace{1mm}b \right |}{(a\hspace{1mm}+\hspace{1mm}b)/2}]\times 100$
$Percent\hspace{1mm} difference = [\dfrac{\left | 400 \right |}{600}]\times 100$
$Percent\hspace{1mm} difference = 0.666\times 100 = 66.7\%$
• Let $a = 600$ and $b = 1800$
$a\hspace{1mm}-\hspace{1mm}b = 600\hspace{1mm} – \hspace{1mm}1800 = – 1200$
$| a \hspace{1mm}-\hspace{1mm} b |= | -1200 | = 1200$
$\dfrac{(a\hspace{1mm}+\hspace{1mm}b)}{2} = \dfrac{(600\hspace{1mm}+\hspace{1mm}800)}{2} = \frac{2400}{2} = 1200$
$Percent\hspace{1mm} difference = [\dfrac{\left | a\hspace{1mm}-\hspace{1mm}b \right |}{a+b/2}]\times 100$
$Percent\hspace{1mm} difference = [\dfrac{\left | 1200 \right |}{1200}]\times 100$
$Percent\hspace{1mm} difference = 1\times 100 = 100\%$
• Let $a = 6000$ and $b = 2000$
$a\hspace{1mm}-\hspace{1mm}b = 6000\hspace{1mm} – \hspace{1mm}2000 = 4000$
$| a\hspace{1mm} – \hspace{1mm}b |= | 4000 | = 4000$
$d\frac{(a\hspace{1mm}+\hspace{1mm}b)}{2} = \dfrac{(6000\hspace{1mm}+\hspace{1mm}2000}{2} = \dfrac{8000}{2} = 4000$
$Percent\hspace{1mm} difference = [\dfrac{\left | a\hspace{1mm}-\hspace{1mm}b \right |}{(a\hspace{1mm}+\hspace{1mm}b)/2}]\times 100$
$Percent\hspace{1mm} difference = [\dfrac{\left | 4000 \right |}{4000}]\times 100$
$Percent\hspace{1mm} difference = 1\times 100 = 100\%$
Example 4: Adam has scored 300 goals in his entire football career while Steve has scored 100 goals. Calculate the percentage difference of goals between these two players
Solution:
Let $a = 300$ and $b = 100$
$a\hspace{1mm}-\hspace{1mm}b = 300\hspace{1mm} – \hspace{1mm}100 = -200$
$| a\hspace{1mm} – \hspace{1mm}b |= | -200 | = 200$
$\dfrac{(a\hspace{1mm}+\hspace{1mm}b)}{2} = \dfrac{(100\hspace{1mm}+\hspace{1mm}300)}{2}= \dfrac{400}{2} = 200$
$Percent\hspace{1mm} difference = [\dfrac{\left | a\hspace{1mm}-\hspace{1mm}b \right |}{(a\hspace{1mm}+\hspace{1mm}b)/2}]\times 100$
$Percent\hspace{1mm} difference = [\dfrac{\left | 200 \right |}{200}]\times 100$
$Percent\hspace{1mm} difference = 1\times 100 = 100\%$
If we analyze example 3 and the last two rows of the table in example number 2, we can clearly see that if one number is 3 times greater than the other number, the percent difference is always 100%. Let us prove this in the following example.
Example 5: Prove that when $a = 3b$, the percent difference equals $100\%$.
Solution:
$Percent\hspace{1mm} difference = [\dfrac{\left | a\hspace{1mm}-\hspace{1mm}b \right |}{(a\hspace{1mm}+\hspace{1mm}b)/2}]\times 100$
When percent difference is $= 100\%$
$| a \hspace{1mm}-\hspace{1mm} b |= \dfrac{(a\hspace{1mm}+\hspace{1mm}b)}{2}$
$2\times (a\hspace{1mm}-\hspace{1mm}b) = a\hspace{1mm}+\hspace{1mm}b$
$2a\hspace{1mm} -\hspace{1mm}2b = a\hspace{1mm} + \hspace{1mm}b$
$a = b\hspace{1mm} +\hspace{1mm}2b$
$a =3b$
### Practice Questions
1. Annie is $25$ years old, and her friend Naila is $13$ years old. Which of the following shows the percent difference of age between these two friends?
2. Allan and his friend Mike are both athletes and do running practice daily to compete for the upcoming Olympics events. Allan and Mike run for a distance of $20$ and $30$KM daily. Which of the following shows the percent difference of distance covered by these two friends?
3. Building “A” has a height of $250$ ft, and building “B” has a height of $700$ ft. Which of the following shows the percent difference of height between these two buildings?
4. Michael and Oliver recently joined a new organization as HR manager and deputy manager, respectively. Michael worked for $280$ hrs, and Oliver worked for $200$ hrs during their first month of the job. Which of the following shows the percent difference of job hours of these two friends?
5. Harold read $24$ books while Robin read $36$ books. Which of the following shows the percent difference of the number of books read between the two friends?
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Interesting Math Problem Math Problems Home | Home | Send Feedback
An election, with a twins twist
A city council consists of 10 members, from whom will be selected 4 officers: President, Vice-President, Secretary, and Ombudsman. But two of the members of the council are identical twins, and no one can tell them apart! If the twins are "interchangeable", whether elected to an office or not, in how many ways can the positions be filled ?
(Historical note: In 2006-2007, the identical twin Kaczynski brothers were President and Prime Minister of Poland.)
Normal straightforward solution:
Case 0:
If neither twin is elected, we have 8 * 7 * 6 * 5 = 1680 arrangements.
Case 1:
If one twin is elected, we have 4 positions for the twin * 8 * 7 * 6 ways to fill the other positions = 1344 arrangements.
Case 2:
If both twins are elected, we have 4C2 = 6 choices of positions for them * 8 * 7 = 6 * 56 = 336 arrangements.
Total of all three cases: 3360.
More interesting solution:
Total, unrestricted arrangements = 10 * 9 * 8 * 7 = 5040.
Since 1680 have neither twin (as found at left), that leaves 3360.
For each of those, write the names of those elected
in order by office. If only one twin is elected, add the other's name
as a fifth name, so both are always listed.
Suppose the twins are named Alice and Beth.
In half the arrangements, Alice's name is first,
and in the other half, Beth's name is first.
Since it doesn't matter which one comes first
(for either of two offices, or one elected and the other not),
toss out the half where Beth's name is before Alice's.
That leaves 3360 / 2 = 1680 for these cases.
The total then is 1680 "no twin" cases + 1680 "one or two twin" cases,
and again we have the total of 3360.
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Equivalent Resistance
Spring 2019
Equivalent Resistance
Use Series, Parallel, and Ohm's law to find Equivalent Resistance
Learn It!
Pre-Requisite Knowledge
Goal
Find the equivalent resistance.
This group of resistors will act like one equivalent resistor. What is its resistance?
Part 1
Redraw The Circuit
+
1A
Circuit transformation
Now our circuit is nicely transformed. After doing this a few times, you will be able to do all this in your head. It just takes a little practice.
+
!
Are the legal moves 'real'
Part 2
Finding Equivalent Resistance
Use parallel and series formulas to decompose circuits one piece at a time.
+
2A
Reduce two series resistors on the right
$R_{eq1} = R_1 + R_2 = 4+2 = 6$
Now our circuit is nicely transformed. After doing this a few times, you will be able to do all this in your head. It just takes a little practice.
+
2B
Reduce two parallel resistors on the left
$R_{eq2} = \dfrac{1}{\dfrac{1}{R_1} + {\dfrac{1}{R_2}}} = \dfrac {1} {\dfrac{1}{20} + \dfrac{1}{30}} = 12$
Apply the parallel resistance formula! Note that our answer is reasonable: a parallel configuration reduces resistance.
+
2C
Reduce the series resistors on the left
$R_{eq3} = 8 +12 = 20$
Apply the series equivalent resistance formula.
+
2D
Combine three parallel resistors
$R_{eq} = \dfrac{1}{\dfrac{1}{20} + \dfrac{1}{10} + \dfrac{1}{6}} = 3.157 \Omega$
$R_{eq} = \dfrac{1}{\dfrac{1}{20} + \dfrac{1}{10} + \dfrac{1}{6}} = 3.157 \Omega$
Apply the parallel resistance formula to three resistors. We have succeeded! The whole mess we started with behaves as though it were a single resistor with a resistance of about $3.1 \Omega$.
+
!
Can all circuits be broken down with series and parallel?
|
# Dividing 3-Digit Numbers by 2-Digit Numbers
## Before we begin:
Before proceeding with this lesson, it would be helpful for you ;
Let's divide 672 by 42.
We start from the largest place values. How many times does 42 fit into 67? You have to approach this with a bit of estimation. If there were two, it would be 40 + 40 = 80 (in fact, the actual total would be even more), which is more than 67, so we can say there's one.
I'm writing 1 over the 7.
I'm multiplying 1 by 42 and subtracting the result from 67.
I'm bringing down 2 next to 25.The number becomes 252.
Now I need to divide 252 by 42, similar to before, I should think about how many times 42 fits into 252.
If you're having trouble figuring out how many there are, try to estimate by rounding the numbers.
When rounding, you don't necessarily have to take rounding rules into account. For example, 252 is rounded to 250, but thinking of it as 260 is not that important, because your aim is to find an approximate result, you don't necessarily have to make the best estimate. You can try, and depending on your result, you can give one more or one less and recalculate.
We estimated that there are six. We need to multiply 6 by 42 and look at our result. The result can be larger than 252 (in this case, we need to try a smaller number), it can be smaller (in this case, we need to try a larger number), or if we are lucky, we can find exactly 252.
6 times is a perfect fit ! so write this to the quotient, above the number 2.
I need to multiply 6 by 42 and subtract it from 252 to find out how much more I need to divide, or to see if I have a remainder left.
I have no remainder, the result is 16.
our teachers have taught you all these steps, or you have learned them through YouTube channels or websites, now let's examine it in a bit more detail.
# Actually, what is happening
What does 'how many 42s are in 67' mean? Is 67 just the number we know, or does it represent something else?
67 does not represent the 67 we know, the number 672 is a number composed of 6 hundreds, 7 tens, and 2 ones. If we break down the hundreds, one hundred means ten tens, so 6 hundreds would be 60 tens, and I already have 7 tens, so I can think of it as having a total of 67 tens,
I'm distributing 67 tens into 42 buckets.. one ten falls into each bucket, so when I write 1 in the quotient, I'm not writing it to another digit, I'm writing it to the tens place.
I multiply 1 by 42 and subtract it from 67 to find out how many tens I have left.I have 25 left..
25 tens cannot be distributed equally into 42 buckets.That's why I need to convert these 25 tens to ones. 25 tens make 250 ones, in fact, this number you see doesn't represent 25, it represents 250.
I had two more ones on hand, and when I bring down those two ones, bingo! My count becomes 252, so I hit two birds with one stone. I'm both taking into account the two ones and converting the 25 tens into ones.
Now I need to distribute these 252 ones into 42 buckets again..
You need to use estimation to find out how many can go into each bucket. This depends on your skill and how much practice you have done. If you practice a lot, your estimation ability will improve.
You can make the best and fastest estimates by rounding the numbers. As I said before, you don't have to stick to the rounding rules, it's an estimate after all! You can estimate a little more or less. Estimation should not be an extra problem for you to overcome.
We estimated six, and when we tried it, we saw that it fit exactly. I'm writing 6 in the ones place in the quotient because I have six ones in each of my buckets.
Since I've distributed the ones, to find out how many ones I have left, I'm multiplying 6 by 42 to calculate the distributed ones. Distributing 6 ones into each of 42 buckets means 6 * 42 = 252 ones have been distributed.
Since I've distributed the ones, to find out how many ones I have left, I'm multiplying 6 by 42 to calculate the distributed ones. Distributing 6 ones into each of 42 buckets means 6 * 42 = 252 ones have been distributed.
# Alternative Ways
In fact, when dividing 672 by 42, we are looking for the answer to the question of how many 42s there are in 672. You can think of this division as taking large or small bites from a big meal, piece by piece.
Let's try to estimate by multiplying 42 by 10, 100, or 1000. For example, multiplying by 10 makes sense, as multiplying by 100 would give 4200, which is too much. Why am I trying to estimate by taking 10, 100, or 1000? Because multiplying by these numbers is simple, so we can estimate easily.
Twenty times will also be too much, it would result in a number in the 800s (400+400), whereas the number I want to divide starts with 600, so ten times seems to be the ideal figure.
Let's take another bite, 10 will be too much, let's give it 5.
total 16 times..
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## Spanning trees
Use determinant to calculate the number of spanning trees of the following graphs:
• #### Solution
The Laplace’s matrix of the graph is
$$L_G=\begin{pmatrix} 4 & -1 & -1 & -1 & -1 \\ -1 & 4 & -1 & -1 & -1 \\ -1 & -1 & 3 & -1 & 0 \\ -1 & -1 & -1 & 3 & 0 \\ -1 & -1 & 0 & 0 & 2 \\ \end{pmatrix}$$
By the theorem on the number of spanning trees
$$\kappa(G)=\det L_G^{11}= \begin{vmatrix} 4 & -1 & -1 & -1 \\ -1 & 3 & -1 & 0 \\ -1 & -1 & 3 & 0 \\ -1 & 0 & 0 & 2 \\ \end{vmatrix} = \begin{vmatrix} 4 & -1 & -1 & -1 \\ -1 & 3 & -1 & 0 \\ -1 & -1 & 3 & 0 \\ 7 & -2 & -2 & 0 \\ \end{vmatrix} = \begin{vmatrix} -1 & 3 & -1 \\ -1 & -1 & 3 \\ 7 & -2 & -2 \\ \end{vmatrix}$$
$$= \begin{vmatrix} -1 & 3 & -1 \\ 0 & -4 & 4 \\ 0 & 19 & -9 \\ \end{vmatrix} =-4 \begin{vmatrix} -1 & 1 \\ 19 & -9 \\ \end{vmatrix} =-4(9-19)=40$$.
It is possible to verify the result by a listing of all spanning trees: $$K_4$$ has 16 spanning trees and each has two options to join the top vertex (the left or the right edge) – in total 32 possibilities. Otherwise the spanning tree contains both edges incident with the top vertex and then we have 4 options that contain the bottom edge an 4 that do not.
The graph has 40 spanning trees.
• #### Solution
Laplace’s matrix: $$L_G= \begin{pmatrix} 5 & -2 & 0 & -1 & 0 & -2 \\ -2 & 5 & -2 & 0 & -1 & 0 \\ 0 & -2 & 5 & -2 & 0 & -1 \\ -1 & 0 & -2 & 5 & -2 & 0 \\ 0 & -1 & 0 & -2 & 5 & -2 \\ -2 & 0 & -1 & 0 & -2 & 5\\ \end{pmatrix}$$
Number of spanning trees: $$\kappa(G)=|L_G^{11}|= \begin{vmatrix} 5 & -2 & 0 & -1 & 0 \\ -2 & 5 & -2 & 0 & -1 \\ 0 & -2 & 5 & -2 & 0 \\ -1 & 0 & -2 & 5 & -2 \\ 0 & -1 & 0 & -2 & 5\\ \end{vmatrix} = 960$$
|
How to Merging of Exponents and Roots
Merging Exponents and Roots
In this section, we shall use the properties of integral exponents to motivate definitions for the use of rational numbers as exponents. These definitions will tie together the concepts of exponent and root.
Let’s consider the following comparisons.
From our study of radicals we know that
$(\sqrt{5})^2=5$
If $(b^n)^m=b^{mn}$ is to hold when n equals a rational number of the form $\frac{1}{p}$, where p is a positive integer greater than one, then
$(5^{\frac{1}{2}})^2=5^{2(\frac{1}{2}}=5^1=5$
It would seem reasonable to make the following definition.
Definition 1 :
If b is a real number, n a positive integer greater than one, and $\sqrt[n]{b}$ exists, then $b^{\frac{1}{n}}=\sqrt[n]{b}$.
Definition 1 states that : $b^{\frac{1}{n}}$ means the nth root of b. We shall assume that b and n are chosen so that $\sqrt[n]{b}$ exists.
For example, $(-25)^{\frac{1}{2}}$ is not meaningful at this time because $\sqrt{-25}$ is not a real number.
Consider the following examples that illustrate the use of definition 1.
$25^{\frac{1}{2}}=\sqrt{25}=5$
$8^{\frac{1}{3}}=\sqrt[3]{8}=2$
$(-27)^{\frac{1}{3}}=\sqrt[3]{-27}=-3$
$16^{\frac{1}{4}}=\sqrt[4]{16}=2$
$(\frac{36}{49})^{\frac{1}{2}}=\sqrt{\frac{36}{49}}=\frac{6}{7}$
The following definition provides the basic for the use of all rational numbers as exponents.
Definition 2 :
If $\frac{m}{n}$ is a rational number, where n is a positive integer greater than one, and b is a real number such that $\sqrt[n]{b}$ exists, then
$b^{\frac{m}{n}}=\sqrt[n]{b^m}=(\sqrt[n]{b})^m$
In definition 2, notice that denominator of the exponent is the index of the radical and the numerator of the exponent is either the exponent of the radicand or the exponent of the root.
Whether we use the form $\sqrt[n]{b^m}\text{ or }(\sqrt[n]{b})^m$ for computational purposes depends somewhat on the magnitude of the problem. Let’s use both forms on two problems to illustrate this point.
$8^{\frac{2}{3}}=\sqrt[3]{8^2}=\sqrt[3]{64}=4$
Or
$8^{\frac{2}{3}}=(\sqrt[3]{8})^2=2^2=4$
$27^{\frac{2}{3}}=\sqrt[3]{27^2}=\sqrt[3]{729}=9$
Or
$27^{\frac{2}{3}}=(\sqrt[3]{27})^2=3^2=9$
To compute $8^{\frac{2}{3}}$, either form seems to work about as well as the other one. However, to compute $27^{\frac{2}{3}}$, it should be obvious that $(\sqrt[3]{27})^2$ is much easier to handle than $\sqrt[3]{27^2}$.
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# Special Products
Special Products are the products of binomials that can be simplified further than regular products. A majority of these products can be expanded and simplified using the FOIL method covered in Expanding Polynomials.
## Distributive Law
Distributive Law involves products expressed in the form:
$$a(x + y)$$
These products can be expanded and simplified using the following process:
$$= (a)(x) + (a)(y)$$
$$= ax + ay$$
Example
Expand and simplify the expression $$5(a + 7)$$.
$$= (5)(a) + (5)(7)$$
$$= 5a + 35$$
Therefore, we can determine that $$5(a + 7)$$ expanded and simplified is $$5a + 35$$.
## Difference of 2 Squares
The Difference of Squares involves products expressed in the form:
$$(x + y)(x - y)$$
These products can be expanded and simplified using the following process:
$$= (x)(x) + (x)(-y) + (y)(x) + (y)(-y)$$
$$= x^2 - xy + xy - y^2$$
$$= x^2 - y^2$$
Example
Simplify the expression $$(6g + 7h)(6g - 7h)$$.
We can easily simplify the expression using the formula $$x^2 - y^2$$:
$$= (6g)^2 - (7h)^2$$
$$= 36g^2 - 49h^2$$
Therefore, we can determine that $$(6g + 7h)(6g - 7h)$$ simplified is $$36g^2 - 49h^2$$.
## Square of a Sum
The Square of a Sum involves products expressed in the form:
$$(x + y)^2$$
These products can be expanded and simplified using the following process:
$$= (x + y)(x + y)$$
$$= (x)(x) + (x)(y) + (y)(x) + (y)(y)$$
$$= x^2 + 2xy + y^2$$
Example
Expand and simplify the expression $$(4t + 3s)^2$$.
We can easily simplify the expression using the formula $$x^2 + 2xy + y^2$$:
$$= (4t)^2 + 2(4t)(3s) + (3s)^2$$
$$= 16t^2 + 24st + 9s^2$$
Therefore, we can determine that $$(4t + 3s)^2$$ expanded and simplified is $$16t^2 + 24st + 9s^2$$.
## Square of a Difference
The Square of a Difference involves products expressed in the form:
$$(x - y)^2$$
These products can be expanded and simplified using the following process:
$$= (x - y)(x - y)$$
$$= (x)(x) + (x)(-y) + (-y)(x) + (-y)(-y))$$
$$= x^2 -2xy + y^2$$
Example
Expand and simplify the expression $$(5q - 8)^2$$.
We can easily simplify the expression using the formula $$x^2 -2xy + y^2$$:
$$= (5q)^2 + (2)(5q)(8) + (8)^2$$
$$= 25q^2 - 80q + 64$$
Therefore, we can determine that $$(5q - 8)^2$$ expanded and simplified is $$25q^2 - 80q + 64$$.
Expand and simplify the following expressions:
$$(5x^2 + 7y^2)^2$$
$$(x - 3)^2 - (x + 3)(x - 3)$$
$$(3x^2 + 5x - 1)^2$$
$$(2x - 3)^3$$
The side length of a square is represented by $$x\;[cm]$$. The length of a rectangle is $$3\;[cm]$$ greater than the side length of the square. The width of the rectangle is $$3\;[cm]$$ less than the side length of the square. Which figure has the greater area and by how much?
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## Monday, September 25, 2017
### Middle School Math Solutions – Expand Calculator, Distributive Law
The distributive law helps with multiplication problems by breaking down large numbers into smaller numbers. In this blog post, we will talk about the distributive law and how to use it.
The law says that multiplying a number by a group of numbers added together is the same as multiplying each separately. What does this mean? Here we can see it in a formula:
You can see that the multiplication of a has been distributed among the sum of b and c.
Let’s see some examples of how to use the distributive law.
Expand 4(2x+5)
1. Define the values for a, b, c
a=4, b=2x, c=5
Refer to the formula to see what these values are.
2. Plug these values into the distributive law formula
4(2x+5)=4∙2x+4∙5
4∙2x+4∙5=8x+20
Expand 5(10-9p)
1. Define the values for a, b, c
a=5, b=10, c=-9p
2. Plug these values into the distributive law formula
5(10-9p)=5∙10+5∙(-9p)
5∙10+5∙(-9p)=50-45p
In this last example, we will see another application of the distributive law.
5(106)
Just by looking at this problem, this might be difficult to calculate quickly. So let’s use the distributive law.
1. Make 106 the sum of two numbers
106=100+6
2. Plug this in for 106
5(100+6)
3. Define values a, b, c
a=5, b=100, c=6
4. Plug these into the distributive law formula
5(100+6)=5∙100+5∙6
5∙100+5∙6=500+30=530
As you can see, the distributive law is very handy. Therefore, I highly suggest you memorize this formula and become familiar with how to use it. You can do so by doing as many examples as you can. For practice examples and more help, check out Symbolab’s Practice
Until next time,
Leah
## Tuesday, September 19, 2017
### High School Solutions – Functions Calculator, Inverse
Last blog posts, we focused on how to find the domain and range of functions. In this blog post, we will discuss inverses of functions and how to find the inverse of a function.
An inverse of a function f(x), denoted f^(-1)(x), is a function that reverses or undoes f(x). What does that mean? This means that the domain or inputs of a function is the range or output of the function’s inverse and the range or outputs of a function is the domain or inputs of the function’s inverse. If f(x)=y, then f^(-1)(y)=x. Let’s see some pictures to better understand.
It is important to note that some functions have more than one inverse. For example, quadratic equations have two inverses because the negative and positive value for an input goes to the same y value (For x^2+4, when x = 1 and x = -1 , we get y = 5). When you find the inverse of quadratics, you’ll notice you get two inverses, one is a postive square root and the other is a negative square root. We will see an example of this later in the post.
Now, that you’ve got the concept of what the inverse of a function is, we will see the steps on how to find the inverse of a function.
Steps to find the inverse of a function:
1. Replace y for f(x)
2. Solve for x
3. Substitute y = x
4. If you want to check the function, then f(f^(-1) (x))=x and f^(-1) (f(x))=x
The steps are pretty simple to remember and follow. Now, let’s see some examples.
Find the inverse of f(x)=3x+5
1. Replace y for f(x)
y=3x+5
2. Solve for x
y=3x+5
y-5=3x
\frac{y-5}{3}=x
3. Substitute y = x
\frac{x-5}{3}=y=f^(-1) (x)
4. Check to make sure it is correct (f(f^(-1)(x))=x and f^(-1)(f(x))=x)
f^(-1)(\f(x))=\frac{(3x+5)-5}{3}=\frac{3x}{3}=x
f(f^(-1)(x))=3(\frac{x-5}{3})+5=x-5+5=x
Find the inverse of f(x)=\sqrt{x+3}
1. Replace y for f(x)
y=\sqrt{x+3}
2. Solve for x
y=\sqrt{x+3}
y^2=x+3
y^2-3=x
3. Substitute y = x
x^2-3=y=f^(-1)(x)
4. Check to make sure it is correct (f(f^(-1)(x))=x and f^(-1)(f(x))=x)
f^(-1)(\f(x))=(\sqrt{x+3})^2-3=x+3-3=x
f(f^(-1)(x))=\sqrt{(x^2-3)+3}=\sqrt{x^2}=x
Last example (
Find the inverse of f(x)=2x^2-2
1. Replace y for f(x)
y=2x^2-2
2. Solve for x
y+2=2x^2
\frac{y+2}{2}=x^2
±\sqrt{\frac{y+2}{2}}=x
3. Substitue y = x
±\sqrt{\frac{x+2}{2}}=y=f^(-1)(x)
4. Check to make sure it is correct (f(f^(-1)(x))=x and f^(-1)(f(x))=x)
You can do this step on your own. We will skip it to save some time.
As you can see, finding the inverse of a function is pretty simple. It is easy to make algebraic answers, so make sure you check your answer to see if it is correct. With practice, you’ll be able to master this topic easily. For more help or practice on this topic, check out Symbolab’s Practice.
Until next time,
Leah
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»»» 8 Circle Theorems Visualised and Explained – GCSE Maths
# 8 Circle Theorems Visualised and Explained – GCSE Maths
Let’s learn 8 circle theorems – prepare for GCSE maths and A-level maths.
## What is a Circle Theorem?
A circle theorem is a mathematical statement that describes a property related to the geometry of a circle and the lines, angles, and other shapes associated with it.
These theorems are proven statements that reveal the relationships between different parts of a circle and provide a way to solve geometric problems without the need for direct measurement.
## 1. Angle at the Centre Theorem
This theorem states that the angle subtended at the centre of a circle by two given points on the circumference is precisely twice the angle subtended at the circumference by those same points. If A, B, and C are points on the circumference of a circle with the centre O, and ∠AOB is an angle at the centre, while ∠ACB is an angle at the circumference, then the theorem can be expressed mathematically as:
AOB=2×∠ACB
## 2. Angles in the Same Segment Theorem
The theorem asserts that angles subtended by the same arc at the circumference of a circle are equal. In essence, if two angles are inscribed on the circumference of a circle and subtended by the same arc, they will be congruent.
This theorem states that the sum of the opposite angles of a cyclic quadrilateral is 180 degrees. The theorem is significant because it applies to any quadrilateral where all four vertices lie on the circumference of a circle.
If we have a cyclic quadrilateral ABCD, with the vertices A, B, C, and D resting on the circumference of a circle, the theorem can be expressed as:
A+∠C=180∘ , ∠B+∠D=180∘
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Maths Tutor
Student at UNIVERSITY OF WARWICK
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## 4. Angle in a Semicircle Theorem
The theorem posits that the angle inscribed in a semicircle is a right angle. That is to say, if an angle is subtended at a point on the circumference of a circle by a diameter of that circle, then the angle is always 90 degrees.
If A, P, and B are points on a circle such that AB is a diameter and P lies on the circumference, then the angle ∠APB inscribed in the semicircle is a right angle.
APB=90∘
The theorem states that a tangent to a circle is perpendicular to the radius drawn to the point of tangency. In simpler terms, the angle formed between a tangent line and a radius at the point where the tangent meets the circle is always a right angle (90 degrees).
If ST is a tangent to a circle at point C, and OC is the radius drawn to the point of tangency C, then the theorem can be mathematically expressed as:
OCT=90∘
## 6. Two Tangents from a Point Theorem
This theorem asserts that if two tangents are drawn to a circle from a single external point, then these two tangents are equal in length. Additionally, the line segment from the external point to the circle’s centre bisects the angle formed between the two tangents.
If PA and PB are two tangents to a circle from an external point P, and O is the centre of the circle, then the theorem can be mathematically stated as:
PA=PB
Moreover, if OP is the line segment from the external point to the circle’s centre, then:
APB=2×∠AOB
## 7. Chord Bisects Diameter
The theorem posits that if a chord is perpendicular to a diameter of a circle at its midpoint, then it bisects the diameter, and conversely, if a chord bisects a diameter, then it must be perpendicular to that diameter.
So, If BC is perpendicular to OD, then: BD=DC and OD bisects BC.
## 8. Alternate Segment Theorem
This theorem states that the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment of the circle. The alternate segment is defined by the chord and the arc opposite to the point of contact of the tangent.
∠B=∠D
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## Self Testing – Circle Theorems
### 1. In a circle, an angle at the circumference subtended by an arc is 35∘. What is the measure of the angle at the centre subtended by the same arc? A) 17.5∘ B) 35∘ C) 70∘ D) 140∘
C) 70∘
According to the Angle at the Centre Theorem, the angle at the centre is twice the angle at the circumference. Therefore, if the angle at the circumference is 35∘, the angle at the centre is 2×35∘=70∘.
### 2. A triangle is inscribed in a semicircle with one of its sides as the diameter. If the diameter is 10 cm, what is the measure of the angle opposite this side? A) 30∘ B) 45∘ C) 60∘ D) 90∘
D) 90∘
The Angle in a Semicircle Theorem states that an angle inscribed in a semicircle is a right angle. Therefore, the angle opposite the diameter is 90∘.
### 3. In a circle, two angles are subtended by the same arc and are 45∘ and x∘ respectively. What is the value of x? A) 22.5∘ B) 45∘ C) 60∘ D) 90∘
B) 45∘
The Angles in the Same Segment Theorem tells us that angles in the same segment are equal. Therefore, x=45∘
### 4. In a circle, a tangent forms an angle of 30∘ with a chord. What is the angle between the chord and the radius at the point of contact? A) 30∘ B) 60∘ C) 90∘ D) 120∘
C) 90∘
The angle between a tangent and a chord is equal to the angle in the alternate segment. However, the angle between a chord and the radius at the point of contact is always 90∘ according to the Tangent-Radius Theorem.
### 5. In a cyclic quadrilateral, one of the angles is 70∘70∘. What is the measure of its opposite angle? A) 70∘ B) 110∘ C) 130∘ D) 180∘
B) 110∘
The Cyclic Quadrilateral Theorem states that the sum of the opposite angles of a cyclic quadrilateral is 180∘. So, the opposite angle is 180∘−70∘=110∘
### 6. A chord is 6 cm long and bisects the diameter of a circle. If the radius of the circle is 5 cm, what is the length of each segment of the diameter? A) 2.5 cm B) 5 cm C) 7.5 cm D) 10 cm
B) 55 cm
If a chord bisects the diameter, it divides the diameter into two equal segments. The radius being 5 cm means each segment of the diameter is also 5 cm.
### 7. In a circle, two chords of equal length subtend angles of x∘ and 40∘ at the center. What is the value of x? A) 20∘ B) 40∘ C) 60∘ D) 80∘
B) 40∘
According to the Equal Chords, Equal Angles Theorem, equal chords subtend equal angles at the center. Therefore, x=40∘.
### 8. If a tangent is drawn to a circle of radius 4 cm, what is the angle formed between the radius and the tangent? A) 30∘ B) 45∘ C) 60∘ D) 90∘
D) 90∘
The Tangent-Radius Theorem states that the angle between a tangent to a circle and the radius drawn to the point of tangency is always 90∘.
### 9. Two tangents are drawn from a point P to a circle, touching the circle at A and B. If PA=7 cm, what is the length of PB? A) 3.5 cm B) 7 cm C) 14 cm D) 28 cm
B) 7 cm
The Two Tangents from a Point Theorem states that two tangents drawn from the same external point are equal in length. Hence, PB=7 cm.
### 10. Problem: In a circle, two chords AB and CD intersect at point E. If AE=3 cm, EB=4 cm, and CE=6 cm, what is the length of ED? A) 0.5 cm B) 2 cm C) 4 cm D) 8 cm
B) 22 cm
According to the Chord Distances Theorem, AE×EB=CE×ED. Therefore, 3×4=6×ED. Solving this gives ED=2 cm.
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### GCSE Maths Formulae Students Need to Learn – Explained with Visuals
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# Area Measure, Introduction
In 2 dimensions, area is a measure of the amount of space on a flat surface, enclosed within a certain boundary.
The simplest 2D shape to consider when thinking about area measure, is probably a square or a rectangle.
The rectangle above, is 3cm high, and 4cm wide.
It is made up of 12 squares, all of which are 1cm high, and 1cm wide.
Instead of counting the squares though, any time you need to work out the area of a simple square or rectangle.
One can multiply the width by the height.
So for the rectangle above:
AREA = 3 × 4 = 12
The area of the whole rectangle is 12 square centimeters.
More examples of the area of a square and rectangle can be seen on the Area of a Rectangle/Square page.
It's the case that many of the most common plane shapes in Math have standard formulas to establish the area, making life much easier in many instances.
## Area Measure, Dimensions
It's important to always remember that when establishing the area of shapes with multiple sides such as a square or rectangle, the dimensions have to be the same on all sides.
For example a rectangle can be seen as:
To work out the area here in square centimeters, one would have to convert the width from millimeters to centimeters first, before multiplying.
20 millimeters is 2 centimeters.
AREA = 4 × 2 = 8, 8 square centimeters.
## Notation
It's also important to know the correct notation for area values.
For the shapes above, we just wrote square centimetres. However, instead of writing square centimetres, a shorter notation of cm² can be used.
Likewise if the area was in metres, m² is used.
The unit being measured, with 2 to the top right.
Example
What is the area of a rectangular garden that is 10 metres wide, and 6 metres long.
Solution
10 × 6 = 60
The area of the rectangular garden is 60m
².
## Area Units
Similar to length, area can be measured in a range of different units.
Above, we just looked at centimetres and metres, but there are other values of area measure.
Rooms in house building plans are often measured in square feet, ft².
Large land areas are usually measured in square miles, mile², or square kilometres, km².
Each time though, it’s the same principal as the first example on this page.
How many smaller squares fill up the larger surface area. But this number often doesn't turn out to be a nice round number.
› › Area Measure
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# Correctly Solve “Must Be True” Questions
by on January 9th, 2013
One of the more-feared types of GMAT Problem Solving questions is the “must be true” question with three statements; these questions often look like:
If , which of the following must be true?
I. x > y
II. x^2 > y^2
III. x/7 is an integer
A) I only
B) II only
C) III only
D) I and III
E) I, II, and III
These problems can be daunting, mainly because:
1. There are multiple problems in one, as you have three relationships you have to deal with.
2. It can be difficult to know if something is “always true” – how many possibilities do you try before you conclude that it is always true?
But rest assured – there is an efficient method for solving these questions that puts all the common GMAT trap answers firmly on your side by doing what human beings do best: be critical!
The opposite of “Must Be True” is “Could be False”, so instead of trying to prove that something is always true, you can use process of elimination by trying to find one situation in which each of the statements could be false. And the best way to do this is to go on the attack – use all the “weird” numbers that tend to trap you, as your weapons against the test. “Weird” numbers like negatives, 0, and fractions tend to give the alternative answer, so you should keep those in mind as weapons that allow you to attack that idea that a statement must be true.
In the above question, for example, your goal should be to try to disprove each statement. If you can find just one set of numbers x and y for which x is not greater than y but , you can confidently eliminate statement I. So try to attack, and use the GMAT’s tendencies against it. “Equal to” is not the same thing as “greater than”, so you actually do not have to find a case where x is less than y if you can just make them equal. With that in mind, try using the one all time “game changer” number, 0. If x and y each equal 0, the given statement is true (0 does equal 0), but both statements I and II are not, as x equals y and equals . So by going on the attack and using strange numbers to your advantage, you can quickly eliminate two statements at once. And look now at the answer choices – there isn’t a choice for “none of the above”…so the answer simply must be C.
More important than this example is the set of takeaways, so for Must Be True questions, remember:
1. Go on the attack and try to find a situation for each statement that it is not true. It is almost always easier to find one example of a “false” than it is to systematically prove that it’s always true.
2. To effectively attack, consider those “weird” numbers like negatives, nonintegers, and 0. Many fear these types of numbers as traps…but they are also your weapons against the test.
3. Consider the layout of the answer choices as an asset, too – with three statements and only five answer choices, the test cannot ask you about every possible combination, so sometimes you can save the “hardest” statement for last and end up not even having to deal with it because you have eliminated the other answer choices.
So don’t fear “Must Be True” questions – with some technique, strategy, and practice these questions that many feel must be feared can actually become those that you must get right.
Plan on taking the GMAT soon? We run a free online GMAT prep seminar every couple of weeks and be sure to check out our blog for up to date GMAT and MBA admissions advice.
• The answer C is correct only if it is mentioned in the question that x and y are integers and that is not the case here. For example, x can be 24.5 and y can be 10. This combination satisfies the equation in the question. But 24.5 is not divisible by 7.
• x is equal to 1/20 and y is equal to 1/49, then "C" is not the right answer.
• @Nithiya is 100% correct... I was also thinking in terms of fraction and found that if we consider x and y as fractions C is also not correct.
If x = 1/{49(p)}
y = 1/{20(p)}
where p is any number so it can give us many fraction values for which x/7 is not an integer.
So please mention in the question that x, y belongs to integer.
• None of the following must be true
• 20x = 49y, which is x/y = 49/20, which is x/y >1 which says x>y , but then x2 >Y2 so A and B can be true, so the question has a problem !!
• I think its A, becuase for -x>-y, x2 need not be greater than y eg: -2>-3, but then 4<9; so A is the answer 100%
• I is the correct answer
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# 2017 AMC 10A Problems/Problem 4
## Problem
Mia is "helping" her mom pick up $30$ toys that are strewn on the floor. Mia’s mom manages to put $3$ toys into the toy box every $30$ seconds, but each time immediately after those $30$ seconds have elapsed, Mia takes $2$ toys out of the box. How much time, in minutes, will it take Mia and her mom to put all $30$ toys into the box for the first time?
$\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5$
## Solution
Every $30$ seconds, $3$ toys are put in the box and $2$ toys are taken out, so the number of toys in the box increases by $3-2=1$ every $30$ seconds. Then after $27 \times 30 = 810$ seconds (or $13 \frac{1}{2}$ minutes), there are $27$ toys in the box. Mia's mom will then put the remaining $3$ toys into the box after $30$ more seconds, so the total time taken is $27\times 30+30=840$ seconds, or $\boxed{(\textbf{B})\ 14}$ minutes.
Note: During the last time Mia's mom will complete picking up the 30 toys(before Mia can take 2 out) which is the reason that you calculate up to 27 and then the rest.
## Video Solutions
https://youtu.be/str7kmcRMY8 -Video Solution by TheBeautyOfMath
~savannahsolver
2017 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
|
# Examples On Integration By Parts Set-3
Go back to 'Indefinite Integration'
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## Integration by parts examples
Example – 45
Evaluate the following integrals.
