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# Mathematics - Permutation (Ordered Combination)
A permutation is a selection of elements from a set where the order of selection matters.
Order matters means that the selections AB and BA are considered as two differents selections.
If the order does not matter such as in a lottery, the selections AB and BA would be considered as a the same selection and is known as combination
## Example
• with the set {1,2,3}
• with a selection without_repetition (where the elements are deleted from the set when selected)
• the possible permutations are:
• 1 2 3
• 1 3 2
• 2 1 3
• 2 3 1
• 3 1 2
• 3 2 1
If the order would have not matter, all this selections would have been seen as a single combination 1 2 3 (because the order in the selections does not matter)
## Type
There are basically two types of permutation:
• With Repetition is Allowed (ie AA): lock, pin, surrogate identifier
• Without Repetition (ie the pair AA is not allowed): the first three people in a running race (a participant can't be first and second), all queue problems.
### With repetition
• When a set has n different elements,
• For each choice, there is n choice,
For a set of length n and a selection of length r (known as k for combination), the number of permutation $_nP_r \text{ (n permute r)}$ is: $$_nP_r = n_1 \times n_2 \dots \times n_r = n^r$$
### Without repetition
• When a set has n different elements,
• For the choice i, there is n-i+1 choice,
For a set of length n and a selection of length k, the number of permutation is: $$_nP_k = \frac{n!}{(n-k)!}$$
The division (n-k)! is a trick to be able to generalize the calculation with factorial
Example: if for a set of 10 element, with a selection of 3 elements, we can choose for:
• the first choice: 10 elements (we delete the selected element of the set)
• the second choice: 9 elements (we delete the selected element of the set)
• the third choice: 8
In total, we have: $$10 * 9 * 8 \text{ permutations}$$
To generalize with the factorial, we add the rest of the factorial serie in the dividend and divisor:
$$10 \times 9 \times 8 \times \frac{ 7 \times 5 \times 4 \times 3 \times 2 \times 1}{7 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{10!}{(10-3)!} = \frac{n!}{(n-k)!}$$
## Documentation / Reference
Discover More
Collisions of Hash or Identifier Generation
A collision happens when a function produces the same result while it should be unique. In a hash function, it happens when two different inputs produces the same hash In a identifier generator, it...
Mathematics - Combination (Binomial coefficient|n choose k)
A combination is a selection of elements from a set where the order of selection does not matter. Order doesn't matter means that the selections AB and BA are considered a single combination (a single...
Mathematics - Exponentiation (square, cube) - Power
Exponentiation is a binary operation involving two numbers: the base (b) the exponent (n) (or index or power). In text notation or computer language, generally the exponentiation operator... |
Question
# The marked price of two articles together is Rs.6.000. The sales tax on article A is 8% and that on article B is 10%. If on selling both the articles, the total sales tax collected is Rs.552, find the sum of the marked price of each of the articles and B.
Hint: Assume a marked price of A and according to solve the question by finding the percentage of sales tax collected. It is given in the question that the total sales tax is 552.
To find the percentage the formula is: $\dfrac{\text{given}\;\text{amount}}{\text{total}\;\text{amount}}\times 100$
And to find the amount from a given percentage the formula is: $\dfrac{\text{amount}\;\text{in}\;\text{percentage}(a{\scriptstyle{}^{0}/{}_{0}})\times \text{total}\;\text{amount}\;\text{of}\;\text{that}\;\text{percentage}}{100}$
Complete step by step solution: Given that,
The total marked price of 2 articles together is 6000,
Sales tax on A is 8%
Sales tax on B is 10%
The total sales tax on selling both the articles = 552
then, we have to calculate the marked price of each article.
Therefore, we will assume the marked price as some variable.
Let the marked price of A= Rs x.
and the marked price of B = Rs. (6000 − x)
Sales tax on A = 8% of x
= Rs $\left( \dfrac{\dfrac{2}{{{8}}}}{\dfrac{{1}{0}{0}}{25}}\times x \right)=\dfrac{2x}{25}$
Sales tax on B = 10% of (6000- x )
$=\dfrac{6000-x}{100}\times 1{0}$
= Rs $\left( \dfrac{6000-x}{10} \right)$
Now
According to the question
$\dfrac{2x}{5}+\dfrac{6000-x}{10}=552$ ①
[The Sum of the sales tax = 552 (given)]
Solving $eq$ ⑩
$\Rightarrow \dfrac{8x+60000-10x}{100}=552$
$=60000-2x=55200$
$\Rightarrow 2x=4800$
$\Rightarrow x=2400$
Therefore the marked price of A = Rs 2400
and the marked price of B = Rs (6000-x)
= Rs (6000-2400 )
= Rs 3600
Note: In this type of question do not assume 2 variables. With one variable we can solve this question. Here we were dealing with tax which is paid annually. It’s important to understand the rate of interest. If we are dealing with interest then it can be monthly also. So, while solving such a question please keep our eyes on this. |
# Irrational Exponents
Irrational Exponents
Before we begin this lesson on irrational exponents, let's have a quick recap about the concept of exponents. Did you know that the first modern use of the word exponent was noticed in "Arithemetica Integra," written by an English author and mathematician Micheal Stifel in 1544. This concept has been in use for years and is still considered very important today.
We will begin this lesson with an overview of irrational exponents and then proceed to solving them. Let's begin!
## Lesson Plan
1 What do you Mean by Irrational Exponents? 2 Tips and Tricks 3 Important Notes on Irrational Exponents 4 Solved Examples on Irrational Exponents 5 Interactive Questions on Irrational Exponents
## What do you Mean by Irrational Exponents?
Before we learn about irrational exponents, let's go through the topic of exponents one more time.
Exponents show the number of times a number is replicated in multiplication.
For example, $$4^2 = 4 \times 4 = 16$$
Here, the exponent 2 is a whole number.
Irrational exponent is given as the exponent which is an irrational number and it cannot be expressed in $$\frac{p}{q}$$ form.
Look at the figure shown at the beginning of this page to understand how irrational exponents are represented.
For the irrational exponent $$2^\sqrt{2}$$ as $$\sqrt{2} \approx 1.414$$, then $$2^\sqrt{2} \approx 2^1.414$$
The decimal number 1.414 can be written as, $$\frac{1414}{1000}$$, hence $$2^\sqrt{2} = 2^{\frac{1414}{1000}}$$
This shows, we can raise 2 to the power 1414 and take the 1000th root of the resulting value.
This is just an approximation, and not the actual value of our exponential term. To that, one could reply: alright, then lets calculate it more accurately, by taking a better approximation of $$\sqrt{2} \approx 1.41421$$
So, $$2^\sqrt{2} = 2^{\frac{141421}{100000}}$$
Once again, we know how to interpret and calculate the right hand side term. If you were to now say: this is still an approximation, then one could in turn say: we can take an even better approximation of square root of 2 and find the value of this term even more accurately. And we could keep doing this and come closer and closer to the actual value of this term.
## How do you Simplify Irrational Exponents?
We can simplify irrational exponents using the law of exponents.
Hence the irrational exponents rules are same as the law of exponents.
They are:
• Law of Product
• Law of Quotient
• Law of Zero Exponent
• Law of Negative Exponent
• Law of Power of a Power
• Law of Power of a Product
• Law of Power of a Quotient
### Examples
Example 1: Solve $$3^\sqrt{2} \times 3^\sqrt{2}$$
Solution: As per the law of product we get,
$$3^{\sqrt{2}+\sqrt{2}} = 3^{2\sqrt{2}} = (3^2)^\sqrt{2} = (9)^\sqrt{2}$$
$$\therefore$$ $$3^\sqrt{2} \times 3^\sqrt{2} = (9)^\sqrt{2}$$
Example 2: Solve $$(4^\sqrt{2})^3$$
Solution: As per the law of power of a power we get,
$$(4^\sqrt{2})^3 = (4^3)^\sqrt{2} = (64)^\sqrt{2}$$
$$\therefore$$ $$(4^\sqrt{2})^3 = 64^\sqrt{2}$$
Example 3: Solve $$\dfrac{7^{\sqrt{3}+2}}{49}$$
Solution: 49 can be written as $$7^2$$
Hence, on substitution of the value we get,
$$\dfrac{7^{\sqrt{3}+2}}{7^2}$$
The numerator can be written as,
$$7^{\sqrt{3}+2} = 7^{\sqrt{3}} \times 7^2$$
The irrational exponent will hence be written as,
$$\dfrac{7^{\sqrt{3}} \times 7^2}{7^2} = 7^{\sqrt{3}}$$
$$\therefore$$ $$\dfrac{7^{\sqrt{3}+2}}{49} = 7^{\sqrt{3}}$$
Tips and Tricks
• The exponential term x raised to the power $$\sqrt{y}$$ is mathematically well-defined. To calculate its value, we could take better and better rational approximations of $$\sqrt{y}$$ for the exponent, and come closer and closer to the actual value of this exponential term.
• An exponent can be an arbitrary real number hence, no matter whether exponent is an integer, a non-integer rational number, or an irrational number it is possible to interpret and calculate that term.
## What Is the Difference Between Rational and Irrational Exponents?
The differences between rational and irrational exponents are,
• Rational exponents can be expressed in $$\dfrac{p}{q}$$ form while irrational exponents cannot be expressed in $$\dfrac{p}{q}$$ form.
• Irrational exponents are non repeating or infinite decimals while rational exponents are rational numbers.
• The value of an irrational exponent when calculated is approximate in nature while the value of rational exponent is exact.
Important Notes
• Irrational exponents follow all the laws of exponents which are given as:
1. $$a^m \times a^n = a^{m+n}$$
2. $$a^m \div a^n = a^{m-n}$$
3. $$a^0 = 1$$
4. $$a^(-m) = \frac{1}{a^{m}}$$
5. $$(a^m)^n = a^{mn}$$
6. $$(ab)^m = a^{m}b^{m}$$
7. $$(\frac{a}{b})^m = \frac{a^m}{b^m}$$
## Solved Examples
Example 1
Solve the below given question.
$$(3^\sqrt{2})^\sqrt{2}$$
Solution
As per the Law of Power of a Product, we get $$(a^m)^n = a^{mn}$$.
Applying the Law of Power of a Product we get,
$$(3^\sqrt{2})^\sqrt{2} = 3^{\sqrt{2} \times \sqrt{2}}$$
As $$(\sqrt{2})^2 = 2$$
Hence, the question is written as,
$$3^{\sqrt{2} \times \sqrt{2}} = 3^2 = 9$$
$$\therefore$$ Answer is 9
Example 2
Help Nicolas in simplifying the irrational exponents given below.
$$\dfrac{3^{\sqrt{2}+2}}{3^{\sqrt{2}-2}}$$
Solution
Nicolas will apply the Law of Quotient for the simplification of exponents.
He will write the exponents as,
$$\dfrac{3^{\sqrt{2}+2}}{3^{\sqrt{2}-2}} = (3)^{(\sqrt{2}+2)-(\sqrt{2}-2)}$$
As, the negative sign reverses all the signs within the brackets,
Hence, the exponents will now become,
$$(3)^{\sqrt{2}+2-\sqrt{2}+2} = (3^4) = 81$$
$$\therefore$$ The answer is 81
## Interactive Questions
Here are a few activities for you to practice.
## Let's Summarize
We hope you enjoyed learning about irrational exponents and irrational exponent definition with the simulations and practice questions. Now, you will be able to easily solve problems on simplifying irrational exponents, multiplying irrational exponents, rational and irrational exponents by using irrational exponents calculator and irrational exponent rules.
The mini-lesson targeted the fascinating concept of irrational exponents. The math journey around polynomial expressions starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.
At Cuemath, our team of math experts are dedicated to make learning fun for our favorite readers, the students! Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic. Be it problems, online classes, videos, or any other form of relation, it’s the logical thinking and smart learning approach that we at Cuemath believe in.
## FAQs on Irrational Exponents
### 1. What is the exponent of a power?
The number which raised to the base of a power is known as exponent of a power.
For example, for in the power $$5^8$$, 8 is the exponent.
### 2. What are 7 rules of exponents?
There are 7 rules of exponents,
• Law of Product, $$a^m \times a^n = a^{m+n}$$
• Law of Quotient, $$a^m \div a^n = a^{m-n}$$
• Law of Zero Exponent, $$a^0 = 1$$
• Law of Negative Exponent, $$a^(-m) = \frac{1}{a^{m}}$$
• Law of Power of a Power, $$(a^x)^y = a^{xy}$$
• Law of Power of a Product, $$(ab)^m = a^{m}b^{m}$$
• Law of Power of a Quotient, $$(\frac{a}{b})^m = \frac{a^m}{b^m}$$
### 3. Are exponents rational numbers?
Exponents can be rational and irrational numbers.
### 4. Does the power rule work for irrational exponents?
Yes, the power rule work for irrational exponents.
### 5. Can an irrational number raised to an irrational power be rational?
Yes, an irrational number raised to an irrational power can be rational.
### 6. How do you solve irrational exponents?
The irrational exponents can be solved using the laws of exponents.
### 7. Is root 2 rational or irrational?
Root 2 is irrational.
### 8. Is "e" a rational or irrational number?
"e" is an irrational number.
### 9. How do you prove a root is irrational?
A root cannot be expressed in the $$\dfrac{p}{q}$$ form, hence they are irrational.
### 10. Are logarithms exponents?
Yes logarithms are exponents. |
# Lesson Video: Linear Approximation Mathematics
In this video, we will use derivatives to find the equation of the line that approximates the function near a certain value and use differentials to approximate the change in the function.
16:15
### Video Transcript
In this video, weβll learn how we can use derivatives to find the equation of the line that approximates a function near a certain value. By this stage, you should feel confident finding the derivative of the function and the application of this. In this lesson, weβre going to investigate the application of the tangent line to the function. And how it allows us to approximate more complicated functions. Weβll then look at a number of examples of this application with varying degrees of difficulty and consider the geometrical interpretation.
Consider the tangent to the curve π¦ equals π₯ squared at the point with Cartesian coordinates one, one. We can see that the tangent lies close to the curve near the point of tangency, near one, one. If we zoom in on the graph and its tangent at the point of tangency, we see that thereβs a brief interval to either side for which the π¦ values along the tangent give a good approximation to the π¦ values of our curve. And the more we magnify the graph near the point where it is differentiable, the flatter the graph appears and the more it will resemble its tangent. We can use this fact to develop a formula that can be used to give approximations for a function π of π₯.
Remember, the formula for the equation of a straight line with a gradient of π and which passes through the points π₯ one π¦ one is π¦ minus π¦ one equals π times π₯ minus π₯ one. We also know that the derivative of a function π of π₯, π prime of π₯, tells us the gradient of the curve at a given point. In particular, it tells us the gradient of the tangent line to the curve at that point. We can say then that the tangent line to a function π of π₯ at a point π₯ equals π where π is differentiable passes through the point π, π of π. And the gradient at this point will be given by π prime of π.
Substituting these values into our formula for the equation of a straight line and we get π¦ minus π of π equals π prime of π times π₯ minus π. We add π of π to both sides of our equation. And we get π¦ equals π of π plus π prime of π times π₯ minus π as the equation for the tangent line. Letβs formalise this a bit. If π is differentiable at π₯ equals π, then the equation of the tangent line π of π₯ is given as π of π plus π prime of π times π₯ minus π. We call this the linear approximation to the function at π₯ equals π. Letβs have a look at the application of our definition.
Find the linear approximation of the function π of π₯ equals π₯ cubed minus π₯ squared plus three at π₯ equals negative two.
Here, we have a polynomial function for which weβre being asked to find the linear approximation at π₯ equals negative two. Remember, if π is differentiable at π₯ equals π, then the equation of the tangent line and the equation that can be used to find a linear approximation to the function at π₯ equals π is given by π of π₯ equals π of π plus π prime of π times π₯ minus π. Letβs break this down piece by piece.
In this example, we want to find the linear approximation at π₯ equals negative two. So we let π equal negative two. Then π of π is π of negative two. And we can find this value by substituting π₯ is equal to negative two into the function π₯ cubed minus π₯ squared plus three. π of negative two is, therefore, negative two cubed minus negative two squared plus three which is negative nine. Next, weβre looking to find π prime of π.
So weβll begin by finding π prime of π₯, thatβs the derivative of our function, and evaluating that at π₯ equals negative two. The first derivative of our function with respect to π₯ is three π₯ squared minus two π₯. And this means the first derivative evaluated at negative two is given by three times negative two squared minus two times negative two which is 16. But what about this last part, π₯ minus π? What do we know here? Well, we know that π is negative two. So this becomes π₯ minus negative two which is equal to π₯ plus two.
We can substitute each part into our equation for π of π₯. And we get negative nine plus 16 times π₯ plus two. Distributing our expression, we get negative nine plus 16π₯ minus 32. And then we simplify fully. And we see that the linear approximation of the function π of π₯ equals π₯ cubed minus π₯ squared plus three at π₯ equals negative two is π of π₯ equals 16π₯ plus 23.
This demonstrates a very simple example for how to find the linear approximation of a function. It is also important to realise that we can use this same process for more complicated functions by applying the rules for differentiation. Letβs see what that might look like.
What is the tangent line approximation π of π₯ of the square root of one minus π₯ near π₯ equals zero? Remember, if π is differentiable at π, then the equation for the tangent line approximation π of π₯ is given by π of π plus π prime of π times π₯ minus π. Weβll look at our example piece by piece. But first letβs find π of π. Our function π of π₯ is the square root of one minus π₯. And weβre finding the tangent line approximation near π₯ equals zero. So weβre going to let π be equal to zero. This means, in our expression, π of π is going to be π of zero. And we can evaluate this by substituting π₯ equals zero into our function. And we get the square root of one minus zero or the square root of one which is simply one.
The next part weβre interested in is π prime of π. π prime of π₯ is the derivative of π with respect to π₯. So weβre going to need to differentiate the square root of one minus π₯ with respect to π₯. We need to spot here that this is a function of a function or a composite function. And we can apply the chain rule. This says that if π¦ is a function in π’ and π’ itself is a function in π₯, then dπ¦ by dπ₯ is the same as dπ¦ by dπ’ times dπ’ by dπ₯. If we say π¦ is the function the square root of one minus π₯, we can let π’ be equal to one minus π₯ and π¦ be equal to the square root of π’ which Iβve written as π’ to the power of one-half.
dπ’ by dπ₯, the derivative of one minus π₯ with respect to π₯, is simply negative one. And the derivative of π¦ with respect to π’ is half times π’ to the power of one-half minus one which is negative one-half. So the derivative of the square root of one minus π₯ with respect to π₯ is a half times π’ to the negative a half multiplied by negative one. Replacing π’ with one minus π₯ and we see that the derivative of the square root of one minus π₯ with respect to π₯ is negative a half times one minus π₯ to the power of negative one-half. Note, at this stage, that we could have used the general power rule here. And thatβs just a special case of the chain rule.
So since we now know π prime of π₯, we can evaluate π prime of π. Thatβs π prime of zero. So weβre going to substitute zero into our formula for the derivative of our function. Itβs negative a half times one minus zero to the power of negative a half which is negative one-half. The final part of our tangent line approximation that weβre interested in is π₯ minus π. And since π is zero, this becomes π₯ minus zero which is just π₯.
Substituting all of this into our formula and we see the π of π₯ equals one plus negative a half times π₯. And this simplifies to one minus π₯ over two.
So weβve seen so far how the chain rule can be used alongside the formula for the tangent line approximation. We can even use the tangent line approximation when dealing with trigonometric functions.
Find the linear approximation of the function π of π₯ equals π₯ sin of π₯ at π₯ equals two π.
Remember, if π is differentiable at π₯ equals π, then the equation for the tangent line approximation is given by π of π₯ equals π of π plus π prime of π times π₯ minus π. In this example, we can let π be equal to two π. Weβre going to need to evaluate π of π and π prime of π. Letβs begin with π of π. In this case, thatβs π of two π. So we substitute π₯ is equal to two π into π₯ sin π₯. And we get two π times sin of two π. We should know that sin of two π is equal to zero. So π of two π is two π times zero which is just zero.
Now, π prime of π is going to require a little more work. Weβll find the derivative of our function. Thatβs the derivative of π₯ sin π₯ with respect to π₯, noticing that we have a function which is itself the product of two differentiable functions. Weβll, therefore, use the product rule. This says that, for two differentiable functions π’ and π£, the derivative of their product is π’ times dπ£ by dπ₯ plus π£ times dπ’ by dπ₯. For our function, weβll let π’ be equal to π₯ and π£ be equal to sin π₯.
Weβre going to need to differentiate each of these with respect to π₯. dπ’ by dπ₯ is one. And here, we recall the derivative of sin π₯ with respect to π₯ is cos of π₯. And we substitute these into our formula for the product rule. And we see that the derivative π prime of π₯ is equal to π₯ times cos of π₯ plus sin π₯ times one. Thatβs π₯ cos π₯ plus sin π₯. To find π prime of two π, weβll evaluate this when π₯ is equal to two π. That gives us two π times cos of two π plus sin of two π. We already said that sin of two π is zero. And cos of two π is one. So π prime of two π is two π times one plus zero which is simply two π.
Letβs substitute everything we now have into the tangent line approximation formula. π of π is zero. π prime of π is two π. And π₯ minus π is π₯ minus two π. We distribute our parentheses. And we see that the linear approximation of our function π of π₯ equals π₯ sin π₯ at π₯ equals two π is two ππ₯ minus four π squared.
In our next two examples, weβll look at how we can use the linear approximation for a function to approximate values.
By finding the linear approximation of the function π of π₯ equals π₯ to the power of four at a suitable value of π₯, estimate the value of 1.999 to the power of four.
Weβre told to use the linear approximation of the function π of π₯ equals π₯ to the power of four. So we begin by finding the linear approximation, sometimes called the tangent line approximation. This says that if π is differentiable at some point π₯ equals π, then the equation that can be used to find a linear approximation to the function at π₯ equals π is π of π plus π prime of π times π₯ minus π. In this example, weβre going to try to approximate the value of 1.999 to the power of four. This is going to be very close in value to two to the power of four.
So in our linear approximation, weβre going to let π be equal to two. This means that π of π becomes π of two. And we substitute π₯ equals two into our function to get two to the power of four which is 16. Next, we find π prime of π. First, of course, weβre going to need to find an expression for the derivative of π₯ to the power of four. So we begin by differentiating π₯ to the power of four with respect to π₯. And we get four π₯ cubed. This now means that π prime of π becomes π prime of two which becomes four times two cubed. Two cubed is eight. So this is four times eight which is 32.
We substitute everything we now have into our tangent line approximation formula. And we get π of π₯ equals 16 plus 32 times π₯ minus two. And when we distribute our parentheses we see π of π₯ is equal to 32π₯ minus 48. We can use this to approximate the value of 1.999 to the power of four. We need to substitute π₯ equals 1.999. And when we do, we get 32 times 1.999 minus 48 which is 15.968. An estimation for the value of 1.999 to the power of four is, therefore, 15.968. Now, if we were to type 1.999 to the power of four into our calculator, weβd get 15.96802399 and so on. So we can see that this is a very good estimate. And thatβs because 1.999 is fairly close to two.
How do we use this to evaluate, say, 2.3 to the power of four? Our answer might have been a little bit further out.
By finding the linear approximation of the function π of π₯ equals π to the power of π₯ at a suitable value of π₯, estimate the value of π to the power of 0.1.
Weβre told to use the linear approximation of the function π of π₯ equals π to the power of π₯. So we recall the formula. If π is differentiable at π₯ equals π, then the equation that can be used to find the linear approximation to the function at π₯ equals π is π of π₯ equals π of π plus π prime of π times π₯ minus π. In this example, weβre trying to approximate the value of π to the power of 0.1. This is going to be close to the value of π to the power of zero. So we let π be equal to zero. This means π of π is equal to π of zero. And substituting zero into our function π of π₯ equals π to the power of π₯ gives π to the power of zero which is one.
Next, we find π prime of π. First, of course, we need to find an expression for the derivative of our function. So we different π to the power of π₯ with respect to π₯. The first derivative of π to the power of π₯ is π to the power of π₯. So π prime of π becomes π prime of zero which is π to the power of zero. And once again, thatβs one. Substituting what we know into our formula for our tangent line approximation and we see that π of π₯ is equal to one plus one times π₯ minus zero. And that simplifies to π₯ plus one.
Weβll use this to approximate the value of π to the power of 0.1 by finding π of 0.1. Thatβs 0.1 plus one which is 1.1. And an estimate to the value of π to the 0.1 is 1.1. And if we type this into our calculator, π to the 0.1 is 1.10517 and so on. Thatβs very close in value to our estimation. And thatβs because 0.1 is fairly close to zero. Had we tried the larger value, our number might not have been so accurate. Letβs check that.
For example, π of 0.3 is 0.3 plus one. So according to our approximation, π to the 0.3 is approximately 1.3. Typing π to the 0.3 into our calculator and we get 1.349858808, still not a bad approximation but not quite as close as π to the power of 0.1.
In this video, weβve learnt that we can use derivatives to find a tangent line approximation that can be used to approximate a function near a given value. If π is differentiable at π₯ equals π, then the equation that can be used to find a linear approximation to the function at this point is π of π₯ equals π of π plus π prime of π times π₯ minus π. We also saw that the closer the value for π₯ is to the value of π, the more accurate the approximations will be. |
# How to Find the Greatest Common Factor: 2 Methods
By Suvo
How do you find the Greatest Common Factor of a set of numbers? Read on for an explanation and two methods of finding the GCF.
## What is the GCF?
As the name indicates, Greatest Common Factor (GCF) is the common and largest factor of the given set of numbers (for which GCF is to be calculated). To find out the GCF,
1. Factorize the numbers.
2. Compare the factors.
3. Find all the common prime factors and multiply them.
## Example
The following example uses 24 and 30.
1. Factorize both numbers and write them on board:
• 24 = 2 x 2 x 2 x 3
• 30 = 3 x 2 x 5
2. Find the common factors of the two numbers.
• 2 and 3.
3. Because both 30 and 24 have 2 and 3 as a prime factor, multiply these two numbers together.
4. Write the product of 2 and 3 as 6. This is the GCF of 24 and 30.
## Another Method
Please understand another method of finding the Greatest Common Factor:
1. Take the smallest number (say A) among the numbers for which GCF needs to find out.
2. Divide any one of the other number (say B) by this number (A).
3. Find reminder (say C) of the division.
4. Divide the divider (A) of the earlier division by the remainder (C).
5. Find remainder (say D).
6. Divide C by D.
7. Repeat step 6 until there is no remainder.
8. The last divider for which there is no remainder is the GCF of the two numbers.
See the figure below for better clarity of the process.
If you have more than two numbers, then first find out the GCF of any two numbers and use that GCF to divide any one of the remaining numbers and so on to find final GCF.
Once you are clear with this method of finding GCF, explain it by taking example of 24 and 30. Like below:
As you can see, both methods produce the same result. |
# RBSE Solutions for Class 9 Maths Chapter 10 Area of Triangles and Quadrilaterals Ex 10.3
RBSE Solutions for Class 9 Maths Chapter 10 Area of Triangles and Quadrilaterals Ex 10.3 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 10 Area of Triangles and Quadrilaterals Ex 10.3.
Board RBSE Class Class 9 Subject Maths Chapter Chapter 10 Chapter Name Area of Triangles and Quadrilaterals Exercise Ex 10.3 Number of Questions Solved 15 Category RBSE Solutions
## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 10 Area of Triangles and Quadrilaterals Ex 10.3
Question 1.
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm², then ar (∆ABC) = 24 cm². Is this statement true or false. Give reason for your answer.
Solution.
ABCD is a parallelogram and X is the mid-point of AB. Draw the diagonal AC.
We know that diagonal of (RBSESolutions.com) a parallelogram divides it into two triangles of equal areas.
So, ar (||gm ABCD) = ar (∆ADC) x 2
= 2 x 24 = 48 cm²
Since, CX is the median of ∆ABC
∴ ar (∆BCX) = $$\frac { 1 }{ 2 }$$ ar (∆ABC)
= $$\frac { 1 }{ 2 }$$ x 24 = 12 cm²
Now,
ar (∆XCD) = ar(||gm ABCD) – ar (∆BXC)
= 48 – 12 = 36 cm²
Hence, the given statement is false.
Question 2.
PQRS is a rectangle inscribed in a quadrant (RBSESolutions.com) of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar (∆PAS) = 30 cm².
Solution.
In ∆PQR,
PR² = PQ² + RQ²
(by Pythagoras theorem)
=> PQ² = PR² – RQ² = (13)² – (5)²
(given PS = 5 cm, PR = 13 cm)
=> PQ = √169-25 = √144 = 12 cm
Now, area of rectangle PQRS
= 12 x 5 = 60 cm²
Area of ∆PSQ = $$\frac { 1 }{ 2 }$$ x 60 = 30 cm²
(Since, diagonal of a rectangle bisects it into two triangles of equal areas)
Now, A is any point on PQ. So, area of ∆APS depends on position of A. If A is at point Q, then area will be 30 cm².
∴ ar (∆PAS) ≤ 30 cm².
Hence proved.
Question 3.
PQRS is a parallelogram, whose area is 180 cm² and A is any point (RBSESolutions.com) on the diagonal QS. Then, the area of ∆ASR = 90 cm². This statement is true or false. Why?
Solution.
This statement is False.
Given, area of parallelogram PQRS is 180 cm² and QS is its diagonal which divides it into two triangles of equal areas.
∴ Area of ∆PQS = 90 cm²
Now, A is any point on QS.
∴ Area of ∆ASR < Area of ∆SRQ.
Question 4.
∆ABC and ∆BDE are two equilateral triangles such (RBSESolutions.com) that D is the mid-point of BC, then
ar (∆BDE) = $$\frac { 1 }{ 4 }$$ ar (∆ABC).
Solution.
Given: In a ∆ABC, AD is the median and E is mid-point of AD.
To prove: ar (∆BDE) = $$\frac { 1 }{ 4 }$$ ar (∆ABC).
Proof: ∵ AD is the median of ∆ABC
∴ ar (∆ABD) = $$\frac { 1 }{ 2 }$$ ar (∆ABC) …(i)
Reason: Median divides the (RBSESolutions.com) triangle into two triangles equal in area.
Also in ∆BAD, BE is median.
∴ ar (∆BED) = $$\frac { 1 }{ 2 }$$ ar (∆ABD) …(ii)
Using (i) and (ii), we get
area (∆BED) = $$\frac { 1 }{ 2 }$$ X $$\frac { 1 }{ 2 }$$ ar (∆ABC)
Yes, the given statement is true.
Question 5.
In figure, ABCD and EFGD are (RBSESolutions.com) two parallelograms and G is the mid-point of CD, then
ar (∆DPC) = $$\frac { 1 }{ 2 }$$ ar (EFGD)
Solution.
Given: ABCD and EFGD are ||gm and G is mid-point of CD.
To prove: ar (∆DPC) = $$\frac { 1 }{ 2 }$$ ar (EFGD)
Proof: ar (∆DPC) = $$\frac { 1 }{ 2 }$$ ar (||gm ABCD)
(∵ ∆PDC and parallelogram ABCD are on the the same base DC and between the same parallels AB and CD)
∵ G is mid-point of DC.
∴DG = $$\frac { 1 }{ 2 }$$DC
∴ar (EFGD) = $$\frac { 1 }{ 2 }$$ ar (ABCD) …(ii)
=> ar (∆DPC) = ar (EFGD)
The given statement is false.
Question 6.
In a trapezium ABCD, AB || DC and L is the (RBSESolutions.com) mid-point of BC. Through L a line PQ is drawn parallel to AD which meets DC produced at Q.
Prove that ar (ABCD) = ar (APQD).
Solution.
In ∆LPB and ∆CLQ
∠1 = ∠2
(Vertically opposite angles)
∠3 = ∠4 (alternate angles)
and BL = LC
(as L is mid-point of BC)
∴∆LPB ≅ ∆LQC
(by ASA congruency rule)
⇒ ar (∆LPB) + ar (∆DCLP)
⇒ ar (∆CLQ) + ar (∆DCLP)
⇒ ar (∆BCD) = ar (∆PQD)
Hence proved.
Question 7.
If the mid-point of the sides of (RBSESolutions.com) a quadrilateral are joined in order, then prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (see figure).
[Hint: Join BD and draw perpendicular from A on BD]
Solution.
Given: ABCD is (RBSESolutions.com) a quadrilateral and P, F, R and S are mid-points of the sides BC, CD, AD and AB respectively and PFRS is a parallelogram.
To prove:
ar (||gm PFRS) = $$\frac { 1 }{ 2 }$$ar (quad. ABCD)
Construction: Join BD and BR.
Proof: Median BR divides ∆BDA into two triangles of equal areas.
ar (∆BRA) = $$\frac { 1 }{ 2 }$$ ar (∆BDA) ..(i)
Similarly, median RS divides ∆BRA into (RBSESolutions.com) two triangles of equal areas.
ar (∆ASR) = $$\frac { 1 }{ 2 }$$ ar (∆BRA) …(ii)
From equation (i) and (ii), we get
ar (∆ASR) = $$\frac { 1 }{ 4 }$$ ar (∆BDA) …(iii)
Similarly,
ar (∆CFP) = $$\frac { 1 }{ 4 }$$ ar (∆BCD) …(iv)
On adding equations (iii) and (iv), we get
ar (∆ASR) + ar (∆CFP) = $$\frac { 1 }{ 4 }$$ [ar (∆BDA) + ar (∆BCD)
ar (∆ASR) + ar (∆CFP) = $$\frac { 1 }{ 4 }$$ ar (quadrilateral ABCD) …(v)
Similarly, ar (∆DRF) + ar (∆BSP) = $$\frac { 1 }{ 2 }$$ ar (quadrilateral ABCD) …(vi)
On adding (v) and (vi), we get
ar (∆ASR) + ar (∆CFP) + ar (∆DRF) + ar (ABSP)
= $$\frac { 1 }{ 2 }$$ ar (quadrilateral ABCD) …..(vii)
But
ar (∆ASR) + ar (∆CFP) + ar (∆DRF) + ar (∆BSP) + ar (||gm PFRS)
On subtracting equation (vii) from equation (viii), we get
ar (||gm PFRS) = $$\frac { 1 }{ 2 }$$ ar
Hence proved.
Question 8.
A man walks 10 m towards (RBSESolutions.com) East and then he walks 30 m towards North. Find his distance from the starting point.
Solution.
Question 9.
A ladder is placed in such a way that its foot is (RBSESolutions.com) at a distance of 7 m from the wall. If its other end reaches a window, a height of 24 m, then find the length of the ladder.
Solution.
In ∆ABC,
AC² = AB² + BC²
(Using Baudhayan theorem)
AC² = (24)² + (7)²
= 576 + 49 = 625
=> AC = √625 = 25 m
Hence, length of the ladder = 25 m.
Question 10.
Two poles stand vertically on (RBSESolutions.com) a level ground. Their height are 7 m and 12 m respectively. If the distance between their feets is 12 m. Find the distance between their tops.
Solution.
Here AB = 7 m and CD = 12 m
CP = (12 – 7) m = 5 m
∴In right angled ∆APC
AC² = AP² + PC²
=> AC² = (12 m)² + (5 m)²
=> AC² = 144 m² + 25 m²
=> AC = √169 m
=> AC = 13 m
Hence, distance between their tops is 13 m.
Question 11.
Find the length of the altitude and area of (RBSESolutions.com) an equilateral triangle of side ‘a’
Solution.
∵ ABC is an equilateral triangle.
∴ AB = BC = CA = a
Question 12.
Find the length of the diagonal of (RBSESolutions.com) a square whose each side is of 4 m.
Solution.
Length of diagonal of a square = √2a
= √2×4
= 4√2 m
Question 13.
In an equilateral triangle ABC, AD is perpendicular to BC, prove that 3AB² = 4AD².
Solution.
Hence proved.
Question 14.
O is a point inside (RBSESolutions.com) the rectangle ABCD. Prove that OB² + OD² = OA² + OC².
Solution.
Given: O is a point inside the rectangle ABCD.
To prove: OB² + OD² = OA² + OC².
Construction: Join O, with the vertices A, B, C and D of rectangle ABCD. Now draw (RBSESolutions.com) a parallel line LM through O which meets AB and DC at M and L respectively.
Proof: In ∆OMB
OB² = OM² + MB² = OM² + CL²
(by Baudhayan theorem)
Since, ABCD is a rectangle and ML ⊥ AB
MB = CL and in ∆ODL
OD² = OL² + DL²
= OL² + AM² (∵ DL = AM)
∴ OB² + OD² = OM² + CL² + OL² + AM²
= (OM² + AM²) + (CL² + OL²)
= OA² + OC²
Hence, OB² + OD² = OA² + OC²
Hence proved.
Question 15.
In an obtuse angled triangle ABC, ∠C is an obtuse angle and AD ⊥ BC which (RBSESolutions.com) meets BC produced at D. Then prove that AB² = AC² + BC² + 2BC • CD.
Solution.
Using Baudhayan theorem
or AB² = AD² + (DC + CB)²
or AB² = AD² + DC² + BC² + 2BC • DC
or AB² = (AD² + DC²) + BC² + 2BC • CD
=> AB² = AC² + BC² + 2BC • CD |
# Harmonic Warping
Harmonic Warping of Blue Wash
Fields: Calculus and Fractals
Image Created By: Paul Cockshott
Website: Fractal Art
Harmonic Warping of Blue Wash
This image is a tiling based on harmonic warping operations. These operations take a source image and compress it to show the infinite tiling of the source image within a finite space.
# Basic Description
This image is an infinite tiling. If you look closely at the edges of the image, you can see that the tiles become smaller and smaller and seem to fade into the edges. The border of the image is infinite so that the tiling continues unendingly and the tiles become eternally smaller.
The for this tiling is another image that is mathematically interesting and is also featured on this website. See Blue Wash for more information about how the source image was created.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *Single Variable Calculus
Harmonic Warping Equation
To create this image, a harmon [...]
Harmonic Warping Equation
To create this image, a harmonic warping operation was used to map the infinite tiling of the source image onto a finite plane. This operation essentially took the entire infinite plane and squashed it into a square. This type of operation can be called a distance compressing warp.
The equations used to perform the harmonic warp is show in a graph to the right and is as follows, where (x,y) is a coordinate on the Euclidean plane tiling and (d(x), d(y)) is a coordinate on the non-Euclidean square tiling
$d(x) = 1 - \frac{1}{1+x}$
$d(y) = 1 - \frac{1}{1+y}$
You can observe for both of these equations that as x and y go to infinity, d(x) and d(y) both approach a limit of 1.
The graph to the right shows clearly that d(x) approaches 1 as x goes to infinity. Mathematically:
$\lim_{x \rightarrow \infty}d(x) = 1 - \frac{1}{1+x}$
$\lim_{x \rightarrow \infty}d(x) = 1 - \frac{1}{1+\infty}$
$\lim_{x \rightarrow \infty}d(x) = 1 - \frac{1}{\infty}$
$\lim_{x \rightarrow \infty}d(x) = 1 - 0$
$\lim_{x \rightarrow \infty}d(x) = 1$
Since d(x) and d(y) approach 1 as x and y go to infinity, the square plane that the infinite tiling is mapped to must be a unit square (that is its dimensions are 1 unit by 1 unit). Since the unit square fits an infinite tiling within its finite border, the square is not a traditional Euclidean plane. As the tiling approaches the border of the square, distance within the square increases non-linearly. In fact, the border of the square is infinite because the tiling goes on indefinitely.
Here is another example of this type of tiling contained in a square using the Union Flag:
## Polar Harmonic Warping
A polar tiling
A tiling similar to the one mentioned above can be performed in polar coordinates. Polar tilings are infinite Euclidean tilings condensed into a finite circular space with an infinite border. The harmonic warping operations used to map the infinite Euclidean plane onto the circle are performed based on the radius of the circle being 1 unit.
## Four Infinite Poles
Another tiling that is possible involves designating the four cardinal poles of the circular border as infinite areas. The tiling then becomes very similar to a tiling done in the Poincaré Disk Model representing hyperbolic geometry. If the center of the circle of this type of tiling corresponds to the Euclidean origin where the x and y axes meet, then we can see that the tiling is consist with the source image. In the Poincaré Disk Model, straight lines are represented as circles and perpendicular lines still intersect at an angle of 90 degrees. A break down of the four infinite pole tiling of the Union Flag to the right shows that in each tile, the cross of the Union Flag still intersects with the x axes and the "X" of the Union Flag collapse towards each of the four poles.
Comparing the Different Types of Tilings
Saint Andrew's Flag Saint George's Flag Original Flag Rectangular Tiling Polar Tiling Four Infinite Poles Tiling
# Teaching Materials
There are currently no teaching materials for this page. Add teaching materials.
# About the Creator of this Image
Paul Cockshott is a computer scientist and a reader at the University of Glasgow. The various math images featured on this page emerged from his research dealing with digital image processing. |
# 2.8: Right Triangle Trigonometry
Difficulty Level: At Grade Created by: CK-12
## The Pythagorean Theorem
Reducing Radicals - Students may or may not have spent much time reducing radicals in previous math courses. You may need to review how to do this and how to perform operations with radicals. Students may need to see more than one way to reduce radicals. The most common is to take out the greatest factor which is a perfect square, but this can be difficult for many students with weak mental math skills. Another method is to make a factor tree and find the prime factorization of the number and identify doubles that can be taken out. There is not necessarily a “best” way, it is more important to figure out which way your students can best reduce radicals correctly.
Method 1: \begin{align*}\sqrt{72}=\sqrt{36 \times 2}=6\sqrt{2}\end{align*}
Method 2: \begin{align*}\sqrt{72}=\sqrt{9 \times 8}=\sqrt{3 \times 3 \times 2 \times 2 \times 2}=3 \times 2\sqrt{2}=6\sqrt{2}\end{align*}
The following are some examples that will help students review the basic properties of radicals and practice reducing radicals.
Example 1: \begin{align*}\sqrt{112}\end{align*}
Answer: \begin{align*}\sqrt{16 \times 7}=4\sqrt{7}\end{align*}
Students may not recognize right away that 16 is a factor of 112. This problem can also be solved by completely factoring 112 (method 2).
Example 2: \begin{align*}4\sqrt{192}\end{align*}
Answer: \begin{align*}4 \sqrt{64 \times 3}=4 \times 8 \sqrt{3}=32 \sqrt{3}\end{align*}
Example 3: \begin{align*}2\sqrt{5}+\sqrt{45}\end{align*}
Answer: \begin{align*}2\sqrt{5}+3\sqrt{5}=5\sqrt{5}\end{align*}
Review with students that in order to add or subtract radicals, the radical must be identical. Sometimes they can meet this condition by simplifying one or both terms and sometimes they just won’t be able to add them together.
Example 4: \begin{align*}\sqrt{6} \times \sqrt{18}\end{align*}
Answer: \begin{align*}\sqrt{108}=\sqrt{36 \times 3}=6\sqrt{3}\end{align*}
Students may have forgotten that radicals can be multiplied or divided to form a new radical which they may then be able to reduce.
Example 5: \begin{align*}\left(\sqrt{11}\right)^2\end{align*}
Review the definition of a square root with students to help them understand this one. The square root and the square cancel each other out - they are inverse operations.
Skipping Around - Not all texts present material in the same order, and many instructors have a preferred way to develop concepts that is not always the same as the one used in the text. The Pythagorean theorem is frequently moved from place to place. You should follow the link given in the text to see additional proofs of the Pythagorean Theorem. Proofs are hard for most students to understand. It is important to choose one that the students can feel good about. Don’t limit the possibilities to these two, research other methods, and pick the one that is most appropriate for your class. Or better yet, pick the best two or three. Different proofs will appeal to different students.
The Height Must be Measured Along a Segment That is Perpendicular to the Base - When given an isosceles triangle where the altitude is not explicitly shown, student will frequently try to use the length of one of the sides of the triangle for the height. Tell them that they must find the length of the altitude that is perpendicular to the segment that’s length is being used for the base in the formula \begin{align*}A=\frac{1}{2} bh\end{align*}. Sometimes they do not know what to do, and are just trying something, which is, in a way, admirable. The more common explanation though is that they forget. The students have been using this formula for years, they think they know this material, so they just plug and chug, not realizing that the given information has changed. Remind the students that now that they are in Geometry class, there is an extra step. The new challenge is to find the height, and then they can do the easy part and plug it into the formula.
Derive the Distance Formula - After doing an example with numbers to show how the distance formula is basically just the Pythagorean theorem, use variables to derive the distance formula. Most students will understand the proof if they have seen a number example first. Point out to the students that the number example was inductive reasoning, and that the proof was deductive reasoning. Taking the time to do this is a good review of logic and algebra as well as great proof practice.
Squaring a Negative - One of the most common errors students make when using the distance formula is that they use their calculator incorrectly when squaring negative numbers. Remind students that when they square a negative number, the result is positive. Describe the difference between \begin{align*}-8^2\end{align*} and \begin{align*}(-8)^2\end{align*} and review the correct of order of operations. You may also want to stress the connection between the distance formula and the Pythagorean theorem and encourage them to think of the difference between the \begin{align*}x\end{align*}’s and \begin{align*}y\end{align*}’s as lengths and therefore they can just use the absolute value of the difference (i.e. ignore the negative).
## Converse of the Pythagorean Theorem
Acute and Obtuse Triangles - Many students have trouble remembering that the inequality with the greater than is true when the triangle is acute, and that the equation with the less than is true for obtuse triangles. It seems backwards to them. One way to present this relationship is to compare the longest side and the angle opposite of it. In a right triangle, the equation has an equal sign; the hypotenuse is the perfect size. When the longest side of the triangle is shorter than what it would be in a right triangle, the angle opposite that side is also smaller, and the triangle is acute. When the longest side of the triangle is longer than what is would be in a right triangle, the angle opposite that side is also larger, and the triangle is obtuse.
Is It Really a Triangle? - I have found that once students start using the Pythagorean theorem to determine whether lengths form a right, acute or obtuse triangle that they forget completely that the sum of two sides must be greater than the remaining side in order for a triangle to exist at all. The following example illustrates this misunderstanding. You may want to put it on the board and ask your students what kind of triangle is formed.
Example: What kind of triangle is formed by lengths 3, 4, 7?
Answer: None! There is no triangle at all. \begin{align*}3+4=7\end{align*}.
If students used the Pythagorean theorem and didn’t check to make sure there was a triangle at all then they would have said that the triangle is obtuse. This in incorrect.
## Using Similar Right Triangles
Separate the Three Triangles - The altitude from the right angle of a triangle divides the triangle into two smaller right triangles that are similar to each other, and to the original triangle. All the relationships among the segments in this figure are based on the similarity of the three triangles. Many students have trouble rotating shapes in their minds, or seeing individual polygons when they are overlapping. It is helpful for these students to draw the triangles separately and oriented in the same direction. After going through the process of turning and redrawing the triangles a few times, they will remember how the triangles fit together, and this step will no longer be necessary.
Color-Coded Flashcards - It is difficult to describe in words which segments to use in the geometric mean to find the desired segment. Labeling the figure with variables and using a formula is the standard method. The relationship is easier to remember if the labeling of the triangles is kept the same every time the figure is drawn. The students need to remember the location of the segments relative to each other. Making color-coded pictures or flashcards will be helpful. For each relationship the figure should be drawn on both sides of the card. The segment whose measure is to be found should be highlighted in one color on the front, and on the back, the two segments that need to be used in the geometric mean should be highlighted with two different colors. Using two colors on the back is important because the segments often overlap. Making these cards will be helpful even if the students never use them. Those that have trouble remembering the relationship will use these cards frequently as a reference.
Add a Step and Find the Areas - The exercises in this section have the students find the base or height of triangles. They have all the information that they need to also calculate the areas of these triangles. Students need practice with multi-step problems. Having them find the area will help them think through a more complex problem, and give them practice laying out organized work for calculations that are more complex. Chose to extend the assignment or not based on how well the students are doing with the material, and how much time there is to work on this section.
## Special Right Triangles
Memorize These Ratios - There are some prevalent relationships and formulas in mathematics that need to be committed to long term memory, and the ratios made by the sides of these two special right triangles are definitely among them. Students will use these relationships not only in the rest of this class, but also in trigonometry, and in other future math classes. Students are expected to know these relationships, so the sooner they learn to use them and commit them to memory, the better off they will be.
Two is Greater Than the Square Root of Three - One way that students can remember the ratios of the sides of these special right triangles, is to use the fact that in a triangle, the longest side is opposite the largest angle, and the shortest side is opposite the smallest angle. At this point in the class, students know that the hypotenuse is the longest side in a right triangle. What sometimes confuses them is that in the 30-60-90 triangle, the ratio of the sides is \begin{align*}1:2:\sqrt{3}\end{align*}, and if they do not really think about it, they sometimes put the \begin{align*}\sqrt{3}\end{align*} as the hypotenuse because it might seem bigger than 2. Using the opposite relationship is a good method to use when working with these triangles. Just bring to the students’ attention that \begin{align*}2>\sqrt{3}\end{align*}.
How Do I Find the Short Leg Again? - While students may quickly memorize the two special right triangle ratios, they may have trouble applying the ratios to find the unknown sides. One way to help students with this process is to have them write the ratios with a variable. For example the 30-60-90 triangle ratio would be \begin{align*}x:2x:x\sqrt{3}\end{align*}. Next, have them identify which side they are given and use the appropriate part of the ratio to determine \begin{align*}x\end{align*} and the other side.
Example 1: Find the other two sides in a 30-60-90 triangle given that the hypotenuse is 8.
Answer: First set \begin{align*}2x=8\end{align*} and solve to get \begin{align*}x=4\end{align*} which is the short leg. The long leg is then \begin{align*}4\sqrt{3}\end{align*}.
Example 2: Find the length of a leg in an isosceles right triangle with hypotenuse 3.
Answer: The ratio for an isosceles right triangle or 45-45-90 triangle is \begin{align*}x:x:x\sqrt{2}\end{align*}. Since we are given the hypotenuse here, \begin{align*}x\sqrt{2}=3\end{align*}. Now we must solve for \begin{align*}x\end{align*} as shown below.
\begin{align*}\frac{x\sqrt{2}}{\sqrt{2}} &= \frac{3}{\sqrt{2}}\\ x &= \frac{3}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{3\sqrt{2}}{2}\end{align*}
Rationalizing the Denominator - Sometimes student will not recognize that \begin{align*}\frac{1}{\sqrt{2}}\end{align*} and \begin{align*}\frac{\sqrt{2}}{2}\end{align*} are equivalent. Most likely, they learned how to rationalize denominators in algebra, but it is nice to do a short review before using these types of ratios in special right triangles.
Derive with Variables - The beginning of the last chapter offers students a good amount of experience with ratios. If they did well on those sections, it would benefit them to see the derivation of the ratios done with variable expressions. It would give them practice with a rigorous derivation, review and apply the algebra they have learned, and help them see how the triangles can change in size.
Exact vs. Decimal Approximation - Many students do not realize that when they enter \begin{align*}\sqrt{2}\end{align*} into a calculator and get 1.414213562, that this decimal is only an approximation of \begin{align*}\sqrt{2}\end{align*}. They also do not realize that when arithmetic is done with an approximation, that the error usually grown. If 3.2 is rounded to 3, the error is only 0.2, but if the three is now multiplied by five, the result is 15, instead of the 16 it would have been if original the original number had not been rounded. The error has grown to 1.0. Most students find it more difficult to do operations with radical expressions than to put the numbers into their calculator. Making them aware of error magnification will motivate them to learn how to do operations with radicals. In the last step, it may be nice to have a decimal approximation so that the number can be easily compared with other numbers. It is always good to have an exact form for the answer so that the person using your work can round the number to the desired degree of accuracy.
## Tangent, Sine and Cosine
Trig Thinking - Students sometimes have a difficult time understanding trigonometry when they are first introduced to this new branch of mathematics. It is quite a different way of thinking when compared to algebra or even geometry. Let them know that as they begin their study of trigonometry in the next few sections the calculations won’t be difficult, the challenge will be to understand what is being asked. Sometimes students have trouble because they think it must be more difficult than it appears to be. Most students find they like trigonometry once they get the feel of it.
Ratios for a Right Angle - Students will sometimes try to take the sine, cosine or tangent of the right angle in a right triangle. They should soon see that something is amiss since the opposite leg is the hypotenuse. Let them know that there are other methods of finding the tangent of angles \begin{align*}90^\circ\end{align*} or more. The triangle based definitions of the trigonometric functions that the students are learning in this chapter only apply to angles in the interval \begin{align*}0^\circ .
The Ratios of an Angle - The sine, cosine, and tangent are ratios that are associated with a specific angle. Emphasize that there is a pairing between an acute angle measure, and a ratio of side lengths. Sine, cosine, and tangent is best described as functions. If the students’ grasp of functions is such that introducing the concept will only confuse matters, the one-to-one correspondence between acute angle and ratio can be taught without getting into the full function definition. When students understand this, they will have an easier time using the notation and understanding that the sine, cosine, and tangent for a specific angle are the same, no matter what right triangle it is being used because all right triangles with that angle will be similar.
Use Similar Triangles - Many students have trouble understanding that the sine, cosine, and tangent of a specific angle measure do not depend on the size of the right triangle used to take the ratio. Take some time to go back and explain why this is true using what the students know about similar triangle. It will be a great review and application. Remind the students that if the right triangles have one set of congruent acute angles, then they are similar by the AA Triangle Similarly Postulate. Once the triangles are known to be similar it follows that their sides are proportional. It may be a good activity to have students make right triangles with a particular acute angle measure and compare the ratios of the sides. They should see that no matter how big or small they made their triangle, they get the same ratios as their classmates. You could also refer to the special right triangles to make this connection. Explaining that the names sine, cosine and tangent were given to these ratios and the values were recorded in tables by angle measure may help them understand the idea a little better. You may even want to show them a trig table of values and explain that their calculator is simply looking up a value in a table when they type in the trig function and a particular angle measure.
Here is a Sketchpad activity that may further enhance student understanding:
1. Students can construct similar right triangles using dilation from the transformation menu.
2. After choosing a specific angle they should measure the corresponding angle in all the triangles. Each of these measurements should be equal.
3. The legs of all the right triangles can be measured.
4. Then the tangents can be calculated.
5. Student should observe that all of ratios are the same.
Trig Errors are Hard to Catch - The math of trigonometry is, at this point, not difficult. Not much computation is necessary to chose two numbers and put them in a ratio. What students need to be aware of is how easy it is to make a little mistake and not realize that there is an error. When solving an equation the answer can be substituted back into the original equation to be checked. The sine and cosine for acute angles do not have a wide range. It is extremely easy to mistakenly use the sine instead of the cosine in an application and the difference often is small enough to seem reasonable, but still definitely wrong. Ask the student to focus on accuracy as they work with these new concepts. Remind them to be slow and careful.
SOHCAHTOA - This pneumonic device has been around for a while because it helps students keep the ratios straight. Another way to write it that makes it even more clear is: \begin{align*}S \frac{O}{H} \ C \frac{A}{H} \ T \frac{O}{A}\end{align*}.
Something to Consider - Ask the students to combine their knowledge of side-angle relationships in a triangle with the definition of sine. How does the length of the hypotenuse compare to the lengths of the legs of a right triangle? What does that mean about the types of numbers that can be sine ratios? With leading questions like these students should be able to see that the sine and cosine ratios for an acute angle will always be less than one. This type of analysis will prepare them for future math classes and increase their analytical thinking skills. It will also be a good review of previous material and help them check there work when they first start writing sine and cosine ratios.
Two-Step Problems - Having the students write sine, cosine, and tangent ratios as part of two-step problems will help them connect the new material that they have learned to other geometry they know. They will remember it longer, and be better able to see where it can be applied.
Example: \begin{align*}\Delta ABC\end{align*} is a right triangle with the right angle at vertex \begin{align*}C\end{align*}. \begin{align*}AC=3 \ cm,BC=4 \ cm\end{align*}. What is the sine of \begin{align*}\angle A\end{align*}?
Answer: \begin{align*}AB=5 \ cm\end{align*} by the Pythagorean theorem, therefore \begin{align*}\sin A = \frac{4}{5}\end{align*}.
Note: The sine of an angle does not have units. The units will cancel out in the ratio.
## Inverse Trigonometric Ratios
Regular or Arc - Students will sometimes be confused about when to use the regular trigonometric function and when to use the inverse. They understand to concepts, but do not want to go through the entire thought process each time they must make the decision. I give them this short rule of thumb to help them remember: When looking for a ratio or side length, use regular and when looking for an angle use arc. They can associate “angle” and “arc” in their minds. Use the alliteration.
It may also help to explain that the inverse or “arc” trigonometric function does the reverse of the original. Each of the original three trig functions essentially go to a table, look up an angle measure and find the corresponding ratio. The “arc” functions will look up the ratio in the table and give back the corresponding angle measure. Understanding what it is that the calculator does when they use these functions may help reinforce student understanding.
Which Trig Ratio - A common mistake students make when using the inverse trigonometric functions to find angles in right triangles is to use the wrong function. They may use arcsine instead of arccosine for example. There is a process that students can use to reduce the number of these kinds of errors.
1. First, the students should mark the angle whose measure is to be found. With the angle in question highlighted, it is easier for the students to see the relationship the sides have to that angle. It is fun for the students to use colored pencils, pens, or highlighters.
2. Next, the students should look at the sides with known side measures and determine their relationship to the angle. They can make notes on the triangle, labeling the hypotenuse, the adjacent leg and the opposite let. If they are having trouble with this I have them look for the hypotenuse first and always highlight it green, then they can decide between opposite and adjacent for the remaining two sides.
3. Now, they need to look at the two sides they have chosen, and decide if they need to use sine, cosine, or tangent.
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# A plane has a take off speed of $88.3 m/s$ and requires 1365 m to reach that speed. Determine the acceleration of the plane.(A) $3.86m/{s^2}$ (B) $2.86m/{s^2}$(C) $2.8m/{s^2}$(D) $2.6m/{s^2}$
Last updated date: 11th Sep 2024
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Hint:Given problem is straight forward formula based problem. If initial velocity, final velocity and distance or displacement is given and we have to find acceleration then use Newton’s III equation of motion which is
${v^2} = {u^2} + 2as$
Where
u $=$ Initial velocity
v $=$ Final velocity
a $=$ Acceleration
s $=$ Displacement
Given that initial velocity of plane is $u = 0 m/s$ final velocity of plane $v = 88.3 m/s$
Distance travelled by plane
$s = 1365 m$
Let a be the acceleration of the plane.
So, using Newton’s III equation of motion
$\Rightarrow {v^2} = {u^2} + 2as$
$\Rightarrow {(88.3)^2} = {(0)^2} + 2a(1365)$
$\Rightarrow 7796.89 = 0 + (2730)a$
$\Rightarrow (2730)a = 7796.89$
$\Rightarrow a = \dfrac{{7796.89}}{{2730}}$
$\Rightarrow a = 2.865$
$\therefore a \simeq 2.86m/{s^2}$
Hence, the acceleration of plane is $2.86m/{s^2}$
So, option B is the correct answer.
Note: In many problems, students may get confused about where and which equation of Newton’s we should apply. So, If acceleration, time and initial velocity is given and we have to find final velocity of the body then we should apply Newton’s I equation.If displacement is asked then we should apply Newton’s II equation.If acceleration is asked and time is not given then we should apply Newton’s III equation. |
Using polar coordinates, there is an alternate way to define the conics. Rectangular coordinates place the most importance on the location of the center of the conic, but polar coordinates place more importance on where the focus of a conic is. In certain situations, this makes more sense (the reflective property of a parabola depends more on the location of the focus than the center).
Now we will define a conic this way: a conic is a set of points such that the distance between a point on the conic and a fixed point is related to the distance from that point to a fixed line by a constant ratio. The fixed point is the focus, and the fixed line is the directrix. This constant ratio is the eccentricity e of the conic. e tells us which kind of conic it is. If 0 < e < 1, the conic is an ellipse. If e = 1, the conic is a parabola. If e > 1, the conic is a hyperbola.
In a polar equation for a conic, the pole is the focus of the conic, and the polar axis lies along the positive x-axis, as is conventional. Let p be the distance between the focus (pole) and the directrix of a given conic. Then the polar equation for a conic takes one of the following two forms:
r =
r =
When r = , the directrix is horizontal and p units above the pole; the axis, major axis, or transverse axis of the conic (depending on which type it is) is vertical, on the line θ = .
When r = , the directrix is horizontal and p units below the pole; the "main" axis (term varies depending on which type of conic it is) is vertical, on the line θ = .
When r = , the directrix is vertical and p units to the right of the pole; the axis is horizontal, on the line θ = 0.
When r = , the directrix is vertical and p units to the left of the pole; the axis is horizontal, on the line θ = 0.
This information is enough to analyze any conic in polar form. First, find e and decide which type of conic it is. Then, based on the form of the conic, decide where the directrix is and find p. Finally, plugging in different values for θ based on whether the main axis of the conic is vertical or horizontal, you can find the vertices of the conic, and find values for a, b, and c. |
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# In the earlier exercise, a population and sample size are given. Find the mean and standard deviation of the sampling distribution of sample means and compare them with the mean and standard deviation of the population.The minutes of overtime reported by each of the three executives at a corporation are 90,120 , and 210. Use a sample size of 3.
Solution
Verified
Step 1
1 of 5
Given: $90, 120, 210$
$N*=\text{Number of samples}=27$
We need to determine all possible samples of size $3$ from the given population.
Next, we also need to determine the mean of each sample, which is the sum of the two data values divided by $3$.
$\begin{array}{ c c c | c } \text{1st data value} & \text{2nd data value} & \text{3rd data value} &\text{Mean} \\ \hline 90& 90& 90& \color{#4257b2}(90+90+90)/3=\color{black}90 \\ 90 & 90 & 120 & 100 \\ 90 & 90 & 210 & 130 \\ 90 & 120 & 90 & 100 \\ 90 & 120 & 120 & 110 \\ 90 & 120 & 210 & 140 \\ 90 & 210 & 90 & 130 \\ 90 & 210 & 120 & 140 \\ 90 & 210 & 210 & 170 \\ 120 & 90 & 90 & 100 \\ 120 & 90 & 120 & 110 \\ 120 & 90 & 210 & 140 \\ 120 & 120 & 90 & 110 \\ 120 & 120 & 120 & 120 \\ 120 & 120 & 210 & 150 \\ 120 & 210 & 90 & 140 \\ 120 & 210 & 120 & 150 \\ 120 & 210 & 210 & 180 \\ 210 & 90 & 90 & 130 \\ 210 & 90 & 120 & 140 \\ 210 & 90 & 210 & 170 \\ 210 & 120 & 90 & 140 \\ 210 & 120 & 120 & 150 \\ 210 & 120 & 120 & 180 \\ 210 & 210 & 90 & 170 \\ 210 & 210 & 120 & 180 \\ 210 & 210 & 210 & 210 \end{array}$
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# 1-4 Solving Inequalities - PowerPoint PPT Presentation
1-4 Solving Inequalities. Big Idea: -Solve equations and inequalities. Solving and Graphing Inequalities. As with an equation, the solutions of an inequality are the numbers that make it true. The properties for solving inequalities are similar to the properties for solving equations.
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1-4 Solving Inequalities
Big Idea:
-Solve equations and inequalities.
As with an equation, the solutions of an inequality are the numbers that make it true.
The properties for solving inequalities are similar to the properties for solving equations.
The exception occurs when you multiply or divide each side by a negative.
MULTIPLYING OR DIVIDING BOTH SIDES BY A NEGATIVE REVERSES THE INEQUALITY SYMBOL!
3x ≥ 15
Multiplying
Ex 1: solve and Graph each solution.
A) -2x < 3(x – 5)
B) 7x > 7(2 + x)
Compound Inequality: a pair of inequalities joined by “and” or “or”.
Ex: -1 < x and x ≤ 3 same as -1 <x ≤ 3
x < 2 or x ≥ 5
To solve a compound inequality containing “and”, find all values of the variable that make both inequalities true.
-Name a student that is a girl and wearing red
-Name a teacher that is female and is short
-Find x such that x > 2 and x ≤ 5.
Ex 2: values of the variable that make at least one of the inequalities true.Graph the solution.
A) 2x – 1 < 3x and x > 4x – 9.
B) 3x + 9 < -3 or -2x + 1 < 5 values of the variable that make at least one of the inequalities true..
• Classwork/Homework: values of the variable that make at least one of the inequalities true.
• Page 24 #3-11, 17-28
Essential values of the variable that make at least one of the inequalities true.Question:
What are the similarities and differences between inequalities and equations? |
Finitary Relation
Get Finitary Relation essential facts below. View Videos or join the Finitary Relation discussion. Add Finitary Relation to your PopFlock.com topic list for future reference or share this resource on social media.
Finitary Relation
In mathematics, a finitary relation over sets is a subset of the Cartesian product ; that is, it is a set of n-tuples consisting of elements xi in Xi.[1][2][3][4] Typically, the relation describes a possible connection between the elements of an n-tuple. For example, the relation "x is divisible by y and z" consists of the set of 3-tuples such that when substituted to x, y and z, respectively, make the sentence true.
The non-negative integer n giving the number of "places" in the relation is called the arity, adicity or degree of the relation. A relation with n "places" is variously called an n-ary relation, an n-adic relation or a relation of degree n. Relations with a finite number of places are called finitary relations (or simply relations if the context is clear). It is also possible to generalize the concept to infinitary relations with infinite sequences.[5]
An n-ary relation over sets is an element of the power set of .
0-ary relations count only two members: the one that always holds, and the one that never holds. This is because there is only one 0-tuple, the empty tuple . They are sometimes useful for constructing the base case of an induction argument.
Unary relations can be viewed as a collection of members (such as the collection of Nobel laureates) having some property (such as that of having been awarded the Nobel prize).
Binary relations are the most commonly studied form of finitary relations. When X1 = X2 it is called a homogeneous relation, for example:
• Equality and inequality, denoted by signs such as = and < in statements such as "", or
• Divisibility, denoted by the sign | in statements such as "13|143".
Otherwise it is a heterogeneous relation, for example:
## Example
Consider the ternary relation R "x thinks that y likes z" over the set of people }, defined by:
}.
