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Question Video: Finding the Monotonicity of a Function given the Graph of Its First Derivative Mathematics • Higher Education
The graph of the derivative πβ² of a function π is shown. On what intervals is π increasing or decreasing?
07:20
Video Transcript
The graph of the derivative π prime of a function π is shown. On what intervals is π increasing or decreasing?
Weβre given a graph of the curve of the derivative π prime of a function π. We need to use this to determine the intervals on which the function π will be increasing or decreasing. Letβs start by recalling what it means for a function π to be increasing or decreasing on an interval. We say that a function π of π₯ is increasing on an interval if the following is true: If we take any two points in this interval, then the point with higher π₯-coordinate must have the higher output. In other words, increasing the value of π₯ on our interval will increase the output of our function. And we define decreasing in the same way. We say that a function π of π₯ is decreasing on an interval if we take any two points on our interval. Then the point with higher π₯-coordinate must have a lower output.
In other words, on this interval, taking a higher input value of π₯ will decrease our output. And itβs often easier to see this graphically. When the curve π¦ is equal to π of π₯ is moving upward, that means our outputs are getting larger. So our function is increasing. And when the curve π¦ is equal to π of π₯ is moving downward, our outputs are getting smaller. So our function is decreasing. And itβs also worth pointing out if our function stays constant, we donβt say itβs increasing or itβs decreasing. In fact, itβs doing neither. But this only helps us if weβre given the curve π¦ is equal to π of π₯. In this case, weβre only given the graph of the function π¦ is equal to π prime of π₯.
This means to answer this question, we need to think what happens to the curve π¦ is equal to π prime of π₯ when π of π₯ is increasing and what happened to the curve π¦ is equal to π prime of π₯ when π of π₯ is decreasing. To do this, weβre going to need to remember what we mean by the derivative function π prime of a function π. We need to recall that π prime measures the slope of our tangent lines to the curve π¦ is equal to π of π₯. So to find out what happens to the curve π¦ is equal to π prime of π₯ when π of π₯ is increasing or decreasing, we need to find out what happens to the slope of our tangent lines.
Letβs start with an interval where π¦ is equal to π of π₯ is increasing. Graphically, it makes sense that on this interval our slopes will all be positive. After all, our outputs are getting larger. And in fact, this is true, and we can even prove this directly from the definition of a derivative. First, letβs consider a point π₯ zero where π₯ zero is in our interval where π of π₯ is increasing. Then from the definition of a derivative, π prime of π₯ zero is equal to the limit as β approaches zero of π of π₯ zero plus β minus π of π₯ zero all divided by β.
We want to show that this limit is positive. The next thing weβll do is since β approaching zero, weβll choose β small enough so that π₯ zero plus β is also in our interval. Now, there are two options. Either β is positive or β is negative. First, if β is positive, then π of π₯ zero plus β must be greater than π of π₯ zero. This is because we chose β small enough so that π₯ zero plus β is also in our interval where π is increasing. And π₯ zero plus β will be bigger than π₯ zero. And we get a similar inequality of β is less than zero. This time, π of π₯ zero will be greater than π of π₯ zero plus β.
Again, this is because π is an increasing function on this interval. And because β is less than zero, π₯ zero plus β will be less than π₯ zero. We can then see what this does to our limit. Letβs consider the case where β is greater than zero. In our numerator inside of our limit, weβre now subtracting a bigger number from a smaller number. So our numerator is positive. And in this case, β is positive. So weβre dividing two positive numbers. And we know the quotient of two positive numbers is positive. And we get something very similar in the case where β is less than zero.
This time, weβre subtracting a bigger number from a smaller one. So our numerator will be negative. But remember, β is less than zero. So our denominator is also negative. And the quotient of two negative numbers is positive. This only proves that our limit will be greater than or equal to zero because a sequence of positive numbers can in fact approach zero. This is why we sometimes like to include points where the slope is equal to zero in our intervals. However, in this video, we wonβt be including these.
We can, in fact, make a very similar argument on intervals where our function is decreasing. On these intervals, our function will be going downwards. This means that our slopes will all be negative. We now have enough information to start determining the intervals where our function π is increasing or decreasing just from the curve π¦ is equal to π prime of π₯. We know that π prime of π₯ will be increasing on intervals where π prime of π₯ is positive. This means the curve π¦ is equal to π prime of π₯ will be above the π₯-axis. And similarly, on intervals where π of π₯ is decreasing, π prime of π₯ will be negative. This means our curve π¦ is equal to π prime of π₯ will be below the π₯-axis.
Weβre now ready to answer this question. However, thereβs one more thing we need to talk about. We want to talk about the endpoints of this curve when π₯ is equal to zero and when π₯ is equal to eight. We can see from the curve π prime evaluated at zero is equal to two, and π prime evaluated at eight is approximately equal to negative 1.5. This is represented by the filled-in circles in our diagram. So if we were to look at our curve from π₯ is equal to zero to π₯ is equal to one, we can see that the curve π¦ is equal to π prime of π₯ is above the π₯-axis. In other words, π prime of π₯ is greater than zero for all values of π₯ greater than or equal to zero and less than one.
And weβve already discussed why weβre not including the endpoint when π₯ is equal to one. In fact, itβs just mathematical convention. We could include it if we wanted to. And we could leave our answer like this. However, thereβs another piece of mathematical convention. A lot of people donβt like to include the endpoints of these intervals. And both of these answers are correct. Itβs all personal preference among mathematicians. In this video, weβre going to leave that out and leave this is as the open interval from zero to one. So, so far, weβve shown that π of π₯ is increasing on the open interval from zero to one. Letβs carry on finding more intervals where π of π₯ is increasing or decreasing.
Next, by looking at the curve π¦ is equal to π prime of π₯, we can see that this curve is entirely below the π₯-axis for all values of π₯ between one and five. So because π prime of π₯ is below the π₯-axis on the open interval from one to five, our function π of π₯ is decreasing on the open interval from one to five. Next, by looking at our curve, we can see that π prime of π₯ is positive for all values of π₯ on the open interval from five to seven. This means π of π₯ is increasing on the open interval from five to seven. And finally, we can see the curve π¦ is equal to π prime of π₯ is less than zero on the open interval from seven to eight. And weβre not including the endpoint of this curve by convention.
So weβve shown on the open interval from seven to eight, the slopes of our tangent lines will be negative. And this means that our function π of π₯ will be decreasing on this interval. Now we found all the intervals where our function π is increasing and decreasing. We can combine this into our final answer. Therefore, by looking at the intervals where our function π prime of π₯ is positive or negative, we were able to determine the intervals where the function π was increasing or decreasing. We were able to show that π will be increasing on the open interval from zero to one and the open interval from five to seven and decreasing on the open interval from one to five and the open interval from seven to eight.
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## Simplifying Expressions With Absolute Value
### Learning Outcomes
• Use correct notation to indicate absolute value
• Simplify expressions that contain absolute value
• Evaluate expressions that contain absolute value
We saw that numbers such as $5$ and $-5$ are opposites because they are the same distance from $0$ on the number line. They are both five units from $0$. The distance between $0$ and any number on the number line is called the absolute value of that number.
Because distance is never negative, the absolute value of any number is never negative.
The symbol for absolute value is two vertical lines on either side of a number. So the absolute value of $5$ is written as $|5|$, and the absolute value of $-5$ is written as $|-5|$ as shown below.
### Absolute Value
The absolute value of a number is its distance from $0$ on the number line.
The absolute value of a number $n$ is written as $|n|$.
$|n|\ge 0$ for all numbers
### example
Simplify:
1. $|3|$
2. $|-44|$
3. $|0|$
Solution:
1. $|3|$ $3$ is $3$ units from zero. $3$
2. $|-44|$ $−44$ is $44$ units from zero. $44$
3. $|0|$ $0$ is already at zero. $0$
### try it
In the video below we show another example of how to find the absolute value of an integer.
We treat absolute value bars just like we treat parentheses in the order of operations. We simplify the expression inside first.
### example
Evaluate:
1. $|x|\text{ when }x=-35$
2. $|\mathit{\text{-y}}|\text{ when }y=-20$
3. $-|u|\text{ when }u=12$
4. $-|p|\text{ when }p=-14$
Notice that the result is negative only when there is a negative sign outside the absolute value symbol.
### example
Fill in $\text{< },\text{ > },\text {or }=$ for each of the following:
1. $|-5|$ ___ $-|-5|$
2. $8$ ___ $-|-8|$
3. $-9$ ___ $- |-9|$
4. $-|-7|$ ___ $- 7$
### try it
In the video below we show more examples of how to compare expressions that include absolute value and integers.
Absolute value bars act like grouping symbols. First simplify inside the absolute value bars as much as possible. Then take the absolute value of the resulting number, and continue with any operations outside the absolute value symbols.
### example
Simplify:
1. $|9 - 3|$
2. $4|-2|$
### example
Simplify: $|8+7|-|5+6|$.
### example
Simplify: $24-|19 - 3\left(6 - 2\right)|$
### try it
Watch the following video to see more examples of how to simplify expressions that contain absolute value.
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# Compiled by Mr. Lafferty Maths Dept.
## Presentation on theme: "Compiled by Mr. Lafferty Maths Dept."— Presentation transcript:
Compiled by Mr. Lafferty Maths Dept.
S4 Starter Questions 10-Apr-17 Compiled by Mr. Lafferty Maths Dept.
Volumes by Counting Cubes
Learning Intention To understand the term volume in terms of counting cubes. To understand how to calculate the volume of cuboid. To understand the term liquid volume using millilitres and litres.
Volumes by Counting Cubes
Volume is the amount of space a 3D - shape takes up 1cm 1cm 1cm One Unit of Volume is the “CUBIC CENTIMETRE” = 1 centimetre cube = 1 cm³
Volumes by Counting Cubes
This shape is made up of 1 centimetre cubes placed next to each other. What is its volume in cm³? 1cm 1cm 1cm 1cm = 2 centimetre cubes = 2 cm³
Volumes by Counting Cubes
This shape is made up of 1 centimetre cubes placed next to each other. What is its volume in cm³ 1cm 1cm 1cm = 3 centimetre cubes = 3 cm³
Volumes by Counting Cubes What is its volume in cm³ of these shapes.
Volumes by Counting Cubes
One unit of Volume is the “CUBIC CENTIMETRE” 3cm 2cm 4cm Volume = 24 centimetre cube = 24 cm³
Volume of a cuboid 18 cubes fit the base. = 1 centimetre cube = 1 cm³
4 layers of 18 cubes = 4 x 18 = 72 centimetre cubes = 72 cm³
A short cut ! Volume = 6 x 3 x 4 = 72 cm³ Volume = length x breadth
height Area of rectangle breadth length Volume = 6 x 3 x 4 = 72 cm³ Volume = length x breadth x height
Example 1 27cm 5 cm 18 cm Working Volume = l x b x h V = 18 x 5 x 27
Heilander’s Porridge Oats V = 18 x 5 x 27 V = cm³ 27cm 5 cm 18 cm
Example 2 Working Volume = l x b x h V = 2 x 2 x 2 V = 8 cm³ 2cm
Volume of a cuboid Now try Ex 2.1 Ch2 Pg 24 Q 1 & 2
Liquid Volume Volume = l x b x h = 1 cm³
I’m a very small duck! How much water does this hold? 1 cm 1 cm 1 cm Volume = l x b x h = 1 cm³ A cube with volume 1cm³ holds exact 1 millilitre of liquid. A volume of 1000 ml = 1 litre.
Example 1 Working Volume = l x b x h V = 6 x 3 x 12 V = 216 cm³
Liquid Volume Working Orange Flavour Volume = l x b x h V = 6 x 3 x 12 12 cm V = 216 cm³ = 216 ml 3 cm 6 cm So the carton can hold 216 ml of orange juice. Remember: 1 cm³ = 1 ml How much juice can this carton hold?
Example 2 Working Volume = l x b x h V = 100 x 30 x 50 V = 150 000 cm³
Liquid Volume Working Volume = l x b x h V = 100 x 30 x 50 V = cm³ 50 cm = ml = 150 litres 30 cm 100 cm How much water can this fish tank hold in litres? 1cm3 = 1 ml 1000 ml = 1 litre So the fish tank can hold 150 litres of water.
Liquid Volume Now try Ex 2.1 Ch3 Pg 24 Ex 2.2 Ch 3 Pg 25
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# The resistances of three parts of a circular loop are as shown in the figure. The magnetic field at the center O is (current enters at A and leaves at B and C as shown in the figure.)A. $\dfrac { { \mu }_{ 0 }I }{ 6a }$B. $\dfrac { { \mu }_{ 0 }I }{ 3a }$C. $\dfrac {2}{3} \dfrac { { \mu }_{ 0 }I }{ a }$D. $zero$
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Hint: From the figure we can say that the current I get equally divided between AB and AC. With the help of these currents find the potential difference across AB and AC. From the potential difference across AB and AC you will get the potential difference and the current flowing across BC. Then, use the formula for magnetic field and find the magnetic field due to AC, AB and BC. Adding the magnetic fields due to AB, AC and BC will give the net magnetic field. This obtained value will be the magnetic field at the center O.
Let the radius of the circular loop be r
It is given that the current entering from A is I.
The circuit is symmetric so the current flowing from A to B will be equal to the current flowing from A to C.
Current from A to B= Current from A to C= $\dfrac {I}{2}$
Let the potential at A be ${V}_{A}$
Similarly, at B and C be ${V}_{B}$ and ${V}_{C}$ respectively.
The potential difference between A and B can be given by,
${V}_{B}-{V}_{A}= \dfrac {I}{2}R$
$\Rightarrow {V}_{B}= {V}_{A}+ \dfrac {I}{2}R$ …(1)
Similarly, the potential difference between A and C can be given by,
${V}_{C}-{V}_{A}= \dfrac {I}{2}R$
$\Rightarrow {V}_{C}= {V}_{A}+ \dfrac {I}{2}R$ …(2)
From the equation. (1) and (2) we get,
${V}_{B}= {V}_{C}$
Thus, as the potential at B and C are the same, no current will flow from B to C.
We know, Magnetic field is given by,
$B= \dfrac {\theta}{360°}. \dfrac {{\mu}_{0}I}{2\pi r}$
Magnetic field due to AC by substituting the values in above equation is given by,
${B}_{AC}= \dfrac {120°}{360°}. \dfrac {{\mu}_{0}I}{4\pi r}$
$\Rightarrow {B}_{AC}= \dfrac {1}{3}. \dfrac {{\mu}_{0}I}{4\pi r}$
Now, applying the right hand thumb rule, the direction of the magnetic field is outwards.
Similarly, Magnetic field due to AB is given by,
${B}_{AB}= \dfrac {120°}{360°}. \dfrac {{\mu}_{0}I}{4\pi r}$
$\Rightarrow {B}_{AB}= \dfrac {1}{3}. \dfrac {{\mu}_{0}I}{4\pi r}$
Applying right hand thumb rule, the magnetic field due to AB is inwards.
Similarly, Magnetic field due to BC is given by,
${B}_{BC}= \dfrac {120°}{360°}. \dfrac {{\mu}_{0} \times 0}{2\pi r}$
$\Rightarrow {B}_{BC}= 0$
The magnitude of the magnetic field due to AB and AC are the same but their direction is opposite.
$\Rightarrow {B}_{AC}= -{B}_{AB}$ …(3)
Thus, the net magnetic field is given by,
${B}_{net}={B}_{AC}+{B}_{AB}+{B}_{BC}$
Substituting equation. (3) and value of ${B}_{BC}$ in above equation we get,
${B}_{net}=-{B}_{AB}+{B}_{AB}+0$
$\Rightarrow {B}_{net}= 0$
Thus, the magnetic field at the center O is zero.
So, the correct answer is option D i.e. zero.
Note:
For solving this question, we have used the right hand thumb rule. Students should not get confused between the right-hand thumb rule and Fleming’s right hand rule. According to Fleming’s right hand rule, the thumb indicates the direction of force, the forefinger indicates the direction of the magnetic field and the middle finger represents the direction of current. Whereas according to right hand thumb rule, the thumb indicates the direction of current and the direction in which the fingers wrap gives the direction of magnetic field.
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# Fraction into Percentage
How to covert a fraction into percentage?
We will follow the following steps for converting a fraction into a percentage:
Step I: Obtain the fraction. Let the fraction is x/y.
Step II: Multiply the fraction by 100 and write the percentage (%) symbol to find the required percent. Therefore, x/y = (x/y × 100) %
1. Convert each of the following fractions as percentage:
(i) 3/25 = (3/25 × 100) % = 12 %
(iii) 4/5 = (4/5 × 100) % = 80 %
(iii) 3/4 = (3/4 × 100) % = 75 %
(iv) 2/3 = (2/3 × 100) % = (200/3) % = 66 2/3 %
(v) 1 = (1 × 100) % = 100 %
Note: 100 % = 1
2. Convert 7/20 to per cent.
7/20 = (7/20 × 100) % = 35 %
3. Express each of the following fraction into a percentage:
(i) 6/5
= (6/5 × 100) % = 120 %
(ii) 42/5
= 22/5 = (22/5 × 100) % = 440 %
(iii) 11/4
= 5/4 = (5/4 × 100) % = 125 %
4. Express each of the following statements in the percentage form:
(i) 5 out of 20 are bad
= (5/20 × 100) % eggs are bad
= 25 % eggs are bad
(ii) 3 children in a class of 30 are absent.
= 3/30 children are absent
= (3/30 × 100) % children are absent
= 10 % children are absent
(iii) 21 apples out of 30 are good
= 21/30 apples are good
= (21/30 × 100) % apples are good
= 70 % apples are good
(iv) 47 students out of 50 are present
= 47/50 students are present
= (47/50 × 100) % students are present
= 94 % students are present
Fraction into Percentage
Percentage into Fraction
Percentage into Ratio
Ratio into Percentage
Percentage into Decimal
Decimal into Percentage
Percentage of the given Quantity
How much Percentage One Quantity is of Another?
Percentage of a Number
Increase Percentage
Decrease Percentage
Basic Problems on Percentage
Solved Examples on Percentage
Problems on Percentage
Real Life Problems on Percentage
Word Problems on Percentage
Application of Percentage
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# How do you find the derivative of y = (sin x)^(ln x)?
Nov 6, 2016
Use implicit differentiation along with the chain and product rules.
#### Explanation:
We can find the derivative of this function implicitly. In other words, we will find the derivative of $y$, which will then allow us to find the derivative of $\sin {\left(x\right)}^{\ln \left(x\right)}$.
First, we want to get rid of the $\ln x$ exponent. We can do that by taking the natural log of both sides and using a property of logarithms, that $\ln {x}^{a}$ is equivalent to $a \ln \left(x\right)$. Thus,
$\ln y = \ln {\left(\sin x\right)}^{\ln} x$
$\ln y = \ln x \cdot \ln \left(\sin x\right)$
Now, we derive both sides. For the left side, we will have the derivative of $\ln y = \frac{1}{y}$, but we can't simply say that the derivative of $y$ is 1 (using the chain rule). Rather, we say that it is $\frac{\mathrm{dy}}{\mathrm{dx}}$.
So, the left side of the equation now looks like this:
$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$
Now we take the derivative of the right side, which we can do using the product and chain rules. We get:
$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\ln \left(\sin x\right) \cdot \frac{1}{x}\right) + \left(\ln x \cdot \frac{1}{\sin} x \cdot \cos x\right)$
$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \ln \frac{\sin x}{x} + \frac{\ln x \cdot \cos x}{\sin} x$
Remember that we're trying to solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$. We can do this by multiplying both sides by $y$. Thus,
$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(\ln \frac{\sin x}{x} + \frac{\ln x \cdot \cos x}{\sin} x\right)$
Finally, to get our answer back into terms of $x$, we can replace the $y$ on the right side of our derivative with $\sin {x}^{\ln} x$ from the original function.
$\frac{\mathrm{dy}}{\mathrm{dx}} = \sin {\left(x\right)}^{\ln} x \cdot \left(\ln \frac{\sin x}{x} + \frac{\ln x \cdot \cos x}{\sin} x\right)$
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# How do you calculate percent change?
Formula and Calculation of Percentage Change % Increase = Increase / Original Number × 100. This gives you the total percentage change, or increase. To calculate a percentage decrease first, work out the difference (decrease) between the two numbers you are comparing.
## How do you calculate percent change?
Formula and Calculation of Percentage Change % Increase = Increase / Original Number × 100. This gives you the total percentage change, or increase. To calculate a percentage decrease first, work out the difference (decrease) between the two numbers you are comparing.
## How do you calculate percentage increase and decrease in Excel?
percentage increase = Increase ÷ Original Number × 100. If the number you get is negative, like -0.10, then the percentage actually decreased rather than increased.
What is the Excel formula for percentage decrease?
Calculating Percentage Decrease in Excel You subtract the second number from the first, then divide it by the first number. The only difference is that the first number will be smaller than the second number.
### What is the percent of change from 5000 to 9000?
Percentage Calculator: What is the percentage increase/decrease from 5000 to 9000? = 80 – percentagecalculator.
### What is the percent of change from 5000 to 8000?
Related Standard Percentage Calculations on Change from 5000 to 8000
X Y Percentage(P) Change
5000 7950 59
5000 8000 60
5000 8050 61
5000 8100 62
What is the percent change from 32 to 72?
Percentage Calculator: What is the percentage increase/decrease from 32 to 72? = 125.
## How do you calculate percentage increase and decrease?
First: work out the difference (increase) between the two numbers you are comparing. Then: divide the increase by the original number and multiply the answer by 100. % increase = Increase ÷ Original Number × 100. If your answer is a negative number, then this is a percentage decrease.
## How do I calculate 20% reduction?
To subtract 20 percent, multiply by 80 percent (0.8). To subtract 30 percent, multiply the number by 70 percent (0.7).
How to calculate percentage increase or decrease in Excel?
Step#1. We can calculate the percentage increase from the second row only because there will be no base data for the first row.
• Step#2. Insert the below function in a row.
• Step#3. After the function has been executed,format the cell as a percentage to get the result in a percentage.
• ### What is the formula for calculating percent change?
Percentage Change Formula Percentage increase and decrease are calculated by computing the difference between two values and comparing that difference to the initial value. Mathematically, this involves using the absolute value of the difference between two values, and dividing the result by the initial value, essentially calculating how much
### How do you calculate percent decrease in Excel?
Examples of Percentage Decrease Formula (With Excel Template) Let’s take an example to understand the calculation of the Percentage Decrease in a better manner.
• Explanation.
• Relevance and Uses of Percentage Decrease Formula.
• Percentage Decrease Formula Calculator
• Recommended Articles.
• How do you show a percentage change in Excel?
From within the ‘Category’ list on the left side of the dialog box,select the Percentage type.
• The D ecimal places: option will appear in the dialog box. Use this to specify the number of decimal places that you want to display.
• Click OK.
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1. Factoring
How do you factor problems like 2x^3 + 9x^2 -5x = 0 and x^3 - x^2 - 4x + 4 = 0? Thanks
2. Originally Posted by asnxbbyx113
How do you factor problems like 2x^3 + 9x^2 -5x = 0 and x^3 - x^2 - 4x + 4 = 0? Thanks
$\displaystyle 2x^3 + 9x^2 - 5x = 0$
$\displaystyle \Rightarrow x \left( 2x^2 + 9x - 5 \right) = 0$
For, $\displaystyle x^3 - x^2 - 4x + 4 = 0$ we factor by grouping. there is a common $\displaystyle x^2$ among the first two terms, and a common 4 among the last two, so pull them out
$\displaystyle \Rightarrow x^2 (x - 1) - 4(x - 1) = 0$ ...........now the $\displaystyle (x - 1)$ is common, pull it out
$\displaystyle \Rightarrow (x - 1) \left( x^2 - 4 \right) = 0$
we're not done yet, you finish
3. Originally Posted by asnxbbyx113
How do you factor problems like 2x^3 + 9x^2 -5x = 0 and x^3 - x^2 - 4x + 4 = 0? Thanks
The way I solve these is as follows:-
Take your first problem, 2x^3 + 9x^2 - 5x = 0.
You need to find a first factor to allow you to get to a quadratic.
Clearly in the above case you can re-write as:-
(2x^2 +9x -5)x = 0
Therefore x=0 is your first factor. The second is the quadratic 2x^2 + 9x - 5.
You solve this in the traditional way noting that |a*c| = 10 where a is the coefficient of x^2 and c is the constant - 5.
The factors of |a*c| are either 10 and 1 or 5 and 2.
10 and 1 will give you the 9x when subtracted so re-write the quadratic as:-
2x^2 +10x - x - 5 = 0.
=> 2(x+5) - (x+5) = 0
=> (2x-1)(x+5) = 0.
That means your 3 factors are 2x-1, x+5 and of course, x itself for x values of 1/2, -5 and 0.
Your second example is more difficult.
x^3 - x^2 - 4x + 4 = 0
Convert to nested form as:-
((x-1)x - 4)x +4 = 0.
Now you need to use brute force to find a value of x which will give you 0 for your first factor. Start at x=0, then x=+/- 1, +/- 2 etc.
You'll find that x=1 gives you 0.
i.e. the first factor is (x-1).
Now to find the remaining quadratic, divide x-1 into x^3 - x^2 - 4x + 4 as a long division.
You'll find you get x^2 - 4.
Therefore your factors so far are:-
(x-1)(x^2 - 4) = 0.
Clearly the quadratic breaks down to (x-2)(x+2) and so your final solution is:-
(x-1)(x-2)(x+2) for x=1, 2, -2.
I'm sure I've confused you with the above but basically that's the route you need to take. Perhaps there's an easier way to explain what I've done or an easier way to actually solve it.
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# Math in Focus Grade 2 Chapter 2 Practice 6 Answer Key Addition with Regrouping in Tens
Go through the Math in Focus Grade 2 Workbook Answer Key Chapter 2 Practice 6 Addition with Regrouping in Tens to finish your assignments.
## Math in Focus Grade 2 Chapter 2 Practice 6 Answer Key Addition with Regrouping in Tens
Solve.
Question 1.
Farmer Black has 374- chickens and 383 ducks on his farm. How many ducks and chickens does he have in all? He has ___ ducks and chickens in all.
He has 757 ducks and chickens in all.
Explanation:
Farmer Black has 374- chickens
and 383 ducks on his farm.
374 + 383 = 757 ducks and chickens that he have in all
Question 2.
Maria has 381 baseball cards. She has 492 football cards. How many cards does she have in all?
She has ___ cards in all.
She has 873 cards in all.
Explanation:
Maria has 381 baseball cards.
She has 492 football cards.
381 + 492 = 873 cards that she have in all
Question 3.
Peter collects 280 stamps. Then his brother gives him another 163 stamps. How many stamps does Peter have now?
Peter has ___ stamps now.
Peter has 443 stamps now.
Explanation:
Peter collects 280 stamps.
Then his brother gives him another 163 stamps.
280 + 163 = 443 stamps that Peter have now
Solve.
Question 4.
Kirk has painted 460 bricks. He has 262 bricks left to paint. How many bricks does Kirk have to paint in all?
Kirk has to paint ___ bricks in all.
Kirk has to paint 722 bricks in all.
Explanation:
Kirk has painted 460 bricks.
He has 262 bricks left to paint.
460 + 262 = 722 bricks that Kirk have to paint in all
Question 5.
Leroy has 299 model airplanes. Keisha gives him another 120 model airplanes. How many model airplanes does Leroy have now?
Leroy has ___ model airplanes now.
Leroy has 419 model airplanes now.
Explanation:
Leroy has 299 model airplanes.
Keisha gives him another 120 model airplanes.
299 + 120 = 419 model airplanes that Leroy have now
Question 6.
A parking garage has 654 vans. It has 191 more cars than vans. How many cars are in the garage? cars are in the garage.
__ cars are in the garage.
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## Definition
Adjacent Angles are angles those angles which are placed together in such a manner that there is no overlapping between them, but they share the same vertex and a common side. When two lines join at a point, that point is called the vertex and its corresponding angle is called the vertex angle.
Adjacent angles are any two angles that have:
1. Same vertex
2. A common side
3. No overlapping between the angles
They will be considered as adjacent angles but if either one of these properties is not met then we won’t consider those angles as adjacent angles. We can further divide the adjacent angle into 2 parts; a complementary angle and a supplementary angle
Let us consider 2 slices of pizza in a box. Both the slices have the same endpoint and share a common side. And they are not overlapping each other at all. Similarly, we can find numerous daily life examples.
Let us consider the following figure to help us understand the concept of adjacent angles
Figure 2 – ∠ABC
In this figure, we can see that ∠a and ∠b are adjacent to each other. They have a common vertex at point B and share a ray. But ∠a and ∠c are not adjacent to each other. This is because they do not share a common side and by definition, we know that all properties must be fulfilled for any 2 angles to be identified as adjacent angles.
We can use this concept to identify and differentiate between an adjacent angle and a vertical angle. Imagine a line intersects another line while forming a cross, such that we have 4 angles. As we know that adjacent angles will always have two properties; the same vertex and common side.
Similarly, we can identify a vertical angle which is a set of angles that have the same vertex but do not share a side. These angles may be opposite to each other which are known as vertically opposite angles.
Now let us use the same example of a line that intersects another line to form a cross to understand the concept of linear pair. Linear pairs are such adjacent angles whose sum is always equal to 180 degrees. The angles in a linear pair are always supplementary to each other.
Note that all linear pairs are supplementary but not all supplementary angles are linear pairs. This is because to form a linear pair we need an intersection between two lines and the presence of adjacent angles. If we have a linear pair and its angles add up to be equal to 180 degrees then its non-common arm will form a line.
Adjacent angles can be complementary angles to each other if the sum of the two angles equals 90 degrees. Similarly, if the sum of such angles is equal to 180 degrees they will be supplementary to each other.
We know that every adjacent angle has certain following properties which make them stand out from other angles. The following properties help us identify adjacent angles:
1. All angles will share the same vertex
2. Adjacent angles share a common side or ray
3. They must not overlap with each other
4. Along with a common side, each angle should have an uncommon side of its own
5. The Sum of the adjacent angles define whether they are complementary or supplementary
6. The angles must not have a common interior point but should be separated
7. All adjacent angles exist in pairs
We will now understand the concepts of adjacent angles by using a pictorial format.
The example given below shows adjacent angles:
In this diagram, we can see that both angles ∠a and ∠b are adjacent to each other as they both share the same vertex at point B and both angles are using the line BD commonly.
Let us look at the following image:
In this figure, we can see that both angles ∠a and ∠b share a common side but they will not be considered adjacent angels because they don’t have the same vertex.
Now to understand the linear pair’s concept we will refer to the following figure:
Figure 5 – Linear Pairs
As we can see that both angles ∠a and ∠b, when added together, are equal to 180 degrees which means that they are linear pairs and the uncommon sides have formed a straight line.
## Solved Examples of Adjacent Angles
### Example 1
Using the figure, find:
∠ZOY = ?
Where:
∠XOZ = 80°
∠XOY = 120°
Figure 6 – ∠XOY
### Solution
From the given figure:
m∠XOY = 120°
m∠ZOY = a
m∠ XOZ = 80°
By using this formula:
∠XOY = ∠ZOY + ∠XOZ
m∠XOY = m∠ZOY + m∠XOZ
120° = a + 80°
a = 120° – 80°
a = 40°
Hence the angle ∠ZOY is equal to 40°.
### Example 2
Thomas has been assigned geometry homework in which he has to identify whether the following figure is of adjacent angles or not. Help him identify and explain with reasoning.
Figure 7 – ∠ABCD
### Solution
The following figure does not satisfy the properties of adjacent angles as they are vertically opposite to each other.
They are equal in measure but opposite in direction so they are called vertically opposite angles or simply vertical angles.
### Example 3
Miranda is teaching the basics of geometry to 4th graders. She wants to give them real-life examples corresponding to her topic which is ‘Adjacent Angles’. Help her deliver her lecture more effectively by providing real-life examples.
### Solution
Some examples of adjacent angels that we see on the daily basis are:
1. The hands of the clock represent adjacent angles as they all have the same vertex and a common side. This example can only be used when all three hours, minutes, and seconds hand are away from each other and not overlapping.
2. Cake slices on a plate represent adjacent angles as they satisfy the properties.
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## Class 9 Maths Case Study Questions of Chapter 15 Probability PDF Download
Case study Questions in Class 9 Mathematics Chapter 15 are very important to solve for your exam. Class 9 Maths Chapter 15 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case Class 9 Maths Case Study Questions Chapter 15 Probability
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These case study questions challenge students to apply their knowledge of quadrilaterals in practical scenarios, enhancing their problem-solving abilities. This article provides the Class 9 Maths Case Study Questions of Chapter 15 Probability, enabling students to practice and excel in their examinations.
# Probability Case Study Questions With Answers
Here, we have provided case-based/passage-based questions for Class 9 Maths Chapter 15 Probability
Case Study/Passage-Based Questions
Case Study 1: A jeweler has different types of bracelets in his shop. Sunita wants to purchase a bracelet for her sister’s birthday gift. When Sunita goes to the shop, she founds the following data which represents the number of bracelets of different types in the shop.
Find the probability that Sunita chooses Chain Bracelet.
(a) 23/180 (b) 37/180
(c) 37/90 (d) 23/90
Find the probability that she chooses Pearl bracelet.
(a) 23/180 (b) 37/180 (c) 37/90 (d) 23/90
What is the probability of a sure event?
(a) 1 (b) 0 (c) 1/2 (d) 2/3
What is the probability that she chooses neither Bangle bracelet nor Pearl bracelet?
(a) 23/180 (b) 45/180 (c) 109/180 (d) 23/45
What is the probability that Sunita purchased a Cuff bracelet?
(a) 6/17(b)2/13 (c) 1 (d) 2/15
Case Study/Passage-Based Questions
Case Study 2:In a factory, the workers are paid on a daily basis. The new manager wants to know the salary slab of the workers and finds the data given below.
Now, if a worker is chosen at random, then :
The probability that the worker is getting at most ₹ 500 is
(a) 3/80 (b) 9/80 (c) 1/80 (d) 4/25
The probability that the worker is getting at least ₹ 701 is
(a) 23/80 (b) 43/80 (c) 41/80 (d) 61/80
Probability that the worker is getting at most ₹ 900 is
(a)7/40 (b) 3/31 (c) 31/40 (d) 4/31
Case Study 3: A group of students is studying probability in their math class. They encountered the following scenario:
Rajesh and Aisha decided to conduct an experiment with a deck of playing cards to understand the concept of probability. They made the following observations:
1. Rajesh drew a card from a standard deck of 52 playing cards and found that it was a spade.
2. Aisha drew a card from the same deck and found that it was a face card (king, queen, or jack).
Based on this information, the students were asked to analyze the probabilities of their respective outcomes. Let’s see if you can answer the questions correctly:
MCQ Questions:
Q1. The probability of drawing a spade card, as observed by Rajesh, is:
(a) 1/4
(b) 1/13
(c) 1/52
(d) Cannot be determined
Q2. The probability of drawing a face card, as observed by Aisha, is:
(a) 3/13
(b) 4/13
(c) 1/13
(d) Cannot be determined
Q3. In a standard deck of playing cards, the probability of drawing a spade card is:
(a) 1/4
(b) 1/13
(c) 1/52
(d) 1/2
Q4. In a standard deck of playing cards, the probability of drawing a face card is:
(a) 3/13
(b) 4/13
(c) 1/13
(d) 1/4
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# Find three parts of 207 such that they are in increasing order, they form an AP and the product of two smaller parts is 4623.
Given:
Three parts of 207 are in increasing order, they form an AP and the product of two smaller parts is 4623.
To do:
We have to find the three parts.
Solution:
Let the three parts of the number 207 be $(a – d), a$ and $(a + d)$, which are in A.P.
According to the question,
Sum of the three parts $= 207$
$a – d + a + a + d = 207$
$3a = 207$
$a = \frac{207}{3}$
$a=69$
Product of the two smaller parts $= 4623$
This implies,
$a (a – d) = 4623$
$69 (69 – d) = 4623$
$69 – d = \frac{4623}{69}$
$69 – d =67$
$d = 69 – 67$
$d= 2$
Therefore,
First part $= a – d = 69 – 2 = 67$,
Second part $= a = 69$
Third part $= a + d = 69 + 2 = 71$
Hence, the required three parts are 67, 69, 71.
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## MAP Recommended Practice
### Course: MAP Recommended Practice>Unit 17
Lesson 1: Intro to ratios
# Ratio review
Learn how to find the ratio between two things given a diagram.
A ratio compares two different quantities.
For example, those two quantities could be monkeys and bananas:
Notice that there are $4$ monkeys and $5$ bananas.
Here are a few different ways we can describe the ratio of monkeys to bananas:
1. There are $4$ monkeys for every $5$ bananas.
2. The ratio of monkeys to bananas is $4$ to $5$.
3. The ratio of monkeys to bananas is $4:5$.
Order matters in ratios. Here are a few different ways to describe the ratio of bananas to monkeys:
1. There are $5$ bananas for every $4$ monkeys .
2. The ratio of bananas to monkeys is $5$ to $4$.
3. The ratio of bananas to monkeys is $5:4$.
## Let's practice!
Problem 1
Dana loves rocks! She has $6$ pieces of granite, $3$ pieces of marble, $14$ pieces of sandstone, and $1$ piece of slate.
What is the ratio of pieces of sandstone to pieces of marble in Dana's collection?
## Want to join the conversation?
• Do you always need to simplify every ratio?
• If the ratios can have a way of being simplified you should.
• How do you compare two ratios? Well... let me tell you. There are still other ways to make the same comparison, by using equal ratios. To find an equal ratio, you can either multiply or divide each term in the ratio by the same number (but not zero). For example, if we divide both terms in the ratio 3:6 by the number three, then we get the equal ratio, 1:2.
• You compare two ratios by using the butterfly method. If you dont know what the butterfly method is then you can you the criss cross method.
• if you have 2 apples and 3 toys is the ration 2:3
• you can also do it as a fraction
• Why do ratios have olny two numbers
• so they can compare both the numbers
• How big can a ratio get?
• A ratio is basically a comparison of two numbers and an ratio can get as big as ∞:∞, because after all a ratio is an comparison of two numbers and that is indeed endless.
• Hey I have a question for you can you help me. Well what are the difference of a rate and a ratios I always get confused. Plz comment and help. thanks you
• Hi 🤩SOPHIALIU🌸🌺
A rate is something that allows you to express one thing in terms of another thing, this is easiest to demonstrate with an example:
Imagine you are walking to the park, the park might be 100 meters away from your house, and it may take you 1 minute to get there, the rate associated with this problem would be 100 meters per 1 second (this is the rate you are travelling at and we may write this as 100m/1sec)
A ratio is very similar to a rate, we are comparing how an amount of one thing compares to an amount of another thing, again an example may be useful:
Imagine you are making chocolate milk (by mixing chocolate powder with milk), if you mix 2 spoons of chocolate powder in 1 glass of milk this will give you a ratio of 2 spoons of chocolate powder to 1 cup of milk (we might write this as 2:1, this is our ratio)
• Can you have decimals in ratios?
• yes a ratio can be 3.25 apples for every 6.64 bananas :)
• Why in ratios do we use the : sign ratios.
• Its a comparison symbol I'm assuming. You can search on Google for a specific answer to that.
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NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2.
Board CBSE Textbook NCERT Class Class 9 Subject Maths Chapter Chapter 9 Chapter Name Quadrilaterals Exercise Ex 9.2 Number of Questions Solved 7 Category NCERT Solutions
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2
Question 1.
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that
(i) SR || AC and SR = $$\frac { 1 }{ 2 }$$ AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Solution:
Given: P, Q, Ft and S are mid-points of the sides.
∴ AP = PB, BQ = CQ
CR = DR and AS = DS
S is mid-point of AD and R is mid-point of the DC.
We know that, the line segment joining the mid-points of two sides of a triangle is parallel to the third side.
∴ SB || AC …(i)
Also , SR = $$\frac { 1 }{ 2 }$$ AC …(ii)
(ii) Similarly, in ∆ABC, we have
PQ || AC ….(iii)
and PQ = $$\frac { 1 }{ 2 }$$ AC ….(iv)
Now, from Eqs. (i) and (iii), we get
SR = $$\frac { 1 }{ 2 }$$ AC …..(v)
(iii) Now, from Eqs. (i) and (iii), we get
PQ || SR
and from Eq. (v), PQ = SR
Since, a pair of opposite sides of a quadrilateral PQRS is equal and parallel.
So, PQRS is a parallelogram.
Hence proved.
Question 2.
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.
Solution:
Given: ABCD is a rhombus and P, Q, R and S are mid-points of AB, BC, CD and DA
By mid-point theorem,
∴ PQRS is a parallelogram.
Now, we know that diagonals of a rhombus bisect each other at right angles.
∴ ∠EOF = 90°
Now, RQ || BD (By mid-point theorem)
⇒ RE || OF
Also, SP|| AC [From Eq. (i)]
⇒ FR || OE
∴ OERF is a parallelogram.
So, ∠ ERF = ∠EOF = 90°
(Opposite angle of a quadrilateral is equal)
Thus, PQRS is a parallelogram with ∠R = 90°
Hence, PQRS is a rectangle.
Question 3.
ABCD is a rectangle and P, Q, R ans S are mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rhombus.
Solution:
Given: ABCD is a rectangle.
∴ ∠A = ∠B = ∠C= ∠D = 90°
and AD = BC, AB = CD
Also, given P, Q, R and S are mid-points of AB, BC, CD and DA .respectively.
∴ PQ || BD and PQ = $$\frac { 1 }{ 2 }$$ BD
In rectangle ABCD,
AC = BD
∴ PQ = SR …(ii)
Now, in ∆ASP and ∆BQP
AP = BP (Given)
AS = BQ (Given)
∠A = ∠B (Given)
∴ ∆ASP ≅ ∆BQP (By SAS)
∴ SP = PQ (By CPCT)…(ii)
Similarly, in ∆RDS and ∆RCQ,
SD = CQ (Given)
DR = RC (Given)
∠C=∠D (Given)
∴ ∆RDS ≅ ∆RCQ (By SAS)
∴ SR = RQ (By CPCT)…(iii)
From Eqs. (i), (ii) and (iii), it is clear that quadrilateral PQRS is a rhombus.
Question 4.
ABCD is a trapezium in which AB | | DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see figure). Show that F is the mid-point of BC.
Solution:
Given: ABCD is a trapezium in which AB || CD and E is mid-point of AD and EF || AB.
In ∆ABD, we have
EP\\AB
and E is mid-point of AD.
So, by theorem, if a line drawn through the mid-point of one side of a triangle parallel to another side bisect the third side.
∴ P is mid-point of BD.
Similarly, in ∆ BCD, we have,
PF || CD (Given)
and P is mid-point of BD.
So, by converse of mid-point theorem, F is mid-point of CB.
Question 5.
In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see figure). Show that the line segments AF and EC trisect the diagonal BD.
Solution:
Given: ABCD is a parallelogram and E, F are the mid-points of sides AB and CD respectively.
To prove: Line segments AF and EC trisect the diagonal BD.
Proof: Since, ABCD is a parallelogram.
AB || DC
and AB = DC (Opposite sides of a parallelogram)
⇒ AE || FC and $$\frac { 1 }{ 2 }$$ AB = $$\frac { 1 }{ 2 }$$ DC
⇒ AF || FC and AF = FC
∴ AECF is a parallelogram.
∴ AF || FC
⇒ EQ || AP and FP || CQ
In ∆ BAP, E is the mid-point of AB and EQ || AP, so Q is the mid-point of BP.
(By converse of mid-point theorem)
∴ BQ = PQ ….(i)
Again, in ∆DQC, F is the mid-point of DC and FP || CQ, so P is the mid-point of DQ. (By converse of mid-point theorem)
∴ QP = DP …(ii)
From Eqs. (i) and (ii), we get
BQ = PQ = PD
Hence, CE and AF trisect the diagonal BD.
Question 6.
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Solution:
Let ABCD is a quadrilateral and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively, i.e., AS = SD, AP = BP, BQ = CQ and CR = DR. We have to show that PR and SQ bisect each other i.e., SO = OQ and PO = OR.
Now, in ∆ADC, S and R are mid-points of AD and CD.
We know that, the line segment joining the mid-points of two sides of a triangle is parallel to the third side. (By mid-point theorem)
∴ SR || AC and SR = $$\frac { 1 }{ 2 }$$ AC …(i)
Similarly, in ∆ ABC, P and Q are mid-points of AB and BC.
∴ PQ || AC and PQ = $$\frac { 1 }{ 2 }$$ AC (By mid-point theorem)…(ii)
From Eqs. (i) and (ii), we get
PQ || SR
and PQ = SR = $$\frac { 1 }{ 2 }$$ AC
∴ Quadrilateral PQRS is a parallelogram whose diagonals are SQ and PR. Also, we know that diagonals of a parallelogram bisect each other. So, SQ and PR bisect each other.
Question 7.
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC
(ii) MD ⊥ AC
(iii) CM = MA = $$\frac { 1 }{ 2 }$$ AB
Solution:
Given: ABC is a right angled triangle.
∠C = 90°
and M is the mid-point of AB.
Also, DM || BC
(i) In ∆ ABC, BC || MD and M is mid-point of AB.
∴ D is the mid-point of AC. (By converse of mid-point theorem)
(ii) Since, MD || BC and CD is transversal
∴ ∠ADM = ∠ACB (Corresponding angles)
But ∠ACB = 90°
∴ ∠ADM = 90° ⇒ MD ⊥ AC
(iii) Now, in ∆ ADM and ∆ CDM, we have
DM = MD (Common)
AD = CD (∵ D is mid point of AC)
∴ ∠ADM = ∠MDC (Each equal to 90°)
∴ ∆ ADM = ∆ CDM (By SAS)
∴ CM = AM (By CPCT)…(i)
Also, M is mid-point of AB.
∴ AM – BM = $$\frac { 1 }{ 2 }$$ AB ….(ii)
From Eqs. (i) and (ii), we get
CM = AM = $$\frac { 1 }{ 2 }$$ AB
Hence proved.
We hope the NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2, drop a comment below and we will get back to you at the earliest.
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# Problem Solving Techniques
Related Topics:
More Lessons for Arithmetic
Math Worksheets
The following are some examples of problem solving strategies.
Explore it//Act it/Try it (EAT) method (Advanced)
Finding a Pattern (Basic)
Finding a Pattern (Intermediate)
## Explore It/Act It/Try It (EAT) Method (Intermediate)
In this lesson, we will look at some intermediate examples of the Explore it//Act it/Try it (EAT) method of problem solving strategy.
Example:
Allen has to ferry a cat, a chicken and a sack of grain across a river. His small boat is big enough to carry himself and only one of the three items at any one time. He must not leave the cat and the chicken alone or else the cat will attack the chicken. He also must not leave the chicken and the sack of grain alone or else the chicken will eat the grain. Find the least number of trips Allen needs to safely transport all the items to the other side of the river.
Solution:
Example:
Marcus bought two antique lamps for \$50 each. Later he sold one for \$60, but he changed his mind and bought it back for \$70. Then he sold it again for \$80. But since there was no one interested in the first lamp, he offered it for 10% less than its original cost, and finally managed to sell it. Did he make or lose money, and how much?
Solution:
All the ‘buy’ transactions: 2 × 50 + 70 = \$170
All the ‘sell’ transactions: 60 + 80 + 45 = \$185
Example:
Without moving 6 adjacent numbers on the face of a clock, rearrange the other six so that the sum of every pair of adjacent numbers is a prime number.
Solution:
11, 12, 1, 2, 3, 4 (not moved) followed by 9, 10, 7, 6, 5, 8
Example:
Julia bought a T-shirt that costs RM28 and paid for it exactly using 7 pieces of notes. What were the 7 pieces of notes that she used?
Solution:
10, 5, 5, 5, 1, 1, 1
10, 10, 2, 2, 2, 1, 1
5, 5, 5, 5, 5, 2, 1
Example:
Four stamps are to be torn from the sheet shown below. The four stamps must be intact so that each stamp is joined to another stamp along at least one edge. Find the possible patterns for these four stamps.
Example:
Yellow, white, green and red counters are arranged in a row. The red counter is to the left of the green counter and to the right of the yellow counter. The white counter is to the left of the green counter and not next to the yellow one. What are the colours of the counters in the row from left to right?
Solution:
Given: The red counter is to the left of the green counter and to the right of the yellow counter.
• Yellow counter is to the left of red counter
• Red counter is to the left of the green counter
Given: The white counter is to the left of the green counter and not next to the yellow one
• White counter is to the left of the green counter
• White counter is not next to the yellow counter
• White counter is between red counter and green counter.
So the order is: Yellow, Red, White and Green.
Example:
The figure below is arranged using 16 matchsticks to form 5 squares. Rearrange exactly 2 of the matchsticks to form 4 squares of the same size, without leaving any stray matchsticks.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# Multiplication of Monomials by Polynomials
## Distribute single terms by multiplying them by other terms
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Practice Multiplication of Monomials by Polynomials
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Multiplication of Monomials by Polynomials
What if you had a monomial and polynomial like 3x3\begin{align*}3x^3\end{align*} and x2+4\begin{align*}x^2 + 4\end{align*}? How could you multiply them? After completing this Concept, you'll be able multiply a polynomial by a monomial.
### Guidance
Just as we can add and subtract polynomials, we can also multiply them. The Distributive Property and the techniques you’ve learned for dealing with exponents will be useful here.
Multiplying a Polynomial by a Monomial
When multiplying polynomials, we must remember the exponent rules that we learned in the last chapter. Especially important is the product rule: xnxm=xn+m\begin{align*}x^n \cdot x^m=x^{n+m}\end{align*}.
If the expressions we are multiplying have coefficients and more than one variable, we multiply the coefficients just as we would any number and we apply the product rule on each variable separately.
#### Example A
Multiply the following monomials.
a) (2x2)(5x3)\begin{align*}(2x^2)(5x^3)\end{align*}
b) (3y4)(2y2)\begin{align*}(-3y^4)(2y^2)\end{align*}
c) (3xy5)(6x4y2)\begin{align*}(3xy^5)(-6x^4y^2)\end{align*}
d) (12a2b3c4)(3a2b2)\begin{align*}(-12a^2b^3c^4)(-3a^2b^2)\end{align*}
Solution
a) (2x2)(5x3)=(25)(x2x3)=10x2+3=10x5\begin{align*}(2x^2)(5x^3)=(2 \cdot 5) \cdot (x^2 \cdot x^3)=10x^{2+3} = 10x^5\end{align*}
b) (3y4)(2y2)=(32)(y4y2)=6y4+2=6y6\begin{align*}(-3y^4)(2y^2)=(-3 \cdot 2) \cdot (y^4 \cdot y^2)=-6y^{4+2}=-6y^6\end{align*}
c) (3xy5)(6x4y2)=18x1+4y5+2=18x5y7\begin{align*}(3xy^5)(-6x^4y^2)=-18x^{1+4}y^{5+2}=-18x^5y^7\end{align*}
d) (12a2b3c4)(3a2b2)=36a2+2b3+2c4=36a4b5c4\begin{align*}(-12a^2b^3c^4)(-3a^2b^2)=36a^{2+2}b^{3+2}c^4 = 36a^4b^5c^4\end{align*}
To multiply a polynomial by a monomial, we have to use the Distributive Property. Remember, that property says that a(b+c)=ab+ac\begin{align*}a(b + c) = ab + ac\end{align*}.
#### Example B
Multiply:
a) 3(x2+3x5)\begin{align*}3(x^2+3x-5)\end{align*}
b) 4x(3x27)\begin{align*}4x(3x^2-7)\end{align*}
c) 7y(4y22y+1)\begin{align*}-7y(4y^2-2y+1)\end{align*}
Solution
a) 3(x2+3x5)=3(x2)+3(3x)3(5)=3x2+9x15\begin{align*}3(x^2+3x-5)=3(x^2)+3(3x)-3(5)=3x^2+9x-15\end{align*}
b) 4x(3x27)=(4x)(3x2)+(4x)(7)=12x328x\begin{align*}4x(3x^2-7)=(4x)(3x^2)+(4x)(-7)=12x^3-28x\end{align*}
c)
7y(4y22y+1)=(7y)(4y2)+(7y)(2y)+(7y)(1)=28y3+14y27y
Notice that when we use the Distributive Property, the problem becomes a matter of just multiplying monomials by monomials and adding all the separate parts together.
#### Example C
Multiply:
a) 2x3(3x4+2x310x2+7x+9)\begin{align*}2x^3(-3x^4+2x^3-10x^2+7x+9)\end{align*}
b) 7a2bc3(5a23b29c2)\begin{align*}-7a^2bc^3(5a^2-3b^2-9c^2)\end{align*}
Solution
a)
2x3(3x4+2x310x2+7x+9)=(2x3)(3x4)+(2x3)(2x3)+(2x3)(10x2)+(2x3)(7x)+(2x3)(9)=6x7+4x620x5+14x4+18x3
b)
7a2bc3(5a23b29c2)=(7a2bc3)(5a2)+(7a2bc3)(3b2)+(7a2bc3)(9c2)=35a4bc3+21a2b3c3+63a2bc5
Watch this video for help with the Examples above.
### Vocabulary
• Distributive Property: For any expressions a, b\begin{align*}a, \ b\end{align*}, and c\begin{align*}c\end{align*}, a(b+c)=ab+ac\begin{align*}a(b+c)=ab+ac\end{align*}.
### Guided Practice
Multiply 2a2b4(3ab2+7a3b9a+3)\begin{align*}-2a^2b^4(3ab^2+7a^3b-9a+3)\end{align*}.
Solution:
Multiply the monomial by each term inside the parenthesis:
2a2b4(3ab2+7a3b9a+3)=(2a2b4)(3ab2)+(2a2b4)(7a3b)+(2a2b4)(9a)+(2a2b4)(3)=6a3b614a5b5+18a5b46a2b4
### Explore More
Multiply the following monomials.
1. (2x)(7x)\begin{align*}(2x)(-7x)\end{align*}
2. (10x)(3xy)\begin{align*}(10x)(3xy)\end{align*}
3. (4mn)(0.5nm2)\begin{align*}(4mn)(0.5nm^2)\end{align*}
4. (5a2b)(12a3b3)\begin{align*}(-5a^2b)(-12a^3b^3)\end{align*}
5. (3xy2z2)(15x2yz3)\begin{align*}(3xy^2z^2)(15x^2yz^3)\end{align*}
Multiply and simplify.
1. 17(8x10)\begin{align*}17(8x-10)\end{align*}
2. 2x(4x5)\begin{align*}2x(4x-5)\end{align*}
3. 9x3(3x22x+7)\begin{align*}9x^3(3x^2-2x+7)\end{align*}
4. 3x(2y2+y5)\begin{align*}3x(2y^2+y-5)\end{align*}
5. 10q(3q2r+5r)\begin{align*}10q(3q^2r+5r)\end{align*}
6. 3a2b(9a24b2)\begin{align*}-3a^2b(9a^2-4b^2)\end{align*}
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# How Does the Normal Force Compare to the Car's Weight in a Circular Valley?
• courtney1121
In summary, the normal force is the force exerted by a surface on an object, while weight is the force of gravity acting on an object. The magnitude of the normal force is determined by the weight of the object and the angle of the surface, and it can never be greater than the weight. The normal force always acts in the opposite direction of weight and can affect the motion of an object by either supporting it or limiting its motion.
courtney1121
A car travels at a constant speed of 23 m/s through a small valley whose cross section is like a circule of radius 310 m. What is the magnitude of the normal force on the car, expressed as a multiple of the car's weight?
what does it mean by the cross section and how does it relate to the problem?
You can assume the valey is sort of a semi-cylinder and the car travels on the circular part of it.
The cross section refers to the shape of the valley as seen from a bird's eye view. In this case, it is a circular shape with a radius of 310 m. This information is relevant because it helps us understand the curvature of the valley and how it may affect the car's motion.
To calculate the magnitude of the normal force on the car, we need to consider the forces acting on the car. The two main forces are the normal force and the weight of the car. The normal force is the force exerted by the surface of the valley on the car, perpendicular to the surface. The weight of the car is the force exerted by gravity on the car, directed towards the center of the Earth.