(a) \begin{align}\int {\sqrt {{x^2} + x + 1} \,\,dx} \end{align} (b) \begin{align}\int {\sqrt {1 - 2x - {x^2}} \,\,dx}\end{align} (c) \begin{align}\int {\sqrt {{x^2} + 3x + 2} \,\,dx}\end{align} (d) \begin{align}\int {x\,\,{e^x}\cos x\,\,dx} \end{align}
Solution: (a) We can write $${x^2} + x + 1$$ as
\begin{align} & {x^2} + x + 1 = \frac{3}{4} + {\left( {x + \frac{1}{2}} \right)^2}\\ &\qquad\qquad\;\;= {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + {\left( {x + \frac{1}{2}} \right)^2}\end{align}
Thus, this integral is of the standard form (26):
$I = \frac{1}{2}\left\{ {\left( {x + \frac{1}{2}} \right)\sqrt {{x^2} + x + 1} + \frac{3}{4}\ln \left| {\left( {x + \frac{1}{2}} \right) + \sqrt {{x^2} + x + 1} } \right|} \right\} + C$
(b) We can write $$1 - 2x - {x^2}$$ upon rearrangement as
\begin{align} 1 - 2x - {x^2} = 2 - (1 + 2x + {x^2})\\ = {\left( {\sqrt 2 } \right)^2} - {\left( {x + 1} \right)^2}\end{align}
This integral therefore of the standard form (27) :
$I = \frac{1}{2}\left\{ {(x + 1)\sqrt {1 - 2x - {x^2}} + 2{{\sin }^{ - 1}}\left( {\frac{{x + 1}}{{\sqrt 2 }}} \right)} \right\} + C$
(c) Upon rearrangement, $${x^2} + 3x + 2$$ can be written as
${x^2} + 3x + 2 = {\left( {x + \frac{3}{2}} \right)^2} - \frac{1}{4}\\ \qquad\qquad\qquad\quad = {\left( {x + \frac{3}{2}} \right)^2} - {\left( {\frac{1}{2}} \right)^2}$
Thus, this integral is of the standard form (28):
$I = \frac{1}{2}\left\{ {\left( {x + \frac{3}{2}} \right)\sqrt {{x^2} + 3x + 2} - \frac{1}{4}\ln \left| {\left( {x + \frac{3}{2}} \right) + \sqrt {{x^2} + 3x + 2} } \right.} \right\} + C$
(d) From the standard form (30), we know the integral of $${{e}^{x}}\cos x$$ :
$\int {{e^x}\cos x\,\,dx = \frac{1}{2}{e^x}\left( {\sin x + \cos x} \right)} + C$
Thus, we can apply integration by parts on the expression $$x\,{{e}^{x}}\cos x,taking\;x$$, taking as the first function:
\begin{align}& I = \int {\mathop x\limits_{{\rm{Ist}}} ({e^x}\mathop {\cos x}\limits_{{\rm{IInd}}} )dx} \\ &= \frac{1}{2}x{e^x}(\sin x + \cos x) - \int {1 \cdot \left( {\frac{1}{2}{e^x}(\sin x + \cos x)} \right)dx} \\& = \frac{1}{2}x{e^x}(\sin x + \cos x) - \frac{1}{2}\left\{ {{e^x}\sin x\,\,dx + \int {{e^x}\cos x\,\,dx} } \right\}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\mathop \nearrow \limits_{{\rm{standard form (29)}}} \mathop \nearrow \limits_{{\rm{standard form (30)}}} \\& = \frac{1}{2}x{e^x}(\sin x + \cos x) - \frac{{{e^x}}}{4}\left\{ {(\sin x - \cos x) + (\cos x + \sin x)} \right\} + C\\& = \frac{1}{2}x{e^x}(\sin x + \cos x) - \frac{1}{2}{e^x}\sin x + C\end{align}
_____________________________________________________________________
We now consider an integral of the form $$\int {L(x)\sqrt {Q(x)} \,\,dx}$$ where $$L(x)$$ is a linear factor and $$Q(x)$$ is a quadratic factor. As described earlier, we can always find constants $$\alpha \,\,{\rm{and }}\beta$$ such that
$L(x) = \alpha Q'(x) + \beta$
so that
\begin{align}& I = \int {L(x)\sqrt {Q(x)} \,\,dx}\\&= \alpha \int {Q'(x)\sqrt {Q(x)} \,\,dx + \beta \int {\sqrt {Q(x)} \,\,dx} } \\& = \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{I_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;{I_2}\end{align}
$${I_1}$$ can be solved by the substitution $$Q(x) = t$$ so that $$Q'(x)dx = dt$$ and $${I_1}$$ reduces to $$\alpha \int {\sqrt t \,\,dt}$$ whereas $${I_2}$$ is of the standard form (26), (27) or (28) depending on the form of $$Q(x)$$ .
Example – 46
Evaluate the following integrals:
(a) $$\int {(x + 3)\sqrt {{x^2} + 2x + 3} } \,\,dx$$ (b) \begin{align}\int {\frac{x}{{x - \sqrt {{x^2} - 1} }}} \,\,dx\end{align}
Solution : (a) The derivative of $${x^2} + 2x + 3 \;is \;2x + 3$$ . Thus, we can find $$\alpha \,\,{\rm{and }}\beta$$ such that
\begin{align}&\qquad (x + 3) = \alpha (2x + 3) + \beta\\& \Rightarrow\quad \alpha =\frac{1}{2},\,\,\,\beta =\frac{3}{2} \\& \Rightarrow\quad I = \frac{1}{2}\int {(2x + 3)\sqrt {{x^2} + 2x + 3} } \,\,dx + \frac{3}{2}\int {\sqrt {{x^2} + 2x + 3} } \,\,dx\\ &\qquad\quad = \mathop {\frac{1}{2}\int {\sqrt t } \,\,dt\,\,\,\,\,\,}\limits_{({\text{where }}t{\text{ = }}{x^{\rm{2}}} + 2x + 3)} + \mathop {\frac{3}{2}\int {\sqrt {{{(x + 1)}^2} + 2} } \,\,dx}\limits_{({\text{of the standard form (26)}})} \\&\qquad\quad = \frac{{{t^{3/2}}}}{3} + \frac{3}{4}\left\{ {(x + 1)\sqrt {{x^2} + 2x + 3} + 2\ln \left| {(x + 1) + \sqrt {{x^2} + 2x + 3} } \right|} \right\} + C\\ &\qquad\quad= \frac{{{{({x^2} + 2x + 3)}^{3/2}}}}{3} + \frac{3}{4}\left\{ {(x + 1)\sqrt {{x^2} + 2x + 3} + 2\ln \left| {(x + 1) + \sqrt {{x^2} + 2x + 3} } \right|} \right\} + C\end{align}
(b) We first rationalize the denominator:
\begin{align}& I = \int {\frac{x}{{x - \sqrt {{x^2} - 1} }}\,\,dx}\\& = \int {x(x + \sqrt {{x^2} - 1} )\,\,dx} \\& = \int {{x^2}dx + \mathop {\int {x\sqrt {{x^2} - 1} } \,\,dx}\limits_{{\text{(substitute }}{x^{\text{2}}}{\text{-1 = }}t{\text{)}}} } \\& = \frac{{{x^3}}}{3} + \frac{1}{2}\int {\sqrt t \,\,dt} \\& = \frac{{{x^3}}}{3} + \frac{{{t^{3/2}}}}{3} + C\\& = \frac{{{x^3}}}{3} + \frac{{{{({x^2} - 1)}^{3/2}}}}{3} + C\end{align}
Indefinite Integration
grade 11 | Questions Set 1
Indefinite Integration
Indefinite Integration
grade 11 | Questions Set 2
Indefinite Integration
|
# Real Zeros of Polynomials
Document Sample
``` Real Zeros of
Polynomials
Section 3.4
College Algebra, MATH 171
Mr. Keltner
Fundamental
Theorem of Algebra
• The Fundamental Theorem of
Algebra, first proved by Carl Johann
Friedrich Gauss, tells us that:
– An nth-degree polynomial will have at most n
real zeros.
– Including the complex number system (i’s),
he notes that every nth-degree polynomial will
have precisely n real zeros.
– This leads into the Linear Factorization
Theorem, mentioned on pg. 298, which
incorporates the idea of complex zeros of
polynomial functions.
Zeros of a
Polynomial
• Zeros of a function are also known as:
– Roots
– X-intercepts (where y = 0)
• The zeros of a function may be
substituted into the original equation
and make it true.
• The degree of the equation (or
function) will be the MOST amount of
zeros we will have for our equation,
including complex zeros and those
factors with multiplicity.
A visual example
• Before we get to the • From its factored form,
next theorem, look at we see that P(x) has
an example of where it zeros at 2, 3, and -4.
is applied. • When it is expanded (or
multiplied out), note
• Consider the polynomial that the constant term
function of 24 will come from
P(x) = (x - 2)(x - 3)(x + 4) multiplying (-2)(-3)(4).
• If we multiplied out its • This means the zeros of
factors, we get the polynomial are all
P(x) = x3 - x2 - 14x + 24. factors of the constant
term.
Rational Zeros
Theorem
• This theorem says that if we have a
polynomial function, like
P(x) = anxn + . . . + a1x + a0
and if this function has integer
coefficients, then each of its rational
zeros can be written in the form:
a
p factor of constant coefficient0
a
q factor of leading coefficient n
Example 1
• List the possible rational zeros of
the function using the Rational
Zero Test.
P(x) = 2x3 + x2 - 13x + 6
Must…find…zeros!
• Very similar to what we did in Section 3.3 if
• In this section, if we find one zero, it will help
us reduce the polynomial to find the others.
– This is so that the polynomial doesn’t look as large.
Example 2
• Find all real zeros of
P(x) = x4 - 5x3 - 5x2 + 23x + 10.
– Start by finding all the possible rational zeros of the
function that can be written in the form p/q.
– Once you have found a zero by synthetic division, you can
use the answer from that to try and factor or use synthetic
division again.
– You can always apply the Quadratic Formula if you have
difficulty factoring a 2nd-degree polynomial of the form ax2
+ bx + c, which can be set equal to zero and solved.
• To solve an equation of the form ax2 + bx + c = 0, use
b b 2 4 ac
x
2a
Example 3
• Find all real zeros of
f(x) = 9x4 + 3x3 - 30x2 + 6x + 12.
Example 3:
Narrowing Down
Options
• The function f(x) = 9x4 + 3x3 - 30x2 + 6x + 12
has possible zeros at:
1, 1/3, 1/9, 2, 2/3, 2/9, 3, 3/3, 3/9, 4, 4/3,
4/9, 6, 6/3, 6/9, 12, 12/3, and 12/9.
• By narrowing down the values that are repeated
(or reduce so that they are equal to another
possible zero), we condense our list of possible
zeros to:
1, 1/3, 1/9, 2, 2/3, 2/9, 3, 4, 4/3, 4/9, 6,
and 12.
That’s almost half the amount we had before!
Example 3:
Narrowing Down MORE
Options
f(x) = 9x4 + 3x3 - 30x2 + 6x + 12
• Using the graph
QuickTime™ and a
PNG decompressor of f(x), what x-
are neede d to se e this picture. value do you see
that is a root of
the function?
When we find one of our possible roots that works, we
can use synthetic division to help factor the rest of the
polynomial.
This can help us find the other roots of the function.
Complex Conjugates
Theorem
• When we investigate complex
conjugates, we will notice that
their product eliminates the i
terms.
• If our polynomial function has a
zero in the form a + bi, then its
conjugate, a - bi, is also a zero of
the function.
Example 4
• Find all real zeros of
f(x) = x3 - x + 6.
Package Design
• Package Design is an
occupation that deals with
polynomials regularly.
• This field of study strives
to make a package that
does two things:
– Uses as little material as
possible
– Holds as much of the
product as possible
Example 5:
Package Design
• From the diagram below for a Special K box
that is unfolded from an 18” by 36” piece of
cardboard, calculate the dimensions for a box
that maximizes its volume.
35-2x 35-2x
1” x 2 x 2
x
2
18-x
x
2
Assessment
Pgs. 308-309:
#’s 5-70, multiples of 5
```
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# Angle trisection
Angle trisection
The problem of trisecting the angle is a classic problem of compass and straightedge constructions of ancient Greek mathematics.Two tools are allowed
# An un-marked straightedge, "and"
# a compass,
Problem: construct an angle one-third a given arbitrary angle.
With such tools, it is generally impossible. This requires taking a cube root, impossible with the given tools; see below.
A common misunderstanding
It is common to hear "It is impossible to trisect an angle! "Q.E.D." Leaving aside the lack of proof, this statement is false: it is only impossible to solve "in general" using only an un-marked straightedge and a compass, i.e. it may be done by using other tools, "and", some angles may be trisected with a straightedge and a compass.
Perspective and relationship to other problems
[
Bisection of arbitrary angles has long been solved.] Using only an unmarked straightedge and a compass, Greek mathematicians found means to divide a line into an arbitrary set of equal segments, to draw parallel lines, to bisect angles, to construct many polygons, and to construct squares of equal or twice the area of a given polygon.
Three problems proved elusive, specifically:
* Trisecting the angle,
* Doubling the cube, "and"
* Squaring the circle
Angles may not in general be trisected
Denote the rational numbers $mathbb\left\{Q\right\}$.
Note that a number constructible in one step from a field $K$ is a solution of a second-order polynomial; again, see constructible number. Note also that $pi/3$ radians (60 degrees, written 60°) is constructible.
However, the angle of $pi/3$ radians (60 degrees) cannot be trisected. Note $cos\left(pi/3\right) = cos\left(60^circ\right) = 1/2$.
If 60° could be trisected, the minimal polynomial of $cos\left(20^circ\right)$ over $mathbb\left\{Q\right\}$ would be of second order. Note the trigonometric identity $cos\left(3alpha\right) = 4cos^\left\{3\right\}\left(alpha\right) - 3cos\left(alpha\right)$. Now let $y = cos\left(20^circ\right)$.
By the above identity, $cos\left(60^circ\right) = 1/2 = 4y^\left\{3\right\} - 3y$. So $4y^\left\{3\right\} - 3y - 1/2 = 0$. Multiplying by two yields $8y^\left\{3\right\} - 6y - 1 = 0$, or $\left(2y\right)^\left\{3\right\} - 3\left(2y\right) - 1 = 0$. Now substitute $x = 2y$, so that $x^\left\{3\right\} - 3x - 1 = 0$. Let $p\left(x\right) = x^\left\{3\right\} - 3x - 1$.
The minimal polynomial for "x" (hence $cos\left(20^circ\right)$) is a factor of $p\left(x\right)$. If $p\left(x\right)$ has a rational root, by the rational root theorem, it must be 1 or −1, both clearly not roots. Therefore $p\left(x\right)$ is irreducible over $mathbb\left\{Q\right\}$, and the minimal polynomial for $cos\left(20^circ\right)$ is of degree 3.
So an angle of $60^circ = pi/3$ radians cannot be trisected.
ome angles may be trisected
However, some angles may be trisected. Given angle $heta$, angle $3 heta$ trivially trisects to $heta$. More notably, $2pi/5$ radians (72°) may be constructed, and may be trisected. [http://www.physicsforums.com/showthread.php?t=160571] Also there are angles, while non-constructable, but (if somehow given) are trisectable, for example $3pi/7$. [Five copies of $3pi/7$ combine to make $15pi/7$ which is a full circle plus $pi/7$.]
One general theorem
Again, denote the rational numbers $Q$:
Theorem: The angle $heta$ may be trisected if and only if $q\left(t\right) = 4t^\left\{3\right\}-3t-cos\left(3 heta\right)$ is reducible over the field extension $Q\left(cos\left( heta\right)\right)$.
Proof. The proof would take us afield, but it may be derived from the above trig identity.cite book | last = Stewart | first = Ian | authorlink = Ian Stewart (mathematician)|Ian Stewart | title = "Galois Theory" | publisher = Chapman and Hall Mathematics | date = 1989 | pages = pg. 58 | doi = | isbn = 0412345501]
Means to trisect angles by going outside the Greek framework
Origami
Trisection, like many constructions impossible by ruler and compass, can easily be accomplished by the more powerful (but physically easy) operations of paper folding, or origami. Huzita's axioms (types of folding operations) can construct cubic extensions (cube roots) of given lengths, whereas ruler-and-compass can construct only quadratic extensions (square roots). See mathematics of paper folding.
Auxiliary curve
There are certain curves called trisectrices which, if drawn on the plane using other methods, can be used to trisect arbitrary angles. [ [http://www.jimloy.com/geometry/trisect.htm#curves Trisection of an Angle ] ]
With a marked ruler
Another means to trisect an arbitrary angle by a "small" step outside the Greek framework is via a ruler with two marks a set distance apart. The next construction is originally due to Archimedes, called a "Neusis construction", i.e., that uses tools other than an "un-marked" straightedge.
This requires three facts from geometry (at right):
# Any full set of angles on a straight line add to 180°,
# The sum of angles of any triangle is 180°, "and",
# Any two equal sides of an isosceles triangle meet the third in the same angle.
Look to the diagram at right; note angle "a" left of point "B". We trisect angle "a".
First, a ruler has two marks distance "AB" apart. Extend the lines of the angle and draw a circle of radius "AB".
"Anchor" the ruler at point "A", and move it until one mark is at point "C", one at point "D", i.e., "CD = AB". A radius "BC" is drawn as obvious. Triangle "BCD" has two equal sides, thus is isosceles.
That is to say, line segments "AB", "BC", and "CD" all have equal length. Segment "AC" is irrelevant.
Now: Triangles "ABC" and "BCD" are isosceles, thus by Fact 3 each has two equal angles. Now re-draw the diagram, and label all angles:
Hypothesis: Given "AD" is a straight line, and "AB", "BC", and "CD" are all equal length,
Conclusion: angle $b = \left(1/3\right) a$.
Proof:
Steps:
# From Fact 1) above, $e + c = 180$°.
# Looking at triangle "BCD", from Fact 2) $e + 2b = 180$°.
# From the last two equations, $c = 2b$.
# From Fact 2), $d + 2c = 180$°, thus $d = 180$°$- 2c$, so from last, $d = 180$°$- 4b$.
# From Fact 1) above, $a + d + b = 180$°, thus $a + \left(180$°$- 4b\right) + b = 180$°.
Clearing, $a - 3b = 0$, or $a = 3b$, and the theorem is proved.
Again: this construction stepped outside the framework of allowed constructions by using a marked straightedge. There is an unavoidable element of inaccuracy in placing the straightedge.
With a string
Hutcheson published an article in Mathematics Teacher, vol. 94, No. 5, May, 2001 that used a string instead of a compass and straight edge. A string can be used as either a straight edge (by stretching it) or a compass (by fixing one point and identifying another), but can also wrap around a cylinder, the key to Hutcheson's solution.
Hutcheson constructed a cylinder from the angle to be trisected by drawing an arc across the angle, completing it as a circle, and constructing from that circle a cylinder on which a, say, equilateral triangle was inscribed (a 360-degree angle divided in three). This was then "mapped" onto the angle to be trisected, with a simple proof of similar triangles.
For the detailed proof and its generalization, see the article cited: Mathematics Teacher, vol. 94, No. 5, May, 2001, pp. 400-405.
There are other constructions (references).
*Bisection
*Constructible number
*Constructible polygon
*Doubling the cube
*Euclidean geometry
*Galois theory
*History of geometry
*Intercept theorem
*List of geometry topics
*Morley's trisector theorem
*Neusis construction
*Squaring the circle
*Tomahawk (geometric shape)
*Trisectrix
Notes
External references
* [http://mathworld.wolfram.com/AngleTrisection.html MathWorld site]
* [http://mathworld.wolfram.com/GeometricProblemsofAntiquity.html Geometric problems of antiquity, including angle trisection]
* [http://www-history.mcs.st-andrews.ac.uk/HistTopics/Trisecting_an_angle.html Some history]
* [http://www.uwgb.edu/dutchs/PSEUDOSC/trisect.HTM One link of marked ruler construction]
* [http://www.cut-the-knot.org/pythagoras/archi.shtml Another, mentioning Archimedes]
* [http://www.jimloy.com/geometry/trisect.htm A long article with many approximations & means going outside the Greek framework]
* [http://www.geom.uiuc.edu/docs/forum/angtri/ Geometry site]
Other means of trisection
* [http://www.uwgb.edu/dutchs/PSEUDOSC/trisect.HTM Trisecting via] an "Archimedean Spiral"
* [http://xahlee.org/SpecialPlaneCurves_dir/ConchoidOfNicomedes_dir/conchoidOfNicomedes.html Trisecting via] the "Conchoid of Nicomedes"
* [http://www.sciencenews.org/articles/20070602/mathtrek.asp sciencenews.org site] on using origami
* [http://www.song-of-songs.net/Star-of-David-Flower-of-Life.html Hyperbolic trisection and the spectrum of regular polygons]
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• Trisection — de l angle La trisection de l angle est un problème classique de mathématiques. C est un problème géométrique, faisant partie des trois grands problèmes de l Antiquité, avec la quadrature du cercle et la duplication du cube. Ce problème consiste… … Wikipédia en Français
• Trisection de l’angle — Trisection de l angle La trisection de l angle est un problème classique de mathématiques. C est un problème géométrique, faisant partie des trois grands problèmes de l Antiquité, avec la quadrature du cercle et la duplication du cube. Ce… … Wikipédia en Français
• trisection — [ trisɛksjɔ̃ ] n. f. • 1691; de tri et section ♦ Géom. Division d une grandeur en trois parties égales. La trisection de l angle. ● trisection nom féminin Division d un ensemble, d une grandeur en trois parties égales. Détermination d un secteur… … Encyclopédie Universelle
• Trisection de l'angle — La trisection de l angle est un problème classique de mathématiques. C est un problème géométrique, faisant partie des trois grands problèmes de l Antiquité, avec la quadrature du cercle et la duplication du cube. Ce problème consiste à diviser… … Wikipédia en Français
• angle — [ ɑ̃gl ] n. m. • XIIe; lat. angulus 1 ♦ Cour. Saillant ou rentrant formé par deux lignes ou deux surfaces qui se coupent. ⇒ arête, coin, encoignure, renfoncement. À l angle de la rue. Former un angle, être en angle. La maison qui fait l angle,… … Encyclopédie Universelle
• Trisection — Tri*sec tion, n. [Cf. F. trisection.] The division of a thing into three parts, Specifically: (Geom.) the division of an angle into three equal parts. [1913 Webster] … The Collaborative International Dictionary of English
• Trisection — (v. lat.), Theilung des Winkels in drei gleiche Theile; ähnlich ist die Multisection, d.i. Theilung in irgend viel gleiche Theile. Vgl. Azumar, Trisection de l angle, Par. 1809 … Pierer's Universal-Lexikon
• trisection — (tri sè ksion) s. f. Division d une chose en trois parties. Terme de géométrie. Division en trois parties égales. • La trisection de l angle, problème fameux chez les anciens, et qui les a beaucoup exercés, FONTEN. Viviani.. Terme de… … Dictionnaire de la Langue Française d'Émile Littré
• TRISECTION — s. f. (On prononce l S fortement. ) T. de Géom. Division d une chose en trois parties égales. Il se dit principalement de La division d un angle en trois angles égaux. La trisection de l angle … Dictionnaire de l'Academie Francaise, 7eme edition (1835)
• TRISECTION — n. f. T. de Géométrie Division d’une chose en trois parties égales. Il se dit principalement de la Division d’un angle en trois angles égaux. La trisection de l’angle … Dictionnaire de l'Academie Francaise, 8eme edition (1935)
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Solutions for Session 9, Part A
See solutions for Problems: A1 | A2 | A3 | A4 | A5 | A6| A7
Problem A1
a. The area is 3/12 (or 1/4) of a square mile. b. The area is 6/15 (or 2/5) of a square mile. c. The area is 8/36 (or 2/9) of a square mile.
Problem A2 If the factors are each less than 1, the product cannot be larger than its factors. The part that overlaps the horizontally shaded area and the vertically shaded area is the area of the product, and each of the original areas will be larger than the area shaded by both. Note that this only works with "proper" fractions, where the numerator is less than the denominator (and therefore the fraction is less than 1).
Problem A3
In essence, you are looking to find the length of a rectangle whose one dimension is 3/4 and whose area is equal to 2/3 -- in other words, 3/4 • x = 2/3, or 2/3 3/4 = x.
You can start by marking 2/3 vertically. Next, mark 3/4 horizontally. Then subdivide everything into ninths. Then rearrange the top (purple) pieces from the original 2/3 area to fit the 3/4 height. The result will be the purple rectangle, whose length is 8/9: 2/3 3/4 = 8/9
Problem A4 Since we model division as the reverse of multiplication, the first fraction is represented as an overlap of two areas. It will be smaller than the quotient for the same reason that the multiplication area is smaller than either of the fractions multiplied. In some cases, the first fraction will be smaller than both the quotient and second fraction. (The illustration in the solution to Problem A3 is an example of such a case.) If the first fraction happens to be larger than the second one in the division problem, the result will be an improper fraction (where the numerator is larger than the denominator).
Problem A5 All the answers are the same, 2, because the units are the same for each division problem.
Problem A6 The original question is "How many 3/4s are there in 3/5?" The common denominator is 20: 3/4 = 15/20, and 3/5 = 12/20. The question is now "How many 15/20s are there in 12/20?," which is easier to answer because the units are the same: 12/20 15/20 = 12 15 = 12/15 = 4/5. The answer is 4/5.
Problem A7 Both 0.6 and 0.2 can be expressed in units of tenths: "How many 2/10s are there in 6/10?" The answer to this question must be the same as "How many 2s are there in 6?" The answer to both questions will always be 3, regardless of the particular units involved.
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# Speed Distance Time Word Problems with Solutions
Last updated date: 21st Feb 2024
Total views: 122.7k
Views today: 2.22k
## Introduction
Speed, distance, and time are the three main pillars behind mathematics and physics. Whenever you are presented with a question related to any kind of transportation, you should immediately relate to these concepts. The train speed, the relation between the time taken by two cars at different speeds or even a simple question about a person walking from one place to another can be answered using simple formulas.
## What is Speed?
When you think of speed, you must think of how fast? Speed is all about calculating the rate at which something can be accomplished.
A Device Used to Measure Speed Known as a ‘Speedometer’
## What is Distance?
Distance in this concept refers to the distance travelled. It means how far? It can be measured in meters, for small distances and kilometres, for larger distances.
Distance is the Measurement From One Place to Another
## What is Time?
Time is the answer to the question, how long? The best way to measure time is using the ‘world clock.’ In ancient times, time was measured with the help of the sun and the moon, and other celestial objects.
Time as Shown by a Clock
## Relation Between Speed Distance and Time
There is a very simple mathematical relation between speed, distance and time.
$\text{speed=}\frac{\text{distance}}{\text{time}}$
$\text{time=}\frac{\text{distance}}{\text{speed}}$
$\text{distance=speed }\!\!\times\!\!\text{ time}$
These relations can be memorised using the triangle method. See the following image to understand the use of this method.
Explaining the Triangle Method
## Speed Distance Time Word Problems With Solutions
Q1. A train is travelling at a speed of 160 km/hour. It takes 15 hours to cover the distance from city A to city B. Find the distance between the two cities.
Solution:
Speed= 160 km/hr
Time= 15 hours
Using the formula, $\text{distance=speed }\!\!\times\!\!\text{ time}$
Distance= $160\times 15=2400km$
Answer: The distance between city A and city B is 2400km
Q2. A car travelling from city A to city C completes the journey in 3 hours, whereas a person travelling on a bike completes the journey in 5 hours. What is the speed of the man on the bike if the car is travelling at 45 kmph?
Solution:
Find the distance using the information on the car
Using the formula, $\text{distance=speed }\!\!\times\!\!\text{ time}$
Speed= 45km/hour
Time= 3 hours
$\therefore \text{distance}=45\times 3=135km$
Using the distance, we can find the speed of the bike
Formula used, $\text{speed=}\frac{\text{distance}}{\text{time}}$
Time taken by bike is 5 hours
$\therefore \text{speed}=\frac{135}{5}=27$
Answer: The bike is moving at a speed of 27 kmph.
Q3. A person travels at a speed of 15 kmph from point A to point B, which are 25 km from each other. Another person is travelling from point C to point B, 44 km from each other. Both people start their journey at the same time; the second person arrives at the point 2 hours after the first. What is the speed at which the second person was travelling?
Solution:
Speed of person 1 = 15km/hour
Distance travelled by person 1= 25 km/hour
Time taken by first-person = $\frac{\text{distance}}{\text{speed}}=\frac{25}{15}=\frac{5}{3}$hours
Note: While solving the questions, keep the answers in fractions until you reach the final answer to make calculations easier.
Time taken by second person =$\frac{5}{3}+2=\frac{11}{3}$hours
Distance travelled by second person= 44 km
$\therefore \text{speed}=\frac{\text{distance}}{\text{time}}=\frac{44}{\frac{11}{3}}=\frac{44\times 3}{11}=4\times 3=12$kmph
Answer: The second person was travelling at a speed of 12kmph.
Q4. Raju is travelling from one station to another 560km away in a train which is moving at a speed of 125kmph. Monu is travelling to the same destination at a speed of 80kmph. Monu starts his journey 3 hours after Raju, from 120 km closer to the destination. How long will Raju have to wait at the train station for Monu to arrive?
Solution:
Distance travelled by Raju = 560km
Speed of Raju= 125km/hour
∴ Time taken by Raju to reach the station = $\frac{\text{distance}}{\text{speed}}=\frac{560}{125}=\frac{112}{25}$hours
Distance travelled by Monu = $560-120=440$km
The speed at which Monu travelled = 80kmph
Time taken by Monu to complete journey = $\frac{440}{80}=\frac{22}{4}=\frac{11}{2}$hours
Total time taken by Monu = $\frac{11}{2}+3=\frac{17}{2}$hours
Therefore, the time Raju had to wait = difference between the two times taken
$\therefore \frac{17}{2}-\frac{112}{25}=\frac{17\times 25}{50}-\frac{112\times 2}{50}=\frac{201}{50}=4.02$hours
Answer: Raju had to wait for 4.02 hours at the train station
Convert 4.02 hours to hours and minutes
$\therefore 0.02=\frac{2}{100}\times 60=1.2$minutes
$\therefore 0.2=\frac{2}{10}\times 60=12$seconds
Therefore, to be precise, Raju had to wait at the train station for 4 hours, 1 minute and 12 seconds.
## FAQs on Speed Distance Time Word Problems with Solutions
1. Is there any other formula to relate the speed distance and time?
No, there is only one relation that one must remember to solve all speed distance time problems. The triangle method is the easiest and fastest way to remember these relations.
2. What are other speed measures, and how to convert them to km/hour?
The other speed measures include meters per second, denoted as m/sec or m/s and miles per hour, denoted as mph.
To convert m/sec into km/hour, one must use the following formula—
$\frac{m}{s}\times \frac{5}{18}=\frac{km}{hr}$ ($\because$1 m/s = 3.6 km/h)
To convert mph into kmph one must use the following relation—
1 mile= approximately 1.609 km
1 km= 0.6214 miles
3. What is the device that records distance called?
The device that calculates the distance travelled is known as an odometer.
4. What other factors need to be considered when calculating real-life problems?
Simple formulas are not useful when solving speed distance time problems in real life since many other factors affect the outcome. One must consider air drag, frictional forces, repulsion, wear and tear and, most importantly, gravitational forces when solving these real-life problems. The formulas of relation, however, remain the same.
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# Converting Complex Numbers From Rectangular Form to Trigonometric - Concept 8,925 views
Both the trigonometric form and the rectangular form are useful ways to describe complex numbers, and so we must understand how to convert from rectangular form to trigonometric form. To convert from rectangular form to trigonometric form we need to calculate the modulus and the angle of the position vector. It is also important to be able to convert from trigonometric to rectangular form.
We're converting from rectangular form to trigonometric form and we're starting with the complex number z equals negative root 2 plus i times root 2. Now, first of all, what does it mean to convert to trigonometric form?
Well, I have my number in rectangular form, so it's in a+bi form. This is the a value, this is the b value. The a's the real part of the imaginary number and root 2 is the the imaginary part. Now if I were to plot this, you'd have a is negative and b is positive so you'd be somewhere in the second quadrant. And I want to choose theta, right? When i get into trigonometric form, theta is the argument of the complex number. I want to choose theta so that it's between 0 and 2 pi. So I'd want to choose this theta and r is the distance between the point and the origin. That's the absolute value of the number. r is the easiest to get because we have this formula r equals the square root of a squared plus b squared, so let's get r first.
The square root of negative root 2 squared plus positive root 2 squared. So this is going to end up being 2 plus 2, 4. Root 4 which is 2, so r=2.
Next, let's try to find theta. That's the harder thing to find but we can use these 2 conversions to find theta. I'm going to use the fact that cosine theta would equal a over r. So cosine theta equals a over r. a is negative root 2 and r is 2. And we also use the fact that sine theta is b over r. Sine theta is b over r. b is root 2. r is 2.
So we need to find, figure out what theta is going to have a cosine of negative root 2 over 2 and a sine of root 2 over 2. Now the numbers root 2 over 2, you may remember are the sine and cosine of pi over 4, pi over 4's our reference angle. That means this angle is pi over 4 and that means this angle is going to be 3 pi over 4. So theta is 3 pi over 4. That's our argument of z. And so we put it into trig form. r times the quantity cosine of theta plus i sine theta. So z=r which is 2 cosine of 3 pi over 4 plus i times the sine of 3 pi over 4.
Don't forget when you're in trig form this number will always be the same for both cosine and sine, right? This is the argument of the complex number. This is our final answer for the trig form of the number.
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Math resources Algebra Math equations
Solve equations with fractions
# Solve equations with fractions
Here you will learn about how to solve equations with fractions, including solving equations with one or more operations. You will also learn about solving equations with fractions where the unknown is the denominator of a fraction.
Students will first learn how to solve equations with fractions in 7th grade as part of their work with expressions and equations and expand that knowledge in 8th grade.
## What are equations with fractions?
Equations with fractions involve solving equations where the unknown variable is part of the numerator and/or denominator of a fraction.
The numerator (top number) in a fraction is divided by the denominator (bottom number).
To solve equations with fractions, you will use the “balancing method” to apply the inverse operation to both sides of the equation in order to work out the value of the unknown variable.
The inverse operation of addition is subtraction.
The inverse operation of subtraction is addition.
The inverse operation of multiplication is division.
The inverse operation of division is multiplication.
For example,
\begin{aligned} \cfrac{2x+3}{5} \, &= 7\\ \colorbox{#cec8ef}{$\times \, 5$} \; & \;\; \colorbox{#cec8ef}{$\times \, 5$} \\\\ 2x+3&=35 \\ \colorbox{#cec8ef}{$-\,3$} \; & \;\; \colorbox{#cec8ef}{$- \, 3$} \\\\ 2x & = 32 \\ \colorbox{#cec8ef}{$\div \, 2$} & \; \; \; \colorbox{#cec8ef}{$\div \, 2$}\\\\ x & = 16 \end{aligned}
## Common Core State Standards
• Grade 7: Expressions and Equations (7.EE.A.1)
Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.
• Grade 8: Expressions and Equations (8.EE.C.7)
Solve linear equations in one variable.
• Grade 8: Expressions and Equations (8.EE.C.7b)
Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
## How to solve equations with fractions
In order to solve equations with fractions:
1. Identify the operations that are being applied to the unknown variable.
2. Apply the inverse operations, one at a time, to both sides of the equation.
3. When you have the variable on one side, you have the final answer.
4. Check the answer by substituting the answer back into the original equation.
## Solve equations with fractions examples
### Example 1: equations with one operation
Solve for x \text{: } \cfrac{x}{5}=4 .
1. Identify the operations that are being applied to the unknown variable.
The unknown is x.
Looking at the left hand side of the equation, the x is divided by 5.
\cfrac{x}{5}
2Apply the inverse operations, one at a time, to both sides of the equation.
The inverse of “dividing by 5 ” is “multiplying by 5 ”.
You will multiply both sides of the equation by 5.
3When you have the variable on one side, you have the final answer.
4Check the answer by substituting the answer back into the original equation.
You can check the answer by substituting the answer back into the original equation.
\cfrac{20}{5}=20\div5=4
### Example 2: equations with one operation
Solve for x \text{: } \cfrac{x}{3}=8 .
Identify the operations that are being applied to the unknown variable.
Apply the inverse operations, one at a time, to both sides of the equation.
Write the final answer, checking that it is correct.
### Example 3: equations with two operations
Solve for x \text{: } \cfrac{x \, + \, 1}{2}=7 .
Identify the operations that are being applied to the unknown variable.
Apply the inverse operations, one at a time, to both sides of the equation.
Write the final answer, checking that it is correct.
### Example 4: equations with two operations
Solve for x \text{: } \cfrac{x}{4}-2=3 .
Identify the operations that are being applied to the unknown variable.
Apply the inverse operations, one at a time, to both sides of the equation.
Write the final answer, checking that it is correct.
### Example 5: equations with three operations
Solve for x \text{: } \cfrac{3x}{5}+1=7 .
Identify the operations that are being applied to the unknown variable.
Apply the inverse operations, one at a time, to both sides of the equation.
Write the final answer, checking that it is correct.
### Example 6: equations with three operations
Solve for x \text{: } \cfrac{2x-1}{7}=3 .
Identify the operations that are being applied to the unknown variable.
Apply the inverse operations, one at a time, to both sides of the equation.
Write the final answer, checking that it is correct.
### Example 7: equations with the unknown as the denominator
Solve for x \text{: } \cfrac{24}{x}=6 .
Identify the operations that are being applied to the unknown variable.
Apply the inverse operations, one at a time, to both sides of the equation.
Write the final answer, checking that it is correct.
### Example 8: equations with the unknown as the denominator
Solve for x \text{: } \cfrac{18}{x}-6=3 .
Identify the operations that are being applied to the unknown variable.
Apply the inverse operations, one at a time, to both sides of the equation.
Write the final answer, checking that it is correct.
### Teaching tips for solving equations with fractions
• When students first start working through practice problems and word problems, provide step-by-step instructions to assist them with solving linear equations.
• Introduce solving equations with fractions with one-step problems, then two-step problems, before introducing multi-step problems.
• Students will need lots of practice with solving linear equations. These standards provide the foundation for work with future linear equations in Algebra I and II.
• Provide opportunities for students to explain their thinking through writing. Ensure that they are using key vocabulary, such as, absolute value, coefficient, equation, common factors, inequalities, simplify, etc.
### Easy mistakes to make
• The solution to an equation can be any type of number
The unknowns do not have to be integers (whole numbers and their negative opposites). The solutions can be fractions or decimals. They can also be positive or negative numbers.
• The unknown of an equation can be on either side of the equation
The unknown, represented by a letter, is often on the left hand side of the equations; however, it doesn’t have to be. It could also be on the right hand side of an equation.
• When multiplying both sides of an equation, multiply each and every term
When multiplying each side of the equation by a number, it is a common mistake to forget to multiply every term.
For example,
Solve:
\cfrac{x}{2}+3=9
Here, the + 3 was not multiplied by 2, resulting in the incorrect answer.
This person has correctly multiplied each term by the denominator.
• Lowest common denominator (LCD)
It is common to get confused between solving equations involving fractions and adding and subtracting fractions. When adding and subtracting, you need to work out the lowest/least common denominator (sometimes called the least common multiple or LCM). When you solve equations involving fractions, multiply both sides of the equation by the denominator of the fraction.
### Practice solve equations with fractions questions
1. Solve: \cfrac{x}{6}=3
x=9
x=36
x=12
x=18
You will multiply both sides of the equation by 6, because the inverse of “dividing by 6 ” is “multiplying by 6 ”.
The final answer is x = 18.
You can check the answer by substituting the answer back into the original equation.
\cfrac{18}{6}=18 \div 6=3
2. Solve: \cfrac{x \, + \, 4}{2}=7
x=18
x=10
x=26
x=30
First, clear the fraction by multiplying both sides of the equation by 2.
Then subtract 4 from both sides.
The final answer is x = 10.
You can check the answer by substituting the answer back into the original equation.
\cfrac{10 \, + \, 4}{2}=\cfrac{14}{2}=14 \div 2=7
3. Solve: \cfrac{x}{8}-5=1
x=40
x=64
x=48
x=56
First, add 5 to both sides of the equation.
Then multiply both sides of the equation by 8.
The final answer is x = 48.
You can check the answer by substituting the answer back into the original equation.
\cfrac{48}{8}-5=48 \div 8-5=1
4. Solve: \cfrac{3x \, + \, 2}{4}=2
x=4
x=16
x=12
x=2
First, multiply both sides of the equation by 4.
Next, subtract 2 from both sides.
Finally, divide both sides by 3.
The final answer is x = 2.
You can check the answer by substituting the answer back into the original equation.
\cfrac{3 \, \times \, 2+2}{4}=\cfrac{6 \, + \, 2}{4}=\cfrac{8}{4}=8 \div 4=2
5. Solve: \cfrac{4x}{7}-2=6
x=11
x=14
x=7
x=10
First, add 2 to both sides of the equation.
Then multiply both sides of the equation by 7.
Finally, divide both sides by 4.
The final answer is x = 14.
You can check the answer by substituting the answer back into the original equation.
\cfrac{4 \, \times \, 14}{7}-2=\cfrac{56}{7}-2=56 \div 7-2=6
6. Solve: \cfrac{42}{x}=7
x=5
x=294
x=7
x=6
You need to multiply both sides of the equation by x.