R can be represented equivalently by the following table:
Relation R "x thinks that y likes z"
P P P
Alice Bob Denise
Charles Alice Bob
Charles Charles Alice
Denise Denise Denise
Here, each row represents a triple of R, that is it makes a statement of the form "x thinks that y likes z". For instance, the first row states that "Alice thinks that Bob likes Denise". All rows are distinct. The ordering of rows is insignificant but the ordering of columns is significant.[1]
The above table is also a simple example of a relational database, a field with theory rooted in relational algebra and applications in data management.[6] Computer scientists, logicians, and mathematicians, however, tend to have different conceptions what a general relation is, and what it is consisted of. For example, databases are designed to deal with empirical data, which is by definition finite, whereas in mathematics, relations with infinite arity (i.e., infinitary relation) are also considered.
## Definitions
When two objects, qualities, classes, or attributes, viewed together by the mind, are seen under some connexion, that connexion is called a relation.
The first definition of relations encountered in mathematics is:
Definition 1. -- An n-ary relation R over sets is a subset of the Cartesian product .[1]
The second definition of relations makes use of an idiom that is common in mathematics, stipulating that "such and such is an n-tuple" in order to ensure that such and such a mathematical object is determined by the specification of mathematical objects with n elements. In the case of a relation R over n sets, there are things to specify, namely, the n sets plus a subset of their Cartesian product. In the idiom, this is expressed by saying that R is a -tuple.
Definition 2. -- An n-ary relation R over sets is an -tuple where G is a subset of the Cartesian product called the graph of R.
As a rule, whatever definition best fits the application at hand will be chosen for that purpose, and if it ever becomes necessary to distinguish between the two definitions, then an entity satisfying the second definition may be called an embedded or included relation.
Both statements in R (under the first definition) and in G (under the second definition) read "x1, ..., xn are R-related" and are denoted using prefix notation by and using postfix notation by . In the case where R is a binary relation, those statements are also denoted using infix notation by .
The following considerations apply under either definition:
• The set Xi is called the ith domain of R.[1] Under the first definition, the relation does not uniquely determine a given sequence of domains. In the case where R is a binary relation, X1 is also called simply the domain or set of departure of R, and X2 is also called the codomain or set of destination of R.
• When the elements of Xi are relations, Xi is called a nonsimple domain of R.[1]
• The set of all xi in Xi for which there exists in such that is called the ith domain of definition or active domain of R.[1] In the case where R is a binary relation, its first domain of definition is also called simply the domain of definition or active domain of R, and its second domain of definition is also called the codomain of definition or active codomain of R.
• When the ith domain of definition of R is equal to Xi, R is said to be total on Xi. In the case where R is a binary relation, when R is total on X1, it is also said to be left-total or serial, and when R is total on X2, it is also said to be right-total or surjective.
• When for all x and y in ?i ? I Xi and for all z in ?i ? J Xi where } is a partition of }, if the components of x and z are R-related and the components of y and z are R-related then , R is said to be unique on , and is called a primary key[1] of R. In the case where R is a binary relation, when R is unique on {X1}, it is also said to be left-unique or injective, and when R is unique on {X2}, it is also said to be right-unique or functional.
• When all Xi are the same set X, it is simpler to refer to R as an n-ary relation over X, called a homogeneous relation. Otherwise R is called a heterogeneous relation.
• When any of Xi is empty, the defining Cartesian product is empty, and the only relation over such a sequence of domains is the empty relation . Hence it is commonly stipulated that all of the domains be nonempty.
Let a Boolean domain B be a two-element set, say, }, whose elements can be interpreted as logical values, typically and . The characteristic function of R, denoted by ?R, is the Boolean-valued function , defined by if and otherwise.
In applied mathematics, computer science and statistics, it is common to refer to a Boolean-valued function as an n-ary predicate. From the more abstract viewpoint of formal logic and model theory, the relation R constitutes a logical model or a relational structure, that serves as one of many possible interpretations of some n-ary predicate symbol.
Because relations arise in many scientific disciplines, as well as in many branches of mathematics and logic, there is considerable variation in terminology. Aside from the set-theoretic extension of a relational concept or term, the term "relation" can also be used to refer to the corresponding logical entity, either the logical comprehension, which is the totality of intensions or abstract properties shared by all elements in the relation, or else the symbols denoting these elements and intensions. Further, some writers of the latter persuasion introduce terms with more concrete connotations (such as "relational structure" for the set-theoretic extension of a given relational concept).
## History
The logician Augustus De Morgan, in work published around 1860, was the first to articulate the notion of relation in anything like its present sense. He also stated the first formal results in the theory of relations (on De Morgan and relations, see Merrill 1990).
Charles Peirce, Gottlob Frege, Georg Cantor, Richard Dedekind and others advanced the theory of relations. Many of their ideas, especially on relations called orders, were summarized in The Principles of Mathematics (1903) where Bertrand Russell made free use of these results.
In 1970, Edgar Codd proposed a relational model for databases, thus anticipating the development of data base management systems.[1]
## References
1. Codd, Edgar Frank (June 1970). "A Relational Model of Data for Large Shared Data Banks" (PDF). Communications of the ACM. 13 (6): 377-387. doi:10.1145/362384.362685. Retrieved .
2. ^ "The Definitive Glossary of Higher Mathematical Jargon -- Relation". Math Vault. 2019-08-01. Retrieved .
3. ^ "Relation - Encyclopedia of Mathematics". www.encyclopediaofmath.org. Retrieved .
4. ^ "Definition of n-ary Relation". cs.odu.edu. Retrieved .
5. ^ Nivat, Maurice (1981). Astesiano, Egidio; Böhm, Corrado (eds.). "Infinitary relations". Caap '81. Lecture Notes in Computer Science. Springer Berlin Heidelberg. 112: 46-75. doi:10.1007/3-540-10828-9_54. ISBN 978-3-540-38716-9.
6. ^ "Relations -- CS441" (PDF). www.pitt.edu. Retrieved .
7. ^ De Morgan, A. (1858) "On the syllogism, part 3" in Heath, P., ed. (1966) On the syllogism and other logical writings. Routledge. P. 119,
## Bibliography
• Codd, Edgar Frank (1990). The Relational Model for Database Management: Version 2 (PDF). Boston: Addison-Wesley. ISBN 978-0201141924.
• Bourbaki, N. (1994) Elements of the History of Mathematics, John Meldrum, trans. Springer-Verlag.
• Carnap, Rudolf (1958) Introduction to Symbolic Logic with Applications. Dover Publications.
• Halmos, P.R. (1960) Naive Set Theory. Princeton NJ: D. Van Nostrand Company.
• Lawvere, F.W., and R. Rosebrugh (2003) Sets for Mathematics, Cambridge Univ. Press.
• Lewis, C.I. (1918) A Survey of Symbolic Logic, Chapter 3: Applications of the Boole--Schröder Algebra, via Internet Archive
• Lucas, J. R. (1999) Conceptual Roots of Mathematics. Routledge.
• Maddux, R.D. (2006) Relation Algebras, vol. 150 in 'Studies in Logic and the Foundations of Mathematics'. Elsevier Science.
• Merrill, Dan D. (1990) Augustus De Morgan and the logic of relations. Kluwer.
• Peirce, C.S. (1870), "Description of a Notation for the Logic of Relatives, Resulting from an Amplification of the Conceptions of Boole's Calculus of Logic", Memoirs of the American Academy of Arts and Sciences 9, 317-78, 1870. Reprinted, Collected Papers CP 3.45-149, Chronological Edition CE 2, 359-429.
• Peirce, C.S. (1984) Writings of Charles S. Peirce: A Chronological Edition, Volume 2, 1867-1871. Peirce Edition Project, eds. Indiana University Press.
• Russell, Bertrand (1903/1938) The Principles of Mathematics, 2nd ed. Cambridge Univ. Press.
• Suppes, Patrick (1960/1972) Axiomatic Set Theory. Dover Publications.
• Tarski, A. (1956/1983) Logic, Semantics, Metamathematics, Papers from 1923 to 1938, J.H. Woodger, trans. 1st edition, Oxford University Press. 2nd edition, J. Corcoran, ed. Indianapolis IN: Hackett Publishing.
• Ulam, S.M. and Bednarek, A.R. (1990), "On the Theory of Relational Structures and Schemata for Parallel Computation", pp. 477-508 in A.R. Bednarek and Françoise Ulam (eds.), Analogies Between Analogies: The Mathematical Reports of S.M. Ulam and His Los Alamos Collaborators, University of California Press, Berkeley, CA.
• Ulam, S.M. (1990) Analogies Between Analogies: The Mathematical Reports of S.M. Ulam and His Los Alamos Collaborators in A.R. Bednarek and Françoise Ulam, eds., University of California Press.
• Roland Fraïssé (2000) [1986] Theory of Relations, North Holland |
# ISEE Middle Level Math : Numbers and Operations
## Example Questions
### Example Question #61 : Numbers And Operations
Natasha is going Christmas shopping for the twelve people on her list. On average, how much money can she spend per gift if she has a $225.00 budget? Possible Answers: Correct answer: Explanation: Divide$225 by 12:
Answer: On average, Natasha can spend \$18.75 per gift.
### Example Question #61 : Numbers And Operations
Which of the following is a factor of 72?
None of these
Explanation:
Factors can be multiplied to get a certain number; when a certain number is divided by a factor, the result is a whole number.
When 18 and 4 are multiplied, the result is 72; thus, 18 and 4 are both factors of 72.
When 72 is divided by the other answer choices, a whole number does NOT result.
### Example Question #63 : Numbers And Operations
There are 15 animals in a pet store. The only animals are dogs and cats. There are twice as many dogs as cats. How many dogs are there?
Explanation:
In this problem, the sum of the cats and dogs must equal 15. If there are 10 dogs, there must be 5 cats because there are twice as many dogs as cats. (Two times 5 is 10.)
Given that 10 plus 5 is 15, 10 is the correct answer.
### Example Question #64 : Numbers And Operations
A dog has a litter of 6 puppies. The average weight of the puppies is 7 pounds. After one week, half of the puppies have gained one pound. What is the new average weight of the puppies after one week?
Explanation:
If half of the 6 puppies gain one pound, that means that 3 puppies will gain one pound. This means that 3 total pounds will be gained among the 6 puppies.
We can assume that the weight of each puppy was originally 7 pounds, since the average was 7. Three of the puppies have gained a pound, meaning they will weigh 8 pounds.
Weights of the puppies after one week: 7, 7, 7, 8, 8, 8
Find the new average by summing the individual weights and dividing by the number of puppies.
### Example Question #65 : Numbers And Operations
Brett buys a shirt. The original price was , but the shirt is on sale for off. Additionally, Brett has a off coupon that he uses after the off is applied. What is the price of the shirt in dollars, before tax?
Explanation:
If the original price of the shirt was , but it is on sale for off, then that means the shirt will be discounted by because of is . . Additionally, with the off coupon, the final price would be .
### Example Question #66 : Numbers And Operations
Sweaters are each, but they are on sale: when you buy one, you get one free. Rebecca buys one sweater and gets one free. At checkout, she then presents a coupon for an additional off. What is the average cost of one sweater in dollars (before tax)?
Explanation:
If Rebecca buys a sweater that is part of a buy one, get one free sale, that means that she will get two sweaters for a total of . A off discount will mean that Rebecca will save off of her total because of is .
Therefore, the total she will pay is equal to
The average price of a sweater will therefore be .
### Example Question #67 : Numbers And Operations
of the students in a classroom of students are boys. of the girls in this classroom wear their hair in pigtails. How many girls wear their hair in pigtails?
Explanation:
If of the students in a classroom of students are boys, that means that students are boys, since . (Alternatively, you could figure this out by realizing that since of is and , of must equal , which is .) If there are boys in the classroom, there must be girls, since .
Given that of the girls wear their hair in pigtails, and of is , it follows that there are girls in the classroom who wear their hair in pigtails.
### Example Question #68 : Numbers And Operations
If the sum of 3 consecutive numbers is 111, what is the value of the middle number?
Explanation:
The sum of . The easiest way to discover the middle number of this set is to divide 111 by 3:
Now that you have the middle number, simply add one to it and subtracy one from it to get your consecutive numbers:
### Example Question #69 : Numbers And Operations
3 brothers weigh 80 pounds, 120 pounds, and 135 pounds. The smallest brother gained 15 pounds over the past year, while the largest brother lost 5 pounds. The middle brother's weight did not change. Based on the brothers' new weights, what is their average weight?
Explanation:
If 3 brothers weigh 80 pounds, 120 pounds and 135 pounds, and the smallest brother gained 15 pounds over the past year, while the largest brother lost 5 pounds, the new weights of the brothers are:
95 pounds, 120 pounds, and 130 pounds. (The middle brother's weight did not change.)
Given that the average is found by dividing the sum of the numbers in a list by the number of items in that list, the following equation will be used:
Therefore, the new average is 115 pounds.
### Example Question #70 : Numbers And Operations
If 25 percent of a number is 3, then what would two thirds of the same number be equal to? |
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# Inverse Trigonometric Functions Help
(not rated)
By — McGraw-Hill Professional
Updated on Sep 12, 2011
## Introduction to Inverse Trigonometric Functions
Figure 6.14 shows the graphs of each of the six trigonometric functions. Notice that each graph has the property that some horizontal line intersects the graph at least twice. Therefore none of these functions is invertible. Another way of seeing this point is that each of the trigonometric functions is 2π-periodic (that is, the function repeats itself every 2π units: f ( x + 2π) = f ( x )), hence each of these functions is not one-to-one.
If we want to discuss inverses for the trigonometric functions, then we must restrict their domains (this concept was introduced in Subsection 1.8.5). In this section we learn the standard methods for performing this restriction operation with the trigonometric functions
### Inverse Sine Function
Consider the sine function with domain restricted to the interval [− π/2, π/2] (Fig. 6.15). We use the notation Sin x to denote this restricted function. Observe that
Fig. 6.14
Fig. 6.15
on the interval (− π/2, π/2). At the endpoints of the interval, and only there, the function Sin x takes the values −1 and +1. Therefore Sin x is increasing on its entire domain. So it is one-to-one. Furthermore the Sine function assumes every value in the interval [−1, 1]. Thus Sin : [− π/2, π/2] → [−1, 1] is one-to-one and onto; therefore f ( x ) = Sin x is an invertible function.
We can obtain the graph of Sin −1 x by the principle of reflection in the line y = x (Fig. 6.16). The function Sin −1 : [−1, 1] → [−π/2, π/2] is increasing, one-to-one, and onto.
Fig. 6.16
### Inverse Cosine Function
The study of the inverse of cosine involves similar considerations, but we must select a different domain for our function. We define Cos x to be the cosine function restricted to the interval [0, ]. Then, as Fig. 6.17 shows, g ( x ) = Cos x is a one-to-one function. It takes on all the values in the interval [−1, 1]. Thus Cos : [0, ] → [−1, 1] is one-to-one and onto; therefore it possesses an inverse.
We reflect the graph of Cos x in the line y = x to obtain the graph of the function Cos −1 . The result is shown in Fig. 6.18.
Calculate
Fig. 6.17
Fig. 6.18
#### Solution 1
We have
Notice that even though the sine function takes the value at many different values of the variable x , the function Sine takes this value only at x = π/3. Similar comments apply to the other two examples.
We also have
We calculate the derivative of f ( t ) = Sin −1 t by using the usual trick for inverse functions. The result is
The derivative of the function Cos −1 t is calculated much like that of Sin −1 t. We find that
#### Example 2
Calculate the following derivatives:
#### Solution 2
We have
You Try It : Calculate ( d/dx )Cos −1 [ x 2 + x ]. Also calculate ( d/dx ) Sin −1 × [ln x − x 3 ].
#### Example 3
Calculate each of the following derivatives:
#### Solution 3
We have
You Try It : Calculate ( d/dx ) ln[Cos −1 x ] and ( d/dx ) exp[Sin −1 x ].
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# This absolute value inequalities lesson uses the weight gain formula as an example
Subjects:
Math, P.E. & Health
10
Title – Absolute Value Inequalities
By – Glen Cotant
Primary Subject – Math
Secondary Subject – PE & Health
Standard:
CA STD 1.0
Total Recall (Warm-up): (5 minutes approx.)
Exercise from yesterday’s lesson on inequalities.
Solve for x and graph the solution for this inequality:
200 < 2000 – 0.5x < 500
(A: 3600
>
x
>
3000 )
Direct Instruction: (10 minutes approx.)
The inequality in today’s “Total Recall” is a formula for
caloric intake with a desire for a weight gain
.
200 < 2000 – 0.5x < 500
2000
is the amount of calories taken in (
what you eat
). This varies.
0.5x
represents a half hour of some exercise ‘x’ burning off calories.
• Brisk jogging ( running ) burns off 800 calories per hour.
• Walking burns off 500 calories per hour.
• What exercise would burn 3000 to 3600 calories per hour?
(None; maximum healthy exercise is around 800 to 1000 calories per hour.)
200 to 500
is the range of the final daily caloric outcome, and since it is positive, it represents a weight gain.
Q:
What would have to change for this to become a weight loss situation?
Q:
What different types of workouts could you add to the list?
Q:
What if you performed two different workouts during the day, one for ½ hour and one for ¾ hour? How would you express that in an inequality?
Direct Instruction: (10 minutes approx.)
Another way to describe a calorie gain between 200 to 500 calories would be:
350 ± 150 calories
The ideal is the middle of the range, and the tolerance is the variation. Here the ideal is 350 and the tolerance is 150. Using absolute value, there is an inequality to describe our initial compound inequality:
| (2000 – 0.5x) – 350 | < 150
Let’s break this down and see if it is the same as our initial compound inequality. The difference of the ACTUAL and IDEAL can be a positive or a negative value, because of the absolute value symbol.
Therefore, we have two cases:
1.
{(2000 – 0.5x) – 350} is a negative value
(
the absolute value will change it to a positive value, but taking away the absolute value – its going to be a negative value
):
-{(2000 – 0.5x) – 350} < 150
and if we multiply both sides by -1, the inequality direction changes and we have:
{(2000 – 0.5x) – 350} > -150
2.
{(2000 – 0.5x) – 350} is a positive value
(
the absolute value doesn’t change it so we end up with
):
{(2000 – 0.5x) – 350} < 150
Putting these both together in a compound inequality:
-150 < {(2000 – 0.5x) – 350} < 150
and if we add 350 to both sides of the inequality, we get:
200 < (2000 – 0.5x) < 500
which is what we started with!
Practice: (5 minutes approx.)
The format:
|ACTUAL – IDEAL| < TOLERANCE
is used extensively in quality control for production manufacturing. In a bolt making factory, the size of the bolt ideally is 1/4″ and each bolt is measured for accuracy to be up to plus or minus 0.005″ or it is rejected. Write an absolute value inequality for the testing of these bolts:
|x – 0.25| < .005
where x is the measure of each bolt.
Direct Instruction: (5 minutes approx.)
Here is a summary of the various absolute value inequalities: (
Copy in notebooks
).
Abs. Value Ineq. Compound Form Graph of Solution |ax + b| < c -c < ax + b < c
Example: |3x + 5| <10
-10 < 3x + 5 < 10
-5 < x < 5/3 |ax + b| < c
-c < ax + b < c |
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Surds calculator that allows you to list the first power somewhere in the problem and make sure it sense! Calculator will show you the step-by-step solution. step 2: Plug the coefficients value positive. \ ( -3 + \frac { 1 } { 2 } x\ ) steps using the square root of (! Check the answer is simple: you arrive the quadratic formula that provides the solution ( s ) a! Calculator Type your algebra problem into the formula or calculator above of other that... Equation this is a quadratic equation solver lets you see all the workings in a by! Eg: enter 3x+2=14, Hit the calculate button or press enter key proceed... Your due diligence as you try to find a calculator used when computing square roots square root equation calculator with steps... 1 } { 2 } x\ ) equation calculator an is an arithmetic calculator and volume... Perfect Combination be certain to do your due diligence as you try to find the discriminant of the equation applying... 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Discriminant ) when necessary here square root equation calculator with steps a perfect square factors of quadratic equations equation calculator - find square roots -... Part of the equation has no real solution. sure to simplify all radical expressions rationalize. And it shows all the steps and graphs of the form: with \ b^2! But I do n't know how to find the square root on each of. That I have two roots or zeroes namely ; Root1 and Root2, this app featured! After decimal point, pair Them from left to right ) … this polynomial is considered to have perfect... And practice number is between the\: square\: x^2-x-6 \$ use this equation calculator an an... Is 4, as 4 x 4 = 0 please use at your own,. |
# Simplifying Trigonometric Expressions using Sum and Difference Formulas
## Simplify sine, cosine, and tangent of angles that are added or subtracted.
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Simplifying Trig Expressions using Sum and Difference Formulas
As Agent Trigonometry you are given this clue: \begin{align*}\sin(\frac{\pi}{2} - x)\end{align*}. How could you simplify this expression to make solving your case easier?
### Simplifying Trigonometric Expressions
We can also use the sum and difference formulas to simplify trigonometric expressions.
The \begin{align*}\sin a = -\frac{3}{5}\end{align*} and \begin{align*}\cos b =\frac{12}{13}\end{align*}. \begin{align*}a\end{align*} is in the \begin{align*}3^{rd}\end{align*} quadrant and \begin{align*}b\end{align*} is in the \begin{align*}1^{st}\end{align*}. Let's find \begin{align*}\sin(a+b)\end{align*}.
First, we need to find \begin{align*}\cos a\end{align*} and \begin{align*}\sin b\end{align*}. Using the Pythagorean Theorem, missing lengths are 4 and 5, respectively. So, \begin{align*}\cos a=-\frac{4}{5}\end{align*} because it is in the \begin{align*}3^{rd}\end{align*} quadrant and \begin{align*}\sin b = \frac{5}{13}\end{align*}. Now, use the appropriate formulas.
\begin{align*}\sin (a+b) &=\sin a \cos b + \cos a \sin b \\ &= -\frac{3}{5}\cdot \frac{12}{13}+-\frac{4}{5}\cdot \frac{5}{13} \\ &= -\frac{56}{65}\end{align*}
Now, using the information from the previous problem above, let's find \begin{align*}\tan (a+b)\end{align*}.
From the cosine and sine of \begin{align*}a\end{align*} and \begin{align*}b\end{align*}, we know that \begin{align*}\tan a=\frac{3}{4}\end{align*} and \begin{align*}\tan b=\frac{5}{12}\end{align*}.
\begin{align*}\tan (a+b) &=\frac{\tan a +\tan b}{1-\tan a \tan b} \\ &= \frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot\frac{5}{12}} \\ &= \frac{\frac{14}{12}}{\frac{11}{16}}=\frac{56}{33}\end{align*}
Finally, let's simplify \begin{align*}\cos (\pi - x)\end{align*}.
Expand this using the difference formula and then simplify.
\begin{align*}\cos (\pi - x) &=\cos \pi \cos x +\sin \pi \sin x \\ &=-1\cdot \cos x +0\cdot \sin x \\ &=-\cos x\end{align*}
### Examples
#### Example 1
Earlier, you were asked to simplify \begin{align*}\sin(\frac{\pi}{2} - x)\end{align*}.
You can expand the expression using the difference formula and then simplify.
\begin{align*}\sin(\frac{\pi}{2} - x)=\sin \frac{\pi}{2} \cos x - \cos \frac{\pi}{2} \sin x \\ &=1\cdot \cos x - 0\cdot \sin x \\ &=\cos x\end{align*}
#### Example 2
Using the information from the first problem above (where we found \begin{align*}\sin (a+b)\end{align*}), find \begin{align*}\cos(a-b)\end{align*}.
\begin{align*}\cos(a-b) &=\cos a \cos b + \sin a \sin b \\ &=-\frac{4}{5}\cdot \frac{12}{13}+-\frac{3}{5}\cdot\frac{5}{13} \\ &=-\frac{63}{65}\end{align*}
#### Example 3
Simplify \begin{align*}\tan (x+\pi)\end{align*}.
\begin{align*}\tan (x+\pi)&=\frac{\tan x +\tan \pi}{1-\tan x \tan \pi} \\ &=\frac{\tan x +0}{1-\tan 0} \\ &=\tan x\end{align*}
### Review
\begin{align*}\sin a =-\frac{8}{17}, \pi \le a < \frac{3\pi}{2}\end{align*} and \begin{align*}\sin b =-\frac{1}{2}, \frac{3\pi}{2}\le b <2\pi\end{align*}. Find the exact trig values of:
1. \begin{align*}\sin (a+b)\end{align*}
2. \begin{align*}\cos (a+b)\end{align*}
3. \begin{align*}\sin (a-b)\end{align*}
4. \begin{align*}\tan (a+b)\end{align*}
5. \begin{align*}\cos (a-b)\end{align*}
6. \begin{align*}\tan (a-b)\end{align*}
Simplify the following expressions.
1. \begin{align*}\sin (2\pi-x)\end{align*}
2. \begin{align*}\sin \left(\frac{\pi}{2}+x\right)\end{align*}
3. \begin{align*}\cos (x+\pi)\end{align*}
4. \begin{align*}\cos \left(\frac{3\pi}{2}-x\right)\end{align*}
5. \begin{align*}\tan(x+2\pi)\end{align*}
6. \begin{align*}\tan(x-\pi)\end{align*}
7. \begin{align*}\sin \left(\frac{\pi}{6}-x\right)\end{align*}
8. \begin{align*}\tan \left(\frac{\pi}{4}+x\right)\end{align*}
9. \begin{align*}\cos \left(x-\frac{\pi}{3}\right)\end{align*}
Determine if the following trig statements are true or false.
1. \begin{align*}\sin(\pi - x)=\sin (x-\pi)\end{align*}
2. \begin{align*}\cos(\pi - x)=\cos (x-\pi)\end{align*}
3. \begin{align*}\tan(\pi - x)=\tan (x-\pi)\end{align*}
### Answers for Review Problems
To see the Review answers, open this PDF file and look for section 14.13.
### Notes/Highlights Having trouble? Report an issue.
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FUNDAMENTALS OF ALGEBRA
POLYNOMIALS
A variable is a letter that can represent any number from a given set of real numbers. The common variables used are x, y, z etc. When we use addition, subtraction, multiplication, division, powers, roots on the variables, we get algebraic expressions. Examples : 2x 2 @ 3x + 2, 2 p y + x2 ,
w w w w w w w
xf+fff ffff ffff ffff fy etc are algebraic expressions. x @y
A monomial is a real number or a real number multiplied with variables with powers as 2f f whole numbers. 6x 2 , 2y, @ 5, @ fx 3 y 2 z are examples of monomials. 3
Sum of two or more monomials is called a polynomial. [Sum of two monomials is called a binomial.] 1f 1f f f Example: fx 2 @ 3x + 2 is a polynomial which has three terms fx 2 , @ 3x and 2 A 3 3
COMBINING ALGEBRAIC EXPRESSIONS: Addition and Subtraction :
To combine polynomials, we use the same properties of real number operations (see the chapter : Sets, Numbers, Operations, Properties. The terms in which the same variables are raised to the same powers are the like terms. 1f f Among the terms 3x 2 , @ 4xy 2 , 2xy, @ fxy 3 z, @ x 2 , 3xy 3 z, 10xy, 7xy 2 the like terms are 2 1f f 3x 2 and @ x 2 , @ 4xy 2 and 7xy 2 , 2xy and 10xy, @ fxy 3 z and 3xy 3 z A 2 To add or subtract two polynomials, we combine the like terms by using the Distributive Property. Example : 3x 2 + 5x 2 = 3 + 5 x 2 = 8x 2
` a
If there is a minus sign before a parentheses, while removing the parentheses, signs of all the terms inside the parentheses are changed.
@ 2x 2 @ 3x + 4 = @ 2x 2 + 3x @ 4
b c
Example : (a) Find the sum of 2x 3 @ 3x 2 + x @ 5 and x 3 + 2x 2 @ 2x + 1 (b) Subtract x 3 + 2x 2 @ 2x + 1 from 2x 3 @ 3x 2 + x @ 5 Solution : b c b c ` a a 2x 3 @ 3x 2 + x @ 5 + x 3 + 2x 2 @ 2x + 1
b c b c `
b c b
b
c
b
c
c
= 2x 3 + x 3 + @ 3x 2 + 2x 2 + x @ 2x + @ 5 + 1
a `
= 3x 3 @ x 2 @ x @ 4 c b c ` a b 3 b 2x @ 3x 2 + x @ 5 @ x 3 + 2x 2 @ 2x + 1
a
= b 3 @ 3x 2 + x @ 5 @ x 3 @ c 2 + 2x @ 1 2x 2x c b ` a ` a = 2x 3 @ x 3 + @ 3x 2 @ 2x 2 + x + 2x + @ 5 @ 1 = x 3 @ 5x 2 + 3x @ 6
Multiplication :
To find the product of polynomials we use the Distributive property and laws of exponents. Example : Find the product 3x 2 @ 5 x 3 + 2x 2 @ 2x + 1
b cb c
Solution : b cb c 3x 2 @ 5 x 3 + 2x 2 @ 2x + 1
b c
= 3x 2 x 3 + 2x 2 @ 2x + 1 @ 5 x 3 + 2x 2 @ 2x + 1
b
= 3x 5 + 6x 4 @ 6x 3 + 3x 2 @ 5x 3 @ 10x 2 + 10x @ 5 Distributive property = 3x 5 + 6x 4 @ 11x 3 @ 7x 2 + 10x @ 5 Combining the like terms
c
Distributive property
SOME SPECIAL PRODUCT FORMULAS:
1A A + B A @B = A @B 2 A A + B = A + 2AB + B
` ` ` ` a2 a3
2
`
a`
a
2
2
2 2 2 3 3
3 A A @ B = A @ 2AB + B
a2 a3
2 3 2 3 2
4 A A + B = A + 3A B + 3AB + B
2
5 A A @ B = A @ 3A B + 3AB @ B 6 A A + B A @ AB + B = A + B
` `
2 2 3
7 A A @ B A + AB + B = A @ B
2 2 3
ab
ab
c
c
3
3
Example : Find 2x @ 3 and 2 + 3y 2 @ 3y
` b cb
a2
c
Solution :
= 2x @ 2.2x A 3 + 3 = 4x 2 @ 12x + 9
b ` a2
`
2x @ 3
a2
2
taking A = 2x,B = 3 and using A @ B = A @ 2AB + B
` a2
2
2
2 + 3y 2 @ 3y
2
cb
= 2 @ 3y
b
= 4 @ 9 y2
c2
c
taking A = 2,B = 3y and using A + B A @ B = A @ B
` a` a
2
2
FACTORING
Factoring is the opposite process of finding products. Here a polynomial is to be written as the product of two or more simpler polynomials. If we multiply 2x @ 3 ` with 4x + 5` we get , ` a` a a a 2x @ 3 4x + 5 = 2x 4x + 5 @ 3 4x + 5
= 8x 2 + 10x @ 12x @ 15 = 8x 2 @ 2x @ 15 On the reverse, if we are `asked to factor 8x 2 @ 2x @ 15, we have to write ` a a 8x 2 @ 2x @ 15 = 2x @ 3 4x + 5 . ` a ` a Here, 2x @ 3 and 4x + 5 are the factors of 8x 2 @ 2x @ 15.
To factor a polynomial, we use the following methods wherever applicable :
Factoring out common factors:
If all the terms of a polynomial have one or more common factors, we take out the greatest common factor and use the Distributive Property. Example : Factor a each polynomial/expression : ` a 5x 2 y @ 10xy
` a
2
Solution : ` a 2 a 5x y = 5 Bx Bx By 10xy = 5 B2 Bx By
Hence
b 6ab c 3 @ 18a 2 c 2 + 12ac 2 ` a` a` a ` a c 2x + 4 2x @ 4 @ 3 2x @ 4
The greatest common factor is 5 Bx By or 5xy
` a
b 6ab c 3 = 2 B3 Ba Bb Bb Bc Bc Bc
2
5x 2 y @ 10xy = 5xy Bx @ 5xy B2 ` a = 5xy x @ 2
18a 2 c 2 = 2 B3 B3 Ba Ba Bc Bc 12ac 2 = 2 B2 B3 Ba Bc Bc The geatest common factor is 2 B3 Ba Bc Bc or 6ac 2 Hence b c 2 2 6ab c 3 @ 18a 2 c 2 + 12ac 2 = 6ac 2 b c @ 3a + 2
` a`
Hence,
c 2x + 4 2x @ 4 @ 3 2x @ 4 ` a the greatest common factor is 2x @ 4
a` a ` a ` a` a `
2x + 4 2x @ 4 @ 3 2x @ 4 = 2x @ 4
a ` `
= 2x @ 4 2x + 1
a`
aB`
2x + 4 @ 3
a a
C
Factoring by using the Product Formulas given above:
See the right hand sides of all the special product formulas given above. The left hand sides are in the factor form. If any expression is in the form of the right hand sides, we can use the reverse of that. Example :aFactor ` a 4x 2 @ 49
` a ` a
b 27x 3 + c 3 c 4a 3 @ 32b
3
Solution :
` a ` a
a 4x 2 @ 49 = 2x @ 7 ` a` a = 2x + 7 2x @ 7
` a2
2
using the reverse of A + B A @ B = A @ B
` a` a
2
2
b 27x 3 + c 3 = 3x + c 3
` ` ` a3 aB` ab a2
3
using the reverse of A + B A @ AB + B = A + B
C ` ab
2 2
= 3x + c 3x @ 3x A c + c 2 = 3x + c 9x 2 @ 3cx + c 2
c
c
3
3
` a
c 4a 3 @ 32b = 4 a 3 @ 8b
3
= 4 a @ 2b a 2 + a A 2b + 2b = 4 a @ 2b a 2 + 2ab + 4b
` ab ` `
2
b
aB
c
using the reverse of
a2C
`
A @ B A + AB + B = A @ B
ab
2 2
c
3
3
c
Factoring by grouping :
Example` : a Factor a 2x 3 + x 2 @ 6x @ 3
` a
b @ 6x 3 @ 2x 2 + 3x + 1
Solution : b c ` ` a 3 a a 2x + x 2 @ 6x @ 3 = 2x 3 + x 2 @ 6x + 3
Group terms
a
= x 2x + 1 @ 2 2x + 1 = x 2 @ 2 2x + 1
b ` 2 b c` a ` a c ` a
factor out common factors
` a
b @ 6x 3 @ 2x 2 + 3x + 1 = @ 6x 3 + 2x 2 + 3x + 1 = @ 2x 2 + 1 3x + 1
b ` c` `
= @ 2x 2 3x + 1 + 1 3x + 1
a
a
a
A rational expression is an expression of the type
` a
Pffff xf ffff fff ` polynomials and Q x ≠ 0. Also, fffafis to be in the simplest form (i.e. after canceling Q x out any common factor other than 1 or -1).
2 1 x+1 ffff 3xffffffff ffffffff ffff fffffffff ffffffff ffff fffffffff fffffff f @x @2 b c Examples : ffff, fffffffff, ffffffffare rational expressions. x @1 x @2 x x3@ 1
` a
Pffff x ffff ffff ffff ` a where Q x
` a
P x andQ x are
` a
` a
Some important operations of Rational Expressions: b c These are similar to operations with fractions
If A, B, C, D are polynomials, then ACf Af fff ff fff f ff ff f = 1A BC B Af Cf ACf ff ff fff f ff fff f f ff ff 2A A = B D BD Af Cf ADf ff ff fff f ff fff f f ff ff 3A D = B D BC Af Bf Affff +B ff ff fffff f f fffff f f fffff 4A + = C C C Af Bf Affff @B ff ff fffff f f fffff f f fffff 5A @ = C C C Af Cf ff ff f ff f f 6 A = if and only if AD = BC B D
Example : Simplify 2 ` a xffffffff @x @2 ffffffff ffffffff a ffffffff x2@ 1
Solution : a x2@ 1
2 ` a xffffffff @x @2 ffffffff ffffffff ffffffff
` a ffff fffff x 3 ffff ffff ffff ffff ffff ffff
b
x+1
@
x @2
x @1 x + 1 xffff @2 ffff ffff = fffff x @1 ` a ` a 3 x @2 x x+1 ` a ffff fffff fffffffffff fffffffffff 3 ffff fx ff fffffffffff fffffffffff ffff ffff fffffffffff fffffffffff ffff ffff fffffffffff fffffffffff f a` a@` a` a b @ =` x @2 x + 1 x + 1 x @2 x + 1 x @2 ` a ` a 3ffffffffffff+fff x fff fff x f 1 fff@ffffffffffff ffffff@ffffffff ffff2 ff x ffffff a` a = ` x @2 x + 1
2 3xffff@ffffff ffffffffffff ffffffx ffff f@ 6 fff @ x ` a` a = fffffffffffff x @2 x + 1 2 @ffffffffff fffffffffff fx f+fff@ 6 f ff f 2x fff a` a = `fffffffffff x @2 x + 1
` a` a xfffffffffff @2 x + f fffffffffff fffffffff1f fffffffffff a` a =`
Example: Simplify 2 ` a xfffff xfffffffff ` a ffff ffffff fffff 2 3 @ 3x @ 4 1 fffff fffffffff f@ff ff 4 f fffffffff fff fffffffff ffff ffffff fffff ffff ffffff fffff ffff ffffff ffff a 2 D 2 b @` a2 + 2 x+1 x @1 x @ 4 x + 5x + 6 x+1 Solution:
2 ` a xfffff xfffffffff @4 @ 3x @ 4 fffff fffffffff fffff fffffffff ffff fffffffff
D x 2 @ 4 x 2 + 5x + 6 2 xfffff xfffffffff + 5x + 6 f@ff fffffffff ff 4 f fffffffff fff = fffff fffffffff B 2 2 x @ 4 x @ 3x @ 4 ` a` a` a x @4 x + 2 x + 3 fffffffffffffffffffffff fffffffffffffffffffffff fffffffffffffffffffffff ffffffffffffffffffffff a` a` a` a =` x + 2 x @2 x @4 x + 1 x+3 fffffffffff fffffffffff fffffffffff a` a = `fffffffffff x @2 x + 1 ` a ffff ffffff fffff 2 f 1 ffff ffffff f3fff ffff ffff f ff b ffff ` ffffaff fffff @ + 2 fff 2 x+1 x @1 x+1
a
1 3 f f fffffffffff ffff fffffffffff fffffffffff ffff ffff2 fffff ffff ffffffffff fffffffffff a a+ a` a = ffff@ `fffff`ffffff `fffffffffff x+1 x+1 x+1 x + 1 x @1 ` a` a 1 @f 3 +1 f+ 1 x @ 1 ffffffffff f ffff fffffffffff ffff fffffffffff fffffffffff xffff fffffffffff xffff ffff x ffffffffff fffffffffff ffff fffffffffff ffff ffff fffffffffff ffff2ffffff ffff fffffffffff ffff ff1f fffffffffff ffff a` a@ ` a` aB a` aB = B` +` x + 1 x + 1 x @1 x + 1 x + 1 x @1 x + 1 x @1 x + 1 ` a` a ` a ` a x ff1 fffffffffffffffffffffffff fffffffffffffffffffffffffffff fffffffffffffffffffffffffffff ffffffffffffffffffffffffffff f+ f x @ 1 @ 2 x @ 1 + 3 x + 1 ` a` a` a = x @1 x + 1 x + 1
2 xfffffffff+ffffffffff @ f @ ff f 2 + 3x f 3 fff1ffffffffffff+ff ffffff2x fffffffffff ` a` a` a = ffffffffffffffffffff x @1 x + 1 x + 1
x 2 + x +ff fffffffffffffffff ffffffffffffffff ffffffffff4 fffff a` ` a = `ffffffffffaffffff x @1 x + 1 x + 1
Some reminders in algebra:
1A 2A 3A 4A
`
a + b ≠ a2 + b
a2
2
as
wwww wwww wwww wwww wwww wwww wwww www qa 2 + b 2 ≠ a +
`
a + b = a + b a + b = a 2 + 2ab + b
a2 ` a` a
2
b
as squaring both sides we get as as as
a2 + b ≠ a + b
2
1f 1f ffff 1 f f ffff f f ffff f f ffff + ≠ a b a+b affff +b ffff ffff ffff ≠b a
1f 1f 1f bf 1f af ff ff bffff +a f f f f f f b f a f fffff f f f f f f ff ff ffff f f f f f f ff ff ffff f f + = B + B = + = a b a b b a ab ab ab affff af bf +b bf ffff f f ffff f f ffff f f f f f = + =1+ a a a a
wwww wwww wwww wwww www www www www www www wwww wwww wwww www www www q a 2 A b 2 = q` a A ba2 = ab
`
a2
f
but
abf ff ff ff f =a b
g
wwww wwww www www www www www www qa 2 A b 2 = a A b 5A
EXPONENTS & RADICALS: These are discussed in the chapter Exponents and Logarithm. |
# Square Root (Sqrt) Decomposition Algorithm
• Difficulty Level : Medium
• Last Updated : 21 Dec, 2022
Square Root Decomposition Technique is one of the most common query optimization techniques used by competitive programmers. This technique helps us to reduce Time Complexity by a factor of sqrt(N)
The key concept of this technique is to decompose a given array into small chunks specifically of size sqrt(N)
Follow the below steps to solve the problem:
• We have an array of n elements and we decompose this array into small chunks of size sqrt(N)
• We will be having exactly sqrt(N) such chunks provided that N is a perfect square
• Therefore, now our array of N elements is decomposed into sqrt(N) blocks, where each block contains sqrt(N) elements (assuming the size of the array is a perfect square)
• Let’s consider these chunks or blocks as an individual array each of which contains sqrt(N) elements and you have computed your desired answer(according to your problem) individually for all the chunks
• Now, you need to answer certain queries asking you the answer for the elements in the range l to r(l and r are starting and ending indices of the array) in the original n-sized array
Naive Approach: To solve the problem follow the below idea:
Simply iterate over each element in the range l to r and calculate its corresponding answer. Therefore, the Time Complexity per query will be O(N)
Efficient Approach(Sqrt Decomposition Trick): To solve the problem follow the below idea:
As we have already precomputed the answer for all individual chunks and now we need to answer the queries in range l to r. Now we can simply combine the answers of the chunks that lie in between the range l to r in the original array. So, if we see carefully here we are jumping sqrt(N) steps at a time instead of jumping 1 step at a time as done in the naive approach. Let’s just analyze its Time Complexity and implementation considering the below problem:
Problem :
Given an array of n elements. We need to answer q
queries telling the sum of elements in range l to
r in the array. Also the array is not static i.e
the values are changed via some point update query.
Range Sum Queries are of form : Q l r
where l is the starting index r is the ending index
Point update Query is of form : U idx val
where idx is the index to update val is the
updated value
Below is the illustration of the above approach:
Let us consider that we have an array of 9 elements: A[] = {1, 5, 2, 4, 6, 1, 3, 5, 7}
1. Let’s decompose this array into sqrt(9) blocks, where each block will contain the sum of elements lying in it. Therefore now our decomposed array would look like this:
2. Till now we have constructed the decomposed array of sqrt(9) blocks and now we need to print the sum of elements in a given range.
So first let’s see two basic types of overlap that a range query can have on our array:
### Range Query of type 1 (Given Range is on Block Boundaries) :
In this type the query, the range may totally cover the continuous sqrt blocks. So we can easily answer the sum of values in this range as the sum of completely overlapped blocks.
So the answer for the above query in the described image will be: ans = 11 + 15 = 26
Time Complexity: O(sqrt(N)). In the worst case, our range can be 0 to N-1(where N is the size of the array and assuming N to be a perfect square). In this case, all the blocks are completely overlapped by our query range. Therefore, to answer this query we need to iterate over all the decomposed blocks for the array and we know that the number of blocks = sqrt(N). Hence, the complexity for this type of query will be O(sqrt(N)) in the worst case.
### Range Query of type 2 (Given Range is NOT on boundaries):
We can deal with these types of queries by summing the data from the completely overlapped decomposed blocks lying in the query range and then summing elements one by one from the original array whose corresponding block is not completely overlapped by the query range.
So the answer for the above query in the described image will be: ans = 5 + 2 + 11 + 3 = 21
Time Complexity: O(sqrt(N)). Let’s consider a query [l = 1 and r = n-2] (n is the size of the array and has 0-based indexing). Therefore, for this query exactly ( sqrt(n) – 2 ) blocks will be completely overlapped whereas the first and last blocks will be partially overlapped with just one element left outside the overlapping range. So, the completely overlapped blocks can be summed up in ( sqrt(n) – 2 ) ~ sqrt(n) iterations, whereas the first and last blocks are needed to be traversed one by one separately. But as we know that the number of elements in each block is at max sqrt(n), to sum up, the last block individually we need to make,
(sqrt(n)-1) ~ sqrt(n) iterations and same for the last block.
So, the overall Complexity = O(sqrt(n)) + O(sqrt(n)) + O(sqrt(n)) = O(3*sqrt(N)) = O(sqrt(N))
### Update Queries(Point update):
In this query, we simply find the block in which the given index lies, then subtract its previous value and add the new updated value as per the point update query.
Time Complexity: O(1)
Below is the implementation of the above approach:
## C++
`// C++ program to demonstrate working of Square Root``// Decomposition.``#include ``using` `namespace` `std;` `#define MAXN 10000``#define SQRSIZE 100` `int` `arr[MAXN]; ``// original array``int` `block[SQRSIZE]; ``// decomposed array``int` `blk_sz; ``// block size` `// Time Complexity : O(1)``void` `update(``int` `idx, ``int` `val)``{`` ``int` `blockNumber = idx / blk_sz;`` ``block[blockNumber] += val - arr[idx];`` ``arr[idx] = val;``}` `// Time Complexity : O(sqrt(n))``int` `query(``int` `l, ``int` `r)``{`` ``int` `sum = 0;`` ``while` `(l < r and l % blk_sz != 0 and l != 0) {`` ``// traversing first block in range`` ``sum += arr[l];`` ``l++;`` ``}`` ``while` `(l + blk_sz - 1 <= r) {`` ``// traversing completely overlapped blocks in range`` ``sum += block[l / blk_sz];`` ``l += blk_sz;`` ``}`` ``while` `(l <= r) {`` ``// traversing last block in range`` ``sum += arr[l];`` ``l++;`` ``}`` ``return` `sum;``}` `// Fills values in input[]``void` `preprocess(``int` `input[], ``int` `n)``{`` ``// initiating block pointer`` ``int` `blk_idx = -1;` ` ``// calculating size of block`` ``blk_sz = ``sqrt``(n);` ` ``// building the decomposed array`` ``for` `(``int` `i = 0; i < n; i++) {`` ``arr[i] = input[i];`` ``if` `(i % blk_sz == 0) {`` ``// entering next block`` ``// incrementing block pointer`` ``blk_idx++;`` ``}`` ``block[blk_idx] += arr[i];`` ``}``}` `// Driver code``int` `main()``{`` ``// We have used separate array for input because`` ``// the purpose of this code is to explain SQRT`` ``// decomposition in competitive programming where`` ``// we have multiple inputs.`` ``int` `input[] = { 1, 5, 2, 4, 6, 1, 3, 5, 7, 10 };`` ``int` `n = ``sizeof``(input) / ``sizeof``(input[0]);` ` ``preprocess(input, n);` ` ``cout << ``"query(3,8) : "` `<< query(3, 8) << endl;`` ``cout << ``"query(1,6) : "` `<< query(1, 6) << endl;`` ``update(8, 0);`` ``cout << ``"query(8,8) : "` `<< query(8, 8) << endl;`` ``return` `0;``}`
## Java
`// Java program to demonstrate working of``// Square Root Decomposition.``import` `java.util.*;` `class` `GFG {` ` ``static` `int` `MAXN = ``10000``;`` ``static` `int` `SQRSIZE = ``100``;` ` ``static` `int``[] arr = ``new` `int``[MAXN]; ``// original array`` ``static` `int``[] block`` ``= ``new` `int``[SQRSIZE]; ``// decomposed array`` ``static` `int` `blk_sz; ``// block size` ` ``// Time Complexity : O(1)`` ``static` `void` `update(``int` `idx, ``int` `val)`` ``{`` ``int` `blockNumber = idx / blk_sz;`` ``block[blockNumber] += val - arr[idx];`` ``arr[idx] = val;`` ``}` ` ``// Time Complexity : O(sqrt(n))`` ``static` `int` `query(``int` `l, ``int` `r)`` ``{`` ``int` `sum = ``0``;`` ``while` `(l < r && l % blk_sz != ``0` `&& l != ``0``) {`` ``// traversing first block in range`` ``sum += arr[l];`` ``l++;`` ``}`` ``while` `(l + blk_sz - ``1` `<= r) {`` ``// traversing completely`` ``// overlapped blocks in range`` ``sum += block[l / blk_sz];`` ``l += blk_sz;`` ``}`` ``while` `(l <= r) {`` ``// traversing last block in range`` ``sum += arr[l];`` ``l++;`` ``}`` ``return` `sum;`` ``}` ` ``// Fills values in input[]`` ``static` `void` `preprocess(``int` `input[], ``int` `n)`` ``{`` ``// initiating block pointer`` ``int` `blk_idx = -``1``;` ` ``// calculating size of block`` ``blk_sz = (``int``)Math.sqrt(n);` ` ``// building the decomposed array`` ``for` `(``int` `i = ``0``; i < n; i++) {`` ``arr[i] = input[i];`` ``if` `(i % blk_sz == ``0``) {`` ``// entering next block`` ``// incrementing block pointer`` ``blk_idx++;`` ``}`` ``block[blk_idx] += arr[i];`` ``}`` ``}` ` ``// Driver code`` ``public` `static` `void` `main(String[] args)`` ``{` ` ``// We have used separate array for input because`` ``// the purpose of this code is to explain SQRT`` ``// decomposition in competitive programming where`` ``// we have multiple inputs.`` ``int` `input[] = { ``1``, ``5``, ``2``, ``4``, ``6``, ``1``, ``3``, ``5``, ``7``, ``10` `};`` ``int` `n = input.length;` ` ``preprocess(input, n);` ` ``System.out.println(``"query(3, 8) : "` `+ query(``3``, ``8``));`` ``System.out.println(``"query(1, 6) : "` `+ query(``1``, ``6``));`` ``update(``8``, ``0``);`` ``System.out.println(``"query(8, 8) : "` `+ query(``8``, ``8``));`` ``}``}` `// This code is contributed by PrinciRaj1992`
## Python 3
`# Python 3 program to demonstrate working of Square Root``# Decomposition.``from` `math ``import` `sqrt` `MAXN ``=` `10000``SQRSIZE ``=` `100` `arr ``=` `[``0``]``*``(MAXN) ``# original array``block ``=` `[``0``]``*``(SQRSIZE) ``# decomposed array``blk_sz ``=` `0` `# block size` `# Time Complexity : O(1)` `def` `update(idx, val):`` ``blockNumber ``=` `idx ``/``/` `blk_sz`` ``block[blockNumber] ``+``=` `val ``-` `arr[idx]`` ``arr[idx] ``=` `val` `# Time Complexity : O(sqrt(n))` `def` `query(l, r):`` ``sum` `=` `0`` ``while` `(l < r ``and` `l ``%` `blk_sz !``=` `0` `and` `l !``=` `0``):` ` ``# traversing first block in range`` ``sum` `+``=` `arr[l]`` ``l ``+``=` `1` ` ``while` `(l ``+` `blk_sz ``-` `1` `<``=` `r):` ` ``# traversing completely overlapped blocks in range`` ``sum` `+``=` `block[l``/``/``blk_sz]`` ``l ``+``=` `blk_sz` ` ``while` `(l <``=` `r):` ` ``# traversing last block in range`` ``sum` `+``=` `arr[l]`` ``l ``+``=` `1` ` ``return` `sum` `# Fills values in input[]` `def` `preprocess(``input``, n):` ` ``# initiating block pointer`` ``blk_idx ``=` `-``1` ` ``# calculating size of block`` ``global` `blk_sz`` ``blk_sz ``=` `int``(sqrt(n))` ` ``# building the decomposed array`` ``for` `i ``in` `range``(n):`` ``arr[i] ``=` `input``[i]`` ``if` `(i ``%` `blk_sz ``=``=` `0``):` ` ``# entering next block`` ``# incrementing block pointer`` ``blk_idx ``+``=` `1` ` ``block[blk_idx] ``+``=` `arr[i]` `# Driver code` `# We have used separate array for input because``# the purpose of this code is to explain SQRT``# decomposition in competitive programming where``# we have multiple inputs.``input` `=` `[``1``, ``5``, ``2``, ``4``, ``6``, ``1``, ``3``, ``5``, ``7``, ``10``]``n ``=` `len``(``input``)` `preprocess(``input``, n)` `print``(``"query(3,8) : "``, query(``3``, ``8``))``print``(``"query(1,6) : "``, query(``1``, ``6``))``update(``8``, ``0``)``print``(``"query(8,8) : "``, query(``8``, ``8``))` `# This code is contributed by Sanjit_Prasad`
## C#
`// C# program to demonstrate working of``// Square Root Decomposition.``using` `System;` `class` `GFG {`` ``static` `int` `MAXN = 10000;`` ``static` `int` `SQRSIZE = 100;` ` ``static` `int``[] arr = ``new` `int``[MAXN]; ``// original array`` ``static` `int``[] block`` ``= ``new` `int``[SQRSIZE]; ``// decomposed array`` ``static` `int` `blk_sz; ``// block size` ` ``// Time Complexity : O(1)`` ``static` `void` `update(``int` `idx, ``int` `val)`` ``{`` ``int` `blockNumber = idx / blk_sz;`` ``block[blockNumber] += val - arr[idx];`` ``arr[idx] = val;`` ``}` ` ``// Time Complexity : O(sqrt(n))`` ``static` `int` `query(``int` `l, ``int` `r)`` ``{`` ``int` `sum = 0;`` ``while` `(l < r && l % blk_sz != 0 && l != 0) {`` ``// traversing first block in range`` ``sum += arr[l];`` ``l++;`` ``}`` ``while` `(l + blk_sz - 1 <= r) {`` ``// traversing completely`` ``// overlapped blocks in range`` ``sum += block[l / blk_sz];`` ``l += blk_sz;`` ``}`` ``while` `(l <= r) {`` ``// traversing last block in range`` ``sum += arr[l];`` ``l++;`` ``}`` ``return` `sum;`` ``}` ` ``// Fills values in input[]`` ``static` `void` `preprocess(``int``[] input, ``int` `n)`` ``{`` ``// initiating block pointer`` ``int` `blk_idx = -1;` ` ``// calculating size of block`` ``blk_sz = (``int``)Math.Sqrt(n);` ` ``// building the decomposed array`` ``for` `(``int` `i = 0; i < n; i++) {`` ``arr[i] = input[i];`` ``if` `(i % blk_sz == 0) {`` ``// entering next block`` ``// incrementing block pointer`` ``blk_idx++;`` ``}`` ``block[blk_idx] += arr[i];`` ``}`` ``}` ` ``// Driver code`` ``public` `static` `void` `Main(String[] args)`` ``{` ` ``// We have used separate array for input because`` ``// the purpose of this code is to explain SQRT`` ``// decomposition in competitive programming where`` ``// we have multiple inputs.`` ``int``[] input = { 1, 5, 2, 4, 6, 1, 3, 5, 7, 10 };`` ``int` `n = input.Length;` ` ``preprocess(input, n);` ` ``Console.WriteLine(``"query(3, 8) : "` `+ query(3, 8));`` ``Console.WriteLine(``"query(1, 6) : "` `+ query(1, 6));`` ``update(8, 0);`` ``Console.WriteLine(``"query(8, 8) : "` `+ query(8, 8));`` ``}``}` `// This code is contributed by 29AjayKumar`
## Javascript
``
Output
```query(3,8) : 26
query(1,6) : 21
query(8,8) : 0```
Time Complexity: O(N)
Auxiliary Space: O(MAXN), since MAXN extra space has been taken, where MAXN is the maximum value of N
Note: The above code works even if N is not a perfect square. In this case, the last block will contain even less number of elements than sqrt(N), thus reducing the number of iterations.
Let’s say n = 10. In this case, we will have 4 blocks first three blocks of size 3, and the last block of size 1.
This article is contributed by Nitish Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.
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# What is an angle?
An angle is the figure formed by two rays sharing a common endpoint, called the vertex of the angle.
We can classify angles according to their size and position.
## Angles according to their size
The following are the types of angles according to their size:
1. Acute Angle
2. Obtuse Angle
3. Right Angle
4. Straight Angle
5. Reflex Angle
6. Revolution
7. Complementary Angles
8. Supplementary Angles
Let’s discuss each of the angles in detail:
### Acute Angle
If an angle is greater than 0 degree but less than 90 degrees, then we call that angle as acute angle.
The following is an acute angle.
Example: 30 degrees, 70 degrees etc. are acute angle
### Obtuse Angle
An angle whose measure is greater than 90 degrees but less than 180 degrees is called an obtuse angle.
Thus, it is between 90 degrees and 180 degrees. The following is an obtuse angle.
Example: 150 degrees, 120 degrees etc.
### Right Angle
An angle whose measure is 90 degrees is called a right angle.
A right angle is marked on the diagram as a small square.
### Straight angle
An angle whose measure is 180 degrees. A straight angle looks like a straight line.
The following is a straight angle.
### Reflex angle
An angle whose measure is greater than 180 degrees but less than 360 degrees.
The following is a reflex angle.
### Revolution
A revolution is an angle that equals exactly 360º.
### Complementary angles
If sum of two angles are 90 degree then they are complementary angles.
For example: Angle measure of 65° and 25° are complementary angles.
As, 65 + 25 = 90°
But angle 65 and angle 25 do not have to be adjacent to be complementary as long as they add up to 90 degrees.
### Supplementary angles
If sum of two angles are 180 degree then they are supplementary angles.
Angle measure of 145° and 35° are supplementary angles.
As, 145 + 35= 180
Try This:
1. What is the measure of a right angle?
2. Name the following angles:
a) 160 degrees
b) 75 degrees
c) Sum of angles 110 and 70
1. 90 degree angle
2. a) Obtuse angle
b) Acute angle
c) Supplementary angles
## Angles according to their position
The following are the types of angles according to their position:
2. Vertically opposite Angles
3. Angles between parallel lines:
3. (a) Interior Angles
3. (b) Exterior Angles
3. (c) Alternate Interior Angles
3. (d) Alternate Exterior Angles
3. (e) Corresponding Angles
Let’s discuss each of them in detail:
Angle with a common vertex and one common side.
In the following figure ∠ 1 and ∠ 2, are adjacent angles.
### Vertically Opposite Angles
Angles that have a common vertex and whose sides are formed by the same lines. Vertically Opposite angles are always equal.
The following figure represents vertically opposite angles.
### Angle formed by Parallel lines
When two parallel lines are crossed by a third line (Transversal), 8 angles are formed.
Take a look at the following figure
We will now discuss the angles formed by the parallel lines:
### Interior Angles
In the adjoining figure, ∠3, ∠4, ∠5, ∠6 are interior angles.
### Exterior Angles
In the adjoining figure, ∠1, ∠2, ∠7, ∠8 are exterior angles.
### Alternate Interior Angles
Pairs of interior angles on opposite sides of the transversal are Alternate Interior Angles.
∠3 and ∠6are alternate interior angles.
∠4 and ∠5 are also alternate interior angles.
### Alternate Exterior Angles
Pairs of exterior angles on opposite sides of the transversal.
∠1 and ∠8 are alternate exterior angles.
∠ 2 and ∠7 are alternate exterior angles.
### Corresponding angles
Pairs of angles in matching corners are called Corresponding Angles.
∠1 and ∠5 are corresponding angles.
∠2 and ∠6 are corresponding angles.
∠3 and ∠7 are corresponding angles
∠4 and ∠8 are corresponding angles.
Try this:
1. Write 1 pair of alternate exterior angle.
2. Write 1 pair of corresponding angle.
1. ∠ a and ∠ h
2. ∠ b and ∠ f
Now try it yourself! Should you still need any help, click here to schedule live online session with e Tutor! |
# A straight line moves so that the sum of the reciprocals of its intercepts made on axes is constant. Show that the line passes through a fixed point. [ Hint : $\large\frac{x}{a}$$+ \large\frac{y}{b}$$=1$ where $\large\frac{1}{a}$$+\large\frac{1}{b} = constant = \large\frac{1}{k} (say ). This implies that \large\frac{k}{a}$$+\large\frac{k}{b}$$=1 \Rightarrow line passes through the fixed point (k, k).] ## 1 Answer Comment A) Need homework help? Click here. Toolbox: • Equation of a line in its intercept form is \large\frac{x}{a}$$+\large\frac{y}{b}$$=1 Step 1 : Let the equation of the line be \large\frac{x}{a}$$+\large\frac{y}{b}$$=1 Its intercepts on the x and y axes are a and b respectively. It is given that \large\frac{1}{a}$$+\large\frac{1}{b}$ = constant = $k$ ( assume )
$\therefore \large\frac{1}{ka}$$+ \large\frac{1}{kb}$$=1$
$\large\frac{\Large\frac{1}{k}}{a}$$+\large\frac{\Large\frac{1}{k}}{b}$$=1$
Hence $\large\frac{1}{k}$$,\large\frac{1}{k} satisfies the equation. \large\frac{x}{a}$$+\large\frac{y}{b}$$=1 Hence the line passes through the fixed point. \bigg( \large\frac{1}{k}$$, \large\frac{1}{k} \bigg)$ |
# The elevation angle of the bird
CB is the line of sight. Angle of elevation (EDC), from the top of the tower. AC is the highest point that the balcony is. It is the height at which the boy’s DA stands. In DBCD BCD, The angle at which the deformation at point B. In DCDE the well-known D is opposite to the side CE It is recognized to be the opposite of the side CD.1
The height of the the balcony AC = as well as its distance from the ground to the floor on the floor AB + CD. Therefore, what is the trigonometry equation which can be used to determine the three variables? Choose tan D or D, as their ratios involve CD in addition to CE. According to the data given the trigonometry equation can be applied since it can include both undiscovered and known amounts.1
In calculating the distance of the tower, or any other thing, one must take into consideration that the size of the kid and include in the final result out of the trigonometry. Examples of Problems. With the help of the following example this idea will be better understand. Problem 1. An angle of depression.1 A pole sits horizontally on the plan. Imagine a situation like in the following figure 4. the subject is looking at an object from an elevated balcony.
From a place at the top of the plane which is 12 m from the bottom of the pole the elevation angle from the top of the pole is 30deg. The ball’s line of view is below the horizontal line.1 Find the size that the pole is.
Its angle with respect to the horizontal level and the the horizontal level is referred to as"the angle of depression" . Solution: The angle at which the point is depressed that is on this object will be the angle that lies between the horizontal line and the line of sight when the point is below the horizontal level.1 Begin by drawing a simple diagram of the issue in the following manner: In the figure above the person standing at point C, is looking up in B. In this diagram, BC represents the height of the electric pole. CB is the line of sight.
CAB or A indicates the angle that defines the elevation of the tower’s top.1 the tower. AC is the highest point that the balcony is. In DABC, it is CAB that represents right angle, and 12m is 12m.
In DBCD BCD, The angle at which the deformation at point B. In DABC CB, it is required to be calculated i.e. what is the diameter of the pole. The height of the the balcony AC = as well as its distance from the ground to the floor on the floor AB + CD.1 To solve the problem apply trigonometry ratios tan A or cot A because they require certain sides in proportions. According to the data given the trigonometry equation can be applied since it can include both undiscovered and known amounts.
Now, i.e. or. Examples of Problems. The height of the survey is 43.1 Problem 1. Problem 2: A boy serves two clouds from a particular place. A pole sits horizontally on the plan. The elevation angle of the clouds is between thirty degrees and forty-five degrees. From a place at the top of the plane which is 12 m from the bottom of the pole the elevation angle from the top of the pole is 30deg.1
If the cloud’s height from the surface is the same , and the distance between clouds is 300m, then figure out the size that the cloud is. Find the size that the pole is. Solution The first step is to draw an easy diagram of the given problem using the below. Solution: In this illustration, CE and BD represents the height of the clouds .1 Begin by drawing a simple diagram of the issue in the following manner: DAB and EAC represent the angle of the height of clouds from point A. In this diagram, BC represents the height of the electric pole. When using DABD, DBA is the right angle, when the height of the cloud is h and trigonometry uses the ratio to A i.e.1 CAB or A indicates the angle that defines the elevation of the tower’s top. the tower. or the equation AB = or AB = (Since that tan 45deg is 1,) In DACE that is ACE.
In DABC, it is CAB that represents right angle, and 12m is 12m. ACE is the correct angle in the case that the elevation of cloud CE is h, then using trigonometry ratio , it is: i.e.1 or i.e. In DABC CB, it is required to be calculated i.e. what is the diameter of the pole. AC = H3. To solve the problem apply trigonometry ratios tan A or cot A because they require certain sides in proportions.
From the figure above 4. Now, i.e. or. AC equals AB + BC Based on the following formula: BC = 300.1 The height of the survey is 43. So, h3 = + 300 i.e. Problem 2: A boy serves two clouds from a particular place. Therefore, the maximum height that the cloud rises to is 410.96 millimeters.
The elevation angle of the clouds is between thirty degrees and forty-five degrees. Problem 3. If the cloud’s height from the surface is the same , and the distance between clouds is 300m, then figure out the size that the cloud is.1 The elevation angle of the bird who was sitting on an oak branch, from the place on the groundthat sits 60 meters away from the bottom of the tree, is 60deg. Solution The first step is to draw an easy diagram of the given problem using the below.
Calculate the length of the tree. (Take 3 = 1.73). In this illustration, CE and BD represents the height of the clouds .1 Solution: First rough a basic diagram of the issue as follows: DAB and EAC represent the angle of the height of clouds from point A. In the above diagram, AB represents the distance between the point of the ground and the foot of the tree i.e. 60 meters. When using DABD, DBA is the right angle, when the height of the cloud is h and trigonometry uses the ratio to A i.e.1 BC is the height of the tree. or the equation AB = or AB = (Since that tan 45deg is 1,) In DACE that is ACE. Let’s take h as the height of the tree. in DABC, ABC is the right angle, while the angle of the elevation of the tree is called B i.e.
60deg. using trigonometry ratios tan A and a result, The height of the tree will be 103.8 meters.1 ACE is the correct angle in the case that the elevation of cloud CE is h, then using trigonometry ratio , it is: i.e. or i.e. Problem 4. AC = H3. The angle of deformation of a bicycle located in a park on top of a 45m tall building is 30 degrees.