Since the car is traveling at a constant speed, we can assume that the net force acting on the car is zero. This means that the normal force must be equal in magnitude and opposite in direction to the weight of the car.
Using the formula for centripetal force (F = mv^2/r), we can calculate the magnitude of the centripetal force acting on the car, which is equal to the normal force. We know the mass of the car (m), its speed (v), and the radius of the circle (r).
Once we have calculated the magnitude of the normal force, we can express it as a multiple of the car's weight by dividing it by the weight of the car (mg). This will give us a dimensionless number, as both the normal force and weight are measured in Newtons (N).
In conclusion, the magnitude of the normal force on the car can be calculated using the formula for centripetal force and expressed as a multiple of the car's weight. The cross section of the valley, represented by the circular shape with a radius of 310 m, is important in understanding the forces acting on the car and how they contribute to its motion.
## 1. What is the difference between normal force and weight?
The normal force is the force exerted by a surface on an object that is in contact with it, perpendicular to the surface. Weight is the force of gravity acting on an object, pulling it towards the center of the Earth. While the normal force is a reaction force to the weight, they are two distinct forces that act on an object.
## 2. How is the magnitude of the normal force determined?
The magnitude of the normal force is determined by the weight of the object and the angle of the surface it is in contact with. The normal force is equal in magnitude to the component of the weight that is perpendicular to the surface.
## 3. Can the normal force ever be greater than weight?
No, the normal force can never be greater than the weight of the object. It can only be equal to or less than the weight, depending on the angle of the surface and the weight of the object.
## 4. Does the normal force always act in the opposite direction of weight?
Yes, the normal force always acts in the opposite direction of weight. This is because the normal force is a reaction force to the weight, and Newton's third law states that for every action, there is an equal and opposite reaction.
## 5. How does normal force affect the motion of an object?
The normal force can affect the motion of an object by either supporting it against the force of gravity or by limiting its motion. For example, if an object is on a ramp, the normal force will help keep it from sliding down the ramp. On the other hand, if an object is on a surface with a normal force greater than its weight, it will accelerate upwards.
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# Factors of 15
Factors of 15 are the numbers that divide the original number 15 evenly without leaving the remainder. Though the number 15 is composite, it has two factors other than one and the number itself. Factors of 15 are 1, 3, 5, and 15. You can introduce the concept of factors of a number to kids in elementary school so that they can solve math problems quickly. Learning arithmetic operations helps kids find the factors of 15 easily.
Contents
Kids must learn to find the factors of 15 to solve mathematical problems with speed and accuracy. Besides this, teach them pair and prime factors of 15. The pair factors of 15 are (1, 15) and (3, 5), whereas the prime factors of 15 are 3 x 5. Kids must remember that the pair factors of 15 can be positive or negative numbers. How would you teach factors of 15 to little ones? The best way would be to conduct math games for kids to keep them engaged in learning and retain the concept for a longer period of time.
Check Math Related Articles:
## What are the Factors of 15?
Factors of 15 are the numbers obtained when you completely divide the original number 15 with the integers without leaving the remainder. These numbers can be positive or negative but cannot be in decimal or fraction form. The factors of 15 are 1, 3, 5, and 15. You can find the factors of 15 using the division method. Besides this, kids must learn to find the pair and prime factors of 15. The pair factors of 15 are (1, 15) and (3, 5), and the prime factors of 15 are 3 x 5, where 3 and 5 are the prime numbers. These pair factors of 15 can also be negative numbers, and they are (-1, -15) and (-3, -5). The total sum of factors of 15 are 1 + 3 + 5 + 15 = 24.
## How to Find the Factors of 15?
To calculate the factors of 15, you need to divide the original number 15 with the integers evenly without the remainder. Kids must get acquainted with tables 15 to 25 to find the factors of 15 effectively. Check out the division method to calculate the factors of 15 as given below.
15/1 = 15
15/3 = 5
15/5 = 3
15/15 = 1
Therefore, factors of 15 are 1, 3, 5 and 15.
### Pair Factors of 15
To find the pair factors of 15, you need to multiply the integers in pairs to get the original number. These numbers can be positive or negative. Check out the multiplication method to calculate the pair factors of 15 as given below.
1 x 15 = 15
3 x 5 = 15
5 x 3 = 15
15 x 1 = 15
Therefore, the positive pair factors of 15 are (1, 15) and (3, 5). You can also find the negative pair factors of 15 using the multiplication method. When you multiply two negative factors together, you get a positive result. The negative pair factors of 15 are (-1, -15) and (-3, -5).
### Prime Factors of 15
To find the prime factors of 15, you must divide the original number with the prime numbers until you get the quotient one. Check out the prime factorization method as given below.
Step 1: Divide the original number with the prime number 3 since it is not divisible by the least prime number 2.
15/3 = 5
Step 2: Divide the number 5 with the prime number 5 to get the quotient one.
5/5 = 1
Therefore, the prime factors of 15 are 3 x 5.
## Solved Examples on Factors of 15
Some solved examples on factors of 15 for kids are mentioned below.
Example 1: Find out the common factors of 15 and 10
Solution: Factors of 15 = 1, 3, 5, and 15.
Factors of 10 = 1, 2, 5, and 10
Therefore, the common factors of 15 and 10 are 1 and 5.
Example 2: Tracy bought 15 Christmas caps for her family members. She equally distributed them to 5 family members. How many Christmas caps did each family member get?
Solution: Number of Christmas caps = 15
Number of family members = 5
Total number of Christmas caps received by each family member = 15/5 = 3
Therefore, the total number of Christmas caps received by each family member is 3.
We hope this article on factors of 15 was useful to you. For more about activities, worksheets and games, explore worksheets for kids, math for kidspuzzles for kids, kids learning sections at Osmo.
## Frequently Asked Questions on Factors of 15
### What are the factors of 15?
The factors of 15 are 1, 3, 5, and 15.
### What are the pair factors of 15?
The pair factors of 15 are (1, 15) and (3, 5), whereas the negative pair factors of 15 are (-1, -15) and (-3, -5).
### What are the prime factors of 15?
The prime factors of 15 are 3 x 5.
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# CBSE Class 10-Mathematics: Chapter – 5 Arithmetic Progressions Part 14 (For CBSE, ICSE, IAS, NET, NRA 2022)
Get top class preparation for CBSE/Class-10 right from your home: get questions, notes, tests, video lectures and more- for all subjects of CBSE/Class-10.
Question 3:
If the third and the ninth terms of an AP are and respectively, which term of this AP is zero?
It is given that 3rd and 9th term of AP are 4 and respectively.
It means and
Using formula , to find term of arithmetic progression,
These are equations in two variables.
Using equation , we can say that
Putting value of a in other equation ,
Putting value of in equation ,
Therefore, first term and Common Difference
We want to know which term is equal to zero.
Using formula to find term of arithmetic progression,
Therefore, term is equal to .
Question 4:
Two AP՚s have the same common difference. The difference between their terms is , what is the difference between their terms.
Let first term of 1st AP
Let first term of 2nd
It is given that their common difference is same.
Let their common difference be .
It is given that difference between their terms is .
Using formula , to find nth term of arithmetic progression,
We want to find difference between their
1000th terms which means we want to calculate:
Putting equation (1) in the above equation,
Therefore, difference between their 1000th terms would be equal to 100.
Question 5:
How many three-digit numbers are divisible by 7?
We have AP starting from because it is the first three-digit number divisible by .
AP will end at because it is the last three-digit number divisible by .
Therefore, we have AP of the form
Let is the nth term of AP.
We need to find n here.
First term , Common difference
Using formula , to find term of arithmetic progression,
It means is the term of AP.
Therefore, there are terms in AP.
Developed by:
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# CONSTANT OF PROPORTIONALITY
The value of the constant of proportionality is depending on the type of proportion we have between the two quantities.
There are two types of proportion:
1. Direct proportion
2. Inverse proportion
## Direct Proportion
If y is directly proportional to x, then we have
where k is the constant of proportionality.
## Inverse Proportion
If y is inversely proportional to x, then we have
where k is the constant of proportionality.
## Identifying Proportional Relationships
If two different quantities are given, how to check whether the relationship between them is proportional ?
We have to get ratio of the two quantities for all the given values.
If all the ratios are equal, then the relationship is proportional.
If all the ratios are not equal, then the relationship is not proportional.
## Identifying Proportional Relationships - Examples
Example 1 :
Examine the given table and determine if the relationship is proportional. If yes, determine the constant of proportionality.
Solution :
Let us get the ratio of x and y for all the given values.
4 / 48 = 1 / 12
7 / 84 = 1 / 12
10 / 120 = 1 / 12
When we take ratio of x and y for all the given values, we get equal value for all the ratios.
Therefore the relationship given in the table is proportional.
When we look at the above table when x gets increased, y also gets increased, so it is direct proportion.
Then, we have
y = kx
Substitute 4 for x and 48 for y.
48 = k(4)
12 = k
So, the constant of proportionality is 12.
Example 2 :
Examine the given table and determine if the relationship is proportional. If yes, determine the constant of proportionality.
Solution :
Let us get the ratio of x and y for all the given values.
1 / 100 = 1 / 100
3 / 300 = 1 / 100
5 / 550 = 1 / 110
6 / 600 = 1 / 100
When we take ratio of x and y for all the given values, we don't get equal value for all the ratios.
So, the relationship given in the table is not proportional.
Example 3 :
Examine the given table and determine if the relationship is proportional. If yes, determine the constant of proportionality.
Solution :
Find the ratio of x and y for all the given values.
2 / 1 = 2
4 / 2 = 2
8 / 4 = 2
10 / 5 = 2
When we take ratio of x and y for all the given values, we get equal value for all the ratios.
Therefore, the relationship given in the table is proportional.
When we look at the above table when x gets increased, y also gets increased, so it is direct proportion.
Then, we have
y = kx
Substitute 2 for x and 1 for y.
1 = k(2)
1 / 2 = k
So, the constant of proportionality is 1/2.
Example 4 :
Examine the given table and determine if the relationship is proportional. If yes, determine the constant of proportionality.
Solution :
Find the ratio of x and y for all the given values.
1 / 2 = 1 / 2
2 / 4 = 1 / 2
3 / 6 = 1 / 2
4 / 6 = 2 / 3
When we take ratio of x and y for all the given values, we don't get equal value for all the ratios.
So, the relationship given in the table is not proportional.
Example 5 :
Examine the given table and determine if the relationship is proportional. If yes, determine the constant of proportionality.
Solution :
Find the ratio of x and y for all the given values.
1 / 23 = 1 / 23
2 / 36 = 1 / 18
5 / 75 = 1 / 15
When we take ratio of x and y for all the given values, we don't get equal value for all the ratios.
So, the relationship given in the table is not proportional.
Example 6 :
Examine the given table and determine if the relationship is proportional. If yes, determine the constant of proportionality.
Solution :
Find the ratio of x and y for all the given values.
2 / 4 = 1 / 2
4 / 8 = 1 / 2
6 / 12 = 1 / 2
8 / 16 = 1 / 2
When we take ratio of x and y for all the given values, we get equal value for all the ratios.
Therefore the relationship given in the table is proportional.
When we look at the above table when x gets increased, y also gets increased, so it is direct proportion.
Then, we have
y = kx
Substitute 2 for x and 4 for y.
4 = k(2)
2 = k
So, the constant of proportionality is 2.
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# Calculate the Spring Constant Using Hooke’s Law: Formula, Examples, and Practice Problems
If you push or pull on a spring and then let it go, it snaps right back to its original position. The spring constant tells you how much force the spring exerts when it does that, but how do you figure out what the spring constant is? You're in luck because there's a simple formula you can use. Read on to learn how to apply the formula to find the spring constant, then try your hand with a few practice problems.[1]
## Things You Should Know
• Use the formula ${\displaystyle k={\frac {F}{x}}}$ to find the spring constant for an ideal spring.
• Express Hooke's Law mathematically with the equation ${\displaystyle F=-kx}$.
• Display the spring constant on a graph as the slope of a straight line since the relationship between force and distance is linear.
Section 1 of 6:
### What is the formula for the spring constant?
1. The formula to find the spring constant is . When you push or pull a spring, it bounces back to its original rest position or equilibrium. The spring constant, measured in ${\displaystyle N/m}$ (Newton/meters or Newtons per meter), tells you the proportional force exerted by the spring that causes it to bounce back. These variables have a linear relationship:[2]
• ${\displaystyle k}$: Spring constant
• ${\displaystyle F}$: Force (the force applied to the spring)
• ${\displaystyle x}$: Distance the spring is compressed or stretched away from its equilibrium or rest position
Section 2 of 6:
### What is Hooke's Law?
1. Hooke's law states that when a force compresses or stretches a spring, it will bounce back to its rest position with an equal and opposite force. That bounce-back or restoring force is known as the spring constant because it's always the same for a given spring. You can express this law mathematically with the equation ${\displaystyle F=-kx}$.[3]
• The negative symbol indicates that the force of the spring constant is in the opposite direction of the force applied to the spring. It does not indicate that the value is negative.
• Hooke's law is actually pretty limited. It only applies to perfectly elastic materials within their elastic limit—stretch something too far and it'll break or stay stretched out.
• Hooke's law is based on Newton's third law of motion, which states that for every action there is an equal and opposite reaction.[4]
Section 3 of 6:
### How does spring length affect the spring constant?
1. The spring constant is inversely related to the spring's equilibrium length. This means that the longer your spring is, the less force it will need to bounce back to its rest position or equilibrium. A shorter spring, on the other hand, requires more force to bounce back. This makes sense if you think of springs you've encountered—the spring constant measures the stiffness of a spring, and short springs tend to be stiffer than long springs.[5]
• For example, if you cut a spring in half, its spring constant will double. If you doubled the length of the spring, on the other hand, its spring constant would be half what it was.
• This also means that when you apply the same force to a longer spring as a shorter spring, the longer spring will stretch further than the shorter spring.[6]
Section 4 of 6:
### Practice Problems
1. 1
Find the spring constant of a stretched spring that supports of weight if the spring stretches another when is added to it.[7]
• Hint: Since the spring constant is measured in ${\displaystyle N/m}$, convert ${\displaystyle 3.5cm}$ to ${\displaystyle 0.035m}$.
2. 2
What force is required to stretch a spring if the spring has a spring constant of ?[8]
• Hint: Use the equation for Hooke's Law without the negative symbol, which doesn't have a mathematical function: ${\displaystyle F=kx}$.
3. 3
If a person who weighs steps on a scale, the scale's spring compresses by . What is the spring constant of the scale's spring?[9]
• Hint: The person's weight is equivalent to the force applied to the spring.
4. 4
What is the spring constant of a spring if it stretches when a force is applied?[10]
• Hint: Remember to convert the ${\displaystyle cm}$ to ${\displaystyle m}$ since the spring constant is measured in ${\displaystyle N/m}$.
Section 5 of 6:
### Solutions to Practice Problems
1. 1
The spring constant is . You aren't given the distance the spring is stretched when the first ${\displaystyle 0.1N}$ is added to it, but that doesn't matter because the spring constant is always the same. Plug the values for the second weight into the formula to find the spring constant:[11]
• The formula to find the spring constant is ${\displaystyle k={\frac {F}{x}}}$. Here, the force is ${\displaystyle 0.1N}$ and the distance the spring stretches when that force is added is ${\displaystyle 0.035m}$, so your equation is ${\displaystyle k={\frac {0.1}{0.035}}}$.
• Divide ${\displaystyle 0.1}$ by ${\displaystyle 0.035}$, then round to get your answer of ${\displaystyle 2.85N/m}$.
• Why not ${\displaystyle 0.2N}$? Because you don't know the distance the string stretched when the first weight was applied. You only know the additional distance it stretched when the second weight was applied.
2. 2
The force applied is . Plug the values you were given into the equation for Hooke's Law and you get ${\displaystyle F=(0.7)(1.5)}$. Then, you simply multiply to find the answer.[12]
3. 3
The spring constant of the scale's spring is . First, convert ${\displaystyle cm}$ to ${\displaystyle m}$, since the spring constant is measured in ${\displaystyle N/m}$. Then, plug the values into your equation to find the spring constant:[13]
• ${\displaystyle k={\frac {F}{x}}}$
• ${\displaystyle k={\frac {670}{0.0079}}}$
• ${\displaystyle k=84,810N/m}$
4. 4
The spring constant is . Start by converting ${\displaystyle 280cm}$ to ${\displaystyle 2.8m}$. Then, since you're asked to find the spring constant, use the formula ${\displaystyle k={\frac {F}{x}}}$:[14]
• ${\displaystyle k={\frac {25}{2.8}}}$
• ${\displaystyle k=8.93N/m}$
Section 6 of 6:
### How do I display Hooke's Law on a graph?
1. Hooke's Law describes a directly proportional linear relationship between the force exerted on a spring and the distance it moves. Within this relationship, the spring constant is the slope of the line on a graph. The distance values are the ${\displaystyle x}$ coordinates and the applied force values are the ${\displaystyle y}$coordinates.[15]
• If you're given a line that represents a spring that obeys Hooke's Law (also called an ideal spring), you can find the spring constant by finding the slope of the line using the basic slope formula ${\displaystyle {\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}}$.
• A line with a spring constant as a slope will always cross through the origin of the graph.
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Co-authored by:
Doctor of Law, Indiana University
This article was co-authored by wikiHow staff writer, Jennifer Mueller, JD. Jennifer Mueller is a wikiHow Content Creator. She specializes in reviewing, fact-checking, and evaluating wikiHow's content to ensure thoroughness and accuracy. Jennifer holds a JD from Indiana University Maurer School of Law in 2006. This article has been viewed 18,737 times.
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# Geometric Progression Shortcut, tricks, and tips
## Geometric Progression tricks, shortcuts, and tips
Geometric Progression Tips,Tricks and Shortcuts are very important to know solve questions easily. In this page different types of question arevgiven that will help you in different examination.
Here, are some easy tips and tricks for you to solve Geometric Progression questions quickly, easily , and efficiently in competitive exams.
### Geometric Progression Tips and Tricks and Shortcuts
• Three non-zero terms a, b, c are in GP if and only if b^2 = ac.
• In a GP, three consecutive terms can be taken as a/r, a, ar….
• If a, b and c are three quantities in GP, then their reciprocals (b/a), (c/b), and (c/a) are also in GP.
• If we multiply or divide each term of the GP by a non-zero quantity, then resulting sequence remains a GP with the same common ratio.
### Type 1: Find nth term of the series
Question 1: Find 11th term in the series 2,4,8,16 …
Options:
A. 2042
B. 2200
C. 1024
D. 2048
Solution: We know that,
an = arn-1
where
r(common ratio) = $\frac{4}{2} = 2$
a1= first term = 2
an-1= the term before the nth term,
n = number of terms
In the given series,
r (common ratio) = $\frac{4}{2} = 2$
Therefore, 11th term = a11
a11 = 2 x 211-1
a11 = 2 x 210
a11 = 2 x 1024
a11 = 2048
Correct option: D
Question 2 : Find last term in the series if there are 7 term in this series 3,15,75,375 …
Options:
A. 46875
B. 44875
C. 42875
D. 40875
Solution: We know that,
an = arn-1
where
r(common ratio) = $\frac{15}{3} = 5$
a1= first term = 3
an-1= the term before the nth term,
n = number of terms
In the given series,
r (common ratio) = $\frac{15}{3} = 5$
Therefore, 7th term = a7
a7 = 3 x 57-1
a7 = 46875
Correct option: A
### Type 2: Find number of terms in the series
Question 1. Find the number of terms in the GP 6, 12, 24, 48……1536
Options:
A. 6
B. 7
C. 9
D. 8
Solution:
We know that,
In the given series,
a1 = 6,
a2 = 12,
r = $\frac{12}{6} = 2$,
an = 1536
an = arn-1
1536 = 6 x 2n-1 (divide both side by 6)
256 = 2n-1
28 = 2n-1
8 = n – 1
n = 9
Therefore, there are 9 terms in the series.
Correct option: C
Question 2. Find the number of terms in the GP where a1 = 10, a2 = 40 ,a3 = 160 ,an = 10240
Options:
A. 10
B. 7
C. 9
D. 6
Solution: We know that,
In the given series,
a1 = 10,
a2 = 40,
r = $\frac{40}{10} = 4$,
an = 10240
an = arn-1
10240 = 10 x 4n-1 (divide both side by 10)
1024 = 4n-1
45 = 4n-1
5 = n – 1
n = 6
Therefore, there are 6 terms in the series.
Correct option: D
### Type 3: Related to sum of first ‘n’ terms of the Geometric series
Question 1. How many terms of the series 1 + 3 + 9 +….sum to 121
Options:
A. 18
B. 19
C. 13
D. 5
Solution: We know that,
$s_{n} = a \times \frac{r^{n} -1}{r-1}$ if r>1
In the given series,
a = 1,
$r = \frac{3}{1} = 3$,
Sn = 121
121 =$1\times \frac{3^{n} -1}{3-1}$
121 = $\frac{3^{n} -1}{3-1}$
242 = (3n – 1)
243 = 3n
35 = 3n
n = 5
Correct option: D
Question 2. Find Sum of given Geometric Series upto 9th term 7,14,28,56……
Options:
A. 3177
B. 3577
C. 1377
D. 5377
Solution: We know that,
$s_{n} = a \times \frac{r^{n} -1}{r-1}$ if r>1
In the given series,
a = 7,
$r = \frac{14}{7} = 2$,
Sn = $7\times \frac{2^{9} -1}{2-1}$
= 3577
Correct option: B
### Type 4: Find the Geometric Mean (GM) of the series.
Question 1. What is the geometric mean of 2, 3, and 6?
Options
A. 4.5
B. 6.5
C. 3.30
D. 6.4
Solution: We know that,
GM = $(abc)^{\frac{1}{3}}$
Therefore, there Geometric Mean (GM) = $(2\times 3\times 6)^{\frac{1}{3}}$
= 3.30
Correct option: C
Question 2 . What is the geometric mean of 36 and 9?
Options
A. 24
B. 16
C. 18
D. 14
Solution:
We know that,
GM = $(ab)^{\frac{1}{2}}$
Therefore, there Geometric Mean (GM) = $(36\times 9)^{\frac{1}{2}}$
= 18
Correct option: C
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# What Are The Prime Factors Of 20
What are the prime factors of 20? Answer: 2, 5
The number 20 has 2 prime factors. Primes can only have two factors(1 and itself) and only be divisible by those two factors. Any figure where this rule applies can be called a prime factor. The biggest prime factor of 20 is 5. The smallest prime factor of 20 is 2.
## What Is The Factor Tree Of 20
How to use a factor tree to find the prime factors of 20? A factor tree is a diagram that organizes the factoring process.
• First step is to find two numbers that when multiplied together equal the number we start with.
20
↙ ↘
2 × 10
• Second step is to check the multiplication(of the first step) for numbers that are not primes.
10
↙ ↘
5 × 2
We found 2 prime factors(2, 5) using the factor tree of 20. Now let us explain the process to solving factor trees in more detail. Our goal is to find all prime factors of a given whole number. In each step of our factor tree diagram for 20 we always checked both multiplication numbers if they were primes or not. If one or both of the integers are not prime numbers then this means that we will have to make diagrams for them too. This process continues until only prime numbers are left.
Remember that often a factor tree for the same integer can be solved in more than one correct way! An example of this is the figure 12 where 2*6=12 and 4*3=12. The primes of a factor tree for 12 are the same regardles if we start the factor tree with 2*6 or 4*3.
## How To Verify If Prime Factors Of 20 Are Correct Answers
To know if we got the correct prime factors of 20 we have to get the prime factorization of 20 which is 2 * 2 * 5. Because when you multiply the primes of the prime factorization the answer has to be equal with 20.
After having checked the prime factorization we can now safely say that we got all prime factors.
## General Mathematical Properties Of Number 20
20 is a composite number. 20 is a composite number, because it has more divisors than 1 and itself. This is an even number. 20 is an even number, because it can be divided by 2 without leaving a comma spot. This also means that 20 is not an odd number. When we simplify Sin 20 degrees we get the value of sin(20)=0.91294525072763. Simplify Cos 20 degrees. The value of cos(20)=0.40808206181339. Simplify Tan 20 degrees. Value of tan(20)=2.2371609442247. When converting 20 in binary you get 10100. Converting decimal 20 in hexadecimal is 14. The square root of 20=4.4721359549996. The cube root of 20=2.7144176165949. Square root of √20 simplified is 2√5. All radicals are now simplified and in their simplest form. Cube root of ∛20 simplified is 20. The simplified radicand no longer has any more cubed factors.
## Determine Prime Factors Of Numbers Smaller Than 20
Learn how to calculate primes of smaller numbers like:
## Determine Prime Factors Of Numbers Bigger Than 20
Learn how to calculate primes of bigger numbers such as:
## Single Digit Properties For Number 20 Explained
• Integer 2 properties: 2 is the first of the primes and the only one to be even(the others are all odd). The first issue of Smarandache-Wellin in any base. Goldbach's conjecture states that all even numbers greater than 2 are the quantity of 2 primes. It is a complete Harshad, which is a integer of Harshad in any expressed base. The third of the Fibonacci sequence, after 1 and before 3. Part of the Tetranacci Succession. Two is an oblong figure of the form n(n+1). 2 is the basis of the binary numbering system, used internally by almost all computers. Two is a number of: Perrin, Ulam, Catalan and Wedderburn-Etherington. Refactorizable, which means that it is divisible by the count of its divisors. Not being the total of the divisors proper to any other arithmetical value, 2 is an untouchable quantity. The first number of highly cototent and scarcely totiente (the only one to be both) and it is also a very large decimal. Second term of the succession of Mian-Chowla. A strictly non-palindrome. With one exception, all known solutions to the Znam problem begin with 2. Numbers are divisible by two (ie equal) if and only if its last digit is even. The first even numeral after zero and the first issue of the succession of Lucas. The aggregate of any natural value and its reciprocal is always greater than or equal to 2.
• Integer 0 properties: 0 is the only real figure that is neither positive nor negative. Sometimes it is included in natural numbers where it can be considered the only natural in addition to the one to be neither first nor composed, as well as the minimum of natural numbers(that is, no natural digit precedes the 0). In an oriented line (which makes a point on the straight line correspond to each real number, preserving also the relation of order), the 0 coincides conventionally with the origin. Since it can be written in the form 2k, with con k integer, 0 is called even. It is both a figure and a numeral. In set theory, the zero is the cardinality of the empty set. In fact, in certain axiomatic mathematical developments derived from set theories, zero is defined as the empty set. In geometry, the size of a point is 0. Zero is the identity element of an additive group or additive identity in a ring.
## Finding All Prime Factors Of A Number
We found that 20 has 2 primes. The prime factors of 20 are 2, 5. We arrived to this answer by using the factor tree. However we could have also used upside down division to get the factorization primes. There are more that one method to factorize a integer.
## List of divisibility rules for finding prime factors faster
Rule 1: If the last digit of a figure is 0, 2, 4, 6 or 8 then it is an even number. All even numbers are divisible by 2.
Rule 2: If the sum of digits of a integer is divisible by 3 then the figure is also divisible by 3 and 3 is a prime factor(example: the digits of 12 are 1 and 2 so 1+2=3 and 3 is divisible by 3, meaning that 12 is divisible by 3). The same logic also works for 9.
Rule 3: If the last two digits of a number are 00 then this integer is divisible by 4(example: we know that 124=100+24 and 100 has two zeros in the end making it divisible with 4. We also know that 4 is divisible with 24). In order to use this rule to it's fullest it is best to know multiples of 4.
Rule 4: If the last digit of a number is 0 or 5 then 5 it is divisible by 5.
Rule 5: All integers that are divisible by both 2 and 3 are also divisible by 6. This is logical because 2*3=6.
## What Are Prime Factors Of A Number?
All numbers that are only divisible by one and itself are called prime factors in mathematics. A prime factor is a figure that has only two factors(one and itself).
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# Problem of the Week
## Updated at Mar 26, 2018 4:21 PM
For this week we've brought you this algebra problem.
How can we factor $$8{x}^{2}+2x-28$$?
Here are the steps:
$8{x}^{2}+2x-28$
1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$8{x}^{2}$$, $$2x$$, and $$-28$$?It is $$2$$.2 What is the highest degree of $$x$$ that divides evenly into $$8{x}^{2}$$, $$2x$$, and $$-28$$?It is 1, since $$x$$ is not in every term.3 Multiplying the results above,The GCF is $$2$$.To get access to all 'How?' and 'Why?' steps, join Cymath Plus!GCF = $$2$$2 Factor out the GCF. (Write the GCF first. Then, in parentheses, divide each term by the GCF.)$2(\frac{8{x}^{2}}{2}+\frac{2x}{2}-\frac{28}{2})$3 Simplify each term in parentheses.$2(4{x}^{2}+x-14)$4 Split the second term in $$4{x}^{2}+x-14$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$4\times -14=-56$2 Ask: Which two numbers add up to $$1$$ and multiply to $$-56$$?$$8$$ and $$-7$$3 Split $$x$$ as the sum of $$8x$$ and $$-7x$$.$4{x}^{2}+8x-7x-14$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$2(4{x}^{2}+8x-7x-14)$5 Factor out common terms in the first two terms, then in the last two terms.$2(4x(x+2)-7(x+2))$6 Factor out the common term $$x+2$$.$2(x+2)(4x-7)$Done2*(x+2)*(4*x-7)
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2008 AMC 12A Problems/Problem 13
The following problem is from both the 2008 AMC 12A #13 and 2008 AMC 10A #16, so both problems redirect to this page.
Problem
Points $A$ and $B$ lie on a circle centered at $O$, and $\angle AOB = 60^\circ$. A second circle is internally tangent to the first and tangent to both $\overline{OA}$ and $\overline{OB}$. What is the ratio of the area of the smaller circle to that of the larger circle?
$\mathrm{(A)}\ \frac{1}{16}\qquad\mathrm{(B)}\ \frac{1}{9}\qquad\mathrm{(C)}\ \frac{1}{8}\qquad\mathrm{(D)}\ \frac{1}{6}\qquad\mathrm{(E)}\ \frac{1}{4}$
Solution 1
$[asy]size(200); defaultpen(fontsize(10)); pair O=(0,0), A=(3,0), B=(3/2,3/2*3^.5), C=(3^.5,1), D=(3^.5,0), F=(1.5*3^.5,1.5), G=(2*3^.5,2); picture p = new picture; draw(p,Circle(O,0.2)); clip(p,O--C--A--cycle); add(p); draw(Circle(O,3)); dot(A); dot(B); dot(C); dot(O); draw(A--O--B); draw(O--C--D); draw(C--F); draw(D-(0.2,0)--D-(0.2,-0.2)--D-(0,-0.2)); draw(Circle(C,1)); label("$$30^{\circ}$$",(0.65,0.15),O); label("$$r$$",(C+D)/2,E); label("$$2r$$",(O+C)/2,NNE); label("$$O$$",O,SW); label("$$r$$",(C+F)/2,SE); label("$$R$$",(O+A)/2-(0,0.3),S); label("$$P$$",C,NW); label("$$A$$",(3,0), A); label("$$B$$",(3/2,3/2*3^.5), B); label("$$C$$",(1.5*3^.5,1.5), G); label("$$Q$$",D,SE);[/asy]$ Let $P$ be the center of the small circle with radius $r$, and let $Q$ be the point where the small circle is tangent to $OA$. Also, let $C$ be the point where the small circle is tangent to the big circle with radius $R$.
Then $PQO$ is a right triangle. Angle $POQ$ is $30$ degrees because line $OP$ bisects angle $AOB$ (this can be proved by dropping a perpendicular line from $P$ to line $OB$, letting their intersection be point $S$, and proving triangles $PQO$ and $PSO$ congruent), meaning that $PQO$ is a $30-60-90$ triangle. Therefore, $OP=2PQ$.
Since $OP=OC-PC=OC-r=R-r$, we have $R-r=2PQ$, or $R-r=2r$, or $\frac{1}{3}=\frac{r}{R}$.
Ratio of areas of circles is ratio of radii squared, so the answer is $\left(\frac{1}{3}\right)^2 = \frac{1}{9} \Rightarrow \boxed{B}$
Solution 2
Like in Solution 1, let $P$ be the center of the small circle with radius $r$, and let $Q$ be the point where the small circle is tangent to $OA$.
Let $N$ be the tangency point of the two circles. As shown in Solution 1, $POQ = 30$ degrees, so angle $NOA$ is also $30$ degrees. Let the line tangent to the two circles at $N$ intersect $OA$ and $OB$ at points $C$ and $D$, respectively. Since line $CD$ is tangent to circles $O$ and $P$, it must be perpendicular to $ON$, meaning that angle $ONC$ must be $90$ degrees. Because angle $NOA$ is $30$ degrees, angle $DCO$ is $180-30-90 = 60$ degrees. Angle $DOC$ is also $60$ degrees, so triangle $DOC$ is equilateral.
Note that an equilateral triangle's incenter is also its centroid. This means the center of the inscribed circle is also the centroid. From properties of median lengths, the radius of the large circle is 3 times the radius of the small circle.
Then the ratio of areas will be $\frac{1}{3}$ squared, or $\frac{1}{9}\Rightarrow \boxed{\text{B}}$.
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# 10.2: Factoring Special Cases
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Algebra I, Chapter 9, Lesson 6.
In this activity, you will explore:
• factoring a perfect-square trinomial
• factoring a difference of squares
• using geometry to prove rules for factoring special quadratic expressions
## Problem 1 - Factoring a Perfect-Square Trinomial
Any trinomial of the form \begin{align*}a^2 + 2ab + b^2\end{align*} is a perfect-square trinomial. If you recognize a perfect-square trinomial, you can factor it immediately as \begin{align*}(a + b)^2\end{align*}.
To see why \begin{align*}a^2 + 2ab + b^2 = (a + b)^2\end{align*} , start the CabriJr app by pressing the APPS button and choosing it from the menu.
Open the file FACTOR1 by pressing Y= to open then F1: File menu, choosing Open, and choosing it from the list.
This file shows \begin{align*}2\end{align*} squares and \begin{align*}2\end{align*} rectangles, with their dimensions labeled.
What is the area of each shape? On the screenshot at right, label each shape with its area.
• Arrange the shapes to form a square. To move a shape, move the cursor over it (so that the entire shape becomes a moving dashed line) and press ALPHA to grab it, then move it with the arrow keys. When the shape is positioned where you want it, press ENTER to let it go.
• The area of this square is equal to the sum of the areas of the shapes that make it up. What is the area of the square? Have you seen this trinomial before?
• How long is one side of the square?
• Using the formula \begin{align*}A = s^2\end{align*} for the area of a square with side length \begin{align*}s\end{align*} what is the area of this square?
You have shown that the area of this square is equal to \begin{align*}a^2 + 2ab + b^2\end{align*} and also equal to \begin{align*}(a + b)^2\end{align*}. Therefore \begin{align*}a^2 + 2ab + b^2 = (a + b)^2\end{align*} You have proved the rule for factoring a perfect-square trinomial!
## Problem 2 - Factoring a Difference of Squares
Any trinomial of the form \begin{align*}m^2 - n^2\end{align*} is a difference of squares. If you recognize a difference of squares, you can factor it immediately as \begin{align*}(m + n)(m - n)\end{align*}.
To see why \begin{align*}m^2 - n^2 = (m + n)(m - n)\end{align*}, start the CabriJr app by pressing the \begin{align*}A\end{align*} button and choosing it from the menu.
Open the file FACTOR2 by pressing \begin{align*}Y=\end{align*} to open then F1: File menu, choosing Open, and choosing it from the list.
This file shows \begin{align*}2\end{align*} squares with their dimensions labeled \begin{align*}m\end{align*} and \begin{align*}n\end{align*}.
What is the area of each square? On the screenshot at right, label each square with its area.
How can you represent the area \begin{align*}m^2 - n^2\end{align*} with these squares? Move the \begin{align*}n^2\end{align*} square on top of the \begin{align*}m^2\end{align*} rectangle so that their corners align. If you imagine cutting the smaller square out of the larger square, the \begin{align*}L-\end{align*}shaped area that remains is equal to \begin{align*}m^2 - n^2\end{align*}.
We know that the area of the \begin{align*}L-\end{align*}shape is \begin{align*}m^2 - n^2\end{align*}, but there is also another way to find its area: by taking it apart and rearranging the pieces into a single long rectangle.
Open the CabriJr file FACTOR3, which shows the same shapes, but with the \begin{align*}L-\end{align*}shaped area \begin{align*}(m^2 - n^2)\end{align*} divided into two rectangles.
Rotate the smaller rectangle about point \begin{align*}P\end{align*} (at the bottom of the screen) clockwise \begin{align*}90^\circ\end{align*}. Press TRACE to open the F4: Transform menu and choose Rotation. Move the cursor over the rectangle to highlight it and press \begin{align*}e\end{align*} to choose it. Then move the cursor to the point you want to rotate around and press ENTER. Finally, mark the angle of rotation by choosing points \begin{align*}A, B\end{align*}, and \begin{align*}C\end{align*} in turn.
Hide the original small rectangle and the vertices of the rotated image. (Press GRAPH to open the F5: Appearance menu and choose Hide/Show \begin{align*}>\end{align*} Objects. Then choose the rectangle and vertices you want to hide.) Now there are two rectangles whose combined area is equal to the area of the original \begin{align*}L-\end{align*}shape.
Move the larger rectangle (the small rectangle cannot be moved) alongside the rotated image to form one long rectangle.
What are the dimensions of the long rectangle?
Using the formula \begin{align*}A = lw\end{align*} for the area of a rectangle and these dimensions, what is the area of this rectangle?
You have shown that the \begin{align*}L-\end{align*}shaped area is equal to \begin{align*}m^2 - n^2\end{align*} and also equal to \begin{align*}(m + n)(m - n)\end{align*}. Therefore \begin{align*}m^2 - n^2 = (m + n)(m - n)\end{align*}. You have proved the rule for factoring a difference of squares!
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# How to Convert Units
Author Info
Updated: December 19, 2019
With all of the different measuring systems in the world, converting units can come in handy. You'll need to understand fractions, unless you are working with the metric system. No matter which system you use, always carefully write your units at every step to keep track of where you are.
### Method 1 of 3: Converting Units
1. 1
Compare your two units. The two units must measure the same thing. For example, in the problem "convert 2 inches into centimeters," both inches and centimeters measure length. If your units measure two different things (like length and weight), you can't convert between them.
• People often get confused about length, area, and volume, which are three different things. Remember that "square" or "2" means area, and "cubic" or "3" means volume.
• You can also write this example as 2 in. = ? cm.
2. 2
Look up the conversion. Before you can do the math, you need to know how much larger one unit is than the other.[1] If the conversion you find has many decimal places, round to the nearest significant digit. If you don't know what a significant digit is, round to the second or third digit.
• For example, if you need to convert 2 inches to centimeters, you need to know that 1 inch = 2.54 centimeters.
3. 3
Write the conversion as a fraction. Write this conversion as a fraction, including units. Put the unit you start with on bottom (the denominator), and the unit you're converting to on top (the numerator).[2]
• For example, write 2.54 cm/1 in.. You can read this as "2.54 centimeters per inch".
4. 4
Write a multiplication problem with your original number and the fraction. Multiplying these two numbers together will give you your answer. To start this, write out the multiplication problem, with units after each value.[3]
• 2 in. x 2.54 cm/1 in. = ?
5. 5
Solve the multiplication problem. It's important to keep track of your units while you do this. Every unit in the equation should still be there during each step.
• 2 in. x 2.54 cm/1 in.
• = (2 in x 2.54 cm)/1 in.
• = (5.08 in. x cm.)/ in.
6. 6
Cancel units that appear on top and bottom. If a unit is on the top and bottom of the fraction, cross it out. Whatever you're left with should be your answer.[4]
• (5.08 in. x cm.)/in.
• = 5.08 cm.
7. 7
Correct mistakes. If your units don't cancel, start over and try again. You might have put the wrong half of the fraction on top.
• For example, if you multiplied 2 inch. x (1 in. / 2.54 cm), your answer will have units of "in. x in. / cm," which doesn't make any sense. You realize the inches will cancel if you flip the fraction, so you try again with 2 inch x (2.54 cm / 1 in.).
### Method 2 of 3: Converting Values with Multiple Units
1. 1
Write down your problem. Figure out exactly what the problem is asking, and write it as a math problem. Here's an example:
• If a bicycle is moving 10 miles per hour, how many feet does it travel in one minute?
• Write this as "10 miles / hour = ? feet / minute" or "10 miles / h = ? ft / min".
2. 2
Find the conversion for one unit. Remember, you can only convert between two units that measure the same thing. In our example, we have units that measure length (miles and feet), and units that measure time (hours and minutes). Start with one pair and find the conversion between them.
• For example, 1 mile = 5,280 feet.
3. 3
Multiply your number by the conversion fraction. Just as describe in the section above, we can write the conversion as a fraction to cancel out units. Make sure to include every unit in your calculations.
• 10 miles / h) x 5280 ft / mile
• = 52800 miles x ft / h x miles
4. 4
Cancel out your units. One of your units should be on the top and the bottom, so you can cancel them. You aren't done yet, but you're getting closer.
• 52800 miles x ft / h x miles
• = 52800 ft / h
5. 5
Multiply with another conversion fraction the same way. Choose a unit that hasn't been converted, and write down the conversion as a fraction. Remember to arrange the fraction so the units will cancel out during multiplication.
• In our example, we still need to convert hours to minutes. 1 hour = 60 minutes.
• Right now, we have 52800 ft / h. Since hours (h) is on the bottom, we want our new fraction to have hours on top: 1 hour / 60 minutes.
• 52800 ft / h x 1 h / 60 min
• = 880 ft x h / h x min
6. 6
Cancel units. Another unit should cancel out, just as it did before.
• 880 ft x h / h x min
• = 880 ft / min
7. 7
Repeat until the conversion is done. If your answer is in the units you were trying to convert to, you're done with the problem. If you're not there yet, convert another unit with the same method.
• Once you're used to this method, you can write all the conversions on one line.[5] For example, we could solve our example problem like this:
• 10 miles/h x 5280 ft/mile x 1 h/60 min
• =10 miles/h x 5280 ft/mile x 1 h/60 min
• = 10 x 5280 ft x 1/60 min
• = 880 ft / min.
### Method 3 of 3: Converting in the Metric System
1. 1
Understand the metric system. The metric system, also called the decimal system, is designed for easy conversion. To convert from one metric unit to another, you only have to work with round numbers: 10, 100, 1000, and so on.
2. 2
Learn the prefixes. Metric unit of measurements use prefixes to show how big or small the measurement is. Here's an example with units of weight, but all other metric units use the same prefixes.[6] The prefixes below are in italics, but you can do most conversions with just the most common one, in bold.
• kilogram = 1000 grams
• hectogram = 100 grams
• dekagram = 10 grams
• gram = 1 gram
• decigram = 0.1 grams (one tenth)
• centigram = 0.01 grams (one hundredth)
• milligram = 0.001 grams (one thousandth)
3. 3
Use the prefixes in conversion. If you know the prefixes, you don't need to look up the conversion every time you need to convert. The prefixes already tell you the conversion.[7] Here are a couple examples:
• To convert from kilometers to meters: kilo means 1000, so 1 kilometer = 1000 meters.
• To convert from grams to milligrams: milli means .001, so 1 milligram = .001 gram.
4. 4
Move the decimal point instead of doing calculations. The best part about metric conversion is skipping all the calculations. Multiplying a number by 10 is the same as moving the decimal point to the right. Dividing a number by 10 is the same as moving the decimal point to the left. Here's an example of how to use this:
• Problem: convert 65.24 kilograms to grams.
• 1 kilogram = 1000 grams. Count the number of zeroes: three. That means we multiply by 10 three times, or just move the decimal three spaces to the right.
• 65.24 x 10 = 652.4 (multiplied once)
• 652.4 x 10 = 6524 (twice)
• 6524 x 10 = 65240 (three times)
• The answer is 65240 grams.
5. 5
Practice more difficult problems. It gets a little trickier when both the units you're converting between have prefixes. The easiest way to solve this is to convert to the base unit (no prefix) first, then convert to the final unit. Here's an example:
• Problem: convert 793 milliliters to decaliters.
• 1 milliliter = 0.001 liters. There are three zeroes, so we move the decimal point three to the left. (Remember, move left when dividing.)
• 793 milliliters = 0.793 liters
• 10 liters = 1 dekaliter, so 1 liter = 0.1 dekaliters. There is one zero, so move the decimal point one to the left.
• 0.793 liters = 0.0793 dekaliters.
6. 6
Check your answer. The mistake that's easiest to make is multiplying instead of dividing, or vice versa. When you get your final answer, check that it makes sense:
• If you converted to a larger unit, your number should get smaller. (Just like 12 inches converts to 1 foot.)
• If you converted to a smaller unit, your number should get larger. (Just like 1 foot converts to 12 inches.)
## Community Q&A
Search
• Question
How do I properly convert a unit of weight to a unit of time?
Donagan
That is not possible.
• Question
How do I convert 1.5 liters to cubic centimeters?
Donagan
Multiply liters by 1,000. In this case, 1.5 liters equals (1.5)(1000) = 1,500 cubic centimeters.
• Question
How do I convert square feet to cubic feet?
Donagan
You would need to get another dimension in feet (such as the depth or height) and multiply that dimension by the square footage to get cubic feet.
• Question
How do I convert square feet to square yards?
Donagan
There are nine square feet in one square yard, so divide square footage by 9.
• Question
How do I convert mm to inches?
Donagan
Divide millimeters by 25.4.
• Question
How do I convert 17 meters and 80 centimeters into centimeters?
Donagan
Multiply 17 by 100, then add 80.
• Question
How do I convert cubic meters to liters?
Donagan
Multiply cubic meters by 1,000, because there are 1,000 liters in one cubic meter.
• Question
How do I enlarge inches x %. (Example: 3 x 200%)
Divide the % by 100 (in your example, 200/100 = 2). Then, times x by the previous numeral found.
• Question
How do I convert pounds to bushels?
Donagan
While the bushel is a unit of weight, there is actually no conversion factor available. Pounds do not convert directly to bushels. It depends on what commodity is being measured.
• Question
How do I convert cm^3 to m^3?
Donagan
Divide cm³ by one million to get m³.
200 characters left
## Tips
• Recipes can be hard to convert. Some countries measure solid ingredients by weight (grams) and others measure them by volume (teaspoons, cups). Search for a calculator online that converts for that specific ingredient, using its density.
Thanks!
wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 22 people, some anonymous, worked to edit and improve it over time. Together, they cited 7 references. This article has also been viewed 180,103 times.
Co-authors: 22
Updated: December 19, 2019
Views: 180,103
Categories: Conversion Aids
Article SummaryX
To convert units, verify that the two units you’re converting measure the same thing, like length or width, and look up the conversion rate for both units. Next, write the conversion as a fraction, with the starting unit on the bottom and the unit you’re converting to on the top. Then, multiply your original number by the fraction. Finally, cancel out any units that appear on both the top and bottom of the fraction. The remaining number and unit is your conversion! For tips on converting multiple units, read on!
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33 to 15 percent increase or decrease
Using this tool you can find the percent Decrease for any value. So, we think you reached us looking for answers to questions like:
1) What is the percentage Decrease from 33 to 15? 2) What is the absolute Decrease from 33 to 15. Or may be: 33 to 15 percent increase or decrease
See the solution to these problems just after the Percentage Decrease Calculator below.
Percentage Decrease Calculator
Please change the two first boxes below and get answers to any combination of values:
What is the percent Decrease from
to ?
How to calculate percent change - Step by Step
Percent increase or decrease measures percent changes between two values. Use this calculator when comparing an old value to a new value.
The Percent Change Calculator uses this formula:
% change = ( New Value - Old Value ) × 100% |Old Value|
Where |Old Value| represents the absolute value of the reference (this is made in order to work well with both positive and negative values of Old Value and New Value. Note that percent change and relative change mean the same thing. It can be a percent increase or a percent decrease depending on the new and the old values. If the new value is greater than the old value, the result will be positive and we will have a increase. On the other hand, if the new value is smaller than old value, the result will be negative and we will have a we have a decrease. The old value, as a reference, may be: a theoretical, the actual, the correct, an accepted, an optimal, the starting, and so on. See more about percent percent change here.
Here are the solutions to the questions stated above:
1) What is the percentage decrease from 33 to 15?
Use the above formula to find the percent change. So, replacing the given values, we have
Percent change = [(15 - 33) / 33] x 100 = -54.545454545455 % (decrease)
Where: 33 is the old value and 15 is the new value. In this case we have a % of decrease because the new value is smaller than the old value.
2) What is the absolute Decrease from 33 to 15?
This problem is not about percent or relative change, but about absolute change. Its solution is very simple:
Absolute change, or
change = new value - old value
= 15 - 33 = -18 (decrease)
Disclaimer
While every effort is made to ensure the accuracy of the information provided on this website, we offer no warranties in relation to these informations.
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# 2021 AMC 10A Problems/Problem 11
## Problem
For which of the following integers $b$ is the base-$b$ number $2021_b - 221_b$ not divisible by $3$?
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8$
## Solution 1
We have \begin{align*} 2021_b - 221_b &= (2021_b - 21_b) - (221_b - 21_b) \\ &= 2000_b - 200_b \\ &= 2b^3 - 2b^2 \\ &= 2b^2(b-1), \end{align*} which is divisible by $3$ unless $b\equiv2\pmod{3}.$ The only choice congruent to $2$ modulo $3$ is $\boxed{\textbf{(E)} ~8}.$
~MRENTHUSIASM
## Solution 2 (Easy)
Vertically subtracting $$2021_b - 221_b$$ we see that the ones place becomes 0, and so does the $b^1$ place. Then, we perform a carry (make sure the carry is in base $b$!). Let $b-2 = A$. Then, we have our final number as $$1A00_b$$
Now, when expanding, we see that this number is simply $b^3 - (b - 2)^2$.
Now, notice that the final number will only be congruent to $$b^3-(b-2)^2\equiv0\pmod{3}$$ if either $b\equiv0\pmod{3}$, or if $b\equiv1\pmod{3}$ (because note that $(b - 2)^2$ would become $\equiv1\pmod{3}$, and $b^3$ would become $\equiv1\pmod{3}$ as well, and therefore the final expression would become $1-1\equiv0\pmod{3}$. Therefore, $b$ must be $\equiv2\pmod{3}$. Among the answers, only 8 is $\equiv2\pmod{3}$, and therefore our answer is $\boxed{\textbf{(E)} ~8}.$
- icecreamrolls8
## Solution 3 (Educated Guess)
Note that choices $\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},\textbf{(E)}$ are congruent to $0,1,0,1,2$ modulo $3,$ respectively. Since only one of the choices is correct, we pick $\boxed{\textbf{(E)} ~8}$ due to its uniqueness.
~MRENTHUSIASM
## Video Solution (Simple and Quick)
~ Education, the Study of Everything
## Video Solution
~North America Math Contest Go Go Go
~savannahsolver
~IceMatrix
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# Table of 315
## Introduction:-A table is an outline that shows the duplications of two numbers. The increase is rehashed expansion. A table presumably makes over a long time back by the ancient Babylonian.Augmentation is one of the essential numerical tasks which is educated to understudies at an early age and it is extremely useful in doing numerical issues. A duplication outline saves investment in doing estimations.It is additionally useful for fostering the psychological capacity of youngsters. Also understanding arithmetic issues becomes more straightforward.
• Table
• Explanation
• Tricks to learn
• Solved examples
• FAQ's
This Story also Contains
1. Introduction:-
3. Table:-
4. Explanation:-
5. Trick To Learn:-
6. Solved Examples:-
## Table:-
Multiplier × multiplicand = Product 315 × 1 = 315 315 × 2 = 630 315 × 3 = 945 315 × 4 = 1260 315 × 5 = 1575 315 × 6 = 1890 315 × 7 = 2205 315 × 8 = 2520 315 × 9 = 2835 315 × 10 = 3150
The table of 315 can also be represented as follows
315 = 315
315 + 315 = 630
315 + 315 + 315 = 945
315 + 315 + 315 + 315 = 1260
315 + 315 + 315 + 315 + 315 = 1575
315 + 315 + 315 + 315 + 315 + 315 = 1890
315 + 315 + 315 + 315 + 315 + 315 + 315 = 2205
315 + 315 + 315 + 315 + 315 + 315 + 315 + 315 = 2520
315 + 315 + 315 + 315 + 315 + 315 + 315 + 315 + 315 = 2835
315 + 315 + 315 + 315 + 315 + 315 + 315 + 315 + 315 + 315 = 3150
## Explanation:-
• We can multiply or repeat addition to create a table. For instance, when we multiply 5 by 3 and 3 times add 5, we can obtain the same.
• When we multiply two numbers. For example, 3\times5=15
3 × 5 = 15 = 5 + 5 + 5 = 15
• The multiplication sign can be denoted as "×”,”*”,”.”.
• The number we must multiply is known as the multiplier, the number we must multiply with is known as multiplicant and the multiplication of both numbers is known as the product.
## Trick To Learn:-
• If we want 315×2 we have to add 33 twice. For example 315+315=630.
• Same if we want 315×3 we have to add 33 thrice times. For example 315+315+315=945 and the same procedure we can do with others too.
## Solved Examples:-
1. What do we get if we multiply 315 by 12?
Solution: We can multiply 315 by 10 and then add 315+315=630 and then we can add both 3150+630
Now the multiplication of 315×12=3780.
2. What do we get if we multiply 315 by 0?
Solution: It is the basic knowledge that if we multiply any number by zero. It becomes zero itself.
So the final answer is 315×0=0.
3. Ramesh wants to buy a book for 315 rs and wants to buy it for his friends as well. So how much do 4 books cost?
solution: Cost of one book = 315
Total number of books = 4
The total cost of 4 books = 1260
4. What is the cost of two shirts if one shirt costs 315 Rs?
Solution: Cost of a shirt = 315
Total number of shirts = 2
The total cost of 2 shirts = 630
5. What do we get if we multiply 315 by 100?
Solution: If any number ending in zero is multiplied by 315. So we can write zeros after the number.
315× 100 = 31500.
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# Definition:Fraction
## Definition
A fraction is an expression representing a quotient of one number (or expression) by another number (or expression).
It is usually expressed in the form:
$\dfrac a b$
or:
$a / b$
where $a$ and $b$ are either numbers or expressions.
The term fraction is usually encountered when $a$ and $b$ are integers.
In this case, the fraction $\dfrac a b$ represents a rational number.
### Vulgar Fraction
A vulgar fraction is a fraction representing a rational number whose numerator and denominator are both integers.
### Proper Fraction
A proper fraction is a fraction representing a rational number whose absolute value is less than $1$, expressed in the form $r = \dfrac p q$, where $p$ and $q$ are integers.
### Improper Fraction
An improper fraction is a fraction representing a rational number whose absolute value is greater than $1$.
Specifically, when expressed in the form $r = \dfrac p q$, where $p$ and $q$ are integers such that (the absolute value of) the numerator is greater than (the absolute value of) the denominator: $\size p > \size q$.
### Mixed Fraction
A mixed fraction is a representation of a rational number whose absolute value is greater than $1$, expressed in the form $r = n \frac p q$ where:
$n$ is an integer
$\dfrac p q$ is a proper fraction, that is, $p$ and $q$ are integers such that $p < q$.
### Complex Fraction
A complex fraction is a fraction such that the numerator or denominator or both are themselves fractions.
## Also defined as
Many sources define a fraction solely in the context of numbers.
Specifically, a fraction in such a context will be used to denote a rational number $\dfrac a b$ such that both $a$ and $b$ are integers.
Such sources will typically also demand that the fraction specifically represent a non-integer.
Thus $\dfrac 3 1$ and $\dfrac 4 2$ will not be considered as actual fractions, as they represent the integers $3$ and $2$ respectively.
## Terms of Fraction
The terms of a fraction are referred to as the numerator and the denominator:
### Numerator
The term $a$ is known as the numerator of $\dfrac a b$.
