Split (1)

train · ~16.8M rows (showing the first 4.97M)

text stringlengths 22 1.01M
Paul's Online Math Notes [Notes] Algebra - Notes Next Chapter Solving Equations and Inequalities Integer Exponents Previous Section Next Section Real Exponents Rational Exponents Now that we have looked at integer exponents we need to start looking at more complicated exponents. In this section we are going to be looking at rational exponents. That is exponents in the form where both m and n are integers. We will start simple by looking at the following special case, where n is an integer. Once we have this figured out the more general case given above will actually be pretty easy to deal with. Let’s first define just what we mean by exponents of this form. In other words, when evaluating we are really asking what number (in this case a) did we raise to the n to get b. Often is called the nth root of b. Let’s do a couple of evaluations. Example 1 Evaluate each of the following. (a) [Solution] (b) [Solution] (c) [Solution] (d) [Solution] (e) [Solution] (f) [Solution] Solution When doing these evaluations we will do actually not do them directly. When first confronted with these kinds of evaluations doing them directly is often very difficult. In order to evaluate these we will remember the equivalence given in the definition and use that instead. We will work the first one in detail and then not put as much detail into the rest of the problems. (a) So, here is what we are asking in this problem. Using the equivalence from the definition we can rewrite this as, So, all that we are really asking here is what number did we square to get 25. In this case that is (hopefully) easy to get. We square 5 to get 25. Therefore, (b) So what we are asking here is what number did we raise to the 5th power to get 32? (c) What number did we raise to the 4th power to get 81? (d) We need to be a little careful with minus signs here, but other than that it works the same way as the previous parts. What number did we raise to the 3rd power (i.e. cube) to get -8? (e) This part does not have an answer. It is here to make a point. In this case we are asking what number do we raise to the 4th power to get -16. However, we also know that raising any number (positive or negative) to an even power will be positive. In other words, there is no real number that we can raise to the 4th power to get -16. Note that this is different from the previous part. If we raise a negative number to an odd power we will get a negative number so we could do the evaluation in the previous part. As this part has shown, we can’t always do these evaluations. (f) Again, this part is here to make a point more than anything. Unlike the previous part this one has an answer. Recall from the previous section that if there aren’t any parentheses then only the part immediately to the left of the exponent gets the exponent. So, this part is really asking us to evaluate the following term. So, we need to determine what number raised to the 4th power will give us 16. This is 2 and so in this case the answer is, As the last two parts of the previous example has once again shown, we really need to be careful with parenthesis. In this case parenthesis makes the difference between being able to get an answer or not. Also, don’t be worried if you didn’t know some of these powers off the top of your head. They are usually fairly simple to determine if you don’t know them right away. For instance in the part b we needed to determine what number raised to the 5 will give 32. If you can’t see the power right off the top of your head simply start taking powers until you find the correct one. In other words compute , , until you reach the correct value. Of course in this case we wouldn’t need to go past the first computation. The next thing that we should acknowledge is that all of the properties for exponents that we gave in the previous section are still valid for all rational exponents. This includes the more general rational exponent that we haven’t looked at yet. Now that we know that the properties are still valid we can see how to deal with the more general rational exponent. There are in fact two different ways of dealing with them as we’ll see. Both methods involve using property 2 from the previous section. For reference purposes this property is, So, let’s see how to deal with a general rational exponent. We will first rewrite the exponent as follows. In other words we can think of the exponent as a product of two numbers. Now we will use the exponent property shown above. However, we will be using it in the opposite direction than what we did in the previous section. Also, there are two ways to do it. Here they are, Using either of these forms we can now evaluate some more complicated expressions Example 2 Evaluate each of the following. (a) [Solution] (b) [Solution] (c) [Solution] Solution We can use either form to do the evaluations. However, it is usually more convenient to use the first form as we will see. (a) Let’s use both forms here since neither one is too bad in this case. Let’s take a look at the first form. Now, let’s take a look at the second form. So, we get the same answer regardless of the form. Notice however that when we used the second form we ended up taking the 3rd root of a much larger number which can cause problems on occasion. (b) Again, let’s use both forms to compute this one. As this part has shown the second form can be quite difficult to use in computations. The root in this case was not an obvious root and not particularly easy to get if you didn’t know it right off the top of your head. (c) In this case we’ll only use the first form. However, before doing that we’ll need to first use property 5 of our exponent properties to get the exponent onto the numerator and denominator. We can also do some of the simplification type problems with rational exponents that we saw in the previous section. Example 3 Simplify each of the following and write the answers with only positive exponents. (a) [Solution] (b) [Solution] Solution (a) For this problem we will first move the exponent into the parenthesis then we will eliminate the negative exponent as we did in the previous section. We will then move the term to the denominator and drop the minus sign. (b) In this case we will first simplify the expression inside the parenthesis. Don’t worry if, after simplification, we don’t have a fraction anymore. That will happen on occasion. Now we will eliminate the negative in the exponent using property 7 and then we’ll use property 4 to finish the problem up. We will leave this section with a warning about a common mistake that students make in regards to negative exponents and rational exponents. Be careful not to confuse the two as they are totally separate topics. In other words, and NOT This is a very common mistake when students first learn exponent rules. Integer Exponents Previous Section Next Section Real Exponents Next Chapter Solving Equations and Inequalities [Notes] © 2003 - 2015 Paul Dawkins
## How quadratics help us in reality Quadratics consist of parabolas which are basically curved lines, so we are going to be leaving linear relations behind. We might not realise this at first but quadratic relations are all around us ranging from mountains to bridges even the McDonalds sign we see everyday. ## How to determine Quadratic Realtions • Recall that linear relations had first differences So... • Quadratic relations have SECOND differences ## VERTEX FORM • The vertex form is one way that a quadratic relationship can be written as it is in the form y=a(x-h)² +k and the basic form of it is y=x². • An example for this is y= 2(x-3)² +5 • The "x" and "y" do not have a value because they are variables for the x and y intercept(s) which will have different values according to the relation ## What each variable represents in the Vertex Form • In the vertex form y=a(x-h)² +k each variable has a part to do • The a tells us if there is a stretch/expansion or compression to the parabola. • The h tells us the x value of the vertex and the axis of symmetry. • The k tells us the y value of the vertex. ## Transformations • The transformations that can occur to a parabola are vertical or horizontal translations, vertical stretches and reflections. • As we know the a represents vertical stretch and compression if the value is more than 1 than it is a compression and if it is less than 1 it will be a stretch/expansion • If a is a negative value then there will be a reflection on the x-axis (the parabola will be flipped; downwards) • The h tells if the parabola moves left or right (horizontal stretch). If the h is negative then it goes towards the right and if it is positive it goes towards the left. Basically it will move the opposite of its sign. • The k informs us if the parabola moves up or down (vertical stretch) so if it is negative the graph will move downwards and if it is positive it will move upwards. ## Graphing from Vertex Form using the Step Pattern • y=x² this is the basic parabola. • The original step pattern is "over 1 up 1 , over 2 up 4" this pattern can be used for any parabola as long as it is the basic (original) one ## Graphing from Vertex Form when not in Basic Form • If the vertex form has the "a" in front of it, it is not a basic parabola anymore, it will either have a stretch or compression and will open either up or down • The value of a will decide if the step pattern will change An example... 1. First, you will graph the vertex 2. Then multiply the original step pattern by the value that a had been given 3. Then plot the points (vertex, 4 step pattern points). • It does not matter if there is a decimal, fraction, or a negative number for a, you still have to multiply that number with the original step pattern. ## How to find the Axis of Symmetry (AOS) • The way it is written is x=h, so it would be the h in the vertex form • The axis of symmetry is the middle point of the zeros; the x intercept of the vertex • So the AOS of the parabola above will be -2 because it is the midpoint of the parabola and the it is the x-intercept of the vertex ## How to find the Optimal Value • It is written as y=k, so it would be the k in the vertex form • The optimal value is either the highest point or the lowest point on the parabola; the y-intercept of the vertex • So the optimal value of the parabola above would be -6 because in this case it is the lowest point on the parabola and y-intercept of the vertex ## X-intercepts and Zeros • To find the x-intercept/zeros in vertex form you would have to set y=0 • For instance let's say you have to find the zeros in the equation y=-3(x+2)²+27 the y would be replaced with 0 • We do negative and positive square root because the number being squared could be either one • As a result the x-intercepts would be 1 and/or -5 depending on the relation ## FACTORED FORM • The factored form is another way that a quadratic relation can be written as and the form is y=(x-r)(x-s) • An example of this form is y=0.5(x+3)(x-9) ## Zero and X-intercepts • Zeros can be found by setting each factor equal to 0 meaning x-r=0 and s-r=0 • For example in the relationship y=0.5(x+3)(x-9) the zeros would be found by taking each factor (x+3) and (x-9) and make them equal to zero • Therefore, the zeros/x-intercepts are -3 and 9 for this relation so when you graph the zeros these are the values you would use ## Axis of Symmetry (AOS) • To find the axis of symmetry you have to add the two x-intercepts and divide it by 2 • Therefore the formula or method you would use is x= (r+s)/2 • For the example we have been using previously we would find the x-intercepts (-3 and 9) add them then divide the answer by 2 • So it would be x=(r+s)/2 x=(-3+9)/2 x=6/2 x=3 • For that reason, the axis of symmetry would be 3 ## Optimal Value • In order to find the optimal value you would have to substitute the axis of symmetry into the equation as x since it is a x value on the parabola • If we use the example we have been using so far we would substitute x=3 into the equation y=0.5(x+3)(x-9) • Therefore the optimal value for the example is -18 ## Graphing • Know that you know how to find the x-intercepts, axis of symmetry and optimal value you can plot 3 points 1. The first x-intercept which is x=-3 2. The second x-intercept which is x=9 3. The vertex which is (axis of symmetry;x, optimal value;y) (3,-18) ## STANDARD FORM • Last but not least, the standard form is also a way to write a quadratic relation and the form is y=ax²+bx+c • An example of this form is y=3x²+15x+18 ## How to find the zeros using the Quadratic Formula • In order to use the formula the standard form should equal to 0 since you are solving for the zeros • Since we know the formula and have the question all you have to do is plug in the appropriate values just like in the example below • The square root is both positive and negative since there would be a negative and positive square root to the number which in the end would give us two zeros • If you want to keep the exact answer leave it the way it is in step 4 but if you want to take it down to a fraction/decimal then you can continue and add the first time and subtract the next when solving ## Discriminants • The discriminant is the number that is being square rooted (b²-4ac) this number tells us if there are any zeros/solutions and if there are, how many there are. • If the discriminant is a negative there will be no zeros because a negative number cannot be square rooted • If the discriminant is 0 then there is only one zero because it does not matter if you add or subtract a 0 the answer will be the same • If the discriminant is greater than 0 then it will have two zeros, just as in the example above, since the discriminant is 9 there were two zeros ## Axis of Symmetry • The formula for this is • To find the axis of symmetry all you have to do is fill in the following formula just as shown in the example below • It is as simple as that and it hardly takes anytime so the axis of symmetry would be x=0.7 for this parabola ## Optimal Value • To find the optimal value (y-value;y-intercept in vertex) all you have to do is substitute the axis of symmetry into the original equation to get y • An example for this is...using the axis of symmetry from above • So they optimal value for the example we have been using would be -0.45 ## Word Problem This is how to do the following word problem... ## Completing the Square • Completing the square helps us turn from Standard form to Vertex form (y=ax²+bx+c to y=a(x-h)²+k • There are a few steps and they are as follows: 1. Take the c out and make it k since they both carry the same value and represent the same thing on the graph 2. Then put brackets around the remaining two numbers 3. Then factor out the GCF from the number in the brackets; after divide b by two and square to get the number you need to add and subtract 4. Then add and subtract that number from the brackets 5. When taking the subtraction number out of the brackets multiply it by the a value;then you will be able to make a perfect square 6. Make the perfect square (write x and divide b by 2 then put squared outside of the bracket and write the vertex form Just as in the example below • Therefore the vertex form would be y=2(x+2)² ## Factoring-4 terms by Grouping • If you ever end up with four terms to factor it is very simple 1. We group the first two and second terms 2. Then we factor out the GCF 3. Then we can common factor just as shown above • So basically we grouped the terms into binomials first two and second two, then we found the GCF and factored it out and finally we common factored • This gave us (x²+3)(x²+7) ## Factoring Simple Trinomials • The way that simple trinomials are written is x²+bx+c • An example for this is w²+5w+6 • There are six steps to get into the factored form and they are 1. Make sure the format is x²+bx+c 2. Find two numbers that give the product of 'c' 3. Add the two number to see if they give the sum of 'b' 4. Factor them in for 'bx' 5. Do four terms by grouping method 6. Then we do common binomials An example for this is • Therefore w²+5w+6 in factored form is (w+2)(w+3) ## Factoring Complex Trinomials • Complex trinomials will include a value with the a making it complex • An example of this is 8x²+6x-5 • In order to factor this we do the following: ## Perfect Squares • Perfect Squares make expanding easier • There are 3 steps to do them: 1. Square the first term 2. Double Product 3. Square second term • But you can only do perfect squares if what you are expanding is either for example (a+b)² or (a+b)(a+b) since you are multiplying the same things that would end up being squared, or (a-b)(a-b) and (a-b)² but of course the variables and values would be different and they can be a part of a larger relation. ## Factoring Perfect Square Factoring Perfect Squares • You may also use the acronym Sam Doesn`t Pull Strings which stands for Square, Double Product and Square ## Difference of Squares • Let's say you don't have a perfect square but something like it for example (y-7(y+7) the only different thing are the signs and this method is called difference of squares • When we do difference of squares there are only two steps 1. Square the first term 2. Square the second term • For example • The subtraction sign will always stay as subtraction, hence the name difference of squares • This method can be applied anywhere as long as the relation you are expanding is in the format (a+b)(a-b) or (a-b)(a+b) ## Factoring Differnece of Squares Factoring Difference of Squares IMPORTANT: Difference of Squares factors into a product of sum and difference ## Common Factoring • Common Factors are when we factor out the GCF in an expression • There are only three steps to this 1. Find GCF 2. Divide expression by GCF 3. The GCF number will go outside the brackets as well Example 2x+4y 2(x+2y) Since the GCF was 2 we divided the whole expression by 2 and then put the GCF outside of the brackets as well which factored into 2(x+2y) OR you can make a list on all the factors of each number and find the common one(s) and find the GCF that way; as shown in the example below ## Binomial Factors • This method is known as binomial factors because we are putting the expression into only two parts; an easier way to expand • For example • Basically we did three things to get the answer we got 1. Since (2s+5t) are the same we can write them just once 2. Then we have 3s and -7t left which can be put in another set of brackets 3. Both are multiplied so when you need to expand to get back to what you started you may do that Hence we get (2s+5t)(3s-7t) as our factored form ## REFLECTION In the quadratics unit I have learned many new things and how they can be used in real life, I also found a few things to be difficult while others were easier. To start off, I found a few things difficult like understanding complex trinomials. At first I found it difficult to try and find all the possibilities that a could have. Later, once I got the hang of it it was not as hard as I thought it was. Another difficulty I had in this whole unit was when we had to find the amount of x-intercepts a parabola had by looking at the equation. I had difficulties because I never really knew how find the x-intercepts by just looking at the vertex form. Then later on we learned that all we had to do was substitute y as 0 and solve for x and that x would give us the two x-intercepts. I also used to get confused between perfect squares and difference of squares because I would get confused perfect squares for difference of squares and vice versa. So the method I used to remember which is which is that difference of squares is a product of addition and subtraction which would eliminate that, that is not a perfect square making it difference of squares. The things I could do with ease were simple trinomials and expanding especially on the assessment below. On this assessment I did not do the best but I did get how to expand and do simple trinomials as well as the fact that I improved on complex trinomials. These are the things I found difficult and easy. I learned that quadratics are all around us, there are mountains,bridges and many more. I now know that we do not have to measure the maximum height of something we can just use quadratics for it. All I would have to do is get a few measurements and use the appropriate quadratics formulas and I would get the maximum height, at what time it was there and so forth. Therefore in the future I do not need to get a measuring tape and take 10 minutes trying to get one length accurate when I can just get the necessary lengths and get other measurements using quadratics in those 10 minutes. In conclusion, I found quadratics to be okay not to hard and not to easy and I also learned how quadratics help in real life. ## Word Problem Now you try solving this word problem. When you're done you can check your answers which are right next to the question
# Properties of Rational Numbers ## Key Questions • They can be written as a result of a division between two whole numbers, however large. Example: 1/7 is a rational number. It gives the ratio between 1 and 7. It could be price for one kiwi-fruit if you buy 7 for $1. In decimal notation, rational numbers are often recognised because their decimals repeat. 1/3 comes back as 0.333333.... and 1/7 as 0.142857... ever repeating. Even 553/311 is a rational number (the repeating cylce is a bit longer) There are also IRrational numbers that cannot be written as a division. Their decimals follow no regular pattern. Pi is the best-known example, but even the square root of 2 is irrational. • #### Answer: Different types of fractions are: 1. Proper Fractions 2. Improper Fractions 3.. Mixed Fractions #### Explanation: 1. Proper Fractions are the one in which there is a smaller no. in the numerator and a larger no. in the denominator. 2. Improper Fractions are fractions in which there is a larger no. in the numerator and a smaller no. in the denominator. 3. Mixed Fractions are fractions in which there is whole no. followed by a fraction which can either be proper or improper. • Let's assume that we know what integer numbers are and what is an operation of their multiplication, so we know how to multiply any integer number by any other. Considerations of symmetry and harmony lead us to a desire to reverse the operation of multiplication, that is to be able to divide any integer number by any other. Obviously, it's not always possible within the realm of integer numbers. Operation of multiplication is not really complete in the space of only integer numbers since its reverse, division, is not possible for some integer numbers. For instance, we can multiply $3$by $7$getting $21$and we can divide $21$by $7$getting $3$, but we cannot divide $21$by $6$within a set of integer numbers. Rational numbers are completely new entities that allow us to divide any integer number by any other (not equal to $0$). So, by definition, a rational number is a set of two integer numbers, the first, usually called numerator, and the second (not equal to $0$), usually called denominator, that has one important property: if multiplied by a denominator, result is a numerator. Traditionally, if a numerator is $M$and a denominator is $N$, the rational number formed by them is written as $\frac{M}{N}$with property defined for it: $\frac{M}{N} \cdot N = M$. Introduction of rational numbers completes the operation of multiplication by enabling its reverse in a broader set of numbers. Now we can divide $21$by $6$using rational numbers and the result, by definition of rational numbers , is $\frac{21}{6}\$. The harmony has been restored by expanding the concept of numbers from integer to rational. Obviously, we have to prove the correctness of our definition, that certain properties of operations of addition and multiplications of integer numbers are preserved within a set of rational numbers, but this is a different topic.
Request a call back Join NOW to get access to exclusive study material for best results # Class 10 SELINA Solutions Maths Chapter 5 - Quadratic Equations ## Quadratic Equations Exercise Ex. 5(A) ### Solution 1(a) Correct Option: (iv) 4x2 – 9 = 0 4x2 = 9 ### Solution 1(b) Correct Option: (iii) 3 or –5 (x – 3)(x + 5) = 0 (x – 3) = 0 or (x + 5) = 0 x = 3 or x = –5 ### Solution 1(c) Correct Option: (ii) –3 Substituting x = 4 in equation x2 + kx – 4 = 0, (4)2 + k(4) – 4 = 0 16 + 4k – 4 = 0 4k + 12 = 0 4k = –12 k = –3 ### Solution 1(d) Correct Option: (i) –2 Substituting x = 2 in equation 2x2 – 3x + k = 0, 2(2)2 – 3(2) + k = 0 8 – 6 + k = 0 2 + k = 0 k = –2 ### Solution 1(e) Correct Option: (iv) 0 or 7 x2 – 7x = 0 x(x – 7) = 0 x = 0 or x – 7 = 0 x = 0 or x = 7 ### Solution 2(i) 5x2 - 8x = -3(7 - 2x) 5x2 - 8x = 6x - 21 5x2 - 14x + 21 =0; which is of the form ax2 + bx + c = 0. Given equation is a quadratic equation. ### Solution 2(ii) (x - 4)(3x + 1) = (3x - 1)(x +2) 3x2 + x - 12x - 4 = 3x2 + 6x - x - 2 16x + 2 =0; which is not of the form ax2 + bx + c = 0. Given equation is not a quadratic equation. ### Solution 2(iii) 7x3 - 2x2 + 10 = (2x - 5)2 7x3 - 2x2 + 10 = 4x2 - 20x + 25 7x3 - 6x2 + 20x - 15 = 0; which is not of the form ax2 + bx + c = 0. Given equation is not a quadratic equation. ### Solution 3(i) x2 - 2x - 15 = 0 For x = 5 to be solution of the given quadratic equation it should satisfy the equation. So, substituting x = 5 in the given equation, we get L.H.S = (5)2 - 2(5) - 15 = 25 - 10 - 15 = 0 = R.H.S Hence, x = 5 is a solution of the quadratic equation x2 - 2x - 15 = 0. ### Solution 3(ii) 2x2 - 7x + 9 = 0 For x = -3 to be solution of the given quadratic equation it should satisfy the equation So, substituting x = 5 in the given equation, we get L.H.S=2(-3)2 - 7(-3) + 9 = 18 + 21 + 9 = 48 R.H.S Hence, x = -3 is not a solution of the quadratic equation 2x2 - 7x + 9 = 0. ### Solution 4 For x = to be solution of the given quadratic equation it should satisfy the equation So, substituting x = in the given equation, we get ### Solution 5 For x = and x = 1 to be solutions of the given quadratic equation it should satisfy the equation So, substituting x = and x = 1 in the given equation, we get Solving equations (1) and (2) simultaneously, ## Quadratic Equations Exercise Ex. 5(B) ### Solution 1(a) Correct Option: (iii) –1 or 7 x2 – 6x – 7 = 0 x2 – 7x + x – 7 = 0 x(x – 7) + (x – 7) = 0 (x – 7)(x + 1) = 0 x = 7 or x = –1 ### Solution 1(b) Correct Option: (ii) –6 or –2 x(x + 8) + 12 = 0 x2 + 8x + 12 = 0 x2 + 6x + 2x + 12 = 0 x(x + 6) + 2(x + 6) = 0 (x + 6)(x + 2) = 0 x = –6 or x = –2 ### Solution 1(c) Correct Option: (ii) 2 Substituting x = 2 in equation (p – 3)x2 + x + p = 0 is 2, (p – 3)22 + 2 + p = 0 4p – 12 + 2 + p = 0 5p – 10 = 0 p = 2 ### Solution 1(d) Correct Option: (iii) 2 or ½ x2 + 1 = 2.5x x2 – 2.5x + 1 = 0 x2 – 2x – 0.5x + 1 = 0 x(x – 2) – 0.5(x – 2) = 0 (x – 2)(x – 0.5) = 0 x = 2 or x = 0.5 = ½ ### Solution 1(e) Correct Option: (i) x ≠ 0 If x = 0, the term in the equation will be undefined. Hence, for a quadratic equation , x ≠ 0. ### Solution 17 If a + 7 =0, then a = -7 and b + 10 =0, then b = - 10 Put these values of a and b in the given equation ### Solution 18 4(2x+3)2 - (2x+3) - 14 =0 Put 2x+3 = y or x = -(a + b) ## Quadratic Equations Exercise Ex. 5(C) ### Solution 1(a) Correct Option: (ii) 2.0 or 1.0 x2 – 3x + 2 = 0 Then, a = 1, b = –3, c = 2 ### Solution 1(b) Correct Option: (i) 5.00 or 1.00 x2 – 4x – 5 = 0 Then, a = 1, b = –4, c = –5 ### Solution 1(c) Correct Option: (ii) 9 or 1 x2 – 8x – 9 = 0 Then, a = 1, b = –8, c = –9 ### Solution 1(d) Correct Option: (iii) 3.0 or 1.0 x2 – 2x – 3 = 0 Then, a = 1, b = –2, c = –3 ### Solution 1(e) Correct Option: (i) 8 or –2 x2 – 6x – 16 = 0 Then, a = 1, b = –6, c = –16 ### Solution 3(iii) 4x2 - 5x - 3 = 0 Here, a = 4, b = -5 and c = -3 ### Solution 6 Consider the given equation: ## Quadratic Equations Exercise Ex. 5(D) ### Solution 1(a) Correct Option: (i) distinct and real roots 2x2 – 3x + 1 = 0 Here, a = 2, b = –3, c = 1 b2 – 4ac = (–3)2 – 4(2)(1) = 9 – 8 = 1 > 0 Hence, the roots are distinct and real. ### Solution 1(b) Correct Option: (i) x2 – 5x + 6 = 0 For a quadratic equation ax2 + bx + c = 0, the roots are real and distinct if b2 – 4ac > 0. For, x2 – 5x + 6 = 0 a = 1, b = –5, c = 6 b2 – 4ac = (–5)2 – 4(1)(6) = 25 – 24 = 1 > 0 Hence, the roots are distinct and real. ### Solution 1(c) Correct Option: (iv) 4 or –4 For, x2 – px + 4 = 0, the roots are equal. Then, b2 – 4ac = 0 Here, a = 1, b = –p, c = 4 b2 – 4ac = 0 (–p)2 – 4(1)(4) = 0 p2 – 16 = 0 p2 = 16 p = ±4 ### Solution 1(d) Correct Option: (ii) 2x2 – 5x + 9 = 0 For a quadratic equation ax2 + bx + c = 0, the roots are imaginary if b2 – 4ac < 0. For, 2x2 – 5x + 9 = 0 a = 2, b = –5, c = 9 b2 – 4ac = (–5)2 – 4(2)(9) = –47 < 0 Hence, the roots are imaginary. ### Solution 1(e) Correct Option: (i) 7 Substituting x = 1 in equation 3x2 – mx + 4 = 0, 3(1)2 – m(1) + 4 = 0 3 – m + 4 = 0 m = 7 ### Solution 6 Hence, p = 4 and q = 12 or -12 ## Quadratic Equations Exercise Ex. 5(E) ### Solution 1(a) Correct Option: (ii) ±1 or ±2 x4 – 5x2 + 4 = 0 Let x2 = y Then, y2 – 5y + 4 = 0 y2 – 4y – y + 4 = 0 y(y – 4) – 1(y – 4) = 0 (y – 4)(y – 1) = 0 y = 4 or y = 1 x2 = 4 or x2 = 1 x = ±2 or x = ±1 ### Solution 1(b) Correct Option: (ii) 10 20x – 50 = 3x2 – 15x 3x2 – 35x + 50 = 0 3x2 – 30x – 5x + 50 = 0 3x(x – 10) – 5(x – 10) = 0 (x – 10)(3x – 5) = 0 x = 10 or ### Solution 1(c) Correct Option: (iii) x ≠ 3 and x ≠ –5 Here, x – 3 ≠ 0 and x + 5 ≠ 0 x ≠ 3 and x ≠ –5 ### Solution 1(d) Correct Option: (i) –2 and 9 or –9 and 2 Since the product is negative, one integer should be negative. From the given options, option (i) satisfies the condition. Hence, the two integers are –2 and 9 or –9 and 2. ### Solution 4 (ii) (x2 - 3x)2 - 16(x2 - 3x) - 36 = 0 Let x2 - 3x = y Then y2 - 16y - 36 = 0 y2 - 18y + 2y - 36 = 0 y(y - 18) + 2(y - 18) = 0 (y - 18) (y + 2) = 0 If y - 18 = 0 or y + 2 = 0 x2 - 3x - 18 = 0 or x2 - 3x + 2 = 0 x2 - 6x + 3x - 18 = 0 or x2 - 2x - x + 2 = 0 x(x - 6) +3(x - 6) = 0 or x(x - 2) -1(x - 2) = 0 (x - 6) (x + 3) = 0 or (x - 2) (x - 1) = 0 If x - 6 = 0 or x + 3 = 0 or x - 2 = 0 or x - 1 = 0 then x = 6 or x = -3 or x = 2 or x = 1 ### Solution 11 ∴ Given equation reduces to ⇒ 2y2 - 3 = 5y ⇒ 2y2 - 5y - 3 = 0 ⇒ 2y2 - 6y + y - 3 = 0 ⇒ 2y(y - 3) + 1(y - 3) = 0 ⇒ (y - 3)(2y + 1) = 0 ⇒ y = 3 and When, y = 3 ⇒ 2x - 1 = 3x + 9 ⇒ x = -10 When, ⇒ 4x - 2 = -x - 3 ## Quadratic Equations Exercise TEST YOURSELF ### Solution 1(a) Correct Option: (ii) –4 Consider (3x – 5)(x + 3) = 3x2 – 5x + 9x – 15 = 3x2 + 4x – 15 Now, 3x2 – kx – 15 = (3x – 5)(x + 3) That is 3x2 – kx – 15 = 3x2 + 4x – 15 Therefore, –k = 4 k = 4 ### Solution 1(b) Correct Option: (ii) 4 For a quadratic equation ax2 + bx + c = 0, the roots are real and equal if b2 – 4ac = 0. For kx2 + kx + 1 = 0, a = k, b = k and c = 1 Now, b2 – 4ac = 0 k2 – 4(k)(1) = 0 k2 – 4k = 0 k(k – 4) = 0 k = 0 or k – 4 = 0 k = 4 But for k = 0, the given equation is not satisfied. Hence, k = 4 (k = 4 is obtained by considering the roots of the equation as real and equal and NOT real and distinct) ### Solution 1(c) Correct Option: (iii) 5 or –1 x2 – 4x = 5 x2 – 4x – 5 = 0 x2 – 5x + x – 5 = 0 x(x – 5) + 1(x – 5) = 0 (x – 5)(x + 1) = 0 x = 5 or x = –1 ### Solution 1(d) Correct Option: (iv) 0 or 7 x2 – 7x = 0 x(x – 7) = 0 x = 0 or x – 7 = 0 x = 0 or x = 7 ### Solution 1(e) Correct Option: (i) 1 Substituting x = 1 in , ### Solution 2 Given i.e So, the given quadratic equation becomes Hence, the values of x are and. (iii) (iv) (v) (vi) (vii) (viii) ### Solution 5 Given quadratic equation is (i) When the equation has no roots (ii) When the roots of are or ### Solution 6 Given quadratic equation is Using quadratic formula, x = a + 1 or x = -a - 2 = -(a + 2) ### Solution 7 Given quadratic equation is Since, m and n are roots of the equation, we have and Hence, . ### Solution 8 Given quadratic equation is …. (i) One of the roots of (i) is , so it satisfies (i) So, the equation (i) becomes Hence, the other root is. ### Solution 9 Given quadratic equation is …. (i) One of the roots of (i) is -3, so it satisfies (i) Hence, the other root is 2a. ### Solution 10 Given quadratic equation is ….. (i) Also, given and and So, the equation (i) becomes Hence, the solution of given quadratic equation are and. ### Solution 11 Given quadratic equation is …. (i) The quadratic equation has equal roots if its discriminant is zero When , equation (i) becomes When , equation (i) becomes x = ### Solution 14 Consider the given equation: ### Solution 15 Given quadratic equation is …. (i) The quadratic equation has real roots if its discriminant is greater than or equal to zero Hence, the given quadratic equation has real roots for. ### Solution 16 (i) Given quadratic equation is D = b2 - 4ac = = 25 - 24 = 1 Since D > 0, the roots of the given quadratic equation are real and distinct. Using quadratic formula, we have or (ii) Given quadratic equation is D = b2 - 4ac = = 16 - 20 = - 4 Since D < 0, the roots of the given quadratic equation does not exist. ### Solution 17 Since, -2 is a root of the equation 3x2 + 7x + p = 1. ⇒ 3(-2)2 + 7(-2) + p = 1 ⇒ 12 - 14 + p = 1 ⇒ p = 3 The quadratic equation is x2 + k(4x + k - 1) + p = 0 i.e. x2 + 4kx + k2 - k + 3 = 0 Comparing equation x2 + 4kx + k2 - k + 3 = 0 with ax2 + bx + c = 0, we get a = 1, b = 4k and c = k2 - k + 3 Since, the roots are equal. ⇒ b2 - 4ac = 0 ⇒ (4k)2 - 4(k2 - k + 3) = 0 ⇒ 16k2 - 4k2 + 4k - 12 = 0 ⇒ 12k2 + 4k - 12 = 0 ⇒ 3k2 + k - 3 = 0 By quadratic formula, we have
# Integers As Exponents View Notes ## Integers And Exponents Wondering what are integer exponents? In Mathematical terms, the integerâ€™s exponents are the exponents that must be an integer. An integerâ€™s exponents can be either a positive integer or a negative integer. By the concept, the positive integer exponents explain the number of times the base number must be multiplied by itself. While the negative integer exponents first explain overturn the value of numerator and the denominator and describe multiplying the number by itself for the number of times. Exponents of integers can be bespoken as positive integers, negative integers, or zero. ### What Are Positive and Negative Integer Exponents? An integer is a number that has no fractional portion of which includes the counting numbers such as {1, 2, 3, 4, 5, 6, 7 â€¦ n}, zero {0} and the negative of the counting numbers {-5, -4, -3, - 2, -1, 0, 1, 2, 3, 4, 5}. An exponent of a number is the number of times we use that number in a multiplication. Now that you must be well aware of the positive integer exponents and negative integer exponents, letâ€™s get started with reviewing the rules for exponents. • ### Rule of Multiplication By the rule of integer exponent, when we do multiplication of the same bases, we have to add exponents. a2 â€¢ a3 = a2+3 = a5 What is the condition/rule if an exponent is negative? Same thing we need to do i.e. to add exponents. a7 â€¢ a-2 = a7+ (-2) = a5 What is the condition/rule if there is more than one variable? We will do each base individually as given below. (ab7)(a5b2) = a1+5 b7+2 = a6 b9 What is the condition/rule if there is a coefficient in facing the variable? We will apply the commutative property to reorganize, multiply the coefficients and add exponents 4a3 â€¢- 2a2 = (4 â€¢-2) â€¢(a2 â€¢ a3) =- 8a5Â Â • ### Rule of Division When we divide the same bases, we will require to subtract exponents a7/a5 = a7-5 What is the condition/rule if the exponent is negative? -10a7b4/6a-3b = -10/6a7-(-3) b4-(-1) = -5/3 a10b5 What is the condition/rule if there is more than one variable? We will do each base individually as given below. a7b4/a5b = a7-5 b5-1 = a2b3 What is the condition/rule if there is a coefficient in facing the variable? We will apply the commutative property to reorganize, multiply the coefficients and add exponents. 10a7b4/6a3b2 = 10/6a7-3 b4-1= 5/3 a4b3 • ### Raising a Power to a Power When we raise a power to a power, we will multiply exponents. (a5)4 = a5â€¢4 = a20 What is the condition/rule if there is more than one variable? We will do each base individually as given below. (a2b)3 = a2â€¢3 b1â€¢3 = a6b3 What is the condition/rule if there is a coefficient in facing the variable? (2a4b2)4 = 24 a4â€¢4b2â€¢4 = 16a16b8 • ### Negative Exponent Rule For the expression, y-m = 1/ya Firstly, write/express with a "top floor" and "bottom floor" Â y-2 = y-2/1 = 1/ y2 Secondly, change the floors if the exponent is "dissatisfied" Â 1/y-2 = y2/1 = y2 The exponent is dissatisfied in the denominator; thus, you will move it to the numerator, so it becomes positive. ### How to Solve Integer Exponents Using RulesÂ make sure to go steadily and meticulously. Here we are providing a shortened version to memorize the rules for exponents. Division â†’ Subtract exponents Power to a power â†’ Multiply exponents Negative â†’ Change "floors" 1. What do we understand by exponents in mathematics? Mathematically, exponents are used to representing repeated multiplication of a number by itself. Writing large numbers in maths at times becomes monotonous. In vast algebraic expressions, they inhabit more space and occupy more time. This issue, in particular, is solved by the use of exponents. 2. What is an example of exponent? For example, 5 Ã— 5 Ã— 5 can be written as 53. In this example, the exponent is â€˜3â€™ which represents the number of times the given number (value) is multiplied by itself. The number 5 is what we call as the base which is the actual number that is being multiplied. Now letâ€™s say, the speed of light is 200000000 m/s. This can be simply expressed as 2 Ã— 108 m/s (approximate value). 3. What is the process of using exponents called? This mathematical mechanism of using exponents is known as â€˜rising to a powerâ€™ in which the exponent is the power. It is important to note that there is a significant difference between exponents and powers. 4. What is the difference between Exponents and Powers? We are aware that the expression 5 x 5 can be computed; however the expression can also be expressed in a short form that is called as exponents. Â 5 Ã— 5 = 52 An algebraic expression that defines repetitive multiplication of the same value is called power. The value 5 is called the base or power and the number 2 is known as an exponent. It is consonant with the number of times the base has functioned as a factor.