Then you divide both sides by 7.
The final answer is x = 6.
You can check the answer by substituting the answer back into the original equation.
\cfrac{42}{6}=42 \div 6=7
## Solve equations with fractions FAQs
Do I still follow the order of operations when solving equations with fractions?
Yes, you still follow the order of operations when solving equations with fractions. You will start with any operations in the numerator and follow PEMDAS (parenthesis, exponents, multiply/divide, add/subtract), followed by any operations in the denominator. Then you will solve the rest of the equation as usual.
## Still stuck?
At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts.
Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence.
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# Intro in Binary Representations and Bit Manipulation
First of all, what is a bit (short for binary digit)? You can think of it as an atom which can be used to build information. In fact, a bit has only two unique states: zero or one. Sometimes you can find another definition like “yes/no”, “on/off” or “true/false”. But, all this stuff boils down to the same thing.
To represent more meaningful information other than zero or one we need a
bunch of bits that can be grouped together. For example, eight bits are able to produce 256 (2^8) unique numbers:
`11110100 = 244 or 10000100 = 132`
### Conversion to base 10 and vice versa
To convert a binary number to base 10 you can use a very simple method which multiplies bits in respective positions with 2 to the power of their position. For example, binary number 10000110 can be converted as follows:
```10000110 =
1 * 2^7 +
0 * 2^6 +
0 * 2^5 +
0 * 2^4 +
0 * 2^3 +
1 * 2^2 +
1 * 2^1 +
0 * 2^0 = 134```
A repeated division and reminder can be used to convert decimal to binary number. The method contains four steps:
1. Divide the decimal by two.
2. Append the reminder as the most significant bit.
3. Go to step 1 if decimal number is greater than zero, otherwise go to 4.
4. End.
Example:
```134 / 2 = 67 reminder 0
67 / 2 = 33 reminder 1
33 / 2 = 16 reminder 1
16 / 2 = 8 reminder 0
8 / 2 = 4 reminder 0
4 / 2 = 2 reminder 0
2 / 2 = 1 reminder 0
1 / 2 = 1/2 reminder 1```
Up to now we have been talking about only integer numbers. So, what can we do to convert the fractional part of the number to binary representation? The fractional part of the binary number works on the same principle as the fractional part in base 10. In base 10 all numbers followed the decimal point are multiplied by 10 to the negative power of the number’s position. For instance:
```123.456 =
1 * 10^2 +
2 * 10^1 +
3 * 10^0 +
4 * 10^-1 +
5 * 10^-2 +
6 * 10^-3```
So we can use the same principle to convert 2.75 to the binary number:
```10.11 =
1 * 2^1 +
0 * 2^0 +
1 * 2^-1 +
1 * 2^-2```
### Bitwise Operations
Basically, it’s not good idea to use bitwise operations in routine applications because it makes code less readable and clear. But, in case if memory consumption and speed matters you can consider this as an optimization.
Another application of bitwise operations are graphics, encryption and compression.
Btw, here goes my utility class for bit manipulations: https://github.com/alexvolov/algorithms-and-data-structures/blob/master/src/main/java/com/alexvolov/ads/common/BitManipulation.java
###### NOT
The unary bitwise complement operator “~” inverts a bit pattern, making every “0” a “1” and every “1” a “0”. For instance:
`~10101011 = 01010100`
###### AND
The AND operator works just like multiplication and it also known as logical conjunction. It outputs 1 only if all inputs are 1.
A B Result
0 0 0
1 0 0
0 1 0
1 1 1
###### OR
The OR operator outputs 1 if at least one input is 1. This is why it’s also known as logical disjunction operator.
A B Result
0 0 0
1 0 1
0 1 1
1 1 1
###### XOR
The XOR operator or exclusive disjunction outputs 1 only if both inputs don’t match. Actually it’s the same as adding modulo 2.
A B Result
0 0 0
1 0 1
0 1 1
1 1 0
Some properties of XOR operation:
```A^B = B^A
A^B^A = B```
###### Shifting
The bitwise left shift operator moves all bits in the given number to the left and inserts 0 to vacated bit positions.
For instance, we can perform a left shift by 2 bits:
`10010101 << 2 = 01010100`
Shifting a number to the left by n bits has the effect of multiplying it by 2^n.
Likewise, the bitwise right shift operator moves all bits of the given number to the right.
Example,
`10010101 >>> 2 = 00100101`
Moving bits to the right is equivalent of dividing the number by 2^n.
In other words, bitwise shifting is an efficient way of performing multiplication or division by powers of two.
### Two’s complement
Two’s complement is just a good and convenient way to represent negative numbers in base 2. If you are a Java developer you probably know that Integer values in java has 32 bits length. The leftmost bit is used for sign (in some sources it’s called the sign bit). In case the number is negative the sign bit will be equal to 1, otherwise to 0. Unfortunately we can’t simply toggle the sign bit to make a number positive or negative because in this case we will have two zero’s. One negative zero and one positive zero which is not possible in math.
To solve this problem with zero’s we can use two’s complement. For instance, let’s say we have 8 bits word and we want to represent a negative five.
The positive five in binary will look like this:
Please note that the left most bit is sign bit. To convert it to negative we need to do two things:
1. Invert all bits:
That’s all. Now try to convert zero using the above algorithm and result will always be zero. Fascinating, wright?
Moreover, in case if you need to convert the negative five to positive five you can use the same algorithm.
### Intro to IEEE 754
IEEE standard for floating point computations was established by the Institute of Electrical and Electronics Engineers (IEEE) in 1985.
A real number can be represented as follows:
N = a * q^n
Where a is significand, q is a base and n is an exponent. As a result, the same number can be represented in many ways. For example, number 120:
```120 = 120 * 10^0
120 = 12 * 10^1
120 = 1.2 * 10^2
...```
All these representations are absolutely correct. But, to reduce the number of the representations to one a normalised notation is used. In this notation significand is always greater or equal to 1 and less than the base. In the above example the only normalised notation for 120 will be 1.2 * 10^2 because 1 <= 1.2 < 10.
Now let’s try to represent a binary number in normalised form. For instance, number 01111011 will look like 1.111011 * 2^6 and it might be represented in IEEE format in following way:
As you can see there are 32 bits in total. The most significant bit is reserved for sign. The sign bit is set to zero if number is positive, otherwise to 1.
Eight bits are allocated for exponent. But, it’s not that simple. Instead of using two’s complement it was decided to add 127 to the exponent and store it as unsigned number. If the stored value is greater than the bias, that means the value of the exponent is positive, if it’s lower than the bias, it’s negative, if it’s equal, it’s zero. In my example the exponent is equal to 10000101 = 01111111 + 110.
The rest of the bits are reserved for fractional part of mantissa. In other words, the integral part of the mantissa almost always (except zero) equals to 1 and we don’t need to store it.
Please note that in my post I described 32 bit IEEE 754 which is equivalent to float in Java. But, there is 64 bit IEEE 754 that is known as double in Java.
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# Differentiate g(x)=x^3 * cos^2 x.Differentiate g(x)=x^3 * cos^2 x.
sciencesolve | Certified Educator
You need to implicit differentiate the function with respect to x, using the rules of differentiation, such that:
`(dg(x))/(dx) = (d(x^3))/(dx)*cos^2 x + x^3*(d(cos^2 x))/(dx)`
`(dg(x))/(dx) = 3x^2*cos^2 x + x^3*2cos x*(d(cos x))/(dx)`
`(dg(x))/(dx) = 3(x*cos x)^2 + 2x^2*(x*cos x)*(-sin x)`
Factoring out `x*cos x` yields:
`(dg(x))/(dx) = (x*cos x)*(3(x*cos x) - 2x*(x*sin x))`
Factoring out x yields:
`(dg(x))/(dx) = (x^2*cos x)*(3(cos x) - 2(x*sin x))`
Hence, differentiating the given function, yields `(dg(x))/(dx) = (x^2*cos x)*(3(cos x) - 2(x*sin x).`
giorgiana1976 | Student
To calculate the first derivative of the given function, we'll use the product rule and the chain rule:
g'(x) = x^3 * (cos x)^2
We'll have 2 functions f and h:
(f*h)' = f'*h + f*h'
We'll put f = x^3 => f' = 3x^2
We'll put h = (cos x)^2 => h' = 2(cos x)*(cos x)'
h' = -2(sin x)*(cos x)
h' = - sin 2x
We'll substitute f,h,f',h' in the expression of (f*h)':
(f*h)' = 3x^2*(cos x)^2 - x^3*(sin 2x)
We'll factorize by x^2:
g'(x) = x^2[3*(cos x)^2 - x*(sin 2x)]
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# Chapter 12: Angles
In this chapter you will work with angles in different 2D shapes.
An angle is the amount of turn between two straight lines that have a common end point. The common end point is called the vertex of the angle.
The size of an angle does not depend on the length of the arms. The size of the angle depends on the amount of turn between the two arms.
In the diagram below, both angles have the same size.
angle An angle is the amount of turn between two straight lines that have a common end point
vertex The vertex of an angle is common end point where the two arms of the angle meet.
## 12.1 Sum of angles in polygons
A 2D shape has interior angles on the inside of the shape, as well as exterior angles on the outside of a shape.
interior angle Interior angles are the angles formed formed by the sides of a shape, on the inside of the shape.
exterior angle Exterior angles are outside the shape. They are created by extending one of the sides of the shape. The side of the shape and the extension must form a straight line.
In this chapter you will focus on the sum of the interior angles of different polygons.
### Sum of interior angles of a triangle
A triangle is a polygon with three sides.
Here is an activity for you to do:
Draw a big triangle $PQR$ on a sheet of paper, with angles of any size. Label the angles on the inside of the triangle as shown below. Cut out the triangle.
Tear or cut the triangle apart so that you have the three angles on three separate pieces of paper. Arrange the three angles next to each other. What do you notice?
The three angles of the triangle form a straight line. This means that the interior angles of a triangle add up to $180^{\circ}$.
We write:
In geometry we always give reasons for our answers, so we add the reason between brackets:
### Scalene triangles
One of the types of triangles we will work with is the scalene triangle. In a scalene triangle, all the interior angles are different in size, and all the sides are different in size.
As for any other triangle, the interior angles of a scalene triangle add up to $180^{\circ}$.
scalene triangle A scalene triangle has three unequal angles and three unequal sides.
### Worked example 12.1: Calculating an interior angle of a scalene triangle
Determine the size of $y$, giving a reason for your answer.
1. Step 1:Think of the geometric statement or fact that you will use, and write it down. Give the reasons between brackets.
2. Step 2: Substitute the given values and variables from the diagram into your statement.
\begin{align} \hat {P}+ \hat {Q} + \hat {R} &= 180^{\circ} \quad (\angle \text{ sum in } \triangle) \\ y + 50^{\circ} + 80^{\circ} & = 180^{\circ}\\ \end{align}
3. Step 3: Solve the equation.
\begin{align} \hat {P} + \hat {Q} + \hat {R}& = 180^{\circ} \quad (\angle \text{ sum in } \triangle) \\ y + 50^{\circ} + 80^{\circ} & = 180^{\circ}\\ y + 130^{\circ} & = 180^{\circ} \\ y + 130^{\circ} - 130^{\circ} & = 180^{\circ} - 130^{\circ} \\ \therefore y & = 50^{\circ} \\ \end{align}
4. Step 4: Check your solution.
A mental calculation will show that the sum of the three interior angles of $\triangle PQR$ adds up to $180^{\circ}$:
\begin {align} \hat {P}+ \hat {Q} + \hat {R} & = 50^{\circ} + 50^{\circ} + 80^{\circ}\\ &= 180^{\circ}\\ \end {align}
### Isosceles triangles
An isosceles triangle has two equal sides. As for all other triangles, all the interior angles of an isosceles triangle add up to $180^{\circ}$.
We need to know two geometry facts about isosceles triangles:
1. If two sides of a triangle are equal, then the angles opposite those sides will be equal.
In $\triangle DEF$, the two sides $DE$ and $DF$ are marked equal.
This means we can write the following geometric statement with its reason:
2. If two angles in a triangle are equal, then the sides opposite those angles will be equal.
In $\triangle STU$, the two angles $\hat T$ and $\hat U$ are marked equal.
This means we can write the following geometric statement with its reason:
isosceles triangle An isosceles triangle has two equal sides. The angles opposite the equal sides are equal.
### Worked example 12.2: Calculating interior angles of an isosceles triangle
Determine the values of $y$ and $x$, giving reasons for your answer.
1. Step 1: Always begin a solution with a geometric statement and a reason.
The information on the diagram shows that $BC$ = $AC$.
So, the angles opposite these sides will be equal.
Write down:
2. Step 2: Determine the value of $y$.
\begin{align} {\hat {B}} &{= \hat {A} \quad (\angle \text{s opp equal sides}) } \\ {\therefore y} & {= 70^{\circ}}\\ \end {align}
3. Step 3: Determine the value of $x$.
\begin{align} {\hat {A} + \hat {B} + \hat {C}} & { = 180^{\circ} \quad (\angle \text{ sum in } \triangle) } \\ {70^{\circ} + 70^{\circ} + x} & {= 180^{\circ}}\\ {x + 140^{\circ}} & {= 180^{\circ}} \\ {x + 140^{\circ} - 140^{\circ}} & {= 180^{\circ} - 140^{\circ}} \\ {\therefore x} & {= 40^{\circ}} \\ \end{align}
4. Step 4: Check your solution.
A mental calculation will show that the sum of the three interior angles of $\triangle ABC$ adds up to $180^{\circ}$:
\begin{align} \hat {A} + \hat {B} + \hat {C} & = 70^{\circ} + 70^{\circ} + 40^{\circ} \\ &= 180^{\circ} \\ \end{align}
Summary of the properties of triangles
• The three angles of any triangle always always add up to $180^{\circ}$.
• A scalene triangle has three angles of different sizes.
• An isosceles triangle has two equal sides. The two angles opposite the equal sides are equal.
• An equilateral triangle has three equal sides and three equal angles. Each angle is equal to $60^{\circ}$.
• A right-angled triangle has one angle of $90^{\circ}$.
### Exercise 12.1: Find the size of angles of a triangle
1. Determine the value of $x$, giving a reason for your answer.
\begin{align} \hat {Q} + \hat {R} + \hat {P} & = 180^{\circ} \quad (\angle \text{ sum in } \triangle) \\ x + 55^{\circ} + 65^{\circ} & = 180^{\circ} \\ x + 120^{\circ} & = 180^{\circ} \\ x + 120^{\circ} - 120^{\circ} & = 180^{\circ} - 120^{\circ} \\ \therefore x & = 60^{\circ} \\ \end{align}
2. Determine the value of $h$, giving a reason for your answer.
\begin{align} \hat {A} + \hat {B} + \hat {C} & = 180^{\circ} \quad (\angle \text{ sum in } \triangle) \\ 40^{\circ} + 100^{\circ} + h & = 180^{\circ} \\ h + 140^{\circ} & = 180^{\circ} \\ h + 140^{\circ} - 140^{\circ} & = 180^{\circ} - 140^{\circ} \\ \therefore h & = 40^{\circ} \\ \end{align}
3. In the triangle below, the interior angles are given as $\hat A$ = $33^{\circ}$, $\hat B$ = $m$ and $\hat C$ = $92^{\circ}$. Calculate the size of the missing angle represented by $m$.
\begin{align} \hat {A} + \hat {B} + \hat {C} & = 180^{\circ} \quad (\angle \text{ sum in } \triangle) \\ 33^{\circ} + m + 92^{\circ} + h & = 180^{\circ} \\ m + 125^{\circ} & = 180^{\circ} \\ m + 125^{\circ} - 125^{\circ} & = 180^{\circ} - 125^{\circ} \\ \therefore m & = 55^{\circ} \\ \end{align}
4. Find the value of $x$ in the diagram below, giving reasons for your answer.
\begin{align} \hat {A} + \hat {B} + \hat {C} & = 180^{\circ} \quad (\angle \text{ sum in } \triangle) \\ 90^{\circ} + x + 23^{\circ} & = 180^{\circ} \\ x + 113^{\circ} & = 180^{\circ} \\ x + 113^{\circ} - 113^{\circ} & = 180^{\circ} - 113^{\circ} \\ \therefore x & = 67^{\circ} \\ \end{align}
5. Find the value of $k$ in the diagram below, giving reasons for your answer.
All three sides are marked equal, so $\triangle XYZ$ is an equilateral triangle
\begin{align} k &= 180^{\circ} \div 3 \quad (\angle \text{s in equilateral } \triangle) \\ & = 60^{\circ} \\ \end{align}
6. In the triangle below $\angle M$ = $59^{\circ}$. It is also given that $NM$ = $NP$. Calculate the size of the missing angle represented by $z$.
\begin{align} \hat {P} & = \hat {M} \quad (\angle s \text{ opp equal sides })\\ \therefore z & = 59^{\circ}\\ \end{align}
7. Determine the value of $y$ in the diagram below, giving reasons for your answer.
\begin{align} \hat {J} + \hat {K} + \hat {L} & = 180^{\circ} \quad (\angle \text{ sum in} \triangle) \\ 47^{\circ} + 47^{\circ} + y & = 180^{\circ} \\ y + 94^{\circ} & = 180^{\circ} \\ y + 94^{\circ} - 94^{\circ} & = 180^{\circ} - 94^{\circ} \\ \therefore y & = 86^{\circ} \\ \end{align}
8. Determine the value of $y$ in the diagram below, giving a reason for your answer.
The information on the diagram shows that $QP$ = $RP$. So, the angles opposite these sides will be equal.
\begin{align} \hat {Q} &= \hat {R} \quad (\angle \text{s opp equal sides}) \\ \therefore \hat {Q} &= 57^{\circ} \\ \hat {P} + \hat {Q} + \hat {R} & = 180^{\circ} (\angle \text{ sum in } \triangle) \\ y + 57^{\circ} + 57^{\circ} & = 180^{\circ} \\ y + 114^{\circ} & = 180^{\circ} \\ y + 114^{\circ} - 114^{\circ} & = 180^{\circ} - 114^{\circ} \\ \therefore y & = 66^{\circ} \\ \end{align}
9. Determine the value of $z$ in the diagram below, giving reasons for all statements.
If two sides of a triangle are equal, then the angles opposite those sides will be equal.
\begin{align} \hat {C} & = \hat {A} \quad (\angle \text{s opp equal sides}) \\ \therefore \hat {C} & = 52^{\circ}\\ \hat {A} + \hat {B} + \hat {C} & = 180^{\circ} (\angle \text{ sum in } \triangle) \\ 52^{\circ} + z + 52^{\circ} & = 180^{\circ} \\ z + 104^{\circ} & = 180^{\circ} \\ z + 104^{\circ} - 104^{\circ} & = 180^{\circ} - 104^{\circ} \\ \therefore z & = 76^{\circ} \\ \end{align}
### Sum of interior angles of a quadrilateral
We are also going to look at the properties of the interior angles of quadrilaterals, so that we can find missing angles.
A quadrilateral is a flat shape with four straight sides and four angles. Examples of quadrilaterals are squares, rectangles, rhombuses, parallelograms and kites.
quadrilateral A quadrilateral is a flat shape with four straight sides and four angles.
We know that a rectangle has four right angles. This means that the sum of the interior angles of a rectangle is $4 \times 90^{\circ}$.
So, the sum of the interior angles of a rectangle is $360^{\circ}$.
Similarly, the sum of the interior angles of a square will also be $360^{\circ}$.
What about other quadrilaterals?
For example, what is the sum of the interior angles of quadrilateral $CBDA$ given below?
You can divide quadrilateral $CBDA$ into two triangles:
• The sum of the interior angles of $\triangle ACB$ = $180^{\circ}$.
• The sum of the interior angles of $\triangle ABD$ = $180^{\circ}$.
Therefore, the sum of the interior angles of quadrilateral $CBDA$ = $360^{\circ}$.
Any quadrilateral can be divided into two triangles. So, the sum of the interior angles of a quadrilateral is equal to the sum of the interior angles of two triangles.
\begin{align} \text{Sum of interior angles of quadrilateral} &= \text {sum of interior angles of two triangles}\\ &= 2 \times 180^{\circ}\\ &= 360^{\circ}\ \end{align}
### Worked example 12.3: Calculating interior angles of a quadrilateral
In quadrilateral $QRST$ below, $\hat{Q}$ = $x$, $\hat{R}$ = $110^{\circ}$, $\hat{S}$ = $104^{\circ}$ and $\hat{T}$ = $47^{\circ}$.
Determine the value of $x$, giving a reason for your answer.
1. Step 1: Remember that the interior angles of a quadrilateral add up to $360^{\circ}$.
You do not need to show this step in your working out. It is just here so that you can see where the fact comes from.
2. Step 2: Write down a geometric statement with a reason.
3. Step 3: Solve the equation.
\begin{align} x + 110^{\circ} + 104^{\circ} + 47^{\circ} & = 360^{\circ} \quad ( \text{sum of } \angle \text{s in quad} )\\ x + 261^{\circ} & = 360^{\circ} \\ x + 261^{\circ} - 261^{\circ} & = 360^{\circ} - 261^{\circ} \\ \therefore x & = 99^{\circ} \\ \end{align}
### Exercise 12.2: Calculate interior angles of a quadrilateral
1. Draw the following quadrilaterals: a square, a parallelogram and a kite.
Divide each quadrilateral into two triangles.
What conclusion can you make about the sum of the interior angles of a quadrilateral?
Conclusion: A square, a parallelogram and a kite are all quadrilaterals. They can all be dvided into two triangles. Therefore, the sum of the interior angles of a quadrilateral is $360^{\circ}$.
2. In quadrilateral $QRST$ below, $\hat{Q}$ = $84^{\circ}$, $\hat{R}$ = $x$, $\hat{S}$ = $105^{\circ}$ and $\hat{T}$ = $65^{\circ}$.
Determine the value of $x$, giving a reason for your answer.
\begin{align} 84^{\circ} + x + 105^{\circ} + 65^{\circ} & = 360^{\circ} \quad ( \text{sum of }\angle \text{s in quad} ) \\ x + 254^{\circ} & = 360^{\circ} \\ x + 254^{\circ} - 254^{\circ} & = 360^{\circ} - 254^{\circ} \\ \therefore x & = 106^{\circ} \\ \end{align}
3. In quadrilateral $QRST$ below, $\hat{Q}$ = $x$, $\hat{R}$ = $212^{\circ}$, $\hat{S}$ = $50^{\circ}$ and $\hat{T}$ = $48^{\circ}$.
Determine the value of $x$, giving a reason for your answer.
\begin{align} x + 212^{\circ} + 50^{\circ} + 48^{\circ} & = 360^{\circ} \quad ( \text{sum of } \angle \text{s in quad} ) \\ x + 310^{\circ} & = 360^{\circ} \\ x + 310^{\circ} - 310^{\circ} & = 360^{\circ} - 310^{\circ} \\ \therefore x & = 50^{\circ} \\ \end{align}
4. In quadrilateral $ABCD$:
• $AB \parallel CD$
• $AD \parallel BC$
1. What type of quadrilateral is $ABCD$?
$ABCD$ has two pairs of parallel sides, so $ABCD$ is a parallelogram.
1. Find the value of $w$, giving reasons for your answer.
In a parallelogram the opposite angles are equal.
So, $\hat {C}$ = $\hat {A}$
\begin{align} \hat {A} + \hat {B} + \hat {C} + \hat {D} & = 360^{\circ} \quad ( \text{sum of } \angle \text{s in quad} ) \\\ 120^{\circ} + \hat {B} + 120^{\circ} + \hat {D} & = 360^{\circ} \\ \hat {B} + \hat {D} + 240^{\circ} & = 360^{\circ} \\ \hat {B} + \hat {D} + 240^{\circ} - 240^{\circ} & = 360^{\circ} - 240^{\circ} \\ \hat {B} + \hat {D} & = 120^{\circ} \\ \hat {B} & = \hat {D} (\text{opp } \angle \text{s of parallelogram}) \\ \therefore \hat {D} & = 120^{\circ} \div 2 \\ &= 60^{\circ} \\ \therefore w & = 60^{\circ} \\ \end{align}
### Sum of interior angles of other polygons
A polygon is a closed shape made of straight lines. In a closed shape, the sides are made using one line that follows another line until you get back to where you started, and there are no openings.
polygon A polygon is a closed shape made of straight lines.
We name polygons by the number of sides they have.
Number of sides Name of polygon
3 triangle
4 quadrilateral
5 pentagon
6 hexagon
7 heptagon
8 octagon
9 nonagon
10 decagon
We can determine the sum of the interior angles of any polygon by dividing the polygon into triangles.
A pentagon is a polygon with five sides.
Draw lines from one vertex of the pentagon and divide the pentagon into triangles, as shown below:
The pentagon can be divided into three triangles.
\begin{align} \text{Sum of interior angles of pentagon} & = \text{ sum of interior angles of three } \triangle \text{s} \\ & = 3 \times 180^{\circ} \\ & = 540^{\circ}\\ \end{align}
### Worked example 12.4: Finding the sum of the interior angles of a polygon by dividing into triangles
Determine the sum of the interior angles of the polygon by dividing it into triangles.
1. Step 1: Count the number of sides and identify the polygon.
This polygon has 6 sides, so it is a hexagon.
2. Step 2: Draw lines from one vertex and divide the polygon into triangles. Write down the number of triangles.
The polygon can be divided into four triangles.
3. Step 3: Determine the sum of the interior angles of the polygon.
\begin{align} \text{Sum of interior angles of hexagon} & = \text{ sum of interior angles of four } \triangle \text{s} \\ & = 4 \times 180^{\circ} \\ & = 720^{\circ}\\ \end{align}
We have seen that:
• a polygon with 4 sides can be divided into 2 triangles
• a polygon with 5 sides can be divided into 3 triangles
• a polygon with 6 sides can be divided into 4 triangles.
This pattern will continue, so:
• a polygon with 7 sides can be divided into 5 triangles
• a polygon with 8 sides can be divided into 6 triangles
• a polygon with 9 sides can be divided into 7 triangles
• a polygon with 10 sides can be divided into 8 triangles, and so on.
There is a pattern when we divide a polygon into triangles: the number of triangles is always 2 less than the number of sides of the polygon.
The sum of the interior angles of a triangles is $180^{\circ}$.
$180^{\circ}$ = $90^{\circ}$ + $90^{\circ}$.
So, we can also say that $180^{\circ}$ is the sum of two right angles.
A table shows the pattern very clearly:
Number of sides Number of triangles Number of right angles
4 2 4
5 3 6
6 4 8
7 5 10
8 6 12
9 7 14
10 8 16
$\vdots$ $\vdots$ $\vdots$
$n$ $n - 2$ $2n - 4$
This means we can use a formula to calculate the sum of the interior angles of a polygon:
A right angle is $90^{\circ}$, so we can write:
### Worked example 12.5: Finding the sum of the interior angles of a polygon using a formula
Identify the polygon below and determine the sum of the interior angles by using a formula.
1. Step 1: Count the number of sides and identify the polygon.
The diagram in this question shows a polygon with 5 sides. This polygon is called a pentagon.
2. Step 2: Write down the formula to determine the sum of the interior angles of the polygon.
Sum of interior angles = $(2n - 4) \text { right angles}$
3. Step 3: Substitute the correct value for $n$.
A pentagon has 5 sides, so $n = 5$.
4. Step 4: Solve the equation.
\begin{align} \text{Sum of interior } \angle \text{s of pentagon} & = (2 \times 5 - 4) \text { right angles} \\ & = (10 - 4) \text { right angles} \\ & = 6 \times \text {right angles} \\ & = 6 \times 90^{\circ} \\ \therefore \text{Sum of interior } \angle \text{s of pentagon} & = 540^{\circ} \\ \end{align}
So, this polygon is a pentagon and the sum of the interior angles of a pentagon is $540^{\circ}$.
Keep the rules of BODMAS in mind when you have to simplify the following calculation: $2 \times 7 - 4$
• You have to multiply first: $2 \times 7 - 4$ = $14 - 4$
• Subtraction follows after multiplication: $14 - 4$ = $10$
### Exercise 12.3: Find the sum of interior angles of polygons
1. Identify the polygon below and determine the sum of the interior angles by using the formula.
The diagram shows a polygon with 6 sides. This polygon is called a hexagon.
A hexagon has 6 sides, so $n = 6$.
\begin{align} \text{Sum of interior } \angle \text{s of hexagon} & = (2 \times 6 - 4) \text { right angles} \\ & = (12 - 4) \text { right angles} \\ & = 8 \times \text {right angles} \\ & = 8 \times 90^{\circ} \\ \therefore \text{Sum of interior } \angle \text{s of hexagon} & = 720^{\circ} \\ \end{align}
2. Identify the polygon, and determine the sum of the interior angles, using a formula.
A polygon with 7 sides is called a heptagon.
\begin{align} \text{Sum of interior } \angle \text{s of heptagon} & = (2 \times 7 - 4) \text { right angles} \\ & = (14 - 4) \text { right angles} \\ & = 10 \times \text {right angles} \\ & = 10 \times 90^{\circ} \\ \therefore \text{Sum of interior } \angle \text{s of heptagon} & = 900^{\circ} \\ \end{align}
3. Identify the polygon and use the formula to determine the sum of the interior angles.
A polygon with 8 sides is called an octagon.
\begin{align} \text{Sum of interior } \angle \text{s of octagon} & = (2 \times 8 - 4) \text { right angles} \\ & = (16 - 4) \text { right angles} \\ & = 12 \times \text {right angles} \\ & = 12 \times 90^{\circ} \\ \therefore \text{Sum of interior } \angle \text{s of octagon} & = 1,080^{\circ} \\ \end{align}
4. Identify the polygon and use the formula to determine the sum of the interior angles.
A polygon with 10 sides is called a decagon.
\begin{align} \text{Sum of interior } \angle \text{s of decagon} & = (2 \times 10 - 4) \text { right angles} \\ & = (20 - 4) \text { right angles} \\ & = 16 \times \text {right angles} \\ & = 16 \times 90^{\circ} \\ \therefore \text{Sum of interior } \angle \text{s of decagon} & = 1,440^{\circ} \\ \end{align}
5. Identify the polygon and use the formula to determine the sum of the interior angles.
A polygon with 4 sides is called a quadrilateral.
\begin{align} \text{Sum of interior } \angle \text{s of quadrilateral} & = (2 \times 4 - 4) \text { right angles} \\ & = (8 - 4) \text { right angles} \\ & = 4 \times \text {right angles} \\ & = 4 \times 90^{\circ} \\ \therefore \text{Sum of interior } \angle \text{s of quadrilateral} & = 360^{\circ} \\ \end{align}
## 12.2 Angles of elevation and depression
In this section you are going to learn about two new types of angles. These angles depend on the way that we look at things.
### Recognising angles of elevation and angles of depression
Angles of elevation and angles of depression are angles that are formed by looking upwards or downwards from a horizontal line.
If the line of sight is upward from the horizontal line, the angle is an angle of elevation. For example, when you sit outside and look up at a bird, the position of the bird is higher than your position. So you are looking at the bird at an angle of elevation.
If the line of sight is downward from the horizontal line, the angle is an angle of depression. For example, when you stand on a cliff and look down at a friend on the road below, the position of your friend is lower than your position. So you are looking at an angle of depression.
It is important to realise that the angle of elevation is equal to the angle of depression if the two angles are taking place between the same horizontal lines. This is because the horizontal lines are parallel. When two parallel lines are intersected by a transversal, the alternate angles are equal.
angle of elevation An angle of elevation is the angle formed when the line of sight is upward from a horizontal line.
angle of depression An angle of depression is the angle formed when the line of sight is downward from a horizontal line.
### Worked example 12.6: Recognising angles of elevation and angles of depression
Wole is standing outside his house. He sees an aeroplane as well as a cat.
Show the angle of elevation and the angle of depression in the diagram.
1. Step 1: Draw the horizontal line.
Draw a horizontal line at the line of sight (from Wole's eyes).
2. Step 2: Mark the angle of elevation.
An angle of elevation is formed when the line of sight is upward from the horizontal line.
3. Step 3: Mark the angle of depression.
An angle of depression is formed when the line of sight is downward from the horizontal line.
### Exercise 12.4: Recognise angles of elevation and angles of depression
1. Yuwa is sitting outside the classroom. She looks at the top of flag pole.
Is the angle marked $x$ an angle of elevation or an angle of depression? Explain your answer.
The angle marked $x$ is an angle of elevation.
Yuwa is looking upwards from the horizontal line.
2. Habib is standing on a cliff next to a lake. He sees a fishing boat.
Is the angle marked $\alpha$ an angle of elevation or an angle of depression? Explain your answer.
The angle marked $\alpha$ is an angle of depression.
Habib is looking downwards from the horizontal line.
3. A bird is sitting at point $B$ on the roof of a building. Which angle in the diagram is an angle of depression? Which angle in the diagram is an angle of elevation?
$D \hat{B}C$ is an angle of depression.
$D \hat{B}E$ is an angle of elevation.
4. In the diagram below $AC$ is a tree. The angle of elevation from $D$ to $A$ is $x$.
Which other angle in the diagram is also equal to $x$? Give a reason for your answer.
$B \hat{A} D$ = $x$.
Reason: $AB \parallel CD$ because they are both horizontal lines.
If two parallel lines are intersected by a transversal, the alternate angles are equal.
## 12.3 Practical applications
It is not always possible to measure the actual size of an object, for example the height of a tall tree. Then we can make use of a scale drawing where we also include an angle of elevation.
We need to know what the ratio is in which we have to reduce (or enlarge) the real life object to do a scale drawing. This ratio is called the scale of the drawing.
For example, a scale of $1 : 5$ means that 1 unit on the scale drawing represents 5 units in real life.
scale drawing A scale drawing is a diagram of a real object where every measurement is reduced (or enlarged) in the same proportion.
scale The scale of a scale drawing tells us what ratio to use to reduce (or enlarge) the size of the real life object to make the scale drawing.
### Worked example 12.7: Using scale in scale drawings
Idara wants to know the height of a tree that is standing across the road from her house. The tree is too high to measure, but she knows that the angle of elevation from the road to the top of the tree is $35^{\circ}$. The distance between Idara and the base of the tree is 10 metres.
Make use of your knowledge of scale drawing and angles of elevation to calculate the height of the tree. Use a scale of $1 : 100$.
1. Step 1: Fill in all the given information on a rough sketch.
It is a good idea to label the vertices of the triangle. It is easier to work with a named triangle, for example $\triangle ABC$, when you have to do calculations.
2. Step 2: Use the given scale to calculate the length of the line segment on the scale drawing.
1 metre = 100 cm, so 10 metres = 1,000 cm
\begin{align} \text {Length BC in real life} & = \text{1,000 cm}\\ \text {Length of BC on scale drawing} & = \text{1,000 cm} \div 100\\ &= \text{10 cm} \end{align}
3. Step 3: Make an accurate scale drawing.
Use the following steps:
• Construct line segment $BC$ = 10 cm.
• Construct $\hat {B}$ = $35^{\circ}$.
• Construct $AC \perp BC$.
It is possible that the diagram given above does not display correctly on your device. The accurate scale drawing should have the given measurements.
4. Step 4: Measure the line segment representing the height of the tree.
Line segment $AC$ represents the height of the tree.
$AC$ = 7 cm
5. Step 5: Calculate the height of the tree in real life.
\begin{align} \text {Length of AC on scale drawing} & = \text{7 cm}\\ \text {Length of AC in real life} & = \text{7 cm} \times 100\\ &= \text{700 cm}\\ &= \text {7 metres} \end{align}
The height of the tree is 7 metres.
When we do scale drawings, we have to measure the lengths of line segments on diagrams accurately. It is possible that the given diagrams do not display correctly on your device. If so, you will not be able to measure the lengths of the line segments as expected.
To assist you, the expected measurements are given in each question in the exercise below. You may use these measurements to do the necessary calculations.
### Exercise 12.5: Use scale in scale drawings
1. The distance from where Uyai is standing to the foot of a pole is 420 cm. The angle of elevation from $A$ to the top of the pole is $65^{\circ}$.
Make a scale drawing to determine the height of the pole. Use a scale of $1 : 100$.
\begin{align} \text {Length AC in real life} & = \text{420 cm}\\ \text {Length of AC on scale drawing} & = \text{420 cm} \div 100\\ &= \text{4.2 cm} \end{align}
Construct a scale drawing, using the following steps:
• Construct $CA$ = 4.2 cm.
• Construct $\hat A$ = $65^{\circ}$.
• Construct $CB \perp CA$.
Measure the length of $BC$.
Expected measurement for $BC$: $BC$ = 9 cm
\begin{align} \text {Length of BC on scale drawing} & = \text{9 cm}\\ \text {Length of BC in real life} & = \text{9 cm} \times 100\\ &= \text{900 cm}\\ &= \text {9 metres} \end{align}
The height of the pole is 9 metres.
2. Yabani is standing on a bridge (point $P$). He looks downward at an angle of $55^{\circ}$ to his friend who is swimming in the river below (at point $Q$). How far is his friend from him if the distance from $P$ to $R$ is 40 metres in real life?
Draw a scale drawing, using a scale of $1 : 1,000$.
1 metre = 100 cm, so 40 metres = 4,000 cm
\begin{align} \text {Length PR in real life} & = \text{4,000 cm}\\ \text {Length of PR on scale drawing} & = \text{4,000 cm} \div 1,000\\ &= \text{4 cm} \end{align}
Construct a scale drawing, using the following steps:
• Construct $PR$ = 4 cm.
• Construct $R \hat{P} Q$ = $55^{\circ}$.
• Construct $RQ \perp PR$.
Measure the length of $PQ$.
Expected measurement for $PQ$: $PQ$ = 7 cm
\begin{align} \text {Length of PQ on scale drawing} & = \text{7 cm}\\ \text {Length of PQ in real life} & = \text{7 cm} \times 1,000\\ &= \text{7,000 cm}\\ &= \text {70 metres} \end{align}
The distance between Yabani and his friend is 70 metres.
3. Two buildings are 20 metres apart from each other. The angles of elevation and depression from the roof of the one building are shown in the diagram below.
If $EC$ is the height of the second building), make use of a scale drawing to determine this height. Use a scale of $1 : 400$.
If the buildings are 20 metres apart, it means that $BD$ = 20 m also.
1 metre = 100 cm, so 20 metres = 2,000 cm
\begin{align} \text {Length BD in real life} & = \text{2,000 cm}\\ \text {Length of BD on scale drawing} & = \text{2,000 cm} \div 400\\ &= \text{5 cm} \end{align}
Construct a scale drawing of $\triangle EBC$ using the following steps:
• Construct $BD$ = $\text{5 cm}$.
• Construct $E \hat {B} D$ = $34^{\circ}$.
• Construct $D \hat {B} C$ = $62^{\circ}$.
• Construct $EDC \perp BD$.
Measure the length of $EC$ on the scale drawing.
Expected measurement for $EC$: $EC$ = 12.7 cm
\begin{align} \text {Length of EC on scale drawing} & = \text{12.7 cm}\\ \text {Length of EC in real life} & = \text{12.7 cm} \times 400\\ &= \text{5,080 cm}\\ &= \text {50.8 metres} \end{align}
The height of the second building is 50.8 metres.
## 12.4 Summary
• An angle is the amount of turn between two straight lines that have a common end point. The common end point is called the vertex of the angle.
• A 2D shape has interior angles as well as exterior angles.
• Interior angles are inside the shape.
• Exterior angles are outside the shape.
• The sum of the interior angles of a triangle is $180^{\circ}$.
• The sum of the interior angles of a quadrilatral is $360^{\circ}$.
• Polygons are named according to the number of sides they have:
Number of sides Name of polygon
3 triangle
4 quadrilateral
5 pentagon
6 hexagon
7 heptagon
8 octagon
9 nonagon
10 decagon
• We can determine the sum of the interior angles of any polygon by dividing the polygon into triangles. This table shows the pattern:
Number of sides Number of triangles Number of right angles
3 1 2
4 2 4
5 3 6
6 4 8
7 5 10
8 6 12
9 7 14
10 8 16
$\vdots$ $\vdots$ $\vdots$
$n$ $n - 2$ $2n - 4$
• The formula to calculate the sum of the interior angles of a polygon is: sum of interior angles of a polygon with $n$ sides $=$ $(2n - 4)$ right angles
• An angle of elevation is the angle formed when the line of sight is upward from a horizontal line.