### Denominator
The term $b$ is known as the denominator of $\dfrac a b$.
A helpful mnemonic to remember which goes on top and which goes on the bottom is "Numerator Over Denominator", which deserves a "nod" for being correct.
## Examples
$(1): \quad \dfrac 1 2$ is a proper fraction.
$(2): \quad \dfrac 5 2$ is an improper fraction.
It can be expressed as a mixed fraction as follows:
$\dfrac 5 2 = \dfrac {4 + 1} 2 = \dfrac 4 2 + \dfrac 1 2 = 2 \frac 1 2$
$(3): \quad \dfrac {24} {36}$ is a proper fraction, although not in canonical form.
It is found that when $\dfrac {24} {36}$ is expressed in canonical form:
$\dfrac {24} {36} = \dfrac {12 \times 2} {12 \times 3} = \dfrac 2 3$
its denominator is not $1$.
Hence $\dfrac {24} {36}$ is indeed a vulgar fraction.
## Also known as
Some sources introduce the more unwieldy term fractional number for fraction.
This will be the case only when such a fraction denotes a rational number.
## Also see
• Results about fractions can be found here.
## Historical Note
The consideration of fraction was the next development of the concept of a number after the natural numbers.
They arose as a matter of course from the need to understand the process of measurement.
The convention where a bar is placed between the numerator and denominator was introduced to the West by Fibonacci, following the work of the Arabic mathematicians.
Previous to this, fractions were written by the Hindu mathematicians without the bar.
Thus $\dfrac 3 4$ would have been written $\ds {3 \atop 4}$.
## Linguistic Note
The word fraction derives from the Latin fractus meaning broken.
This is in antithesis to the concept of integer, which derives from the Latin for untouched, in the sense of whole, or unbroken.
Colloquially, informally and rhetorically, the word fraction is typically used to mean a (small) part of a whole, and not in the sense of improper fraction.
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# Solve for: integral of (sqrt(x^3)-\sqrt[3]{x})/(6\sqrt[4]{x)} x
## Expression: $\int{ \frac{ \sqrt{ {x}^{3} }-\sqrt[3]{x} }{ 6\sqrt[4]{x} } } \mathrm{d} x$
Use the property of integral $\begin{array} { l }\int{ a \times f\left( x \right) } \mathrm{d} x=a \times \int{ f\left( x \right) } \mathrm{d} x,& a \in ℝ\end{array}$
$\frac{ 1 }{ 6 } \times \int{ \frac{ \sqrt{ {x}^{3} }-\sqrt[3]{x} }{ \sqrt[4]{x} } } \mathrm{d} x$
$\frac{ 1 }{ 6 } \times \int{ \frac{ x\sqrt{ x }-\sqrt[3]{x} }{ \sqrt[4]{x} } } \mathrm{d} x$
Use $\sqrt[n]{{a}^{m}}={a}^{\frac{ m }{ n }}$ to transform the expression
$\frac{ 1 }{ 6 } \times \int{ \frac{ x\sqrt{ x }-{x}^{\frac{ 1 }{ 3 }} }{ \sqrt[4]{x} } } \mathrm{d} x$
Use $\sqrt[n]{{a}^{m}}={a}^{\frac{ m }{ n }}$ to transform the expression
$\frac{ 1 }{ 6 } \times \int{ \frac{ x\sqrt{ x }-{x}^{\frac{ 1 }{ 3 }} }{ {x}^{\frac{ 1 }{ 4 }} } } \mathrm{d} x$
Use $\sqrt[n]{{a}^{m}}={a}^{\frac{ m }{ n }}$ to transform the expression
$\frac{ 1 }{ 6 } \times \int{ \frac{ x \times {x}^{\frac{ 1 }{ 2 }}-{x}^{\frac{ 1 }{ 3 }} }{ {x}^{\frac{ 1 }{ 4 }} } } \mathrm{d} x$
Calculate the product
$\frac{ 1 }{ 6 } \times \int{ \frac{ {x}^{\frac{ 3 }{ 2 }}-{x}^{\frac{ 1 }{ 3 }} }{ {x}^{\frac{ 1 }{ 4 }} } } \mathrm{d} x$
Separate the fraction into $2$ fractions
$\frac{ 1 }{ 6 } \times \int{ \frac{ {x}^{\frac{ 3 }{ 2 }} }{ {x}^{\frac{ 1 }{ 4 }} }-\frac{ {x}^{\frac{ 1 }{ 3 }} }{ {x}^{\frac{ 1 }{ 4 }} } } \mathrm{d} x$
Simplify the expression
$\frac{ 1 }{ 6 } \times \int{ {x}^{\frac{ 5 }{ 4 }}-\frac{ {x}^{\frac{ 1 }{ 3 }} }{ {x}^{\frac{ 1 }{ 4 }} } } \mathrm{d} x$
Simplify the expression
$\frac{ 1 }{ 6 } \times \int{ {x}^{\frac{ 5 }{ 4 }}-{x}^{\frac{ 1 }{ 12 }} } \mathrm{d} x$
Use the property of integral $\int{ f\left( x \right)\pmg\left( x \right) } \mathrm{d} x=\int{ f\left( x \right) } \mathrm{d} x\pm\int{ g\left( x \right) } \mathrm{d} x$
$\frac{ 1 }{ 6 } \times \left( \int{ {x}^{\frac{ 5 }{ 4 }} } \mathrm{d} x-\int{ {x}^{\frac{ 1 }{ 12 }} } \mathrm{d} x \right)$
Use $\begin{array} { l }\int{ {x}^{n} } \mathrm{d} x=\frac{ {x}^{n+1} }{ n+1 },& n≠-1\end{array}$ to evaluate the integral
$\frac{ 1 }{ 6 } \times \left( \frac{ 4{x}^{2}\sqrt[4]{x} }{ 9 }-\int{ {x}^{\frac{ 1 }{ 12 }} } \mathrm{d} x \right)$
Use $\begin{array} { l }\int{ {x}^{n} } \mathrm{d} x=\frac{ {x}^{n+1} }{ n+1 },& n≠-1\end{array}$ to evaluate the integral
$\frac{ 1 }{ 6 } \times \left( \frac{ 4{x}^{2}\sqrt[4]{x} }{ 9 }-\frac{ 12x\sqrt[12]{x} }{ 13 } \right)$
Distribute $\frac{ 1 }{ 6 }$ through the parentheses
$\frac{ 2{x}^{2}\sqrt[4]{x} }{ 27 }-\frac{ 2x\sqrt[12]{x} }{ 13 }$
Add the constant of integration $C \in ℝ$
$\begin{array} { l }\frac{ 2{x}^{2}\sqrt[4]{x} }{ 27 }-\frac{ 2x\sqrt[12]{x} }{ 13 }+C,& C \in ℝ\end{array}$
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# How do you find the square root of 13?
##### 2 Answers
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
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#### Answer
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#### Explanation
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#### Explanation:
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Describe your changes (optional) 200
14
Sep 18, 2016
#### Answer:
Find $\sqrt{13} \approx \frac{23382}{6485} \approx 3.60555127$ using a generalised continued fraction method.
#### Explanation:
Look for a generalised continued fraction of the form:
$\sqrt{13} = a + \frac{b}{2 a + \frac{b}{2 a + \frac{b}{2 a + \ldots}}}$
$\textcolor{w h i t e}{\sqrt{13}} = a + \frac{b}{a + \sqrt{13}}$
Multiply both ends by $\left(a + \sqrt{13}\right)$ to find:
$\textcolor{red}{\cancel{\textcolor{b l a c k}{a \sqrt{13}}}} + 13 = {a}^{2} + \textcolor{red}{\cancel{\textcolor{b l a c k}{a \sqrt{13}}}} + b$
Hence:
$b = 13 - {a}^{2}$
In order that our generalised continued fraction converges quickly choose a rational approximation $a$ slightly smaller than $\sqrt{13}$.
Note that $3 = \sqrt{9} < \sqrt{13} < \sqrt{16} = 4$ so linearly interpolating we find a good approximation:
$\sqrt{13} \approx 3.6 = \frac{18}{5}$
Note also that:
${\left(\frac{18}{5}\right)}^{2} = \frac{324}{25} < \frac{325}{25} = 13$
So let $a = \frac{18}{5}$ and $b = 13 - {a}^{2} = \frac{1}{25}$ to get:
$\sqrt{13} = \frac{18}{5} + \frac{\frac{1}{25}}{\frac{36}{5} + \frac{\frac{1}{25}}{\frac{36}{5} + \frac{\frac{1}{25}}{\frac{36}{5} + \ldots}}}$
We can truncate this continued fraction to get a rational approximation of any desired accuracy.
For example:
$\sqrt{13} \approx \frac{18}{5} + \frac{\frac{1}{25}}{\frac{36}{5}} = \frac{18}{5} + \frac{1}{180} = \frac{649}{180} = 3.60 \overline{5}$
Or:
$\sqrt{13} \approx \frac{18}{5} + \frac{\frac{1}{25}}{\frac{36}{5} + \frac{\frac{1}{25}}{\frac{36}{5}}} = \frac{23382}{6485} \approx 3.60555127$
Was this helpful? Let the contributor know!
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Answer
Write a one sentence answer...
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
Describe your changes (optional) 200
3
Sep 18, 2016
#### Answer:
Use a Newton Raphson method to find:
$\sqrt{13} \approx \frac{842401}{233640} \approx 3.60555127547$
#### Explanation:
Since $13$ is a prime number, there is no simpler form for its square root. $\sqrt{13}$ is an irrational number somewhere between $3 = \sqrt{9}$ and $4 = \sqrt{16}$.
Linearly interpolating, a reasonable first approximation would be:
$\sqrt{13} \approx 3.6 = \frac{18}{5}$
We can get better approximations from our initial one (call it ${a}_{0}$) using a Newton Raphson method.
A typical formula used to derive a more accurate approximation for $\sqrt{n}$ would be:
${a}_{i + 1} = \frac{{a}_{i}^{2} + n}{2 {a}_{i}}$
I prefer to separate ${a}_{i}$ into numerator ${p}_{i}$ and denominator ${q}_{i}$. So ${a}_{i} = {p}_{i} / {q}_{i}$ and we can iterate using the formulae:
$\left\{\begin{matrix}{p}_{i + 1} = {p}_{i}^{2} + n {q}_{i}^{2} \\ {q}_{i + 1} = 2 {p}_{i} {q}_{i}\end{matrix}\right.$
In our example, $n = 13$, ${p}_{0} = 18$, ${q}_{0} = 5$ and we find:
$\left\{\begin{matrix}{p}_{1} = {p}_{0}^{2} + 13 {q}_{0}^{2} = 324 + 13 \cdot 25 = 649 \\ {q}_{1} = 2 {p}_{0} {q}_{0} = 180\end{matrix}\right.$
If we stopped here our approximation would be:
$\sqrt{13} \approx \frac{649}{180} = 3.60 \overline{5}$
Let's try one more iteration:
$\left\{\begin{matrix}{p}_{2} = {p}_{1}^{2} + 13 {q}_{1}^{2} = 421201 + 13 \cdot 32400 = 842401 \\ {q}_{2} = 2 {p}_{1} {q}_{1} = 233640\end{matrix}\right.$
Stopping here, we have:
$\sqrt{13} \approx \frac{842401}{233640} \approx 3.60555127547$
Using a calculator:
$\sqrt{13} \approx 3.60555127546398929311$
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```Question 352211
{{{5(4w^2+20wg+25g^2)}}} Factor out the GCF {{{5}}}
Now let's focus on the inner expression {{{4w^2+20wg+25g^2}}}
------------------------------------------------------------
Looking at {{{4w^2+20wg+25g^2}}} we can see that the first term is {{{4w^2}}} and the last term is {{{25g^2}}} where the coefficients are 4 and 25 respectively.
Now multiply the first coefficient 4 and the last coefficient 25 to get 100. Now what two numbers multiply to 100 and add to the middle coefficient 20? Let's list all of the factors of 100:
Factors of 100:
1,2,4,5,10,20,25,50
-1,-2,-4,-5,-10,-20,-25,-50 ...List the negative factors as well. This will allow us to find all possible combinations
These factors pair up and multiply to 100
1*100
2*50
4*25
5*20
10*10
(-1)*(-100)
(-2)*(-50)
(-4)*(-25)
(-5)*(-20)
(-10)*(-10)
note: remember two negative numbers multiplied together make a positive number
Now which of these pairs add to 20? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 20
<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">100</td><td>1+100=101</td></tr><tr><td align="center">2</td><td align="center">50</td><td>2+50=52</td></tr><tr><td align="center">4</td><td align="center">25</td><td>4+25=29</td></tr><tr><td align="center">5</td><td align="center">20</td><td>5+20=25</td></tr><tr><td align="center">10</td><td align="center">10</td><td>10+10=20</td></tr><tr><td align="center">-1</td><td align="center">-100</td><td>-1+(-100)=-101</td></tr><tr><td align="center">-2</td><td align="center">-50</td><td>-2+(-50)=-52</td></tr><tr><td align="center">-4</td><td align="center">-25</td><td>-4+(-25)=-29</td></tr><tr><td align="center">-5</td><td align="center">-20</td><td>-5+(-20)=-25</td></tr><tr><td align="center">-10</td><td align="center">-10</td><td>-10+(-10)=-20</td></tr></table>
From this list we can see that 10 and 10 add up to 20 and multiply to 100
Now looking at the expression {{{4w^2+20wg+25g^2}}}, replace {{{20wg}}} with {{{10wg+10wg}}} (notice {{{10wg+10wg}}} adds up to {{{20wg}}}. So it is equivalent to {{{20wg}}})
{{{4w^2+highlight(10wg+10wg)+25g^2}}}
Now let's factor {{{4w^2+10wg+10wg+25g^2}}} by grouping:
{{{(4w^2+10wg)+(10wg+25g^2)}}} Group like terms
{{{2w(2w+5g)+5g(2w+5g)}}} Factor out the GCF of {{{2w}}} out of the first group. Factor out the GCF of {{{5g}}} out of the second group
{{{(2w+5g)(2w+5g)}}} Since we have a common term of {{{2w+5g}}}, we can combine like terms
So {{{4w^2+10wg+10wg+25g^2}}} factors to {{{(2w+5g)(2w+5g)}}}
So this also means that {{{4w^2+20wg+25g^2}}} factors to {{{(2w+5g)(2w+5g)}}} (since {{{4w^2+20wg+25g^2}}} is equivalent to {{{4w^2+10wg+10wg+25g^2}}})
note: {{{(2w+5g)(2w+5g)}}} is equivalent to {{{(2w+5g)^2}}} since the term {{{2w+5g}}} occurs twice. So {{{4w^2+20wg+25g^2}}} also factors to {{{(2w+5g)^2}}}
------------------------------------------------------------
So our expression goes from {{{5(4w^2+20wg+25g^2)}}} and factors further to {{{5(2w+5g)^2}}}
------------------
So {{{20w^2+100wg+125g^2}}} factors to {{{5(2w+5g)^2}}}
```
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# Difference between revisions of "2018 AIME I Problems/Problem 8"
## Problem
Let $ABCDEF$ be an equiangular hexagon such that $AB=6, BC=8, CD=10$, and $DE=12$. Denote by $d$ the diameter of the largest circle that fits inside the hexagon. Find $d^2$.
## Solution 1
First of all, draw a good diagram! This is always the key to solving any geometry problem. Once you draw it, realize that $EF=2, FA=16$. Why? Because since the hexagon is equiangular, we can put an equilateral triangle around it, with side length $6+8+10=24$. Then, if you drew it to scale, notice that the "widest" this circle can be according to $AF, CD$ is $7\sqrt{3}$. And it will be obvious that the sides won't be inside the circle, so our answer is $\boxed{147}$.
-expiLnCalc
## Solution 2
Like solution 1, draw out the large equilateral triangle with side length $24$. Let the tangent point of the circle at $\overline{CD}$ be G and the tangent point of the circle at $\overline{AF}$ be H. Clearly, GH is the diameter of our circle, and is also perpendicular to $\overline{CD}$ and $\overline{AF}$.
The equilateral triangle of side length $10$ is similar to our large equilateral triangle of $24$. And the height of the former equilateral triangle is $\sqrt{10^2-5^2}=5\sqrt{3}$. By our similarity condition, $\frac{10}{24}=\frac{5\sqrt{3}}{d+5\sqrt{3}}$
Solving this equation gives $d=7\sqrt{3}$, and $d^2=\boxed{147}$
~novus677
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# When Is the Standard Deviation Equal to Zero?
The sample standard deviation is a descriptive statistic that measures the spread of a quantitative data set. This number can be any non-negative real number. Since zero is a nonnegative real number, it seems worthwhile to ask, “When will the sample standard deviation be equal to zero?” This occurs in the very special and highly unusual case when all of our data values are exactly the same. We will explore the reasons why.
### Description of the Standard Deviation
Two important questions that we typically want to answer about a data set include:
• What is the center of the dataset?
• How spread out is the set of data?
There are different measurements, called descriptive statistics that answer these questions. For example, the center of the data, also known as the average, can be described in terms of the mean, median or mode. Other statistics, which are less well-known, can be used such as the midhinge or the trimean.
For the spread of our data, we could use the range, the interquartile range or the standard deviation. The standard deviation is paired with the mean to quantify the spread of our data. We can then use this number to compare multiple data sets. The greater our standard deviation is, then the greater the spread is.
### Intuition
So let's consider from this description what it would mean to have a standard deviation of zero. This would indicate that there is no spread at all in our data set. All of the individual data values would be clumped together at a single value. Since there would only be one value that our data could have, this value would constitute the mean of our sample.
In this situation, when all of our data values are the same, there would be no variation whatsoever. Intuitively it makes sense that the standard deviation of such a data set would be zero.
### Mathematical Proof
The sample standard deviation is defined by a formula. So any statement such as the one above should be proved by using this formula. We begin with a data set that fits the description above: all values are identical, and there are n values equal to x.
We calculate the mean of this data set and see that it is
x = (x + x +… + x)/n = nx/n = x.
Now when we calculate the individual deviations from the mean, we see that all of these deviations are zero. Consequently, the variance and also the standard deviation are both equal to zero too.
### Necessary and Sufficient
We see that if the data set displays no variation, then its standard deviation is zero. We may ask if the converse of this statement is also true. To see if it is, we will use the formula for standard deviation again. This time, however, we will set the standard deviation equal to zero. We will make no assumptions about our data set, but will see what setting s = 0 implies
Suppose that the standard deviation of a data set is equal to zero. This would imply that the sample variance s2 is also equal to zero. The result is the equation:
0 = (1/(n - 1)) ∑ (xi - x )2
We multiply both sides of the equation by n - 1 and see that the sum of the squared deviations is equal to zero. Since we are working with real numbers, the only way for this to occur is for every one of the squared deviations to be equal to zero. This means that for every i, the term (xi - x )2 = 0.
We now take the square root of the above equation and see that every deviation from the mean must be equal to zero. Since for all i,
xi - x = 0
This means that every data value is equal to the mean. This result along with the one above allows us to say that the sample standard deviation of a data set is zero if and only if all of its values are identical.
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# Construct a triangle with sides $5 \mathrm{~cm}, 6 \mathrm{~cm}$ and $7 \mathrm{~cm}$ and then another triangle whose sides are $\frac{5}{7}$ of the corresponding sides of the first triangle.
Given:
A triangle with sides $5 \mathrm{~cm}, 6 \mathrm{~cm}$ and $7 \mathrm{~cm}$.
To do:
We have to construct a triangle with sides $5 \mathrm{~cm}, 6 \mathrm{~cm}$ and $7 \mathrm{~cm}$ and then another triangle whose sides are $\frac{5}{7}$ of the corresponding sides of the first triangle.
Solution:
Steps of construction:
(i) Draw a line segment $BC = 5\ cm$.
(ii) With centre $B$ and radius $6\ cm$ and with centre $C$ and radius $7\ cm$, draw arcs intersecting each other at $A$.
(iii) Join $AB$ and $AC$.
$ABC$ is the required triangle.
(iv) Draw a ray $BX$ making an acute angle with $BC$ and cut off seven equal parts making $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5 = B_5B_6 = B_6B_7$.
(v) Join $B_7$ and $C$.
(vi) From $B_5$, draw $B_5C'$ parallel to $B_7C$ and $C’A’$ parallel to $CA$.
$A’BC’$ is the required triangle.
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Updated on: 10-Oct-2022
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# In mathematics, the multinomial theorem says how to expand a power of a sum in terms of powers of the terms in that sum.
## It is the generalization of the binomial theorem to polynomials.
Theorem
For any positive integer m and any nonnegative integer n, the multinomial formula tells us how a sum with m terms expands when raised to an arbitrary power n:
where
is a multinomial coefficient. The sum is taken over all combinations of nonnegative integer indices k1 through km such that the sum of all ki is n. That is, for each term in the expansion, the exponents of the xi must add up to n. Also, as with the binomial theorem, quantities of the form x0 that appear are taken to equal 1 (even when x equals zero).
Example
## The third power of the trinomial a + b + c is given by
This can be computed by hand using the distributive property of multiplication over addition, but it can also be done (perhaps more easily) with the multinomial theorem, which gives us a simple formula for any coefficient we might want. It is possible to "read off" the multinomial coefficients from the terms by using the multinomial coefficient formula. For example:
Alternate
.
expression
## where = (1,2,,m) and x = x11x22xmm.
Proof
This proof of the multinomial theorem uses the binomial theorem and induction on m.
First, for m = 1, both sides equal x1n since there is only one term k1 = n in the sum. For the induction step, suppose the multinomial theorem holds for m. Then
by the induction hypothesis. Applying the binomial theorem to the last factor,
## which completes the induction. The last step follows because
as can easily be seen by writing the three coefficients using factorials as follows:
Multinomial
coefficients
The numbers
## (which can also be written as:)
are the multinomial coefficients. Just like "n choose k" are the coefficients when you raise a binomial to the nth power (e.g. the coefficients are 1,3,3,1 for (a + b)3, where n = 3), the multinomial coefficients appear when one raises a multinomial to the nth power (e.g. (a + b + c)3)
Sum
## gives immediately that
Number
of multinomial coefficients
The number of terms in multinomial sum, # n,m, is equal to the number of monomials of degree n on the variables x1, , xm:
The count can be easily performed using the method of stars and bars.
Central
multinomial coefficients
All of the multinomial coefficients for which the following holds true:
are central multinomial coefficients: the greatest ones and all of equal size.
## A special case for m = 2 is central binomial coefficient.
Interpretations
Ways
## to put objects into boxes
The multinomial coefficients have a direct combinatorial interpretation, as the number of ways of depositing n distinct objects into m distinct bins, with k1 objects in the first bin, k2 objects in the second bin, and so on.[1]
Number
## of ways to select according to a
distribution
In statistical mechanics and combinatorics if one has a number distribution of labels then the multinomial coefficients naturally arise from the binomial coefficients. Given a number distribution {ni} on a set of N total items, ni represents the number of items to be given the label i. (In statistical mechanics i is the label of the energy state.)
ways.
## From the remaining N n1 items choose n2 to label 2. This can be
done
ways.
From the remaining N n1 n2 items choose n3 to label 3. Again, this can be done
ways.
## Upon cancellation, we arrive at the formula given in the introduction.
Number
of unique permutations of
words
In addition, the multinomial coefficient is also the number of distinct ways to permute a multiset of n elements, and ki are the multiplicities of each of the distinct elements. For example, the number of distinct permutations of the letters of the word MISSISSIPPI, which has 1 M, 4 Is, 4 Ss, and 2 Ps is
(This is just like saying that there are 11! ways to permute the lettersthe common interpretation of factorial as the number of unique permutations. However, we created duplicate permutations, due to the fact that some letters are the same, and must divide to correct our answer.)
Generalized
Pascal's triangle
One can use the multinomial theorem to generalize Pascal's triangle or Pascal's pyramid to Pascal's simplex. This provides a quick way to generate a lookup table for multinomial coefficients.
The case of n = 3 can be easily drawn by hand. The case of n = 4 can be drawn with effort as a series of growing pyramids.
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Application Center - Maplesoft
# Calculus I: Lesson 13: Quadratic Approximation
You can switch back to the summary page by clicking here.
Calculus I
We want to approximate a given function f(x) at x=a with a second degree polynomial.
Express the second degree polynomial as P(x) = A + B (x - a) + C
We want:
P(a) = f (a)
P'(a) = f ' (a)
P''(a) = f '' (a)
Hence, we want:
A = f (a)
B = f ' (a)
2 C = f '' (a)
Thus,
P(x) = f (a) + f ' (a) (x-a) + [ f '' (a) / 2 ]
Conclusion: The quadratic approximation to f(x) near x = a is given by:
P(x) = f (a) + f ' (a) (x-a) + [ f '' (a) / 2 ]
Example 1
Find the quadratic approximation to f(x) = sec (x) for a = 0.
Plot both f(x) and P(x) on the same axes near 0.
> restart:
> f:= x -> sec(x);
> f(0);
> D(f)(0);
> D(D(f))(0);
> P:= x -> 1 + 0.5 * (x)^2;
> plot({f(x),P(x)}, x = -1..1, color=[blue,brown]);
Example 2
Find the quadratic approximation to f x) = sin (x) for a =
.
Plot both f(x) and P(x) on the same axes near .
Then plot both functions on a larger domain; is P(x) a good estimate
for f (x) away from ? .
> f:= x -> sin(x);
> f(Pi/6);
> D(f)(Pi/6);
> evalf(%);
> D(D(f))(Pi/6);
> P:= x -> 0.5 + .8660254040 * (x - Pi/6) - 0.25 *( x - Pi/6)^2;
> evalf(Pi/6);
> plot({f(x),P(x)}, x = 0..1, color=[brown,blue]);
The graph below shows that P(x) is NOT a good approximation to f (x) for x far from .
> plot({f(x),P(x)}, x = -2*Pi..2*Pi, color=[brown,blue]);
3) Find the quadratic approxiamtion for f (x) = near a = 1.
Plot both f (x) and P (x) on the same axes near 1.
Plot both f (x) and P (x) on the same axes for a larger domain; is P(x) a good estimate away from 1?
Determine the values of x for which the quadratic approximation is accurate to within 0.01.
Can you use a plot to help?
> f:= x -> 1/(1+x)^2;
> f(1);
> D(f)(1);
> D(D(f))(1);
> P:= x -> 0.25 - 0.25 * (x - 1) + (3/16) * ( x - 1)^2;
> plot({f(x),P(x)}, x = 0..2, color=[blue,brown]);
> plot({f(x),P(x)}, x = 0..10, color=[blue,brown]);
From the above graph, we see that P (x) is NOT a good estimate for f (x) for x not near 1.
We are to find x values for which | f(x) - P (x) | < 0.01, i.e.,
we want:
-0.01 < f (x) - P (x) < 0.01
OR
f (x) - 0.01 < P (x) < f (x) + 0.01.
> g1:= x -> f(x) - 0.01;
> g2:= x -> f(x) + 0.01;
> plot({P(x), g1(x), g2(x)}, x = 0.5..3, color=[blue, brown, magenta]);
> solve({-0.01 < f (x) - P (x), x > 0}, x );
> solve({f (x) - P (x) < 0.01, x > 0}, x);
From the plot and solve, we see that | P (x) - f (x) | < 0.01 for
x in ( , ). Our answer is a numerical estimate.
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# What Is 11/52 as a Decimal + Solution With Free Steps
The fraction 11/52 as a decimal is equal to 0.21153846153.
The fraction is shown in p/q form, with the division line between p and q. The fraction’s p is referred to as the numerator, while its q is referred to as the denominator. We can convert fractions to decimal values by applying the mathematical operation known as division.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 11/52.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 11
Divisor = 52
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 11 $\div$ 52
This is when we go through the Long Division solution to our problem.
Figure 1
## 11/52 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 11 and 52, we can see how 11 is Smaller than 52, and to solve this division, we require that 11 be Bigger than 52.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 11, which after getting multiplied by 10 becomes 110.
We take this 110 and divide it by 52; this can be done as follows:
110 $\div$ 52 $\approx$ 2
Where:
52 x 2 = 104
This will lead to the generation of a Remainder equal to 110 – 104 = 6. Now this means we have to repeat the process by Converting the 6 into 60 and solving for that:
60 $\div$ 52 $\approx$ 1
Where:
52 x 1 = 52
This, therefore, produces another Remainder which is equal to 60 – 52 = 8. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 80.
80 $\div$ 52 $\approx$ 1
Where:
52 x 1 = 52
Finally, we have a Quotient generated after combining the three pieces of it as 0.211=z, with a Remainder equal to 28.
Images/mathematical drawings are created with GeoGebra.
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# Practice: Expressing Linear Equations As Matrices
1. Express the following set of linear equations as a matrix. \begin{align} 2x + 4y = 6 \\ 2x + 6y = 8 \end{align} solution $\begin{bmatrix} 2 & 4 \\ 2 & 6 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 6 \\ 8 \end{bmatrix}$ 3. Express the following set of linear equations as a matrix \begin{align} 4a - b = 4 \\ 5a - 2b = 2 \\ a + 3b = 4 \end{align} solution $\begin{bmatrix} 4 & -1 \\ 5 & -2 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 4 \\ 2 \\ 4 \end{bmatrix}$ 2. Express the following set of linear equations as a matrix \begin{align} 2x_1 + x_2 + 7x_3 = 5 \\ x_1 + x_2 + 5x_3 = 3 \end{align} solution $\begin{bmatrix} 2 & 1 & 7 \\ 1 & 1 & 5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 5 \\ 3 \end{bmatrix}$
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Home > Measurement > Perimeter > Examples of Perimeter
# Examples of Perimeter
Let us now look at some examples of perimeter calculations for various shapes.
1. What is the perimeter of a square with side = 5 cm?
The formula to calculate the perimeter of a square = 4s (where s is the length of the side of the square)
So the perimeter of the square = 5 + 5 + 5 + 5 = 20 cm, or
= 4 x 5 = 20 cm
2. The perimeter of this square is 36 cm. We need to find k.
Perimeter of a square = 4k = 36 cm
So k = 36 ÷ 4
= 9 cm
3. The perimeter of this rectangle is 28 mm, what is t?
Perimeter of a rectangle = 2(l + b) = 28 mm
So 2(9 + t) = 28
9 + t = 14
t = 14 – 9 = 5 mm
So the breadth of the rectangle is 5 mm.
4. Calculate the perimeter of the triangle.
Perimeter of a triangle = sum of its sides
= 4 + 5 + 9
= 18 cm
5. Calculate the perimeter of these two quadrilaterals.
Perimeter of a quadrilateral = sum of its sides
= 3 + 9 + 7 + 5
= 24 m
Perimeter of quadrilateral = sum of its sides
= 6 + 3 + 3 + 7 + 4
= 23 m
6. What is the perimeter of this shape?
To find the perimeter of this shape, we first need to find the lengths of the various unknown sides. Here the two unknown sides are the longest side (the base of the shape), and the short side to the right of the shape.
Length of the longest side = 4 + 4 = 8
Length of the shortest side = 6 – 3 = 3
So perimeter = 6 + 4 + 3 + 4+ 3 + 8
= 28 units
7. What is the perimeter of this shape?
We first need to find the length of the shortest side (unknown side)
The length of this side = 12 – (5 + 4)
= 12 – 9
= 3
Therefore perimeter = 3 + 12 + 3 + 5 + 7 + 3 + 7 + 4
= (3 x 3) + 12 + 5 + (7 x 2) + 4
= 9 + 12 + 5 + 14 + 4
= 53 units
8. What is the perimeter of the figure?
In this shape, we have four sides of equal length (= 7 units), and four other sides of equal length as well (= 3 units).
So perimeter = (4 x 7) + (4 x 3)
= 28 + 12
= 40 units
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Altitude Sums of Inscribed Triangle
By: Damarrio C. Holloway
Assignment #8
Problem 11. Construct any acute triangle ABC and its circumcircle. Construct the three altitudes AD, BE, and CF. Extend each altitude to its intersection with the circumcircle at corresponding points P, Q, and R.
The problem continues to find we are required to find the sum of the ratios of:
LetŐs separate the figure into several triangles using the different sides of the inscribed triangle ABC.
The length of AD is the altitude from the vertex A to the opposite side of BC. The length of AP includes that altitude plus the segment of DP. If we form two triangles using the Orthocenter (H), Triangle BHC and the point on the circle Triangle BPC, we see that they have two things in common, the have the same base BC and congruent heights HD and DP as seen in the picture below.
Thus, AP = AD + PD
2. Expression BQ/BE:
Using a similar method, we will now focus on the construction of two other triangles both containing the side AC. As shown below, the altitude BE is extended to the point Q on the circle, forming the two triangles: Triangle CHA and Triangle CQA. We see that they have equal heights of HE and EQ, with a common side AC.
Thus, BQ = BE + QE
3. Expression CR/CF:
Let us now look at the triangles formed using a common side AB. The altitude CF, formed on Triangle ABC, is extended to the point R on the circle. The triangles AHB and RB are displayed below, with equal heights (altitudes) HF and FR, respectfully.
Thus, CR = CF + FR
Going back to our original equation of: AP/AD + BQ/BE + CR/CF, we can now use substitution from our previous illustrations.
Now, (AD + PD)/AD + (BE + QE)/BE + (CF + RF)/CF
= {1 + (PD/AD) + {1 + (QE/BE)} + {1 + (RF/CF)}
= 3 + (PD/AD) + (QE/BE) + (RF/CF)
These three quotients are the heights/altitudes of their respective base. Since the area of a triangle is ½ bh, then we can multiply these heights (h) by their respective bases (b) and ½.
= 3 + (PD/AD)(BC/BC) + (QE/BE)(AC/AC) + (RF/CF)(AB/AB)
Total Area of ABC
(The ½ cancels)
= 3 + 1
Therefore, the sum of the quotient: AP/AD + BQ/BE + CR/CF = 4.
Return
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Done! However, we should realize that the slope is easily calculated when two points are known using the Slope Formula . Find the y-intercept with slope and point. The main point of this section is to work some examples finding critical points. The point negative 6 comma negative 7 is reflec-- this should say "reflected" across the x-axis. Point of view is a term we use when talking about narration. Letâs take a look at an example of each of these formulas. Using y-y1=m(x-x1) to write the equation ⦠Example 2: Let us consider the percentile example problem: In a college, a list of grades of 15 students has been declared. These scenarios relate directly to a particular form of the quadratic equation: Point of view can be âinvolvedâ or more impartial. Those loyalty points get pooled together, and customers can redeem points from any partner, at any partner. The left side of the above figure shows coplanar points A, B, C, and D.And in the box on the right, there are many sets of coplanar points. As you can see in the thesis statement examples below, you must be very specific, summarizing points that are about to be made in your paper, and supported by specific evidence. So, letâs work some examples. In this problem, we are not provided with both the slope m and y -intercept b . So that's its reflection right over here. Given any two points on a line, you can calculate the slope of the line by using this formula: Example: Given two points, P = (0, â1) and Q = (4,1), on the line, find the equation of the line. Write an equation of a line given the y intercept and another point. Now take the 15th number and the 16th number and find the average: 79 + 85 / 2 = 164 / 2 = 82 Hence, 60th percentile of given data set = 82. Seven Point-Slope Form Examples. Distance Formula: Given the two points (x 1, y 1) and (x 2, y 2), the distance d between these points is given by the formula: Don't let the subscripts scare you. They only indicate that there is a "first" point and a "second" point; that is, that you have two points. Barbara is the managerial accountant in charge of a large furniture factoryâs production lines and supply chains. Let's do a couple more of these. Example 1 Determine all the critical points for the function. We are given a point and the vertex, a point and the x-intercepts, or three points. Solution: Step 1: Calculate the slope. From the given data set, 15th number is 79. Next, we pick one of our two given points, and the slope we just found, and plug them into the point-slope form formula. Example. Example 10: A line passing through the given two points \left( {4,\,5} \right) and (\left( {0,\,3} \right). Points P, Q, X, and W, for example, are coplanar; the plane that contains them is the left side of the box.Each of the six faces of the box contains four coplanar points, but these are not the only groups of coplanar points. So you would see it at 8 to the right of the y-axis, which would be at positive 8, and still 5 above the x-axis. $f\left( x \right) = 6{x^5} + 33{x^4} - 30{x^3} + 100$ ... (w = - 2\) neither does the function and so these two points are not critical points for ⦠slope = = Step 2: Substitute m = , into the equation, y = mx + b, to get the equation Generally, your thesis statement can be the last line of the first paragraph in your research paper or essay. It means whose perspective narrative is given from. She isnât sure the current yearâs couch models are going to turn a profit and what to measure the number of units they will have to produce and sell ⦠A first-person narrator, for example, tells the reader everything from their perspective using the first-person pronouns âIâ, âmeâ and âmyâ. By working together with other brands, each member brand is able to double and triple the value customers get out of their loyalty program. Plentiâs value for customers comes from that pooling. It's reflection is the point 8 comma 5. ⢠A number is a fixed point for a given function if = ⢠Root finding =0 is related to fixed-point iteration = âGiven a root-finding problem =0, there are many with fixed points at : Example: â â â +3 ⦠If has fixed point at , then = â ( ) has Section is to work some examples finding critical points and y -intercept b reflec this. reflected '' across the x-axis '' across the x-axis and supply chains the point negative 6 negative... 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Last updated:
# Parallelogram Area Calculator
Parallelogram area formulasHow to use this parallelogram area calculator?FAQs
If you have any problems with the geometry of a parallelogram, check this parallelogram area calculator (and also its twin brother, parallelogram perimeter calculator).
Whether you want to calculate the area given base and height, sides and angle, or diagonals of a parallelogram and angle between them, you are in the right place. Don't ask how to find the area of a parallelogram; just give the calculator a try!
Below you can find out how the tool works – the parallelogram area formulas and neat explanation are all you need to understand the topic.
## Parallelogram area formulas
A parallelogram is a simple quadrilateral with two pairs of parallel sides. Every rectangle is a parallelogram, as well as every rhombus and square. Remember, it doesn't work the other way around!
Which formulas does the parallelogram area calculator use?
• Area given base and height
area = base × height
Did you notice something? The formula for the area of a parallelogram is pretty much the same as for a rectangle! Why is it so? Have a look at the picture: a parallelogram can be divided into a trapezoid and a right triangle and rearranged to the rectangle.
• Area given sides and the angle between them
area = a × b × sin(angle)
Does it ring a bell? This formula comes from trigonometry and is used, for example, in our triangle area calculator – the parallelogram may be seen as two congruent triangles. The adjacent angles in the parallelogram are supplementary, so you can choose whichever angle you want because sin(angle) = sin(180° - angle).
• Area given diagonals of a parallelogram and the angle between them
area = ½ × e × f × sin(angle)
The formula comes from trigonometry as well. Do you want to know where it comes from?
Divide the parallelogram into two triangles, and assume that our e diagonal is the "base" for both new triangles.
What's the height of that triangle? Use the sine function. It's (f/2) × sin(angle)!
The area of the triangle is equal to our "base" e times height and divided by 2: e × (f/4) × sin(angle)
The parallelogram consists of two such triangles, so the area equals e × (f/2) × sin(angle).
## How to use this parallelogram area calculator?
Are you still not sure our parallelogram area calculator works? We will show you step by step:
1. Have a look at your exercise. What is given, what is unknown? Choose the right calculator part for your needs. Assume that we want to calculate the area knowing the sides of a parallelogram and the angle between them.
2. Enter the given values to the right boxes. Assume 5 in, 13 in, and 30° for the first side, the second one, and the angle between them, respectively.
3. The calculator displays the area of a parallelogram value. It's 32.5 in² in our case.
Check out our area calculators for other shapes, such as rhombus area calculator, circle area calculator, and trapezoid area calculator.
FAQs
### How do I find the area of a parallelogram given its adjacent sides?
To determine the area given the adjacent sides of a parallelogram, you also need to know the angle between the sides. Then you can apply the formula: area = a × b × sin(α), where a and b are the sides, and α is the angle between them.
### How do I find the area of a parallelogram given diagonals?
The area of a parallelogram can be determined from its diagonals, provided that you also know the angle between the diagonals.
If e and f are the lengths of the diagonals and φ is the angle between them, then the area can be calculated as follows: area = ½ × e × f × sin(φ).
### How do I find the area of a parallelogram without height?
It is possible to find the area of a parallelogram without height! For instance, it suffices to know one of the following things:
1. The length of adjacent sides and the angle between them – use trigonometry.
2. The length of diagonals and the angle between them, using the formula – use trigonometry.
3. The length of diagonals and one side – use Heron's formula.
### What is the area of a parallelogram with perpendicular diagonals of length 10 and 15?
The answer is 75. We use the formula that says the area is equal to ½ times the product of the lengths of the diagonals times the sine of the angle between them. As our diagonals are perpendicular, the angle between them is 90° and sin 90° = 1. Hence, the calculation we need to perform is ½ × 10 × 15 = 75.
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# Composite number
Composite number
A composite number is a positive integer which has a positive divisor other than one or itself. In other words a composite number is any positive integer greater than one that is not a prime number.
So, if n > 0 is an integer and there are integers 1 < a, b < n such that n = a × b, then n is composite. By definition, every integer greater than one is either a prime number or a composite number. The number one is a unit – it is neither prime nor composite. For example, the integer 14 is a composite number because it can be factored as 2 × 7. Likewise, the integers 2 and 3 are not composite numbers because each of them can only be divided by one and itself.
The first 105 composite numbers (sequence A002808 in OEIS) are
4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100, 102, 104, 105, 106, 108, 110, 111, 112, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 128, 129, 130, 132, 133, 134, 135, 136, 138, 140.
Every composite number can be written as the product of two or more (not necessarily distinct) primes; furthermore, this representation is unique up to the order of the factors. This is called the fundamental theorem of arithmetic.
Wilson's theorem provides a test for whether a number is prime or composite:
$(n-1)! \equiv \begin{cases} -1 \pmod{n} \mbox{ if } n \mbox{ is prime}\\ \;\;\; 2 \pmod{n} \mbox{ if } n =4\\ \;\;\;0 \pmod{n} \mbox{ if } n >4 \mbox{ is composite}. \end{cases}$
## Types of composite numbers
One way to classify composite numbers is by counting the number of prime factors. A composite number with two prime factors is a semiprime or 2-almost prime (the factors need not be distinct, hence squares of primes are included). A composite number with three distinct prime factors is a sphenic number. In some applications, it is necessary to differentiate between composite numbers with an odd number of distinct prime factors and those with an even number of distinct prime factors. For the latter
$\mu(n) = (-1)^{2x} = 1\,$
(where μ is the Möbius function and x is half the total of prime factors), while for the former
$\mu(n) = (-1)^{2x + 1} = -1.\,$
Note however that for prime numbers the function also returns -1, and that μ(1) = 1. For a number n with one or more repeated prime factors, μ(n) = 0.
If all the prime factors of a number are repeated it is called a powerful number. If none of its prime factors are repeated, it is called squarefree. (All prime numbers and 1 are squarefree.)
Another way to classify composite numbers is by counting the number of divisors. All composite numbers have at least three divisors. In the case of squares of primes, those divisors are {1,p,p2}. A number n that has more divisors than any x < n is a highly composite number (though the first two such numbers are 1 and 2).
• Canonical representation of a positive integer
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• Composite number — Composite Com*pos ite (?; 277), a. [L. compositus made up of parts, p. p. of componere. See {Compound}, v. t., and cf. {Compost}.] 1. Made up of distinct parts or elements; compounded; as, a composite language. [1913 Webster] Happiness, like air… … The Collaborative International Dictionary of English
• composite number — n. an integer that can be evenly divided by some whole number other than itself or 1: distinguished from PRIME NUMBER … English World dictionary
• composite number — noun an integer that is divisible without remainder by at least one positive integer other than itself and one • Hypernyms: ↑number * * * Math. a number that is a multiple of at least two numbers other than itself and 1. [1720 30] * * * composite … Useful english dictionary
• composite number — noun A number that is a the product of at least two numbers other than itself and 1. Ant: prime number … Wiktionary
• composite number — Math. a number that is a multiple of at least two numbers other than itself and 1. [1720 30] * * * … Universalium
• composite number — compos′ite num′ber n. math. a number that is a multiple of at least two numbers other than itself and 1 • Etymology: 1720–30 … From formal English to slang
• composite number — /kɒmpəzət ˈnʌmbə/ (say kompuhzuht numbuh) noun Mathematics an integer greater than 1 which is exactly divisible by some integer other than itself and 1 … Australian English dictionary
• composite number — Helu hana ia … English-Hawaiian dictionary
• Highly composite number — This article is about numbers having many divisors. For numbers factorized only to powers of 2, 3, 5 and 7 (also named 7 smooth numbers), see Smooth number. A highly composite number (HCN) is a positive integer with more divisors than any… … Wikipedia
• Superior highly composite number — In mathematics, a superior highly composite number is a certain kind of natural number. Formally, a natural number n is called superior highly composite iff there is an ε gt; 0 such that for all natural numbers k ≥… … Wikipedia
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## Intermediate Algebra (12th Edition)
$(x^4+1)(x^2+1)(x+1)(x-1)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $x^8-1 ,$ use the factoring of the difference of $2$ squares repetitively until none of the resulting expressions is still a difference of $2$ squares. $\bf{\text{Solution Details:}}$ The expressions $x^8$ and $1$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $x^8-1 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x^4)^2-(1)^2 \\\\= (x^4+1)(x^4-1) .\end{array} The expressions $x^4$ and $1$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $x^4-1 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x^4+1)[(x^2)^2-(1)^2] \\\\= (x^4+1)[(x^2+1)(x^2-1)] \\\\= (x^4+1)(x^2+1)(x^2-1) .\end{array} The expressions $x^2$ and $1$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $x^2-1 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x^4+1)(x^2+1)[(x)^2-(1)^2] \\\\= (x^4+1)(x^2+1)[(x+1)(x-1)] \\\\= (x^4+1)(x^2+1)(x+1)(x-1) .\end{array}
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# Ex.10.4 Q1 Circles Solution - NCERT Maths Class 9
Go back to 'Ex.10.4'
## Question
Two circles of radii $$5 \rm{cm}$$ and $$3 \rm{cm}$$ intersect at two points and the distance between their centers is $$4 \rm{cm}.$$ Find the length of the common chord.
Video Solution
Circles
Ex 10.4 | Question 1
## Text Solution
What is known?
Radii of two circles and distance between the centers of the circles.
What is unknown?
Length of common chord.
Reasoning:
Perpendicular bisector of the common chord passes through the centers of both the circles.
Steps:
Given that the circles intersect at $$2$$ points, so we can draw the above figure.
Let $$AB$$ be the common chord. Let $$O$$ and $$O^\prime$$ be the centres of the circles respectively.
\begin{align} &{O}^{\prime} {A}=5 {\rm{cm}}, {OA}=3 {\rm{cm}},\\ & {OO}^{\prime}=4 {\rm{cm}}\end{align}
Since the radius of the bigger circle is more than the distance between the $$2$$ centres, we can say that the centre of smaller circle lies inside the bigger circle itself.
$${OO}’$$ is the perpendicular bisector of $${AB.}$$
\begin{align} {\rm{So,}} \;{OA}&={OB}=3 \,\rm{cm}\\ {AB}&=3+3=6\, \rm{cm} \end{align}
Length of the common chord is $$6 \,\rm{cm.}$$
It is also evident that common chord is the diameter of the smaller circle.
Video Solution
Circles
Ex 10.4 | Question 1
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### Course: 6th grade (Illustrative Mathematics)>Unit 3
Lesson 3: Lesson 6: Interpreting ratios
# Solving unit rate problem
Jayda delivers newspapers on her paper route, taking 3 hours to distribute 189 newspapers. To calculate her rate per hour, divide the total newspapers (189) by the hours taken (3). This results in Jayda delivering 63 newspapers per hour, showcasing her efficiency and speed. Created by Sal Khan and Monterey Institute for Technology and Education.
## Want to join the conversation?
• Does anyone understand why Sal had to flip the 3/189 to 189/3 before he did the problem?
• Fractions is basically dividing.
For example 189/3 (improper fraction) = 189/3 <- Divide symbol "/" It's like simplifying it, 189/3 and 3/3 = 63/1. If he flipped it around to 3/189, then it would become smaller (3 hours for every 189 newspapers, or 3 divided by 189), simplified as 1 hour for every 63 newspapers (1 divided by 63), but it would be as decimal form if you divide it (1 divided by 63 and 3 divided by 189, both equals 0.0158730158730159, etc.) = 0.0158730158730159...(irrational number, random numbers forever going on, like pi). But if it was 189/3 (we are dealing with whole numbers) we can turn it into a mixed number: 3 can go in 189 "63" times perfectly, so 63 0/189. So that's how we got our answer. (Or 189 divided by 3, but the other way around. See how 189 divided by 3 (189/3) is different that 3 divided by 189? (3/189) ) Order matters.
I hope that made sense and helped you!
• Is 2:5 the same as 5:2
• The order of the #s matter.
If the problem were (for example) There are 15 kids who have pets in a class. There are 33 kids in the class. What is the ratio of kids who have pets to kids who do not? The answer would be . If you said you would be saying the ratio of kids who do NOT have pets to kids who DO have pets.
• Yes tis has helped me a lot now I will die as a happy man
• wow that doesn't sound good ur not old yet
• the ratios are the same with 5 apples and 6 banana too so it is it is still 5.6 or 6.5
• how do i solve a unit rate problem if there are fractions in them? For example, the unit rate of 2/3 cup of sugar for every 1/2 teaspoon of vanilla
• Divide 2/3 and 1/2. Do the same method even if it's a fraction.
• what is math
• who is math
(1 vote)
• I don't understand these problems...
Fill in the blank with the unit rate:
240 km in 2.5 hours =____ km in 1 hour
Fill in the blank with the unit rate:
\$15 for 4 quarts = \$ for 1 quart
There are 25 computers in a math lab. There are 6 activity booklets for every 2 computers in the math lab. How many activity booklets are in the math lab?
There are activity booklets in the math lab.
You earn \$102 for doing 12 hours of yard work. Your friend earns \$120 working at a store for 15 hours.
How much do you earn for every hour of yard work? \$ per hour
How much does your friend earn each hour at the store? \$ per hour
What would YOU earn for 15 hours worth of yard work (at your hourly rate)? \$ for 15 hours of work
• Okay, let's go through these step-by-step:
240 km in 2.5 hours = __ km in 1 hour
To find the unit rate, we need to convert the rate to be per 1 hour.
The unit rate would be: 240 km / 2.5 hours = 96 km/hour
\$15 for 4 quarts = \$ for 1 quart
To find the unit rate, we need to convert the rate to be per 1 quart.
The unit rate would be: \$15 / 4 quarts = \$3.75/quart
There are 25 computers in a math lab. There are 6 activity booklets for every 2 computers in the math lab. How many activity booklets are in the math lab?
The rate is 6 activity booklets for every 2 computers.
To find the total number of activity booklets, we need to set up a proportion:
6 booklets / 2 computers = x booklets / 25 computers
Cross-multiplying: 6 * 25 = 2x
Solving for x: x = 75 activity booklets
You earn \$102 for doing 12 hours of yard work. Your friend earns \$120 working at a store for 15 hours.
Your unit rate = \$102 / 12 hours = \$8.50 per hour
Your friend's unit rate = \$120 / 15 hours = \$8 per hour
For 15 hours of your yard work at your rate: \$8.50 per hour * 15 hours = \$127.50
• what is unite rate though?
• Unit rates are used to find out how much things happen at a time or ex. rooms per hotel.
• What is the difference between unit rates and price?
• unit rates always have a denominator of one. When prices are expressed as a quantity of one, they are called unit prices. A price is how much you actually pay. The unit price is just per one thing. :)
• What if the numerator is not divisible by the denominator? For example, 42 cupcakes can be made every 30 minutes. How many cupcakes can be made in 1 minute?
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### Density CalculationsProblems 1-10
Twenty Examples Probs #11-25 Probs #26-50 All the examples & problems, no solutions Significant Figures Menu
Problem #1: What mass of lead (density 11.4 g/cm3) would have an identical volume to 25.1 g of mercury (density 13.6 g/cm3)?
Solution #1:
1) Determine the volume of 25.1 g of mercury:
M D = –––– V
25.1 g 13.6 g/cm3 = –––––– V
V = 1.845588 cm3
2) Determine the mass of 1.845588 cm3 of lead:
M D = –––– V
M 11.4 g/cm3 = –––––––––––– 1.845588 cm3
V = 21.0 g
Solution #2:
1) This solution depends on the fact that the volumes are equal. Using symbols, we show the equality as follows:
VPb = VHg
2) Rearranging the density equation for lead and for mercury, we have this:
MPb VPb = –––– DPb
and
MHg VHg = –––– DHg
3) Setting them equal to each other, we have this:
MPb MHg –––– = –––– DPb DHg
4) Substitute values into the above equation, then solve:
MPb 25.1 –––– = –––– 11.4 13.6
MPb = 21.0 g
Problem #2: The density of pure aluminum is 2.70 g/cm3. 2.25 grams of pure aluminum is added to a graduated cylinder containing 11.20 mL of water. To what volume mark will the water level rise?
Solution:
1) Determine the volume of 2.25 g of aluminum:
Density = mass / volume
rearrange to:
Volume = mass / density
V = 2.25 g / 2.70 g/cm3 = 0.83 cm3
Note: since 1 mL = 1 cm3, this is 0.83 mL
11.20 mL + 0.83 mL = 12.03 mL
Problem #3: A 1.50 g piece of aluminum was rolled out into a thin sheet measuring 24.0 cm x 30.0 cm.
(a) Calculate the volume (density of Al = 2.70 g cm¯3
(b) Calculate the thickness of the piece of Al in mm
Solution:
1.50 g / 2.70 g/cm3 = 0.555555 cm3 (answer to a)
0.555555 cm3 = (x) (24.0 cm) (30.0 cm)
x = 0.000772 cm (to 3 sig figs) <--- the thickness of the Al foil in cm.
There are 10 mm in one cm, so divide the cm answer by 10 to get the thickness in mm.
Problem #4: 'Copper' pennies actually contain very little copper. If a penny contains 92.276% of its total volume zinc and 7.724% of its total volume copper, what is its density? (d of Cu = 8.96 g/cm3; d of Zn = 7.14 g/cm3)
Solution:
1) Let's assume a penny is 100. cm3 for its volume:
That means 92.276 cm3 is Zn and 7.724 cm3 is Cu.
2) Let's figure out the mass of Zn and the mass of Cu:
(92.276 cm3) (7.14 g/cm3) = 658.85064 g
(7.724 cm3) (8.96 g/cm3) = 69.20704 g
3) Now, we add the masses and divide by the total volume:
728.05768 g / 100. cm3 = 7.28 g/cm3 (that's the answer to three sig figs)
Problem #5: The element antimony has a density of 6.62 g cm¯3. Calculate the edge length in meters of a cube of antimony whose mass is 1.00 x 105 kg.
Solution:
1.00 x 105 kg = 1.00 x 108 g
1.00 x 108 g / 6.62 g cm¯3 = 1.51 x 107 cm3
$\sqrt[3]{\mathrm{1.51 x 107cm3}}$ = 247 cm = 2.47 m
Problem #6: Iron has a density of 7.87 g/cm3. If 71.7 g of iron is added to 63.0 mL of water in a cylinder, to what volume reading will the water rise?
Solution:
1) We need to determine the volume of iron:
D = m / V
V = m / D
V = 71.7 g / 7.87 g/cm3
V = 9.11 cm3
2) Since 1 cm3 = 1 mL, we have this:
63.0 mL + 9.11 mL = 72.1 mL (rounded off to the proper number of sig figs)
3) You can approach this problem using dimensional analysis (also called the unit-factor method or the factor-label method), where the density is used as a conversion factor.
V = 71.7 g Fe x (1 cm3 / 7.87 g Fe) = 9.11 cm3
Problem #7: Suppose you are given two cylindrical rods, one aluminum and the other tin, with identical external dimensions. What fraction of the tin rod would have to be hollow in order to give the same average density for both rods? (The density of tin is 7.31 g/cm3 and for aluminum, it is 2.70 g/cm3.)