From the figure above 4. Which is the length between the bike and the ground (in millimeters)?1 AC equals AB + BC Based on the following formula: BC = 300. Below is a basic illustration of the issue. So, h3 = + 300 i.e.
In the above image, AB represents the distance between the foundation on the construction and that of the bicycle. Therefore, the maximum height that the cloud rises to is 410.96 millimeters.1 AC is the high point of the building i.e. 45 meters. Problem 3. In DBCD BCD, BCD is the right-angled angle and The angle that the depression has is called C i.e. 30deg. The elevation angle of the bird who was sitting on an oak branch, from the place on the groundthat sits 60 meters away from the bottom of the tree, is 60deg.1
By using trigonometry ratio, C is DBCD. Calculate the length of the tree. (Take 3 = 1.73). In this case, AC is BD while AB equals CD. i.e. Solution: First rough a basic diagram of the issue as follows:
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# Unit 8 Family Materials
Probability and Sampling
### Probability and Sampling
Here are the video lesson summaries for Grade 7, Unit 8: Probability and Sampling. Each video highlights key concepts and vocabulary that students learn across one or more lessons in the unit. The content of these video lesson summaries is based on the written Lesson Summaries found at the end of lessons in the curriculum. The goal of these videos is to support students in reviewing and checking their understanding of important concepts and vocabulary. Here are some possible ways families can use these videos:
• Keep informed on concepts and vocabulary students are learning about in class.
• Watch with their student and pause at key points to predict what comes next or think up other examples of vocabulary terms (the bolded words).
• Consider following the Connecting to Other Units links to review the math concepts that led up to this unit or to preview where the concepts in this unit lead to in future units.
Grade 7, Unit 8: Probability and Sampling
Vimeo
Video 1: Estimating Probabilities (Lessons 1–5)
Video 2: Multi-step Experiments (Lessons 6–10)
Video 3: Sampling (Lessons 11–14)
Video 4: Using Samples (Lessons 15–19)
Video 1
Video 2
Video 3
Video 4
### Probabilities of Single Step Events
This week your student will be working with probability. A probability is a number that represents how likely something is to happen. For example, think about flipping a coin.
• The probability that the coin lands somewhere is 1. That is certain.
• The probability that the coin lands heads up is $$\frac12$$, or 0.5.
• The probability that the coin turns into a bottle of ketchup is 0. That is impossible.
Sometimes we can figure out an exact probability. For example, if we pick a random date, the chance that it is on a weekend is $$\frac{2}{7}$$, because 2 out of every 7 days fall on the weekend. Other times, we can estimate a probability based on what we have observed in the past.
People at a fishing contest are writing down the type of each fish they catch. Here are their results:
• Person 1: bass, catfish, catfish, bass, bass, bass
• Person 2: catfish, catfish, bass, bass, bass, bass, catfish, catfish, bass, catfish
• Person 3: bass, bass, bass, catfish, bass, bass, catfish, bass, catfish
1. Estimate the probability that the next fish that gets caught will be a bass.
2. Another person in the competition caught 5 fish. Predict how many of these fish were bass.
3. Before the competition, the lake was stocked with equal numbers of catfish and bass. Describe some possible reasons for why the results do not show a probability of $$\frac12$$ for catching a bass.
Solution:
1. About $$\frac{15}{25}$$, or 0.6, because of the 25 fish that have been caught, 15 of them were bass.
2. About 3 bass, because $$\frac35 = 0.6$$. It would also be reasonable if they caught 2 or 4 bass, out of their 5 fish.
3. There are many possible answers. For example:
• Maybe the lures or bait they were using are more likely to catch bass.
• With results from only 25 total fish caught, we can expect the results to vary a little from the exact probability.
### Probabilities of Multi-step Events
To find an exact probability, it is important to know what outcomes are possible. For example, to show all the possible outcomes for flipping a coin and rolling a number cube, we can draw this tree diagram:
The branches on this tree diagram represent the 12 possible outcomes, from “heads 1” to “tails 6.” To find the probability of getting heads on the coin and an even number on the number cube, we can see that there are 3 ways this could happen (“heads 2”, “heads 4”, or “heads 6”) out of 12 possible outcomes. That means the probability is $$\frac{3}{12}$$, or 0.25.
A board game uses cards that say “forward” or “backward” and a spinner numbered from 1 to 5.
1. On their turn, a person picks a card and spins the spinner to find out which way and how far to move their piece. How many different outcomes are possible?
2. On their next turn, what is the probability that the person will:
1. get to move their piece forward 5 spaces?
2. have to move their piece backward some odd number of spaces?
Solution:
1. There are 10 possible outcomes (“forward 1”, “forward 2”, “forward 3”, “forward 4”, “forward 5”, “backward 1”, “backward 2”, “backward 3”, “backward 4”, or “backward 5”).
1. $$\frac{1}{10}$$ or 0.1, because “forward 5” is 1 out of the 10 possibilities.
2. $$\frac{3}{10}$$ or 0.3, because there are 3 such possibilities (“backward 1”, “backward 3”, or “backward 5”)
### Sampling
This week your student will be working with data. Sometimes we want to know information about a group, but the group is too large for us to be able to ask everyone. It can be useful to collect data from a sample (some of the group) of the population (the whole group). It is important for the sample to resemble the population.
• For example, here is a dot plot showing a population: the height of 49 plants in a sprout garden.
• This sample is representative of the population, because it includes only a part of the data, but it still resembles the population in shape, center, and spread.
• This sample is not representative of the population. It has too many plant heights in the middle and not enough really short or really tall ones.
A sample that is selected at random is more likely to be representative of the population than a sample that was selected some other way.
A city council needs to know how many buildings in the city have lead paint, but they don’t have enough time to test all 100,000 buildings in the city. They want to test a sample of buildings that will be representative of the population.
1. What would be a bad way to pick a sample of the buildings?
2. What would be a good way to pick a sample of the buildings?
Solution:
1. There are many possible answers.
• Testing all the same type of buildings (like all the schools, or all the gas stations) would not lead to a representative sample of all the buildings in the city.
• Testing buildings all in the same location, such as the buildings closest to city hall, would also be a bad way to get a sample.
• Testing all the newest buildings would bias the sample towards buildings that don’t have any lead paint.
• Testing a small number of buildings, like 5 or 10, would also make it harder to use the sample to make predictions about the entire population.
2. To select a sample at random, they could put the addresses of all 100,000 buildings into a computer and have the computer select 50 addresses randomly from the list. Another possibility could be picking papers out a bag, but with so many buildings in the city, this method would be difficult.
### Using Samples
We can use statistics from a sample (a part of the entire group) to estimate information about a population (the entire group). If the sample has more variability (is very spread out), we may not trust the estimate as much as we would if the numbers were closer together. For example, it would be easier to estimate the average height of all 3-year olds than all 40-year olds, because there is a wider range of adult heights.
We can also use samples to help predict whether there is a meaningful difference between two populations, or whether there is a lot of overlap in the data.
Students from seventh grade and ninth grade were selected at random to answer the question, “How many pencils do you have with you right now?” Here are the results:
4 1 2 5 2 1 1 2 3 3
9 4 1 14 6 2 0 8 2 5
1. Use the sample data to estimate the mean (average) number of pencils carried by:
1. all the seventh grade students in the whole school.
2. all the ninth grade students in the whole school.
2. Which sample had more variability? What does this tell you about your estimates in the previous question?
3. A student, who was not in the survey, has 5 pencils with them. If this is all you know, can you predict which grade they are in?
Solution:
1. Since the samples were selected at random, we predict they will represent the whole population fairly well.
1. About 2.4 pencils for all seventh graders, because the mean of the sample is $$(4+1+2+5+2+1+1+2+3+3) \div 10$$ or 2.4 pencils.
2. About 5.1 pencils for all ninth graders, because the mean of the sample is $$(9+4+1+14+6+2+0+8+2+5) \div 10$$ or 5.1 pencils.
2. The survey of ninth graders had more variability. Those numbers were more spread out, so I trust my estimate for seventh grade more than I trust my estimate for ninth grade.
3. There are many possible answers. For example:
• Since they only asked 10 students from each grade, it is hard to predict. It would help if they could ask more students.
• The student is probably in ninth grade, because 5 is closer to the sample mean from ninth grade than from seventh grade.
• The student could possibly be in seventh grade, because at least one student in seventh grade has 5 pencils. |
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## Solving exponential equations
Do you need help with your math homework? Are you struggling to understand concepts how to Solve exponential equations? There are a few steps to solving log equations. First, identify what the base of the log is. This can usually be done by looking at the coefficients in front of the log. Once the base is identified, use the following properties to solve the equation: - if the base is the same on both sides, then the arguments must be equal - if the base is different on both sides, then you can use algebra to solve for the argument - if there is only
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Review In mathematics, substitution is a method for solving equations in which one variable is replaced with an expression in terms of other variables. The expression obtained after substituting the variable is then solved for the other variable. For example, in the equation `x + 3 = 5`, one could substitute `x` with `5 - 3` to obtain `5 - 3 + 3 = 5`, which simplifies to `5 = 5`. From this, we can see that `x = 2
To solve for absolute value, you need to find the distance of a number from zero on a number line. To do this, you take the number's distance from zero on the number line and make it positive. This will give you the absolute value of the number.
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# How do you rationalize the denominator and simplify 30/sqrt18?
Mar 21, 2016
$5 \sqrt{2}$
#### Explanation:
To rationalise the denominator multiply numerator and denominator by $\sqrt{18}$
$\text{-------------------------------------------------------}$
Note : $\sqrt{a} \times \sqrt{a} = a \text{ eliminate the radical }$
example : $\sqrt{100} \times \sqrt{100} = 10 \times 10 = 100$
Also : $\sqrt{a} \times \sqrt{b} \Leftrightarrow \sqrt{a} b$
$\text{-----------------------------------------------------------}$
$\Rightarrow \frac{30}{\sqrt{18}} \times \frac{\sqrt{18}}{\sqrt{18}} = \frac{30 \sqrt{18}}{18} = \frac{5 \sqrt{18}}{3}$
now $\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3 \sqrt{2}$
hence : $\frac{5 \sqrt{18}}{3} = \frac{5 \times 3 \sqrt{2}}{3} = \frac{5 \cancel{3} \sqrt{2}}{\cancel{3}}$
$\Rightarrow \frac{30}{\sqrt{18}} = 5 \sqrt{2}$ |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Unknown Measures of Similar Figures
## Use ratios and proportions to solve for missing lengths in similar figures.
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Unknown Measures of Similar Figures
In the field behind the museum, Mrs. Gilson showed the students a replica of a sculpture known as the “Angel of the North.” The students went outside to have a better look at the huge structure.
“It is huge,” Carmen said to Henry.
“Yup, I wonder how tall it is?” he whispered.
“You can figure that out quite easily with math,” Mrs. Gilson said overhearing the conversation.
“How can I do that?” Henry smirked.
“How tall are you?”
Henry looked at Carmen and then back at Mrs. Gilson.
“I have an idea,” he said smiling.
Do you have an idea what Henry is thinking about? If you pay attention to this Concept, you will know how to figure out how tall the statue is. We will come back to Henry, Carmen and the sculpture at the end of the Concept.
### Guidance
We can use a scale factor to help us to determine unknown measures. We don’t use the scale factor alone we apply it to the proportion. If we know the length of a side in one figure, we can use the scale factor to find the measure of the corresponding side in a similar figure. Let’s see how this works.
Side a in triangle ABC\begin{align*}ABC\end{align*} corresponds to side x\begin{align*}x\end{align*} in the smaller triangle XYZ\begin{align*}XYZ\end{align*}. Side x\begin{align*}x\end{align*} is 4 meters long and the scale factor is 6. What is the measure of side a\begin{align*}a\end{align*}?
We have been told that two sides, a\begin{align*}a\end{align*} and x\begin{align*}x\end{align*}, correspond in a small triangle and a large one. If we know the length of one and the scale factor, we can find the length of the other. Side x\begin{align*}x\end{align*} is 4 meters long, and the scale factor tells us that side a\begin{align*}a\end{align*} will be six times longer. Let’s write this out and solve.
side x×scale factor4×624 m=side a=side a=side a
Side a must have a length of 24 meters.
We can check by setting up the ratio that compares the lengths of the two sides. If the scale factor is 6, then our work is accurate.
ax=244=6\begin{align*}\frac{a}{x} = \frac{24}{4} = 6\end{align*}
Now we know that our work is accurate.
Use the scale factor of the similar figures below to find the measure of KJ\begin{align*}KJ\end{align*}.
Now the first thing that we can do is to set up a proportion to solve for the missing side. Remember that a proportion is two equal ratios. We can set up and compare the corresponding sides.
Here is our proportion.
KJ5=64\begin{align*}\frac{KJ}{5} = \frac{6}{4}\end{align*}
Our proportion is written so that the corresponding sides form the two ratios of the proportion. We can say that KJ\begin{align*}KJ\end{align*} is our unknown in this proportion.
Do you remember how to solve proportions?
We can see a clear relationship between five and four, so we need to use cross products.
KJ×45×64KJ=4KJ=30=30
Now we can solve the equation for KJ\begin{align*}KJ\end{align*} by dividing both sides of the equation by 4.
30÷4KJ=7.5=7.5
The side length of KJ\begin{align*}KJ\end{align*} is 7.5.
Sometimes, you can figure out missing side lengths just by looking at the given measures. In the last example it was too tricky because we couldn’t see the relationship between 5 and 4. However this isn’t always the case. Always look at the diagram of the figures and see if you can determine the missing length without measuring.
Look at these two rectangles. First, look and see if we can figure out the relationship between the two figures. To do this, we compare the side lengths of each part of the two figures.
We need to figure out the measurement of side GH\begin{align*}GH\end{align*} in the second rectangle.
You can see that the measurements in the second rectangle are half as big as the measurements in the first. Also, you know that the opposite sides of a rectangle are congruent. Therefore, the missing side length is 4.
Working in this way can often save you some time!
You may be surprised how often we use similar figures that are related by a scale factor. Maps, architectural blueprints, and diagrams are just some examples. In most of these cases, the scale factor is given so that we know how to enlarge the items in the drawing to their real sizes. Take a look at the floor plan below. It shows where the furniture is located in a living room.
The size of everything in the drawing has been made smaller from a real size by the scale factor. What is the scale factor for the floor plan?
It tells us that one inch in the drawing is equal to two feet in actual size. Therefore, if we know the size in inches of any object in the floor plan, we can find its actual size in feet. Let’s give it a try.
How many feet long is the sofa?
Let’s find the sofa on the floor plan. Then we can use a ruler to find its length in inches. How many inches long is the drawing of the sofa? The sofa in the floor plan is 2 inches long. Imagine this is like knowing the length of one side in a similar figure. Now we need to use the scale factor as we would to find the length of the corresponding side in a similar figure (in this case the “corresponding side” is the actual sofa). We simply multiply the length we know by the scale factor:
sofa drawing×scale factor2 inches×2=actual sofa size=4 feet
The sofa is four feet long.
How long is the fireplace?
Use a ruler to measure the fireplace in the drawing. It is 2.5 inches long. We multiply this by the scale factor to find the length in feet.
fireplace drawing×scale factor2.5 inches×2=actual fireplace length=5 feet
The real length of the fireplace is 5 feet.
We can also reverse the process to take an actual size and reduce it.
Now it's time for you to practice by solving proportions.
#### Example A
34=x12\begin{align*}\frac{3}{4} = \frac{x}{12}\end{align*}
Solution: x=9\begin{align*}x = 9\end{align*}
#### Example B
36=1x\begin{align*}\frac{3}{6} = \frac{1}{x}\end{align*}
Solution: x=2\begin{align*}x = 2\end{align*}
#### Example C
58=1x\begin{align*}\frac{5}{8} = \frac{1}{x}\end{align*}
Solution: 1.6\begin{align*}1.6\end{align*}
Here is the original problem once again.
In the field behind the museum, Mrs. Gilson showed the students a replica of a sculpture known as the “Angel of the North.” The students went outside to have a better look at the huge structure.
“It is huge,” Carmen said to Henry.
“Yup, I wonder how tall it is?” he whispered.
“You can figure that out quite easily with math,” Mrs. Gilson said overhearing the conversation.
“How can I do that?” Henry smirked.
“How tall are you?”
Henry looked at Carmen and then back at Mrs. Gilson.
“I have an idea,” he said smiling.
Let’s think about how Henry and Carmen could figure out the height of the statue. We know that Henry is five feet tall and that his shadow is half as long as he is tall. Now we can write a ratio to compare Henry’s height to his shadow’s length.
Henrys heightShadows length=5 feet2.5 feet\begin{align*}\frac{Henry' s \ height}{Shadow' s \ length} = \frac{5 \ feet}{2.5 \ feet}\end{align*}
Next, we figure out the height of the statue. Henry and Carmen figure out very quickly that they need to figure out the length of the shadow of the statue to figure out the height of the statue. Once they know the length of the shadow, they can use proportional reasoning and indirect measurement to figure out the statue’s height.
Approximating 1 foot using a length a little longer than Henry’s sneaker, they measure 3212\begin{align*}32 \frac{1}{2}\end{align*} feet. It is not an exact measure, but they feel that it is very close.
Now they wrote the following proportion.
5 ft2.5 ft=x32.5 ft\begin{align*}\frac{5 \ ft}{2.5 \ ft} = \frac{x}{32.5 \ ft}\end{align*}
Taking out a notebook, Carmen cross multiplies to solve the proportion.
5(32.5)162.5x=2.5x=2.5x=65
The sculpture is approximately 65 feet tall.
After completing their work, Henry and Carmen check out their answer with the curator of the museum. The statue is actually 65.6 feet tall. Their work was very close to accurate! Indirect measurement was very useful!!
### Vocabulary
Similar Figures
figures that have the same angle measures but not the same side lengths.
Indirect Measurement
using relationships between side lengths to figure out missing measures.
Scale Factor
the proportional relationship between two side lengths.
Proportions
two equal ratios.
### Guided Practice
Here is one for you to try on your own.
Chris is making a drawing of his school and the grounds around it. The basketball court is 75 feet long and 40 feet wide. If Chris uses a scale factor in which 1 inch equals 10 feet, what should the dimensions of the basketball court be in his drawing?
First of all, what do we need to find?
We need to know the dimensions (length and width) that the small version of the basketball court should be.
What information have we been given?
We know the actual size of the basketball court, and we know the scale factor Chris is using for his drawing. We can set up an equation to find the drawn dimensions. We’ll have to find the length first and then the width.
drawing length×scale factordrawing length×10drawing lengthdrawing length=actual basketball court length=75 feet=75÷10=7.5 inches
The length of the basketball court in Chris’s drawing should be 7.5 inches.
Now let’s use the same process to find the width Chris should draw.
drawing width×scale factordrawing width×10drawing widthdrawing width=actual basketball court width=40 feet=40÷10=4 inches
Great! Now we know that Chris should represent the basketball court as a 4 by 7.5 inch rectangle on his drawing.
### Practice
Directions: Solve each problem.
1. Side m\begin{align*}m\end{align*} in triangle LMN\begin{align*}LMN\end{align*} corresponds to side c\begin{align*}c\end{align*} in the smaller triangle BCD\begin{align*}BCD\end{align*}. Side m\begin{align*}m\end{align*} is 12 cm long and the scale factor is 4. What is the measure of side c\begin{align*}c\end{align*}?
2. Side q\begin{align*}q\end{align*} in triangle PQR\begin{align*}PQR\end{align*} corresponds to side y\begin{align*}y\end{align*} in the smaller triangle XYZ\begin{align*}XYZ\end{align*}. Side y\begin{align*}y\end{align*} is 8 inches long and the scale factor is 7. What is the measure of side q\begin{align*}q\end{align*}?
Directions: Solve each proportion for the missing side length.
3. 710=x30\begin{align*}\frac{7}{10} = \frac{x}{30}\end{align*}
4. 1.53=x6\begin{align*}\frac{1.5}{3} = \frac{x}{6}\end{align*}
Directions: Now use the scale factor to create a new ratio.
5. 13\begin{align*}\frac{1}{3}\end{align*}, scale factor 4
6. 85\begin{align*}\frac{8}{5}\end{align*}, scale factor 5
7. 93\begin{align*}\frac{9}{3}\end{align*}, scale factor 3
Directions: Find the scale factor of the similar figures below and then use it to find the measure of LO\begin{align*}LO\end{align*}.
8. Use the scale factor of the similar figures below to find the measure of JK\begin{align*}JK\end{align*}.
Directions: Use the map below and a ruler to answer the questions that follow.
9. How far does Delia live from her school?
10. How far is it from the library to the park?
11. How far does Delia live from City Hall?
12. Delia drew another point to show the police station on her map. She drew it 1.5 inches away from the City Hall. What is the actual distance between the police station and City Hall?
13. How far does Delia live from the park?
14. How far does she live from the library?
15. What is the farthest that Delia will travel to any one item on her map?
### Vocabulary Language: English
Indirect Measurement
Indirect Measurement
Indirect measurement is the process of using the characteristics of similar triangles to measure distances.
Scale Factor
Scale Factor
A scale factor is a ratio of the scale to the original or actual dimension written in simplest form.
Similar
Similar
Two figures are similar if they have the same shape, but not necessarily the same size. |
## Engage NY Eureka Math 7th Grade Module 3 Lesson 9 Answer Key
### Eureka Math Grade 7 Module 3 Lesson 9 Example Answer Key
Example 1.
Fred and Sam are a team in the local 138.2 mile bike-run-athon. Fred will compete in the bike race, and Sam will compete in the run. Fred bikes at an average speed of 8 miles per hour and Sam runs at an average speed of 4 miles per hour. The bike race begins at 6:00 a.m., followed by the run. Sam predicts he will finish the run at 2:33 a.m. the next morning.
a. How many hours will it take them to complete the entire bike-run-athon?
From 6:00 a.m. to 2:00 a.m. the following day is 20 hours.
33 minutes in hours is $$\frac{33}{60}$$=$$\frac{11}{20}$$=0.55, or 0.55 hours.
Therefore, the total time it will take to complete the entire bike-run-athon is 20.55 hours.
b. If t is how long it takes Fred to complete the bike race, in hours, write an expression to find Fred’s total distance.
d=rt
d=8t
The expression of Fred’s total distance is 8t.
c. Write an expression, in terms of t to express Sam’s time.
Since t is Fred’s time and 20.55 is the total time, then Sam’s time would be the difference between the total time and Fred’s time. The expression would be 20.55-t.
d. Write an expression, in terms of t, that represents Sam’s total distance.
d=rt
d=4(20.55-t)
The expressions 4(20.55-t) or 82.2-4t is Sam’s total distance.
e. Write and solve an equation using the total distance both Fred and Sam will travel.
8t+4(20.55-t)=138.2
8t+82.2-4t=138.2
8t-4t+82.2=138.2
4t+82.2=138.2
4t+82.2-82.2=138.2-82.2
4t+0=56
($$\frac{1}{4}$$)(4t)=($$\frac{1}{4}$$)(56)
t=14
Fred’s time: 14 hours
Sam’s time: 20.55-t=20.55-14=6.55
6.55 hours
f. How far will Fred bike, and how much time will it take him to complete his leg of the race?
8(14)=112
Fred will bike 112 miles and will complete the bike race in 14 hours.
g. How far will Sam run, and how much time will it take him to complete his leg of the race?
4(20.55-t)
4(20.55-14)
4(6.55)
26.2
Sam will run 26.2 miles, and it will take him 6.55 hours.
→ How do you find the distance traveled?
→ Multiply the rate of speed by the amount of time.
→ Model how to organize the problem in a distance, rate, and time chart.
→ Explain how to write the equation to have only integers and no decimals. Write the equation.
→ Since the decimal terminates in the tenths place, if we multiply every term by 10, the equation would result with only integer coefficients. The equation would be 40t+822=1382.
Example 2.
Shelby is seven times as old as Bonnie. If in 5 years, the sum of Bonnie’s and Shelby’s ages is 98, find Bonnie’s present age. Use an algebraic approach.
x+5+7x+5=98
8x+10=98
8x+10-10=98-10
8x=88
($$\frac{1}{8}$$)(8x)=($$\frac{1}{8}$$)(88)
x=11
Bonnie’s present age is 11 years old.
→ The first step we must take is to write expressions that represent the present ages of both Bonnie and Shelby. The second step is to write expressions for future time or past time, using the present age expressions. How would the expression change if the time were in the past and not in the future?
→ If the time were in the past, then the expression would be the difference between the present age and the amount of time in the past.
### Eureka Math Grade 7 Module 3 Lesson 9 Opening Exercise Answer Key
Heather practices soccer and piano. Each day she practices piano for 2 hours. After 5 days, she practiced both piano and soccer for a total of 20 hours. Assuming that she practiced soccer the same amount of time each day, how many hours per day, h, did Heather practice soccer?
h: hours per day that soccer was practiced
5(h+2)=20
5h+10=20
5h+10-10=20-10
5h=10
($$\frac{1}{5}$$)(5h)=($$\frac{1}{5}$$)(10)
h=2
Heather practiced soccer for 2 hours each day.
Over 5 days, Jake practices piano for a total of 2 hours. Jake practices soccer for the same amount of time each day. If he practiced piano and soccer for a total of 20 hours, how many hours, h, per day did Jake practice soccer?
h: hours per day that soccer was practiced
5h+2=20
5h+2-2=20-2
5h=18
($$\frac{1}{5}$$)(5h)=(18)($$\frac{1}{5}$$)
h=3.6
Jake practiced soccer 3.6 hours each day.
### Eureka Math Grade 7 Module 3 Lesson 9 Problem Set Answer Key
Question 1.
A company buys a digital scanner for $12,000. The value of the scanner is 12,000(1-$$\frac{n}{5}$$) after n years. The company has budgeted to replace the scanner when the trade-in value is$2,400. After how many years should the company plan to replace the machine in order to receive this trade-in value?
12,000(1-$$\frac{n}{5}$$)=2,400
12,000-2,400n=2,400
-2,400n+12,000-12,000=2,400-12,000
-2,400n=-9,600
n=4
They will replace the scanner after 4 years.
Question 2.
Michael is 17 years older than John. In 4 years, the sum of their ages will be 49. Find Michael’s present age.
x represents Michael’s age now in years.
x+4+x-17+4=49
x+4+x-13=49
2x-9=49
2x-9+9=49+9
2x=58
($$\frac{1}{2}$$)(2x)=($$\frac{1}{2}$$)(58)
x=29
Michael’s present age is 29 years old.
Question 3.
Brady rode his bike 70 miles in 4 hours. He rode at an average speed of 17 mph for t hours and at an average rate of speed of 22 mph for the rest of the time. How long did Brady ride at the slower speed? Use the variable t to represent the time, in hours, Brady rode at 17 mph.
The total distance he rode: 17t+22(4-t)
The total distance equals 70 miles:
17t+22(4-t)=70
17t+88-22t=70
-5t+88=70
-5t+88-88=70-88
-5t=-18
t=3.6
Brady rode at 17 mph for 3.6 hours.
Question 4.
Caitlan went to the store to buy school clothes. She had a store credit from a previous return in the amount of $39.58. If she bought 4 of the same style shirt in different colors and spent a total of$52.22 after the store credit was taken off her total, what was the price of each shirt she bought? Write and solve an equation with integer coefficients.
t: the price of one shirt
4t-39.58=52.22
4t-39.58+39.58=52.22+39.58
4t+0=91.80
($$\frac{1}{4}$$)(4t)=($$\frac{1}{4}$$)(91.80)
t=22.95
The price of one shirt was $22.95. Question 5. A young boy is growing at a rate of 3.5 cm per month. He is currently 90 cm tall. At that rate, in how many months will the boy grow to a height of 132 cm? Answer: Let m represent the number of months. 3.5m+90=132 3.5m+90-90=132-90 3.5m=42 ($$\frac{1}{3.5}$$)(3.5m)=($$\frac{1}{3.5}$$)(42) m=12 The boy will grow to be 132 cm tall 12 months from now. Question 6. The sum of a number, $$\frac{1}{6}$$ of that number, 2 $$\frac{1}{2}$$ of that number, and 7 is 12 $$\frac{1}{2}$$. Find the number. Answer: Let n represent the given number. n+$$\frac{1}{6}$$ n+(2 $$\frac{1}{2}$$)n+7=12 $$\frac{1}{2}$$ n(1+$$\frac{1}{6}$$+$$\frac{5}{2}$$)+7=12 $$\frac{1}{2}$$ n($$\frac{6}{6}$$+$$\frac{1}{6}$$+$$\frac{15}{6}$$)+7=12 $$\frac{1}{2}$$ n($$\frac{22}{6}$$)+7=12 $$\frac{1}{2}$$ 1$$\frac{1}{3}$$ n+7-7=12 $$\frac{1}{2}$$-7 1$$\frac{1}{3}$$ n+0=5 $$\frac{1}{2}$$ 1$$\frac{1}{3}$$ n=5 $$\frac{1}{2}$$ $$\frac{3}{11}$$∙1$$\frac{1}{3}$$ n=$$\frac{3}{11}$$∙1$$\frac{1}{2}$$ 1n=$$\frac{3}{2}$$ n=1 $$\frac{1}{2}$$ The number is 1 $$\frac{1}{2}$$. Question 7. The sum of two numbers is 33 and their difference is 2. Find the numbers. Answer: Let x represent the first number, then 33-x represents the other number since their sum is 33. x-(33-x)=2 x+(-(33-x))=2 x+(-33)+x=2 2x+(-33)=2 2x+(-33)+33=2+33 2x+0=35 2x=35 $$\frac{1}{2}$$∙2x=$$\frac{1}{2}$$∙35 1x=$$\frac{35}{2}$$ x=17 $$\frac{1}{2}$$ 33-x=33-(17 $$\frac{1}{2}$$)=15 $$\frac{1}{2}$$ {17 $$\frac{1}{2}$$,15 $$\frac{1}{2}$$} Question 8. Aiden refills three token machines in an arcade. He puts twice the number of tokens in machine A as in machine B, and in machine C, he puts $$\frac{3}{4}$$ of what he put in machine A. The three machines took a total of 18,324 tokens. How many did each machine take? Answer: Let A represent the number of tokens in machine A. Then $$\frac{1}{2}$$ A represents the number of tokens in machine B, and $$\frac{3}{4}$$ A represents the number of tokens in machine C. A+$$\frac{1}{2}$$ A+$$\frac{3}{4}$$ A=18,324 $$\frac{9}{4}$$ A=18,324 A=8,144 Machine A took 8,144 tokens, machine B took 4,072 tokens, and machine C took 6,108 tokens. Question 9. Paulie ordered 250 pens and 250 pencils to sell for a theatre club fundraiser. The pens cost 11 cents more than the pencils. If Paulie’s total order costs$42.50, find the cost of each pen and pencil.
Let l represent the cost of a pencil in dollars. Then, the cost of a pen in dollars is l+0.11.
250(l+l+0.11)=42.5
250(2l+0.11)=42.5
500l+27.5=42.5
500l+27.5+(-27.5)=42.5+(-27.5)
500l+0=15
500l=15
$$\frac{500l}{500}$$=$$\frac{15}{500}$$
l=0.03
A pencil costs $0.03, and a pen costs$0.14.
Question 10.
A family left their house in two cars at the same time. One car traveled an average of 7 miles per hour faster than the other. When the first car arrived at the destination after 5 $$\frac{1}{2}$$ hours of driving, both cars had driven a total of 599.5 miles. If the second car continues at the same average speed, how much time, to the nearest minute, will it take before the second car arrives?
Let r represent the speed in miles per hour of the faster car, then r-7 represents the speed in miles per hour of the slower car.
5 $$\frac{1}{2}$$ (r)+5 $$\frac{1}{2}$$ (r-7)=599.5
5 $$\frac{1}{2}$$ (r+r-7)=599.5
5 $$\frac{1}{2}$$ (2r-7)=599.5
1$$\frac{1}{2}$$ (2r-7)=599.5
$$\frac{2}{11}$$∙$$\frac{11}{2}$$ (2r-7)=$$\frac{2}{11}$$∙599.5
1∙(2r-7)=$$\frac{1199}{11}$$
2r-7=109
2r-7+7=109+7
2r+0=116
2r=116
$$\frac{1}{2}$$∙2r=$$\frac{1}{2}$$∙116
1r=58
r=58
The average speed of the faster car is 58 miles per hour, so the average speed of the slower car is 51 miles per hour.
distance=rate∙time
d=51∙5 $$\frac{1}{2}$$
d=51∙$$\frac{11}{2}$$
d=280.5
The slower car traveled 280.5 miles in 5 $$\frac{1}{2}$$ hours.
d=58∙5 $$\frac{1}{2}$$
d=58∙$$\frac{11}{2}$$
d=319
OR
599.5-280.5 = 319
The faster car traveled 319 miles in 5 $$\frac{1}{2}$$ hours.
The slower car traveled 280.5 miles in 5 $$\frac{1}{2}$$ hours. The remainder of their trip is 38.5 miles because
319-280.5=38.5.
distance=rate∙time
38.5=51 (t)
$$\frac{1}{51}$$ (38.5)=$$\frac{1}{51}$$ (51)(t)
$$\frac{38.5}{51}$$=1t
$$\frac{77}{102}$$=t
This time is in hours. To convert to minutes, multiply by 60 minutes per hour.
$$\frac{77}{102}$$∙60=$$\frac{77}{51}$$∙30=$$\frac{2310}{51}$$≈45
The slower car will arrive approximately 45 minutes after the first.
Question 11.
Emily counts the triangles and parallelograms in an art piece and determines that altogether, there are 42 triangles and parallelograms. If there are 150 total sides, how many triangles and parallelograms are there?
If t represents the number of triangles that Emily counted, then 42-t represents the number of parallelograms that she counted.
3t+4(42-t)=150
3t+4(42+(-t))=150
3t+4(42)+4(-t)=150
3t+168+(-4t)=150
3t+(-4t)+168=150
-t+168=150
-t+168-168=150-168
-t+0=-18
-t=-18
-1∙(-t)=-1∙(-18)
1t=18
t=18
There are 18 triangles and 24 parallelograms.
Note to the Teacher: Problems 12 and 13 are more difficult and may not be suitable to assign to all students to solve independently.
Question 12.
Stefan is three years younger than his sister Katie. The sum of Stefan’s age 3 years ago and $$\frac{2}{3}$$ of Katie’s age at that time is 12. How old is Katie now?
If s represents Stefan’s age in years, then s+3 represents Katie’s current age, s-3 represents Stefan’s age 3 years ago, and s also represents Katie’s age 3 years ago.
(s-3)+($$\frac{2}{3}$$)s=12
s+(-3)+$$\frac{2}{3}$$ s=12
s+$$\frac{2}{3}$$ s+(-3)=12
$$\frac{3}{3}$$ s+$$\frac{2}{3}$$ s+(-3)=12
$$\frac{5}{3}$$ s+(-3)=12
$$\frac{5}{3}$$ s+(-3)+3=12+3
$$\frac{5}{3}$$ s+0=15
$$\frac{5}{3}$$ s=15
$$\frac{3}{5}$$∙$$\frac{5}{3}$$ s=$$\frac{3}{5}$$∙15
1s=3∙3
s=9
Stefan’s current age is 9 years, so Katie is currently 12 years old.
Question 13.
Lucas bought a certain weight of oats for his horse at a unit price of $0.20 per pound. The total cost of the oats left him with$1. He wanted to buy the same weight of enriched oats instead, but at $0.30 per pound, he would have been$2 short of the total amount due. How much money did Lucas have to buy oats?
The difference in the costs is $3.00 for the same weight in feed. Let w represent the weight in pounds of feed. 0.3w-0.2w=3 0.1w=3 $$\frac{1}{10}$$ w=3 10∙$$\frac{1}{10}$$ w=10∙3 1w=30 w=30 Lucas bought 30 pounds of oats. Cost=unit price×weight Cost=($0.20 per pound)∙(30 pounds)
Cost=$6.00 Lucas paid$6 for 30 pounds of oats. Lucas had $1 left after his purchase, so he started with$7.
### Eureka Math Grade 7 Module 3 Lesson 9 Exit Ticket Answer Key
Question 1.
Brand A scooter has a top speed that goes 2 miles per hour faster than Brand B. If after 3 hours, Brand A scooter traveled 24 miles at its top speed, at what rate did Brand B scooter travel at its top speed if it traveled the same distance? Write an equation to determine the solution. Identify the if-then moves used in your solution.
x: speed, in mph, of Brand B scooter
x+2: speed, in mph, of Brand A scooter
d=rt
24=(x+2)(3)
24=3(x+2)
Possible solution 1:
24 = 3(x+2)
8=x+2
8-2=x+2-2
6=x
If-then Moves: Divide both sides by 3.
Subtract 2 from both sides.
Possible solution 2:
24=3(x+2)
24=3x+6
24-6=3x+6-6
18=3x+0
($$\frac{1}{3}$$)(18)=($$\frac{1}{3}$$)(3x)
6=x
If-then Moves: Subtract 6 from both sides.
Multiply both sides by $$\frac{1}{3}$$.
Question 2.
At each scooter’s top speed, Brand A scooter goes 2 miles per hour faster than Brand B. If after traveling at its top speed for 3 hours, Brand A scooter traveled 40.2 miles, at what rate did Brand B scooter travel if it traveled the same distance as Brand A? Write an equation to determine the solution and then write an equivalent equation using only integers.
x: speed, in mph, of Brand B scooter
x+2: speed, in mph, of Brand A scooter
d=rt
40.2=(x+2)(3)
40.2=3(x+2)
Possible solution 1:
40.2=3(x+2)
13.4=x+2
134=10x+20
134-20=10x+20-20
114=10x
($$\frac{1}{10}$$)(114)=($$\frac{1}{10}$$)(10x)
11.4=x
Possible solution 2:
40.2=3(x+2)
40.2=3x+6
402=30x+60
402-60=30x+60-60
342=30x
($$\frac{1}{30}$$)(342)=($$\frac{1}{30}$$)(30x)
11.4=x
Brand B’s scooter travels at 11.4 miles per hour. |
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Let's learn some vocabulary for quadratic functions. First, vertex. What is a vertex of a quadratic function? Well,the word vertex means a peak. So the vertex of a quadratic function is the peak point on a parabola. This means for a parabola opening down, the vertex is the highest point. Now for a parabola opening up, the vertex is the lowest point. So in this case, we have a parabola opening up, so the vertex is the lowest point, righthere. quadratic functions
Sitting at x equals 1, y equals negative 9. So the vertex is at 1 and negative 9. Next, the axis of symmetry. The axis of symmetry of a quadratic function is a vertical line that divides the parabola into two equal halves, like this. And the axis of symmetry must always pass through the vertex. The symmetry behavior can be best understood when we fold the parabola in half along the axis of symmetry. Then the right half of the parabola will completely overlap the left half of the parabola. Can you see that? Let me show you. Here we have a parabola, just like this one, and the axis of symmetry runs down the middle of a the parabola.
Now, I'm going to fold the graph along the axis of symmetry.Can you see that? The two halves of the parabola completely overlap each other. Yeah? Cool. Now let's determine the equation of the axis of symmetry. For a vertical line that passes through x equals 1, the equation of the line issimply x equals 1. Because for every point on this vertical line, the coordinate is always 1. So in this case, the equation of the axis of symmetry is equaled 1 quadratic functions
Quadratic equation . Next, y-intercept. The y-intercept is the point where the parabola intersects the y-axis. So in this case, that's right here and the y-intercept is negative 8. Next, x-intercepts. The x-intercepts are the points where the parabola intersects the x-axis. So in this case, the parabola intersects the x axis at two points. One at x equals negative 2 another one at x equals 4.Next, quadratic functions domain. By definition, domain is the set of x values for which a function is defined. So as we can see, a parabola is defined for all x values. Let me show you.This point, the x value is positive 1. This point x value is positive 2. This point x value is positive 3. This point x value is positive 4,
and thispoint x value is positive 5. So a parabola is defined for all positive integers forx. Now if you take any points in between, now see, these points the x values are positive decimal numbers, right? Yeah. So a parabola can take on any positive values for x, whether they are integers or whether they are decimal numbers. Parabolacan take on any positive numbers for x. Now, guess what? Parabola can also take onany negative values for x. For example, this point x value is negative 1.This point x value is negative 2. This point x value is negative 3. You see, negative integers. quadratic functions
Quadratic equation Now, for any point in between them, for any points between the negative integers, this point the x values are all negative decimal numbers.So in conclusion, parabola can take on any positive values for x. Parabola can alsotake on any negative values for x and parabola can even take on zero for x.That would be this point, for the y intercept, the x value is zero, right?So therefore, the domain for a quadratic function is all real numbers. And we writea mathematical notation like this
Quadratic equation , x is all real numbers. This symbol means "is"and this symbol here, represents, see this symbol here is an R with an extra bar, represents "all real numbers." In fact, the domain for any quadratic function is always x is all real numbers. Next, range. quadratic functions By definition, range is theset of y values for which a function is defined. So here, the parabola only goesas low as right here, where the y value is negative 9. However, you will go ashigh it likes and therefore we say the range . . . Range talks about y right? Sothe range is y must be negative 9 and anything above.
So the range is y is greater than or equal to negative 9. Does this parabola have a minimum value ora maximum value? Well first of all, in math, whenever we talk about the value of a function, we are always referring to the y value. Remember that. The value of function always refers to they value. Now, since the parabola can go as high asit likes, there's no maximum value. However, a parabola can only go as low asright here, where the y value is negative 9. Therefore, we say this parabola has a minimum value of negative 9.
Hey there! My name is Mayank, a professional Blogger, Graphic Designer, Coder as well as Content Creator from Uttar Pradeh, India. I love to Code and create interesting things while playing with it. |
## CONTENTS
As students’ progress to Grade 10, they familiarize themselves further with mathematical concepts while also initiating the process of their college applications, shortlisting career options.
## Numbers
The study of whole numbers and their properties is called Number theory. Number theory is a large and interesting area in mathematics, including studying Prime numbers, rational numbers and so on. Number theory is important because it helps you to understand and master how the numbers function which helps with logical reasoning skills.
Factors
Prime factorization
Greatest common factor
Least common multiple
GCF and LCM: word problems
Square roots
Cube roots
Classify rational and irrational numbers
Classify numbers
Compare and order rational numbers
Absolute value and opposites
Number lines
Convert between decimals and fractions
## Exponents
The exponent of a number tells us how many times to multiply that number with itself. It is written as a small number to the right and above the base number.
Exponents with integer bases
Exponents with decimal and fractional bases
Negative exponents
Multiplication with exponents
Division with exponents
Multiplication and division with exponents
Power rule
Evaluate expressions using properties of exponents
Identify equivalent expressions involving exponents
## Scientific notation
Scientific notation is a way of representing a number where that number is written in two parts: just the digits with the decimal point placed after the first digit, followed by power of 10.
Convert between standard and scientific notation
Compare numbers written in scientific notation
Multiply numbers written in scientific notation
Divide numbers written in scientific notation
The symbol √ is called radical in mathematics. Radical expression is any expression containing this symbol. The √ symbol is used to indicate square root or nth root of a number. Understanding radical expressions are important for future math works and also for real lives problems such as calculating inflation and interest.
Simplify radical expressions by rationalizing the denominator
Simplify radical expressions using the distributive property
## Rational exponents
Using rational numbers as exponents. A rational exponent represents both an integer exponent and an nth root. The root is found in the denominator and the integer exponent is found in the numerator. In other words, a rational exponent is an exponent that is a fraction.
Evaluate rational exponents
Multiplication with rational exponents
Division with rational exponents
Power rule with rational exponents
Simplify expressions involving rational exponents I
Simplify expressions involving rational exponents II
## Coordinate plane
The plane containing X axis and Y axis is called coordinate plane. Cartesian coordinated can be used to pinpoint where we are on a map or graph. We can mark a point on a graph by how far along and how far up it is, the point (10,6) is 10 units along and 6 units up. Coordinate plane is exciting and important for learning math and it has important use in real life like mapping an area or arranging furniture in your room.
Coordinate plane review
Midpoints
Distance between two points
## Solve equations
Solving an equation is the process of finding a value (or values) that we can put in place of a variable which makes the equation true. Solving an equation is like solving a puzzle which means there are things we can (an cannot) do.
Model and solve equations using algebra tiles
Write and solve equations that represent diagrams
Solve one-step linear equations
Solve two-step linear equations
Solve equations with variables on both sides
Solve equations: complete the solution
Find the number of solutions
Create equations with no solutions or infinitely many solutions
Solve linear equations: word problems
Solve linear equations: mixed review
## Single-variable inequalities
In mathematics sometimes we only know that something is greater or smaller than. Inequality tells us about the relative size of two values. A single-variable inequality is a mathematical statement that relates a linear expression as either less than or greater than another. Learning equations and inequalities helping to get ready for more advanced math problems.
Graph inequalities
Write inequalities from graphs
Identify solutions to inequalities
Solve one-step linear inequalities: addition and subtraction
Solve one-step linear inequalities: multiplication and division
Solve one-step linear inequalities
Graph solutions to one-step linear inequalities
Solve two-step linear inequalities
Graph solutions to two-step linear inequalities
Graph solutions to advanced linear inequalities
Graph compound inequalities
Write compound inequalities from graphs
Solve compound inequalities
Graph solutions to compound inequalities
## Data and graphs
A collection of facts, such as numbers, measurements or observations is called data. We can create a table with the data. A diagram of values, usually shown as lines is called graph. Understanding data and the appropriate graph related to it can help interpreting data.
Interpret bar graphs, line graphs and histograms
Create bar graphs, line graphs and histograms
Interpret circle graphs
Interpret stem-and-leaf plots
Interpret box-and-whisker plots
Interpret a scatter plot
Scatter plots: line of best fit
## Relations and functions
A relation is a set of inputs and outputs, while a function is a relation with one output for each input. It works like a machine that takes something in (input) and at the end gives us something back (output). F(x) is the traditional way of expressing functions. Each function has three parts the Input, the Relationship and the Output.
Relations: convert between tables, graphs, mappings and lists of points
Domain and range of relations
Identify independent and dependent variables
Identify functions
Identify functions: vertical line test
Find values using function graphs
Evaluate a function
Evaluate a function: plug in an expression
Complete a function table from a graph
Complete a function table from an equation
Interpret the graph of a function: word problems
## Direct and inverse variation
A relationship between two variables in which one is a constant multiple of the other is called direct variation. The statement “Y varies directly as X” means that when X increases, Y increases by the same factor.
Identify proportional relationships
Find the constant of variation
Graph a proportional relationship
Write direct variation equations
Write and solve direct variation equations
Identify direct variation and inverse variation
Write inverse variation equations
Write and solve inverse variation equations
## Linear functions
A function related an input to an output. It works like a machine that takes something in (input) and at the end gives us something back (output). F(x) is the traditional way of expressing functions. Each function has three parts the Input, the Relationship and the Output. Linear equations are equations that make a straight line when graphed.
Identify linear functions
Find the slope of a graph
Find the slope from two points
Find a missing coordinate using slope
Slope-intercept form: find the slope and y-intercept
Slope-intercept form: graph an equation
Slope-intercept form: write an equation from a graph
Slope-intercept form: write an equation
Slope-intercept form: write an equation from a table
Slope-intercept form: write an equation from a word problem
Write linear functions to solve word problems
Complete a table and graph a linear function
Compare linear functions: graphs, tables and equations
Write equations in standard form
Standard form: find x- and y-intercepts
Standard form: graph an equation
Equations of horizontal and vertical lines
Graph a horizontal or vertical line
Point-slope form: graph an equation
Point-slope form: write an equation
Point-slope form: write an equation from a graph
Slopes of parallel and perpendicular lines
Write an equation for a parallel or perpendicular line
Find the distance between a point and a line
Find the distance between two parallel lines
Transformations of linear functions
## Systems of linear equations
Two or more equations containing common variables is called the system of equations. A system of equations in which every equation is linear is called system of linear equations. For any linear system, there are three possible outcomes: there is only one solution, there are infinitely solutions or there are no solutions at all. If the number of equations is more than the variables the system is called overdetermined, while if the variables are more than the equations the system is called underdetermined.
Is (x, y) a solution to the system of equations?
Solve a system of equations by graphing
Solve a system of equations by graphing: word problems
Find the number of solutions to a system of equations by graphing
Find the number of solutions to a system of equations
Classify a system of equations by graphing
Classify a system of equations
Solve a system of equations using substitution
Solve a system of equations using substitution: word problems
Solve a system of equations using elimination
Solve a system of equations using elimination: word problems
Solve a system of equations using any method
Solve a system of equations using any method: word problems
## Monomials
A polynomial with just one term is called a Monomial.
Identify monomials
Multiply monomials
Divide monomials
Multiply and divide monomials
Powers of monomials
## Polynomials
The sum or difference of terms which have variables raised to positive integer powers and which have coefficients. A polynomial can have constants, variables and exponents, but never division by variable. Even though the poly- means many the polynomials terms should be finite.
Polynomial vocabulary
Model polynomials with algebra tiles
Add and subtract polynomials using algebra tiles
Multiply a polynomial by a monomial
Multiply two polynomials using algebra tiles
Multiply two binomials
Multiply two binomials: special cases
Multiply polynomials
## Factoring
Finding what to multiply to get an expression is called factoring or factorizing. Factoring is like splitting an expression into a multiplication of simpler expressions. Factoring is important because it helps us to simplify expressions.
GCF of monomials
Factor out a monomial
Factor by grouping
Factor polynomials
An equation where the highest exponent of the variable is two is called quadratic equation. In other words, an equation includes only second-degree polynomials. The quadratic equation is used to find the curve on a Cartesian plane.
Complete a function table: quadratic functions
Solve a quadratic equation using square roots
Solve a quadratic equation using the zero product property
Solve a quadratic equation by factoring
Complete the square
Solve a quadratic equation by completing the square
Using the discriminant
## Parabolas
Parabola is a special curve, shaped like an arch. Any point of a parabola is at an equal distance from a fixed point (the focus), and a fixed straight line (the directrix). For formal definition we say: for a given point, called focus, and a given line not through the focus, called directrix, a parabola is the locus of points such that the distance of the focus equals the distance to the directrix. Not all U-shaped curves are parabola, a parabola should satisfy the conditions listed above. Parabola has many important applications in real life, such as designing automobiles headlights to calculating the path of a ballistic missile.
Identify the direction a parabola opens
Find the vertex of a parabola
Find the axis of symmetry of a parabola
Write equations of parabolas in vertex form from graphs
Write equations of parabolas in vertex form using properties
Graph parabolas
A function related an input to an output. It works like a machine that takes something in (input) and at the end gives us something back (output). F(x) is the traditional way of expressing functions. Each function has three parts the Input, the Relationship and the Output. Linear equations are equations that make a straight line when graphed.
Identify linear, quadratic and exponential functions from graphs
Identify linear, quadratic and exponential functions from tables
Write linear, quadratic and exponential functions
Linear functions over unit intervals
Exponential functions over unit intervals
Describe linear and exponential growth and decay
## Rational functions and expressions
A function that be written as a polynomial divided by a polynomial is a rational function. It is rational because one is divided by the other, like a ratio.
Rational functions: asymptotes and excluded values
Simplify complex fractions
Simplify rational expressions
Multiply and divide rational expressions
Divide polynomials
Solve rational equations
## Circles in the coordinate plane
A circle is a two-dimensional shape made by drawing a curve that is always the same distance from a center. In other words, a circle is the locus of all points that are a fixed distance from a given point.
Find the centre of a circle
Find the radius or diameter of a circle
Write equations of circles in standard form from graphs
Write equations of circles in standard form using properties
Graph circles
## Logic
One area of mathematics that has its roots deep in philosophy is the study of logic. Mathematical log helps to detect whether a statement is valid or invalid.
Identify hypotheses and conclusions
Counterexamples
Truth tables
Truth values
Conditionals
Negations
Converses, inverses and contrapositives
Biconditionals
## Measurement
Measurement is the process of assigning a number to a physical property, like length, area, mass, etc.
Estimate metric measurements
Convert rates and measurements: metric units
Estimate imperial measurements of length
Convert rates and measurements: imperial units of length
Unit rates
Unit prices with imperial length conversions
Imperial mixed units of length
Convert between metric and imperial units of length
Scale drawings: word problems
## Constructions
Construction in Geometry means to draw shapes, angles or lines accurately. These constructions use only compass, straightedge and a pencil. Construction is important in geometry because it allows you to draw lines, angles, and polygons with the simplest tools at you hand.
Construct an angle bisector
Construct a congruent angle
## Two-dimensional figures
Two-dimensional geometry or plane geometry is about flat shapes like triangles and circles. Two-dimensional figures have only two dimensions such as width and height but no thickness. It also known as “2D”.
Polygon vocabulary
Interior angles of polygons
Perimeter
Area and perimeter in the coordinate plane I
Area and perimeter in the coordinate plane II
Area and circumference of circles
Area of compound figures
Area between two shapes
Area and perimeter of similar figures
Identify trapezoids
Properties of parallelograms
Properties of rhombuses
## Three-dimensional figures
Having three dimensions such as Height, Width and Depth, like any real-world object is a three-dimensional figure. Three-dimensional geometry is about solid shapes like spheres or cubes. It is also known as “3D”.
Three-dimensional figure vocabulary
Parts of three-dimensional figures
Nets and drawings of three-dimensional figures
Introduction to surface area and volume
Surface area of prisms and cylinders
Surface area of pyramids and cones
Volume of prisms and cylinders
Volume of pyramids and cones
Surface area and volume of spheres
Introduction to similar solids
Surface area and volume of similar solids
Surface area and volume review
Cross-sections of three-dimensional figures
Solids of revolution
## Congruent figures
When we change a shape by using Turn, flip, slide or resize it is called transformation. If one shape can become another using Turn, Flip or Slide then shapes are congruent. Congruence keeps the size, area, angles and line lengths of the shape.
Congruent line segments
Congruence statements and corresponding parts
Solve problems involving corresponding parts
Identify congruent figures
SSS and SAS Theorems
ASA and AAS Theorems
SSS, SAS, ASA and AAS Theorems
SSS Theorem in the coordinate plane
Congruency in isosceles and equilateral triangles
Hypotenuse-Leg Theorem
## Triangles
Triangle is a polygon with 3 sides.
Classify triangles
Triangle Angle-Sum Theorem
Midsegments of triangles
Triangles and bisectors
Identify medians, altitudes, angle bisectors and perpendicular bisectors
Angle-side relationships in triangles
Triangle Inequality Theorem
## Similarity
When we change a shape by using Turn, flip, slide or resize it is called transformation. Two shapes are similar when one can become the other after a resize, flip, slide or turn. When two shapes are similar then the corresponding angles are equal, and the lines are in proportion. This can make life a lot easier when solving geometry problems.
Identify similar figures
Similarity ratios
Similarity statements
Side lengths and angle measures in similar figures
Similar triangles and indirect measurement
Perimeters of similar figures
Similarity rules for triangles
Similar triangles and similarity transformations
Similarity of circles
Triangle Proportionality Theorem
Areas of similar figures
## Right triangles
Right triangles are important in geometry due to their properties and the use of Pythagoras theorem.
Pythagorean Theorem
Converse of the Pythagorean theorem
Pythagorean Inequality Theorems
Special right triangles
## Trigonometry
Trigonometry is the study of triangles with emphasis on calculations involving the lengths of sides and the angles. Trigonometry is based on six functions, namely Sine, Cosine, Tangent, Cotangent, Secant, and Cosecant. Trigonometry also includes study of the properties of these functions and their graphs. Trigonometry is an important field in mathematics with a lot of applications in real life problems, like calculating the waves and tides in oceans, creating maps, satellite systems and a lot more.
Trigonometric ratios: sin, cos and tan
Trigonometric ratios: csc, sec and cot
Find trigonometric functions of special angles
Find trigonometric functions using a calculator
Inverses of trigonometric functions
Trigonometric ratios: find a side length
Trigonometric ratios: find an angle measure
Solve a right triangle
## Transformations
When we change a shape by using Turn, flip, slide or resize it is called transformation. Transformation is an operation or operations that alter the form of a figure. Transformations are important in geometry and advanced math works.
Translations: graph the image
Translations: find the coordinates
Translations: write the rule
Reflections: graph the image
Reflections: find the coordinates
Rotations: graph the image
Rotations: find the coordinates
Classify congruence transformations
Compositions of congruence transformations: graph the image
Transformations that carry a polygon onto itself
Congruence transformations: mixed review
Dilations: graph the image
Dilations: find the coordinates
Dilations: scale factor and classification
Dilations and parallel lines
## Probability
Probability is the chance of something happening or how likely it is that some event will happen. Probability is a number between 0 (not happening) to 1 (certainly happening).
Theoretical and experimental probability
Compound events: find the number of outcomes
Independent and dependent events
Factorials
Permutations
Counting principle
Permutation and combination notation
Geometric probability
## Statistics
Statistics is a branch of mathematics dealing with the collection, analysis, interpretation, and presentation of data. Statistics is a strong tool in everyday life to get answers about data and make concrete decisions.
Mean, median, mode and range
Quartiles
Identify biased samples
Mean absolute deviation
Variance and standard deviation |
# 10 circular pieces of paper each of radius 1 cm have been cut out from a piece of paper having a shape of an equilateral triangle. What should be the minimum area of the equilateral triangle?
##### 2 Answers
Jun 8, 2016
$A = 6 \left(3 + 2 \sqrt{3}\right)$[${\text{cm}}^{2}$]
#### Explanation:
The circular pieces of paper aligned in rows, like
1 over
2 over
3 over
4
configure the skeleton of an equilateral triangle.
The external equilateral triangle must have a minimum side length of $6 r$ plus an addition of
$2 \times \frac{r}{\tan \left({30}^{\circ}\right)} = 2 \times \sqrt{3} \times r$.
So each side must have a length of $l = \left(6 + 2 \times \sqrt{3}\right) \times r$ or $l = \lambda r$. For the equilateral triangle the height is $h = \frac{\sqrt{3}}{2} l$ so the area is given by
$A = \frac{1}{2} l \times h = \frac{1}{2} \frac{\sqrt{3}}{2} {\left(\lambda r\right)}^{2} = 6 \left(3 + 2 \sqrt{3}\right) {r}^{2}$
now if $r = 1$[$\text{cm}$] then $A = 6 \left(3 + 2 \sqrt{3}\right)$[${\text{cm}}^{2}$]
Jun 10, 2016
$= 6 \left(3 + 2 \sqrt{3}\right) c {m}^{2}$
#### Explanation:
After cutting out 10 circular pieces the triangular paper will have structure as shown below
$r = 1 c m \to \text{Radius of each circular piece of paper}$
$B T = \text{projection of BO} = x$
$\angle O B T = \frac{1}{2} \angle A B C = \frac{1}{2} \times {60}^{\circ} = {30}^{\circ}$
$\frac{x}{r} = \cot {30}^{\circ} = \sqrt{3}$
$\implies x = \sqrt{3} r$
$\text{Each side of the } \Delta A B C = 6 r + 2 x = 6 r + 2 \sqrt{3} r = \left(6 + 2 \sqrt{3}\right) c m$
${\text{Area " Delta ABC=sqrt3/4*"side}}^{2} = \frac{\sqrt{3}}{4} {\left(6 + 2 \sqrt{3}\right)}^{2} c {m}^{2}$
$= \sqrt{3} {\left(3 + \sqrt{3}\right)}^{2} c {m}^{2}$
$= \sqrt{3} \left(12 + 6 \sqrt{3}\right) c {m}^{2}$
$= \left(18 + 12 \sqrt{3}\right) c {m}^{2}$
$= 6 \left(3 + 2 \sqrt{3}\right) c {m}^{2}$ |
# 10.6 Parametric equations (Page 5/6)
Page 5 / 6
## Verbal
What is a system of parametric equations?
A pair of functions that is dependent on an external factor. The two functions are written in terms of the same parameter. For example, $\text{\hspace{0.17em}}x=f\left(t\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=f\left(t\right).$
Some examples of a third parameter are time, length, speed, and scale. Explain when time is used as a parameter.
Explain how to eliminate a parameter given a set of parametric equations.
Choose one equation to solve for $\text{\hspace{0.17em}}t,\text{\hspace{0.17em}}$ substitute into the other equation and simplify.
What is a benefit of writing a system of parametric equations as a Cartesian equation?
What is a benefit of using parametric equations?
Some equations cannot be written as functions, like a circle. However, when written as two parametric equations, separately the equations are functions.
Why are there many sets of parametric equations to represent on Cartesian function?
## Algebraic
For the following exercises, eliminate the parameter $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ to rewrite the parametric equation as a Cartesian equation.
$\left\{\begin{array}{l}x\left(t\right)=5-t\hfill \\ y\left(t\right)=8-2t\hfill \end{array}$
$y=-2+2x$
$\left\{\begin{array}{l}x\left(t\right)=6-3t\hfill \\ y\left(t\right)=10-t\hfill \end{array}$
$\left\{\begin{array}{l}x\left(t\right)=2t+1\hfill \\ y\left(t\right)=3\sqrt{t}\hfill \end{array}$
$y=3\sqrt{\frac{x-1}{2}}$
$\left\{\begin{array}{l}x\left(t\right)=3t-1\hfill \\ y\left(t\right)=2{t}^{2}\hfill \end{array}$
$\left\{\begin{array}{l}x\left(t\right)=2{e}^{t}\hfill \\ y\left(t\right)=1-5t\hfill \end{array}$
$x=2{e}^{\frac{1-y}{5}}\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}y=1-5ln\left(\frac{x}{2}\right)$
$\left\{\begin{array}{l}x\left(t\right)={e}^{-2t}\hfill \\ y\left(t\right)=2{e}^{-t}\hfill \end{array}$
$\left\{\begin{array}{l}x\left(t\right)=4\text{log}\left(t\right)\hfill \\ y\left(t\right)=3+2t\hfill \end{array}$
$x=4\mathrm{log}\left(\frac{y-3}{2}\right)$
$\left\{\begin{array}{l}x\left(t\right)=\text{log}\left(2t\right)\hfill \\ y\left(t\right)=\sqrt{t-1}\hfill \end{array}$
$\left\{\begin{array}{l}x\left(t\right)={t}^{3}-t\hfill \\ y\left(t\right)=2t\hfill \end{array}$
$x={\left(\frac{y}{2}\right)}^{3}-\frac{y}{2}$
$\left\{\begin{array}{l}x\left(t\right)=t-{t}^{4}\hfill \\ y\left(t\right)=t+2\hfill \end{array}$
$\left\{\begin{array}{l}x\left(t\right)={e}^{2t}\hfill \\ y\left(t\right)={e}^{6t}\hfill \end{array}$
$y={x}^{3}$
$\left\{\begin{array}{l}x\left(t\right)={t}^{5}\hfill \\ y\left(t\right)={t}^{10}\hfill \end{array}$
${\left(\frac{x}{4}\right)}^{2}+{\left(\frac{y}{5}\right)}^{2}=1$
$\left\{\begin{array}{l}x\left(t\right)=3\mathrm{sin}\text{\hspace{0.17em}}t\hfill \\ y\left(t\right)=6\mathrm{cos}\text{\hspace{0.17em}}t\hfill \end{array}$
${y}^{2}=1-\frac{1}{2}x$
$\left\{\begin{array}{l}x\left(t\right)=\mathrm{cos}\text{\hspace{0.17em}}t+4\\ y\left(t\right)=2{\mathrm{sin}}^{2}t\end{array}$
$\left\{\begin{array}{l}x\left(t\right)=t-1\\ y\left(t\right)={t}^{2}\end{array}$
$y={x}^{2}+2x+1$
$\left\{\begin{array}{l}x\left(t\right)=-t\\ y\left(t\right)={t}^{3}+1\end{array}$
$\left\{\begin{array}{l}x\left(t\right)=2t-1\\ y\left(t\right)={t}^{3}-2\end{array}$
$y={\left(\frac{x+1}{2}\right)}^{3}-2$
For the following exercises, rewrite the parametric equation as a Cartesian equation by building an $x\text{-}y$ table.
$\left\{\begin{array}{l}x\left(t\right)=2t-1\\ y\left(t\right)=t+4\end{array}$
$\left\{\begin{array}{l}x\left(t\right)=4-t\\ y\left(t\right)=3t+2\end{array}$
$y=-3x+14$
$\left\{\begin{array}{l}x\left(t\right)=2t-1\\ y\left(t\right)=5t\end{array}$
$\left\{\begin{array}{l}x\left(t\right)=4t-1\\ y\left(t\right)=4t+2\end{array}$
$y=x+3$
For the following exercises, parameterize (write parametric equations for) each Cartesian equation by setting $x\left(t\right)=t$ or by setting $\text{\hspace{0.17em}}y\left(t\right)=t.$
$y\left(x\right)=3{x}^{2}+3$
$y\left(x\right)=2\mathrm{sin}\text{\hspace{0.17em}}x+1$
$\left\{\begin{array}{l}x\left(t\right)=t\hfill \\ y\left(t\right)=2\mathrm{sin}t+1\hfill \end{array}$
$x\left(y\right)=3\mathrm{log}\left(y\right)+y$
$x\left(y\right)=\sqrt{y}+2y$
$\left\{\begin{array}{l}x\left(t\right)=\sqrt{t}+2t\hfill \\ y\left(t\right)=t\hfill \end{array}$
For the following exercises, parameterize (write parametric equations for) each Cartesian equation by using $x\left(t\right)=a\mathrm{cos}\text{\hspace{0.17em}}t$ and $\text{\hspace{0.17em}}y\left(t\right)=b\mathrm{sin}\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ Identify the curve.