• An angle of depression is the angle formed when the line of sight is downward from a horizontal line.
• The angle of elevation is equal to the angle of depression, because the horizontal lines are parallel. When two parallel lines are intersected by a transversal, the alternate angles are equal.
• We can make use of a scale drawing and an angle of elevation or an angle of depression to solve practical problems.
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# Solve this
Question:
If $\cos \theta=\frac{-\sqrt{3}}{2}$ and $\theta$ lies in Quadrant III, find the value of all the other five trigonometric functions.
Solution:
Given: $\cos \theta=\frac{-\sqrt{3}}{2}$
Since, θ is in IIIrd Quadrant. So, sin and cos will be negative but tan will be positive
We know that,
$\cos ^{2} \theta+\sin ^{2} \theta=1$
Putting the values, we get
$\left(-\frac{\sqrt{3}}{2}\right)^{2}+\sin ^{2} \theta=1$ [given]
$\Rightarrow \frac{3}{4}+\sin ^{2} \theta=1$
$\Rightarrow \sin ^{2} \theta=1-\frac{3}{4}$
$\Rightarrow \sin ^{2} \theta=\frac{4-3}{4}$
$\Rightarrow \sin ^{2} \theta=\frac{1}{4}$
$\Rightarrow \sin \theta=\sqrt{\frac{1}{4}}$
$\Rightarrow \sin \theta=\pm \frac{1}{2}$
Since, $\theta$ in III $^{\text {rd }}$ quadrant and $\sin \theta$ is negative in III $^{\text {rd }}$ quadrant
$\therefore \sin \theta=-\frac{1}{2}$
Now,
$\tan \theta=\frac{\sin \theta}{\cos \theta}$
Putting the values, we get
$\tan \theta=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$
$=-\frac{1}{2} \times\left(-\frac{2}{\sqrt{3}}\right)$
$=\frac{1}{\sqrt{3}}$
Now,
$\operatorname{cosec} \theta=\frac{1}{\sin \theta}$
Putting the values, we get
$\operatorname{cosec} \theta=\frac{1}{-\frac{1}{2}}$
$=-2$
Now,
$\sec \theta=\frac{1}{\cos \theta}$
Putting the values, we get
$\sec \theta=\frac{1}{-\frac{\sqrt{3}}{2}}$
$=-\frac{2}{\sqrt{3}}$
Now,
$\cot \theta=\frac{1}{\tan \theta}$
Putting the values, we get
$\cot \theta=\frac{1}{\frac{1}{\sqrt{3}}}$
$=\sqrt{3}$
Hence, the values of other trigonometric Functions are:
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Short Blocks
# Maths Year 1/2 Autumn Addition & Subtraction (A)
Each unit has everything you need to teach a set of related skills and concepts.
First time using Hamilton Maths?
The PowerPoint incorporates step-by-step teaching, key questions, an in-depth mastery investigation, problem-solving and reasoning questions - in short, everything you need to get started.
All the other resources are there to support as-and-when required. Explore at your leisure - and remember that we are always here to answer your questions.
## Unit 1 Partition numbers; learn number bonds (suggested as 5 days)
### Objectives
Partition numbers and learn number bonds
Unit 1: ID# 12174
Y1: Partition to create number bonds
National Curriculum
Y1 Hamilton Objectives
7. Know number bonds to 10, e.g. 5 + 5, 6 + 4, etc. Also know what is left if objects are taken from 10, e.g. 10 fingers, fold down 4, leaves 6 standing.
8. Begin to know pairs which make 5, 6, 7, 8, 9 and 20.
12. Solve missing number problems and addition/subtraction problems in number stories.
Y2 Hamilton Objectives
7. Know securely number pairs for all the numbers up to and including 20.
14. Recognise that addition and subtraction are inverse operations; use addition to check subtractions and solve missing number problems.
### Planning and Activities
Day 1 Teaching
Use five pegs on a hanger and a cloth to model different additions to partition 5.
Further Teaching with Y2
Use a number balance to model ‘balancing sums’ to 7, 8 and 9, e.g. 5 + = 6 + 1
Group Activities: T with Y2
Y1 -- Make ‘snakes’ using five cubes in two colours. Record the additions.
Y2 -- Use number balances to create ‘balancing sums’.
Day 2 Teaching
Use six pegs on a hanger and a cloth to model different additions and subtractions. Use a 20-bead frame to show linked pairs to 10 and 20, e.g. 5 + = 10 and 15 + = 20.
Group Activities: T with Y1
Use the ‘Domino loop’ in-depth problem-solving investigation below as today’s group activity.
Or, use these activities:
Y1 -- Solve a missing eggs problem … or … Make pairs to 6 using egg boxes and modelling clay eggs.
Y2 -- Explore pairs of numbers that add to 20.
Day 3 Teaching
Use a stick of cubes and bar models to show pairs to 10 and 20.
Further Teaching with Y1
Break sticks of 10 in different places; write matching additions.
Group Activities: T with Y1
Y1 -- Explore pairs to 10, making ‘snakes’ of cubes in two colours or by matching sticks of 10 cubes to bar models.
Y2 -- List all possible additions and subtractions for 10 or 20 to go with a number story.
Day 4 Teaching
Use fingers to find pairs to 10 and record additions.
Further Teaching with Y2
Use a 20-bead frame to explore complements to 20, first adding to the next 10.
Group Activities: T with Y2
Y1 -- Explore complements to 10 using sticks of cubes and number cards, or by playing number bond snap.
Y2 -- Take a card and say what needs to be added to make 20; record the matching addition and subtraction.
Day 5 Teaching
Partition 10 pegs in two colours on a hanger and record the related additions. Hide some pegs and record missing number sentences. Children say how many are hidden.
Further Teaching with Y2
Write number sentences with symbols standing for mystery numbers. Children work out what numbers they represent.
Group Activities: T with Y2
Y1 -- Place 10 spots on ladybirds in different ways. Explore all possibilities.
Y2 -- Use number facts to work out how many cubes are hidden under a cup or in a number sentence under sticky notes.
### You Will Need
• Pegs in a variety of colours (plus 6 in same colour)
• Coat hanger and cloth
• Number balance and number weights
• Coloured interlocking cubes
• Eggs made of modelling clay or real hardboiled eggs, Baskets, Egg box
• Long coloured strips of paper
• Sticky notes (two different colours)
• Modelling clay and counters
• 20-bead strings, Sand timer (e.g. 5 minute)
• ‘Slidy box' cards
• ‘Pairs to 10: cubes’ (see resources) or interconnecting cubes in the same colours
• Additional activity sheets (see resources)
• A3 paper
• Flipchart
• 0–20 number cards
• Strips of paper and plastic cups
### Short Mental Workouts
Day 1
Count on and back to 20
Day 2
Count on to 100
Day 3
Numbers to 100
Day 4
Count in tens
Day 5
Number facts up to 10
### Worksheets
Day 1
Y1: Partition five biscuits onto two plates in different ways.
Y2: Fill in missing sums with totals of 7, 8 and 9. Draw weights to create balancing sums.
Day 2
Y1: Colour eggs in boxes and write the matching addition and subtraction bond to 6.
Y2: Partition 10. Find the missing numbers for the totals of 10 and 20.
Day 3
Y1: How many are needed to make 10?
Y2: Complete bar models for pairs to 10 and 20.
Day 4
Y1: Make pairs of balloons that add up to 10.
Y2: Draw bar models for pairs to 20.
Day 5
Y1: Use pairs to work out how many pegs are hidden, or cross out pegs to compete subtraction facts.
Y2: Use number facts to work out mystery numbers.
### Mastery: Reasoning and Problem-Solving
Y1
• Write the number of blank bricks in these diagrams.
Diagram1:
6 // // // //
Diagram 2:
7 // // // //
Diagram 3:
5 // //
• The crocodile has 6 fish. He eats 4. How many are left?
• The mouse has 5 bits of cheese. She finds 2 more. How many has she now?
• Write all the different ways of making 7, e.g. 1 and 6. Make sure you have them all!
• Fill in the missing numbers.
5 + ☐ = 10
7 + ☐ = 10
2 + ☐ = 10
Y2
• Write the pairs of possible missing numbers:
20 11 ? ?
• How many different ways are there of writing numbers in the two boxes in this sentence?
☐ + ◯ = 10
What if the two boxes = 9? Or = 8?
• You have 21 cards: numbers 0 to 20. You can create how many pairs of numbers making 20? Write these. Write the pair you cannot create.
In-depth Investigation: Domino Loop
Children make loops of dominoes such that meeting ends have a given total number of spots.
### Extra Support
Y1
Give Me Five
Stick Sums
Y2
Ten Totally Rocks
Finding pairs with a total of 10
## Unit 2 Add by counting on in 1s or 10s (suggested as 3 days)
### Objectives
Add by counting on in 1s or 10s
Unit 2: ID# 12190
Y1: Add small numbers by counting on
Y2: Add 10, 11, 20 and 21
National Curriculum
Y1: Addition and subtraction (i) (iii)
Y2: Addition and subtraction (i) (iii)
Y1 Hamilton Objectives
10. Recognise the + and – and = signs, and use these to read and write simple additions and subtractions.
11. Add small numbers by counting on.
Y2 Hamilton Objectives
10. Add a 2-digit number and 10s.
### Planning and Activities
Day 1 Teaching
Use ‘spider counting’ to count in 10s, then to add 10 to a number on the 100-square.
Further Teaching with Y1
Demonstrate how to add by counting on, using soft toys getting onto a ‘bus’.
Group Activities: T with Y1
Use the ‘Dotty triangle corners’ in-depth problem-solving investigation below as today’s group activity.
Or, use these activities:
Y1 – Add by counting on, to a maximum of 10 or 11.
Y2 -- Add 10, then 20 to 2-digit numbers on a 100-square.
Day 2 Teaching
Model counting on 1, 2, 3, 4 and 5 more, from five fingers.
Further Teaching with Y2
Count on from 7p in 10s using 10p coins. Add 10p to prices.
Group Activities: T with Y2
Y1 -- Add by counting on, either with a snake of cubes, or spots on dominoes. Order dominoes.
Y2 -- Make 2-digit prices less than 90p, adding 10p.
Day 3 Teaching
Count on 1, 2 and 3 from numbers on a 20-bead bar.
Further Teaching with Y2
Add 10, then 11 to numbers on a 100-square. Add 20, then 21.
Group Activities: T with Y2
Y1 -- Add 1, 2 or 3 to numbers to 20 by counting on.
Y2 -- Add 10 and 11, then 20 and 21 to numbers between 20 and 70 on a 100-square. Some children will focus on adding 11p to 2-digit prices.
### You Will Need
• ‘1-100 grid ’ (see resources)
• Plastic spider and 10 soft toys
• A3 paper and two large spotty dice
• 10s and 1s place value dice or cards
• Sticky notes and sets of dominoes
• Coins (1p, 2p, 5p, 10p) and a tin
• Five items with 2-digit prices less than 90p
• Coloured interlocking cubes
• Strips of coloured paper
• 20-bead bar and dice with faces labelled +1, +2 and +3
• 1–6 spotty dice and a dice labelled 1, 1, 1, 2, 2, 2
• Five items with price tags
### Short Mental Workouts
Day 1
Count on
Day 2
Pairs to 10
Day 3
Finger pairs to 10
### Worksheets
Day 1
Y1: Count on to find the total of two dice.
Y2: Use addition to fill in vertical strips from a 100-square.
Day 2
Y1: Count on to find the number of fingers.
Y2: Add 10p to cake prices.
Day 3
Y2: Add 10 and 11. Add 10 and 11, 20 and 21.
### Mastery: Reasoning and Problem-Solving
Y1
• Draw 5 faces on a bus. Let 2 or 3 people get on. How many on the bus now? Write the addition.
• What is 5 add 4?
• Use a 0 to 10 number line to help you complete the sums below:
Point at the first number and count on.
5 + 3 = ☐
6 + 1 = ☐
4 + 4 = ☐
7 + 2 = ☐
• 5 little frogs
Sitting on a log
On hop 3
• How many can there be?
Y2
• What number adds to 63 to make 83?
What number adds to 29 to make 40?
What number adds to 50 to make 71?
• Write the missing numbers in the sequences:
13 ☐ 53 ☐ 93 113
42 ☐ 64 75 ☐ ☐
• Complete this grid:
+ 31 ? 11 ? 58 ? 43 ?
• Write the missing numbers.
32 + ☐ = 43
57 + ☐ = 78
In-depth Investigation: Dotty Triangle Corners
Children use trial and improvement to find different ways of making a given total by adding three numbers.
### Extra Support
Y1
What's Next Norris?
Saying the next number after any number from 1 to 9
Room For One More!
Adding 1 to any number from 1 to 9, by saying the next number
Y2
Spider's Workout
Adding and subtracting 10 using a 100-square
Easy Come, Easy Go
Adding and subtracting 10, using coins
## Unit 3 Counting back; understand + and – (suggested as 5 days)
### Objectives
Subtract by counting back; understand + and –
Unit 3: ID# 12206
Y1: Count back to subtract, inverse operations, word problems.
Y2: Count back 2, 10, 11, 20, 21, inverse operations, word problems.
National Curriculum
Y1: Addition and subtraction (i) (iii) (iv)
Y2: Addition and subtraction (i) (iii) (v)
Y1 Hamilton Objectives
10. Recognise the + and – and = signs, and use these to read and write simple additions and subtractions.
11. Add small numbers by counting on and subtract small numbers by counting back.
12. Solve addition/subtraction problems in number stories.
Y2 Hamilton Objectives
10. Add a 2-digit number and 10s.
11. Count back in 1s or 10s or use number facts to take away.
14. Recognise that addition and subtraction are inverse operations; use addition to check subtractions and solve missing number problems.
15. Solve problems involving addition and subtraction of numbers, quantities and measures, using recall of number facts and appropriate models and images.
### Planning and Activities
Day 1 Teaching
Count on and back on a 1–20 number track. Point to a number: Y1 Say the number that is 1 less; Y2 Say the number that is 2 less.
Further Teaching with Y1
Using a bus context, model counting back 1 and 2 to subtract from numbers to 10.
Group Activities: T with Y1
Use the in-depth problem-solving investigations ‘Number Balance’ (Y1) or ‘Getting the Balance’ (Y2) from NRICH as today’s group activity.
Or, use these activities:
Y1 – Count back to find how many dogs are left in the rescue centre, or are left on the bus.
Y2 -- Subtract 2 from 2-digit numbers using bead strings or the 100-square.
Day 2 Teaching
Start with 10 pegs on a coat hanger. Keep removing one or two pegs, recording subtractions until no pegs are left.
Further Teaching with Y2
Use ‘spider counting’ to count back in 10s on a 100-square. Subtract 10, then 20 from 2-digit numbers.
Group Activities: T with Y2
Y1-- Make a tower of cubes and take away the number on a dice. Record the number sentence.
Y2 -- Subtract 10 to make 2-digit numbers on a 100-square. Subtract 10, then 20 by filling in ‘strips’ from a 100-square.
Day 3 Teaching
Place five cubes in a hat. Roll a large dice, sides labelled +1, +2, +2, –1, –2 and –2. Children add and subtract accordingly, with
Y2s -- Recording number sentence
Further Teaching with Y2
Subtract 10, then 11 from numbers on a 100-square. Subtract 20, then 21.
Group Activities: T with Y2
Y1 -- Game with towers of cubes. Take away and add 1 and 2 to see who can reach 10 or 0 cubes first.
Y2 -- Subtract 10 and 11 from numbers between 40 and 100 on a 100-square. Some children will also subtract 20 and 21.
Day 4 Teaching
Group Activities: T with Y1
Y1 -- Explore the relationship between adding and subtracting by calculating how much money is left when a child spends some and then earns some back.
Y2 -- Work out what subtraction needs to be done to ‘undo’ an addition (+11, +21) and vice versa. Some children will write addition and subtraction chains,
Day 5 Teaching
Read a word problem about dogs in a rescue centre. Model how to use fingers or cubes to represent numbers in the story. Write a matching calculation and repeat with other stories.
Group Activities: T with Y1 or Y2
Y1 -- Explore practical number stories using equipment to represent numbers; create ‘leaf posters’ to show number sentences.
Y2 -- Choose to add or subtract to solve ‘worm’ number stories.
### You Will Need
• ‘1–20 track’ (see resources)
• ‘Count back bus’ (see resources)
• How many dogs? (cut out images so that you can subtract from any number between 5 and 10 - see resources)
• Additional activity sheets (see resources)
• Two dice, one labelled 1, 1, 1, 2, 2, 2 and one labelled 2, 2, 2, 3, 3, 3
• 5 red pegs, 5 yellow pegs and coat hanger
• Number cards 5–10 and 5–15
• 10s and 1s place value dice or cards
• Flipchart, Sticky notes
• Interlocking cubes, a hat and plastic spider
• Large dice, labelled +1, +2, +2, –1, –2, –2
• A3 paper, Strips of paper
• 20-bead bar and ten £1 coins
• Leaves
### Short Mental Workouts
Day 1
Count on to 100
Day 2
Count on and back in 10s
Day 3
Count back
Day 4
Count on and back in 10s
Day 5
Number bonds to 10
### Worksheets
Day 1
Y1: Subtract 1 or 2 frogs. Subtract 1 or 2 using the context of people getting off a bus.
Y2: Subtract 2 on a beaded line. Subtract 2 to fill in ladders.
Day 2
Y1: Work out how tall the tower is after some cubes have fallen off.
Y2: Use subtraction to fill in vertical strips from a 100-square.
Day 3
Y1: Add/subtract 1, 2 or 3 cubes to/from a bag.
Y2: Subtract 10 and 11, 20 and 21.
Day 4
Y1: Add then subtract 2 or 3 cubes to/from towers.
Y2: Add then subtract 10, 11, 20 and 21.
Day 5
Y1: Dog addition and subtraction number stories.
Y2: Garden addition and subtraction number stories.
### Mastery: Reasoning and Problem-Solving
Y1
• Set up using small world play equipment.
There are 7 ducks in the pond. If 2 get out, how many are left?
Write a matching number sentence. If 3 more get in, how many ducks are on the pond? Write a matching number sentence.
If 2 get out, how many are left on the pond? Write a matching number sentence.
• Draw a number track from 1 to 10.
1 2 3 4 5 6 7 8 9 10
Place your counter on a number. Write it. Now hop along 2. Where do you end up? Write the addition. Repeat.
Place your counter on a number. Write it. Now hop back 2. Where do you end up? Write the subtraction. Repeat
• Write the missing numbers:
7 – 3 = ☐
5 + ☐ = 6
☐ + 2 = 5
Y2
• What number is subtracted from 83 to give 73?
What number is subtracted from 50 to give 29?
What number is subtracted from 48 to give 37?
• Write the missing numbers in the sequences:
76, 56, ☐, ☐
☐, 85, 74, ☐, ☐
☐, ☐, 45, ☐, ☐, 39
• Write the missing numbers:
76 – ☐ = 55
☐ – 11 = 47
• How many times can 21 be subtracted from 100 before you get a number smaller than 21?
In-depth Investigation
Y1: Number Balance
Can you hang weights in the right place to make the equaliser balance? Number Balance from nrich.maths.org.
Y2: Getting the Balance
If you hang two weights on one side of this balance, in how many different ways can you hang three weights on the other side for it to be balanced? Getting the Balance from nrich.maths.org.
### Extra Support
Y1
What's Before Beryl?
Saying the number before any number up to 6, then 10
Dig for Acorns
Subtracting 1 from any number from 2 to 6, and then from 2 to 10, by saying the number which comes before
Y2
Spider's Workout
Adding and subtracting 10 using a 1-100 grid
Easy Come, Easy Go
Adding and subtracting 10 using coins
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Get ICSE Solutions for Class 10 Mathematics Chapter 5 Linear Inequations (Solving Linear Inequations in One Variable) for ICSE Board Examinations on ICSESolutions.com. We provide step by step Solutions for ICSE Mathematics Class 10 Solutions Pdf. You can download the Class 10 Maths ICSE Textbook Solutions with Free PDF download option.
## Linear Inequations (Solving Linear Inequations in One Variable) Class 10 Maths ICSE Solutions
### Formulae
Two permissible rules:
If the same number or expression is added to or subtracted from both sides of an inequation, the resulting inequation has the same solution (or solutions) as the original.
2. Multiplication – Division Rule:
(i) If both sides of an inequation are multiplied or divided by the same positive number, the resulting inequation has the same solution (or solutions) as the original.
(ii) If both sides of an inequation are multiplied or divided by the same negative number, the resulting inequation has the same solution (or solutions) as the original if the symbol of the inequality is reversed.
Thus, the only difference between solving a linear equation and solving an inequation concerns multiplying or dividing both sides by a negative number. Therefore, always reverse the symbol of an inequation when multiplying or dividing by a negative number.
3. Properties of absolute values:
### Determine the Following
Question 1. Give that x ∈ I. Solve the inequation and graph the solution on the number line:
Question 2. Graph the solution set for each inequality:
Question 3. Solve the given inequation and graph the solution on the number line
Question 4. Given that x ∈ R, solve the following inequality and graph the solution on the number line:
Question 5. Given:
Question 6. Solve 2 ≤ 2x – 3 ≤ 5, x ∈ R and mark it on a number line.
Question 7. For each inequality, determine which of the given numbers are in the solution set:
Question 8. Graph the solution sets of the following inequalities:
Question 9. Solve the equation and represent the solution set on the number line.
Question 10. Solve the following inequation and represent the solution set on the number line:
Question 11. Solve the following in equalities and graph their solution set
Question 12. Solve the following inequation and graph the solution set,
Question 13. Solve the following inequation and graph the solution on the number line.
Question 14. Solve the following inequalities and represent the solution on a number line:
Question 15. Solve the following inequalities and represent the solution set on a number line:
Question 16. Solve the following inequation, write the solution set and represent it on the number line:
Question 17. Find the values of x, which satisfy the inequation
Question 18. Solve the following inequalities in the given universal set:
Question 19. Find the solution set of the following inequalities and draw the graph of their solutions sets:
Question 20. Solve the following inequalities and graph their solution set:
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### Hex
Explain how the thirteen pieces making up the regular hexagon shown in the diagram can be re-assembled to form three smaller regular hexagons congruent to each other.
### Transformation Game
Why not challenge a friend to play this transformation game?
### Growing Rectangles
What happens to the area and volume of 2D and 3D shapes when you enlarge them?
# Who is the fairest of them all ?
##### Age 11 to 14Challenge Level
Isabel from St Andrew's Scots School in Argentina used Geogebra to investigate the problem. This is Isabel's work, with some teacher comments.
Robert from Kings Ely said this very concisely:
The vector for the flags is $\frac{n-1}{n}$ of the vector of centres.
See if you can follow this explanation of why this is the result using this diagram:
Suppose the first scale factor is $k$ (so the second one is $\frac{1}{k}$). Let $\mathbf{x}$ denote the vector from the first centre of enlargement to the second. Then the required transformation is a translation by the vector $\frac{k-1}{k}\mathbf{x}$.
To see this, consider a single point on the flag. (If we show that the required transformation for a single point is the given translation, then the same will apply to the flag.) Let $\mathbf{a}$ denote the vector from the first centre of enlargement to the point. Then the vector from the first centre of enlargement to the image of the point under the first enlargement will be $k\mathbf{a}$. The vector from the second centre of enlargement to this image will be $k\mathbf{a}- \mathbf{x}$, so the vector from the second centre of enlargement to the final image will be $\frac{1}{k}\left(k\mathbf{a}-\mathbf{x}\right) =\mathbf{a}-\frac{1}{k}\mathbf{x}$. So the vector from the initial point to the final image will be $-\mathbf{a}+\mathbf{x}+\mathbf{a}- \frac{1}{k}\mathbf{x}=\frac{k-1}{k}\mathbf{x}$, as required.
Here is the diagram with k=2:
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# s36 - = 1/G Pythagoras’s Theorem gives the relationships...
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Solution to Problem36 Congratulations to this week’s winner Mike Fitzpatrick Correct solutions were also received from Ray Kremer, William Webb, Greg Falcon, and Jeff Dowin. Two partial and one incorrect solution were also received. Greg Falcon and William Webb provided very different solutions. First I’ll sketch Greg Falcon’s solution. [IMAGE] In the diagram on the left F is the distance along the wall from the top of the ladder to the top of the box and G is the horizontal distance from the bottom of the ladder to the right hand edge of the box. The point P is halfway down the ladder, that is, 3.5 feet from both top and bottom, and D is the distance from P to the upper right hand corner of the box, marked C. The three triangles are all similar, which gives the relationship F
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Unformatted text preview: = 1/G. Pythagoras’s Theorem gives the relationships 1 + F 2 = (3.5 + D) 2 and 1 + G 2 = (3.5 - D) 2 F 2 = (4.5 + D)(2.5 + D) and G 2 = (4.5 - D)(2.5 - D) Multiply these last equations together and use F = 1/G to get 1 = (20.25 - D 2 )(6.25 - D 2 ) which yields a quadratic equation in D 2 . The rest is straightforward. The height of the ladder is [IMAGE] William Webb’s solutions is trigonometric in nature. Start by labeling the lower right hand angle A. The distance from C to B is csc(A) while that from C to T is sec(A). He now uses various trig identities to show that sin(2A) is a root of the quadratic equation 49x 2- 4x - 4 = 0. The rest follows easily. to this page....
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# Which shape has the greatest number of lines of symmetry Brainly?
## Which shape has the greatest number of lines of symmetry Brainly?
Answer: square has the greatest number of lines of symmetry in these following.
Which shape has the greatest number of lines of symmetry besides a circle?
A circle has infinite lines of symmetry. Likewise, a triangle has three lines of symmetry, while rectangle and square have four such lines which divide them into identical parts.
### How do you find the number of lines of symmetry?
Look at the shape. To find a line of symmetry, look for two halves of the polygon that are mirror images of each other. The right half of the shape is the mirror image of the left half. So, the vertical line that passes through the middle is a line of symmetry.
What is the greatest number of lines of symmetry that can be drawn on the figure?
2 Answers By Expert Tutors. You can draw 4 lines of symmetry: 1 through each diagonal, one cutting in half vertically, and another cutting in half horizontally. Nitin P.
## What is a symmetry line?
A line of symmetry is a line that cuts a shape exactly in half. This means that if you were to fold the shape along the line, both halves would match exactly. Equally, if you were to place a mirror along the line, the shape would remain unchanged. A square has 4 lines of symmetry, as shown below.
Which shape has only one line of symmetry?
Kite Rhombus (all sides equal length) 1 Line of Symmetry 2 Lines of Symmetry
### What shape has no line of symmetry?
scalene triangle
A scalene triangle – A scalene triangle has no line symmetry because it has all sides unequal in length. A parallelogram – A parallelogram has opposite sides equal and parallel but a parallelogram has no line of symmetry. A trapezium – In a trapezium, only two opposite sides are parallel to each other.
What is symmetry and example?
Symmetry is an attribute where something is the same on both sides of an axis. An example of symmetry is a circle that is the same on both sides if you fold it along its diameter.
## Which shape has no line of symmetry?
A scalene triangle, parallelogram, and a trapezium are three examples of shapes with no line of symmetry.
What shape has 2 line of symmetry?
rectangle
A rectangle has two lines of symmetry. It has rotational symmetry of order two.
### Which is a line of symmetry?
What shapes have two lines of symmetry?
Two Lines of Symmetry: Some figure can be divided into two equal parts with two lines. These shapes are said to have Two Lines of Symmetry. The rectangle is an example of Two Line of Symmetry. A Rectangle can be divided vertically, horizontal or diagonally to get the two symmetrical parts. A rhombus also has two lines of symmetry.
## How many lines of symmetry does a line have?
You can make only four lines of symmetry: a vertical line in the middle of the square, a horizontal line in the middle of the square, a line along one diagonal,
How many lines of symmetry does a rectangle?
Rectangles are geometric shapes with two lines of symmetry. One line of symmetry divides a rectangle in half horizontally, and the other line divides the shape in half vertically. Rectangles also have four right angles and two pairs of lines, with the lines in each pair having equal lengths.
### What is a line of symmetry?
Line of symmetry. A symmetrical figure has a line of symmetry (or mirror symmetry or reflection symmetry). If there is a line about which it can be folded so that the two equal parts of the figure coincide, that line is called the line of symmetry or line of reflection.
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## TRICKY MATHS RIDDLES WITH ANSWERS
This blog explains about TRICKY MATHS RIDDLES WITH ANSWERS . It is explained clearly well below :
1. If 6 + 4 = 210
9 + 2 = 711
? + ? = 113
Then what is the number ?
It traces the pattern that ( a – b ) followed by (a + b )
( 6 – 4 )( 6 + 4 ) = 210
( 9 – 2 ) ( 9 + 2 ) = 711
So ( 7 – 6 ) ( 7 + 6 ) = 113
2 . What will be the next number in the given sequence ?
0 1 3 2 6 3 9 4 12 5 ?
Let us consider the numbers to be “ abcdefghijkl”
Here the numbers “ a , c , e , g , i , k ” are consecutive numbers and the numbers “ b , d , f , h , j , l ” are multiples of 3 .
So the next number will be 15
3. If 1 + 4 = 5
2 + 5 = 12
3 + 6 = 21
Then 8 + 11 = ?
It is of the form ( X * Y ) + X
( 1 * 4 ) + 1 = 5
( 2 * 5 ) + 2 = 12
( 3 * 6 ) + 3 = 21
So , ( 8 * 11 ) + 8 = 96
4 . What will be the next number in the sequence ?
70 68 64 58 50 ?
It follows the pattern that the difference between the consecutive numbers are multiples of 2
70 – 68 = 2 ,
68 – 64 = 4 ,
64 – 58 = 6 ,
58 – 50 = 8 ,
50 – x = 10
So x = 40
5. If 1 1 1 1 = R
2 2 2 2 = T
3 3 3 3 = E
Then 4 4 4 4 = ?
The sum of 1 1 1 1 = 4 (FOUR ) = R
2 2 2 2 = 8 ( EIGHT ) = T
3 3 3 3 = 12 ( TWELVE ) = E
4 4 4 4 = 16 (SIXTEEN ) = N
6. If 2 x – 4 / y = 10
2 x + 2 / y = 4
Then what will be the value of x and y ?
Solution :
Multiplying by y we get ,
2 x y – 4 = 10 y
2 x y + 2 = 4 y
( – 6 ) = 6 y
• Y = -1
Then 2 x(- 1) – 4 = 10(-1)
( – 2 x ) = 4 – 10
– 2 x = – 6
X = 3
Thus the values of x and y are 3 and -1
7 . Using only addition how will add eight 8 ‘s and you get 1000 ?
888
88
8
8
( + ) 8
__________
1000
8 . It is a 9 letter word 123456789
A ) If you lose it you will die
B ) If you have 234 then you can 1234
C ) 56 is a disease
D ) 89 indicates the exact time and location
E ) 5 and 9 are the same alphabets , 2 and 7 are same and 3 and 8 are the same alphabets .
ANSWER : A ) HEARTBEAT
B) EAR => HEAR
C) TB
D) AT
E ) 5 = 9 = T , 2 = 7 = E and 3 = 8 = A
9 . It is a 6 letter word
A ) First three letters is the largest thing in the world
B ) 234 indicates an human organ
C ) 345 indicates a part of a circle
D ) The whole word plays the vital key role in the internet today
ANSWER : A ) SEA
B ) EAR
C ) ARC
D ) SEARCH
10 . If 2651 = 86
3342 = 66
9017 = 98
then 6113 = ?
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Contents
# Shortest route With Graph Coverage
## Traversable graphs
A traversable graph is one that can be drawn without taking a pen from the paper and without retracing the same edge. In such a case the graph is said to have an Eulerian trail.
If we try drawing the three graphs shown above we find:
· It is impossible to draw Graph 1 without either taking the pen off the paper or re-tracing an edge
· We can draw Graph 2, but only by starting at either A or D – in each case the path will end at the other vertex of D or A
· Graph 3 can be drawn regardless of the starting position and you will always return to the start vertex.
What is the difference between the three graphs?
In order to establish the differences, we must consider the order of the vertices for each graph. We obtain the following:
When the order of all the vertices is even, the graph is traversable and we can draw it. When there are two odd vertices we can draw the graph but the start and end vertices are different. When there are four odd vertices the graph can’t be drawn without repeating an edge.
To draw the graph with odd vertices, edges need to be repeated. To find such a trail we have to make the order of each vertex even.
In graph 1 there are four vertices of odd order and we need to pair the vertices together by adding an extra edge to make the order of each vertex four. We can join AB and CD, or AC and BD, or AD and BC.
In each case the graph is now traversable.
A minimum spanning tree of a network always has one edge fewer than the number of vertices in the graph (trivially). It may not be unique, but is the shortest length of arcs (edges) required to link every single node (vertex).
## Kruskal’s algorithm:
To find a minimum spanning tree for a network with n edges.
Step 1: Choose the unused edge with the lowest value.
Step 3: If there are n-1 edges in your tree, stop. If not, go to step 1.
NOTE: Ensure you do not create a cycle – if choosing an arc would produce a cycle, don’t choose it.
## Prim’s algorithm:
To find a minimum spanning tree for a network with n edges.
Step 1: From a start vertex draw the lowest valued edge to start your tree. (Any vertex can be chosen as the start vertex; however, it will always be given in an exam question.)
Step 2: From any vertex on your tree, add the edge with the lowest value.
Step 3: If there are n-1 edges in your tree, you have finished. If not, go to step 2.
NOTE: Ensure you do not create a cycle – if choosing an arc would produce a cycle, don’t choose it.
Kruskal’s algorithm adds edges regardless of their connectedness to the previously selected edges. Prim’s limits your choice to connected edges.
## Shortest path problem (Dijkstra’s algorithm) Dijkstra’s algorithm
Used for finding the shortest path through a network.
Step 1: Label the start vertex as 0.
Step 2: Box this number (permanent label).
Step 3: Label each vertex that is connected to the start vertex with its distance (temporary label).
Step 4: Box the smallest number.
Step 5: From this vertex, consider the distance to each connected vertex.
Step 6: If a distance is less than the distance already in this vertex, cross out the distance and write in the new distance. If there was no distance at the vertex, write down the new distance.
Step 7: Repeat from step 4 until the destination vertex is boxed.
Note: At step 4, the ‘smallest number’ refers to the smallest available temporary label – this does not need to be directly connected to the latest vertex to be permanently labelled.
Note: in some networks, called directed networks, sometimes edges will be unidirectional (eg, oneway streets). To apply Dijkstra’s algorithm in these cases, only consider edges leading away from the vertex.
If there are multiple start points, then we apply Dijkstra’s algorithm from the end point until we have reached each of the starting points. In this way we can find the shortest route.
Limitations: If we use Dijkstra’s algorithm on a network containing an edge that has a negative value it does not work.
To find the route of the shortest path using Dijkstra’s algorithm:
As well as listing temporary values, we put a letter after each value, which indicates the preceding vertex on the route. We find the route by backtracking through the network from the finishing point.
To find second shortest path, store each path in a priority queue sorted by length of the path.
Maintain a k size priority queue, if kth shortest path is to be calculated.
## A* Search Algorithm:
In computer science<http://en.wikipedia.org/wiki/Computer_science>, A* is a computer algorithm<http://en.wikipedia.org/wiki/Computer_algorithm> that is widely used in pathfinding<http://en.wikipedia.org/wiki/Pathfinding> and graph traversal<http://en.wikipedia.org/wiki/Graph_traversal>, the process of plotting an efficiently traversable path between points, called nodes. Noted for its performance<http://en.wikipedia.org/wiki/Computer_performance> and accuracy, it enjoys widespread use. However, in practical travel-routing systems, it is generally outperformed by algorithms which can pre-process the graph to attain better performance.
A* uses a best-first search<http://en.wikipedia.org/wiki/Best-first_search> and finds a least-cost path from a given initial node<http://en.wikipedia.org/wiki/Node_(graph_theory)> to one goal node<http://en.wikipedia.org/wiki/Goal_node> (out of one or more possible goals). As A* traverses the graph, it follows a path of the lowest expected total cost or distance, keeping a sorted priority queue<http://en.wikipedia.org/wiki/Priority_queue> of alternate path segments along the way.
It uses a knowledge-plus-heuristic<http://en.wikipedia.org/wiki/Heuristic> cost function of node x (usually denoted f(x)) to determine the order in which the search visits nodes in the tree. The cost function is a sum of two functions:
· the past path-cost function, which is the known distance from the starting node to the current node x (usually denoted g(x))
· A future path-cost function, which is an admissible<http://en.wikipedia.org/wiki/Admissible_heuristic> “heuristic estimate” of the distance from x to the goal (usually denoted h(x)).
The h(x) part of the f(x) function must be an admissible heuristic<http://en.wikipedia.org/wiki/Admissible_heuristic>; that is, it must not overestimate the distance to the goal. Thus, for an application like routing<http://en.wikipedia.org/wiki/Routing>, h(x) might represent the straight-line distance to the goal, since that is physically the smallest possible distance between any two points or nodes.
Like breadth-first search<http://en.wikipedia.org/wiki/Breadth-first_search>, A* is complete and will always find a solution if one exists.
Special cases:
Dijkstra’s algorithm<http://en.wikipedia.org/wiki/Dijkstra%27s_algorithm>, as another example of a uniform-cost search algorithm, can be viewed as a special case of A* where h(x) = 0 for all x. General depth-first search<http://en.wikipedia.org/wiki/Depth-first_search> can be implemented using the A* by considering that there is a global counter C initialized with a very large value. Every time we process a node we assign C to all of its newly discovered neighbours. After each single assignment, we decrease the counter C by one. Thus the earlier a node is discovered, the higher its h(x) value. It should be noted, however, that both Dijkstra’s algorithm and depth-first search can be implemented more efficiently without including an h(x) value at each node.
A* is admissible<http://en.wikipedia.org/wiki/Admissible_heuristic> and considers<http://en.wikipedia.org/wiki/Consistent_heuristic> fewer nodes than any other admissible search algorithm with the same heuristic. This is because A* uses an “optimistic” estimate of the cost of a path through every node that it considers—optimistic in that the true cost of a path through that node to the goal will be at least as great as the estimate. But, critically, as far as A* “knows”, that optimistic estimate might be achievable.
When A* terminates its search, it has found a path whose actual cost is lower than the estimated cost of any path through any open node. But since those estimates are optimistic, A* can safely ignore those nodes. In other words, A* will never overlook the possibility of a lower-cost path and so is admissible.
## Chinese postman problem
Dijkstra’s algorithm finds the shortest path between two vertices. The Chinese postman problem is an attempt to find the shortest path which traverses every edge. The obvious application would be a postman who needs to traverse each road in a given area.
To find a minimum Chinese postman route we must walk along each edge at least once and in addition we must also walk along the least pairings of odd vertices on one extra occasion.
An algorithm for finding an optimal Chinese postman route is:
Step 1 List all odd vertices.
Step 2 List all possible pairings of odd vertices.
Step 3 For each pairing find the edges that connect the vertices with the minimum weight.
Step 4 Find the pairings such that the sum of the weights is minimised.
Step 5 On the original graph add the edges that have been found in Step 4.
Step 6 The length of an optimal Chinese postman route is the sum of all the edges added to the total found in Step 4.
Step 7 A route corresponding to this minimum weight can then be easily found.
Step 1 The odd vertices are A and H.
Step 2 There is only one way of pairing these odd vertices, namely AH.
Step 3 The shortest way of joining A to H is using the path AB,
BF, FH, a total length of 160.
Step 4 Draw these edges onto the original network.
Step 5 The length of the optimal Chinese postman route is the sum of all the edges in the original network, which is 840 m, plus the answer found in Step 4, which is 160 m.
Hence the length of the optimal Chinese postman route is 1000 m.
Step 6 One possible route corresponding to this length is ADCGHCABDFBEFHFBA, but many other possible routes of the same minimum length can be found.
How does this problem differ from the problem of the travelling salesman?
The travelling salesman needs to visit every node (or vertex) whereas the Postman needs to travel along each arc (or edge). Therefore it is required that the network is traversable (Eulerian).
## Travelling salesman problem
Where the Chinese postman was trying to minimise the distance travelled while traversing every edge, the travelling salesman needs to minimise the distance travelled while visiting every vertex.