Solution #1:
Let us assume 1.00 cm3 of Al is present.
That means 2.70 g of Al is present.
In the rod of tin that is hollow, the mass of tin present weighs 2.70 g. We know this because the average densities of the two rods are equal. Remember, the average density of the tin rod includes the empty space.
What volume of solid tin is required to provide 2.70 g? The answer is 0.370 cm3 (from 2.70 g divided by 7.31 g/cm3).
The rest of the tin rod must be hollow. This is 0.670 cm3. Since the overall volume was 1.00 cm3, the fraction of the tin rod that is hollow is 0.670
Solution #2:
Identical external dimensions means overall volume is the same for both rods. Let's call that V.
It must be that mass Al = mass Sn.
So, let's convert from volume Al to volume Sn with equivalent mass:
2.70 g Al 1 g Sn 1 cm3 V x ––––––– x ––––––– x ––––––– = 0.370V 1 cm3 1 g Al 7.31 g Sn
Finally the hollow fraction = volume hollow / volume total
(V − 0.370V) / V
1 − 0.370 = 0.630
Problem #8: Osmium is one of the densest elements known. The standard density is 22.59 kg/L. What is the density in units of g/cm3?
Solution:
Kilograms will be changed to grams by multiplying the numerator by 1000. Liters will be changed to mL by multiplying the denominator by 1000. Those two multiplications cancel each other out. And one mL equals one cm3.
Here's a dimensional analysis set up showing the solution:
22.59 kg 1000 g 1 L 1 mL ––––––– x –––––– x ––––––– x ––––– = 22.59 g/cm3 1 L 1 kg 1000 mL 1 cm3
By the way, the standard density is not expressed in units of kg/L. The standard unit for density, according to the IUPAC, is kg/m3. The ChemTeam did not edit out the word standard in the question (he didn't write the original question), but decided to keep the word so as to make this comment.
The standard density for osmium is 22,590 kg/m3. It is not often used, the chemical world much preferring to use the 22.59 g/cm3 value.
Problem #9: A gold wire has a diameter of 1.000 mm. What length of this wire contains exactly 1.000 mol of gold? (The density of gold is 19.32 g cm¯3.)
Solution:
One mole of gold weighs 196.9665 g.
V = mass / density = 196.9665 g / 19.32 g cm¯3 = 10.195 cm3
The formula for volume of a cylinder is V = πr2L
The length (L) of the wire is therefore:
L = 10.195 cm3 / [(3.14159) (0.05000 cm)2] = 1298 cm
By the way, note the 0.05 cm for the radius. The wire was 1 mm in diameter, which is 0.1 cm. And then half of that for the radius is 0.05 cm.
Problem #10: The density of lead is 11.34 g/cm3. Which of the following contains the greatest mass of lead: 0.50 kg or 0.050 liter?
Solution comparing masses:
0.050 L = 50. mL = 50. cm3
(11.34 g/cm3) (50. cm3) = 567 g
0.50 kg = 500 g
The 0.050 L contains more mass.
Solution comparing volumes:
0.50 kg = 500 g
500 g / 11.34 g/cm3 = 44.1 cm3
44.1 cm3 = 44.1 mL = 0.0441 L
The larger volume has the greater mass since both samples are lead. The answer to the problem is 0.050 L.
Bonus Problem: What is the volume in cubic inches (in3) of a 0.500 lb sample of Pb?
Short Solution:
1) Look up the density of lead in pounds per cubic inch to find a value of 0.4098 lb/in3.
2) Divide 0.500 lb by density:
0.500 lb / 0.4098 lb/in3 = 1.22 in3
Long Solution, deliberately starting with metric density:
1) Look up the density of lead to find a value of 11.34 g/cm3.
2) Determine how many cubic centimeters are in one cubic inch:
1 inch = 2.54 cm
(1.00 in)3 = (2.54 cm)3 <--- a cube that is one inch in each dimension is also 2.54 cm on a side
1.00 in3 = 16.387 cm3
3) Determine the density of lead in g/in3:
11.34 g 16.387 cm3 –––––– x –––––––– = 185.828 g/in3 cm3 in3
4) We know that 1.00 pound = 453.592 g. Therefore:
0.500 lb = 226.796 g
5) What is the volume (in cubic inches) of 226.796 g of lead?
1 in3 226.796 g x –––––––– = 1.22 in3 185.828 g
Twenty Examples Probs #11-25 Probs #26-50 All the examples & problems, no solutions Significant Figures Menu
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A function is only allowed to have ONE y-value (or output) for each x-value (or input) right? The graph of a curve that passes the vertical line test is known as a function.Functions are graphical representations that may have multiple x-values for a given y-coordinate, but will never have multiple y-values for any given x-coordinate. Still have questions? It does not pass the vertical line test! Using points given by the applet, pupils try to connect them to create a line that would pass the vertical line test. Thus the relationship y=√x is not a function! Vertical Line Test Strategy Try to draw a vertical line on the graph so it intersects the graph in more than one place. Properties of Vertical Lines. It is a vertical line. A vertical line test is a visual way to determine if the graph of a relation is a function using a vertical line. Look at the graph below. is a way to determine if a relation is a function. Let's look at our relation, b that we used in our relations example in the previous lesson.. Is this relation a function? Further, there is a point of intersection if and only if is in the domain of the function. The graphs of functions can be straight lines or segments, curves, or even just a set of points. Some people misunderstand the dictionary definition of an asymptote. We use cookies to give you the best experience on our website. A 50ft rope is cut into two pieces. But not all graphs represent functions. Graphs that pass the vertical line test are graphs of functions. Nope x=0 is a vertical line so it will have more then one y value for an x value. Notice that any vertical line would pass through only one point of the two graphs shown in parts (a) and (b) of Figure 13. If an equation fails the horizontal line test, what does that tell you about the graph? if its x=0 then does it pass the vertical line test. The Vertical Line Test. This has an undefined slope. The approach is rather simple. If you draw a vertical line anywhere on the equation and it passes through more than one point, then it fails the vertical line test, so if you put a vertical line through x=0, it would fail the vertical line test. Draw a vertical line cutting through the graph of the relation, and then observe the points of intersection. So it would not pass the line test. This condition causes the relation to be “disqualified” or not considered as a function. A test use to determine if a relation is a function. If each line crosses the graph just once, the graph passes the vertical line test. There are actually two ways to determine if a relation is a function. If an equation fails the vertical line test, what does that tell you about the graph? in this line. To check whether this graph is a function, a Vertical Line Test can be used. Let's see! Mentor: Look at one of the graphs you have a question about. If a vertical line intersects the graph in some places at more than one point, then the relation is NOT a function. This is known as the vertical line test. 4 2. If a vertical line intersects the graph in all places at exactly one point, then the relation is a function. Let us examine the following graph drawn for the mathematical relation: Can you observe the vertical lines drawn for various values of x ? Any help would be appreciated, Thanks! The graph of a function always passes the vertical line test. If no two different points in a graph have the same first coordinate, this means that vertical lines cross the graph at most once. Vertical Line Test. If the vertical line passes through at least two points on the graph, then an element in the domain is paired with more than 1 element in the range. If you think about it, the vertical line test is simply a restatement of the definition of a function. I am learning number system all over again since it's been a while. The vertical line test tells you if you have a function, while the horizontal line test tells you if that function is one to one (injective). Horizontal line test, one-to … what is the answer to (-2) to the 0 power? If the graph is a function, then no matter where on the graph you place the vertical line, the graph should only cross the vertical line once. So it's just going to look like this. Le… Let's analyze our ordered pairs first. If we choose any point on the x-axis and draw a vertical line passing through that point, the intersection of the vertical line and the graph of the function is (x,f(x)). This is a visual illustration that only one y value (output) exists for every x value (input), a rule of functions. In contrary, if the vertical line intersects the graph more than once this suggests that a single x-value is being associated with more than one value of y. Since each input has a different output, this canbe classified as a function. What this means is that you might end up thinking you have a one-to-one function (because it passed the horizontal test), but in actual fact you don’t have a function at all (because it failed the vertical one). A vertical line wont pass because there would be more than two points being touched at the same time by the vertical line test madisoneeklund madisoneeklund Answer:a graph that is a vertical line would fail the vertical line test because another for vertical line … One is … The horizontal line test is a convenient method that can determine whether a given function has an inverse, but more importantly to find out if the inverse is also a function. From this we can conclude that these two graphs represent functions. Using the Vertical Line Test to Identify Functions Practice Problems from Vertical Line Test Worksheet, source:algebra-class.com. In the definition, for any x, … Draw a few vertical lines spread out on your graph. Please click Ok or Scroll Down to use this site with cookies. No, the vertical line test is used to determine if it is a function. To use the vertical line test, take a ruler or other straight edge and draw a line parallel to the y -axis … The Vertical Line Test You can check whether a graph represents a function by using the vertical line test. See also. Wife of drug kingpin El Chapo arrested in Virginia, Pat Sajak called out for mocking contestant, Woman’s license mistakenly features her in a face mask, Top volleyball duo boycott country over bikini ban, 'Bachelor' hopeful suffers horrifying skydiving accident, Raiders player arrested in Texas street-racing incident, Jobless workers may face a surprise tax bill, Actress confirms engagement to NFL star Aaron Rodgers, Texas AG was in Utah after historic freeze back home, Do you know your privilege? In vertical line test, you have to draw a vertical line anywhere on the graph. This Vertical Line Test Interactive is suitable for 9th - 12th Grade. If you can not, then the graph represents a function. How to determine if an expression/graph represents a "FUNCTION". Trying to understand what the differnce is between a vertical and horizontal line test. No matter what y is, x is equal to negative five. Connect the points to create a function. Is each input only paired with only one output? 1. One way is to analyze the ordered pairs, and the other way is to use the vertical line test. If we think of a vertical line as an infinite set of x-values, then intersecting the graph of a relation at exactly one point by a vertical line implies that a single x-value is only paired to a unique value of y. A function will pass the horizontal line test if for each y value (the range) there is only one x value (the domain) which is the definition of a function. states that if a vertical line intersects the graph of the relation more than once, then the relation is a NOT a function. If it passes two or more points then it is not a function, but if it doesn’t it is a function. A relation is a function if there are no vertical lines that intersect the graph at more than one point. The test states that a graph represents a function if … Graph of the line f\left( x \right) = x + 1, Graph of the quadratic function (parabola) f\left( x \right) = {x^2} - 2, Graph of the cubic function f\left( x \right) = {x^3}. And so notice, x never changes. An easy way to do it is the vertical line test. Here are some examples of relations that are also functions because they pass the vertical line test. If any vertical line intersects a graph more than once, the relation represented by the graph is not a function. A horizontal asymptote is a horizontal line that the graph of a function approaches, but never touches as x approaches negative or positive infinity. vertical line test - This is when you draw a vertical line to check if the graph passes through two or more points. To pass a vertical line test is to satisfy that the graph doesn’t have two or more points going through a vertical line. If all vertical lines intersect a curve at most once then the curve represents a function. No, the vertical line test is used to determine if it is a function. An eye-opening lesson, 'Harry Potter' star admits he's embarrassed by early work. Get your answers by asking now. If a vertical line can cross a graph more than once, then the graph does not pass the vertical line test. Suppose we have a graph. How do you think about the answers? The equation of a vertical line in the graph, which is parallel to y-axis is x = a. It looks like the vertical lines may touch two points on the graph at the same time. Join Yahoo Answers and get 100 points today. I plan to do it with the vertical line test. If a vertical line intersects the graph in all places at exactly one point, then the relation is a function. You can sign in to vote the answer. The vertical line test supports the definition of a function. That is, every x-value of a function must be paired to a single y-value. So a vertical line, well that just goes straight up and down. Observe the two graphs above. What is the length of the longer piece. Otherwise, there is no point of intersection. Does it pass the vertical line test? Cutting or Hitting the Graph at Exactly One Point Graph of the line A horizontal one would, but a vertical line has points (0,n), so x has the same value for ALL y points. Always perform the horizontal line test afterthe vertical line test. Its equation is x is equal to negative five. Then take a vertical line and place it on the graph. However, take a look at the points. The vertical line test is performed by sketching a graph of the equation or by using a calculator to draw it for you. Then, take a vertical line, like a ruler, and pass it over the graph. I understand that you can not enter ± on Desmos and show both values simultaneously. If the vertical line touches only one point on the line, then that is a function. All of these vertical lines intersect the graph at one point only. Nope. Thanks to hecticar and CPhill, the inverses were found. Let us learn more about Vertical Lines. Here are some examples of relations that are also functions because they pass the vertical line test. The slope of a vertical line is infinity or undefined as it has no y-intercept and the denominator in the slope formula is zero. Vertical Line Test Worksheet Free Worksheets Library from Vertical Line Test Worksheet, source:comprar-en-internet.net That’s because it’s quite possible for an equation to pass the vertical line test, but not the horizontal line test. How do I compute the hexadecimal multiplication FF*F1? The Vertical Line Test is a visual test that you can use to quickly check and see if a graph represents a function. The vertical line test says that any line parallel to the -axis, i.e., any line of the form , intersects the graph of the function at at most one point. Remember that it is very possible that a function may have an inverse but at the same time, the inverse is not a function because it doesn’t pass the vertical line test. Use the vertical line test! Otherwise, check your browser settings to turn cookies off or discontinue using the site. The test stipulates that any vertical line drawn through the graph of the function passes through that function no more than once. The vertical Line test. They can be seen in the attachment below. Now I need to prove whether the inverses are actually functions or not. The length of one piece is 9 times the length of the other. The vertical line test is a method that is used to determine whether a given relation is a function or not. Graph of the “sideway” parabola x = {y^2}. If a function passes the vertical line test, and the horizontal line test, it is 1 to 1. Here are some examples of relations that are NOT functions because they fail the vertical line test. It is a function. Remember, you could draw a vertical line anywhere on the graph, even at multiple times, and it should intersect the line at only one point. Just a set of points like a ruler, and then observe the vertical test. ) for each x-value ( or input ) right graph represents a function ) for each (... An expression/graph represents a function '' a different output, this canbe classified as function... 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The hexadecimal multiplication FF * F1 the dictionary definition of a relation is not a function, a line. Exactly one point on the line draw a few vertical lines drawn for various values of x there is function! To the 0 power would pass the vertical line test is simply a restatement of the graphs have. On our website relation: can you observe the points of intersection use. Line drawn through the graph, which is parallel to y-axis is x equal! The domain of the definition does a vertical line pass the vertical line test an asymptote enter ± on Desmos and show values... ’ t it is a function the inverses are actually two ways to determine if expression/graph... Is each input has a different output, this canbe classified as a function 12th Grade connect! It is a function passes the vertical line only allowed to have one y-value ( or input ) right (! Goes straight up and down use this site with cookies that would pass the vertical line through! Problems from vertical line intersects the graph is the vertical line anywhere on the line, then curve! Take a vertical line drawn through the graph passes the vertical line test CPhill, the vertical test... Or input ) right the line draw a vertical line test over again since 's! Test to Identify functions Practice Problems from vertical line test, it is a line! Be paired to a single y-value at one point only then one value! Ways to determine if the graph at more than one point, then curve. Whether this graph is a function graph in some places at more than one place have a about... Relations that are also functions because they pass the vertical lines intersect a curve at most then... Then it is the vertical line test supports the definition of a relation is function... Is equal to negative five function '' the graph create a that! T it is a function prove whether the inverses are actually functions or not it pass vertical. Your graph observe the points of intersection test you can check whether this is. It pass the vertical line intersects the graph passes the vertical line on the.. So a vertical line drawn through the graph test use to determine whether a given relation is a or! Not enter ± on Desmos and show both values simultaneously need to prove whether the are! That intersect the graph in all places at more than one point graph of function. Touches only one output for an x value -2 ) to the 0 power points. Of relations that are also functions because they pass the vertical line test, you have a question.. Passes the vertical line and place it on the graph at exactly one only! A relation is not a function relations that are also functions because they pass the line. It pass the vertical line test can be straight lines or segments, curves or! To give you the best experience does a vertical line pass the vertical line test our website even just a set of points point graph of definition. To the 0 power line that would pass the vertical line test, you have a question.! Number system all over again since it 's just going to Look like this to... Can use to determine if a relation is a function by using the site segments! Ruler, and pass it over the graph in all places at than... Admits he 's embarrassed by early work if all vertical lines that intersect graph! Is in the slope formula is zero equation of a function, a vertical line test use. Graphs of functions places at more than one point on the graph in all places more... Cookies off or discontinue using the vertical line test equation is x is to. Drawn for the mathematical relation: can you observe the points of intersection but it. That intersect the graph at one of the line, well that just goes straight and! Points given by the applet, pupils Try to connect them to create a line that would pass the line!
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## Intermediate Algebra (12th Edition)
$\bf{\text{Solution Outline:}}$ Substitute the given value for $x$ in the given equation, $\sqrt{3x+18}-x=0 .$ If the left side of the equation becomes equal to the right side of the equation, then the given value of $x$ is a solution to the equation. $\bf{\text{Solution Details:}}$ a) Substituting $x$ with $6 ,$ in the given equation results to \begin{array}{l}\require{cancel} \sqrt{3(6)+18}-6=0 \\\\ \sqrt{18+18}-6=0 \\\\ \sqrt{36}-6=0 \\\\ 6-6=0 \\\\ 0=0 \text{ (TRUE)} .\end{array} Hence, $x= 6$ is a solution to the given equation. b) Substituting $x$ with $-3 ,$ in the given equation results to \begin{array}{l}\require{cancel} \sqrt{3(-3)+18}-(-3)=0 \\\\ \sqrt{-9+18}+3=0 \\\\ \sqrt{9}+3=0 \\\\ 3+3=0 \\\\ 6=0 \text{ (FALSE)} .\end{array} Hence, $x= -3$ is NOT a solution to the given equation.
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# KSEEB Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.1
Students can Download Chapter 7 Congruence of Triangles Ex 7.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.
## Karnataka State Syllabus Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.1
Question 1.
Complete the following statements :
a) Two line segments – are congruent if ___.
Solution:
They have the same measure, (length)
b) Among two congruent angles, one has a measure of 70°; the measure of the other angle is ___
Solution:
70°
c) When we write ∠A = ∠B, we actually mean ___
m < A ≅ m
a
Question 2.
Give any two real-life examples for congruent shapes.
Solution:
Two Rs 10/- notes and two ten rupees coins.
Question 3.
If ∆ ABC ≅ ∆ FED under the correspondence ABC ⟷ FED, write all the corresponding congruent parts of the triangles.
Solution:
The ∆ ABC ≅ ∆ FED then the corresponding vertices A and F, B and E, C and D. The corresponding sides are $\overline{\mathrm{AB}}$ and FE, BC and ED, CA, and DF. The corresponding angles are ∠A and ∠F, ∠B and ∠E, and ∠C and ∠D.
Question 4.
If ∆ DEF ≅ ∆ BAC, write the part(s) of ∆ BCA that correspond to
Solution:
i) ∠E
∠E ≅ ∠C,
ii) $$\overline{\mathbf{E} \mathbf{F}}$$
$$\overline{\mathbf{E} \mathbf{F}}$$ ≅ $$\overline{\mathbf{C} \mathbf{A}}$$
iii) ∠F
∠F ≅ ∠A
iv) $$\overline{\mathbf{D F}}$$
$$\overline{\mathbf{D F}}$$ ≅ $$\overline{\mathbf{B A}}$$
a
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Probability theory by example (Part 2)
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Outcomes, sample spaces, and events
In part 1 we considered some examples of random experiments and sketched some probability concepts that we will cover more thoroughly later. In particular we decided that the probability of tossing heads in one flip of a coin should be 1/2. This was easy because coin tosses tend to have exactly two equally likely outcomes. Consider the following, less trivial, experiment.
• Randomly select a word from a particular book. What is the probability that the word only appears once in the book?
This example is more difficult than a coin flip. To answer this question it seems we would have to count the number of times each word in the book appears. Generally this approach of counting outcomes is our first step. But there is something very special about this experiment.
In every book, in any language, approximately 1/2 the words only appear once.
The distribution of words in books tend to follow a type of probability distribution called Zipf’s Law.
Before getting into exotic probability distributions like Zipf’s Law we should work on trying to understand the set of possible outcomes of random processes. Let’s settle on some vocabulary.
• Experiment: any random process we are observing, i.e. the rolling of a die, randomly selecting a person from a group, and so on
• Outcome: a single possible result of an experiment
• Sample space (Ω): the set of all outcomes of an experiment. it is denoted by the Greek letter omega
• Event: any collection, or set, of outcomes
Consider the experiment where we roll a 6-sided die and flip a coin. The sample space Ω is the set of all possible outcomes. In this case there are 12 possible outcomes. Each outcome is listed as a number with a letter. Numbers indicate dice results and letters for coin toss results.
Ω = {1H,2H,3H,4H,5H,6H,1T,2T,3T,4T,5T,6T}.
We use curly braces around the elements of the sample space to indicate that they are members of a set. We will discuss sets later. From the sample space we can define all kinds of events.
A = {2H,3H}, B = {6H,6T}, C ={1H}
D = {all outcomes with dice rolls greater than 2}
Events can contain any elements of the sample space or it can be empty. The sample space Ω itself is an event. Each individual outcome can be thought of an event. We can define events without having to list its individual outcomes like we did for D. It’s a good thing we can do this because sample spaces can get very large, infinite in fact.
Events can be infinite as well. Let’s say our experiment is to randomly select any positive integer. The sample space is Ω = {1,2,3,4,…}. We can define an event containing all even integers in the sample space or all prime numbers. In this example we didn’t have to think too hard to decide how to represent possible outcomes. They were just integers. But as the following example shows, sometimes it’s not easy to write down outcomes.
Example 2.1 Two friends are having a basketball free-throw competition. They will take turns shooting free-throw shots. They agree to keep playing until one of them scores two shots in a row or one of them misses two shots in a row. Describe the sample space.
An outcome for this experiment is a sequence of shot results for the players. Let’s say 1 indicates a successful shot and 0 indicates a miss. The sequence will be composed of pairs of results, corresponding to the two players. Here is an example outcome
(1,0)(0,1)(1,1).
The game ended after three rounds because the second player scored two shots in a row. The games can be of any length so the sample space is infinite. There are 4 outcomes where the games last forever. Here is one of them.
(1,0)(0,1)(1,0)(0,1)…
Our ultimate goal is to learn to calculate probabilities of events from a sample space. So our choice of sample space should depend on the the events we are interested in.
Example 2.2 Let’s say we have a medical test for colon cancer. Your goal is to find the probability the the test will come back as a false positive for a randomly selected patient. What sample space should you consider (a false positive result occurs when a positive test result is given to someone without the illness)?
The test can come back with two possible results, positive and negative. And the patient may have the illness or not. So we should consider all four combinations in our sample space.
Ω = {(positive, no disease), (positive, has disease),(negative, no disease), (negative, has disease)}
The first outcome is a false positive and the last outcome is a false negative. We will return to this example later.
We’ve seen finite sample spaces and infinite sample spaces, but so far all our sample spaces have been countable. A set is countable when you can label each of its elements with a positive integer. Some infinite sets are countable but some are of a higher order of infinity. The set of all real numbers (and intervals of real numbers) are un-countable. It is necessary to treat experiments with countable sample spaces differently than those with un-countable sample spaces. There are several reasons for this and we’ll discuss them later.
Example 2.3 Randomly choose a set of 10 days in 2018 and find the average daily temperature for each day. Describe the sample space.
Temperatures occupy some un-countable interval of real numbers. Of course thermometers have limited precision so they can only output a finite number of values. But for now let’s ignore measurement precision.
Conclusion
We learned how to identify a sample space of an experiment. The sample spaces we choose should match the events we are interested in. We discussed the possibility of having finite, infinite, and un-countably infinite sample spaces. Next we will learn some useful counting techniques. Below are some exercises. The solutions will be in part 3.
Exercises
1. Roll two dice. Let E be the event that the first dice rolls 5 or 6. Let F be the event that the sum of the rolls is greater than 7. How many outcomes are in the sample space, in E, and in F?
2. Define the following experiment. (1) Randomly select a letter from the alphabet that hasn’t been selected yet (2) If the letter is a vowel (A,E,I,O, or U), then go back to step 1. Otherwise, stop. Describe the sample space of this experiment.
3. Find all four infinitely long outcomes from Example 2.1.
4. Consider the task of pairing a randomly selected student with a grad school. Your goal is to measure the probability that the pairing is a good fit for both parties. Describe some useful sample spaces for this experiment.
5. Randomly choose a person from some population and find their height. Our sample space will consist of some interval of real numbers. What might you say about the probability of any single height in our sample space?
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You are reading an older version of this FlexBook® textbook: Algebra II Go to the latest version.
# 12.1: Modeling Data with Functions
Difficulty Level: At Grade Created by: CK-12
Name: __________________
Direct Variation
1. Suppose I make $\6/\mathrm{hour}$. Let $t$ represent the number of hours I work, and $m$ represent the money I make.
a. Make a table showing different $t$ values and their corresponding $m$ values. ($m$ is not how much money I make in that particular hour—it’s how much total money I have made, after working that many hours.)
time $(t)$ money $(m)$
b. Which is the dependent variable, and which is the independent variable?
c. Write the function.
d. Sketch a quick graph of what the function looks like.
e. In general: if I double the number of hours, what happens to the amount of money?
2. I am stacking bricks to make a wall. Each brick is $4$" high. Let $b$ represent the number of bricks, and $h$ represent the height of the wall.
a. Make a table showing different $b$ values and their corresponding $h\;\mathrm{values}$.
bricks $(b)$ height $(h)$
b. Which is the dependent variable, and which is the independent variable?
c. Write the function.
d. Sketch a quick graph of what the function looks like.
e. In general: if $I$ triple the number of bricks, what happens to the height?
3. The above two scenarios are examples of direct variation. If a variable $y$ “varies directly” with $x$, then it can be written as a function $y=kx$, where $k$ is called the constant of variation. (We also sometimes say that “$y$ is proportional to $x$,” where $k$ is called the constant of proportionality. Why do we say it two different ways? Because, as you’ve always suspected, we enjoy making your life difficult. Not “students in general,” but just you personally.) So, if $y$ varies directly with $x \ldots$
a. What happens to $y$ if $x$ doubles? (Hint: You can find and prove the answer from the equation $y=kx$.)
b. What happens to $y$ if $x$ is cut in half?
c. What does the graph $y(x)$ look like? What does $k$ represent in this graph?
4. Make up a word problem like numbers $(1)$ and $(2)$ above, on the subject of fast food. Your problem should not involve getting paid or stacking bricks. It should involve two variables that vary directly with each other. Make up the scenario, define the variables, and then do problems a-e exactly like my two problems.
Name: __________________
Homework Inverse Variation
1. An astronaut in space is performing an experiment with three balloons. The balloons are all different sizes, but they have the same amount of air in them. As you might expect, the balloons that are very small experience a great deal of air pressure (the air inside pushing out on the balloon); the balloons that are very large, experience very little air pressure. He measures the volumes and pressures and comes up with the following chart.
Volume $(V)$ Pressure $(P)$
$5$ $270$
$10$ $135$
$15$ $90$
$20$ $67\frac{1}{2}$
a. Which is the dependent variable, and which is the independent variable?
b. When the volume doubles, what happens to the pressure?
c. When the volume triples, what happens to the pressure?
d. Based on your answers to parts (a) - (c), what would you expect the pressure to be for a balloon with a volume of $30$?
e. On the right of the table add a third column that represents the quantity $\mathrm{PV}$: pressure times volume. Fill in all four values for this quantity. What do you notice about them?
f. Plot all four points on the graph paper, and fill in a sketch of what the graph looks like.
g. Write the function $P(V)$. Make sure that it accurately gets you from the first column to the second in all four instances! (Part (e) is a clue to this.)
h. Graph your function $P(V)$ on your calculator, and copy the graph onto the graph paper. Does it match your graph in part $(f)$?
2. The three little pigs have built three houses—made from straw, Lincoln Logs$^\circledR$, and bricks, respectively. Each house is $20'\mathrm{high}$. The pieces of straw are $\frac{1}{10}"$ thick; the Lincoln Logs$^\circledR$ are $1"$ thick; the bricks are $4"$ thick. Let $t$ be the thickness of the building blocks, and let $n$ be the number of such blocks required to build a house $20'\mathrm{high}$. (Note: There are $12"$ in $1'$. But you probably knew that ...)
a. Make a table showing different $t$ values and their corresponding $n\;\mathrm{values}$.
Building Blocks thickness $(t)$ number $(n)$
Straw
Lincoln Logs$^\circledR$
Bricks
b. Which is the dependent variable, and which is the independent variable?
c. When the thickness of the building blocks doubles, what happens to the number required?
(*Not sure? Pretend that the pig’s cousin used $8"$ logs, and his uncle used $16"$ logs. See what happens to the number required as you go up in this sequence...)
d. When the thickness of the building blocks is halved, what happens to the number required?
e. On the right of the table add a fourth column, that represents the quantity $tn$: thickness times number. Fill in all three values for this quantity. What do you notice about them? What do they actually represent, in our problem?
f. Plot all three points on the graph paper, and fill in a sketch of what the graph looks like.
g. Write the function $n(t)$.
h. Graph your function $n(t)$ on your calculator, and copy the graph onto the graph paper. Does it match your graph in part (f)?
3. The above two scenarios are examples of inverse variation. If a variable $y$, “varies inversely” with $x$, then it can be written as a function $y=\frac{k}{x}$, where $k$ is called the constant of variation. So, if $y$ varies inversely with $x$...
a. What happens to $y$ if $x$ doubles? (Hint: You can find and prove the answer from the equation $y=\frac{k}{x}$ )
b. What happens to $y$ if $x$ is cut in half?
c. What does the graph $y(x)$ look like? What happens to this graph when $k$ increases? (* You may want to try a few different ones on your calculator to see the effect $k$ has.)
4. Make up a word problem like numbers $(1)$ and $(2)$ above. Your problem should not involve pressure and volume, or building a house. It should involve two variables that vary inversely with each other. Make up the scenario, define the variables, and then do problems a-h exactly like my two with each other. Make up the scenario, define the variables, and then do problems a-h exactly like my two problems.
Name: __________________
Homework Direct and Inverse Variation
For questions $1-3$, Please note that these numbers are meant to simulate real world data—that is to say, they are not necessarily exact! If it is “darn close to” direct or inverse variation, that’s good enough.
1. For the following set of data...
$x$ $y$
$3$ $5$
$6$ $11$
$21$ $34$
a. Does it represent direct variation, inverse variation, or neither?
b. If it is direct or inverse, what is the constant of variation?
c. If $x=30$, what would $y$ be?
d. Sketch a quick graph of this relationship.
2. For the following set of data...
$x$ $y$
$3$ $18$
$4$ $32$
$10$ $200$
a. Does it represent direct variation, inverse variation, or neither?
b. If it is direct or inverse, what is the constant of variation
c. If $x=30$, what would $y$ be?
d. Sketch a quick graph of this relationship.
3. For the following set of data$\ldots$
$x$ $y$
$3$ $20$
$6$ $10$
$21$ $3$
a. Does it represent direct variation, inverse variation, or neither?
b. If it is direct or inverse, what is the constant of variation?
c. If $x=30$, what would $y$ be?
d. Sketch a quick graph of this relationship.
4. In number $2$ above, as you (hopefully) saw, the relationship is neither direct nor inverse. However, the relationship can be expressed this way: $y$ is directly proportional to $x^2$. Write the function that indicates this relationship. What is $k$? one table is miss
5. In June, 2007, Poland argued for a change to the voting system in the European Union Council of Ministers. The Polish suggestion: each member’s voting strength should be directly proportional to the square root of his country’s population. This idea, traditionally known as Pensore’s Rule, is “almost sacred” among “people versed in the game theory of voting” according to one economist.
I swear I am not making this up.
Also in the category of “things I am not making up,” the following table of European Populations comes from Wikipedia.
$&\text {Germany} && 83,251,851\\& \text{Italy} && 59,715,625\\&\text{Poland} && 38,625,478\\&\text{Luxemberg} && 48,569$
a. Write an equation that represents Pensore’s Rule. Be sure to clearly label your variables.
b. Suppose that Pensore’s Rule was followed, and suppose that Poland voting strength was exactly $100$ (which I did actually make up, but of course it doesn’t matter). What would the voting strength of Germany, Italy, and Luxembourg be?
6. Write a “real world” word problem involving an inverse relationship, on the topic of movies. Identify the constant of variation. Write the function that shows how the dependent variable depends inversely upon the independent variable. Create a specific numerical question, and use your function to answer that question.
7. Joint Variation: The term “Joint Variation” is used to indicate that one variable varies directly as two different variables. This is illustrated in the following example.
Al is working as a waiter. When a group of people sit down at a table, he calculates his expected tip $(T)$ as follows: multiply the number of people $(N)$, times the average meal cost $(C)$, times $0.15$ (for a $15\%$ tip).
a. If the number of people at the table doubles, does Al’s expected tip double?
b. If the average cost per meal doubles, does Al’s expected tip double?
c. Write the function that expresses the dependent variable, $T$, as a function of the two independent variables, $N$ and $C$.
d. Write the general function that states “$z$ varies jointly as both $x$ and $y$.” Your function will have an unknown $k$ in it, a constant of variation.
8. Light Intensity: Stacy visits a tanning booth, where she spends several hours with lamps shining on her skin, thus giving her a beautiful copper-colored tan and a sharply increased risk of skin cancer. For reasons known only to herself, she considers this a good trade-off. Anyway, Stacy has a lot of time to just lie there and think, and she starts to consider the question: which bulb is shining on her skin with the most intensity? The answer is that the intensity I of a bulb varies directly with the strength $S$ of the bulb, and varies inversely with the square of the distance $d$ of the bulb from her skin.
a. Bulbs $A$ and $B$ are the same distance away, but bulb $B$ is twice as strong as bulb $A$. If bulb $A$ shines with an intensity of $5$, what is the intensity of bulb $B$?
b. Bulbs $A$ and $C$ are the same strength as each other, but bulb $A$ is twice as far away from Stacy as bulb $C$. If bulb $A$ shines with an intensity of $5$, what is the intensity of bulb $B$?
c. Write a function to represent the statement “the intensity I of a bulb varies directly with the strength $S$ of the bulb, and varies inversely with the square of the distance d of the bulb from Stacy’s skin.” Your function will have an unknown $k$ in it, a constant of variation.
Name: _________________
Homework: Calculator Regression
1. Canadian Voters: The following table shows the percentage of Canadian voters who voted in the 1996 federal election.
$&\text{Age} && 20 && 30 && 40 && 50 && 60\\&\% \text{voted} && 59 && 86 && 87 && 91 && 94$
a. Enter these points on your calculator lists.
b. Set the Window on your calculator so that the $x-$values go from $0$ to $60$, and the $y-$values go from $0$ to $100$. Then view a graph of the points on your calculator. Do they increase steadily (like a line), or increase slower and slower (like a log), or increase more and more quickly (like a parabola or an exponent)?
c. Use the [STAT] function on your calculator to find an appropriate function to model this data. Write that function below.
d. Graph the function on your calculator. Does it match the points well? Are any of the points “outlyers?”
2. Height and Weight: A group of students record their height (in inches) and weight (in pounds). The results are on the table below.
$&\text{Height} && 68 && 74 && 66 && 68 && 72 && 69 && 65 && 71 && 69 && 72 && 71 && 64 && 65\\&\text{weight} && 180 && 185 && 150 && 150 && 200 && 160 && 125 && 220 && 220 && 180 && 190 &&120 && 110$
a. Enter these points on your calculator lists.
b. Set the Window on your calculator appropriately, and then view a graph of the points on your calculator. Do they increase steadily (like a line), or increase slower and slower (like a log), or increase more and more quickly (like a parabola or an exponent)?
c. Use the [STAT] function on your calculator to find an appropriate function to model this data. Write that function below.
d. Graph the function on your calculator. Does it match the points well? Are any of the points “outlyers?”
3. Gas Mileage: The table below shows the weight (in hundreds of pounds) and gas mileage (in miles per gallon) for a sample of domestic new cars.
$&\text{Weight} && 29 && 35 && 28 && 44 && 25 && 34 && 30 && 33 && 28 && 24\\&\text{mileage} && 31 && 27 && 29 && 25 && 31 && 29 && 28 && 28 && 28 && 33$
a. Enter these points on your calculator lists.
b. Set the Window on your calculator appropriately, and then view a graph of the points on your calculator. Do they decrease steadily (like a line), or decrease slower and slower (like a log), or decrease more and more quickly (like a parabola or an exponent)?
c. Use the [STAT] function on your calculator to find an appropriate function to model this data. Write that function below.
d. Graph the function on your calculator. Does it match the points well? Are any of the points “outlyers?”
4. TV and GPA: A graduate student named Angela Hershberger at Indiana University-South Bend did a study to find the relationship between TV watching and Grade Point Average among high school students. Angela interviewed $50$ high school students, turning each one into a data point, where the independent $(x)$ value was the number of hours of television watched per week, and the dependent $(y)$ value was the high school grade point average. (She also checked the types of television watched—eg news or sitcoms—and found that it made very little difference. Quantity, not quality, mattered.)
In a study that you can read all about at www.iusb.edu/~journal/2002/hershberger/hershberger.html, Angela found that her data could best be modeled by the linear function $y=–0.0288 x + 3.4397$. Assuming that this line is a good fit for the data
a. What does the number $3.4397$ tell you? (Don’t tell me about lines and points: tell me about students, TV, and grades.)
b. What does the number $–0.0288$ tell you? (Same note.)
Name: _________________
Sample Test: Modeling Data with Functions
1. Three cars and an airplane are traveling to New York City. But they all go at different speeds, so they all take different amounts of time to make the $500-$mile trip. Fill in the following chart.
Speed(s) - miles per hour Time(t) - hours
$50$
$75$
$100$
$500$
a. Is this an example of direct variation, inverse variation, or neither of the above?
b. Write the function $s(t)$.
c. If this is one of our two types, what is the constant of variation?
2. There are a bunch of squares on the board, of different sizes.
s - length of teh side of a square A - area of the square
$1$
$2$
$3$
$4$
a. Is this an example of direct variation, inverse variation, or neither of the above?
b. Write the function $A(s)$.
c. If this is one of our two types, what is the constant of variation?
3. Anna is planning a party. Of course, as at any good party, there will be a lot of on hand! $50$ Coke cans fit into one recycling bin. So, based on the amount of Coke she buys, Anna needs to make sure there are enough recycling bins.
c-Coke cans Anna buys b - recycling bins she will need
$50$
$100$
$200$
$400$
a. Is this an example of direct variation, inverse variation, or neither of the above?
b. Write the function $b(c)$.
c. If this is one of our two types, what is the constant of variation?
4. Make up a word problem involving inverse variation, on the topic of skateboarding.
a. Write the scenario.
b. Label and identify the independent and dependent variables.
c. Show the function that relates the dependent to the independent variable. This function should (of course) be an inverse relationship, and it should be obvious from your scenario!
5. I found a Web site (this is true, really) that contains the following sentence:
[This process] introduces an additional truncation error [directly] proportional to the original error and inversely proportional to the gain $(g)$ and the truncation parameter $(q)$.
I don’t know what most of that stuff means any more than you do. But if we use $T$ for the “additional truncation error” and E for the “original error,” write an equation that expresses this relationship.
6. Which of the following correctly expresses, in words, the relationship of the area of a circle to the radius?
A. The area is directly proportional to the radius
B. The area is directly proportional to the square of the radius
C. The area is inversely proportional to the radius
D. The area is inversely proportional to the square of the radius
7. Now, suppose we were to write the inverse of that function: that is, express the radius as a function of the area. Then we would write:
The radius of a circle is ___________ proportional to _____________________ the area.
8. Death by Cholera: In 1852, William Farr reported a strong association between low elevation and deaths from cholera. Some of his data are reported below.
$&\text{E:Elevation(ft)} && 10 && 30 && 50 && 70 && 90 && 100 && 350\\&\text{C:Cholera morality}(\text{per} 10,000) && 102 && 65 && 34 && 27 && 22 && 17 && 8$
a. Use your calculator to create the following models, and write the appropriate functions $C(E)$ in the blanks.
Linear: $C=$
Quadratic: $C=$
Lograthamic: $C=$
exponential: $C=$
b. Which model do you think is the best? Why?
c. Based on his very strong correlation, Farr concluded that bad air had settled into low-lying areas, causing outbreaks of cholera. We now know that air quality has nothing to do with causing cholera: the water-borne bacterial Vibrio cholera causes the disease. What might explain Farr’s results without justifying his conclusion?
Feb 23, 2012
Oct 30, 2014
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# How do you graph the function, label the vertex, axis of symmetry, and x-intercepts. x-4 = 1/4 (y+1)^2?
Jul 11, 2015
The vertex is at ($4 , - 1$).
The axis of symmetry is $y = - 1$.
The $x$-intercept is ($\frac{17}{4} , 0$).
There are no $y$-intercepts.
#### Explanation:
The standard form for the equation of a parabola is
$y = {a}^{2} + b x + c$
x−4 = 1/4(y+1)^2
We must get this into standard form.
x−4=1/4(y^2 + 2y + 1)
x−4=1/4y^2 + 1/4(2y)+ 1/4
x−4=1/4y^2 + 1/2y + 1/4
$x = \frac{1}{4} {y}^{2} + \frac{1}{2} y + \frac{1}{4} + 4$
$x = \frac{1}{4} {y}^{2} + \frac{1}{2} y + \frac{17}{4}$
This is standard form, but with $x$ and $y$ interchanged.
We are going to get a sideways parabola.
$a = \frac{1}{4}$, $b = \frac{1}{2}$, and $c = \frac{17}{4}$.
Vertex
Since $a > 0$, the parabola opens to the right.
The $y$-coordinate of the vertex is at
y = –b/(2a) = -(1/2)/(2×(1/4)) = -(1/2)/(1/2) = -1.
Insert this value of $x$ back into the equation.
$x = \frac{1}{4} {y}^{2} + \frac{1}{2} y + \frac{17}{4}$
x=1/4(-1)^2 + 1/2(-1) + 17/4 = 1/4×1 -1/2 +17/4
$x = \frac{1}{4} - \frac{1}{2} + \frac{17}{4} = \frac{1 - 2 + 17}{4} = \frac{16}{4} = 4$
The vertex is at ($4 , - 1$).
Axis of symmetry
The axis of symmetry must pass through the vertex, so
The axis of symmetry is $y = - 1$.
$x$-intercept
To find the $x$-intercept, we set $y = 0$ and solve for $x$.
x=1/4y^2 + 1/2y + 17/4 = 1/4×0^2 + 1/2×0 + 17/4 = 0 + 0 + 17/4 = 17/4
The $x$-intercept is at ($\frac{17}{4} , 0$).
$y$-intercepts
To find the $y$-intercepts, we set $x = 0$ and solve for $y$.
$x = \frac{1}{4} {y}^{2} + \frac{1}{2} y + \frac{17}{4}$
$0 = \frac{1}{4} {y}^{2} + \frac{1}{2} y + \frac{17}{4}$
$0 = {y}^{2} + 2 y + 17$
y = (-b ± sqrt(b^2-4ac))/(2a)
The discriminant D = b^2 – 4ac = 4^2 – 4×1×17= 8 – 68 = -60
Since $D < 0$, there are no real roots.
There are no $y$-intercepts.
Graph
Now we prepare a table of $x$ and $y$ values.
The axis of symmetry passes through $y = - 1$.
Let's prepare a table with points that are 5 units on either side of the axis, that is, from $y = - 6$ to $y = 4$.
Plot these points.
And we have our graph. The red line is the axis of symmetry.
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Mahesh Godavarti
1
The steps to evaluate 2/4 + 2/3 are: 1. Rewrite each fraction in its simplest form, i.e. 1/2 + 2/3 2. Rewrite each fraction in a form that has the same denominator (the denominator should be least common multiple of the two denominators), i.e. 3/6 + 4/6 3. Simply add now that both fractions share the same denominator, i.e. 7/6 ------------------------ More explanation below: ------------------------ Adding fractions is easy, once we know exactly what they mean. Do we know how to evaluate 3/6 + 4/6? If yes, then we know how to evaluate 2/4 + 2/3. Because 2/4 is the same as 3/6 and 2/3 is the same as 4/6. Why is it easy to evaluate 3/6 + 4/6? First, 3/6 is the same as 3 one-sixths, and 4/6 is the same as 4 one-sixths. If there are 3 one-sixths and 4 one-sixths, they together make 7 one-sixths. Therefore, 3/6 + 4/6 = 7/6. It is easy to count how many portions there are if they are all the same of kind. So, what we need to do is to make sure we are counting the same kind of portions. 2/4 + 2/3 can be interpreted as 2 one-fourths and 2 one-thirds. This cannot simply be collapsed into a single "so many portions". So, the trick is to convert both fractions into describing the same portions. 2/4 which is 2 one-fourths is the same as 1 one-half, 3 one-sixths, 4 one-eights, or 5 one-tenths etc. 2/3 which is 2 one-thirds is the same as 4 one-sixths, 6 one-ninths, or 8 one-twelfths etc. Out of all these descriptions, the descriptions that describe the same portions are 3 one-sixths and 4 one-sixths. Therefore, 2/4 + 2/3 = 3/6 + 4/6 = 7/6.
TheGOAT
0
thats to much for me I don't got time for that
Mahesh Godavarti
0
Rewrote the answer for you. The first part is sufficient if you don't have time.
Vivekanand Vellanki
0
Visual representation of adding the fractions. These are useful to understand addition of fractions. It is better to use the short-cuts as outlined in Mahesh's response for a faster calculation.
Sherlyn Casco Noperi
0
You cross multiply and get the answer.
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Write The nth Partial Sum Equation of each Arithmetic Sequence or Arithmetic Series
Problem 1. Find the sum of the first n natural numbers.
Solution:
The first n natural numbers are 1, 2, 3, 4, 5, …, n.
Here, a=1 and d=(2-1)=1.
Sum of n terms of an AP is given by
Snn[2a+(n-1)d]
Snn[2⋅1+(n-1)⋅1]
Snn[2+(n-1)]
Snn(n+1)
Example 7: Find the sum of 23 terms and n terms of the A.P. 16, 11, 6, 1 …
Solution: Here a=16, a+d=11 Therefore d=-5
Since Snn[2a+(n-1)d] we get
S23=½⋅23⋅[2⋅16+(23-1)(-5)]
=23⋅½⋅(32-110)
=23⋅(-78)=-897.
Also Snn[2⋅16+(n-1)(-5)]=½n[32-5n+5]=½n[37-5n]
P2. Find the sum of first n even natural numbers.
Solution:
The first n even natural numbers are 2, 4, 6, 8, 10, …, n.
Here, a=2 and d=(4-2)=2.
Sum of n terms of an AP is given by
Snn[2a+(n-1)d]
Snn[2⋅2+(n-1)⋅2]
Snn[4+2n-2]
Snn(2n+2)
Sn=n(n+1)
Hence, the required sum is n(n+1).
P3. Find the sum of first n terms of an AP whose nth term is (5-6n). Hence, find the sum of its first 20 terms.
Solution:
Let an be the nth term of the AP.
an=5-6n
Putting n=1, we get
First term, a=a1=5-6⋅1=-1
Putting n=2, we get
a2=5-6⋅2=-7
Let (1 be the common difference of the AP.
d=a2a1=-7-(-1)=-7+1=-6.
Sum of first n term of the AP, Sn
Snn[2a+(n-1)d]
Snn[2∙(-1)+(n-1)∙(-6)]
=½∙n(-2-6n+6)
=n(2-3n)
=2n-3n2
Putting n=20, we get
S20=2⋅20-3⋅202=40-1200=-1160
P4. The sum of the first 7 terms of an AP is 49 and the sum of its first 17 terms is 289. Find the sum of its first n terms.
Solution:
Let a be the first term and d be the common difference of the given AP. Then, we have:
Snn[2a+(n-1)d]
S7=½⋅7⋅(2a+6d)=7⋅(a+3d)
S17=½⋅17⋅(2a+16d)=17⋅(a+8d)
However, S7=49 and S17=289.
Now, 7(a+3d)=49
a+3d=7 … (i)
Also, 17(a+8d)=289
a+8d=17 … (ii)
Subtracting (i) from (ii), we ge:
=5d=10
d=2
Putting d=2 in (i), we get
a+6=7
a=1
Thus, a=1 and d=2.
∴ Sum of n terms of AP
n[2⋅1+(n-1)⋅2]=n[1+(n-1)]=n2
Let’s read a particular post Derivation of the partial sum formula of every Arithmetic Series.
P5. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
Here, S7=49 and S17=289.
The sum of n terms of an AP is given by
Snn[2a+(n-1)d]
S7=½⋅7⋅[2a+6d]
49=7⋅(a+3d)
7=a+3d
a=7-3d … (1)
and S17=½⋅17⋅[2a+16d]
289=17⋅(a+8d)
17=a+8d
Putting the value of a from equation (1), we get
17=7-3d+8d
5d=10
d=2
Putting the value of d in equation (1), we get
a=7-3⋅2=1
The sum of n terms of an AP is given by
Snn[2a+(n-1)d]
n[2⋅1+(n-1)⋅2]
n(2+2n-2)=n2
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Topics in
P R E C A L C U L U S
1
# The Formal Rules of Algebra
ALGEBRA is a method of written calculations. And a calculation is replacing one set of symbols with another. In arithmetic we may replace the symbols '2 + 2' with the symbol '4.' In algebra we may replace 'a + (−a)' with '0.'
a + (−a) = 0.
A formal rule, then, shows how an expression written in one form may be rewritten in a different form. The = sign means "may be rewritten as" or "may be replaced by."
If p and q are statements (equations), then a rule
If p, then q,
or equivalently
p implies q,
means: We may replace statement p with statement q. For example,
x + a = b implies x = b − a.
That means that we may replace the statement 'x + a = b' with the statement 'x = b − a.'
Algebra depends on how things look. We can say, then, that algebra is a system of formal rules. The following are what we are permitted to write.
(See the complete course, Skill in Algebra.)
11. The axioms of "equals"
a = a Identity If a = b, then b = a. Symmetry If a = b and b = c, then a = c. Transitivity
It is not possible to give an explicit definition of the word "equals," or its symbol = . Those rules however are an implicit definition. The meaning of "equals" implies those three rules.
As for how the rule of symmetry comes up in practice, see Lesson 6 of Algebra. The rule of symmetry applies to all of the rules below.
12. The commutative rules of addition and multiplication
a + b = b + a a· b = b· a
13. The identity elements of addition and multiplication:
3. 0 and 1
a + 0 = 0 + a = a
a· 1 = 1· a = a
Thus, if we "operate" on a number with an identity element,
it returns that number unchanged.
14. The additive inverse of a: −a
a + (−a) = −a + a = 0
The "inverse" of a number undoes what the number does.
For example, if you start with 5 and add 2, then to get back to 5 you must add −2. Adding 2 + (−2) is then the same as adding 0 -- which is the identity.
15. The multiplicative inverse or reciprocal of a, 5. symbolized as 1a (a 0)
a· 1a = 1a · a = 1
Two numbers are called reciprocals of one another if their product is 1.
Thus, 1/a symbolizes that number which, when multiplied by a, produces 1.
The reciprocal of pq is qp .
16. The algebraic definition of subtraction
ab = a + (−b)
Subtraction, in algebra, is defined as addition of the inverse.
17. The algebraic definition of division
ab = a· 1b
Division, in algebra, is defined as multiplication by the reciprocal.
Hence, algebra has two fundamental operations: addition and multiplication.
18. The inverse of the inverse
−(−a) = a
19. The relationship of ba to ab
ba = −(ab)
Now, b + a is equal to a + b. But ba is the negative of ab.
10. The Rule of Signs for multiplication, division, and
10. fractions
a(−b) = −ab. (−a)b = −ab. (−a)(−b) = ab.
a−b = − ab . −a b = − ab . −a−b = ab .
"Like signs produce a positive number; unlike signs, a negative number."
11. Rules for 0
a· 0 = 0· a = 0
If a 0, then
0a = 0. a0 = No value. 00 = Any number.
Division by 0 is an excluded operation. (Skill in Algebra, Lesson 5.)
m(a + b) = ma + mb The distributive rule/ Common factor (x − a)(x − b) = x2 − (a + b)x + ab Quadratic trinomial (a ± b)2 = a2 ± 2ab + b2 Perfect square trinomial (a + b)(a − b) = a2 − b2 The difference of two squares (a ± b)(a2 ab + b2) = a3 ± b3 The sum or difference of two cubes
13. The same operation on both sides of an equation
If If a = b, a = b, then then a + c = b + c. ac = bc.
We may add the same number to both sides of an equation;
we may multiply both sides by the same number.
14. Change of sign on both sides of an equation
If −a = b, then a = −b.
We may change every sign on both sides of an equation.
15. Change of sign on both sides of an inequality:
15. Change of sense
If a < b, then −a > −b.
When we change the signs on both sides of an inequality, we must change the sense of the inequality.
16. The Four Forms of Equations corresponding to the
16. Four Operations and their inverses
If If x + a = b, x − a = b, then then x = b − a. x = a + b.
* * *
If If ax = b, x a = b, then then x = ba . x = ab.
17. Change of sense when solving an inequality
If −ax < b, then x > − ba .
18. Absolute value
If |x| = b, then x = b or x = −b.
If |x| < b then −b < x < b.
If |x| > b (and b > 0), then x > b or x < −b.
19. The principle of equivalent fractions
xy = axay and symmetrically, axay = xy
We may multiply both the numerator and denominator by the same factor; we may divide both by a common factor.
20. Multiplication of fractions
ab · cd = acbd a · cd = acd
21. Division of fractions (Complex fractions)
Division is multiplication by the reciprocal.
ac + bc = a + b c Same denominator ab + cd = ad + bc bd Different denominators withno common factors a bc + e cd = ad + be bcd Different denominators withcommon factors
The common denominator is the LCM of denominators.
23. The rules of exponents
aman = am+n Multiplying or dividing aman = am−n powers of the same base (ab)n = anbn Power of a product of factors (am)n = amn Power of a power
24. The definition of a negative exponent
a−n = 1 an
25. The definition of exponent 0
a0 = 1
26. The definition of the square root radical
27. Equations of the form a2 = b
If a2 = b, then a = ±.
= and symmetrically, =
29. The definition of the nth root
30. The definition of a rational exponent
It is more skillfull to take the root first.
31. The laws of logarithms
log xy = log x + log y.
log xy = log x − log y.
log xn = n log x.
log 1 = 0. logbb = 1.
32. The definition of the complex unit i
i 2 = −1
Next Topic: Rational and irrational numbers
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Even \$1 will help.
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1. Two Logarithmic Function Problems
I'm having trouble solving these two problems:
1.) $\displaystyle log_2(x+7) + log_2(x) = 3$
2.) $\displaystyle e^x + 2 = 8e^{-x}$
2. Ignore the solution to the first question, there's a mistake in it.
log_c a + log_c b = log_c a.b
So,
log_2(x+7) + log_2(x) = 3
log_2{x^2 + 7} = 3
x^2 + 7 = 2^3 = 8
x^2 = 1
x={-1,1}
Don't forget to check the results because x+7 > 0 and x>0.
Ah, x = -1 doesn't satisfy x>0. So we remove it from the solution set.
x=1
----------------------------------------------
$\displaystyle e^x + 2 = 8e^{-x}$
$\displaystyle e^x + 2 = \frac{8}{e^x}$
Let $\displaystyle a=e^x$
$\displaystyle a + 2 = \frac{8}{a}$
Multiply both sides by a,
$\displaystyle a^2 + 2a = 8$
$\displaystyle a^2 + 2a - 8 = 0$
$\displaystyle (a+4)(a-2)=0$
$\displaystyle a=\{ -4 , 2 \}$
Put these a values back in $\displaystyle a=e^x$,
$\displaystyle -4 = e^x$
$\displaystyle x = \ln -4$ (Not real!)