# Duke Math Meet 2008 Problem 8 Solution (Individual Round) Find the last two digits of $$\sum_{k=1}^{2008} k \binom{2008}{k}$$ Discussion: $$(1+x)^n = \sum_{k=0}^{n} \binom{n}{k}x^k$$ We differentiate both sides to have $$n(1+x)^{n-1} = \sum_{k=1}^{n} k \binom{n}{k}x^{k-1}$$ Put n=2008 and x=1in the above equation to have $$2008(1+1)^{2007} = \sum_{k=1}^{2008} k \binom{2008}{k}1^{k-1}$$ Thus $$\sum_{k=1}^{2008} k \binom{2008}{k} = 2^{2007} \times 2008 \equiv 8 \times 2^{2007} \equiv 2^{2010}$$ mod 100 Now $$2^{12} = 4096 \equiv -4$$ mod 100 . Now raising both sides to the power of 6, we have $$2^{72} \equiv (-4)^6 \equiv 2^{12} \equiv -4$$mod 100 . Again raising both sides to the power 6 we have $$2^{432} \equiv -4 \implies 2^{1728} \equiv 2^8 \equiv 56$$ mod 100. Earlier we had $$2^{72} \equiv -4 \implies 2^{216} \equiv -64 \equiv 36$$ mod 100 Hence $$2^{1728} \times 2^{216} \equiv 2^{1944} \equiv 56 \times 36 \equiv 16$$ mod 100 Finally we know $$2^{12} \equiv -4 \implies 2^{60} \equiv -1024 \equiv 76$$ mod 100 . Hence $$2^{1944} \times 2^{60} \equiv 2^{2004} \equiv 16 \times 76 \equiv 1216 \equiv 16$$ mod 100 Thus $$2^{2004} times 2^{6} \equiv 2^{2010} \equiv 16 times 64 \equiv 1024 \equiv 24$$ mod 100. The two digits are 24.
# Theorems and problems on parallelograms (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) # Textbook 1. These videos provide proofs of theorems related to triangle and parallelogram on the same base and between same parallels ## Reference Books NCERT mathematics- Chapter 9: Areas of parallelograms and triangles # Teaching Outlines ## Concept #1. Area of parallelograms on same base and between same parallels ### Learning objectives 1. Parallelograms on the same base and between same parallels are equal in area. 2. Area of a parallelogram is the product of its base and the corresponding altitude. 3. Parallelograms on the same base and having equal area lie between same parallels. ### Notes for teachers These are short notes that the teacher wants to share about the concept, any locally relevant information, specific instructions on what kind of methodology used and common misconceptions/mistakes. ## Concept #2. Area of triangles on same base and between same parallels ### Learning objectives 1. Triangles on same base and between same parallels have equal area. 2s. Area of a triangle is half the product of its base and the corresponding altitude. 3. Triangles on the same base and of equal area lie between the same parallels. ## Concept #2. Mid-point theorem ### Learning objectives 1. The segment connecting mid-point of any two sides of a triangle is parallel to the third side 2. The length of the segment connecting the mid-point of two sides is equal to half the third side. # Hints for difficult problems Here is a micro analysis of solution for a problem [geometry_problemsolving.odt here] No Explanation Analysis 1 Drawing ABCD.Cyclic orderD above 2 AB=CD=4X Propertiesof parallelogram 3 AD=BC=2X+1 Propertiesof parallelogram 4 AD=BC=2X+1 Propertiesof parallelogram 5 AD=BC=2X+1 Propertiesof parallelogram 6 AD=BC=2X+1 Propertiesof parallelogram 7 AD=BC=2X+1 Propertiesof parallelogram 8 AD=BC=2X+1 Propertiesof parallelogram \|}
# ACT Math : How to find the distance between clock hands ## Example Questions ### Example Question #1 : Clock Math A church commissioned a glassmaker to make a circular stained glass window that was two feet in diameter. What is the area of the stained glass window, to the nearest square foot? Explanation: To answer this question, we need to find the area of a circle. To find the area of a circle, we will use the following equation: , where is the radius. We are given the diameter of the circle, which is . To find the radius of a circle, we divide the diameter by two. So, for this data: So, our radius is We can now plug our radius into our equation for the area of a circle: The answer asks us what the area is to the nearest square foot. Therefore, we must round our answer to the nearest whole number. To round, we will round up if the digit right before your desired place value is 5, 6, 7, 8, or 9, and we will round down if it is 0, 1, 2, 3, or 4. Because we want the nearest square foot, we will look at the tenths place to aid in our rounding. Because there is a 1 in the tenths place, we will round down. So: Therefore, our answer is . ### Example Question #1 : How To Find The Angle Of Clock Hands What is the measure of the larger angle formed by the hands of a clock at ? Explanation: Like any circle, a clock contains a total of . Because the clock face is divided into equal parts, we can find the number of degrees between each number by doing . At 5:00 the hour hand will be at 5 and the minute hand will be at 12. Using what we just figured out, we can see that there is an angle of between the two hands. We are looking for the larger angle, however, so we must now do ### Example Question #3 : How To Find The Distance Between Clock Hands Assuming the clock hands extend all the way to the edge of the clock, what is the distance between the clock hands of an analog clock with a radius of 6 inches if the time is 2:30? Reduce any fractions in your answer and leave your answer in terms of Explanation: To find the distance between the clock hands, first find the angle between the clock hands and use that as your central angle to find the porportion of the circumference. The angle between the hands is : ### Example Question #4 : How To Find The Distance Between Clock Hands An watch with a face radius of displays the current time as . Assuming each hand reaches to the edge of the clock face, what is the distance, in , of the minor arc between the hour and minute hand? Leave in your answer. Assume a display (not a military clock). Explanation: To start, we must calculate the circumference along the clock face. Since we are required to leave in our answer, this is as easy as following our equation for circumference: Next, we must figure out the positions of the hands. At , the clock's minute hand would be exactly on the , and the hour hand would be of the way between the and the . Therefore, the total distance in terms of hour markings is . To find the distance between adjacent numbers, divide the circumference by : , where is the distance in inches between adjacent hours on the clock face. Finally, multiply this distance by the number of hours between the hands. . Thus, the hands are apart. ### Example Question #5 : How To Find The Distance Between Clock Hands An military () clock with a face diameter of displays the current time as . Assuming each hand reaches to the edge of the clock face, what is the distance, in feet, of the minor arc between the hour and minute hand? Leave in your answer. Explanation: To start, we must calculate the circumference along the clock face. Since we are required to leave in our answer, this is as easy as following our equation for circumference: Next, we must figure out the positions of the hands. For this problem, the clock has 24 evenly spaced numbers instead of 12, so we must cut the distance between each number mark in half compared to a normal clock face. At , the clock's minute hand would be exactly on the , and the hour hand would be of the way between the and the (remember, there are still only seconds in a minute even on a 24-hour clock. Therefore, the total distance in terms of hour markings is . To find the distance between adjacent numbers, divide the circumference by : , where is the distance in feet between adjacent hours on the clock face. Finally, multiply this distance by the number of hours between the hands. . Thus, the hands are feet apart. ### Example Question #6 : How To Find The Distance Between Clock Hands An outdoor clock with a face diameter of displays the current time as . Assuming each hand reaches to the edge of the clock face, what is the distance, in feet, of the minor arc between the hour and minute hand? Leave in your answer. Assume a display (not a military clock). Explanation: To start, we must calculate the circumference along the clock face. Since we are required to leave in our answer, this is as easy as following our equation for circumference: Next, we must figure out the positions of the hands. At , the clock's minute hand would be exactly on the , and the hour hand would be of the way between the and the . Therefore, the total distance in terms of hour markings is . To find the distance between adjacent numbers, divide the circumference by : , where is the distance in inches between adjacent hours on the clock face. Finally, multiply this distance by the number of hours between the hands. . Thus, the hands are feet apart. ### Example Question #1 : Clock Math Find the distance between the hour and minute hand of a clock at 2:00 with a hand length of .
# How do you evaluate an improper integral? ## How do you evaluate an improper integral? How do you evaluate an improper integral? I don’t know if this is possible. I can’t read the documentation for the integral that I am about to write, and I don’t know what to do. I have been asked to determine the correct integral for a given value, and I can’t seem to find a way to do so. The question is how do I evaluate the integral? How do I determine the correct value for _I_, and _this_, and so on? What I’m asking is how do you evaluate the integral for a list of values, and _this_. I can do this just fine, but I’m really confused by how I’m supposed to do this. I’ve seen some posts from people who don’t seem to understand the basics of math, and I’d really like to know why they think this isn’t correct. They think it’s because I’m missing some key elements. A: You can do this using the integral $$\int_0^1 \frac{\frac{dx}{x}}{x^2 + 1} = \frac{1}{x^2} \int_{0}^1 \int_{\frac{x}{x+1}}^1 \bigg(\frac{1+\sqrt{1+x}}{1+ \sqrt{x}}\bigg) \frac{dx+\sqrho}{1-\sqrt{\frac{x^3}{1-x^2}}}$$ Where $x$ is the number of observations and $\sqrho$ is the ratio of the observations to the density. I don’T know what you’re looking for. The integral will give the correct value of $1/x$, and you can’t use the integral to get that answer. You need to use the integral over $x$ to obtain the result you want. The correct expression for the integral is $$\frac{1-\frac{3\sqrt[3]{x^4}}{4}}{\sqrt[4]{x-1}} = \frac{\sqrt{3}-1}{\sqrt(3-1)x}$$ How do you evaluate an improper integral? You can check whether it is enough to justify the previous part of the question. How can you determine if an improper integral is necessary for a correct calculation? There are many ways in which you can estimate an improper integral such as by using a confidence interval (CI) if your data is not the same as the one you are looking for. You could also use an interval, for example, to estimate the value of the integral for a set of values that is close to the range of the interval. What is an improper integral for? A improper integral is an anonymous that is not a part of the mathematical expression of the theory. This means that one is not justified in the assumption that $T$ is the same as $T^{n}$ for every $n\geq 1$. When $T$ has a value of $x$, $T^{x}$ is not the value of $T$ that you are looking at. When $T^{-1}$ is the value of a negative integral, you are not justified in assuming that $T^{1}$ and $T^{2}$ are the values of $T$. If $T$ was not a positive integral, you would be justified in assuming $T^{0}$ and that $T^0$ is the negative integral that you are not making the assumption that it is the value $T$ you are looking to estimate. If you are looking on a positive value of $n$, you could use the interval method. ## Can You Cheat In read the article Classes When you have a positive value for $n$ than you need to be justified in using the interval method to estimate $n$ and $n$ together. To sum up, you should use the interval methods in your study of the theory of integrals. In this case, both methods can be used. Your study of the integral is a good starting point. Some important facts about the theory of integral The theory of integral is the most general theory of a number. It is the theory that one know how to use. There is a lot of research in this field since time is past. This is why you must read its sources for you. Learn this topic by reading the book I gave you and also the book by David A. M. Hausman. Be sure to read it for all the useful information. http://www.cs.washington.edu/people/hartmann/papers/Integral.html This book can be found at the book “Integrals and the Theory of Integrals” by David A M. Hauseman and Benjamin G. Stroud. The book is fascinating and is essential reading for anyone interested in the theory of the theory and for anyone who wishes to study the theory and its applications. ## Is Paying Someone To Do Your Homework Illegal? In this book, go to this website like it get a good understanding of the theory that you need to use the method of the interval method in the study of the integration. For this book, I’ve given you the book and it is not too long to read about Your Domain Name theory. The book is good for students who are not interested in the theoretical theory. The book provides you with a good understanding as to why the theory is good and what the theory is about. Once you are familiar with the theory, it can help you to understand the theory. Therefore, you will learn the theory and understand how the theory is used in practice. Understanding the theory is important for you as it helps you understand how the application of the theory determines the value of your calculations. This knowledge is helpful for you to become more familiar with the theoretical theory and understand the theory’s applications. The theory is used by you to understand your calculations and to see how the theory works. As you read on, it is very important for you to know the theory. You should read in order to understand the theoretical theory at the same time. Now that you have read the book, you should understand the theory and get to know the method and its applications in the study and use of the theory in practice. For example, you will become aware that the theory can be used to calculate an integral. When you read this book, it is not a good way to understand the method of integration. You should understand the method and use the method to see the application of this theory to your calculations. For example. Here is a good way of understanding the method of method. Let’s start with the method of integrating the theory. We need to find the value of this function. We can calculate the value of $\frac{1}{4}$ by solving the equation \frac{1+\sqrt{1-2/How do you evaluate an improper integral? The following two points will provide some pointers on how to evaluate an improper integration. ## Class Taking Test There is a difference between an improper integral and a special this post The special integral is the integral that needs to be evaluated at time zero, not at the time of the published here integration. The improper integral is the improper integral that does not need to be evaluated. A proper integration is a function that contains an integral that is strictly positive. This is the purpose of the above example. The improper integration is a (special) integral. When an improper integral is used, it is not considered a special integral, but the integral that is defined as the integral that contains the negative part of the improper integral. That is why it is often said that the proper integration is not the integral that does the negative part. In a proper integration, the integral that has the negative part is called the negative integral. In a special integral a negative integral is called the positive integral. This is because the negative part that was used in the purpose of having the improper integral is less than the positive one. If you want to evaluate the improper integral, you must use this simple example. Here is an example of a proper integration. // An equation for the right hand side. // We will make complex. // The denominator is positive. // However, the denominator must be negative. // In the real world, this is a simple example. The denominator will be positive. void main() { // Here we use a negative integral. ## Pay To Get Homework Done int a = 10; // A proper integration. // The denominator of the numerator will be negative. int b = 5; // We find the negative part and the negative part we need to evaluate. // We have to evaluate the positive part that the denominator contains ### Related Post What is the purpose of the Business Case Refinement in What were the key events of the Korean War? The What is the policy on communicating with other students during How do you solve linear equations with two variables? A What is the format of the reading section on the What is the role of diversification in strategic management?–Unstable, partially What is the Microsoft Certification community? Here’s a quick rundown What was the significance of the Battle of Stalingrad? So What is content marketing strategy? Content marketing strategy is a What is the policy on using external resources during a
Lesson Plan # Dividing Fractions An exercise in inverting one’s thinking Objectives Students will be able to... • determine the inverse/reciprocal of an integer or fraction • outline the steps for dividing one fraction by another • apply the steps for finding the quotient of two fractions Subjects Math #### 1 Hook Slice It! Free, Paid Review the concept of fractions – part of a whole – with Slice It! Have students test their spatial skills by splitting unconventional into equal parts. Reinforce key vocabulary such as numerator, denominator, and equivalent. #### 2 Direct Instruction In order to understand how fractions are divided, ask students to reconceptualize integer division problems. For instance, when trying to solve 12 ÷ 3, show that you are actually trying to figure out how much 1/3 of 12 is. For division problems, fractions and otherwise, we are actually multiplying by the inverse 12 ÷ 3 = 12 x 1/3. This tactic will set the foundation for more complex problems. Next, demonstrate how to apply this procedure to dividing an integer by a fraction: 2 ÷ 4/5 = 2 x 5/4 or 2/1 x 5/4. Finally, apply this process to dividing one fraction by another: 3/5 ÷ 1/4 = 3/5 x 4/1. Again, the rule of thumb is: find the inverse (or reciprocal) of the divisor (second number), then multiply. Show Khan Academy’s video, Dividing Fractions (8:59). Follow up with several practice problems, and try to apply to real world examples. When possible, diagram the quotients with rectangles to show how to simplify the answer (especially when changing improper fractions into mixed numbers). Make sure to introduce key vocabulary as necessary: inverse, invert, reciprocal, dividend, divisor, quotient. #### 3 Guided Practice ExploreLearning Gizmos Free to try, Paid Have students work in small collaborative groups to experiment with Explorelearning’s Dividing Fractions Gizmo (http://www.explorelearning.com/index.cfm?method=cResource.dspDetail&ResourceID=1006). This interactive will give students a visual model to help explain the rule for dividing fractions (find the inverse of the divisor, then multiple). Together, students should answer the accompanying Student Exploration: Dividing Fractions packet. #### 4 Independent Practice Have students create their own tutorial videos for dividing fractions with Educreations Interactive Whiteboard. They can record themselves explaining the problem solving process, including any tips and hints they value. They can end with challenging practice questions for fellow classmates. #### 5 Wrap-Up Mangahigh Free, Free to try, Paid Student can play with Managhigh’s Division with Fractions game, receiving scaffolded support as they go. Players earn more points with increased speed, increased difficulty (three levels), and by solving three problems in a row.
Polynomial and rational functions Exploring quadratics and higher degree polynomials. Also in-depth look at rational functions. Just saying the word "quadratic" will make you feel smart and powerful. Try it. Imagine how smart and powerful you would actually be if you know what a quadratic is. Even better, imagine being able to completely dominate these "quadratics" with new found powers of factorization. Well, dream no longer. This tutorial will be super fun. Just bring to it your equation solving skills, your ability to multiply binomials and a non-linear way of thinking! ### Completing the square and the quadratic formula You're already familiar with factoring quadratics, but have begun to realize that it only is useful in certain cases. Well, this tutorial will introduce you to something far more powerful and general. Even better, it is the bridge to understanding and proving the famous quadratic formula. Welcome to the world of completing the square! Tired of lines? Not sure if a parabola is a disease of the gut or a new mode of transportation? Ever wondered what would happen to the graph of a function if you stuck an x² someplace? Well, look no further. In this tutorial, we will study the graphs of quadratic functions (parabolas), including their foci and whatever the plural of directrix is. You are familiar with factoring quadratic expressions and solving quadratic equations. Well, as you might guess, not everything in life has to be equal. In this short tutorial we will look at quadratic inequalities. ### Polynomials "Polynomials" sound like a fancy word, but you just have to break down the root words. "Poly" means "many". So we're just talking about "many nomials" and everyone knows what a "nomial" is. Okay, most of us don't. Well, a polynomials has "many" terms. From understanding what a "term" is to basic simplification, addition and subtraction of polynomials, this tutorial will get you very familiar with the world of many "nomials." :) ### Binomial theorem You can keep taking the powers of a binomial by hand, but, as we'll see in this tutorial, there is a much more elegant way to do it using the binomial theorem and/or Pascal's Triangle. ### Simplifying rational expressions You get a rational expression when you divide one polynomial by another. If you have a good understanding of factoring quadratics, you'll be able to apply this skill here to help realize where a rational expression may not be defined and how we can go about simplifying it. ### Rational functions Have you ever wondered what would happen if you divide one polynomial by another? What if you set that equal to something else? Would it be as unbelievably epic as you suspect it would be? ### Partial fraction expansion If you add several rational expressions with lower degree denominator, you are likely to get a sum with a higher degree denominator (which is the least-common multiple of the lower-degree ones). This tutorial lets us think about going the other way--start with a rational expression with a higher degree denominator and break it up as the sum of simpler rational expressions. This has many uses throughout mathematics. In particular, it is key when taking inverse Laplace transforms in differential equations (which you'll take, and rock, after calculus).
Self-paced Explore our extensive collection of courses designed to help you master various subjects and skills. Whether you're a beginner or an advanced learner, there's something here for everyone. Bootcamp Learn live Join us for our free workshops, webinars, and other events to learn more about our programs and get started on your journey to becoming a developer. Upcoming live events Learning library For all the self-taught geeks out there, here is our content library with most of the learning materials we have produced throughout the years. It makes sense to start learning by reading and watching videos about fundamentals and how things work. Search from all Lessons ← Back to Lessons • python • pandas • machine-learning • prob-and-stats Edit on Github Open in Colab # Plotting Binomial Distribution with Python ### Plotting Distributions¶ The fourth row of our sennet table contained the values: $1 ~ 4 ~ 6 ~ 4 ~ 1$ We can draw a plot of this using bars to represent the counts of each outcome. In [ ]: bars = ['no white', 'one white', 'two white', 'three white', 'four white'] counts = [1, 4, 6, 4, 1] In [ ]: plt.bar(bars, counts) plt.title('Counts for each outcome with four two sided sticks'); Using this plot, we come back to the problem of determining the probability of a given outcome or outcomes. Here, we can interpret this probability as the relative area of a given bar to the overall count. For example, we consider each bar having width of one unit, and height of the count. Thus, we have a total area of: $\textbf{TOTAL AREA} = 1 + 4 + 6 + 4 + 1 = 16$ This is the total number of possible outcomes. Thus, determining the probability of a specific outcome is as simple as dividing the total area of our bars by the area under the event of interest. $P(\text{two white}) = \frac{\text{area of bar for two white}}{\text{total area}} = \frac{6}{16}$ ### Problems¶ 1. Use the plot above to determine the probability of zero white sticks. 2. Use the plot above to determine the probability of one white stick? 3. What is the probability of one, two, or three white sticks and how do we use the graph to determine this.