A minimum spanning tree links all vertices, but a tour for the travelling salesman must start and finish at the same vertex, meaning that, rather than containing one fewer edges than vertices, it contains the same number.
The optimal tour is defined as a tour of the graph with the shortest path. There is no algorithm for calculating this, and brute force calculations would take prohibitively long, even by computer.
Therefore an upper bound is a useful tool.
Any tour that exists is an upper bound. The best upper bound is the lowest upper bound.
Nearest-neighbour algorithm
Used to find an upper bound for the optimal tour.
Step 1: Choose a start vertex.
Step 2: From your current vertex go to the nearest unvisited vertex.
Step 3: Repeat step 2 until all the vertices have been visited.
Note: By starting at different vertices, different upper bounds can be generated.
Just as an upper bound for the length of an optimal tour can be found, a lower bound can be calculated. Since it only shows that the shortest tour cannot be below this value, it does not necessarily mean that any tour exists with a value as low as this lower bound. Conversely to the upper bound, the best lower bound is the greatest lower bound.
Lower bound algorithm
Used to find a lower bound for the optimal tour.
Step 1: Delete a vertex and all edges connected to the vertex.
Step 2: Find a minimum spanning tree for the remaining network.
Step 3: Add the two shortest edges from the deleted vertex.
Note: You can use either Kruskal’s or Prim’s to find the minimum spanning tree, required in step 2.
An incomplete network is one in which not all vertices are linked by edges. This would result in a contradiction should we apply our upper bound or lower bound algorithms.
If any network is incomplete then before any upper or lower bounds are obtained, the network must be made complete. Ensure that the distances between all pairs of vertices are represented by a single edge of minimum length.
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# Classroom activity: Matching criminals - guidance
### Guidance on Matching criminals
Assuming that the same number of people are born on every day of the year and that there are 365 days in a year, the probability of finding someone sharing your birthday is 1/365.
Now suppose you are interested in any two people in a group having matching birthdays. Let's first think about a group that has no matching birthdays. Then the first person in the group can have any of the 365 birthdays. The second person can have 364 out of the 365 birthdays in order to not match the first person. The third person can have 363 out of the 365 birthdays in order to not match either of the first two people, and so on. So that the probability that your group of size N (up to 365 people) has no matching birthdays is:
P(No matching birthdays) = (365/365) x (364/365) x (363/365) x ... x ( (365 - N + 1)/365).
Therefore, the probability that you have at least two people with the same birthday is:
P(At least one match) = 1 - P(No matching birthdays).
The following graph shows the probability of finding a matching pair of birthdays in a group of size N, plotted against N.
The probability of finding a matching pair of birthdays for number of people. Image from Wikipedia.
As shown in the figure, you need 23 people to have a 50:50 chance of two of them sharing a birthday. If you have 100 people, you can be 99.99997% sure that two of them will share a birthday. This result is known as the birthday paradox or birthday problem. It's called a "paradox" because intuitively you might guess that you'd need far more people, for example 365, to have a 50:50 chance that two share the same birthday.
What's the difference between looking for a person to match a given birthday and looking for two people sharing a birthday? In the first case, you're comparing everyone in a group of people to one other person, the one who's got the given birthday. In the second case, you compare everyone in a group to everyone else, so there are far more comparisons and therefore far more opportunities to find a match. For 23 people, looking for someone with a given birthday involves 23 comparisons, but looking for a matching pair involves 253 comparisons.
The birthday paradox helps explain the Arizona controversy. The chance of finding a DNA profile matching a given profile may be very small, but the chance of finding a matching pair of profiles is much larger. Therefore, the fact that the researcher found more matches than expected doesn't in itself prove that the DNA profiling technology is inaccurate.
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# How do you calculate fluid buoyancy?
What are the examples of buoyancy? A boat or a ship floating in the water and the floating of cork in water are examples of buoyancy.
## What are the 3 types of buoyancy?
In a given liquid, the object’s immersed weight is equal to its weight minus the buoyancy. If the density of the object is greater than that of the liquid, it will weigh more than the buoyancy; the immersed weight of this object is positive, and it will sink.
## What is buoyancy example?
Archimedes’ principle states that the buoyant force on a fluid is equal to the weight of the displaced fluid. To calculate the buoyant force, we use the equation buoyant force = density of fluid x volume of displaced fluid x acceleration due to gravity.
## How do you calculate the weight of an object submerged in water?
According to the Archimedes principle, the buoyant force acting on an object immersed in liquid is equal to the weight of liquid it displaces. Hence the buoyant force acting on the given object=20kg×10m/s2=200N.
## How is Archimedes Principle calculated?
The upward force, or buoyant force, that acts on an object in water is equal to the weight of the water displaced by the object. Any object that is in water has some buoyant force pushing up against gravity, which means that any object in water loses some weight.
## What will be the buoyant force when an object is submerged in water and displaced 20 kg of water?
If an object is more dense than water it will sink when placed in water, and if it is less dense than water it will float. Density is a characteristic property of a substance and doesn’t depend on the amount of substance.
## What is the buoyant force of water?
Just divide the mass of the object by the density of water for displaced water.
## What will happen if an object more dense than the fluid is immersed in the fluid?
If the buoyant force is greater than the object’s weight, the object will rise to the surface and float. If the buoyant force is less than the object’s weight, the object will sink.
## How do you find buoyant force without volume?
Buoyancy or a buoyant force can be defined as the tendency of the fluid to exert an upward force on an object, which is wholly or partially immersed in a fluid. The S.I. unit of buoyant force is Newton.
## What happens when the buoyant force is greater than the weight?
The buoyant force on an object is due to the pressure of the fluid on the object. But without gravity there would be no pressure and thus no buoyant force.
## What is the SI unit of buoyancy?
Boat sailing on the river, Iceberg floating on water, A person with a life vest floating on water, Ship floating on the ocean, Helium balloon rising in the air, etc. The buoyant force is proportional to the density.
## Which factors affect buoyancy?
• the density of the fluid,
• the volume of the fluid displaced, and.
• the local acceleration due to gravity.
## Can you have buoyancy without gravity?
Archimedes’ principle states that: “The upward buoyant force that is exerted on a body immersed in a fluid, whether partially or fully submerged, is equal to the weight of the fluid that the body displaces and acts in the upward direction at the center of mass of the displaced fluid”.
## What is a real life example of buoyancy?
Does gravity affect buoyancy? Yes, because buoyancy depends on weight of fluid displaced and we all know weight is a function of gravitational acceleration. Hence in the absence of gravity buoyancy force would be zero.
## What is the law of buoyancy?
If the body is completely submerged, the volume of fluid displaced is equal to the volume of the body. If the body is only partially submerged, the volume of the fluid displaced is equal to the volume of the part of the body that is submerged. Read more about Archimedes’ principle in the fluid mechanics article.
## Does gravity affect buoyancy?
Buoyant force is an upward force responsible for the apparent decrease in the weight of the immersed object. The Formula for Archimedes’ Principle can be given as Fb = ρ x g x V (here, Fb = buoyant force, g = acceleration due to gravity, ρ = density, V = volume.)
## How much fluid does a submerged object displace?
Thus as the density of liquid increases, the buoyant force exerted by it also increases.
## How do you use Archimedes Principle to calculate weight?
Pressure Increases Buoyant Force This pressure is always greater at the bottom than the top, hence the force that creates the displacement, the rising water. Pressure therefore increases buoyant force, which is equal to the weight of the water being dispersed.
## What happens to buoyancy when density increases?
Pascal’s law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container. The pressure at any point in the fluid is equal in all directions.
## Does pressure cause buoyancy?
The buoyant force is guaranteed to be the same for both objects only as long as they are both kept submerged, unless the density (and thus weight) of both objects is the same. Otherwise, the upward buoyant force on each object will depend on the density of each object relative to the density of water.
## What is Pascal’s law in physics?
There is no horizontal buoyant force on a submerged object because there is no difference in pressure horizontally.
## Is buoyant force the same for all objects?
When you float in fresh water, the buoyant force that acts on you is equal to your weight.
## Why is there no horizontal buoyant force on a submerged object?
Why does the buoyant force act upward on an object submerged in water? The pressure upward on the deeper bottom is greater than the downward pressure on the top.
## When you float in fresh water the buoyant force?
Things like ice cubes, oil drops, logs of wood, and paper float on water because they are less dense than water. Hollow objects like balloons, balls, a plastic container, and glass bottles also float because they are filled with air, which is less dense than water.
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# What is (5 -(sqrt)2) (3 + sqrt 2)?
Sep 18, 2015
$13 + 2 \sqrt{2}$
#### Explanation:
First we must expand the equation of $\left(5 - \sqrt{2}\right) \left(3 + \sqrt{2}\right)$. And we will get the equation be like;
$\left(5 - \sqrt{2}\right) \left(3 + \sqrt{2}\right)$
$= 5 \left(3\right) + 5 \left(\sqrt{2}\right) - 3 \left(\sqrt{2}\right) - \sqrt{2} \left(\sqrt{2}\right)$
Note that for $\sqrt{2} \left(\sqrt{2}\right)$, we can calculate it by;
$= \sqrt{2} \times \sqrt{2} = {\left(2\right)}^{\frac{1}{2}} \times {\left(2\right)}^{\frac{1}{2}} = {2}^{\frac{1}{2} + \frac{1}{2}} = {2}^{1} = 2$
And we will get;
$= 15 + 5 \sqrt{2} - 3 \sqrt{2} - 2$
$5 \sqrt{2}$ can subtract $3 \sqrt{2}$ since it has the same base, which is $\sqrt{2}$. Calculate all of it and we will get;
$= 13 + 2 \sqrt{2}$
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Chapter 5 Uniform Circular Motion and Gravitation
# 5.3 Centripetal Force
### Summary
• Calculate friction on a car tire moving in a circle
Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge.
Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: net F = ma. For uniform circular motion, the acceleration is the centripetal acceleration— a = ac. Thus, the magnitude of centripetal force Fc is
F centripetal = F c =m a centripetal = m ac = m v2/r = m ω2 r
You may use whichever expression for centripetal force is more convenient. Centripetal force Fc is always perpendicular to the path and pointing to the centre of curvature, because ac is perpendicular to the velocity and pointing to the centre of curvature.
Note that if you solve the first expression for r, you get
$\boldsymbol{r\:=}$$\boldsymbol{\frac{mv^2}{F_{\textbf{c}}}}.$
This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.
### Example 1: What Coefficient of Friction Do Care Tires Need on a Flat Curve?
(a) Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s.
(b) Assuming an unbanked curve, find the minimum force of friction , also called traction, due to the tires and the road.
Strategy and Solution for (a)
We know that $\boldsymbol{{F}_{\textbf{c}}=\frac{mv^2}{r}}.$Thus,
$\boldsymbol{{F}_{\textbf{c}}\:=}$$\boldsymbol{\frac{mv^2}{r}}$$\boldsymbol{=}$$\boldsymbol{\frac{(900\textbf{ kg})(25.0\textbf{ m/s})^2}{(500\textbf{ m})}}$$\boldsymbol{=\:1125\textbf{ N}}.$
Strategy for (b)
Figure 2 shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case.
Discussion
We could also solve part (a) using the first expression in $\begin{array}{l} \boldsymbol{F_{\textbf{c}}=m\frac{v^2}{r}} \\ \boldsymbol{F_{\textbf{c}}=mr\omega^2} \end{array}$$\rbrace$, because m, v, and r are given.
### TAKE-HOME EXPERIMENT
Ask a friend or relative to swing a golf club or a tennis racquet. Take appropriate measurements to estimate the centripetal acceleration of the end of the club or racquet. You may choose to do this in slow motion.
### PHET EXPLORATIONS: GRAVITY AND ORBITS
Move the sun, earth, moon and space station to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies, and turn off gravity to see what would happen without it!
# Section Summary
• Centripetal force Fc is any force causing uniform circular motion. It is a “centre-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity v and has magnitude
$\boldsymbol{F_{\textbf{c}}=ma_{\textbf{c}}},$
which can also be expressed as
$\boldsymbol{F_{\textbf{c}}=m\frac{v^2}{r}}$
or
$\boldsymbol{F_{\textbf{c}}=mr\omega^2}$
# Conceptual Questions
### Conceptual Questions
1: If you wish to reduce the stress (which is related to centripetal force) on high-speed tires, would you use large- or small-diameter tires? Explain.
2: Define centripetal force. Can any type of force (for example, tension, gravitational force, friction, and so on) be a centripetal force? Can any combination of forces be a centripetal force?
3: If centripetal force is directed toward the centre, why do you feel that you are ‘thrown’ away from the centre as a car goes around a curve? Explain.
4: Race car drivers routinely cut corners as shown in Figure 5. Explain how this allows the curve to be taken at the greatest speed.
5: A number of amusement parks have rides that make vertical loops like the one shown in Figure 6. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if:
(a) The car goes over the top at faster than this speed?
(b)The car goes over the top at slower than this speed?
7: What is the direction of the force exerted by the car on the passenger as the car goes over the top of the amusement ride pictured in Figure 6 under the following circumstances:
(a) The car goes over the top at such a speed that the gravitational force is the only force acting?
(b) The car goes over the top faster than this speed?
(c) The car goes over the top slower than this speed?
8: As a skater forms a circle, what force is responsible for making her turn? Use a free body diagram in your answer.
9: Suppose a child is riding on a merry-go-round at a distance about halfway between its centre and edge. She has a lunch box resting on wax paper, so that there is very little friction between it and the merry-go-round. Which path shown in Figure 7 will the lunch box take when she lets go? The lunch box leaves a trail in the dust on the merry-go-round. Is that trail straight, curved to the left, or curved to the right? Explain your answer.
11: Suppose a mass is moving in a circular path on a frictionless table as shown in figure. In the Earth’s frame of reference, there is no centrifugal force pulling the mass away from the centre of rotation, yet there is a very real force stretching the string attaching the mass to the nail. Using concepts related to centripetal force and Newton’s third law, explain what force stretches the string, identifying its physical origin.
### Problems & Exercises
1: (a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she exert to stay on if she is 1.25 m from its centre?
(b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its center?
(c) Compare each force with her weight.
2: Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg.
7: A large centrifuge, like the one shown in Figure 10(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries.
(a) At what angular velocity is the centripetal acceleration 10 g if the rider is 15.0 m from the centre of rotation?
(b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in Figure 10(b). At what angle θ below the horizontal will the cage hang when the centripetal acceleration is 10 g? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle θ should be.)
9: Modern roller coasters have vertical loops like the one shown in Figure 11. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g?
## Glossary
centripetal force
any net force causing uniform circular motion
ideal banking
the sloping of a curve in a road, where the angle of the slope allows the vehicle to negotiate the curve at a certain speed without the aid of friction between the tires and the road; the net external force on the vehicle equals the horizontal centripetal force in the absence of friction
### Solutions
Problems & Exercises
1: (a) 483 N b) 17.4 N (c) 2.24 times her weight, 0.0807 times her weight
3: 4.14^
5: (a) $$\boldsymbol{24.6\textbf{ m}}$$ (b) $\boldsymbol{36.6\textbf{ m/s}^2}$ (c) $\boldsymbol{a_{\textbf{c}}=3.73\textbf{ g}.}$
This does not seem too large, but it is clear that bobsledders feel a lot of force on them going through sharply banked turns.
7: (a) 2.56 rad/s (b) 5.71o
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36
Q:
# The H.C.F and L.C.M of two numbers are 11 and 385 respectively. If one number lies between 75 and 125 , then that number is
A) 77 B) 88 C) 99 D) 110
Explanation:
Product of numbers = 11 x 385 = 4235
Let the numbers be 11a and 11b . Then , 11a x 11b = 4235 => ab = 35
Now, co-primes with product 35 are (1,35) and (5,7)
So, the numbers are ( 11 x 1, 11 x 35) and (11 x 5, 11 x 7)
Since one number lies 75 and 125, the suitable pair is (55,77)
Hence , required number = 77
Q:
What is a Common Multiple?
If the Multiples for two or more numbers which are Common, then those Multiples are said to be Common Multiples.
For Example :
Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30,...
Multiples of 5 are 5, 10, 15, 20, 25, 30,...
Here common multiples of 3 & 5 upto 30 are 15, 30.
751
Q:
LCD of 12 and 18
A) 36 B) 42 C) 12 D) 6
Explanation:
LCD is nothing but Lowest or Least Common Denominator
Here LCD of 12 and 18 means LCD of two fractions with denominators 12 and 18 respectively.
Therefore, LCM of 12 & 18 = 6 x 3 x 2 = 36
• How to calculate LCD ::
The lowest common denominator or least common denominator (LCD) is the least common multiple (LCM) of the denominators of a set of fractions.
14 1098
Q:
54, 60 LCM is
A) 600 B) 540 C) 60 D) 54
Explanation:
Here the LCM of 54, 60 is
=> 2 x 3 x 3 x 3 x 10
= 6 x 9 x 10
= 54 x 10
540.
17 1370
Q:
If the HCF of 210 and 55 is expressible in the form of 210 x 5 + 55P, then value of P = ?
A) -23 B) 27 C) 16 D) -19
Explanation:
HCF of 210 and 55 is 5
Now, 210x5 + 55P = 5
=> 1050 + 55P = 5
=> 55P = -1045
=> P = -1045/55
=> P = -19.
16 2233
Q:
A heap of coconuts is divided into groups of 2, 3 and 5 and each time one coconut is left over. The least number of Coconuts in the heap is ?
A) 63 B) 31 C) 16 D) 27
Explanation:
To get the least number of coconuts :
LCM = 30 => 30 + 1 = 31
21 3013
Q:
A drink vendor has 368 liters of Maaza, 80 liters of Pepsi and 144 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required ?
A) 47 B) 46 C) 37 D) 35
Explanation:
The number of liters in each can = HCF of 80, 144 and 368 = 16 liters.
Number of cans of Maaza = 368/16 = 23
Number of cans of Pepsi = 80/16 = 5
Number of cans of Sprite = 144/16 = 9
The total number of cans required = 23 + 5 + 9 = 37 cans.
24 2480
Q:
H.C.F of 4 x 27 x 3125, 8 x 9 x 25 x 7 and 16 x 81 x 5 x 11 x 49 is :
A) 360 B) 180 C) 90 D) 120
Explanation:
4 x 27 x 3125 = ;
8 x 9 x 25 x 7 =
16 x 81 x 5 x 11 x 49 =
H.C.F = = 180.
16 1982
Q:
The difference of two numbers is 14. Their LCM and HCF are 441 and 7. Find the two numbers ?
A) 63 and 49 B) 64 and 48 C) 62 and 46 D) 64 and 49
Answer & Explanation Answer: A) 63 and 49
Explanation:
Since their HCFs are 7, numbers are divisible by 7 and are of the form 7x and 7y
Difference = 14
=> 7x - 7y = 14
=> x - y = 2
product of numbers = product of their hcf and lcm
=> 7x * 7y = 441 * 7
=> x * y = 63
Now, we have
x * y = 63 , x - y = 2
=> x = 9 , y = 7
The numbers are 7x and 7y
=> 63 and 49
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# How do you simplify sqrt5(2sqrt5 - sqrt4)?
Apr 1, 2018
$10 - 2 \sqrt{5}$
#### Explanation:
Distribute $\sqrt{5} :$
$\sqrt{5} \left(2 \sqrt{5} - \sqrt{4}\right) = 2 \sqrt{5} \sqrt{5} - \sqrt{4} \sqrt{5}$
Recall that $\sqrt{a} \sqrt{a} = a$
So, we have
$2 \left(5\right) - \sqrt{4} \sqrt{5}$
$\sqrt{4} = 2 ,$ so we end up with
$10 - 2 \sqrt{5}$
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# NYT Digits June 10, 2023 Answers
If you need NYT Digits June 10, 2023 Answers, we have a list of solutions for you!
Digits is a math-based game developed and published by New York Times. Each day, there are five puzzles released, and in each puzzle, players have six numbers that they can add, subtract, multiply, or divide by in order to reach the target number. Once you use a number, you cannot use it again in another operation. If you’ve having trouble solving the NYT Digits puzzles for June 10th, 2023, we have the answers for you!
## NYT Digits, Puzzle 1 Answer for June 10 2023
The target is 55. There are two operations to get to 3-stars for the one puzzle.
• 10 + 1 = 11
• 5 × 11 = 55
## NYT Digits, Puzzle 2 Answer for June 10 2023
The target is 137. There are two operations to get to 3-stars for the two puzzle.
• 15 × 9 = 135
• 135 + 2 = 137
## NYT Digits, Puzzle 3 Answer for June 10 2023
The target is 291. There are three operations to get to 3-stars for the three puzzle.
• 25 + 15 = 40
• 40 × 7 = 280
• 280 + 11 = 291
## NYT Digits, Puzzle 4 Answer for June 10 2023
The target is 315. There are three operations to get to 3-stars for the four puzzle.
• 25 + 3 = 28
• 28 × 11 = 308
• 308 + 7 = 315
## NYT Digits, Puzzle 5 Answer for June 10 2023
The target is 453. There are three operations to get to 3-stars for the five puzzle.
• 23 × 20 = 460
• 460 - 15 = 445
• 445 + 8 = 453
Digits rewards players with up to 3-stars; you’ll get 3-stars for getting to the target exactly, 2-stars if you are within 10 of the target, and 1-star if you are within 25 of the target. Players can start a new operation by selecting a different number, and fractions or negative numbers are not accepted. New puzzles are released daily at midnight.
When working on a target from Digits, one of the first things you can do is to look for factor pairs of the target number, which can sometimes clue you into the direction you need to go. This is often the last operation of the solution, and you’ll use other operations to get to the numbers you need for the final factor pair.
We hope our NYT Digits June 10, 2023 Answer Guide helped you out on today’s puzzles! Check out more of our NYT Digits coverage.
### Christine Mielke
Christine Mielke has been writing content for the web for over 15 years. She is well-known for concise, informative content and her transparency. Christine is a 2011 graduate of Santa Clara University’s JD/MBA program, after having graduated in 2007 from University of California, Irvine with B.A. in Economics and B.A. in Political Science.
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Problem 188
# $\begin{array}{cccccccc} & /0 & 0 & 0 & 0 & 1 & 1 & 1 \mid \\ \text { Let } \mathrm{A}= & /0 & 2 & 6 & 2 & 0 & 0 & 4 \mid . \\ &/0 & 1 & 3 & 1 & 2 & 1 & 2 \mid\end{array}$ Reduce A to the Hermite normal form.
Expert verified
The Hermite normal form of matrix A is: $$HNF(A) = \begin{bmatrix} 0 & 0 & 0 & 0 & 4 & 2 & 0 \\ 0 & 1 & 3 & 1 & -2 & -1 & 2 \end{bmatrix}$$
See the step by step solution
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## Step 1: Ensure positive diagonal values
In the given matrix, the first diagonal element is already positive, so we don't need to make any changes: $$A = \begin{bmatrix} 0 & 2 & 6 & 2 & 0 & 0 & 4 \\ 0 & 1 & 3 & 1 & 2 & 1 & 2 \end{bmatrix}$$
## Step 2: Create zeros below the first nonzero number in a nonzero row
To achieve this, we can perform the following row operation R2 = R2 - R1: $$\begin{bmatrix} 0 & 2 & 6 & 2 & 0 & 0 & 4 \\ 0 & -1 & -3 & -1 & 2 & 1 & -2 \end{bmatrix}$$
## Step 3: Make the first nonzero number in a nonzero row positive
For the second row, R2 = -R2: $$\begin{bmatrix} 0 & 2 & 6 & 2 & 0 & 0 & 4 \\ 0 & 1 & 3 & 1 & -2 & -1 & 2 \end{bmatrix}$$
## Step 4: Create zeros above the first nonzero number in a nonzero row
Perform the row operation R1 = R1 - 2 * R2: $$\begin{bmatrix} 0 & 0 & 0 & 0 & 4 & 2 & 0 \\ 0 & 1 & 3 & 1 & -2 & -1 & 2 \end{bmatrix}$$
## Step 5: Ensure that the first nonzero number in a nonzero row divides all other numbers in the row
The current matrix meets this requirement: $$\begin{bmatrix} 0 & 0 & 0 & 0 & 4 & 2 & 0 \\ 0 & 1 & 3 & 1 & -2 & -1 & 2 \end{bmatrix}$$ Now we have the Hermite normal form of matrix A: $$HNF(A) = \begin{bmatrix} 0 & 0 & 0 & 0 & 4 & 2 & 0 \\ 0 & 1 & 3 & 1 & -2 & -1 & 2 \end{bmatrix}$$
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How do you find the quotient (6t^3-9t^2+6)div(2t-3) using long division?
May 31, 2017
The quotient is $= 3 {t}^{2}$ and the remainder $= 6$
Explanation:
We perform a long division
$\textcolor{w h i t e}{a a a a}$$2 t - 3$$\textcolor{w h i t e}{a a a a}$$|$$6 {t}^{3} - 9 {t}^{2} + 0 t + 6$$\textcolor{w h i t e}{a a a a}$$|$$3 {t}^{2}$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$$6 {t}^{3} - 9 {t}^{2}$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a}$$0 - 0 + 0 t + 6$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a}$$+ 6$
The quotient is $= 3 {t}^{2}$ and the remainder $= 6$
$\frac{6 {t}^{3} - 9 {t}^{2} + 6}{2 t - 3} = 3 {t}^{2} + \frac{6}{2 t - 3}$
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Law of Sines
Introduction: We will show and apply the Law of Sines which is a relationship among the sides and angles of a triangle. Applications will involve solving a triangle, calculating the lengths of sides and the measurements of angles, as well as calculating the area of a triangle.
The Lesson:
Given a triangle ABC with altitude AD observe that
producing the result AD = bsinC.
Also observe that
producing the result AD = csinB.
This gives us csinB = bsinC or equivalently we have .
We can extend this relationship as follows:
This relationship is known as the “Law of Sines.” It can also be written as
The Law of Sines can also be derived from a formula for the area of triangle ABC. The area of a triangle is ½ base x height.
If we use a as the base and the height AD = csinB = bsinC, then we have:
Area of triangle ABC = ½ acsinB = ½ absinC
If we draw an altitude from either B or C, then we can also state that the
Area of triangle ABC = ½ bcsinA
Summarizing,
Area of Triangle ABC = ½ acsinB = ½ absinC = ½ bcsinA.
If we multiply this equation by 2 and then divide through by abc, these area formulas become a statement of the Law of Sines.
Note: The Law of Sines involves a ratio of the sine of an angle to the length of its opposite side. Therefore it will NOT work if no angle of the triangle is known or if an angle is known but its opposite side is not.
Let's Practice:
1. In triangle ABC suppose side c = 12 and angle A = 75º and angle B = 26º. Find the length of side a.
In order to use side c in the Law of Sines, we need to know angle C.
This can be found using the fact that the three angles of ABC have a sum of 180º.
giving us
This answer is reasonable since angle A is a little smaller than angle C. We expect side a to be a little smaller than side c.
1. Suppose triangle ABC has angle A = 67º, angle B = 33º and a = 8. Find the length of side b. Also find the area of this triangle.
From the triangle, we can set up the relationships:
Area = ½ absinC .
We know that angle C is because angle C + 67º + 33º = 180º.
Thus the area of triangle ABC .
Note that in both these examples the measure of two angles is known. While the Law of Sines can be used in situations where only one angle measurement of a triangle is known, it is often more convenient to solve such a triangle by using the Law of Cosines.
Examples
Suppose triangle ABC has angle A = 71º, angle B = 39º, and a = 14. Find the length of side b and find the area of this triangle. What is your answer?
Suppose triangle ABC has angle A = 73º, angle B = 28º, and c = 14. Find the length of side b. What is your answer?
M Ransom
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# Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1
Students can Download Maths Chapter 2 Measurements Ex 2.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1
Question 1.
Fill in the blanks:
(i) The ratio between the circumference and diameter of any circle is _________.
(ii) A line segment which joins any two points on a circle is a ______.
(iii) The longest chord of a circle is _______.
(iv) The radius of a circle of diameter 24 cm is ______.
(v) A part of circumference of a circle is called as _____.
Solution:
(i) π
(ii) chord
(iii) diameter
(iv) 12 cm
(v) an arc
Question 2.
Match the following
Solution:
(i) 3
(ii) 4
(iii) 5
(iv) 2
(v) 1
Question 3.
Find the central angle of the shaded sectors (each circle is divided into equal sectors)
Solution:
Question 4.
For the sectors with given measures, find the length of the arc, area and perimeter, (π = 3.14)
(i) central angle 45°, r = 16 cm
(ii) central angle 120°, d = 12.6 cm
(iii) central angle 60°, r = 36 cm
(iv) central angle 72°, d = 10 cm
Solution:
(i) Central angle 45°, r = 16 cm
Length of the arc l = $$\frac{\theta^{\circ}}{360^{\circ}}$$ × 2πr units
l = $$\frac{45^{\circ}}{360^{\circ}}$$ × 2 × 3.14 × 16 cm
l = $$\frac{1}{8}$$ × 2 × 3.14 × 16 cm
l = 12.56 cm
Area of the sector = $$\frac{\theta^{\circ}}{360^{\circ}}$$ × πr2
A = $$\frac{45^{\circ}}{360^{\circ}}$$ × 3.14 × 16 × 16
A = 100.48 cm2
Perimeter of the sector
P = l + 2r units
P = 12.56 + 2(16) cm
P = 44.56 cm
(ii) Central angle 120°, d = 12.6 cm
∴ r = $$\frac{12.6}{2}$$ cm
r = 6.3 cm
Length of the arc
Area of the sector missing
Perimeter of the sector
P = l + 2r units
P = 6.28 + 2(5) cm
P = 6.28 + 10 cm
P = 16.28 cm
Question 5.
From the measures given below, find the area of the sectors.
Solution:
(i) Area of the sector
A = $$\frac{l r}{2}$$ sq. units
l = 48 m
r = 10 m
= $$\frac{48 \times 10}{2}$$ m2
= 24 × 10 m2
= 240 m2
Area of the sector = 240 m2
(ii) length of the arc l = 12.5 cm
Radius r = 6 cm
Area of the sector
A = $$\frac{l r}{2}$$ sq. units
= $$\frac{12.5 \times 6}{2}$$
= 12.5 × 3 cm2 cm2
= 37.5 cm2
Area of the sector = 37.5 cm2
(iii) length of the arc l = 50 cm
Radius r = 13.5 cm
Area of the sector
A = $$\frac{l r}{2}$$ sq. units
= $$\frac{50 \times 13.5}{2}$$
= 25 × 13.5 cm2 cm2
= 337.5 cm2
Area of the sector = 337.5 cm2
Question 6.
Find the central angle of each of the sectors whose measures are given below (π = $$\frac{22}{7}$$)
Solution:
(i) Radius of the sector = 21 cm
Area of the sector = 462 cm2
$$\frac{l r}{2}$$ = 462
$$\frac{l \times 21}{2}$$ = 462
∴ Central angle of the sector = 120°
(ii) Radius of the sector = 8.4 cm
Area of the sector = 18.48 cm2
(iii) Radius of the sector = 35 m
Length of the arc l = 44 m
Question 7.
Answer the following questions:
(i) A circle of radius 120 m is divided into 8 equal sectors. Find the length of the arc of each of the sectors.
(ii) A circle of radius 70 cm is divided into 5 equal sectors. Find the area of each of the sectors.
Solution:
(i) Radius of the circle r = 120 m
Number of equal sectors = 8
(ii) Radius of the sector r = 70 cm
Number of equal sectors = 5
Note: We can solve this problem using A = $$\frac{1}{n}$$ πr2 sq. units also.
Question 8.
Find the area of a sector whose length of the arc is 50 mm and radius is 14 mm.
Solution:
Length of the arc of the sector l = 50 mm
Radius r = 14 mm
Area of the sector = $$\frac{l r}{2}$$ sq. units = $$\frac{50 \times 14}{2}$$ mm2 = 50 × 7 mm2 = 350 mm2
Area of the sector = 350 mm2
Question 9.
Find the area of a sector whose perimeter is 64 cm and length of the arc is 44 cm.
Solution:
Length of the arc of the sector l = 44 cm
Perimeter of the sector P = 64 cm
l + 2r = 64 cm
44 + 2 r = 64 .
2 r = 64 – 44
2 r = 20
r = $$\frac{20}{2}$$ = 10 cm2
Area of the sector = $$\frac{l r}{2}$$ sq. units = $$\frac{44 \times 10}{2}$$ cm2 = 22 × 10 cm2 = 220 cm2
Area of the sector = 220 cm2
Question 10.
A sector of radius 4.2 cm has an area 9.24 cm2. Find its perimeter
Solution:
Radius of the sector r = 4.2 cm
‘ Area of the sector = 9.24 cm2
$$\frac{l r}{2}$$ = 9.24
$$\frac{l \times 4.2}{2}$$ = 9.24
l × 2.1 = 9.24
l = $$\frac{9.24}{2.1}$$
l = 4.4 cm
Perimeter of the sector = 1 + 2r units = 4.4 + 2(4.2) cm
= 4.4 + 8.4 cm = 12. 8 cm
Perimeter of the sector = 12.8 cm
Question 11.
Infront of a house, flower plants are grown in a circular quadrant shaped pot whose radius is 2 feet. Find the area of the pot in which the plants grow. (π = 3.14)
Solution:
Central angle of the quadrant = 90°
Radius of the circle = 2 feet
Area of the quadrant = 3.14 sq. feet (approximately)
Question 12.
Dhamu fixes a square tile of 30 cm on the floor. The tile has a sector design on it as shown in the figure. Find the area of the sector, (π = 3.14).
Solution:
Side of the square = 30 cm
∴ Radius of the sector design = 30 cm
Area of the quadrant = $$\frac{1}{4}$$ πr2 sq. units
= $$\frac{1}{4}$$ × 3.14 × 30 × 30 cm2
= 3.14 × 15 × 15 cm2
∴ Area of the sector design = 706.5 cm2 (approximately)
Question 13.
A circle is formed with 8 equal granite stones as shown in the figure each of radius 56 cm and whose central angle is 45°. Find the area of each of the granite. (π = $$\frac{22}{7}$$)
Solution:
Number of equal sectors ‘n’ = 8
Radius of the sector ‘r’ = 56 cm
Area of the each sector = $$\frac{1}{n}$$ πr2 sq. units
= $$\frac{1}{8} \times \frac{22}{7}$$ × 56 × 56 cm2 = 1232 cm2
Area of each sector = 1232 cm2 (approximately)
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# Solve for: (p^2-36)/(3p-4) * (3p^2+11p-20)/(p^2+11p+30)
## Expression: $\frac{ {p}^{2}-36 }{ 3p-4 } \times \frac{ 3{p}^{2}+11p-20 }{ {p}^{2}+11p+30 }$
Use ${a}^{2}-{b}^{2}=\left( a-b \right)\left( a+b \right)$ to factor the expression
$\frac{ \left( p-6 \right) \times \left( p+6 \right) }{ 3p-4 } \times \frac{ 3{p}^{2}+11p-20 }{ {p}^{2}+11p+30 }$
Write $11p$ as a difference
$\frac{ \left( p-6 \right) \times \left( p+6 \right) }{ 3p-4 } \times \frac{ 3{p}^{2}+15p-4p-20 }{ {p}^{2}+11p+30 }$
Write $11p$ as a sum
$\frac{ \left( p-6 \right) \times \left( p+6 \right) }{ 3p-4 } \times \frac{ 3{p}^{2}+15p-4p-20 }{ {p}^{2}+6p+5p+30 }$
Factor out $3p$ from the expression
$\frac{ \left( p-6 \right) \times \left( p+6 \right) }{ 3p-4 } \times \frac{ 3p \times \left( p+5 \right)-4p-20 }{ {p}^{2}+6p+5p+30 }$
Factor out $-4$ from the expression
$\frac{ \left( p-6 \right) \times \left( p+6 \right) }{ 3p-4 } \times \frac{ 3p \times \left( p+5 \right)-4\left( p+5 \right) }{ {p}^{2}+6p+5p+30 }$
Factor out $p$ from the expression
$\frac{ \left( p-6 \right) \times \left( p+6 \right) }{ 3p-4 } \times \frac{ 3p \times \left( p+5 \right)-4\left( p+5 \right) }{ p \times \left( p+6 \right)+5p+30 }$
Factor out $5$ from the expression
$\frac{ \left( p-6 \right) \times \left( p+6 \right) }{ 3p-4 } \times \frac{ 3p \times \left( p+5 \right)-4\left( p+5 \right) }{ p \times \left( p+6 \right)+5\left( p+6 \right) }$
Factor out $p+5$ from the expression
$\frac{ \left( p-6 \right) \times \left( p+6 \right) }{ 3p-4 } \times \frac{ \left( p+5 \right) \times \left( 3p-4 \right) }{ p \times \left( p+6 \right)+5\left( p+6 \right) }$
Factor out $p+6$ from the expression
$\frac{ \left( p-6 \right) \times \left( p+6 \right) }{ 3p-4 } \times \frac{ \left( p+5 \right) \times \left( 3p-4 \right) }{ \left( p+6 \right) \times \left( p+5 \right) }$
Cancel out the common factor $p+6$
$\frac{ p-6 }{ 3p-4 } \times \frac{ \left( p+5 \right) \times \left( 3p-4 \right) }{ p+5 }$
Cancel out the common factor $p+5$
$\frac{ p-6 }{ 3p-4 } \times \left( 3p-4 \right)$
Cancel out the common factor $3p-4$
$p-6$
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HomeNCERT SOLUTIONS8th CLASSClass 8 Maths Chapter 3 Understanding Quadrilaterals Solutions
Quadrilaterals
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# Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2
## NCERT Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2 Solutions
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.2
Ex 3.2 Class 8 Maths Question 1.
Find x in the following figures.
Solution:
(a) We know that the sum of all the exterior angles of a polygon = 360°
125° + 125° + x = 360°
⇒ 250° + x = 360°
x = 360° – 250° = 110°
Hence x = 110°
(b) Here ∠y = 180° – 90° = 90°
and ∠z = 90° (given)
x + y + 60° + z + 70° = 360° [∵ Sum of all the exterior angles of a polygon = 360°]
⇒ x + 90° + 60° + 90° + 70° = 360°
⇒ x + 310° = 360°
⇒ x = 360° – 310° = 50°
Hence x = 50°
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Ex 3.2 Class 8 Maths Question 2.
Find the measure of each exterior angle of a regular polygon of
(i) 9 sides
(ii) 15 sides
Solution:
(i) We know the sum of all the exterior angles of polygon = 360°
Measure of each angle of 9 sided regular polygon = $\frac { 360 }{ 9 }$ = 40°
(ii) Sum of all the exterior angles of a polygon = 360°
Measure of each angle of 15 sided regular polygon = $\frac { 360 }{ 15 }$ = 24°
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Ex 3.2 Class 8 Maths Question 3.
How many sides does a regular polygon have if the measure of an exterior angle is 24°?
Solution:
Sum of all exterior angles of a regular polygon = 360°
Number of sides
Hence, the number of sides = 15
Ex 3.2 Class 8 Maths Question 4.
How many sides does a regular polygon have if each of its interior angles is 165°?
Solution:
Let re be the number of sides of a regular polygon.
Sum of all interior angles = (n – 2) × 180°
and, measure of its each angle
Hence, the number of sides = 24
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Ex 3.2 Class 8 Maths Question 5.
(a) Is it possible to have a regular polygon with measure of each exterior angle a is 22°?
(b) Can it be an interior angle of a regular polygon? Why?
Solution:
(a) Since, the sum of all the exterior angles of a regular polygon = 360° which is not divisible by 22°.
It is not possible that a regular polygon must have its exterior angle 22°.
(b) Sum of all interior angles of a regular polygon of side n = (n – 2) × 180°
not a whole number.
Since number of sides cannot be in fractions.
It is not possible for a regular polygon to have its interior angle = 22°.
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Ex 3.2 Class 8 Maths Question 6.
(a) What is the minimum interior angle possible for a regular polygon? Why?
(b) What is the maximum exterior angle possible for a regular polygon?
Solution:
(a) Sum of all interior angles of a regular polygon of side n = (n – 2) × 180°
The measure of each interior angle
The minimum measure the angle of an equilateral triangle (n = 3) = 60°.