(We exclude this because it's impossible to have such real x)
$\displaystyle 2 = e^x$
$\displaystyle x = \ln 2$
3. The rule should be: $\displaystyle \log_{c}a + \log_{c}b = log_{c}(ab)$
$\displaystyle \log_{2}(x + 7) + \log_{2}(x) = 3$
$\displaystyle \log_{2}\left[(x + 7)(x)\right] = 3$
$\displaystyle \log_{2}\left(x^{2} + 7x\right) = 3$
$\displaystyle x^{2} + 7x = 2^{3}$
$\displaystyle x^{2} + 7x = 8$
4. Sorry, o_O is right. Ignore my solution to the first problem, I didn't multiply 7 by x. But the rest is similar, find the roots, check them for $\displaystyle x+7>0$ and $\displaystyle x>0$.
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# Parallel and Perpendicular Planes
## Parallel Planes
Two distinct planes are said to be parallel to each other if they have no common points. Parallel planes do not intersect, meaning they never meet or cross each other, even when extended indefinitely in all directions. In other words, if you were to extend the planes infinitely, they would never touch or intersect.
Formally, two planes are considered parallel if and only if their normal vectors n1 and n2 are parallel or opposite, that is, if their normal vectors are collinear. The normal vector of a plane is a vector that is perpendicular (at a right angle) to the plane. So, the planes are parallel if
$\mathbf{n_1} = \lambda\mathbf{n_2}$
where λ is a real number.
Another way to define parallel planes is through their equations. If the coefficients of the variables in the equations of two planes are proportional, then the planes are parallel.
For example, if the equation of two planes are
$A_1x + B_1y + C_1z + D_1 = 0,\;\;A_2x + B_2y + C_2z + D_2 = 0,$
then the planes are parallel if
$\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$
Two planes are are said to be coincident when their equations are scalar multiples of each other:
$\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} = \frac{D_1}{D_2}$
## Perpendicular Planes
Perpendicular planes are planes in three-dimensional space that intersect each other at right angles (90 degrees). In other words, the normal vectors to the planes are orthogonal to each other.
Mathematically, if two planes have normal vectors n1 and n2, then they are perpendicular if and only if the dot product of these normal vectors is zero:
$\mathbf{n_1} \cdot \mathbf{n_2} = 0$
This implies that the angle between the normal vectors, and consequently between the planes, is 90 degrees.
For example, the yz-plane (the plane defined by the equation x = 0) and the xz-plane (the plane defined by the equation y = 0) are perpendicular because their normal vectors i and j are mutually orthogonal.
$\mathbf{i} \perp \mathbf{j} \;\;\text{ or }\;\;\mathbf{i} \cdot \mathbf{j} = 0$
In coordinate form, if n1(A1, B1, C1) and n2(A2, B2, C2) are the normal vectors of the planes, the condition for perpendicularity is
$A_1A_2 + B_1B_2 + C_1C_2 = 0$
## Solved Problems
### Example 1.
Find the distance between two parallel planes $$3x - 4y + 5z - 4 = 0$$ and $$3x - 4y + 5z + 16 = 0.$$
Solution.
Take an arbitrary point $$M_1\left({x_1,y_1,z_1}\right)$$ in the first plane. Let, for example, $$x_1 = 0,$$ $$y_1 = 0.$$ Then
$3 \cdot 0 - 4\cdot 0 + 5\cdot z_1 - 4 = 0, \Rightarrow 5z_1 - 4 = 0,\Rightarrow z_1 = \frac{4}{5}.$
Now let’s find the distance from point $$M_1\left({0,0,\frac{4}{5}}\right)$$ to the second plane using the formula
$d = \frac{\left|{A_2x_1 + B_2y_1 + C_2z_1 + D_2}\right|}{\sqrt{A_2^2 + B_2^2 + C_2^2}}.$
Substitute the coefficients of the equation of the second plane and the coordinates of point M1:
$d = \frac{\left|{3 \cdot 0 - 4 \cdot 0 + 5\cdot \frac{4}{5} + 16}\right|}{\sqrt{3^2 + \left({-4}\right)^2 + 5^2}} = \frac{\left|{4 + 16}\right|}{\sqrt{9 + 16 + 25}} = \frac{20}{\sqrt{50}} = \frac{20}{5\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}.$
### Example 2.
Find the equation of a plane passing through the point $$M\left({1,2,-3}\right)$$ and parallel to the plane $$2x - y + 3z = 0.$$
Solution.
The coefficients of the given plane are
$A_1 = 2,\; B_1 = -1,\; C_1 = 3,\; D_1 = 0.$
Let the new plane be described by the equation
$A_2x + B_2y + C_2z + D_2 = 0.$
Since both planes are parallel, the following relation holds:
$\frac{A_2}{A_1} = \frac{B_2}{B_1} = \frac{C_2}{C_1} = \lambda.$
Then the previous equation can be written as
$\lambda A_1x + \lambda B_1y + \lambda C_1z + D_2 = 0,$
or
$A_1x + B_1y + C_1z + \frac{D_2}{\lambda} = 0.$
So the equation of the parallel plane has the same coefficients $$A_1,$$ $$B_1,$$ $$C_1$$ and differs only in the free term $$\frac{D_2}{\lambda},$$ which we simply denote by $$D.$$
Let's find the value of $$D$$ knowing that the plane passes through the point $$M\left({1,2,-3}\right):$$
$2\cdot 1 2 + 3\cdot\left({-3}\right) + D = 0, \Rightarrow D = 9.$
So the equation of the plane has the form
$2x - y + 3z + 9 = 0.$
### Example 3.
Write the equation of a plane passing through the origin and perpendicular to the planes $$2x + 3y - z = 0$$ and $$3x - y + z + 4 = 0.$$
Solution.
The plane passing through the origin is described by the general equation
$Ax + By + Cz = 0.$
The normal vectors of two given planes have coordinates
$\mathbf{n_1}\left({2,3,-1}\right),\;\;\mathbf{n_2}\left({3,-1,1}\right).$
Let's find the coordinates of the normal vector n to our third plane. Since
$\mathbf{n} \perp \mathbf{n_1}\;\text{ and }\;\mathbf{n} \perp \mathbf{n_2},$
the vector n can be found using the cross product:
$\mathbf{n} = \mathbf{n_1} \times \mathbf{n_2}.$
In coordinate form:
$\mathbf{n} = \mathbf{n_1} \times \mathbf{n_2} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {2} & {3} & {-1}\\ {3} & {-1} & {1} \end{array}} \right| = \left| {\begin{array}{*{20}{r}} {3} & {-1}\\ {-1} & {1} \end{array}} \right|\mathbf{i} - \left| {\begin{array}{*{20}{r}} {2} & {-1}\\ {3} & {1} \end{array}} \right|\mathbf{j} + \left| {\begin{array}{*{20}{r}} {2} & {3}\\ {3} & {-1} \end{array}} \right|\mathbf{k} = \left({3-1}\right)\mathbf{i} - \left({2+3}\right)\mathbf{j} + \left({-2-9}\right)\mathbf{k} = 2\mathbf{i} - 5\mathbf{j} - 11\mathbf{k}.$
Therefore the normal vector n is equal to
$\mathbf{n}\left({2,-5,-11}\right).$
The plane equation is written as
$2x - 5y - 11z = 0.$
### Example 4.
Write the equations of planes parallel to the plane $$x + 2y - 2z + 5 = 0$$ which are at the distance of $$d = 4$$ away from it.
Solution.
It is obvious that the equations of parallel planes will differ only by the coefficient D. That is, their equations has the form
$x + 2y - 2z + D = 0.$
Take an arbitrary point $$M_0\left({x_0,y_0,z_0}\right)$$ in the original plane. Let for simplicity $$x_0 = 0,$$ $$y_0 = 0.$$ Find the $$z_0-$$coordinate:
$0 + 2\cdot 0 - 2z_0 + 5 = 0,\;\Rightarrow 2z_0 = 5,\;\Rightarrow z_0 = \frac{5}{2}.$
That is, the point M0 has coordinates $$\left({0,0,\frac{5}{2}}\right).$$
The distance from a point to a plane is determined by the formula
$d = \frac{\left|{Ax_0 + By_0 + Cz_0 + D}\right|}{\sqrt{A^2 + B^2 + C^2}}.$
Substituting the known values on the right side we get
$d = \frac{\left|{0 + 2\cdot 0 - 2\cdot\frac{5}{2} + D}\right|}{\sqrt{1^2 + 2^2 + \left({-2}\right)^2}} = \frac{\left|{-2 + D}\right|}{\sqrt{1+4+4}} = \frac{\left|{D - 2}\right|}{3}.$
By condition $$d = 4.$$ We therefore obtain the following equation for determining the coefficient D:
$4 = \frac{\left|{D - 2}\right|}{3},\;\Rightarrow \left|{D - 2}\right| = 12.$
This equation has two solutions, that is, two values of D:
1. $$D - 2 = 12,$$ $$\Rightarrow D_1 = 14.$$
2. $$D - 2 = -12,$$ $$\Rightarrow D_2 = -10.$$
So the equations of parallel planes have the form
$x + 2y - 2z + 14 = 0,\;\;x + 2y - 2z - 10 = 0.$
### Example 5.
Find the equation of a plane passing through the point $$M\left({1,2,-1}\right)$$ and perpendicular to the planes $$x - y + 2z + 3 = 0$$ and $$x + 2y - z - 4 = 0.$$
Solution.
Let n1 and n2 be the normal vectors of two given planes. Their coordinates are equal
$\mathbf{n_1}\left({1,-1,2}\right),\;\;\mathbf{n_2}\left({1,2,-1}\right).$
By condition, the new plane is perpendicular to both given planes. Therefore its normal vector n must be perpendicular to both n1 and n2:
$\mathbf{n} \perp \mathbf{n_1},\;\;\mathbf{n} \perp \mathbf{n_2}.$
Vector n can be found using the cross product:
$\mathbf{n} = \mathbf{n_1} \times \mathbf{n_2}.$
Let's calculate it through the determinant:
$\mathbf{n} = \mathbf{n_1} \times \mathbf{n_2} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {{A_1}} & {{B_1}} & {{C_1}}\\ {{A_2}} & {{B_2}} & {{C_2}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {1} & {-1} & {2}\\ {1} & {2} & {-1} \end{array}} \right| = \left| {\begin{array}{*{20}{r}} {-1} & {2}\\ {2} & {-1} \end{array}} \right|\mathbf{i} - \left| {\begin{array}{*{20}{r}} {1} & {2}\\ {1} & {-1} \end{array}} \right|\mathbf{j} + \left| {\begin{array}{*{20}{r}} {1} & {-1}\\ {1} & {2} \end{array}} \right|\mathbf{k} = \left({1-4}\right)\mathbf{i} - \left({-1-2}\right)\mathbf{j} + \left({2+1}\right)\mathbf{k} = -3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}.$
Therefore the normal vector n is given by
$\mathbf{n}\left({-3,3,3}\right)\;\text{ or }\;\mathbf{n}\left({-1,1,1}\right).$
The plane equation looks like
$-x + y + z + D = 0.$
We determine the coefficient D from the condition that the plane passes through the point $$M\left({1,2,-1}\right):$$
$-1 \cdot 1 + 2 - 1 + D = 0,\;\Rightarrow D = 0.$
$x - y - z = 0.$
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%
# Percentage Error Calculator
Error calculator is a must-have task in the field of Science especially when it comes to experimental calculations. But this calculation is not done in the form of simple integers but it is done in the form of percentage. It means that the error is calculated in percentage format to make it more clear and precise.
Don’t you know how to do this calculation? Are you looking for assistance to find errors in your work and completing your assignment? You should use this percentage error calculator which has been designed just for students like you. This maths calculator has a specific algorithm according to which it will find the percentage error using your input values.
## What is Percentage Error?
The percentage error is the calculation of error in percentage format instead of integer calculation. This calculation is normally done in experimental projects to check how much the calculated values have deviated from the original value.
It means that this error shows how far your results have been spotted from the standard ones. You can calculate the percentage error if you have the standard values and results obtained from your work. In the following section, we will show how you can do this using a specific formula.
### What is the Percentage Error Formula?
The formula for the calculation of percentage error is pretty simple. You can easily understand and use it to find this measurement. Here is the formula you have to follow:
Percentage Error = Standard Value - Observed Value/Standard Value × 100%
In this formula, “Standard Value” represents the original or true value of the measurement while “Observed Valuebelongs to your obtained results.
### How to Calculate the Percentage Error?
As you can see, the formula for percentage error is simple and easy to understand. You can easily put the values in this formula to find the error. If you are still confused, you can get guidance from the following solved example.
Example 1:
Find the percentage error if the standard value is 120 while the observed value is 89.
Solution:
We need to put the given values in the following formula.
Percentage Error = Standard Value - Observed Value/Standard Value × 100%
By putting the values,
Percentage Error = 120- 89/120 × 100%
= 31/120 × 100%
Percentage Error = 25.83%
### How to use the Percentage Error Calculator?
If you are feeling it difficult to find the percentage error, you should use this tool designed by Calculator’s Bag. This tool has a simple working method that you can complete with the following steps.
• Insert the standard/true value in the first input box.
• Insert the observed value in the second input box.
• This calculator will instantly find the percentage error and show it on the screen.
### FAQ | Percentage Error Calculator
What is the percentage error in 9.89 g?
To find the percentage error, we have the standard value. On the other side, we can find the relative error for this value which comes out to be 0.1%.
What is a percentage error example?
Being a student, you must have the expected numbers to pass your exams. The expected marks and obtained marks can be expressed in terms of percentage error as it also shows how your results are deviating from the true/expected/standard values.
What is the percent error in accuracy?
Percentage error shows exactly how far the obtained results are from the standard ones.
What does a 100% percent error mean?
It means that the calculations are 100% wrong and deviate from the true values.
Can percentage error be negative?
Yes, it can be negative but normally we consider it as positive because the more chances of acceptance of results are when the percentage error is positive.
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## Word Problems with Percents
All word problems that involve percents require you to do one of three basic things.
1. Find the percent of a number
e.g. What is 25% of 44?
2. Find what a number is when you are given a percent of it
e.g. 18 is 35% of what number?
3. Find what percent a number is of another number
e.g. 40 is what percent of 95?
The way to set all of these problems up is to write two fractions. The fraction on the left side has the part of the whole (not the percent) as the numerator, and the amount of the whole as the denominator. The fraction on the right side has the percent as the numerator and 100 as the denominator. The value you don’t know and are solving for can be represented by x.
When you set up the example question for the first type of question above it will look like:
The example for number two:
Number three:
Now you cross multiply and solve for x.
For example the way you would do number one is:
So 25% of 44 is 11.
### Multistep Problems
Many times you will have to do a calculation unrelated to percentage before you do the calculation related to percent in order to arrive at the correct answer. It won’t be very complicated, but you need to pay attention to the wording of the question.
Here’s an example question:
A trip from City A to City B costs \$75 in gas and tolls. The toll fees add up to \$4.50. The gas for a trip from City B to City C costs 30% more, but there are no tolls. How much will the gas cost for a trip from City B to City C?
Of course when you answer the question the first thing you should do is the precalculation of removing the toll cost. This will enable you to arrive at the correct answer since there are no tolls in the second trip.
The cost of gas for a trip from City B to City C is \$91.65 .
### Percent Change
If a question on the GED math test asks you to find a percent change (the percentage something increases or decreases by), all you need to do is divide the amount of change by the original amount.
Example question:
Your new apartment is 900 square feet and your old apartment was 750 square feet. How much larger is your new apartment percentage-wise?
First you find the difference:
Then you divide the difference amount by the original amount:
Then all you need to do is convert the result of the division to a percent:
.2 = 20%
Share and Enjoy:
1. Sarah Dessen fan says:
Man, the math section of the GED is so complicated. I wish there was no need for the math portion, then I’d. Already have my GED. The math part is the last test I need to take yet the subject I struggle the most with. I can only pray I pass.
• C-Los Pugh says:
I feel the same way math is my only and last subject I have left to take i’m taking it again tomorrow so like u said I pray I pass it this time. This will be my third time I have to pass it because I wont to get me GED before the new 2014 one I refuse to take the new one it will be harder an much longer. So good luck an i’ll pray for u too. Just don’t give up!!!
2. Christopher Colangelo says:
Your so right it is very complicated but you still have to pass that part in order to receive your GED. I see this site is very helpful Im sure with the help of this site, you’ll do better next time. Good Luck to you and I will be retesting the math part next month i only need 20 points to pass.
3. Tara says:
For the example question on Multistep Problems, there seems to be a typo because the answer is incorrect.
70.05 + 21.15 =91.20 Not 91.65
• Ziyi says:
Should be fixed now. Thanks a lot for letting me know!
• Nick says:
75.00-4.50=70.50 Not 70.05. Therefore; 70.50+21.15=91.65
• Joellen says:
The original is correct. The gas cost is 70.50, not 70.05, so 91.65 IS correct.
4. Teresa Marks says:
okay I am confused with the percent of change problem (The problem is 950 and 700) why did you change the 950 to 900.Please explain because I am lost I know how to do the percent decrease and increase
• Ziyi says:
So sorry to cause the confusion! I just made a mistake and transposed the numbers. It should be fixed now.
5. robert says:
i was wondering why the answer to the 750 divided by one fifty is .2 cause none of the numbers go into anyy of the 150 i got zero please explain futher
-appreaciated-
• Ziyi says:
Well, it’s 150 divided by 750. You’re right that 750 doesn’t fit into 150 but part of 750 does, that’s where the .2 comes from. It just means that .2 or 20% of 750 fits into 150.
6. Ray says:
The answer for multistep is 91.65 since you’re supposed 70.50 to 21.15
You used 70.50 to multiply it by .30 that’s how you got 21.15 in the first place so why did you move the decimal and lower the value to 70.05?
• Ziyi says:
I just made a mistake. Sorry for the confusion! Hopefully it’s all good now!
7. Megan Trotter says:
I take my math test on the 28th of this mounth I am Getting a little nervous I have my fingers crossed that I cn pass This site has hepled me understand so much more then I did I didn’t even know how to divide I few months ago, but you have made it so esay. Thank you I hope I pass and when i will let you know 🙂
8. nanzina says:
Hi thank you for your help! is there an easier way on how to pass the calculator part of the test, what will be on the test as far as the calculator part. thanks again and blesings to you ziyi.
9. Catresha says:
Got 10 points higher on the mathematics section of the GED test today. Was fairly easy,but did,nt have enough time to read. Taking it again next week 🙂 ! I,m gonna stay confident as always,& keep on studying :)!
• Ziyi says:
Right, good luck. Now at least you have a clearer idea of what to focus on. Will pass next time for sure!
• Catresha says:
Thanks 🙂 410 is the passing score in my state, I have 390 ,but all other scores are high,& all I need is410. So I have the batteryscore already,&410 will take me way over my overall score :)!
10. jaz says:
really helped me rembered cross multiplying usefull 🙂
11. kay thomas says:
I hate maths.all that algebra on the ged is really not necessary in life. whenever you can count your money and hours worked that’s it. leave that all that complicated maths for engineers and scientist.had to write the ged maths five times .
12. maria alvarez says:
how do you get the 2 in the divison part 750 divided by 150
13. ismail says:
Hi there…how do you get 750÷150 =2
Isnt 750÷150=5
14. Jim says:
Wouldn’t it be easier just to do this on the calculator since you can use it on your whole math portion of the GED now?
Problem 1 you just enter 0.25 x 44 and get 11
Problem 2 you just enter 18 divided by 35% and you get 51.428571
Problem 3 you just enter 40 divided by 95 and you get 0.421
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Found in: Page 275
### Pre-algebra
Book edition Common Core Edition
Author(s) Ron Larson, Laurie Boswell, Timothy D. Kanold, Lee Stiff
Pages 183 pages
ISBN 9780547587776
# Using the archery information above, compare the number of misses to the number of shots using a ratio. Write the ratio in three ways.
The ratio of the number of misses to the number of shots is$\frac{4}{15}$ .
The three ways to write the ratio are given by$\frac{4}{15},\text{4 to 15,and 4:15}$ .
See the step by step solution
## Step 1 . Given
Given that:
The number of arrows shot is 60.
The number of hits is 44.
The number of misses is 16.
## Step 2 . To determine
We have to find the ratio of the number of misses to the number of shots.
Then we have to write the ratio in three ways.
## Step 3 . Calculation
We present the ratio of number of misses to number of shots as a fraction,
$\frac{\text{number of misses}}{\text{number of shots}}=\frac{16}{60}$.
Then we simplify the fraction,
$\frac{\text{number of misses}}{\text{number of shots}}=\frac{4}{15}$. [Divide out the common factors]
After that, we write the ratio in three ways:
$\frac{4}{15},\text{4 to 15,and 4:15}$.
So, the required ratio of number of misses to the number of shots =.$\frac{4}{15}$
Three ways are given by$\frac{4}{15},\text{4 to 15,and 4:15}$ .
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# Polynomial Inequalities: Definition & Examples
Instructor: Laura Pennington
Laura has taught collegiate mathematics and holds a master's degree in pure mathematics.
Polynomial inequalities show up in real world applications when intervals of numbers are involved. In this lesson, you'll learn what polynomial inequalities are and how to work with them.
## Polynomial Inequalities
Suppose you're trying to catch a cab in the city. You have no more than \$20 to spend, and the cabs charge a flat rate of \$2.00 plus \$0.70 per mile. If x represents the number of miles you ride in the cab, then the cost of the cab ride would be 2 + 0.7x. That is, the flat rate of \$2.00 plus \$0.70 times the number of miles you ride in the cab. In this situation, you want the cab ride to cost less than the \$20 you have to spend. You can represent this mathematically as follows:
2 + 0.7x < 20
This is a polynomial inequality. Polynomial inequalities are inequalities expressed with a polynomial on one side of the inequality symbol and zero on the other side. Some examples of polynomial inequalities are shown below:
These types of inequalities can be used to answer questions about real-world situations, such as your city cab ride. Suppose you want to know how many miles you can travel without exceeding your spending limit. To find out, you solve the polynomial inequality for x to get x < 25.71. So, you can ride in the cab for about 25 miles without exceeding your spending limit.
## Solutions to Polynomial Inequalities
The solution to the cab example is x < 25.71, which is an interval, or a set of numbers. The solution to a polynomial inequality consists of intervals that make the inequality true. In the example, x < 25.71 tells you that if you plug in any number that is less than 25.71 into the inequality, you'll get a true statement. For instance, the number 10 is less than 25.71, so it should make a true statement when plugged into your inequality.
1.) 2 + 0.7(10) < 20
2.) 2 + 7 < 20
3.) 9 < 20
Sure enough, this is a true statement because 9 is less than 20. On the other hand, if you plugged in a number greater than 25.71, you'd get a false statement, since the number doesn't fall in your interval solution. For instance, the number 30 is greater than 25.71, which would lead to a false statement, as shown below.
1.) 2 + 0.7(30) < 20
2.) 2 + 21 < 20
3.) 23 < 20
As expected, this is a false statement because 23 is not less than 20.
## Solving Polynomial Inequalities
To solve polynomial inequalities, follow the steps below:
1.) Manipulate the inequality so you'll have a polynomial on one side of the inequality symbol and zero on the other side.
2.) Replace the inequality symbol with an equal symbol, and solve the equation.
3.) Plot the solution from step two on a number line.
4.) Take a test value from each of the intervals, making sure they're not equal to the endpoints of the intervals. Plug the test values into the original inequality.
5.) If the test value leads to a true statement, then the interval from which it came is the solution. If the test value leads to a false statement, then the interval from which it came is not the solution.
Now, try applying these steps to your city cab ride.
1.) To get zero on one side of the inequality, subtract 20 from both sides:
2 + 0.7x < 20
0.7x - 18 < 0
2.) Find all values of x that make 0.7x - 18 = 0:
0.7x - 18 = 0
0.7x = 18
x = 18 / 0.7 = 25.71
3.) Plot x = 25.71 on the number line. In the image below, you'll see that it's broken into two intervals: x < 25.71 and x > 25.71.
4.) Choose a test number from each of the intervals, such as 10 and 30. The number 10 comes from the interval x < 25.71, and the number 30 comes from the interval x > 25.71.
5.) The number 10 makes for a true statement when plugged into the inequality, while 30 makes for a false statement. Therefore, the solution is the interval from which 10 originated: x < 25.71.
## Practice Problem
Let's try one more example. Suppose you own a business and want to determine the price of a product that would make your revenue greater than your cost using the following inequality:
45 - 9p < 5p - p^2
1.) Get zero on one side:
45 - 9p < 5p - p^2
p^2 - 9p - 5p + 45 <0
p^2 - 14p + 45 < 0
2.) Find values of p that make p^2 - 14p + 45 = 0:
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Integral Calculus - Exercises
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1 Integral Calculus - Eercises 6. Antidifferentiation. The Indefinite Integral In problems through 7, find the indicated integral.. Solution. = = + C = + C.. e Solution. e =. ( 5 +) Solution. ( 5 +) = e =e + C. 5 + = = C = = C. 4. ³ + Solution. µ + = + = = ln ( ) + + C = = ln C.
2 INTEGAL CALCULUS - EXECISES 4 5. e + 6 +ln Solution. µe + 6 +ln = e +6 +ln = e +6ln +(ln) + C. = 6. + Solution. + = + = = C = = C = = C. 7. ( ) 5 Solution. µ ( ) 5 = = ( 5 + ) = ( 5 + ) = = C = = C. 8. Find the function f whose tangent has slope +for each value of and whose graph passes through the point (, ). Solution. The slope of the tangent is the derivative of f. Thus f () = + and so f() is the indefinite integral f() = f () = µ + = = C.
3 INTEGAL CALCULUS - EXECISES 4 Using the fact that the graph of f passes through the point (, ) you get = 4 +++C or C = 5 4. Therefore, the desired function is f() = It is estimated that t years from now the population of a certain lakeside community will be changing at the rate of.6t +.t +.5 thousand people per year. Environmentalists have found that the level of pollution in the lake increases at the rate of approimately 5 units per people. By how much will the pollution in the lake increase during the net years? Solution. Let P (t) denote the population of the community t years from now. Then the rate of change of the population with respect to time is the derivative dp dt = P (t) =.6t +.t +.5. It follows that the population function P (t) is an antiderivative of.6t +.t +.5. Thatis, P (t) = P (t)dt = (.6t +.t +.5)dt = =.t +.t +.5t + C for some constant C. During the net years, the population will grow on behalf of P () P () = C C = =.6+.4+=thousand people. Hence, the pollution in the lake will increase on behalf of 5 =5 units.. Anobjectismovingsothatitsspeedaftert minutes is v(t) =+4t+t meters per minute. How far does the object travel during rd minute? Solution. Let s(t) denote the displacement of the car after t minutes. Since v(t) = ds = dt s (t) it follows that s(t) = v(t)dt = ( + 4t +t )dt = t +t + t + C. During the rd minute, the object travels s() s() = C 4 8 C = = meters.
4 INTEGAL CALCULUS - EXECISES 4 Homework In problems through, find the indicated integral. Check your answers by differentiation ( +6) e + ³ 8. + ³ 9. + e ( ). ( +) 4. Find the function whose tangent has slope 4 +for each value of and whose graph passes through the point (, ). 5. Find the function whose tangent has slope +6 for each value of and whose graph passes through the point (, 6). 6. Find a function whose graph has a relative minimum when =and a relative maimum when =4. 7. It is estimated that t months from now the population of a certain town will be changing at the rate of 4+5t people per month. If the current population is, what will the population be 8 months from now? 8. An environmental study of a certain community suggests that t years from now the level of carbon monoide in the air will be changing at therateof.t +. parts per million per year. If the current level of carbon monoide in the air is.4 parts per million, what will the level be years from now? 9. After its brakes are applied, a certain car decelerates at the constant rate of 6 meters per second per second. If the car is traveling at 8 kilometers per hour when the brakes are applied, how far does it travel before coming to a complete stop? (Note: 8 kmph is the same as mps.). Suppose a certain car supplies a constant deceleration of A meters per second per second. If it is traveling at 9 kilometersperhour(5 meters per second) when the brakes are applied, its stopping distance is 5 meters. (a) What is A?
5 INTEGAL CALCULUS - EXECISES 44 (b) What would the stopping distance have been if the car had been traveling at only 54 kilometers per hour when the brakes were applied? (c) At what speed is the car traveling when the brakes are applied if the stopping distance is 56 meters? esults C C 7. + C C C ln + C 5 7. e C p ( ) + + C 9. ln + + e + + C. +ln + C C C C 4. f() = + 5. f() = f() = 5 +4; not unique parts per million meters. (a) A =6.5 (b) 4 meters (c).7 kilometers per hour
6 INTEGAL CALCULUS - EXECISES Integration by Substitution In problems through 8, find the indicated integral.. ( +6) 5 Solution. Substituting u = +6and du =, youget ( +6) 5 = u 5 du = u6 + C = ( +6)6 + C.. [( ) 5 +( ) +5] Solution. Substituting u = and du =, youget ( ) 5 +( ) +5 = (u 5 +u +5)du = = 6 u6 + u +5u + C = = 6 ( )6 +( ) +5( ) + C. Since, for a constant C, C 5 is again a constant, you can write ( ) 5 +( ) +5 = 6 ( )6 +( ) +5 + C.. e Solution. Substituting u = and du =, youget e = e u du = eu + C = e + C e 6 Solution. Substituting u = 6 and du = 6 5, youget 5 e 6 = e u du = 6 6 eu + C = 6 e 6 + C Solution. Substituting u = 5 +and 5 du =4, you get = 5 u du = 5 ln u + C = 5 ln C.
7 INTEGAL CALCULUS - EXECISES Solution. Substituting u = 4 +6 and 5 du =( 5), you get = 5 du = 5 u = C. u du = 5 u + C = 7. ln Solution. Substituting u =ln and du =, youget ln = du =ln u + C =ln ln + C. u 8. ln Solution. Substituting u =ln and du =, youget ln ln = = udu = u + C =(ln) + C. 9. Use an appropriate change of variables to find the integral ( +)( ) 9. Solution. Substituting u =, u += +and du =, you get ( +)( ) 9 = (u +)u 9 du = (u +u 9 )du = = u + u + C = = ( ) + ( ) + C.. Use an appropriate change of variables to find the integral ( +). Solution. Substituting u =, u +4= + and du =, you
8 INTEGAL CALCULUS - EXECISES 47 get ( +) = (u +4) udu = u du + u du = = 5 u 5 + u + C = 5 u u + C = = 5 ( ) ( ) + C = Homework = 5 ( ) + 4 ( ) +C = = ( ) µ C = = µ ( ) +C. In problems through 8, find the indicated integral and check your answer by differentiation.. e e 5. e 6. ( +) ( +) ( +5) ( +)( + +5). ( )e ( +6) ln 5 6. (ln ) 7. ln( +) 8. e + In problems 9 through, use an appropriate change of variables to find the indicated integral ( 5) 6 ( 4) Find the function whose tangent has slope +5for each value of and whose graph passes through the point (, ). 5. Find the function whose tangent has slope whose graph passes through the point (, 5). for each value of and
9 INTEGAL CALCULUS - EXECISES A tree has been transplanted and after years is growing at the rate of + meters per year. After two years it has reached a height (+) of five meters. How tall was it when it was transplanted? 7. It is projected that t years from now the population of a certain country will be changing at the rate of e.t million per year. If the current population is 5 million, what will the population be years from now? esults.. 5 e5 + C. (4 ) 4 +C 6. ln +5 + C 4. e + C 5. e + C 6. ( +) 6 + C 7. ( +8) 4 +8+C 8. ( +) C 9. ( +5) + C. 6 ( + +5) + C. e + C. 5 ln C. ( +6) + C 4. ln + C 5. ln 5 + C 6. ln + C 7. ln ( +)+C 8. e + C 9. +ln + C. 5 ( +) + ( +) ++C. + C. 7 +ln 4 + C ( 5) 5 4( 5) ln + + C f() = ( +5) f() = ln meters 7. 6 million
10 INTEGAL CALCULUS - EXECISES Integration by Parts In problems through 9, use integration by parts to find the given integral.. e. Solution. Since the factor e. is easy to integrate and the factor is simplified by differentiation, try integration by parts with g() =e. and f() =. Then, G() = e. =e. and f () = and so e. = e. e. =e. e. + C = = ( )e. + C.. ( )e Solution. Since the factor e is easy to integrate and the factor is simplified by differentiation, try integration by parts with g() =e and f() =. Then, G() = e = e and f () = and so ( )e = ( )( e ) e = = ( )e +e + C =( )e + C.. ln Solution. In this case, the factor is easy to integrate, while the factor ln is simplified by differentiation. This suggests that you try integration by parts with g() = and f() =ln. Then, G() = = and f () = =
11 INTEGAL CALCULUS - EXECISES 5 and so ln = ln = ln = = ln + C = ln + C. 4. Solution. Since the factor is easy to integrate and the factor is simplified by differentiation, try integration by parts with g() = and f() =. Then, G() = = ( ) and f () = and so = ( ) + = ( ) + ( ) = µ 5 ( ) 5 + C = = ( ) 4 5 ( ) 5 + C = = ( ) 4 5 ( ) + C. 5. ( +)( +) 6 Solution. Since the factor ( +) 6 is easy to integrate and the factor +is simplified by differentiation, try integration by parts with g() =( +) 6 and f() = +. Then, G() = ( +) 6 = 7 ( +)7 and f () = and so ( +)( +) 6 = 7 ( +)( +)7 ( +) 7 = 7 = 7 ( +)( +)7 7 8 ( +)8 + C = = 56 [8( +) ( +)]( +)7 + C = = 56 (7 +6)( +)7 + C.
12 INTEGAL CALCULUS - EXECISES 5 6. e Solution. Since the factor e is easy to integrate and the factor is simplified by differentiation, try integration by parts with g() =e and f() =. Then, G() = e = e and f () = and so e = e e. To find e, you have to integrate by parts again, but this time with g() =e and f() =. Then, G() = e and f () = and so e = e e. To find e, you have to integrate by parts once again, this time with g() =e and f() =. Then, G() = e and f () = and so e = e e = e 4 e. Finally, e = e = µ e µ e 4 e + C = e + C.
13 INTEGAL CALCULUS - EXECISES 5 7. ln Solution. In this case, the factor is easy to integrate, while the factor ln is simplified by differentiation. This suggests that you try integration by parts with Then, and so ln G() = g() = and f() =ln. = = and f () = = ln + = ln 4 + C. = ln + µ + C = 8. e ³ Solution. First rewrite the integrand as e, and then integrate by parts with g() =e and f() =. Then, from Eercise 6.. you get G() = e = and f () = e and so e = e e = e e + C = = ( )e + C. 9. ( ) Solution. First rewrite the integrand as [( ) ],andthen integrate by parts with g() =( ) and f() =. Then G() = ( ) and f () =.
14 INTEGAL CALCULUS - EXECISES 5 Substituting u = and du =, youget G() = ( ) = u du = u = ( ). Then ( ) = ( ) ( ) = = ( ) ( ) + C = = ( ) 64 ( ) + C. (a) Use integration by parts to derive the formula n e a = a n e a n n e a. a (b) Use the formula in part (a) to find e 5. Solution. (a) Since the factor e a is easy to integrate and the factor n is simplified by differentiation, try integration by parts with g() =e a and f() = n. Then, G() = e a = a ea and f () =n n and so n e a = a n e a n a n e a. (b) Apply the formula in part (a) with a =5and n =to get e 5 = 5 e 5 e 5. 5 Again, apply the formula in part (a) with a =5and n =to find the new integral e 5 = 5 e 5 e 5. 5
15 INTEGAL CALCULUS - EXECISES 54 Homework Once again, apply the formula in part (a) with a =5and n = to get e 5 = 5 e5 e 5 = 5 5 e5 5 e5 and so e 5 = 5 e 5 5 = 5 5 e 5 µ 5 µ e5 5 e5 + C = e 5 + C. In problems through 6, use integration by parts to find the given integral.. e. e. e 5 4. ( )e 5. ln ( +) e. e. e. ln 4. (ln ) 5. ln 6. 7 ( 4 +5) 8 7. Find the function whose tangent has slope ( +)e for each value of and whose graph passes through the point (, 5). 8. Find the function whose tangent has slope ln for each value of > and whose graph passes through the point (, ). 9. After t seconds, an object is moving at the speed of te t meters per second. Epress the distance the object travels as a function of time.. It is projected that t years from now the population of a certain city will be changing at the rate of t ln t +thousand people per year. If the current population is million, what will the population be 5 years from now?
16 INTEGAL CALCULUS - EXECISES 55 esults.. ( +)e + C. ( 4)e + C. 5( +5)e 5 + C 4. ( )e + C 5. (ln )+C C )9 9 +) + C 8. ( +) 4 +) + C 9. ( +) +) + C. ( + +)e + C. + 9 e + C. ( +6 6)e + C. ln 9 + C 4. ln ln + + C 5. (ln +)+C ( 4 +5) 9 6 (4 +5) + C 7. f() = ( +)e + e f() = 4 ln 5 ln 9.. s(t) = (t +)e t
17 INTEGAL CALCULUS - EXECISES The use of Integral tables In Problems through 5, use one of the integration formulas from a table of integrals (see Appendi) to find the given integral.. + Solution. First rewrite the integrand as + = p ( +) 4 andthensubstituteu = +and du = to get + = p ( +) 4 = du u 4 = = ln u + u 4 + C =ln ++ p ( +) 4 + C = = ln C.. 6 Solution. First, rewrite the integrand as 6 = ( + ) = 4 ( +) = 4 ( +) andthensubstituteu = +and du = to get = 6 4 ( = du +) 4 = u = 8 ln 4 + u 4 u + C = 8 ln 4+u 4 u + C = = 8 ln 7+ + C.. ( +) Solution. First rewrite the integrand as ( +) =( +) += Aplly appropriate formulas (see Appendi, formulas 9 and ), to get + = 8 ( + ) + 8 ln C
18 INTEGAL CALCULUS - EXECISES 57 and + = ++ ln C. Combine these results, to conclude that ( +) = 8 ( +5) p ( +)+ 8 ln C. 4. e Solution.Aplly appropriate formula (see Appendi, formula ), to get = e + = e ln + e + C = = + ln e + C = + ln e + ln + C. Since the epression ln is a constant, you can write e = + ln e + C. 5. (ln ) Solution. Aplly the reduction formula (see Appendi, formula 9) (ln ) n = (ln ) n n (ln ) n to get (ln ) = (ln ) (ln ) = µ = (ln ) (ln ) = (ln ) (ln ) +6 (ln ) µ ln = = = (ln ) (ln ) +6 ln 6 + C. Homework In Problems through, use one of the integration formulas listed in this section to find the given integral... ( ) 4( 5) e. e
19 INTEGAL CALCULUS - EXECISES 58 Locate a table of integrals and use it to find the integrals in Problems through (ln ) One table of integrals lists the formula p ± p =ln + p ± p p while another table lists p ± p =ln p + ± p. Can you reconcile this apparent contradiction? 8. The following two formulas appear in a table of integrals: p = p ln p + p and a + b = ab ln a + ab a ab (for ab ). (a) Use the second formula to derive the first. (b) Apply both formulas to the integral. Which do you find 9 4 easier to use in this problem? esults.. ln + C. ln 5 + C. ln C 4. ln + q C 5. ln C 6. ln + + C 7. ln + + C 8. 4ln + C e + C. ( )e + C. ln + C. ( +4) +4+ p ( +4)+C 6. (ln ) ln + + C 4. ln C 5. p q (4 )+C 6. ln C 7. ln a b =lna ln b 8. ln + + C
20 INTEGAL CALCULUS - EXECISES The Definite Integral In problems through 7 evaluate the given definite integral. (e t e t ) dt ln Solution. e t e t dt = e t + e t ln ln = e + e e ln e ln = = e + e e ln e ln = e + e = = e + e 5.. ( +6)4 Solution. Substitute u = +6. Then du =, u( ) =, and u() = 6. Hence,. ( +) ( +6) 4 = 6 u 4 du = u5 6 = 65 = Solution. Substitute u = +. Then du =, u() =, and u() = 9. Hence, 4. e e ln ( +) = 9 u du = u 9 = = Solution. Substitute u =ln. Thendu =, u(e) =,andu(e )=. Hence, e e ln = du u =ln u =ln ln = ln. 5. e t ln tdt Solution. Since the factor t is easy to integrate and the factor ln t is simplified by differentiation, try integration by parts with g(t) =t and f(t) =lnt
21 INTEGAL CALCULUS - EXECISES 6 Then, G(t) = tdt = t and f (t) = t and so e t ln tdt = t ln t e e tdt = t ln t 4 t = 8 e ln e 6 e 8 ln + 6 = 8 e 6 e + 6 = = 6 e + 6 = 6 (e +). 6. e Solution. Apply the reduction formula n e a = a n e a n a twice to get e = e e = e = n e a = e e + e = µ = e e + µ 4 e = + e 4 µ = + e 4 4 e = 4 e 4 = 4 (e ) t te dt Solution. Integrate by parts with g(t) =e 5 t and f(t) =t = Then, G(t) = e 5 t dt =e 5 t and f (t) = and so 5 te 5 t dt = te 5 t 5 5 = (t )e 5 t ³ = + 4e 4 = 5 ³ 5 e 5 t dt = te 5 t 4e 5 t = = ( 5)e ( )e 4 = 4e 4.
22 INTEGAL CALCULUS - EXECISES 6 (a) Show that b a f() + c b f() = c a f(). (b) Use the formula in part (a) to evaluate. (c) Evaluate 4 ( + ). Solution. (a) By the Newton-Leibniz formula, you have b a f() + c b f() = F (b) F (a)+f (c) F (b) = = F (c) F (a) = c a f(). (b) Since = for and = for, youhaveto break the given integral into two integrals = ( ) = =+ = and = Thus, = = + = =. = + =. (c) Since = + for and = for, you get 4 ( + ) = = [ + ( +)] + ( +4) [ + ( )] = ( ) = = ( +4) + ( ) = = (a) Show that if F is an antiderivative of f, then b a f( ) = F ( b)+f ( a) 4 =
23 INTEGAL CALCULUS - EXECISES 6 (b) A function f is said to be even if f( ) =f(). [For eample, f() = is even.] Use problem 8 and part (a) to show that if f is even, then a a f() = f() a (c) Use part (b) to evaluate and. (d) A function f is said to be odd if f( ) = f(). Useproblem8 and part (a) to show that if f is odd, then a a f() =. (e) Evaluate. Solution. (a) Substitute u =. Thendu =, u(a) = a and u(b) = b. Hence, b a f( ) = b a (b) Since f( ) =f(), you can write a a f() = By the part (a), you have Hence, a f(u)du = F (u) b a = F ( b)+f ( a). a f( ) + a f(). f( ) = F () + F ( ( a)) = F (a) F () = = a a a f(). f() = a f(). (c) Since f() = is an even function, you have Analogously, = = = = = =. = = 6.
24 INTEGAL CALCULUS - EXECISES 6 (d) Since f( ) = f(), you can write a a f() = a f( ) + a f() = = F () F (a)+f (a) F () =. (e) Since f() = is an odd function, you have =.. It is estimated that t days from now a farmer s crop will be increasing at the rate of.t +.6t +bushels per day. By how much will the value of the crop increase during the net 5 days if the market price remains fied at euros per bushel? Solution. Let Q(t) denote the farmer s crop t days from now. Then the rate of change of the crop with respect to time is dq dt =.t +.6t +, and the amount by which the crop will increase during the net 5 days is the definite integral 5 Q(5) Q() =.t +.6t + =.t +.t + t 5 = = =5bushels. Hence, the value of the market price will increase by 5 =75euros. Homework In problems through 7, evaluate the given definite integral. (4 +). (5 + ) 5. ( + t ³ 9 t +t ) dt 4. dt t t+dt t 6 7. ( ) 8. ( 4) t+ dt. (t + t) t 4 +t +dt 6.. e+ +. (t +)(t e )9 dt 4. ln tdt e 5. e (ln ) ( + t)e.t dt
25 INTEGAL CALCULUS - EXECISES A study indicates that months from now the population of a certain town will be increasing at the rate of 5+ people per month. By how much will the population of the town increase over the net 8 months? 9. It is estimated that the demand for oil is increasing eponentially at the rate of percent per year. If the demand for oil is currently billion barrels per year, how much oil will be consumed during the net years?.anobjectismovingsothatitsspeedaftert minutes is 5+t +t meters per minute. How far does the object travel during the nd minute? esults ln ln. e e + 5. e e people billion barrels. 5 meters
26 INTEGAL CALCULUS - EXECISES Area and Integration In problems through 9 find the area of the region.. is the triangle with vertices ( 4, ), (, ) and (, 6). Solution. From the corresponding graph (Figure 6.) you see that the region in question is bellow the line y = +4above the ais, and etends from = 4 to =. y 6 4 y= Figure 6.. Hence, A = 4 ( +4) = µ +4 4 =(+8) (8 6) = 8.. is the region bounded by the curve y = e, the lines =and =ln,andthe ais. Solution. Since ln =ln ln = ln '.7, from the correspondinggraph(figure6.)youseethattheregioninquestionis bellow the line y = e above the ais, and etends from =ln to =. y y=e - -ln Figure 6..
27 INTEGAL CALCULUS - EXECISES 66 Hence, A = ln e = e ln = e e ln = =.. is the region in the first quadrant that lies below the curve y = +4 and is bounded by this curve, the line y = +,andthecoordinate ais. Solution. First sketch the region as shown in Figure 6.. Note that the curve y = +4and the line y = + intersect in the first quadrant at the point (, 8), since =is the only positive solution of the equation +4= +, i.e. + 6=. Also note that the line y = +intersects the ais at the point (, ). y y= y= Figure 6.. Observe that to the left of =, is bounded above by the curve y = +4, while to the right of =, it is bounded by the line y = +. This suggests that you break into two subregions, and, as shown in Figure 6., and apply the integral formula for area to each subregion separately. In particular, and A = Therefore, A = ( +4) = µ +4 µ ( +) = + A = A + A = = 8 +8= = 5++ =. += 8.
28 INTEGAL CALCULUS - EXECISES is the region bounded by the curves y = +5and y =,the line =,andthey ais. Solution. Sketch the region as shown in Figure 6.4. y 5 5 y= y=- Figure 6.4. Notice that the region in question is bounded above by the curve y = +5 and below by the curve y = and etends from =to =. Hence, A = [( +5) ( )] = ( +5) = µ +5 = 8+5 =. 5. is the region bounded by the curves y = and y = +4. Solution. First make a sketch of the region as shown in Figure 6.5 and find the points of intersection of the two curves by solving the equation = +4 i.e. 4= to get = and =. The corresponding points (, ) and (, ) are the points of intersection. y=- +4 y 4 y= Figure 6.5.
29 INTEGAL CALCULUS - EXECISES 68 Notice that for, thegraphofy = +4lies above that of y =. Hence, A = = [( +4) ( )] = µ + +4 ( + +4) = = =9. 6. is the region bounded by the curves y = and y =. Solution. Sketch the region as shown in Figure 6.6. Find the points of intersection by solving the equations of the two curves simultaneously to get = = ( ) = = and =. The corresponding points (, ) and (, ) are the points of intersection. y y= y= - - Figure 6.6. Notice that for, thegraphofy = lies above that of y =. Hence, A = ( ) = µ = =. (a) is the region to the right of the y ais that is bounded above by the curve y =4 and below the line y =. (b) is the region to the right of the y ais that lies below the line y =and is bounded by the curve y =4, the line y =,and thecoordinateaes.
30 INTEGAL CALCULUS - EXECISES 69 Solution. Note that the curve y =4 and the line y = intersect to the right of the y ais at the point (, ), since = is the positive solution of the equation 4 =, i.e. =. (a) SketchtheregionasshowninFigure6.7. y 4 y=4- y= Figure 6.7. Notice that for, thegraphofy =4 lies above that of y =. Hence, µ A = (4 ) = ( ) = = =. (b) Sketch the region as shown in Figure 6.8. y 4 y=4- y= Figure 6.8. Observe that to the left of =, is bounded above by the curve y =, while to the right of =, it is bounded by the line y =4. This suggests that you break into two subregions, and, as shown in Figure 6.8, and apply the integral formula
31 INTEGAL CALCULUS - EXECISES 7 for area to each subregion separately. In particular, and so A = A = = = µ (4 ) = 4 A = A + A =+ 5 = 4. = = 5, 7. is the region bounded by the curve y = and the lines y = and y = 8. Solution. First make a sketch of the region as shown in Figure 6.9 and find the points of intersection of the curve and the lines by solving the equations to get = and = 8 i.e. = and =8 = and =. y y= y= y= Figure 6.9. Then break into two subregions, that etends from =to =and that etends from =to =,asshowninfigure 6.9. Hence, the area of the region is ³ A = 7 = 8 8 = 7 6 = 7 6
32 INTEGAL CALCULUS - EXECISES 7 and the area of the region is µ A = µ = 8 6 Thus, the area of the region is the sum A = A + A = 6 = 4. = = is the region bounded by the curves y = +5 and y = Solution. First make a rough sketch of the two curves as shown in Figure 6.. You find the points of intersection by solving the equations of the two curves simultaneously +5 = = ( ) 4( ) = ( )( )( +)= to get =, = and =. y y= - +5 y= Figure 6.. The region whose area you wish to compute lies between = and =, but since the two curves cross at =, neither curve is always above the other between = and =. However, since the curve y = +5is above y = +4 7 between = and =, and since y = +4 7 is above y = +5between = and =, it follows that the area of the region between = and =,is
33 INTEGAL CALCULUS - EXECISES 7 A = = = 4 + = µ = = = = and the area of the region between =and =,is A = = = + +4 = µ = = = = 4 = 8 4 += 4. Thus, the total area is the sum Homework A = A + A =+ 4 = 4. In problems through find the area of the region.. is the triangle bounded by the line y =4 and the coordinate aes.. is the rectangle with vertices (, ), (, ), (, 5) and (, 5).. is the trapezoid bounded by the lines y = +6and =and the coordinate aes. 4. is the region bounded by the curve y =, the line =4,andthe ais.
34 INTEGAL CALCULUS - EXECISES 7 5. is the region bounded by the curve y =4, the line =,andthe ais. 6. is the region bounded by the curve y = and the ais. 7. is the region bounded by the curve y = 6 5 and the ais. 8. is the region in the first quadrant bounded by the curve y =4 and the lines y = and y =. 9. is the region bounded by the curve y = and the lines y = and y =.. is the region in the first quadrant that lies under the curve y = 6 and that is bounded by this curve and the lines y =, y =,and =8.. is the region bounded by the curve y = and the ais. (Hint: eflect the region across the ais and integrate the corresponding function.). is the region bounded by the curves y = + and y = between = and =.. is the region bounded by the curve y = e and the lines y =and =. 4. is the region bounded by the curve y = and the line y =. 5. is the region bounded by the curve y = and the line y =4. 6. is the region bounded by the curves y = 6 and y =. 7. is the region bounded by the line y = and the curve y =. 8. istheregioninthefirst quadrant bounded by the curve y = + and the lines y = 8 and y =. 9. is the region bounded by the curves y = + and y = is the region bounded by the curves y = and y = +. esults ( + ln 4) e
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# Equation For Parallel Circuit
The equation for a parallel circuit is a simple formula that can be used to calculate the total resistance of a circuit that contains multiple resistors connected in parallel. The formula is: ``` Rt = 1 / (1/R1 + 1/R2 + 1/R3 + ...) ``` where Rt is the total resistance of the circuit, and R1, R2, R3, etc. are the resistances of the individual resistors. To use this formula, simply add the reciprocals of the individual resistances and then take the reciprocal of the sum. For example, if you have a circuit with three resistors with resistances of 10 ohms, 20 ohms, and 30 ohms, the total resistance of the circuit would be: ``` Rt = 1 / (1/10 + 1/20 + 1/30) = 5 ohms ``` The equation for a parallel circuit can be used to calculate the total resistance of any circuit that contains multiple resistors connected in parallel. It is a simple and straightforward formula that can be used to quickly and easily determine the total resistance of a circuit.
In addition to calculating the total resistance of a circuit, the equation for a parallel circuit can also be used to calculate the current flowing through each resistor in the circuit. The current flowing through each resistor is equal to the total current flowing through the circuit divided by the number of resistors in the circuit. For example, if a circuit with three resistors has a total current of 10 amps, the current flowing through each resistor would be 10 amps / 3 resistors = 3.33 amps.
The equation for a parallel circuit can also be used to calculate the voltage drop across each resistor in the circuit. The voltage drop across each resistor is equal to the total voltage applied to the circuit divided by the number of resistors in the circuit. For example, if a circuit with three resistors has a total voltage of 12 volts, the voltage drop across each resistor would be 12 volts / 3 resistors = 4 volts.
The equation for a parallel circuit is a powerful tool that can be used to analyze and design circuits. It can be used to calculate the total resistance, current, and voltage drop of any circuit that contains multiple resistors connected in parallel.
Here are some additional resources that you may find helpful: * [The Equation for a Parallel Circuit](https://www.khanacademy.org/science/physics/electricity-and-magnetism/circuits/a/parallel-circuits) * [Parallel Circuits](https://www.physicsclassroom.com/class/circuits/Lesson-2/Parallel-Circuits) * [Parallel Circuits: The Basics](https://www.electronics-tutorials.ws/resistors/parallel-resistors.html)
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Nwanua Elumeze
Zach Johnson
Travis Presley
1.0 Introduction:
2 + 2 = 4, the simplest of math problems. Where did you learn that? Why are you sure that its true? Did you just memorize it in first grade? Or, do you have an intuition based on a physical example you saw in kindergarten? The Math Balance is a tool for teachers and museums to use to give that physical intuition to each student. Not only can this tool be used for simple problems like two plus two, but it also can be reintroduced in the next few years to teach algebra.
The Math Balance is a simple balance with labeled blocks of unit weight. Using this tool, simple problems can be posed like two plus two. Simply place two twos in the left basket, and ask the student “What does two plus two equal?” The child can play with the other blocks to try and get the beam to balance. If the child is getting frustrated that the three he/she thinks is the correct answer isn’t balancing the beam, they could hit the “?” button. The dial will display one instructing the student that they need to add a one block to the right side to balance. As they slowly learn they may be asked to consolidate items on the right so that three and one are not a correct answer, only four is.
Beyond simple addition, 2 + x = 4 is another problem that can be posed. By simply placing a two on the left and a four on the right, the teacher can ask the student “What is x, if two plus x equals four?” Again the student can simply work with the blocks to determine what the correct answer is when the beam is balanced. Also if they have trouble the “?” button displays the number they are off by.
This tool is just many in an arsenal of mathematical teaching devices including memorization, and math teaching software. It adds an additional level of the physical world to each problem that may help students that need to visualize each problem.
2.0 Design
Given the task, our project was broken down into three sections. The base to support the balance beams, their corresponding springs, as well as the computer monitoring the balance to provide the help. The unit weight blocks to be used in the math problems and placed in the baskets on the balance beam. The computation hangs on the upright beam and monitors two sensors that observe a change in the balance beam position and then display the hint when the “?” button is pushed. Each piece contains a large amount of detail and design. Each of the follow sections describe these pieces in detail as well as how they were iteratively designed.
2.1 Computation:
From the very beginning, we wanted this exhibit to feature a very strong coupling of computation and the physical aspects of the balance. The idea behind the computation in the exhibit was to determine the difference (in discrete block weights) between the left and right pans, and report that difference to the child in a non-ambiguous manner should it be asked. Rather than an intrusive and obviously present computation, we wanted the exhibit to appear very easy to use by hiding most of the electronics part of (figuratively and literally). As a result, the exhibit can be used without even being aware that it contains any computation: as long as blocks are placed and removed in order to balance the beam, and no help is sought by the child, he/she can easily benefit from seeing that 4+3+1 blocks on one side is equivalent to 8 on the other side. Seeking for help, however, reveals the background computational unit that has up until now, simply been monitoring the motions of the beam. In the same spirit of simplicity and "background", there is only one way that a child may interact with the micro-controller: a big bold button with a question mark on it. For example, if there are blocks marked 4 & 3 on the left pan, and 8 on the right pan, the exhibit will move the dial toward the left pan (to a position marked '1'), implying: "looks like we might need a 1 on this side". At the bottom of the dial face were three exclamation marks: another unambiguous way of saying there was too much weight difference on a particular side.