$\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1$
$\frac{{x}^{2}}{16}+\frac{{y}^{2}}{36}=1$
$\left\{\begin{array}{l}x\left(t\right)=4\mathrm{cos}\text{\hspace{0.17em}}t\hfill \\ y\left(t\right)=6\mathrm{sin}\text{\hspace{0.17em}}t\hfill \end{array};\text{\hspace{0.17em}}$ Ellipse
${x}^{2}+{y}^{2}=16$
${x}^{2}+{y}^{2}=10$
$\left\{\begin{array}{l}x\left(t\right)=\sqrt{10}\mathrm{cos}t\hfill \\ y\left(t\right)=\sqrt{10}\mathrm{sin}t\hfill \end{array};\text{\hspace{0.17em}}$ Circle
Parameterize the line from $\text{\hspace{0.17em}}\left(3,0\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\left(-2,-5\right)\text{\hspace{0.17em}}$ so that the line is at $\text{\hspace{0.17em}}\left(3,0\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=0,\text{\hspace{0.17em}}$ and at $\text{\hspace{0.17em}}\left(-2,-5\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=1.$
Parameterize the line from $\text{\hspace{0.17em}}\left(-1,0\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\left(3,-2\right)\text{\hspace{0.17em}}$ so that the line is at $\text{\hspace{0.17em}}\left(-1,0\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=0,\text{\hspace{0.17em}}$ and at $\text{\hspace{0.17em}}\left(3,-2\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=1.$
$\left\{\begin{array}{l}x\left(t\right)=-1+4t\hfill \\ y\left(t\right)=-2t\hfill \end{array}$
Parameterize the line from $\text{\hspace{0.17em}}\left(-1,5\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\left(2,3\right)$ so that the line is at $\text{\hspace{0.17em}}\left(-1,5\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=0,\text{\hspace{0.17em}}$ and at $\text{\hspace{0.17em}}\left(2,3\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=1.$
Parameterize the line from $\text{\hspace{0.17em}}\left(4,1\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\left(6,-2\right)\text{\hspace{0.17em}}$ so that the line is at $\text{\hspace{0.17em}}\left(4,1\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=0,\text{\hspace{0.17em}}$ and at $\text{\hspace{0.17em}}\left(6,-2\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}t=1.$
$\left\{\begin{array}{l}x\left(t\right)=4+2t\hfill \\ y\left(t\right)=1-3t\hfill \end{array}$
## Technology
For the following exercises, use the table feature in the graphing calculator to determine whether the graphs intersect.
yes, at $t=2$
For the following exercises, use a graphing calculator to complete the table of values for each set of parametric equations.
$\left\{\begin{array}{l}{x}_{1}\left(t\right)=3{t}^{2}-3t+7\hfill \\ {y}_{1}\left(t\right)=2t+3\hfill \end{array}$
$t$ $x$ $y$
–1
0
1
$\left\{\begin{array}{l}{x}_{1}\left(t\right)={t}^{2}-4\hfill \\ {y}_{1}\left(t\right)=2{t}^{2}-1\hfill \end{array}$
$t$ $x$ $y$
1
2
3
$t$ $x$ $y$
1 -3 1
2 0 7
3 5 17
$\left\{\begin{array}{l}{x}_{1}\left(t\right)={t}^{4}\hfill \\ {y}_{1}\left(t\right)={t}^{3}+4\hfill \end{array}$
$t$ $x$ $y$
-1
0
1
2
## Extensions
Find two different sets of parametric equations for $\text{\hspace{0.17em}}y={\left(x+1\right)}^{2}.$
Find two different sets of parametric equations for $\text{\hspace{0.17em}}y=3x-2.$
Find two different sets of parametric equations for $\text{\hspace{0.17em}}y={x}^{2}-4x+4.$
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael |
Comment on page
# 15. 3 Sum
## Solution
This builds off of the Two Sum problem that uses the two pointer solution. I found these posts helpful to understand 3Sum:
Below are two implementations. The first uses a set to handle duplicates results. The other handles duplicates results by skipping them using the two pointer trick.
### Set solution
The first step is to sort the array. This will let us skip duplicate numbers in the nums array. We don't need to process duplicate numbers as we already found the solution(s) for that number. The sorted array will also allow us to look for solution items.
We want to find three numbers that sum to zero: a + b + c = 0. We can set 'a' to be the current item when iterating through the array. Then we need to find b,c that when added with 'a' will equal zero. Since 'a' is fixed for the iteration, we can then use a start and an end pointer to scan the rest of the array to see if there are b,c values that will satisfy the condition.
If the result of a + b + c > 0 then the end pointer will be decremented in an attempt to find zero in the next iteration. If the sum in less than zero then the start pointer will be incremented.
When a solution is found, a + b + c = 0, then we add it to the set. We continue looking in that iteration for other b,c that may satisfy the condition.
Example:
nums = [-1,0,1,2,-1,-4]
nums sorted = [-4, -1, -1, 0, 1, 2]
The below table shows the first two iterations. The first iteration does not find a solution. The second iteration finds a solution.
iteration
a
start index
end index
b
c
0
-4
1
5
-1
2
2
5
-1
2
3
5
0
2
4
5
1
2
1
-1
2
5
-1
2
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
result = set()
for i in range(len(nums) - 2):
a = nums[i]
# skip duplicate numbers
if i > 0 and a == nums[i - 1]:
continue
start = i + 1
end = len(nums) - 1
while start < end:
b = nums[start]
c = nums[end]
if a + b + c == 0:
start += 1
end -= 1
elif a + b + c > 0:
end -= 1
else:
start += 1
return list(map(list, result))
### Skipping "Two pointer trick" Solution
The main algorithm is the same as above. However, instead of using a set to store the result, an array is used to store the result. Duplicate results are avoided by skipping duplicate numbers. In the outer loop we skip duplicate numbers. But in the inner while loop we can skip duplicate numbers as well. We do this by incrementing the start pointer past any duplicates. We also decrement the end pointer before any duplicates as well.
Example
nums = [-1,0,1,2,-1,-4, 2]
nums sorted = [-4, -1, -1, 0, 1, 2, 2]
In the second iteration, when the index is 1, and a = -1, the algorithm will find a solution of [-1, -1, 2]. In order to not get a duplicate result, start will increment past the second -1. The start index will be 3. The end pointer will decrement to before the first 2.The end index will be 4.
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
result = []
for i in range(len(nums) - 2):
a = nums[i]
# skip duplicate numbers
if i > 0 and a == nums[i - 1]:
continue
start = i + 1
end = len(nums) - 1
while start < end:
b = nums[start]
c = nums[end]
summ = a + b + c
if summ == 0:
result.append([a, b, c])
# the while loops below are what lets us use an array rather than a set to store the result (since
# duplicates are not allowed). Here duplicate numbers are skipped by.
while start < end and b == nums[start]:
start += 1
while start < end and c == nums[end]:
end -= 1
elif summ > 0:
end -= 1
else:
start += 1
return result
## Test Cases
input
result
[-1, 0, 1, 2, -1, -4]
[[-1, -1, 2], [-1, 0, 1]]
[]
[]
[0]
[]
[-25, -10, -7, -3, 2, 4, 8, 10]
[-10, 2, 8], [-7, -3, 10] |
# Difference between revisions of "2004 AIME I Problems/Problem 5"
## Problem
Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted questions worth a total of 500 points. Alpha scored 160 points out of 300 points attempted on the first day, and scored 140 points out of 200 points attempted on the second day. Beta who did not attempt 300 points on the first day, had a positive integer score on each of the two days, and Beta's daily success rate (points scored divided by points attempted) on each day was less than Alpha's on that day. Alpha's two-day success ratio was 300/500 = 3/5. The largest possible two-day success ratio that Beta could achieve is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m+n$?
## Solution
Let $q$ be the number of questions Beta takes on day 1 and $a$ be the number he gets right. Let $b$ be the number he gets right on day 2.
These inequalities follow: $$\frac{a}{q} < \frac{160}{300} = \frac{8}{15}$$ $$\frac{b}{500-q} < \frac{140}{200} = \frac{7}{10}$$ Solving for a and b and adding the two inequalities: $$a + b < \frac{8}{15}q + (350 - \frac{7}{10}q)$$ $$a + b < 350 - \frac{1}{6}q$$ From here, we see the largest possible value of $a+b$ is $349$.
Checking our conditions, we know that $a$ must be positive so therefore $q$ must be positive. A quick check shows that for $2\le q \le 5$, $q$ follows all the conditions and results in $a+b=349$.
This makes Beta's success ratio $\frac{349}{500}$. Thus, the answer is $m+n = 349 + 500 = \boxed{849}$.
## Simple Solution
We see that we want Beta to have more points where there is a higher Alpha success rate (that way, the score is shifted more towards the higher Alpha score). With that in mind, .7 > 8/15. Thus, we might as well put as many points as possible into the second day.
That said, we need Beta to have a positive integer score on either end per requirements. The best way is for Beta to go 1/2 on the first day. That way, we "waste" the least points- both in the numerator and denominator.
Thereafter, we see that, letting $x$ be the number of points Beta gained on the second day, $x/498 < 7/10$; thus $max(x) = 348$.
Aha. 348+1 = 349. |
Home | | Maths 10th Std | Surface Area of Right Circular Cone
# Surface Area of Right Circular Cone
A right circular cone is a solid generated by the revolution of a right angled triangle about one of the sides containing the right angle as axis.
Right Circular Cone
Observe the given figures in Fig.7.9 and identify which solid shape they represent?
These objects resemble the shape of a cone.
## Definition :
A right circular cone is a solid generated by the revolution of a right angled triangle about one of the sides containing the right angle as axis.
## Formation of a Right Circular Cone – Demonstration
In Fig. 7.10, if the right triangle ABC revolves about AB as axis, the hypotenuse AC generates the curved surface of the cone represented in the diagram. The height of the cone is the length of the axis AB, and the slant height is the length of the hypotenuse AC.
## Surface area of a right circular cone
Suppose the surface area of the cone is cut along the hypotenuse AC and then unrolled on a plane, the surface area will take the form of a sector ACD, of which the radius AC and the arc CD are respectively the slant height and the circumference of the base of the cone.
Here the sector of radius ‘l ’ and arc length ‘s’ will be similar to a circle of radius l.
## (i) Curved surface area
Curved Surface Area of the cone = Area of the Sector = πrl sq. units.
C.S.A. of a right circular cone = πrl sq. units.
Derivation of slant height ‘l
ABC is a right angled triangle, right angled at B. The hypotenuse, base and height of the triangle are represented by l, r and h respectively.
Now, using Pythagoras theorem in ΔABC,
## (ii) Total surface area
Total surface area of a cone = C.S.A. + base area of the cone
= πrl + πr2 (since, the base is a circle)
T.S.A. of a right circular cone = πr (l + r ) sq. units.
### Example 7.5
The radius of a conical tent is 7 m and the height is 24 m. Calculate the length of the canvas used to make the tent if the width of the rectangular canvas is 4 m?
### Solution
Let r and h be the radius and height of the cone respectively.
Given that, radius r =7 m and height h = 24 m
Hence,
C.S.A. of the conical tent = πrl sq. units
Area of the canvas = (22/7) ×7 ×25 = 550 m2
Now, length of the canvas = Area of the canvas / width = 550/4 = 137.5 m
Therefore, the length of the canvas is 137.5 m
### Example 7.6
If the total surface area of a cone of radius 7cm is 704 cm2, then find its slant height.
### Solution
Given that, radius r = 7 cm
Now, total surface area of the cone = πr (l + r ) sq. units
T.S.A. = 704 cm2
704 = (22/7) ×7 (l + 7)
32 = l + 7 implies l = 25 cm
Therefore, slant height of the cone is 25 cm.
### Example 7.7
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and base is hollowed out (Fig.7.13). Find the total surface area of the remaining solid.
### Solution
Let h and r be the height and radius of the cone and cylinder.
Let l be the slant height of the cone.
Given that, h = 2.4 cm and d = 1.4 cm ; r = 0.7 cm
1= 2.5 cm
Area of the remaining solid = 2πrh + πrl + πr2 sq. units
= πr (2h + l + r )
= (22/7) × 0.7 × [(2 × 2.4) + 2.5 + 0.7]
= 17.6
Therefore, total surface area of the remaining solid is 17.6 m2
Tags : Definition, Formula, Solved Example Problems | Mensuration | Mathematics , 10th Mathematics : UNIT 7 : Mensuration
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
10th Mathematics : UNIT 7 : Mensuration : Surface Area of Right Circular Cone | Definition, Formula, Solved Example Problems | Mensuration | Mathematics |
# 31700 in Words
31700 in words can be written as Thirty-one Thousand Seven Hundred. The fundamental concepts in Mathematics like counting or count can be learnt efficiently here. If you buy gold worth Rs. 31700, then you can say that “I bought gold ornaments for Thirty-one Thousand Seven Hundred Rupees”. To write numbers in words, the English alphabet is used. The numbers in words concept is explained here in a simple way to improve the conceptual knowledge of students. The 31700 can be read as “Thirty-one Thousand Seven Hundred” in English.
31700 in words Thirty-one Thousand Seven Hundred Thirty-one Thousand Seven Hundred in Numbers 31700
## How to Write 31700 in Words?
Students can learn about the expanded form and the place value chart of 31700. Five digits are present in the number 31700. With the help of the place value chart given below, students will be able to understand the concepts with ease.
Ten Thousands Thousands Hundreds Tens Ones 3 1 7 0 0
31700 can be written in expanded form as:
3 x Ten Thousand + 1 x Thousand + 7 × Hundred + 0 × Ten + 0 × One
= 3 x 10000 + 1 x 1000 + 7 × 100 + 0 × 10 + 0 × 1
= 30000 + 1000 + 700
= 31700
= Thirty-one Thousand Seven Hundred
Hence, 31700 in words is written as Thirty-one Thousand Seven Hundred.
31700 is a natural number that precedes 31701 and succeeds 31699.
31700 in words – Thirty-one Thousand Seven Hundred
Is 31700 an odd number? – No
Is 31700 an even number? – Yes
Is 31700 a perfect square number? – No
Is 31700 a perfect cube number? – No
Is 31700 a prime number? – No
Is 31700 a composite number? – Yes
## Frequently Asked Questions on 31700 in Words
Q1
### How to write 31700 in words?
31700 can be written in words as “Thirty-one Thousand Seven Hundred”.
Q2
### How to write Thirty-one Thousand Seven Hundred in numbers?
Thirty-one Thousand Seven Hundred in numbers can be written as 31700.
Q3
### Is 31700 an odd or even number?
31700 is an even number as it is completely divisible by 2. |
# The result of 17 times 15x minus 12y.
Title: Understanding the Difference Between 17 times 15x and 12y
Introduction:
Welcome to Warren Institute's blog, where we delve into the fascinating world of Mathematics education. In today's article, we will explore the concept of the difference between "seventeen times 15x" and "twelve y." This topic often confuses students, but fear not! We will break it down step by step and provide you with a clear understanding of the distinctions between the two expressions. So, grab your pens and get ready to unravel the mysteries behind these mathematical expressions!
Tags: Mathematics education, difference, 17 times 15x, 12y, understanding, expressions
## The Difference of Seventeen Times 15x and 12y
1. What is the expression "seventeen times 15x and 12y"?
The expression "seventeen times 15x and 12y" refers to the mathematical operation of multiplying seventeen by the sum of 15x and 12y. This expression involves both variables (x and y) and constants (17, 15, and 12). It represents a combination of terms that can be simplified or evaluated further.
2. How to simplify the expression?
To simplify the expression "seventeen times 15x and 12y," we can first distribute the multiplication of seventeen to both 15x and 12y. This results in (17 * 15x) + (17 * 12y). We then simplify each term individually by performing the multiplication. The simplified expression becomes 255x + 204y.
3. What is the difference between seventeen times 15x and 12y?
The difference between seventeen times 15x and 12y refers to subtracting the expression 12y from the expression 17 times 15x. Mathematically, it can be represented as (17 * 15x) - 12y. To calculate the difference, we first distribute the multiplication of seventeen to 15x and then subtract 12y from the result. The simplified expression becomes 255x - 12y.
4. How can the concept be applied in Mathematics education?
Understanding the difference between expressions like seventeen times 15x and 12y is essential in Mathematics education. It allows students to develop skills in simplifying expressions, distributing multiplication, and performing subtraction. This concept is fundamental in algebraic manipulations and lays the foundation for more complex topics such as solving equations and inequalities. By grasping this concept, students can enhance their problem-solving abilities and analytical thinking in various mathematical situations.
### How do I simplify the expression "the difference of seventeen times 15x and 12y" in terms of Mathematics education?
To simplify the expression "the difference of seventeen times 15x and 12y" in terms of Mathematics education, we can start by writing it as an algebraic expression: 17(15x) - 12y. This means we need to multiply 17 by 15x and then subtract 12y from the result.
### What strategies can I use to solve problems involving expressions like "the difference of seventeen times 15x and 12y" in Mathematics education?
One strategy to solve problems involving expressions like "the difference of seventeen times 15x and 12y" is to translate the verbal expression into a mathematical equation. In this case, the expression can be written as (17 * 15x) - 12y. Then, you can simplify the equation by multiplying and subtracting the terms accordingly. Remember to follow the order of operations and combine like terms if possible.
### Can you provide an example of a word problem that requires finding the difference of seventeen times 15x and 12y in Mathematics education?
Sure! An example of a word problem in Mathematics education that requires finding the difference of seventeen times 15x and 12y could be:
Problem: John has 17 times as many apples as Sarah has oranges. If Sarah has 15 apples and John has 12 oranges, what is the difference between seventeen times 15x and 12y?
Solution: To find the difference between seventeen times 15x and 12y, we first need to calculate the values of 17 times 15x and 12y.
17 times 15x = 255x (seventeen multiplied by fifteen times x)
12y remains unchanged.
Therefore, the difference between seventeen times 15x and 12y is 255x - 12y.
### How does understanding the concept of "the difference of seventeen times 15x and 12y" contribute to overall Mathematical fluency in Mathematics education?
Understanding the concept of "the difference of seventeen times 15x and 12y" contributes to overall mathematical fluency in mathematics education by enhancing students' ability to interpret and simplify algebraic expressions. It helps students develop skills in identifying key operations (subtraction in this case), applying order of operations, and simplifying complex expressions. This understanding facilitates problem-solving, critical thinking, and the application of mathematical concepts in various contexts.
### Are there any real-life applications where knowing how to calculate the difference of seventeen times 15x and 12y is useful in Mathematics education?
Yes, there are real-life applications where knowing how to calculate the difference of seventeen times 15x and 12y is useful in Mathematics education. This skill is particularly applicable in algebraic expressions, equations, and problem-solving scenarios. It helps students develop their understanding of variables, coefficients, and simplifying expressions. Additionally, it lays the foundation for more advanced mathematical concepts such as factoring, solving equations, and graphing.
In conclusion, the difference between seventeen times 15x and twelve times 12y plays a crucial role in Mathematics education. By understanding the concept of variables and their representations, students can develop their problem-solving skills and critical thinking abilities. It is important to emphasize the significance of utilizing proper mathematical notation, such as 17 * 15x and 12 * 12y, to accurately represent and manipulate these equations. Mastery of this mathematical concept will lay a solid foundation for students' future mathematical endeavors and enable them to tackle more complex problems with confidence.
See also Power up your circuits with a 6V to 100V DC voltage boost converter circuit - Electronics!
If you want to know other articles similar to The result of 17 times 15x minus 12y. you can visit the category General Education.
Michaell Miller
Michael Miller is a passionate blog writer and advanced mathematics teacher with a deep understanding of mathematical physics. With years of teaching experience, Michael combines his love of mathematics with an exceptional ability to communicate complex concepts in an accessible way. His blog posts offer a unique and enriching perspective on mathematical and physical topics, making learning fascinating and understandable for all.
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$\text{Set Builder Notation: } \left\{ x|x\le-8 \right\} \\\text{Interval Notation: } \left( -\infty,-8 \right]$
$\bf{\text{Solution Outline:}}$ Use the properties of inequality to solve the given inequality, $7\le1-\dfrac{3}{4}x .$ Write the answer in both set-builder notation and interval notation. $\bf{\text{Solution Details:}}$ Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 7\le1-\dfrac{3}{4}x \\\\ 4(7)\le4\left( 1-\dfrac{3}{4}x \right) \\\\ 28\le4-3x \\\\ 3x\le4-28 \\\\ 3x\le-24 \\\\ x\le-\dfrac{24}{3} \\\\ x\le-8 .\end{array} Hence, the solution set is \begin{array}{l}\require{cancel} \text{Set Builder Notation: } \left\{ x|x\le-8 \right\} \\\text{Interval Notation: } \left( -\infty,-8 \right] .\end{array} |
## Greatest Common Factor (GCF)
Finding a greatest common factor (GCF) is the process of identifying the what numbers and variables a group of terms has in common - in other words, what do they share?
In some cases you will "factor it out" and in other cases your teacher will simply ask you to identify it. Finding a GCF is an important skill to learn because it should always be your first step when factoring. For more information about factoring, try these pages: Factoring in a factor tree.
Before looking at some examples, let's break down the phrase:
Greatest = Largest Common = Shared Factor = Factored Piece
Example: Factor the GCF
Example: 3x2 + 6x
Step 1: Factor each term completely. Help with this
3xx + 3•2•x
Step 2: Find all factors that are in common (the same in all terms)
3xx + 3•2•x
Step 3: Pull out the GCF and then divide every term by it
Please remember to put parentheses around the terms, with the GCF on the outside. Once you get more advanced, you will probably be able to do the division in your head.
Step 4: Simplify each term (perform the division)
3x(x + 2)
More examples of the Greatest Common Factor
1) 5x3 - 125x The GCF is 5x. Take that out, and then divide each term 5x(x2 - 25) Now factor the binomial in parentheses farther. 2) 4x3 + 6x + 2x2 The GCF is 2x. Again, take that out. 2x(2x2 + 3 + x) 3) 6x2y + 9xy2 The GCF is 3xy. Notice each term has at least one x and y. Once you are able to factor the greatest common factor of a polynomial, you will need to learn other factoring techniques. Click here to find out how to factor trinomials and binomials. Of course, you can learn the basics of how to factor at our factoring main page. We know that factoring can be an intimidating concept for the beginning algebra student, but don't worry, it only takes a little practice.
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Center of Mass and External Forces
0 pts ended
x.øi5ng>Learning Goal: Understand that, for many purposes, a system can be treated as a point-like particle with its mass concentrated at the center of mass.
A complex system of objects, both point-like and extended ones, can often be treated as a point particle, located at the system's center of mass. Such an approach can greatly simplify problem solving.
Before you use the center of mass approach, you should first understand the following terms:
System: Any collection of objects that are of interest to you in a particular situation. In many problems, you have a certain freedom in choosing your system. Making a wise choice for the system is often the first step in solving the problem efficiently.
Center of mass: The point that represents the "average" position of the entire mass of a system. To find the coordinate of the center of mass of a system in a given frame of reference, the following definition can be used:
In this definition, each is the mass of one of the objects within the system and is the corresponding -coordinate of that object.
Of course, and can be found in a similar way.
Internal force: Any force that results from an interaction between the objects inside your system. As we will show, the internal forces do not affect the motion of the system's center of mass.
External force: Any force acting on an object inside your system that results from an interaction with an object outside your system.
Consider a system of two blocks that have masses and :
Assume that the blocks are point-like particles and have coordinates and as shown. In this problem, the blocks' motion is restricted along the x-axis.
A)
Find the x-coordinate of the center of mass of the system.
Express your answer in terms of , , and .
D)
Let us assume that the blocks are in motion, and the -components of their velocities at a certain moment are and , respectively.
Find the -component of the velocity of the center of mass at that moment. Remember that, in general,
Express your answer in terms of , , and .
F)
Let us assume that the blocks are in motion, and the x-components of their momenta at a certain moment are and , respectively.
Find the x-component of the velocity of the center of mass at that moment.
Express your answer in terms of , , and .
H)
Let us assume that the blocks are accelerating, and the -xcomponents of their accelerations at a certain moment are and , respectively.
Find the x-component of the acceleration of the center of mass at that moment. Remember that, in general,
Express your answer in terms of , , and .
We will now consider the effect of external and internal forces on the acceleration of the center of mass.
I)
Consider the same system of two blocks. An external force is now acting on block . No forces are applied to block .
Find the acceleration of the center of mass of the system.
Express your answer in terms of the x-component of the force, and .
J)
Consider the same system of two blocks. Now, there are two forces involved. An external force is acting on block and another external force is acting on block .
Find the acceleration of the center of mass of the system.
Express your answer in terms of the x-components (not magnitudes!) and of the forces, and .
L)
Consider the same system of two blocks. Now, there are two internal forces involved. An internal force is applied to block by block and another internal force is applied to block by block .
Find the acceleration of the center of mass of the system.
Express your answer in terms of the x-components (not magnitudes!) and of the forces, and .
ðx.øi5·Sx.øi5
• Understand that, for many purposes, a system can be treated as a point-like particle with its mass concentrated at the center of mass.
A complex system of objects, both point-like and extended ones, can often be treated as a point particle, located at the system's center of mass. Such an approach can greatly simplify problem solving.
Before you use the center of mass approach, you should first understand the following terms:
System: Any collection of objects that are of interest to you in a particular situation. In many problems, you have a certain freedom in choosing your system. Making a wise choice for the system is often the first step in solving the problem efficiently.
Center of mass: The point that represents the "average" position of the entire mass of a system. To find the coordinate of the center of mass of a system in a given frame of reference, the following definition can be used:
In this definition, each is the mass of one of the objects within the system and is the corresponding -coordinate of that object.
Of course, and can be found in a similar way.
Internal force: Any force that results from an interaction between the objects inside your system. As we will show, the internal forces do not affect the motion of the system's center of mass.
External force: Any force acting on an object inside your system that results from an interaction with an object outside your system.
Consider a system of two blocks that have masses and :
Assume that the blocks are point-like particles and have coordinates and as shown. In this problem, the blocks' motion is restricted along the x-axis.
A)
Find the x-coordinate of the center of mass of the system.
Express your answer in terms of , , and .
D)
Let us assume that the blocks are in motion, and the -components of their velocities at a certain moment are and , respectively.
Find the -component of the velocity of the center of mass at that moment. Remember that, in general,
Express your answer in terms of , , and .
F)
Let us assume that the blocks are in motion, and the x-components of their momenta at a certain moment are and , respectively.
Find the x-component of the velocity of the center of mass at that moment.
Express your answer in terms of , , and .
H)
Let us assume that the blocks are accelerating, and the -xcomponents of their accelerations at a certain moment are and , respectively.
Find the x-component of the acceleration of the center of mass at that moment. Remember that, in general,
Express your answer in terms of , , and .
We will now consider the effect of external and internal forces on the acceleration of the center of mass.
I)
Consider the same system of two blocks. An external force is now acting on block . No forces are applied to block .
Find the acceleration of the center of mass of the system.
Express your answer in terms of the x-component of the force, and .
J)
Consider the same system of two blocks. Now, there are two forces involved. An external force is acting on block and another external force is acting on block .
Find the acceleration of the center of mass of the system.
Express your answer in terms of the x-components (not magnitudes!) and of the forces, and .
L)
Consider the same system of two blocks. Now, there are two internal forces involved. An internal force is applied to block by block and another internal force is applied to block by block .
Find the acceleration of the center of mass of the system.
Express your answer in terms of the x-components (not magnitudes!) and of the forces, and .
ðx.øi5·Sx.øi5
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# 3.3 Incommensurability and Irrationality
## ODDS AND EVENS
• The side and the diagonal of a square are incommensurable.
To recap, the Greek concept of magnitude was somewhat tied to what a person could measure in the real world. The Pythagoreans believed that all magnitudes in nature could be represented through arithmos, the intrinsic properties of whole numbers. This means that for any two magnitudes, one should always be able to find a fundamental unit that fits some whole number of times into each of them (i.e., a unit whose magnitude is a whole number factor of each of the original magnitudes)—an idea known as commensurabilty. Hipassus argued against this idea by demonstrating that for some magnitudes this simply isn't the case—they are incommensurable. Although his original argument is lost to the ages, the following proof, which uses algebraic notions that would have been unfamiliar to the Greeks, gives a sense of the discovery that changed Greek mathematics forever.
Let's imagine a square with a side of length a and diagonal of length b.
If these lengths are commensurable, as Pythagoras and his followers believed (without proof), then there is a common unit u such that a = mu and b = nu for some whole numbers m and n. We can assume that m and n are not both even (for if they were, it would indicate that the common unit could instead be u, and we would simply make that adjustment). So, we can safely assume that at least one of these numbers is odd.
Applying Pythagoras' theorem to the triangle formed in the square, we have:
a2 + a2 = b2
That is,
2a2 = b2
or, substituting our common unit expressions for the two lengths,
2m2u2 = n2u2
We know that our common unit, u can't be zero, so we can cancel the u2 term from both sides of the equation, leaving:
2m2 = n2
Obviously, n2 is even, because it is equal to some number, m2, multiplied by two. If n2 is even, then n must be even also (if n were an odd number, then n2 would be odd). We can express the even number n as two times some number.
n = 2w
Substituting this expression for n into the preceding equation gives us:
(2w)2 = 2m2
4w2 = 2m2
m2 = 2w2
This reveals that m2 is a multiple of two, that is, an even number. Consequently, as we reasoned before, m must also be even, and we can write:
m = 2h
Now we have found a contradiction! Remember, we assumed at the beginning
that either m or n was odd, yet we have just shown that both have to be even. This logical contradiction proves that there is no common unit, u, that fits a whole number of times into both a and b—therefore, a and b, the lengths of the side and diagonal of a square, are incommensurable.
## THE INFINITE CHASE
• Incommensurable quantities are not rationally related, because this logically leads to an infinite regress.
What does incommensurability have to do with infinity? A contemporary of Hipassus, Theodorus of Cyrene, proved the incommensurability of the side and diagonal of a square by showing that no matter how small of a unit one uses to measure the side and the diagonal, it will never fit a whole number of times into both. In fact, selecting smaller and smaller units merely leads to an infinite regression of triangles. Theodorus' approach is illuminating in that it is more in line with how the Greeks thought about mathematics than the previous demonstration of incommensurability.
To get a sense of Theodorus' proof, let's again focus on the isosceles right (also commonly called a "45°-45°-90°") triangle formed by two sides and the connecting diagonal of a square. Designating this as triangle ABC, with legs of length a and hypotenuse length b, let's once more assume that there is a fundamental unit of measurement capable of representing the lengths of both a side and the hypotenuse in whole number multiples; that is:
a = mu and b = nu
Along the hypotenuse of the right triangle, we can measure a length equal to the side's length and construct a new 45°-45°-90° triangle CDE as shown:
Without too much difficulty, we can show that all three segments, BE, ED, and DC are congruent. (We won't go through the proof, but you would begin by constructing a line from A to E and showing that the two triangles ABE and ADE are congruent.) and that each of these lengths is b-a, again a whole number of copies of u.
Thus, from any 45°-45°-90° triangle with sides whose measure is a multiple of u, we can construct a smaller 45°-45°-90° triangle with sides whose measure is also a multiple of u. We can keep doing this for a number of iterations.
Eventually, however, we will obtain a 45°-45°-90° triangle so small that the length of each of its sides is less than u, which can't be—u was supposed to be the fundamental unit! We might be tempted to think that perhaps u was too big to be the fundamental unit. Using a smaller unit, however, would only delay the inevitable fact that at some point we will reach a triangle with sides whose lengths are shorter than our fundamental unit. Choosing ever smaller units leads to ever smaller "terminal" triangles for as along as we care to continue the process, another example of potential infinity. Our beginning assumption that there was a common unit of measure leads to an absurdity.
We have seen two different ways of demonstrating that the diagonal of a square is incommensurable with its side length. It is not uncommon today to calculate that if the side length of the square is 1 unit long, then its diagonal is units long. The Greeks, themselves, may not have agreed that something such as this is a number. Recall that the Pythagoreans viewed numbers as discrete collections of atom-like units. This view of numbers requires that we have a whole number "counter" to determine the size of the collection and a whole number "namer" to sit in the denominator of the ratio and designate the size of the unit. However, poses a problem because it is not useful in this method; it does not allow us to use whole numbers to serve as "counters" and "namers." This concept put the Pythagoreans in a bind, because it demonstrates that the length of the diagonal of a unit square cannot be a number; consequently, "all is not number." If we insist that such a number must exist because it measures a magnitude that actually does exist, then it is clear that we do not know what a number really is. We shall return to this problem a bit later in the text. The incommensurability argument essentially shows that there are no whole numbers m and n such that =. We call quantities like these, "irrational," and we have seen that their existence is fundamentally linked to a manifestation of infinity (the infinite regress of Theodorus' proof, for instance.) In the previous section, we saw that any rational number can be written as a repeating decimal and vice versa. However, it doesn't take much thought to conceive of a decimal that does not repeat any finite digit sequence and does not end, such as:
0.101001000100001…
Putting aside for a moment the question of whether or not something like this actually exists, we can say at least that this thing cannot be rational, because if it was it would repeat itself, which it is clearly not going to do. Its decimal expansion extends to infinity with no repetitive elements. This brings us to the point that any non-repeating decimal is non-rational, or irrational. It can also be shown that, like the , any square root of a number that is not a square number, will also be irrational. Values such as , , , etc., are all irrational.
Shortly after Hipassus made his arguments for incommensurability, which would lead to the discovery of irrational quantities, an Eleatic philosopher, Zeno, would also show the absurdity of a world in which there were fundamental smallest units of space and time. Recall that the Eleatics held beliefs somewhat diametrically opposed to those of the Pythagoreans—that multiplicity, the idea that the universe is composed of fundamental parts—is ridiculous. They believed in continuous magnitudes in which any perceived boundaries were illusions. This idea is somewhat similar to the concept that "all is one." Similarly to Hipassus' argument for incommensurable magnitudes, Zeno would show that treating a line as a multitude of individual points was philosophically contradictory. These ideas would force thinkers to confront notions of actual infinity—an infinity contained in a limited space—which would prove to be both a powerful concept and a troublesome idea in mathematics. |
NCERT Textbook - Linear Inequalities
# NCERT Textbook - Linear Inequalities | NCERT Textbooks (Class 6 to Class 12) - CTET & State TET PDF Download
``` Page 1
5 Chapter
vMathematics is the art of saying many things in many
different ways. – MAXWELLv
5.1 Introduction
In earlier classes, we have studied equations in one variable and two variables and also
solved some statement problems by translating them in the form of equations. Now a
natural question arises: ‘Is it always possible to translate a statement problem in the
form of an equation? For example, the height of all the students in your class is less
than 160 cm. Your classroom can occupy atmost 60 tables or chairs or both. Here we
get certain statements involving a sign ‘<’ (less than), ‘>’ (greater than), ‘=’ (less than
or equal) and = (greater than or equal) which are known as inequalities.
In this Chapter, we will study linear inequalities in one and two variables. The
study of inequalities is very useful in solving problems in the field of science, mathematics,
statistics, economics, psychology, etc.
5.2 Inequalities
Let us consider the following situations:
(i) Ravi goes to market with ` 200 to buy rice, which is available in packets of 1kg. The
price of one packet of rice is ` 30. If x denotes the number of packets of rice, which he
buys, then the total amount spent by him is ` 30x. Since, he has to buy rice in packets
only, he may not be able to spend the entire amount of ` 200. (Why?) Hence
30x < 200 ... (1)
Clearly the statement (i) is not an equation as it does not involve the sign of equality.
(ii) Reshma has ` 120 and wants to buy some registers and pens. The cost of one
register is ` 40 and that of a pen is ` 20. In this case, if x denotes the number of
registers and y, the number of pens which Reshma buys, then the total amount spent by
her is ` (40x + 20y) and we have
40x + 20y = 120 ... (2)
LINEAR INEQUALITIES
2024-25
Page 2
5 Chapter
vMathematics is the art of saying many things in many
different ways. – MAXWELLv
5.1 Introduction
In earlier classes, we have studied equations in one variable and two variables and also
solved some statement problems by translating them in the form of equations. Now a
natural question arises: ‘Is it always possible to translate a statement problem in the
form of an equation? For example, the height of all the students in your class is less
than 160 cm. Your classroom can occupy atmost 60 tables or chairs or both. Here we
get certain statements involving a sign ‘<’ (less than), ‘>’ (greater than), ‘=’ (less than
or equal) and = (greater than or equal) which are known as inequalities.
In this Chapter, we will study linear inequalities in one and two variables. The
study of inequalities is very useful in solving problems in the field of science, mathematics,
statistics, economics, psychology, etc.
5.2 Inequalities
Let us consider the following situations:
(i) Ravi goes to market with ` 200 to buy rice, which is available in packets of 1kg. The
price of one packet of rice is ` 30. If x denotes the number of packets of rice, which he
buys, then the total amount spent by him is ` 30x. Since, he has to buy rice in packets
only, he may not be able to spend the entire amount of ` 200. (Why?) Hence
30x < 200 ... (1)
Clearly the statement (i) is not an equation as it does not involve the sign of equality.
(ii) Reshma has ` 120 and wants to buy some registers and pens. The cost of one
register is ` 40 and that of a pen is ` 20. In this case, if x denotes the number of
registers and y, the number of pens which Reshma buys, then the total amount spent by
her is ` (40x + 20y) and we have
40x + 20y = 120 ... (2)
LINEAR INEQUALITIES
2024-25
90 MATHEMATICS
Since in this case the total amount spent may be upto ` 120. Note that the statement (2)
consists of two statements
40x + 20y < 120 ... (3)
and 40x + 20y = 120 ... (4)
Statement (3) is not an equation, i.e., it is an inequality while statement (4) is an equation.
Definition 1 Two real numbers or two algebraic expressions related by the symbol
‘<’, ‘>’, ‘=’ or ‘=’ form an inequality.
Statements such as (1), (2) and (3) above are inequalities.
3 < 5; 7 > 5 are the examples of numerical inequalities while
x < 5; y > 2; x = 3, y = 4 are some examples of literal inequalities.
3 < 5 < 7 (read as 5 is greater than 3 and less than 7), 3 < x < 5 (read as x is greater
than or equal to 3 and less than 5) and 2 < y < 4 are the examples of double inequalities.
Some more examples of inequalities are:
ax + b < 0 ... (5)
ax + b > 0 ... (6)
ax + b = 0 ... (7)
ax + b = 0 ... (8)
ax + by < c ... (9)
ax + by > c ... (10)
ax + by = c ... (11)
ax + by = c ... (12)
ax
2
+ bx + c = 0 ... (13)
ax
2
+ bx + c > 0 ... (14)
Inequalities (5), (6), (9), (10) and (14) are strict inequalities while inequalities (7), (8),
(11), (12), and (13) are slack inequalities. Inequalities from (5) to (8) are linear
inequalities in one variable x when a ? 0, while inequalities from (9) to (12) are linear
inequalities in two variables x and y when a ? 0, b ? 0.
Inequalities (13) and (14) are not linear (in fact, these are quadratic inequalities
in one variable x when a ? 0).
In this Chapter, we shall confine ourselves to the study of linear inequalities in one
and two variables only.
2024-25
Page 3
5 Chapter
vMathematics is the art of saying many things in many
different ways. – MAXWELLv
5.1 Introduction
In earlier classes, we have studied equations in one variable and two variables and also
solved some statement problems by translating them in the form of equations. Now a
natural question arises: ‘Is it always possible to translate a statement problem in the
form of an equation? For example, the height of all the students in your class is less
than 160 cm. Your classroom can occupy atmost 60 tables or chairs or both. Here we
get certain statements involving a sign ‘<’ (less than), ‘>’ (greater than), ‘=’ (less than
or equal) and = (greater than or equal) which are known as inequalities.
In this Chapter, we will study linear inequalities in one and two variables. The
study of inequalities is very useful in solving problems in the field of science, mathematics,
statistics, economics, psychology, etc.
5.2 Inequalities
Let us consider the following situations:
(i) Ravi goes to market with ` 200 to buy rice, which is available in packets of 1kg. The
price of one packet of rice is ` 30. If x denotes the number of packets of rice, which he
buys, then the total amount spent by him is ` 30x. Since, he has to buy rice in packets
only, he may not be able to spend the entire amount of ` 200. (Why?) Hence
30x < 200 ... (1)
Clearly the statement (i) is not an equation as it does not involve the sign of equality.
(ii) Reshma has ` 120 and wants to buy some registers and pens. The cost of one
register is ` 40 and that of a pen is ` 20. In this case, if x denotes the number of
registers and y, the number of pens which Reshma buys, then the total amount spent by
her is ` (40x + 20y) and we have
40x + 20y = 120 ... (2)
LINEAR INEQUALITIES
2024-25
90 MATHEMATICS
Since in this case the total amount spent may be upto ` 120. Note that the statement (2)
consists of two statements
40x + 20y < 120 ... (3)
and 40x + 20y = 120 ... (4)
Statement (3) is not an equation, i.e., it is an inequality while statement (4) is an equation.
Definition 1 Two real numbers or two algebraic expressions related by the symbol
‘<’, ‘>’, ‘=’ or ‘=’ form an inequality.
Statements such as (1), (2) and (3) above are inequalities.
3 < 5; 7 > 5 are the examples of numerical inequalities while
x < 5; y > 2; x = 3, y = 4 are some examples of literal inequalities.
3 < 5 < 7 (read as 5 is greater than 3 and less than 7), 3 < x < 5 (read as x is greater
than or equal to 3 and less than 5) and 2 < y < 4 are the examples of double inequalities.
Some more examples of inequalities are:
ax + b < 0 ... (5)
ax + b > 0 ... (6)
ax + b = 0 ... (7)
ax + b = 0 ... (8)
ax + by < c ... (9)
ax + by > c ... (10)
ax + by = c ... (11)
ax + by = c ... (12)
ax
2
+ bx + c = 0 ... (13)
ax
2
+ bx + c > 0 ... (14)
Inequalities (5), (6), (9), (10) and (14) are strict inequalities while inequalities (7), (8),
(11), (12), and (13) are slack inequalities. Inequalities from (5) to (8) are linear
inequalities in one variable x when a ? 0, while inequalities from (9) to (12) are linear
inequalities in two variables x and y when a ? 0, b ? 0.
Inequalities (13) and (14) are not linear (in fact, these are quadratic inequalities
in one variable x when a ? 0).
In this Chapter, we shall confine ourselves to the study of linear inequalities in one
and two variables only.
2024-25
LINEAR INEQUALITIES 91
5.3 Algebraic Solutions of Linear Inequalities in One Variable and their
Graphical Representation
Let us consider the inequality (1) of Section 6.2, viz, 30x < 200
Note that here x denotes the number of packets of rice.
Obviously, x cannot be a negative integer or a fraction. Left hand side (L.H.S.) of this
inequality is 30x and right hand side (RHS) is 200. Therefore, we have
For x = 0, L.H.S. = 30 (0) = 0 < 200 (R.H.S.), which is true.
For x = 1, L.H.S. = 30 (1) = 30 < 200 (R.H.S.), which is true.
For x = 2, L.H.S. = 30 (2) = 60 < 200, which is true.
For x = 3, L.H.S. = 30 (3) = 90 < 200, which is true.
For x = 4, L.H.S. = 30 (4) = 120 < 200, which is true.
For x = 5, L.H.S. = 30 (5) = 150 < 200, which is true.
For x = 6, L.H.S. = 30 (6) = 180 < 200, which is true.
For x = 7, L.H.S. = 30 (7) = 210 < 200, which is false.
In the above situation, we find that the values of x, which makes the above
inequality a true statement, are 0,1,2,3,4,5,6. These values of x, which make above
inequality a true statement, are called solutions of inequality and the set {0,1,2,3,4,5,6}
is called its solution set.
Thus, any solution of an inequality in one variable is a value of the variable
which makes it a true statement.
We have found the solutions of the above inequality by trial and error method
which is not very efficient. Obviously, this method is time consuming and sometimes
not feasible. W e must have some better or systematic techniques for solving inequalities.
Before that we should go through some more properties of numerical inequalities and
follow them as rules while solving the inequalities.
Y ou will recall that while solving linear equations, we followed the following rules:
Rule 1 Equal numbers may be added to (or subtracted from) both sides of an equation.
Rule 2 Both sides of an equation may be multiplied (or divided) by the same non-zero
number.
In the case of solving inequalities, we again follow the same rules except with a
difference that in Rule 2, the sign of inequality is reversed (i.e., ‘<‘ becomes ‘>’, =’
becomes ‘=’ and so on) whenever we multiply (or divide) both sides of an inequality by
a negative number. It is evident from the facts that
3 > 2 while – 3 < – 2,
– 8 < – 7 while (– 8) (– 2) > (– 7) (– 2) , i.e., 16 > 14.
2024-25
Page 4
5 Chapter
vMathematics is the art of saying many things in many
different ways. – MAXWELLv
5.1 Introduction
In earlier classes, we have studied equations in one variable and two variables and also
solved some statement problems by translating them in the form of equations. Now a
natural question arises: ‘Is it always possible to translate a statement problem in the
form of an equation? For example, the height of all the students in your class is less
than 160 cm. Your classroom can occupy atmost 60 tables or chairs or both. Here we
get certain statements involving a sign ‘<’ (less than), ‘>’ (greater than), ‘=’ (less than
or equal) and = (greater than or equal) which are known as inequalities.
In this Chapter, we will study linear inequalities in one and two variables. The
study of inequalities is very useful in solving problems in the field of science, mathematics,
statistics, economics, psychology, etc.
5.2 Inequalities
Let us consider the following situations:
(i) Ravi goes to market with ` 200 to buy rice, which is available in packets of 1kg. The
price of one packet of rice is ` 30. If x denotes the number of packets of rice, which he
buys, then the total amount spent by him is ` 30x. Since, he has to buy rice in packets
only, he may not be able to spend the entire amount of ` 200. (Why?) Hence
30x < 200 ... (1)
Clearly the statement (i) is not an equation as it does not involve the sign of equality.
(ii) Reshma has ` 120 and wants to buy some registers and pens. The cost of one
register is ` 40 and that of a pen is ` 20. In this case, if x denotes the number of
registers and y, the number of pens which Reshma buys, then the total amount spent by
her is ` (40x + 20y) and we have
40x + 20y = 120 ... (2)
LINEAR INEQUALITIES
2024-25
90 MATHEMATICS
Since in this case the total amount spent may be upto ` 120. Note that the statement (2)
consists of two statements
40x + 20y < 120 ... (3)
and 40x + 20y = 120 ... (4)
Statement (3) is not an equation, i.e., it is an inequality while statement (4) is an equation.
Definition 1 Two real numbers or two algebraic expressions related by the symbol
‘<’, ‘>’, ‘=’ or ‘=’ form an inequality.
Statements such as (1), (2) and (3) above are inequalities.
3 < 5; 7 > 5 are the examples of numerical inequalities while
x < 5; y > 2; x = 3, y = 4 are some examples of literal inequalities.
3 < 5 < 7 (read as 5 is greater than 3 and less than 7), 3 < x < 5 (read as x is greater
than or equal to 3 and less than 5) and 2 < y < 4 are the examples of double inequalities.
Some more examples of inequalities are:
ax + b < 0 ... (5)
ax + b > 0 ... (6)
ax + b = 0 ... (7)
ax + b = 0 ... (8)
ax + by < c ... (9)
ax + by > c ... (10)
ax + by = c ... (11)
ax + by = c ... (12)
ax
2
+ bx + c = 0 ... (13)
ax
2
+ bx + c > 0 ... (14)
Inequalities (5), (6), (9), (10) and (14) are strict inequalities while inequalities (7), (8),
(11), (12), and (13) are slack inequalities. Inequalities from (5) to (8) are linear
inequalities in one variable x when a ? 0, while inequalities from (9) to (12) are linear
inequalities in two variables x and y when a ? 0, b ? 0.
Inequalities (13) and (14) are not linear (in fact, these are quadratic inequalities
in one variable x when a ? 0).
In this Chapter, we shall confine ourselves to the study of linear inequalities in one
and two variables only.
2024-25
LINEAR INEQUALITIES 91
5.3 Algebraic Solutions of Linear Inequalities in One Variable and their
Graphical Representation
Let us consider the inequality (1) of Section 6.2, viz, 30x < 200
Note that here x denotes the number of packets of rice.
Obviously, x cannot be a negative integer or a fraction. Left hand side (L.H.S.) of this
inequality is 30x and right hand side (RHS) is 200. Therefore, we have
For x = 0, L.H.S. = 30 (0) = 0 < 200 (R.H.S.), which is true.
For x = 1, L.H.S. = 30 (1) = 30 < 200 (R.H.S.), which is true.
For x = 2, L.H.S. = 30 (2) = 60 < 200, which is true.
For x = 3, L.H.S. = 30 (3) = 90 < 200, which is true.
For x = 4, L.H.S. = 30 (4) = 120 < 200, which is true.
For x = 5, L.H.S. = 30 (5) = 150 < 200, which is true.
For x = 6, L.H.S. = 30 (6) = 180 < 200, which is true.
For x = 7, L.H.S. = 30 (7) = 210 < 200, which is false.
In the above situation, we find that the values of x, which makes the above
inequality a true statement, are 0,1,2,3,4,5,6. These values of x, which make above
inequality a true statement, are called solutions of inequality and the set {0,1,2,3,4,5,6}
is called its solution set.
Thus, any solution of an inequality in one variable is a value of the variable
which makes it a true statement.
We have found the solutions of the above inequality by trial and error method
which is not very efficient. Obviously, this method is time consuming and sometimes
not feasible. W e must have some better or systematic techniques for solving inequalities.
Before that we should go through some more properties of numerical inequalities and
follow them as rules while solving the inequalities.
Y ou will recall that while solving linear equations, we followed the following rules:
Rule 1 Equal numbers may be added to (or subtracted from) both sides of an equation.
Rule 2 Both sides of an equation may be multiplied (or divided) by the same non-zero
number.
In the case of solving inequalities, we again follow the same rules except with a
difference that in Rule 2, the sign of inequality is reversed (i.e., ‘<‘ becomes ‘>’, =’
becomes ‘=’ and so on) whenever we multiply (or divide) both sides of an inequality by
a negative number. It is evident from the facts that
3 > 2 while – 3 < – 2,
– 8 < – 7 while (– 8) (– 2) > (– 7) (– 2) , i.e., 16 > 14.
2024-25
92 MATHEMATICS
Thus, we state the following rules for solving an inequality:
Rule 1 Equal numbers may be added to (or subtracted from) both sides of an inequality
without affecting the sign of inequality.
Rule 2 Both sides of an inequality can be multiplied (or divided) by the same positive
number. But when both sides are multiplied or divided by a negative number, then the
sign of inequality is reversed.
Now, let us consider some examples.
Example 1 Solve 30 x < 200 when
(i) x is a natural number, (ii) x is an integer.
Solution We are given 30 x < 200
or
30 200
30 30
x
<
(Rule 2), i.e., x < 20 / 3.
(i) When x is a natural number, in this case the following values of x make the
statement true.
1, 2, 3, 4, 5, 6.
The solution set of the inequality is {1,2,3,4,5,6}.
(ii) When x is an integer, the solutions of the given inequality are
..., – 3, –2, –1, 0, 1, 2, 3, 4, 5, 6
The solution set of the inequality is {...,–3, –2,–1, 0, 1, 2, 3, 4, 5, 6}
Example 2 Solve 5x – 3 < 3x +1 when
(i) x is an integer, (ii) x is a real number.
Solution We have, 5x –3 < 3x + 1
or 5x –3 + 3 < 3x +1 +3 (Rule 1)
or 5x < 3x +4
or 5x – 3x < 3x + 4 – 3x (Rule 1)
or 2x < 4
or x < 2 (Rule 2)
(i) When x is an integer, the solutions of the given inequality are
..., – 4, – 3, – 2, – 1, 0, 1
(ii) When x is a real number, the solutions of the inequality are given by x < 2,
i.e., all real numbers x which are less than 2. Therefore, the solution set of
the inequality is x ? (– 8, 2).
We have considered solutions of inequalities in the set of natural numbers, set of
integers and in the set of real numbers. Henceforth, unless stated otherwise, we shall
solve the inequalities in this Chapter in the set of real numbers.
2024-25
Page 5
5 Chapter
vMathematics is the art of saying many things in many
different ways. – MAXWELLv
5.1 Introduction
In earlier classes, we have studied equations in one variable and two variables and also
solved some statement problems by translating them in the form of equations. Now a
natural question arises: ‘Is it always possible to translate a statement problem in the
form of an equation? For example, the height of all the students in your class is less
than 160 cm. Your classroom can occupy atmost 60 tables or chairs or both. Here we
get certain statements involving a sign ‘<’ (less than), ‘>’ (greater than), ‘=’ (less than
or equal) and = (greater than or equal) which are known as inequalities.
In this Chapter, we will study linear inequalities in one and two variables. The
study of inequalities is very useful in solving problems in the field of science, mathematics,
statistics, economics, psychology, etc.
5.2 Inequalities
Let us consider the following situations:
(i) Ravi goes to market with ` 200 to buy rice, which is available in packets of 1kg. The
price of one packet of rice is ` 30. If x denotes the number of packets of rice, which he
buys, then the total amount spent by him is ` 30x. Since, he has to buy rice in packets
only, he may not be able to spend the entire amount of ` 200. (Why?) Hence
30x < 200 ... (1)
Clearly the statement (i) is not an equation as it does not involve the sign of equality.
(ii) Reshma has ` 120 and wants to buy some registers and pens. The cost of one
register is ` 40 and that of a pen is ` 20. In this case, if x denotes the number of
registers and y, the number of pens which Reshma buys, then the total amount spent by
her is ` (40x + 20y) and we have
40x + 20y = 120 ... (2)
LINEAR INEQUALITIES
2024-25
90 MATHEMATICS
Since in this case the total amount spent may be upto ` 120. Note that the statement (2)
consists of two statements
40x + 20y < 120 ... (3)
and 40x + 20y = 120 ... (4)
Statement (3) is not an equation, i.e., it is an inequality while statement (4) is an equation.
Definition 1 Two real numbers or two algebraic expressions related by the symbol
‘<’, ‘>’, ‘=’ or ‘=’ form an inequality.
Statements such as (1), (2) and (3) above are inequalities.
3 < 5; 7 > 5 are the examples of numerical inequalities while
x < 5; y > 2; x = 3, y = 4 are some examples of literal inequalities.
3 < 5 < 7 (read as 5 is greater than 3 and less than 7), 3 < x < 5 (read as x is greater
than or equal to 3 and less than 5) and 2 < y < 4 are the examples of double inequalities.
Some more examples of inequalities are:
ax + b < 0 ... (5)
ax + b > 0 ... (6)
ax + b = 0 ... (7)
ax + b = 0 ... (8)
ax + by < c ... (9)
ax + by > c ... (10)
ax + by = c ... (11)
ax + by = c ... (12)
ax
2
+ bx + c = 0 ... (13)
ax
2
+ bx + c > 0 ... (14)
Inequalities (5), (6), (9), (10) and (14) are strict inequalities while inequalities (7), (8),
(11), (12), and (13) are slack inequalities. Inequalities from (5) to (8) are linear
inequalities in one variable x when a ? 0, while inequalities from (9) to (12) are linear
inequalities in two variables x and y when a ? 0, b ? 0.
Inequalities (13) and (14) are not linear (in fact, these are quadratic inequalities
in one variable x when a ? 0).
In this Chapter, we shall confine ourselves to the study of linear inequalities in one
and two variables only.
2024-25
LINEAR INEQUALITIES 91
5.3 Algebraic Solutions of Linear Inequalities in One Variable and their
Graphical Representation
Let us consider the inequality (1) of Section 6.2, viz, 30x < 200
Note that here x denotes the number of packets of rice.
Obviously, x cannot be a negative integer or a fraction. Left hand side (L.H.S.) of this
inequality is 30x and right hand side (RHS) is 200. Therefore, we have
For x = 0, L.H.S. = 30 (0) = 0 < 200 (R.H.S.), which is true.
For x = 1, L.H.S. = 30 (1) = 30 < 200 (R.H.S.), which is true.
For x = 2, L.H.S. = 30 (2) = 60 < 200, which is true.
For x = 3, L.H.S. = 30 (3) = 90 < 200, which is true.
For x = 4, L.H.S. = 30 (4) = 120 < 200, which is true.
For x = 5, L.H.S. = 30 (5) = 150 < 200, which is true.
For x = 6, L.H.S. = 30 (6) = 180 < 200, which is true.
For x = 7, L.H.S. = 30 (7) = 210 < 200, which is false.
In the above situation, we find that the values of x, which makes the above
inequality a true statement, are 0,1,2,3,4,5,6. These values of x, which make above
inequality a true statement, are called solutions of inequality and the set {0,1,2,3,4,5,6}
is called its solution set.
Thus, any solution of an inequality in one variable is a value of the variable
which makes it a true statement.
We have found the solutions of the above inequality by trial and error method
which is not very efficient. Obviously, this method is time consuming and sometimes
not feasible. W e must have some better or systematic techniques for solving inequalities.
Before that we should go through some more properties of numerical inequalities and
follow them as rules while solving the inequalities.
Y ou will recall that while solving linear equations, we followed the following rules:
Rule 1 Equal numbers may be added to (or subtracted from) both sides of an equation.
Rule 2 Both sides of an equation may be multiplied (or divided) by the same non-zero
number.
In the case of solving inequalities, we again follow the same rules except with a
difference that in Rule 2, the sign of inequality is reversed (i.e., ‘<‘ becomes ‘>’, =’
becomes ‘=’ and so on) whenever we multiply (or divide) both sides of an inequality by
a negative number. It is evident from the facts that
3 > 2 while – 3 < – 2,
– 8 < – 7 while (– 8) (– 2) > (– 7) (– 2) , i.e., 16 > 14.
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92 MATHEMATICS
Thus, we state the following rules for solving an inequality:
Rule 1 Equal numbers may be added to (or subtracted from) both sides of an inequality
without affecting the sign of inequality.
Rule 2 Both sides of an inequality can be multiplied (or divided) by the same positive
number. But when both sides are multiplied or divided by a negative number, then the
sign of inequality is reversed.
Now, let us consider some examples.
Example 1 Solve 30 x < 200 when
(i) x is a natural number, (ii) x is an integer.
Solution We are given 30 x < 200
or
30 200
30 30
x
<
(Rule 2), i.e., x < 20 / 3.
(i) When x is a natural number, in this case the following values of x make the
statement true.
1, 2, 3, 4, 5, 6.
The solution set of the inequality is {1,2,3,4,5,6}.
(ii) When x is an integer, the solutions of the given inequality are
..., – 3, –2, –1, 0, 1, 2, 3, 4, 5, 6
The solution set of the inequality is {...,–3, –2,–1, 0, 1, 2, 3, 4, 5, 6}
Example 2 Solve 5x – 3 < 3x +1 when
(i) x is an integer, (ii) x is a real number.
Solution We have, 5x –3 < 3x + 1
or 5x –3 + 3 < 3x +1 +3 (Rule 1)
or 5x < 3x +4
or 5x – 3x < 3x + 4 – 3x (Rule 1)
or 2x < 4
or x < 2 (Rule 2)
(i) When x is an integer, the solutions of the given inequality are
..., – 4, – 3, – 2, – 1, 0, 1
(ii) When x is a real number, the solutions of the inequality are given by x < 2,
i.e., all real numbers x which are less than 2. Therefore, the solution set of
the inequality is x ? (– 8, 2).
We have considered solutions of inequalities in the set of natural numbers, set of
integers and in the set of real numbers. Henceforth, unless stated otherwise, we shall
solve the inequalities in this Chapter in the set of real numbers.
2024-25
LINEAR INEQUALITIES 93
Example 3 Solve 4x + 3 < 6x +7.
Solution We have, 4x + 3 < 6x + 7
or 4x – 6x < 6x + 4 – 6x
or – 2x < 4 or x > – 2
i.e., all the real numbers which are greater than –2, are the solutions of the given
inequality. Hence, the solution set is (–2, 8).
Example 4 Solve
5 2
5
3 6
– x x
– =
.
Solution We have
5 2
5
3 6
– x x
– =
or 2 (5 – 2x) = x – 30.
or 10 – 4x = x – 30
or – 5x = – 40, i.e., x = 8
Thus, all real numbers x which are greater than or equal to 8 are the solutions of the
given inequality, i.e., x ? [8, 8).
Example 5 Solve 7x + 3 < 5x + 9. Show the graph of the solutions on number line.
Solution We have 7x + 3 < 5x + 9 or
2x < 6 or x < 3
The graphical representation of the solutions are given in Fig 5.1.
Fig 5.1
Example 6 Solve
3 4 1
1
2 4
x x - +
= -
. Show the graph of the solutions on number line.
Solution We have
3 4 1
1
2 4
x x - +
= -
or
3 4 3
2 4
x x - -
=
or 2 (3x – 4) = (x – 3)
2024-25
```
## NCERT Textbooks (Class 6 to Class 12)
681 docs|672 tests
## FAQs on NCERT Textbook - Linear Inequalities - NCERT Textbooks (Class 6 to Class 12) - CTET & State TET
1. What is a linear inequality?
Ans. A linear inequality is an inequality in which the highest power of the variable is 1. It can be written in the form ax + b < c or ax + b > c, where a, b, and c are real numbers.
2. How do you graph a linear inequality?
Ans. To graph a linear inequality, first, graph the corresponding linear equation. Then, choose a test point not on the line and substitute its coordinates into the inequality. If the inequality is true, shade the region that contains the test point. If it is false, shade the other region.
3. What is the solution set of a linear inequality?
Ans. The solution set of a linear inequality is the set of all values of the variable that satisfy the inequality. It can be represented graphically as the shaded region on the graph or algebraically as a range of values for the variable.
4. How do you solve a system of linear inequalities?
Ans. To solve a system of linear inequalities, graph each inequality on the same coordinate plane. The solution is the overlapping region or the intersection of all the shaded regions. It represents the values that satisfy all the inequalities simultaneously.
5. Can a linear inequality have no solution?
Ans. Yes, a linear inequality can have no solution. This occurs when the shaded regions of the inequalities do not overlap or intersect. In other words, there are no values of the variable that satisfy all the given inequalities simultaneously.
## NCERT Textbooks (Class 6 to Class 12)
681 docs|672 tests
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# 11.4: Volume of Prisms and Cylinders
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Find the volume of prisms and cylinders.
## Review Queue
1. Define volume in your own words.
2. What is the surface area of a cube with 3 inch sides?
3. A regular octahedron has 8 congruent equilateral triangles as the faces.
1. If each edge is 4 cm, what is the slant height for one face?
2. What is the surface area of one face?
3. What is the total surface area?
Know What? Let’s fill the pool it with water. The shallow end is 4 ft. and the deep end is 8 ft. The pool is 10 ft. wide by 25 ft. long. How many cubic feet of water is needed to fill the pool?
## Volume of a Rectangular Prism
Volume: The measure of how much space a three-dimensional figure occupies.
Another way to define volume would be how much a three-dimensional figure can hold. The basic unit of volume is the cubic unit: cubic centimeter , cubic inch , cubic meter , cubic foot .
Volume of a Cube Postulate: .
What this postulate tells us is that every solid can be broken down into cubes. For example, if we wanted to find the volume of a cube with 9 inch sides, it would be .
Volume Congruence Postulate: If two solids are congruent, then their volumes are congruent.
These prisms are congruent, so their volumes are congruent.
Example 1: Find the volume of the right rectangular prism below.
Solution: Count the cubes. The bottom layer has 20 cubes, or , and there are 3 layers. There are 60 cubes. The volume is also .
Each layer in Example 1 is the same as the area of the base and the number of layers is the same as the height. This is the formula for volume.
Volume of a Rectangular Prism: .
Example 2: A typical shoe box is 8 in by 14 in by 6 in. What is the volume of the box?
Solution: We can assume that a shoe box is a rectangular prism.
## Volume of any Prism
Notcie that is equal to the area of the base of the prism, which we will re-label .
Volume of a Prism: .
” is not always going to be the same. So, to find the volume of a prism, you would first find the area of the base and then multiply it by the height.
Example 3: You have a small, triangular prism shaped tent. How much volume does it have, once it is set up?
Solution: First, we need to find the area of the base.
Even though the height in this problem does not look like a “height,” it is, according to the formula. Usually, the height of a prism is going to be the last length you need to use.
## Oblique Prisms
Recall that oblique prisms are prisms that lean to one side and the height is outside the prism. What would be the volume of an oblique prism? Consider to piles of books below.
Both piles have 15 books, which means they will have the same volume. Cavalieri’s Principle says that leaning does not matter, the volumes are the same.
Cavalieri’s Principle: If two solids have the same height and the same cross-sectional area at every level, then they will have the same volume.
If an oblique prism and a right prism have the same base area and height, then they will have the same volume.
Example 4: Find the area of the oblique prism below.
Solution: This is an oblique right trapezoidal prism. Find the area of the trapezoid.
## Volume of a Cylinder
If we use the formula for the volume of a prism, , we can find the volume of a cylinder. In the case of a cylinder, the base is the area of a circle. Like a prism, Cavalieri’s Principle holds.
Volume of a Cylinder: .
Example 5: Find the volume of the cylinder.
Solution: If the diameter is 16, then the radius is 8.
Example 6: Find the volume of the cylinder.
Solution:
Example 7: If the volume of a cylinder is and the height is 4 in, what is the radius?
Solution: Solve for .
Example 8: Find the volume of the solid below.
Solution: This solid is a parallelogram-based prism with a cylinder cut out of the middle.
The total volume is .
Know What? Revisited Even though it doesn’t look like it, the trapezoid is the base of this prism. The area of the trapezoids are .
## Review Questions
• Question 1 uses the volume formula for a cylinder.
• Questions 2-4 are similar to Example 1.
• Questions 5-18 are similar to Examples 2-6.
• Questions 19-24 are similar to Example 7.
• Questions 25-30 are similar to Example 8.
1. Two cylinders have the same surface area. Do they have the same volume? How do you know?
2. How many one-inch cubes can fit into a box that is 8 inches wide, 10 inches long, and 12 inches tall? Is this the same as the volume of the box?
3. A cereal box in 2 inches wide, 10 inches long and 14 inches tall. How much cereal does the box hold?
4. A can of soda is 4 inches tall and has a diameter of 2 inches. How much soda does the can hold? Round your answer to the nearest hundredth.
5. A cube holds . What is the length of each edge?
6. A cube has sides that are 8 inches. What is the volume?
7. A cylinder has and the radius is 4 cm. What is the volume?
8. A cylinder has a volume of . If the height is 6 ft., what is the diameter?
Use the right triangular prism to answer questions 9 and 10.
1. What is the length of the third base edge?
2. Find the volume of the prism.
3. Fuzzy dice are cubes with 4 inch sides.
1. What is the volume of one die?
2. What is the volume of both dice?
4. A right cylinder has a 7 cm radius and a height of 18 cm. Find the volume.
Find the volume of the following solids. Round your answers to the nearest hundredth.
Algebra Connection Find the value of , given the surface area.
1. The area of the base of a cylinder is and the height is 6 in. Find the volume.
2. The circumference of the base of a cylinder is and the height is 15 cm. Find the volume.
3. The lateral surface area of a cylinder is and the circumference is . What is the volume of the cylinder?
The bases of the prism are squares and a cylinder is cut out of the center.