(b) From part (a) we can conclude that the maximum exterior angle of a regular polygon = 180° – 60° = 120°.
# Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3
## NCERT Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3 Solutions
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3
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Ex 3.3 Class 8 Maths Question 1.
Given a parallelogram ABCD. Complete each statement along with the definition or property used.
(i) AD = …………
(ii) ∠DCB = ………
(iii) OC = ………
(iv) m∠DAB + m∠CDA = ……..
Solution:
(i) AD = BC [Opposite sides of a parallelogram are equal]
(ii) ∠DCB = ∠DAB [Opposite angles of a parallelogram are equal]
(iii) OC = OA [Diagonals of a parallelogram bisect each other]
(iv) m∠DAB + m∠CDA = 180° [Adjacent angles of a parallelogram are supplementary]
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Ex 3.3 Class 8 Maths Question 2.
Consider the following parallelograms. Find the values of the unknowns x, y, z.
Solution:
(i) ABCD is a parallelogram.
∠B = ∠D [Opposite angles of a parallelogram are equal]
∠D = 100°
⇒ y = 100°
∠A + ∠B = 180° [Adjacent angles of a parallelogram are supplementary]
⇒ z + 100° = 180°
⇒ z = 180° – 100° = 80°
∠A = ∠C [Opposite angles of a ||gm]
x = 80°
Hence x = 80°, y = 100° and z = 80°
(ii) PQRS is a parallelogram.
∠P + ∠S = 180° [Adjacent angles of parallelogram]
⇒ x + 50° = 180°
x = 180° – 50° = 130°
Now, ∠P = ∠R [Opposite angles are equal]
⇒ x = y
⇒ y = 130°
Also, y = z [Alternate angles]
z = 130°
Hence, x = 130°, y = 130° and z = 130°
(iii) ABCD is a rhombus.
[∵ Diagonals intersect at 90°]
x = 90°
Now in ∆OCB,
x + y + 30° = 180° (Angle sum property)
⇒ 90° + y + 30° = 180°
⇒ y + 120° = 180°
⇒ y = 180° – 120° = 60°
y = z (Alternate angles)
⇒ z = 60°
Hence, x = 90°, y = 60° and z = 60°.
(iv) ABCD is a parallelogram
∠A + ∠B = 180° (Adjacent angles of a parallelogram are supplementary)
⇒ x + 80° = 180°
⇒ x = 180° – 80° = 100°
Now, ∠D = ∠B [Opposite angles of a |jgm]
⇒ y = 80°
Also, z = ∠B = 80° (Alternate angles)
Hence x = 100°, y = 80° and z = 80°
(v) ABCD is a parallelogram.
∠D = ∠B [Opposite angles of a ||gm]
y = 112°
x + y + 40° = 180° [Angle sum property]
⇒ x + 112° + 40° = 180°
⇒ x + 152° = 180°
⇒ x = 180° – 152 = 28°
z = x = 28° (Alternate angles)
Hence x = 28°, y = 112°, z = 28°.
Ex 3.3 Class 8 Maths Question 3.
Can a quadrilateral ABCD be a parallelogram if
(i) ∠D + ∠B = 180°?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) ∠A = 70° and ∠C = 65°?
Solution:
(i) For ∠D + ∠B = 180, quadrilateral ABCD may be a parallelogram if following conditions are also fulfilled.
(a) The sum of measures of adjacent angles should be 180°.
(b) Opposite angles should also be of same measures. So, ABCD can be but need not be a parallelogram.
(ii) Given: AB = DC = 8 cm, AD = 4 cm, BC = 4.4 cm
In a parallelogram, opposite sides are equal.
Here AD ≠ BC
Thus, ABCD cannot be a parallelogram.
(iii) ∠A = 70° and ∠C = 65°
Since ∠A ≠ ∠C
Opposite angles of quadrilateral are not equal.
Hence, ABCD is not a parallelogram.
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Ex 3.3 Class 8 Maths Question 4.
Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Solution:
ABCD is a rough figure of a quadrilateral in which m∠A = m∠C but it is not a parallelogram. It is a kite.
Ex 3.3 Class 8 Maths Question 5.
The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD is parallelogram such that
m∠B : m∠C = 3 : 2
Let m∠B = 3x° and m∠C = 2x°
m∠B + m∠C = 180° (Sum of adjacent angles = 180°)
3x + 2x = 180°
⇒ 5x = 180°
⇒ x = 36°
Thus, ∠B = 3 × 36 = 108°
∠C = 2 × 36° = 72°
∠B = ∠D = 108°
and ∠A = ∠C = 72°
Hence, the measures of the angles of the parallelogram are 108°, 72°, 108° and 72°.
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Ex 3.3 Class 8 Maths Question 6.
Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram in which
∠A = ∠B
We know ∠A + ∠B = 180° [Sum of adjacent angles = 180°]
∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = 90°
Thus, ∠A = ∠C = 90° and ∠B = ∠D = 90°
[Opposite angles of a parallelogram are equal]
Ex 3.3 Class 8 Maths Question 7.
The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.
Solution:
∠y = 40° (Alternate angles)
∠z + 40° = 70° (Exterior angle property)
⇒ ∠z = 70° – 40° = 30°
z = ∠EPH (Alternate angle)
In ∆EPH
∠x + 40° + ∠z = 180° (Adjacent angles)
⇒ ∠x + 40° + 30° = 180°
⇒ ∠x + 70° = 180°
⇒ ∠x = 180° – 70° = 110°
Hence x = 110°, y = 40° and z = 30°.
Ex 3.3 Class 8 Maths Question 8.
The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)
Solution:
(i) GU = SN (Opposite sides of a parallelogram)
⇒ y + 7 = 20
⇒ y = 20 – 7 = 13
Also, ON = OR
⇒ x + y = 16
⇒ x + 13 = 16
x = 16 – 13 = 3
Hence, x = 3 cm and y = 13 cm.
Ex 3.3 Class 8 Maths Question 9.
In the above figure both RISK and CLUE are parallelograms. Find the value of x.
Solution:
Here RISK and CLUE are two parallelograms.
∠1 = ∠L = 70° (Opposite angles of a parallelogram)
∠K + ∠2 = 180°
Sum of adjacent angles is 180°
120° + ∠2 = 180°
∠2 = 180° – 120° = 60°
In ∆OES,
∠x + ∠1 + ∠2 = 180° (Angle sum property)
⇒ ∠x + 70° + 60° = 180°
⇒ ∠x + 130° = 180°
⇒ ∠x = 180° – 130° = 50°
Hence x = 50°
Ex 3.3 Class 8 Maths Question 10.
Explain how this figure is a trapezium. Which of its two sides are parallel?
Solution:
∠M + ∠L = 100° + 80° = 180°
∠M and ∠L are the adjacent angles, and sum of adjacent interior angles is 180°
KL is parallel to NM
Hence KLMN is a trapezium.
Ex 3.3 Class 8 Maths Question 11.
Find m∠C in below figure if $\bar { AB }$ || $\bar { DC }$
Solution:
Given that $\bar { AB }$ || $\bar { DC }$
m∠B + m∠C = 180° (Sum of adjacent angles of a parallelogram is 180°)
120° + m∠C = 180°
m∠C = 180° – 120° = 60°
Hence m∠C = 60°
Ex 3.3 Class 8 Maths Question 12.
Find the measure of ∠P and ∠S if $\bar { SP }$ || $\bar { RQ }$ in figure, is there any other method to find m∠P?)
Solution:
Given that ∠Q = 130° and ∠R = 90°
$\bar { SP }$ || $\bar { RQ }$ (given)
∠P + ∠Q = 180° (Adjacent angles)
⇒ ∠P + 130° = 180°
⇒ ∠P = 180° – 130° = 50°
and, ∠S + ∠R = 180° (Adjacent angles)
⇒ ∠S + 90° = 180°
⇒ ∠S = 180° – 90° = 90°
Alternate Method:
∠Q = 130°, ∠R = 90° and ∠S = 90°
We know that
∠P + ∠Q + ∠R + ∠Q = 360° (Angle sum property of quadrilateral)
⇒ ∠P + 130° + 90° + 90° = 360°
⇒ ∠P + 310° = 360°
⇒ ∠P = 360° – 310° = 50°
Hence m∠P = 50°
# Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4
## NCERT Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4 Solutions
NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.4
Ex 3.4 Class 8 Maths Question 1.
State whether True or False.
(a) All rectangles are squares.
(b) All rhombuses are parallelograms.
(c) All squares are rhombuses and also rectangles.
(d) All squares are not parallelograms.
(e) All kites are rhombuses.
(f) All rhombuses are kites.
(g) All parallelograms are trapeziums.
(h) All squares are trapeziums.
Solution:
(a) False
(b) True
(c) True
(d) False
(e) False
(f) True
(g) True
(h) True
Ex 3.4 Class 8 Maths Question 2.
Identify all the quadrilaterals that have
(a) four sides of equal length
(b) four right angles
Solution:
(a) Squares and rhombuses.
(b) Rectangles and squares.
Ex 3.4 Class 8 Maths Question 3.
Explain how a square is
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle
Solution:
(i) Square is a quadrilateral because it is closed with four line segments.
(ii) Square is a parallelogram due to the following properties:
(a) Opposite sides are equal and parallel.
(b) Opposite angles are equal.
(iii) Square is a rhombus because its all sides are equal and opposite sides are parallel.
(iv) Square is a rectangle because its opposite sides are equal and has equal diagonal.
Ex 3.4 Class 8 Maths Question 4.
Name the quadrilaterals whose diagonals
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal
Solution:
(i) Parallelogram, rectangle, square and rhombus
(ii) Square and rhombus
(iii) Rectangle and square
Ex 3.4 Class 8 Maths Question 5.
Explain why a rectangle is a convex quadrilateral.
Solution:
In a rectangle, both of its diagonal lie in its interior. Hence, it is a convex quadrilateral.
Ex 3.4 Class 8 Maths Question 6.
ABC is a right-angled triangle and O is the midpoint of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).
Solution:
Since the right-angled triangle ABC makes a rectangle ABCD by the dotted lines.
Therefore OA = OB = OC = OD [Diagonals of a rectangle are equal and bisect each other]
Hence, O is equidistant from A, B and C.
# Extra Questions
## Extra Questions Maths Chapter 3
Extra Questions for Class 8 Maths Chapter 3 Understanding Quadrilaterals
### Understanding Quadrilaterals Class 8 Extra Questions Very Short Answer Type
Question 1.
In the given figure, ABCD is a parallelogram. Find x.
Solution:
AB = DC [Opposite sides of a parallelogram]
3x + 5 = 5x – 1
⇒ 3x – 5x = -1 – 5
⇒ -2x = -6
⇒ x = 3
Question 2.
In the given figure find x + y + z.
Solution:
We know that the sum of all the exterior angles of a polygon = 360°
x + y + z = 360°
Question 3.
In the given figure, find x.
Solution:
∠A + ∠B + ∠C = 180° [Angle sum property]
(x + 10)° + (3x + 5)° + (2x + 15)° = 180°
⇒ x + 10 + 3x + 5 + 2x + 15 = 180
⇒ 6x + 30 = 180
⇒ 6x = 180 – 30
⇒ 6x = 150
⇒ x = 25
Question 4.
The angles of a quadrilateral are in the ratio of 2 : 3 : 5 : 8. Find the measure of each angle.
Solution:
Sum of all interior angles of a quadrilateral = 360°
Let the angles of the quadrilateral be 2x°, 3x°, 5x° and 8x°.
2x + 3x + 5x + 8x = 360°
⇒ 18x = 360°
⇒ x = 20°
Hence the angles are
2 × 20 = 40°,
3 × 20 = 60°,
5 × 20 = 100°
and 8 × 20 = 160°.
Question 5.
Find the measure of an interior angle of a regular polygon of 9 sides.
Solution:
Measure of an interior angle of a regular polygon
Question 6.
Length and breadth of a rectangular wire are 9 cm and 7 cm respectively. If the wire is bent into a square, find the length of its side.
Solution:
Perimeter of the rectangle = 2 [length + breadth]
= 2[9 + 7] = 2 × 16 = 32 cm.
Now perimeter of the square = Perimeter of rectangle = 32 cm.
Side of the square = $\frac { 32 }{ 4 }$ = 8 cm.
Hence, the length of the side of square = 8 cm.
Question 7.
In the given figure ABCD, find the value of x.
Solution:
Sum of all the exterior angles of a polygon = 360°
x + 70° + 80° + 70° = 360°
⇒ x + 220° = 360°
⇒ x = 360° – 220° = 140°
Question 8.
In the parallelogram given alongside if m∠Q = 110°, find all the other angles.
Solution:
Given m∠Q = 110°
Then m∠S = 110° (Opposite angles are equal)
Since ∠P and ∠Q are supplementary.
Then m∠P + m∠Q = 180°
⇒ m∠P + 110° = 180°
⇒ m∠P = 180° – 110° = 70°
⇒ m∠P = m∠R = 70° (Opposite angles)
Hence m∠P = 70, m∠R = 70°
and m∠S = 110°
Question 9.
In the given figure, ABCD is a rhombus. Find the values of x, y and z.
Solution:
AB = BC (Sides of a rhombus)
x = 13 cm.
Since the diagonals of a rhombus bisect each other
z = 5 and y = 12
Hence, x = 13 cm, y = 12 cm and z = 5 cm.
Question 10.
In the given figure, ABCD is a parallelogram. Find x, y and z.
Solution:
∠A + ∠D = 180° (Adjacent angles)
⇒ 125° + ∠D = 180°
⇒ ∠D = 180° – 125°
x = 55°
∠A = ∠C [Opposite angles of a parallelogram]
⇒ 125° = y + 56°
⇒ y = 125° – 56°
⇒ y = 69°
∠z + ∠y = 180° (Adjacent angles)
⇒ ∠z + 69° = 180°
⇒ ∠z = 180° – 69° = 111°
Hence the angles x = 55°, y = 69° and z = 111°
Question 11.
Find x in the following figure. (NCERT Exemplar)
Solution:
In the given figure ∠1 + 90° = 180° (linear pair)
∠1 = 90°
Now, sum of exterior angles of a polygon is 360°, therefore,
x + 60° + 90° + 90° + 40° = 360°
⇒ x + 280° = 360°
⇒ x = 80°
### Understanding Quadrilaterals Class 8 Extra Questions Short Answer Type
Question 12.
In the given parallelogram ABCD, find the value of x andy.
Solution:
∠A + ∠B = 180°
3y + 2y – 5 = 180°
⇒ 5y – 5 = 180°
⇒ 5y = 180 + 5°
⇒ 5y = 185°
⇒ y = 37°
Now ∠A = ∠C [Opposite angles of a parallelogram]
3y = 3x + 3
⇒ 3 × 37 = 3x + 3
⇒ 111 = 3x + 3
⇒ 111 – 3 = 3x
⇒ 108 = 3x
⇒ x = 36°
Hence, x = 36° and y – 37°.
Question 13.
ABCD is a rhombus with ∠ABC = 126°, find the measure of ∠ACD.
Solution:
∠ABC = ∠ADC (Opposite angles of a rhombus)
∠ADC = 126°
∠ODC = $\frac { 1 }{ 2 }$ × ∠ADC (Diagonal of rhombus bisects the respective angles)
⇒ ∠ODC = $\frac { 1 }{ 2 }$ × 126° = 63°
⇒ ∠DOC = 90° (Diagonals of a rhombus bisect each other at 90°)
In ΔOCD,
∠OCD + ∠ODC + ∠DOC = 180° (Angle sum property)
⇒ ∠OCD + 63° + 90° = 180°
⇒ ∠OCD + 153° = 180°
⇒ ∠OCD = 180° – 153° = 27°
Hence ∠OCD or ∠ACD = 27°
Question 14.
Find the values of x and y in the following parallelogram.
Solution:
Since, the diagonals of a parallelogram bisect each other.
OA = OC
x + 8 = 16 – x
⇒ x + x = 16 – 8
⇒ 2x = 8
x = 4
Similarly, OB = OD
5y + 4 = 2y + 13
⇒ 3y = 9
⇒ y = 3
Hence, x = 4 and y = 3
Question 15.
Write true and false against each of the given statements.
(a) Diagonals of a rhombus are equal.
(b) Diagonals of rectangles are equal.
(c) Kite is a parallelogram.
(d) Sum of the interior angles of a triangle is 180°.
(e) A trapezium is a parallelogram.
(f) Sum of all the exterior angles of a polygon is 360°.
(g) Diagonals of a rectangle are perpendicular to each other.
(h) Triangle is possible with angles 60°, 80° and 100°.
(i) In a parallelogram, the opposite sides are equal.
Solution:
(a) False
(b) True
(c) False
(d) True
(e) False
(f) True
(g) False
(h) False
(i) True
Question 16.
The sides AB and CD of a quadrilateral ABCD are extended to points P and Q respectively. Is ∠ADQ + ∠CBP = ∠A + ∠C? Give reason.
(NCERT Exemplar)
Solution:
Join AC, then
∠CBP = ∠BCA + ∠BAC and ∠ADQ = ∠ACD + ∠DAC (Exterior angles of triangles)
Therefore,
∠CBP + ∠ADQ = ∠BCA + ∠BAC + ∠ACD + ∠DAC
= (∠BCA + ∠ACD) + (∠BAC + ∠DAC)
= ∠C + ∠A
### Understanding Quadrilaterals Class 8 Extra Questions Higher Order Thinking Skills (HOTS)
Question 17.
The diagonal of a rectangle is thrice its smaller side. Find the ratio of its sides.
Solution:
Let AD = x cm
diagonal BD = 3x cm
In right-angled triangle DAB,
AD2 + AB2 = BD2 (Using Pythagoras Theorem)
x2 + AB2 = (3x)2
⇒ x2 + AB2 = 9x2
⇒ AB2 = 9x2 – x2
⇒ AB2 = 8x2
⇒ AB = √8x = 2√2x
Required ratio of AB : AD = 2√2x : x = 2√2 : 1
Question 18.
If AM and CN are perpendiculars on the diagonal BD of a parallelogram ABCD, Is ∆AMD = ∆CNB? Give reason. (NCERT Exemplar)
Solution:
In triangles AMD and CNB,
AD = BC (opposite sides of parallelogram)
∠AMB = ∠CNB = 90°
∠ADM = ∠NBC (AD || BC and BD is transversal.)
So, ∆AMD = ∆CNB (AAS)
## NCERT Solutions for Class 8 Maths
### లాజికల్ రీజనింగ్ అండ్ డేటా ఇంటర్ప్రెటేషన్ 10
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# Parabolas Warm Up Lesson Presentation Lesson Quiz
## Presentation on theme: "Parabolas Warm Up Lesson Presentation Lesson Quiz"— Presentation transcript:
Parabolas Warm Up Lesson Presentation Lesson Quiz
Holt McDougal Algebra 2 Holt Algebra 2
Warm Up 1. Given , solve for p when c = Find each distance.
2. from (0, 2) to (12, 7) 13 3. from the line y = –6 to (12, 7) 13
Objectives Write the standard equation of a parabola and its axis of symmetry. Graph a parabola and identify its focus, directrix, and axis of symmetry.
Vocabulary focus of a parabola directrix
In Chapter 5, you learned that the graph of a quadratic function is a parabola. Because a parabola is a conic section, it can also be defined in terms of distance.
A parabola is the set of all points P(x, y) in a plane that are an equal distance from both a fixed point, the focus, and a fixed line, the directrix. A parabola has a axis of symmetry perpendicular to its directrix and that passes through its vertex. The vertex of a parabola is the midpoint of the perpendicular segment connecting the focus and the directrix.
The distance from a point to a line is defined as the length of the line segment from the point perpendicular to the line. Remember!
Example 1: Using the Distance Formula to Write the Equation of a Parabola
Use the Distance Formula to find the equation of a parabola with focus F(2, 4) and directrix y = –4. PF = PD Definition of a parabola. Distance Formula. Substitute (2, 4) for (x1, y1) and (x, –4) for (x2, y2).
Example 1 Continued Simplify. Square both sides. (x – 2)2 + (y – 4)2 = (y + 4)2 (x – 2)2 + y2 – 8y + 16 = y2 + 8y + 16 Expand. Subtract y2 and 16 from both sides. (x – 2)2 – 8y = 8y Add 8y to both sides. (x – 2)2 = 16y Solve for y.
Check It Out! Example 1 Use the Distance Formula to find the equation of a parabola with focus F(0, 4) and directrix y = –4. PF = PD Definition of a parabola. Distance Formula Substitute (0, 4) for (x1, y1) and (x, –4) for (x2, y2).
Check It Out! Example 1 Continued
Simplify. x2 + (y – 4)2 = (y + 4)2 Square both sides. x2 + y2 – 8y + 16 = y2 + 8y +16 Expand. Subtract y2 and 16 from both sides. x2 – 8y = 8y x2 = 16y Add 8y to both sides. Solve for y.
Previously, you have graphed parabolas with vertical axes of symmetry that open upward or downward. Parabolas may also have horizontal axes of symmetry and may open to the left or right. The equations of parabolas use the parameter p. The |p| gives the distance from the vertex to both the focus and the directrix.
Example 2A: Writing Equations of Parabolas
Write the equation in standard form for the parabola. Step 1 Because the axis of symmetry is vertical and the parabola opens downward, the equation is in the form y = x2 with p < 0. 1 4p
Step 3 The equation of the parabola is . y = – x2
Example 2A Continued Step 2 The distance from the focus (0, –5) to the vertex (0, 0), is 5, so p = –5 and 4p = –20. Step 3 The equation of the parabola is y = – x2 1 20 Check Use your graphing calculator. The graph of the equation appears to match.
Example 2B: Writing Equations of Parabolas
Write the equation in standard form for the parabola. vertex (0, 0), directrix x = –6 Step 1 Because the directrix is a vertical line, the equation is in the form The vertex is to the right of the directrix, so the graph will open to the right.
Step 2 Because the directrix is x = –6, p = 6 and 4p = 24.
Example 2B Continued Step 2 Because the directrix is x = –6, p = 6 and 4p = 24. Step 3 The equation of the parabola is x = y2 1 24 Check Use your graphing calculator.
Check It Out! Example 2a Write the equation in standard form for the parabola. vertex (0, 0), directrix x = 1.25 Step 1 Because the directrix is a vertical line, the equation is in the form of The vertex is to the left of the directrix, so the graph will open to the left.
Check It Out! Example 2a Continued
Step 2 Because the directrix is x = 1.25, p = –1.25 and 4p = –5. Step 3 The equation of the parabola is Check Use your graphing calculator.
Check It Out! Example 2b Write the equation in standard form for each parabola. vertex (0, 0), focus (0, –7) Step 1 Because the axis of symmetry is vertical and the parabola opens downward, the equation is in the form
Check It Out! Example 2b Continued
Step 2 The distance from the focus (0, –7) to the vertex (0, 0) is 7, so p = –7 and 4p = –28. Step 3 The equation of the parabola is Check Use your graphing calculator.
The vertex of a parabola may not always be the origin
The vertex of a parabola may not always be the origin. Adding or subtracting a value from x or y translates the graph of a parabola. Also notice that the values of p stretch or compress the graph.
Example 3: Graphing Parabolas
Find the vertex, value of p, axis of symmetry, focus, and directrix of the parabola Then graph. y + 3 = (x – 2)2. 1 8 Step 1 The vertex is (2, –3). Step , so 4p = 8 and p = 2. 1 4p 8 =
Example 3 Continued Step 3 The graph has a vertical axis of symmetry, with equation x = 2, and opens upward. Step 4 The focus is (2, –3 + 2), or (2, –1). Step 5 The directrix is a horizontal line y = –3 – 2, or y = –5.
Check It Out! Example 3a Find the vertex, value of p, axis of symmetry, focus, and directrix of the parabola. Then graph. Step 1 The vertex is (1, 3). Step , so 4p = 12 and p = 3. 1 4p 12 =
Check It Out! Example 3a Continued
Step 3 The graph has a horizontal axis of symmetry with equation y = 3, and opens right. Step 4 The focus is (1 + 3, 3), or (4, 3). Step 5 The directrix is a vertical line x = 1 – 3, or x = –2.
Check It Out! Example 3b Find the vertex, value of p axis of symmetry, focus, and directrix of the parabola. Then graph. Step 1 The vertex is (8, 4). Step , so 4p = –2 and p = – 1 4p 2 = –
Check It Out! Example 3b Continued
Step 3 The graph has a vertical axis of symmetry, with equation x = 8, and opens downward. Step 4 The focus is or (8, 3.5). Step 5 The directrix is a horizontal line or y = 4.5.
Light or sound waves collected by a parabola will be reflected by the curve through the focus of the parabola, as shown in the figure. Waves emitted from the focus will be reflected out parallel to the axis of symmetry of a parabola. This property is used in communications technology.
Example 4: Using the Equation of a Parabola
The cross section of a larger parabolic microphone can be modeled by the equation What is the length of the feedhorn? x = y2. 1 132 The equation for the cross section is in the form x = y2, 1 4p so 4p = 132 and p = 33. The focus should be 33 inches from the vertex of the cross section. Therefore, the feedhorn should be 33 inches long.
The equation for the cross section is in the form
Check It Out! Example 4 Find the length of the feedhorn for a microphone with a cross section equation x = y2. 1 44 The equation for the cross section is in the form x = y2, 1 4p so 4p = 44 and p = 11. The focus should be 11 inches from the vertex of the cross section. Therefore, the feedhorn should be 11 inches long.
Lesson Quiz 1. Write an equation for the parabola with focus F(0, 0) and directrix y = 1. 2. Find the vertex, value of p, axis of symmetry, focus, and directrix of the parabola y – 2 = (x – 4)2, 1 12 then graph. vertex: (4, 2); focus: (4,5); directrix: y = –1; p = 3; axis of symmetry: x = 4
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Ian's Shoelace Site – 2 Trillion Lacing Methods?
# 2 Trillion Lacing Methods?
On an average shoe with six pairs of eyelets, there are almost 2 Trillion ways to feed a shoelace through those twelve eyelets! Impossible? This page shows the maths behind that extraordinary number.
## Shoe Lacing Mathematics
It hardly seems possible that there could be quite that many ways to feed a shoelace through twelve eyelets! So let's look at the mathematics:
• Feed through one of 12 eyelets from either inside or outside. That's 24 possible ways to start.
• Continue through one of 11 remaining eyelets from either inside or outside (×22 more ways).
• Then 10 remaining eyelets (×20 more ways). We've only gone through three eyelets and we're already up to 24×22×20 = 10,560 ways!
• By the time we reach the last eyelet (×2 more ways), the possible ways have multiplied to 24×22×20×18×16×14×12×10×8×6×4×2 ways, a staggering total of 1,961,990,553,600.
That's almost 2 TRILLION possibilities!
In more general mathematical notation, if “n” is the total number of eyelets, then the formula for the total possible paths through those eyelets is:
paths = n!×2n
Thus for the above example of six pairs of eyelets = twelve eyelets, the total number of paths is: 12!×212 = 1,961,990,553,600.
### Duplicate Paths
This number of possible paths can be halved for those that are mirror-images of other paths. Similarly, the number can be halved again for those that follow the identical path from opposite ends. (Actually, it's a bit more than half because some opposite-end duplicates have already been eliminated as mirror-image duplicates.) Even so, this still results in at least 500 billion unique paths.
In addition to the basic mathematics of passing through the eyelets, we can multiply by the many different ways the shoelaces can be crossed or interwoven prior to passing through those eyelets (as shown at right), and multiply again if we allow the laces to either pass through any eyelet more than once or skip any eyelet, and even more if we use two or more laces per shoe. This results in almost infinite possibilities, limited mainly by the length of the shoelaces.
### Real-World Constraints
In the real world, however, we can impose some sensible constraints, such as:
• The lace should generally start and finish from the top pair of eyelets.
• The lace should pass through each eyelet only once.
• Each eyelet should contribute to pulling together the sides of the shoe.
• The lacing should not be too difficult to tighten or loosen.
• Any pattern formed should be relatively stable.
• Ignore irrelevant variations (eg. changing the direction through a single eyelet).
• Above all, the finished result should be visually pleasing.
So how many possible ways are there to lace a shoe with twelve eyelets if we DO take into account some or all of the above constraints? This requires far more complicated maths than the simple multiplications above. For example:
The above combinatorial equation came from research by Australian mathematician Burkard Polster, who, in December 2002, caused a sudden worldwide surge of scientific and academic interest in the mathematics of shoelacing following the publication of an article in the respected journal “Nature”.
Although not quoted in the Nature article, Polster's calculation for the number of real-world lacing methods for a typical shoe with 12 eyelets came to 43,200.
### Conclusion
I'm sure that the number of Shoe Lacing Methods on this website is destined to grow as I discover more worthwhile methods from the thousands of possibilities that I haven't yet explored.
Burkard Polster's 2002 article in Nature spawned a number of articles, one of the most informative and readable of which is reproduced here on the San Francisco Chronicle's website. In Oct-2009, Burkard re-visited the subject in his Maths Masters column in The Age newspaper.
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# 6.2. Sets and Relations¶
## 6.2.1. Set Notation¶
The concept of a set in the mathematical sense has wide application in computer science. The notations and techniques of set theory are commonly used when describing and implementing algorithms because the abstractions associated with sets often help to clarify and simplify algorithm design.
A set is a collection of distinguishable members or elements. The members are typically drawn from some larger population known as the base type. Each member of a set is either a primitive element of the base type or is a set itself. There is no concept of duplication in a set. Each value from the base type is either in the set or not in the set. For example, a set named $\mathbf{P}$ might consist of the three integers 7, 11, and 42. In this case, $\mathbf{P}$’s members are 7, 11, and 42, and the base type is integer.
The following table shows the symbols commonly used to express sets and their relationships.
Table 6.2.1
$\begin{split}\begin{array}{l|l} \{1, 4\}& \mbox{A set composed of the members 1 and 4}\\ \{\mathsf{x}\, |\, \mathsf{x}\ \mbox{is a positive integer}\}& \mbox{A set definition using a set former}\\ &\qquad \mbox{Example: the set of all positive integers}\\ \mathsf{x} \in \mathbf{P}&\mathsf{x}\ \mbox{is a member of set}\ \mathbf{P}\\ \mathsf{x} \notin \mathbf{P}&\mathsf{x}\ \mbox{is not a member of set}\ \mathbf{P}\\ \emptyset&\mbox{The null or empty set}\\ |\mathbf{P}|& \mbox{Cardinality: size of set}\ \mathbf{P} \mbox{or number of members for set}\ \mathbf{P}\\ \mathbf{P}\,\subseteq\,\mathbf{Q}, \mathbf{Q}\,\supseteq\,\mathbf{P}& \mbox{Set}\ \mathbf{P}\ \mbox{is included in set}\ \mathbf{Q},\\ &\qquad \mbox{set}\ \mathbf{P}\ \mbox{is a subset of set}\ \mathbf{Q},\\ &\qquad \mbox{set}\ \mathbf{Q}\ \mbox{is a superset of set}\ \mathbf{P}\\ \mathbf{P}\,\cup\,\mathbf{Q} & \mbox{Set Union: all elements appearing in} \ \mathbf{P}\ \mbox{OR}\ \mathbf{Q}\\ \mathbf{P}\,\cap\,\mathbf{Q} & \mbox{Set Intersection: all elements appearing in}\ \mbox{P} \ \mbox{AND}\ \mathbf{Q}\\ \mathbf{P}\,-\,\mathbf{Q} & \mbox{Set difference: all elements of set} \ \mathbf{P}\ \mbox{NOT in set}\ \mathbf{Q}\\ \mathbf{P}\,\times\,\mathbf{Q} & \mbox{Set (Cartesian) Product: yields a set of ordered pairs}\\ \end{array}\end{split}$
Here are some examples of this notation in use. First define two sets, $\mathbf{P}$ and $\mathbf{Q}$.
$\mathbf{P} = \{2, 3, 5\}, \qquad \mathbf{Q} = \{5, 10\}.$
$|\mathbf{P}| = 3$ (because $\mathbf{P}$ has three members) and $|\mathbf{Q}| = 2$ (because $\mathbf{Q}$ has two members). Both of these sets are finite in length. Other sets can be infinite, for example, the set of integers.
The union of $\mathbf{P}$ and $\mathbf{Q}$, written $\mathbf{P} \cup \mathbf{Q}$, is the set of elements in either $\mathbf{P}$ or $\mathbf{Q}$, which is {2, 3, 5, 10}. The intersection of $\mathbf{P}$ and $\mathbf{Q}$, written $\mathbf{P} \cap \mathbf{Q}$, is the set of elements that appear in both $\mathbf{P}$ and $\mathbf{Q}$, which is {5}. The set difference of $\mathbf{P}$ and $\mathbf{Q}$, written $\mathbf{P} - \mathbf{Q}$, is the set of elements that occur in $\mathbf{P}$ but not in $\mathbf{Q}$, which is {2, 3}. Note that $\mathbf{P} \cup \mathbf{Q} = \mathbf{Q} \cup \mathbf{P}$ and that $\mathbf{P} \cap \mathbf{Q} = \mathbf{Q} \cap \mathbf{P}$, but in general $\mathbf{P} - \mathbf{Q} \neq \mathbf{Q} - \mathbf{P}$. In this example, $\mathbf{Q} - \mathbf{P} = \{10\}$. Finally, the set {5, 3, 2} is indistinguishable from set $\mathbf{P}$, because sets have no concept of order. Likewise, set {2, 3, 2, 5} is also indistinguishable from $\mathbf{P}$, because sets have no concept of duplicate elements.
The set product or Cartesian product of two sets $\mathbf{Q} \times \mathbf{P}$ is a set of ordered pairs. For our example sets, the set product would be
$\{(2, 5),\ (2, 10),\ (3, 5),\ (3, 10),\ (5, 5),\ (5, 10)\}.$
The powerset of a set $\mathbf{S}$ (denoted $2^S$) is the set of all possible subsets for $\mathbf{S}$. Consider the set $\mathbf{S} = \{ a, b, c \}$. The powerset of $\mathbf{S}$ is
$\{ \emptyset,\ \{a\},\ \{b\},\ \{c\},\ \{a, b\}, \ \{a, c\},\ \{b, c\},\ \{a, b, c\}\}.$
A collection of elements with no order (like a set), but with duplicate-valued elements is called a bag 1. To distinguish bags from sets, we will use square brackets [] around a bag’s elements. For example, bag [3, 4, 5, 4] is distinct from bag [3, 4, 5], while set {3, 4, 5, 4} is indistinguishable from set {3, 4, 5}. However, bag [3, 4, 5, 4] is indistinguishable from bag [3, 4, 4, 5].
A sequence is a collection of elements with an order, and which may contain duplicate-valued elements. A sequence is also sometimes called a tuple or a vector. In a sequence, there is a 0th element, a 1st element, 2nd element, and so on. We will use angle brackets $\langle\rangle$ to enclose the elements of a sequence. For example, $\langle3, 4, 5, 4\rangle$ is a sequence. Note that sequence $\langle3, 5, 4, 4\rangle$ is distinct from sequence $\langle3, 4, 5, 4\rangle$, and both are distinct from sequence $\langle3, 4, 5\rangle$.
1
The object referred to here as a bag is sometimes called a multilist. But, the term multilist also refers to a list that may contain sublists.
### 6.2.1.1. Relations¶
A relation $R$ over set $\mathbf{S}$ is a set of ordered pairs from $\mathbf{S}$. As an example of a relation, if $\mathbf{S}$ is $\{a, b, c\}$, then
$\{ \langle a, c\rangle, \langle b, c\rangle, \langle c, b\rangle \}$
is a relation, and
$\{ \langle a, a\rangle, \langle a, c\rangle, \langle b, b\rangle, \langle b, c\rangle, \langle c, c\rangle \}$
is a different relation. If tuple $\langle x, y\rangle$ is in relation $R$, we may use the infix notation $xRy$. We often use relations such as the less than operator ($<$) on the natural numbers, which includes ordered pairs such as $\langle1, 3\rangle$ and $\langle2, 23\rangle$, but not $\langle3, 2\rangle$ or $\langle2, 2\rangle$. Rather than writing the relationship in terms of ordered pairs, we typically use an infix notation for such relations, writing $1<3$.
Define the properties of relations as follows, with $R$ a binary relation over set $\mathbf{S}$.
• $R$ is reflexive if $aRa$ for all $a \in \mathbf{S}$.
• $R$ is irreflexive if $aRa$ is not true for all $a \in \mathbf{S}$.
• $R$ is symmetric if whenever $aRb$, then $bRa$, for all $a, b \in \mathbf{S}$.
• $R$ is antisymmetric if whenever $aRb$ and $bRa$, then $a = b$, for all $a, b \in \mathbf{S}$.
• $R$ is transitive if whenever $aRb$ and $bRc$, then $aRc$, for all $a, b, c \in \mathbf{S}$.
As examples, for the natural numbers, $<$ is irreflexive (because $aRa$ is never true), antisymmetric (because there is no case where $aRb$ and $bRa$), and transitive. Relation $\leq$ is reflexive, antisymmetric, and transitive. Relation $=$ is reflexive, symmetric (and antisymmetric!), and transitive. For people, the relation “is a sibling of” is symmetric and transitive. If we define a person to be a sibling of themself, then it is reflexive; if we define a person not to be a sibling of themself, then it is not reflexive.
## 6.2.2. Equivalence Relations¶
$R$ is an equivalence relation on set $\mathbf{S}$ if it is reflexive, symmetric, and transitive. An equivalence relation can be used to partition a set into equivalence classes. If two elements $a$ and $b$ are equivalent to each other, we write $a \equiv b$. A partition of a set $\mathbf{S}$ is a collection of subsets that are disjoint from each other and whose union is $\mathbf{S}$. An equivalence relation on set $\mathbf{S}$ partitions the set into disjoint subsets whose elements are equivalent. The UNION/FIND algorithm efficiently maintains equivalence classes on a set. One application for such disjoint sets computing a minimal cost spanning tree.
Example 6.2.1
For the integers, $=$ is an equivalence relation that partitions each element into a distinct subset. In other words, for any integer $a$, three things are true.
1. $a = a$,
2. if $a = b$ then $b = a$, and
3. if $a = b$ and $b = c$, then $a = c$.
Of course, for distinct integers $a$, $b$, and $c$ there are never cases where $a = b$, $b = a$, or $b = c$. So the requirements for symmetry and transitivity are never violated, and therefore the relation is symmetric and transitive.
Example 6.2.2
If we clarify the definition of sibling to mean that a person is a sibling of themself, then the sibling relation is an equivalence relation that partitions the set of people.
Example 6.2.3
We can use the modulus function to define an equivalence relation. For the set of integers, use the modulus function to define a binary relation such that two numbers $x$ and $y$ are in the relation if and only if $x \bmod m = y \bmod m$. Thus, for $m = 4$, $\langle1, 5\rangle$ is in the relation because $1 \bmod 4 = 5 \bmod 4$. We see that modulus used in this way defines an equivalence relation on the integers, and this relation can be used to partition the integers into $m$ equivalence classes. This relation is an equivalence relation because
1. $x \bmod m = x \bmod m$ for all $x$;
2. if $x \bmod m = y \bmod m$, then $y \bmod m = x \bmod m$; and
3. if $x \bmod m = y \bmod m$ and $y \bmod m = z \bmod m$, then $x \bmod m = z \bmod m$.
## 6.2.3. Partial Orders¶
A binary relation is called a partial order if it is antisymmetric and transitive. If the relation is reflexive, it is called a non-strict partial order. If the relation is irreflexive, it is called a strict partial order. The set on which the partial order is defined is called a partially ordered set or a poset. Elements $x$ and $y$ of a set are comparable under a given relation $R$ if either $xRy$ or $yRx$. If every pair of distinct elements in a partial order are comparable, then the order is called a total order or linear order.
Example 6.2.4
For the integers, relations $<$ and $\leq$ define partial orders. Operation $<$ is a total order because, for every pair of integers $x$ and $y$ such that $x \neq y$, either $x < y$ or $y < x$. Likewise, $\leq$ is a total order because, for every pair of integers $x$ and $y$ such that $x \neq y$, either $x \leq y$ or $y \leq x$.
Example 6.2.5
For the powerset of the integers, the subset operator defines a partial order (because it is antisymmetric and transitive). For example, $\{1, 2\}\subseteq\{1, 2, 3\}$. However, sets {1, 2} and {1, 3} are not comparable by the subset operator, because neither is a subset of the other. Therefore, the subset operator does not define a total order on the powerset of the integers.