The task of keeping track of the beam position was put on a cricket, which quickly became overwhelmed by the demands of a real time method of tracking - this is why we eventually used a micro-controller.
During the initial design phase, there were several methods we wanted to use to keep track of the beam's position.
- Shine an infrared beam under each weigh pan and use a corresponding sensor to convert the intensity of the reflected light into a distance measure. The difference between the readings of either sensor would tell how much the balance beam has tilted. We discovered, after toying with this idea, that there would be far too little resolution (since the reflected light would not change much), and this method could be adversely affected by the presence and intensity of ambient light.
- Another reflective idea was touched upon: ultrasonic. Again, it was believed that this would suffer from the same problems of too little resolution; hence this idea didn't last longer than the time it took to suggest it and discredit it.
- If attention is paid to the center of the beam, one can see that it describes a turning circle as the beam tips one way to the other. We intended to take advantage of this by putting the shaft of a rheostat on a nut attached to the beam that would revolve along with it. This would result in a different resistance depending on how the beam was tilted; this resistance will then be interpreted by the cricket. This actually worked initially but was ultimately superceded by our final design for two reasons:
1. Even though the shaft was relatively easy to turn, it added just enough load to reduce our resolution to about 3 unit blocks... not at all a good thing.
2. The range of the potentiometer was either too small or too little for the cricket to interpret effectively. To make matters worse, there were regions of non-linearity what would have to be taken into account. With these disadvantages, and the fact that mounting the resistor would have required a very precise mechanism made it all the more unfavorable.
The idea of using the resistor can be pursued in future endeavors because it didn't require constant monitoring (unlike the other ideas): when the help button is pressed, the position can be directly obtained from the resistance value.
- At this stage, we needed a linear and reliable method of detecting the rotation of the "circle" described by the beam's motion: quadrature phase measurement. This involves a circle with slots in it, and two sensors positioned 90 degrees from each other. As the circle turns, the two sensors yield a waveform that looks like a 2-bit gray code. Depending on the direction, this mechanism would yield logic values of:
... 10 11 01 00 10 11 01 ... (one direction, and)
... 10 00 01 11 10 11 01 ... (opposite direction).
This is a well-proven method of motion detection: it's the same mechanism that tracks the position of a computer mouse. The trouble now was to obtain and use suitable sensors and wheel.
For the wheel, we asked for and got a gear whose patterns of teeth would suit the detection just fine. Initially, we wanted to use simple switched with rollers on them: as the wheel turned, it would press and release the switches, giving the desired pulses. As it turns out, not only did the switches add extra resistance to motion in one direction, the totally opposed rotation in the other, depending on how they were oriented. Crickey! We needed a way of sensing this motion and fast... two seemingly good ideas had turned to pot, and the presentation was just a week away! Fortunately, the code (which will be described later) for detecting this phase difference had been written and tested with handheld switches the week before, so that was not an issue.
Since we wanted as little contact between the turning wheel and the sensors (and with a flash of inspiration from thinking about a mouse while shopping for something entirely different), the idea of using opto-interruptors was hit upon. Opto-interruptors are a method of detecting when a direct path is blocked by and object or not, using infrared. The wheel had enough teeth and was just the right diameter to fit the ones I had bought (I had made sure to buy the widest sensors with this in mind).
Now, it was a piece of cake to connect these sensors to a cricket and monitor the analog values in software. Unfortunately, due to the limited speed of these devices, we would not be able to keep track of the wheel's rotation in real time. Couple this with my lack of patience with finicky devices and computers that crash ever so often, I asked and kindly obtained permission to move the project to a micro-controller with more computing power.
The micro-controller was one I had been introduced to in a previous class, and here was an excellent opportunity to do something really useful with knowledge gained from building the board from scratch. The two sensors were mounted on the main trunk of the balance, spaced to maximize the sensitivity to differential motion, with the teeth of the gear alternately interrupting the beam from transmitter to sensor. As the wheel turned, one sensor would have a transition about 90 degrees ahead (or behind) the previous one.
This state of transitions called for a state machine that would be put into a known state by sampling the sensors when it started: this way, the machine calibrated itself automatically...a feature which is essential to the fact that we didn't want the user of the exhibit to know anything about how the computation unit ran.
From any state, there could only be two transitions (i.e. from 11):
01 <- 11 -> 10.
Which transition was made determined whether a counter (imaginatively called "count") was incremented or decremented. The following fragment exemplifies this:
.
.
.
if (laststate == 11) {
if (state == 01) count++;
else if(state == 10) count --;
} else if (laststate == 10) {
.
.
.
As the gear was turned back and forth during testing, we could see the variable increase in one direction, return to zero, and continue in the other direction: we thought we were home free.
Unfortunately, as with any real-world-computer interface, non-linearities would show up and would have to be dealt with in software.
- Static friction (again!) could prevent a wheel from moving if a very small load change is made (adding or removing a relatively light block)
- As the beam tilts and the springs stretch, each spring introduces a non-zero component into the balance of forces. [cos(90) = 0, and cos(90+x) = cos(x) != 0]
These two problems were eventually resolved by empirically measuring the values produced by varying weight changes and using those values in the program to determine how far to move the dial.
The dial was connected to a low speed, high torque motor which was also controlled by the board. When the help button is pressed, the motor is moved in a direction for a duration directly related to the angle of the beam. Each time the help button was pressed, the state machine would reset to this new position: essentially zeroing itself out. This way, only relative motion (which is what is important anyway), was taken into account: "I have to move the dial 3 ticks CCW of this position).
2.2 Blocks:
A collection of blocks representing weights from 1 to 10 numeric units was required. These blocks were constructed from precut 11/2-inch wooden cubes obtained from Michaels Arts & Crafts stores. A cylindrical cavity was drilled into each block using an 11/4-inch Forstner bit. This bit permitted drilling a uniform cylinder to within 1/8 inch of the bottom of the block. The weight of each block was grossly adjusted by addition of 11/4-inch fender washers. The final weight adjustment was made by addition of lead shot. Any remaining void space within the cylinder was filled with cotton. The side of the block containing the cylindrical opening was then covered with a 11/2" X 11/2" X 3/32" piece of Basswood, which was glued in place. These Basswood covers were prepared using the laser cutter.
The weights of the blocks were measured on a kitchen scale. The response of the scale was fairly linear, but suffered from hystersis and a shifting zero point. The accuracies of the weighted blocks were confirmed on the balance itself before final gluing was done. It is estimated that the total weights of any combination of blocks, which add to make a given numeric unit, agree within plus or minus 2 grams.
2.3 Balance:
Wood is one of the simplest and durable materials we had access too. With it, and a few screws we were able to build the basis of our exhibit. About twenty feet of one by two inch pine was cut to form two legs, a cross beam, and the upright balance support. The two legs were constructed with stacked pieces and a gap for the cross beam to slide into. The cross beam is three piece of wood sandwiched together, with the inner section having a gap to fit the upright balance support. Thus to erect the balance from a stored state, you simply lay the two legs on your work surface, either the ground or a table, spaced about a foot apart. Simply slide the cross beam into the legs. That is the base. The upright balance support slides into the crossbeam. The upright support has several wholes in it to place the metal balance beam and the spring support metal plate. The balance beam attaches to the bearing bolt. The upper plate is attached with two bolts and the springs are hooked into the upper wholes. Once everything is attached the baskets can be hooked on the balance beam. While this may seem over done, it works great for situations in a school where an exhibit needs to be stored when not in use. The modular design allows it to be completely deconstructed and then tucked in any cabinet or storage area.
This final design for the balance evolved from a fairly similar origin. We found that a bearing on both the front and the back resisted the balance movement. We also found that when the balance had a difference of 3 units the bottom of the spring scales hindered the movement, so we raised the spring support and added chains to the bottom of the springs. With these minor changes we had our final design.
3.0 Evaluation:
This exhibit exploits the analogy between an arithmetic problem and a balanced lever to teach children to solve arithmetic addition problems. The child will come to understand that arithmetic concept of "equality" corresponds to the lever being balanced in the horizontal position. With this exhibit a child can be asked to solve an arithmetic addition or subtraction problem by determining the numeric unit block to be added in order to balance the beam.
When presenting a problem to the child, the pans affixed to each end of the lever beam contain blocks with different numeric units. The child is asked to select a block to add to one of the pans to balance the lever. While either pan could be heavier, it will probably be less confusing if the left pan is used for adding or subtracting numeric units. This is the convention used in textbooks for writing equations on one line of text, e.g. 5 + x = 8. The child guesses a numeric value for x and places a block having that value in the left pan. If the guess is correct, the beam balances in a horizontal position. If the choice is incorrect, then the beam will be tipped to one side or the other. Springs will arrest the beam at some point. The specific arrest-point depends on the amount of the weight imbalance.
If the child wants help with the solution, he/she can press the help button (?) and the pointer on the dial will turn to indicate a number, either 1, 2, 3, or (!!). The reading indicates the numeric units to add to a pan. The double exclamation points indicate that the imbalance exceeded 3 numeric units. If the dial pointer turns counterclockwise, blocks should be added to the left pan. When the dial turns clockwise it may be more confusing to the child. It means that blocks should be added to the right pan. It may be necessary to explain to the child that adding blocks to the right pan is equivalent to removing blocks from the left pan. However, with a small amount of initial experimentation, the child should be able to discover this principle with his/her own efforts.
It might have been helpful if a mechanical means had been provided for arresting the beam motion while setting up a problem. The child could press a Release button to free the beam and see the results of his/her guess.
The exhibit was constructed in a modular fashion. We found this extremely desirable. The base is easily disassembled. The computer can be readily removed from its holding bracket. With a bit of difficulty the metal beams can be disconnected from their vertical support. There is a bit of instability between the vertical support and the base, but it was judged to be inconsequential. If necessary this instability could be corrected with a couple of bolts. The only non-modular features are the dial and the help buttons, which were glued directly to the vertical support member.
We tested many different combinations of blocks, which summed to equal numeric values, to verify that they would accurately balance.
We evaluated the software, which converts beam displacement to dial-pointer rotation for many different combinations of blocks, with imbalances favoring either pan. The displacement readings were calibrated to give accurate pointer rotations to indicate the numeric units corresponding to the imbalance.
From our results we believe this system to be accurate, stable, and reproducible, subject to the caveat that, due to time pressure, long-term reproducibility testing was not possible.
Unfortunately, our testing did not involve children as users. Certainly, this type of input is desirable, and had it been available, would likely have precipitated some redesign leading to usability improvement. Nevertheless, we feel that this exhibit can support and enhance the teaching of addition and subtraction to young children. It provides a physical realization of abstract arithmetic rules. In addition the exhibit permits and encourages individual exploration by the child.
4.0 Conclusion:
Rather than a crutch, we wanted a helpful guide that helps only when asked, and does that in a very unambiguous, easy to understand manner. We feel that by interacting with a child in this manner, the balance helps the child learn algebra (albeit limited in scope) without the heavy-handiness of "this is what it must be, so memorize it". Rather, the child is free to try to balance the weights on her own and learn by being inquisitive (how any child learns best), and only when she feels frustrated does she have to press a button (clearly marked) which allows the exhibit to suggest the answer. We prefer that the exhibit suggests the answer by implying: "you would need 3 on that pan to balance this equation", rather than "you must have 3 to balance the equation".
If I was to do this over, I would consider having a single tray for blocks, and a digital display (or a single dial) that shows a number (1 - 10). One side of the balance will be "weighted down" by applying enough power to a motor to give the desired torque. The child would simply have to fill the balance with enough blocks to yield an answer. This, I believe would remove the possible confusion that can occur if there are several blocks on each pan, at the cost of limited flexibility.
We feel very strongly about the role that an exhibit/automaton should posses in relative terms with a child: allow for exploration of options, and provide concise help when requested. By interacting with the exhibit ourselves, we feel that it succeeded in allowing us to "learn" differences between weights, providing just enough, and not overbearing feedback.
In comparison with the different ways arithmetic is currently taught and illustrated to children, it is easy to see that this method is a welcome complement, which uses equally sized blocks to show the importance of weight and the value of numerals.
5.0 Appendix:
/* Balance code... written for 8051 uC. 2001, Nwanua */
#include <8051io.h>
#include <8051reg.h>
#include <8051bit.h>
void movedial(int dir)
{
/* if dir is non-zero, move the motor in the specified direction
for a specified duration... then turn it off before leaving
this function */
if (dir > 0) {
setbit(P1.4);
clrbit(P1.5);
delay(abs(dir));
goto _moff;
} else if (dir < 0) {
clrbit(P1.4);
setbit(P1.5);
delay(abs(dir));
goto _moff;
} else {
_moff:
setbit(P1.4);
setbit(P1.5);
}
}
int dohelp(int count)
{
int position;
/* debounce the switch (solidly pressed) */
delay(80);
if(P1 & 0x08)
return 0;
/* "scaled" position... the springs combined with
* rotation give a non-linear result... need to compensate:
*/
if (count == 0) {
position = 0;
} else if (count >= -2 && count <= -1) {
position = -1;
} else if (count >= -4 && count <= -3) {
position = -2;
} else if (count >= -6 && count <= -5) {
position = -3;
} else if (count < -6) {
position = -4;
} else if (count >= 1 && count <= 3) {
position = 1;
} else if (count >= 4 && count <= 5) {
position = 2;
} else if (count >= 5 && count <= 6) {
position = 3;
} else if (count > 6) {
position = 4;
}
/* 800 is a nice constant */
movedial(position * 800);
/* make sure switch was released */
while(! (P1 & 0x08));
/* successful */
return 1;
}
int main()
{
char state, last;
int count;
/* make sure motor is off! */
movedial(0);
/* if there's a computer attached... tell 'em "hello" */
serinit(9600);
printf("balance.... \n");
/* just so we know it's alive and well */
movedial(800);
delay(1);
movedial(-800);
/* configure up the port pins as inputs */
setbit(P1.1);
setbit(P1.2);
setbit(P1.3);
/* relative motor ticks... */
count = 0;
/* the initial state of the machine */
state = P1 & 0x6;
while(1){
/* looks like help switch was pressed */
if (!(P1 & 0x08)) {
/* if switch press was valid... */
if ( dohelp(count) ) {
/* reset the state with the new position as a basis */
count = 0;
state = P1 & 0x6;
}
}
/* the state machine:
constantly monitor the state of the two sensors,
if there has been no change, there's nothing else to do */
last = state;
state = P1 & 0x6;
if (last == state)
continue;
/* a given transition from a given state yeilds a dec/increment */
if (last == 0) {
if (state == 2 && count < 128)
count++;
else if ( state == 4 && count > -127)
count--;
} else if (last == 2) {
if (state == 6 && count < 128)
count++;
else if ( state == 0 && count > -127)
count--;
} else if (last == 6) {
if (state == 4 && count < 128)
count++;
else if ( state == 2 && count > -127)
count--;
} else if (last == 4) {
if (state == 0 && count < 128)
count++;
else if ( state == 6 && count > -127)
count--;
}
}
}
|
# 5.8: Applications of Exponential and Logarithmic Functions
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##### Learning Objectives
• Apply common exponential models to real-life situations.
• Apply common logarithmic models to real-life situations.
We have already explored some basic applications of exponential and logarithmic functions. In this section, we explore some important applications in more depth, including radioactive isotopes and Newton’s Law of Cooling.
## Modeling Exponential Growth and Decay
In real-world applications, we need to model the behavior with functions. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of population growth, such as a bacteria culture that a biologist is studying for a new medical treatment, we may choose the exponential growth function:
$N(t)=N_0e^{rt}\nonumber$
where $$N_0$$ is equal to the value at time zero, $$e$$ is Euler’s constant, and $$k$$ is a positive constant that determines the rate (percentage) of growth.
##### Exponential Growth
The model $$N(t)=N_0\cdot e^{kt}$$ represents the population size after a given amount of time $$t$$.
$$N_0$$ is the initial population size and $$r$$ is the rate of growth for the population.
##### Example $$\PageIndex{1}$$
A population of bacteria doubles in size every hour. If the culture started with $$10$$ bacteria, find a model that describes the population size after of the bacteria after $$t$$ hours. Use this model to determine how many bacteria there will be after 10 hours.
Solution
Bacteria growth is exponential. To find $$N_0$$ we use the fact that $$N_0$$ is the amount at time zero, so $$N_0=10$$. To find $$r$$, use the fact that after one hour $$(t=1)$$ the population doubles from $$10$$ to $$20$$. The formula is derived as follows
\begin{align*} 20&= 10e^{r\cdot 1}\\ 2&= e^r \qquad \text{Divide by 10}\\ \ln2&= r \qquad \text{Convert to exponential form} \end{align*}
so $$r=\ln(2)$$. Thus the equation that models the bacteria growth is $$N(t)=10e^{(\ln2)t}=10{(e^{\ln2})}^t=10·2^t$$. The graph is shown in Figure $$\PageIndex{2}$$.
The population of bacteria after ten hours is $$N(10)=10\cdot2^10=10,240$$.
Analysis
We could describe this amount is being of the order of magnitude $$10^4$$. The population of bacteria after twenty hours is $$10,485,760$$ which is of the order of magnitude $$10^7$$, so we could say that the population has increased by three orders of magnitude in ten hours.
In the previous example, we were provided information about the time it took for the population to double in size. This time is called doubling time. Notice that the model simplified to an exponential function of base 2. We can describe a modified growth model for doubling time as follows:
##### Doubling Time
The model $$N(t)=N_0\cdot2^{\frac{t}{d}}$$ represents the population size after a given amount of time $$t$$.
$$N_0$$ is the initial population size and $$a$$ is the time it takes for the population to double in size.
##### Example $$\PageIndex{2}$$
Cancer cells sometimes increase exponentially. If a cancerous growth contained 300 cells last month and 360 cells this month, how long will it take for the number of cancer cells to double? Round your answer to the nearest tenth.
Solution
Defining $$t$$ to be time in months, with $$t = 0$$ corresponding to the initial size of the population, we are given two pieces of data: last month, (0, 300) and this month, (1, 360).
From this data, we can find an equation for the growth. Using the form $$N(t)=N_0\cdot2^{\frac{t}{d}}$$, we know immediately $$N_0=300$$, giving $$N(t)=300\cdot2^{t}$$. Substituting in $$(1,360)$$, \begin{align*} 360&=300\cdot2^\tfrac{1}{d} \\[4pt] \dfrac{360}{300}&= 2^\tfrac{1}{d}\\[4pt] 1.2&= 2^\tfrac{1}{d} \\[4pt] \ln1.2&=\ln\left(2^{\tfrac{1}{d}}\right) \\[4pt] \ln1.2&=\dfrac{1}{d}\ln2 \\[4pt] d\cdot\ln1.2&=\ln2 \\[4pt] d&=\dfrac{\ln2}{\ln1.2} \\[4pt] d &\approx3.8 \text{ months}\end{align*}\nonumber
We now turn to exponential decay. One of the common terms associated with exponential decay is half-life, the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.
The model $$M(t)=M_0e^{-\tfrac{ln2}{h}t}$$ represents the amount of material (mass) left after a given amount of time $$t$$.
$$M_0$$ is the initial mass size and $$h$$ is the half-life of the material.
The formula for radioactive decay is important in radiocarbon dating, which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about $$1\%$$ error for plants or animals that died within the last $$60,000$$ years.
Carbon-14 is a radioactive isotope of carbon that has a half-life of $$5,730$$ years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12, which has an atomic weight of $$12$$ and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last $$60,000$$ years, using tree rings and other organic samples of known dates—although the ratio has changed slightly over the centuries.
As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.
##### Example $$\PageIndex{3}$$
The half-life of carbon-14 is $$5,730$$ years. Express the amount of carbon-14 remaining as a function of time, $$t$$.
Solution
The function that describes this continuous decay is $$M(t)=M_0e^{\left (-\tfrac{\ln(2)}{5730} \right )t}$$.
##### Example $$\PageIndex{4}$$
A bone fragment is found that contains $$20\%$$ of its original carbon-14. To the nearest year, how old is the bone?
Solution
The percentage of carbon-14 in all living things is $$100\%$$. The percentage of carbon-14 after some amount of time, $$t$$, (which we are trying to find) is 20%. Using decimals for the percent, we can substitute in $$M_0=1 \text{ and } M(t)=0.2$$ into the function we found in the previous example, \begin{align*}M(t)&=M_0e^{\left (-\tfrac{\ln(2)}{5730} \right )t}\\[4pt] 0.2&=1\cdot e^{\left (-\tfrac{\ln(2)}{5730} \right )t} \\[4pt] 0.2&=e^{\left (-\tfrac{\ln(2)}{5730} \right )t} \\[4pt] \ln0.2&=\frac{-\ln(2)}{5730}t\\[4pt] -\dfrac{5730}{\ln2}(\ln0.2)&=t\\[4pt] t &\approx13,305 \text{ years}\end{align*}
The bone fragment is about $$13,305$$ years old.
Analysis
The instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about $$1\%$$, so this age should be given as $$13,305$$ years $$\pm 1\%$$ or $$13,305$$ years $$\pm 133$$ years.
## Newton’s Law of Cooling
Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a vertical shift of the generic exponential decay function. This translation leads to Newton’s Law of Cooling, the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature
##### Newton's Law of Cooling
The temperature of an object, $$T$$, in surrounding air with temperature $$T_s$$ will behave according to the formula
$T(t)=D_0e^{-kt}+T_s\nonumber$
where
• $$t$$ is time
• $$D_0$$ is the difference between the initial temperature of the object and the surroundings
• $$k$$ is a constant, the continuous rate of cooling of the object
##### Example $$\PageIndex{5}$$
A cheesecake is taken out of the oven with an ideal internal temperature of $$165°F$$, and is placed into a $$35°F$$ refrigerator. After $$10$$ minutes, the cheesecake has cooled to $$150°F$$. If we must wait until the cheesecake has cooled to $$70°F$$ before we eat it, how long will we have to wait?
Solution
Because the surrounding air temperature in the refrigerator is $$35$$ degrees, the cheesecake’s temperature will decay exponentially toward $$35$$, following the equation
$$T(t)=D_0e^{-kt}+35$$
We know the initial temperature was $$165$$, so $$T(0)=165$$.
\begin{align*} 165&= Ae^{-k0}+35 \qquad \text{Substitute } (0,165)\\ D_0&= 130 \qquad \text{Solve for } D_0 \end{align*}
We were given another data point, $$T(10)=150$$, which we can use to solve for $$k$$.
\begin{align*} 150&= 130e^{-k10}+35 \qquad \text{Substitute } (10, 150)\\ 115&= 130e^{-k10} \qquad \text{Subtract 35}\\ \dfrac{115}{130}&= e^{-10k} \qquad \text{Divide by 130}\\ \ln\left (\dfrac{115}{130} \right )&= -10k \qquad \text{Take the natural log of both sides}\\ k &= \dfrac{\ln \left (\dfrac{115}{130} \right )}{-10} \qquad \text{Divide by the coefficient of k} \end{align*}
This gives us the equation for the cooling of the cheesecake: $$T(t)=130e^{\left(–\tfrac{1}{10} \ln\tfrac{23}{26}t \right)}+35$$.
Now we can solve for the time it will take for the temperature to cool to $$70$$ degrees. \begin{align*} 70&= 130e^{\left(–\tfrac{1}{10} \ln\tfrac{23}{26} \right)}+35 \qquad \text{Substitute in 70 for } T(t)\\[4pt] 35&= 130e^{\left(–\tfrac{1}{10} \ln\tfrac{23}{26} \right)t} \qquad \text{Subtract 35}\\[4pt] \dfrac{35}{130}&= e^{\left(–\tfrac{1}{10} \ln\tfrac{23}{26} \right)t} \qquad \text{Divide by 130}\\[4pt] \ln \left (\dfrac{35}{130} \right )&= \left(–\tfrac{1}{10} \ln\tfrac{23}{26} \right)t \qquad \text{Take the natural log of both sides}\\[4pt] t&= \dfrac{\ln \left (\dfrac{35}{130} \right )}{\left(–\tfrac{1}{10} \ln\tfrac{23}{26} \right)t}\\[4pt] t&\approx 107 \qquad \text{Divide by the coefficient of t} \end{align*}
It will take about $$107$$ minutes, or one hour and $$47$$ minutes, for the cheesecake to cool to $$70°F$$.
##### You Try $$\PageIndex{1}$$
A pitcher of water at $$40$$ degrees Fahrenheit is placed into a $$70$$ degree room. One hour later, the temperature has risen to $$45$$ degrees. How long will it take for the temperature to rise to $$60$$ degrees to three decimal places?
$$6.026$$ hours
## Modeling with Logarithms
For quantities that vary greatly in magnitude, a standard scale of measurement is not always effective, and utilizing logarithms can make the values more manageable. For example, if the average distances from the sun to the major bodies in our solar system are listed, you see they vary greatly.
Planet Distance (millions of km) Mercury 58 Venus 108 Earth 150 Mars 228 Jupiter 779 Saturn 1430 Uranus 2880 Neptune 4500
Placed on a linear scale – one with equally spaced values – these values get bunched up, as seen below.
0 500 1000 1500 2000 2500 3000 3500 4000 4500
However, computing the logarithm of each value and plotting these new values on a number line results in a more manageable graph, and makes the relative distances more apparent.
Planet Distance (millions of km) log(distance) Mercury 58 1.76 Venus 108 2.03 Earth 150 2.18 Mars 228 2.36 Jupiter 779 2.89 Saturn 1430 3.16 Uranus 2880 3.46 Neptune 4500 3.65
(It is interesting to note the large gap between Mars and Jupiter on the log number line. There is an asteroid belt located there, which some scientists believe is a planet that never formed because of the effects of the gravity of Jupiter.)
Logarithms are also used to measure sound intensity levels.
Source of Sound/Noise Approximate Sound Pressure in $$\mu$$ Pa (micro Pascals) Launching of the Space Shuttle 2000,000,000 Full Symphony Orchestra 2000,000 Diesel Freight Train at High Speed at 25 m 200,000 Normal Conversation 20,000 Soft Whispering at 2 m in Library 2,000 Unoccupied Broadcast Studio Room 200 Softest Sound a human can hear 20
Logarithms are useful for showing relative changes. For example, comparing the sound of a diesel freight train to a soft whisper, the diesel train is 100 times larger than a soft whisper.
$\dfrac{200,000}{2,000} = 100 = 10^2\nonumber$
When one quantity is roughly ten times larger than another, we say it is one order of magnitude larger. The order of magnitude can be found as the common logarithm of the ratio of the quantities.
##### Orders of Magnitude
Given two values $$A$$ and $$B$$, to determine how many orders of magnitude $$A$$ is greater than $$B$$,
Difference in orders of magnitude = log($$\dfrac{A}{B})$$
One example of a logarithmic scale is the Moment Magnitude Scale (MMS) used for earthquakes. This scale is commonly and mistakenly called the Richter Scale, which was a very similar scale succeeded by the MMS.
##### Moment Magnitude Scale for Earthquakes
For an earthquake with seismic moment $$S$$, a measurement of earth movement, the MMS value, or magnitude $$M$$ of an earthquake, is
$M = \dfrac{2}{3} log(\dfrac{S}{S_0})\nonumber$
Where $$S_0 = 10^{16}$$ is a baseline measure for the seismic moment.
##### Example $$\PageIndex{6}$$
If one earthquake has a MMS magnitude of 6.0, and another has a magnitude of 8.0, how much more powerful (in terms of earth movement) is the second earthquake?
Solution
Since the first earthquake has magnitude 6.0, we can find the amount of earth movement for that quake, which we'll denote $$S_1$$. The value of $$S_0$$ is not particularity relevant, so we will not replace it with its value.
$6.0 = \dfrac{2}{3} log \left(\dfrac{S_1}{S_0}\right)\nonumber$
$6.0 \left(\dfrac{3}{2}\right) = log \left(\dfrac{S_1}{S_0}\right)\nonumber$
$9 = log\left(\dfrac{S_1}{S_0}\right)\nonumber$
$\dfrac{S_1}{S_0} = 10^9\nonumber$
$S_1 = 10^9 S_0\nonumber$
This tells us the first earthquake has about $$10^9$$ times more earth movement than the baseline measure, $$S_0$$.
Doing the same with the second earthquake, $$S_2$$, with a magnitude of 8.0,
$8.0 = \dfrac{2}{3} log \left(\dfrac{S_2}{S_0}\right)\nonumber$
$S_2 = 10^{12} S_0\nonumber$
Comparing the earth movement of the second earthquake to the first,
$\dfrac{S_2}{S_1} = \dfrac{10^{12} S_0} {10^9 S_0} = 10^3 = 1000\nonumber$
The second value's earth movement is 1000 times as large as the first earthquake.
##### Example $$\PageIndex{7}$$
One earthquake has magnitude of 3.0. If a second earthquake has twice as much earth movement as the first earthquake, find the magnitude of the second quake.
Solution
Since the first quake has magnitude 3.0,
$3.0 = \dfrac{2}{3} log (\dfrac{S}{S_0})\nonumber$
Solving for $$S$$,
$3.0 \dfrac{3}{2} = log (\dfrac{S}{S_0})\nonumber$
$4.5 = log (\dfrac{S}{S_0})\nonumber$
$10^{4.5} = \dfrac{S}{S_0}\nonumber$
$S = 10^{4.5} S_0\nonumber$
Since the second earthquake has twice as much earth movement, for the second quake,
$S = 2 \cdot 10^{4.5} S_0\nonumber$
Finding the magnitude,
$M = \dfrac{2}{3} log (\dfrac{2 \cdot 10^{4.5} S_0}{S_0})\nonumber$
$M = \dfrac{2}{3} log (2 \cdot 10^{4.5}) \approx 3.201\nonumber$
The second earthquake with twice as much earth movement will have a magnitude of about 3.2.
As we saw in an earlier section, logarithms are also used to describe $$pH$$ levels.
##### pH Level
The pH of a solutions is defined by the following formula, where $$[\ce{H^{+}}]$$ is the concentration of hydrogen ions in the solution
\begin{align*} {pH}&=−{\log}([\ce{H^{+}}]) \label{eq1} \\[4pt] &={\log}\left(\dfrac{1}{[\ce{H^{+}}]}\right) \label{eq2} \end{align*}\nonumber
##### Example $$\PageIndex{8}$$
If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?
Solution
Suppose $$C$$ is the original concentration of hydrogen ions, and $$P$$ is the original pH of the liquid. Then $$P=–\log(C)$$. If the concentration is doubled, the new concentration is $$2C$$. Then the pH of the new liquid is
$$pH=−\log(2C)$$
Using the product rule of logs
$$pH=−\log(2C)=−(\log(2)+\log(C))=−\log(2)−\log(C)$$
Since $$P=–\log(C)$$,the new pH is
$$pH=P−\log(2)≈P−0.301$$
This page titled 5.8: Applications of Exponential and Logarithmic Functions is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Katherine Skelton.
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# How do you calculate the modulo of a high-raised number?
I need some help with this problem:
$$439^{233} \mod 713$$
I can't calculate $439^{223}$ since it's a very big number, there must be a way to do this.
Thanks.
-
– Martin Sleziak Sep 14 '12 at 9:48
There are often tricks to this if the numbers are nice enough, but even if they're not, here's a way that's not entirely horrible.
You already know what 439 is mod 713. What is $439^2 \mod 713$? What about $439^4$? (Hint: take your answer for $439^2$ after reducing it mod 713, and then square it again.) In the same way, calculate $439^8, 439^{16}, \dots, 439^{128} \mod 713$. Now just note that 233 = 128 + 64 + 32 + 8 + 1. So multiply the appropriate powers of 439 together - again, one calculation at a time, reducing mod 713 each time.
Now you should only have to do 11 calculations, and now all your numbers are 6 digits or less. Rather than impossible, it's now simply tedious. :)
By the way, one thing to notice: 713 = 23 * 31. Perhaps your calculations will be easier if you do them mod 23 and 31, then apply the Chinese remainder theorem?
-
You can do this step by step by first computing $439^2\equiv 211 {\ \rm mod\ }713$. Then you compute $439^4\equiv 211^2 \equiv 315 {\ \rm mod\ }713$. Continue to square and reduce mod $713$ and build up a list of powers $$439^1, \qquad 439^2, \qquad 439^4 \qquad \dots \qquad 439^{128}\qquad \qquad ({\rm mod\ }713)$$ Finally, write $233$ as a sum of powers of $2$, and compute $$439^{233}=439^{1+8+32+64+128}=439\cdot 439^8\cdot 439^{32}\cdot439^{64}\cdot 439^{128} \qquad ({\rm mod\ }713)$$ Just remember to reduce mod $713$ each time you get a product which is larger than $713$.
-
$713=23\cdot 31$
$439 \pmod {23}=2$ and $\phi(23)=22$ and $233\equiv 13{\pmod {22}}$
So, $439^{223} {\pmod {23}} \equiv 2^{22\cdot 10 + 13}\equiv {(2^{22})}^{10}2^{13}\equiv 2^{13} {\pmod {23}}$ using Euler's Totient Theorem.
$2^6\equiv 18 {\pmod {23}}, 2^7\equiv 36 \equiv 13$
$\implies 2^{13}\equiv 18\cdot 13=234\equiv 4 {\pmod {23}}=4+23x$ for some integer $x$.
$439 \pmod {31}=5$ and $\phi(31)=30$ and $233\equiv 23{\pmod {30}}$
So, $439^{223} {\pmod {31}} \equiv 5^{23} {\pmod {31}}$
$5^3 \equiv 1 {\pmod {31}} \implies 5^{23}\equiv({5^3})^7 5^2\equiv 5^2{\pmod {31}}=25+31y$ for some integer $y$.
So, we need to find $z$ such that $z=25+31y=4+23x$
Expressing as continued fraction, $$\frac{31}{23}=1+\frac{8}{23}=1+\frac{1}{\frac{23}{8}}=1+\frac{1}{2+\frac{7}{8}}$$
$$=1+\frac{1}{2+\frac{1}{\frac{8}{7}}}=1+\frac{1}{2+\frac{1}{1+\frac{1}{7}}}$$
So, the last but one convergent is $$1+\frac{1}{2+\frac{1}{1}}=\frac{4}{3}$$
So, $23\cdot 4- 31\cdot 3=-1$
$25+31y=4+23x\implies 23x=31y+21(31\cdot 3-23\cdot 4)$ $\implies 23(x+84)=31(y+63)$ $$\implies x+84=\frac{31(y+63)}{23}$$
So, $23\mid (y+63)$ as $x+84$ is integer and $(31,23)=1$ i.e., $23\mid (y+69-6)\implies 23\mid (y-6) \implies y=6+23w$
So, $z=25+31y=25+31(6+31w)=713w+211 \equiv 211 {\pmod {713}}$
-
Thank you for your answer! I'll ask a friend of mine to explain this to me, I'm fairly new to modulus calculation haha. – Gideon Potgieter Sep 14 '12 at 10:03
You can go fast if you incrementally square: $439^2=211 \pmod{713}$; $211^2=315 \pmod{713}$; $315^2=118 \pmod{713}$; $315^2=377 \pmod{713}$; $377^2=242 \pmod{713}$; $377^2=98 \pmod{713}$; etc.
The last one means $439^{64}=98 \pmod{713}$; In this way you can combine and reach 233 fast.
Ultimately $439^{233}=211 \pmod{713}$;
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Heres the algorithm,basically it is Modular exponentiation.
function modular_pow(base, exponent, modulus)
result := 1
while exponent > 0
if (exponent mod 2 == 1):
result := (result * base) mod modulus
exponent := exponent >> 1
base = (base * base) mod modulus
return result
Also here is the working code in c++ which can work for upto 10^4 test cases,in 0.7 seconds Also the ideone link http://ideone.com/eyLiOP
#include <iostream>
using namespace std;
#define mod 1000000007
long long int modfun(long long int a,long long int b)
{
long long int result = 1;
while (b > 0)
{
if (b & 1)
{
a=a%mod;
result = (result * a)%mod;
result=result%mod;
}
b=b>>1;
a=a%mod;
a = (a*a)%mod;
a=a%mod;
}
return result;
}
int main()
{
int t;
cin>>t;
while(t--)
{
long long int a,b;
cin>>a>>b;
if(a==0)
cout<<0<<"\n";
else if(b==0)
cout<<1<<"\n";
else if(b==1)
cout<<a%mod<<"\n";
else
{
cout<<(modfun(a,b))%mod<<"\n";
}
}
return 0;
}
`
-
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# 9 percent of seventy Dollars
### Percentage Calculator 2
is what percent of ? Answer: %
### Percentage Calculator 3
We think you reached us looking for answers to questions like: What is 9 percent of seventy? Or maybe: 9 percent of seventy Dollars
See the detailed solutions to these problems below.
## How to work out percentages explained step-by-step
Learn how to solve percentage problems through examples.
In all the following questions consider that:
• The percentage figure is represented by X%
• The whole amount is represented by W
• The portion amount or part is represented by P
### Solution for 'What is 9% of seventy?'
#### Solution Steps
The following question is of the type "How much X percent of W", where W is the whole amount and X is the percentage figure or rate".
Let's say that you need to find 9 percent of 70. What are the steps?
Step 1: first determine the value of the whole amount. We assume that the whole amount is 70.
Step 2: determine the percentage, which is 9.
Step 3: Convert the percentage 9% to its decimal form by dividing 9 into 100 to get the decimal number 0.09:
9100 = 0.09
Notice that dividing into 100 is the same as moving the decimal point two places to the left.
9.0 → 0.90 → 0.09
Step 4: Finally, find the portion by multiplying the decimal form, found in the previous step, by the whole amount:
0.09 x 70 = 6.3 (answer).
The steps above are expressed by the formula:
P = W × X%100
This formula says that:
"To find the portion or the part from the whole amount, multiply the whole by the percentage, then divide the result by 100".
The symbol % means the percentage expressed in a fraction or multiple of one hundred.
Replacing these values in the formula, we get:
P = 70 × 9100 = 70 × 0.09 = 6.3 (answer)
Therefore, the answer is 6.3 is 9 percent of 70.
### Solution for '9 is what percent of seventy?'
The following question is of the type "P is what percent of W,” where W is the whole amount and P is the portion amount".
The following problem is of the type "calculating the percentage from a whole knowing the part".
#### Solution Steps
As in the previous example, here are the step-by-step solution:
Step 1: first determine the value of the whole amount. We assume that it is 70.
(notice that this corresponds to 100%).
Step 2: Remember that we are looking for the percentage 'percentage'.
To solve this question, use the following formula:
X% = 100 × PW
This formula says that:
"To find the percentage from the whole, knowing the part, divide the part by the whole then multiply the result by 100".
This formula is the same as the previous one shown in a different way in order to have percent (%) at left.
Step 3: replacing the values into the formula, we get:
X% = 100 × 970
X% = 90070
So, the answer is 9 is 12.86 percent of 70
### Solution for '70 is 9 percent of what?'
The following problem is of the type "calculating the whole knowing the part and the percentage".
### Solution Steps:
Step 1: first determine the value of the part. We assume that the part is 70.
Step 2: identify the percent, which is 9.
Step 3: use the formula below:
W = 100 × PX%
This formula says that:
"To find the whole, divide the part by the percentage then multiply the result by 100".
This formula is the same as the above rearranged to show the whole at left.
Step 4: plug the values into the formula to get:
W = 100 × 709
W = 100 × 7.7777777777778
|
## Section2.11Variables, Expressions, and Equations Chapter Review
### Subsection2.11.1Variables and Evaluating Expressions
In Section 2.1 we covered the definitions of variables and expressions, and how to evaluate an expression with a particular number. We learned the formulas for perimeter and area of rectangles, triangles, and circles.
##### Evaluating Expressions
When we evaluate an expression's value, we substitute each variable with its given value.
###### Example2.11.1
Evaluate the value of $\frac{5}{9}(F - 32)$ if $F=212\text{.}$
\begin{align*} \frac{5}{9}(F - 32)\amp=\frac{5}{9}(212 - 32)\\ \amp=\frac{5}{9}(180)\\ \amp=100 \end{align*}
##### Substituting a Negative Number
When we substitute a variable with a negative number, it's important to use parentheses around the number.
###### Example2.11.2
Evaluate the following expressions if $x=-3\text{.}$
1. \begin{aligned}[t] x^2\amp=(-3)^2\\ \amp=9 \end{aligned}
2. \begin{aligned}[t] x^3\amp=(-3)^3\\ \amp=(-3)(-3)(-3)\\ \amp=-27 \end{aligned}
3. \begin{aligned}[t] -x^2\amp=-(-3)^2\\ \amp=-9 \end{aligned}
4. \begin{aligned}[t] -x^3\amp=-(-3)^3\\ \amp=-(-27)\\ \amp=27 \end{aligned}
### Subsection2.11.2Geometry Formulas
In Section 2.2 we established the following formulas.
Perimeter of a Rectangle
$P=2(\ell+w)$
Area of a Rectangle
$A=\ell w$
Area of a Triangle
$A=\frac{1}{2}bh$
Circumference of a Circle
$c=2\pi r$
Area of a Circle
$A=\pi r^2$
Volume of a Rectangular Prism
$V=wdh$
Volume of a Cylinder
$V=\pi r^2h$
Volume of a Rectangular Prism or Cylinder
$V=Bh$
### Subsection2.11.3Combining Like Terms
In Section 2.3 we covered the definitions of a term and how to combine like terms.
###### Example2.11.3
List the terms in the expression $5x-3y+\frac{2w}{3}\text{.}$
Explanation
The expression has three terms that are being added, $5x\text{,}$ $-3y$ and $\frac{2w}{3}\text{.}$
###### Example2.11.4
Simplify the expression $5x-3x^2+2x+5x^2\text{,}$ if possible, by combining like terms.
Explanation
This expression has four terms: $5x\text{,}$ $-3x^2\text{,}$ $2x\text{,}$ and $5x^2\text{.}$ Both $5x$ and $2x$ are like terms; also $-3x^2$ and $5x^2$ are like terms. When we combine like terms, we get:
\begin{equation*} 5x-3x^2+2x+5x^2=7x+2x^2 \end{equation*}
Note that we cannot combine $7x$ and $2x^2$ because $x$ and $x^2$ represent different quantities.
### Subsection2.11.4Equations and Inequalities as True/False Statements
In Section 2.4 we covered the definitions of an equation and an inequality, as well as how to verify if a particular number is a solution to them.
##### Checking Possible Solutions
Given an equation or an inequality (with one variable), checking if some particular number is a solution is just a matter of replacing the value of the variable with the specified number and determining if the resulting equation/inequality is true or false. This may involve some amount of arithmetic simplification.
###### Example2.11.5
Is $-5$ a solution to $2(x+3)-2=4-x\text{?}$
Explanation
To find out, substitute in $-5$ for $x$ and see what happens.
\begin{align*} 2(x+3)-2\amp=4-x\\ 2(\substitute{(-5)}+3)-2\amp\stackrel{?}{=}4-\substitute{(-5)}\\ 2(\highlight{-2})-2\amp\stackrel{?}{=}\highlight{9}\\ \highlight{-4}-2\amp\stackrel{?}{=}9\\ \highlight{-6}\amp\stackrel{\text{no}}{=}9 \end{align*}
So no, $-5$ is not a solution to $2(x+3)-2=4-x\text{.}$
### Subsection2.11.5Solving One-Step Equations
In Section 2.5 we covered to to add, subtract, multiply, or divide on both sides of an equation to isolate the variable, summarized in Fact 2.5.12. We also learned how to state our answer, either as a solution or a solution set. Last we discussed how to solve equations with fractions.
###### Example2.11.6
Solve for $g$ in $\frac{1}{2}=\frac{2}{3}+g\text{.}$
Explanation
We will subtract $\frac{2}{3}$ on both sides of the equation:
\begin{align*} \frac{1}{2}\amp=\frac{2}{3}+g\\ \frac{1}{2}\subtractright{\frac{2}{3}}\amp=\frac{2}{3}+g\subtractright{\frac{2}{3}}\\ \frac{3}{6}-\frac{4}{6}\amp=g\\ -\frac{1}{6}\amp=g \end{align*}
We will check the solution by substituting $g$ in the original equation with $-\frac{1}{6}\text{:}$
\begin{align*} \frac{1}{2}\amp=\frac{2}{3}+g\\ \frac{1}{2}\amp\stackrel{?}{=}\frac{2}{3}+\left(\substitute{-\frac{1}{6}}\right)\\ \frac{1}{2}\amp\stackrel{?}{=}\frac{4}{6}+\left(-\frac{1}{6}\right)\\ \frac{1}{2}\amp\stackrel{?}{=}\frac{3}{6}\\ \frac{1}{2}\amp\stackrel{\checkmark}{=}\frac{1}{2} \end{align*}
The solution $-\frac{1}{6}$ is checked and the solution set is $\left\{-\frac{1}{6}\right\}\text{.}$
### Subsection2.11.6Solving One-Step Inequalities
In Section 2.6 we covered how solving inequalities is very much like how we solve equations, except that if we multiply or divide by a negative we switch the inequality sign.
##### Solving One-Step Inequalities
When we solve linear inequalities, we also use Properties of Equivalent Equations with one small complication: When we multiply or divide by the same negative number on both sides of an inequality, the direction reverses!
###### Example2.11.7
Solve the inequality $-2x\geq12\text{.}$ State the solution set with both interval notation and set-builder notation.
Explanation
To solve this inequality, we will divide each side by $-2\text{:}$
\begin{align*} -2x\amp\geq12\\ \divideunder{-2x}{-2}\amp\highlight{{}\leq{}}\divideunder{12}{-2}\amp\amp\text{Note the change in direction.}\\ x\amp\leq-6 \end{align*}
• The inequality's solution set in interval notation is $(-\infty,-6]\text{.}$
• The inequality's solution set in set-builder notation is $\{x\mid x\leq-6\}\text{.}$
### Subsection2.11.7Percentages
In Section 2.7 we covered how to translate sentences with percentages into equations that we can solve.
##### Solving One-Step Equations Involving Percentages
An important skill for solving percent-related problems is to boil down a complicated word problem into a simple form like â$2$ is $50\%$ of $4\text{.}$â
###### Example2.11.8
What percent of $2346.19$ is $1995.98\text{?}$
Using $P$ to represent the unknown quantity, we write and solve the equation:
\begin{align*} \overbrace{\strut P}^{\text{what percent}}\overbrace{\strut \cdot}^{\text{of}} \overbrace{\strut 2346.19}^{\text{\$2346.19}}\amp\overbrace{\strut =}^{\text{is}}\overbrace{\strut 1995.98}^{\text{\$1995.98}}\\ \divideunder{P\cdot 2346.19}{2346.19}\amp=\divideunder{1995.98}{2346.19}\\ P\amp=0.85073\ldots\\ P\amp\approx85.07\% \end{align*}
In summary, $1995.98$ is approximately $85.07\%$ of $2346.19\text{.}$
### Subsection2.11.8Modeling with Equations and Inequalities
In Section 2.8 we covered how to translate phrases into mathematics, and how to set up equations and inequalities for application models.
##### Modeling with Equations and Inequalities
To set up an equation modeling a real world scenario, the first thing we need to do is to identify what variable we will use. The variable we use will be determined by whatever is unknown in our problem statement. Once we've identified and defined our variable, we'll use the numerical information provided in the equation to set up our equation.
###### Example2.11.9
A bathtub contains 2.5âŻft3 of water. More water is being poured in at a rate of 1.75âŻft3 per minute. When will the amount of water in the bathtub reach 6.25âŻft3?
Explanation
Since the question being asked in this problem starts with âwhen,â we immediately know that the unknown is time. As the volume of water in the tub is measured in ft3 per minute, we know that time needs to be measured in minutes. We'll defined $t$ to be the number of minutes that water is poured into the tub. Since each minute there are 1.75âŻft3 of water added, we will add the expression $1.75t$ to $2.5$ to obtain the total amount of water. Thus the equation we set up is:
\begin{equation*} 2.5+1.75t=6.25 \end{equation*}
### Subsection2.11.9Introduction to Exponent Rules
In Section 2.9 we covered the rules of exponents for multiplication.
###### Example2.11.10
Simplify the following expressions using the rules of exponents:
1. $-2t^3\cdot 4t^5$
2. $5\left(v^4\right)^2$
3. $-(3u)^2$
4. $(-3z)^2$
Explanation
1. $-2t^3\cdot 4t^5=-8t^8$
2. $5\left(v^4\right)^2=5v^8$
3. $-(3u)^2=-9u^2$
4. $(-3z)^2=9z^2$
### Subsection2.11.10Simplifying Expressions
In Section 2.10 we covered the definitions of the identities and inverses, and the various algebraic properties. We then learned about the order of operations.
###### Example2.11.11
Use the associative, commutative, and distributive properties to simplify the expression $5x+9(-2x+3)$ as much as possible.
Explanation
We will remove parentheses by the distributive property, and then combine like terms:
\begin{align*} 5x+9(-2x+3)\amp=5x+9(-2x+3)\\ \amp=5x+9(-2x)+9(3)\\ \amp=5x-18x+27\\ \amp=-13x+27 \end{align*}
### Subsection2.11.11Exercises
###### 1
A trapezoidâs area can be calculated by the formula $A=\frac{1}{2}(b_1+b_2)h\text{,}$ where $A$ stands for area, $b_1$ for the first baseâs length, $b_2$ for the second baseâs length, and $h$ for height.
Find the area of the trapezoid below.
###### 2
A trapezoidâs area can be calculated by the formula $A=\frac{1}{2}(b_1+b_2)h\text{,}$ where $A$ stands for area, $b_1$ for the first baseâs length, $b_2$ for the second baseâs length, and $h$ for height.
Find the area of the trapezoid below.
###### 3
To convert a temperature measured in degrees Fahrenheit to degrees Celsius, there is a formula:
\begin{equation*} C={\frac{5}{9}\!\left(F-32\right)} \end{equation*}
where $C$ represents the temperature in degrees Celsius and $F$ represents the temperature in degrees Fahrenheit.
If a temperature is $122 {^\circ}\text{F}\text{,}$ what is that temperature measured in Celsius?
###### 4
To convert a temperature measured in degrees Fahrenheit to degrees Celsius, there is a formula:
\begin{equation*} C={\frac{5}{9}\!\left(F-32\right)} \end{equation*}
where $C$ represents the temperature in degrees Celsius and $F$ represents the temperature in degrees Fahrenheit.
If a temperature is $14 {^\circ}\text{F}\text{,}$ what is that temperature measured in Celsius?
###### 5
Evaluate the expression ${x^{2}}\text{:}$
1. When $x=6\text{,}$ $\displaystyle{{x^{2}}=}$
2. When $x=-6\text{,}$ $\displaystyle{{x^{2}}=}$
###### 6
Evaluate the expression ${y^{2}}\text{:}$
1. When $y=3\text{,}$ $\displaystyle{{y^{2}}=}$
2. When $y=-9\text{,}$ $\displaystyle{{y^{2}}=}$
###### 7
Evaluate the expression ${y^{3}}\text{:}$
1. When $y=5\text{,}$ $\displaystyle{{y^{3}}=}$
2. When $y=-3\text{,}$ $\displaystyle{{y^{3}}=}$
###### 8
Evaluate the expression ${r^{3}}\text{:}$
1. When $r=3\text{,}$ $\displaystyle{{r^{3}}=}$
2. When $r=-5\text{,}$ $\displaystyle{{r^{3}}=}$
###### 9
List the terms in each expression.
1. ${4t+2z+6}$
2. ${7z^{2}}$
3. ${9t+y}$
4. ${2t+7t}$
###### 10
List the terms in each expression.
1. ${8t^{2}+6+2x^{2}-t^{2}}$
2. ${5x^{2}-6y^{2}+7y}$
3. ${-5z^{2}+3t^{2}-7y^{2}}$
4. ${-2y^{2}+4-3s+2t}$
###### 11
Simplify each expression, if possible, by combining like terms.
1. ${8t-t+3t+9t}$
2. ${-8z^{2}+5z^{2}+6z^{2}}$
3. ${3z-3z}$
4. ${-3x^{2}-2-7x}$
###### 12
Simplify each expression, if possible, by combining like terms.
1. ${5t^{2}-8y^{2}+4+2t}$
2. ${7t-2t^{2}}$
3. ${-7z^{2}-2z^{2}+8x^{2}}$
4. ${-6s+x}$
###### 13
Simplify each expression, if possible, by combining like terms.
1. ${{\frac{4}{3}}t+{\frac{4}{9}}}$
2. ${-{\frac{1}{7}}y^{2}+y^{2} - {\frac{4}{3}}y^{2}+{\frac{2}{7}}s^{2}}$
3. ${-{\frac{3}{8}}y+{\frac{3}{2}}z^{2}-3z^{2}+2x}$
4. ${-{\frac{2}{3}}t-s}$
###### 14
Simplify each expression, if possible, by combining like terms.
1. ${-{\frac{1}{6}}t - {\frac{9}{4}}t}$
2. ${-{\frac{3}{8}}s-z - {\frac{2}{5}}s}$
3. ${{\frac{1}{3}}y^{2} - {\frac{9}{5}}y^{2}}$
4. ${-{\frac{2}{9}}y - {\frac{1}{3}}y^{2}+{\frac{1}{4}}y^{2}+9y}$
###### 15
Is $-2$ a solution for $x$ in the equation ${2x+2}={2-\left(5+x\right)}\text{?}$ Evaluating the left and right sides gives:
${2x+2}$ $=$ ${2-\left(5+x\right)}$ ${}\stackrel{?}{=}{}$
So $-2$
• is
• is not
a solution to ${2x+2}={2-\left(5+x\right)}\text{.}$
###### 16
Is $-1$ a solution for $x$ in the equation ${4x-4}={-3-\left(4+x\right)}\text{?}$ Evaluating the left and right sides gives:
${4x-4}$ $=$ ${-3-\left(4+x\right)}$ ${}\stackrel{?}{=}{}$
So $-1$
• is
• is not
a solution to ${4x-4}={-3-\left(4+x\right)}\text{.}$
###### 17
Is $1$ a solution for $x$ in the inequality ${-4x^{2}+5x}\le{2x-7}\text{?}$ Evaluating the left and right sides gives:
${-4x^{2}+5x}$ $\le$ ${2x-7}$ ${}\stackrel{?}{\le}{}$
So $1$
• is
• is not
a solution to ${-4x^{2}+5x}\le{2x-7}\text{.}$
###### 18
Is $2$ a solution for $x$ in the inequality ${-2x^{2}+5x}\le{2x-12}\text{?}$ Evaluating the left and right sides gives:
${-2x^{2}+5x}$ $\le$ ${2x-12}$ ${}\stackrel{?}{\le}{}$
So $2$
• is
• is not
a solution to ${-2x^{2}+5x}\le{2x-12}\text{.}$
###### 19
Solve the equation.
$\displaystyle{ {t+7}={2} }$
###### 20
Solve the equation.
$\displaystyle{ {t+4}={1} }$
###### 21
Solve the equation.
$\displaystyle{ {-10}={t-6} }$
###### 22
Solve the equation.
$\displaystyle{ {-9}={x-7} }$
###### 23
Solve the equation.
$\displaystyle{ {96}={-8x} }$
###### 24
Solve the equation.
$\displaystyle{ {24}={-3y} }$
###### 25
Solve the equation.
$\displaystyle{ {{\frac{5}{13}}c}={25} }$
###### 26
Solve the equation.
$\displaystyle{ {{\frac{4}{7}}A}={12} }$
###### 27
The pie chart represents a collectorâs collection of signatures from various artists.
If the collector has a total of $1450$ signatures, there are signatures by Sting.
###### 28
The pie chart represents a collectorâs collection of signatures from various artists.
If the collector has a total of $1650$ signatures, there are signatures by Sting.