1. Find the volume of the prism.
2. Find the volume of the cylinder in the center.
3. Find the volume of the figure.
This is a prism with half a cylinder on the top.
1. Find the volume of the prism.
2. Find the volume of the half-cylinder.
3. Find the volume of the entire figure.
1. The amount a three-dimensional figure can hold.
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Date Created:
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CK.MAT.ENG.SE.1.Geometry-Basic.11.4 |
INTRODUCTION
In previous classes, we have studided certain constructions using a ruler and a compass. There are : bisecting an angle, drawing the perpendicular bisector of a line segment, construction of some triangle, etc. We have also given their justifications. Here, we shall study some more constructions taking into use the above pervious knowledge. Also, we shall give mathematical reasoning underlying these constructions.
Divison of A line segment :
In order to divide a line segment internally in a given ratio m : n, where both m and n are positive integers, we follow the following steps :
Steps of construction :
(i) Draw a line segment AB of given length by using a ruler.
(ii) Draw any ray AX making a suitable acute angle with AB.
(iii) Along AX draw (m + n) arcs intersecting the rays AX at A1, A2 ............, Am, Am+1, ........., Am + n such that AA1 = A1A2 =...............= Am+n–1 Am+n
(iv) Join B Am+n
(v) Through the point Am draw a line parallel to Am+n B by making AAm P = AAm+n B.
Suppose this line meets AB at point P.
The point P so obtained is the required point which divides AB internally in the ratio m : n.
Illustration
Divide a line segment of length 12 cm internally in the ratio 3 : 2.
Solution
Steps of construction :
(i) Draw a line segment AB = 12 cm by using a ruler.
(ii) Draw a ray making a suitable acute angle BAX with AB.
(iii) Along AX, draw 5 ( = 3 + 2) arcs intersecting the rays AX at A1, A2, A3, A4 and A5 such that
AA1 = A1A2 = A2A3 = A3A4 = A4A5 |
Utah Math 2 - 2020 Edition
Lesson
A quadrilateral is a four-sided 2D figure. By showing that quadrilaterals have specific properties, we can narrow it down to the most specific type of quadrilateral.
### The hierarchy of parallelograms
We define a parallelogram as any quadrilateral whose opposite side lengths are parallel. In some instances, a parallelogram can also be a rhombus, rectangle or square, depending on what knowledge we know about the parallelogram.
Parallelogram
Properties of parallelograms:
• Opposite sides are parallel
• Opposite sides are congruent
• Opposite angles are congruent
• Consecutive angles are supplementary
• The diagonals bisect each other
Bisecting diagonals
We define a rhombus as a parallelogram whose all four sides are congruent.
Rhombus
Properties of a rhombus:
• Four congruent sides
• Diagonals are perpendicular
• Each diagonal bisects a pair of opposite angles
We define a rectangle as a parallelogram whose interior angles are right angles.
Rectangle
Properties of rectangles:
• Four right angles
• Diagonals are congruent
• Diagonals bisect each other
And lastly, we define a square as a parallelogram whose all four sides are congruent and and all interior angles are right angles.
Square
Properties of squares:
• Four congruent sides and four congruent angles
• A special case of both a rectangle and a rhombus
• Diagonals are perpendicular
If we can prove that a parallelogram possesses one of these necessary conditions, then we can prove that it is a rhombus, rectangle or square respectively.
#### Exploration
Consider the following theorem.
If a parallelogram has one pair of consecutive sides that are congruent, then the parallelogram is a rhombus.
To begin proving the above theorem, we can first label the parallelogram by its set of vertices, $PQRS$PQRS.
Parallelogram $PQRS$PQRS
From the given information, we know that a pair of consecutive sides are congruent. So let's choose $\overline{PQ}$PQ and $\overline{SP}$SP. Ultimately we could have chosen another pair, but the proof would be the same. We write this as a geometric statement $\overline{PQ}\cong\overline{SP}$PQSP.
Parallelogram $PQRS$PQRS with congruent sides
We also know that the $PQRS$PQRS is a parallelogram, and one feature of the parallelogram is that opposite sides are congruent. We can state this as $\overline{PQ}\cong\overline{RS}$PQRS and $\overline{SP}\cong\overline{QR}$SPQR. Congruence is transitive, meaning that if two things are congruent to the same thing, they are also congruent. This means that $\overline{PQ}\cong\overline{QR}$PQQR and $\overline{SP}\cong\overline{RS}$SPRS and more generally all the sides are congruent. By definition, a rhombus is a parallelogram with all sides congruent. So $PQRS$PQRS is a rhombus and we are finished with the proof.
We can formalize the above steps into a two-column proof where each line contains a geometric statement in the left column and a corresponding reason in the right column.
### Two-column proof
Given the parallelogram $PQRS$PQRS, and the fact that $\overline{PQ}\cong\overline{SP}$PQ≅SP, prove that $PQRS$PQRS is a rhombus. Statements Reasons $PQRS$PQRS is a parallelogram and $\overline{PQ}\cong\overline{SP}$PQ≅SP Given $\overline{PQ}\cong\overline{RS}$PQ≅RS and $\overline{SP}\cong\overline{QR}$SP≅QR If a quadrilateral is a parallelogram, then its opposite sides are congruent. $\overline{PQ}\cong\overline{QR}\cong\overline{RS}\cong\overline{SP}$PQ≅QR≅RS≅SP Transitive property of congruence $PQRS$PQRS is a rhombus Definition of a rhombus
The final line contains the statement that the quadrilateral is a rhombus, which is what we wanted to prove.
#### Practice questions
##### question 1
In order to show that a given parallelogram is a rectangle, which of the following must be proved?
1. The parallelogram has four congruent sides.
A
The parallelogram has four right angles.
B
The parallelogram has opposite sides congruent.
C
The parallelogram has opposite sides parallel.
D
The parallelogram has opposite angles congruent.
E
The parallelogram has four congruent sides.
A
The parallelogram has four right angles.
B
The parallelogram has opposite sides congruent.
C
The parallelogram has opposite sides parallel.
D
The parallelogram has opposite angles congruent.
E
##### question 2
Given that $PQRS$PQRS is a parallelogram, and $\overline{PR}$PR and $\overline{QS}$QS are perpendicular, prove that $PQRS$PQRS is a rhombus.
### The hierarchy of trapezoids
We define a trapezoid as any quadrilateral with one pair of opposite sides that are parallel.
A special type of a trapezoid, called an isosceles trapezoid, occurs when the two legs of the trapezoid are congruent and hence the diagonals are congruent.
Isosceles trapezoid
There are a number of theorems that explain equivalent ways of identifying whether a trapezoid is isosceles. We call these special theorems criteria.
1. If a trapezoid has congruent base angles, then it is an isosceles trapezoid.
Trapezoid with base angles congruent
1. If a trapezoid has opposite angles that are supplementary, then it is an isosceles trapezoid.
Trapezoid with $\angle1$∠1 and $\angle3$∠3 are supplementary, and $\angle2$∠2 and $\angle4$∠4 are supplementary
1. If a trapezoid has congruent diagonals, then it is an isosceles trapezoid.
Trapezoid with congruent diagonals
#### Exploration
Consider the following theorem.
If a trapezoid has congruent base angles, then it is an isosceles trapezoid.
To begin proving the above theorem, we can first label the trapezoid by its set of vertices, $ABCD$ABCD. Let $\angle ADC$ADC and $\angle BCD$BCD be congruent.
Trapezoid $ABCD$ABCD
We next construct the pair of altitudes $\overline{AP}$AP and $\overline{BQ}$BQ, which form the pair of triangles $\triangle APD$APD and $\triangle BQC$BQC. If we can show that these triangles are congruent, then we can show that $\overline{AD}$AD is congruent to $\overline{BC}$BC, which will mean that $ABCD$ABCD is an isosceles trapezoid.
Trapezoid $ABCD$ABCD with altitudes drawn
Firstly we are given that $\angle ADC$ADC and $\angle BCD$BCD are congruent. Since altitudes create right angles with the base of a trapezoid, $\angle APD$APD and $\angle BQC$BQC are right angles. All right angles are congruent, so we have that $\angle APD$APD and $\angle BQC$BQC are congruent. The altitudes themselves are congruent, so $\overline{AP}$AP is congruent to $\overline{BQ}$BQ. This means that $\triangle APD\cong\triangle BQC$APDBQC are congruent by angle-angle-side congruence theorem.
It follows that the two sides $\overline{AD}$AD and $\overline{BC}$BC are congruent because they are corresponding parts of the two congruent triangles. Hence the trapezoid $ABCD$ABCD is isosceles, by definition.
We can formalize the above steps into a two-column proof where each line contains a geometric statement in the left column and a corresponding reason in the right column.
Statements Reasons $ABCD$ABCD is a trapezoid, $\overline{AP}$AP and $\overline{BQ}$BQ are altitudes, and $\angle ADC\cong\angle BCD$∠ADC≅∠BCD Given $\angle APD$∠APD and $\angle BQC$∠BQC are right angles Definition of an altitude $\angle APD\cong\angle BQC$∠APD≅∠BQC All right angles are congruent. $\overline{AP}\cong\overline{BQ}$AP≅BQ Altitudes of a trapezoid are congruent $\triangle APD\cong\triangle BQC$△APD≅△BQC Angle-angle-side congruence theorem $\overline{AD}\cong\overline{BC}$AD≅BC Corresponding parts of congruent triangles are congruent (CPCTC) $ABCD$ABCD is an isosceles trapezoid Definition of an isosceles trapezoid
Below is a summary of the criteria for a trapezoid to be an isosceles trapezoid.
Criteria for isosceles trapezoids
A trapezoid is an isosceles trapezoid if any of the following are true:
• The base angles are congruent.
• The opposite angles are supplementary.
• The diagonals are congruent.
#### Practice Questions
##### Question 1
In order to show that a given trapezoid is isosceles, which of the following must be proved?
1. The trapezoid has diagonals that bisect each other.
A
The trapezoid has diagonals that are perpendicular.
B
The trapezoid has diagonals that are congruent.
C
The trapezoid has diagonals that are perpendicular bisectors of each other.
D
The trapezoid has diagonals that bisect each other.
A
The trapezoid has diagonals that are perpendicular.
B
The trapezoid has diagonals that are congruent.
C
The trapezoid has diagonals that are perpendicular bisectors of each other.
D
##### Question 2
Given that $ABCD$ABCD is a trapezoid with altitudes $\overline{AP}$AP and $\overline{BQ}$BQ and that $\angle ADC$ADC is congruent to $\angle BCD$BCD, prove that $ABCD$ABCD is an isosceles trapezoid.
### Outcomes
#### II.G.CO.11
Prove theorems about parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals.
#### II.G.SRT.5
Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. |
# 1990 AHSME Problems/Problem 20
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
$[asy] pair A = (0,0), B = (7,4.2), C = (10, 0), D = (3, -5), E = (3, 0), F = (7,0); draw(A--B--C--D--cycle,dot); draw(A--E--F--C,dot); draw(D--E--F--B,dot); markscalefactor = 0.1; draw(rightanglemark(B, A, D)); draw(rightanglemark(D, E, C)); draw(rightanglemark(B, F, A)); draw(rightanglemark(D, C, B)); MP("A",(0,0),W); MP("B",(7,4.2),N); MP("C",(10,0),E); MP("D",(3,-5),S); MP("E",(3,0),N); MP("F",(7,0),S); [/asy]$
In the figure $ABCD$ is a quadrilateral with right angles at $A$ and $C$. Points $E$ and $F$ are on $\overline{AC}$, and $\overline{DE}$ and $\overline{BF}$ are perpendicual to $\overline{AC}$. If $AE=3, DE=5,$ and $CE=7$, then $BF=$
$\text{(A) } 3.6\quad \text{(B) } 4\quad \text{(C) } 4.2\quad \text{(D) } 4.5\quad \text{(E) } 5$
## Solution
Label the angles as shown in the diagram. Since $\angle DEC$ forms a linear pair with $\angle DEA$, $\angle DEA$ is a right angle.
$[asy] pair A = (0,0), B = (7,4.2), C = (10, 0), D = (3, -5), E = (3, 0), F = (7,0); draw(A--B--C--D--cycle,dot); draw(A--E--F--C,dot); draw(D--E--F--B,dot); markscalefactor = 0.075; draw(rightanglemark(B, A, D)); draw(rightanglemark(D, E, A)); draw(rightanglemark(B, F, A)); draw(rightanglemark(D, C, B)); draw(rightanglemark(D, E, C)); draw(rightanglemark(B, F, C)); MP("A",(0,0),W); MP("B",(7,4.2),N); MP("C",(10,0),E); MP("D",(3,-5),S); MP("E",(3,0),N); MP("F",(7,0),S); [/asy]$
Let $\angle DAE = \alpha$ and $\angle ADE = \beta$.
Since $\alpha + \beta = 90^\circ$, and $\alpha + \angle BAF = 90^\circ$, then $\beta = \angle BAF$. By the same logic, $\angle ABF = \alpha$.
As a result, $\triangle AED \sim \triangle BFA$. By the same logic, $\triangle CFB \sim \triangle DEC$.
Then, $\frac{BF}{AF} = \frac{3}{5}$, and $\frac{CF}{BF} = \frac{5}{7}$.
Then, $7CF = 5BF$, and $5BF = 3AF$.
By the transitive property, $7CF = 3AF$. $AC = AF + CF = 10$, and plugging in, we get $CF = 3$.
Finally, plugging in to $\frac{CF}{BF} = \frac{5}{7}$, we get $BF = 4.2$ $\fbox{C}$ |
## Pages
### Trigonometric multiple-angle formulas
In our previous post, we show a compass-and-straightedge construction of a regular pentagon based on the following trigonometric formula $$\cos{\frac{\pi}{5}} = \frac{1 + \sqrt{5}}{4}.$$
We derive this formula of $\cos{\frac{\pi}{5}}$ by observing that $\cos{\frac{2 \pi}{5}} = -\cos{\frac{3 \pi}{5}}$ and then applying the trigonometric formulas for double angle and triple angle:
$$\cos{2 x} = 2 \cos^2{x} - 1,$$ $$\cos{3 x} = 4 \cos^3{x} - 3 \cos{x}$$ to set up a cubic equation for $\cos{\frac{\pi}{5}}$.
It seems a good occasion now for us to learn about trigonometric multiple-angle formulas. In this post, we will show how to derive formulas for $\sin{nx}$$\cos{nx}$$\tan{nx}$ and $\cot{nx}$ using de Moivre's identity of the complex numbers.
de Moivre's identity for complex numbers
Complex numbers have the form $$a + ib$$ where $a$ and $b$ are two real numbers; $a$ is called the real part and $ib$ is called the imaginary part. Here are a few examples of complex numbers $$3+2i, ~~ 5-3i, ~~ 2i+7, ~~ 8i, ~~ 4i-2, ~~ 5, ~~ i+1, \dots$$ We do addition, subtraction, multiplication and division for complex numbers exactly the same as we do normal arithmetic for real numbers, except that we need to remember $$i^2 = -1, ~~ i^3 = -i, ~~ i^4 = -i^2 = 1, ~~ i^5 = i, ~~i^6 = i^2 = -1, \dots$$
• Addition and Subtraction $$(a + i b) + (c + i d) = (a+c) + i (b+d) ,$$ $$(a + i b)- (c + i d) = (a-c) + i (b - d).$$
• Multiplication $$(a + i b)(c + i d) = ac + i ad + i bc + i^2 bd = (ac - bd) + i (bc + ad ) .$$
• Division we need to use the identity $$(a + i b)(a - ib ) = a^2 - i^2 b^2 = a^2 + b^2$$ and so $$\frac{c + i d}{a + i b} = \frac{(c + i d)(a - ib)}{(a + ib)(a - ib)} = \frac{(ac + bd) + i(ad - bc)}{a^2 + b^2} = \frac{ac + bd}{a^2 + b^2} + i \frac{ad - bc}{a^2 + b^2}.$$
• Complex conjugate $$\overline{a + i b} = a - ib, ~~~~\overline{a- ib} = a + ib.$$
• Absolute value $$|a + ib| = \sqrt{a^2 + b^2}.$$
There is a very important fact about complex numbers, that is, any complex number $z$ can be written in a trigonometric form as follows $$z = |z| (\cos{\phi} + i ~\sin{\phi}).$$
This trigonometric form is very convenient when we do multiplication and division. This is because $$(\cos{\alpha_1} + i \sin{\alpha_1})(\cos{\alpha_2} + i \sin{\alpha_2})$$ $$= (\cos{\alpha_1} \cos{\alpha_2} - \sin{\alpha_1} \sin{\alpha_2}) + i (\sin{\alpha_1} \cos{\alpha_2} + \cos{\alpha_1} \sin{\alpha_2})$$ $$= \cos{(\alpha_1 + \alpha_2)} + i \sin{(\alpha_1 + \alpha_2)}.$$
So if we have two complex numbers $$z_1= |z_1|(\cos{\alpha_1} + i \sin{\alpha_1}),$$ $$z_2= |z_2|(\cos{\alpha_2} + i \sin{\alpha_2}),$$ then we have the following simple formulas for multiplication and division $$z_1 z_2 = |z_1| |z_2| (\cos{(\alpha_1 + \alpha_2)} + i \sin{(\alpha_1 + \alpha_2)}),$$ $$\frac{z_1}{z_2}= \frac{|z_1|}{|z_2|} (\cos{(\alpha_1 - \alpha_2)} + i \sin{(\alpha_1 - \alpha_2)}).$$
In a special case when $\alpha_1 = \alpha_2$, then we have $$z^2= |z|^2 (\cos{2 \alpha} + i \sin{2 \alpha}).$$
By applying the multiplication formula repeatedly, we derive the following formula $$z^n= |z|^n (\cos{n \alpha} + i \sin{n \alpha}).$$
When $|z|=1$ we obtain
de Moivre's identity:
$$(\cos{\alpha} + i \sin{\alpha})^n = \cos{n \alpha} + i \sin{n \alpha}.$$
From de Moivre's identity, if we use binomial theorem to expand the left hand side, and compare with the right hand side, we can derive formulas for $\cos{n \alpha}$ and $\sin{n \alpha}$ !
Trigonometric multiple-angle formulas
Trigonometric formulas for double angle
Using de Moivre's identity for $n=2$, we have $$\cos{2 x} + i \sin{2 x} = (\cos{x} + i \sin{x})^2$$ $$= \cos^2{x} + 2 \cos{x} (i \sin{x}) + (i \sin{x})^2$$ $$= \cos^2{x} -\sin^2{x} + i 2 \sin{x} \cos{x}$$
We derive the following
formula for $\cos{2 x}$
$$\cos{2 x} = \cos^2{x} -\sin^2{x} = \cos^2{x} - ( 1 - \cos^2{x}) = 2 \cos^2{x} - 1$$
formula for $\sin{2 x}$ $$\sin{2 x} = 2 \sin{x} \cos{x}$$
formula for $\tan{2 x}$ $$\tan{2x} = \frac{\sin{2x}}{\cos{2x}} = \frac{2 \sin{x} \cos{x}}{\cos^2{x} -\sin^2{x}} = \frac{2 \sin{x} \cos{x} / \cos^2{x}}{(\cos^2{x} -\sin^2{x}) / \cos^2{x}}= \frac{2 \tan{x}}{1 - \tan^2{x}}$$
formula for $\cot{2 x}$ $$\cot{2x} = \frac{\cos{2x}}{\sin{2x}} = \frac{\cos^2{x} -\sin^2{x}}{2 \sin{x} \cos{x}} = \frac{(\cos^2{x} -\sin^2{x}) / \sin^2{x}}{2 \sin{x} \cos{x} / \sin^2{x}} = \frac{\cot^2{x} - 1}{2 \cot{x}}$$
Trigonometric formulas for triple angle
Using de Moivre's identity for $n=3$, we have $$\cos{3 x} + i \sin{3 x} = (\cos{x} + i \sin{x})^3$$ $$= \cos^3{x} + 3 \cos^2{x} (i \sin{x}) + 3 \cos{x} (i \sin{x})^2 + (i \sin{x})^3$$ $$= \cos^3{x} - 3 \cos{x} \sin^2{x} + i (3 \cos^2{x} \sin{x} - \sin^3{x})$$
Therefore,
formula for $\cos{3 x}$ $$\cos{3 x} = \cos^3{x} - 3 \cos{x} \sin^2{x} =\cos^3{x} - 3 \cos{x} (1 - \cos^2{x}) =4 \cos^3{x} - 3 \cos{x}$$
formula for $\sin{3 x}$ $$\sin{3 x} = 3 \cos^2{x} \sin{x} - \sin^3{x} =3 (1 - \sin^2{x}) \sin{x} - \sin^3{x} =3 \sin{x} - 4 \sin^3{x}$$
formula for $\tan{3 x}$ $$\tan{3 x} = \frac{\sin{3 x}}{\cos{3 x}} = \frac{3 \cos^2{x} \sin{x} - \sin^3{x}}{\cos^3{x} - 3 \cos{x} \sin^2{x}}= \frac{3 \tan{x} - \tan^3{x}}{1 - 3 \tan^2{x} }$$
formula for $\cot{3 x}$ $$\cot{3 x} = \frac{\cos{3 x}}{\sin{3 x}} = \frac{\cos^3{x} - 3 \cos{x} \sin^2{x}}{3 \cos^2{x} \sin{x} - \sin^3{x}}= \frac{\cot^3{x} - 3 \cot{x}}{3 \cot^2{x} - 1}$$
Using de Moivre's identity for $n=4$, we have $$\cos{4 x} + i \sin{4 x} = (\cos{x} + i \sin{x})^4$$ $$= \cos^4{x} + 4 \cos^3{x} (i \sin{x}) + 6 \cos^2{x} (i \sin{x})^2 + 4 \cos{x} (i \sin{x})^3 + (i \sin{x})^4$$ $$= \cos^4{x} - 6 \cos^2{x} \sin^2{x} + \sin^4{x} + i (4 \cos^3{x} \sin{x} - 4 \cos{x} \sin^3{x})$$ Therefore,
formula for $\cos{4 x}$ $$\cos{4 x} = \cos^4{x} - 6 \cos^2{x} \sin^2{x} + \sin^4{x} = \cos^4{x} - 6 \cos^2{x} (1-\cos^2{x}) + (1- \cos^2{x})^2$$ $$= 8 \cos^4{x} - 8 \cos^2{x} + 1$$
formula for $\sin{4 x}$ $$\sin{4 x} = 4 \cos^3{x} \sin{x} - 4 \cos{x} \sin^3{x} = 4 \cos{x} (1 - \sin^2{x}) \sin{x} - 4 \cos{x} \sin^3{x}$$ $$= 4 \cos{x} (\sin{x} - 2 \sin^3{x})$$
formula for $\tan{4 x}$ $$\tan{4x} = \frac{\sin{4 x}}{\cos{4 x}} = \frac{4 \cos^3{x} \sin{x} - 4 \cos{x} \sin^3{x}}{ \cos^4{x} - 6 \cos^2{x} \sin^2{x} + \sin^4{x}} = \frac{4 \tan{x} - 4 \tan^3{x}}{ 1 - 6 \tan^2{x} + \tan^4{x}}$$
formula for $\cot{4 x}$ $$\cot{4 x} = \frac{\cos{4 x}}{\sin{4 x}} = \frac{\cos^4{x} - 6 \cos^2{x} \sin^2{x} + \sin^4{x}}{4 \cos^3{x} \sin{x} - 4 \cos{x} \sin^3{x}}= \frac{\cot^4{x} - 6 \cot^2{x} + 1}{4 \cot^3{x} - 4 \cot{x} }$$
By now, you probably know how to derive the general formulas. Please continue with the cases for $n=5,6,7$. Using these formulas, it is possible to calculate the explicit values of $\sin{\frac{\pi}{7}}$, $\cos{\frac{\pi}{7}}$, $\sin{\frac{\pi}{9}}$, $\cos{\frac{\pi}{9}}$, ...
Let us stop here for now. In the next post, we will return to the topic of compass-and-straightedge construction. Hope to see you again then.
Homework.
1. Continue with $n=5,6,7, \dots$ and establish formulas for $\sin{nx}$, $\cos{nx}$, $\tan{nx}$, $\cot{nx}$.
2. Find explicit formulas for $$\sin{\frac{\pi}{7}}, \sin{\frac{2 \pi}{7}}, \sin{\frac{3 \pi}{7}}, \dots$$ $$\cos{\frac{\pi}{7}}, \cos{\frac{2 \pi}{7}}, \cos{\frac{3 \pi}{7}}, \dots$$
3. Find explicit formulas for $$\sin{\frac{\pi}{9}}, \sin{\frac{2 \pi}{9}}, \sin{\frac{3 \pi}{9}}, \dots$$ $$\cos{\frac{\pi}{9}}, \cos{\frac{2 \pi}{9}}, \cos{\frac{3 \pi}{9}}, \dots$$
4. Search on Google http://google.com about Chebyshev polynomial, find out the relationship between Chebyshev polynomial and the trigonometric multiple-angle formulas. |
# Real numbers, sequences and limits: part IV
Representation of Q on the Number Line.
We have seen earlier that every element of can be represented on a straight line with a certain point representing 0. Positive integers are represented by points to the right of this marked point at equal lengths, and negative integers are similarly represented by points to the left of this.
We can use the same straight line to represent the newly constructed rational numbers as follows.
We know that we can divide a line segment, using a straight edge and compass, into q equal parts. So, the line segment between 0 to 1 can be divided into q equal parts for every positive integer q.
The points of division will now represent the numbers $\frac{1}{q}, \frac{2}{q}, \ldots , \frac{q-1}{q}$. It is now clear that any rational number of the form $\frac{p}{q}$ can be represented on the straight line. We have seen that between any two distinct rational numbers r and $r^{'}$, no matter how close they are, we can always find another rational number between them, example, $\frac{r+r^{"}}{2}$. Geometrically, this means that between any two distinct points on the line representing rational numbers, there is a point between them representing a rational number.
One can ask: does every point on the line correspond to a rational number? Note that we also have a point on the line representing length of the diagonal of a square of side 1. But, we shall show in the next blog that the length of the diagonal of unit square does not correspond to a rational number.
To prepare the ground for such discussions, let us start with a few definitions:
A subset $A \subset Q$ is said to be bounded below if there exists a rational number $\alpha$ such that
$\alpha \leq \beta$ for every $\beta \in A$
and $\alpha$ is called a lower bound of A. Similarly, $A \subset Q$ is said to be bounded above if there is a $\gamma \in Q$ such that $\beta \leq \gamma$, and $\gamma$ is called an upper bound of A. A set which is bounded above and also bounded below is called a bounded set. Observe that if a set has a lower bound $\alpha$ then it has many more like $\alpha -1$, $\alpha -2, \ldots$. Similarly, a set that is bounded above has many upper bounds. If there is a least element of all these upper bounds, then we call it the least upper bound of the set. But, it is not true that every set of rational numbers which is bounded above has a least upper bound. We can say similar things about sets which are bounded below. Some sets may have least upper bound. For example, the set of negative rational numbers is certainly bounded above and it is easy to see that 0 is the least upper bound.
More later,
Nalin Pithwa
# Real Numbers, Sequences and Series: part I:
Natural Numbers.
Natural numbers are perhaps, the earliest mathematical notions of man. The difference between a pack of wolves and one wolf, a swarm of bees and one bee, a school of fish and a single fish, a heap of stones and a single stone, etc., may have led primitive man tbo systematize the idea of numbers. The fact that in Sanskrti (and in some other ancient Indo-European languages) there are usages where one object, two objects and many objects are distinguished, example, narah, nari, naraah — indicates man’s attempt to distinguish betweeh numbers. Heaps of stones found in caves, where early man lived, and the discovery of animal bones on which notches have been cut in regular series of five are indicative of devices by the human beings of yore to keep count of their livestock. That is how they understood plurality.
But, what do we understand by numbers? Is number the same thing as plurality ? We often use expressions like “there are five fingers on my hand”, “Pandavas were five brothers”, “there are five books on this table”, “there are five mangoes in the basket”, etc. Observe what we are talking about. In the first case, we were talking about a set of fingers on my hand, in the next about a set of brothers, next about a set of books on a table, next about a set of mangoes. But to each of these sets we attached the common attribute “five”. Why was that? There must have been something common in all these sets to justify our attaching the common attribute. The commonality between them is that the elements of one set can be put in one-to-one correspondence with the elements of the other sets under discussion.
We say that two sets A and B are equinumerous (or of the same cardinality) if there is a map
$f: A \rightarrow B$
which is one-to-one and onto, that is, a bijection. One sees that if A and B are equinumerous and the sets B and C are also equinumerous, then so are A and C. One can easily observe that being equinumerous is an equivalence relation among sets.
This relation decomposes any class of sets into disjoint equivalence classes of sets. Two sets are in the same equivalence class if they are equinumerous. The equivalence class is characterized by a common property of its members, that is, any two of them are equinmerous. This characterizing property is what we would like to call the number of elements of the set belonging to the equivalence class. Since equivalence classes, into which we have classified the sets, are disjoint, the characterizing property associated with disjoint classes are distinct. “Commonness”, “common property” or ‘characteristics” is too imprecise an idea to be handled easily; some authors have defined the number of elements of a set as the equinumerous class to which it belongs. Thus, the number “two” would mean the class of all pairs, the number “three” the class of all triplets, …and so on. But, this too has some foundational problems which are hard to circumvent. We do not go into the details of such things here.
Instead, we adopt the axioms devised by Peano for the natural numbers N with the following properties:
1) There is a map (called the successor map) from N to N, sending an element n to $n^{+}$ (called the successor of n), which is injective, that is, $n \neq m \Longrightarrow n^{+} \neq m^{+}$
2) There is an element 1 that is not a successor of any element (that is, 1 is not in the image of any successor map, that is, $1 \neq n^{+}$ for all $n \in N$). Also, every other element of N is the successor of some element of N.
3) Suppose that $A \subseteq N$ and that (a) $1 \in A$ (b) $n^{+} \in A$ whenever $n \in A$. Then, $A=N$, that is, the only subset of N which contains 1 and the successor of each of its elements in N itself.
The third axiom is known as the Principle of Mathematical Induction.
More later,
Nalin Pithwa |
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# Combine Volumes of Prisms
Sep 27, 2022
### Key Concepts
• Composite figures
• The volume of the composite figure
## 11.3 Combine volumes of Prisms
The shape and size of a kitchen storage are shown in the figure. Ria’s mother wants to find the volume to determine how much storage space is available. What is the volume?
### 11.3.1 Composite figures
A composite or compound shape is any shape that is made up of two or more geometric shapes.
The image below is an example of composite figures
Two rectangular prisms formed into one rectangular prism.
Example1:
This image below can be separated into two rectangular prisms, as shown.
Identify the measurement for the length, width and height of each prism.
Solution:
Step 1: The length, width and height of prism A.
Length = 4 m
Width =9 m
Height =5 m
Step 2: The length, width and height of prism B.
Length = 10 m
Width =9 m
Height =7 m
Example 2:
Identify the measurement for the length, width and height of each prism.
Solution:
Step1: The length, width and height of the small prism.
Length = 8 m
Width =4 m
Height =2.5 m
Step 2: The length, width and height of the big prism.
Length = 12 m
Width =4 m
Height =15 m
### 11.3.2 Volume of the composite figures
Example 1: Find the volume of composite figures.
Solution:
Step1: Use the formula V= l x w x h to find the volume of each rectangular prism.
Step2: Volume of prism A
= 6 × 4 × 3
= 72
Step 3:
= 8 × 4 × 7
=>32 × 7 = 224
Step 4: Add to find the total volume.
72 + 224 = 296
The volume of the composite figure is 296 cubic meters.
Example 2:
Find the volume of the composite figure.
Solution:
Step1: Use the formula to find the volume of each rectangular prism.
Step2: Volume of prism A
= 10 × 30 × 30
= 9,000
Step3: Volume of prism B
= 20 × 30 × 20
= 12,000
Step 4: Add to find the total volume.
9,000 + 12,000 = 21,000
The volume of the composite figure is 21,000 cubic centimetres.
Example 3:
What is the volume of the composite figure?
Solution:
Step 1: To find the volume of the figure, start by breaking it up into 2 rectangular prisms.
Step 2: Now, find the volume of each prism. Start with the back prism.
The length ℓ is 10 meters. The width w is 5 meters. The height h is 8 meters.
V = 10 × 5 × 8
V = 400
The volume of the back prism is 400 cubic meters.
Step 3: Then, find the volume of the front prism.
The length ℓ is 6 meters. The width w is 2 meters. The height h is
4 meters.
V = 6 × 2 × 4
V = 48
The volume of the front prism is 48 cubic meters.
Step 4: To find the total volume of the figure, add the volumes of the prisms.
Step 5: 400+48=448
The volume of the figure is 448 cubic meters.
## Exercise:
1. Find the length, width and height of rectangular prism 1.
2. Find the volume of rectangular prism 2.
3. Find the volume of the given rectangular prism.
4. Find the volume of the given figure.
5. Find the volume of the rectangular prism.
6. Find the volume of the bottom rectangular prism.
7. Find the volume of the rectangular prism.
### What Have We Learned:
• Understanding how to combine the volume of prisms.
• Understanding cube and rectangular prism.
• How to find the volume of composite figures
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […] |
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# If $z$ and $\overline z$ represent adjacent vertices of a regular polygon of $n$ sides with centre at origin and if $\dfrac{{\operatorname{Im} z}}{{\operatorname{Re} z}} = \sqrt 2 - 1$, then the value of $n$ is equal to A) $2$B) $4$C) $6$D) $8$
Last updated date: 17th Jul 2024
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Hint: First of all we should know that a regular polygon is a polygon that has equal sides and equal angles. In this question we should know that we have complex numbers. We can assume that ${z_1} = x + iy$, then the value of $\overline {{z_1}}$ will be $x - iy$. We know that the angle between two corresponding vertices is $\dfrac{{2\pi }}{n}$.
Complete step by step solution:
Let us assume that ${z_1}$ be the first vertex of the polygon in the first quadrant i.e. ${z_1} = x + iy$, then the $\overline {{z_1}}$ be in the fourth quadrant i.e. $\overline {{z_1}} = x - iy$.
In the above figure we have points in the first quadrant as $(x,y)$ and in the fourth quadrant the coordinates are $(x, - y)$.
Now we have assumed that the angle at the centre in the first quadrant is $\theta$.
So we can write from the question that $\tan \theta = \sqrt 2 - 1$. We know that the angle between two corresponding vertices is $\dfrac{{2\pi }}{n}$. From this we can write $\dfrac{{2\pi }}{n} = 2\theta$, because $\dfrac{\pi }{n} = \theta$ from the above figure.
By multiplying with $\tan$ on both the sides, it gives us $\tan \dfrac{{2\pi }}{n} = \tan 2\theta$. On simplifying this we can write $\tan \dfrac{{2\pi }}{n} = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$.
By putting the value of $\tan \theta = \sqrt 2 - 1$in the right hand side of the equation, we have$\dfrac{{2(\sqrt 2 - 1)}}{{1 - {{(\sqrt 2 - 1)}^2}}}$.
We will solve it now: $\dfrac{{2\sqrt 2 - 2}}{{1 - 2 - 1 + 2\sqrt 2 }} \Rightarrow \dfrac{{2\sqrt 2 - 2}}{{2\sqrt 2 - 2}}$. It gives us the value $\tan \dfrac{{2\pi }}{n} = 1$.
Now we know that the value of $\tan \dfrac{\pi }{4}$ is $1$, so we can write it as $\tan \dfrac{{2\pi }}{n} = \tan \dfrac{\pi }{4} = 1$.
Therefore we can write by eliminating tan, $\dfrac{{2\pi }}{n} = \dfrac{\pi }{4}$. It gives us the value of $n = 4 \times 2 = 8$.
Hence the required answer is (D) $8$.
Note:
We should know that we have used the trigonometric identity in the above question i.e. $\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$. We have also applied the algebraic identity of difference square formula which is ${(a - b)^2} = {a^2} + {b^2} - 2ab$. Before solving this kind of question we should have a clear knowledge of the trigonometric identities and their functions. |
# Introduction to Writing Algebraic Terms
Introduction to Writing Algebraic Terms
• We use letters in algebra when we don’t know the value of something.
• These letters can be called variables.
• We don’t know how much each apple costs, so we say each apple costs ‘a’.
• The total cost of three of these apples is a + a + a, which is 3 lots of a.
• We don’t write the multiplication sign in algebra because we use the letter ‘x’ a lot which looks similar.
• We write a + a + a as 3a, which means that there are 3 a’s.
• We always write the number in front of the letter.
We do not write multiplication signs with algebra.
We write how many of each letter there are with a number in front of the letter.
• Here is b + b + b.
• We have 3 b’s in total.
• We can write b + b + b as 3b.
• We don’t write 3 × b because we don’t write multiplication signs in algebra.
• We always write the number in front of the letter, so we write 3b and not b3.
# Introduction to Writing Algebra
## What are Algebraic Terms?
Algebraic terms are the different letters or numbers in an expression that are separated by plus or minus signs. A single term can even contain multiple different letters and numbers multiplied together. As long as there is not a plus or minus sign, then it is an algebraic term.
The letters in a term are called variables and the numbers in front of them are called coefficients.
For example, in the expression: c + c + c + c, we have 4 algebraic terms. Each letter ‘c’ is separated by a plus sign, so each letter ‘c’ is a separate term.
In most cases, the algebraic terms will be different letters, such as in the next example below.
It is important to note that even numbers on their own can be algebraic terms. It is not necessary to have an algebraic variable present for it to be classed as a term. In this example of d – 3 + e, we have 3 terms.
The first term is ‘d’, the second term is ‘3’ and the third term is ‘e’.
We can see that in this example, the terms are separated by both plus and minus signs.
Remember, if there is not a plus or minus sign separating the letters or numbers, then it is one single term.
For example in the expression: 2a – 4cd there are 2 terms. The first term is ‘2a’ and the second term is ‘4cd’.
## How to Write Algebraic Terms
When writing algebraic terms use the following rules:
• Do not write multiplication signs.
• Write the number in front of the letters in each term.
• If there is only one of a certain algebraic term, do not write the number 1 in front of it. Simply write the letters only.
For example, here is the expression 𝑥 + 𝑥. There are two 𝑥 terms.
𝑥 + 𝑥 is the same as 2 × 𝑥.
The multiplication sign of ‘×’ and the variable ‘𝑥’ look very similar. 𝑥 is a letter that is commonly used in algebra. For this reason, we do not use the multiplication sign in algebra. Removing the multiplication sign also keeps the algebraic expressions shorter and simpler to work with.
Instead of 2 × 𝑥 we write 2𝑥.
2𝑥 means that we have 2 × 𝑥, or 2 x’s.
An important rule used when writing algebraic terms is to avoid using the number 1 in front of an algebraic variable if there is only 1 of them.
For example, here is the variable 𝑥.
Although we only have 1 𝑥, we simply write 𝑥.
We do not write 1𝑥.
The reason we do not write 1 in front of algebraic terms is because we don’t need to. It is quicker and easier, especially in longer algebraic expressions to avoid using the number 1. It is easier to write 𝑥 than it is to write 1𝑥.
This is just a simple example and when it comes to more complex expressions and equations, it can be helpful to remove unnecessary information.
In this example, we have y + y + y + y + y.
We have 5 ‘y’ terms, which is 5 lots of y, or 5 × y.
We do not write the multiplication symbol and so, y + y + y + y + y = 5y.
5y simply means that we have 5 y terms.
We can see that it is quicker to write 5y than it is to write y + y + y + y + y.
We can do this because all of the terms are y terms. There is no other letter in the expression.
Counting up the same letter terms is known as collecting like terms and is explored in a different lesson Collecting Like Terms.
Here is another example of collecting like terms.
In this expression, we have b + b + b.
We have 3 b terms, so we write this as 3b.
Remember that we do not write the multiplication sign and we always write the number in front of the letters.
We write 3b, not b3.
The number that comes in front of the algebraic variables (letters) is called the coefficient.
In the example of 5ab, 5ab is one term. ‘5’ is the coefficient and both ‘a’ and ‘b’ are the variables.
## Writing the Letter 𝑥 in Algebra
In this example we have: 𝑥 + 𝑥 + 𝑥 + 𝑥.
We have 4 𝑥’s in total or 4 × 𝑥.
Remember, we do not write the multiplication sign so we simply write 𝑥 + 𝑥 + 𝑥 + 𝑥 as 4𝑥.
𝑥 is a letter commonly used in algebra. You may notice that 𝑥 is written differently to the letter x on your keyboard.
In algebra, 𝑥 is written differently to the usual letter x. The letter x looks similar to the multiplication sign, ×. The symbol of 𝑥 is used instead so that it is clearer that we are not doing a multiplication.
To write the algebra 𝑥, write a letter c and then write a letter c backwards just before it.
The algebra 𝑥 is more curved than the standard letter x.
## What are Algebraic Variables?
Variables are the letters used in algebra that hold the place of numbers. They are used when the value of something is unknown or can vary, hence the name variable.
For example if the price of an apple is $2, then we can work out the price of three apples by multiplying$2 by 3.
$2 × 3 =$6. The total cost would be $6. However, you might pick up the three apples when shopping and not know what price they are. You may only find out the price when someone tells you as you go to pay. Because we do not know the price of each apple, we can use an algebraic variable. We can use the letter ‘a’ to represent the price of an individual apple. Any letter can be used, however I chose ‘a’ because it is apples that we are buying. If we were buying bananas, I could still use ‘a’ but I would probably choose ‘b’ to help remind myself what the letter represents. We have 3 apples, so the total price is a + a + a. We have 3 apples, so the total price is 3 × a. Remember that we do not use multiplication signs in algebra and so we would just write this as 3a. We say that the total price is 3a. When teaching or introducing algebra for the first time, people can be uncomfortable with not having a numerical answer. It helps to explain how the answers in these cases will look like and what they mean. The answer in this case is 3a. If we later find out the price of an apple is$4, then we can use the algebraic term of ‘3a’ to work out the price.
If a = $4, then 3a means 3 ×$4.
If the price of one apple is $4, then the total price of 3 apples would be$12.
The letter ‘a’ is used to hold the value of the apple until we know what it is.
Now try our lesson on Negative Numbers on a Number Line where we introduce negative numbers.
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# SPECIAL RIGHT TRIANGLES
Special right triangles are the triangles whose angle measures are
45° - 45° - 90°
or
30° - 60° - 90°
## Theorems about Special Right Triangles
45° - 45° - 90° Triangle Theorem :
In a 45° - 45° - 90° triangle, the hypotenuse is √2 times as long as each leg.
It has been illustrated in the diagram shown below.
30° - 60° - 90° Triangle Theorem :
In a 30° - 60° - 90° triangle, the hypotenuse is twice as long as the shorter leg, and the longer leg is √3 times as long as the shorter leg.
It has been illustrated in the diagram shown below.
## Finding the Hypotenuse in a 45° - 45° - 90° Triangle
Example 1 :
Find the length of the hypotenuse in the triangle shown below.
Solution :
By the Triangle Sum Theorem, the measure of the third angle is 45°.
The triangle is a 45°-45°-90° right triangle, so the length x of the hypotenuse is √2 times the length of a leg.
By 45°-45°-90° Triangle Theorem, we have
Hypotenuse = √2 ⋅ leg
In any right angled triangle, the side which is opposite the right angle is hypotenuse. In the triangle shown above, x represents hypotenuse. Because, the side x represents is opposite the right angle.
Substitute.
x = √2 ⋅ 3
Simplify.
x = 3√2
So, the length of the hypotenuse is 3√2.
## Finding a Leg in a 45° - 45° - 90° Triangle
Example 2 :
Find the value of x in the triangle shown below.
Solution :
Because the triangle is an isosceles right triangle, its base angles are congruent. The triangle is a 45°-45°-90°right triangle, so the length of the hypotenuse is √2 times the length of a leg.
By 45°-45°-90° Triangle Theorem, we have
Hypotenuse = √2 ⋅ leg
Substitute.
5 = √2 ⋅ x
Divide each side by √2.
5/√2 = x
To rationalize denominator in 5/√2, multiply numerator and denominator by √2.
(√2/√2) ⋅ (5/√2) = x
5√2/2 = x
## Side Lengths in a 30° - 60° - 90° Triangle
Example 3 :
Find the value of s and x in the triangle shown below.
Solution :
Finding the value of s :
Because the triangle is a 30°-60°-90° triangle, the longer leg is √3 times the length s of the shorter length.
By 30°-60°-90° Triangle Theorem, we have
Longer leg = √3 ⋅ shorter leg
Substitute.
5 = √3 ⋅ s
Divide each side by √3.
5/√3 = s
To rationalize denominator in 5/√3, multiply numerator and denominator by √3.
(√3/√3) ⋅ (5/√3) = s
5√3/3 = s
Finding the value of x :
By 30°-60°-90° Triangle Theorem, the length x of the hypotenuse is twice the length s of the shorter leg.
So, we have
Hypotenuse = ⋅ shorter leg
Substitute.
x = 2(5√3/3)
x = 10√3/3
## Using Special Right Right Triangles in Real Life
Example 4 :
A tipping platform is a ramp used to unload trucks, as shown in the picture below. How high is the end of an 70 foot ramp when it is tipped by a 30° angle ? by a 45° angle ?
Solution :
Part I :
Let h be the height of the ramp.
When the angle of elevation is 30°, we get a 30°-60°-90° special right triangle.
The height h of the ramp is the length of the shorter leg of the 30°-60°-90° triangle. And also, the length of the hypotenuse is 70 feet.
By 30°-60°-90° Triangle Theorem, we have
Hypotenuse = 2 ⋅ Shorter length
70 = 2 ⋅ h
Divide each side by 2.
35 = h
Part II :
When the angle of elevation is 45°, we get a 45°-45°-90° special right triangle.
By 45°-45°-90° Triangle Theorem, we have
Hypotenuse = √2 ⋅ Leg
70 = √2 ⋅ h
Divide each side by √2.
70/√2 = h
Use calculator to approximate.
49.5 ≈ h
When the angle of elevation is 30°, the ramp height is 35 feet. When the angle of elevation is 45°, the ramp height is about 49 feet 6 inches.
Example 5 :
The road sign is shaped like an equilateral triangle. If the length of each side is 36 inches, estimate the area of the sign by finding the area of the equilateral triangle.
Solution :
First find the height h of the triangle by dividing it into two 30°-60°-90° triangles. The length of the longer leg of one of these triangles is h. The length of the shorter leg is 18 inches.
By 30°-60°-90° Triangle Theorem, we have
h = √3 ⋅ 18
h = 18√3
Use h = 18√3 to find the area of the equilateral triangle.
Area = 1/2 ⋅ b ⋅ h
Substitute.
= 1/2 ⋅ 36 ⋅ 18√3
Use calculator to approximate.
561.18 square inches
Kindly mail your feedback to v4formath@gmail.com
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Unit 5 Section 3 - 5.3 Matrices and Determinants Objectives After completing this section you should be able to solve systems of linear equations using
# Unit 5 Section 3 - 5.3 Matrices and Determinants Objectives...
• Notes
• 19
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5.3 Matrices and Determinants Objectives: After completing this section, you should be able to solve systems of linear equations using the following methods: 1. Matrices and Gaussian elimination 2. Matrices and Gauss-Jordan elimination 3.Cramer’s ruleIn Sections 1 and 2 we learned how to solve systems of linear equations in two and three variables using elimination by addition and its variant, Gaussian elimination. We can organize more efficiently our work for solving linear systems using these methods, and two other methods you will see later, by introducing a mathematical form called a matrix. Corresponding to a linear system of equations is an augmented matrixwhich contains all of the information in the system without the x, y, and variable labels to carry around. In this section we will revisit Gaussian elimination but this time using augmented matrices. Keep This in Mind A matrixis a rectangular array of numbers written within brackets that contains rows and columns. If m and nare positive integers, an m xn(read “m byn”) matrix is a rectangular array of numbers mnmmnnaaaaaaaaa212222111211with mrows and n columns in which each number ijais referred to as an entrythe matrix in the ithrow and jth column. The expression m xnis called the sizedimension of the matrix. A matrix with nrows and ncolumns is called a square matrix of order n. A matrix with only one is called a column matrix, and a matrix with only one row is called a row matrixRow 1 Row 2 Row Col. 1 Col. 2 Col. z of or . m n
Unit 5. Systems of Equations and Inequalities Section 3 page 2 _____________________________________________________________________________________ Example 5.3.1 Below are two matrices: 7 0 2 4 3 1 A 2 5 0 3 8 6 1 0 15 1 0 4 B We usually name a matrix by a single capital letter, such as A , B, C, and so on. Matrix A above has six entries arranged in two rows and three columns, while matrix B has 12 entries arranged in three rows and four columns. So we call A as a 2 x 3 matrix and B as a 3 x 4 matrix. In matrix A we have 7 23 a and 3 12 a , while in matrix B we have 6 23 b and . 3 31 b Matrices Associated with Systems of Linear Equations Consider the following system of linear equations in the workstation problem of Example 5.1.2 in Section1: . 16 38 3 2 y x y x Related to this system are the following matrices: 1 1 3 2 16 38 16 1 1 38 3 2 The augmented matrix is what we need for Gaussian elimination. It is ca lled “ augmented ” because it includes the right-hand side constants of the equations. The other matrices will be used later in this section. The augmented matrix contains the essential parts of the linear system both the coefficients and the constants. The vertical bar is included only as a visual aid to help us separate the coefficients from the constants. |
What is 680/1000 as a decimal?
Decimals and fractions are two ways you can write the same number. A decimal is written when in two parts, its whole numbers part and its decimals part – with a decimal point in between. In this article, you’ll learn how to convert 680/1000 to a decimal using the division method.
## Solution: 680/1000 as a decimal is 0.68
Methods
### Method 1 – Explanation using the division method:
A fraction is usually split into two parts: the first part is the number on top, called the numerator; and the second part is the number on the bottom, called the denominator. These are both separated by a line called the “divisor line”. We can use the division method help to solve this question: to get a decimal, simply divide the numerator 680 by the denominator 1000 (which you can enter in any calculator):
680 (numerator) ÷ 1000 (denominator) = 0.68
And finally, you get 0.68 as your answer when you convert 680/1000 to a decimal.
### Method 2 – Explanation using the factors of 10 method:
Sometimes, dividing a number by a power of 10 (like 10, 100, 1000 etc.) is easy because all you need to do is move the invisible decimal place left by how many zeros you have in the power of 10. But for this question, the denominator, 1000, isn’t a power of 10.
However, we can change the fraction: if the denominator is a factor of a power of 10, like 2, 4, 5, 8, 10, 16, 20, 25, 40, 50 and so on, then we can convert the fraction into one that is easy to calculate.
To do this, we need to make the denominator to a power of 10 - to make 1000 a power of 10, we need to multiply by 1 (1000 * 1 = 1000). We multiplied the denominator by 1, so now we need to also multiply the numerator by 1 too:
$\frac{680 * 1}{1000 * 1} = \frac{680}{1000} = 0.68$
And finally, you get 0.68 as your answer when you convert 680/1000 to a decimal.
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# Logical Negation
This is where things get a bit weird. Binary has only two digits: a 1 and a 0, nothing else. So how do you represent a negative number? It turns out we have two ways of doing this.
## One's Complement
In decimal you sutract a number (let's say it's `y`) from another number (let's say it's `x`) by inverting it and then peforiming addition:
``````z = x + -y
``````
Inverting a number in decimal is signified by using the `-` sign operator. This equation is a bit noisy, however, so we tend to write it in the simpler form `x-y`. We can do the exact same in binary, sort of, by inverting (or "flipping") a binary number to get it's complement.
Flipping is a straightforward thing: whenever you see a 1 you replace it with a 0 and vice-versa. That's what one's complement is: inverting and then signing a number. To "sign" a number in binary means setting the left-most bit to either a 0 (positive) or 1 (negative).
The one's complement of a binary 1 (`01`), therefore, is `110`. It looks a bit weird and takes some getting used to, but it works.
We can now perform subtraction. Let's try 1 - 1 in binary. Remember, the left-most digit is the sign:
``````001
110
---
111
``````
Our answer is 111, which is the binary representation of zero. Which is weird because the binary representation of zero is also `000` and, as it turns out, there are two representations of zero in one's complement. Which sucks and is confusing.
There's a mathematical reason why which is in the video.
## Two's Complement
We need to do a little more work to get everything line up the way we want, which is where two's complement comes in. It's the same as one's complement, but you just add one to get over that negative zero thing. Sounds kind of goofy, but it works:
``````001
110
001
---
000
``````
You'll notice here that there's a carry as the last operation and that, if we're sticking to the rules, the answer should really be `1000`. Why isn't it?
## No Last Carry in Two's Complement
The simple reason is that we've already accounted for the carry in two's complement by adding one. Nice and tidy. So: if there's a carry, you can just ignore it. |
Month: April 2019
# Convert Quadrantal Bearings to Whole Circle Bearing -Example
## Convert Quadrantal Bearings to Whole Circle Bearing
Convert the following quadrantal bearing into whole circle bearing:
1. N 30°30՛ E
2. S 50° E
3. S 20°45՛ W
4. N 25° W
Solution:
How to draw the line for a Q.B ??
For Example: N 30°30՛ E, S 20°45՛ W & for N 25° W.
In case of N 30°30՛ E, the line will be drawn starting from the North line up to 30°30՛ towards the East line.
In case of S 20°45՛ W, the line will be drawn starting from the South line up to 20°45՛ towards the West line.
In case of N 25° W, the line will be drawn starting from the North line up to 25° towards the west line.
Now, the whole circle bearing can be written directly as below:
1. Line OB, W.C.B = 30°30՛
2. Line OC, W.C.B =180° – 50° = 130°
3. Line OD, W.C.B = 180° + 20°45՛ = 200°45՛
4. Line OE, W.C.B = 360° – 25° = 335°
Convert Whole Circle Bearing to Reduced Bearing – || Rules & Example ||
How To Calculate True Bearing From Magnetic Bearing
# Convert The Whole Circle Bearing to Reduced Bearing – || Rules and Example ||
## Convert Whole Circle Bearing to Reduced Bearing Rules
1. When Whole Circle Bearing(W.C.B) lies between 0° to 90°.
• Reduced Bearing(R.B) = W.C.B
2. When W.C.B lies between 90° to 180°.
• R.B = (180° – W.C.B)
3. When W.C.B lies between 180° to 270°.
• R.B = (W.C.B – 180°)
4. When Whole Circle Bearing(W.C.B) lies between 270° to 360°.
• Reduced Bearing(R.B) = (360° – W.C.B)
### Example of Convert Whole Circle Bearing to Reduced Bearing
Example:
Convert the following whole circle bearing to reduced bearing(with quadrant)
a. 30°
b. 160°
c. 200°
d. 315°
Solution:
For 30°, Reduced Bearing(R.B) = 30° NE
For 160°, R.B = (180° – 160°) = 20° SE
For 200°, R.B = (200° – 180°) = 20° SW
For 315°, R.B = (360° – 315°) = 45° NW. Where, NE, SE, SW, NW is the quadrant.
Uses Of Surveying
Convert Quadrantal Bearings to Whole Circle Bearing
# Numerical Example For Error Due To Wrong Measuring Scale – Survey Engineering
## Numerical Example For Error Due To Wrong Measuring Scale
Following numerical example shows that how to calculate correct length and area, if errors occur due to wrong measuring scale:
### Numerical Example
A surveyor measured the distance between two points marked on the plan drawn to a scale of 1cm = lm (R.F. = 1: 100) and found it to be 60m. Later he detected that he used a wrong scale of 1 cm = 50 cm (R.F. = 1: 50) for the measurement.
(a) Determine the correct length.
(b) What would be the correct area if the measured area is 70 m²?
### Mathematical Solution
a. Correct length = (R.F. of the wrong scale/R.F. of the correct scale) ⨯ measured length = [(1/50)/(1/100)] ⨯ 60 = 120m.
b. Correct area = (R.F. of the wrong scale/R.F. of the correct scale)² ⨯ measured area = [(1/50)/(1/100)]² ⨯ 70 = 280 m².
# 7 Uses Of Surveying
## 7 Uses Of Surveying
Surveying may be used for the following various purpose:
### 1. Engineering Surveying
Engineering surveying is used for the design and construction of a new project such as road, railway, building, city, towns etc.
### 2. Topographical Surveying
To prepare a topographical map, which shows the hills, rivers, valleys, forest, village, towns, etc. of a country.
### 3. City Surveying
City surveying, which is used to locate the streets, water supply, sanitary system etc.
Cadastral surveying, which is conducted in order to produce plans of property boundaries for legal purpose.
### 5. Astronomic Surveying
Astronomic surveying is used to determining latitudes, longitudes, azimuths, local time etc. for the various places on the earth.
### 6. Geological Surveying
To prepare a geological map, which shows the information about different strata of the earth surface for the purpose of the geological studies.
### 7. Photogrammetric Surveying
Photogrammetric surveying is used for obtaining topographical details of large areas which are difficult to access.
Preliminary Survey
Final Location And Detailed Survey
Reconnaissance Survey
# Quantity Of Soil Sample Required For Different Test
The following tests and required sample weight are:
## i) Soil identification, natural moisture content test, mechanical analysis and index properties chemical test
⇛ For Cohesive Soils – Required sample weight is 1 kg
⇛ For Sand and gravels – Required sample weight 3 kg
## ii) Compaction tests
⇛ For Cohesive Soils and Sands – 12.5 kg
⇛ For Gravelly Soils – 25 kg
## iii) Comprehensive examination materials including soil stabilization
⇛ For Cohesive Soils and Sands – 25 to 50 kg
⇛ For Gravelly Soils – 50 to 100 kg
Types of Soil Sample
Different Grading Zone Value of Fine Aggregates(sand) As Per IS: 383
# Method Of Providing Superelevation
## Method Of Providing Superelevation
Introduction of superelevation on a horizontal curve of a road is an important feature in road construction. Superelevation is provided in the following two methods.
1. Elimination of the crown of the cambered section.
2. Rotation of pavement to attain full superelevation.
Read More: Purposes of Providing Superelevation
### 1. Elimination of The Crown of The Cambered Section
In this method, the outer half of the camber is gradually decreased. This may be done by two methods.
In the first method, the outer half of the camber is rotated about the crown at the desired rate such that the surface falls on the same plane as the inner half.
In the second method, the crown is progressively shifted outwards. This method is not usually adopted.
### 2. Rotation of Pavement To Attain Full Superelevation
In this stage, superelevation is gradually provided over the full width of the carriageway so that the required superelevation is available at the beginning of the circular curve. The different method employed for attaining the superelevation is as follows:
#### A. Revolving Pavement About The Center Line
In this method the surface of the road is rotated about the center line of the carriageway, gradually lowering the inner edge and rising the upper edge. The level of the center line is kept constant. This method is widely used.
#### B. Revolving Pavement About The Inner Edge
In this method, the surface of the road is rotated about the inner edge, raising the center and outer edge.
#### C. Revolving Pavement About The Outer Edge
In this method, the surface of the road is rotated about the outer edge depressing the center and inner edge.
# 5 Types of Road Tar
Tar is the viscous liquid obtained by the destructive distillation of coal or wood in the absence of. Generally, the colour of the road tar is black and it contains 72 to 96% bituminous contents.
There are 5 types(grades) of road tars viz RT-1, RT-2, RT-3, RT-4, RT-5 based on their viscosity and other properties.
RT-1 has the lowest viscosity and used for surface painting under exceptionally cold weather. The specific gravity of this road tar is 1.26.
RT-2 is recommended for standard surface painting under normal Indian climate condition. Specific gravity 1.26
RT-3 may be used for surface painting, renewal coats and premixed top course and light carpets. The specific gravity of this tar is 1.28
RT-4 is generally used for premixing tarmacadam in base course. Specific gravity 1.28
RT-5 has the highest viscosity and may be used for grouting purpose. Sp.g 1.28.
4 Steps in Bituminous Road Construction
# Factors Affecting The Thickness Of A Load-Bearing Wall
## Factors Affecting The Thickness Of A Load-Bearing Wall
The thickness of a load-bearing wall shall be sufficient to ensure that in worst condition loading, the stresses developed are within the safe limits.
The thickness of a non-load-bearing wall should be sufficient to ensure stability and strength against lateral forces.
The thickness of a wall depends on the following factors:
2. The overall height of the wall.
3. The height of the wall between floors.
4. The overall length of the wall and the centre-to-centre distances between lateral supports.
5. The crushing strength of bricks to be used and the strength of the mortar to be applied.
Types of Joints in Stone Masonry
# Prevention of Dampness
## Prevention of Dampness
There are various methods of damp proofing. Adoption of any specific method depends upon various factors such as
―nature of the surface,
―location of the structure,
―sub-soil water table,
―the amount of dampness to be handled, etc.
Following are the methods or measures adopted to prevent dampness:
1) Surface treatment.
2) Integral waterproofing.
3) Incorporation of a membrane between the source of dampness and the part of the building adjacent to it.
4) Construction of cavity wall.
5) Guniting.
9 Effects of Dampness
8 Causes of Dampness
# 9 Effects of Dampness
## 9 Effects of Dampness
Following 9 harmful effects of dampness are:
1. The materials used in the construction of a building are seriously affected by dampness.
2. The plaster becomes softened and may crumble due to dampness.
3. Unpleasant white patches are formed on the damp surface.
4. The appearance of distemper is badly affected by dampness.
5. Timber may decay, warp and buckle due to dampness.
6. The metallic fittings may corrode.
7. The electrical fittings may get damaged.
8. Dampness promotes the growth of termites and mosquitoes breeding.
9. Dam building creates unhealthy conditions for occupants. |
The addition of numbers is pretty simple but the addition of the same numbers with exponents is not an easy task. For students, it may be hard because those exponents need to be solved first. With the help of adding an exponents calculator, this task can be done quickly within seconds. It is a simple tool that has been programmed for this purpose particularly. All students with any level can use this tool to add numbers with exponents. This tool will perform the calculation quickly within a glimpse and show you the output.
It will perform the conversion of exponents in numbers at the backend and then add them to show you the results. You can use this math calculator even if you haven’t experienced working with such tools. Isn’t it a good way to add exponents quickly without errors?
## Exponents Definition
In Mathematics, the term exponent is used to represent the power of a number or variable. It means the digits or alphabets that will be written as the power of any base will be called exponents.
For example, in 23, the digit “3” is an exponent as it is the power of “2”. In Mathematics, adding exponents is not possible unless you have solved them according to the base involved. It makes the overall process of addition hard for the students struggling with Mathematical problems
### What is meant by adding exponents?
Adding exponents is the process to add different numbers with different powers. It means neither the base is same nor the powers. To solve such questions, you need to remove the exponents by solving them using a general Mathematical problem.
The addition of exponents doesn’t include any specific formula. It includes a specific method that you need to follow for accurate adding. Every Adding Exponents Calculator will also follow the same procedure we are going to share here.
• Dissolve the exponent of the first number by multiplication
• Dissolve the exponent of the second number by multiplication
• Add the final numbers to get the output of the complete problem
To make the process of adding exponents understandable, let us share an example here.
Example 1:
Add 43and 34 to find the final output.
Solution:
To find the answer to the addition of exponents, we need to first solve them separately to dissolve their exponents.
43 = 4 x 4 x 4 = 64
Similarly,
34 = 3 x 3 x 3 x 3 = 81
Now, we need to add the results we have found from the above conversion.
64 + 81 = 145
### How to use Adding Exponents Calculator?
This adding exponents calculator by Calculator’s Bag has a simple interface. You can use this tool by following these steps: You can use us to add an exponent's calculator in various mathematical and other scientific principles to instantly determine the value in natural numbers.
### FAQs | Adding Exponents Calculator
Adding decimals is a little complex process as compared to the addition of integers. But overall it is an easy process to add decimals.
What are the four rules of decimals?
To add exponents, we need to first dissolve the exponents by multiplying the base with itself.
Can you add different exponents together?
If the base of exponents is the number, we can add them using the above method. But if the base is an alphabet, we won’t be able to add or subtract them.
What is the exponent rule?
The exponent rule varies with the operation involved. It means the rule for adding an exponent is different from the rule used for multiplication.
What are the rules for exponents in fractions?
The rule for exponent in fractions states that the numerator of the fraction will be the power and the denominator will act like the root of the number or equation.
How do you add and subtract exponents with different bases?
If the base of exponents is a number, we need to solve them first for addition and subtraction. It will not be possible to add or subtract if the base is an alphabet instead of a number. |
# How to Graph Quadratic Inequalities? (+FREE Worksheet!)
Want to know how to graph Quadratic inequalities? You can do it in a few simple and easy steps.
## Step by step guide to Graphing Quadratic inequalities
• A quadratic inequality is in the form $$y>ax^2+bx+c$$ (or substitute $$<,≤,$$ or $$≥$$ for $$>$$).
• To graph a quadratic inequality, start by graphing the quadratic parabola. Then fill in the region either inside or outside of it, depending on the inequality.
• Choose a testing point and check the solution section.
### Graphing Quadratic inequalities – Example 1:
Sketch the graph of $$y>3x^2$$.
Answer:
First, graph $$y=3x^2$$
Since the inequality sing is $$>$$, we need to use dash lines.
Now, choose a testing point inside the parabola. Let’s choose $$(0,2)$$.
$$y>3x^2→2>3(0)^2→2>0$$
This is true. So, inside the parabola is the solution section.
The Absolute Best Books to Ace Pre-Algebra to Algebra II
### Graphing Quadratic inequalities – Example 2:
Sketch the graph of $$y>2x^2$$.
Answer:
First, graph $$y=2x^2$$
Since the inequality sign is $$>$$, we need to use dash lines.
Now, choose a testing point inside the parabola. Let’s choose $$(0,2)$$. $$y>2x^2→2>2(0)^2→2>0$$
This is true. So, inside the parabola is the solution section.
The Best Book to Help You Ace Pre-Algebra
$14.99 Satisfied 92 Students ## Exercises for Graphing Quadratic inequalities ### Sketch the graph of each function. • $$\color{blue}{y<-2x^2}$$ • $$\color{blue}{y≥4x^2}$$ • $$\color{blue}{y<-2x^2}$$ • $$\color{blue}{y≥4x^2}$$ The Greatest Books for Students to Ace the Algebra ## Related to This Article ### More math articles ### What people say about "How to Graph Quadratic Inequalities? (+FREE Worksheet!) - Effortless Math: We Help Students Learn to LOVE Mathematics"? No one replied yet. X 45% OFF Limited time only! Save Over 45% SAVE$40
It was $89.99 now it is$49.99 |
Back
What Does it Mean to Subtract a Negative Number?