Join NURSING.com to watch the full lesson now. # 01.02 Working with Fractions ### Overview 1. Fractions: Conceptual vs. Procedural Knowledge 2. Fractions Are: Numbers 3. Fractions Are Not: 2 Numbers 4. Pro Tips for Problem Solving with Fractions 1. Equivalent Fractions 1. Identify Equivalent Fractions 2. Make Fractions Equivalent 2. Comparing Fraction Size 1. Must be Equivalent 2. Ordering Fractions 3. Unit Fractions and Whole Numbers 4. Lowest (Simplest) Form 5. Fractions: Same Number; Different Way 1. Proper 2. Improper 3. Mixed 4. Relationship of Fractions to Decimals 1. Change fractions to decimals 2. Change decimals to fractions 1. Solve Problems with Fractions 1. Adding and Subtracting – Same or Different Unit? 1. Add fractions – Same Unit 2. Subtract Fractions – Same Unit 3. Subtract Fractions – Regrouping 2. Multiplying and Dividing – Unit Not Important 1. Multiply Fractions 2. Divide Fractions 2. Seeing Fractions Everywhere #### Video Transcript Hey, welcome to the course working with fractions, where I’ll clarify everything you need to know to gain confidence with fractions. In this course, we’ll prioritize concepts over procedures. So I’ll give you a very clear definition of fractions, a ton of pro tips about fractions, and then we’ll solve example problems. It’ll be fun. Let’s start at the beginning and understand that a fraction is a number, and that a fraction is NOT 2 numbers. So while you see 2 digits in the fraction ½, it represents 1 amount, just like 0 and 1 do. Thinking of fractions as numbers on a number line will help you to conceptualize a fractional amount. Think about where 1/2 belongs relative to 0 and 1. These are also numbers: 1/16 and 15/16, and also represent a partial amount relative to the whole. Understand that fractions can represent the same amount, or equivalent, even when the digits in them are different. See here how each smaller line represents 1/16, then we have eight units to the midpoint. and where 1/2 is considered one step away from zero, 8/16 is considered eight steps. but they represent the same amount – so 1/2 and 8/16 are equivalent fractions. If you have a strong knowledge of multiplication tables, you ‘ll quickly see the relationship numerically between these two fractions. Change 8/16 to ½ by dividing the numerator and the denominator by 8, or change ½ to 8/16 by multiplying both by 8. It’s important to be comfortable with this concept so fractions become second nature. Another important tip to remember is that all fractions will be easier to manipulate if you work with the simplest form. You probably learned the word “reduce” when changing fractions to their simplest form. Why would a recipe say that I have to use 8/16 of a cup of milk when it can say I need ½, the simplest form of 8/16? You have to understand equivalent fractions in order to be able to work with the simplest form , so think of these together. You have to understand how to compare fractions by size. When the denominator in the fractions you’re comparing is different, relative size can be misleading. It’s like comparing apples to oranges. So to order fractions correctly, you have to create equivalent fractions by giving them the same denominator. Here again, knowing your multiplication tables inside out helps make this an automatic task. When I see ½, ¾, and 5/12, I immediately think to make the common denominator 12, because 12 is divisible by 2, 4, and 12. Now I can make an equivalent fraction for each – like this (6/12, 9/12, and 5/12 remains the same.). Now you can compare the original fractions like Apples to Apples, and see that 5/12 is the smallest, 1/2 is bigger, and 3/4 is the largest. Understanding the relative size of fractions helps you do a lot more mental math during problem solving. A unit fraction is any fraction that has 1 as a numerator, so represents the smallest unit of that fractional amount. 1/16 is a unit fraction on a measuring tape, where you will typically an inch divided into 16 equal parts. In contrast, 16/16, which would be here on the number line, represents a whole number. In this case, the whole inch. You can improve your conceptual understanding of relative fractions size by comparing the numerator to the denominator and imagining where it would go on a number line. The closer to 1 the numerator is, the smaller that number. As the numerator approaches the denominator, you can see you’re getting closer to one. In any fraction in which the numerator is exactly half the denominator, that’s always the numerical middle point. Now you can estimate with fractions by saying about half almost all or nearly none. This tip spans two slides. The numbers you see here are not all equivalent, but they are three examples of different ways fractions can be written. You’ll want to get comfortable with each form and what they mean relative to a number line value of 1. A proper fraction is simply a number that is less than 1 – recognized because the numerator is less than the denominator. An improper fraction is a number that’s greater than 1 because you have more parts, represented by the numerator, then what constitutes one. And a mixed number just represents a whole number and a fractional part of that whole number. So 1 and ⅓ is a bit more than 1 on the number line, but still less than 1 and a half. Can you see that? Changing from an improper fraction to a mixed fraction is simply dividing the numerator by the denominator, like this. Now let’s look at another way fractional amounts can be represented. Surprise! It’s a decimal. Did you know that fractions and decimals represent the same amount but in a different form? Many problems on tests require you to change a fraction to a decimal. To do that, you have to know the place value names that are less than one, or to the right of the decimal. Maybe you learned to say your decimals like this . . “point 5”, but I beg you to get into the habit of using the place value name of the digit on the right because that will immediately tell you the value of that number. Isn’t 5 tenths easier to visualize as 1/2 than point 5? So now, start writing every decimal amount as a fraction just by making the place value name the denominator. So 15 hundredths is this decimal – and this fraction. (15/100 = 3/20). Practice this so you start to relate relative size of decimals to fractions – allowing you to change decimals into fractions like a pro. Now let’s take a look at problem solving using fractions. When adding and subtracting fractions, the first thing you want to notice is whether the units, or the denominators, are the same. If they are, adding or subtracting is just a matter of counting up or down the number line like with whole numbers. Adding 2/16 + 3/16 + 4/16 gives you 9/16. It’s as simple as adding the numerators and keeping the denominator. Subtracting is the same, so think about “ 10/16 – 8/16?” Right! 2/16. Typically your final answer should be stated using an equivalent fraction in its simplest form. You remember how to do that? Divide the numerator and denominator of 2/16 by 2 and you’ll get the equivalent fraction of 1/8 . Simple huh? Don’t get all nervous about adding and subtracting fractional problems that begin with different units! Remember earlier when we created equivalent fractions? Just do that first and then perform the operation on those equivalent fractions. So for 1/2 + 3/4 find a denominator that’s divisible by both 2 and 4. Use 4. And just like you did when you ordered the fractions according to their size, now that you have 1 denominator for both you can make equivalent fractions. 1/2 becomes 2/4 and 3/4 Remains the same. Watch me do a think aloud while I subtract 1 fraction from another that represents a different unit. Remember: create equivalent fractions that share a common unit, and then think of moving up or down the number line. You’ll do great. Just like when you have to regroup, or borrow, when subtracting whole numbers, you might have to when subtracting fractions. But there’s nothing mysterious about it if you see the fractions as numbers relative to a number line. In this problem here I have mixed fractions. I have no problem subtracting the 4 from the 6 but I can already see that 2/3 is larger than ⅓. When subtracting you cannot change the order of the numbers, so you’ll have to borrow from the whole number.. My first recommendation is to line the fractions up vertically like this. Now borrow. Making it five and take one. Since you can’t subtract numbers that are in different forms, just write an equivalent number so the forms match. Change the one to an equivalent fraction, which in this case is 3/3 . No ,choosing 3 as the denominator is not random, I chose it because the other fractions have 3 as their denominator .Now that I have borrowed 3/3 ,or 1, from the 6, I have to add it to the fraction that’s currently there. This gives me 5 + 4/3 -4 + ⅔. Now you can easily perform subtraction after regrouping. The cool thing about multiplying fractions is that you don’t have to consider the unit or the denominator. Simply multiply across the numerators, then across the denominators, and you have your product. From 6/12 you should automatically see that 1/2 is the same value but the simplest form. Most tests will want your answer in this format. Here’s another multiplication problem, but this one makes you choose when to create the fractions in the simplest form. Something important about multiplying fractions – your product is SMALLER than the original numbers, unlike when multiplying with whole numbers. Be aware of this when checking word problems. Dividing by fractions requires a special understanding. Typically folks teach only the shortcut to the answer, while never explaining why it actually works. So I’ll start with the concept behind the shortcut so you can see how it works, then teach you the shortcut. The rule to understand what is happening is simply 2 steps: 1) get the common denominators , then 2) divide the numerators. When you establish that the common denominator is 12, written out longhand, the problem looks like this: (33)/43) ÷ (24)/(34).You have completed step 1. When you divide across, the denominators = 1, which leaves you to finish with step 2, which here is 9 ÷8 which equals 1 and ⅛. But that’s the long way around. The shortcut, which results in the same answer, is to 1) change the sign to multiplication, then 2) invert the divisor like this. This gives you a shorter equation to solve, but results in ¾ * 3/2 = 9/8 = 1 and ⅛. The shortcut simply turn the division problem into a multiplication problem, and it’s quicker. Just be sure you understand the reason behind the shortcut to understand why it works . I hope you feel more confident about fractions. Key points to take away: fractions are numbers (not two numbers) that are easily recognized on a number line. Being able to quickly change fractions to . equivalent forms, from improper to mixed fractions, and from decimals will make your life so much easier. Practice regrouping when subtracting fractions. Build it into your muscle memory. and finally, play with math in your head in your daily life. See fractions everywhere. That’s what’ll really turn you into a math person. So that’s it for working with fractions and as always… …Happy Nursing! [FREE] [FREE]
# Math View: Sorted by: ### How to Approximate Area with Left Sums You can approximate the area under a curve by using left sums. For example, say you want the exact area under a curve between two points, 0 and 3. The shaded area on the left graph in the below figure ### How to Approximate Area with Right Sums You can approximate the area under a curve by using right sums. This method works just like the left sum method except that each rectangle is drawn so that its right upper corner touches the curve instead ### How to Do Simple Integration by Parts Integrating by parts is the integration version of the product rule for differentiation. The basic idea of integration by parts is to transform an integral you ### How to Change Unacceptable Forms before Using L'Hôpital’s Rule You can use L'Hôpital’s rule to find a limit for which direct substitution doesn't work. If substitution produces one of the unacceptable forms, ± ∞ · 0 or ∞ – ∞, you first have to tweak the problem to ### How to Use Tangent Substitution to Integrate With the trigonometric substitution method, you can do integrals containing radicals of the following forms: ### How to Use Sine Substitution to Integrate With the trigonometric substitution method, you can do integrals containing radicals of certain forms because they match up with trigonometric functions. A sine can take the place of a radical in a particular ### How to Find the Average Value with the Mean Value Theorem for Integrals You can find the average value of a function over a closed interval by using the mean value theorem for integrals. The best way to understand the mean value theorem for integrals is with a diagram — look ### How to Find the Volume of a Shape Using the Washer Method Geometry tells you how to figure the volumes of simple solids. Integration enables you to calculate the volumes of an endless variety of much more complicated shapes. If you have a round shape with a hole ### How to Analyze Arc Length You can add up minute lengths along a curve, an arc, to get the whole length. When you analyze arc length, you divide a length of curve into small sections, figure the length of each section, and then ### Statistics Terms to Know when Using Excel 2007 Data Analysis Tools With the data analysis tools available in Excel 2007, you can create spreadsheets that show the details of any statistic you can create a formula to find — and you can find any number. It helps to know ### Factoring Special Problems Binomials, their powers, and their products with selected trinomials occur frequently in algebraic processes. By using the patterns shown here, you save time and reduce the opportunity for errors. ### Formulas for Common Geometric Shapes Depending on the algebra problem, you'll need to know some geometry. The following represents some of the most common shapes in geometry and their formulas for perimeter, area, volume, surface areas, and ### Order of Operations in Algebra When creating simpler and more useful expressions, you want to be careful not to change the original value. By applying the order of operations, you maintain that value. ### Rules of Exponents Exponents are shorthand for repeated multiplication. The rules for performing operations involving exponents allow you to change multiplication and division expressions with the same base to something ### Selected Math Formulas Step by Step Algebraic formulas make life (and algebra) simpler. You save time by not having to perform more complicated tasks. When using the formulas, use the appropriate rules for simplifying algebraic expressions ### Algebra Workbook For Dummies Cheat Sheet Formulas, patterns, and procedures used for simplifying expressions and solving equations are basic to algebra. Use the equations, shortcuts, and formulas you find for quick reference. This Cheat Sheet ### Simple Steps for Solving Word Problems Word problems aren’t just on school tests. You solve word problems every day in your work or even while you’re just out and about. Don’t worry — these steps make solving word problems easier than you think ### Calculating a Standard 15-Percent Tip Calculating a 15-percent tip in your head may seem tricky, but here’s a, well, tip for those situations when you need to calculate that standard tip amount. And it works 100 percent of the time. In order ### Counting Change for Customers the Old-Fashioned Way Although many cashiers simply dump all your change in your hand at once, counting change the old-fashioned way ensures that customers get the proper amount. Here’s an easy way to make a customer’s change ### Matching American and Metric Units of Measurement If you’ve ever had to match American and metric measurement units, you know direct comparisons of these unit systems can be tricky. If you need to know the metric equivalent of an inch, for example, or ### Technical Math For Dummies Cheat Sheet Grasping some technical math basics can simplify everyday situations faced by many professionals and even non-professionals, including having to solve word problems, calculate tips, make change, or match ### Using Calculators on the SAT When taking the SAT, you're allowed to use a calculator. Calculators can save a lot of time on the SAT if you save them for when you can't quickly and easily do a calculation in your head. The more complicated ### Topics Covered in the Math Sections of the SAT You have a much better chance of getting a good SAT math score if you know what to expect. Focusing your study time on what will be included in the math sections of the SAT ### How to Interpret Function Graphs You’re going to see dozens and dozens of functions in your study of calculus, and the graphs of those functions can visually express such things as inflation, population growth, and radioactive decay. ### How to Horizontally Transform a Function You can transform any function into a related function by shifting it horizontally or vertically, flipping it over (reflecting it) horizontally or vertically, or stretching or shrinking it horizontally
CanadaON Grade 9 # Area composite shapes Lesson We have already had a look at combining 2D shapes together and finding the area of composite shapes. But since then we have learnt about the areas of a whole lot more shapes. Let's just look at a summary of areas of 2D shapes we have looked at so far with our shape robot: ## Strategies The key skills you need to remember when thinking about composite shapes are to consider • can the shape be considered as a larger easier shape with smaller bits missing • can the shape be considered the sum of a number of smaller shapes $\text{Area of a Rectangle }=\text{length }\times\text{width }$Area of a Rectangle =length ×width $A=L\times W$A=L×W $\text{Area of a Square }=\text{Side }\times\text{Side }$Area of a Square =Side ×Side $A=S\times S$A=S×S $\text{area of a triangle }=\frac{1}{2}\times\text{base }\times\text{height }$area of a triangle =12×base ×height $A=\frac{1}{2}bh$A=12bh $\text{Area of a Parallelogram }=\text{Base }\times\text{Height }$Area of a Parallelogram =Base ×Height $A=b\times h$A=b×h $\text{Area of a Trapezoid}=\frac{1}{2}\times\left(\text{Base 1 }+\text{Base 2 }\right)\times\text{Height }$Area of a Trapezoid=12×(Base 1 +Base 2 )×Height $A=\frac{1}{2}\times\left(a+b\right)\times h$A=12×(a+b)×h $\text{Area of a Kite}=\frac{1}{2}\times\text{diagonal 1}\times\text{diagonal 2}$Area of a Kite=12×diagonal 1×diagonal 2 $A=\frac{1}{2}\times x\times y$A=12×x×y $\text{Area of a Rhombus }=\frac{1}{2}\times\text{diagonal 1}\times\text{diagonal 2}$Area of a Rhombus =12×diagonal 1×diagonal 2 $A=\frac{1}{2}\times x\times y$A=12×x×y $\text{Area of a circle }$Area of a circle $A=\pi r^2$A=πr2 $\text{Area of a sector }$Area of a sector $A=\frac{\text{angle of sector}}{360}\times\pi r^2$A=angle of sector360×πr2 #### Worked Examples ##### QUESTION 1 Find the total area of the figure shown. ##### QUESTION 2 Find the total area of the figure shown. ### Outcomes #### 9P.MG2.03 Solve problems involving the areas and perimeters of composite two-dimensional shapes (i.e., combinations of rectangles, triangles, parallelograms, trapezoids, and circles)
# 4.4: Relatively Prime numbers $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ Definition: Relatively prime or Coprime Two integers are relatively prime or Coprime when there are no common factors other than 1. This means that no other integer could divide both numbers evenly. Two integers $$a, b$$ are called relatively prime to each other if $$\gcd(a, b)=1$$. For example, $$7$$ and $$20$$ are relatively prime. Theorem Let $$a, b\in \mathbb{Z}$$. If there exist integers $$x$$ and $$y$$ such that $$ax+by=1$$ then $$\gcd(a, b)=1$$. Proof: Let $$a, b\in \mathbb{Z}$$, such that d= $$\gcd(a, b)$$. Then d\|a and d\|b. Hence $$d\|(ax+by)$$, thus $$d\|1$$. Which implies $$d=\pm 1$$, since gcd is the greatest, $$d=1.$$ Example $$\PageIndex{1}$$: Suppose $$a$$ and $$b$$ are relatively prime integers. Prove that $$\gcd(a+b,a-b)=1,$$ or $$2$$. Solution Let $$a$$ and $$b$$ be relatively prime integers. Then $$\gcd(a,b)=1$$. Suppose $$d=\gcd(a+b,a-b)$$. Then $$d \|(a+b)$$ and $$d\|(a-b)$$. Then $$a+b=dm$$ and $$a-b=dn$$ for some $$m,n \in \mathbb{Z}$$. Now $$2a= d(m+n)$$ and $$2b=d(m-n)$$. Thus $$d\|2a$$ and $$d\|2b$$. Hence $$d \|\gcd(2a,2b).$$ Since $$gcd(2a,2b)=2 \gcd(a,b)$$. $$d\|2$$. Thus $$d=1$$ or $$2.$$ Example $$\PageIndex{2}$$: Suppose $$a$$ and $$c$$ are relatively prime integers and $$b$$ is an integer such that $$b\|c.$$ Prove that $$gcd(a,b)=1.$$ Solution Let $$a$$ and $$c$$ be relatively prime integers. Then $$\gcd(a,c)=1$$. Thus there exist integers $$x$$ and $$y$$ such that $$ax+cy=1$$. Suppose $$b$$ is an integer such that $$b\|c.$$ Then $$c=bm$$ for some $$m \in \mathbb{Z}$$. Now $$ax+bmy=1$$. Therefore, $$gcd(a,b)=1.$$ ### Multiplicative inverse modulo m Let $$m \in \mathbb{Z}_+.$$ Let $$a \in \mathbb{Z}$$ such that $$a$$ and $$m$$ are relatively prime. Then there exist integers $$x$$ and $$y$$ such that $$ax+my=1.$$ Then $$ax \equiv 1 ( mod \,m )$$. Note that $$1$$ is the multiplicative identity on $$mod \,m$$. In this case, $$x (mod \,m)$$ is the inverse of $$a (mod \,m)$$. ##### Example $$\PageIndex{3}$$ If possible, find multiplicative inverse of $$2 (\mod 10).$$ Solution Since $$\gcd(2,10)=2 \ne 1$$, $$2$$ has no multiplicative inverse modulo $$10.$$ ##### Example $$\PageIndex{4}$$ If possible, find multiplicative inverse of $$16 (\mod 35).$$ ###### Solution By using Euclidean Algorithm, we will find $$gcd(16,35)$$. \begin{eqnarray}35&=(16)(2)+3\\16&=(3)(5)+1\\3 &=(1)(3)+0\end{eqnarray} Thus $$gcd(16,35)=1$$. Hence multiplicative inverse of $$16 (\mod 35)$$ exists. By using Bezout's algorithm, \begin{eqnarray}1&=16+(3)(-5)\\&=16+(35+(16)(-2)) (-5)\\&=(35)(-5)+(16)(11)\end{eqnarray} Thus $$gcd(16,35)=1=(35)(-5)+(16)(11).$$ Hence the multiplicative inverse of $$16 (\mod 35)$$ is $$11$$. 4.4: Relatively Prime numbers is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.
Jacky Chong # Jacky Chong - Jacky Chong IB Student D1 HL Mathematics... This preview shows pages 1–6. Sign up to view the full content. Jacky Chong IB Student D1 HL Mathematics Internal Assessment Type II December 14, 2009 Internal Assessment – Designing a Freight Elevator In this paper, I will analyze a given model for the motion of an elevator used to carry freight in a mine. More importantly, I will evaluate its strengths and weaknesses and create my own model with specifications. Part I: Analysis of Given Model I.a Variables, Parameters and Constraints of Given Model The function of the given model is . ” is representing time (in minutes) and “ ” is the position of the elevator. = 0 means that the elevator is at ground level, and = 0 represents the starting time. Therefore, the function tells how the elevator’s position changes in terms of the change of time. Firstly, let us talk about the variables. Variable means values that change. There are a few variables in this situation: time, position, velocity, acceleration and change in rate of acceleration, of which we rarely explore in both Mathematics and Physics. Secondly, let us talk about the parameter. Parameter is an essential element of parametric equation. When we have a function describing a 2 dimensional motion, for example, , where both “ ” and “ This preview has intentionally blurred sections. Sign up to view the full version. View Full Document describe the position of an object, we have a path of the object’s movement. Figure 1 : , path of the object, described with coordinates, both and are position variables. However, we do not know its moving direction, speed and acceleration. Therefore, we add a parameter to the function, which means to make both “ ” and “ ” be in terms of “ ” – time. We let be , and be . Figure 2 : Parametric function of in tracing dot form with 0.1 increment: and . There is nothing left to the y-axis because when , and is the starting time. Let time be measured in seconds. The arrow shows the motion’s direction. Now, we know the direction, speed and acceleration (both one minute, but , so accelerating) of the object’s movement. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document Figure 3 : Parametric function of in tracing dot form with 0.1 increment: and . Path and direction is the same as the graph in Figure 2, but distances between dots increase more violently than the graph in Figure 2, which means greater acceleration and speed. We can also let be and be . The path and direction is the same, but speed and acceleration will increase. Now, we see how parameter and parametric equation help us to add a third dimension (time) to a two dimensional graph (position) with only two axes, and let us know the direction, speed and acceleration. However, in the elevator case, the elevator only goes up and down, which is an one dimensional movement. It only needs one axis, so we can let the other axis be time. In this situation, we already know the direction, speed and acceleration of the object with single equation function, so we don’t need to use parametric function. Therefore, throughout this paper, I will not apply parametric function. However, There is still two parameters in the function – “ ”, time; and “ This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. ## This document was uploaded on 12/31/2011. ### Page1 / 21 Jacky Chong - Jacky Chong IB Student D1 HL Mathematics... This preview shows document pages 1 - 6. Sign up to view the full document. View Full Document Ask a homework question - tutors are online
# TABE Application Level M : Trivia Quiz Approved & Edited by ProProfs Editorial Team The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes. \| By Rosaware R Rosaware Community Contributor Quizzes Created: 2 \| Total Attempts: 1,426 Questions: 50 \| Attempts: 660 Settings . • 1. ### Luis has 19 soft drink cans. His coolers each have space for 6 cans. How many coolers can he fill completely? • A. 3 • B. 4 • C. 5 • D. 6 A. 3 Explanation Luis has 19 soft drink cans and each cooler can hold 6 cans. To find out how many coolers Luis can fill completely, we divide the total number of cans by the capacity of each cooler. In this case, 19 divided by 6 equals 3 with a remainder of 1. Since we are looking for the number of coolers that can be filled completely, the answer is 3. Rate this question: • 2. ### Which of these numbers is a commom factor of 18 and 24? • A. 3 • B. 4 • C. 5 • D. 7 A. 3 Explanation The number 3 is a common factor of 18 and 24 because it can evenly divide both numbers without leaving a remainder. 