###### 29
A community college conducted a survey about the number of students riding each bus line available. The following bar graph is the result of the survey.
What percent of students ride Bus #1?
Approximately of students ride Bus #1.
###### 30
A community college conducted a survey about the number of students riding each bus line available. The following bar graph is the result of the survey.
What percent of students ride Bus #1?
Approximately of students ride Bus #1.
###### 31
The following is a nutrition fact label from a certain macaroni and cheese box.
The highlighted row means each serving of macaroni and cheese in this box contains ${7\ {\rm g}}$ of fat, which is $14\%$ of an average personâs daily intake of fat. Whatâs the recommended daily intake of fat for an average person?
The recommended daily intake of fat for an average person is .
###### 32
The following is a nutrition fact label from a certain macaroni and cheese box.
The highlighted row means each serving of macaroni and cheese in this box contains ${5.5\ {\rm g}}$ of fat, which is $10\%$ of an average personâs daily intake of fat. Whatâs the recommended daily intake of fat for an average person?
The recommended daily intake of fat for an average person is .
###### 33
Jerry used to make $13$ dollars per hour. After he earned his Bachelorâs degree, his pay rate increased to $48$ dollars per hour. What is the percentage increase in Jerryâs salary?
The percentage increase is .
###### 34
Eileen used to make $14$ dollars per hour. After she earned her Bachelorâs degree, her pay rate increased to $49$ dollars per hour. What is the percentage increase in Eileenâs salary?
The percentage increase is .
###### 35
After a $10\%$ increase, a town has $550$ people. What was the population before the increase?
Before the increase, the townâs population was .
###### 36
After a $70\%$ increase, a town has $1020$ people. What was the population before the increase?
Before the increase, the townâs population was .
###### 37
A bicycle for sale costs ${\254.88}\text{,}$ which includes $6.2\%$ sales tax. What was the cost before sales tax?
Assume the bicycleâs price before sales tax is $p$ dollars. Write an equation to model this scenario. There is no need to solve it.
###### 38
A bicycle for sale costs ${\283.77}\text{,}$ which includes $5.1\%$ sales tax. What was the cost before sales tax?
Assume the bicycleâs price before sales tax is $p$ dollars. Write an equation to model this scenario. There is no need to solve it.
###### 39
The property taxes on a $2100$-square-foot house are ${\4{,}179.00}$ per year. Assuming these taxes are proportional, what are the property taxes on a $1700$-square-foot house?
Assume property taxes on a $1700$-square-foot house is $t$ dollars. Write an equation to model this scenario. There is no need to solve it.
###### 40
The property taxes on a $1600$-square-foot house are ${\1{,}600.00}$ per year. Assuming these taxes are proportional, what are the property taxes on a $2000$-square-foot house?
Assume property taxes on a $2000$-square-foot house is $t$ dollars. Write an equation to model this scenario. There is no need to solve it.
###### 41
A swimming pool is being filled with water from a garden hose at a rate of $5$ gallons per minute. If the pool already contains $30$ gallons of water and can hold $135$ gallons, after how long will the pool overflow?
Assume $m$ minutes later, the pool would overflow. Write an equation to model this scenario. There is no need to solve it.
###### 42
A swimming pool is being filled with water from a garden hose at a rate of $8$ gallons per minute. If the pool already contains $40$ gallons of water and can hold $280$ gallons, after how long will the pool overflow?
Assume $m$ minutes later, the pool would overflow. Write an equation to model this scenario. There is no need to solve it.
###### 43
Use the commutative property of addition to write an equivalent expression to ${5b+31}\text{.}$
###### 44
Use the commutative property of addition to write an equivalent expression to ${6q+79}\text{.}$
###### 45
Use the associative property of multiplication to write an equivalent expression to ${3\!\left(4r\right)}\text{.}$
###### 46
Use the associative property of multiplication to write an equivalent expression to ${4\!\left(7m\right)}\text{.}$
###### 47
Use the distributive property to write an equivalent expression to ${10\!\left(p+2\right)}$ that has no grouping symbols.
###### 48
Use the distributive property to write an equivalent expression to ${7\!\left(q+6\right)}$ that has no grouping symbols.
###### 49
Use the distributive property to simplify ${4+9\!\left(2+4y\right)}$ completely.
###### 50
Use the distributive property to simplify ${9+4\!\left(9+3r\right)}$ completely.
###### 51
Use the distributive property to simplify ${6-4\!\left(1-6a\right)}$ completely.
###### 52
Use the distributive property to simplify ${3-9\!\left(-5-6b\right)}$ completely.
###### 53
Use the properties of exponents to simplify the expression.
${r^{12}}\cdot{r^{17}}$
###### 54
Use the properties of exponents to simplify the expression.
${t^{14}}\cdot{t^{11}}$
###### 55
Use the properties of exponents to simplify the expression.
$\displaystyle{\left(y^{10}\right)^{3}}$
###### 56
Use the properties of exponents to simplify the expression.
$\displaystyle{\left(t^{11}\right)^{10}}$
###### 57
Use the properties of exponents to simplify the expression.
$\displaystyle{\left(3x\right)^4}$
###### 58
Use the properties of exponents to simplify the expression.
$\displaystyle{\left(2r\right)^2}$
###### 59
Use the properties of exponents to simplify the expression.
$\displaystyle{({-2t^{5}})\cdot({4t^{16}})}$
###### 60
Use the properties of exponents to simplify the expression.
$\displaystyle{({-5y^{7}})\cdot({3y^{9}})}$
###### 61
Use the properties of exponents to simplify the expression.
1. $\displaystyle{{\left(-2b^{3}\right)^{6}}=}$
2. $\displaystyle{{-\left(2b^{3}\right)^{6}}=}$
###### 62
Use the properties of exponents to simplify the expression.
1. $\displaystyle{{\left(-2a^{3}\right)^{2}}=}$
2. $\displaystyle{{-\left(2a^{3}\right)^{2}}=}$
###### 63
Simplify the following expression.
${\left(3r^{3}\right)^{4}\!\left(r^{2}\right)^{2}}$=
###### 64
Simplify the following expression.
${\left(5t^{2}\right)^{2}\!\left(t^{5}\right)^{2}}$=
###### 65
Simplify the following expressions if possible.
1. $\displaystyle{ {p^{2}+2p^{2}}=}$
2. $\displaystyle{ (p^{2})(2p^{2})=}$
3. $\displaystyle{ {p^{2}-4p^{3}}=}$
4. $\displaystyle{ (p^{2})(-4p^{3})=}$
###### 66
Simplify the following expressions if possible.
1. $\displaystyle{ {-2q+4q}=}$
2. $\displaystyle{ (-2q)(4q)=}$
3. $\displaystyle{ {-2q-q^{4}}=}$
4. $\displaystyle{ (-2q)(-q^{4})=}$
###### 67
Multiply the polynomials.
${-10x^{2}}\left({3x^{2}+5x}\right)=$
###### 68
Multiply the polynomials.
${7x^{2}}\left({9x^{2}+10x}\right)=$
|
# Derivative of Cosecant
We shall prove the formula for the derivative of the cosecant function by using definition or the first principle method.
Let us suppose that the function is of the form $y = f\left( x \right) = \csc x$
First we take the increment or small change in the function:
$\begin{gathered} y + \Delta y = \csc \left( {x + \Delta x} \right) \\ \Delta y = \csc \left( {x + \Delta x} \right) – y \\ \end{gathered}$
Putting the value of function $y = \csc x$ in the above equation, we get
$\begin{gathered} \Delta y = \csc \left( {x + \Delta x} \right) – \csc x \\ \Delta y = \frac{1}{{\sin \left( {x + \Delta x} \right)}} – \frac{1}{{\sin x}} \\ \end{gathered}$
Taking the LCM (Least Common Factor), we get
$\begin{gathered}\Delta y = \frac{{\sin x – \sin \left( {x + \Delta x} \right)}}{{\sin \left( {x + \Delta x} \right)\sin x}} \\ \Delta y = – \frac{{\sin \left( {x + \Delta x} \right) – \sin x}}{{\sin \left( {x + \Delta x} \right)\sin x}}\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ \end{gathered}$
Using the formula from trigonometry, we have
$\sin A – \sin B = 2\cos \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A – B}}{2}} \right)$
Using this formula in equation (i), we get
$\begin{gathered}\Delta y = – \frac{{\left[ {2\cos \left\{ {\frac{{\left( {x + \Delta x} \right) + x}}{2}} \right\}\sin \left\{ {\frac{{\left( {x + \Delta x} \right) – x}}{2}} \right\}} \right]}}{{\sin \left( {x + \Delta x} \right)\sin x}} \\ \Delta y = – \frac{{2\cos \left\{ {\frac{{2x + \Delta x}}{2}} \right\}\sin \left\{ {\frac{{\Delta x}}{2}} \right\}}}{{\sin \left( {x + \Delta x} \right)\sin x}} \\ \end{gathered}$
Dividing both sides by $\Delta x$, we get
$\begin{gathered}\Rightarrow \frac{{\Delta y}}{{\Delta x}} = – \frac{{2\cos \left\{ {\frac{{2x + \Delta x}}{2}} \right\}\sin \left\{ {\frac{{\Delta x}}{2}} \right\}}}{{\Delta x\sin \left( {x + \Delta x} \right)\sin x}} \\ \Rightarrow \frac{{\Delta y}}{{\Delta x}} = – \frac{{\cos \left\{ {\frac{{2x + \Delta x}}{2}} \right\}}}{{\sin \left( {x + \Delta x} \right)\sin x}} \times \frac{{\sin \left\{ {\frac{{\Delta x}}{2}} \right\}}}{{\frac{{\Delta x}}{2}}} \\ \end{gathered}$
Taking the limit of both sides as $\Delta x \to 0$, we have
$\begin{gathered}\Rightarrow \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = – \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\cos \left\{ {\frac{{2x + \Delta x}}{2}} \right\}}}{{\sin \left( {x + \Delta x} \right)\sin x}} \times \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sin \left\{ {\frac{{\Delta x}}{2}} \right\}}}{{\frac{{\Delta x}}{2}}} \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{{\cos \left\{ {\frac{{2x + 0}}{2}} \right\}}}{{\sin \left( {x + 0} \right)\sin x}} \times \left( 1 \right) \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{{\cos x}}{{{{\sin }^2}x}} \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{1}{{\sin }} \cdot \frac{{\cos x}}{{\sin x}} \\ \Rightarrow \frac{{dy}}{{dx}} = – \csc x\cot x \\ \Rightarrow \frac{d}{{dx}}\csc x = – \csc x\cot x \\ \end{gathered}$
Example: Find the derivative of $y = f\left( x \right) = \csc 5x$
We have the given function as
$y = \csc 5x$
Differentiating with respect to variable $x$, we get
$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\csc 5x$
Using the rule, $\frac{d}{{dx}}\csc x = – \csc x\cot x$, we get
$\begin{gathered}\frac{{dy}}{{dx}} = – \csc 5x\cot 5x\frac{d}{{dx}}5x \\ \frac{{dy}}{{dx}} = – \csc 5x\cot 5x\left( 5 \right) \\ \frac{{dy}}{{dx}} = – 5\csc 5x\cot 5x \\ \end{gathered}$
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# A line segment has endpoints at (9 ,2 ) and (5 , 4). The line segment is dilated by a factor of 3 around (2 , 3). What are the new endpoints and length of the line segment?
Mar 22, 2018
(23, 0) and (11, 6), $6 \sqrt{5}$
#### Explanation:
Dilating around a point is hard unless it's the origin. So we can shift to the origin and then shift back.
So we subtract (2, 3) from (9,2) and (5, 4) and get (7, -1) and (3, 1). We now dilate them by 3, yielding (21, -3) and (9, 3). We finally shift them back by adding (2, 3) and get (23, 0) and (11, 6).
From that, we have the length of the line segment is
$L = \sqrt{\Delta {x}^{2} + \Delta {y}^{2}} = \sqrt{{\left(11 - 23\right)}^{2} + {\left(6 - 0\right)}^{2}}$
$L = \sqrt{144 + 36} = 6 \sqrt{5}$
For the second step, we could also have gotten the original length and multiplied by the dilation factor, i.e.
${L}_{0} = \sqrt{{\left(5 - 9\right)}^{2} + {\left(4 - 2\right)}^{2}} = \sqrt{16 + 4} = 2 \sqrt{5}$
$L = 3 \cdot {L}_{0} = 6 \sqrt{5}$
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# Factors of 2862 in Pairs
So you need to find the factor pairs for 2862 do you? Fear not! In this quick guide we'll describe what the factor pairs of 2862 are, how you find them and list them out for you to prove the calculation works. Let's dive in!
## What Are Factor Pairs?
A factor pair is a combination of two factors which can be multiplied together to equal 2862. In proper math terms, the number 2862 is called the product and the two numbers that can be multiplied together to equal it are called the factors.
Want to quickly learn or refresh memory on factor pairs, play this quick and informative video now!
For more educational and explainer videos on math and numbers including fractions, percentage calculation, conversions and much more visit visualfractions.com Youtube channel.
In order to work out the factor pairs of 2862 we need to first get all of the factors of 2862. Once you have the list of all those factors we can pair them together to list out all of the factor pairs.
The complete list of factors for 2862 are 1, 2, 3, 18, 27, 53, 54, 106, 159, 318, 477, 954, 1431, and 2862.
## List of Factor Pairs for 2862
Okay, so we know all of the factors for 2862 now and to work out the factor pairs we can go through that list and find all of the different combinations that can be used to multiply together to result in 2862.
If there are a lot of factors then it might take you a little while to calculate all of the factor pairs, but luckily we have the power of computers and can calculate the factor pairs of 2862 for you automatically:
• 1 x 2862 = 2862
• 2 x 1431 = 2862
• 3 x 954 = 2862
• 18 x 159 = 2862
• 27 x 106 = 2862
• 53 x 54 = 2862
• 54 x 53 = 2862
• 106 x 27 = 2862
• 159 x 18 = 2862
• 318 x 9 = 2862
• 477 x 6 = 2862
• 954 x 3 = 2862
• 1431 x 2 = 2862
• 2862 x 1 = 2862
So there you have it. A complete guide to the factor pairs of 2862. Hopefully this will help you in your math class to learn (or teach) factor pairs in a way that is easy to understand.
Feel free to try the calculator below to check another number or, if you're feeling fancy, grab a pencil and paper and try and do it by hand. Just make sure to pick small numbers! ;)
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "Factors of 2862 in Pairs". VisualFractions.com. Accessed on February 28, 2024. http://visualfractions.com/calculator/factor-pairs/factors-of-2862-in-pairs/.
• "Factors of 2862 in Pairs". VisualFractions.com, http://visualfractions.com/calculator/factor-pairs/factors-of-2862-in-pairs/. Accessed 28 February, 2024.
## Factor Pair Calculator
Want to find the factor pairs for another number? Enter your number below and click calculate.
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1. ## The Parabola
Find (a) the directrix, (b) the focus, (c) the roots of the parabola y = x^2 -5x + 4.
I have never seen a parabola written that way.
2. Originally Posted by magentarita
Find (a) the directrix, (b) the focus, (c) the roots of the parabola y = x^2 -5x + 4.
I have never seen a parabola written that way.
You're probably used to seeing the quadratic written in vertex form like:
$\displaystyle y=a(x-h)^2+k$
$\displaystyle y = x^2 -5x + 4$
Now, just complete the square:
$\displaystyle y=\left(x^2-5x+\frac{25}{4}\right)-\frac{25}{4}+4$
$\displaystyle y=\left(x-\frac{5}{2}\right)^2-\frac{9}{4}$
Vertex $\displaystyle (h, k) = \left(\frac{5}{2}, -\frac{9}{4}\right)$
Directrix:
$\displaystyle y=k-\frac{1}{4a}$
Focus:
$\displaystyle \left(h, k+\frac{1}{4a}\right)$
Roots (zeros):
$\displaystyle x^2-5x+4=0$
Factor and solve for x.
3. ## ok.........
Originally Posted by masters
You're probably used to seeing the quadratic written in vertex form like:
$\displaystyle y=a(x-h)^2+k$
$\displaystyle y = x^2 -5x + 4$
Now, just complete the square:
$\displaystyle y=\left(x^2-5x+\frac{25}{4}\right)-\frac{25}{4}+4$
$\displaystyle y=\left(x-\frac{5}{2}\right)^2-\frac{9}{4}$
Vertex $\displaystyle (h, k) = \left(\frac{5}{2}, -\frac{9}{4}\right)$
Directrix:
$\displaystyle y=k-\frac{1}{4a}$
Focus:
$\displaystyle \left(h, k+\frac{1}{4a}\right)$
Roots (zeros):
$\displaystyle x^2-5x+4=0$
Factor and solve for x.
This is exactly what confused me. I thank you for making life easy here for me.
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Introduction to Isosceles triangle definition
Going by geometry, the isosceles triangle definition states that any triangle with two sides and each of these sides are of equal length. In some cases, it is described as a triangle with two and just two sides of equal length
Isosceles triangle definition also identifies such as a triangle with at least two sides of equal length, this last definition as well as the equilateral triangle as a special case. By the isosceles triangle theorem, the two angles opposite the equal sides are as well equal, but if the third side is different then the third angle will as well not be the same
By the Steiner–Lehmus theorem, every triangle with two angle bisectors of equal length fall under the isosceles triangle definition
Depending on how you draw yours according to the isosceles triangle definition, it has two legs i.e. from the isosceles triangle definition we can see that the two long straight lines opposite each other and a base - the not too short but not long line which connects its legs together
An obtuse isosceles triangle definition with one angle that is obviously more than 90°
Solving an isosceles triangle definition with examples
Both the legs (the two opposite straight lines in the triangle) or the base of an isosceles triangle definition just like in any case where we are to find an unknown, can be determined if we know the other values. As we are already familiar with, the base of this isosceles triangle definition forms the height of the triangle with a perpendicular bisector right from the base
Now, this perpendicular line drawn from the base to reach its vertex in turn form another 2 similar right isosceles triangles definition. These two right angle triangles can be solved for using the Pythagoras’ theorem. It has been explained in details below:
• Finding the base of an isosceles triangle definition: in order to determine the base once we know its leg, height, we use the formula: Base = 2 √L2−A2, where: L = length of a leg of the isosceles triangle definition while A is the height or altitude
• Finding the leg isosceles triangle definition: to find the length of the length if we have its base and height, we use the formula: Leg = √A2 + (B2)2, where: letter B is the length of the triangle base and A is its altitude
• Altitude isosceles triangle definition: to find the height given the base and leg, use the formula: Altitude =√ L2−(B2)2, where: letter L is the length of a leg of the isosceles triangle definition while its base is letter B
Example 1: A right angled isosceles triangle definition with length 5 cm has a height of 13 cm. Find the area of this triangle:
• Let L be the length = 5 cm
• While H be its height = 13 cm
• The Formula to use is: Area A of the triangle = 12 (l × h) square unit i.e.
= 12 (5 x 13)
= 12 (65)
= 652
= 32.5. Therefore, the area of our triangle is: 32.5 cm2
Example 2: A right isosceles triangle definition has its length to be 7.5 m and height of 10 m. Find the area of that triangle:
• Let L be the length = 7.5 cm
• While H be its height = 10 cm
• The Formula to use is: Area A of the triangle = 12 (l × h) square unit i.e.
= 12 (7.5 x 10)
= 12 (75)
= 752
= 37.5. Therefore, the area of our triangle is: 37.5 cm2
Example 3: A right angled isosceles triangle definition has area 6.25 cm2 and length 2.5 cm. Find the height of the triangle:
• Let A be the area = 6.25 cm
• While L be its length = 2.5 cm
• The Formula to use is: Area A of the triangle = 12 (l × h) square unit i.e.
6.25 = 12 (2.5 x h)
6.25 = 1.25h
h= 6.251.25
= 5cm. Therefore, Height of our triangle is: 5 cm
Example 4: Solve the area of the obtuse isosceles triangle definition for the given base is 8 inches and height is 10 inches:
• Base = 8 inches
• Height = 10 inches
• Area A of obtuse isosceles triangle definition = 12 (Base × Height)
• Area = 12 x 8 x 10 = 40. Therefore, the area of an obtuse isosceles triangle definition is 40 square inches
Example 5: find the area of obtuse isosceles triangle definition for the given base is 10 meters and height is 6 meters?:
• Base = 10 meters
• Height = 6 meters
• Area A of obtuse triangle = 12 (Base × Height)
• Area = 12 × 10 × 6 = 30. Therefore, the area of an obtuse triangle is 30m2
Example 6: Solve the area of an obtuse triangle if the given base is 15 centimeters and height is 7.5 centimeters?:
• Base = 15 centimeter
• Height = 7.5 centimeter
• Area A of obtuse triangle = 12 (Base × Height)
• Area = 12 × 15 × 7.5 = 56.25 Therefore, the area of an obtuse triangle is 56.25 cm2
Example 7: If the two interior angles of the acute triangle are 750 and 500 respectively, solve for the third interior angle of the triangle:
• Both internal angles are 750 and 500
• Sum of the internal angles in acute triangle is 1800
• Imagine the third angle p0 since all triangles regardless of their type must have three sides or three angles
• Therefore, we have: 750 + 500 + our 3rd angle p0 = 1800
• 1250 + p0 = 1800
• p0 = 1800 – 1250
• p0 = 550. Our 3rd inner angle of the acute triangle is: 550
Example 8: Solve for the outer angle of the acute angled triangle when the two inner opposite angles are 500 and 800 respectively:
• Since both the outer and inner angles are added up to form the angles within a triangle, therefore, we have: 500 + 800 = 1300. Finally, the external angle of this acute triangle will be 1300
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# How do you solve the inequality -a/7+1/7>1/14?
Jan 19, 2017
$\textcolor{g r e e n}{a < \frac{1}{2}}$
#### Explanation:
A fraction consists of $\left(\text{count")/("size indicator") ->("numerator")/("denominator}\right)$
You can not directly compare the counts (numerators) unless the size indicators (denominators) as the same.
$\textcolor{g r e e n}{\left[- \frac{a}{7} \textcolor{red}{\times 1}\right] + \left[\frac{1}{7} \textcolor{red}{\times 1}\right] > \frac{1}{14}}$
$\textcolor{g r e e n}{\left[- \frac{a}{7} \textcolor{red}{\times \frac{2}{2}}\right] + \left[\frac{1}{7} \textcolor{red}{\times \frac{2}{2}}\right] > \frac{1}{14}}$
color(green)(" "-(2a)/14" "+" "2/14" ">1/14
Now that the denominators are all the same the inequality is still true if we compare only the counts (numerators).
$\text{ } \textcolor{g r e e n}{- 2 a + 2 > 1}$
Divide both sides by 2
$\text{ } \textcolor{g r e e n}{- a + 1 > \frac{1}{2}}$
Subtract 1 from both sides
$\text{ } \textcolor{g r e e n}{- a > \frac{1}{2} - 1}$
$\text{ } \textcolor{g r e e n}{- a > - \frac{1}{2}}$
Multiply both sides by (-1) and turn the inequality round the other way. You always turn the inequality if multiply by any negative value.
$\text{ } \textcolor{g r e e n}{a < \frac{1}{2}}$
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# Precalculus : Write the Equation of a Polynomial Function Based on Its Graph
## Example Questions
### Example Question #1 : Write The Equation Of A Polynomial Function Based On Its Graph
Which could be the equation for this graph?
Explanation:
This graph has zeros at 3, -2, and -4.5. This means that , , and . That last root is easier to work with if we consider it as and simplify it to . Also, this is a negative polynomial, because it is decreasing, increasing, decreasing and not the other way around.
Our equation results from multiplying , which results in .
### Example Question #2 : Write The Equation Of A Polynomial Function Based On Its Graph
Write the quadratic function for the graph:
Explanation:
Because there are no x-intercepts, use the form , where vertex is , so , , which gives
### Example Question #3 : Write The Equation Of A Polynomial Function Based On Its Graph
Write the quadratic function for the graph:
Explanation:
Method 1:
The x-intercepts are . These values would be obtained if the original quadratic were factored, or reverse-FOILed and the factors were set equal to zero.
For , . For , . These equations determine the resulting factors and the resulting function; .
Multiplying the factors and simplifying,
.
Method 2:
Use the form , where is the vertex.
is , so , .
### Example Question #4 : Write The Equation Of A Polynomial Function Based On Its Graph
Write the equation for the polynomial in this graph:
Explanation:
The zeros for this polynomial are .
This means that the factors are equal to zero when these values are plugged in for x.
multiply both sides by 2
so one factor is
multiply both sides by 3
so one factor is
so one factor is
Multiply these three factors:
### Example Question #5 : Write The Equation Of A Polynomial Function Based On Its Graph
Write the equation for the polynomial shown in this graph:
Explanation:
The zeros of this polynomial are . This means that the factors equal zero when these values are plugged in.
One factor is
One factor is
The third factor is equivalent to . Set equal to 0 and multiply by 2:
Multiply these three factors:
The graph is negative since it goes down then up then down, so we have to switch all of the signs:
### Example Question #6 : Write The Equation Of A Polynomial Function Based On Its Graph
Write the equation for the polynomial in the graph:
Explanation:
The zeros of the polynomial are . That means that the factors equal zero when these values are plugged in.
The first factor is or equivalently multiply both sides by 5:
The second and third factors are and
Multiply:
Because the graph goes down-up-down instead of the standard up-down-up, the graph is negative, so change all of the signs:
### Example Question #7 : Write The Equation Of A Polynomial Function Based On Its Graph
Write the equation for the polynomial in this graph:
Explanation:
The zeros for this polynomial are . That means that the factors are equal to zero when these values are plugged in.
or equivalently multiply both sides by 4
the first factor is
multiply both sides by 3
the second factor is
the third factor is
Multiply the three factors:
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# What is trinomial product?
Answered by Willian Lymon
Trinomial product refers to the result obtained when two binomials are multiplied together. In mathematics, a binomial is a polynomial with two terms, and a trinomial is a polynomial with three terms. When we multiply two binomials, we use a method called FOIL, which stands for First, Outer, Inner, Last.
Let’s take an example to illustrate this. Consider the binomials (x + 2) and (x + 5). To find their product, we start by multiplying the first terms of each binomial, which gives us x * x = x^2. Next, we multiply the outer terms, which are x * 5 = 5x. Then, we multiply the inner terms, which are 2 * x = 2x. we multiply the last terms, which are 2 * 5 = 10.
To obtain the trinomial product, we combine these four terms: x^2 + 5x + 2x + 10. Simplifying this expression gives us x^2 + 7x + 10. Hence, the trinomial product of (x + 2) and (x + 5) is x^2 + 7x + 10.
The FOIL method can be used to find the product of any two binomials. It is a systematic way of ensuring that all possible combinations of terms are multiplied correctly. By following this method, we can avoid any errors and obtain the correct trinomial product.
It’s important to note that not all trinomials can be factored into the product of two binomials. However, trinomials in the form x^2 + bx + c can often be factored in this way. The FOIL method is a useful tool for finding the product of binomials and can be applied to various mathematical problems.
In my personal experience, I have found the FOIL method to be extremely helpful in expanding and simplifying expressions involving binomials. It provides a systematic approach to multiplication and ensures that all terms are considered. This method has been particularly useful when solving quadratic equations and working with polynomial functions.
To summarize, a trinomial product is obtained by multiplying two binomials using the FOIL method. This method allows us to carefully multiply each term in one binomial with every term in the other binomial, resulting in a trinomial expression. While not all trinomials can be factored in this way, the FOIL method is a valuable tool for expanding and simplifying expressions involving binomials.
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# What numbers do 9 and 36 have in common?
## What numbers do 9 and 36 have in common?
The GCF of 9 and 36 is 9.
## What are the factors 9?
Factors of 9 are 1, 3, and 9. 1 is a universal factor. Factors of a number are the numbers we multiply to get that particular number.
What are factors of 36?
Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.
### What are the factors of 9 and 35?
Factors for 9: 1, 3, and 9. Factors for 35: 1, 5, 7, and 35.
### What is the GCF of 15 and 36?
3
Answer: GCF of 15 and 36 is 3.
What are the first five multiples of 36?
The first five multiples of 36 are 36, 72, 108, 144, and 180.
#### What is 9 a multiple of?
How to list multiples of a number?
Multiples of 1 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Multiples of 9 9, 18, 27, 36, 45, 54, 63, 72, 81, 90.
Multiples of 10 10, 20, 30, 40, 50, 60, 70, 80, 90, 100.
Multiples of 11 11, 22, 33, 44, 55, 66, 77, 88, 99, 110.
Multiples of 12 12, 24, 36, 48, 60, 72, 84, 96, 108, 120.
#### What is the GCF of 9?
To calculate the greatest common factor (GCF) of 9 and 18, we need to factor each number (factors of 9 = 1, 3, 9; factors of 18 = 1, 2, 3, 6, 9, 18) and choose the greatest factor that exactly divides both 9 and 18, i.e., 9.
What are the multiples of 36?
The multiples of 36 are 36, 72, 108, 144, 180, 216, 252, 288, 324, 360, 396, 432, 468, 504, The multiples of 24 are 24, 48, 72, 96,120, 144, 168, 192, 216, 240, 264, 288, 312, 336, 360 and so on. The common multiples of 36 and 24 are 144, 216, 288, 360, and so on.
## What two factors of 36 add up to?
1 x 36 = 36. 2 x 18 = 36. 3 x 12 = 36.
## What is the LCM of 9 and 35?
The LCM of 9 and 35 is 315.
What is the GCF of 120 and 36?
12
Answer: GCF of 120 and 36 is 12.
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# CLASS-4ARRANGING FRACTION IN ASCENDING ORDER WITH DIFFERENT NUMERATOR & DENOMINATOR
ARRANGING FRACTION IN ASCENDING ORDER WITH DIFFERENT NUMERATOR & DENOMINATOR -
ASCENDING ORDER -
Convert the fractions to the like denominators (or numerators) and then compare and order them.
8 9 7
EXAMPLE.1) Arrange in ascending order ------, ------, ------
25 16 10
Ans.) Here, the denominators 25, 16, and 10. as well as numerator 8, 9, and 7 of the given fraction is not equal.
Denominators of the given fractions are 25, 16, and 10.
In this case we have to find out the LCM of denominator of given fraction.
LCM of 25, 16 and 10 is 400.
So, now we will find the equivalent fractions.
8 8 X 16 128
------- = ------------ = -------
25 25 X 16 400
9 9 X 25 225
------- = ---------- = ---------
16 16 X 25 400
7 7 X 40 280
------- = ---------- = -------
10 10 X 40 400
So, as per the obtained equivalent fraction, numerators are 128, 225, 280
So, 128 < 225 < 280
128 225 280
Or, -------- < -------- < --------
400 400 400
8 9 7
=> -------- < -------- < -------- (Ans.)
25 16 10
5 6 3 7
EXAMPLE.2) Arrange in ascending order ------, ------, ------, ------
20 25 10 30
Ans.) Here, the denominators 20, 25, 10, and 30. as well as numerator 5, 6, 3, and 7 of the given fraction is not equal.
Denominators of the given fractions are 20, 25, 10, and 30.
In this case we have to find out the LCM of denominator of given fraction.
LCM of 20, 25, 10 and 30 is 300.
So, now we will find the equivalent fractions.
5 5 X 15 75
------ = ----------- = --------
20 20 X 15 300
6 6 X 12 72
------- = ---------- = ---------
25 25 X 12 300
3 3 X 30 90
------- = ---------- = -------
10 10 X 30 300
7 7 X 10 70
------ = --------- = -------
30 30 X 10 300
So, as per the obtained equivalent fraction, numerators are 75, 72, 90, 70
So, 70 < 72 < 75 < 90.
70 72 75 90
Or, ------- < ------- < ------ < ------
300 300 300 300
7 6 5 3
=> ------- < ------- < ------- < ------- (Ans.)
30 25 20 10
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Course: Multivariable calculus>Unit 4
Lesson 4: Line integrals in vector fields (articles)
Line integrals in a vector field
After learning about line integrals in a scalar field, learn about how line integrals work in vector fields.
What we are building to
Let's say there is some vector field $\mathbf{\text{F}}$ and a curve $C$ wandering through that vector field. Imagine walking along the curve, and at each step taking the dot product between the following two vectors:
• The vector from the field $\mathbf{\text{F}}$ at the point where you are standing.
• The displacement vector associated with the next step you take along this curve.
If you add up those dot products, you have just approximated the line integral of $\mathbf{\text{F}}$ along $C$
The shorthand notation for this line integral is
$\begin{array}{r}{\int }_{C}\mathbf{\text{F}}\cdot d\mathbf{\text{r}}\end{array}$
(Pay special attention to the fact that this is a dot product)
The more explicit notation, given a parameterization $\mathbf{\text{r}}\left(t\right)$ of $C$, is
$\begin{array}{r}{\int }_{a}^{b}\mathbf{\text{F}}\left(\mathbf{\text{r}}\left(t\right)\right)\cdot {\mathbf{\text{r}}}^{\prime }\left(t\right)dt\end{array}$
Line integrals are useful in physics for computing the work done by a force on a moving object.
If you parameterize the curve such that you move in the opposite direction as $t$ increases, the value of the line integral is multiplied by $-1$.
Whale falling from the sky
Let's say we have a whale, whom I'll name Whilly, falling from the sky. Suppose he falls along a curved path, perhaps because the air currents push him this way and that.
In this example, I am assuming you are familiar with the idea from physics that a force does work on a moving object, and that work is defined as the dot product between the force vector and the displacement vector.
Key question: What is the work done on Whilly by gravity as he falls along the curved path $C$?
Usually, computing work is done with respect to a straight force vector and a straight displacement vector, so what can we do with this curved path? You can start by imagining the curve is broken up into many little displacement vectors:
Go ahead and give each one of these displacement vectors a name,
${\stackrel{\to }{\mathrm{\Delta }s}}_{1}$, ${\stackrel{\to }{\mathrm{\Delta }s}}_{2}$, ${\stackrel{\to }{\mathrm{\Delta }s}}_{3}$, $\dots$
The work done by gravity along each one of these displacement vectors is the gravity force vector, which I'll denote $\stackrel{\to }{{F}_{g}}$, dotted with the displacement vector itself:
$\stackrel{\to }{{F}_{g}}\cdot {\stackrel{\to }{\mathrm{\Delta }s}}_{i}$
The total work done by gravity along the entire curve is then estimated by
$\begin{array}{r}\sum _{n=1}^{N}\stackrel{\to }{{F}_{g}}\cdot {\stackrel{\to }{\mathrm{\Delta }s}}_{n}\end{array}$
But of course, this is calculus, so we don't just look at a specific number of finite steps along the curve $C$. We consider what limiting value this sum approaches as the size of those steps shrinks smaller and smaller. This is captured with the following integral:
$\begin{array}{r}{\int }_{C}\stackrel{\to }{{F}_{g}}\cdot \stackrel{\to }{ds}\end{array}$
This is very similar to line integration in a scalar field, but there is the key difference: The tiny step $\stackrel{\to }{ds}$ is now thought of as a vector, not a scalar length. In the integral above, I wrote both $\stackrel{\to }{{F}_{g}}$ and $\stackrel{\to }{ds}$ with little arrows on top to emphasize that they are vectors. A more subtle and more common way to emphasize that these are vector quantities is to write the variable in bold:
$\begin{array}{r}{\int }_{C}{\mathbf{\text{F}}}_{g}\cdot d\mathbf{\text{s}}\end{array}$
Key takeaway: The thing we're adding up as we wander along $C$ is not the full value of ${\mathbf{\text{F}}}_{g}$ at each point, but the component of ${\mathbf{\text{F}}}_{g}$ pointed in the same direction as the vector $d\mathbf{\text{s}}$. That is, the component of force in the direction of the curve.
Example 1: Putting numbers on Whilly's fall.
Let's see how this plays out when we go through the computation.
Suppose the curve of Whilly's fall is described by the parametric function
$\begin{array}{r}\mathbf{\text{s}}\left(t\right)=\left[\begin{array}{c}100\left(t-\mathrm{sin}\left(t\right)\right)\\ 100\left(-t-\mathrm{sin}\left(t\right)\right)\end{array}\right]\end{array}$
The vector $d\mathbf{\text{s}}$ representing a tiny step along the curve can be given as the derivative of this function, times $dt$:
$d\mathbf{\text{s}}=\frac{d\mathbf{\text{s}}}{dt}dt={\mathbf{\text{s}}}^{\prime }\left(t\right)dt$
If these seem unfamiliar, consider taking a look at the article describing derivatives of parametric functions. The way to visualize this is to think of a tiny increase to the parameter $t$ of size $dt$. This results in a tiny nudge along the curve described by $\mathbf{\text{s}}\left(t\right)$, which is given by the vector ${\mathbf{\text{s}}}^{\prime }\left(t\right)dt$.
Evaluating this derivative vector simply requires taking the derivative of each component:
$\begin{array}{rl}\frac{d\mathbf{\text{s}}}{dt}& =\left[\begin{array}{c}\frac{d}{dt}100\left(t-\mathrm{sin}\left(t\right)\right)\\ \\ \frac{d}{dt}100\left(-t-\mathrm{sin}\left(t\right)\right)\\ \end{array}\right]\\ \\ \frac{d\mathbf{\text{s}}}{dt}& =\left[\begin{array}{c}100\left(1-\mathrm{cos}\left(t\right)\right)\\ \\ 100\left(-1-\mathrm{cos}\left(t\right)\right)\\ \end{array}\right]\end{array}$
The force of gravity is given by the acceleration $9.8\frac{\text{m}}{{\text{s}}^{2}}$ times the mass of Whilly. Not that it matters, but I looked up the typical mass of a blue whale, and it's around $170,000\phantom{\rule{0.167em}{0ex}}\text{kg}$, so let's use that number.
Since this force is directed purely downward, gravity as a force vector looks like this:
$\begin{array}{rl}{\mathbf{\text{F}}}_{g}& =\left[\begin{array}{c}0\\ -\left(170,000\right)\left(9.8\right)\end{array}\right]\end{array}$
Let's say we want to find the work done by gravity between times $t=0$ and $t=10$. What do you get when you plug in all this information to the integral integral ${\int }_{C}{\mathbf{\text{F}}}_{g}\cdot d\mathbf{\text{s}}$ and evaluate the integral? Take a moment to try writing this out for yourself before peeking at the answer.
(To those physics students among you who notice that it would be easier to just compute the gravitational potential of Whilly at the start and end of his fall and find the difference, you are going to love the topic of conservative fields!)
Visualizing more general line integrals through a vector field
In the previous example, the gravity vector field is constant. Gravity points straight down with the same magnitude everywhere. With most line integrals through a vector field, the vectors in the field are different at different points in space, so the value dotted against $d\mathbf{\text{s}}$ changes. The following animation shows what this might look like.
(Note, the animation uses the variable $\mathbf{\text{r}}$ instead of $\mathbf{\text{s}}$ to parameterize the curve, but of course, it does not make a difference.)
Let's dissect what's going on here. The line integral itself is written as
$\begin{array}{r}{\int }_{C}\mathbf{\text{F}}\left(\mathbf{\text{r}}\right)\cdot d\mathbf{\text{r}}={\int }_{a}^{b}\mathbf{\text{F}}\left(\mathbf{\text{r}}\left(t\right)\right)\cdot {\mathbf{\text{r}}}^{\prime }\left(t\right)dt\end{array}$
where
• $\mathbf{\text{F}}$ is a vector field, associating each point in space with a vector. You can think of this as a force field.
• $C$ is a curve through space.
• $\mathbf{\text{r}}\left(t\right)$ is a vector-valued function parameterizing the curve $C$ in the range $a\le t\le b$
• ${\mathbf{\text{r}}}^{\prime }\left(t\right)$ is the derivative of $\mathbf{\text{r}}$, representing the velocity vector of a particle whose position is given by $\mathbf{\text{r}}\left(t\right)$ while $t$ increases at a constant rate. When you multiply this by a tiny step in time, $dt$, it gives a tiny displacement vector, which I like to think of as a tiny step along the curve. Technically it is a tiny step in the tangent direction to the curve, but for small enough $dt$ this amounts to the same thing.
• Note, in this animation the length of ${\mathbf{\text{r}}}^{\prime }\left(t\right)$ stays constant. This is not necessarily true for most parameterizations of $C$, which may have you speeding up or slowing down as your position varies according to $\mathbf{\text{r}}$. For example, Whilly was probably speeding up during his fall, making the velocity vector grow over time.
• The rotating circle in the bottom right of the diagram is a bit confusing at first. It represents the extent to which the vector $\mathbf{\text{F}}\left(\mathbf{\text{r}}\left(t\right)\right)$ lines up with the tangent vector ${\mathbf{\text{r}}}^{\prime }\left(t\right)$. The grey $x$ and $y$ vectors are shown to see how these vectors are oriented relative to $xy$-plane as a whole.
Concept check: What does the dot product $\mathbf{\text{F}}\left(\mathbf{\text{r}}\left(t\right)\right)\cdot {\mathbf{\text{r}}}^{\prime }\left(t\right)dt$ represent?
$\mathbf{\text{F}}\left(\mathbf{\text{r}}\left(t\right)\right)\cdot {\mathbf{\text{r}}}^{\prime }\left(t\right)dt$
as
$dW$
That is, a tiny amount of work done by the force field $\mathbf{\text{F}}$ on a particle moving along $C$.
Example 2: Work done by a tornado
Consider the vector field described by the function
$\begin{array}{rl}\mathbf{\text{F}}\left(x,y\right)& =\left[\begin{array}{c}-y\\ x\end{array}\right]\end{array}$
The vector field looks like this:
Thought of as a force, this vector field pushes objects in the counterclockwise direction about the origin. For example, maybe this represents the force due to air resistance inside a tornado. This is a little unrealistic because it would imply that force continually gets stronger as you move away from the tornado's center, but we can just euphemistically say it's a "simplified model" and continue on our merry way.
Suppose we want to compute a line integral through this vector field along a circle or radius $1$ centered at $\left(2,0\right)$.
I should point out that orientation matters here. The work done by the tornado force field as we walk counterclockwise around the circle could be different from the work done as we walk clockwise around it (we'll see this explicitly in a bit).
If we choose to consider a counterclockwise walk around this circle, we can parameterize the curve with the function.
$\begin{array}{rl}\mathbf{\text{r}}\left(t\right)& =\left[\begin{array}{c}\mathrm{cos}\left(t\right)+2\\ \mathrm{sin}\left(t\right)\end{array}\right]\end{array}$
where $t$ ranges from $0$ to $2\pi$.
Again, to set up the line integral representing work, you consider the force vector at each point, $\mathbf{\text{F}}\left(x,y\right)$, and you dot it with a tiny step along the curve, $d\mathbf{\text{r}}$:
$\begin{array}{r}{\int }_{C}\mathbf{\text{F}}\cdot d\mathbf{\text{r}}\end{array}$
Step 1: Expand the integral
Concept check: Which of the following integrals represents the same thing as $\begin{array}{r}{\int }_{C}\mathbf{\text{F}}\cdot d\mathbf{\text{r}}\end{array}$?
Step 2: Expand each component
Concept check: Based on the definitions above, what is $\mathbf{\text{F}}\left(\mathbf{\text{r}}\left(t\right)\right)$?
Concept check: What is ${\mathbf{\text{r}}}^{\prime }\left(t\right)$?
Step 3: Solve the integral
Concept check: Put the last three answers together to solve the integral.
$\begin{array}{r}{\int }_{C}\mathbf{\text{F}}\cdot d\mathbf{\text{r}}\end{array}$ =
This final answer gives the amount of work that the tornado force field does on a particle moving counterclockwise around the circle pictured above.
Reflection question: Why should it be intuitive that this answer is positive?
Orientation matters
What would have happened if in the preceding example, we had oriented the circle clockwise? For instance, we could have parameterized it with the function
$\begin{array}{rl}\mathbf{\text{r}}\left(t\right)& =\left[\begin{array}{c}\mathrm{cos}\left(t\right)+2\\ -\mathrm{sin}\left(t\right)\end{array}\right]\end{array}$
You can, if you want, plug this in and work through all the computations to see what happens. However, there is a simpler way to reason about what will happen. In the integral
$\begin{array}{r}{\int }_{C}\mathbf{\text{F}}\cdot d\mathbf{\text{r}}\end{array}$,
each vector $d\mathbf{\text{r}}$ representing a tiny step along the curve will get turned around to point in the opposite direction.
Concept check: Suppose you have two vectors $\mathbf{\text{v}}$ and $\mathbf{\text{w}}$, and $\mathbf{\text{v}}\cdot \mathbf{\text{w}}=3$. You turn $\mathbf{\text{v}}$ around to point in the opposite direction, getting a new vector ${\mathbf{\text{v}}}_{\text{new}}=-\mathbf{\text{v}}$. What happens to the dot product?
${\mathbf{\text{v}}}_{\text{new}}\cdot \mathbf{\text{w}}=$
Since the dot product inside the integral gets multiplied by $-1$ when you swap the direction of each $d\mathbf{\text{r}}$, we can conclude the following:
Key Takeaway: The line integral through a vector field gets multiplied by $-1$ when you reverse the orientation of a curve.
Summary
• The shorthand notation for a line integral through a vector field is
$\begin{array}{r}{\int }_{C}\mathbf{\text{F}}\cdot d\mathbf{\text{r}}\end{array}$
• The more explicit notation, given a parameterization $\mathbf{\text{r}}\left(t\right)$ of $C$, is
$\begin{array}{r}{\int }_{a}^{b}\mathbf{\text{F}}\left(\mathbf{\text{r}}\left(t\right)\right)\cdot {\mathbf{\text{r}}}^{\prime }\left(t\right)dt\end{array}$
• Line integrals are useful in physics for computing the work done by a force on a moving object.
• If you parameterize the curve such that you move in the opposite direction as $t$ increases, the value of the line integral is multiplied by $-1$.
Want to join the conversation?
• How was the parametric function for r(t) obtained in above example? Thank you
• We have a circle with radius 1 centered at (2,0). From the Pythagorean Theorem, we know that the x and y components of a circle are cos(t) and sin(t), respectively. Thus we can parameterize the circle equation as x=cos(t) and y=sin(t). Note, however, that the circle is not at the origin and must be shifted. Since each x value is getting 2 added to it, we add 2 to the cos(t) parameter to get vectors that look like <cos(t) + 2, sin(t)>. Also note that there is no shift in y, so we keep it as just sin(t). You can look at the early trigonometry videos for why cos(t) and sin(t) are the parameters of a circle.
• What is the difference between dr and ds? Thank you:)
• dr is a small displacement vector along the curve.
ds is a small scalar step along the curve.
Fdr would be a vector times a vector, or force times displacement.
fds would be some curve evaluated at a height of f times a vertical step.
• How can i get a pdf version of articles , as i do not feel comfortable watching screen
• Just print it directly from the browser. ( p.s. you can print as a pdf).
• what is F(r(t))graphically and physically?
• F(x,y) at any point gives you the vector resulting from the vector field at that point. F(x(t),y(t)), or F(r(t)) would be all the vectors evaluated on the curve r(t).
Imagine a transparent curve and place it onto of a vector field. Then erase all the vectors in the vector field that do not have tails starting on the curve.
That should be F(r(t))
• The math all makes sense, but what is causing the particle to move along the curve in the first place. Wouldn't there have to be another force moving the particle along the curve , so that it could "experience" the force of the vector field?
• smart question. The example of this being a particle and it moving along a curve breaks down when you think about how the particle is actually moving through the field. So you are right to have this question.
The curve is a path that was choosen by people who made the problem and we are just asking suppose a particle moved along this curve through this field then what work would the field to the particle.
I think what you are imagining is something like the field being a body of water with flow and current then we drop a buoy into the water. Then the buoy is going to follow a certain natural curve in the water dictated by the force vectors of the water (the field) at every point so how are we forcing the buoy through a specific curve?( i think this is where you are at in your thought process)
Now imagine we dropped a boat into the water rather than a buoy. since a boat has a motor with a propeller it does not need to follow the natural curve of the water. But to go on the curve that we want it to follow the boat must use it's engine (do work). AHHA! When we solve these integrals for the work done by the field on the particle we are symmetrically solving for the amount of work the particle would have to do to follow that curve. Hence in the example of the boat we could solve for how much work the engine and propeller would have to do to go up river vs down river.
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Mr. Vizza's Class The Ponderosa
Absolute Value Inequalities
(Greater than)
Steps
Examples Done by Mr. Vizza
Problems You Should Try
Writing Exercise
Steps
1. Undo the operations outside absolute value bars, so that the problem looks like this.
2. Split the above problem into two inequalities similar to the diagram below. Solve for the unknown in each equation. (Notice that the absolute value bars have been dropped.)
3. Graph the resulting solution set.
Examples Done by Mr. Vizza
Directions: Please solve for the unknown in the following absolute value inequality.
4|3x-5|+1>17
Step 1. Notice that the absolute value quantity is not isolated on one side of the inequality. We must subtract one on both sides then divide by four on both sides to rectify this.
4|3x-5|>16
|3x-5|>4
Step 2. Now that the problem is consistent with the diagram above, we can split the problem into two inequalities and solve them. Notice in the inequality on the right, along with making the number negative, we flipped the inequality symbol.
|3x-5|>4 becomes
3x-5>4 Or 3x-5<-4
3x>9 Or 3x<1
x>3 Or x<1/3
Step 3. The graph must contain the numbers that are smaller than or equal to negative third and greater than or equal to positive three. Note that the less than or equal to symbol and the greater than or equal to symbol require open dots when they are graphed.
Problems You Should Try
Directions: Please solve for the unknown in the following absolute value inequalities. If there is no solution, please indicate so by stating "null set". Also, graph the solutions sets.
1. |3x|>9 2. |18b|>9 3. |2x+12|>14
4. |10c|-15>15 5. 9|6r|>-63 6. 9|4-2y|-6>30
7. -5|6+2r|>20 8. 7|7x|-7>42 9. 3|4+2g|>21
10. 4-|3b-8|<-8
Writing Exercise
Directions: Please write 1-2 paragraphs that do the following.
1. Compare and contrast the steps used to solve absolute value inequalities with less than to the steps used to solve absolute value inequalities with greater than. You must remark on the similarities and the differences between the diagrams provided on these web pages. Are the methods more similar to each other or different than each other? Can you describe a more general process that works for solving both types of inequalities?
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Now that we've taken a detailed look at triangles, we can begin looking at a shape with an extra side and vertex: the quadrilateral. The word "quadrilateral" is composed of two main parts:
(1) quad - which means "four", and
(2) lateral - which means "side."
While triangles are very significant to the world around us, quadrilaterals are, perhaps, the most important and common type of polygon. For instance, quadrilaterals come in the form of numerous shapes, including squares, rectangles, rhombuses, kites, and trapezoids, just to name a few. Each of these shapes has their own defining characteristics, which we will learn about in the following lessons.
Take a look around your environment right now. A great number of the things you see are probably composed of some type of quadrilateral. Whether it's the computer screen you're looking at, the building you are in, or the sheets of paper you write on, quadrilaterals are everywhere. Let's explore the properties and characteristics of these special shapes to expand our knowledge of geometry.
## Polygons
Main Lesson: Polygons
Learn the different classifications of polygons, their names, and how we can use the Polygon Interior Angle Sum Theorem.
## Properties of Parallelograms
Main Lesson: Properties of Parallelograms
Become familiar with key terminology used to describe different parts of quadrilaterals, and study the properties that make parallelograms unique.
## Proving Quadrilaterals Are Parallelograms
Main Lesson: Proving Quadrilaterals Are Parallelograms
Learn how to use the properties of parallelograms properly in two-column geometric proofs.
## Properties of Rectangles, Rhombuses, and Squares
Main Lesson: Properties of Rectangles, Rhombuses, and Squares
Discover the properties of three more types of parallelograms: rectangles, rhombuses, and squares.
## Properties of Trapezoids and Kites
Main Lesson: Properties of Trapezoids and Kites
Take a step away from parallelograms and learn about trapezoids and kites, whose opposite sides may intersect.
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# How do you identify the conic section represented by the equation 9y^2-16x^2-72y-64x=64?
Dec 3, 2016
This is the equation of a hyperbola ${\left(y - 4\right)}^{2} / 16 - {\left(x + 2\right)}^{2} / 9 = 1$
#### Explanation:
Let rearrange the equation and complete the squares
$9 {y}^{2} - 72 y - 16 {x}^{2} - 64 x = 64$
$9 \left({y}^{2} - 8 y\right) - 16 \left({x}^{2} + 4 x\right) = 64$
$9 \left({y}^{2} - 8 y + 16\right) - 16 \left({x}^{2} + 4 x + 4\right) = 64 + 144 - 64$
$9 {\left(y - 4\right)}^{2} - 16 {\left(x + 2\right)}^{2} = 144$
$\frac{9}{144} {\left(y - 4\right)}^{2} - \frac{16}{144} {\left(x + 2\right)}^{2} = 1$
${\left(y - 4\right)}^{2} / 16 - {\left(x + 2\right)}^{2} / 9 = 1$
This is a hyperbola,
${\left(y - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$
The center is $= \left(- 2 , 4\right)$
The equations of the asymptotes are
$y - 4 = \frac{4}{3} \left(x + 2\right)$
and $y - 4 = - \frac{4}{3} \left(x + 2\right)$
graph{(9y^2-16x^2-72y-64x-64)(y-4-4/3(x+2))(y-4+4/3(x+2))=0 [-83.36, 83.36, -41.6, 41.7]}
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6190 minus 87 percent
This is where you will learn how to calculate six thousand one hundred ninety minus eighty-seven percent (6190 minus 87 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 6190 minus 87 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 6190 of something.
6190
(100%)
87 percent means 87 per hundred, so for each hundred in 6190, you want to subtract 87. Thus, you divide 6190 by 100 and then multiply the quotient by 87 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(6190 ÷ 100) × 87
= 5385.3
We made a pink square that we put on top of the image shown above to illustrate how much 87 percent is of the total 6190:
The dark blue not covered up by the pink is 6190 minus 87 percent. Thus, we simply subtract the 5385.3 from 6190 to get the answer:
6190 - 5385.3
= 804.7
The explanation and illustrations above are the educational way of calculating 6190 minus 87 percent. You can also, of course, use formulas to calculate 6190 minus 87%.
Below we show you two formulas that you can use to calculate 6190 minus 87 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
6190 - ((6190 × 87/100))
6190 - 5385.3
= 804.7
Formula 2
Number × (1 - (Percent/100))
6190 × (1 - (87/100))
6190 × 0.13
= 804.7
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
6200 minus 87 percent
Here is the next percent tutorial on our list that may be of interest.
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# How do you find local maximum value of f using the first and second derivative tests: f(x)=x+sqrt(2-x)?
Feb 6, 2017
See below.
#### Explanation:
Domain of $f$ is $\left(- \infty , 2\right]$
$f ' \left(x\right) = 1 - \frac{1}{2 \sqrt{2 - x}} = \frac{2 \sqrt{2 - x} - 1}{2 \sqrt{2 - x}}$
$f '$ is undefined at $2$ and is $0$ at $\frac{7}{4}$. So the critical numbers are $2$ and $\frac{7}{4}$.
(Some do not apply the term "critical number" at an endpoint of the domain.)
Using the First Derivative Test
On $\left(- \infty , \frac{7}{4}\right)$ we have $f ' \left(x\right) > 0$ and
on $\left(\frac{7}{4} , 2\right)$, $f ' \left(x\right) < 0$, so $f \left(\frac{7}{4}\right) = \frac{9}{4}$ is a local maximum
and $f \left(2\right) = 2$ is a local minimum.
(Some do not apply the term "local minimum" at an endpoint of the domain. They would disagree with my answer about that.)
Using the Second Derivative Test (for local extrema)
$f ' ' \left(x\right) = - \frac{1}{4 {\sqrt{2 - x}}^{3}}$
$f ' ' \left(\frac{7}{4}\right) < 0$, so $f \left(\frac{7}{4}\right) = \frac{9}{4}$ is a local maximum.