> ... Math > Addition and Subtraction > What Does it Mean to Subtract a Negative Number?
Subtracting a negative number is the same as adding a positive number.
Example: $$-5 - (-2) = -3$$
Subtract a negative number
### Practice Problems
Evaluate the following expression:
1. $$6 - (- 3) = ?$$ (
Solution
Solution: $$9$$
Details:
Let’s explore the patterns:
$$6 - (1) = 5$$
$$6 - (0) = 6$$
$$6 - (-1) = 7$$
$$6 - (-2) = 8$$
$$\color{red} 6 - (-3) = 9$$
Subtracting a negative number from a positive number result in a positive answer. This problem can also be rewritten as $$6 + 3 = 9$$.
)
2. $$4 - (- 4) = ?$$ (
Solution
Solution: $$8$$
Details:
Let’s explore the patterns:
4 $$- (1) = 3$$
4 $$- (0) = 4$$
4 $$- (-1) = 5$$
4 $$- (-2) = 6$$
4 $$- (-3) = 7$$
$$\color{red} 4 - (-4) = 8$$
Subtracting a negative number from a positive number results in a positive answer. This problem can also be rewritten as $$4 + 4 = 8$$.
)
3. $$3 - (- 5) = ?$$ (
Video Solution
Solution: 8
Details:
(Video Source | Transcript)
)
4. $$4 - (- 8) = ?$$ (
Video Solution
Solution: 12
Details:
(Video Source | Transcript)
)
5. $$10 - (- 19) = ?$$ (
Solution
Solution:
29
)
6. $$48 - (- 16) = ?$$ (
Solution
Solution:
64
)
## Need More Help?
1. Study other Math Lessons in the Resource Center.
2. Visit the Online Tutoring Resources in the Resource Center. |
You are on page 1of 4
(a) 0th iteration
(b) 1st iteration
(c) 2nd iteration
(d) 20th iteration
Figure 1: Sierpi´
nski Carpet Fractal.
Problems of the day:
1
Assuming that the area of the black square in (a) is equal to 9.
(a) How could we calculate the area in each of the drawings (b) – (d)?
(b) Write a general rule for the area after n iterations. What is the area after
(c) If you have extra time, consider what else you might measure that is also
a function of n?
2
Now, we are ready for an application that relates the concept of fractals (Figure 1) to the concept of fractional exponents. First let’s clarify what mathematicians mean when we use the word dimension:
(a) In the rectangle below, draw one square with unit side length (l = 1).
Next to it, draw a two-times scaled (s = 2) version of the square with
twice the side length (l = 2).
(b) How many copies (c) of the first square are needed to construct the second
square? How many are needed for a three-times scaled (s = 3) square?
(c) We have been working with two dimensional (d = 2) objects. How would
your answer to part (b) change for one dimensional (d = 1) line segments?
Three dimensional (d = 3) solids? Points (d = 0)? Answer the question
by completing the table below.
shape
dimension
scaling
copies
(d)
factor (s)
(c)
square
2
1
1
square
2
2
square
2
3
segment
1
1
segment
1
2
cube
3
1
cube
3
2
point
0
1
point
0
2
(d) Can you come up with a relationship between dimension (d), copies (c),
and scaling factor (s)?
(e) Challenge! Using your relationship in (d), can you determine the dimension of the object in Figure 1?
he dimension, d, of a fractal depends upon its construction rule. As you
can imagine, this structural parameter also critically influences a physical object’s function. Engineers and Nature (via natural selection)1 have
repeatedly capitalized on this strategy. For example, the many different
length scales of a cell phone fractal antenna absorb a wider range of electromagnetic
frequencies using only a fraction of the footprint.
T
(a) Fractal
(b) Cell phone
Figure 2: Fractal in technology.
Design question! If you were an engineer and you were considering fractal materials, what variable would you likely want to know the most, the material’s surface
area, volume, or fractal dimension? Why?
Bonus Challenge! Make your own fractal with a tunable dimension, d; 0 < d < 1?
1
There are many naturally occurring instances of fractal geometry in nature, including the
branching on trees, sponges and internal organs such as the lungs, neurons and kidneys |
Home PHYSICS TOPIC 4: FORCES IN EQUILIBRIUM ~ PHYSICS FORM 2
# TOPIC 4: FORCES IN EQUILIBRIUM ~ PHYSICS FORM 2
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### FORCES IN EQUILIBRIUM
#### The Effects of Turning Forces
in motion or making it move if it was at rest.
If a body under the action of a net external force is allowed to rotate about a pivot, the body will tend to turn in the direction of the applied force.
Examples of turning effect of force:
A person pushing a swing will make the swing rotate about its pivot.
A worker applies force to a spanner to rotate a nut.
A person removes a bottle’s cork by pushing down the bottle opener’s lever.
Force is applied to a door knob and the door swings open about its hinge.
A driver can turn a steering wheel by applying force on its rim.
### The Moment of Force
Moment of force about a point is the turning effect of the force about that point. It is obtained as the product of applied force and the perpendicular distance from the point of the application of the force.
Therefore, moment of a force depends on two factors namely;
1. The applied force.
2. The perpendicular distance from the point of action of the force to the turning point(fulcrum).
The change of state of a body can appear in several forms and the most common form is the turning effect which is referred to as moment of a force.
The unit for the moment of a force is Newton-meter (Nm).
### The turning effect can bring about two kinds of moments;
Clockwise moment -caused by the forces which tend to turn the body in a clockwise direction.
Anticlockwise moment -caused by the forces which tend to turn the body in an anticlockwise direction.
Consider the diagram below,
The above uniform rod AB is balanced about the turning point (fulcrum) f. The weight W1 acting at point A tends to turn the rod in anticlockwise direction. The weight W2 acting at point B and the weight of the rigid body W0 tends to move the rod in clockwise direction. Hence, we have two anticlockwise moments due to weight W2 and weight of the rod W0.
### The Principle of Moments
The Principle of moments states that:
” If a body is in equilibrium under action of forces which lie on one plane, then, the sum of the clockwise moments is equal to the sum of the anticlockwise moments about any point in that plane.”
The equilibrium of state of a body is brought about by two conditions;
1. The sum of anticlockwise moments is equal to sum of clockwise moments about the turning point.
2. The sum of upward forces is equal to sum of downward forces acting on a body.
If the data for the weight (force) W, and distance d of the weight from the fulcrum can be obtained then the moment created about the fulcrum is obtained by multiplying the weight with the respective distance i.e Wd. According to the principle of moments, at equilibrium the sum of clockwise moments (Wd) and that of anticlockwise moments (Wd) must be equal. to make it simple the below table can be used to record and compare the moments;
W1(g) W2(g) d1(cm) d2(cm) W1d1(gcm) W2d2(gcm)
Materials and apparatus:Meter rule, several different weights, inelastic cotton thread, knife edge and a marker pen.
Procedures
Balance the meter rule horizontally on a knife edge.
Mark a balance point as C. Use the marker pen to do that.
Suspend a meter rule from a fixed axis through C. Suspend unequal weights W₁ and W₂ from the meter rule by using thin cotton threads.
Adjust the distance d₁ and d₂ of the weights W₁ and W₂ from C until the meter rule balance.
Repeat the experiment five times using different values of W₁ and W₂.Record the results on the table as shown below.
W₁(g) W₂(g) d₁(cm) d₂(cm) W₁ d₁ (gcm) W₂ d₂ (gcm)
Observation:In each case it will be found that W₁ d₁ is equal to W₂ d₂.
The Principle of Moment in Daily Life
### Moment of a force is used in the following:
Is applied by a hand to unscrew a stopper on the bottle.
Is applied by a spanner to unscrew a nut on a bottle.
Turning a steering wheel of a car.
### Centre of Gravity
Center of gravity of a body is the point at which the weight of a body appears to be concentrated. OR center of gravity of a body is the point of application of the resultant force due to the earth attraction on the body.The center of gravity of a regular body is found to be at its geometrical center.
Example 1
1. Centre of gravity of a uniform meter rule is at the 50cm mark.
2. Centre of gravity of a circular object is at its center.
The center of gravity of irregular bodies can be found experimentally.
### Types of Equilibrium
The Condition for Equilibrium
Equilibrium is simply the state of balance of a body. A body is said to be balanced if its center of gravity is directly above the point of support.
The body is more stable when its center of gravity is closer to its base and if it has a wide base.
The body can be in motion while in equilibrium state. Equilibrium of this type is known as dynamic equilibrium.
A body in state of rest while in equilibrium is under static equilibrium while the one which is rotating but still in equilibrium is said to be under rotational equilibrium.
Stable, Unstable and Neutral Equilibrium
There are three types of equilibrium, namely:
1. Stable equilibrium
2. Unstable equilibrium
3. Neutral equilibrium
Stable equilibrium:A body is said to be in stable equilibrium if is given with small displacement the center of gravity will be raised and the body returns to its original position after displacement.
Unstable equilibrium:A body is said to be in unstable equilibrium if when given a small displacement the center of gravity will be lowered and the body doesn’t returns to its original position after displacement.
Neutral equilibrium:A body is said to be in neutral equilibrium when a small displacement doesn’t alter the position of the center of gravity; the body is at rest in whichever position it is placed, eg, rolling a sphere or a barrel.
Conditions of Stable, Unstable and Neutral Equilibrium in Daily Life
Application of turning effect in daily life
Is applied by a hand to unscrew a stopper on the bottle.
Is applied by a spanner to unscrew a nut on a bottle.
Turning a steering wheel of a car.
Exercise 1
.The moment of a force about a point is 1120 Nm.If the magnitude of the force is 5600N,find the perpendicular distance between the point and the line of action of the force |
What is the perimeter of a triangle with corners at (2 ,6 ), (4 ,5 ), and (3 ,1 )?
Jan 26, 2018
See a solution process below:
Explanation:
The perimeter of a triangle is:
$p = {s}_{1} + {s}_{2} + {s}_{3}$
Where: ${s}_{1}$, ${s}_{2}$ and ${s}_{3}$ are the lengths of the sides of the triangle.
We need to find the distance between each of the points in the problem. The formula for calculating the distance between two points is:
$d = \sqrt{{\left(\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}\right)}^{2} + {\left(\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}\right)}^{2}}$
Distance Between (2, 6) and (4, 5) is:
${s}_{1} = \sqrt{{\left(\textcolor{red}{4} - \textcolor{b l u e}{2}\right)}^{2} + {\left(\textcolor{red}{5} - \textcolor{b l u e}{6}\right)}^{2}}$
${s}_{1} = \sqrt{{2}^{2} + {\left(- 1\right)}^{2}}$
${s}_{1} = \sqrt{4 + 1}$
${s}_{1} = \sqrt{5}$
Distance Between (2, 6) and (3, 1) is:
${s}_{2} = \sqrt{{\left(\textcolor{red}{3} - \textcolor{b l u e}{2}\right)}^{2} + {\left(\textcolor{red}{1} - \textcolor{b l u e}{6}\right)}^{2}}$
${s}_{2} = \sqrt{{1}^{2} + {\left(- 5\right)}^{2}}$
${s}_{2} = \sqrt{1 + 25}$
${s}_{2} = \sqrt{26}$
Distance Between (4, 5) and (3, 1) is:
${s}_{3} = \sqrt{{\left(\textcolor{red}{3} - \textcolor{b l u e}{4}\right)}^{2} + {\left(\textcolor{red}{1} - \textcolor{b l u e}{5}\right)}^{2}}$
${s}_{3} = \sqrt{{\left(- 1\right)}^{2} + {\left(- 4\right)}^{2}}$
${s}_{3} = \sqrt{1 + 16}$
${s}_{3} = \sqrt{17}$
The Perimeter of the Triangle is:
$p = {s}_{1} + {s}_{2} + {s}_{3}$
$p = \sqrt{5} + \sqrt{26} + \sqrt{17}$
If you need a single number answer:
$p = 11.458$ rounded to the nearest thousandth. |
# Equivalent Ratios
Equivalent ratios : A ratio obtained by multiplying or dividing the numerator and denominator of a given ratio by the same number is called an equivalent-ratios.
Example : Consider the ratio 6 : 4,
6/4 = (6 x2)/(4 x 2) = 12/8
6/4 = (6 x3)/(4 x 3) = 18/12
6/4 = (6 x4)/(4 x 4) = 24/16
And so on.
All these are equivalent-ratio of 6/4.
If a : b and c : d are two equivalent-ratios, we write a/b = c/d
Solved examples :
1) Find two equivalent-ratios of 6 : 15.
Solution :
6/15 = (6 x2)/(15 x 2) = 12/30
6/15 = (6 ÷ 3)/(15 ÷ 3) = 2/5
So, 12 : 30 and 2 : 5 are the two equivalent-ratios of 6 : 15.
--------------------------------------------------------------------
2) Fill in the blank.
Solution :
In order to find the first missing number, consider the denominator 21 and 3.
Think of a number in such a way that when we divide 21 with that number we will get 3.
21 ÷ ( ) = 3 So that number must be 7. So divide 14 with 7 that gives us the 1st missing number.
∴ 1st missing number = 14 ÷ 7 = 2
So the 2nd ratio is 2/3.
For 2nd missing number, consider
As we know that when we multiply 2 with 3 we get 6 so with that same number multiply 3 we will get the 2nd missing number.
∴ 2nd missing number = 3 x 3 = 9
So the 3rd ratio is 6/9.
--------------------------------------------------------------------
3) Find the value of ‘a’ and ‘b’ in
Solution :
Consider the 1st two ratios,
12/20 = a/5 ⇒ 12 x 5 = 20 x a
60 = 20a
∴ a = 60/20
∴ a = 3
For finding b, consider 1st and 3rd ratio.
12/20 = 9/b ⇒ 12 x b = 20 x 9
12b = 180
∴ b = 180/12
∴ b = 15.
Ratio - Proportion
Ratio and Proportion
Ratio in the simplest form
Comparison of ratios
Equivalent ratios
Proportion
Continued Proportion
From equivalent-ratios to number system |
Easy Math Quiz On Word Problems! Trivia
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| By DanielCarig
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Community Contributor
Quizzes Created: 205 | Total Attempts: 120,021
Questions: 15 | Attempts: 648
Settings
.
• 1.
How many CD racks does it take to hold 300 CDs if each rack holds 75 CDs?
• A.
Three
• B.
Four
• C.
Six
• D.
Two
B. Four
Explanation
Each CD rack can hold 75 CDs. To find the number of CD racks needed to hold 300 CDs, we divide 300 by 75, which equals 4. Therefore, four CD racks are needed to hold 300 CDs.
Rate this question:
• 2.
How many six packs of soda do you need to buy to have 54 cans of soda?
• A.
Three
• B.
Five
• C.
Nine
• D.
Seven
C. Nine
Explanation
To have a total of 54 cans of soda, you would need to buy nine six packs of soda. Each six pack contains six cans, so by buying nine six packs, you would have a total of 54 cans.
Rate this question:
• 3.
What is the total price of a \$20 item if the sales tax is 7%?
• A.
20.07
• B.
22.89
• C.
21.4
• D.
21
C. 21.4
Explanation
The total price of a \$20 item with a 7% sales tax can be calculated by multiplying the item's price by the tax rate and adding the result to the original price. In this case, 7% of \$20 is \$1.40, so the total price would be \$20 + \$1.40 = \$21.40.
Rate this question:
• 4.
How many days are there in 40 weeks?
• A.
47
• B.
280
• C.
375
• D.
440
B. 280
Explanation
There are 7 days in a week. To find the number of days in 40 weeks, we multiply 7 by 40, which gives us 280. Therefore, the correct answer is 280.
Rate this question:
• 5.
How fast is a car averaging if it goes 200 miles in 4 hours?
• A.
70 mph
• B.
40 mph
• C.
60 mph
• D.
50 mph
D. 50 mpH
Explanation
The car is averaging 50 mph because it traveled 200 miles in 4 hours. To find the average speed, we divide the total distance traveled by the total time taken. In this case, 200 miles divided by 4 hours equals 50 mph.
Rate this question:
• 6.
How many pieces of gum would you need to give 33 kids two pieces each?
• A.
64
• B.
67
• C.
66
• D.
65
C. 66
Explanation
To give 33 kids two pieces of gum each, you would need a total of 66 pieces of gum. This is because you multiply the number of kids (33) by the number of pieces each kid needs (2), which equals 66.
Rate this question:
• 7.
If one scoop of drink mix makes one quart of lemonade, how much mix do you need for a gallon?
• A.
One Scoop
• B.
Three Scoops
• C.
Four Scoops
• D.
Two Scoops
C. Four Scoops
Explanation
To make one quart of lemonade, one scoop of drink mix is needed. Since there are 4 quarts in a gallon, you would need 4 times the amount of mix. Therefore, to make a gallon of lemonade, you would need four scoops of the drink mix.
Rate this question:
• 8.
If a gas tank holds 15 gallons, and has 2.6 gallons now, how much do you need to fill it up?
• A.
17.6
• B.
12.4
• C.
2.6
• D.
9
B. 12.4
Explanation
To fill the gas tank, you need to add the amount of gas that is missing. Since the tank can hold 15 gallons and currently has 2.6 gallons, you need to add 12.4 gallons to fill it up completely. Therefore, the correct answer is 12.4.
Rate this question:
• 9.
If you are leaving at 9 PM for a destination that is about 6 hours away, what time will you arrive?
• A.
12 AM
• B.
6 AM
• C.
3 AM
• D.
10 PM
C. 3 AM
Explanation
If you leave at 9 PM and the destination is about 6 hours away, you will arrive at 3 AM. This is because if you subtract 6 hours from 9 PM, you get 3 AM.
Rate this question:
• 10.
If you divide your 8-piece shrimp cocktail amongst 4 people, how many shrimp does each person get?
• A.
3
• B.
8
• C.
4
• D.
2
D. 2
Explanation
If you divide the 8-piece shrimp cocktail amongst 4 people, each person will get 2 shrimp. This is because when you divide 8 by 4, the result is 2. Therefore, each person will receive 2 shrimp.
Rate this question:
• 11.
A plane departs at 5:30 pm. At what time will it land if the flight takes eleven hours and a half?
• A.
4:30am
• B.
5:30am
• C.
5:30pm
• D.
5:00am
D. 5:00am
Explanation
The flight duration is given as eleven hours and a half. If the plane departs at 5:30 pm, adding eleven and a half hours to this time will give us the landing time. Therefore, the plane will land at 5:00 am.
Rate this question:
• 12.
If a burger has 300 calories, fries have 250 and soda has 200, how many calories is your meal?
• A.
750
• B.
650
• C.
450
• D.
550
A. 750
Explanation
The meal consists of a burger with 300 calories, fries with 250 calories, and soda with 200 calories. To find the total number of calories in the meal, we add these three values together: 300 + 250 + 200 = 750. Therefore, the correct answer is 750.
Rate this question:
• 13.
If there are 48 cookies in a box and 3 cookies make one serving, how many servings are in a box?
• A.
24
• B.
16
• C.
11
• D.
22
B. 16
Explanation
There are 48 cookies in a box, and 3 cookies make one serving. To find the number of servings in a box, we need to divide the total number of cookies by the number of cookies in one serving. Therefore, 48 divided by 3 equals 16 servings.
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• 14.
If hot dogs come in 10-packs and buns in 8-packs, how many buns should you buy to use up both?
• A.
20
• B.
40
• C.
24
• D.
18
B. 40
Explanation
To use up both the hot dogs and buns, we need to find the least common multiple (LCM) of 10 and 8. The LCM of 10 and 8 is 40. This means that if we buy 40 buns, we will have enough to use up both the 10-pack of hot dogs and the 8-pack of buns.
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• 15.
You have won \$50,000 in the lottery. Taxes on income are 20%. How much is owed in taxes?
• A.
10000
• B.
5000
• C.
50000
• D.
1000
A. 10000
Explanation
The taxes owed on the \$50,000 lottery win can be calculated by multiplying the amount won by the tax rate. In this case, the tax rate is 20%, so the taxes owed would be \$50,000 * 0.20 = \$10,000.
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# Math Tip Of The Day: GCF Part 1
In mathematics, you sometimes have to find the GCF (greatest common factor). This concept is used once students have learned multiplication, and is also used in Algebra when factoring polynomials. Part 1 of this math tip will focus simply on finding the GCF of whole numbers. Part 2 will extend to factoring polynomials.
First of all, what is a factor? A factor is a number that is multiplied to get another number or product. So, the factors of 2 are 1 and 2, since they are the only numbers that can multiply to arrive at a product of 2.
Anytime that I’ve had to find the GCF, or the largest common factor of two numbers, I’ve made a T-table to assist me. For example, if you need to find the GCF of the numbers 30 and 75, you can set up a T-table to do so. See the diagram below.
I list the numbers that I need the GCF for at the top of the table. Underneath each number, I list all of the factors. For 30, I know that 1 x 30, 2 x 15, 3 x 10 and 5 x 6 all equal 30. So I list them in pairs. I did the same for the number 75. Then I circle all the factors that both numbers have in common. Once all common factors are circled, I find the largest of those numbers and place a check mark. This indicates the GCF for those numbers.
It’s really not that difficult to find the GCF, but there are some things to keep in mind. Look at the following tips:
• 1 and the number itself will always be factors. So if you can’t think of any other factors, at least you’ll have those two.
• The number 1 will more than likely NOT be the GCF. All numbers have a factor of 1. The only way that 1 will be the GCF is if both numbers are prime numbers (which means they have no other factors than 1 and themselves).
• Multiplication is key. The better you are at multiplication, the better you will be at finding the factors of numbers.
• When in doubt, try out every possible whole number. Divide the number by 1, then 2, then 3 and so on until you’ve found all possible factors.
So, feeling confident that you can find the GCF? Try these examples.
1. 24 and 28
2. 25 and 55
**Stay tuned for Part 2 of this Math Tip!! |
Classification
Classification means to assort the items of a given group on the basis of certain common qualities (characteristics) they possess and then spot the stranger (odd man) out. We are required to select the figure which differs from all other figures in the given set.
Bases of Classification
The figures may broadly be classified on the following bases:
1. SFR (Same figure rotates)
The four similar figures are actually the rotated forms of the same figure. For example, in the figures below, it is only the rotation of the same figure in four figures while one of them is a different figure.
Example:
Solution: 3
1. Number of lines or components
The figures may be classified on the basis of the number of lines or the number of components. For example, all figures (except the odd man out) may be made of the same number of components; they all may be made of only even number of (or odd number of) lines or the sum (or difference) of the number of lines may be a fixed number etc.
Example:
Solution: 5; all others are divided in six segments.
1. Origin of Lines
Apart from the number of lines, the origin of lines is also an important consideration for some of the figures. Some figures are made of several components, an important component of which could be a line. The point from where it starts is an important factor to consider and it may be the basis on which a figure may be differentiated from the remaining four.
Example:
Solution: 5; all figures are made of two circles of different sizes, each circle having a line within it. One of these lines is vertical and the other horizontal. But in (5), both the lines are horizontal.
1. Comparative positions of the elements
A figure consists of several components. Now it may be that in the figures all the components may have a definite position in relation to each other and this could be the basis on which any four figures could be said to be similar and the fifth could be said to be different.
Example:
Solution: 4; in each figure there are two identical shapes in rotated form.
1. Other considerations
i. Interior-exterior consideration:
When a figure is made of more than one elements/components, it is naturally likely that some of the elements may lie within (interior) some other elements while some others may lie outside (exterior) the other elements.
Example:
Solution: 5; All other figures exhibit non-parallelism except figure 5.
SYNTHESIS OF FIGURES
In a given group of figures, each figure consists of several components and these components, their number, their positioning; their shape and size etc are formed according to a definite rule. If we are able to spot the logic that is the basis of the synthesis of figures we may be able to detect the figure.
It may be noted that we have already seen cases where the figures were created and classified on the basis of comparative shapes, sizes and positions of components or on the basis of number or origin of lines etc. Even after these considerations have been made, it does not remain very easy to detect the logic or the algorithm behind the making of the figures. However, with a little bit of foresight and experience, such problems can easily be tackled as the following examples suggest. |
Math Focus 3
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Chapter 7: Fractions Planning a Vegetable Garden
INTRODUCTION
When you are trying to grow a garden, insects can be big problem — some of them eat your plants! If you are trying to grow food for you to eat, these insects (or pests) become an even bigger problem. Luckily, you can plant certain plants together to help keep away pests. This can help when you are planning a garden.
Imagine that you are planting a vegetable garden in the shape of a square. You want to plant four different types of vegetables. Two of the vegetables will be tomatoes and carrots. You would like the other vegetables you plant to keep pests out of the garden. Make a plan for your garden where 2/6 of the garden is for carrots, 2/6 of the garden is for tomatoes, 1/6 of the garden is for one other vegetable, and 1/6 is for the other vegetable.
THE PROCESS
1. Visit Gardening for Your Health and scroll down to the “Health for your garden” section to find out what 2 other vegetables you should plant beside carrots and tomatoes to help keep pests out of your garden.
2. Use Fraction Pieces to design your garden. Click “Squares” at the top of the screen and “Show Labels” at the bottom. Then click on the yellow rectangles to add pieces to the window. Drag the pieces onto the square to divide it into 6 pieces. Choose a colour for each vegetable. To change the colour of a rectangle, click on the rectangle and then one of the colours shown at the bottom of the screen. Remember to place the vegetables that keep pests away beside the vegetables that the insects might eat.
3. Draw your design on paper and colour and label each vegetable in the garden. Include the fraction for each part.
4. How do the fractions tell you what vegetables you will have more of to eat?
5. How could you change your design if you wanted to grow carrots in 3/6 of your garden? Use Fraction Pieces to make a model of your new design. Then draw the design on paper, including labels and fractions. (Remember, you have to keep the pests off your other plants too!)
6. Use the fractions to compare the parts of your garden design from step 5.
Websites:
Fraction Pieces
Materials:
pencil
pencil crayons
grid paper
ASSESSMENT
Mathematical Processes Work meets standard of excellence Work meets standard of proficiency Work meets acceptable standard Work does not yet meet acceptable standard Communication • uses effective and specific mathematical language, symbols, and conventions to enhance communication about fractions • organizes and displays results in effective and clear ways that enhance interpretation • uses appropriate and correct mathematical language, symbols, and conventions to support communication about fractions • organizes and displays results in appropriate and reasonably clear ways that assist interpretation • uses mathematical language, symbols, and conventions to partially support communication about fractions • organizes and displays results in somewhat appropriate and partially clear ways that make inferring necessary by the reader • uses mathematical and nonmathematical language and conventions incorrectly and/or inconsistently which interfere with communication about fractions • organizes and displays results in haphazard and/or unclear ways that impede interpretation Connections • makes insightful connections between real-world contexts and mathematical ideas • makes meaningful connections between real-world contexts and mathematical ideas • makes simple connections between real-world contexts and mathematical ideas • makes minimal or weak connections between real-world contexts and mathematical ideas Visualization • uses visual representations insightfully to foster/demonstrate a thorough understanding of fractions • uses visual representations meaningfully to foster/demonstrate a reasonable understanding of fractions • uses visual representations simply to foster/demonstrate a basic understanding of fractions • uses visual representations poorly to foster/demonstrate an incomplete understanding of fractions |
CATEGORIES:
# Decimal to Unsigned Binary Conversion
Converting from decimal to unsigned binary is a little more complicated, but it still isn't too difficult. Once again, there is a well-defined process. To begin with, it is helpful to remember the powers of 2 that correspond to each bit position in the binary numbering system. These were presented in Figure 4 for the powers of 20 up to 27. What we need to do is separate the decimal value into its power of 2 components. The easiest way to begin is to find the largest power of 2
that is less than or equal to our decimal value. For example if we were converting 7510 to binary, the largest power of 2 less than or equal to 7510 is 26 = 64.
The next step is to place a 1 in the location corresponding to that power of 2 to indicate that this power of 2 is a component of our original decimal value. Next, subtract this first power of 2 from the original decimal value. In our example, that would give us 7510 – 6410 = 1110. If the result is not equal to zero, go back to the first step where we found the largest power of 2 less than or equal to the new decimal value. In the case of our example, we would be looking for the largest power of 2 less than or equal to 1110 which would be 23 = 8. When the result of the subtraction reaches zero, and it eventually will, then the conversion is complete. Simply place 0's in the bit positions that do not contain 1's. Figure 6 illustrates this process using a flowchart.
If you get all of the way to bit position zero and still have a non-zero result, then one of two things has happened. Either there was an error in one of your subtractions or you did not start off with a large enough number of bits. Remember that a fixed number of bits, n, can only represent an integer value up to 2n – 1. For example, if you are trying to convert 31210 to unsigned binary, eight bits will not be enough because the highest value eight bits can represent is 28 – 1 = 25510. Nine bits, however, will work because its maximum unsigned value is 29 – 1 = 51110.
Figure 1.6Decimal to Unsigned Binary Conversion Flow Chart
Example1Convert the decimal value 13310 to an 8 bit unsigned binary number.
Solution Since 13310 is less than 28 – 1 = 255, 8 bits will be sufficient for this conversion. Using Figure 4, we see that the largest power of 2 less than or equal to 13310 is 27 = 128. Therefore, we place a 1 in bit position 7 and subtract 128 from 133.
Bit position
133 – 128 = 5
Our new decimal value is 5. Since this is a non-zero value, our next step is to find the largest power of 2 less than or equal to 5. That would be 22 = 4. So we place a 1 in the bit position 2 and subtract 4 from 5.
Bit position
5 – 4 = 1
Our new decimal value is 1, so find the largest power of 2 less than or equal to 1. That would be 20 = 1. So we place a 1 in the bit position 0 and subtract 1 from 1.
Bit position
1 – 1 = 0
Since the result of our last subtraction is 0, the conversion is complete. Place zeros in the empty bit positions.
Bit position
And the result is:
13310 = 100001012
Example 2 Convert the decimal number 99 to its binary equivalent:
Divide 99 by 2. The quotient is 49 with a remainder of 1; indicate the 1 on the right.
Divide 49 by 2(the quotient from the previous division). The quotient is 24 with a remainder of 1, indicated on the right.
Divide 24 by 2. The quotient is 12 with a remainder of 0, as indicated .
Divide 12 by 2. The quotient is 6 with a remainder of 0, as indicated.
Divide 6 by 2. The quotient is 3 with a remainder of 0, as indicated.
Divide 3 by 2. The quotient is 1 with a remainder of 1, as indicated.
Divide 1 by 2. The quotient is 0 with a remainder of 1, as indicated. Since the quotient is 0, stop here.
The base 2 number is the numeric remainder reading from the last division to the first |
Web Results
www.math.com/school/subject2/lessons/S2U2L3GL.html
We have learned that, in in an algebraic expression, letters can stand for numbers. When we substitute a specific value for each variable, and then perform the operations, it's called evaluating the expression. Let's evaluate the expression 3y + 2y when 5 = y. Click on the steps to see how it's done.
www.mathplanet.com/education/pre-algebra/introducing-algebra/evaluate-expressions
To evaluate an algebraic expression, you have to substitute a number for each variable and perform the arithmetic operations. In the example above, the variable x is equal to 6 since 6 + 6 = 12. If we know the value of our variables, we can replace the variables with their values and then evaluate the expression.
Evaluating Problem Solving in Mathematics Effective assessment of problem solving in math requires more than a look at the answers students give. Teachers need to analyze their processes and get students to communicate their thinking. WALTER SZETELA AND CYNTHIA NICOL In its Curriculum and Evaluation Standards for School Mathematics,
www.mathpapa.com/calc/tutorial/evaluate-expressions
Learn how to use the Algebra Calculator to evaluate expressions. Example Problem Evaluate the expression 2x for x=3. How to Evaluate the Expression in Algebra Calculator. First go to the Algebra Calculator main page. Type the following: First type the expression 2x. Then type the @ symbol. Then type x=3. Try it now: 2x @ x=3 Clickable Demo
Learn to evaluate expressions in word problems to gain new information. ... Math · Algebra I · Algebra foundations · Evaluating expressions word problems. Evaluating expressions with variables word problems. CCSS Math: 6.EE.A.2, 6.EE.A.2c.
A local hospital is holding a raffle as a fundraiser. The individual cost of participating in the raffle is given by the following expression-- 5t plus 3, or 5 times t plus 3, where t represents the number of tickets someone purchases. Evaluate the expression when t is equal to 1, t is equal to 8, and t is equal to 10.
www.math10.com/en/problem-solver
Basic Math Plan. Basic Math Solver offers you solving online fraction problems, metric conversions, power and radical problems. You can find area and volume of rectangles, circles, triangles, trapezoids, boxes, cylinders, cones, pyramids, spheres. You can simplify and evaluate expressions, factor/multiply polynomials, combine expressions. |
# How do you divide (6x^4+6x^3-12x^2-7x-7)/(x-2) ?
Aug 6, 2016
I got:
$= 6 {x}^{3} + 18 {x}^{2} + 24 x + 41 + \frac{75}{x - 2}$
A fairly straightforward way to do it is synthetic division.
For this, you use the coefficients of each term. It's like regular long division, but faster/more compact.
$6 {x}^{4} + 6 {x}^{3} - 12 {x}^{2} - 7 x - 7$
$\implies$ $6 \text{ "6" "-12" "-7" } - 7$
The factor divided by is as if you were setting it equal to $0$ and solved for it. If we have $x - 2$, the factor in the upper-left square is $2$, because $x - 2 = 0 \implies x = 2$.
So, we begin with:
$\textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{\underline{2} |} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{- 12} , \textcolor{b l a c k}{- 7} , \textcolor{b l a c k}{- 7}\right) , \left(\textcolor{b l a c k}{+} , \textcolor{b l a c k}{\underline{\text{ "),color(black)(ul" "),color(black)(ul" "),color(black)(ul" "),color(black)(ul" }}}\right) , \left(\textcolor{b l a c k}{} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{}\right)\right]}$
Then, the basic steps are:
1. Bring down the first polynomial term.
2. Multiply by the divisor, place under the next polynomial term, moving left to right.
3. Add the current polynomial term and the current product from step 2, place underneath.
4. Repeat steps 2 and 3 until you've reached the last column and added the terms together already.
So, we would get:
$\textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{\underline{2} |} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{- 12} , \textcolor{b l a c k}{- 7} , \textcolor{b l a c k}{- 7}\right) , \left(\textcolor{b l a c k}{+} , \textcolor{b l a c k}{\underline{\text{ "),color(black)(ul" "),color(black)(ul" "),color(black)(ul" "),color(black)(ul" }}}\right) , \left(\textcolor{b l a c k}{} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{}\right)\right]}$
$\textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{\underline{2} |} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{- 12} , \textcolor{b l a c k}{- 7} , \textcolor{b l a c k}{- 7}\right) , \left(\textcolor{b l a c k}{+} , \textcolor{b l a c k}{\underline{\text{ "),color(black)(ul"12"),color(black)(ul" "),color(black)(ul" "),color(black)(ul" }}}\right) , \left(\textcolor{b l a c k}{} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{18} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{}\right)\right]}$
$\textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{\underline{2} |} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{- 12} , \textcolor{b l a c k}{- 7} , \textcolor{b l a c k}{- 7}\right) , \left(\textcolor{b l a c k}{+} , \textcolor{b l a c k}{\underline{\text{ "),color(black)(ul"12"),color(black)(ul"36"),color(black)(ul" "),color(black)(ul" }}}\right) , \left(\textcolor{b l a c k}{} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{18} , \textcolor{b l a c k}{24} , \textcolor{b l a c k}{} , \textcolor{b l a c k}{}\right)\right]}$
$\textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{\underline{2} |} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{- 12} , \textcolor{b l a c k}{- 7} , \textcolor{b l a c k}{- 7}\right) , \left(\textcolor{b l a c k}{+} , \textcolor{b l a c k}{\underline{\text{ "),color(black)(ul"12"),color(black)(ul"36"),color(black)(ul"48"),color(black)(ul" }}}\right) , \left(\textcolor{b l a c k}{} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{18} , \textcolor{b l a c k}{24} , \textcolor{b l a c k}{41} , \textcolor{b l a c k}{}\right)\right]}$
$\textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{\underline{2} |} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{- 12} , \textcolor{b l a c k}{- 7} , \textcolor{b l a c k}{- 7}\right) , \left(\textcolor{b l a c k}{+} , \textcolor{b l a c k}{\underline{\text{ "),color(black)(ul"12"),color(black)(ul"36"),color(black)(ul"48"),color(black)(ul"82}}}\right) , \left(\textcolor{b l a c k}{} , \textcolor{b l a c k}{6} , \textcolor{b l a c k}{18} , \textcolor{b l a c k}{24} , \textcolor{b l a c k}{41} , \textcolor{b l a c k}{75}\right)\right]}$
Now just re-assign the coefficients to the polynomial solution. Our answer had to have gone down one degree from the quartic that we started with.
Therefore, our answer is a cubic:
$= q \left(x\right) + r \left(x\right)$
$= \textcolor{b l u e}{\stackrel{q \left(x\right) , \text{quotient")overbrace(6x^3 + 18x^2 + 24x + 41) + stackrel(r(x), "remainder}}{\overbrace{\frac{75}{x - 2}}}}$ |
# In the beginning, there was the empty set
What would you say if I told you that in math everything is a set? That, whatever mathematical object you name, it can be defined using sets? And not sets, that contains some special ingredient somewhere deep inside - but by basically wrapping up empty sets and merging it with more wrapped empty sets in various degree of wrapping?
Because that’s how things are. If you are curious how to do it, let us take a small walk through the foundations of the set theory!
## Let’s start with some axioms
If we want to build up the whole universe with sets, we need to set up some ground rules to follow. Otherwise, we wouldn’t know what is a set and what is allowed and what is not.
To be honest, we don’t define a set. It just is. A thing that contains other things… or not. That’s basically what we can tell about a set: does it contains something or not.
Set(1, 2, 3).contains(2) // true
Set(1, 2, 3).contains(4) // false
### Axiom of extensionality
So, the first thing we can do with sets is taking them and checking if they are the same thing. Therefore we say, that two sets are equal (and they are the same set) if they contain the same elements.
$\forall_{X,Y} \forall_{z} [(z \in X \iff z \in Y) \implies X = Y]$
(For all possible (sets) $X$, $Y$, for all possible (elements) $z$ , $z$ belongs to $X$ if and only if $z$ belongs $Y$, implies that (set) $X$ is equal to $Y$).
The important takeaway here is that sets equality completely ignores the order. Because, you know, sets are unordered, you would have to define order (which we’ll do, another day).
Set(1, 2, 3) ==
Set(1, 3, 2) ==
Set(2, 1, 3) ==
Set(2, 3, 1) ==
Set(3, 1, 2) ==
Set(3, 2, 1)
Also, another property that is implied is that sets don’t allow you to repeat elements.
Set(1, 2, 3, 2) == Set(1, 2, 3)
### Axiom of empty set
There is an empty set ($\emptyset$).
$\exists_{\emptyset}\forall_x\neg(x \in \emptyset)$
If we want to build set(s) up, we need to start up with something. The empty set is that something.
The classical set theory does it in another way - uses axiom schema of specification, which tells that for each set you can define a subset. (It’s called schema because it has to be defined for each set in separation, but there is a template for it where you just need to fix the set).
### Axiom of pairing
If you have (sets) $x$ and $y$, you can create a set, that contains them both.
$\forall_x\forall_y\exists_A x \in A \land y \in A$
Set(x, y)
If $x=y$, then we are effectively building a $x$ singleton ($\{x\}$).
Set(x, x) == Set(x)
### Axiom of union
If you have two sets, you can combine/sum/union them into one (a sum/union set).
$\forall_X\forall_Y\exists_Z a \in X \lor a \in Y \iff a \in Z$
(For each $X$, $Y$ exists $Z$ that contains elements of both of them.)
Set(1) ++ Set(2) == Set(1, 2)
We can write it dawn using set sum operator
$Z = X \cup Y$
(Notice the similarity between $\lor$ and $\cup$ - it helps to remember what is sum $\cup$ and what is intersection $\cap$).
Together with the axiom of pairing it lets you build up a set out of any elements you want: you just made singletons or sets of pairs out of them and then unionize the resulting sets until you get the set you want.
Set(1, 2) ++ Set(3, 3) == Set(1, 2, 3)
### Axiom of regularity (axiom of foundation)
If we said, that set can contain itself, or some other set that would contain such set, things would get pretty hairy, pretty soon - it would easily lead to the paradox of set of all sets (Russell’s paradox).
To prevent that, we require that each set as well as each of its elements (if there are some; such elements are also sets) would be disjoint (share no common elements).
$\forall_X[\exists_A A \in X \implies \exists_{Y \in X} \neg(\exists_z z \in X \land z \in Y)]$
x = Set(x) // not allowed!
### Axiom schema of replacement
Let’s say we want to have a pair of things and we want to remember which one is first and which one is second. We can define it like this
$(a, b) = \{ \{a\}, \{a, b\} \}$
An inner set containing two elements defines our 2 elements and inner set with one element shows which one is first. This is Kuratowski’s ordered pair/tuple definition.
Now, if we had a set of such pairs, and we knew that there are no two pairs that have the same first element, then we could call such set a function.
$y = f(x) \iff (x, y) \in f$
The set of first elements would be a function’s domain, while the set of second values would be the function’s image or codomain.
Axiom schema of replacement is a way of stating that if way of creating function $F$ for set $A$ (domain), then its image $F[A]$ is also a set. It express it using logical predicate where $y = f(x) \land x \in A$ is denoted as $\phi(x,y,A)$ ($y$ is a value returned for an argument $x$ belonging to a domain $A$).
$\forall_A ( [ \forall_{x \in A} \exists!_y \phi(x, y, A) ] \implies \exists_B \forall_y [y \in B \iff \exists_{x \in A} \phi(x, y, A) ] )$
### Axiom of infinity
Let’s define $S(w) = w \cup \{w\}$ (union of $w$ set with $w$ singleton - achievable thanks to axiom of pairing and ). Then we can build a tower of $S$es:
$\emptyset, S(\emptyset), S(S(\emptyset)), ...$
Axiom of infinity says, that we can do this infinitely many times and obtain the final (infinite) set:
$\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}, \{\emptyset, \{\emptyset\}, \{\emptyset,\{\emptyset\}\}\}, ...\}$
(These are called von Neumann ordinals and we get back to them later).
To allow the existence of such set it’s enough if we allow that
$\exists_X [\emptyset \in X \land \forall_y(y\in X\implies S(y) \in X)]$
Of course, this set $X$ can contain more elements, than required above (von Neuman set is just a minimal example).
Having one infinite set, we can create other infinite sets just by applying other axioms on it.
### Axiom of power set
Let’s say we have a set $A = \{a,b,c\}$. It has 8 possible subsets:
$\emptyset, \{a\}, \{b\}, \{a\}, \{a,b\}, \{a,c\}, \{b,c\}, \{a,b,c\}$
Set of all possible subsets is called A’s power set. We call it $P(A)$. The axiom of power set states that for every set $A$ there exists a power set $P(A)$.
$\forall_A\exists_{P(A)}\forall_x[x\subseteq A\implies x\in P(A)]$
If our $A$ set has $n$ elements (or a cardinality of $n$, $n = \|A\|$), then the power set will have
• (first element yes/no) $\times$ (second element yes/no) $\times$ $...$ $\times$ (n-th element yes/no)
• $2 \times 2 \times ... \times 2$ ($n$ times)
• $2^n$
elements ($\|P(A)\| = 2^n = 2^{\|A\|}$). Its cardinality is $2$ power $A$’s cardinality thus the name.
### Axiom of choice
We defined a sum set and a tuple. So for completeness, let’s define a product set.
If we have sets $A$ and $B$ we can define a Cartesian product of $A \times B$ as a set of all pairs where the first element belongs to $A$ and second to $B$.
$A \times B = \{ (a,b): a \in A \land b \in B \}$
If we generalize tuple (e.g. by stating that $(a, b, c) = (a, (b, c))$, $(a, b, (c, d)) = (a, b, c, d)$ and so on) we can also generalize Cartesian product
$X_1 \times ... \times X_n = \{(x_1, ..., x_n): x_1 \in X_1, ..., x_n \in X_n \}$
It looks like product set’s cardinality should be the result of multiplication (product) of cardinalities of multiplied sets. So, if all of the sets are non-empty the cardinality of a product should also be non-zero and the set itself non-empty.
Now, if we looked at it from the other way round, it means that if the product is non-empty $X_1, ... X_n$ we can always pick n-tuple, that contains one element from each set (basically any element of a product set will do). We can also think of such tuple as a function from $X_i$ to $x_i \in X_i$. We would call such function a selector.
The axiom of choice basically tells, that for any set of non-empty sets you can find a selector.
$\forall_X [ \emptyset \not\in X \implies \exists f: X \rightarrow \bigcup X\quad \forall A\in X\ \land (f(A)\in A)]$
(For all sets of sets $X$, if $X$ doesn’t have an empty set, then there is a function $f$ from $X$ to sum of all X elements, such that for each set $A \in X$ it returns value belonging to $A$).
The axiom of choice is sometimes replaced by well-ordering theorem or Kuratowski–Zorn lemma.
With that, we have some foundations to build on.
## Peano arithmetics or how natural numbers came to be
Minimal requirements for building up natural numbers were defined in the 19th century by Giuseppe Peano. He presented that to the define numbers we need 2 things:
• something that will serve as a $0$,
• something that will return a value $n+1$ when we pass it $n$ - a successor function or $succ$.
So, in Peano arithmetic, to build up a $n$ number we need to apply $succ$ $n$ times:
$1 = succ(0)$ $2 = succ(1) = succ(succ(0))$ $3 = succ(2) = succ(succ(1)) = succ(succ(succ(0)))$
and so on.
So we could define natural numbers as a set made of $0$ as well as all results of applying $succ$ to other natural numbers. We did something similar with axiom of infinity - von Neuman set starts with an empty set and then created its successor with $s_{\emptyset} = S(\emptyset)$, then pass $s_{s_{\emptyset}}$ to $S$ and so on. It perfectly fits the requirement of Peano arithmetic! As a matter of the fact, these are called von Neumann ordinals and are one way of implementing it in set theory. Other would be Zermelo ordinals.
To make it a true arithmetic, we need some operations, e.g.:
• addition:
$n + 0 = n$ $n + succ(m) = succ(n) + m$
We basically move $succ$ from one side to the other, until one side is $0$ - then we have a result.
$1 + 2 = succ(0) + succ(succ(0)) =$ $= succ(succ(0)) + succ(0) =$ $= succ(succ(succ(0))) + 0 = 3$
• multiplication:
$n \times 0 = 0 \times n = 0$ $n \times 1 = 1 \times n = n$ $n \times succ(m) = n + (n \times m)$
Here, we basically unroll multiplication into a sequence of additions. And we have the addition already defined.
$2 \times 2 = succ(succ(0)) \times succ(succ(0)) =$ $= succ(succ(0)) + (succ(succ(0)) \times succ(0)) =$ $= succ(succ(0)) + succ(succ(0)) =$ $=succ(succ(succ(0))) + succ(0) =$ $=succ(succ(succ(succ(0)))) + 0 =$ $=succ(succ(succ(succ(0)))) = 4$
• comparison:
$0 < succ(n)$ $succ(n) > 0$ $0 = 0$ $succ(n) < succ(m) \iff n < m$ $succ(n) > succ(m) \iff n > m$ $succ(n) = succ(m) \iff n = m$
If you have $0$ of either side, it is obvious. If not, remove $succ$ from both sides and try again.
These definitions assume, that just as you are able to wrap things up, you can also unwrap, then and check if they match some pattern. In functional programming, it is called a pattern matching. In mathematics, I never saw any name for it as it is a granted property of the world.
## Integers, anyone?
OK, so we defined an infinite set of natural numbers $\N = {0, 1, 2, …}$. It has addition, multiplication, and comparison, but it doesn’t have subtraction. At least one that would be defined for all possible arguments. If we define it like:
$a - b = c \iff a = b + c$
and then we’ll just try to find 3 matching the equation, then:
$2 - 1 = x \iff 1 + x = 2$
has one solution for $x = 1$. But:
$1 - 2 = ?$
What would be the meaning of that? (I saw a definition of a subtraction in lambda calculus, that would return 0 - because it has to return something, but we would like something better than that).
So, let’s upgrade our numbers. Let’s write all pairs of natural numbers and draw them as points on a graph:
^
|(0,3) (1,3) (2,3) (3,3)
+
|(0,2) (1,2) (2,2) (3,2)
+
|(0,1) (1,1) (2,1) (3,1)
+
|(0,0) (1,0) (2,0) (3,0)
+------+------+------+------>
If we assume that $(a, b)$ represents $b - a$, then we’ll have the result defined for all tuples lying on diagonal $(n,n)$ and below it. We’ll also see, we can group tuples with the same result.
$I_0 = \{(0,0), (1,1), (2,2), (3,3), ...\}$ $I_1 = \{(1,0), (2,1), (3,2), (4,3), ...\}$ $I_2 = \{(2,0), (3,1), (4,2), (5,3), ...\}$
In general, if for any $(n, m)$ where $n,m \in \mathbb{N}$, $(n+1,m+1)$ would have the same subtraction, and so it will belong to the same group. If we connect the dots in the same group we get:
^
|(0,3) (1,3) (2,3) (3,3)
+ .*
| .*
|(0,2) (1,2) (2,2) (3,2)
+ .* .*
| .* .*
|(0,1) (1,1) (2,1) (3,1)
+ .* .* .*
| .* .* .*
|(0,0) (1,0) (2,0) (3,0)
+------+------+------+------>
Now, let’s say from now on we’ll identify a natural number with the group to in which it is a result of subtraction. As a matter of the fact, what we did here is creation of an equivalent class - we partitioned the whole set of pairs to smaller disjoints subsets, that altogether cover the whole original set $\mathbb{N} \times \mathbb{N}$, where each pair is identical to the other in some regard (here: same result of subtraction). Because they are the same (in a way), we might not distinguish between them, and even identify the whole set with them. And (as long as we won’t violate the property used to partition them), we can translate the operation on elements of the set to the operation on a whole set. E.g. if we add any pairs which represent 1 to any pair which represent 2, we receive a pair that must represent 3. Of course, as long as we define that addition properly.
So, from now on:
• addition:
$n + m = I_n + I_m = \{ (a+c, b+d): (a, b) \in I_n \land (c,d) \in I_m \}$
• multiplication:
$n \times m = I_n \times I_m = \{ (bd + ac, bc + ad): (a, b) \in I_n \land (c,d) \in I_m \}$
(Try to calculate $(b-a)(d-c)$ to check where it came from).
What we might have not paid attention so far, is that from the very beginning such definition defines classes of equivalence for all possible pairs - even those, that were not handled by the original definition of subtraction.
^
|(0,3) (1,3) (2,3) (3,3)
+ .* .* .*
| .* .* .*
|(0,2) (1,2) (2,2) (3,2)
+ .* .* .*
| .* .* .*
|(0,1) (1,1) (2,1) (3,1)
+ .* .* .*
| .* .* .*
|(0,0) (1,0) (2,0) (3,0)
+------+------+------+------>
As you probably guessed, these new numbers are negative numbers we missed.
^
|(0,3) (1,3) (2,3) (3,3)
-3+ .* .* .*
| .* .* .*
|(0,2) (1,2) (2,2) (3,2)
-2+ .* .* .*
| .* .* .*
|(0,1) (1,1) (2,1) (3,1)
-1+ .* .* .*
| .* .* .*
|(0,0) (1,0) (2,0) (3,0)
+------+------+------+------>
0 1 2 3
So, now we can reliably subtract things. But we still cannot reliably divide them. For that, we’ll need rational numbers.
## Rationals - still talking about tuples
Rational numbers are fractions. Dividend, divisor and a fraction bar between them.
$\frac{p}{q}: p, q \in \mathbb{Z} \land q \neq 0$
At this point, you should know, what we’re going to do: use tuples!
$\frac{p}{q} = (p, q): p, q \in \mathbb{Z} \land b \neq 0$
Again, we will introduce a equality classes, stating that:
$(p, q) = (n \times q, n \times q) \land n \in \mathbb{Z}-\{0\}$
meaning that you can reduce a fraction and it is still the same fraction.
Also, for compatibility: $\frac{p}{1} = p, p \in \mathbb{Z}$
Then, we can define operations:
• addition:
$(a, b) + (c, d) = (ad + bc, bd)$
• multiplication:
$(a, b) \times (c, d) = (ab, cd)$
• subtraction:
$(a, b) - (c, d) = (a, b) + (-c, d) = (a, b) + (c, -d)$
• division:
$(a, b) / (c, d) = (ad, bc)$
At this point, I’m skipping showing how comparison work, or things like neutral elements, but the curious reader has all the tools to figure them out.
## Real numbers - finally, something made up!
So far everything could be solved with tricks like wrapping things up in a specific way. With real numbers, things are more difficult. In a way, as for something based in eponymous reality, they are the first thing we cannot easily construct from smaller things.
This indirectly became an issue for Pythagoreans, when they found out that numbers like $\sqrt{2}$ or $\pi$ cannot be constructed from (finite amount of) rational numbers, no matter how you arrange them. It offended their sense of mathematical aesthetic. Even millenniums later we call such numbers irrationals. Yet, we need them to be able to describe positions and sizes in (among others) Euclidean spaces.
There are 2 popular ways to define real numbers using rationals: Dedekind’s cuts and Cauchy’s sequences.
Dedekind’s cuts uses bounds or partitions to define numbers. E.g. I am looking for the smallest number that, when squared, would be greater or equal to 2. In rational numbers, there is no such smallest element. We can get closer and closer to something that squared becomes 2, but it’s always slightly bigger.
Then we can state, that on e.g. numerical line we can reach such a point, and we need a good name for it. Since no rational fit it, we can name it $\sqrt{2}$ and call it a day. Every time we will see $\sqrt{2}$ we would be able to think, that this is a placeholder for something that fulfills the condition, that this is the smallest number that squared is greater or equal to 2 (or it’s a least upper bound of $i$ such that $i^2 >= 2$).
Problem with such definition is that we aren’t constructing anything, we just list the requirements, give a name to something that fulfills them, and from now on pretend that this is a thing (and, since it’s mathematics, we can often get away with it if the definition is not contradictory).
Cauchy’s sequences are sequences where elements become closer to each other as the sequence progresses. E.g.
$4, -4, 2, -2, 1, -1, \frac{1}{2}, -\frac{1}{2}, \frac{1}{4}, -\frac{1}{4}, ...$
We can easily design a sequence, that will approach something that is not a rational number. E.g.
$\frac{2}{1}, \frac{14}{10}, \frac{141}{100}, \frac{1414}{1000}, ...$
could be a Cauchy’s sequence of rationals that gets closer and closer to $\sqrt{2}$.
Then, we could identify each sequence, with a number it is approaching and this way we would design real numbers. An infinite sequence is always something that we can construct (as opposed to hand-waving with partitions).
How to operate on such sequences? Well, if you add, subtract, multiply or divide values on corresponding positions in 2 sequences, you should get a sequence that approaches the result of addition, subtraction, multiplication or division.
For comparison, you can subtract one sequence from the other and check the sign.
Now, you can construct and use every real number.
## Beyond real
So far everything was more or less convoluted way of wrapping up empty set and sets of the empty set, and sets of sets of the empty set, etc. We build this way natural numbers, integers, rationals and reals, tuples and functions. Is everything made up of sets? Well, in set theory, basically yes.
Vectors and matrices could be defined as functions from elements coordinates to the element (sets). Or tuples (vectors) and tuples of tuples with matching sizes (matrices). Complex numbers are, well, tuples of real and imaginary part. Also coordinates on a complex plane. Quaternions? Matrices. Algebras are sets of values with operations on these values (functions, thus also sets). Graphs? A tuples of vertices’ set and nodes’ set.
At some point, mathematicians introduced the idea of a class - a collection of objects defined by properties objects must have. They did it because sets have to be constructible from bottom-up (one way or the other), which excludes e.g. ability to build the set of all sets, which might be limiting. The class of all sets is a thing. Also, an example of a proper class - a class which is not a set.
However, besides proper classes (and proper classes of proper classes, etc), everything based on a set theory operates on sets. And each of these sets originates from an empty set wrapped in different ways. Lambda calculus needs only $\alpha$-congruences and $\beta$-reductions to implement everything, category theory has objects and arrows. And a set theory achieves it all with the empty sets and wrapping. |
# Factoring Quadratic Equations (a = 1 Case)
Topic: Factoring Quadratic Equations When a = 1
Introduction
Another factoring method for quadratic equations will be shown here. We deal with the case of in . If then another method may be needed. (Also if then we no longer have a quadratic equation.)
It is assumed that the reader is familiar with common factoring.
The Factoring Method
To illustrate the method, an example will be used.
Suppose that we are given . We need two numbers and which are factors of 4 and satisfy = 4 and . The two numbers which fit that criteria are 1 and 4 (or 4 and 1) since and .
Using the numbers 1 and 4 we can factor into . The equation follows the factored form format of .
To check that is indeed the factored form of , we use the FOIL method when multiplying binomials.
There are times when . As an example, the factored form of is .
The General Method
Given a quadratic equation of the form , we can factor it into the form . To determine what and , we seek two numbers and such that (since here and .
The examples below will illustrate how the general method works.
Examples
Example One
Factor .
Two factors of 12 which multiply to 12 are 1 and 12, 2 and 6, 3 and 4 (you can inculde the reverse ordered pairs too). Out of the pairs which multiply to 12, 3 and 4 sum to 7. The quadratic equation would be factored as .
Example Two
Factor .
Factors of 99 are 1 and 99, 3 and 33, 9 and 11. The two numbers which multiply together to get 99 and add together to get 20 are 9 and 11. The factored form of would be .
Example Three
Factor by factoring out as a common factor first.
From factoring out the (-1), we have a factorable quadratic in the brackets. The only factors of 2 are 1 and 2. We can factor into .
Example Four
Factor .
In this equation we have , and in . Unlike examples one and two, we have to be careful with signs. I prefer to ignore the negative and look at the factors of .
Factors of 6 are 1 and 6, 2 and 3. Now, we consider the signs. One of the two numbers in the factor pair is negative. The pair 6 and 1 would not work as 6 – 1 = 5 and 1 – 6 = -5. With 2 and 3, we have 2 – 3 = -1 and 3 – 2 = 1. We go with 2 and -3.
The factored form of is .
Practice Problems
Here are some practice problems to build understanding.
1) Factor .
2) Factor .
3) Factor .
4) Factor .
5) Factor .
6) In , factor a (-1) first and then factor using the method described in this page.
7) Using common factoring first, factor .
1)
2)
3)
4)
5)
6)
7) |
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# Class 10 NCERT Solutions- Chapter 11 Constructions – Exercise 11.1
### Question 1. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Solution:
Steps of construction:
To divide the line segment of 7.6 cm in the ratio of 5 : 8.
Step 1. Draw a line segment AB of length 7.6 cm.
Step 2. Draw a ray AC which forms an acute angle with the line segment AB.
Step 3. Mark the points = 13 as (5+8=13) points, such as A1, A2, A3, A4 …….. A13, on the ray AC such that it becomes AA1 = A1A2 = A2A3 and such like this.
Step 4. Now join the line segment and the ray, BA13.
Step 5. Hence, the point A5, construct a line parallel to BA13 which makes an angle equal to ∠AA13B.
Step 6. Point A5 intersects the line AB at point X.
Step 7. X is that point which divides line segment AB into the ratio of 5:8.
Step 8. Thus, measure the lengths of the line AX and XB. Hence, it measures 2.9 cm and 4.7 cm respectively.
Justification:
The construction can be justified by proving that
From construction, we have A5X || A13B. By the Basic proportionality theorem for the triangle AA13B, we will get
….. (1)
By the figure we have constructed, it can be seen that AA5 and A5A13 contains 5 and 8 equal divisions of line segments respectively.
Thus,
… (2)
Comparing the equations (1) and (2), we get
Thus, Justified.
### Question 2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.
Solution:
Steps of Construction:
Step 1. Draw a line segment XY which measures 4 cm, So XY = 4 cm.
Step 2. Taking point X as centre, and construct an arc of radius 5 cm.
Step 3. Similarly, from the point Y as centre, and draw an arc of radius 6 cm.
Step 4. Thus, the following arcs drawn will intersect each other at point Z.
Step 5. Now, we have XZ = 5 cm and YZ = 6 cm and therefore ΔXYZ is the required triangle.
Step 6. Draw a ray XA which will make an acute angle along the line segment XY on the opposite side of vertex Z.
Step 7. Mark the 3 points such as X1, X2, X3 (as 3 is greater between 2 and 3) on line XA such that it becomes XX1 = X1X2 = X2X3.
Step 8. Join the point YX3 and construct a line through X2 which is parallel to the line YX3 that intersect XY at point Y’.
Step 9. From the point Y’, construct a line parallel to the line YZ that intersect the line XZ at Z’.
Step 10. Hence, ΔXY’Z’ is the required triangle.
Justification:
The construction can be justified by proving that
From the construction, we get Y’Z’ || YZ
∴ ∠XY’Z’ = ∠XYZ (Corresponding angles)
In ΔXY’Z’ and ΔXYZ,
∠XYZ = ∠XY’Z (Proved above)
∠YXZ = ∠Y’XZ’ (Common)
∴ ΔXY’Z’ ∼ ΔXYZ (From AA similarity criterion)
Therefore,
…. (1)
In ΔXXY’ and ΔXXY,
∠X2XY’ =∠X3XY (Common)
From the corresponding angles, we get,
∠AA2B’ =∠AA3B
Thus, by the AA similarity criterion, we get
ΔXX2Y’ and XX3Y
So,
Therefore, ……. (2)
From the equations (1) and (2), we obtain
It is written as
XY’ =
Y’Z’ =
XZ’=
Therefore, justified.
### Question 3. Construct a triangle with sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle
Solution:
Steps of construction:
Step 1. Construct a line segment XY =5 cm.
Step 2. By taking X and Y as centre, and construct the arcs of radius 6 cm and 5 cm respectively.
Step 3. These two arcs will intersect each other at point Z and hence ΔXYZ is the required triangle with the length of sides as 5 cm, 6 cm, and 7 cm respectively.
Step 4. Construct a ray XA which will make an acute angle with the line segment XY on the opposite side of vertex Z.
Step 5. Pinpoint the 7 points such as X1, X2, X3, X4, X5, X6, X7 (as 7 is greater between 5 and 7), on the line XA such that it becomes XX1 = X1X2 = X2X3 = X3X4 = X4X5 = X5X6 = X6X7
Step 6. Join the points YX5 and construct a line from X7 to YX5 that is parallel to the line YX5 where it intersects the extended line segment XY at point Y’.
Step 7. Now, construct a line from Y’ the extended line segment XZ at Z’ that is parallel to the line YZ, and it intersects to make a triangle.
Step 8. Hence, ΔXY’Z’ is the needed triangle.
Justification:
The construction can be justified by proving that
XY’ =
Y’Z’ =
XZ’=
By the construction, we have Y’Z’ || YZ
Therefore,
∠XY’Z’ = ∠XYZ {Corresponding angles}
In ΔXY’Z’ and ΔXYZ,
∠XYZ = ∠XY’Z {As shown above}
∠YXZ = ∠Y’XZ’ {Common}
Therefore,
ΔXY’Z’ ∼ ΔXYZ { By AA similarity criterion}
Therefore,
…. (1)
In ΔXX7Y’ and ΔXX5Y,
∠X7XY’=∠X5XY (Common)
From the corresponding angles, we will get,
∠XX7Y’=∠XX5Y
Hence, By the AA similarity criterion, we will get
ΔXX2Y’ and XX3Y
Thus,
Hence, ……. (2)
From the equations (1) and (2), we obtain
It can be also shown as
XY’ =
Y’Z’ =
XZ’=
Thus, justified.
### Question 4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are times the corresponding sides of the isosceles triangle
Solution:
Steps of construction:
Step 1. Construct a line segment YZ of 8 cm.
Step 2. Now construct the perpendicular bisector of the line segment YZ and intersect at the point A.
Step 3. Taking the point A as centre and draw an arc with the radius of 4 cm which intersect the perpendicular bisector at the point X.
Step 4. Join the lines XY and XZ and the triangle is the required triangle.
Step 5. Construct a ray YB which makes an acute angle with the line YZ on the side opposite to the vertex X.
Step 6. Mark the 3 points Y1, Y2 and Y3 on the ray YB such that YY1 = Y1Y2 = Y2Y3
Step 7. Join the points Y2Z and construct a line from Y3 which is parallel to the line Y2Z where it intersects the extended line segment YZ at point Z’.
Step 8. Now, draw a line from Z’ the extended line segment XZ at X’, that is parallel to the line XZ, and it intersects to make a triangle.
Step 9. Hence, ΔX’YZ’ is the required triangle.
Justification:
The construction can be justified by proving that
X’Y =
YZ’ =
X’Z’=
By the construction, we will obtain X’Z’ || XZ
Therefore,
∠ X’Z’Y = ∠XZY {Corresponding angles}
In ΔX’YZ’ and ΔXYZ,
∠Y = ∠Y (common)
∠X’YZ’ = ∠XZY
Therefore,
ΔX’YZ’ ∼ ΔXYZ {By AA similarity criterion}
Hence,
Thus, the corresponding sides of the similar triangle are in the same ratio, we get
Thus, justified.
### Question 5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
Solution:
Steps of construction:
Step 1. Construct a ΔXYZ with base side YZ = 6 cm, and XY = 5 cm and ∠XYZ = 60°.
Step 2. Construct a ray YA that makes an acute angle with YZ on the opposite side of vertex X.
Step 3. Mark 4 points (as 4 is greater in 3 and 4), such as Y1, Y2, Y3, Y4, on line segment YA.
Step 4. Join the points Y4Z and construct a line through Y3, parallel to Y4Z intersecting the line segment YZ at Z’.
Step 5. Construct a line through Z’ parallel to the line XZ which intersects the line XY at X’.
Step 6. Therefore, ΔX’YZ’ is the required triangle.