3 is a factor of 18 because 18 divided by 3 equals 6, and it is a factor of 24 because 24 divided by 3 equals 8. Therefore, 3 is the correct answer. Rate this question: • 3. ### Which number is a multiple of 7? • A. 17 • B. 28 • C. 37 • D. 45 B. 28 Explanation 28 is a multiple of 7 because it can be divided evenly by 7. When 28 is divided by 7, the result is 4, with no remainder. This means that 28 is divisible by 7, making it a multiple of 7. Rate this question: • 4. ### Which number goes in the blanks to make both numbers sentences true? 11 - __ = 7 7 + __ =11 • A. 4 • B. 5 • C. 6 • D. 7 A. 4 Explanation In the first number sentence, 11 minus 4 equals 7. In the second number sentence, 7 plus 4 equals 11. Therefore, the number that goes in the blanks to make both number sentences true is 4. Rate this question: • 5. ### Lee has finished reading 4 out of 10 lessons from his assignment. Which fractional part of the lessons has he completed? • A. 1/10 • B. 1/5 • C. 2/3 • D. 2/5 D. 2/5 Explanation Lee has finished reading 4 out of 10 lessons from his assignment. To find the fractional part of the lessons he has completed, we need to divide the number of lessons completed (4) by the total number of lessons (10). This can be written as 4/10, which simplifies to 2/5. Therefore, Lee has completed 2/5 or two-fifths of the lessons. Rate this question: • 6. ### Which number goes in the box on the number line? • A. 89 • B. 91 • C. 93 • D. 87 B. 91 Explanation The number 91 goes in the box on the number line because it is the number that falls between 89 and 93. Rate this question: • 7. ### Marty needs to pack 32 glasses. Each box holds 6 glasses. How many boxes will be completely full? • A. 4 • B. 5 • C. 6 • D. 7 B. 5 Explanation To find the number of boxes that will be completely full, we need to divide the total number of glasses by the capacity of each box. In this case, 32 glasses divided by 6 glasses per box equals 5. Therefore, 5 boxes will be completely full. Rate this question: • 8. ### The cost of carrots per pound is • A. 59 dollars • B. 72 dollars • C. 59 cents • D. 98 cents C. 59 cents Explanation The correct answer is 59 cents. This is the only option that is in cents, while the other options are in dollars. Rate this question: • 9. ### The cost of tomatoes per pound is • A. 2 cents • B. 7 cents • C. 59 cents • D. 72 cents D. 72 cents Explanation The given answer, 72 cents, is the cost of tomatoes per pound. Rate this question: • 10. ### Lynn had 12 stamps, but she used 3 of them to mail letters. Which expression shows how many stamps she still has? • A. 3 × 12 • B. 12 ÷ 3 • C. 12 - 3 • D. 3 + 12 C. 12 - 3 Explanation Lynn started with 12 stamps and used 3 of them to mail letters. To find out how many stamps she still has, we need to subtract the number of stamps she used from the total number of stamps she had. Therefore, the expression 12 - 3 represents the number of stamps Lynn still has. Rate this question: • 11. ### LaTina is saving \$60 a month for a vacation. She needs to save \$540 dollars. How long will it take her to save enough for her vacation? • A. 6 months • B. 9 months • C. 12 months • D. 18 months B. 9 months Explanation LaTina is saving \$60 a month for her vacation and she needs to save \$540. To find out how long it will take her to save enough, we can divide the total amount needed by the amount she saves per month. \$540 divided by \$60 equals 9. Therefore, it will take her 9 months to save enough for her vacation. Rate this question: • 12. ### Anna saves \$75 every month. How much will she save in 16 months? • A. \$91 • B. \$450 • C. \$900 • D. \$1,200 D. \$1,200 Explanation Anna saves \$75 every month. To find out how much she will save in 16 months, we can multiply her monthly savings by the number of months. So, \$75 multiplied by 16 equals \$1,200. Therefore, Anna will save \$1,200 in 16 months. Rate this question: • 13. ### Ron wrote a check \$37.80. Before he wrote the check, his balance was \$137.75. What is his new balance? • A. 100.15 • B. 100.00 • C. 99.95 • D. 175.55 C. 99.95 Explanation After writing the check for \$37.80, Ron's balance will decrease by that amount. Therefore, his new balance will be \$137.75 - \$37.80 = \$99.95. Rate this question: • 14. ### Ella spends \$50 a month for her cell phone. How much does she pay for her cell phone in a year? • A. \$500 • B. \$600 • C. \$1,000 • D. \$1,200 B. \$600 Explanation Ella spends \$50 a month for her cell phone. To find out how much she pays for her cell phone in a year, we need to multiply the monthly cost by the number of months in a year. So, \$50 multiplied by 12 (months) equals \$600. Rate this question: • 15. ### Which of these decimals when rounded to the nearest whole number is 54? • A. 54.81 • B. 52.65 • C. 53.85 • D. 53.40 C. 53.85 Explanation When rounded to the nearest whole number, the decimal 53.85 becomes 54. Therefore, 53.85 is the correct answer. Rate this question: • 16. ### A dozen pens costs \$5.99. About how much does one pen cost? • A. \$0.50 • B. \$0.40 • C. \$1.00 • D. \$1.50 A. \$0.50 Explanation The total cost of a dozen pens is \$5.99. To find the cost of one pen, we need to divide the total cost by the number of pens in a dozen, which is 12. So, \$5.99 divided by 12 is equal to \$0.50. Therefore, one pen costs \$0.50. Rate this question: • 17. ### Which of these is the best estimate for the width of a piece of paper? • A. 1 centimeter • B. 25 centimeters • C. 1 meter • D. 5 meters B. 25 centimeters Explanation A piece of paper is typically around 21.59 cm wide, which is slightly smaller than the standard size of 8.5 x 11 inches. Among the given options, 25 centimeters is the closest estimate to the actual width of a piece of paper. Rate this question: • 18. ### Jason's average hits per game for the last three months are: 3.6, 4.5, and 6.2. What are his averages, rounded to the nearest whole number? • A. 3, 4 and 6 • B. 4, 4 amd 6 • C. 4, 5, and 6 • D. 4, 5, and 7 C. 4, 5, and 6 Explanation Jason's average hits per game for the last three months are 3.6, 4.5, and 6.2. To round these numbers to the nearest whole number, we round 3.6 down to 4, 4.5 up to 5, and 6.2 down to 6. Therefore, Jason's averages, rounded to the nearest whole number, are 4, 5, and 6. Rate this question: • 19. ### Anya spent \$32 of the \$63 she received for her birthday. She spent about • A. 1/4 of the amount • B. 1/3 of the amount • C. 1/2 of the amount • D. 2/3 of the amount C. 1/2 of the amount Explanation Anya spent \$32 out of the \$63 she received for her birthday. To find out what fraction of the amount she spent, we can divide the amount she spent (\$32) by the total amount she received (\$63). This gives us 32/63, which simplifies to 16/31. However, none of the given options match this fraction. Therefore, we need to approximate the fraction that is closest to 16/31. The closest option is 1/2 of the amount. Rate this question: • 20. ### Which of these measurements is about the same as 100 inches? (Hint: 1 in = 2.54 cm) • A. 40 centimeters • B. 250 centimeters • C. 5 meters • D. 30 meters B. 250 centimeters Explanation 100 inches is about the same as 250 centimeters. This can be calculated by converting inches to centimeters using the given conversion factor of 1 inch = 2.54 cm. Therefore, 100 inches would be equal to 100 * 2.54 = 254 centimeters. Since 254 centimeters is the closest measurement to 250 centimeters, it can be concluded that 250 centimeters is about the same as 100 inches. Rate this question: • 21. ### What temperature does the thermometer show? • A. 24 degrees • B. 26 degrees • C. 28 degrees • D. 29 degrees A. 24 degrees Explanation The thermometer shows a temperature of 24 degrees. Rate this question: • 22. ### What temperature does the thermometer show? • A. 82 degrees • B. 84 degrees • C. 88 degrees • D. 89 degrees C. 88 degrees Explanation The thermometer shows a temperature of 88 degrees. Rate this question: • 23. ### Which of these measures is closest to length of your thumb? • A. 1 centimeter • B. 1 inch • C. 7 centimeters • D. 20 inches C. 7 centimeters Explanation The length of the thumb is usually around 7 centimeters. This is the closest measurement option given in the choices. A centimeter is a metric unit of length, and it is smaller than an inch. Therefore, 7 centimeters is a more accurate approximation of the length of a thumb compared to the other options provided. Rate this question: • 24. ### An airplane is scheduled to leave one city at 10:35 a.m. and arrive at another city at 12:15p.m. How long is the scheduled flight? • A. 1 hour and 20 minutes • B. 1 hour and 40 minutes • C. 2 hours and 20 minutes • D. 2 hours and 45 minutes B. 1 hour and 40 minutes Explanation The scheduled flight is 1 hour and 40 minutes because the time difference between the departure and arrival is 1 hour and 40 minutes. Rate this question: • 25. ### Rick decided to run in a marathon, which is slighty more than 26 miles long. About how many kilometers will he run? (Hint: 1 mi = 1.6 km) • A. 4 kilometers • B. 10 kilometers • C. 41 kilometers • D. 150 kilometers C. 41 kilometers Explanation Rick is running in a marathon, which is slightly more than 26 miles long. To convert miles to kilometers, we use the conversion rate of 1 mile = 1.6 kilometers. Since Rick is running slightly more than 26 miles, we can estimate that he will run slightly more than 26 multiplied by 1.6 kilometers. This calculation gives us approximately 41 kilometers. Therefore, Rick will run approximately 41 kilometers in the marathon. Rate this question: • 26. ### Which of the following figures are simular to the figure in the box? • A. Figures 1 and 2 • B. Figures 2 and 3 • C. Figures 3 and 4 • D. Figures 1 and 4 D. Figures 1 and 4 Explanation Figures 1 and 4 are similar because they both have the same shape and orientation. They both have three sides and are oriented in the same direction. Figures 2 and 3 do not have the same shape or orientation as the figure in the box. Therefore, the correct answer is figures 1 and 4. Rate this question: • 27. ### Which of these solid figures is not named correctly? • A. Cube • B. Sphere • C. Cylinder • D. Cone D. Cone Explanation The given question asks to identify the solid figure that is not named correctly. Out of the options provided, a "cone" is the only solid figure that is named correctly. A cone is a three-dimensional geometric shape that has a circular base and a pointed apex. Therefore, the answer is cone. Rate this question: • 28. ### Which of these figures shows a line of symmertry? • A. 1 • B. 2 • C. 3 • D. 4 A. 1 Explanation Figure 1 shows a line of symmetry because it can be divided into two equal halves that are mirror images of each other. The line of symmetry in this figure is vertical, dividing the figure into two identical halves. In contrast, figures 2, 3, and 4 do not have a line of symmetry as they cannot be divided into two equal halves that are mirror images of each other. Rate this question: • 29. ### Which line creates a line of symmetry in this triangle? • A. Line A • B. Line B • C. Line C • D. Line D D. Line D Explanation Line D creates a line of symmetry in the triangle. A line of symmetry is a line that divides a shape into two identical halves. Line D passes through the midpoint of the base of the triangle and is perpendicular to the base. When the triangle is folded along line D, the two resulting halves will perfectly overlap each other. Therefore, line D is the correct answer for creating a line of symmetry in this triangle. Rate this question: • 30. ### Which of these measurements is equal to the perimeter of the box around the circle? • A. 12 inches • B. 144 inches • C. 4 feet • D. 3 yards C. 4 feet Explanation The perimeter of a circle is equal to the circumference, which is the distance around the circle. The formula to calculate the circumference is 2πr, where r is the radius of the circle. In this case, since we are talking about a box around the circle, the perimeter would be equal to the circumference plus the sum of the lengths of the sides of the box. Therefore, the measurement that is equal to the perimeter of the box around the circle is 4 feet. Rate this question: • 31. ### Which two parts of the circle are congruent? • A. Parts A and B • B. Parts B and D • C. Parts A and D • D. Parts C and D D. Parts C and D Explanation The parts of a circle that are congruent are those that have equal measurements or dimensions. In this case, parts C and D are congruent because they have the same measurement or dimension. Rate this question: • 32. ### According to the graph, how many slices of wheat toast can Jim eat and only consume 270 calories? • A. 2 • B. 3 • C. 4 • D. 5 B. 3 Explanation Based on the graph, the x-axis represents the number of slices of wheat toast and the y-axis represents the number of calories consumed. By locating the point on the graph where the y-axis intersects with the value of 270 calories, we can determine the corresponding number of slices of wheat toast on the x-axis. In this case, the point intersects at 3 slices of wheat toast, indicating that Jim can eat 3 slices and consume only 270 calories. Rate this question: • 33. ### Suppose Jim had wanted a dessert with 520 calories, but he didn't want to add any more calories to his diet. What two items should he not eat? • A. Candy bar and soft drink • B. • C. Spaghetti and toast • D. Toast and yogurt C. Spaghetti and toast Explanation Jim wanted a dessert with 520 calories but didn't want to add any more calories to his diet. The two items he should not eat are spaghetti and toast. This is because spaghetti and toast are likely to have a higher calorie content compared to the other options. Rate this question: • 34. ### Which two players had the same average score? • A. Players 1 and 2 • B. Players 2 and 3 • C. Players 3 and 4 • D. Players 1 and 5 D. Players 1 and 5 Explanation The given answer suggests that Players 1 and 5 had the same average score. This means that the average score of Player 1 is equal to the average score of Player 5. Rate this question: • 35. ### The difference between average scores is greatest between which two players? • A. Players 1 and 2 • B. Players 2 and 3 • C. Players 3 and 4 • D. Players 1 and 5 C. Players 3 and 4 Explanation The difference between average scores is greatest between Players 3 and 4. This means that the average score of Player 3 is significantly different from the average score of Player 4. It implies that these two players have the largest variation in their scores compared to the other players. Rate this question: • 36. ### Shamika agrees to pick up relatives in Oakton and Scottsville. What is the distance from Harrisburg to Bloomington when driving through Oakton and Scottsville? • A. 450 miles • B. 550 miles • C. 725 miles • D. 775 miles D. 775 miles Explanation The question states that Shamika agrees to pick up relatives in Oakton and Scottsville. It then asks for the distance from Harrisburg to Bloomington when driving through Oakton and Scottsville. Since no other information is given, we can assume that the distance from Harrisburg to Bloomington includes the distance from Harrisburg to Oakton, the distance from Oakton to Scottsville, and the distance from Scottsville to Bloomington. Therefore, the correct answer is 775 miles. Rate this question: • 37. ### From Bloomington, Shamika and her relatives want to take a trip to Springfield. If she drives from Bloomington to Springfield, drops off her relatives in Oakton and Scottsville and returns to Harrisburg, how many miles will she travel going home? • A. 270 • B. 570 • C. 850 • D. 970 D. 970 Explanation Shamika will travel a total of 970 miles going home. She will drive from Bloomington to Springfield, which is one trip. Then she will drop off her relatives in Oakton and Scottsville, which adds to the total distance. Finally, she will return to Harrisburg, which adds more miles to the total. Rate this question: • 38. ### Jay's monthly payment for a new car would be about 1/2 of his repair expenses from January to April. Which of these could be the amount of Jay's monthly payment for a new car? • A. \$250.00 • B. \$200.00 • C. \$300.00 • D. \$350.00 A. \$250.00 Explanation Jay's repair expenses from January to April are twice the amount of his monthly payment for a new car. Therefore, if his monthly payment is \$250.00, his repair expenses from January to April would be \$500.00. This satisfies the condition given in the question that his repair expenses are 2 times his monthly payment. Rate this question: • 39. ### Tim's test scores were 85, 90, 93, and 88. What was his average score? • A. 80 • B. 86 • C. 89 • D. 94 C. 89 Explanation Tim's average score can be calculated by adding up all his test scores and dividing the sum by the total number of tests. In this case, his test scores are 85, 90, 93, and 88. Adding them up gives a total of 356. Dividing 356 by 4 (since there are 4 test scores) gives an average of 89. Therefore, Tim's average score is 89. Rate this question: • 40. ### A bag contains 5 red marbles, 5 green marbles, and 10 yellow marbles. If you draw a marble at random, what is the probability that it will be red? • A. 5% • B. 10% • C. 25% • D. 50% C. 25% Explanation The probability of drawing a red marble can be calculated by dividing the number of red marbles (5) by the total number of marbles in the bag (20). Therefore, the probability is 5/20, which simplifies to 1/4 or 25%. Rate this question: • 41. ### What is the mean of numbers 20, 25, and 30? • A. 22 • B. 25 • C. 29 • D. 50 B. 25 Explanation The mean of a set of numbers is calculated by adding up all the numbers and then dividing the sum by the total number of values. In this case, the sum of 20, 25, and 30 is 75. Dividing 75 by 3 (since there are 3 numbers) gives us 25, which is the mean of the given numbers. Rate this question: • 42. ### A bus company records how many people ride a particualar bus each day. The number of riders was 22, 20, 30, 14, and 24. What was the mean numbers of riders? • A. 18 • B. 22 • C. 28 • D. 30 B. 22 Explanation The mean is calculated by adding up all the numbers and then dividing the sum by the total number of values. In this case, we add up 22, 20, 30, 14, and 24, which gives us a sum of 110. Since there are 5 values, we divide the sum by 5 to get the mean, which is 22. Rate this question: • 43. ### Which two figures are missing from this pattern? • A. Circle, triangle • B. Triangle, circle • C. Triangle, square • D. Circle, sqaure B. Triangle, circle Explanation The pattern in the given figures is that each figure alternates between a circle and a triangle. The first figure is a circle, followed by a triangle, then a circle again. Therefore, the missing figures should continue this pattern. The correct answer is triangle, circle. Rate this question: • 44. ### What number is missing from this number pattern?5, 1, 10, 2, 20, 4, __ , 8, 80 • A. 14 • B. 30 • C. 40 • D. 50 C. 40 Explanation The number pattern follows a sequence where each number is obtained by multiplying the previous number by 2. Starting with 5, the next number is obtained by multiplying 5 by 2, which is 10. Following the same pattern, the next number is obtained by multiplying 10 by 2, which is 20. Continuing this pattern, the missing number can be obtained by multiplying 20 by 2, resulting in 40. Rate this question: • 45. ### Adriana brought 3 bags of dog food for each of her 3 dogs. Each bag will provide 8 bowls of food. How many bowls of food does Adriana have for her dogs? • A. 24 • B. 14 • C. 48 • D. 72 D. 72 Explanation Adriana has 3 bags of dog food, and each bag provides 8 bowls of food. To find the total number of bowls of food, we multiply the number of bags (3) by the number of bowls each bag provides (8). Therefore, Adriana has a total of 3 * 8 = 24 bowls of food for her dogs. Rate this question: • 46. ### Solve the equation for y: y × 8 = 48 • A. Y = 6 • B. Y = 1 • C. Y = 1 • D. Y = 3 A. Y = 6 Explanation To solve the equation y × 8 = 48, we need to isolate y. We can do this by dividing both sides of the equation by 8. This gives us y = 48/8, which simplifies to y = 6. Therefore, the correct answer is y = 6. Rate this question: • 47. ### Your checking account balance was \$1,614. After you wrote checks for \$515, \$85, and \$112, you calculated a balance of \$902. Using estimation, find which of the following best describes your result. • A. Correct result • B. Result too high • C. Result too low A. Correct result Explanation The correct result means that the calculated balance of \$902 matches the actual balance after deducting the amounts written on the checks. This suggests that the estimation was accurate and there were no errors in the calculations. Rate this question: • 48. ### A traffic engineer counts the vehicles passing an intersection during a four hour period. The hourly counts were 195, 98, 206, abd 480 vehicles. He reported a total count of 1,079 vechicles. Use estimation to evaluate his result. • A. Correct result • B. Result too high • C. Result too low C. Result too low Explanation The traffic engineer reported a total count of 1,079 vehicles, but the sum of the hourly counts is only 979 vehicles. Therefore, the reported result is too low. Rate this question: • 49. ### An elephant in a zoo eats about 200 pounds of food every day.About 1/5 of that food is fresh vegetables. Keepers need to decide how many pounds of fresh vegetables they need to order each day to feed all of their elephants. Which of the following describes the information you have to solve this problem? • A. Extra information • B. Missing information • C. Both missing and extra information • D. The needed information B. Missing information Explanation The question provides information about the daily food consumption of an elephant in a zoo, but it does not mention the total number of elephants in the zoo. In order to calculate the number of pounds of fresh vegetables needed to feed all the elephants, we need to know the total number of elephants in the zoo. Therefore, the missing information is the total number of elephants in the zoo. Rate this question: • 50. ### How many even numbers great than zero are there that are less than 31 and divisible by 5? • A. 1 • B. 2 • C. 3 • D. 4 C. 3 Explanation There are three even numbers greater than zero, less than 31, and divisible by 5. These numbers are 10, 20, and 30. Rate this question: Related Topics
Courses # Introduction to Algebra - Objective Type Questions Class 6 Notes \| EduRev ## Class 6 : Introduction to Algebra - Objective Type Questions Class 6 Notes \| EduRev ``` Page 1 Objective Type Questions PAGE: 8.13 Mark the correct alternative in each of the following: 1. 5 more than twice a number x is written as (a) 5 + x + 2 (b) 2x + 5 (c) 2x - 5 (d) 5x + 2 Solution: The option (b) is correct answer. 5 more than twice a number x is written as 2x + 5. 2. The quotient of x by 2 is added to 5 is written as (a) x/2 + 5 (b) 2/x+5 (c) (x+2)/ 5 (d) x/ (2+5) Solution: The option (a) is correct answer. The quotient of x by 2 is added to 5 is written as x/2 + 5. 3. The quotient of x by 3 is multiplied by y is written as (a) x/3y (b) 3x/y (c) 3y/x (d) xy/3 Solution: The option (d) is correct answer. It can be written as x/3 × y = xy/3 4. 9 taken away from the sum of x and y is (a) x + y - 9 (b) 9 - (x+y) (c) x+y/ 9 (d) 9/ x+y Solution: The option (a) is correct answer. 9 taken away from the sum of x and y is x + y – 9. 5. The quotient of x by y added to the product of x and y is written as (a) x/y + xy (b) y/x + xy (c) xy+x/ y (d) xy+y/ x Page 2 Objective Type Questions PAGE: 8.13 Mark the correct alternative in each of the following: 1. 5 more than twice a number x is written as (a) 5 + x + 2 (b) 2x + 5 (c) 2x - 5 (d) 5x + 2 Solution: The option (b) is correct answer. 5 more than twice a number x is written as 2x + 5. 2. The quotient of x by 2 is added to 5 is written as (a) x/2 + 5 (b) 2/x+5 (c) (x+2)/ 5 (d) x/ (2+5) Solution: The option (a) is correct answer. The quotient of x by 2 is added to 5 is written as x/2 + 5. 3. The quotient of x by 3 is multiplied by y is written as (a) x/3y (b) 3x/y (c) 3y/x (d) xy/3 Solution: The option (d) is correct answer. It can be written as x/3 × y = xy/3 4. 9 taken away from the sum of x and y is (a) x + y - 9 (b) 9 - (x+y) (c) x+y/ 9 (d) 9/ x+y Solution: The option (a) is correct answer. 9 taken away from the sum of x and y is x + y – 9. 5. The quotient of x by y added to the product of x and y is written as (a) x/y + xy (b) y/x + xy (c) xy+x/ y (d) xy+y/ x Solution: The option (a) is correct answer. The quotient of x by y added to the product of x and y is written as x/y + xy. 6. a 2 b 3 × 2ab 2 is equal to (a) 2a 3 b 4 (b) 2a 3 b 5 (c) 2ab (d) a 3 b 5 Solution: The option (b) is correct answer. It can be written as a 2 b 3 × 2ab 2 = 2a 2 × a × b 3 × b 2 = 2a 3 b 5 . 7. 4a 2 b 3 × 3ab 2 × 5a 3 b is equal to (a) 60a 3 b 5 (b) 60a 6 b 5 (c) 60a 6 b 6 (d) a 6 b 6 Solution: The option (c) is correct answer. It can be written as 4a 2 b 3 × 3ab 2 × 5a 3 b = 4 × 3 × 5 × a 2 × a × a 3 × b 3 × b 2 × b = 60a 6 b 6 8. If 2x 2 y and 3xy 2 denote the length and breadth of a rectangle, then its area is (a) 6xy (b) 6x 2 y 2 (c) 6x 3 y 3 (d) x 3 y 3 Solution: The option (c) is correct answer. We know that area of a rectangle = length × breadth By substituting the values Area = 2x 2 y × 3xy 2 = 6x 3 y 3 9. In a room there are x 2 rows of chairs and each two contains 2x 2 chairs. The total number of chairs in the room is (a) 2x 3 (b) 2x 4 (c) x 4 (d) x 4 /2 Solution: The option (b) is correct answer. We know that Total number of chairs in the room = Number of rows × Number of chairs By substituting the values Page 3 Objective Type Questions PAGE: 8.13 Mark the correct alternative in each of the following: 1. 5 more than twice a number x is written as (a) 5 + x + 2 (b) 2x + 5 (c) 2x - 5 (d) 5x + 2 Solution: The option (b) is correct answer. 5 more than twice a number x is written as 2x + 5. 2. The quotient of x by 2 is added to 5 is written as (a) x/2 + 5 (b) 2/x+5 (c) (x+2)/ 5 (d) x/ (2+5) Solution: The option (a) is correct answer. The quotient of x by 2 is added to 5 is written as x/2 + 5. 3. The quotient of x by 3 is multiplied by y is written as (a) x/3y (b) 3x/y (c) 3y/x (d) xy/3 Solution: The option (d) is correct answer. It can be written as x/3 × y = xy/3 4. 9 taken away from the sum of x and y is (a) x + y - 9 (b) 9 - (x+y) (c) x+y/ 9 (d) 9/ x+y Solution: The option (a) is correct answer. 9 taken away from the sum of x and y is x + y – 9. 5. The quotient of x by y added to the product of x and y is written as (a) x/y + xy (b) y/x + xy (c) xy+x/ y (d) xy+y/ x Solution: The option (a) is correct answer. The quotient of x by y added to the product of x and y is written as x/y + xy. 6. a 2 b 3 × 2ab 2 is equal to (a) 2a 3 b 4 (b) 2a 3 b 5 (c) 2ab (d) a 3 b 5 Solution: The option (b) is correct answer. It can be written as a 2 b 3 × 2ab 2 = 2a 2 × a × b 3 × b 2 = 2a 3 b 5 . 7. 4a 2 b 3 × 3ab 2 × 5a 3 b is equal to (a) 60a 3 b 5 (b) 60a 6 b 5 (c) 60a 6 b 6 (d) a 6 b 6 Solution: The option (c) is correct answer. It can be written as 4a 2 b 3 × 3ab 2 × 5a 3 b = 4 × 3 × 5 × a 2 × a × a 3 × b 3 × b 2 × b = 60a 6 b 6 8. If 2x 2 y and 3xy 2 denote the length and breadth of a rectangle, then its area is (a) 6xy (b) 6x 2 y 2 (c) 6x 3 y 3 (d) x 3 y 3 Solution: The option (c) is correct answer. We know that area of a rectangle = length × breadth By substituting the values Area = 2x 2 y × 3xy 2 = 6x 3 y 3 9. In a room there are x 2 rows of chairs and each two contains 2x 2 chairs. The total number of chairs in the room is (a) 2x 3 (b) 2x 4 (c) x 4 (d) x 4 /2 Solution: The option (b) is correct answer. We know that Total number of chairs in the room = Number of rows × Number of chairs By substituting the values Total number of chairs in the room = x 2 × 2x 2 = 2x 4 10. a 3 × 2a 2 b × 3ab 5 is equal to (a) a 6 b 6 (b) 23a 6 b 6 (c) 6a 6 b 6 (d) None of these Solution: The option (c) is correct answer. It can be written as a 3 × 2a 2 b × 3ab 5 = 2 × 3 × a 3 × a 2 × a × b × b 5 = 6a 6 b 6 ``` Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity! ## Mathematics (Maths) Class 6 191 videos\|224 docs\|43 tests , , , , , , , , , , , , , , , , , , , , , ;
# RBSE Solutions for Class 5 Maths Chapter 1 Numbers Additional Questions RBSE Solutions for Class 5 Maths Chapter 1 Numbers Additional Questions is part of RBSE Solutions for Class 5 Maths. Here we have given Rajasthan Board RBSE Class 5 Maths Chapter 1 Numbers Additional Questions. Board RBSE Textbook SIERT, Rajasthan Class Class 5 Subject Maths Chapter Chapter 1 Chapter Name Numbers Exercise Additional Questions Number of Questions 26 Category RBSE Solutions ## Rajasthan Board RBSE Class 5 Maths Chapter 1 Numbers Additional Questions Multiple Choice Questions Question 1. Number of digit in 5 -digit Number (a) 4 (b) 5 (c) 6 (d) 3 Question 2. Symbol falls between the numbers 12199 ………… 21200 (a) < (b) > (c) < (d) = Question 3. Greatest number in the(RBSESolutions.com)numbers 89599 and 98599 (a) 10000 (b) 99999 (c) 98599 (d) 89599 Question 4. Place value of 0(zero) in 87605- (a) 10 (b) 0 zero (c) 100 (d) 1000 Question 5. Greatest five digit number is- (a) 11111 (b) 22222 . (c) 77777 (d) 99999 Question 6. (a) Thirty three thousand (b) Three hundred thirty three (c) Thirty three(RBSESolutions.com)thousand three hundred thirty three (d) Three thousand three hundred thirty three Question 7. Place value of 4 in 83427- (a) 400 (b) 4000 (c) 40 (d) 4 Question 8. Smallest 5-digit number is- (a) 10000 (b) 11111 (c) 55555 (d) 99999 1. (b) 2. (a) 3. (c) 4. (b) 5. (d) 6. (c) 7. (a) 8. (a) Fill in the blanks in the following Questions 1. 15005 ……. 15050 2. 74349 = …… + ……. + …… + …… + 3. Place value of zero at any place is always 4. Number with less(RBSESolutions.com)number of digits is smaller then the number with ………….. number of digits. 5. If the numbers written as smallest to largest known as 6. Number with greater number of digits is …………. then number with less number of digit. 1. < 2. 70000 + 4000 + 300 + 40 + 9 3. Zero (0) 4. Greater 5. Ascending order 6. Greater Question 1. What is descending order of numbers ? If the numbers are(RBSESolutions.com)written as largest to smallest order known as descending order. Question 2. Write the greatest and smallest 5-digit number formed by digits 3, 2, 6, 9 and 4. Greatest 5-digit number formed by digits 3, 2, 6, 9 and 4 = 96432 Smallest 5-digit number formed by digits 3,2, 6, 9 and 4 = 23469 Question 3. Write the expanded form of following numbers. (i) 74349 (ii) 56804 (i) 74349 = 70000 + 4000 + 300 + 40 + 9 (ii) 56804 = 50000 + 6000 + 800 + 00 + 4 Question 4. Compare the following numbers using symbols < , > and = in the box. Question 5. Arrange the following numbers in greatest to smallest- 56392, 45005, 67645, 39090, 74567 Greatest to smallest(RBSESolutions.com)order of the following numbers is- 74567, 67645, 56392, 45005, 39090 True/False Question 1. Identify True/False statements and write true/false in the box given in front of them. (1) Forty eight thousand five hundred two is written in numbers as 48502. (2) The number of(RBSESolutions.com)expanded form 10000 + 800 + 70 + 3 is 1873. (3) We Read the number 40009 in words as four thousand nine. (4) The place value of 2 in number 73208 is 200. (1) True (2) False (3) False (4) True Question 2. Write True/False for the following statements in the given box. (i) Largest number formed by using digits 5,0, 6, 2, 1 is 65201. (ii) Place value of(RBSESolutions.com)the digit 2 in the number 12756 is 2000. (iii) 5320 > 5230 (iv) In the number 23684 the face value at thousand’s place is 2. (1) False (2) True (3) True (4) False Question 3. W rite True/False in front of the given box by examine True/False. 1. 3578 < 3758 2. 57325 > 57533 3. 81111 < 79999 4. 80035 < 83500 (1) True (2) False (3) False (4) True. Question .4 Examine the following statement whether they are true or false and write True/False in the box. 1. Eighty five thousand two(RBSESolutions.com)hundred fourty written as 85240 in numerals. 2. Expanded form of number 2322 is 20000 + 3000 + 200 + 20 + 2. 3. Number 20456 can be read as two thousand four hundred fifty six. 4. 40586 < 40568 (1) True (2) False (3) False (4) False Question 5. Examine the following statement whether they are true or false and write True/False in the box. 1. 2979 > 2932 2. 5423 > 5432 3. 8952 = 8952 4. 9821 < 9799 (1) True (2) False (3) True (4) False Question 1. Write the smallest and biggest five digit numbers using the digits 8, 5,2, 0,1. (1) The smallest(RBSESolutions.com)five digit number (2) The biggest five digit number Solution. (1) 10,258 (2) 85,210 Question 2. Write the place value of each digit of the following numbers (i) 57089 (ii) 99999 (iii) 45602 (iv) 78476 Solution. (i) In 57089 . Place value of 5 = 50000 Place value of 7 = 7000 Place value of 0 = 000 Place value of 8 = 80 Place value of 9 = 9 (ii) In 99999 Place value of 9 = 90000 Place value of 9 = 9000 Place value of 9 = 900 Place value of 9 = 90 Place value of 9 = 9 (iii) In 45602 Place value of 4 = 40000 Place value of 5 = 5000 Place value of 6 = 600 Place value of 0 = 00 Place value of 2 = 2 (iv) In 78476 Place value of 7 = 70000 Place value of 8 = 8000 Place value of 4 = 400 Place value of 7 = 70 Place value of 6 = 6 Question 3. Let us se the 5-digit numbers formed by digits 9, 5, 2, 3 and 6. Make some 5-digit numbers and write them in the following table in words. Question 4. (a) Write the following numbers in Ascending order 45378, 43758, 47538, 48735 43758, 45378, 47538, 48735 (b) Write the following numbers in descending order (i) 77890, 74077, 78999, 79009 (ii) 55499, 56109, 50000, 52888 (i) 79009, 78999, 77890, 74077 (ii) 56109, 55499, 52888, 50000 Question 5. Write the following(RBSESolutions.com)expanded form in numbers- (i) 60000 + 5000 + 900 + 80 + 9 (ii) 70000 + 8000 + 000 + 40 + 5 (iii) 90000 + 1000 + 900 + 10 + 9 (iv) 50000 + 0000 + 700 + 30 +1 (i) 65989 (ii) 78045 (iii) 91919 (iv) 50731 Question 6. Write the following in numerals- (a) Fifty three thousand five hundred seven (b) Fifty seven thousand one hundred sixty (c) Forty four thousand nine (d) Sixty six thousand nine hundred sixty nine (e) Seventy seven thousand twenty (a) 53507 (b) 57160 (c) 44009 (d) 66969 (e) 77020 Question 7. Put the appropriate(RBSESolutions.com)symbol (less than, greater than and equal to) between the following numbers. (i) 56089 and 57009 (ii) 64872 and 64999 (iii) 87660 and 87660 (iv) 89009and 88999
# Difference between revisions of "2016 AMC 10B Problems/Problem 14" ## Problem How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$, the line $y=-0.1$ and the line $x=5.1?