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# How do you solve the following system: 5x+y=-7, 6x + 7y = -9 ?
Jul 2, 2017
See a solution process below:
#### Explanation:
Step 1) Solve the first equation for $y$:
$5 x + y = - 7$
$- \textcolor{red}{5 x} + 5 x + y = - \textcolor{red}{5 x} - 7$
$0 + y = - 5 x - 7$
$y = - 5 x - 7$
Step 2) Substitute $\left(- 5 x - 7\right)$ for $y$ in the second equation and solve for $x$:
$6 x + 7 y = - 9$ becomes:
$6 x + 7 \left(- 5 x - 7\right) = - 9$
$6 x + \left(7 \cdot - 5 x\right) - \left(7 \cdot 7\right) = - 9$
$6 x + \left(- 35 x\right) - 49 = - 9$
$6 x - 35 x - 49 = - 9$
$\left(6 - 35\right) x - 49 = - 9$
$- 29 x - 49 = - 9$
$- 29 x - 49 + \textcolor{red}{49} = - 9 + \textcolor{red}{49}$
$- 29 x - 0 = 40$
$- 29 x = 40$
$\frac{- 29 x}{\textcolor{red}{- 29}} = \frac{40}{\textcolor{red}{- 29}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 29}}} x}{\cancel{\textcolor{red}{- 29}}} = - \frac{40}{29}$
$x = - \frac{40}{29}$
Step 3) Substitute $- \frac{40}{29}$ for $x$ in the solution to the first equation at the end of Step 1 and calculate $y$:
$y = - 5 x - 7$ becomes:
$y = \left(- 5 \times - \frac{40}{29}\right) - 7$
$y = \frac{200}{29} - 7$
$y = \frac{200}{29} - \left(\frac{29}{29} \times 7\right)$
$y = \frac{200}{29} - \frac{203}{29}$
$y = - \frac{3}{29}$
The solution is: $x = - \frac{40}{29}$ and $y = - \frac{3}{29}$ or $\left(- \frac{40}{29} , - \frac{3}{29}\right)$
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The gradient for this radius is $$m = \frac{5}{3}$$. Calculate the coordinates of $$P$$ and $$Q$$. Example: Find the outer intersection point of the circles: (r 0) (x − 3) 2 + (y + 5) 2 = 4 2 (r 1) (x + 2) 2 + (y − 2) 2 = 1 2. This gives the points $$F(-3;-4)$$ and $$H(-4;3)$$. The product of the gradient of the radius and the gradient of the tangent line is equal to $$-\text{1}$$. A circle has a center, which is that point in the middle and provides the name of the circle. \end{align*}. Substitute the $$Q(-10;m)$$ and solve for the $$m$$ value. We’ll use the point form once again. The tangent to a circle equation x2+ y2=a2 at (x1, y1) isxx1+yy1= a2 1.2. In our crop circle U, if we look carefully, we can see a tangent line off to the right, line segment FO. Example: Find equations of the common tangents to circles x 2 + y 2 = 13 and (x + 2) 2 + (y + 10) 2 = 117. Determine the gradient of the tangent to the circle at the point $$(2;2)$$. Determine the coordinates of $$M$$, the mid-point of chord $$PQ$$. The required equation will be x(4) + y(-3) = 25, or 4x – 3y = 25. Solved: In the diagram, point P is a point of tangency. After having gone through the stuff given above, we hope that the students would have understood "Find the equation of the tangent to the circle at the point ". Circle centered at any point (h, k), ( x – h) 2 + ( y – k) 2 = r2. \end{align*}. Here is a crop circle with three little crop circles tangential to it: [insert cartoon drawing of a crop circle ringed by three smaller, tangential crop circles]. Find the gradient of the radius at the point $$(2;2)$$ on the circle. At this point, you can use the formula, \\ m \angle MJK= \frac{1}{2} \cdot 144 ^{\circ} \\ m \angle ... Back to Circle Formulas Next to Arcs and Angles. &= \sqrt{(-4 -(-10))^{2} + (-2 - 10)^2} \\ Because equations (3) and (4) are quadratic, there will be as many as 4 solutions, as shown in the picture. Points of tangency do not happen just on circles. Get help fast. Equation of the two circles given by: (x − a) 2 + (y − b) 2 = r 0 2 (x − c) 2 + (y − d) 2 = r 1 2. Solution: Intersections of the line and the circle are also tangency points.Solutions of the system of equations are coordinates of the tangency points, Several theorems are related to this because it plays a significant role in geometrical constructionsand proofs. The equations of the tangents are $$y = -5x - 26$$ and $$y = - \frac{1}{5}x + \frac{26}{5}$$. In geometry, a tangent of a circle is a straight line that touches the circle at exactly one point, never entering the circle’s interior. x 2 + y 2 = r 2. Determine the gradient of the radius $$OQ$$: Substitute $$m_{Q} = - \frac{1}{5}$$ and $$Q(1;5)$$ into the equation of a straight line. The point P is called the point … The centre of the circle is $$(-3;1)$$ and the radius is $$\sqrt{17}$$ units. The points on the circle can be calculated when you know the equation for the tangent lines. Specifically, my problem deals with a circle of the equation x^2+y^2=24 and the point on the tangent being (2,10). We can also talk about points of tangency on curves. by this license. Find a tutor locally or online. Finally we convert that angle to degrees with the 180 / π part. The radius of the circle $$CD$$ is perpendicular to the tangent $$AB$$ at the point of contact $$D$$. Here are the circle equations: Circle centered at the origin, (0, 0), x2 + y2 = r2. How to determine the equation of a tangent: Determine the equation of the tangent to the circle $$x^{2} + y^{2} - 2y + 6x - 7 = 0$$ at the point $$F(-2;5)$$. This means a circle is not all the space inside it; it is the curved line around a point that closes in a space. Determine the gradient of the radius. m_r & = \frac{y_1 - y_0}{x_1 - x_0} \\ The Tangent Secant Theorem explains a relationship between a tangent and a secant of the same circle. Show that $$S$$, $$H$$ and $$O$$ are on a straight line. Use the distance formula to determine the length of the radius: Write down the general equation of a circle and substitute $$r$$ and $$H(2;-2)$$: The equation of the circle is $$\left(x + 4\right)^{2} + \left(y - 8\right)^{2} = 136$$. This means that AT¯ is perpendicular to TP↔. To do that, the tangent must also be at a right angle to a radius (or diameter) that intersects that same point. Determine the gradient of the radius $$OT$$. \begin{align*} The tangent lines to circles form the subject of several theorems and play an important role in many geometrical constructions and proofs. If $$O$$ is the centre of the circle, show that $$PQ \perp OH$$. A circle with centre $$(8;-7)$$ and the point $$(5;-5)$$ on the circle are given. Notice that the diameter connects with the center point and two points on the circle. One circle can be tangent to another, simply by sharing a single point. Identify and recognize a tangent of a circle, Demonstrate how circles can be tangent to other circles, Recall and explain three theorems related to tangents. Setting each equal to 0 then setting them equal to each other might help. Embedded videos, simulations and presentations from external sources are not necessarily covered c 2 = a 2 (1 + m 2) p 2 /16 = 16 (1 + 9/16) p 2 /16 = 16 (25/16) p 2 /16 = 25. p 2 = 25(16) p = ± 20. Point of tangency is the point where the tangent touches the circle. The two vectors are orthogonal, so their dot product is zero: Get better grades with tutoring from top-rated professional tutors. The second theorem is called the Two Tangent Theorem. The points will be where the circle's equation = the tangent's … Complete the sentence: the product of the $$\ldots \ldots$$ of the radius and the gradient of the $$\ldots \ldots$$ is equal to $$\ldots \ldots$$. Determine the equations of the two tangents to the circle, both parallel to the line $$y + 2x = 4$$. 1.1. To determine the coordinates of $$A$$ and $$B$$, we substitute the straight line $$y = - 2x + 1$$ into the equation of the circle and solve for $$x$$: This gives the points $$A(-4;9)$$ and $$B(4;-7)$$. equation of tangent of circle. Substitute the straight line $$y = x + 4$$ into the equation of the circle and solve for $$x$$: This gives the points $$P(-5;-1)$$ and $$Q(1;5)$$. &= \sqrt{36 \cdot 2} \\ The gradient for the tangent is $$m_{\bot} = \frac{3}{2}$$. More precisely, a straight line is said to be a tangent of a curve y = f(x) at a point x = c if the line passes through the point (c, f(c)) on the curve and has slope f '(c), where f ' is the derivative of f. That distance is known as the radius of the circle. This point is called the point of tangency. A line that joins two close points from a point on the circle is known as a tangent. The word "tangent" comes from a Latin term meaning "to touch," because a tangent just barely touches a circle. If (2,10) is a point on the tangent, how do I find the point of tangency on the circle? In geometry, a circle is a closed curve formed by a set of points on a plane that are the same distance from its center O. So, if you have a graph with curves, like a parabola, it can have points of tangency as well. Solution : Equation of the line 3x + 4y − p = 0. Here a 2 = 16, m = −3/4, c = p/4. Where it touches the line, the equation of the circle equals the equation of the line. We won’t establish any formula here, but I’ll illustrate two different methods, first using the slope form and the other using the condition of tangency. The tangent to a circle equation x2+ y2=a2 at (a cos θ, a sin θ ) isx cos θ+y sin θ= a 1.4. Let the gradient of the tangent at $$Q$$ be $$m_{Q}$$. & = \frac{5 - 6 }{ -2 -(-9)} \\ From the graph we see that the $$y$$-coordinate of $$Q$$ must be positive, therefore $$Q(-10;18)$$. The tangent to the circle at the point $$(5;-5)$$ is perpendicular to the radius of the circle to that same point: $$m \times m_{\bot} = -1$$. We think you are located in Let the point of tangency be ( a, b). Let the two tangents from $$G$$ touch the circle at $$F$$ and $$H$$. Find the equation of the tangent at $$P$$. \begin{align*} Example 3 : Find the value of p so that the line 3x + 4y − p = 0 is a tangent to x 2 + y 2 − 64 = 0. &= - 1 \\ \begin{align*} The tangent line $$AB$$ touches the circle at $$D$$. I need to find the points of tangency on a circle (x^2+y^2=100) and a line y=5x+b the only thing I know about b is that it is negative. Tangent to a Circle A tangent to a circle is a straight line which touches the circle at only one point. Equation of the circle x 2 + y 2 = 64. The equation for the tangent to the circle at the point $$Q$$ is: The straight line $$y = x + 2$$ cuts the circle $$x^{2} + y^{2} = 20$$ at $$P$$ and $$Q$$. Let's look at an example of that situation. m_{PQ} \times m_{OM} &= - 1 \\ Notice that the line passes through the centre of the circle. Determine the gradient of the tangent to the circle at the point $$(5;-5)$$. &= \sqrt{36 + 144} \\ This line runs parallel to the line y=5x+7. EF is a tangent to the circle and the point of tangency is H. Tangents From The Same External Point. Tangent to a Circle. m_{OM} &= \frac{1 - 0}{-1 - 0} \\ In other words, we can say that the lines that intersect the circles exactly in one single point are Tangents. Point Of Tangency To A Curve. Plot the point $$P(0;5)$$. We have already shown that $$PQ$$ is perpendicular to $$OH$$, so we expect the gradient of the line through $$S$$, $$H$$ and $$O$$ to be $$-\text{1}$$. A tangent to a circle is a straight line that touches the circle at one point, called the point of tangency. This means we can use the Pythagorean Theorem to solve for AP¯. Given a circle with the central coordinates $$(a;b) = (-9;6)$$. D(x; y) is a point on the circumference and the equation of the circle is: (x − a)2 + (y − b)2 = r2 A tangent is a straight line that touches the circumference of a circle at … We can also talk about points of tangency on curves. Is this correct? Creative Commons Attribution License. where r is the circle’s radius. The line joining the centre of the circle to this point is parallel to the vector. \end{align*}. At the point of tangency, a tangent is perpendicular to the radius. &= \sqrt{(-6)^{2} + (-6)^2} \\ The solution shows that $$y = -2$$ or $$y = 18$$. to personalise content to better meet the needs of our users. It states that, if two tangents of the same circle are drawn from a common point outside the circle, the two tangents are congruent. & = - \frac{1}{7} &= \sqrt{180} The gradient of the radius is $$m = - \frac{2}{3}$$. The equation of tangent to the circle{x^2} + {y^2} The condition for the tangency is c 2 = a 2 (1 + m 2) . The tangent to a circle equation x2+ y2=a2 for a line y = mx +c is y = mx ± a √[1+ m2] &= \sqrt{(12)^{2} + (-6)^2} \\ Determine the equation of the tangent to the circle at the point $$(-2;5)$$. The key is to find the points of tangency, labeled A 1 and A 2 in the next figure. 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Question Video: Finding the Monotonicity of a Function given Its Derivative Graph | Nagwa Question Video: Finding the Monotonicity of a Function given Its Derivative Graph | Nagwa
# Question Video: Finding the Monotonicity of a Function given Its Derivative Graph Mathematics • Third Year of Secondary School
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The graph of the derivative 𝑓′ of a function 𝑓 is shown. On what intervals is 𝑓 increasing or decreasing?
01:51
### Video Transcript
The graph of the derivative 𝑓 prime of a function 𝑓 is shown. On what intervals is 𝑓 increasing or decreasing?
To answer this question, we need to recall the link between whether a function is increasing or decreasing and its first derivative. Formally, a function is increasing on an interval 𝐼 if 𝑓 of 𝑥 one is less than 𝑓 of 𝑥 two for all pairs of 𝑥-values, 𝑥 one and 𝑥 two, with 𝑥 one less than 𝑥 two in the interval 𝐼. In practical terms though, this just means that the graph of the function is sloping upwards. And so its first derivative which, remember, is the slope function of the curve is positive. On the other hand, a function is decreasing on an interval 𝐼 if 𝑓 of 𝑥 one is greater than 𝑓 of 𝑥 two for all 𝑥 one less than 𝑥 two in the interval 𝐼, which in practical terms just means the line is sloping downwards. And so the first derivative, 𝑓 prime of 𝑥, is negative.
To determine the intervals on which any function is increasing or decreasing then, we just need to consider the sign of its first derivative. So the function 𝑓 will be increasing when the graph of its first derivative 𝑓 prime is above the 𝑥-axis. From the given figure, we see that this is true on the open interval, one to five. 𝑓 will be decreasing when the graph of its first derivative is below the 𝑥-axis. From the figure, we see that this is true on two open intervals, the interval zero, one and the interval five, six. So we can conclude then that 𝑓 is increasing on the open interval one to five and decreasing on the open intervals zero to one and five to six.
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# How do you solve 32x^2-3x-14=(2x-1)^2?
May 20, 2018
$x = \frac{5}{7} \mathmr{and} x = - \frac{3}{4}$
#### Explanation:
$32 {x}^{2} - 3 x - 14 = {\left(2 x - 1\right)}^{2} \text{ } \leftarrow$ remove the brackets
$32 {x}^{2} - 3 x - 14 = 4 {x}^{2} - 4 x + 1 \text{ } \leftarrow$ make it $= 0$
$32 {x}^{2} - 4 {x}^{2} - 3 x + 4 x - 14 - 1 = 0$
$28 {x}^{2} + x - 15 = 0 \text{ } \leftarrow$ factorise the quadratic
$\left(7 x - 5\right) \left(4 x + 3\right) = 0 \text{ } \leftarrow$ solve each factor set $= 0$
If $7 x - 5 = 0 \Rightarrow \text{ } x = \frac{5}{7}$
If $4 x + 3 = 0 \Rightarrow \text{ } x = - \frac{3}{4}$
May 20, 2018
$x = \frac{5}{7} , x = - \frac{3}{4}$
#### Explanation:
$32 {x}^{2} - 3 x - 14 = {\left(2 x - 1\right)}^{2}$ expand the right side,
$32 {x}^{2} - 3 x - 14 = 4 {x}^{2} - 4 x + 1$
$32 {x}^{2} - 4 {x}^{2} - 3 x + 4 x - 14 - 1 = 0$ shift them to the left side
$28 {x}^{2} + x - 15 = 0$
by using the equation $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$a = 28 , b = 1 , c = - 15$
sub a,b,c into the equation$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \left(28\right) \left(- 15\right)}}{2 \left(28\right)}$
$x = \frac{- 1 \pm \sqrt{1 + 1680}}{56}$
$x = \frac{- 1 \pm \sqrt{1681}}{56}$
$x = \frac{- 1 \pm 41}{56}$
$x = \frac{- 1 + 41}{56}$ or $x = \frac{- 1 - 41}{56}$
$x = \frac{40}{56}$ or $x = \frac{- 42}{56}$
$x = \frac{5}{7}$ or $x = \frac{- 3}{4}$
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# 2005 AIME I Problems/Problem 10
## Problem
Triangle $ABC$ lies in the Cartesian Plane and has an area of 70. The coordinates of $B$ and $C$ are $(12,19)$ and $(23,20),$ respectively, and the coordinates of $A$ are $(p,q).$ The line containing the median to side $BC$ has slope $-5.$ Find the largest possible value of $p+q.$
## Solution
$[asy]defaultpen(fontsize(8)); size(170); pair A=(15,32), B=(12,19), C=(23,20), M=B/2+C/2, P=(17,22); draw(A--B--C--A);draw(A--M);draw(B--P--C); label("A (p,q)",A,(1,1));label("B (12,19)",B,(-1,-1));label("C (23,20)",C,(1,-1));label("M",M,(0.2,-1)); label("(17,22)",P,(1,1)); dot(A^^B^^C^^M^^P);[/asy]$
### Solution 1
The midpoint $M$ of line segment $\overline{BC}$ is $\left(\frac{35}{2}, \frac{39}{2}\right)$. The equation of the median can be found by $-5 = \frac{q - \frac{35}{2}}{p - \frac{39}{2}}$. Cross multiply and simplify to yield that $-5p + \frac{39 \cdot 5}{2} = q - \frac{35}{2}$, so $q = -5p + 107$.
### Solution 2
Use determinants to find that the area of $\triangle ABC$ is $\frac{1}{2} \begin{vmatrix}p & 12 & 23 \\ q & 19 & 20 \\ 1 & 1 & 1\end{vmatrix} = 70$ (note that there is a missing absolute value; we will assume that the other solution for the triangle will give a smaller value of $p+q$, which is provable by following these steps over again). We can calculate this determinant to become $140 = \begin{vmatrix} 12 & 23 \\ 19 & 20 \end{vmatrix} - \begin{vmatrix} p & q \\ 23 & 20 \end{vmatrix} + \begin{vmatrix} p & q \\ 12 & 19 \end{vmatrix}$ $\Longrightarrow 140 = 240 - 437 - 20p + 23q + 19p - 12q$ $= -197 - p + 11q$. Thus, $q = \frac{1}{11}p - \frac{337}{11}$.
Setting this equation equal to the equation of the median, we get that $\frac{1}{11}p - \frac{337}{11} = -5p + 107$, so $\frac{56}{11}p = \frac{107 \cdot 11 + 337}{11}$. Solving produces that $p = 15$. Substituting backwards yields that $q = 32$; the solution is $p + q = 047$.
### Solution 3
Using the equation of the median, we can write the coordinates of $A$ as $(p,\ -5p + 107)$. The equation of $\overline{BC}$ is $\frac{20 - 19}{23 - 12} = \frac{y - 19}{x - 12}$, so $x - 12 = 11y - 209$. In general form, the line is $x - 11y + 197 = 0$. Use the equation for the distance between a line and point to find the distance between $A$ and $BC$ (which is the height of $\triangle ABC$): $\frac{|1(p) - 11(-5p + 107) + 197|}{1^2 + 11^2} = \frac{|56p - 980|}{\sqrt{122}}$. Now we need the length of $BC$, which is $\sqrt{(23 - 12)^2 + (20 - 19)^2} = \sqrt{122}$. The area of $\triangle ABC$ is $70 = \frac{1}{2}bh = \frac{1}{2}\left(\frac{|56p - 980|}{\sqrt{122}}\right) \cdot \sqrt{122}$. Thus, $|28p - 490| = 70$, and $p = 15,\ 20$. We are looking for $p + q = -4p + 107 = 47,\ 27$. The maximum possible value of $p + q = 47$.
### Solution 4
Let $A'$ be the point $(17, 22)$, which lies along the line through $M$ of slope $-5$. The area of triangle $A'BC$ can be computed in a number of ways (one possibility: extend $A'B$ until it hits the line $y = 19$, and subtract one triangle from another), and each such calculation gives an area of 14. This is $\frac{1}{5}$ of our needed area, so we simply need the point $A$ to be 5 times as far from $M$ as $A'$ is. Thus $A = \left(\frac{35}{2}, \frac{39}{2}\right) \pm 5\left(-\frac{1}{2}, \frac{5}{2}\right)$, and the sum of coordinates will be larger if we take the positive value, so $A = \left(\frac{35}{2} - \frac{5}2, \frac{39}{2} + \frac{25}{2}\right)$ and the answer is $\frac{35}{2} - \frac{5}2 + \frac{39}{2} + \frac{25}{2} = 047$.
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• Surface Area of A Cube (Definition, Formula & Examples)
# Surface Area of A Cube (Definition, Formula & Examples)
September 20, 2022
The surface area of a 3D shape (solid object) is a measure of the total area that the surface of the object occupies. The surface area of a cube is the total area covered by all six faces of the cube. The total surface area of cube can be calculated if we calculate the area of the two bases and the area of the four lateral (side) faces.
It’s important to know and understand the surface area as it helps you to know the amount of sheet of paper required to wrap a cube, paint the surfaces of the cube, etc.
A cube has some characteristic features that are used to find the surface area of cube such as all edges of a cube being equal and all faces of a cube being square. Let’s learn how to find the surface area of cube and its uses.
## What is the Surface Area of a Cube?
The surface area of a cube is the sum of the area of the bases($2$ bases) and the area of lateral faces($4$ lateral faces) of the cube. Since all six faces of the cube are made up of squares of the same dimensions then the total surface area of cube will be numerically equal to six times the area of one face.
The surface area of cube is measured as the “number of square units” ($cm^{2}$, $m^{2}$, $in^{2}$, $ft^{2}$, etc.). There are two types of surface areas of a cube
• Lateral Surface Area
• Total Surface Area
### Lateral Surface Area of Cube
The lateral surface area of cube refers to the total area covered by the side or lateral faces of a cube. There are four lateral faces in a cube, so to calculate LSA, we find the sum of the areas of these $4$ square faces.
If $a$ is the length of the edge (side) of a cube, then the area of one face (square) is $a \times a = a^{2}$ sq units.
Therefore, the lateral surface area of cube(LSA) is $4 \times a^{2} = 4a^{2}$ sq units.
### Lateral Surface Area of a Cube Formula
The formula for lateral surface area of cube (LSA) = $4a^{2}$.
### Total Surface Area of Cube
The total surface area of cube refers to the total area covered by all the faces of a cube. There are six square faces in a cube, so to calculate TSA, we find the sum of the areas of these $6$ faces.
If $a$ is the length of the edge (side) of a cube, then the area of one face (square) is $a \times a = a^{2}$ sq units.
Therefore, the total surface area of cube (TSA) is $6 \times a^{2} = 6a^{2}$ sq units.
### Lateral Surface Area of a Cube Formula
The formula for total surface area of cube = $6a^{2}$.
### Examples
Ex 1: Find the area of the metal sheet required to make a cubical box of side length $4 cm$.
.Length of side of cubical box = $a = 4 cm$
Amount of metal sheet required to make a cubical box = Total Surface Area of Cube = $6a^{2}$
$6a^{2} = 6 \times 4^{2} = 6 \times 16 = 96 cm^{2}$
Therefore, the area of the metal sheet required to make a cubical box of side length $4 cm$ is $96 cm^{2}$.
Ex 2: Find the ratio of the lateral surface area and total surface area of cube.
Let the side of a cube be $a$ units
Lateral surface area of cube = $4a^{2}$ sq units
Total surface area of cube = $6a^{2}$ sq units
The ratio of LSA to TSA is $4a^{2} : 6a^{2} = \frac {4a^{2}}{6a^{2}} = \frac {4}{6} = \frac {2}{3} = 2 : 3$.
Ex 3: Find the length of the edge of the cube whose total surface area is $384 cm^{2}$.
TSA of cube = $384 cm^{2}$
$6a^{2} = 384 => a^{2} = \frac {384}{6} => a^{2} = 64 => a = 8 cm$
Length of the edge of a cube = $8 cm$
## Difference Between Cube and Cuboid
Although cube and cuboid are similar $3D$ objects, there are few differences between these two. Following are the differences between a cube and a cuboid.
## Practice Problems
1. If the length of the edge of a cube is $5 cm$, calculate its
• lateral surface area
• total surface area
2. The surface area of a cube is 150 feet square. What is the length of the cube?
3. If the total surface area of cube is $96 cm^{2}$, find its lateral surface area.
4. A cube of edge $2$ cm is divided into cubes of edge $1$ cm.
• How many cubes will be made?
• Find the total surface area of the larger cube.
• Find the total surface area of the smaller cubes.
• Which of the above two surface areas is larger?
5. A solid cube of length $10 m$ is to be painted on its $6$ faces. If the painting rate is ₹$45$ per square metre, find the total cost of painting the cube.
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### Precalculus Mathematics for Calculus
Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508
# Irrigation An irrigation system uses a straight sprinkler pipe long that pivots around a central point as shown. Because of an obstacle the pipe is allowed to pivot through ${\mathbf{280}}^{\mathbf{°}}$ only. Find the area irrigated by this system.
The area irrigated by this system is $\mathbf{219}\mathbf{,}\mathbf{911}{\mathbf{\text{ft}}}^{\mathbf{2}}$.
See the step by step solution
## Step 1. Given information.
The area A of a sector with central angle of radians is:
$A=\frac{1}{2}{r}^{2}\theta$
So use above formula to find the area irrigated by this system.Given:Central angle $\left(\theta \right)={280}^{°}$
radius $\left(r\right)=300ft$
To convert into radians, multiply the angle by $\frac{\pi }{180}$
## Step 2. Finding the area irrigated by the system.
To convert $\left(\theta \right)$ into radians, multiply by $\frac{\pi }{180}$:
$\theta ={280}^{°}×\frac{\pi }{{180}^{°}}\text{rad}\phantom{\rule{0ex}{0ex}}=\frac{14\pi }{9}\text{rad}$
The area irrigated by this system is:
$A=\frac{1}{2}{r}^{2}\theta$
Put given values in the above formula, we get:
$A=\frac{1}{2}×{300}^{2}×\frac{14\pi }{9}$
Putting $\pi =3.14$ in the above formula, we get:
$A=\frac{1}{2}×{300}^{2}×\frac{14×3.14}{9}\phantom{\rule{0ex}{0ex}}=\frac{14×3.14×300×300}{18}\phantom{\rule{0ex}{0ex}}=\frac{3956400}{18}\phantom{\rule{0ex}{0ex}}\approx 219911.485751\phantom{\rule{0ex}{0ex}}\approx 219,911{\text{ft}}^{2}$
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Sherpa
Maths
>
GCSE
Rates of Change
Question
How do you find the rate of change?
2 years ago
·
171 Replies
·
6744 views
V
Vickie Shanahan
L
Leeland
The rate of change is how something is changing. Imagine you have a toy car that moves on a track. The rate of change tells you how fast the car is moving or if it's going faster or slower.
To find the rate of change, you can look at two things: how much something changes and how much time it takes. For example, if the car moves 10 centimeters in 2 seconds, the rate of change is 5 centimeters per second because it moves 5 centimeters each second.
You can also think of the rate of change as a slope on a hill. If the hill is steep, it means things are changing quickly. If it's not very steep, things are changing slowly.
So, the rate of change is like a measurement that tells you how fast something is changing or how steep the change is. It can help you understand how things are moving or growing.
D
Dewain Sewell
Distance traveled divided by time it takes. On a graph that would be calculating the difference in the change in y and the change in x
Z
Zeid Al-Arqan
(Change in y) divided by (change in x) for the rate of change in a graph but for a problem solving question its the the change divided by the original
H
Haseeb Ullah Khan
Consider the following equation
y=mx+c
The changing quantity is represented by the letter "y" called the independent variable. It is dependent upon a variable called the dependent variable denoted by 'x'. Find the value of "y" at two values of x. let Y1 be the output of X1 and Y2 be the output of X2. The rate of change of y will be calculated as follows:
dy/dx= (Y2-Y1)/(X2-X1)
M
Michael Welch
Rate of change is found through differentiation, this in its simplest form can be found by reducing the power of the function by one and timesing through by the power you have reduced, if it is a constant function you could simply find the gradient.
F
Farooq
The rate of change (ROC) is the speed at which a variable changes over a specific period of time
F
Finn Mortimore
If we have a graph, the rate of change at a particular point is the gradient of the tangent at that point - this is known as the derivative. Given some values, we could approximate the rate of change by calculating the gradient of the graph of those values at that point i.e. rise over run
D
This is an easy thing to find out. If you have a straight line on a X-Y graph, the formula you would use is y=mx+c. The rate of change is m in this formula is is how steep the line is . To calculate m for this formula you choose two separate points on the line (taking note of their X-Y coordinates). Then you take the difference between the y coordinates then take the difference between the x coordinates. Then you take these two numbers and divide the difference between the y coordinates by the difference between the x coordinates. This will give you the steepness of the line which is m in the formula of a straight line.
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The rate of change can be found by calculating the difference in the values of a quantity over a certain period of time or another variable. It is often calculated using the formula:
Rate of change = (Change in quantity) / (Change in time or another variable)
For example, to find the rate of change of distance with respect to time, you would divide the change in distance by the change in time. This gives you the speed or velocity.
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A
The rate of change refers to what degree a variable relates to another.
Considering two points connected by a line on an XY graph. The rate of change of the variable Y in relation to X can be calculated by finding the gradient. The gradient = the change in Y, divided by, the change in X.
For example, given a line graph of distance travelled against time, the average gradient between two points along the graph can be found by: gradient = distance1-distance2/time1-time2. This also represents the speed between time1 and time2, as speed is defined as the rate of change of distance.
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Experienced Physics and Mathematics tutor from beginner to degree
The Rate of change (ROC) is the increase or decrease of one value related to another value.
The speed of a car for instance is the distance covered divided by the time taken.
In mathematics this can be the gradient of a graph in general terms.
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S
Sam Morris
To find the average rate of change of a function over an interval, you can use the formula:
Average Rate of Change = (Change in Y) / (Change in X) = (ΔY) / (ΔX)
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Hi Vickie,
The rate of change is usually obtained if you divide by time.
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Rojan Santhakrishnan
Derivative with respect to another variable
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We can find the rate of change by calculating the gradient of the curve or line. We identify two points on the line. Next we find the difference in the y-axis points and divide by the difference in the x-axis points. E.g. (Y2 - Y1)/(X2 - X1)
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# Sector of a Circle: Definition, Formula, Area, Perimeter, Examples
Home » Math Vocabulary » Sector of a Circle: Definition, Formula, Area, Perimeter, Examples
## What Is a Sector of a Circle?
A sector of a circle is a portion or part of a circle that is composed of an arc and its two radii. You can compare the sector of a circle to the shape of a pizza slice. A sector is formed when two radii of the circle meet at both ends of the arc. An arc is simply a portion of the circumference of the circle.
### Sector of a Circle Definition
The definition of the sector of a circle in geometry can be given as the part of the circle enclosed by two radii and an arc of the circle. The arc of the circle is a part of the boundary/circumference of the circle.
Two radii meet at the center of the circle to form two sectors.
• Minor sector
• Major sector
### Minor Sector
A sector of a circle is called the minor sector if the minor arc of the circle is a part of its boundary. It is the sector with a smaller area. The angle of a minor sector is less than 180 degrees.
### Major Sector
A sector is called the major sector if the major arc of the circle is a part of its boundary.
It is the sector with the greater area. The angle of a major sector is greater than 180 degrees.
## Sector of a Circle Formulas
Let’s learn how to find the area of a sector of a circle. The formula for determining the area of a sector is given in two ways, with an angle and without an angle.
### Area of a Sector Formula: When Angle Is Given
If the radius of a circle is given as “r” and the angle of the sector is given as . This angle is made by the two radii at the center.
As we know, for a complete circle, the angle made at the center is equal to 2 or $360^\circ$.
• If is measured in degrees, then “the area of a sector of a circle formula” is given by
Area of sector $= \frac{\theta}{360^\circ} \times \pi r^2$
• If is measured in radians, then “the area of a sector of a circle formula” is given by
Area of sector $= \frac{1}{2} \times \theta \times r^2$
• Perimeter of sector $=2r + \frac{\theta}{360} \times 2\pi r$
### Area of Sector Formula: When Angle Is Not Given
How can we find the area of a sector of a circle when the central angle is not given? Let’s find out.
If l is the length of the arc, r is the radius of the circle and θ is the angle subtended at the center, then the angle is expressed in terms of l and r as
$\theta = \frac{l}{r}$, where is in radians.
If the angle of the sector $=2\pi$, the area of the sector (full circle) is $r^2$.
Similarly, for the angle $= 1$, the area of the sector $= \frac{\pi r^2}{2\pi} = \frac{r^2}{2}$
Thus, for the angle , area of the sector $\theta = \frac{r^2}{2} = \frac{l}{r}\times\frac{r^2}{2} = \frac{lr}{2}$
• Area of the sector without an angle $= \frac{lr}{2}$.
• Perimeter of sector $= 2r + l$
### Arc Length of a Sector Formula
The length of the arc “l” of the sector with angle is given by:
$l = \frac{\theta}{360}\times2\pi r$ …when $\theta$ is given in degrees
$l = \theta r$ …when$\theta$ is given in radians
## Facts about Sector of a Circle
• A section or part of a circle involved by two radii with a central angle $90^\circ$ is called a quadrant.
• A section or part of a circle involved by two radii with a central angle of $180^\circ$ is called a semicircle.
• The combination of any two hands (minute hands and hour hands or hour hands and second hands or minute hands and second hands) of a circular analog clock form sectors.
## Conclusion
In this article, we learned about the sector of a circle, minor and major sector, the sector formula for area, perimeter and arc length with and without angle. Now, let us look at some solved examples and practice questions.
## Solved Examples On Sector of a Circle
1. Calculate the area of the sector.
Solution:
The radius of sector $= r = 6$ inches
Angle of sector $= \theta = 60^\circ$
The area of sector $= \frac{\theta}{360^\circ}\times\pi r^2 = \frac{60^\circ}{360^\circ}\times3.14\times6^2 = 18.84$ sq. in.
1. Find the area of a sector of a circular region whose central angle is 3 radians with a radius of 5 feet.
Solution:
The radius of sector $= r = 5$ feet
Angle of sector $= \theta = 3$ radians
If is measured in radians, then
The area of the sector $= \frac{\theta}{2}\times r^2 = \frac{3}{2}\times5^2 =37.5$ sq. feet.
1. Find the central angle of a sector (in degrees) which has a 25 sq. yard area and a radius of 4 yards. Use $\pi = 3.14$.
Solution:
Radius of sector $= r = 4$ yards
Area of sector $= 25$ sq. yards
If is measured in degrees, then
Area of the sector $= \frac{\theta}{360^\circ} \times \pi r^2$
$25 = \frac{\theta}{360^\circ}\times3.14\times4^2$
$\theta = \frac{25\times360}{3.14\times4^2} = 179.14^0$
1. Find the perimeter of the sector shown below.
Solution:
The radius of sector $= r = 8$ inches
Angle of sector $= \theta = 115^\circ$
The perimeter of sector $= 2r + \frac{\theta}{360}\times2\pi r$
$=(2\times8) + \frac{115^\circ}{360^\circ}\times(2\times3.14\times8)$
$=16 + (\frac{1}{4}\times23.14\times8)$
$=16 + 12.56$
$=28.56$ in
1. Find the area and perimeter of a sector with a radius of 10 feet and an arc length of 12.56 feet.
Solution:
The radius of sector $= r = 10$ feet
Arc length $= l = 12.56$ feet
Area of the sector without an angle $= \frac{lr}{2} = \frac{12.56\times10}{2}=62.8$ sq. feet
Perimeter of sector $= 2r + l = 2(10) + 12.56 = 32.56$ feet.
1. Find the arc length of a sector having a radius of 5 feet and a central angle of $120^\circ$.
Solution:
The radius of sector $= r = 5$ feet
Angle of sector $= \theta = 120^\circ$
The length of the arc “l” of the sector with angle is given by;
$l = \frac{\theta \pi r}{180} = \frac{120\times3.14\times5}{180} = 10.47$ feet
## Practice Problems On Sector of a Circle
1
### The sector of a circle is formed by two ____ and an arc.
chords
diameters
tangents
CorrectIncorrect
The sector of a circle is formed by two radii and an arc.
2
### The central angle of the minor sector is _________.
less than 270 degrees
less than 180 degrees
more than 270 degrees
more than 180 degrees
CorrectIncorrect
Correct answer is: less than 180 degrees
The central angle of the minor sector is less than 180 degrees.
3
### For the quadrant of a circle, the central angle is _________.
$90^\circ$
$180^\circ$
$60^\circ$
$45^\circ$
CorrectIncorrect
Correct answer is: $90^\circ$
The quadrant of a circle can be a sector of a circle with a central angle of $90^\circ$.
4
### If$\theta$is measured in radians, then the formula for the area of a sector of a circle is equal to ____________.
Area of sector $=\frac{\theta}{180^\circ}\times\pi r^2$
Area of sector $=\frac{\theta}{360^\circ}\times\pi r^2$
Area of sector $=\theta r^2$
Area of sector $=\frac{\theta}{2}\times r^2$
CorrectIncorrect
Correct answer is: Area of sector $=\frac{\theta}{2}\times r^2$
If $\theta$ is measured in radians, then
Area of sector $= \frac{\theta}{2}\times r^2$
5
### The area of the quadrant of a circle is equal to _____ of that of the circle.
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{4}$
$\frac{1}{8}$
CorrectIncorrect
Correct answer is: $\frac{1}{4}$
The area of the quadrant of a circle is equal to one-fourth, i.e., $\frac{1}{4}$ of that of the circle.
## Frequently Asked Questions On Sector of a Circle
Arc represents the part of the circumference. The sector of a circle is a part of the circle that is enclosed by two radii and an arc of the circle as a part of its boundary.
The area of the sector of a circle is the area of the part of a circle composed of an arc and two radii.
The perimeter of a sector is formed by two radii and an arc.
Perimeter of the sector $= 2r + l = 2r + \frac{\theta}{360} \times 2\pi r$, where $r =$ radius of the circle, $l =$ arc length, $\theta =$ angle of the sector.
There are two types of a sector, which can be categorized as a minor or a major part.
Area of the sector without an angle $= \frac{lr}{2}$, where $r =$ radius of the circle, $l =$ arc length.
The arc length of a sector of a circle can be found using the formula:
$l = \frac{\theta}{360} \times 2\pi r$ or,
$l = \frac{\theta \pi r}{180}$. Where, $r =$ radius of the circle, $l =$ arc length, $\theta =$ angle of the sector.
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1. Chapter 11 Class 7 Perimeter and Area
2. Concept wise
3. Circumference of circle
Transcript
Ex 11.3, 4 A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs Rs 4 per meter. (๐๐๐๐ ๐=22/7) Since the gardener makes two rounds of the fence Thus, Length of rope required = 2 ร Circumference of circular garden Diameter = 21 m Radius = r = ๐ท๐๐๐๐๐ก๐๐/2 = 21/2 m Circumference of circular garden = 2๐r = 2 ร 22/7 ร 21/2 = 22/7 ร 21 = 22 ร 3 = 66 m Now, Length of rope required = 2 ร Circumference of circular garden = 2 ร 66 = 132 m Now, we also need to find cost of rope, Given, Cost of 1 m rope = Rs 4 Cost of 132 m rope = Rs 4 ร 132 = Rs 528 โด It would cost Rs 528 to fence the garden
Circumference of circle
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This is a Clilstore unit. You can .
# Trigonometry
In this video I want to give you the basics of Trigonometry. It sounds like a very complicated topic but you're going to see this is just the study of the ratios of sides of Triangles. The "Trig" part of "Trigonometry" literally means Triangle and the "metry" part literally means Measure. So let me just give you some examples here. I think it'll make everything pretty clear.