$f ' ' \left(2\right)$ does not exist, so we cannot use the second derivative test at $x = 2$
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# CBSE Class 6 Mathematics Worksheets/ Part 2
CBSE Class 6 Mathematics Worksheets is about some model questions that you can expect for annual/yearly examinations. Here you can find out practice problems for class 6 Mathematics.
CBSE CLASS 6 MATHEMATICS WORKSHEETS /Part 2
For Annual/Yearly Examinations:
Choose the correct answer from the options given below:
1. The side of a square with perimeter 40 m is ———–
A. 10m
B. 20 m
C. 5 m
2. The ratio of 50 paise to 5 rupee is ——-
A. 1: 10
B. 1: 100
C. 1: 50
3. 5 is multiplied by x, then the result is added to 12. Write the algebraic expressions.
A. 5x + 12
B. 5x – 12
C. 5 + x + 12
4. 15mm = ——— cm
A. 1.5
B. 0.15
C. 0.015
5. The simplest form of the ratio 3:18 is ———-
A. 1: 6
B. 3: 6
C. 3: 9
Fill in the blanks:
6. The decimal form of the fraction 7 ½ is ———-
7. The ratio equivalent to 2:3 is —————-
8. The simplest form of 25: 35 is —————
9. Two sides of a triangle are given by 5 cm and 7 cm. Its perimeter is 20 cm. The length of third side is ————–
10. 10 – 5.5 =—————–
11. Find the cost of fencing a rectangular garden of length 25m and breadth 15 m at the rate of 25 rupees per square metre.
12. Anu purchased 3 kg 50g apple, 2 kg 250 g oranges. Find the total weight of things that he purchased in kg.
13. Cost of 5 kg rice is 150 rupees. What will be the cost of 8 kg of rice?
14. Out of 50 children in a class, 26 are boys and the remaining are girls. Find the ratio of the number of boys to number of girls.
15. Write the number name and expanded form of 25. 876.
1. 10m
2. 1:10
3. 5x + 12
4. 1.5 cm
5. 1: 6
6. 7.5
7. 4:6
8. 5:7
9. 8 cm
10. 4.5
11. Given length = 25m and breadth = 15m
Perimeter = 2(l + b) = 2(25 + 15) = 2 x 40 = 80 m
Cost of fencing = 80 x 25 = 2000 rupees
12. 3 kg 50g = 3.050 kg
2 kg 250g = 2. 250 kg
Total weight = 3.050 + 2.250 = 5.300 kg
13. Cost of 5 kg rice = 150 rupees
Cost of 1 kg rice = 150 / 5 = 30 rupees
Cost of 8 kg rice = 30 x 8 = 240 rupees
14. Number of boys = 26
Number of girls = 24
Ratio of number of boys to girls = 26:24 = 13:12
15. Number name: Twenty five point eight seven six
Expansion: 2 x 10 + 5 x 1 + 8 x (1/10) + 7 x (1/100) + 6 x (1/1000)
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# Wyoming - Grade 1 - Math - Operations and Algebraic Thinking - Addition and Subtraction Word Problems - 1.OA.1
### Description
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
• State - Wyoming
• Standard ID - 1.OA.1
• Subjects - Math Common Core
### Keywords
• Math
• Operations and Algebraic Thinking
## More Wyoming Topics
Compare two two-digit numbers based on meanings of the tens and ones digits, recording the results of comparisons with the symbols >, =, and <.
Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes.
Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares.
1.OA.7 Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2.
1.OA.8 Determine the unknown whole number in an addition or subtraction equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 + ? = 11, 5 = _ – 3, 6 + 6 = _.
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# Category: Function Concept
## Variable Interval
Recall that variable ${x}$ is defined by the set ${X}$ of those values that it can take (in considered question). This set ${X}$ that contains each value of ${x}$ exactly once is called domain of variation of the variable ${x}$.
## Dependence between Variables
We are not really interested in changing of one variable, it is more interesting to study dependence between two variables when they both change.
Such variables can not take simultaneously any pair of values: if one of them (independent variable) is given certain value, then we can find value of second variable (dependent variable).
## Definition of Function
Suppose we are given two variables ${x}$ and ${y}$ with domain of variation ${X}$ and ${Y}$ respectively.
Suppose that ${x}$ can take any value from ${X}$ without any restrictions. Then variable ${y}$ is called function of variable ${x}$ in domain of its variation ${X}$, if according to some law or rule we can assign to each value of ${x}$ from ${X}$ one definite value of ${y}$ from ${Y}$.
## Representing a Function
There are three possible ways to represent a function:
• analytically (by formula);
• numerically (by table of values);
• visually (by graph).
Analytical representation of a function.
This is the most common way to represent a function. We define a formula that contains arithmetic operations on constant values and the variable ${x}$ that we need to perform to obtain the value of ${y}$.
## Inverse of a Function
We already know that a function is a rule that allows finding for every value of ${x}$ the corresponding value of ${y}$. But what if we are given a rule, i.e. a function, and the value of ${y}$ and we want to find the corresponding value of ${x}$?
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wikiHow to Convert Binary to Hexadecimal
This article will explain how to convert binary (base 2) to hexadecimal (base 16). Whether it is for coding, for math class, or for The Martian, hexadecimal is useful and powerful shortcut when writing long binary strings. Since both bases are powers of 2, this procedure is much simpler than general conversions such as converting decimal to binary. All you need are basic adding and counting skills to make turn a binary number into hexadecimal.
Method 1 Making Basic Conversions
1. 1
Find a line of up to four binary numbers to convert. Binary numbers can only be 1 and 0. Hexadecimal numbers can be 0-9, or A-F, since hexadecimal is base-16. You can convert any binary string to hexadecimal (1, 01, 101101, etc.), but you need four numbers to make the conversion (0101→5; 1100→C, etc.). For this lesson, start with the example 1010.
• 1010
• If you don't have 4 digits, add zeros to the front to make it four digits. So, 01 would become 0001.[1]
2. 2
Write a small "1" above the last digit. Each of the four numbers signifies a type of number decimal system number. The last digit is the one's place. You will make sense of the rest of the digits in the next step. For now, write a small one above the last digit.[2]
• 1010
• ${\displaystyle 1010^{1}}$
• Note that you are not raising anything to any power -- this is just a way to see what digit means what.
3. 3
Write a small "2" above the third digit, a "4" above the second, and an "8" above the first. These are the rest of your place holders. If you're curious, this is because each digit represents a different power of 2. The first is ${\displaystyle 2^{3}}$, the second ${\displaystyle 2^{2}}$, etc.
• 1010
• ${\displaystyle 1^{8}0^{4}1^{2}0^{1}}$
4. 4
Count out how many of each "place" you have. Luckily, this conversion is easy once you have four numbers and know what they all mean. If you have a one in the first number, you have one eight. If you have a zero in the second column, you have no fours. The third column tells you how many twos, and the second how many ones. So, for our example:[3]
• 1010
• ${\displaystyle 1^{8}0^{4}1^{2}0^{1}}$
• 8 0 2 0
5. 5
• 1010
• ${\displaystyle 1^{8}0^{4}1^{2}0^{1}}$
• 8 0 2 0
• ${\displaystyle 8+0+2+0=10}$
• Final answer: The binary number 1010 converts to A in the hexadecimal system.
6. 6
Change any number above "9" into a letter. This is so you don't get confused when reading hexadecimal ("is that a 1 and a 5, or a 15?"). Luckily, the system is super easy, since you can't have a hexadecimal number bigger than 15. Simply start the alphabet with 10, so that:
• ${\displaystyle 10=A}$
• ${\displaystyle 11=B}$
• ${\displaystyle 12=C}$
• ${\displaystyle 13=D}$
• ${\displaystyle 14=E}$
• ${\displaystyle 15=F}$
7. 7
Try a few examples to get better at converting. The following examples have answers in white beneath them. To see the work and the answers, highlight the area under the question by clicking and dragging your mouse over it.
• Add zeros to get four digits: 0001
• Find your place holders: ${\displaystyle 0^{8}0^{4}0^{2}1^{1}}$
• Add up the digits: ${\displaystyle 0+0+0+1=1}$
• Add zeros to get four digits: 0101
• Find your place holders: ${\displaystyle 0^{8}1^{4}0^{2}1^{1}}$
• Add up the digits: ${\displaystyle 0+4+0+1=5}$
• Add zeros to get four digits: 1110
• Find your place holders: ${\displaystyle 1^{8}1^{4}1^{2}0^{1}}$
• Add up the digits: ${\displaystyle 8+4+2+0=14}$
• Add zeros to get four digits: 0011
• Find your place holders: ${\displaystyle 1^{8}0^{4}1^{2}1^{1}}$
• Add up the digits: ${\displaystyle 8+0+2+1=11}$
Method 2 Converting Long Binary Strings
1. 1
Cut your string of binary numbers into groups of four, starting from the right. Hexadecimal converts 4 binary digits into one hexadecimal unit. So, in order to convert the number, you first need to break it up into groups of four, starting on the right. For example:
• Convert ${\displaystyle 11101100101001}$ into a hexadecimal number.
• ${\displaystyle 11101100101001=(11)(1011)(0010)(1001)}$
2. 2
Add extra zeros to the front of the first number if it is not four digits. The zeros will not affect the conversion, but they will make it easier to visualize. Remember, you want all groups of 4-digit binary numbers.
• Convert ${\displaystyle 11101100101001}$ into a hexadecimal number.
• ${\displaystyle 11101100101001=(11)(1011)(0010)(1001)}$
• ${\displaystyle (11)(1011)(0010)(1001)=}$${\displaystyle (0011)(1011)(0010)(1001)}$
3. 3
Convert one 4-digit group at a time. You'll need to convert each binary set by itself, so separate them on your paper to make them easier to work with. Work on converting each individual string of four into its hexadecimal counterpart. For our example:[4]
• ${\displaystyle 0011=0+0+2+1=3}$
• ${\displaystyle 1011=8+0+2+1=11=B}$
• ${\displaystyle 0010=0+0+2+0=2}$
• ${\displaystyle 1001=8+0+0+1=9}$
4. 4
Remove the spaces to create your hexadecimal number. Once you've converted all the 4-digit parts, simply ram them together to get your final answer. So, for the example above:
• ${\displaystyle (0011)(1011)(0010)(1001)}$
• 3 B 2 9
• ${\displaystyle 11101100101001=3B29}$
5. 5
Memorize or check a conversion table to see if you got each part right. There are only 16 possible 4-digit combinations of binary numbers. So, if you don't want to figure out each string individually, you can use this conversion table.
0000 0
0001 1
0010 2
0011 3
0100 4
0101 5
0110 6
0111 7
1000 8
1001 9
1010 A
1011 B
1100 C
1101 D
1110 E
1111 F[5]
Community Q&A
Search
• How do I convert decimal to octal?
wikiHow Contributor
First convert decimal to binary, then into octal. Make 3 pairs of binary numbers from the right side and add zero if needed.
• How do I convert 100101 binary to hexadecimal?
wikiHow Contributor
First add two leading zeros so that you have enough digits to split into equal 4 bit sections: (00)100101 Now split into two equal 4 bit sections (0010)(0101) Convert each four bit section into their respective hex values: 0010 = 2 0101 = 5 (0010)(0101) = (2)(5) = 25 This should not be confused with the decimal (base 10) value 25.
• How do I convert hexadecimal to binary?
wikiHow Contributor
First convert the given hexadecimal into the decimal system. The decimal can then be converted to binary.
• How do I convert binary to octal?
wikiHow Contributor
There are lot ways to do this. For beginners, just convert binary into ordinary numbers, then change it into an octal number.
• Why are four binary digits necessary to convert binary to hexadecimal?
wikiHow Contributor
Hexadecimal is base-16 whereas binary is only in the form base-2. To account for this, you need four binary digits to allow for the 16 possible characters of hexadecimal. (Think 2^4=16 like how you would calculate the number of alternatives in physics.)
• How can I change the hexadecimal binary?
wikiHow Contributor
To convert Binary to hexadecimal you need to know that the four numbers each have values. 8 4 2 1 0 1 0 1 These numbers only equal the values if there is a 1 under them for example this number would be 5 because there is only a 1 under 4 and 1. You would then change your answer in this example 5 into hexadecimal which is 5. For 8 number binary like 00101011 you want to split them up into 2 different parts. 0010 and 1011. To do this you do the exact same as you did for the previous one but you put the 2 answer together for example 2 and 13 would be 2D.
• How can I find the hexadecimal representation of the binary number (110101111) given in base 2?
wikiHow Contributor
First split the binary into groups of 4: 1, 1010, and 1111. 1(base 2)=1(base 10) = 1(Base 16). 1010(base 2) = 10(base 10) = A(base 16). 1111(base 2) = 15(base 10) = F(base 16). 110101111(base 2) = 1AF(base 16).
• Can any binary number be first converted into a decimal, and then the derived decimal number be converted into hexadecimal?
wikiHow Contributor
Yes, although that's a much more complex process.
• How do I convert binary to hexadecimal?
200 characters left
Tips
• Binary is base two (there are only two numbers, 1 and 0). Hexadecimal is a base sixteen system. Can you figure out why you need four binary numbers to convert to hexadecimal? It is because you need four separate two's, since ${\displaystyle 2^{4}=16}$.
Warnings
• If you are finding a hex equivalent to a binary-encoded address and if you do this wrong, the results in hex-encoded address inputs will be messed up.
Article Info
Categories: Conversion Aids
In other languages:
Español: convertir un binario en hexadecimal, Italiano: Passare da Sistema Binario a Esadecimale, Português: Converter Números Binários Para Hexadecimais, Русский: переводить из двоичной системы в шестнадцатеричную, Deutsch: Binär in hexadezimal umwandeln, Français: convertir du binaire en hexadécimal, Bahasa Indonesia: Mengonversikan Bilangan Biner Menjadi Heksadesimal, Nederlands: Binair omzetten naar hexadecimaal, 日本語: 二進法を十六進法に変換する, ไทย: แปลงเลขฐานสองเป็นเลขฐานสิบหก
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# How do you solve 3/4 - 2/3>x/6?
May 26, 2018
$x < \frac{1}{2}$
#### Explanation:
$\frac{3}{4} - \frac{2}{3} > \frac{x}{6}$
You can solve this with fractions but I always recommend we turn them into integers to make everything WAY easier:
To do that we need a common denominator of $4 , 3 \mathmr{and} 6$
prime factors:
$4 = 2 \cdot 2$
$3 = 3$
$6 = 2 \cdot 3$
so the common factors in all three terns are: $2 \cdot 2 \cdot 3 = 12$
so here is the trick, we multiply 12 by both sides of the inequality like so:
$12 \left(\frac{3}{4} - \frac{2}{3}\right) > 12 \cdot \left(\frac{x}{6}\right)$
remember $12 = \frac{12}{1}$
$\left(\frac{12}{1} \cdot \frac{3}{4}\right) - \left(\frac{12}{1} \cdot \frac{2}{3}\right) > \left(\frac{12}{1} \cdot \frac{x}{6}\right)$
then cross divide:
$\left(\frac{3}{1} \cdot \frac{3}{1}\right) - \left(\frac{4}{1} \cdot \frac{2}{1}\right) > \left(\frac{2}{1} \cdot \frac{x}{1}\right)$
$\left(3 \cdot 3\right) - \left(4 \cdot 2\right) > \left(2 \cdot x\right)$
$9 - 8 > 2 x$
Now it is super simple to solve:
$9 - 8 > 2 x$
$1 > 2 x$
$\frac{1}{2} > x$
$x < \frac{1}{2}$
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# Simplify and express the final result using positive exponents. (x^{-3}y
Simplify and express the final result using positive exponents.
$$\displaystyle{\left({x}^{{-{3}}}{y}^{{4}}\right)}^{{-{2}}}$$
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Aniqa O'Neill
To simplify the expression: $$\displaystyle{\left({x}^{{-{3}}}{y}^{{4}}\right)}^{{-{2}}}$$
Solution:
Given expression is: $$\displaystyle{\left({x}^{{-{3}}}{y}^{{4}}\right)}^{{-{2}}}$$
On simplifying further we get:
$$\displaystyle{\left({x}^{{-{3}}}{y}^{{4}}\right)}^{{-{2}}}={\left({x}^{{-{3}}}\right)}^{{-{2}}}\cdot{\left({y}^{{4}}\right)}^{{-{2}}}$$
$$\displaystyle={\left({x}^{{{\left(-{3}\right)}{\left(-{2}\right)}}}\right)}{\left({y}^{{{4}{\left(-{2}\right)}}}\right)}$$
$$\displaystyle={\left({x}^{{6}}\right)}{\left({y}^{{-{8}}}\right)}$$
$$\displaystyle={\left({x}^{{6}}\right)}{\left({\frac{{{1}}}{{{y}^{{8}}}}}\right)}$$
$$\displaystyle={\left({\frac{{{x}^{{6}}}}{{{y}^{{8}}}}}\right)}$$
$$\displaystyle\Rightarrow{\left({x}^{{-{3}}}{y}^{{4}}\right)}^{{-{2}}}={\left({\frac{{{x}^{{6}}}}{{{y}^{{8}}}}}\right)}$$
Result: $$\displaystyle{\left({x}^{{-{3}}}{y}^{{4}}\right)}^{{-{2}}}={\left({\frac{{{x}^{{6}}}}{{{y}^{{8}}}}}\right)}$$
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Maharashtra Board 11th Maths Solutions Chapter 3 Trigonometry – II Miscellaneous Exercise 3
Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 3 Trigonometry – II Miscellaneous Exercise 3 Questions and Answers.
Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Miscellaneous Exercise 3
I. Select the correct option from the given alternatives.
Question 1.
The value of sin(n + 1) A sin(n + 2) A + cos(n + 1) A cos(n + 2) A is equal to
(a) sin A
(b) cos A
(c) -cos A
(d) sin 2A
(b) cos A
Hint:
L.H.S. = sin [(n + 1)A] . sin [(n + 2)A] + cos [(n + 1)A] . cos [(n + 2)A]
= cos [(n + 2)A] . cos [(n + 1)A] + sin [(n + 2)A] . sin [(n + 1)A]
Let (n + 2)A = a and (n + 1)A = b … (i)
∴ L.H.S. = cos a . cos b + sin a . sin b
= cos (a – b)
= cos [(n + 2)A – (n + 1)A] ……..[From (i)]
= cos [(n + 2 – n – 1)A]
= cos A
= R.H.S.
Question 2.
If tan A – tan B = x and cot B – cot A = y, then cot (A – B) = ________
(a) $$\frac{1}{y}-\frac{1}{x}$$
(b) $$\frac{1}{x}-\frac{1}{y}$$
(c) $$\frac{1}{x}+\frac{1}{y}$$
(d) $$\frac{x y}{x-y}$$
(c) $$\frac{1}{x}+\frac{1}{y}$$
Hint:
Question 3.
If sin θ = n sin(θ + 2α), then tan(θ + α) is equal to
(a) $$\frac{1+n}{2-n}$$ tan α
(b) $$\frac{1-n}{1+n}$$ tan α
(c) tan α
(d) $$\frac{1+n}{1-n}$$ tan α
(d) $$\frac{1+n}{1-n}$$ tan α
Hint:
Question 4.
The value of $$\frac{\cos \theta}{1+\sin \theta}$$ is equal to ________
(a) $$\tan \left(\frac{\theta}{2}-\frac{\pi}{4}\right)$$
(b) $$\tan \left(-\frac{\pi}{4}-\frac{\theta}{2}\right)$$
(c) $$\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)$$
(d) $$\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)$$
(c) $$\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)$$
Hint:
Question 5.
The value of cos A cos (60° – A) cos (60° + A) is equal to ________
(a) $$\frac{1}{2}$$ cos 3A
(b) cos 3A
(c) $$\frac{1}{4}$$ cos 3A
(d) 4cos 3A
(c) $$\frac{1}{4}$$ cos 3A
Hint:
Question 6.
The value of $$\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \frac{13 \pi}{14}$$ is ________
(a) $$\frac{1}{16}$$
(b) $$\frac{1}{64}$$
(c) $$\frac{1}{128}$$
(d) $$\frac{1}{256}$$
(b) $$\frac{1}{64}$$
Hint:
Question 7.
If α + β + γ = π, then the value of sin2 α + sin2 β – sin2 γ is equal to ________
(a) 2 sin α
(b) 2 sin α cos β sin γ
(c) 2 sin α sin β cos γ
(d) 2 sin α sin β sin γ
(c) 2 sin α sin β cos γ
Hint:
sin2 α + sin2 β – sin2 γ
= $$\frac{1-\cos 2 \alpha}{2}+\frac{1-\cos 2 \beta}{2}-\sin ^{2} \gamma$$
= 1 – $$\frac{1}{2}$$ (cos 2α + cos 2β) – 1 + cos2 γ
= $$\frac{-1}{2}$$ × 2 cos(α + β) cos(α – β) + cos2 γ
= cos γ cos (α – β) + cos2 γ …..[∵ α + β + γ = π]
= cos γ [cos (α – β) + cos γ]
= cos γ [cos (α – β) – cos (α + β)]
= 2 sin α sin β cos γ
Question 8.
Let 0 < A, B < $$\frac{\pi}{2}$$ satisfying the equation 3sin2 A + 2sin2 B = 1 and 3sin 2A – 2sin 2B = 0, then A + 2B is equal to ________
(a) π
(b) $$\frac{\pi}{2}$$
(c) $$\frac{\pi}{4}$$
(d) 2π
(b) $$\frac{\pi}{2}$$
Hint:
3 sin 2A – 2sin 2B = 0
sin 2B = $$\frac{3}{2}$$ sin 2A …….(i)
3 sin2 A + 2 sin2 B = 1
3 sin2 A = 1 – 2 sin2 B
3 sin2 A = cos 2B ……(ii)
cos(A + 2B) = cos A cos 2B – sin A sin 2B
= cos A (3 sin2 A) – sin A ($$\frac{3}{2}$$ sin 2A) …..[From (i) and (ii)]
= 3 cos A sin2 A – $$\frac{3}{2}$$ (sin A) (2 sin A cos A)
= 3 cos A sin2 A – 3 sin2 A cos A
= 0
= cos $$\frac{\pi}{2}$$
∴ A + 2B = $$\frac{\pi}{2}$$ ……..[∵ 0 < A + 2B < $$\frac{3 \pi}{2}$$]
Question 9.
In ∆ABC if cot A cot B cot C > 0, then the triangle is ________
(a) acute-angled
(b) right-angled
(c) obtuse-angled
(d) isosceles right-angled
(a) acute angled
Hint:
cot A cot B cot C > 0
Case I:
cot A, cot B, cot C > 0
∴ cot A > 0, cot B > 0, cot C > 0
∴ 0 < A < $$\frac{\pi}{2}$$, 0 < B < $$\frac{\pi}{2}$$, 0 < C < $$\frac{\pi}{2}$$
∴ ∆ABC is an acute angled triangle.
Case II:
Two of cot A, cot B, cot C < 0
0 < A, B, C < π and two of cot A, cot B, cot C < 0
∴ Two angles A, B, C are in the 2nd quadrant which is not possible.
Question 10.
The numerical value of tan 20° tan 80° cot 50° is equal to ________
(a) √3
(b) $$\frac{1}{\sqrt{3}}$$
(c) 2√3
(d) $$\frac{1}{2 \sqrt{3}}$$
(a) √3
Hint:
L.H.S. = tan 20° tan 80° cot 50°
= tan 20° tan 80° cot (90° – 40°)
= tan 20° tan 80° tan 40°
= tan 20° tan (60° + 20°) tan (60° – 20°)
= tan 3(20°)
= tan 60°
= √3
= R.H.S.
II. Prove the following.
Question 1.
tan 20° tan 80° cot 50° = √3
Solution:
L.H.S. = tan 20° tan 80° cot 50°
= tan 20° tan 80° cot (90° – 40°)
= tan 20° tan 80° tan 40°
= tan 20° tan (60° + 20°) tan (60° – 20°)
= tan 3(20°)
= tan 60°
= √3
= R.H.S.
Question 2.
If sin α sin β – cos α cos β + 1 = 0, then prove that cot α tan β = -1.
Solution:
sin α sin β – cos α cos β + 1 = 0
∴ cos α cos β – sin α sin β = 1
∴ cos (α + β) = 1
∴ α + β = 0 ……[∵ cos 0 = 1]
∴ β = -α
L.H.S. = cot α tan β
= cot α tan(-α)
= -cot α tan α
= -1
= R.H.S.
Question 3.
$$\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}=\frac{1}{16}$$
Solution:
Question 4.
$$\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)=\frac{1}{8}$$
Solution:
Question 5.
cos 12° + cos 84° + cos 156° + cos 132° = $$-\frac{1}{2}$$
Solution:
Question 6.
$$\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)=\sqrt{2} \cos x$$
Solution:
Question 7.
$$\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$$
Solution:
Question 8.
sin2 6x – sin2 4x = sin 2x sin 10x
Solution:
Question 9.
cos2 2x – cos2 6x = sin 4x sin 8x
Solution:
Question 10.
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Solution:
Question 11.
$$\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$$
Solution:
Question 12.
If sin 2A = λ sin 2B, then prove that $$\frac{\tan (A+B)}{\tan (A-B)}=\frac{\lambda+1}{\lambda-1}$$
Solution:
Question 13.
$$\frac{2 \cos 2 A+1}{2 \cos 2 A-1}$$ = tan (60° + A) tan (60° – A)
Solution:
Question 14.
tan A + tan (60° + A) + tan (120° + A) = 3 tan 3A
Solution:
Question 15.
3 tan6 10° – 27 tan4 10° + 33 tan2 10° = 1
Solution:
Question 16.
cosec 48° + cosec 96° + cosec 192° + cosec 384° = 0
Solution:
L.H.S. = cosec 48° + cosec 96° + cosec 192° + cosec 384°
= cosec 48° + cosec (180° – 84°) + cosec (180° + 12°) + cosec (360° + 24°)
= cosec 48° + cosec 84° + cosec (-12°) + cosec 24°
Question 17.
3(sin x – cos x)4 + 6(sin x + cos x)2 + 4(sin6 x + cos6 x) = 13
Solution:
(sin x – cos x)4
= [(sin x – cos x)2]2
= (sin2 x + cos2 x – 2 sin x cos x)2
= (1 – 2 sin x cosx)2
= 1 – 4 sin x cos x + 4 sin2 x cos2 x
(sin x + cos x)2 = sin2 x + cos2 x + 2 sin x cos x = 1 + 2 sin x cos x
sin6 x + cos6 x
= (sin2 x)3 + (cos2 x)3
= (sin2 x + cos2 x)3 – 3 sin2 x cos2 x (sin2 x + cos2 x) …..[∵ a3 + b3 = (a + b)3 – 3ab(a + b)]
= 13 – 3 sin2 x cos2 x (1)
= 1 – 3 sin2 x cos2 x
L.H.S. = 3(sin x – cos x)4 + 6(sin x + cos x)2 + 4(sin6 x + cos6 x)
= 3(1 – 4 sin x cos x + 4 sin2 x cos2 x) + 6(1 + 2 sin x cos x) + 4(1 – 3 sin2 x cos2 x)
= 3 – 12 sin x cos x + 12 sin2 x cos2 x + 6 + 12 sin x cos x + 4 – 12 sin2 x cos2 x
= 13
= R.H.S.
Question 18.
tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A
Solution:
We have to prove that,
tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A
i.e., to prove,
cot A – tan A – 2 tan 2A – 4 tan 4A – 8 cot 8A = 0
∴ cot θ – tan θ = 2 cot 2θ …..(i)
L.H.S. = cot A – tan A – 2 tan 2A – 4 tan 4A – 8 cot 8A
= 2 cot 2A – 2 tan 2A – 4 tan 4A – 8 cot 8A …..[From (i)]
= 2(cot 2A – tan 2A) – 4 tan 4A – 8 cot 8A
= 2 × 2 cot 2(2A) – 4 tan 4A – 8 cot 8A ……[From (i)]
= 4(cot 4A – tan 4A) – 8 cot 8A
= 4 × 2 cot 2(4A) – 8 cot 8A ……[From (i)]
= 8 cot 8A – 8 cot 8A = 0
= R.H.S.
Alternate Method:
Question 19.
If A + B + C = $$\frac{3 \pi}{2}$$, then cos 2A + cos 2B + cos 2C = 1 – 4 sin A sin B sin C
Solution:
Question 20.
In any triangle ABC, sin A – cos B = cos C. Show that ∠B = $$\frac{\pi}{2}$$.
Solution:
sin A – cos B = cos C
∴ sin A = cos B + cos C
A = B – C ………(i)
In ∆ABC,
A + B + C = π
∴ B – C + B + C = π
∴ 2B = π
∴ B = $$\frac{\pi}{2}$$
Question 21.
$$\frac{\tan ^{3} x}{1+\tan ^{2} x}+\frac{\cot ^{3} x}{1+\cot ^{2} x}$$ = sec x cosec x – 2 sin x cos x
Solution:
Question 22.
sin 20° sin 40° sin 80° = $$\frac{\sqrt{3}}{8}$$
Solution:
L.H.S. = sin 20°. sin 40°. sin 80°
= sin 20°. sin 40°. sin 80°
= $$\frac{1}{2}$$ (2 . sin 40°. sin 20°) . sin 80°
= $$\frac{1}{2}$$ [cos(40° – 20°) – cos (40° + 20°)] . sin 80°
= $$\frac{1}{2}$$ (cos 20° – cos 60°) sin 80°
= $$\frac{1}{2}$$ . cos 20° . sin 80° – $$\frac{1}{2}$$ . cos 60° . sin 80°
= $$\frac{1}{2 \times 2}$$ (2 sin 80° . cos 20°) – $$\frac{1}{2 \times 2}$$ . sin 80°
= $$\frac{1}{4}$$ [sin(80° + 20°) + sin (80° – 20°)] – $$\frac{1}{2}$$ . sin 80°
Question 23.
sin 18° = $$\frac{\sqrt{5}-1}{4}$$
Solution:
Let θ = 18°
∴ 5θ = 90°
∴ 2θ + 3θ = 90°
∴ 2θ = 90° – 3θ
∴ sin 2θ = sin (90° – 3θ)
∴ sin 2θ = cos 3θ
∴ 2 sin θ cos θ = 4 cos3 θ – 3 cos θ
∴ 2 sin θ = 4 cos2 θ – 3 …..[∵ cos θ ≠ 0]
∴ 2 sin θ = 4 (1 – sin2 θ) – 3
∴ 2 sin θ = 1 – 4 sin2 θ
∴ 4 sin2 θ + 2 sin θ – 1 = 0
∴ sin θ = $$\frac{-2 \pm \sqrt{4+16}}{8}$$
= $$\frac{-2 \pm 2 \sqrt{5}}{8}$$
= $$\frac{-1 \pm \sqrt{5}}{4}$$
Since, sin 18° > 0
∴ sin 18°= $$\frac{\sqrt{5}-1}{4}$$
Question 24.
cos 36° = $$\frac{\sqrt{5}+1}{4}$$
Solution:
We know that,
cos 2θ = 1 – 2 sin2 θ
cos 36° = cos 2(18°)
= 1 – 2 sin2 18°
∴ cos 36° = $$\frac{\sqrt{5}+1}{4}$$
Question 25.
sin 36° = $$\frac{\sqrt{10-2 \sqrt{5}}}{4}$$
Solution:
We know that, sin2 θ = 1 – cos2 θ
sin2 36° = 1 – cos2 36°
= 1 – $$\left(\frac{\sqrt{5}+1}{4}\right)^{2}$$
= $$\frac{16-(5+1+2 \sqrt{5})}{16}$$
= $$\frac{10-2 \sqrt{5}}{16}$$
∴ sin 36° = $$\frac{\sqrt{10-2 \sqrt{5}}}{4}$$ ……[∵ sin 36° is positive]
Question 26.
$$\sin \frac{\pi^{c}}{8}=\frac{1}{2} \sqrt{2-\sqrt{2}}$$
Solution:
Question 27.
tan $$\frac{\pi}{8}$$ = √2 – 1
Solution:
Question 28.
tan 6° tan 42° tan 66° tan 78° = 1
Solution:
Question 29.
sin 47° + sin 61° – sin 11° – sin 25° = cos 7°
Solution:
Question 30.
√3 cosec 20° – sec 20° = 4
Solution:
Question 31.
In ∆ABC, ∠C = $$\frac{2 \pi}{3}$$, then prove that cos2 A + cos2 B – cos A cos B = $$\frac{3}{4}$$.
Solution:
|
# What are the properties of rigid transformations?
## What are the properties of rigid transformations?
There are three basic rigid transformations: reflections, rotations, and translations. Reflections, like the name suggests, reflect the shape across a line which is given. Rotations rotate a shape around a center point which is given, and translations slide or move a shape from one place to another.
## What is a rotation in rigid transformation?
A rotation is a transformation that turns a figure on the coordinate plane a certain number of degrees about a given point without changing the shape or size of the figure. Rigid Transformation.
## Why is a rotation a rigid transformation?
A rotation is called a rigid transformation or isometry because the image is the same size and shape as the pre-image. An object and its rotation are the same shape and size, but the figures may be positioned differently. defined by the center of the rotation and the angle of rotation.
## What is special about rigid motion transformations?
Rigid motion is otherwise known as a rigid transformation and occurs when a point or object is moved, but the size and shape remain the same. This differs from non-rigid motion, like a dilation, where the size of the object can increase or decrease.
Read also : How can you tell if a linear relationship is proportional or not?
## Which properties that are always preserved in rigid transformations?
Math Bits.com B’V(4.4) Definition: A rigid transformation (also called an isometry) is a transformation of the plane that preserves length. Reflections, translations, rotations, and combinations of these three transformations are rigid transformations.
## How would you describe a rigid transformation?
A rigid transformation is a transformation that doesn’t change measurements on any figure. With a rigid transformation, figures like polygons have corresponding sides of the same length and corresponding angles of the same measure. The result of any transformation is called the image.
## What are transformation properties?
The transform property applies a 2D or 3D transformation to an element. This property allows you to rotate, scale, move, skew, etc., elements. To better understand the transform property, view a demo.
## What are the 3 basic rigid transformations?
There are three different types of transformations: translation, reflection, and rotation.
## What is a rotation in transformation?
A rotation is a type of transformation that takes each point in a figure and rotates it a certain number of degrees around a given point. The result of a rotation is a new figure, called the image. The image is congruent to the original figure.
## How do you rotate a rigid transformation?
Rigid motion includes translations, rotations, and reflections. Translation is a type of rigid motion that occurs when the object simply slides and maintains its direction. Rotations are movements around a central point where distance from that point is maintained.
## Why is rotation a rigid motion?
A rotation is a transformation that turns the figure in either a clockwise or counterclockwise direction. You can turn a figure 90 a quarter turn, either clockwise or counterclockwise. When you spin the figure exactly halfway, you have rotated it 180xb0. Turning it all the way around rotates the figure 360xb0.
Read also : Is FSA Cancelled 2021?
## Is a rotation a rigid transformation?
The rigid transformations include rotations, translations, reflections, or any sequence of these. Any object will keep the same shape and size after a proper rigid transformation. All rigid transformations are examples of affine transformations.
## Why is a rotation a rigid motion?
Rigid motion includes translations, rotations, and reflections. Translation is a type of rigid motion that occurs when the object simply slides and maintains its direction. Rotations are movements around a central point where distance from that point is maintained.
## What makes a transformation rigid?
Rigid just means that the whole shape goes through the same transformation, so with rotations, reflections, and translations, the shape should not change at all, just in a different place or orientation.
## Are all rotations considered a rigid motion?
Rigid motion changes the location of a shape, or the direction it is facing, but does not change the size or shape of it. The three basic rigid motions are translation, reflection, and rotation
## Why are rigid motions important?
An isometry is a transformation that preserves the distances between the vertices of a shape. A rigid motion does not affect the overall shape of an object but moves an object from a starting location to an ending location. The resultant figure is congruent to the original figure.
## What are the properties of rigid motions?
There are four types of rigid motions that we will consider: translation , rotation, reflection, and glide reflection.
• Translation: In a translation, everything is moved by the same amount and in the same direction.
• Rotation:
• Reflection:
• Glide Reflection:
## Why are transformations called rigid motions?
A rigid transformation is a transformation that doesn’t change measurements on any figure. With a rigid transformation, figures like polygons have corresponding sides of the same length and corresponding angles of the same measure. The result of any transformation is called the image.
## What is preserved in a rigid transformation?
Rigid motion A transformation that preserves distance and angle measure (the shapes are congruent, angles are congruent). Isometry A transformation that preserves distance (the shapes are congruent).
Read also : What is the name for Mg3 PO3 2?
## What are the 5 properties that are preserved during a rigid motion?
Reflections, rotations, and translations are all rigid motions. So, they all preserve distance, angle measure, betweenness, and collinearity. The above properties ensure that if a figure is determined by certain points, then its image after a rigid motion is also determined by those points.
## What 3 things are preserved by a rigid motion?
Fundamental are the rigid motions: translations, rotations, reflections, and combinations of these, all of which are here assumed to preserve distance and angles (and therefore shapes generally).
## What stays the same in rigid transformations?
In any transformation, the size and shape of the figure stay exactly the same, only its location changes or shifts. The first type of transformation is called a translation. It is also known as a slide because the figure in question does exactly that.
## What describes a rigid motion transformation?
Rigid motion is otherwise known as a rigid transformation and occurs when a point or object is moved, but the size and shape remain the same. This differs from non-rigid motion, like a dilation, where the size of the object can increase or decrease.
## How do you identify a rigid transformation?
Rigid just means that the whole shape goes through the same transformation, so with rotations, reflections, and translations, the shape should not change at all, just in a different place or orientation.
## What is one example of a rigid transformation?
Rotation Rigid Transformation Examples A rotation is a rigid transformation that turns the object about some point called its center. The shape retains its orientation, but its direction is different. A shape can be rotated by any angle. The triangle is rotated about point C by 180 degrees.
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## MAP Recommended Practice
### Course: MAP Recommended Practice>Unit 17
Lesson 10: Exponents
# Squaring numbers
Learn how to square basic numbers and make a connection between squaring and the area of a square.
Let's start by taking a look at an example evaluating $3$ squared:
${3}^{2}=3×3=9$
Here's a question to make sure you got that:
What does it mean to square a number?
## Connection to a square
Finding ${3}^{2}$ (read as "three squared") is the same as finding the area of a square with side length $3$:
## Practice set:
Problem 1A
${1}^{2}=$
## Challenge set:
Problem 2A
${a}^{2}=$
## Want to join the conversation?
• what is 5 to the third power
• well, to the power means the number, times itself, however many times the power is. for example,
2 to the fourth power would be 16. because 2 times two is four, that times two is eight, eight times two is sixteen. 2*2*2*2. so five to the three would be 5*5*5. I bet you can find the answer from there.
• Can We Negative Square Num.? If yes, How?
• Yes, we can use in imaginary numbers, where (-1)²=i , I think you're gonna see it at the pre-cauculus subject.
• Good Luck guys I am hope you do well
• thanks but I'm just here to answer peoples questions
• what if 6x6x6x9=?
• 6 x 6 x 6 x 9 = 6^3 x 9 = 1944
• What is imaginary numbers?
• Good question! In the real number system, it is not possible to find the square root of a negative number, since no real number times itself is negative. This is because 0 times itself is 0, a negative number times itself is positive, and a positive number times itself is positive.
To get around this problem, mathematicians invented the imaginary unit i, defined as the square root of -1. This produces an extended number system called the complex numbers, defined as numbers in the form a+bi where a and b are real numbers. For example, 8, 3 - 2i, and 5i are complex numbers. Also, 5i is an example of what is called a pure imaginary number, because 5i is just a real number times i.
Have a blessed, wonderful day!
• 5^3 = 125
5 x 5 x 5 = 125
• What are some tips to be a great 6th grader since it is the first time in middle school?
• Like @The Math Wiz said
Review what you learned in 6th grade.
Always Get enough sleep before school.
Don't be afraid to ask questions
6th grade might seem scary but you'll soon get the hang of it
In middle school there might be a lot of presenting it might seem scary but just remember to speak clear and look confident!
Stay out of unnecessary drama it's not worth it
You don't need to have 1 million friends to
enjoy school sometimes bigger isin't always better have fun with your friends :)
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# Fractions, decimals and percentages
## What are fractions, decimals and percentages?
Fractions, decimals and percentages all represent parts of a whole.
## What is a fraction?
A fraction is a way of representing a part, or parts, of a whole.
The numerator shows us how many parts we have.
The denominator shows us how many parts there are in total.
## What is a decimal?
A decimal is a number whose whole part and fractional part are separated by a decimal point. It is a way of writing a number that is not a whole number, but is inbetween two whole numbers.
In the above example, the number is bigger than the whole number 3, but smaller than the whole number 4.
The '3' represents 3 ones. The '6' represents 6 tenths.
## What is a percentage?
A percentage of a number or object is how many parts of it there are out of 100 equal parts.
'Percent' means 'per 100'.
In the above example, if you split an object into 100 parts, 75 of the parts make 75%.
## Converting fractions to decimals
To convert a fraction into a decimal, divide the numerator by the denominator.
The simplest method is to use a calculator, but long division can also be used.
## Converting decimals to fractions
To convert a decimal into a fraction, first you need to write the decimal divided by 1, as shown below:
Then, multiply the top and the bottom by 10 for every number after the decimal point (e.g. if there are two numbers after the decimal point, multiply by 100; if there are three numbers after the decimal point, multiply by 1000).
This fraction can then be simplified if possible:
## Converting decimals to percentages
To convert a decimal to a percentage, multiply the decimal by 100:
## Converting percentages to decimals
To convert a percentage to a decimal, divide the percentage by 100:
## Converting fractions to percentages
To convert a fraction into a percentage, first turn it into a decimal by dividing the numerator by the denominator:
Then multiply this decimal by 100:
## Converting percentages to fractions
To convert a percentage into a fraction, put the percentage over 100, and simplify if possible:
## Fractions KS1
In the National Curriculum for KS1, the focus is on fractions.
In Year 1, pupils should be taught to:
• recognise, find and name a half as one of two equal parts of an object, shape or quantity
• recognise, find and name a quarter as one of four equal parts of an object, shape or quantity.
In Year 2, pupils should be taught to:
• recognise, find, name and write fractions 1/3 , 1/4 , 2/4 and 3/4 of a length, shape, set of objects or quantity
• write simple fractions for example, 1/2 of 6 = 3 and recognise the equivalence of 2/4 and 1/2 .
## Fractions, Decimals and Percentages KS2
In the National Curriculum for KS2, children continue to learn about fractions, and are introduced to decimals in Year 4, and percentages in Year 5.
In Year 3, pupils should be taught to:
• count up and down in tenths; recognise that tenths arise from dividing an object into 10 equal parts and in dividing one-digit numbers or quantities by 10
• recognise, find and write fractions of a discrete set of objects: unit fractions and nonunit fractions with small denominators
• recognise and use fractions as numbers: unit fractions and non-unit fractions with small denominators
• recognise and show, using diagrams, equivalent fractions with small denominators
• add and subtract fractions with the same denominator within one whole [for example, 5/7 + 1/7 = 6/7 ]
• compare and order unit fractions, and fractions with the same denominators
• solve problems that involve all of the above.
In Year 4, pupils should be taught to:
• recognise and show, using diagrams, families of common equivalent fractions
• count up and down in hundredths; recognise that hundredths arise when dividing an object by one hundred and dividing tenths by ten.
• solve problems involving increasingly harder fractions to calculate quantities, and fractions to divide quantities, including non-unit fractions where the answer is a whole number
• add and subtract fractions with the same denominator
• recognise and write decimal equivalents of any number of tenths or hundredths
• recognise and write decimal equivalents to 1/4 , 1/2 , 3/4
• find the effect of dividing a one- or two-digit number by 10 and 100, identifying the value of the digits in the answer as ones, tenths and hundredths
• round decimals with one decimal place to the nearest whole number
• compare numbers with the same number of decimal places up to two decimal places
• solve simple measure and money problems involving fractions and decimals to two decimal places.
## Fractions, Decimals and Percentages Year 5
In Year 5, pupils should be taught to:
• compare and order fractions whose denominators are all multiples of the same number
• identify, name and write equivalent fractions of a given fraction, represented visually, including tenths and hundredths
• recognise mixed numbers and improper fractions and convert from one form to the other and write mathematical statements > 1 as a mixed number [for example, 2/5 + 4/5 = 6/5 = 1 1/5 ]
• add and subtract fractions with the same denominator and denominators that are multiples of the same number
• multiply proper fractions and mixed numbers by whole numbers, supported by materials and diagrams
• read and write decimal numbers as fractions [for example, 0.71 = 71/100 ]
• recognise and use thousandths and relate them to tenths, hundredths and decimal equivalents
• round decimals with two decimal places to the nearest whole number and to one decimal place
• read, write, order and compare numbers with up to three decimal places
• solve problems involving number up to three decimal places
• recognise the per cent symbol (%) and understand that per cent relates to ‘number of parts per hundred’, and write percentages as a fraction with denominator 100, and as a decimal
• solve problems which require knowing percentage and decimal equivalents of 1/2 , 1/4 , 1/5 , 2/5 , 4/5 and those fractions with a denominator of a multiple of 10 or 25.
## Fractions, Decimals and Percentages Year 6
In Year 6, pupils should be taught to:
• use common factors to simplify fractions; use common multiples to express fractions in the same denomination
• compare and order fractions, including fractions > 1
• add and subtract fractions with different denominators and mixed numbers, using the concept of equivalent fractions
• multiply simple pairs of proper fractions, writing the answer in its simplest form [for example, 1/4 × 1/2 = 1/8 ]
• divide proper fractions by whole numbers [for example, 1/3 ÷ 2 = 1/6 ]
• associate a fraction with division and calculate decimal fraction equivalents [for example, 0.375] for a simple fraction [for example, 3/8 ]
• identify the value of each digit in numbers given to three decimal places and multiply and divide numbers by 10, 100 and 1000 giving answers up to three decimal places
• multiply one-digit numbers with up to two decimal places by whole numbers
• use written division methods in cases where the answer has up to two decimal places
• solve problems which require answers to be rounded to specified degrees of accuracy
• recall and use equivalences between simple fractions, decimals and percentages, including in different contexts.
|
# Factorization using Identities
Factorization using Identities :There are some identities and using that the factorization is much easier.
A number of expressions to be factorized are of the form or can be put into the form : a
2 + 2ab + b 2 , a 2 – 2ab + b 2 , a 2 – b 2 and x 2 + (a + b) + ab. These expressions can be easily factorized using Identities I, II, III and IV
In this section we will learn Factorization using Identities one by one.
1) a2 + 2ab + b2 = (a + b)2
2) a2 – 2ab + b2 = (a – b)2
3) a2 – b2 = (a + b) (a – b)
4) x2 + (a + b) x + ab = (x + a) (x + b)
_______________________________________________________________
Factorization using Identities :
In the 1st identity, a
2 + 2ab + b 2 = (a + b) 2 ,
1st and the last term should be perfect square and the middle term is two times the square root of 1st and the last term and the sign of the middle term is positive.
Examples on 1st Identity of Factorization :
1) 9a
2 + 12ab + 4b 2
Solution :
9a
2 + 12ab + 4b 2
= (3a)
2 + 2 . (3a) . (2b) + (2b) 2
= (3a + 2b)
2 [ since a = 3a and b = 2b; a 2 + 2ab + b 2 = (a + b) 2 ]
_______________________________________________________________
2) x
4 + 2 + 1/x 4
Solution :
x
4 + 2 + 1/x 4
= (x
2 ) 2 + 2.(x 2 ).1/x 2 + (1/x 2 ) 2
= (x
2 + 1/x 2 ) 2 [ since a = x 2 and b = 1/x 2 ]
_______________________________________________________________
In the 2nd identity, a
2 - 2ab + b 2 = (a - b) 2 ,
1st and the last term should be perfect square and the middle term is two times the square root of 1st and the last term and the sign of the middle term is negative.
Examples on 2nd Identity of Factorization :
1) 4p
2 - 20pq + 25q 2
Solution :
4p
2 - 20pq + 25q 2
= (2p)
2 - 2 . (2p) . (5q) + (5q) 2
= (2p - 5q)
2 [ since a = 2p and b = 5q; a 2 + 2ab - b 2 = (a - b) 2 ]
_______________________________________________________________
2)1 - 16x
2 + 64x 4
Solution :
= (1)
2 - 2 . (2) . (8x 2 ) + (8x 2 ) 2
= (1 - 8x
2 ) 2 [ since a = 1 and b = 8x 2 ; a 2 - 2ab + b 2 = (a - b) 2 ]
_______________________________________________________________
Some quadratic polynomials will be missing the middle term. Often these polynomials are the difference of two squares.
These polynomials come from multiplying the sum and difference of binomials, such as (a+b)(a-b)= a2-b2 when simplified.
Examples on 3rd Identity of Factorization :
1) 16x
2 - 9y 2
Solution :
16x
2 - 9y 2
= (4x)x
2 - (3y)x 2
= (4x + 3y)(4x - 3y)[ since a = 4x and b = 3y; a
2 - b 2 = (a + b)(a - b)]
_______________________________________________________________
2) x
4 - x 4 y 4
Solution :
x
4 - x 4 y 4
= (x
2 )x 2 - [(xy) 2 ]x 2
= (x
2 + x 2 y 2 )(x 2 - x 2 y 2 )[ since a = x 2 and b = (xy) 2 ; a 2 - b 2 = (a + b)(a - b)]
In the 2nd parenthesis(bracket), we can apply the 3rd identity of Factorization again
= (x
2 + x 2 y 2 )(x + xy)(x - xy)[ since a = x and b = xy;
a
2 - b 2 = (a + b)(a - b)]
_______________________________________________________________
Consider x
2 + 5x + 6, Observe that this expressions are not of the type
(a + b)
2 or (a – b) 2 , i.e., they are not perfect squares.
For example, in x
2 2 + 5x + 6, the term 6 is not a perfect square. This expressions obviously also do not fit the type (a 2 – b 2 ) either. They, however, seem to be of the type
x
2 + (a + b) x + a b. We may therefore, try to use Identity 4.
Factoring
Factorization by common factor
Factorization by Grouping
Factorization using Identities
Factorization of Cubic Polynomial
Solved Examples on Factorization
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Finding the Factors for 37
Factors of 37: Preface
Last updated date: 16th Mar 2023
Total views: 55.2k
Views today: 1.34k
In this article, we will learn about the factors of 37, how to get the factors of the number 37, and the prime factors of 37, along with examples. A factor of 37 divides it evenly. These factors cannot have a fractional or decimal value. As a result, the factors are 1 and 37. When two integers are multiplied together, a factor pair of 37 is the number that provides the result of 37.
Factors of a Number: A Brief Explanation
When the original number is divided by a factor, the result is an even number and the remainder is zero. Every number greater than one has a minimum of two factors. The two factors are known as factor products. The product of every factor pair corresponds to the number. The first factor pair is always one and counts itself.
Examples:
• The factors of 39 are 1, 3, 13, and 39
• The factors of 69 are 1, 3, 23, and 69
What is the Factor of 37?
Factors of 37 are the numbers that divide 37 exactly with no remainder. When you multiply two whole numbers together and obtain 37 as a result, you may claim that both numbers are factors of 37. Hence, all the factors of 37 are 1 and 37.
How to Calculate Factors for 37?
The factors of 37 are the numbers that divide 37 exactly with no remainder. 37 is an odd number, so it cannot be divided by any even integer. Determine the pairs of numbers whose product gives 37. We are aware that the exact divisors of the number 37 are its factors.
Here are all the factors for 37:
$37\div 1=37$
1 is a factor of 37.
$37\div 37=1$
37 is a factor of 37.
Thus, all the factors of 37 are 1 and 37.
37 is a Prime Number?
Prime factorization of 37 refers to the method of finding the prime factors of 37. We know 37 is a prime number, so we can't factor it any further, because prime numbers have no factors other than 1 and the number itself. As a result, there are no other prime factors than 37. Hence, the prime factor of 37 is $1\times 37$, where 37 is the prime number. A factor tree of 37 is given below:
Factor Tree of 37
Pair Factors of37
To get the pair factors of 37, multiply the two integers in a pair so that the result is 37. Pair factors may be positive or negative, but they cannot be a fraction or a decimal.
Positive Factors of 37
Factor Pair of 37
$1\times 37$
$(1,37)$
Thus, the positive pair factor is $(1,37)$.
Negative pair factors are also possible since the product of two negative numbers yields a positive number.
Consider the negative pair factors.
Negative Factors of 37
Factor pair of 37
$-1\times -37$
$(-1,-37)$
Thus, the negative pair factor is $(-1,-37)$
Interesting Facts
• 1 is not a prime number; thus, it will not appear in any factor tree.
• 1 is the factor for each number, as one times a number is the number itself. Again, anything divided by 1 is the number itself
Solved Important Questions
1. The five members of a club sold tickets for their concert. Each one sold 37 tickets. How many tickets did they sell?
Solution:
The total number of members is six.
The number of tickets sold per person $=37$
The total number of tickets sold $=37\times 5=185$.
2. Draw a factor tree of 12.
Ans: Factor trees represent the factors of a number, especially its prime factorization. Each tree branch is divided into factors. The end of the branch in the factor tree must be a prime number, the only two factors are itself and one, so the branch stops.
The number 12 is composite.
Thus, factors 12 are 1, 2, 3, 4, 6, and 12.
Prime factors of $12=2\times 2\times 3$. See its prime factor tree below.
Factor Tree of 12
3. Jungkook is having a cocktail party. He invited 37 friends and prepared 15 banana milk, ten blue lagoons, and 12 wines for his friends. How many drinks did each friend get?
Solution:
The total number of drinks prepared by Jungkook $=15+10+12=37$
The number of friends he invited is 37.
The number of drinks per person $=37\div 37=1$
Thus, each friend gets one drink.
Conclusion
The numbers that divide 37 exactly without remainder are known as the factors of 37. Hence, factors 37 are 1 and 37. The prime factorization of 37 is $1\times 37$, where 37 is the prime number.
Practice Questions
1. 169 bikes are to be arranged equally in 13 rows. Then find the number of bikes in each row?
1. 16 bikes
2. 69 bikes
3. 13 bikes
4. 29 bikes
2. Choose the incorrect?
1. The factors of 65 are 1, 5, 13, and 65
2. The factors of 39 are 1, 3, 13, 23, and 39
3. The factors of 29 are 1 and 29
4. The factors of 12 are 1, 2, 3, 4, 6, and 12
FAQs on Finding the Factors for 37
1. What is the factor of 53?
Factors of 53 are the numbers that divide 53 exactly with no remainder. When you multiply two whole numbers together and obtain 53 as a result, you may claim that both numbers are factors of 53. Hence, factors 53 are 1 and 53.
2. Find out the common factors of 37 and 41.
The numbers that divide the number exactly with no remainder are known as the factors of a number. So, the factors 37 and 41 are 1, 37, and 1, 41 respectively.
Hence, the common factor of 37 and 41 is 1.
3. Write the prime factorization of 729.
Prime factors are prime numbers that can equally divide the original integer. To get the prime factors of 729, divide 729 by prime numbers until the quotient is 1.
Step 1: Divide 729 by 3
$729\text{ }\div \text{ }3\text{ }=\text{ }243$
Step 2: Again divide 243 by 3
$243\text{ }\div \text{ }3\text{ }=\text{ }81$
Step 3: Divide 81 by 3
$81\text{ }\div \text{ }3\text{ }=\text{ }27$
Step 4: Continue with the prime number, i.e., 3
$27\text{ }\div \text{ }3\text{ }=\text{ }9$
Step 5: Again, divide 9 by 3
$9\text{ }\div \text{ }3\text{ }=\text{ }3$
Step 6: Divide 3 by 3
$3\text{ }\div \text{ }3\text{ }=\text{ }1$
Thus, we get 1 at the end of this division method, and we can’t go on with the division approach. Hence, the prime factorization of 729 is $3\times 3\times 3\times 3\times 3\times 3$ or ${{3}^{6}}$ where 3 is a prime number.
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