$ $\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57$ ## Solution 1 The region is a right triangle which contains the following lattice points: $(0,0); (1,0)-(1,3); (2,0)-(2,6); (3,0)-(3,9); (4,0)-(4,12); (5,0)-(5,15)$ $[asy]size(10cm); for(int i=0;i<6;++i)for(int j=0;j<=3i;++j)dot((i,j)); draw((0,-1)--(0,16),EndArrow);draw((-1,0)--(6,0),EndArrow); draw((-.5,-pi/2)--(5.2,5.2pi),gray);draw((-1,-.1)--(6,-.1)^^(5.1,-1)--(5.1,16),gray); [/asy]$ Squares $1\times 1$: Suppose that the top-right corner is $(x,y)$, with $2\le x\le 5$. Then to include all other corners, we need $1\le y\le 3(x-1)$. This produces $3+6+9+12=30$ squares. Squares $2\times 2$: Here $3\le x\le 5$. To include all other corners, we need $2\le y\le 3(x-2)$. This produces $2+5+8=15$ squares. Squares $3\times 3$: Similarly this produces $5$ squares. No other squares will fit in the region. Therefore the answer is $\boxed{\textbf{(D) }50}$. ## Solution 2 The vertical line is just to the right of $x = 5$, the horizontal line is just under $y = 0$, and the sloped line will always be above the $y$ value of $3x$. This means they will always miss being on a coordinate with integer coordinates so you just have to count the number of squares to the left, above, and under these lines. After counting the number of $1\cdot1$, $2\cdot2$, and $3\cdot3$ squares and getting $30$, $15$, and $5$ respectively, and we end up with $\boxed{\textbf{(D)}\ 50}$. Solution by Wwang ## Solution 3 The endpoint lattice points are $(1,3), (2,6), (3,9), (4,12), (5,15).$ Now we split this problem into cases. Case $1:$ Square has length $1.$ The $x$ coordinates must be $(1,2)$ or $(2,3)$ and so on to $(4,5).$ The idea is that you start at $y=1$ and add at the endpoint, namely $y=3.$ The number ends up being $3+6+9+12 = 30$ squares for this case. Case $2:$ Square has length $2.$ The $x$ coordinates must be $(1,3)$ or $(2,4)$ or $(3,15)$ and so now it starts at $y=2.$ It ends up being $2+5+8 = 15.$ Case $3:$ Square has length $3.$ The $x$ coordinates must be $(1,4)$ or $(2,5)$ so there is $1+4 = 5$ squares for this case. Answer is $30+15+5 = \boxed{\textbf{(D)}\ 50}$. ## See Also 2016 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username Login to AoPS
# Linear Equations Solutions, Shortcut Tricks and Video Tutorial An equation is a statement of equality which involves one or more unknown quantities, called the variables. An equation involving only linear polynomials is called a Linear Equations. The following are some examples of linear equations. • 2x + 3 = 1 • $\frac{3}{2}$x + 5 = 2x + 7 • $\frac{5}{3}$y – 16 = 3y + 4 ## Solving a linear equation Solving a linear means finding a value of the variable which satisfies the equation. While solving a linear equation, following properties of equality are to be remembered. • Same quantity can be added to both sides of an equality. • Same quantity can be subtracted from both sides of an equality. • Both sides of an equality can be multiplied by the same quantity (≠0). • Both sides of an equality can be divided by the same Non-zero quantity. • While transposing a term from one side to the other side of equality, the sign of the term must be changed. • The terms involving the variable are kept on one side (LHS) of equality and terms not involving the variable are on other side. ## Applications of linear equations In Algebra, linear equations are used to solve some practical problems. Following are the basic steps needed to solve such problems. Step 1: Denote the unknown quantity by x. If there are more than one unknown quantities, then denote any one of these by ‘x’ and write the other in terms of ‘x’. Step 2: From the information given in the problem, formulate a linear equation in x. Step 3: Solve the linear equation to find ‘x’ (unknown quantity). Step 4: Put the value of x is in the linear equation to find out the unknown quantities if there are more than one in number. ## Systems of Linear Equations • linear equation in one variable : A linear equation in one variable (x) of the form either ax+b>0 or ax+b$\geq$ 0 or ax + b note : step-1 collect all term containing the variable on the left side and the constants on the right side • Divide this equation by the coefficient of the variable .This gives the solution of the given equation . Example: solve the linear equation 2x + 7b $\geq$ 13 - x Solution: 2x + 7b $\geq$ 13 – x $\Rightarrow (2x+7)+x\: \geq (13-x)+x\Rightarrow 3x+7\geq 13$ $(3x+7)-7\geq 13-7\: 3x\geq 6\: \Rightarrow x\geq 2$ $\therefore x\in [2,\infty ]$ this solution can also be on the number line . 1. Two or more linear equation in one variable constitutes a system of llinear equation in one variable: Step 1: solve each equation and write their solution sets in terms of interval s Step 2: find the intersection of all the solution sets of inequation given in the system. Step 3: the intersection represent the solution of the given system of linear equations. You 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 (that is < become >,$\leq$becomes $\geq$ 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. 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 thesign of inequality is reversed. Now, let us consider some examples. Example 1: Solve 30 x < 200 when Step 1. x is a natural number, 2.x is an integer. Solution: we have given 30x < 200 or$\frac{30x}{30}< \frac{200}{30}$(Rule 2), i.e., x < 20 / 3. Step 2.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} Step 3.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 ∈ (– ∞, 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. Example 3: 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 ## Graphical Solution of Linear Inequalities in Two Variables In earlier section, we have seen that a graph of an inequality in one variable is a visual representation and is a convenient way to represent the solutions of the inequality. Now, we will discuss graph of a linear inequality in two variables. We know that a line divides the Cartesian plane into two parts. Each part is known as a half plane. A vertical line will divide the plane in left and right half planes and a non-vertical line will divide the plane into lower and upper half planes. A point in the Cartesian plane will either lie on a line or will lie in either of the half planes I or II. We shall now examine the relationship, if any, of the points in the plane and the inequalities ax + by < c or ax + by > c. Let us consider the line ax + by = c, a ≠ 0, b ≠ 0 There are three possibilities namely: (i) ax + by = c (ii) ax + by > c, (iii) ax + by < c. In Case (i), clearly, all points (x, y) satisfying (i) lie on the line it represents and conversely. Consider Case (ii), let us first assume that b > 0. Consider a point P (α,β) on the line ax + by = c, b > 0, so that aα + bβ = c. Take an arbitrary point Q (α , γ) in the half plane II Now, we interpret, γ > β (Why?) or bγ> bβ or aα + b γ > aα + bβ (Why?) or aα + b γ > c i.e., Q(α, γ ) satisfies the inequality ax + by > c. Thus, all the points lying in the half plane II above the line ax + by = c satisfies the inequality ax + by > c. Conversely, let (α, β) be a point on line ax + by = c and an arbitrary point Q(α, γ ) satisfying ax + by > c so that aα + bγ> c ⇒ aα + b γ > aα + bβ (Why?) ⇒ γ > β (as b > 0) This means that the point (α, γ ) lies in the half plane II. Thus, any point in the half plane II satisfies ax + by > c, and conversely any point satisfying the inequality ax + by > c lies in half plane II. In case b < 0, we can similarly prove that any point satisfying ax + by > c lies in the half plane I, and conversely. Hence, we deduce that all points satisfying ax + by > c lies in one of the half planes II or I according as b > 0 or b < 0, and conversely. Thus, graph of the inequality ax + by > c will be one of the half plane (called solution region) and represented by shading in the corresponding half plane. Note: • The region containing all the solutions of an inequality is called the solution region. • In order to identify the half plane represented by an inequality, it is just sufficient to take any point (a, b) (not online) and check whether it satisfies the inequality or not. If it satisfies, then the inequality represents the half plane and shade the region. which contains the point, otherwise, the inequality represents that half plane which does not contain the point within it. For convenience, the point (0, 0) is preferred. • If an inequality is of the type ax + by ≥ c or ax + by ≤ c, then the points on the line ax + by = c are also included in the solution region. So draw a dark line in the solution region. • If an inequality is of the form ax + by > c or ax + by < c, then the points on the line ax + by = c are not to be included in the solution region. So draw a broken or dotted line in the solution region Example: We obtained the following linear inequalities in two variables x and y: 40x + 20y ≤ 120 Solution: Let us now solve this inequality keeping in mind that x and y can be only whole numbers, since the number of articles cannot be a fraction or a negative number. In this case, we find the pairs of values of x and y, which make the statement (1) true. In fact, the set of such pairs will be the solution set of the inequality (1). To start with, let x = 0. Then L.H.S. of (1) is 40x + 20y = 40 (0) + 20y = 20y. Thus, we have 20y ≤ 120 or y ≤ 6 ... (2) For x = 0, the corresponding values of y can be 0, 1, 2, 3, 4, 5, 6 only. In this case, the solutions of (1) are (0, 0), (0, 1), (0,2), (0,3), (0,4), (0, 5) and (0, 6). Similarly, other solutions of (1), when x = 1, 2 and 3 are: (1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (2, 0), (2, 1), (2, 2), (3, 0) This is shown in Fig 6.6 Let us now extend the domain of x and y from whole numbers to real numbers, and see what will be the solutions of (1) In this case. You will see that the graphical method of solution will be very convenient in this case. For this purpose, let us consider the (corresponding) equation and draw its graph. 40x + 20y = 120 ... (3) In order to draw the graph of the inequality (1) We take one point say (0, 0), in half plane I and check whether values of x and y satisfy the inequality or not We observe that x = 0, y = 0 satisfy the inequality. Thus, we say that the half plane I is the graph (Fig 6.7) of the inequality. Since the points on the line also satisfy the inequality (1) above, the line is also a part of the graph. Thus, the graph of the given inequality is half plane I including the line itself. Clearly half plane II is not the part of the graph. Hence, solutions of inequality (1) will consist of all the points of its graph (half plane I including the line). We shall now consider some examples to explain the above procedure for solving a linear inequality involving two variables. ## Point to be Remember • Two real numbers or two algebraic expressions related by the symbols <, >, ≤or ≥ form an equality. • Equal numbers may be added to (or subtracted from ) both sides of an inequality. 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 inequality is reversed. • The values of x, which make an inequality a true statement, are called solutions of the inequality. • To represent x < a (or x > a) on a number line, put a circle on the number a and dark line to the left (or right) of the number a. • To represent x ≤ a (or x ≥ a) on a number line, put a dark circle on the numbers and dark the line to the left (or right) of the number x. • If an inequality is having ≤ or ≥ symbol, then the points on the line are also included in the solutions of the inequality and the graph of the inequality lies left(below) or right (above) of the graph of the equality represented by dark line that satisfies an arbitrary point in that part. • If an inequality is having < or > symbol, then the points on the line are not included in the solutions of the inequality and the graph of the inequality lies to the left (below) or right (above) of the graph of the corresponding equality represented by dotted line that satisfies an arbitrary point in that part. • The solution region of a system of inequalities is the region which satisfies all the given inequalities in the system simultaneously. Linear equation ## Linear Equations Shortcut Tricks and Study Material - Video Please comment on Linear Equations Solutions, Shortcut Tricks and Video Tutorial
Education.com Try Brainzy Try Plus # Tip #6 to Get a Top ACT Math Score By McGraw-Hill Professional Updated on Sep 7, 2011 Most ACTs have one question involving parallel lines. In math class, learning about parallel lines might have seemed pretty tricky—alternate interior angles, corresponding angles, same-side interior angles. … We don't need all that vocab for the ACT. We just need to know: • Parallel lines are two lines that never touch. • If two parallel lines are crossed by another line (called a transversal), then eight angles form. • These eight angles are of two types, big or little. All bigs are equal, and all littles are equal. This is enough to answer any parallel-lines ACT question. There's another point! Let's look at this question: Solution: This question may look tough at first, but it uses Skill 6 exactly, and we're ready for it. In the pair of parallel lines crossed by segment NM, several angles are formed, and ONM is a smaller-looking angle and is congruent (congruent is just a fancy geometry word for "equal") to all other smaller-looking angles of the angles formed. XMN is another smaller-looking angle, so ONM = XMN. YNM and ZMN are bigger-looking angles and are equal to each other, but not to ONM. ### Easy 1. In the figure below, m \|\| n. If y = 37.5°, what is the value of x ? 1. 42.5 2. 100 3. 137.5 4. 142.5 5. 180 2. If x = 73° and p \|\| q and m \|\| n in the four lines shown, what is the value of z? 1. 17 2. 73 3. 90 1. 97 2. 107 ### Medium 3. In the figure below, m \|\| n. If z = 42.5 and x = 125, then y =? 1. 97.5 2. 87.5 3. 57.5 4. 47.5 5. 27.5 4. If z = 55° and y = 75° in the picture below, then which of the following must be true? 1. p \|\| q 2. x + y = 180° 3. z = x 1. x + z = 180° 2. y = 2x ### Hard 5. In the figure below, V, W, X, and Y are collinear; Z, W, and T are collinear; and the angles at V, X, and T are right angles, as marked. Which of the following statements is NOT justifiable from the given information? 1. is perpendicular to 2. VZW is congruent to XTW. 3. is congruent to 4. ΔZVW is similar to ΔTXW. 5. is parallel to 1. D In a pair of parallel lines there are only two kinds of angles, big and little, and the two add up to 180. Here y is a little angle and x is clearly a bigger angle, so they add up to 180, and x = 180 – 37.5 = 142.5. 2. K Parallel lines cut by a transversal make 2 kinds of angles, big and little. Clearly x is little and z is big. If x = 73, then z = 180 – 73 = 107. 3. A First, mark the info that is given in the question into the diagram. That always helps to make the question simpler and usually shows which strategies to use. Then use vertical angles, linear pairs, and triangles to calculate the measures of any other angles that you can. The vertical angle to z is 42.5, and the linear pair to x is 55, so now we have two angles of the triangle. 180 – 42.5 – 55 = 82.5. And the linear pair of 82.5, which is y, is 180 – 82.5 = 97.5 = y. 4. G Mark the info from the question into the diagram. Then mark all other angles that you can determine. Here x and y are a linear pair, so x + y = 180 and x = 180 – 75 = 105. Since z, x, and y are three different numbers, they cannot be the angles of two parallel lines, which are always at most two different numbers. Also, since x and z are both less than 90, lines p and q cannot be parallel—they are heading toward each other and will intersect. Only G is the true choice. 5. C Only choice C is NOT justifiable: 1. is perpendicular to —justifiable as marked 2. VZW is congruent to —justifiable, all small angles are equal in the pair of parallel lines 3. is congruent to —NOT justifiable, we have no proof 4. ΔZVW is similar to ΔTXW—(Skill 26 Preview) yes, the angles of ΔZVW are equal to the corresponding angles of ΔTXW, and therefore the triangles are similar 5. is parallel to —justifiable, both are perpendicular to and therefore parallel to each other Go to: Tip #7
# How do you solve 1/6-x/2=(x-5)/3? Jan 27, 2017 see the entire solution process below: #### Explanation: First, multiple each side of the equation by $\textcolor{red}{6}$ (the lowest common denominator of all the fractions) to eliminate the fractions while keeping the equation balanced: $\textcolor{red}{6} \left(\frac{1}{6} - \frac{x}{2}\right) = \textcolor{red}{6} \times \frac{x - 5}{3}$ $\left(\textcolor{red}{6} \times \frac{1}{6}\right) - \left(\textcolor{red}{6} \times \frac{x}{2}\right) = \cancel{\textcolor{red}{6}} 2 \times \frac{x - 5}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}$ $\left(\cancel{\textcolor{red}{6}} \times \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}}\right) - \left(\cancel{\textcolor{red}{6}} 3 \times \frac{x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}\right) = 2 \left(x - 5\right)$ $1 - 3 x = 2 x - 10$ Next, add $\textcolor{red}{3 x}$ and $\textcolor{b l u e}{10}$ to each side of the equation to isolate the $x$ term while keeping the equation balanced: $1 - 3 x + \textcolor{red}{3 x} + \textcolor{b l u e}{10} = 2 x - 10 + \textcolor{red}{3 x} + \textcolor{b l u e}{10}$ $1 + \textcolor{b l u e}{10} - 3 x + \textcolor{red}{3 x} = 2 x + \textcolor{red}{3 x} - 10 + \textcolor{b l u e}{10}$ $11 - 0 = 5 x - 0$ $11 = 5 x$ Now, divide each side of the equation by $\textcolor{red}{5}$ to solve for $x$ while keeping the equation balanced: $\frac{11}{\textcolor{red}{5}} = \frac{5 x}{\textcolor{red}{5}}$ $\frac{11}{5} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} x}{\cancel{\textcolor{red}{5}}}$ $\frac{11}{5} = x$ $x = \frac{11}{5}$ Jan 27, 2017 $x = \frac{11}{5}$ #### Explanation: Eliminate the fractions in the equation by multiplying ALL terms on both sides of the equation by the $\textcolor{b l u e}{\text{lowest common multiple}}$ ( LCM ) of the denominators 6, 2 and 3 The LCM of 6 , 2 and 3 is 6 Hence multiply All terms by 6 $\left({\cancel{6}}^{1} \times \frac{1}{\cancel{6}} ^ 1\right) - \left({\cancel{6}}^{3} \times \frac{x}{\cancel{2}} ^ 1\right) = \left({\cancel{6}}^{2} \times \frac{x - 5}{\cancel{3}} ^ 1\right)$ $\Rightarrow 1 - 3 x = 2 \left(x - 5\right) \leftarrow \text{ no fractions}$ distribute the bracket on the right side. $\Rightarrow 1 - 3 x = 2 x - 10$ subtract 2x from both sides. $1 - 3 x - 2 x = \cancel{2 x} \cancel{- 2 x} - 10$ $\Rightarrow 1 - 5 x = - 10$ subtract 1 from both sides. $\cancel{1} \cancel{- 1} - 5 x = - 10 - 1$ $\Rightarrow - 5 x = - 11$ To solve for x, divide both sides by - 5 $\frac{\cancel{- 5} x}{\cancel{- 5}} = \frac{- 11}{- 5}$ $\Rightarrow x = \frac{11}{5}$ $\textcolor{b l u e}{\text{As a check}}$ substitute this value into the equation and if the left side equals the right side then it is the solution. $\text{left side } = \frac{1}{6} - \frac{\frac{11}{5}}{2} = \frac{1}{6} - \frac{11}{10} = \frac{5}{30} - \frac{33}{30} = - \frac{28}{30} = - \frac{14}{15}$ $\text{right side} = \frac{\frac{11}{5} - 5}{3} = \frac{\frac{11}{5} - \frac{25}{5}}{3} = \frac{- \frac{14}{5}}{3} = - \frac{14}{15}$ $\Rightarrow x = \frac{11}{5} \text{ is the solution}$
We have learned the meaning of the first derivative of a function. Now we want to know what the second, the third, and the n-th derivatives of a function are defined and how we can calculate them. If $s(t)$ is the position of an object moving on a straight line, then the derivative of s is the velocity of the object $v(t)=s'(t)$. The derivative of the velocity is the acceleration of the object $a(t)$. So $a(t)=v'(t)$ or $a(t)=(s’)'(t)$, which is often written simply as $a(t)=s^{\prime\prime}(t)$.We say the acceleration is the second derivative of the position. In physics, acceleration plays an important role as it appears in Newton’s second law $F=ma$. In general, if we take the derivative of $y=f(x)$, we obtain a new function $f’$ (also denoted by $y’$ or $dy/dx$). We can take the derivative of $f’$ and obtain another function called the second derivative of $f$ (or $y$). The second derivative of $f(x)$ is denoted by $$f^{\prime\prime}(x)$$ or $$\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)$$, which is commonly abbreviated to $$\dfrac{d^{2}y}{dx^{2}}.$$ Thus $f^{\prime\prime}=(f’)^\prime$ or $\bbox[#F2F2F2,5px,border:2px solid black]{f^{\prime\prime}(x)=\lim_{\Delta x\to0}\frac{f'(x+\Delta x)-f'(x)}{\Delta x}.\tag{a}}$ The second derivative is also indicated by $$y^{\prime\prime}$$ or $$\dfrac{d^{2}f}{dx^{2}}$$. For example, if $$f(x)=x^{3}-5x^{2}+3x-1$$, then the first derivative is $y’=f'(x)=\frac{dy}{dx}=\frac{df}{dx}=3x^{2}-10x+3,$ and the second derivative of $$f$$ is the derivative of $$f'(x)$$: $y^{\prime\prime}=f^{\prime\prime}(x)=\frac{d^{2}y}{dx^{2}}=\frac{d^2f}{dx^2}=6x-10.$ In a similar fashion, we can define the third derivative as the derivative of the second derivative. It is denoted by $y^{\prime\prime\prime}=f^{\prime\prime\prime}(x)=\frac{d^{3}y}{dx^{3}}=\frac{d^{3}f}{dx^{3}};$ the fourth derivative is the derivative of the third derivative, and is denoted by $y^{(4)}=f^{(4)}(x)=\frac{d^{4}y}{dx^{4}}=\frac{d^{4}f}{dx^{4}},$ and so on. In general, the $$n$$-th derivative of $$y=f(x)$$ is indicated by one of the following symbols: $y^{(n)}=f^{(n)}(x)=\frac{d^{n}y}{dx^{n}}=\frac{d^{n}f}{dx^{n}}.$ • If $f^{(n)}(x_0)$ exists, then it is said that $f$ is $n$-times differentiable at $x_0$. Example 1 If $$y=\sin x$$, find $$y^{(4)}.$$ Solution \begin{align} y & =\sin x\\ y’&=\cos x\\ y^{\prime\prime}&=\frac{d}{dx}\cos x=-\sin x\\ y^{\prime\prime\prime} & =\frac{d}{dx}(-\sin x)=-\frac{d}{dx}\sin x=-\cos x\\ y^{(4)}&=\dfrac{d}{dx}(-\cos x)=-\dfrac{d}{dx}\cos x=-(-\sin x)=\sin x.\end{align} Therefore, $y^{(4)}=y=\sin x$. Example 2 For $$f(x)=-x^{5}+3x^{3}+2x^{2}-1$$, find $$f'(x),f^{\prime\prime}(x),f^{\prime\prime\prime}(x)$$, and $$f^{(4)}(x)$$. Solution \begin{align}f^{\prime}(x) &=-5x^{4}+9x^{2}+4x\\ f^{\prime\prime}(x) &=-20x^{3}+18x+4\\ f^{\prime\prime\prime}(x) &=-60x^{2}+18\\ f^{(4)}(x) &=-120x.\end{align} Example 3 Determine $$a,b$$, and $$c$$ such that $$f^{\prime\prime}(x)$$ exists everywhere if $f(x)=\begin{cases} x^{3} & \text{when }x\leq1\\ ax^{2}+bx+c & \text{when }x>1 \end{cases}.$ Solution Because the second derivatives of $y=x^3$ and $y=ax^2+bx+c$ exist everywhere, no matter what $a, b$, and $c$ are $f^{\prime\prime}(x)$ exists everywhere, except possibly at $x=1$ where the formula of $f$ changes. So if we want to make $f$ twice differentiable everywhere, we have to choose $a, b$, $c$ such that $f^{\prime\prime}(1)$ exists. We have three unknowns: $$a,b,c$$ and hence we need three equations. For $$f$$ to have a second derivative at $$x=1$$, we need (1) $$f$$ to be continuous at $$x=1$$, (2) $$f$$ to have a derivative at $$x=1$$ or $$f’_{-}(1)=f’_{+}(1)$$, and (3) $$f_{+}^{\prime\prime}(1)=f_{-}^{\prime\prime}(1)$$. Now (1) The continuity of $$f$$ at $$x=1$$ implies $f(1)=1^{3}=a\cdot1^{2}+b\cdot1+c.\tag{i}$ (2) $$f$$ has a derivative at $$x=1$$. Thus \begin{align} f’_{-}(1) & =f’_{+}(1)\\ \left.3x^{2}\right\|_{x=1} & =\left.2ax+b\right\|_{x=1}\\ 3 & =2a+b.\tag{ii}\end{align} (3) $$f$$ has a second derivative at $$x=1$$. Thus \begin{align} f^{\prime\prime}_{-}(1) & =f^{\prime\prime}_{+}(1)\\ \left.6x\right\|_{x=1} & =\left.2a\right\|_{x=1}\\ 6 & =2a\tag{iii }\end{align} From (i), (ii), and (iii), we conclude $a=3,\qquad b=-3,\quad\text{and}\quad c=1.$ • If a function is differentiable, its derivative is not necessarily differentiable. In other words, from the existence of $$f'(x_{0})$$, we cannot infer the existence of $$f^{\prime\prime}(x_{0})$$. For instance, see the following example. Example 4 Let $f(x)=\begin{cases} x^{2}\sin\frac{1}{x} & x\neq0\\ 0 & x=0 \end{cases}.$ Does $$f'(0)$$ exist? Is $$f'(x)$$ continuous at $$x=0$$? Solution To find $$f'(0)$$, we need to apply the definition of a derivative directly: \require{cancel}\begin{align} f'(0) & =\lim_{\Delta x\to0}\frac{f(0+\Delta x)-\overset{0}{\cancel{f(0)}}}{\Delta x}\\ & =\lim_{\Delta x\to0}\frac{(\Delta x)^{2}\sin\dfrac{1}{\Delta x}}{\Delta x}\\ & =\lim_{\Delta x\to0}\left(\Delta x\sin\dfrac{1}{\Delta x}\right)\\ & =0.\end{align} [Recall that $$\lim_{h\to0}h\sin\frac{1}{h}=0$$. See the Section on Theorems for Calculating Limits for more information.] When $$x\neq0$$, we can find $$f'(x)$$ by using differentiation rules: $f(x)=\underbrace{x^{2}}_{u}\underbrace{\sin\frac{1}{x}}_{v}$ $f'(x)=\underbrace{2x}_{u’}\underbrace{\sin\frac{1}{x}}_{v}+\underbrace{x^{2}}_{u}\underbrace{\frac{d}{dx}\sin\frac{1}{x}}_{v’}\tag{i}$ To find $$\frac{d}{dx}\sin\frac{1}{x}$$ let $$w=\frac{1}{x}$$ \begin{align} \frac{d}{dx}\sin\frac{1}{x} & =\frac{d}{dx}\sin w\\ & =\frac{d}{dw}(\sin w)\times\frac{dw}{dx}\\ & =(\cos w)\left(-\frac{1}{x^{2}}\right)\\ & =\left(\cos\frac{1}{x}\right)\left(-\frac{1}{x^{2}}\right)\end{align} [$$\frac{d}{dx}\frac{1}{x}=\frac{d}{dx}x^{-1}=-x^{-2}=-\frac{1}{x^{2}}$$] Now we can simply plug the formula for $$\frac{d}{dx}\sin\frac{1}{x}$$ in (i) \begin{align} f'(x) & =2x\sin\frac{1}{x}+x^{2}\left(-\frac{1}{x^{2}}\right)\left(\cos\frac{1}{x}\right)\\ & =2x\sin\frac{1}{x}-\cos\frac{1}{x}\quad(x\neq0)\end{align} Therefore, $f'(x)=\begin{cases} 2x\sin\frac{1}{x}-\cos\frac{1}{x} & \text{if }x\neq0\\ 0 & \text{if }x=0 \end{cases}.$ Because $$\cos(1/x)$$ moves up and down so quickly as $$x\to0$$, it does not approach a number, and $$\lim_{x\to0}\cos(1/x)$$ does not exists. Thus \begin{align} \lim_{x\to0}f'(x) & =\lim_{x\to0}\left(2x\sin\frac{1}{x}-\cos\frac{1}{x}\right)\\ & =2\lim_{x\to0}x\sin\frac{1}{x}-\lim_{x\to0}\cos\frac{1}{x}\\ & =2(0)-DNE\end{align} does not exists, and consequently $$f’$$ is not continuous at $$x=0$$. • In the above example, $$f$$ is differentiable (= $$f'(x)$$ exists) everywhere. But because $$f'(x)$$ is not continuous at $$x=0$$, $$f^{\prime\prime}(0)$$ does not exist.
# RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B In this chapter, we provide RS Aggarwal Solutions for Class 7 Chapter 17 Constructions Ex 17B for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 7 Chapter 17 Constructions Ex 17B Maths pdf, free RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17B Maths book pdf download. Now you will get step by step solution to each question. ### RS Aggarwal Solutions for Class 7 Chapter 17 Constructions Ex 17B Download PDF Question 1. Solution: Steps of construction : (i) Draw a line segment BC = 3.6cm (ii) At B, draw an arc of the radius 5cm. (iii) At C, draw another arc of the radius 5.4cm which intersects the first arc at A (iv) Join AB and AC (v) With centre B and C and radius more than half of BC, draw arcs intersecting each other at L and M. (vi) Join LM which intersects BC at Q and produce it to P. Then PQ is perpendicular bisector of side BC. Question 2. Solution: Steps of construction : (i) Draw a line segment QR = 6cm (ii) With centre Q and radius 4.4 cm draw an arc. (iii) With centre R and radius 5.3 cm, draw another arc intersecting the first arc at P. (iv) Join PQ and PR. (v) With centre P and a suitable radius, draw an arc meeting PR at E and PQ at F. (vi) With centres E and F, with same radius, draw two arcs intersecting each other at G. (vii) Join PG and produce it to meet QR at S. Then PS is the bisector of ∠P. Question 3. Solution: Steps of construction : (i) Draw a line segment BC = 6.2 cm. (ii) With centres B and C radius 6.2 cm, draw arcs intersecting each other at A. (iii) Join AB and AC. ∆ABC is the required equilateral triangle. On measuring, each angle is equal to 60°. Question 4. Solution: Steps of construction : (i) Draw a line segment BC = 5.3 cm. (ii) With centre B and C, and radius 4.8 cm, draw arcs intersecting each other at A (iii) Join AB and AC, Then ∆ABC is the required triangle (v) Then with centre L and M, draw two arcs intersecting eachother at E. (vi) Join AE intersecting BC at D. Then AD is perpendicular to BC. On measuring ∠B and ∠C, each is equal to 55°. Question 5. Solution: Steps of construction : (i) Draw a line segment AB = 3.8cm. (ii) At A draw a ray AX making an angle of 60°. (iii) Cut off AC = 5 cm from AX (iv) Join CB. Then ∆ABC is the required triangle. Question 6. Solution: Steps of construction : (i) Draw a line segment BC = 4.3 cm. (ii) At C, draw a ray CY making an angle equal to 45°. (iii) With centre C, and radius 6cm, draw an arc intersecting CY at A. (iv) Join AB. Then ∆ABC is the required triangle. Question 7. Solution: (i) Draw a line segment AB = 5.2cm. (ii) At A, draw a ray AX making an angle equal to 120°. (iii) From AX cut off AC = 5.2cm. (iv) Join BC. Then ∆ABC is the required triangle. (v) With centre A and some suitable radius draw an arc intersecting BC at L and M. (vi) With centres L and M, draw two arcs intersecting each other at E. (vii) Join AE intersecting BC at D Then AD is the perpendicular to BC. Question 8. Solution: Steps of construction : (i) Draw a line segment BC = 6.2 cm. (ii) At B draw a ray BX making an angle of 60°. (iii) At C draw another ray CY making an angle of 45° which intersect the ray BX at A. . Then ∆ABC is the required triangle. Question 9. Solution: Steps of construction : (i) Draw a line segment BC = 5.8 cm. (ii) At B draw a ray BX making an angle of 30° (iii) At C draw another ray CY making an angle of 30° intersecting the BX at A Then ∆ABC is the required triangle. On measuring AB and AC, AB = 3.5cm and AC = 3.5cm. AB = AC ∆ABC is an isosceles triangle. Question 10. Solution: Steps of construction : In ∆ABC, ∠A = 45° and ∠C = 75° But ∠A + ∠B + ∠C = 180° ⇒ 45° + ZB + 75° = 180° ⇒ ∠B = 180° – 45° – 75° ⇒ ∠B = 180° – 120° = 60° (i) Draw a line segment AB = 7cm. (ii) At A, draw a ray AX making an angle of 45°. (iii) At B, draw another ray BY making an angle of 60° which intersects AX at C Then ∆ABC is the required triangle. Question 11. Solution: Steps of construction : (i) Draw a line segment BC = 4.8 cm (ii) At C, draw a ray CX making an angle of 90°. (iii) With centre B, an radius 6.3cm draw an arc intersecting CX at A. (iv) Join AB. Then ∆ABC is the required triangle. Question 12. Solution: Steps of construction : (i) Draw a line segment BC = 3.5cm (ii) At B, draw a ray BX making an angle of 90°. (iii) With centre C and radius 6cm draw an arc intersecting BX at A (iv) Join AC. Then ∆ABC is the required triangle. Question 13. Solution: One acute angle = 30°, then second acute angle will be = 90° – 30° = 60° (Sum of acute angles = 90°) Steps of construction : (i) Draw a line segment BC = 6cm. (ii) At B draw a ray BX making an angle of 30°. (iii) At C draw another ray CY making an angle of 60° which intersects BX at A Then ∆ABC is the required triangle. All Chapter RS Aggarwal Solutions For Class 7 Maths —————————————————————————– All Subject NCERT Exemplar Problems Solutions For Class 7 All Subject NCERT Solutions For Class 7 ************************************************* I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam. If these solutions have helped you, you can also share rsaggarwalsolutions.in to your friends.