Justification:
The construction can be justified by proving that
Since here the scale factor is ,
We need to prove
X’Y =
YZ’ =
X’Z’=
From the construction, we will obtain X’Z’ || XZ
In ΔX’YZ’ and ΔXYZ,
Therefore,
∠X’Z’Y = ∠XZY {Corresponding angles}
∠Y = ∠Y {common}
Therefore,
ΔX’YZ’ ∼ ΔXYZ {By AA similarity criterion}
Thus, the corresponding sides of the similar triangle are in the same ratio, we get
Therefore,
Thus, it becomes
Hence, justified.
### Question 6. Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC.
Solution:
To find ∠Z:
Given:
∠Y = 45°, ∠X = 105°
∠X+∠Y +∠Z = 180° {Sum of all interior angles in a triangle is 180°}
105°+45°+∠Z = 180°
∠Z = 180° − 150°
∠Z = 30°
Thus, from the property of triangle, we get ∠Z = 30°
Steps of construction:
Step 1. Construct a ΔXYZ with side measures of base YZ = 7 cm, ∠Y = 45°, and ∠Z = 30°.
Step 2. Construct a ray YA makes an acute angle with YZ on the opposite side of vertex X.
Step 3. Mark 4 points (as 4 is greater in 4 and 3), such as Y1, Y2, Y3, Y4, on the ray YA.
Step 4. Join the points Y3Z.
Step 5. Construct a line through Y4 parallel to Y3Z which intersects the extended line YZ at Z’.
Step 6. Through Z’, construct a line parallel to the line YZ that intersects the extended line segment at Z’.
Step 7. Hence, ΔX’YZ’ is the required triangle.
Justification:
The construction can be justified by proving that
Here the scale factor is , we have to prove
X’Y =
YZ’ =
X’Z’=
From the construction, we obtain X’Z’ || XZ
In ΔX’YZ’ and ΔXYZ,
Therefore.
∠X’Z’Y = ∠XZY {Corresponding angles}
∠Y = ∠Y {common}
Therefore,
ΔX’YZ’ ∼ ΔXYZ {By AA similarity criterion}
Since the corresponding sides of the similar triangle are in the same ratio, it becomes
Therefore,
We get,
Thus, justified.
### Question 7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.
Solution:
Given:
The sides other than hypotenuse are of lengths 4cm and 3cm. Hence, the sides are perpendicular to each other.
Step of construction:
Step 1. Construct a line segment YZ =3 cm.
Step 2. Now measure and draw ∠= 90°
Step 3. Now taking Y as centre and draw an arc with the radius of 4 cm and intersects the ray at the point Y.
Step 4. Join the lines XZ and the triangle XYZ is the required triangle.
Step 5. Construct a ray YA makes an acute angle with YZ on the opposite side of vertex X.
Step 6. Mark 5 such as Y1, Y2, Y3, Y4, on the ray YA such that YY1 = Y1Y2 = Y2Y3= Y3Y4 = Y4Y5
Step 7. Join the points Y3Z.
Step 8. Construct a line through Y5 parallel to Y3Z which intersects the extended line YZ at Z’.
Step 9. Through Z’, draw a line parallel to the line XZ that intersects the extended line XY at X’.
Step 10. Therefore, ΔX’YZ’ is the required triangle.
Justification:
The construction can be justified by proving that
Here the scale factor is , we need to prove
X’Y =
YZ’ =
X’Z’=
From the construction, we obtain X’Z’ || XZ
In ΔX’YZ’ and ΔXYZ,
Therefore,
∠X’Z’Y = ∠XZY {Corresponding angles}
∠Y = ∠Y {common}
Therefore,
ΔX’YZ’ ∼ ΔXYZ {By AA similarity criterion}
Since the corresponding sides of the similar triangle are in the same ratio, it becomes
Therefore,
So, it becomes
Therefore, justified.
My Personal Notes arrow_drop_up
Related Tutorials |
Title
Intermediate Algebra
Tutorial 2: Algebraic Expressions
Answer/Discussion to 1a when y = 2.2 In this problem we will be replacing y with 2.2 and then multiply:
*Replace y with 2.2 *( ) is another way to write multiplication *Multiply numbers together
Answer/Discussion to 1b when In this problem, we will replacing a with 3/5 and b with 2/3 and then adding the two together:
*Replace a with 3/5 and b with 2/3 *Mult. top and bottom of 1st fraction by 3 to get LCD of 15 *Mult. top and bottom of 2nd fraction by 5 to get LCD of 15 *Add the fractions together
Answer/Discussion to 1c You need to know how much area of carpet to purchase to carpet your rectangular floor. You will need to use the algebraic expression lw to find it's area, where l represents the length of the rectangular floor and w is the width of it. If your rectangular floor is 15 feet long and 8 feet wide, find how much area of carpet you would need to cover it. In this problem we will be replacing l with 15 and w with 8 and multiplying the two together:
*Replace l with 15 and w with 8 *Multiply the two numbers together
Answer/Discussion to 2a 15 more than a number. In this problem what mathematical operation would we substitute the phrase 'more than' with? I hope you said addition. Once again, we have an unknown number so we will replace the phrase 'a number' with the variable x. Putting it all together we get:
15 more than a number * 'more than' = + *'a number' = variable x
Answer/Discussion to 2b The quotient of 7 and the product of 3 and a number. First of all the term quotient is going to be replaced with what mathematical operation? If you said division you are correct!! Note how 7 immediately follows the phrase the quotient of, this means that 7 is going to be in the numerator. The phrase that immediately follows the word quotient is going to be in the numerator of it. After the word 'and' you have the phrase the product of 3 and a number. That is the second part of your quotient which means it will go in the denominator. And what operation will we have when we do write that product down below? Multiplication would be the correct response to that. Let's see what we get when we put all of this together:
The quotient of 7 and the product of 3 and a number *'quotient' = division *'product' = multiplication *'a number' = variable x
Answer/Discussion to 2c 17 less than the sum of a number and 8. First of all we have the phrase less than which mathematically translates as what? If you said subtraction you are doing great!! Again, you need to be careful with this phrase - it is very tempting to start off with 17 and put your subtraction sign after the 17 - however, think about it, if you want 17 less than something you are 17 below it, in order to be 17 below something you would have to subtract the 17!!! So you would not have 17 minus but minus 17 as PART of your expression. The other part of the expression involves the phrase 'the sum of', which would be what mathematical operation? Addition would be the correct response to that. And as always, when we have a phrase like 'a number' where we have a unknown number, we replace that with a variable. Let's see what we have:
17 less than the sum of a number and 8 *'less than' = - *'sum' = + *'a number' = variable x
Last revised on June 10, 2011 by Kim Seward. |
Cones - Surface Area and Volume - The Shape of the World - Basic Math and Pre-Algebra
## Basic Math and Pre-Algebra
PART 3. The Shape of the World
CHAPTER 16. Surface Area and Volume
Cones
If cylinders behave a lot like prisms, it’s not surprising that cones have a lot in common with pyramids. Like cylinders, cones have a circle as their base. Like pyramids, their lateral surface slopes to a point. Trying to unroll a cone is a bit trickier than taking the label off a can, as you did with the lateral area of a cylinder. If you try to unroll the cone’s lateral area, you’ll find it’s part of a circle, but how big a part depends on the size of the base circle.
A cone is a solid with a circular base and a lateral surface that slopes to a point.
Surface Area
The surface area of a pyramid is equal to the area of the base plus the combined area of all the surrounding triangles. That lateral area, you learned, was equal to 1/2 x perimeter x slant height. For a cone, the surface area is still the area of the base plus the lateral area. The base is a circle, so the area of the base is πr2. If the slant height is l, the lateral area is a fraction of the area of a circle with area πl2. What fraction? The fraction you create when you put the circumference of the base, 2πr, over the circumference of a circle with radius equal to the slant height. The lateral area is πrl, so the surface area is SA = πr2 + πrl.
To find the surface area of a cone with a radius of 8 cm and a slant height of 10 cm, find the area of the base. πr2 = π(82) = 64π, and then calculate πrl = π(8)(10) = 80π. The surface area is 64π + 80π = 144π square centimeters.
Volume
The volume of a pyramid is one-third of the volume of a prism with the same base and height. Conveniently, the volume of a cone is one-third of the volume of a cylinder with the same base and height. For a cone,
To find the volume of a cone 9 meters high with a base that has a radius of 8 meters, follow the formula.
The volume is 192^ cubic meters, or approximately 603.2 cubic meters.
CHECK POINT
Find the surface area and volume of each cone. SA = πr2 + πrl and
26. A cone with radius 10 cm, height 24 cm, and slant height 26 cm.
27. A cone with diameter 8 inches, height 3 inches, and slant height 5 inches.
28. A cone with circumference 16π cm, height 6 cm, and slant height 10 cm.
29. A cone with radius 12 inches and height 5 inches. (Hint: Use the Pythagorean theorem to find the slant height.)
30. A cone with base area 324π cm and a height of 24 cm.
|
### Archive
Posts Tagged ‘stationary points’
## Stationary Points (Maximum and Minimums) and Differentiation
On a graph a stationary point is any point where the gradient is 0 so where the graph is flat. For example the graph y=x2 has one stationary point at the origin.
### Finding the Stationary Points
We know that stationary point occur when the gradient is 0 so when the derivative of the graph is 0, so in order to find the stationary points we but first differentiate the curve.
For example lets consider the graph $y = 3x^2 + 2x - 7$. We cab differentiate this to find
$\frac{dy}{dx} = 6x + 2$
We must then equate the derivative to 0 and solve the resulting equation. This is because we are trying to find the points where the gradient is zero and these point occur exactly at the solutions of the equation we have formed.
So in our example we form the equation
$6x + 2 = 0$
by equating our expression for $\frac{dy}{dx}$, $6x + 2$, to 0
Solving this equation we find that stationary points occur exactly when
$x = \frac{2}{6} = \frac{1}{3}$
Note that there can be more than solution to this equation, each of which is a valid stationary point.
Finally we should also find the y co-ordinate for the stationary point by putting this value of x into the initial equation. So for this example $y= 3 \cdot \frac{1}{3}^2 + 2 \cdot \frac{1}{3} - 7 = -6$
So the only stationary point is at $(\frac{1}{3},-6)$
### Nature of Stationary Points
The nature of a stationary point simply means what the graph is doing around it and are characterised by the second derivative, $\frac{d^{2}y}{ dx^2}$ (found by differentiating the derivative). There are three types of stationary point:
1. Maximum Points: These are stationary points where the graph is sloping down on either side of the stationary point (a sad face type of curve).
Here ${d^{2}y}{dx^2} < 0$
2. Minimum Points: These are stationary where the graph is sloping upwards on either side of the point (a happy face)
Here ${d^{2}y}{dx^2} > 0$
3. Point of Inflection: Here the direction of the slope of the graph is the same either side of the stationary point, it can be in either direction.
At a point of inflection ${d^{2}y}{dx^2} = 0$ but ${d^{2}y}{dx^2} = 0$ isn’t enough to ensure that a point really is a point of inflection as it could still be a maximum or minimum point
4. Checking the nature of a Stationary Point when ${d^{2}y}{dx^2} = 0$
In this case the easiest thing to do is look a small distance either side of the point and see whether the y value is greater than or less than that of the stationary point. You can then draw yourself a picture to see what it is. For example if they are both greater than the stationary point you know it is a minimum point, but if one is greater and one is less than it is a point of inflection
Warning: checking points either side does not guarantee the correct result as there may be another stationary point or a break in the graph between where you are checking and the stationary point so you should always check using the derivatives if possible |
RATIO AND PROPORTION
Inverse Proportion
(Teacher Copy)
### Activity 12: An Example of Inverse Proportion
• Students use a fulcrum to find inverse proportions.
• To set up and solve inverse proportions.
### Introduction:
Think back to earlier lessons in this unit and answer the following questions:
• What happens to the circumference of a circle when its diameter increases? The circumference increases.
• What happens to the distance traveled by your bicycle when the number of wheel revolutions increases? Distance increases. What happens when the number of pedal revolutions decreases? Distance decreases.
Think about your everyday experiences. In each of the cases below, what happens to the quantity on the right when the quantity on the left increases? And when it decreases?
unit cost total cost time traveled (constant speed) distance traveled distance in feet distance in meters size of a gas tank cost to fill up
Because these are all direct proportions, both either increase together or decrease together. More specifically, they increase or decrease at the same rate.
There are other quantities that are also related in a different way. Consider the following variables:
speed of a car time spent traveling number of workers time needed to finish a task length of a hammer handle effort needed to pull a nail
How are they related? What happens to the quantities on the right when those on the left increase? When one quantity increases, the other decreases.
### Part A
Weight (W2) (unit: grams) 10 20 40 60 80 100 Distance (D2) (unit: mm) 40 20 10 6 2/3 5 4
This table of weight and distance is not of direct proportions because as weight increases, distance decreases (distance would increase if these were directly proportional).
### Part B
In the first table of Part B, W2 X D2 = 400 in each case.
Students should say that as weight increases, distance decreases.
In the second table of Part B, W2 X D2 = 600 in each case.
### Closing Discussion
How can you tell when a relationship between two numbers is a direct proportion, an inverse proportion, or not a proportion at all? If dividing values in corresponding columns of a table always gives the same number, there is a direct proportion. If multiplying values in the corresponding columns of a table always gives the same number, there is a direct proportion. Otherwise, there is no proportion.
Think of a situation where two quantities both increase but are not in direct proportion. Explain. An example of such a proportion is your height over time. Both the amount of time and your height increase, but you do not grow at the same rate over your lifetime.
Think of a situation where one quantity goes up while another goes down but the two quantities are not an inverse proportion. Explain. An example of this type of proportion is your ability to run over a long distance. As the amount of distance increases, you run more and more slowly, but your running speed decreases at a faster rate than the rate of increase in distance.
### Activity 13: Solving Inverse Proportions
• Students set up and solve word problems involving inverse proportions.
To set up and solve inverse proportions.
### Introduction:
At this point, you should have realized that in an inverse proportion, the two terms of each ratio (the numerator and denominator if the ratio is written as a fraction) always multiply to give the same value. Suppose you think it will take you 6 hours to paint your room by yourself without stopping. We can write this as:
The two terms of the fraction (1 and 6) multiply to give you 6. With a friend, it should only take 3 hours. When two people work, you can write
Again, the product of the two terms is 6. This relationship always holds in an inverse proportion, so you can always solve an inverse proportion problem by looking for the number that is the product of the two terms.
Let's try one more example. Suppose Tom and Vickie are working together on a term paper. Tom, who can type 30 words per minute, can type the paper in 2.6 hours. If Vickie, who can type 22 words per minute, decides to help Tom by working on the second part of the paper, how long will it take the two of them working together to type the paper? We can write a fraction representing Tom working alone.
Multiply the two terms of the fraction together. What did you get? When Tom and Vickie work together, their combined typing speed is 52 words per minute. Knowing that the typing speed multiplied by the number of hours must equal 78, find the time needed for Tom and Vickie to type the paper together.
### Part A
Weight (W2) (unit: grams) 10 50 200 1 4 8 Distance (D2) (unit: cm) 40 8 2 400 100 50
### Part B
1. 4.2 hours
2. 15 mph
3. 1.4 hours
4. \$100,750
a. 12.1 hours
b. »7 hours (6.98 hours)
5. 500 cubic feet of water
### Closing Discussion
Look back to question 2 of Part B. Compare the different ways that each group solved the problem. Did all groups get the same answer? Was one way faster? Which of the procedures used is easiest to remember? Answers will vary |
## Labels
### 3º ESO - ECUACIONES DE PRIMER Y SEGUNDO GRADO
Material complementario para repasar el tema de ecuaciones de primer y segundo grado de 3º ESO
## Simultaneous Equations
Simultaneous equations and linear equations, after studying this section, you will be able to:
• solve simultaneous linear equations by substitution
• solve simultaneous linear equations by substitution
• solve simultaneous linear equations by elimination
• solve simultaneous linear equations using straight line graphs
If an equation has two unknowns, such as 2y + x = 20, it cannot have unique solutions. Two unknowns require two equations which are solved at the sametime (simultaneously) − but even then two equations involving two unknowns do not always give unique solutions.
The video below works through examples of simultaneous equations. The step-by-step example shows how to group like terms and then add or subtract to remove one of the unknowns, to leave one unknown to be solved.
This method is called solution by substitution.
It involves what it says − substitution − using one of the equations to get an expression of the form ‘y = …’ or ‘x = …’ and substituting this into the other equation. This gives an equation with just one unknown, which can be solved in the usual way. This value is then substituted in one or other of the original equations, giving an equation with one unknown.
Example
Solve the two simultaneous equations:
2y + x = 8 [1]
1 + y = 2x [2]
from [2] y = 2x -1 ← subtract 1 from each side
Substituting this value for y into [1] gives:
2(2x – 1) + x = 8
4x – 2 + x = 8 ← expand the brackets
5x – 2 = 8 ←tidy up
5x = 10 ←Add 2 to each side
x = 2 ←By dividing both sides by 5 the value of x is found.
Substitute the value of x into y = 2x – 1 gives
y = 4 - 1 = 3
So x = 2 and y = 3
NOTE:
• It is a good idea to label each equation. It helps you explain what you are doing − and may gain you method marks.
• This value of x can be substituted into equation [1] or [2], or into the expression for y: y = 2x − 1.
• Choose the one that is easiest!
• As a check, substitute the values back into each of the two starting equations.
The second method is called solution by elimination.
NOTE:The method is not quite as hard as it first seems, but it helps if you know why it works.
It works because of two properties of equations:
• Multiplying (or dividing) the expression on each side by the same number does not alter the equation.
• Adding two equations produces another valid equation:
e.g. 2x = x + 10 (x = 10) and x − 3 = 7 (x also = 10).
Adding the equations gives 2x + x − 3 = x + 10 + 7 (x also = 10).
The object is to manipulate the two equations so that, when combined, either the x term or the y term is eliminated (hence the name) − the resulting equation with just one unknown can then be solved:
Here we will manipulate one of the equations so that when it is combined with the other equation either the x or y terms will drop out. In this example the x term will drop out giving a solution for y. This is then substituted into one of the otiginal equations.
Label your equations so you know which one your are working with at each stage.
Equation [1] is 2y + x = 8
Equation [2] is 1 + y = 2x
Rearrange one equation so it is similar to the other.
[2] y – 2x = -1
also 2 x [1] gives 4y + 2x = 16 which we call [3]
[2] y – 2x = -1
[3] 4y +2x = 16
[2] + [3] gives 5y = 15
so y = 3
substituting y = 3 into [1] gives 1 + (3) = 2x
so 2x = 4, giving x = 2 and y = 3
Solving simultaneous linear equations using straight line graphs
The 2 lines represent the equations '4x - 6y = -4' and '2x + 2y = 6'.
There is only one point the two equations cross.
Because the graphs of 4x - 6y = 12 and 2x + 2y = 6 are straight lines, they are called linear equations.
### TRABAJAMOS LA AUTOESTIMA I
LA PEQUEÑA LUCIÉRNAGA
Había una vez una comunidad de luciérnagas que habitaba el interior de un gigantesco lampati, uno de los árboles más majestuosos y antiguos de Tailandia. Cada noche, cuando todo se volvía oscuro y apenas se escuchaba el leve murmurar de un cercano río, todas las luciérnagas salían del árbol para mostrar al mundo sus maravillosos destellos. Jugaban a hacer figuras con sus luces, bailando al son de una música inventada para crear un sinfín de centelleos luminosos más resplandeciente que cualquier espectáculo de fuegos artificiales.
Pero entre todas las luciérnagas del lampati había una muy pequeñita a la que no le gustaba salir a volar.
– No, hoy tampoco quiero salir a volar -decía todos los días la pequeña luciérnaga-. Id vosotros que yo estoy muy bien aquí en casita.
Tanto sus padres como sus abuelos, hermanos y amigos esperaban con ilusión la llegada del anochecer para salir de casa y brillar en la oscuridad. Se divertían tanto que no comprendían por qué la pequeña luciérnaga no les quería acompañar. Le insistían una y otra vez, pero no había manera de convencerla. La pequeña luciérnaga siempre se negaba.
-¡Que no quiero salir afuera! -repetía una y otra vez-. ¡Mira que sois pesados!
Toda la colonia de luciérnagas estaba muy preocupada por su pequeña compañera.
-Tenemos que hacer algo -se quejaba su madre-. No puede ser que siempre se quede sola en casa sin salir con nosotros.
-No te preocupes, mujer -la consolaba el padre-. Ya verás como cualquier día de estos sale a volar con nosotros.
Pero los días pasaban y pasaban y la pequeña luciérnaga seguía encerrada en su cuarto.
Una noche, cuando todas las luciérnagas habían salido a volar, la abuela de la pequeña se le acercó y le preguntó con mucha delicadeza:
-¿Qué es lo que ocurre, mi pequeña? ¿Por qué no quieres venir nunca con nosotros a brillar en la oscuridad?
-Es que no me gusta volar-, respondió la pequeña luciérnaga.
-Pero, ¿por qué no te gusta volar ni mostrar tu maravillosa luz? -insistió la abuela luciérnaga.
-Pues… -explicó al fin la pequeña luciérnaga-. Es que para qué voy a salir si nunca podré brillar tanto como la luna. La luna es grande, y muy brillante, y yo a su lado no soy nada. Soy tan diminuta que en comparación parezco una simple chispita. Por eso siempre me quedo en casa, porque nunca podré brillar tanto como la luna.
La abuela había escuchado con atención las razones de su nieta, y le contestó:
-¡Ay, mi niña! hay una cosa de la luna que debería saber y, visto lo visto, desconoces. Si al menos salieras de vez en cuando, lo habrías descubierto, pero como siempre te quedas en el árbol, pues no lo sabes.
-¿Qué es lo que he de saber y no sé? -preguntó con impaciencia la pequeña luciérnaga.
-Tienes que saber que la luna no tiene la misma luz todas las noches -le contestó la abuela-. La luna es tan variable que cada día es diferente. Hay días en los que es grande y majestuosa como una pelota, y brilla sin cesar en el cielo. Pero hay otros días en los que se esconde, su brillo desaparece y el mundo se queda completamente a oscuras.
-¿De veras hay noches en las que la luna no sale? -preguntó sorprendida la pequeña luciérnaga.
-Así es -le confirmó la abuela. La luna es muy cambiante. A veces crece y a veces se hace pequeñita. Hay noches en las que es grande y roja y otras en las que desaparece detrás de las nubes. En cambio tú, mi niña, siempre brillarás con la misma fuerza y siempre lo harás con tu propia luz.
La pequeña luciérnaga estaba asombrada ante tal descubrimiento. Nunca se había imaginado que la luna pudiese cambiar y que brillase o se escondiese según los días.
Y a partir de aquel día, la pequeña luciérnaga decidió salir a volar y a bailar con su familia y sus amigos. Así fue como nuestra pequeña amiguita aprendió que cada uno tiene sus cualidades y por tanto, cada uno debe brillar con su propia luz.
### DECIMAL NUMBERS
Naming decimal places
A decimal number is made of an integer part, placed on the left side of a decimal point, and a fractional part, placed on the right side of a decimal point. As a matter a fact, decimals are numbers which tells us how many parts of a whole we have. We use them to mark measure units of things that are not completely whole.
All the place values of the numbers depend on position on the left or right side of a decimal point. Look at the example with more digits.
Let’s take a look at, for example the number 1,987,654,321.123456,
The first digit before the decimal point represents the ones (number 1),
-the second stands for the tens (number 2), the third for the hundreds (number 3),
-the fourth for the thousands (number 4, after the comma),
-the fifth for the ten thousands (number 5),
-the sixth for the hundred thousands (number 6),
-the seventh for the millions (number 7, after the second comma),
-the eight for the ten millions (number 8),
-the ninth for the hundred millions (number 9) and
-the tenth for the billions (number 1, after the third comma).
All digits after the decimal point are called decimals.
-The first digit represents tenths (number 1),
-the second digit stands for the hundredths (number 2),
-third for the thousandths (number 3),
-fourth for the ten thousandths (number 4),
-fifth for the hundred thousandths (number 5),
-sixth for the millionths (number 6)
There are larger and smaller place values, but these ones are used the most. It doesn’t matter how large the number of digits is, they can be read and understood with ease. Test the knowledge with worksheets.
### Sequences
The general term of a sequence is the expression that represents all the terms of the sequence (there are usually infinitive terms!).The general term is expressed by an algebraic expression that shows therelation between the that term.
A sequenceis increasing if every term is greater the previous one and a sequence is decreasing if every term is less than the previous.
Some sequences are special because they have a singular structure.
The difference of any two successive terms of the sequence is aconstant. They can be defined by the first term and the constant value called common difference. These sequences always increase (or decrease). These sequences are called arithmetic progressions.
A geometric progression is a sequence where every term is generated by multiplying the previous one by a fixed number called the common ratio of the progression.
### TEST DE INTELIGENCIA EMOCIONAL PARA LOS NIÑOS
En cada frase debe evaluar y cuantificar su capacidad en el uso de la habilidad descrita. Antes de responder, intenta pensar en situaciones reales en las que hayas tenido que utilizar dicha habilidad y no pretendas responder de acuerdo a lo que usted crea que sería lo correcto.
No. HABILIDADES PUNTOS 1 Identificar cambios del estímulo fisiológico 2 Relajarse en situaciones de presión 3 Actuar de modo productivo mientras está enojado 4 Actuar de modo productivo mientras está ansioso 5 Tranquilizarse rápidamente mientras está enojado 6 Asociar diferentes indicios físicos con emociones diversas 7 Usar el diálogo interior para controlar estados emocionales 8 Comunicar los sentimientos de un modo eficaz 9 Pensar con sentimientos negativos sin angustiarse 10 Mantenerse en calma cuando es blanco del enojo de otros 11 Saber cuando tiene pensamientos negativos 12 Saber cuando su "discurso interior" es positivo 13 Saber cuando empieza a enojarse 14 Saber como interpreta los acontecimientos 15 Conocer qué sentimientos utiliza actualmente 16 Comunicar con precisión lo que experimenta 17 Identificar la información que influye sobre sus interpretaciones 18 Identificar sus cambios de humor 19 Saber cuándo está a la defensiva 20 Calcular el impacto que su comportamiento tiene en los demás 21 Saber cuando no se comunica con sentido 22 Ponerse en marcha cuando lo desea 23 Recuperarse rápidamente después de un contratiempo 24 Completar tareas a largo plazo dentro del tiempo previsto 25 Producir energía positiva cuando realiza un trabajo poco interesante 26 Abandonar o cambiar hábitos inútiles 27 Desarrollar conductas nuevas y más productivas 28 Cumplir lo que promete 29 Resolver conflictos 30 Desarrollar el consenso con los demás 31 Mediar en los conflictos con los demás 32 Utilizar técnicas de comunicación interpersonal eficaces 33 Expresar los pensamientos de un grupo 34 Influir sobre los demás en forma directa o indirecta 35 Fomentar la confianza con los demás 36 Montar grupos de apoyo 37 Hacer que los demás se sientan bien 38 Proporcionar apoyo y consejo a los demás cuando es necesario 39 Reflejar con precisión los sentimientos de las personas 40 Reconocer la angustia de los demás 41 Ayudar a los demás a controlar sus emociones 42 Mostrar comprensión hacia los demás 43 Entablar conversaciones íntimas con los demás 44 Ayudar a un grupo a controlar sus emociones 45 Detectar incongruencias entre las emociones o sentimientos de los demás y sus conductas
### POLYNOMIALS IDENTITIES
Algebraic Identities
(a + b)(a - b) = a2 - b2
(a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 - 2ab + b2
Example 1: Simplify (3u + 5w)(3u – 5w)
Using the algebraic identities (a + b)(a - b) = a2 - b2, we substitute a for 3u and b for 5w.
(3u + 5w)(3u – 5w)
= (3u)2 – (5w)2
= 9u2 – 25w2
Thus (3u + 5w)(3u – 5w) = 9u2 – 25w2
Example 2 : Using the algebraic identities to simplify (3a + 7b)2
Using (a+b)2 = a2+2ab+b2
In this case we need to substitute 3a for a as well as 7b for b
(3a + 7b)2
= (3a)2 + 2(3a)(7b) + (7b)2
= 9a2+ 42ab + 49b2
Thus (3a + 7b)2 = 9a2+ 42ab + 49b2
Example 3: Simplify (5a – 7b)2
Using (a-b)2 = a2-2ab+b2 we have:
(5a – 7b)2
= (5a)2 – 2(5a) (7b) + (7b)
= 25a2 – 70ab + 49b2.
Thus (5a – 7b)2 = 25a2 – 70ab + 49b2
### PANEL DE LAS EMOCIONES
Si tenemos que definir las emociones nosotros pensamos que son los sentimientos que tenemos desde dentro y que nos hacen estar de una forma o de otra. En el diccionario nos dice que:
La emoción es un sentimiento breve e intenso producido por algo bueno o malo que nos impresiona mucho, una noticia inesperada, un regalo o una película nos puede causar mucha emoción, también las personas y las cosas nos producen emociones.
Vamos a trabajar la alegría, la tristeza, el enfado, el miedo y la satisfacción. Veremos que las emociones se relacionan unas con otras. Unas nos gustan más, otras menos, pero todas son importantes conocerlas, aceptarlas y controlarlas.
1. Vemos una imagen.
2. Pensamos de que emoción puede ser.
3. Luego hablamos de nuestros sentimientos.
4. Descubrimos palabras que están relacionadas con la imagen y con esa emoción.
5. Hablamos de momentos en los que notamos ese sentimiento y qué podemos hacer con esa emoción.
6. Hacemos el cartel de la emoción que hemos trabajado y lo pegamos en nuestro panel.
### DÍA MUNDIAL DE LA BIBLIOTECA
Por iniciativa de la Asociación Española de Amigos del Libro Infantil y Juvenil, desde 1997, en colaboración con el Ministerio de Educación, Cultura y Deporte, cada 24 de octubre se celebra el Día de la Biblioteca, con el objetivo de concienciar a la sociedad de la importancia de la lectura y como homenaje y reconocimiento a la labor de los bibliotecarios/as.
Cada año se encarga a un escritor y a un ilustrador, ambos de reconocido prestigio, la redacción del pregón y el diseño del cartel que se difunde entre todas las bibliotecas de España, asociados e interesados. Este año los seleccionados son el escritor Diego Arboleda, Premio Nacional de Literatura Infantil 2014, y la ilustradora Leticia Ruiz Fernández.
Con este motivo desde la Junta de Andalucía se ha propuesto la semana del 26 al 30 de Octubre, para dar más visibilidad a las bibliotecas de los centros. Nosotros desde el plan de bibliotecas hemos decidido organizar un concurso de carteles y otro de lemas, que se realizaran preferentemente en las horas de tutoría.
X CONCURSO “ LEMAS DE LA BIBLIOTECA” BIBLIOTECA JORDI SIERRA I FABRA
Torre del Mar a 7 de Octubre de 2015, con motivo de la celebración durante la semana del 26 al 30 de Octubre de la semana de la biblioteca se convoca el X concurso, en este caso será de lemas: “lemas de la biblioteca”, con él intentaremos hacerla más visible en nuestro centro y que los alumnos reflexionen sobre la importancia de la lectura.
1.- se desecharán todos aquellos lemas que hagan referencia a violencia de cualquier tipo, tengan contenido sexual u ofensivo y/o denigren la imagen de la persona.
2.- Participantes: cualquier persona miembro de la comunidad educativa que lo desee puede participar.
3.- Se podrán presentar en Word, A4 vertical, se admiten 2 lemas por participante.
4.- Forma de presentación: Correo electrónico, en archivo adjunto o en papel, respetando el mismo formato. Se deberá enviar un mensaje de correo electrónico con el siguiente asunto:
X concurso de la biblioteca Jordi Sierra i Fabra a la dirección
En el mensaje se adjuntará un solo archivo en formato Word (.doc) donde aparecerán el título del relato, el nombre del autor y su fecha de nacimiento así como el curso y clase a que pertenece el participante.
5.- Los trabajos se enviarán antes del 30 de octubre.
7.- El jurado estará compuesto por miembros del proyecto de biblioteca.
8.- El fallo del jurado será inapelable.
XI CONCURSO “ CARTELES DE LA BIBLIOTECA” BIBLIOTECA JORDI SIERRA I FABRA
Torre del Mar a 7 de Octubre de 2015, con motivo de la celebración durante la semana del 26 al 30 de Octubre de la semana de la biblioteca se convoca el X concurso, en este caso será de lemas: “Carteles de la biblioteca”, con él intentaremos hacerla más visible en nuestro centro y que los alumnos reflexionen sobre la importancia de la lectura.
1.- se desecharán todos aquellos lemas que hagan referencia a violencia de cualquier tipo, tengan contenido sexual u ofensivo y/o denigren la imagen de la persona.
2.- Participantes: cualquier persona miembro de la comunidad educativa que lo desee puede participar.
3.- Se podrán presentar en cartulinas o media cartulina, se admitirán 2 por participante.
4.- Forma de presentación, se entregarán en mano al profesor de lengua y por detrás aparecerá:
XI concurso de la biblioteca Jordi Sierra i Fabra.
Nombre y apellidos :
Curso:
5.- Los trabajos se entregarán antes del 30 de octubre. |
# How do you combine -4( - 8x - 8) + 12x?
Feb 5, 2018
See below...
#### Explanation:
First we must expand the brackets.
$\therefore - 4 \left(- 8 x - 8\right) = 32 x + 32$
This gives the expression $32 x + 32 + 12 x$
Now we collect the like terms.
$\therefore 32 x + 32 + 12 x = 44 x + 32$
Feb 5, 2018
$44 x + 32$
#### Explanation:
Start off by multiplying out the brackets:
$- 4 \times - 8 x = 32 x$
$- 4 \times - 8 = 32$
Putting these values back in, and replacing the bracket we get...
$32 x + 32 + 12 x$
Combining like terms we get...
$32 x + 12 x = 44 x$
$44 x + 32$ |
# Problems on Work Done in a Given Period of Time
Learn how to solve problems on work done in a given period of time when a person A completes 1/n th part of work in one day, then A will take n days to complete the work.
Now we will apply these in working out various word problems to understand the concept to find work done in a given period of time.
Solved problems on work done in a given period of time:
1. Aden and Brad can do a piece of work in 18 days. Brad can Cody can do it in 24 days and Cody and Aden can do it in 36 days. If Aden, Brad, Cody work together, in how many days will they finish the work?
Solution:
(Brad + Cody)’s 1 day’s work = 1/24
(Cody + Aden)’s 1 day’s work = 1/36
Therefore, 2 (Aden + Brad + Cody)’s 1 day’s work = 1/18 + 1/24 + 1/36
= (4 + 3 + 2)/72
= 9/72
= 1/8
(Aden + Brad + Cody)’s 1 day’s work = 1/2 × 8 = 1/16
Therefore, Aden, Brad, Cody together will complete this work in 16 days.
2. Jack and Max can do a piece of work in 10 days. Max and Troy can do the same work in 12 days and Troy and Jack can do it in 15 days. Find the number of days taken by each to finish it working alone.
Solution:
(Jack + Max)’s 1 day’s work = 1/10
(Max + Troy)’s 1 day’s work = 1/12
(Troy + Jack)’s 1 day’s work = 1/15
Therefore, 2(Jack + Max + Troy)’s 1 day’s work = 1/10 + 1/12 + 1/15
= (6 + 5 + 4)/60
= 15/60
= 1/4
Therefore, (Jack + Max + Troy)’s 1 day’s work = 1/2 × 1/4
= 1/8
Now Jack’s 1 day’s work = (Jack + Max + Troy)’s 1 day’s work – (Max + Troy)’s 1 day’s work.
= 1/8 – 1/12
= (3 – 2)/24
= 1/24
Therefore, Jack alone can complete the work in 25 days.
Now Max’s 1 day’s work = (Jack + Max + Troy)’s 1 day’s work – (Jack + Troy)’s 1 day’s work
= 1/8 – 1/15
= (15 – 8)/120
= 7/120
Therefore, Max alone can complete the work in 120/7 days. = 17.1 days.
Now Troy’s 1 day’s work = (Jack + Max + Troy)’s 1 day’s work – (Jack + Max)’s 1 day’s work
= 1/8 – 1/10
= (5 – 4)/40
= 1/40
Therefore, Troy alone can complete the work in 40 days.
Calculate Time to Complete a Work
Calculate Work Done in a Given Time
Problems on Time required to Complete a Piece a Work
Problems on Work Done in a Given Period of Time
Problems on Time and Work
Pipes and Water Tank
Problems on Pipes and Water Tank
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# How do you change 25 into a fraction?
## How do you change 25 into a fraction?
Answer: 25% as a fraction is 1/4.
### What is 0.26 as a fraction?
13 / 50
Answer and Explanation: 0.26 can be converted to the fraction 13 / 50.
What is 1.5 as a fraction?
3/2
Answer: 1.5 as a fraction is written as 3/2.
How do you turn 25% into a fraction and decimal?
We know that 25% means 25 per 100 or you can say 25 out of 100. So the fraction can be written as, 25%=25100 . Therefore, 25% in decimal is 0.25 and in the fraction is 25100.
## What is .25 written in the simplest fraction form?
The decimal . 25 written in its simplest fraction form is 1/4.
### What is 1.2 As a decimal?
1.2% = . 01. 2 = 0.012 To change a percent to a fraction, Change the percent to a decimal.
What is 1.3 as a fraction?
13/10
Answer: 1.3 as a fraction is 13/10.
What is 1.5 as a fraction or mixed number?
To express 1.5 as a fraction, we need to write the given decimal number in the x/y form, where x and y are the positive integers. Now, we have to multiply and divide 1.5 / 1 by 10. The fraction 15/10 can be further be simplified to 3/2. So, 1.5 as a fraction is equal to 3/2.
## What is 1.25 simplified?
In the fraction 1/25, 1 is the numerator and 25 is the denominator. When you ask “What is 1/25 simplified?”, we assume you want to know how to simplify the numerator and denominator to their smallest values, while still keeping the same value of the fraction. We do this by first finding the greatest common factor of 1 and 25, which is 1.
### How do you write a percentage as a decimal?
Easiest—divide by 100: The simplest way to convert a percentage to a decimal is to divide the number (in percentage format) by 100. Example: Convert 75 percent to decimal format. Divide 75 by 100 (this can also be written as 75 / 100 or 75 ÷ 100) The result is 0.75, meaning that 75 percent is the same as 0.75.
How do you change a mixed fraction into a decimal?
To convert fractions to decimals you just divide the top by the bottom — divide the numerator by the denominator — and if the division doesn’t come out evenly, you can stop after a certain number of decimal places and round off. To convert fractions to decimals: If the fraction is a mixed number, change it to an improper fraction.
How do you convert ratios to decimals?
The first step in converting a ratio to a decimal is to rewrite the ratio as a fraction. To do this, put the first part of the ratio over the second part in fraction form.
Posted In Q&A |
# Reflections on Cartesian Plane
Lesson
## Reflections
As we saw in the previous lesson on Flips, A reflection occurs when we flip an object or shape across a line. Like a mirror, the object is exactly the same size, just flipped in position. So what was on the left may now appear on the right. Every point on the object or shape has a corresponding point on the image, and they will both have the same distance from the reflection line. We can reflect points, lines, polygons on the Cartesian plane by flipping them across an axis line or another line in the plane.
If we reflect horizontally across the $y$y axis, then the y values of the coordinates remain the same and the $x$x values change sign.
In this diagram, the image is reflected across $y$y axis.
Note how the point $\left(-2,1\right)$(2,1) becomes $\left(2,1\right)$(2,1). The $y$y values have not changed and the x values have changed signs.
Similarly the point $\left(-6,3\right)$(6,3) becomes $\left(6,3\right)$(6,3). The $y$y values have not changed and the x values have changed signs.
If we reflect vertically across the $x$x axis, then the $x$x values of the coordinates remain the same and the $y$y values change sign.
In this diagram, the image is reflected across $x$x axis.
Note how the point $\left(4,3\right)$(4,3) becomes $\left(4,-3\right)$(4,3). The $x$x values have not changed and the $y$y values have changed signs.
Similarly the point $\left(0,5\right)$(0,5) becomes $\left(0,-5\right)$(0,5). The $x$x values have not changed and the $y$y values have changed signs.
Have a quick play with this interactive to further consolidate the ideas behind translations on the Cartesian plane.
Let's have a look at these worked examples.
##### Question 1
Consider the point $A\left(7,3\right)$A(7,3).
1. Plot point $A$A on the number plane.
2. Now plot point $A'$A, a reflection of point $A$A across the $x$x-axis.
##### Question 2
Consider the point $A\left(-7,-3\right)$A(7,3).
1. Plot point $A$A on the number plane.
2. Now plot point $A'$A, a reflection of point $A$A across the $y$y-axis.
##### Question 3
Consider the line segment $AB$AB, where the endpoints are $A$A$\left(-4,-2\right)$(4,2) and $B$B$\left(6,7\right)$(6,7).
1. Plot the line segment $AB$AB on the number plane.
2. Now plot the reflection of the line segment $AB$AB across the $x$x-axis.
##### Question 4
Plot the new triangle formed by reflecting the given triangle about the line $x=-3$x=3. |
# Volume
## Number of unit cubes to fill prisms, spheres, pyramids, cylinders, cones, and composite solids.
%
Progress
MEMORY METER
This indicates how strong in your memory this concept is
Progress
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Volume of Solids
The composite solid below is made of a cube and a square pyramid. The length of each edge of the cube is 12 feet and the overall height of the solid is 22 feet. What is the volume of the solid? Why might you want to know the volume of the solid?
### Volume of a Solid
The volume of a solid is the number of unit cubes it takes to fill up the solid.
A prism is a solid with two congruent polygon bases that are parallel and connected by rectangles. Prisms are named by their base shape.
To find the volume of a prism, find the area of its base and multiply by its height.
\begin{align*}Prism: V=A_{Base} \cdot h\end{align*}
A cylinder is like a prism with a circular base.
To find the volume of a cylinder, find the area of its circular base and multiply by its height.
\begin{align*}Cylinder: V=\pi r^2 h \end{align*}
A cone also has a circular base, but its lateral surface meets at a point called the vertex.
To find the volume of a cone, find the volume of the cylinder with the same base and divide by three.
\begin{align*}Cone: V=\frac{\pi r^2 h}{3}\end{align*}
A pyramid is similar to a cone, except it has a base that is a polygon instead of a circle. Like prisms, pyramids are named by their base shape.
To find the volume of a pyramid, find the volume of the prism with the same base and divide by three.
\begin{align*}Pyramid: V= \frac{A_{Base} \cdot h}{3}\end{align*}
A sphere is the set of all points in space equidistant from a center point. The distance from the center point to the sphere is called the radius.
The volume of a sphere relies on its radius.
\begin{align*}Sphere: V= \frac{4}{3} \pi r^3\end{align*}
A composite solid is a solid made up of common geometric solids.
The volume of a composite solid is the sum of the volumes of the individual solids that make up the composite.
Let's look at some problems where we find the volume.
1. Find the volume of the rectangular prism below.
To find the volume of the prism, you need to find the area of the base and multiply by the height. Note that for a rectangular prism, any face can be the “base”, not just the face that appears to be on the bottom.
\begin{align*}Area \ of \ Base &= 4 \cdot 4 = 16 \ in^2 \\ Height &= 5 \ in \\ Volume &=80 \ in^3\end{align*}
2. Find the volume of the cone below.
To find the volume of the cone, you need to find the area of the circular base, multiply by the height, and divide by three.
\begin{align*}Area \ of \ Circle &= 7^2 \pi = 49 \pi \ cm^2 \\ Volume &= \frac{49 \pi \cdot12}{3} = 196 \pi \ cm^3\end{align*}
3. Find the volume of a sphere with radius 4 cm.
The formula for the volume of a sphere is \begin{align*}V= \frac{4}{3} \pi r^3\end{align*}.
\begin{align*}Volume= \frac{4}{3} \pi (4)^3 = \frac{256 \pi}{3} cm^3\end{align*}
### Examples
#### Example 1
Earlier, you were asked what is the volume of the solid, and why might you want to know the volume of the solid.
The composite solid below is made of a cube and a square pyramid. The length of each edge of the cube is 12 feet and the overall height of the solid is 22 feet.
To find the volume of the solid, find the sum of the volumes of the prism (the cube) and the pyramid. Note that since the overall height is 22 feet and the height of the cube is 12 feet, the height of the pyramid must be 10 feet.
\begin{align*}Volume \ of \ Prism(Cube) &=A_{Base} \cdot h=(12 \cdot 12) \cdot 12=1728 \ ft^3 \\ Volume \ of \ Pyramid &=\frac{A_{Base} \cdot h}{3}=\frac{(12 \cdot 12) \cdot 10}{3}=480 \ ft^3 \\ Total \ Volume &=1728+480=2208 \ ft^3\end{align*}
The volume helps you to know how much the solid will hold. One cubic foot holds about 7.48 gallons of liquid. This solid would hold 16,515.84 gallons of liquid!
#### Example 2
The area of the base of the pyramid below is \begin{align*}100 \ cm^2\end{align*}. The height is 5 cm. What is the volume of the pyramid?
\begin{align*}V=100 \ cm^2 \cdot 5 \ cm=500 \ cm^3\end{align*}
#### Example 3
The volume of a sphere is \begin{align*}\frac{500 \pi}{3} \ in^3\end{align*}. What is the radius of the sphere?
\begin{align*}\frac{4}{3}\pi r^3 &=\frac{500 \pi}{3} \\ 4r^3 &=500 \\ r^3 &=125 \\ r &=5 \ in\end{align*}
#### Example 4
The volume of a square pyramid is \begin{align*}64 \ in^3\end{align*}. The height of the pyramid is three times the length of a side of the base. What is the height of the pyramid?
\begin{align*}64=\frac{A_{Base} \cdot h}{3}=\frac{s^2h}{3}=\frac{s^2(3s)}{3}=s^3\end{align*}. Therefore, \begin{align*}s=4 \ in\end{align*} and \begin{align*}h=3(4)=12 \ in\end{align*}.
### Review
Find the volume of each solid or composite solid.
1.
2.
3.
4.
5. The base is an equilateral triangle.
6.
7. Explain why the formula for the volume of a prism involves the area of the base.
8. How is a cylinder related to a prism?
9. How is a pyramid related to a cone?
10. How is a sphere related to a circle?
11. If one cubic centimeter will hold 1 milliliter of water, approximately how many liters of water will the solid in #1 hold? (One liter is 1000 milliliters).
12. If one cubic centimeter will hold 1 milliliter of water, approximately how many liters of water will the solid in #3 hold? (One liter is 1000 milliliters).
13. If 231 cubic inches will hold one gallon of water, approximately how many gallons of water will the solid in #5 hold?
14. The volume of a cone is \begin{align*}125 \pi \ in^3\end{align*}. The height is three times the length of the radius. What is the height of the cone?
15. The volume of a pentagonal prism is \begin{align*}360 \ in^3\end{align*}. The height of the prism is 3 in. What is the area of the pentagon base?
To see the Review answers, open this PDF file and look for section 1.10.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Composite A number that has more than two factors.
Cone A cone is a solid three-dimensional figure with a circular base and one vertex.
Cylinder A cylinder is a solid figure with two parallel congruent circular bases.
Prism A prism is a three-dimensional object with two congruent parallel bases that are polygons.
Pyramid A pyramid is a three-dimensional object with a base that is a polygon and triangular faces that meet at one vertex.
Sphere A sphere is a round, three-dimensional solid. All points on the surface of a sphere are equidistant from the center of the sphere.
Volume Volume is the amount of space inside the bounds of a three-dimensional object. |
# Can 50mm, 13mm and 12mm be a right triangle?
Jun 10, 2015
I do not think so.
#### Explanation:
Try with Pythagora's Theorem:
${50}^{2} = {13}^{2} + {12}^{2}$
$2500 = 169 + 144 = 313$ NO
Jun 10, 2015
It cannot even be a triangle let alone a right angled one.
#### Explanation:
If $a$, $b$ and $c$ are the lengths of the sides of a triangle and $c$ is the largest value, then $a + b \ge c$
$12 + 13 = 25 < 50$
Jun 10, 2015
A triangle with sides 50mm, 13mm, and 12mm can not form a right triangle
#### Explanation:
By the Pythagorean Theorem, to be a right triangle:
the square of the longest side
must be equal to
the sum of the squares of the other two sides
$\textcolor{w h i t e}{\text{XXXX}}$${50}^{2} = 2500$
$\textcolor{w h i t e}{\text{XXXX}}$1${3}^{2} + {12}^{2} = 169 + 144 = 313$
${50}^{2} \ne {13}^{2} + {12}^{2}$
Also
Note that no triangle can exist with sides 50mm, 13mm, and 12mm.
Explanation 2:
To form a triangle, every side must be less than the sum of the other two sides.
Picture a line segment of length 50mm with a line segment of 13 mm attached to one end and a line segment of 12 mm attached to the other end. The 13mm and 12mm line segments can not reach far enough to touch each other.
Jun 24, 2015
No, not according to the Pythagorean theorem.
#### Explanation:
If you plug in the side lengths into the Pythagorean theorem, assuming that $50$ is the hypotenuse and that side lengths are $13$ and $12$, you can calculate whether the triangle is a right triangle or not.
${13}^{2} + {12}^{2} = 313$, while $\sqrt{313}$ certainly doesn't equal ${50}^{2}$.
${a}^{2} + {b}^{2} = {c}^{2}$ |
# 797 and Mathematical Ways to Love
Mathwithbaddrawings.com has some thoughtful and entertaining Ways to Tell a Mathematician that you love them.
Artful Maths wrote a post that includes beautiful mathematical origami valentines and a “string art” cardioid that is made with a pencil instead of string.
This puzzle could be another mathematical way to show some love:
Print the puzzles or type the solution on this excel file: 12-factors-795-799
Here are a few facts about the number 797:
797 is a palindrome in three bases:
• 797 BASE 10 because 7(100) + 9(10) + 7(1) = 797
• 565 BASE 12 because 5(144) + 6(12) + 5(1) = 797
• 494 BASE 13 because 4(169) + 9(13) + 4(1) = 797
But there’s one more palindromic fact about the number 797: It is the sum of two square numbers that are also palindromes!
• 797 = 121 + 676. Note that 11² = 121 and 26² = 676.
Since it is the sum of two squares, 797 will also be the hypotenuse a Pythagorean triple:
• 555 – 572 – 797 calculated from 26² – 11², 2(26)(11), 26² + 11².
797 is the sum of three squares seven different ways:
• 28² + 3² + 2² = 797
• 27² + 8² + 2² = 797
• 24² + 14² + 5² = 797
• 24² + 11² + 10² = 797
• 22² + 13² + 12² = 797
• 21² + 16² + 10² = 797
• 20² + 19² + 6² = 797
797 is also the sum of the 15 prime numbers from 23 to 83:
• 23 + 29 + 31 + 37+ 41 + 43 + 47 + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 = 797
Here is the most basic information about the number 797:
• 797 is a prime number.
• Prime factorization: 797 is prime.
• The exponent of prime number 797 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 797 has exactly 2 factors.
• Factors of 797: 1, 797
• Factor pairs: 797 = 1 x 797
• 797 has no square factors that allow its square root to be simplified. √797 ≈ 28.231188.
How do we know that 797 is a prime number? If 797 were not a prime number, then it would be divisible by at least one prime number less than or equal to √797 ≈ 28.2. Since 797 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, or 23, we know that 797 is a prime number.
Here’s another way we know that 797 is a prime number: Since its last two digits divided by 4 leave a remainder of 1, and 26² + 11² = 797 with 26 and 11 having no common prime factors, 797 will be prime unless it is divisible by a prime number Pythagorean triple hypotenuse less than or equal to √797 ≈ 28.2. Since 797 is not divisible by 5, 13, or 17, we know that 797 is a prime number.
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# Table of 49
Table of 49 represents repeated addition of the number 49, to itself. For example, 49 multiplied by 4, represents 49 added four times to itself.
49 x 4 ⇒ 49 + 49 + 49 + 49 ⇒ 196
Thus, we can say that, when 49 is repeatedly added to itself, four times, we get the sum as 196. Also, when 49 is multiplied by 4 the result is 196.
In the same way, we can write the multiplication table of 49 for n natural numbers. Check Tables from 1 to 100 to learn all the tables.
## Multiplication Table of 49
The table of the natural number 49, is provided below up to 20 times. Students can use this table to solve various multiplication problems. Memorise and revise the table for quick calculations.
49 x 1 = 49 49 x 2 = 98 49 x 3 = 147 49 x 4 = 196 49 x 5 = 245 49 x 6 = 294 49 x 7 = 343 49 x 8 = 392 49 x 9 = 441 49 x 10 = 490 49 x 11 = 539 49 x 12 = 588 49 x 13 = 637 49 x 14 = 686 49 x 15 = 735 49 x 16 = 784 49 x 17 = 833 49 x 18 = 882 49 x 19 = 931 49 x 20 = 980
## 49 Times Table
Read the forty-nine times table here.
• One time forty-nine is 49
• Two times forty-nine is 98
• Three times forty-nine is 147
• Four times forty-nine is 196
• Five times forty-nine is 245
• Six times forty-nine is 294
• Seven times forty-nine is 343
• Eight times forty-nine is 392
• Nine times forty-nine is 441
• Ten times forty-nine is 490
Practise writing the 49 times table from 11 to 20 and read them to memorise fast.
## Solved Examples
Q.1: What is the value of 49 times of 5 added to 15?
Solution: 49 times of 5 = 49 x 5 = 245
Now if 15 added, we get;
245 + 15 = 260
Hence, 260 is the required value.
Q.2: Which of these is a multiple of 49?
97, 147, 197, 249, 299, 341, 391
Solution: To check if the given numbers are multiples of 49, we need to divide each of them by 49. If the quotient is a whole number, then it is a multiple of 49.
97 ÷ 49 = 1.97.. (fraction)
147 ÷ 49 = 3 (whole number)
197 ÷ 49 = 4.02.. (fraction)
249 ÷ 49 = 5.08.. (fraction)
299 ÷ 49 = 6.10.. (fraction)
341 ÷ 49 = 6.95.. (fraction)
391 ÷ 49 = 7.97.. (fraction)
We can see, when 147 is divided by 49, we get 3, which is a whole number, but the rest of them are fractions. Therefore, 147 is the multiple of 49.
Q.3: Solve the following: 49 x 11 ÷ 11.
Solution: Given,
49 x 11 ÷ 11
= 49 x (11 ÷ 11) [BODMAS rule]
= 49 x 1
= 49
### Check More Tables
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## Frequently Asked Questions on Table of 49
### What is the Table of 49?
Table of 49 expresses the repeated addition of 49.
49 x 1 = 49
49 x 2 = 49 + 49 = 98
49 x 3 = 49 + 49 + 49 = 147
49 x 4 = 49 + 49 + 49 + 49 = 196
And so on
### What is the value of 49 times of 9?
49 times of 9 = 49 x 9 = 441
### What are the multiples of 49?
The first 10 multiples of 49 are 49, 98, 147, 196, 245, 294, 343, 392, 441 and 490.
### What is the sum of the first five multiples of 49?
The sum of first five multiples of 49 is given by:
49 + 98 + 147 + 196 + 245 = 735
### Is 49 a prime number or a composite number?
49 is a composite number because 49 has more than two factors, they are:
1, 7 and 49. |
# Why Is A Full Circle 360 Degrees, Instead Of Something More Convenient, Like 100?
A full circle is 360 degrees because the Babylonians used the sexagesimal system. It also represents the number of days a year and also because 360 is highly composite.
The other day, I helped my 12-year-old cousin with his math homework. His homework had questions about the area of a circle with a certain radius and circumference. The questions seemed quite simple, and we were able to solve them all in 15 minutes.
What annoyed my cousin all the time, however, was, why was a complete circle 360 degrees? Why not something simpler like 10 degrees or 100 degrees? All calculations would have been so easy; wouldn’t that be more convenient?
## Hypothesis 1: Mathematical Reasons
You may wonder what mathematical reasons there might be for using 360 degrees to represent a complete circle since 360 always seems to cause difficulties in calculating the answers to your math homework.
But actually, it is the ideal solution. A number like 10 or 100 would have been mathematically more uncomfortable.
The number 360 is divisible by any number from 1 to 10, except the 7th. It is actually divided into 24 different numbers: 1,2,3,4,5,6,8,9,10,12,15,18,20,24,30,36,40,45,60,72,90,120,180 and 360 itself. These 24 numbers are called the divisors of the number 360.
This is the highest number of divisors for each positive integer up to its own value of 360. In contrast, the number we often wish would be the value for the full circle—100—has only 9 divisors.
This property of the number 360 makes it a strongly composite number. Numbers are said to be highly composite if they are positive integers with more divisors than any smaller positive integer has. Highly composite numbers are considered good base numbers with which to perform common calculations.
For example, 360 can be divided into two, three and four parts and the resulting number is a whole number. The resulting numbers are 180, 120 and 90.
However, dividing 100 by three doesn’t end in a whole number. Instead, it provides a decimal value of 33.3 recurring, which makes performing calculations difficult. Calculations using 360 actually become pretty simple once you’re smart enough to do them in your head and can put down the calculator.
But is this the whole justification? Was this the only reason why our ancestors decided to break a circle in 360 degrees?
Also Read: Do We Have Any Mathematical Proof That Pi Is Infinite?
## Hypothesis 2: The Length Of A Year
Have you ever wondered why there are exactly 365 days a year? Why not use a more convenient number like 300 or 400?
Okay, there is no mathematical reason for this, but it is just an observation made by our ancestors, and these observations also contributed to a circle being closed by 360 degrees.
The ancient astronomers, especially the Persians and Cappadocians, noticed that it took the sun 365 days to return to exactly the same position.
In other words, the sun advances one degree every day on its elliptical orbit. Persians had a leap month every 6 years to compensate for the 5 extra days. Moreover, the lunar calendar has a total of 355 days, while the solar calendar has 365 days. And, which number sits perfectly between the two and is a composite number?
Yes… 360!
## Hypothesis 3: Historical Reasons
Another theory that suggests why a full circle is 360 degrees comes from the Babylonians. The Sumerians and Babylonians famously used the Sexagesimal number system.
The sexagesimal system is one with a base value of 60, while the current system we use is known as the decimal system and has a base value of 10. So once we reach the 10th number, we start repeating the symbols of earlier numbers from 0 to 9 to form new numbers.
The Babylonians had 60 different symbols to make numbers with. Why should they use 60 again?
Because 60, like 360, is a highly composite number with up to 12 factors. Just as we can count 10 on our fingers for the decimal system, we can also count up to 60.
Start by counting the knuckles of the 4 fingers (not the thumb) on your right hand. 12, right?
Now, on the other hand, raise any of those fingers to remember that you finished one iteration and got the number 12. Now, repeat the same procedure as many times as the number of fingers remaining on the left hand.
The number you will end up with is 12 knuckles x 5 fingers = 60.
### But Anyway… Why 360?
If we were to draw an equilateral triangle with the length of sides equal to the radii of the circle and place one of its vertexes at the center of the circle, then we could fit a total of 6 such equilateral triangles inside a circle. Since the Babylonians used the sexagesimal numeral system, they considered each triangle to have a base value of 60. Thus, 6 triangles x 60 base value again gives us a value of 360.
### The Indian Context
There is even evidence that a circle in the Rigveda from India was divided into 360 parts.
Twelve spokes, one wheel, navels three.
Who can comprehend this?
On it are placed together
three hundred and sixty like pegs.
They shake not in the least.
— Dirghatamas, Rigveda 1.164.48
Next time someone asks you this question, I hope you have something to say!
Also Read: Why Are There 24 Hours In A Day And 60 Minutes In An Hour?
Wanna check how much information you retained from the article?
Can you answer three questions based on the article you just read?
References (click to expand) |
## Exponent Study Guide: Multiplying and Dividing Exponents With the Same Bases
written by: Keren Perles โข edited by: SForsyth โข updated: 1/20/2012
Need a reminder on multiplying and dividing exponents? Follow these simple steps...
• slide 1 of 2
### Multiplying Exponents With the Same Base
Multiplying exponents with the same base is easier to do than it seems, especially assuming that you understand the basic idea of exponents. For example, take the following problem: 3^2 X 3^3. If you understand that 3^2 is the same thing as 3 X 3, and that 3^3 is the same thing as 3 X 3 X 3, then it makes sense that the original problem would be the same thing as (3 X 3) X (3 X 3 X 3). If you remove the parentheses, youโll see that you could write that as 3 X 3 X 3 X 3 X 3 โ or just 3^5. So 3^2 X 3^3 = 3^5.
Instead of writing all of that out each time you want to multiply exponents, just think of it like this: two threes times three threes is just five threes. You could get the same answer just by adding the powers. So thatโs the rule for multiplying exponents โ To multiply exponents with the same base, just add the powers.
• slide 2 of 2
### Dividing Exponents With the Same Base
Dividing exponents with the same is just as easy as multiplying them. For example, take the following problem: 3^3 / 3^2. The numerator (top of the fraction) is 3^3, which is really the same as 3 X 3 X 3. The denominator (bottom of the fraction) is 3^2, which is really the same as 3 X 3. Now think about fractions. If you have 3 X 3 X 3 on the top of the fraction, and 3 X 3 on the bottom of the fraction, you can simplify the fraction by crossing out two threes in the numerator and two threes in the denominator. The result would be 3^1, or just 3.
How could you have gotten that answer more easily? Well, you know that if there are more threes in the numerator, you can simplify the fraction by crossing out the number of threes that are in the denominator. How can you figure out how many threes that would leave in the numerator? Subtract the number of threes in the denominator from the number of threes in the numerator. So thatโs the rule for dividing exponents โ To divide exponents with the same base, just subtract the powers.
#### Math Help for Exponents
Looking for math help for exponents? Whether you're a student, parent, or tutor, this series of articles will explain the basics of how to use exponents correctly. Includes rules for adding, subtracting, multiplying, and dividing exponents, as well as how to use negative exponents. |
# If cot θ=3–√, find the value of cosec2 θ+cot2 θcosec2 θ−sec2 θ.
Question:
If $\cot \theta=\sqrt{3}$, find the value of $\frac{\cos e c^{2} \theta+\cot ^{2} \theta}{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}$.
Solution:
Given: $\cot \theta=\sqrt{3}$
We have to find the value of the expression $\frac{\operatorname{cosec}^{2} \theta+\cot ^{2} \theta}{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}$.
We know that,
$\cot \theta=\sqrt{3} \Rightarrow \cot ^{2} \theta=3$
$\operatorname{cosec}^{2} \theta=1+\cot ^{2} \theta=1+(\sqrt{3})^{2}=4$
$\sec ^{2} \theta=\frac{1}{\cos ^{2} \theta}=\frac{1}{1-\sin ^{2} \theta}=\frac{1}{1-\frac{1}{\operatorname{cosec}^{2} \theta}}=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$
Therefore,
$\frac{\operatorname{cosec}^{2} \theta+\cot ^{2} \theta}{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}=\frac{4+3}{4-\frac{4}{3}}$
$=\frac{21}{8}$
Hence, the value of the given expression is $\frac{21}{8}$. |
# What is an Inequality? An introduction to understanding and graphing inequalities
## An Introduction to Graphing Inequalities
The video is an introduction to what an inequality is. We look at why we need inequalities and how to graph them.
#### Let’s continue this idea by looking at the finer details of graphing inequalities.
Graph: $\color{blue}{x < 4}$
Lets start by graphing some of the numbers that make the statement $x < 4$ true.
These are only a few of the solutions. We use need to use a line to show that all the numbers less than 4 are included. Also by putting an open circle where the start of the solution is (around the 4) we are saying that while $3.99999$ is part of the solution set, 4 is not.
When the the number is not included in the solution set the circle is open.
Graph: $\color{blue}{x \leq 4}$
Again we will use a line to show that all the numbers less than 4 are included, but this time we put a closed circle (which means that we color in the circle) around the number where the start of the solution is. So in this case, the filled in circle around the 4 is telling the reader that 4 is part of the solution set.
When the the number is included in the solution set the circle is filled in.
Graph: $\color{blue}{x > 4}$
Pick a value for x to determine which way to draw the line representing the solution set. Since 6 is greater than 4 the line is drawn to the right.</>
When the the number is not included in the solution set the circle is open. |
# SAT II Math II : Solving Piecewise and Recusive Functions
## Example Questions
### Example Question #271 : Sat Subject Test In Math Ii
Define and as follows:
Evaluate .
Explanation:
by definition.
on the set , so
.
on the set , so
.
### Example Question #2 : Solving Piecewise And Recusive Functions
Define function as follows:
Give the range of .
Explanation:
The range of a piecewise function is the union of the ranges of the individual pieces, so we examine both of these pieces.
If , then . To find the range of on the interval , we note:
The range of this portion of is .
If , then . To find the range of on the interval , we note:
The range of this portion of is
The union of these two sets is , so this is the range of over its entire domain.
### Example Question #1 : Solving Piecewise And Recusive Functions
Define function as follows:
Give the range of .
Explanation:
The range of a piecewise function is the union of the ranges of the individual pieces, so we examine both of these pieces.
If , then
To find the range of on the interval , we note:
The range of on is .
If , then
To find the range of on the interval , we note:
The range of on is .
The range of on its entire domain is the union of these sets, or .
### Example Question #1 : Solving Piecewise And Recusive Functions
Define functions and as follows:
Evaluate .
Undefined
Explanation:
First, we evaluate . Since , the definition of for is used, and
Since
, then
### Example Question #1 : Solving Piecewise And Recusive Functions
Define functions and as follows:
Evaluate Evaluate .
Undefined
Undefined
Explanation:
First, evaluate using the definition of for :
Therefore,
However, is not in the domain of
Therefore, is an undefined quantity.
### Example Question #6 : Solving Piecewise And Recusive Functions
Define functions and as follows:
Evaluate .
Undefined
Explanation:
First, evaluate using the definition of for :
Therefore,
Evaluate using the definition of for :
### Example Question #2 : Solving Piecewise And Recusive Functions
Define functions and as follows:
Evaluate .
Undefined
Explanation:
First we evaluate . Since , we use the definition of for the values in the range :
Therefore,
Since , we use the definition of for the range :
### Example Question #71 : Functions And Graphs
Define two functions as follows:
Evaluate .
Explanation:
By definition,
First, evaluate , using the definition of for nonnegative values of . Substituting for 5:
; evaluate this using the definition of for nonnegative values of :
12 is the correct value.