So let me draw some right triangles, let me just draw one right triangle. So this is a right triangle. When I say it's a right triangle, it's because one of the angles here is 90 degrees. This right here is a right angle. It is equal to 90 degrees. And we will talk about other ways to show the magnitude of angles in future videos. So we have a 90 degree angle. It's a right triangle, let me put some lengths to the sides here. So this side over here is maybe 3. This height right over there is 3. Maybe the base of the triangle right over here is 4. and then the hypotenuse of the triangle over here is 5. You only have a hypotenuse when you have a right triangle. It is the side opposite the right angle and it is the longest side of a right triangle. So that right there is the hypotenuse. You've probably learned that already from geometry.
And you can verify that this right triangle - the sides work out - we know from the Pythagorean theorem, that 3 squared plus 4 squared, has got to be equal to the length of the longest side, the length of the hypotenuse squared is equal to 5 squared so you can verify that this works out that this satisfies the Pythagorean theorem. Now with that out of the way let's learn a little bit of Trigonometry. The core functions of trigonometry, we're going to learn a little more about what these functions mean. There is the sine, the sine function. There is the cosine function, and there is the tangent function. And you write sin, or S-I-N, C-O-S, and "tan" for short. And these really just specify, for any angle in this triangle, it will specify the ratios of certain sides. So let me just write something out.
This is really something of a mnemonic here, so something just to help you remember the definitions of these functions, but I'm going to write down something called "soh cah toa", you'll be amazed how far this mnemonic will take you in trigonometry. We have "soh cah toa", and what this tells us is; "soh" tells us that "sine" is equal to opposite over hypotenuse. It's telling us. And this won't make a lot of sense just now, I'll do it in a little more detail in a second. And then cosine is equal to adjacent over hypotenuse. And then you finally have tangent, tangent is equal to opposite over adjacent.
So you're probably saying, "hey, Sal, what is all this "opposite" "hypotenuse", "adjacent", what are we talking about?" Well, let's take an angle here. Let's say that this angle right over here is theta, between the side of the length 4, and the side of length 5. This is theta. So let figure out the sine of theta, the cosine of theta, and what the tangent of theta are. So if we first want to focus on the sine of theta, we just have to remember "soh cah toa", sine is opposit over hypotonuse, so sine of theta is equal to the opposite - so what is the opposite side to the angle? So this is our angle right here, the opposite side, if we just go to the opposite side, not one of the sides that are kind of adjacent to the angle, the opposite side is the 3, if you're just kind of - it's opening on to that 3, so the opposite side is 3. And then what is the hypotenuse? Well, we already know - the hypotenuse here is 5. So it's 3 over 5. The sine of theta is 3/5. And I'm going to show you in a second, that the sine of theta - if this angle is a certain angle - it's always going to be 3/5.
The ratio of the opposite to the hypotenuse is always going to be the same, even if the actual triangle were a larger triangle or a smaller one. So I'll show you that in a second. So let's go through all of the trig functions. Let's think about what the cosine of theta is. Cosine is adjacent over hypotenuse, so remember - let me label them. We already figured out that the 3 was the opposite side.
This is the opposite side. And only when we're talking about this angle. When we're talking about this angle - this side is opposite to it. When we're talking about this angle, this 4 side is adjacent to it, it's one of the sides that kind of make up - that kind of form the vertex here. So this right here is the adjacent side. And I want to be very clear, this only applies to this angle.
If we're talking about that angle, then this green side would be opposite, and this yellow side would be adjacent. But we're just focusing on this angle right over here. So cosine of this angle - so the adjacent side of this angle is 4, so the adjacent over the hypotenuse, the adjacent, which is 4, over the hypotenuse, 4 over 5. Now let's do the tangent. Let's do the tangent. The tangent of theta: opposite over adjacent. The opposite side is 3. What is the adjacent side? We've already figured that out, the adjacent side is 4. So knowing the sides of this right triangle, we were able to figure out the major trig ratios. And we'll see that there are other trig ratios, but they can all be derived from these three basic trig functions.
Now, let's think about another angle in this triangle, and I'll re-draw it, because my triangle is getting a little bit messy. So I'll re-draw the exact same triangle. The exact same triangle. And, once again, the lengths of this triangle are - we have length 4 there, we have length 3 there, we have length 5 there. In the last example we used this theta. But let's do another angle, let's do another angle up here, and let's call this angle - I don't know, I'll think of something, a random Greek letter. So let's say it's psi. It's, I know, a little bit bizarre. Theta is what you normally use, but since I've already used theta, let's use psi. Or actually - let me simplify it, let me call this angle x. Let's call that angle x. So let's figure out the trig functions for that angle x. So we have sine of x, is going to be equal to what? Well sine is opposite over hypotenuse.
So what side is opposite to x? Well it opens on to this 4, it opens on to the 4. So in this context, this is now the opposite, this is now the opposite side. Remember: 4 was adjacent to this theta, but it's opposite to x. So it's going to be 4 over - now what's the hypotenuse? Well, the hypotenuse is going to be the same regardless of which angle you pick, so the hypotenuse is now going to be 5, so it's 4/5. Now let's do another one; what is the cosine of x? So cosine is adjacent over hypotenuse. What side is adjacent to x, that's not the hypotenuse? You have the hypotenuse here. Well the 3 side, it's one of the sides that forms the vertex that x is at, that's not the hypotenuse, so this is the adjacent side. That is the adjacent. So it's 3 over the hypotenuse, the hypotenuse is 5.
And then finally, the tangent. We want to figure out the tangent of x. Tangent is opposite over adjacent, "soh cah toa", tangent is opposite over adjacent, opposite over adjacent. The opposite side is 4. I want to do it in that blue color. The opposite side is 4, and the adjacent side is 3. And we're done! And in the next video I'll do a ton of more examples of this, just so that we really get a feel for it. But I'll leave you thinking of what happens when these angle start to approach 90 degrees, or how could they even get larger than 90 degrees.
And we'll see that this definition, the "soh cah toa" definition takes us a long way for angles that are between 0 and 90 degrees, or that are less than 90 degrees. But they kind of start to mess up really at the boundries. And we're going to introduce a new definition, that's kind of derived from the "soh cah toa" definition for finding the sine, cosine and tangent of really any angle.
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## Kiss those Math Headaches GOODBYE!
### Posts tagged ‘Math Challenge Problem’
The “fit” for each situation is the following ratio:
(Area of Inner Figure) ÷ (Area of Outer Figure)
For the square peg in a round hole —
Call the radius of the circle r.
Then the diagonal of square “peg” = 2r
Notice that by slicing the square along its diagonal,
we get a 45-45-90 triangle, with the diagonal being
the hypotenuse and the sides being the two equal legs.
Using the proportions in a 45-45-90 triangle,
side of square peg = r times the square root of 2
Multiplying this side of the square by itself gives
us the area of the square, which comes out as:
This being the case,
Area of square is: 2 times radius squared, and
Area of circle is: Pi times radius squared, and so …
Cancelling the value of the radius squared, we get:
Ratio of (Area of square) to (Area of circle) is:
2÷Pi = 0.6366
For the round peg in a square hole —
Call radius of the circle r.
And since the diameter of the circle is the same length as
the side of the square, the side of the square = 2r
Multiplying the side of the square by itself to get the
area of the square, we find that the area of the square
is given by: 4 times radius squared.
This being the case,
Area of circle is: Pi times radius squared
Area of square is: 4 times radius squared, and so …
Ratio of (Area of circle) to (Area of square) is therefore:
Pi ÷ 4 = 0.7854
Of the two ratios, the ratio of the circular peg in a square hole
is greater than that of the square peg in a circular hole.
Therefore we can say that the circular peg in a square hole
provides a better fit than a square peg in a circular hole.
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Home >> CBSE XII >> Math >> Matrices
# Use product $\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$ to solve the system of equations : $\\ x-y+2z=1 \\ 2y-3z=1 \\ 3x-2y+4z=2$
Can you answer this question?
Toolbox:
• If $|A| \neq 0,$ then $'A'$ is a non -singular matrix. Then inverse exists
• $A{-1}=\frac{1}{|A|} adj (A)$
• $Ax=B => x =A^{-1}B$
Step 1:
Let$A=\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$ and $B=\begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$
Now let us find the product AB,by the matrix multiplication.
$AB=\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$
$=\begin{bmatrix} -2-9+12 & 0-2+2 & 1+3-4 \\ 0+18-18 & 0+4-3 & 0-6+6 \\ -6-18+24 & 0-4+4 & 3+6-8 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Hence $A^{-1}=B$
$(ie)\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}^{-1}=\begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$
Step 2:
Now the system of equations of the form $AX=B$
$(ie)\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$
Where $A=\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}\qquad X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}\;and \; B=\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$
we know $A^{-1}B=X$
and $A^{-1}=\begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$
Therefore $X=\begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$
$\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -2+0+2 \\ 9+2-6 \\ 6+1-4 \end{bmatrix}=\begin{bmatrix} 0 \\ 5 \\ 3 \end{bmatrix}$
Hence $x=0,y=5$ and $z=3$
Solution : option C is correct
answered Apr 4, 2013 by
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# Algebra help fractions
Algebra help fractions can be found online or in math books. Our website can solve math word problems.
## The Best Algebra help fractions
There are a lot of Algebra help fractions that are available online. How to solve for roots. There are multiple ways to solve for the roots of a polynomial equation. One way is to use the Quadratic Formula. The Quadratic Formula is: x = -b ± √b² - 4ac/2a. You can use the Quadratic Formula when the highest exponent of your variable is 2. Another way you can solve for the roots is by factoring. You would want to factor the equation so that it is equal to 0. Once you have done that, you can set each factor equal to 0 and solve for your variable. For example, if you had the equation x² + 5x + 6 = 0, you would first want to factor it. It would then become (x + 2)(x + 3) = 0. You would then set each factor equal to zero and solve for x. In this case, x = -2 and x = -3. These are your roots. If you are given a cubic equation, where the highest exponent of your variable is 3, you can use the method of solving by factoring or by using the Cubic Formula. The Cubic Formula is: x = -b/3a ± √(b/3a)³ + (ac-((b) ²)/(9a ²))/(2a). To use this formula, you need to know the values of a, b, and c in your equation. You also need to be able to take cube roots, which can be done by using a graphing calculator or online calculator. Once you have plugged in the values for a, b, and c, this formula will give you two complex numbers that represent your two roots. In some cases, you will be able to see from your original equation that one of your roots is a real number and the other root is a complex number. In other cases, both of your roots will be complex numbers.
The first step in building a better system is to identify the problems that need to be solved. Once you know what the problems are, you can start looking for solutions that will address those problems. One common way to solve a problem is by substituting one thing for another. For example, if the problem with your lawnmower is that it’s too heavy, you could buy a lighter model. If your car breaks down on the way home from work, you could take public transportation instead of driving. By substituting one thing for another, you’re reducing the amount of stress and hassle involved in getting from A to B. But just because one solution works well for one person doesn’t mean it will work equally well for everyone. Substituting one thing for another might be an effective way to solve your own personal problems, but it may not be effective at solving the problems of other people. In other words, the same system might work well for some people but not others. To find out whether your system is working well for everyone, you have to look at all of the different factors that affect each person’s experience of your system: how they use it, what they like or dislike about it, and so on.
This is a free online tool that allows users to create, view and edit equations. It can help with homework assignments and test preparation. It can also be used to solve word problems. The tool provides step-by-step instructions on how to solve an equation or word problem, including the formula and variables. It also provides an estimate of the solution's accuracy. Users can also make their own equations. The tool is available in English, French, Spanish and German. The word equation solver makes it easy for students to practice, understand and create word problems while taking the next step in learning mathematics. This tool provides step-by-step instructions on how to solve an equation or word problem, including the formula and variables.
A solution math example is a type of mathematical example that is used to illustrate a solution to a problem. These examples are usually found in textbooks and other instructional materials. They are designed to help students understand how to solve a particular type of problem.
There are a variety of different types of math problem solving questions that can be asked. Some may ask for the student to find a specific answer, while others may ask for the student to solve a more complex problem. No matter what type of math problem solving question is asked, it is important for the student to take their time and carefully consider the question before attempting to answer it.
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# Question Video: Dividing Multidigit Numbers by 3-Digit Numbers Using Long Division Mathematics
Calculate 64224 ÷ 223.
05:12
### Video Transcript
Calculate 64224 divided by 223.
64224 is the dividend. 223 is the divisor. We start by asking how many times does 223 go into 642. I think 223 is about 220. 642 is about 640. If I think about 220 times three, that will equal 660. And that’s too big. Which means I’ll want to start this problem by trying the number two. I think that 223 will go into 642 two times.
The next step is to multiply 223 times two. And we do that by multiplying two by each term, starting with two times three, which equals six. Two times two equals four. Two times two again also equals four. After that, we subtract 446 from 642. I can’t take six away from two, so I need to borrow from my four, leaving us with three in that place and 12 in the hundreds place. 12 minus six equals six.
Now, I’m trying to subtract four from three, but that also is not possible. So, I borrow from my six, leaving me with five. Carry a one over, and now, in the thousands place, I have 13 minus four, which equals nine. In the ten thousands place, five minus four equals one. Now, we need to bring down the next digit to the right. Bring down our two. And ask, how many times does 223 go into 1962?
I’m wondering what times 220 will equal close to 1900. 220 times eight equals 1760. And that’s pretty close. So, let’s try eight times 223. Eight times three equals 24. Write down your four. Carry your two. Eight times two equals 16. Plus the two we carried over equals 18. Write down your eight. Carry your one. Then, we multiply eight times two again, which equals 16, plus one, 17.
From here, we need to subtract 1784 from 1962. In the tens place, we can’t take four from two. We borrow from our six, leaving us with five hundreds and 12 tens. 12 minus four equals eight. Now, in the hundreds place, we’re trying to subtract eight from five. And that won’t work. So, we borrow from our thousands place. Take one away from 9000, we have 8000 and 15 hundreds. 15 minus eight equals seven. In the thousands place, we now have eight minus seven, which equals one.
In the ten thousands place, one minus one equal zero. We bring down the four in our ones place. And we notice something interesting. The value we were just subtracting is the same value we’re trying to divide 223 into. We already know that eight times 223 equals 1784. We can write down an eight in the quotient. And we know that when you multiply eight times 223, you get 1784. 1784 minus 1784 equal zero, indicating that there’s no remainder. 64224 divided by 223 equals 288.
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# Introduction to Probability
## Probability of an Event
Probabilities are associated with experiments where the outcome is not known in advance or cannot be predicted. For example, if you toss a coin, will you obtain a head or tail? If you roll a die will obtain 1, 2, 3, 4, 5 or 6?
Probability measures and quantifies "how likely" an event, related to these types of experiment, will happen. The value of a probability is a number between 0 and 1 inclusive. An event that cannot occur has a probability (of happening) equal to 0 and the probability of an event that is certain to occur has a probability equal to 1.(see probability scale below).
In order to quantify probabilities, we need to define the sample space of an experiment and the events that may be associated with that experiment.
## Sample Space and Events
The sample space is the set of all possible outcomes in an experiment.
Example 1: If a die is rolled, the sample space S is given by
S = {1,2,3,4,5,6}
Example 2: If two coins are tossed, the sample space S is given by
S = {HH,HT,TH,TT} , where H = head and T = tail.
Example 3: If two dice are rolled, the sample space S is given by
S = { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) }
We define an event as some specific outcome of an experiment. An event is a subset of the sample space.
Example 4: A die is rolled (see example 1 above for the sample space). Let us define event E as the set of possible outcomes where the number on the face of the die is even. Event E is given by
E = {2,4,6}
Example 5: Two coins are tossed (see example 2 above for the sample space). Let us define event E as the set of possible outcomes where the number of head obtained is equal to two. Event E is given by
E = {(HT),(TH)}
Example 6: Two dice are rolled (see example 3 above for the sample space). Let us define event E as the set of possible outcomes where the sum of the numbers on the faces of the two dice is equal to four. Event E is given by
E = {(1,3),(2,2),(3,1)}
## How to Calculate Probabilities?
### 1 - Classical Probability Formula
It is based on the fact that all outcomes are equally likely.
Total number of outcomes in E P(E)= ________________________________________________ Total number of outcomes in the sample space
Example 7: A die is rolled, find the probability of getting a 3.
The event of interest is "getting a 3". so E = {3}.
The sample space S is given by S = {1,2,3,4,5,6}.
The number of possible outcomes in E is 1 and the number of possible outcomes in S is 6. Hence the probability of getting a 3 is P(E) = 1 / 6.
Example 8: A die is rolled, find the probability of getting an even number.
The event of interest is "getting an even number". so E = {2,4,6}, the even numbers on a die.
The sample space S is given by S = {1,2,3,4,5,6}.
The number of possible outcomes in E is 3 and the number of possible outcomes in S is 6. Hence the probability of getting an even number is P(E) = 3 / 6 = 1 / 2.
### 2 - Empirical Probability Formula
It uses real data on present situations to determine how likely outcomes will occur in the future. Let us clarify this using an example
30 people were asked about the colors they like and here are the results:
Color frequency red 10 blue 15 green 5
If a person is selected at random from the above group of 30, what is the probability that this person likes the red color?
Let event E be "likes the red color". Hence
Frequency for red color P(E)= ________________________________________________ Total frequencies in the above table
= 10 / 30 = 1 / 3
Example 8: The table below shows students distribution per grade in a school.
Grade frequency 1 50 2 30 3 40 4 42 5 38 6 50
If a student is selected at random from this school, what is the probability that this student is in grade 3?
Let event E be "student from grade 3". Hence
Frequency for grade 3 P(E)= _______________________________________ Total frequencies
= 40 / 250 = 0.16
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# 1.1 Expressing a remainder as a decimal
To split a prize of £125 between 5 friends you would do this calculation:
• £125 ÷ 5 and get the answer £25.
This is a convenient, exact amount of money. However, often when you perform calculations, especially those involving division, you do not always get an answer that is suitable for the question.
For example, if there were 4 friends who shared the same prize we would do the calculation £125 ÷ 4 and get the answer £31 remainder £1. If we did the same calculation on a calculator you would get the answer £31.25, the remainder has been converted into a decimal. Let’s look at how to express the remainder as a decimal.
You can write one hundred and twenty five pounds in two different ways: £125 or £125.00. Both ways show the same amount but the second way allows you to continue the calculation and express it as a decimal.
Figure 2 Expressed as a decimal: 125 ÷ 4
We can use the same principal with any whole number, adding as many zeros after the decimal point as required. Look at the following example.
A teacher wants to share 35 kg of clay between 8 groups of students. How much clay will each group get?
Figure 3 Expressed as a decimal: 35 ÷ 8
You can see that each group would get 4.375 kg of clay.
## Activity 3: Expressing a remainder as a decimal
Work out the answers to the following without using a calculator.
1. 178 ÷ 4
2. 212 ÷ 5
3. 63 ÷ 8
4. 227 ÷ 4
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 12.6: Addition and Subtraction of Rational Expressions
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
At the end of this lesson, students will be able to:
• Add and subtract rational expressions with the same denominator.
• Find the least common denominator of rational expressions.
• Add and subtract rational expressions with different denominators.
• Solve real-world problems involving addition and subtraction of rational expressions.
## Vocabulary
Terms introduced in this lesson:
least common denominator (LCD)
least common multiple (LCM)
prime factorization
factor completely
equivalent fraction
## Teaching Strategies and Tips
Use the ordinary fractions in Examples 1 and 5 to motivate adding and subtracting variable rational expressions with and without common denominators, respectively.
Draw the analogy between finding the LCM of polynomials and the LCM of integers.
• Use prime factorization of numbers and polynomials.
• In general, the LCM is found by taking each factor to the highest power that it appears in each expression.
In Examples 6, 8, and 9, remind students to distribute the minus to each term in the second rational expression. See also Review Questions 4, 5, 8, 17, 18, 21, 23, 27, and 30.
Draw the analogy between the formula: part of the task completed =\begin{align*}=\end{align*} rate of work \begin{align*}\cdot\end{align*} time spent on the task and the formula: distance =\begin{align*}=\end{align*} rate \begin{align*}\cdot\end{align*} time.
In Review Questions 34-36, encourage students to set up a table similar to the one in Example 12.
• Emphasize that all known and unknown variables for each person or machine can be listed, which makes organizing the given information easy.
• Combining parts of the task completed by each person or machine is a matter of reading across the rows or down the columns depending on how the table is set up.
• Many students find tables useful for work problems; others rely heavily upon it.
In Review Questions 34-36,
• Suggest that students begin by looking at the part of the task completed by each person or machine separately.
• Encourage students to check the reasonableness of their answers. Ask: What kind of answer should we expect based upon the given information?
## Error Troubleshooting
In Review Questions 7 and 8, suggest that students factor out a negative from the second rational expression first.
• In Example 7, the LCM of x4\begin{align*}x-4\end{align*} and 4x=(x4)\begin{align*}4-x=-(x-4)\end{align*} is x4\begin{align*}x-4\end{align*}.
General Tip: Remind students to find the LCD of rational expressions by factoring. Students needlessly use larger common multiples when expressions are not completely factored.
General Tip: Have students leave the LCM in factored form. This makes simplifying and determining excluded values easier.
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# A Survey of Probability Concepts
Chapter 5
McGrawHill/Irwin
TheMcGrawHillCompanies,Inc.2008
GOALS
Define probability.
Describe the classical, empirical, and subjective
approaches to probability.
Explain the terms experiment, event, outcome,
permutations, and combinations.
Define the terms conditional probability and joint
probability.
Calculate probabilities using the rules of addition
and rules of multiplication.
Apply a tree diagram to organize and compute
probabilities.
Calculate a probability using Bayes theorem.
Definitions
A probability is a measure of the likelihood
that an event in the future will happen. It
it can only assume a value between 0 and 1.
A value near zero means the event is not
likely to happen. A value near one means it is
likely.
There are three ways of assigning probability:
classical,
empirical, and
subjective.
Probability Examples
Definitions continued
An
## experiment is the observation of
some activity or the act of taking
some measurement.
An outcome is the particular result
of an experiment.
An event is the collection of one or
more outcomes of an experiment.
## Experiments, Events and Outcomes
Assigning Probabilities
Three approaches to assigning probabilities
Classical
Empirical
Subjective
Classical Probability
## Consider an experiment of rolling a six-sided die. What is the
probability of the event an even number of spots appear face
up?
The possible outcomes are:
## There are three favorable outcomes (a two, a four, and a six) in
the collection of six equally likely possible outcomes.
Events
## are mutually exclusive if the occurrence of any one event
means that none of the others can occur at the same time.
Events are independent if the occurrence of one event does not
affect the occurrence of another.
Events
## are collectively exhaustive
if at least one of the events must
occur when an experiment is
conducted.
Empirical Probability
## The empirical approach to probability is based on what
is called the law of large numbers. The key to
establishing probabilities empirically is that more
observations will provide a more accurate estimate of
the probability.
## Law of Large Numbers
Suppose we toss a fair coin. The result of each toss is either a
head or a tail. If we toss the coin a great number of times, the
probability of the outcome of heads will approach .5. The
following table reports the results of an experiment of flipping a
fair coin 1, 10, 50, 100, 500, 1,000 and 10,000 times and then
computing the relative frequency of heads
## Empirical Probability - Example
On February 1, 2003, the Space Shuttle Columbia
exploded. This was the second disaster in 113 space
missions for NASA. On the basis of this information,
what is the probability that a future mission is
successfully completed?
Number of successful flights
Probability of a successful flight
Total number of flights
111
0.98
113
## If there is little or no past experience or information on which to base a
probability, it may be arrived at subjectively.
## Illustrations of subjective probability are:
1. Estimating the likelihood the New England Patriots will play in the Super Bowl next year.
2. Estimating the likelihood you will be married before the age of 30.
3. Estimating the likelihood the U.S. budget deficit will be reduced by half in the next 10
years.
## Rules for Computing Probabilities
Special Rule of Addition - If two events
A and B are mutually exclusive, the
probability of one or the other events
occurring equals the sum of their
probabilities.
P(A or B) = P(A) + P(B)
## The General Rule of Addition - If A and
B are two events that are not mutually
exclusive, then P(A or B) is given by the
following formula:
P(A or B) = P(A) + P(B) - P(A and B)
What is the probability that a card chosen at
random from a standard deck of cards will be
either a king or a heart?
## P(A or B) = P(A) + P(B) - P(A and B)
= 4/52 + 13/52 - 1/52
= 16/52, or .3077
## The Complement Rule
The complement rule is used to determine the
probability of an event occurring by
subtracting the probability of the event not
occurring from 1.
P(A) + P(~A) = 1
or P(A) = 1 - P(~A).
## Joint Probability Venn Diagram
JOINT PROBABILITY A probability that
measures the likelihood two or more events
will happen concurrently.
The
## special rule of multiplication requires that two events A and B are
independent.
Two
events A and B are independent if the occurrence of one has no effect on the
probability of the occurrence of the other.
This rule is written:
P(A and B) = P(A)P(B)
Multiplication Rule-Example
A survey by the American Automobile association
(AAA) revealed 60 percent of its members made
airline reservations last year. Two members are
selected at random. What is the probability both
Solution:
The probability the first member made an airline reservation last year is .
60, written as P(R1) = .60
The probability that the second member selected made a reservation is
also .60, so P(R2) = .60.
Since the number of AAA members is very large, you may assume that
R1 and R2 are independent.
P(R1 and R2) = P(R1)P(R2) = (.60)(.60) = .36
Conditional Probability
A conditional probability is the
probability of a particular event
occurring, given that another event
has occurred.
The probability of the event A given
that the event B has occurred is
written P(A|B).
## General Multiplication Rule
The general rule of multiplication is used to find the joint probability that two events will occur.
Use the general rule of multiplication to find the joint probability of two events when the
events are not independent.
It states that for two events, A and B, the joint probability that both events will happen is
found by multiplying the probability that event A will happen by the conditional probability
of event B occurring given that A has occurred.
## General Multiplication Rule - Example
A golfer has 12 golf shirts in his closet. Suppose 9 of these shirts are white
and the others blue. He gets dressed in the dark, so he just grabs a shirt
and puts it on. He plays golf two days in a row and does not do laundry.
What is the likelihood both shirts selected are white?
## The event that the first shirt selected is white is W1.
The probability is P(W1) = 9/12
## The event that the second shirt selected is also white is
identified as W2. The conditional probability that the
second shirt selected is white, given that the first shirt
selected is also white, is P(W2 | W1) = 8/11.
## To determine the probability of 2 white shirts being
selected we use formula: P(AB) = P(A) P(B|A)
P(W1 and W2) = P(W1)P(W2 |W1) = (9/12)(8/11) = 0.55
Contingency Tables
A CONTINGENCY TABLE is a table used to classify sample
observations according to two or more identifiable
characteristics
E.g. A survey of 150 adults classified each as to gender and the
number of movies attended last month. Each respondent is
classified according to two criteriathe number of movies
attended and gender.
## Contingency Tables - Example
A sample of executives were surveyed about their loyalty to their company.
One of the questions was, If you were given an offer by another
company equal to or slightly better than your present position, would
you remain with the company or take the other position? The
responses of the 200 executives in the survey were cross-classified
with their length of service with the company.
## What is the probability of randomly selecting an executive who is loyal to
the company (would remain) and who has more than 10 years of
service?
## Contingency Tables - Example
Event A1 happens if a randomly selected executive will remain with
the company despite an equal or slightly better offer from
another company. Since there are 120 executives out of the 200
in the survey who would remain with the company
P(A1) = 120/200, or .60.
Event B4 happens if a randomly selected executive has more than
10 years of service with the company. Thus, P(B4| A1) is the
conditional probability that an executive with more than 10 years
of service would remain with the company. Of the 120
executives who would remain 75 have more than 10 years of
service, so P(B4| A1) = 75/120.
Tree Diagrams
A tree diagram is useful for portraying
conditional and joint probabilities. It is
decisions involving several stages.
A tree diagram is a graph that is helpful in
organizing calculations that involve several
stages. Each segment in the tree is one stage of
the problem. The branches of a tree diagram are
weighted by probabilities.
Tree Diagram
Bayes Theorem
Bayes
## Theorem is a method for
information.
It is computed using the following
formula:
Tree Diagram
## Counting Rules Multiplication
The multiplication formula indicates that if there are m ways of doing one thing and n
ways of doing another thing, there are m x n ways of doing both.
Example: Dr. Delong has 10 shirts and 8 ties. How many shirt and tie outfits does he have?
(10)(8) = 80
## Counting Rules Multiplication:
Example
An automobile dealer
a convertible, a two-door
sedan, or a four-door
of either wire wheel
covers or solid wheel
covers. How many
different arrangements of
models and wheel
covers can the dealer
offer?
Example
## Counting Rules - Permutation
A permutation is any arrangement of r
objects selected from n possible
objects. The order of arrangement is
important in permutations.
Counting - Combination
A combination is the number of
ways to choose r objects from a
group of n objects without regard
to order.
Combination - Example
There are 12 players on the Carolina Forest
Thompson must pick five players among the
twelve on the team to comprise the starting
lineup. How many different groups are
possible?
12!
792
12 C5
5!(12 5)!
Permutation - Example
Suppose that in addition to selecting the group,
he must also rank each of the players in that
starting lineup according to their ability.
12!
95,040
12 P 5
(12 5)!
End of Chapter 5
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Education
### How Do You Find the Radius of a Circle? Exploring Methods for Calculating the Radius of Circles
To find the radius of a circle, measure from the center to the edge. Use the formula C = 2πr to find the circumference. For the area, use A = πr^2. Trigonometry helps relate the radius to other circle properties. Practice with examples to master these methods. By understanding the radius, you can tackle real-world problems and explore geometric principles further.
## The Basics of Circle Radius
Understanding the radius of a circle is essential for various mathematical calculations and geometric applications. The radius is the distance from the center of a circle to any point on its circumference. To find the radius, you can measure the distance from the center to any point on the circle’s edge. This measurement remains constant regardless of where you choose to measure it on the circle.
In geometry, the radius is a fundamental component when calculating the area and circumference of a circle. It’s also crucial in determining the diameter, which is twice the length of the radius. When working with circles, knowing the radius allows you to make precise calculations and solve problems efficiently.
Additionally, the radius plays a significant role in various real-world applications, such as engineering, architecture, and physics. Whether you’re designing a bridge, calculating the area of a circular garden, or analyzing planetary orbits, a solid grasp of the radius is indispensable. By understanding the basics of circle radius, you pave the way for mastering more complex geometric concepts and practical problems.
## Using the Circumference Formula
To calculate the circumference of a circle, you can use the formula C = 2πr, where C represents the circumference and r is the radius. This formula shows that the circumference is equal to two times π times the radius.
By plugging in the value of the radius into this formula, you can find the circumference of the circle. For example, if the radius of a circle is 5 units, you’d calculate the circumference by substituting r = 5 into the formula: C = 2π(5) = 10π units. This means the circumference of the circle with a radius of 5 units is 10π units.
Using the circumference formula is a straightforward way to find the distance around a circle based on its radius. It’s an essential formula in geometry and is often used in various math and real-world applications.
## Applying the Area Formula
When calculating the area of a circle, you can use the formula A = πr^2, where A represents the area and r is the radius. This formula is derived from the fact that the area of a circle is proportional to the square of its radius. By squaring the radius and multiplying it by the constant π (pi), you can find the total space enclosed by the circle.
To apply the area formula, simply substitute the given radius into the equation. For example, if the radius of a circle is 5 units, you’d calculate the area as A = π(5)^2 = 25π square units. Remember to include the units squared since area is a two-dimensional measurement.
Using the area formula is essential in various real-world applications, such as calculating the space within a circular garden, determining the amount of material needed for a circular tablecloth, or estimating the capacity of a circular swimming pool. Mastering this formula will enable you to solve a wide range of problems involving circles efficiently.
Calculating the radius of circles can be effectively achieved by utilizing trigonometry to determine the relationship between the radius and other geometric properties. When dealing with circles, trigonometry comes in handy to find the radius when other parameters are known. By using trigonometric functions such as sine, cosine, and tangent, you can establish connections between the radius, the circle’s diameter, and angles within the circle.
One commonly used method is to consider a right triangle inscribed in a circle. By drawing a radius to one of the points of tangency, you create a right angle. This allows you to apply trigonometric ratios to relate the radius to the other sides of the triangle. For instance, the sine of an angle in the triangle can help you find the radius if the angle and another side length are known.
## Practical Examples and Exercises
In practical applications, understanding how to apply trigonometry to determine circle radii can be honed through engaging in various exercises. By working through practical examples, you can solidify your grasp on calculating circle radii accurately.
For instance, consider a scenario where you need to find the radius of a circular garden. You can measure the garden’s circumference and then use the formula ( r = rac{C}{2\pi} ) to determine the radius.
Another exercise could involve calculating the radius of a pulley system essential in engineering. By applying trigonometric functions such as sine and cosine to the angles involved, you can easily compute the radius of the pulley.
These exercises not only sharpen your trigonometry skills but also demonstrate the real-world applications of finding circle radii. So, grab a pen, solve some circle radius problems, and witness firsthand how trigonometry can unveil the mysteries of circles around you.
### Can the Radius of a Circle Be Negative?
Yes, the radius of a circle cannot be negative since it represents the distance from the center to any point on the circle. It is always a positive value, as it measures a length.
### What Is the Significance of the Radius in Relation to Other Circle Measurements?
Understanding the significance of the radius is crucial in geometry. It’s half the diameter and helps determine the circle’s area and circumference. Knowing the radius lets you calculate various properties of circles accurately, enhancing your mathematical skills.
### Is There a Limit to How Large or Small a Circle’s Radius Can Be?
There’s no strict limit to a circle’s radius size, but practically, it’s constrained by factors like available space or the precision of your tools. A circle’s radius can be as big or small as needed.
### How Does the Radius of a Circle Affect Its Overall Shape and Properties?
When you adjust a circle’s radius, its size and properties change. A larger radius increases the circle’s area and circumference. The radius directly impacts the circle’s curvature and how it interacts with other shapes in geometric calculations and designs.
### Are There Any Real-Life Applications Where Knowing the Radius of a Circle Is Crucial?
In real life, knowing a circle’s radius is crucial in fields like engineering, architecture, and physics. Whether designing a bridge or calculating planetary orbits, understanding the radius helps determine dimensions, forces, and trajectories accurately.
## Conclusion
So there you have it – finding the radius of a circle isn’t as difficult as it may seem. By using formulas like the circumference formula, area formula, and even some trigonometry, you can easily calculate the radius of any circle.
Practice with some examples and exercises to solidify your understanding. With these methods in your toolkit, you’ll be a radius-finding pro in no time!
|
### Ratios and rates
```Ratios
February, 2011
A ___________ is a comparison of two numbers by
division.
A ratio must be expressed in __________ .
Example #1:
There are approximately 600 students at Lewisburg
School. There are 30 teachers. What is the student to
teacher ratio at Lewisburg School?
600/30 or 20/1
3 ways to write a RATIO!
There were 6 chocolate donuts in the box of 12
donuts.
6
Part to Whole = 12
1
= 2
** This can be expressed as: **
1
1:2
1 to 2
2
ALL THREE OF THESE MEAN ~
1 CHOCOLATE DONUT FOR EVERY 2 DONUTS.
Equivalent Ratios
Two ratios are the same when put in simplest form.
Example – Are these two ratios equivalent?:
\$4 for every 16 ounces
\$4/16 oz =
\$10 for every 40 ounces
\$10/40 oz =
These two ratios are equivalent!!
Choose a partner and take turns taking each others pulse for
2 minutes.
1. Count the number of beats for each of you.
2. Write the ratio beats to minutes as a fraction.
1st block
A ___________ is a ratio that compares two different
kinds of units.
Example:
Heartbeat example
Two Units = heartbeats and minutes
A ____________ is a ratio that is simplified so that it has
a denominator of ONE unit.
Examples:
1. I made \$300 for 6 hours of work. Write the unit rate.
300/6 = 50/1
2.
We drove 220 miles on 8 gallons of gas. Write the unit
rate
220/8 = 27.5/1
How to calculate unit rate?
Easiest way is to divide BOTH the numerator AND the
denominator by the denominator.
Example #1:
\$ 15
3 pounds
15 3
33
\$5
1 pound
How to calculate unit rate?
Easiest way is to divide BOTH the numerator AND the
denominator by the denominator.
Example #2:
220 heartbeats
4 min utes
220 4
44
55 heartbeats
1 min ute
Common Unit Rates
Rate
Unit Rate
Common Unit Rates
Rate
Unit Rate
Number of Miles
1 hour
Miles per hour
Number of Miles
1 gallon
Miles per gallon
Number of Dollars
Price per pound
1 pound
Number of Dollars
1 hour
Dollars per
hour
Different Types of Ratios
Different Types of Ratios
Ratios
Comparing
Same
Measures
Part/Whole
(Fraction)
Comparing
Different Type
of Measures
Part/Part
Rate
Different Types of Ratios
Part/Whole
Examples
Ratio
Part/Part
Examples
Ratio
Rate Examples
Number of red roses
to number of flowers
in bouquet
Number of red
roses to number
of yellow roses
Number of roses per
bouquet
Number of footballs
to number of balls in
gym supply
Number of
footballs to
number of soccer
balls
Number of footballs
per class
Number of points
earned to number of
points possible
Number of points
earned to the
number of points
not earned
Number of points
earned per pupil
Number of girls to
number of student in
class
Number of girls
to number of
boys in class
Number of students
per teacher
Rate
Different Types of Ratios
Part/Whole
Examples
Ratio
Part/Part
Examples
Ratio
Rate Examples
Rate
Number of red roses
to number of flowers
in bouquet
6/24
Number of red
roses to number
of yellow roses
6 to 18
Number of roses
per bouquet
12:1
Number of footballs
to number of balls in
gym supply
5/25
Number of
footballs to
number of soccer
balls
5 to 8
Number of
footballs per class
5:5
Number of points
earned to number of
points possible
80/100
Number of points 80 to 20
earned to the
number of points
not earned
Number of points
earned per pupil
1140:15
or
76:1
Number of girls to
number of student in
class
14/22
Number of girls
to number of
boys in class
Number of
students per
teacher
22:1
14 to 8
Grocery Store Comparison – UNIT RATE!
Four
Products:
Find
the best price!!
Let’s try together:
http://math.rice.edu/~lanius/proportions/rate.html
Practice
Small Group:
Number your paper from 1-12
Deck of cards
LEVEL 0: INDIVIDUAL
Deal
out cards evenly to each player
Rotate the card (clockwise) to your partner
Continue until you’ve answered all cards
LEVEL 1: PARTNERS
Check
|
# Math Assignment Help With Angle Between Two Vectors
## 2.7.2.2 Angle between two vectors:
Let the coordinates of any two nonzero vectors u and v. The angle q between them:
u = ai + bj + ck
v = xi + yj + zk
u.v = u v cos q
u.v = a x + b y + c z
therefore,
u v cos q = a x + b y + c z
q = cos-1 o (a x + b y + c z) / ( u v ) p
### 2.7.2.3 Cross Product:
The vector product of two vectors A and B, also called the cross product, is denoted by A X B.
The symbol used to represent cross product is a cross (×). Since this product has both, magnitude and direction, it is also known as the vector product.
A × B = AB sin θ ˆn
The vector ˆn ("n hat") is a unit vector perpendicular to the plane formed by the two vectors. The direction of ˆn is determined by the right hand rule.
The cross product is distributive i.e.
A × (B + C) = (A × B) + (A × C)
but not commutative i.e.
A × B = −B × A
On reversing the order of cross multiplication the direction of the product also get reversed
If we know the components of A and B, we can calculate the components of the vector product, using a procedure similar to that for the scalar product.
The components of
C = A X B
are given by
Cx = Ax + Bx,
Cy = Ay + By,
Cz = Az + Bz
The cross product of any vector with itself is zero .i.e.
A × A = 0
Applying this corollary to the unit vectors means that the cross product of any unit vector with itself is zero.
î × î = ĵ × ĵ = ˆk × ˆk = (1)(1)(sin 0°) = 0
It should be noted that the cross product of any unit vector with other will give a magnitude of one, because
sine90° = 1
The right hand rule for cross multiplication relates the direction of the two vectors with the direction of their product. Since cross multiplication is not commutative.
|
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