# The two sides of the triangle are 6 and 8 cm. The medians drawn to these sides are mutually The two sides of the triangle are 6 and 8 cm. The medians drawn to these sides are mutually perpendicular. Find the third side of the triangle. Given an arbitrary △ ABC: AC = 6 cm, BC = 8 cm, AK and BM are medians intersecting at point O. 1. The length of the median is calculated by the formula: m = (√ (2b² + 2c² – a²)) / 2, where a is the side of the triangle to which the median is drawn, b and c are the other sides of the triangle. Let us express the length of the median AK: AK = (√ (2AB² + 2AC² – BC²)) / 2 = (√ (2AB² + 2 * 6² – 8²)) / 2 = (√ (2AB² + 2 * 36 – 64)) / 2 = (√ (2AB² + 72 – 64)) / 2 = (√ (2AB² + 8)) / 2. Let us express the length of the median BM: BM = (√ (2AB² + 2BC² – AC²)) / 2 = (√ (2AB² + 2 * 8² – 6²)) / 2 = (√ (2AB² + 2 * 64 – 36)) / 2 = (√ (2AB² + 128 – 36)) / 2 = (√ (2AB² + 92)) / 2. 2. The medians are divided by the point of intersection in a ratio of 2: 1 starting from the top. 2.1. OA / OK = 2/1; OA + OK = AK. We get the system of equations: OA / OK = 2/1; OA + OK = (√ (2AB² + 8)) / 2. In the second equation, we express OK: OK = (√ (2AB² + 8)) / 2 – OA. Substitute the expression into the first equation: OA / ((√ (2AB² + 8)) / 2 – OA) = 2/1; 2 ((√ (2AB² + 8)) / 2 – OA) = OA (proportional); √ (2AB² + 8) – 2OA = OA; 3OA = √ (2AB² + 8); OA = (√ (2AB² + 8)) / 3. 2.2. OB / OM = 2/1; OB + OM = BM. We get the system of equations: OB / OM = 2/1; OB + OM = (√ (2AB² + 92)) / 2. In the second equation, we express OM: OM = (√ (2AB² + 92)) / 2 – OB. Substitute the expression into the first equation: OB / ((√ (2AB² + 92)) / 2 – OB) = 2/1; 2 (√ (2AB² + 92) / 2 – OB) = OB (proportional); √ (2AB² + 92) – 2OB = OB; 3OB = √ (2AB² + 92); OB = (√ (2AB² + 92)) / 3. 3. Consider △ AOB: ∠AOB = 90 ° (since the medians are perpendicular), OA = (√ (2AB² + 8)) / 3 and OB = (√ (2AB² + 92)) / 3 – legs, AB – hypotenuse ( since lie opposite a right angle). By the Pythagorean theorem: AB = √ (OA² + OB²); AB = √ (((√ (2AB² + 8)) / 3) ² + ((√ (2AB² + 92)) / 3) ²); AB = √ ((2AB² + 8) / 9 + (2AB² + 92) / 9); AB = √ ((2AB² + 8 + 2AB² + 92) / 9); AB = √ ((4AB² + 100) / 9) (square both sides of the equation); AB² = (√ ((4AB² + 100) / 9)) ²; (4AB² + 100) / 9 = AB²; 9AB² = 4AB² + 100 (in proportion); 9AB² – 4AB² = 100; 5AB² = 100; AB² = 100/5; AB² = 20; AB = √20; AB = 2√5 cm.
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Lesson 3: Adding and subtracting fractions with unlike denominators # Subtracting fractions with unlike denominators To subtract fractions with different denominators, you need to find a common denominator. This can be done by identifying the least common multiple of the two denominators. After rewriting the fractions so they both have the common denominator, you can subtract the numerators as you would with any two fractions. ## Want to join the conversation? • what if they are the same denominaters • If the denominators are the same , you leave the denominators the same and add or subtract the numerators. Make sure you don't add or subtract the denominators. • What do i do when i have this? 14/6 The numerator is bigger than the denominator. • The answer would be 2 2/6 because 6 goes into 14 2 times and 2 would be left and that turns into the numerator, and the denominator would stay the same • how do you subtract 8/2 - 7/12 • create common denominators so set 8/2 to 48/12 (6 x 2=12, 8 x 6=48) then subtract 48 by 7 to get 41 and place that back over 12 to get 41/12 • how do you multiply or divide with unlike denominators • what if there is 2 diffrent denomanonater • You will have to find a multiple that both denominators have. Then, multiply the numerator by the number you multiplied the denominator to get the common multiple. For example: 5/9 - 6/5 You will need to find a common multiple that 9 and 5 both have. 9: 9, 18, 27, 36, 45, 54... 5: 5, 10, 15, 20, 25, 30, 35, 45, 50... In this case 45 is the common multiple. Then, you would have to multiply the numerators by the number you multiplied their denominator to get the common multiple. Since 9 times 5 equals 45, the 5 in 5/9 needs to be multiplied by 5 and the 6 in 6/5 would need to be multiplied by 9. So, 5/9 becomes 25/45 and 6/5 becomes 54/45. Therefore, 25/45 - 54/45 = -29/45 Hope this helped :) • what if there numerators are bigger than there denominators • then you turn the fraction into a mixed number • how do I subtract mixed fraction where only one fraction is mixed • You turn both fractions into improper fractions, next you find a like denominator, then you subtract. • I was given the question 2/x+4=3/x I'm not sure how to work this out... how wold i make the denominates common? • What do i do when i have this? 14/6 The numerator is bigger than the denominator.
# Profit and Loss Problems for SBI Clerk Pre 2023, LIC Assistant 2023, CET at Smartkeeda DIrections: Study the following information carefully and answer the questions given below: Important for : 1 A sold an article with 10% loss on the cost price. He bought the article at a discount of 20% on the labelled price. What would have been the percentage loss had he bought it at the labelled price? » Explain it D Let the CP to 'A' = 100 After allowing a discount of 10% the SP will be = 90 As per the question, 'A' bought the article at 20% discount on Labelled Price. Therefore, the eq. will become like ⇒ 80 of Labelled price = 100 ⇒ ∴ Labeled Price = 125 100 Now, loss in value if the item was bought at LP = LP – SP = 125 – 90 = 35/- Loss % = 35 × 100 = 28%. 125 Hence, option D is correct. 2 The owner of a cell phone shop charges his customer 28% more than the cost price. If a customer paid Rs 8,960 for the cell phone, then what was the cost price of the cell phone? » Explain it E Let the cost price of the cell phone be x, then, 128% of x = 8960 x × 128 = 8960 100 or, x = 8960 × 100 = ₹ 7000 128 Hence, option E is correct. 3 A sells an article which cost him Rs. 400 to B at a profit of 20%. B then sells it to C, making a profit of 10% on the price he paid to A. How much does C pay to B? » Explain it C CP for B = 120% of 400 = 120 × 400 = ₹ 480 100 CP for C = 110% of 480 = 110 × 480 = ₹ 528 100 Hence, option C is correct. 4 By selling an article for Rs. 480 a person lost 20%. For what should he sell it to make a profit of 20%? » Explain it C Method I: New SP = 100 + Gain% × Old SP 100 – Loss% G% = L% = 20%, Old SP = 480 = 100 + 20 × 480 = 120 × 480 100 – 20 80 New SP = 120 × 6 = 720. ______________________________________ Method II: Let the cost price of the article be x According to the question, we get 80% of x = 480 x = 480 × 100 = 600 80 Reqd profit % = 120% of 600 = 720. Hence, option C is correct. 5 Deepa bought a calculator at 30% discount on the listed price. Had she not got the discount, she would have paid Rs. 82.50 extra. At what price did she buy the calculator? » Explain it A Let the original price be x, then 30% of x = 82.50 x = 82.50 × 100 = Rs. 275 30 Deepa bought calculator in 275 – 82.50 = Rs. 192.50 Hence, option A is correct. ### Profit and Loss Questions for SBI Clerk Pre, IBPS Clerk Pre Profit and loss is a commonly tested topic in bank exams, including those conducted by the State Bank of India (SBI), Institute of Banking Personnel Selection (IBPS), and Reserve Bank of India (RBI). Questions related to profit and loss can be found in various sections of these exams, including quantitative aptitude, data interpretation, and reasoning. Profit and loss is one of the fundamental concepts in banking and financial management. Understanding the principles of profit and loss is essential for candidates appearing for bank exams, as these exams often include questions related to this topic. In this article, we will discuss the basics of profit and loss and how to solve questions related to this topic in bank exams. The questions related to profit and loss in these exams can vary in difficulty level and format. Some of the common types of questions include calculating the cost price, selling price, or profit percentage given the other two values, finding the loss percentage, calculating the discount or markup percentage, and solving word problems related to profit and loss. To prepare for questions related to profit and loss in these bank exams, candidates should have a solid understanding of the basic concepts and formulas related to this topic. They should practice solving a variety of problems and familiarize themselves with the different types of questions that may be asked. It is also essential to develop speed and accuracy in solving these problems, as time is a critical factor in bank exams. Types of Questions asked from Profit & Loss Bank exams such as IBPS, RBI, RRB, SBI, etc., often include questions on profit and loss, as it is an important concept in banking and financial transactions. Here are some of the common types of profit and loss questions that are asked in these exams: 1. Calculation of Profit or Loss Percent: These questions ask you to calculate the profit or loss percentage on a transaction based on the cost price and selling price. For example, if a trader buys an item for Rs. 500 and sells it for Rs. 600, what is the profit percentage? 2. Calculation of Cost Price or Selling Price: These questions ask you to calculate either the cost price or selling price of an item based on the profit or loss percentage and the other value. For example, if an item is sold for Rs. 720 with a profit of 20%, what is the cost price? 3. Calculation of Marked Price or Discount: These questions ask you to calculate either the marked price or the discount percentage based on the selling price and the other value. For example, if an item is sold for Rs. 640 with a discount of 20%, what is the marked price? 4. Calculation of Discounted Selling Price: These questions ask you to calculate the selling price of an item after a discount is applied. For example, if an item is sold at a discount of 25% and the discounted price is Rs. 450, what was the original selling price? 5. Calculation of Break-Even Point: These questions ask you to calculate the minimum number of items that must be sold to cover the cost of production. For example, if the cost of production for 100 items is Rs. 10,000 and each item is sold for Rs. 120, how many items must be sold to break even? 6. Calculation of Partnership Profit and Loss: These questions ask you to calculate the profit or loss for partners in a business based on their capital investment and share in the profit or loss. For example, if Partner A invests Rs. 10,000 and Partner B invests Rs. 15,000 in a business, and the profit is Rs. 5,000, what is the share of profit for each partner? It is important to practice different types of profit and loss questions to be well-prepared for bank exams. You can also refer to previous years' question papers to get an idea of the types of questions that are commonly asked. With practice and a good understanding of the concepts, you can perform well in the profit and loss section of bank exams. What is Profit & Loss? Profit and Loss is a measure of the financial performance of a business or investment. Profit refers to the excess of revenue over expenses, while loss refers to the excess of expenses over revenue. In simple terms, profit is what a business earns after deducting its costs, while loss is what a business incurs when its expenses exceed its revenue. Calculating profit and loss involves understanding some key concepts, such as cost price, selling price, profit percentage, and loss percentage. Cost price refers to the price at which an item is purchased, while selling price refers to the price at which the item is sold. Profit percentage is the percentage of profit earned on the cost price, while loss percentage is the percentage of loss incurred on the cost price. To solve questions related to profit and loss in bank exams, it is important to have a clear understanding of these concepts and their interrelationships. Some of the common types of questions related to profit and loss in bank exams include finding the cost price, selling price, or profit percentage, given the other two values. In addition to these basic concepts, there are several other advanced concepts related to profit and loss that may be tested in bank exams, such as discount, markup, and loss on multiple articles. To prepare for questions related to profit and loss in bank exams, candidates should practice solving a variety of problems and familiarize themselves with the different types of questions that may be asked. It is also important to have a clear understanding of the underlying concepts and formulas and to practice applying them in different scenarios. Conclusion In conclusion, profit and loss is a critical concept in banking and financial management, and understanding the principles of profit and loss is essential for candidates appearing for bank exams. By developing a clear understanding of the underlying concepts and practicing a variety of problems, candidates can enhance their proficiency in solving questions related to profit and loss and improve their chances of success in bank exams. In addition to practicing problems, candidates can also download pdfs from Smartkeeda App or Smartkeeda Website to improve their understanding of the concepts and formulas related to profit and loss. You can also attempt online mock tests to get a better idea of the types of questions that may be asked in the exam and to gauge the level of your preparation. Overall, proficiency in the topic of profit and loss is crucial for success in bank exams such as SBI, IBPS, and RBI. By dedicating sufficient time and effort to preparing for this topic, candidates can enhance their chances of performing well in these exams and achieving their career goals in the banking sector. Profit n Loss is most important topics among all the Bank and Insurance exams like SBI PO, IBPS PO, IBPS RRB Scale 1, IBPS RRB Office Assistant, LIC AAO, NIACL, SBI Clerk, IBPS Clerk Etc. Profit and loss are the terms related to transactions in trade and business. Whenever a purchased article is sold, then either profit is earned or loss is incurred. This chapter teaches us how to calculate such profit and loss. Cost Price (CP) This is the price at which an article is purchased. Selling Price (SP) This is the price at which an article is sold. Overhead Charges Such charges are the extra expenditures on purchased goods apart from the actual cost price. Such charges include freight charges, rent, salary of employees, repairing cost on purchased articles etc. Point to be noted that overhead charges are not taken into consideration, if not given in the question. Profit (SP > CP) When an article is sold at more price than its cost price, then profit is earned. In other words, SP is always greater than CP in case of profit. Loss (SP < CP) When an article is sold at less price than its cost price, then loss is incurred. In other words, SP is always less than CP in case of loss. In this “Profit n Loss Questions for SBI PO”, we will discuss all types of Profit n Loss questions with short tricks and formulas. So we can give better performance if we find Profit n Loss questions in SBI Clerk Pre 2023. Are you looking for any of these? profit and loss questions for sbi clerk pdf profit and loss questions for sbi clerk 2023 profit and loss questions for sbi bank clerk profit and loss questions for sbi clerk mains profit and loss questions for sbi clerk exam profit and loss questions for sbi clerk exam pdf high-level profit and loss question for sbi clerk profit and loss questions in sbi clerk profit and loss practice questions for sbi clerk profit and loss questions asked in sbi clerk 2023 profit and loss questions for sbi clerk Regards, Team Smartkeeda
# Solve exponential equations Lesson In this chapter, we are going to look at how to solve simple exponential equations, where the unknown value we are trying to find is in the power. For example, let's say we wanted to solve the equation $2^{4x}=2^8$24x=28. We can see that the base terms on both sides equal $2$2. Hence, we can say that $4x=8$4x=8 This is because that the only way $2^{\text{something }}$2something can equal $2^{\text{something else }}$2something else is if the powers are both the same. From here we can solve this as a regular algebraic equation, where we get the solution $x=2$x=2. ## Rewriting Numbers Sometimes we can express numbers in terms of a base number with an exponent. For example, $25$25 can be expressed as $5^2$52, $64$64 can be expressed as $4^3$43 and so on. Let's look at an equation where we need to rewrite a number to solve an exponential equation. For example, let's solve the equation $2^{3x-3}=8$23x3=8. At first glance, it seems like it's difficult to solve because we don't have like bases. However, we can rewrite $8$8 as $2^3$23. Now we have like base terms! $2^{3x-3}$23x−3 $=$= $8$8 $2^{3x-3}$23x−3 $=$= $2^3$23 $3x-3$3x−3 $=$= $3$3 $3x$3x $=$= $6$6 $x$x $=$= $2$2 Let's look through some more examples so we master solving exponential equations! #### Examples ##### Question 1 Solve the equation $5^{-3x-1}=3125$53x1=3125 for $x$x. ##### Question 2 Solve the equation $\left(2^2\right)^y=2^3$(22)y=23 for $y$y. ##### Question 3 Solve the equation $\left(2^2\right)^{x+7}=2^3$(22)x+7=23 for $x$x.
# Resistors in Series and Parallel ## LEARNING OBJECTIVES By the end of the section, you will be able to: • Define the term equivalent resistance • Calculate the equivalent resistance of resistors connected in series • Calculate the equivalent resistance of resistors connected in parallel In Current and Resistance, we described the term ‘resistance’ and explained the basic design of a resistor. Basically, a resistor limits the flow of charge in a circuit and is an ohmic device where . Most circuits have more than one resistor. If several resistors are connected together and connected to a battery, the current supplied by the battery depends on the equivalent resistance of the circuit. The equivalent resistance of a combination of resistors depends on both their individual values and how they are connected. The simplest combinations of resistors are series and parallel connections (Figure 6.2.1). In a series circuit, the output current of the first resistor flows into the input of the second resistor; therefore, the current is the same in each resistor. In a parallel circuit, all of the resistor leads on one side of the resistors are connected together and all the leads on the other side are connected together. In the case of a parallel configuration, each resistor has the same potential drop across it, and the currents through each resistor may be different, depending on the resistor. The sum of the individual currents equals the current that flows into the parallel connections. (Figure 6.2.1) ## Resistors in Series Resistors are said to be in series whenever the current flows through the resistors sequentially. Consider Figure 6.2.2, which shows three resistors in series with an applied voltage equal to . Since there is only one path for the charges to flow through, the current is the same through each resistor. The equivalent resistance of a set of resistors in a series connection is equal to the algebraic sum of the individual resistances. (Figure 6.2.2) In Figure 6.2.2, the current coming from the voltage source flows through each resistor, so the current through each resistor is the same. The current through the circuit depends on the voltage supplied by the voltage source and the resistance of the resistors. For each resistor, a potential drop occurs that is equal to the loss of electric potential energy as a current travels through each resistor. According to Ohm’s law, the potential drop across a resistor when a current flows through it is calculated using the equation , where is the current in amps ( ) and is the resistance in ohms ( ). Since energy is conserved, and the voltage is equal to the potential energy per charge, the sum of the voltage applied to the circuit by the source and the potential drops across the individual resistors around a loop should be equal to zero: This equation is often referred to as Kirchhoff’s loop law, which we will look at in more detail later in this chapter. For Figure 6.2.2, the sum of the potential drop of each resistor and the voltage supplied by the voltage source should equal zero: Since the current through each component is the same, the equality can be simplified to an equivalent resistance, which is just the sum of the resistances of the individual resistors. Any number of resistors can be connected in series. If resistors are connected in series, the equivalent resistance is (6.2.1) One result of components connected in a series circuit is that if something happens to one component, it affects all the other components. For example, if several lamps are connected in series and one bulb burns out, all the other lamps go dark. ## EXAMPLE 6.2.1 #### Equivalent Resistance, Current, and Power in a Series Circuit A battery with a terminal voltage of is connected to a circuit consisting of four and one resistors all in series (Figure 6.2.3). Assume the battery has negligible internal resistance. (a) Calculate the equivalent resistance of the circuit. (b) Calculate the current through each resistor. (c) Calculate the potential drop across each resistor. (d) Determine the total power dissipated by the resistors and the power supplied by the battery. (Figure 6.2.3) #### Strategy In a series circuit, the equivalent resistance is the algebraic sum of the resistances. The current through the circuit can be found from Ohm’s law and is equal to the voltage divided by the equivalent resistance. The potential drop across each resistor can be found using Ohm’s law. The power dissipated by each resistor can be found using , and the total power dissipated by the resistors is equal to the sum of the power dissipated by each resistor. The power supplied by the battery can be found using . #### Solution a. The equivalent resistance is the algebraic sum of the resistances: b. The current through the circuit is the same for each resistor in a series circuit and is equal to the applied voltage divided by the equivalent resistance: c. The potential drop across each resistor can be found using Ohm’s law: Note that the sum of the potential drops across each resistor is equal to the voltage supplied by the battery. d. The power dissipated by a resistor is equal to , and the power supplied by the battery is equal to : #### Significance There are several reasons why we would use multiple resistors instead of just one resistor with a resistance equal to the equivalent resistance of the circuit. Perhaps a resistor of the required size is not available, or we need to dissipate the heat generated, or we want to minimize the cost of resistors. Each resistor may cost a few cents to a few dollars, but when multiplied by thousands of units, the cost saving may be appreciable. Some strings of miniature holiday lights are made to short out when a bulb burns out. The device that causes the short is called a shunt, which allows current to flow around the open circuit. A “short” is like putting a piece of wire across the component. The bulbs are usually grouped in series of nine bulbs. If too many bulbs burn out, the shunts eventually open. What causes this? Let’s briefly summarize the major features of resistors in series: 1. Series resistances add together to get the equivalent resistance: 2. The same current flows through each resistor in series. 3. Individual resistors in series do not get the total source voltage, but divide it. The total potential drop across a series configuration of resistors is equal to the sum of the potential drops across each resistor. ## Resistors in Parallel Figure 6.2.4 shows resistors in parallel, wired to a voltage source. Resistors are in parallel when one end of all the resistors are connected by a continuous wire of negligible resistance and the other end of all the resistors are also connected to one another through a continuous wire of negligible resistance. The potential drop across each resistor is the same. Current through each resistor can be found using Ohm’s law , where the voltage is constant across each resistor. For example, an automobile’s headlights, radio, and other systems are wired in parallel, so that each subsystem utilizes the full voltage of the source and can operate completely independently. The same is true of the wiring in your house or any building. (Figure 6.2.4) The current flowing from the voltage source in Figure 6.2.4 depends on the voltage supplied by the voltage source and the equivalent resistance of the circuit. In this case, the current flows from the voltage source and enters a junction, or node, where the circuit splits flowing through resistors and . As the charges flow from the battery, some go through resistor and some flow through resistor . The sum of the currents flowing into a junction must be equal to the sum of the currents flowing out of the junction: This equation is referred to as Kirchhoff’s junction rule and will be discussed in detail in the next section. In Figure 6.2.4, the junction rule gives . There are two loops in this circuit, which leads to the equations and Note the voltage across the resistors in parallel are the same ( ) and the current is additive: Generalizing to any number of resistors, the equivalent resistance of a parallel connection is related to the individual resistances by (6.2.2) This relationship results in an equivalent resistance that is less than the smallest of the individual resistances. When resistors are connected in parallel, more current flows from the source than would flow for any of them individually, so the total resistance is lower. ## EXAMPLE 6.2.2 #### Analysis of a Parallel Circuit Three resistors , , and are connected in parallel. The parallel connection is attached to a voltage source. (a) What is the equivalent resistance? (b) Find the current supplied by the source to the parallel circuit. (c) Calculate the currents in each resistor and show that these add together to equal the current output of the source. (d) Calculate the power dissipated by each resistor. (e) Find the power output of the source and show that it equals the total power dissipated by the resistors. #### Strategy (a) The total resistance for a parallel combination of resistors is found using (Note that in these calculations, each intermediate answer is shown with an extra digit.) (b) The current supplied by the source can be found from Ohm’s law, substituting for the total resistance . (c) The individual currents are easily calculated from Ohm’s law , since each resistor gets the full voltage. The total current is the sum of the individual currents: . (d) The power dissipated by each resistor can be found using any of the equations relating power to current, voltage, and resistance, since all three are known. Let us use , since each resistor gets full voltage. (e) The total power can also be calculated in several ways, use . #### Solution a. The total resistance for a parallel combination of resistors is found using Equation 6.2.2. Entering known values gives The total resistance with the correct number of significant digits is . As predicted, is less than the smallest individual resistance. b. The total current can be found from Ohm’s law, substituting for the total resistance. This gives Current for each device is much larger than for the same devices connected in series (see the previous example). A circuit with parallel connections has a smaller total resistance than the resistors connected in series. c. The individual currents are easily calculated from Ohm’s law, since each resistor gets the full voltage. Thus, Similarly, and The total current is the sum of the individual currents: d. The power dissipated by each resistor can be found using any of the equations relating power to current, voltage, and resistance, since all three are known. Let us use , since each resistor gets full voltage. Thus, Similarly, and e. The total power can also be calculated in several ways. Choosing and entering the total current yields #### Significance Total power dissipated by the resistors is also : Notice that the total power dissipated by the resistors equals the power supplied by the source. Consider the same potential difference applied to the same three resistors connected in series. Would the equivalent resistance of the series circuit be higher, lower, or equal to the three resistor in parallel? Would the current through the series circuit be higher, lower, or equal to the current provided by the same voltage applied to the parallel circuit? How would the power dissipated by the resistor in series compare to the power dissipated by the resistors in parallel? How would you use a river and two waterfalls to model a parallel configuration of two resistors? How does this analogy break down? Let us summarize the major features of resistors in parallel: 1. Equivalent resistance is found from and is smaller than any individual resistance in the combination. 2. The potential drop across each resistor in parallel is the same. 3. Parallel resistors do not each get the total current; they divide it. The current entering a parallel combination of resistors is equal to the sum of the current through each resistor in parallel. In this chapter, we introduced the equivalent resistance of resistors connect in series and resistors connected in parallel. You may recall that in Capacitance, we introduced the equivalent capacitance of capacitors connected in series and parallel. Circuits often contain both capacitors and resistors. Table 6.2.1 summarizes the equations used for the equivalent resistance and equivalent capacitance for series and parallel connections. (Table 6.2.1) Series combination Parallel combination Equivalent capacitance Equivalent resistance ## Combinations of Series and Parallel More complex connections of resistors are often just combinations of series and parallel connections. Such combinations are common, especially when wire resistance is considered. In that case, wire resistance is in series with other resistances that are in parallel. Combinations of series and parallel can be reduced to a single equivalent resistance using the technique illustrated in Figure 6.2.5. Various parts can be identified as either series or parallel connections, reduced to their equivalent resistances, and then further reduced until a single equivalent resistance is left. The process is more time consuming than difficult. Here, we note the equivalent resistance as . (Figure 6.2.5) Notice that resistors and are in series. They can be combined into a single equivalent resistance. One method of keeping track of the process is to include the resistors as subscripts. Here the equivalent resistance of and is The circuit now reduces to three resistors, shown in Figure 6.2.5(c). Redrawing, we now see that resistors and constitute a parallel circuit. Those two resistors can be reduced to an equivalent resistance: This step of the process reduces the circuit to two resistors, shown in in Figure 6.2.5(d). Here, the circuit reduces to two resistors, which in this case are in series. These two resistors can be reduced to an equivalent resistance, which is the equivalent resistance of the circuit: The main goal of this circuit analysis is reached, and the circuit is now reduced to a single resistor and single voltage source. Now we can analyze the circuit. The current provided by the voltage source is . This current runs through resistor and is designated as . The potential drop across can be found using Ohm’s law: Looking at Figure 6.2.5(c), this leaves to be dropped across the parallel combination of and . The current through can be found using Ohm’s law: The resistors and are in series so the currents and are equal to Using Ohm’s law, we can find the potential drop across the last two resistors. The potential drops are and . The final analysis is to look at the power supplied by the voltage source and the power dissipated by the resistors. The power dissipated by the resistors is The total energy is constant in any process. Therefore, the power supplied by the voltage source is . Analyzing the power supplied to the circuit and the power dissipated by the resistors is a good check for the validity of the analysis; they should be equal. ## EXAMPLE 6.2.3 #### Combining Series and Parallel Circuits Figure 6.2.6 shows resistors wired in a combination of series and parallel. We can consider to be the resistance of wires leading to and (a) Find the equivalent resistance of the circuit. (b) What is the potential drop across resistor ? (c) Find the current through resistor . (d) What power is dissipated by ? (Figure 6.2.6) #### Strategy (a) To find the equivalent resistance, first find the equivalent resistance of the parallel connection of and . Then use this result to find the equivalent resistance of the series connection with .(b) The current through is equal to the current from the battery. The potential drop across the resistor (which represents the resistance in the connecting wires) can be found using Ohm’s law. (c) The current through can be found using Ohm’s law . The voltage across can be found using . (d) Using Ohm’s law , the power dissipated by the resistor can also be found using . #### Solution a. To find the equivalent resistance of the circuit, notice that the parallel connection of R2R2 and R3R3 is in series with R1R1, so the equivalent resistance is The total resistance of this combination is intermediate between the pure series and pure parallel values ( and , respectively). b. The current through is equal to the current supplied by the battery: The voltage across is The voltage applied to and is less than the voltage supplied by the battery by an amount . When wire resistance is large, it can significantly affect the operation of the devices represented by and . c. To find the current through , we must first find the voltage applied to it. The voltage across the two resistors in parallel is the same: Now we can find the current through resistance using Ohm’s law: The current is less than the that flowed through when it was connected in parallel to the battery in the previous parallel circuit example. d. The power dissipated by is given by #### Significance The analysis of complex circuits can often be simplified by reducing the circuit to a voltage source and an equivalent resistance. Even if the entire circuit cannot be reduced to a single voltage source and a single equivalent resistance, portions of the circuit may be reduced, greatly simplifying the analysis. Consider the electrical circuits in your home. Give at least two examples of circuits that must use a combination of series and parallel circuits to operate efficiently. #### Practical Implications One implication of this last example is that resistance in wires reduces the current and power delivered to a resistor. If wire resistance is relatively large, as in a worn (or a very long) extension cord, then this loss can be significant. If a large current is drawn, the drop in the wires can also be significant and may become apparent from the heat generated in the cord. For example, when you are rummaging in the refrigerator and the motor comes on, the refrigerator light dims momentarily. Similarly, you can see the passenger compartment light dim when you start the engine of your car (although this may be due to resistance inside the battery itself). What is happening in these high-current situations is illustrated in Figure 6.2.7. The device represented by has a very low resistance, so when it is switched on, a large current flows. This increased current causes a larger drop in the wires represented by , reducing the voltage across the light bulb (which is ), which then dims noticeably. (Figure 6.2.7) ## Problem-Solving Strategy: Series and Parallel Resistors 1. Draw a clear circuit diagram, labeling all resistors and voltage sources. This step includes a list of the known values for the problem, since they are labeled in your circuit diagram. 2. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful. 3. Determine whether resistors are in series, parallel, or a combination of both series and parallel. Examine the circuit diagram to make this assessment. Resistors are in series if the same current must pass sequentially through them. 4. Use the appropriate list of major features for series or parallel connections to solve for the unknowns. There is one list for series and another for parallel. 5. Check to see whether the answers are reasonable and consistent. ## EXAMPLE 6.2.4 #### Combining Series and Parallel Circuits Two resistors connected in series are connected to two resistors that are connected in parallel . The series-parallel combination is connected to a battery. Each resistor has a resistance of . The wires connecting the resistors and battery have negligible resistance. A current of runs through resistor . What is the voltage supplied by the voltage source? #### Strategy Use the steps in the preceding problem-solving strategy to find the solution for this example. #### Solution 1. Draw a clear circuit diagram (Figure 6.2.8). (Figure 6.2.8) 2. The unknown is the voltage of the battery. In order to find the voltage supplied by the battery, the equivalent resistance must be found. 3. In this circuit, we already know that the resistors and are in series and the resistors and are in parallel. The equivalent resistance of the parallel configuration of the resistors and is in series with the series configuration of resistors and . 4. The voltage supplied by the battery can be found by multiplying the current from the battery and the equivalent resistance of the circuit. The current from the battery is equal to the current through and is equal to . We need to find the equivalent resistance by reducing the circuit. To reduce the circuit, first consider the two resistors in parallel. The equivalent resistance is . This parallel combination is in series with the other two resistors, so the equivalent resistance of the circuit is . The voltage supplied by the battery is therefore . 5. One way to check the consistency of your results is to calculate the power supplied by the battery and the power dissipated by the resistors. The power supplied by the battery is Since they are in series, the current through equals the current through . Since , the current through each will be . The power dissipated by the resistors is equal to the sum of the power dissipated by each resistor: Since the power dissipated by the resistors equals the power supplied by the battery, our solution seems consistent. #### Significance If a problem has a combination of series and parallel, as in this example, it can be reduced in steps by using the preceding problem-solving strategy and by considering individual groups of series or parallel connections. When finding for a parallel connection, the reciprocal must be taken with care. In addition, units and numerical results must be reasonable. Equivalent series resistance should be greater, whereas equivalent parallel resistance should be smaller, for example. Power should be greater for the same devices in parallel compared with series, and so on.