### Example Question #1 : Solving Piecewise And Recusive Functions
Which of the following would be a valid alternative definition for the provided function?
None of these
Explanation:
The absolute value of an expression is defined as follows:
for
for
Therefore,
if and only if
.
Solving this condition for :
Therefore, for
Similarly,
for
The correct response is therefore |
# Partial Fractions
In this article, we will be learning about partial fractions for proper fractions. More specifically, we will be concentrating on the first two cases listed below.
• Case 1:
Partial fractions with distinct linear factors in the denominator.
• Case 2:
Proper fractions with a repeated linear factor in the denominator.
• Case 3:
Proper fractions with a quadratic factor that cannot be factorised in the denominator.
## What are Partial Fractions?
A proper algebraic fraction may be expressed in two or more partial fractions.
The process from left to right shows the result of subtracting two algebraic fractions.
$\begin{gather} \xrightarrow {\text {Simplification of Algebraic Fractions}} \end{gather}$ \begin{align*} \frac{3}{x+1} -\frac{5}{2x-3}=\frac{x-14}{(x+1)(2x-3)} \end{align*} $\begin{gather} \xleftarrow {\text {Partial Fractions}} \end{gather}$
The reverse process from right to left shows the ‘splitting’ of an algebraic fraction into two or more partial fractions.
## Case 1: Proper Fractions with Distinct Linear Factors in Denominator
An algebraic fraction where the denominator is a product of two distinct linear factors may be expressed in partial fractions as shown below.
\begin{align*} \frac{px+q}{(ax+b)(cx+d)}=\frac{A}{ax+b}+\frac{B}{cx+d} \end{align*}, where $A$ and $B$ are constants.
The denominator is a product of $(ax+b)$ and $(cx+d)$. $(ax+b)$ is one linear factor and $(cx+d)$ is another linear factor.
Question 1:
Express \begin{align*} \frac{(x-14)}{(x+1)(2x-3)} \end{align*} in partial fractions.
Solution:
• Let \begin{align*} \frac{(x-14)}{(x+1)(2x-3)}=\frac{A}{(x+1)}+\frac{B}{(2x-3)} \end{align*}.
We need to find the value of $A$ and of $B$
• Multiply the equation throughout by the denominator $(x + 1)(2x – 3)$
\begin{align*} \frac{(x-14)}{(x+1)(2x+3)} \times {(x+1)(2x-3)} \;=\; \frac{A}{(x+1)} \times {(x+1)(2x-3)} \;+\; \frac {B}{(2x-3)} \times {(x+1)(2x-3)} \end{align*}
• We will be left with the following equation,
\begin{align*} (x-14)=A(2x-3)+B(x+1) \end{align*}
• Then, we need to perform smart substitution. Substitute $x=-1$, so that the$B(x + 1)$ term is eliminated.
So, $x = −1$
\begin{align*} & −1−14 = A(2(−1)−3) \\ & −15 = A(−5) \\ & A = \frac{-15}{-5 } \\ &= 3 \end{align*}
• Then, substitute $x = 1.5$ to eliminate the $A(2x – 3)$ term.
\begin{align} &1.5 − 14 = A(2\times 1.5 − 3) + B(1.5 + 1) \\ &−12.5 = 2.5B \\ & B = −5 \end{align}
• Lastly, substitute the value of $A$ and of $B$ into the first equation.
Therefore, \begin{align*} \frac{(x-14)}{(x+1)(2x+3)}=\frac {3}{(x+1)}-\frac{5}{(2x-3)} \end{align*}.
## Case 2: Proper Fractions with Repeated Linear Factors in the Denominator
An algebraic fraction where the denominator contains a linear factor that is squared may be expressed in partial fractions as shown below.
\begin{align*} \frac {px+q}{(ax+b)^2}=\frac{A}{ax+b}+\frac{B}{(ax+b)^2} \end{align*}, where $A$ and $B$ are constants.
Question 2:
Express \begin{align*} \frac {x}{(x-1)(x+4)^2} \end{align*} in partial fractions.
Solution:
• Let \begin{align*} \frac {x}{(x-1)(x+4)^2}=\frac{A}{(x-1)}+\frac{B}{(x+4)}+\frac{C}{(x+4)^2} \end{align*}
• Multiply the equation throughout by the denominator $(x – 1)(x + 4)^2$
$x = A(x+4) + B(x−1)(x+4) + C(x−1)$
• Substitute $x = −4$,
\begin{align*} &−4 = C(−4−1) \\ &−4 = −5C \\ & C = \frac{4}{5} \end{align*}
• Substitute $x = 1$,
\begin{align*} & 1 = A(1+4)^2+ 0 + 0 \\ &1 = 25A \\ & A = \frac{1}{25} \end{align*}
• For the third substitution, choose any number (or a number that makes the equation easy to manipulate).
Substitute\begin{align*} x = 0, \;C = \frac{4}{5}, \;A = \frac {1}{25}, \end{align*}
\begin{align*} & 0 = \frac{1}{25}\times(0+4)^2 + B(0-1)(0+4) + \frac{4}{5} \times (0-1) \\ & 0 =\frac{16}{25} + B(−4) − \frac{4}{5} \\ & \frac{4}{5} − \frac{16}{25} = − 4B \\ & \frac{4}{25} = − 4B \\ & B = − \frac{1}{25} \end{align*}
• Lastly, substitute the values of $A$, $B$ and $C$ into the first equation.
\begin{align*} \frac{x}{(x-1)(x+4)^2}=\frac{1}{25(x-1)}-\frac{1}{25(x+4)}+\frac{4}{5(x+4)^2} \end{align*}
## Conclusion
In this article, we have covered two cases for splitting an algebraic fraction into partial fractions:
1. A proper algebraic fraction with distinct linear factors
\begin{align*} \frac{px+q}{(ax+b)(cx+d)}=\frac{A}{ax+b}+\frac{B}{cx+d} \end{align*}
1. A proper fraction with repeated linear factors
\begin{align*} \frac {px+q}{(ax+b)^2}=\frac{A}{ax+b}+\frac{B}{(ax+b)^2} \end{align*}
The only difference between these two forms is a square in the denominator. However, please take note of it as the subsequent steps will differ quite a bit!
Keep Learning! Keep Improving!
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Finding Extrema
by Justin Skycak on
Derivatives can be used to find a function's local extreme values, its peaks and valleys.
This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Finding Extrema. In Justin Math: Calculus. https://justinmath.com/finding-local-extrema/
Derivatives can be used to find a function’s local extreme values, its peaks and valleys. At its peaks and valleys, a function’s derivative is either $0$ (a smooth, rounded peak/valley) or undefined (a sharp, pointy peak/valley).
Critical Points
The points at which a function’s derivative is $0$ or undefined, and the function itself exists, are called critical points of the function. We can find the critical points by taking the derivative, noting any singularities, setting the derivative to $0$, and solving.
For example, to find the critical points of the function $f(x)=x \sqrt{1-x^2}$, we start by taking the derivative and simplifying.
\begin{align*} f(x)&=x \sqrt{1-x^2} \\ f'(x)& = \sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}} \\ &= \frac{1-2x^2}{\sqrt{1-x^2}} \end{align*}
The derivative has a singularity when the denominator $\sqrt{1-x^2}$ is $0$, which happens at $x=\pm 1$. The derivative itself is zero when the numerator $1-2x^2$ is $0$, which happens at $x=\pm \sqrt{ \frac{1}{2} }$. The function is defined at all of these x-values, so they all correspond to critical points: $x= \pm 1, \pm \sqrt{ \frac{1}{2} }$.
Classifying Critical Points
Now, how do we tell which critical points correspond to maxima (peaks), and which correspond to minima (valleys)?
It may be tempting to decide whether a critical point is a maximum or minimum by observing whether the resulting function value is large or small. However, it is entirely possible that some local minima may be greater than some local maxima. Think of a mountain range – some valleys may be higher than some peaks.
It may also be possible that some critical points are neither peaks nor valleys, but saddle points on the side of a mountain where the terrain is flat. At saddle points like the one indicated below, the derivative is $0$ but the point is neither a maximum nor a minimum.
First Derivative Test
There are two main methods for determining whether a critical point is a local minimum, local maximum, or neither. One way is to inspect the sign of the derivative on either side of the critical point, which tells whether we are ascending or descending on either side of the critical point.
• • If the derivative is positive to the left of the critical point and negative to the right of the critical point, then we are ascending to a peak and then descending down the peak, which tells us that the critical point is a local maximum.
• • On the other hand, if the derivative is negative to the left of the critical point and positive to the right of the critical point, then we are descending down a valley and then climbing up the valley, which tells us that the critical point is a local minimum.
• • Lastly, if the derivative does not switch sign from the left of the critical point to the right of the critical point, then we are either ascending up the whole way or descending down the whole way, which indicates that the critical point is a saddle point.
This method is called the first derivative test, because it makes use of the first derivative of the function.
Demonstration of First Derivative Test
To use the first derivative test on the critical points $x=\pm 1, \pm \sqrt{ \frac{1}{2} }$ that we found for the function $f(x)=x \sqrt{1-x^2}$, we first split up the number line over the critical points.
The number line splits into $5$ intervals:
\begin{align*} (-\infty,-1) \cup \left( -1,-\sqrt{\frac{1}{2}} \right) \cup \left( -\sqrt{\frac{1}{2}}, \sqrt{\frac{1}{2}} \right) \cup \left( \sqrt{\frac{1}{2}}, 1 \right) \cup (1,\infty) \end{align*}
However, on the intervals $(-\infty,1)$ and $(1,\infty)$ our function $f(x)=x\sqrt{1-x^2}$ is not defined because the argument of the square root becomes negative. We remove these intervals from consideration.
\begin{align*} \left( -1,-\sqrt{\frac{1}{2}} \right) \cup \left( -\sqrt{\frac{1}{2}}, \sqrt{\frac{1}{2}} \right) \cup \left( \sqrt{\frac{1}{2}}, 1 \right) \end{align*}
We want to know whether our function is increasing or decreasing on each of these intervals. To find out this information, we choose a test value in each of the remaining intervals. The actual values of the test values don’t matter, because the derivative maintains the same sign within any given interval. For the sake of example, we choose our test values as, say, $-0.9$, $0$, and $0.9$.
Lastly, we evaluate the sign of the derivative at each of these test values.
\begin{align*} f'(x) &= \frac{1-2x^2}{\sqrt{1-x^2}} \\ f'(-0.9) &= \frac{1-2(-0.9)^2}{\sqrt{1-(-0.9)^2}} = \frac{-}{\sqrt{+}} = - \\ f'(0) &= \frac{1-2(0)^2}{\sqrt{1-(0)^2}} = \frac{+}{\sqrt{+}} = + \\ f'(0.9) &= \frac{1-2(0.9)^2}{\sqrt{1-(0.9)^2}} = \frac{-}{\sqrt{+}} = - \end{align*}
The sign of the derivative at each particular test value tells us the sign of the derivative throughout the interval containing the particular test value. As a result, we know whether the function is increasing or decreasing on each interval, and we can sketch a rough graph of the peaks and valleys of the function.
We see that the function $f(x)=x\sqrt{1-x^2}$ has maxima at $x=-1,\sqrt{ \frac{1}{2} }$ and minima at $x=-\sqrt{ \frac{1}{2} }, 1$.
Second Derivative Test
The other method for classifying a critical point of a function as a maximum or minimum is called the second derivative test, because it makes use of the second derivative of the function.
• • If the second derivative is positive at the critical point, then the function is concave up in the shape of a smile, which means the critical point is a local minimum.
• • If the second derivative is negative at the critical point, then the function is concave down in the shape of a frown, which means the critical point is a local maximum.
• • If the second derivative is $0$ or undefined at the critical point, then we cannot conclude whether the critical point is a local maximum or minimum, and we need to fall back to the first derivative test.
The second derivative test is sometimes inconclusive, but it is mentioned because it is often faster than the first derivative test.
Demonstration of Second Derivative Test
To use the second derivative test on the critical points $x=\pm 1, \pm \sqrt{ \frac{1}{2} }$ that we found for the function $f(x)=x\sqrt{1-x^2}$, we first take the second derivative of the function. We computed the first derivative earlier, so we just have to differentiate once more.
\begin{align*} f'(x) &= \frac{1-2x^2}{\sqrt{1-x^2}} \\ f''(x) &= \frac{-4x\sqrt{1-x^2}+ \frac{x(1-2x^2)}{\sqrt{1-x^2}} }{1-x^2} \\ &= \frac{-4x(1-x^2)+ x(1-2x^2) }{(1-x^2)^\frac{3}{2}} \\ &= \frac{2x^3-3 x}{(1-x^2)^\frac{3}{2}} \end{align*}
We evaluate the sign of the second derivative at each of the critical points.
\begin{align*} f''(-1) &= \frac{-2+3}{(1-1)^\frac{3}{2}} = \frac{+}{0} = \mbox{undefined} \\ f'' \left( - \sqrt{\frac{1}{2}} \right) &= \frac{ \sqrt{\frac{1}{2}} +3 \sqrt{\frac{1}{2}} }{\left( 1- \frac{1}{2} \right)^\frac{3}{2}} = \frac{+}{+} = + \\ f'' \left( \sqrt{\frac{1}{2}} \right) &= \frac{ \sqrt{\frac{1}{2}} -3 \sqrt{\frac{1}{2}} }{\left( 1- \frac{1}{2} \right)^\frac{3}{2}} = \frac{-}{+} = - \\ f''(1) &= \frac{2-3}{(1-1)^\frac{3}{2}} = \frac{-}{0} = \mbox{undefined} \end{align*}
Based on the results of the second derivative test, we see that $x=-\sqrt{ \frac{1}{2} }$ is a minimum, and $x=\sqrt{ \frac{1}{2} }$ is a maximum. The test is inconclusive for $x=-1$ and $x=1$, so we would need to fall back to the first derivative test for these cases.
When to Use Each Test
In general, it’s a good idea to use the first derivative test when the second derivative is more complex than the first derivative, and the second derivative test when the second derivative is less complex than the first derivative.
For example, for polynomial functions, it is usually easiest to use the second derivative test because the second derivative is less complex than the first derivative.
\begin{align*} p(x) &= 2x^3-3x^2-11x+6 \\ p'(x) &= 6x^2 - 6x - 11 \\ p''(x) &= 12x-6 \end{align*}
We find the critical points by solving for where the first derivative is zero.
\begin{align*} 0 &= 6x^2 - 6x - 11 \\ x &= \frac{3 \pm 5\sqrt{3}}{6} \end{align*}
Then, we find the sign of the second derivative at these points.
\begin{align*} p''\left( \frac{3 - 5\sqrt{3}}{6} \right) &= 12 \left( \frac{3 - 5\sqrt{3}}{6} \right) -6 = 2(3 - 5\sqrt{3})-6 = -10 \sqrt{3} < 0 \\ p''\left( \frac{3 + 5\sqrt{3}}{6} \right) &= 12 \left( \frac{3 + 5\sqrt{3}}{6} \right) -6 = 2(3 + 5\sqrt{3})-6 = 10 \sqrt{3} > 0 \end{align*}
The critical point $x=\frac{3-5\sqrt{3}}{6}$ has a negative second derivative, which means the function is concave down and thus the critical point is a maximum. Likewise, the critical point $x=\frac{3+5\sqrt{3}}{6}$ has a positive second derivative, which means the function is concave up and thus the critical point is a minimum.
On the other hand, for the function below, it is easiest to use the first derivative test because the computations for the second derivative will get a bit messy when we use the product rule.
\begin{align*} f(x) &= e^{x^2 + \frac{1}{x}} \\ f'(x) &= \left( 2x - \frac{1}{x^2} \right) e^{x^2 + \frac{1}{x} } \end{align*}
We find the critical points by solving for where the first derivative is zero.
\begin{align*} 0 &= \left( 2x - \frac{1}{x^2} \right) e^{x^2 + \frac{1}{x} } \\ 0 &= 2x - \frac{1}{x^2} \\ 1 &= 2x^3 \\ \sqrt[3]{\frac{1}{2}} &= x \end{align*}
We choose test points $x=-1$ and $x=1$ on each side of our critical point, and evaluate the sign of the first derivative at these points.
\begin{align*} f'(-1) &= (-2-1)e^{1-1} = (-)(+) = - \\ f'(1) &= (2-1)e^{1+1} = (+)(+) = + \end{align*}
The function has a negative derivative to the left of the critical point and a positive derivative to the right of the critical point, which means it is descending to the critical point and then ascending from the critical point. Therefore, the critical point $x=\sqrt[3]{ \frac{1}{2} }$ is a minimum of the function.
Functions Defined on Closed Intervals
Lastly, when a function is defined on a closed interval, we need to use the endpoints as critical points as well, because the derivative isn’t defined there but the function is.
For example, to find the extrema of the function $f(x)=x^2+2x$ with $x \in [-3,5]$, we should also consider $x=-3$ and $x=5$ as critical points, in addition to the point $x=-1$ which makes the derivative $f’(x)=2x+2$ equal to zero.
To apply the first derivative test, we choose a test point $x=-2$ for the interval $[-3,-1)$ and $x=0$ for the interval $(-1,5]$.
\begin{align*} f'(-2) &= 2(-2)+2 = - \\ f'(0) &= 2(0)+2 = + \end{align*}
The function is decreasing from $x=-3$ to $x=-1$, and then increasing from $x=-1$ to $x=5$. Therefore, the function has a minimum at $x=-1$ and maxima at $x=-3$ and $x=5$.
Exercises
For each function, find the critical points and label each critical point as a local maximum, local minimum, or saddle point. (You can view the solution by clicking on the problem.)
\begin{align*}1) \hspace{.5cm} f(x)= x^3-2x+1 \end{align*}
Solution:
\begin{align*} x &= -\sqrt{ \frac{2}{3} } \hspace{.5cm} \text{(max)} \\ x &= \sqrt{ \frac{2}{3} } \hspace{.5cm} \text{(min)} \end{align*}
\begin{align*}2) \hspace{.5cm} f(x)= x^4+10x^3-5 \end{align*}
Solution:
\begin{align*} x &= -\frac{15}{2} \hspace{.5cm} \text{(min)} \\ x &= 0 \hspace{.5cm} \text{(saddle)} \end{align*}
\begin{align*}3) \hspace{.5cm} f(x)= x \ln x \end{align*}
Solution:
\begin{align*} x &= \frac{1}{e} \hspace{.5cm} \text{(min)} \end{align*}
\begin{align*}4) \hspace{.5cm} f(x)= \ln(5x-x^2) \end{align*}
Solution:
\begin{align*} x &= \frac{5}{2} \hspace{.5cm} \text{(max)} \end{align*}
\begin{align*}5) \hspace{.5cm} f(x)= \frac{x^2+1}{x+1} \end{align*}
Solution:
\begin{align*} x &= -1-\sqrt{2} \hspace{.5cm} \text{(max)} \\ x &= -1+\sqrt{2} \hspace{.5cm} \text{(min)} \end{align*}
\begin{align*}6) \hspace{.5cm} f(x)= \frac{1}{x^2} e^{x^2-1} \end{align*}
Solution:
\begin{align*} x &= -1 \hspace{.5cm} \text{(min)} \\ x &= 1 \hspace{.5cm} \text{(min)} \end{align*}
\begin{align*}7) \hspace{.5cm} & f(x) = 4-x^3 \\ &x \in [-1,2] \end{align*}
Solution:
\begin{align*} x &= -1 \hspace{.5cm} \text{(max)} \\ x &= 0 \hspace{.5cm} \text{(saddle)} \\ x &= 2 \hspace{.5cm} \text{(min)} \end{align*}
\begin{align*}8)\hspace{.5cm} &f(x) = x^3 - 4x^2 \\ &x \in [-3,3] \end{align*}
Solution:
\begin{align*} x &= -3 \hspace{.5cm} \text{(min)} \\ x &= 0 \hspace{.5cm} \text{(max)} \\ x &= \frac{8}{3} \hspace{.5cm} \text{(min)} \\ x &= 3 \hspace{.5cm} \text{(max)} \end{align*}
\begin{align*}9) \hspace{.5cm} &f(x) = \frac{x^2}{x+1} \\ &x \in [-5,5] \end{align*}
Solution:
\begin{align*} x &= -5 \hspace{.5cm} \text{(min)} \\ x &= -2 \hspace{.5cm} \text{(max)} \\ x &= 0 \hspace{.5cm} \text{(min)} \\ x &= 5 \hspace{.5cm} \text{(max)} \end{align*}
\begin{align*}10) \hspace{.5cm} f(x) &= \frac{1+\ln x}{x} \\ &x \in \left[ \frac{1}{2}, 10 \right] \end{align*}
Solution:
\begin{align*} x &= \frac{1}{2} \hspace{.5cm} \text{(min)} \\ x &= 1 \hspace{.5cm} \text{(max)} \\ x &= 10 \hspace{.5cm} \text{(min)} \end{align*}
This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Finding Extrema. In Justin Math: Calculus. https://justinmath.com/finding-local-extrema/
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Lesson Explainer: Linear Momentum | Nagwa Lesson Explainer: Linear Momentum | Nagwa
# Lesson Explainer: Linear Momentum Mathematics • Third Year of Secondary School
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In this explainer, we will learn how to calculate the momentum of a particle moving in a straight line using the formula .
Imagine two objects: a truck moving at 30 mph along a road and a paper airplane moving at 2 mph through the air. Which object would require a greater force to stop it in the same amount of time? Intuitively, we know that the truck would require the greater force to stop it because it has a greater mass and it is moving faster. We can say that the truck has a greater momentum. Momentum can be thought of as a measure of how difficult it is to stop an object that is moving.
The two factors that contribute to an objectβs momentum are its mass, , and its velocity, . The greater the mass of the object, the greater its momentum, and similarly the greater the velocity of the object, the greater its momentum. Knowing this, we can now define momentum mathematically.
### Definition: Momentum
The momentum of an object, , is equal to its mass, , multiplied by its velocity, :
Since velocity is a vector quantity and mass is a scalar quantity, momentum is a vector quantity. However, often we just want the magnitude of the momentum, which we can write as
Since mass is a scalar quantity, we can take it outside of the magnitude sign, giving us
On the right-hand side, is just the magnitude of the velocity, or the speed. We can denote the magnitude of the momentum as just and the speed as just , giving us
This is the scalar form of the definition of momentum, and this is what we will end up using most of the time.
So, for example, imagine a bowling ball with a mass of 12 kg moving at a speed of 5 m/s along the lane of a bowling alley. What is the momentum of the bowling ball? We can substitute these values into to get , giving us 60 kgβ
m/s.
This shows us that the standard unit of momentum is kgβ
m/s, or βkilogram-metres per second,β but momentum can be measured in other units; in fact, any unit that is a unit of mass multiplied by a unit of speed is a valid unit of momentum.
Letβs have a look at some example questions.
### Example 1: Finding the Momentum of an Object given Its Speed
Determine the momentum of a car of mass 2.1 tonnes moving at 42 km/h.
The mass of the car is 2.1 tonnes, and the speed of the car is 42 km/h. These are the two factors that affect an objectβs momentum, and we can substitute them straight into the formula to find the momentum:
Here, we have used the symbol for the tonne, which is t. This answer is in units of tonne-kilometres per hour, which is a slightly unusual unit, but still a valid unit of momentum.
### Example 2: Finding the Momentum of a Falling Body after a Given Distance
Calculate the momentum of a stone of mass 520 g after it has fallen 8.1 m vertically downward. Consider the acceleration due to gravity to be .
In this question, we are not told what the speed of the stone is, which we need to know in order to calculate the momentum, so we are going to have to work that out first.
We are told how far the stone falls, and what its acceleration is as it falls. The question says nothing about the initial motion of the stone, so we can assume that initially the stone is not moving. Knowing these three pieces of information, there is a formula that we can use to find the speed of the particle after it has fallen 8.1 mβit is one of the kinematic equations: where is the initial speed of the stone, is the acceleration of the stone, is the distance that the stone moves, and is the final speed of the stone. Since is zero, is also zero, and this formula simplifies to just
We are trying to find , so we want to make the subject of the equation. To do this, we need to square root each side of the equation:
We can now substitute in the values we have for and :
So the stone starts at rest; it then begins to fall at 9.8 m/s2 due to gravity, and by the time it has fallen a distance of 8.1 m, it is moving at a speed of 12.6 m/s.
The other quantity that we need to know in order to calculate the momentum of the stone is its mass, which we were given. The mass was given in grams; letβs convert that to the standard unit of kilograms. There are 1βββ000 g in 1 kg, so 520 g = 0.52 kg.
We can now substitute the values for the mass and speed of the stone into the formula for momentum:
So by the time the stone had fallen a distance of 8.1 m, it had a momentum of 6.552 kgβ
m/s.
### Example 3: Finding the Increase in Momentum of a Body Moving with Uniform Acceleration after a Given Time
A body of mass 17 kg moves in a straight line with constant acceleration of 1.8 m/s2. Its initial velocity is 22.3 m/s. Find the increase in its momentum in the first 5 seconds.
This question asks us to find the increase in momentumβthe change in momentumβover a given time period. In order to do that, we need to work out the difference between its final momentum and its initial momentum. We can express this mathematically as where is the change in momentum, is the final momentum, is the initial momentum, is the final speed, is the initial speed, and is the mass.
We already know the initial speed of the object and its mass; the only thing we do not know is its final speed, so we will have to work that out first.
Since the object undergoes constant acceleration, we can use one of the kinematic equations to find the final speed of the object: where is the acceleration of the object and is how long the object accelerates for. We know all three values that are used on the right-hand side of this equation, so we can use it to find , the final speed:
So we now know the initial speed, the final speed, and the mass of the object. We can substitute these values into our equation from earlier to find the change in momentum:
So, in the first 5 seconds of its motion, the momentum of the object increases by 153 kgβ
m/s.
### Example 4: Finding the Momentum of an Object at a Given Time given Its Displacement with respect to Time
A car of mass 1βββ350 kg moves in a straight line such that at time seconds its displacement from a fixed point on the line is given by . Find the magnitude of the carβs momentum at .
In this question, we are given a function for the position of the car that depends only on time. In order to find the momentum of the car at a particular time, we are going to need to know its velocity at that time. In order to get the velocity at a particular time, we are going to need to get a general function for the velocity of the car over time.
Recall that the velocity of an object, , is defined as the rate of change of the displacement of the object. It is the derivative of the displacement of the object with respect to time:
Since this question is just about an object moving in one dimension, we can just use scalar quantities to represent the velocity, , and the displacement, , giving us
Taking the derivative with respect to of the function for that we were given in the question, we get
We want to know the speed of the car at , so letβs substitute this value for the time into the above equation:
We now know both the speed of the car and the mass of the car, so we can substitute these values into the formula for the momentum of the car:
At , the car has a momentum of 44βββ550 kgβ
m/s.
### Example 5: Finding the Momentum of a Body at a Given Time given an Expression of Its Position with Time
A body of mass 7 kg is moving in a straight line. Its position vector at a time is given by the relation , where is measured in metres and in seconds. Determine the bodyβs momentum after 2 seconds.
In this question, we are given a function for the position of the object that depends only on time. In order to find the momentum of the object at a particular time, we are going to need to know the velocity of the object at that time. In order to get the velocity at a particular time, we are going to need to get a general function for the velocity of the object over time.
The velocity, , of an object is the derivative of its displacement, , with respect to time:
The momentum, , of an object is equal to its mass, , times its velocity:
We now have a formula that will give us the vector momentum of the object at a time . We can substitute the values for the mass of the object and the time into the equation:
The momentum of the object after 2 seconds is kgβ
m/s.
At the start of this explainer, we wrote a mathematical definition of momentum:
If we take the derivative with respect to time of each side of the equation, we can see that the derivative of the momentum with respect to time is equal to the mass of the object, , times the acceleration of the object, . But is also equal to the force on the object, , so
In the case where the acceleration of an object is constant or we know the average acceleration, we can find the change in momentum, , of an object if we know its mass, its acceleration, and the time for which it accelerates, : or, in scalar form,
This formula only works if our acceleration, , is constant. In the case where acceleration changes as time, , changes, we can use the equation we found earlier for the rate of change of momentum with respect to time, and integrate both sides of this formula with respect to . Doing this will give us
On the left, we can see that the integral removes the change in time and all we are left with is a change in momentum. On the right, we can take out of the integral since it is a constant. Let us now consider what this formula would look like if we were trying to find the change in momentum over the time interval : or, in scalar form,
Let us look at an example of how this can be used to find a change in momentum.
### Example 6: Finding the Change in Momentum of a Body given Its Acceleration
A body of mass 5 kg moves along a straight line. At time seconds, its acceleration is given by . Find the change in momentum in the time interval .
We have been given the acceleration, , of a body in terms of time, , and also as a scalar. Therefore, in order to find its change in momentum, we can use the following formula:
From the question, we can see that we have , , , and . Substituting these values into our formula, we have
Using the power rule for integration, we increase the powers of by 1 and then divide each term by the new power of . This gives us
Now, we just need to substitute in our values for and simplify as follows:
We must not forget to include the units, which gives us our solution of
### Key Points
• The momentum, , of an object is the product of its mass, , and its velocity, : .
• Momentum is typically measured in units of kilogram-metres per second, or kgβ
m/s.
• Sometimes, we may have to use the kinematic equations to find the velocity of an object in order to calculate its momentum.
• If we are given a function for the position of an object at a time , we can take the derivative of that function with respect to time to get a function for the velocity and then use that function in calculating the momentum.
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## Law of Exponents - II
Now we can expand the laws of exponents a little bit further. Back in the arithmetic module, we learned about the distributive law. And really, the distributive law is one of the big ones, it’s really one of the big mathematical ideas. And of course, the distributive law says that P to the M plus or minus N, what we can do is just multiply the P separately times each one of those terms.
That is the Distributive Law. Multiplication distributes over addition and subtraction. As it turns out, division also distributes over addition and subtraction. Much in the same way, exponents distribute over multiplication and division. So if I have a times b to the n, or a divided by b to the n, I can distribute the exponent to each factor.
So a times b to the n equals a to the n times b to the n, a divided by b, that fraction to the n, equals a to the n divided by b to the n. So, we can distribute an exponent across multiplication or division. Here’s a very quick numerical example. Suppose we have 18 to the 8th, well we know that we could write 18 as a product, we could write it as its prime factorization.
And of course, the prime factorization of 18 is 2 times 3 squared. So, 18 to the 8th equals 2 times 3 to the 8th, while we can distribute that exponent to each one of those factors,. We’ll get it 2 to the 8th and then we’ll get a 3 squared to the 8th. And for 3 squared to the 8th, of course you’ll use the rule for the product of, for a power to a power, which means multiply the exponents.
And we’ll get 2 to the 8th times 3 to the 16th. So notice it’s very easy to go from the prime factorization to the, from, of the individual number, to the prime factorization of one of its powers. Here’s a practice problem. Pause the video and then we’ll talk about this. Okay, in the numerator, all we’re gonna do is multiply out that 4.
We’re just going to distribute it to each one of those terms. And for each one of those terms, we’re going to have a power to a power, which means multiply the exponents. So we’re gonna wind up with x to the 8th y to the 12th. Then we have the, we have to deal with the division. Well, x to the 8th divided by x to the 5th.
We subtract the exponents that’s going to be x cubed. Y to the 12th divided by y to the negative 5th. That’s going to be 12 minus negative 5 which is 12 plus 5 which is 17 and that’s why we get x cubed y to the 17th. It’s important to be aware of a very common and tempting trap, because it’s close to what is true.
So now we are going to talk about a trap. First of all it’s legal to distribute multiplication over addition and subtraction. That’s 100% legal. It’s legal to distribute exponents over multiplication and division. That’s 100% legal.
But it’s illegal to distribute an exponent over addition and subtraction. So that line, that’s just the distributive log, that’s 100% legal. That’s one of the fundamental patterns in mathematics. This is an also, it is also a version of the distributive law, we are distributing the exponent over multiplication and division, that’s also 100% legal. The thing that is illegal, is distributing the exponent over addition or subtraction.
That is always illegal. And in fact m plus or minus n to the p means that we’re taking that, what’s in the parenthesis, m plus or minus n, and multiplying it by itself p times. So these were variables we’d have to foil out several times. So you’ll never actually have to do that, but it’s just important to keep in mind that, that’s what it would be, not multiplying, not raising the individual terms of those powers.
And I’ll say this is a very tricky one, because even when you understand that this third line is illegal, the human brain’s inborn pattern matching software is tempted to make that mistake again, especially when you are under pressure. So you really need to know this cold so that even when you walk into the test and you’re stressed in the middle of the test you don’t accidentally fall into making this mistake again because it is a very tempting trap.
Again let’s look at all this with numbers. Here is just the ordinary distributive law with numbers. Multiplication distributing over addition. Here is the distributive law with powers. So that exponent distributing over multiplication and division, but it would be illegal if we had 8 plus or minus 5 to the 3rd.
That would not be 8 to the 3rd plus or minus 5 to the 3rd. And one way to see this is to just think let’s just take the subtraction case. If we look at 8 minus 5 to the 3rd, well what is that? Of course that is 3 to the 3rd which is 27. Whereas if we looked at something different, 8 cubed minus 5 cubed. Well 8 cubed as we’ve mentioned in other videos is 512.
5 cubed is 125 and if we subtract them we get 387. And those two are not equal. So in other words, we get two different numerical answers and that’s why we can’t set those things equal. We can do some legal math when sums are differences of power.
First we need to go back to that most impressive pattern, the Distributive Law. Now this is very tricky. When we read this equation from left to right, we say that we are distributing P. When we read this equation from right to left, we say that we are factoring out P. So distributing and factoring out are two sides of the same coin. It’s just a matter of whether we are reading this equation going from left to right or from right to left, but it’s the same fundamental pattern.
It’s also important to remember that any higher power of a base is divisible by any lower power of that same base. Thus, in the sum of a higher power and lower power of the same base. The greatest common factor of the two terms is the lower powner, power, and this can be factored out because the lower power is always a factor of a higher power.
So, for example, 17 to the 30th plus 17 to the 20th. Well, first of all, we know that 17 to the 30th has to be divisible by 17 to the 20th. We know one is a factor of the other. And so 17 to the 20th is the greatest common factor of these two terms. So I’m gonna factor that out.
17 to the 30th, I can write that as 17 to the 20th times 17 to the 10. By the, by the multiplication of powers law, I can write it that way. And, of course, 17 to the 20th, I can write that as 17 to the 20th times 1. I factor out 17 to the 20th, and I get 17 to the 20th times parentheses, 17 to the 10th plus 1. And that is a factored out form of those powers.
Obviously, the powers here are too large to simplify any of these resultant terms, but if the two powers in the sum are closer, sometimes such simplification is easy. So I will say pause the video and see if you can simplify this and then we’ll talk about this. Okay.
3 to the 32 minus 3 to the 28. 3 to the 28th is a factor of 3 to the 32nd. In fact, 3 to the 28th is the greatest common factor of these two terms. So we’re gonna express both of them as products involving 3 to the 28th. So 3 to the 32, we can write that as 3 to the 28th times 3 to the 4th. And of course 3 to the 28th, we can write that as 3 to the 28th times 1.
Factor out 3 to the 28, we get 3 to the 4th minus 1. Now 3 to the 4th, that’s something we can calculate. 3^4. So one way to think about that is if you have memorized 3 to the 4 is 81, also 3 to the 4th is 3 squared square. Well 3 squared is 9 and 9 squared is 81.
So that simplifies to 81. I do 80 min, 81 minus 1. That’s 80. And so this is 80 times 3 to the 28. We don’t often have to solve for something in the exponent, because this usually involves much more advanced ideas than are found on the test.
The test will expect us to know that if the bases are the same, if we have be to the s equals b to the t. Then it must mean that the exponents are equal. That will be very important in the lesson, equations with exponents, which we’ll get to much later in the module. In summary, exponents distribute over multiplication and division, and those are the patterns.
Exponents do not distribute over addition or subtraction. Those are very tempting mistake patterns and we can simplify the sum or difference of powers by factoring out the lower power. And finally, if we have bases are equal and we have a to the m equals a to the n we can equate the exponents.
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# Lesson 6 The Coordinate Plane
Coordinate Plane
y
Axes (plural of axis)
0 x-axis
0 y-axis
x
Coordinate Plane
y
Origin
x
Coordinate Plane
y
II
I
III
IV
x
Coordinates
Plot and label the points:
0 A(4,7)
0 B(-3,5)
0 C(4,-7)
0 D(-6,-3)
each point is graphed.
Coordinates
State the coordinates:
0A
0B
0C
0D
0E
0F
(5,2)
(-8,-5)
(-3,5)
(4,-7)
(0,4)
(-7,0)
C
E
A
F
B
D
Distance in the coordinate plane
Kayleigh walks to a friend’s house by traveling 5 blocks
west and then 12 blocks north. She returns home later
by cutting diagonally through her neighbors’ yards.
What distance did she travel on her voyage home?
13
12
5
a 2 b2 c 2
52 122 c 2
25 144 c 2
169 c 2
169 c 2
13 c
Distance formula
0 The distance between two points x1 , y1 and x2 , y2 on
the coordinate plane (graph) can be found using the
formula:
d
x2 x1 y2 y1
2
2
http://www.bsin.k12.nm.us/schools/BHS/math/joinmath/geometry/GEO1_3.pps#257,4,Slide 4
Practice
Find the distance between R(–2, –6) and S(6, –2) to the nearest tenth.
d=
(x2 – x1)2 + (y2 – y1)2
d=
(6 – (–2))2 + (–2 – (–6))2
d=
82 + (–8)2
d=
64 + 64 =
128
To the nearest tenth, RS = 11.3.
Midpoint Formula
In the coordinate plane, the coordinates of the midpoint
of a segment whose endpoints have coordinates ( x1 , y1 )
and ( x2 , y2 ) are x1 x2 , y1 y2 .
2
2
Example: Find the midpoint between (-2, 5) and (6, 4)
x1 y1
M=
2 6 , 5 4
2
2
M=
4 , 9 = 2, 9
2 2 2
x2 y2
Practice
AB has endpoints (8, 9) and (–6, –3). Find the coordinates of its midpoint M.
(
x1 + x2
2
8 + (–6)
2
=
=1
2
2
,
y1 + y2
2
)
9 + (–3)
6
=
=3
2
2
The coordinates of midpoint M are (1, 3).
Homework
The Coordinate Plane in
Student Practice Packet
(Page 7, #1-13) |
# HBSE 6th Class Maths Solutions Chapter 8 Decimals Intext Questions
Haryana State Board HBSE 6th Class Maths Solutions Chapter 14 Practical Geometry Ex 14.5 Textbook Exercise Questions and Answers.
## 14.5
Try These (Page 211) :
Question 1.
Can you now write the following numbers as decimals ?
Hundreds Tens Ones Tenths (100) (10) (1) (1/10) (i) 5 3 8 1 (ii) 2 7 3 4 (iii) 3 5 4 6
Solution:
(i) 538.1,
(ii) 273.4,
(iii) 354.6.
Question 2.
Write the lengths of Ravi’s and Raju’s pencils in cm using decimals.
Solution:
Length of Ravi’s pencil
= 7 cm 5 mm = 7.5 cm.
Length of Raju’s pencil
= 8 cm 3 mm = 8.3 cm.
Question 3.
Make three more examples similar to the one given in question 1 and solve them.
Solution:
(i) Ram has rupees 40 and seventy-five paise and Shyam has rupees 70 and fifty paise. Express these amounts in decimal fractions.
(ii) The length of a wire is 6 metre and 30 cm. Write this length of wire in metres.
(iii) Pallavi and Ashu measured the lengths of their pencils. Pallavi’s pencil was 6 cm and 4 mm long and Ashu’s pencil was 5 cm 7 mm. Write the lengths of their pencils in cm.
Solution:
(i) Rs. 40 and seventy five paise
= Rs. 40.75
Rs. 70 and fifty paise
= Rs. 70.50
(ii) The length of the wire
= 6 metre and 30 cm = 6.30 m.
(iii) Length of Pallavi’s pencil
= 6 cm and 4 mm = 6.4 cm.
Length of Ashu’s pencil
= 5 cm and 7 mm = 5.7 cm.
Try These (Page 212):
Question.
Represent the following numbers on the number line :
(i) 0.6, (ii) 1.2, (Hi) 2.3.
Solution:
(i) We know that 0.6 is more than zero but less than one. There are 6 tenths in it. Divide the unit length between 0 and 1 into 10 equal parts and take 6 parts as shown below:
(ii) We know that 1.2 is more than one but less than two. There are 2 lengths in it. Divide the unit length between 1 and 2 into 10 the equal parts and take 2 parts as shown below :
(iii) We know that 1.2 is more than one but less than two. There are 2 lengths in it. Divide the unit length between 1 and 2 into 10 the equal parts and take 2 parts as shown below :
Try These (Page 213) :
Question.
Write $$\frac{3}{2}, \frac{4}{5}, \frac{8}{5}$$ in decimal notation.
Solution:
Decimals as Fractions :
Write a decimal number 1.2 as a fraction.
Solution:
1.2 = 1.2 $$=\frac{10}{10}+\frac{2}{10}=\frac{10+2}{10}=\frac{12}{10}=\frac{6}{5}$$
Try These
Question 1.
David was measuring the length of his room. He found that the length of his room is 4 m and 25 cm. He wanted to write the length in metres. Can you help him.
Solution:
4 m and 25 cm = 4 + $$\frac{25}{100}$$ = 4.25 m.
Question 2.
Fill in the blanks :
Ordinary fraction Decimal fraction
Solution:
(a) 0.50,
(b) 0.92.
Question 3.
Write the number in decimal fraction:
Solution:
The number is 2 + $$\frac{4}{10}+\frac{3}{100}$$ = 2 + 0.4 + 0.03 = 2.43.
Try These (Page 224) :
Question.
Compare :
(i) 1.82 and 1.823
(ii) 5.7 and 4.9
(iii) 6.05 and 6.50
(iv) 3.15 and 3.18.
Solution:
Try These (Page 225):
(i) Write 2 rupees 5 paise and 2 rupees 50 paise in decimals.
(it) Write 20 rupees 7 paise and 21 rupees 75 paise in decimals.
Solution:
(i) We know that
100 paise = 1 Re
.’. 1 paise = $$\frac{1}{100}$$ Re = 0.01 Re
.’. 5 paise = $$\frac{5}{100}$$ Re = 0.05 Re
Hence, 2 rupees 5 paise 2.05 Rs.
and 50 paise = $$\frac{50}{100}$$ = 0.5 Re
Hence, 2 rupees 50 paise = 2.50 Rs.
(ii) We know that
100 paise = 1 Re.
1 paise = $$\frac{1}{100}$$ Re = 0.01 Re
7 paise = $$\frac{7}{100}$$ = 0.07 Re
Hence, 20 rupees 7 paise = 20.07 Rs. 75
75 paise = $$\frac{75}{100}$$ = 0.75 Re.
Hence, 21 rupees 75 paise = Rs. 21.75.
Try These (Page 226) :
(1) Can you write 4 mm in cm using decimals ?
(2) How will you write 7 cm 5 mm in cm using decimals ?
Solution:
(1) 4mm = $$\frac{4}{10}$$ cm = 0.4 cm
(2) 7 cm 5 mm = 7 + [atex]\frac{5}{10}[/latex] = 7.5 cm.
Try These (Page 227):
Question.
(a) Can you write 456 g as kg using decimals ?
(b) How will you write 2 kg 9 g in kg using decimals ?
Solution:
(a) We know that
1000 g = 1 kg.
1 kg = $$\frac{1}{1000}$$ kg = 0.001 kg
Hence, 456 g = $$\frac{456}{1000}$$ = 0.456 kg.
(b) We know that
1000 g = 1 kg.
1 kg = $$\frac{1}{1000}$$ kg = 0.001 kg
Hence 2 kg 9 g = 2 + $$\frac{9}{1000}$$
= 2 + 0.009
= 2.009 kg.
Try These (Page 229):
Find: (i) 0.29 + 0.36
(ii) 0.7 + 0.08
(iii) 1.54 + 1.80
(iv) 2.66 + 1.85.
Solution:
Try These {Page 232):
(i) Subtract 1.85 from 5.46
(ii) Subtract 5.25 from 8.28
(iii) Subtract 0.95 from 2.29
(iv) Subtract 2.25 from 5.68.
Solution: |
# 2022 AMC 10A Problems/Problem 1
The following problem is from both the 2022 AMC 10A #1 and 2022 AMC 12A #1, so both problems redirect to this page.
## Problem
What is the value of $$3+\frac{1}{3+\frac{1}{3+\frac13}}?$$ $\textbf{(A)}\ \frac{31}{10}\qquad\textbf{(B)}\ \frac{49}{15}\qquad\textbf{(C)}\ \frac{33}{10}\qquad\textbf{(D)}\ \frac{109}{33}\qquad\textbf{(E)}\ \frac{15}{4}$
## Solution 1
We have \begin{align*} 3+\frac{1}{3+\frac{1}{3+\frac13}} &= 3+\frac{1}{3+\frac{1}{\left(\frac{10}{3}\right)}} \\ &= 3+\frac{1}{3+\frac{3}{10}} \\ &= 3+\frac{1}{\left(\frac{33}{10}\right)} \\ &= 3+\frac{10}{33} \\ &= \boxed{\textbf{(D)}\ \frac{109}{33}}. \end{align*} ~MRENTHUSIASM
## Solution 2
Continued fractions with integer parts $q_i$ and numerators all $1$ can be calculated as \begin{align*} \dfrac{[q_0,q_1,q_2,\ldots,q_n]}{[q_1,q_2,\ldots,q_n]} \end{align*} where \begin{align*} []&=1 \\ [q_0,q_1,q_2,\ldots,q_n]&=q_0[q_1,q_2,\ldots,q_n]+[q_2,\ldots,q_n]\\ \end{align*}
\begin{align*} [3]&=3(1) = 3\\ [3,3]&=3(3)+1=10\\ [3,3,3]&=3(10)+3=33\\ [3,3,3,3]&=3(33)+10=109\\ \dfrac{[q_0,q_1,q_2,\ldots,q_n]}{[q_1,q_2,\ldots,q_n]}&=\dfrac{[3,3,3,3]}{[3,3,3]}\\ &=\boxed{\textbf{(D)}\ \frac{109}{33}} \end{align*} ~lopkiloinm
## Solution 3
Finite continued fractions of form $n+\frac{1}{n+\frac{1}{n+\cdots}}=\frac{x}{y}$ have linear combinations of $x, y$ that solve Pell's Equation. Specifically, the denominator $y$ and numerator $x$ are solutions to the Diophantine equation $(n^2+4)\left(\frac{y}{2}\right)^2-\left(x-\frac{ny}{2}\right)^2=\pm{1}$. So for this problem in particular, the denominator $y$ and numerator $x$ are solutions to the Diophantine equation $13\left(\frac{y}{2}\right)^2-\left(x-\frac{3y}{2}\right)^2=\pm{1}$. That leaves two answers. Since the number of $1$'s in the continued fraction is odd, we further narrow it down to $13\left(\frac{y}{2}\right)^2-\left(x-\frac{3y}{2}\right)^2=-1$, which only leaves us with $1$ answer and that is $(x,y)=(109,33)$ which means $\boxed{\textbf{(D)}\ \frac{109}{33}}$.
~lopkiloinm
(Note: Integer solutions increase exponentially, so our next solution will have a numerator greater than $3^2(109)$. Therefore, when you don't see numerators greater than $3^2(109)$ in the answer choices, this method should be fine.)
## Video Solution 1
~Education, the Study of Everything
~Charles3829
~Math-X |
Lesson Video: Maclaurin Series | Nagwa Lesson Video: Maclaurin Series | Nagwa
# Lesson Video: Maclaurin Series Mathematics • Higher Education
In this video, we will learn how to find Maclaurin series of a function and find the radius of convergence of the series.
13:18
### Video Transcript
In this video, weβll learn the definition of the Maclaurin series and its relationship to the Taylor series. Weβll look at how to apply this definition to represent an arbitrary function as the Maclaurin series and how we can find the radius of convergence of these types of series.
We might recall that if a function π has a power series expansion at π, the Taylor series of a function π about π is given by π of π₯ equals the sum of the πth derivative of π evaluated π over π factorial times π₯ minus π to the πth power for values of π between zero and β. This can be written as π of π plus π prime of π over one factorial times π₯ minus π plus π double prime of π over two factorial times π₯ minus π squared and so on. For the special case where π is equal to zero, the Taylor series becomes the sum from π equals zero to β of the πth derivative evaluated at zero over π factorial times π₯ to the πth power. Now, this case arises frequently enough that itβs given the special name the Maclaurin series. And weβre going to have a look at how we might apply this formula.
Find the Maclaurin series of the function π of π₯ equals π to the power of π₯.
We recall that the Maclaurin series of a function π of π₯ is given by the sum of the πth derivative of π evaluated at zero over π factorial times π₯ to the πth power for values of π between zero and β. So weβll begin by differentiating π of π₯ to evaluate π prime of zero, π double prime of zero, and so on. Ultimately, weβre hoping to find what the πth derivative of our function is at π₯ equals zero. We know that π of π₯ is equal to π to the power of π₯. The derivative of π to the power of π₯ is simply π to the power of π₯. So we find that π prime of π₯ is π to the power of π₯. π double prime of π₯ is the derivative of this, which is also π to the power of π₯. And we can continue this. And weβll see that the πth derivative of π will also be equal to π to the power of π₯.
We need to work out what π of zero, π prime of zero, and so on is. So π of zero is π to the power of zero, which is one. π prime of zero is also π to the power of zero, which is one. And since we know that the πth derivative of π at π₯ is π to the power of π₯, we know that the πth derivative of π at zero is π to the power of zero, which is one. We therefore replace the πth derivative of π evaluated at zero with one. And we find that our summand is one over π factorial times π₯ to the πth power. We can, of course, simplify this to π₯ to the πth power over π factorial. And so we see that the Maclaurin series of the function π of π₯ equals π to the power of π₯ is the sum from π equals zero to β of π₯ to the πth power over π factorial. And this is one plus π₯ plus π₯ squared over two factorial plus π₯ cubed over three factorial and so on.
So what about the radius of convergence of this series? We recall that if a series converges on the open interval from negative π
to π
, its radius of convergence is the number denoted by π
. And of course, a power series always converges absolutely within its radius of convergence. Similarly, the interval of convergence is the open, closed, or semiclosed range of values of π₯, for which the Maclaurin series converges to the value of the function.
So letβs have a look at how we can find the radius of convergence for a Maclaurin series.
A Maclaurin series is given by the sum from π equals one to β of negative one to the power of π plus one times π times π₯ to the power of π minus one. Find the radius of convergence for the series.
We recall that we can use the ratio test to help us find the radius of convergence of a series. The part of the ratio test weβre interested in says, suppose we have the series given by the sum of π π. If the limit as π approaches β of the absolute value of π π plus one over π π is less than one, then the series is absolutely convergent and hence convergent. In our case, we define π π to be equal to negative one to the power of π plus one times π times π₯ to the power of π minus one. This means that π π plus one is negative one to the power of π plus two times π plus one times π₯ to the πth power. And so to find the radius of convergence, we need to work out where the limit as π approaches β of the absolute value of the quotient of these is less than one.
Now, we can actually simplify everything inside our limit. We recall that when we divide two numbers with an equal base, we subtract the exponents. This means negative one to the power of π plus two divided by negative one to the power of π plus one is simply negative one. Similarly, π₯ to the power of π divided by π₯ to the power of π minus one is simply π₯. Letβs rewrite this as the limit as π approaches β of the absolute value of negative one π₯ times π plus one over π. π₯ is independent event. So we can take the absolute value of negative π₯ outside of our limit.
And we can now look to evaluate our limit. We divide each term on the numerator by π. And weβre looking to find the limit as π approaches β of one plus one over π. Well, as π gets larger, one over π gets smaller. And it approaches zero. This means that the limit as π approaches β of the absolute value of one plus one over π is simply one. So we have that the absolute value of negative π₯ times one is less than one. Or indeed, the absolute value of negative π₯ is less than one.
We can rewrite this as the absolute value of negative one times the absolute value of π₯. And then, of course, the absolute value of negative one is simply one. So by the ratio test, the absolute value of π₯ must be less than one for our series to be convergent. And therefore, the radius of convergence is equal to one.
Weβll now look at how we can derive a Maclaurin series for a trigonometric function.
Consider the function π of π₯ equals cos of π₯. What are the first four derivatives of π with respect to π₯? Write the general form for the πth derivative of π with respect to π₯. And hence, derive the Maclaurin series for cos of π₯.
There is a fourth part to this question, which asks us to find the radius π
of convergence of the Maclaurin series for cos of π₯. And weβll consider that at the very end. So we begin by finding the first four derivatives of π with respect to π₯. Weβre told that π of π₯ is equal to cos of π₯. And so we quote the general result for the first derivative of cos of π₯ is negative sin π₯. And therefore, π prime of π₯, the first derivative of π with respect to π₯, is negative sin π₯.
Next, we recall the general result for the derivative of sin π₯. Itβs cos π₯. And this means the derivative of negative sin π₯ must be negative cos of π₯. By returning to the first general result we quoted, weβll be able to differentiate negative cos of π₯. Itβs negative, negative sin of π₯, which is, of course, simply sin of π₯. So the third derivative is sin of π₯. And finally, we obtain the fourth derivative to be equal to cos of π₯. And so we have the first four derivatives. They are negative sin π₯, negative cos π₯, sin π₯, and cos π₯.
The second part of this question asks us to find a general form for the πth derivative of π with respect to π₯. And so we need to spot a pattern with our derivatives. Firstly, we should notice that if we differentiate cos of π₯, we go back to negative sin π₯. And this cycle will continue. Next, we recall that sine and cosine are horizontal translations of one another. And so if we sketch the curve of π¦ equals negative sin of π₯, that, of course, is a reflection on the π₯-axis for the graph of π¦ equals sin π₯. We see that we can write π¦ equals negative sin π₯ as π¦ equals cos of π₯ plus π by two. Itβs a horizontal translation of the graph of π¦ equals cos of π₯, left by π by two radians. So π prime of π₯ is equal to cos of π₯ plus π by two.
Weβll repeat this for the graph of π¦ equals negative cos of π₯. Itβs a reflection again in the π₯-axis of the graph π¦ equals cos of π₯. And it, therefore, also can be represented by a horizontal translation of the graph of π¦ equals cos of π₯ left by π radians. So we can write π double prime of π₯, which was negative cos of π₯, as cos of π₯ plus π. By applying a similar thought process, we find that the third derivative, sin of π₯, can be written as cos of π₯ plus three π by two. And the fourth derivative can be written as cos of π₯ plus two π.
Remember, cos itself is periodic with a period of two π radians. So cos of π₯ plus two π is exactly the same as cos of π₯. And we might be starting to spot a pattern. In fact, if we write the second derivative, cos of π₯ plus π, as cos of π₯ plus two π over two and the fourth derivative as cos of π₯ plus four π over two, we see we can write the πth derivative as cos of π₯ plus ππ over two. And so the general form for the πth derivative of π with respect to π₯ is cos of π₯ plus ππ over two.
The third part of this question asks us to derive the Maclaurin series for cos of π₯. And so we recall this is the sum from π equals zero to β of the πth derivative of π evaluated at zero over π factorial times π₯ to the πth power. Well, in this case, the πth derivative of π evaluated at zero is cos of zero plus ππ over two. When π is zero, we have π of zero, which is cos of zero which is one. When π is one, we have π prime of zero, which is cos of π by two, which is zero. When π is two, we have cos of π, which is negative one. And when π is equal to three, we have cos of three by two, which is once again zero. And of course, as we saw in the second part of this question, this cycle continues.
And this means the first few terms will be one minus π₯ squared over two factorial plus π₯ to fourth power over four factorial minus π₯ to the six over six factorial, which we can write as the sum from π equals zero to β of negative one to the πth power. Remember, this part will just give us the alternating signs. We go from positive to negative and back to positive again. This is all over two π factorial. This bit gives us the even factorials and the denominator times π₯ to the power of two π. And so weβve derived the Maclaurin series for cos of π₯. Itβs the sum from π equals zero to β of negative one to the πth power over two π factorial times π₯ to the power of two π. And, of course, if we chose to alternatively write this as shown, that would be absolutely fine too.
Weβre now going to clear some space and consider the fourth and final part of this question. Weβll keep the Maclaurin series for cos of π₯ on screen as weβre going to use that in a moment. And this question says, what is the radius π
of convergence of the Maclaurin series for cos of π₯? We recall that, in general, thereβs an open interval from negative π
to π
, in which a power series converges. And that number π
is called the radius of convergence. And we can use the ratio test to find this value. The part of the test weβre interested in says that, given a series of sum of π π, if the limit as π approaches β of the absolute value of π π plus one over π π is less than one, then the series is absolutely convergent and hence convergent.
In our case, we let π π be equal to negative one to the πth power times π₯ to the power of two π over two π factorial. Then, π π plus one is negative one to the power of π plus one times π₯ to the power of two times π plus one over two times π plus one factorial. Weβll distribute the parentheses to make the next step easier. Two times π plus one is two π plus two. According to the ratio test, we want to find where the limit as π approaches β of the absolute value of the quotient of these is less than one.
We know that when dividing by a fraction, we simply multiply by the reciprocal of that fraction. So we can say that we want the limit as π approaches β of the absolute value of negative one to the power of π plus one times π₯ to the power of two π plus two over two π plus two factorial times two π factorial over negative one to the πth power times π₯ to the power of two π. And then, we recall that when dividing two numbers with equal basis, we simply subtract their exponents so that π₯ to the power of π divided by π₯ to the power of π is π₯ to the power of π minus π. Weβll use this to simplify negative one to the power of π plus one divided by negative one to the power of π. Itβs simply negative one. Similarly, weβll be able to simplify π₯ to the power of two π plus two divided by π₯ to the power of two π. Itβs π₯ squared.
And what about these factorials? Well, we know that two π plus two factorial is the same as two π plus two times two π plus one times two π times two π minus one and so on. Alternatively, thatβs the same as two π plus two times two π plus one times two π factorial. And that allows us to divide through by two π factorial. When we do, weβre left with two π plus two times two π plus one on the denominator of our fraction. So we want the limit as π approaches β of the absolute value of this fraction to be less than one. We spot that negative one times π₯ squared is independent event. And so we can take the absolute value of negative π₯ squared outside of the limit. And we have the absolute value of negative π₯ squared times the limit as π approaches β of one over two π plus two times two π plus one.
Weβre now ready to evaluate this limit. As π grows larger, one over two π plus two times two π plus one approaches zero. And so, in fact, weβre looking to find the values of π₯ such that the absolute value of negative π₯ squared times zero is less than one. Well, when we multiply the absolute value of negative π₯ squared by zero, weβll always get zero. And this means our Maclaurin series converges for all values of π₯. And we can therefore say that π
equals β or positive β.
In this video, weβve learned that the Maclaurin series is a special case of the Taylor series, where π is equal to zero. Itβs given by the sum from π equals zero to β of the πth derivative of π evaluated at zero over π factorial times π₯ to the πth power. And finally, we saw that if a series converges on the open interval negative π
to π
, its radius of convergence is the number denoted by π
. And we can use the ratio test to find this value.
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# How to Solve a 3×3 Rubik’s Cube: The Ultimate Guide
Baca Cepat
## Unlock the Secrets of the Rubik’s Cube with Our Comprehensive Guide
Welcome to our guide on solving the 3×3 Rubik’s Cube. If you’re reading this, chances are you’ve already spent countless hours struggling to solve this iconic puzzle. Well, fear not! In this article, we’ll be providing you with a step-by-step guide on how to solve the Rubik’s Cube, as well as tips and tricks to help you solve it faster. So, let’s dive in!
## Introduction
First, let’s briefly introduce the Rubik’s Cube. Developed by Ernő Rubik in 1974, the Rubik’s Cube is a 3-dimensional puzzle that consists of 6 faces, each with a different color. The goal is to rearrange the cube so that each face has only one color. Sounds simple, right? Unfortunately, it can be quite challenging, especially for beginners. However, with a bit of practice, anyone can learn how to solve the Rubik’s Cube.
Before we begin, there are a few things you should know. Firstly, there are different methods for solving the Rubik’s Cube. We’ll be teaching you the most popular method, the CFOP method, which involves solving the cube layer by layer. Secondly, these instructions are for right-handed people. If you’re left-handed, you’ll need to mirror the instructions.
Now that we’ve got that out of the way, let’s get started!
### Step 1: The Cross
The first step in solving the Rubik’s Cube is to create a cross on one side of the cube. This may sound simple, but it requires some thought and planning. The goal is to create a cross using the center color on each edge piece.
To create the cross, you’ll want to start with a white cross if possible. Look for the edge pieces that have white as one of their colors and place them in the correct position on the cube. Use the algorithm below to move the pieces:
Move Notation
Front F
Right R
Up U
Left L
Down D
Back B
For example, if the white edge piece is in the top layer but not in the correct position, you’ll need to move it to the correct position using the following algorithm:
R U R’ U’
Repeat this step until you’ve created the cross.
### Step 2: The Corners
Once you’ve created the cross, the next step is to solve the corners. The goal is to place the corners in the correct position and orientation so that the colors on each face match.
The first corner you’ll want to solve is the one that’s directly under the white center. Hold the cube so that the white center is on the bottom and the corner you want to solve is in the top layer. Use the following algorithm to move the corner into place:
U R U’ L’ U R’ U’ L
Repeat this algorithm until the corner is in the correct position and orientation. Then, solve the remaining corners using a similar method.
### Step 3: The Second Layer
With the first layer solved, it’s time to move on to the second layer. The goal of this step is to solve the middle layer of the cube.
To solve the second layer, you’ll need to place the correct edge pieces in the correct position. First, find an edge piece with a white sticker on it and place it in the top layer with the white sticker facing up. Then, look for the corresponding center color on the cube and turn the bottom layer until the edge piece is directly below it. Now, use the following algorithm to move the edge piece into place:
F R U R’ U’ F’
Repeat this step until the second layer is complete.
### Step 4: The Top Layer
The final step in solving the Rubik’s Cube is to solve the top layer. This can be the most challenging step, but with practice, you’ll be able to solve it with ease.
The first step in solving the top layer is to orient the top layer pieces so that they all match. To do this, use the following algorithm:
R U R’ U R U U R’
Once the top layer pieces are oriented correctly, it’s time to permute the corners. To do this, use the following algorithm:
U R U’ L’ U R’ U’ L
Finally, permute the edges using the algorithm below:
F R U R’ U’ F’
## FAQs
### How long does it take to solve a Rubik’s Cube?
The time it takes to solve a Rubik’s Cube varies depending on the person’s skill level. For beginners, it can take anywhere from a few minutes to a few hours. For more experienced cubers, the average solving time is around 20 seconds.
### What is the fastest time for solving a Rubik’s Cube?
The fastest time for solving a Rubik’s Cube is 3.47 seconds, achieved by Yusheng Du from China in November 2018.
### Is it possible to solve the Rubik’s Cube blindfolded?
Yes, it is possible to solve the Rubik’s Cube blindfolded. This is known as blindfolded cubing and involves memorizing the cube’s position and solving it without looking.
### What is the world record for blindfolded Rubik’s Cube solving?
The world record for blindfolded Rubik’s Cube solving is 16.22 seconds, achieved by Marcin Kowalczyk from Poland in May 2019.
### What is the best Rubik’s Cube brand?
There are many different Rubik’s Cube brands on the market, but some of the most popular brands include GAN, Moyu, and QiYi. The best brand for you will depend on your personal preferences.
### What is the hardest Rubik’s Cube to solve?
The Rubik’s Cube with the most significant number of combinations is the 17×17 Rubik’s Cube. It has 66,252,710,885,993,600 possible combinations, making it the most challenging to solve.
### How can I solve a Rubik’s Cube faster?
There are several ways to solve a Rubik’s Cube faster, including learning new algorithms, improving finger speed, and practicing regularly.
### Can I solve a Rubik’s Cube with one hand?
Yes, it is possible to solve the Rubik’s Cube with one hand. Some cubers specialize in one-handed solving and can solve the cube in under 10 seconds.
### What is the Rubik’s Cube world championship?
The Rubik’s Cube World Championship is a biennial competition that brings together the world’s top Rubik’s Cube solvers. Competitors compete in various events, including solving the cube as fast as possible and solving it with one hand or blindfolded.
### Is it possible to solve a Rubik’s Cube with no algorithms?
No, it is not possible to solve the Rubik’s Cube without algorithms. Algorithms are essential to solving the cube efficiently.
### What is the average time to solve a Rubik’s Cube?
The average time to solve a Rubik’s Cube varies depending on the solver’s skill level. For beginners, the average time is around 2-3 minutes, while for more experienced cubers, the average time is around 20 seconds.
### How many combinations are possible on a Rubik’s Cube?
There are over 43 quintillion possible combinations on a Rubik’s Cube.
### What is the Rubik’s Cube world record?
The Rubik’s Cube world record is 3.47 seconds, achieved by Yusheng Du from China in November 2018.
### What is the best lubricant for a Rubik’s Cube?
The best lubricant for a Rubik’s Cube is a silicone-based lubricant. Some popular brands include Traxxas, Lubicle, and Maru.
### What is the best method for solving the Rubik’s Cube?
The best method for solving the Rubik’s Cube is subjective and depends on the solver’s personal preferences. However, the most popular method is the CFOP method, which involves solving the cube layer by layer.
## Conclusion
Congratulations! You’ve now learned how to solve the Rubik’s Cube using the CFOP method. Remember, practice makes perfect, so don’t be discouraged if it takes you a while to solve the cube. With time and dedication, you’ll be able to solve the cube in no time!
Now that you’ve mastered the Rubik’s Cube, why not try competing in a Rubik’s Cube competition? Who knows, you might even set a new record!
Thank you for reading our guide on how to solve the Rubik’s Cube. We hope you found it informative and helpful. If you have any questions or feedback, please don’t hesitate to contact us.
## Closing
The Rubik’s Cube is a truly iconic puzzle that has captivated people of all ages for decades. While it may seem daunting at first, anyone can learn how to solve the Rubik’s Cube with practice and determination.
We hope you’ve enjoyed our guide on how to solve the Rubik’s Cube. Remember, the most important thing is to have fun and enjoy the process. So keep on cubing, and who knows, you might just become the next Rubik’s Cube champion! |
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