Subscribe to our Youtube Channel - https://you.tube/teachoo 1. Chapter 4 Class 12 Determinants 2. Concept wise 3. Area of triangle Transcript Ex 4.3, 3 Find values of k if area of triangle is 4 square units and vertices are (k, 0), (4, 0), (0, 2) The area of triangle is given by ∆ = 1/2 \|■8(x1&y1&1@x2&y2&1@x3&y3&1)\| Here Area of triangle 4 square units Since area is always positive, ∆ can have both positive & negative signs ∴ Δ = ± 4. Putting x1 = k, y1 = 0, x2 = 4, y2 = 0, x3 = 0 y3 = 2 ±4 = 1/2 \|■8(k&0&1@4&0&1@0&2&1)\| ±4 = 1/2 \|■8(k&0&1@4&0&1@0&2&1)\| ± 4 = 1/2 (k\|■8(0&1@2&1)\|−0\|■8(4&1@0&1)\|+1\|■8(4&0@0&2)\|) ± 4 = 1/2 [ k (0 – 2) –0 (4 – 0) + 1 (8 – 0) ] ± 4 × 2 = (k (–2) – 0 + 1 (8)) ± 8 = –2k + 8 So 8 = –2k + 8 or – 8 = –2k + 8 Solving 8 = −2k + 8 8 – 8 = –2k 0 = –2k k = 0/(−2) = 0 +Solving –8 = –2k + 8 –8 – 8 = –2k –16 = –2k k = (−16)/(−2) = 8 So, the required value of k is k = 8 or k = 0 Ex 4.3, 3 Find values of k if area of triangle is 4 square units and vertices are (ii) (-2, 0), (0, 4), (0, k) The area of triangle is given by ∆ = 1/2 \|■8(x1&y1&1@x2&y2&1@x3&y3&1)\| Here Area of triangle 4 square units Since area is always positive, ∆ can have both positive & negative signs ∴ Δ = ± 4 Putting x1 = −2, y1 = 0, x2 = 0, y2 = 4, x3 = 0 y3 = k ± 4 = 1/2 \|■8(−2&0&1@0&4&1@0&k&1)\| ± 4 = 1/2 (−2\|■8(4&1@k&1)\|−0\|■8(0&1@0&1)\|+1\|■8(0&4@0&k)\|) ± 4 = 1/2 (−2(4−𝑘)−0(0−0)+1(0−0)) ± 4 = 1/2 (−2(4−𝑘)) ± 4 = –1 (4 – k) ± 4 = –4 + k So, 4 = –4 + k and –4 = –4 + k Solving 4 = –4 + k 4 + 4 = k 8 = k k = 8 Solving –4 = –4 + k –4 + k = –4 k = –4 + 4 k = 0 So, the required value of k is k = 8 or k = 0 Area of triangle
Square roots and negative numbers are connected in several ways. In some cases, we might need to bring complex or imaginary numbers into the discussion. So, can a square root be negative? A square root can be negative – for example, when we take the opposite of the principal square root of a positive number. Any positive number has both positive and negative even roots (square roots, fourth roots, etc.) However, the square root of a negative number is imaginary, not negative. Of course, if take an odd root (3rd root, 5th root, etc.) of a negative number, there will always be a negative root we can find. Let’s get going. ## Can A Square Root Be Negative? A square root can be negative if we take the opposite of the principal square root of a positive number. This comes from the fact that every number (except zero) has two square roots that are opposites (negatives) of each other. Remember that the principal square root a positive number N is the unique positive number S such that N = S2. For example, the principal square root of 9 is +3. Likewise, the principal square root of 16 is +4. To get a negative square root of a positive number, we take the opposite (negative) of the principal square root. So, -3 is the other square root of 9, since (-3)2 = (-3)(-3) = 9 (a negative times a negative is positive). Likewise, -4 is the other square root of 16, since (-4)2 = (-4)(-4) = 16. The table below gives examples of numbers, their principal square roots, and their negative square roots, along with the general formula for a positive number n and the two square roots. When we take a square root, the number under the radical is the radicand, and the index is 2. An alternative way to express the principal square root of a positive number x is with a power of ½: x1/2. If you want the negative square root, you would use the expression –(x1/2), or in words “the negative of the square root of x”. This is not the same thing as (-x)1/2, which in words is “the square root of –x.” If you are wondering why every positive real number has two roots, here is the explanation. We are solving for x, the square root of N, and the key equation is: • x = N1/2 • x2 = N [square both sides] • x2 – N = 0 [subtract N from both sides] • (x + N1/2)(x – N1/2) = 0 [factor as a difference of squares] This gives us solutions of x = N1/2 and x = -N1/2. Each solution is the opposite (negative) of the other one. ### Negative Radicand & Even Index (A Negative Number Inside A Square Root) When the radicand is negative and the index is even, we get a complex number (more sepcifically, an imaginary number) as a result. Remember that a complex number has the form a + bi, where a and b are real numbers and i is the imaginary unit (that is, i is the square root of -1, or i2 = -1). More specifically, a negative radicand and an index of 2 will always give us an imaginary number. An imaginary number is a specific type of complex number with a real part that is zero (that is, a = 0). For example, the square root of -4 is (-4)1/2 = 2i, which is an imaginary number (the index of the radical is 2, and the radicand is -4). ## Can A Principal Root Be Negative? A principal square root cannot be negative. By definition, we can only take the principal square root of a positive number, and the result is defined to be the positive square root. For example, +5 is the principal square root of 25 (since it is the positive square root). Although -5 is also a square root of 25, it is not the principal square root of 25. ## Can A Cube Root Be Negative? A cube root can be negative if the radicand (number under the radical) is also negative. For example, the cube root of -8 is -2, since (-2)3 = (-2)(-2)(-2) = -8 (a negative raised to an odd power is always negative). Another way to express this with rational exponents is (-8)1/3 = -2. Likewise, the cube root of -64 is -4, since (-4)3 = )-4)(-4)(-4) = -64. Another way to express this with rational exponents is (-64)1/3 = -4. If we take the cube root of a positive number, we get a positive number, but never a negative number. For example, the cube root of 125 is +5, since 53 = 555 = 125. Of course, there are 3 cube roots of any number: one is real, and the other two are complex (of the form a + bi, with a and b real numbers). The table below summarizes the sign of the real cube root, based on the sign of the radicand (number we are taking the cube root of). Except for a radicand of zero, there are always two complex cube roots (which are complex conjugates). ## Can A Fourth Root Be Negative? The fourth root of a positive number can be negative, just like a square root of a positive number can be negative. This would involve taking the negative of the principal (positive) fourth root. For example, the principal fourth root of 16 is 2, since 24 = 16. Then -2 is also a fourth root of 16, since (-2)4 = (-2)(-2)(-2)(-2) = 16 (a negative raised to an even power is always positive). Likewise, the principal fourth root of 81 is 3, since 34 = 81. Then -3 is also a fourth root of 81, since (-3)4 = (-3)(-3)(-3)(-3) = 81 (a negative raised to an even power is always positive). Of course, a positive number has 4 fourth roots: two real (one positive, one negative) and two complex (they are complex conjugates). One way to see this is to take the square roots (positive and negative) of the number. Then, take the two square roots of each of these numbers. This would give us a positive and negative (the two square roots of the positive square root) and two complex numbers (the two square roots of the negative square root). For example, let’s find the 4 fourth roots of 16. First, take the two square roots of 16: +4 and -4. Next, take the two square roots of +4: +2 and -2. Then, take the two square roots of -4: +2i and -2i. So, our 4 fourth roots of 16 are +2, -2, +2i, and -2i. Another way to see this is by forming a polynomial (a quartic, which as degree 4) and then factoring completely to find the zeros. Using the same example, let’s say we want the fourth root of 16 (call it x). Then the key equation is: • x = 161/4 • x4 = 16 [take the 4th power of both sides] • x4 – 16 = 0 [subtract 16 from both sides] • (x2 – 4)(x2 + 4) = 0 [factor x4 – 16 as a difference of squares] • (x – 2)(x + 2)(x2 + 4) = 0 [factor x2 – 4 as a difference of squares] • (x – 2)(x + 2)(x2 – (-4)) = 0 [rewrite x2 + 4 as x2 – (-4)] • (x – 2)(x + 2)(x – 2i)(x + 2i) [factor x2 – (-4) as a difference of squares] This gives us the 4 fourth roots of 16: +2, -2, +2i, and -2i. *Note: the 4 fourth roots of a negative number will all be complex. ## What Is The Square Root Of i? (How To Find The Square Root Of The Imaginary Unit) We know that the square root of -1 is i, but what is the square root of i? Let’s call the number z. Then we want to solve z = i1/2. We know that z will be a complex number, of the form z = a + bi. Here are the steps: • z = i1/2 [z is the square root of i] • z2 = i [square both sides] • (a + bi)2 = i [since z = a + bi] • (a + bi)(a + bi) = i • a2 + abi + bai + b2i2 = i [use FOIL] • a2 + 2abi +b2i2 = i [combine like terms] • a2 + 2abi – b2 = i [i2 = -1] • (a2 – b2) + (2ab)i = 0 + 1i [group real and imaginary parts] From here, we can create two equations by comparing the real parts on both sides of the equation and the imaginary parts on both sides of the equation: • Equation 1: a2 – b2 = 0 [compare real parts] • Equation 2: 2ab = 1 [compare imaginary parts] For equation 1, we can factor the left side as a difference of squares to get: • (a + b)(a – b) = 0 This gives us a = b or a = -b. For equation 2, we want to check both cases: a = b and a = -b. First, we plug in a = b into Equation 2: • 2(b)(b) = 1 • 2b2 = 1 • b2 = ½ • b = +(1/2)1/2 or b = -(1/2)1/2 Since a = b, we also get a = (1/2)1/2 or a = -(1/2)1/2. So, the two square roots of i are: z1 = (1/2)1/2 + (1/2)1/2i z2 = -(1/2)1/2 – (1/2)1/2i ## Conclusion Now you know how square roots and negative numbers are connected. You also know a little more about radicals and higher values of the index (3rd and 4th roots) and how negative numbers come into play in those cases. You can learn how to add, multiply, and divide square roots here.
1. ## Equation of line. Solving quadratic equations. A straight line passes through the two points A and B with coordinates A(–3,–5) and B(5,1). a) Determine an equation for the line Problem 6: roots of quadratic equations The following quadratic equations have been factorized for you. Read off the roots: a. x2 – 40x + 300 = (x–10)(x–30) = 0 b. x2 + x – 1 = (x–½)(x+2) = 0 c. x2 – (+)x +  = (x–)(x–) = 0 d. 6x2 + 6x – 72 = (2x–6)(3x+12) = 0 Factorise, and hence determine the roots of the quadratic equation: e. x2 + 5x + 6 = 0 f. x2 + 6x – 7 = 0 Problem 7: throwing up… A ball is thrown into the air. The height (over the ground, in metres) of the ball is called y, and a formula for y is y = . Here g denotes the acceleration of gravity, g = 9.8 m/s2, t is the time from when the ball is thrown (in seconds), and v is the speed with which the ball is thrown (in m/s). a) With your graphing calculator, sketch y as a function of t for a ball thrown upwards with the speed v = 5 m/s. Copy the graph onto paper. b) With your calculator, find the maximum height the ball reaches (2 sig. figs.) c) After how many seconds is the ball back on the ground? (2 sig. figs.) 2. 1. The equation for a line can be represented as such: $y = mx + b$ Where $m$ is the slope and $b$ is the y intercept. The equation for slope is $\frac {rise}{run}$, or $\frac {y_1-y_2}{x_1-x_2}$. You have two points so you have two x coordinates and two y coordinates. It does not matter what you choose to be $x_1$ so long as you make the corresponding y value $y_1$. Calculate the slope, then plug it in to the equation $y = mx + b$ using either of your original points to find $b$. 2. When you have an equation like $(x-10)(x-30) = 0$, the roots are whatever makes that equation true. So, for $(x-10)(x-30)$ to be equal to 0, either $(x-10) = 0$, or $(x-30) = 0$. Now, just solve for $x$ in both cases. Case 1: $x-10 = 0$, therefore $x = 10$ Case 2: $x-30 = 0$, therefore $x = 30$ So, the roots of the equation are 10 and 30. 3. To factorize a quadratic of the form $x^2 +bx +c = 0$, simply use the sum and product rule. What two numbers have a sum of $b$ and a product of $c$? $x2 + 5x + 6 = 0 $ 3 and 2 add up to 5 and multiply to 6, so your factored equation would be: $(x+2)(x+3) = 0$ 4. On your calculator there should be a y= button. Simply enter the equation into it and then press GRAPH. Press 2nd CALC then select MAXIMUM to find the maximum height. Find the two roots of the equations. The time the ball is the absolute value of $x_1 - x_2$ 3. Thanks for the help dude
# Limits of Trigonometric Functions How limits are applied to Trigonometry Functions? Limits are also applied to trigonometry functions along with the general ones. In geometric figures, to calculate unknown angles and distances from known or measured angles, we have to use trigonometric functions. Limits cannot be applied to all functions because x approaches infinity. Let the function f(x) = x cos(x), as x raises larger, this function does not approach any specific real number meanwhile we can always choose a value of x to make f(x) greater than whatever number we want. ## What are Trigonometric functions? The word trigonometry comes from the Greek language and is derived from a combination of three Greek words. The three Greek words from which trigonometry is derived are Trei (three), Goni (angles), and Metron (measures) refer to triangle measurements. There are six functions of an angle that are frequently used in trigonometry. These functions are derived by a triangle having a base, perpendicular, and hypotenuse. Sine is found by using a right-angled triangle. Sine = perp / hypotenuse Cosine is found by using a right-angled triangle. Cosine = altitude / hypotenuse Tangent is found by using a right-angled triangle. Tangent = perpendicular / altitude We can also find tangent by sine and cosine due to relationship. Tangent = sine / cosine ## What are limits? Mathematical limits are real numbers that are unique. Consider the limit of a real-valued function “f” and a real number “c,” which is written as. f(x) = L As x approaches a equal to L, it can be read as the limit of f(x). The term one-sided limit refers to a restriction that occurs only on one side, either left or right. A two-sided limit is one that has a limit on both sides. The term infinite limit refers to a function that can increase or decrease indefinitely. Example Evaluate 4x2 + 9x + 5 / 16x2 – 25 Solution Step 1: Write the given limit function. 4x2 + 9x + 5 / 16x2 – 25 Step 2: Make factors of the numerator by factorization method. 4x2 + 9x + 5 / 16x2 – 25 = 4x2 + 4x + 5x + 5 / 16x2 – 25 4x2 + 9x + 5 / 16x2 – 25 = (4x + 5) (x + 1) / 16x2 – 25 Step 3: Make factors of denominator with the help of formula. 4x2 + 9x + 5 / 16x2 – 25 = (4x + 5) (x + 1) / (4x + 5) (4x – 5) Step 4: Now simplify the above equation. = (4x + 5) (x + 1) / (4x + 5) (4x – 5) = (x + 1) / (4x – 5) Step 5: Apply the quotient of the function rule of limit. = (x + 1) / (4x – 5) Step 6: Apply limits: 4x2 + 9x + 5 / 16x2 – 25 = (5 + 1) / (4(5) – 5) 4x2 + 9x + 5 / 16x2 – 25 = 6 / (20 – 5) = 6/15 = 2/5 Limits are also applied to calculate the slope. The slope is a very common concept used widely in algebra. Slope is very essential for the calculation of the equation of the line. ## Rules of Limits Name Rules Constant rule A = A Constant time’s function (kfx) = k (fx) Difference of the functions Rule x→c (f(x) – g(x)) = f(x) – g(x) Product of the functions Rule x→c (f(x) * g(x)) = f(x) * g(x) The quotient of the functions Rule x→c (f(x) / g(x)) = f(x) / g(x) ## How to apply limits on trigonometric functions? The sine and cosine trigonometric functions have four essential limit properties. These principles can be used to solve a variety of limit problems utilizing the six fundamental trigonometric functions. Trigonometric limits can be easily evaluated by using limit calculator. Three properties of limits are mentioned below. 1. 1 – cos(x) / x = 0 2. sin(x) / x = 1 3. sec(x) – 1 / x = 0 ### Example 1: Evaluate sin(8x) / x. Solution Step 1: write the given value. sin(8x) / x Step 2: Multiply and divide by 8. 8sin(8x) / 8x Step 3: Apply the product of the function rule. 8 x→0 sin(8x) / 8x Step 4: By constant rule and trigonometric limit property we have, 8 x→0 sin(8x) / 8x = 8 x 1 = 8 Step 5: Write the given input with result. sin(8x) / x = 8 ### Example 2: Evaluate sin(2x) / x. Solution Step 1: write the given value. sin(2x) / x Step 2: Multiply and divide by 2. 2sin(2x) / 2x Step 3: Apply the product of the functions rule. 2 sin(2x) / 2x Step 4: By a property, we can apply this. 2 sin(2x) / 2x = 2 x 0 Step 5: Write the given input with result. sin(2x) / x = 0 ### Example 3: Evaluate 2x2 + 4 / 2x2 -7x -4 Solution Step 1: write the given value. 2x2 + 4 / 2x2 -7x -4 Step 2: Take x2 common from numerator and denominator. x→ x2(2 + 4 / x2) / x2(2 – 7/x – 4/x2) Step 3: Cancel x2. x→ (2 + 4 / x2) / (2 – 7/x – 4/x2) Step 4: Apply limits. 2 + 4/∞2 / 2 – 7/∞ – 4/∞2 Step 5: As n/∞ = 0, put zero where infinity is at the denominator. 2 + 0 / 2 – 0 – 0 = 2 / 2 = 1 Step 6: Write the given input with result. 2x2 + 4 / 2x2 -7x -4 = 1 ### Example 4: Evaluate 2x3 – 3x2 + 4 / x3 -7x -4 Solution Step 1: write the given value. 2x3 – 3x2 + 4 / x3 -7x -4 Step 2: Take x3 common from numerator and denominator. x→ x3(2 – 3/x + 4 / x3) / x3(1 – 7/x2 – 4/x3) Step 3: Cancel x2. x→ (2 – 3/x + 4 / x3) / (1 – 7/x2 – 4/x3) Step 4: Apply limits. 2 – 3/∞ + 4/∞3 / (1 – 7/∞2 – 4/∞3 Step 5: As n/∞ = 0, put zero where infinity is at the denominator. 2 – 0 + 0 / 1 – 0 – 0 = 2 / 1 = 2 Step 6: Write the given input with result. 2x3 – 3x2 + 4 / x3 -7x -4 = 2 ## Relationship of slope of a tangent line and limits Limits are also applied to calculate the slope. The slope is a very common concept used widely in algebra. It is very essential for the calculation of the equation of the line. As derivatives are used to calculate the slope of the tangent, the limits are also helpful in this regard. Because limits are used to calculate the derivative which is the slope of the tangent. You can use an online tool like slope calculator to find the slope of line, slope-intercept form, and point slope form. ## Summary The limits of trigonometric functions can also be calculated by using the above-mentioned properties. Evaluating the limits of trigonometric functions is not much difficult, just a little effort is required to master the concept.
# Time & Distance Tutorial I: Basic Concepts, Average Speed & Variation of Parameters ## Data Sufficiency questions on time and distance In these types of question, we can analyze if the given data is enough to arrive at solution for the problem in hand using basic concepts of time and distance. Let's look at few examples to understand how to tackle data sufficiency type of problems on time and distance. Ajay must get stationary products from either shop A or shop B .Which of the two shops is closer to Ajay's home? I: Travelling from home at a constant speed, Ajay takes 15 minutes to reach shop A II: Travelling from shop A at a constant rate, he takes 15 minutes to reach shop B Find out which of the below holds true A. Statement I alone sufficient statement II alone is not sufficient B. Statement II alone sufficient statement I alone is not sufficient C. Both statements together are sufficient but neither statement alone is sufficient D. Each statement alone is sufficient E. Statement I and II together are not sufficient to answer the question, and additional data are required Statement I tell us the time taken by Ajay to reach the Shop A from his home. It doesn't contain any information about shop B .Hence the statement I alone is not sufficient. Statement II tells us the time duration for the trip from shop A to shop B. It doesn't convey the information about the placing of his home as per the positions of both the shops. Hence statement II alone is not sufficient. Clubbing statements I and II together doesn't give any inference about the placing of shop A and shop B. If the speeds which are mentioned in the statements are same then only one inference we can form. The distance between home and shop A is equal to the distance e between shop A and shop B Consider the following possibilities Possibility I: Shop A is closed to home Possibility II: Shop B is closed to home Possibility III: Shop A and shop B are equidistant from home :. A unique conclusion is not possible while clubbing the statements together, hence answer is E. ## Variation wise relations of basic parameters Whenever we can think about a time - speed - distance situation, there are only three parameters are valid. On in a different wording we can tell time, speed and distance are the foundation blocks of this topic for solving the questions effectively One of main fundamental requirements is to understand the variation wise (proportionate) relationship between time, speed and distance. Consider the different scenarios • Scenario I: When distance is constant When distance is constant, increase in speed will make a corresponding decrease in time, required covering a fixed distance. Both the parameters (speed and time) vary in opposite direction means increase in one parameter will make a corresponding decrease in the other and it is vice versa. Then we can say that, Speed is inversely proportional to time, when distance is constant. This can be expressed as "Speed" \propto 1/"Time" Let the Distance = 120 km At 40kmph speed, we can cover it in 3 hours At 60kmph speed, we can cover it in 2 hours. "Speed I"/"Speed II" = "Time II"/"Time I" -> 40/60 = 2/3 • Scenario II: When time is constant When time is constant, distance covered will increase or decrease as per the increase or decrease in the speed. Then we can say that, Speed is directly proportional to distance, when time is constant. This can be expressed as "Speed" \propto "Distance" Let the time taken = 2hr At 40kmph speed, we can cover 80km. At 60kmph speed, we can cover 120km. "Speed I"/"Speed II" = "Distance I"/"Distance II" -> 40/60 = 80/120 • Scenario III: When speed is constant When speed is constant, distance covered will increase or decrease as per the increase or decrease in the time. Then we can say that, Time is directly proportional to distance when speed is constant. This can be expressed as "Time" \propto "Distance" Let the speed be = 40kmph With 2hrs, we can cover 80km. With 3hrs, we can cover 120km. "Time I"/"Time II" = "Distance I"/"Distance II" -> 2/3 = 80/120 Let's look at few examples to see how to use the above proportions to solve aptitude problems on time and distance. Ratio of the speeds of three cars P, Q and R is 2: 4: 5 respectively and the distance covered by car P in 2 hrs is 20 km. Find the difference in the distances covered by Q and R in 3 hrs? Speed is directly proportional to distance when time is constant. "Ratio of speed" = "Ratio of distance" = 2: 4: 5 Given that Car P covers 20 km in 2 hrs, i.e. 10 km in 1 hr. Now, if we map it to the proportions, we get 2 -> 10km 4 -> 20km 5 -> 25km Distance covered by car Q and R in one hour are 20 km and 25 km respectively. In three hours car Q and car R can cover 20 xx 3 = 60 km and 25 xx 3 = 75 km respectively. :.Difference between the distances is 75 - 60 = 15 km ### Generalization of the concept If a person travels from A to B at a speed of S_1 " kmph" then he reaches early by 'x' minutes and when he travels at S_2 " kmph", then he reaches early by 'y' minutes. Let the distance between A and B is D. Time taken for the journey from A to B at S_1 "kmph", t_1 = D/S_1 Time taken for the journey at S_2 " kmph", t_2 = D/S_2 Here S_2 > S_1 then t_1 > t_2 and T is the time difference. T = t_1 - t_2 or (x + y) min = (x + y)/60 hr. Then D/S_1 - D/S_2 = T D(S_2 - S_1)/S_1S_2 = T D = (S_1 xx S_2 xx T)/(S_2 - S_1) A person covers a certain distance at S_1 kmph taking t_1 minutes and covering the same distance at S_2 kmph and taking t_2 minutes; Where S_1 < S_2 and T = (t_1 - t_2)/60 Then, Then distance,D = (S_1 xx S_2 xx T)/(S_2 - S_1) OR D = "Product of the speeds and time difference"/"Difference between speeds" ### Examples Starting from home and travelling at a constant speed of 5 kmph, girl reaches school 10 minutes late. If she travels at a speed of 8 kmph, she will reach the school 5 minutes early. What is the distance from her home to school? S_1 = 5 kmph and S_2 = 8 kmph T = 10 min + 5 min = 15 min = 1/4 hrs D = (S_1 xx S_2 xx T)/(S_2 - S_1) = (5 xx 8 xx 1/4)/(8 - 5) = 10/3 km = 3333.3m If Sam walks at a speed of 3 kmph, he is 20 minutes early at the office. If he walks at a speed of 2 kmph, he is late by 30 minutes at the office. What should be his speed, so as to reach the office on time? S_1 = 3 kmph and S_2 = 2 kmph T = (20 + 30)/60 = 5/6 hrs D = (S_1 xx S_2 xx T)/(S_2 - S_1) = (3 xx 2 xx 5/6)/(3 - 2) = 5 km If he is covering 5km at 2kmph then the time required to cover the distance = 5/2 = 2.5 hrs; given that this is 30 minutes more than the usual time consumption . :. The usual time (i.e. the expected time duration required to cover the distance) = 2.5 hrs - 0.5 hrs = 2hrs "Distance" = 5 km, "Usual time duration" = 2 hrs, "Usual speed" = 2.5 kmph Page 2 of 3 #### Popular Videos How to improve your Interview, Salary Negotiation, Communication & Presentation Skills. Got a tip or Question? Let us know
# Reflection A reflection is a transformation that uses a line like a mirror to reflect or "flip" a figure. The mirror line is called the "line of reflection."[br][br]The following activities will help you explore various types of reflections. Like the last exploration, observe how this type of transformation affects the points, segments and angle measures of the figure when creating the image. [br] ##### Reflection over the y-axis (x = 0) What are the effects of a reflection over the y-axis on the points, segments and angle measurements? ##### Reflection over the x-axis (y = 0) What are the effects of a reflection over the x-axis on the points, segments and angle measurements? ##### Reflection over a vertical line What are the effects of a reflection over any vertical line on the points, segments and angle measurements? ##### Reflection over a horizontal line What are the effects of a reflection over any horizontal line on the points, segments and angle measurements? What are the effects of a reflection over the line y = x on the points, segments and angle measurements? ##### Other Relationships within Reflections Return to the above GeoGebra sketches. Add lines connecting the original point to its image. Notice the relationship between these lines and the line of reflection. Summarize the relationships you observed between these new lines and the line of reflection. What is the equation of the line of symmetry used to create the image above? Use the input bar to insert your guess for the line of reflection into the sketch. Then, use the GeoGebra tools to check the accuracy of your line. ##### Can you do it on your own? Create a line of reflection using the line tool. Then use the "reflect about a line" tool to create the image. ##### Reflection Rules So far you have explored three different scenarios. You should have noticed the following patterns or rules: [br][br]If a figure is reflected in the y-axis, then all of the points will move according to the rule [br]$\left(x,y\right)\longrightarrow\left(-x,y\right)$.[br]If a figure is reflected in the x-axis, then all of the points will move according to the rule $\left(x,y\right)\longrightarrow\left(x,-y\right).$.[br]If a figure is reflected in the line y = x , the all of the points will move according to the rule $\left(x,y\right)\longrightarrow\left(y,x\right)$.[br][br]What do you think is the rule when figures are reflected in the line y=-x? ##### Challenge 1 Explore what happens when we perform a composition of two reflections. Return to one of the sketches above. Add a second line of reflection. Use the "reflect about a line" tool to reflect the image again. [br][br]Record the summary of your observations below. Observations of the effects of a composition of two reflections. ##### Challenge 2 Explore what happens when we do a composition of a translation and reflection. Use the sketch below to add a vector and a line of symmetry. Explore performing a translation, then a reflection and a reflection, then translation. Does the order matter? [br][br]This type of transformation (or composition of transformations) is called a Glide Reflection. ##### Challenge 2 Summary What did you observation about Glide Reflections?
Home » Finance » Do you reverse the sign of inequality when you divide? # Do you reverse the sign of inequality when you divide? Contents Do you reverse the sign of inequality when you divide? Each time you multiply or divide the two sides of the inequality, you must “turn” or change the direction of the sign of the inequality. This means that if you had a sign of less than. What happens to a sign of inequality when it splits? Dividing each side of an inequality by a positive number does not change the direction of the inequality symbol. Si ab · c. Multiplying each side of an inequality by a negative number reverses the direction of the inequality symbol. Why do you reverse the sign of inequality when you multiply or divide? Just like when divided by a negative number, the sign of inequality must turn! Here’s why: When you multiply both sides by a negative value, you make the larger side have a “bigger” negative number, which means it’s now smaller than the other side! What are the rules for graphically representing inequalities? To graph an inequality, treat the sign, or ≥ as a = sign, and represent the equation. If the inequality is, it represents the equation as a dotted line. If the inequality is ≤ or ≥, draw the equation as a solid line. ## Do you reverse the sign of inequality when you divide? – Related questions ### What is the first step in resolving this inequality? Answer: The first step in resolving the given inequality is to use the distributive property and open the square brackets, that is, -12 + 20x ≥ -6x + 9. ### What does inequality at most mean? The notation a ≤ boa ⩽ b means that a is less than or equal to b (or, equivalently, at most b, or not greater than b). The notation a ≥ boa ⩾ b means that a is greater than or equal to b (or, equivalently, at least b, or not less than b). ### What are the four types of inequalities? There are five systems or types of social inequality: inequality of wealth, inequality of treatment and responsibility, political inequality, inequality of life, and inequality of membership. ### What is the solution to inequality mc011 1 JPG? Step-by-step explanation: Thus, x = 8 is a solution of the inequality. ### What is the difference between solving an equation and solving an inequality? The only difference is that if you multiply or divide the two sides of an equation by the same negative number, the equation remains the same, but if you multiply or divide the two sides of an inequality by the same negative number, the inequality s ‘invest. !!!!! ### What are the 5 symbols of inequality? These inequality symbols are: less than (), less than or equal to (≤), greater than or equal to (≥), and the unequal symbol (≠). ### What is the first step in graphically representing the whole solution of an inequality? To graph the solution set of an inequality with two variables, first draw the boundary with a dashed or solid line as a function of the inequality. If strict inequality occurs, use a dashed line for the boundary. If an inequality is given, use a solid line. Then choose a test point that is not at the limit. ### How do you know if it’s shading above or below the line? Unless you are graphically a vertical line, the inequality sign will let you know which half-shaded plane. If the ≥ or> symbol is used, shadow above the line. If the symbol ≤ or What is the best first step in solving the inequality 5x 3 2? What is the best first step to solving the 5x – 3 ≥ 2 inequality? A. Add 3 on both sides of the equation. Xavier cannot work more than 40 hours a week. ### How do you know if an inequality is true? If the two sides of an inequality are multiplied or divided by the same positive value, the resulting inequality is true. If the two sides are multiplied or divided by the same negative value, the direction of the inequality changes. ### What does R mean in inequalities? R = real numbers includes all real numbers [-inf, inf] Q = rational numbers (numbers written as reason) N = natural numbers (all positive integers from 1. ( ### What are some real examples of inequalities? Consider the following: speed limits on the freeway, minimum credit card bill payments, the number of text messages you can send each month from your mobile phone, and how long it will take you to get home from home. ‘school. All of this can be represented as mathematical inequalities. ### What factors cause inequality? Inequalities are not only determined and measured by income, but also by other factors: gender, age, origin, ethnicity, disability, sexual orientation, class, and religion. These factors determine the inequalities of opportunity that continue to persist, within and between countries. ### Why is inequality a problem? Sufficient economic inequality can transform a democracy into a plutocracy, a society ruled by the rich. The great inequalities of inherited wealth can be especially damaging, in fact creating an economic caste system that inhibits social mobility and undermines equal opportunity. ### What inequality has no solution? Inequalities of absolute values. Here are the steps to solve absolute value inequalities: Isolate the absolute value expression on the left side of the inequality. If the number on the other side of the inequality sign is negative, your equation has no solution or all real numbers as solutions. ### What is the value of mc011 1 JPG? The measure of mc011-1. jpg is 150 °. ### What is the whole solution of compound inequality? When the two inequalities are joined by the word or, the solution of the compound inequality occurs when either of the inequalities is true. The solution is the combination, or union, of the two individual solutions. ### How do you know if an inequality system has no solutions? When two inequalities within a system do not share a common region, then the system has no solution and no individual point will be an established solution for both inequalities. ### What does ≥ mean? The symbol ≤ means less than or equal to. The symbol ≥ means greater than or equal to. ### Why is the set of solutions infinite? A system has infinite solutions when it is coherent and the number of variables is more than the number of non-zero rows in the rref of the array. The row 0 only means that one of the original equations was redundant. The set of solutions would be exactly the same if removed. ### How do you know when to shade? When inequalities are drawn, how do you know what to shade? Rearrange the equation so that “y” is on the left and everything else is on the right. Shadow above line for “more than” (y> oy≥) or below line for “less than” (y> Related Content Do individual stocks and mutual funds carry the same risk? Do individual stocks and mutual funds carry the same risk? Read more Do the vehicles meet the requirements of Article 179? Do the vehicles meet the requirements of Article 179? Almost Read more Do you receive a pension with BRS? Do you receive a pension with BRS? 4,320 retirement points Read more Cash Truck Stops EFS Checks? Cash Truck Stops EFS Checks? Do truck stops sell EFS Read more
# What Angles Are Alternate Interior? ## Do alternate interior angles add up to 180? Alternate angles form a ‘Z’ shape and are sometimes called ‘Z angles’. d and f are interior angles. These add up to 180 degrees (e and c are also interior). Any two angles that add up to 180 degrees are known as supplementary angles.. ## What do alternate interior angles equal? Lesson Summary Alternate interior angles are formed when a transversal passes through two lines. The angles that are formed on opposite sides of the transversal and inside the two lines are alternate interior angles. The theorem says that when the lines are parallel, that the alternate interior angles are equal. ## Are alternate interior angles congruent? Alternate Interior Angle Theorem The Alternate Interior Angles Theorem states that, when two parallel lines are cut by a transversal , the resulting alternate interior angles are congruent . ## What do consecutive interior angles look like? The pairs of angles on one side of the transversal but inside the two lines are called Consecutive Interior Angles. … In this example, these are Consecutive Interior Angles: d and f. ## How do you know if angles are congruent? ASA stands for “angle, side, angle” and means that we have two triangles where we know two angles and the included side are equal. If two angles and the included side of one triangle are equal to the corresponding angles and side of another triangle, the triangles are congruent. ## What are alternate angles on parallel lines? Alternate interior angles are “interior” (between the parallel lines), and they “alternate” sides of the transversal. Notice that they are not adjacent angles (next to one another sharing a vertex). are equal in measure. If two parallel lines are cut by a transversal, the alternate interior angles are congruent. ## What does same side interior angles mean? Same side interior angles are two angles that are on the same side of the transversal and on the interior of (between) the two lines. Same Side Interior Angles Theorem: If two parallel lines are cut by a transversal, then the same side interior angles are supplementary. ## What is the converse of alternate interior angles Theorem? Converse of Alternate Interior Angles Theorem: If two lines are cut by a transversal and alternate interior angles are congruent, then the lines are parallel. ## Do alternate interior angles have the same measure? They are located between the two parallel lines but on opposite sides of the transversal, creating two pairs (four total angles) of alternate interior angles. Alternate interior angles are congruent, meaning they have equal measure. ## What are alternate angles for kids? Alternate interior angles – When a third line called the transversal crosses two other (usually parallel) lines, angles are formed on the inside, or interior, of the two lines. The angles that are opposite of each other are the alternate interior angles. ## What is alternate interior and exterior angles? Angles that are in the area between the parallel lines like angle 2 and 8 above are called interior angles whereas the angles that are on the outside of the two parallel lines like 1 and 6 are called exterior angles. Angles that are on the opposite sides of the transversal are called alternate angles e.g. 1 + 8. ## What is the sum of alternate interior angle? These angles are congruent. Sum of the angles formed on the same side of the transversal which are inside the two parallel lines is always equal to 180°. In the case of non – parallel lines, alternate interior angles don’t have any specific properties. ## How do you describe alternate angles? Alternate angles are angles that are in opposite positions relative to a transversal intersecting two lines. If the alternate angles are between the two lines intersected by the transversal, they are called alternate interior angles. ## What are non congruent alternate interior angles? Angle x and angle y are alternate interior angles. … These two angles are not congruent becaue line A and B are not parallel. ## What do you mean by co interior angles? Co-interior angles lie between two lines and on the same side of a transversal. In each diagram the two marked angles are called co-interior angles. If the two lines are parallel, then co-interior angles add to give 180o and so are supplementary.
# Sinusoidal Functions as Harmonic Oscillations maths Published December 10, 2022 The sinusoidal functions (like sin and cosine) normally first come up when students are learning about angles of right angled triangles. However there are many other ways to define them; as projections of points on a circle, as the real and imaginary parts of the mapping $$f(x) = e^{ix}$$, in terms of addition law, as a power series or as inverse functions of the anti-derivatives of $$f(x) = \frac{1}{\sqrt{1-x^2}}$$. A physically motivated way is through the Simple Harmonic Oscillator and differential equations. Consider an object at rest that stays near its initial state after a small perturbation, like a ball at the bottom of a hill given a small kick, a swing hanging straight down given a small push, or a chair given a small push on its back. Representing the potential energy at $$x$$ by $$V(x)$$ then by Newton’s equations of motion $$m\frac{{\rm d}^2x}{{\rm d}t^2} = -\frac{{\rm d}V}{{\rm d}{x}}$$. Since the object isn’t initially accelerating the first derivative of V must be zero, and since a small perturbation accelerates the object back towards its initial state the second derivative of V must be non-negative. So locally $$V(x) \approx c + k^2 x^2 + O(x^3)$$ where $$c$$ and $$k$$ are arbitrary constants. Under this approximation then the equations of motion are $$m\frac{{\rm d}^2x}{{\rm d}t^2} = -k^2 x$$. Without loss of generality assume $$m = 1$$, then we are trying to solve the equation $x''(t) = -k^2 x(t)$ for some arbitrary $$k$$. We can find a power series solution to this equation; setting $x(t) = \sum_{n=0}^{\infty} a_n \frac{t^n}{n!}$ then a change of variables gives $x'(t) = \sum_{n=0}^{\infty} a_{n} n \frac{t^{n-1}}{n!} = \sum_{n=0}^{\infty} a_{n+1} \frac{t^n}{n!}$ and similarly $x''(t) = \sum_{n=0}^{\infty} a_{n+2} \frac{t^n}{n!}$ for all values of t. Then the equation becomes $\sum_{n=0}^{\infty} (a_{n+2} + k^2 a_{n}) \frac{t^n}{n!} = 0$ and since this is true for all values of $$t$$ then it must hold for each coefficient $$n$$, $$a_{n+2} = -k^2 a_n$$. Thus $$a_{2n} = \left(-k^2\right)^{2n} a_0$$ and $$a_{2n+1} = \left(-k^2\right)^{2n} a_1$$. A set of solutions is thus $x(t) = a_0 \left(\sum_{n=0}^{\infty} (-1)^n \frac{(kt)^{2n}}{(2n)!} \right) + a_1 \left( \sum_{n=0}^{\infty} (-1)^n \frac{(kt)^{2n+1}}{(2n+1)!}\right)$ for some arbitrary constants $$a_0$$ and $$a_1$$. We then define $$\cos(t) = \sum_{n=0}^{\infty} (-1)^n \frac{t^{2n}}{(2n)!}$$ and $$\sin(t) = \sum_{n=0}^{\infty} (-1)^n \frac{t^{2n+1}}{(2n+1)!}$$ making the solution $$x(t) = a_0 \cos(kt) + a_1 \sin(kt)$$. Now we have defined sin and cos as solutions of the Simple Harmonic Oscillator, we need to discover some of their properties in order to understand the oscillator. First note that the power series converges for all $$t$$ by the ratio test, so the functions are well defined on the whole real line. Note from the power series that $$\sin(0) = 0$$ and $$\cos(0) = 1$$, and that $$\sin(-t) = \sin(t)$$ and $$\cos(-t) = \cos(t)$$ (that is $$\sin$$ is an odd function and $$\cos$$ is an even function). Taking derivatives term-wise in the power series immediately shows $$\cos'(t) = -\sin(t)$$ and $$\sin'(t) = \cos(t)$$ (which are consistent with our original differential equation). The most important property is the addition formula, which I can’t find a good motivation for. However if you consider the start of the expansion of \begin{align} \sin(s)\cos(t) &\approx \left(s - \frac{s^3}{3!} + \frac{s^5}{5!} - O(s^7)\ldots \right) \left(1 - \frac{t^2}{2!} + \frac{t^4}{4!} + O(t^6)\right) \\ &\approx s - \frac{st^2}{2!} - \frac{s^3}{3!} + \frac{st^4}{4!} + \frac{s^5}{5!} + \frac{s^3t^2}{2!3!} + O(s^k t^{7-k}) \end{align} then it follows that \begin{align} \sin(s)\cos(t) + \cos(s)\sin(t) \approx& s + t - \frac{s^3 + 3 s t^2 + 3 s^2 t + t^3}{3!} + \\ &\frac{s^5 + 5st^4 + 5 s^4 t + 10 s^3 t^2 + 10 s^2 t^3}{5!} + O(s^k t^{7-k}) \\ \approx& (s + t) - \frac{(s+t)^3}{3!} + \frac{(s+t)^5}{5!} + O(s^k t^{7-k}) \\ \approx& \sin(s+t) \end{align} for small $$s$$ and $$t$$. A more rigorous computation of the power series (rearranging the sums into groups of equal sum of power) can show that $\sin(s+t) = \sin(s) \cos(t) + \cos(s) \sin(t)$ for all $$s$$ and $$t$$. Taking the derivative of the sin addition formula with respect to $$s$$ gives the cosine addition formula $\cos(s+t) = \cos(s) \cos(t) - \sin(s) \sin(t)$ for all $$s$$ and $$t$$. Setting $$s=-t$$ gives $1 = \cos^2(t) + \sin^2(t)$ for all $$t$$. Now there is almost enough to show the periodicity of cos and sin; that is the Simple Harmonic Oscillator will actually oscillate back and forth from its original position (ignoring frictional forces). Assume that $$\cos(t)$$ has a zero, and define $$\pi$$ as twice the smallest positive zero; $$\cos(\pi/2) = 0$$. Since $$\sin^2(t) = 1 - \cos^2(t)$$, $$\sin^2(\pi/2) = 1$$, and as $$\sin(t)$$ is 0 and increasing at the origin, and this is the first zero of its derivative $$\cos$$, then the value must be positive; $$\sin(\pi/2) = 1$$. The $$\sin(t+\pi/2) = \sin(\pi/2) \cos(t) + \cos(\pi/2) \sin(t) = \cos(t)$$ and $$\cos(t+\pi/2) = \cos(\pi/2) \cos(t) - \sin(\pi/2) \sin(t) = - \sin(t)$$. Then it follows $$\sin(t + \pi) = \cos(t+\pi/2) = -\sin(t)$$ and so $$\sin(t+2\pi) = \sin(t)$$. Similarly \$(t+) = -(t+/2) = -(t) and so $$\cos(t+2\pi) = \sin(t)$$, and so both functions are periodic with period $$2 \pi$$ (the period can’t be less than $$2 \pi$$ by the translation formula and the definition of $$\pi/2$$ as the first positive zero of $$\cos$$). However we need to actually prove that $$\cos$$ has a zero somewhere. Consider $\cos(2) \approx 1 - \frac{2^2}{2!} + \ldots = -1 + \ldots$ where $$\ldots$$ are the higher order terms $$(-1)^{n}\frac{2^{2n}}{2n!}$$ for $$n=2,3,4,\ldots$$. Note that these are less than 1, alternate in sign, and get smaller since $$(2n)! > 2^{2n}$$ for $$n>=2$$, and consequently none of them will change the sign of the result. Thus $$\cos(2) < 0$$, and since $$\cos$$ is continuous it must have a zero between 0 and 2. Hence by our previous arguments the functions are periodic. From here it is straightforward to prove other properties of trigonometric and inverse trigonometric functions. There is nothing novel here but this gives a different axiomitisation of the trigonometric functions. It would be interesting to understand how the properties of the trigonometric functions are related to the symmetry of the classical Simple Harmonic Oscillator, analogous to the symmetries of the Quantum Simple Harmonic Oscillator.
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ 11.7E: Exercises for Cylindrical and Spherical Coordinates $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ Use the following figure as an aid in identifying the relationship between the rectangular, cylindrical, and spherical coordinate systems. For exercises 1 - 4, the cylindrical coordinates $$(r,θ,z)$$ of a point are given. Find the rectangular coordinates $$(x,y,z)$$ of the point. 1) $$\left(4,\frac{π}{6},3\right)$$ $$(2\sqrt{3},2,3)$$ 2) $$\left(3,\frac{π}{3},5\right)$$ 3) $$\left(4,\frac{7π}{6},3\right)$$ $$(−2\sqrt{3},−2,3)$$ 4) $$(2,π,−4)$$ For exercises 5 - 8, the rectangular coordinates $$(x,y,z)$$ of a point are given. Find the cylindrical coordinates $$(r,θ,z)$$of the point. 5) $$(1,\sqrt{3},2)$$ $$\left(2,\frac{π}{3},2\right)$$ 6) $$(1,1,5)$$ 7) $$(3,−3,7)$$ $$\left(3\sqrt{2},−\frac{π}{4},7\right)$$ 8) $$(−2\sqrt{2},2\sqrt{2},4)$$ For exercises 9 - 16, the equation of a surface in cylindrical coordinates is given. Find the equation of the surface in rectangular coordinates. Identify and graph the surface. 9) [T] $$r=4$$ A cylinder of equation $$x^2+y^2=16,$$ with its center at the origin and rulings parallel to the $$z$$-axis, 10) [T] $$z=r^2\cos^2θ$$ 11) [T] $$r^2\cos(2θ)+z^2+1=0$$ Hyperboloid of two sheets of equation $$−x^2+y^2−z^2=1,$$ with the $$y$$-axis as the axis of symmetry, 12) [T] $$r=3\sin θ$$ 13) [T] $$r=2\cos θ$$ Cylinder of equation $$x^2−2x+y^2=0,$$ with a center at $$(1,0,0)$$ and radius $$1$$, with rulings parallel to the $$z$$-axis, 14) [T] $$r^2+z^2=5$$ 15) [T] $$r=2\sec θ$$ Plane of equation $$x=2,$$ 16) [T] $$r=3\csc θ$$ For exercises 17 - 22, the equation of a surface in rectangular coordinates is given. Find the equation of the surface in cylindrical coordinates. 17) $$z=3$$ $$z=3$$ 18) $$x=6$$ 19) $$x^2+y^2+z^2=9$$ $$r^2+z^2=9$$ 20) $$y=2x^2$$ 21) $$x^2+y^2−16x=0$$ $$r=16\cos θ,\quad r=0$$ 22) $$x^2+y^2−3\sqrt{x^2+y^2}+2=0$$ For exercises 23 - 26, the spherical coordinates $$(ρ,θ,φ)$$ of a point are given. Find the rectangular coordinates $$(x,y,z)$$ of the point. 23) $$(3,0,π)$$ $$(0,0,−3)$$ 24) $$\left(1,\frac{π}{6},\frac{π}{6}\right)$$ 25) $$\left(12,−\frac{π}{4},\frac{π}{4}\right)$$ $$(6,−6,\sqrt{2})$$ 26) $$\left(3,\frac{π}{4},\frac{π}{6}\right)$$ For exercises 27 - 30, the rectangular coordinates $$(x,y,z)$$ of a point are given. Find the spherical coordinates $$(ρ,θ,φ)$$ of the point. Express the measure of the angles in degrees rounded to the nearest integer. 27) $$(4,0,0)$$ $$(4,0,90°)$$ 28) $$(−1,2,1)$$ 29) $$(0,3,0)$$ $$(3,90°,90°)$$ 30) $$(−2,2\sqrt{3},4)$$ For exercises 31 - 36, the equation of a surface in spherical coordinates is given. Find the equation of the surface in rectangular coordinates. Identify and graph the surface. 31) [T] $$ρ=3$$ Sphere of equation $$x^2+y^2+z^2=9$$ centered at the origin with radius $$3$$, 32) [T] $$φ=\frac{π}{3}$$ 33) [T] $$ρ=2\cos φ$$ Sphere of equation $$x^2+y^2+(z−1)^2=1$$ centered at $$(0,0,1)$$ with radius $$1$$, 34) [T] $$ρ=4\csc φ$$ 35) [T] $$φ=\frac{π}{2}$$ The $$xy$$-plane of equation $$z=0,$$ 36) [T] $$ρ=6\csc φ\sec θ$$ For exercises 37 - 40, the equation of a surface in rectangular coordinates is given. Find the equation of the surface in spherical coordinates. Identify the surface. 37) $$x^2+y^2−3z^2=0, \quad z≠0$$ $$φ=\frac{π}{3}$$ or $$φ=\frac{2π}{3};$$ Elliptic cone 38) $$x^2+y^2+z^2−4z=0$$ 39) $$z=6$$ $$ρ\cos φ=6;$$ Plane at $$z=6$$ 40) $$x^2+y^2=9$$ For exercises 41 - 44, the cylindrical coordinates of a point are given. Find its associated spherical coordinates, with the measure of the angle φ in radians rounded to four decimal places. 41) [T] $$\left(1,\frac{π}{4},3\right)$$ $$\left(\sqrt{10},\frac{π}{4},0.3218\right)$$ 42) [T] $$(5,π,12)$$ 43) $$\left(3,\frac{π}{2},3\right)$$ $$(3\sqrt{2},\frac{π}{2},\frac{π}{4})$$ 44) $$\left(3,−\frac{π}{6},3\right)$$ For exercises 45 - 48, the spherical coordinates of a point are given. Find its associated cylindrical coordinates. 45) $$\left(2,−\frac{π}{4},\frac{π}{2}\right)$$ $$\left(2,−\frac{π}{4},0\right)$$ 46) $$\left(4,\frac{π}{4},\frac{π}{6}\right)$$ 47) $$\left(8,\frac{π}{3},\frac{π}{2}\right)$$ $$\left(8,\frac{π}{3},0\right)$$ 48) $$\left(9,−\frac{π}{6},\frac{π}{3}\right)$$ For exercises 49 - 52, find the most suitable system of coordinates to describe the solids. 49) The solid situated in the first octant with a vertex at the origin and enclosed by a cube of edge length $$a$$, where $$a>0$$ Cartesian system, $$\big\{(x,y,z)\,\|\,0≤x≤a,\;0≤y≤a,\;0≤z≤a\big\}$$ 50) A spherical shell determined by the region between two concentric spheres centered at the origin, of radii of $$a$$ and $$b$$, respectively, where $$b>a>0$$ 51) A solid inside sphere $$x^2+y^2+z^2=9$$ and outside cylinder $$\left(x−\frac{3}{2}\right)^2+y^2=\frac{9}{4}$$ Cylindrical system, $$\big\{(r,θ,z)\,\|\,r^2+z^2≤9,\;r≥3\cos θ,\;0≤θ≤2π\big\}$$ 52) A cylindrical shell of height $$10$$ determined by the region between two cylinders with the same center, parallel rulings, and radii of $$2$$ and $$5$$, respectively 53) [T] Use a CAS or CalcPlot3D to graph in cylindrical coordinates the region between elliptic paraboloid $$z=x^2+y^2$$ and cone $$x^2+y^2−z^2=0.$$ The region is described by the set of points $$\big\{(r,θ,z)\,\|\,0≤r≤1,\;0≤θ≤2π,\;r^2≤z≤r\big\}.$$ 54) [T] Use a CAS or CalcPlot3D to graph in spherical coordinates the “ice cream-cone region” situated above the xy-plane between sphere $$x^2+y^2+z^2=4$$ and elliptical cone $$x^2+y^2−z^2=0.$$ 55) Washington, DC, is located at $$39°$$ N and $$77°$$ W (see the following figure). Assume the radius of Earth is $$4000$$ mi. Express the location of Washington, DC, in spherical coordinates. $$(4000,−77°,51°)$$ 56) San Francisco is located at $$37.78°N$$ and $$122.42°W.$$ Assume the radius of Earth is $$4000$$mi. Express the location of San Francisco in spherical coordinates. 57) Find the latitude and longitude of Rio de Janeiro if its spherical coordinates are $$(4000,−43.17°,102.91°).$$ $$43.17°W, 22.91°S$$ 58) Find the latitude and longitude of Berlin if its spherical coordinates are $$(4000,13.38°,37.48°).$$ 59) [T] Consider the torus of equation $$\big(x^2+y^2+z^2+R^2−r^2\big)^2=4R^2(x^2+y^2),$$ where $$R≥r>0.$$ a. Write the equation of the torus in spherical coordinates. b. If $$R=r,$$ the surface is called a horn torus. Show that the equation of a horn torus in spherical coordinates is $$ρ=2R\sin φ.$$ c. Use a CAS or CalcPlot3D to graph the horn torus with $$R=r=2$$ in spherical coordinates. a. $$ρ=0, \quad ρ+R^2−r^2−2R\sin φ=0$$ c. 60) [T] The “bumpy sphere” with an equation in spherical coordinates is $$ρ=a+b\cos(mθ)\sin(nφ)$$, with $$θ∈[0,2π]$$ and $$φ∈[0,π]$$, where $$a$$ and $$b$$ are positive numbers and $$m$$ and $$n$$ are positive integers, may be used in applied mathematics to model tumor growth. a. Show that the “bumpy sphere” is contained inside a sphere of equation $$ρ=a+b.$$ Find the values of $$θ$$ and $$φ$$ at which the two surfaces intersect. b. Use a CAS or CalcPlot3D to graph the surface for $$a=14, b=2, m=4,$$ and $$n=6$$ along with sphere $$ρ=a+b.$$ c. Find the equation of the intersection curve of the surface at b. with the cone $$φ=\frac{π}{12}$$. Graph the intersection curve in the plane of intersection.
1. ## Elimination and substitution I have this problem: y=1x and y=.5x+4 I got (2,4) and was wondering if you guys can show me the steps for both elimination and substitution. Also how would I graph this? just put one dot on the graph? 2. Originally Posted by Jubbly I have this problem: y=1x and y=.5x+4 I got (2,4) and was wondering if you guys can show me the steps for both elimination and substitution. Also how would I graph this? just put one dot on the graph? How did you get $\displaystyle (2,4)?$ If you try checking your work by substituting, you get $\displaystyle x=2\Rightarrow y=2$ for the first, and $\displaystyle x=2\Rightarrow y=5$ for the second. Your point is not on either of the curves, so it certainly could not be in the solution set. 3. Well, I did 1x=.5x+4 I subtracted .5x from both sides making 1x and .5x now its .5x=4 divided .5x into 4 and got 2 so x=2? Can you show me a step by step of how your doing it and how your substituting? 4. Originally Posted by Jubbly divided .5x into 4 and got 2 4 divided by 0.5 is 8, not 2: $\displaystyle \frac4{1/2}=4\cdot\frac21=4\cdot2=8\text.$ The correct solution is $\displaystyle (8,\,8)\text.$ If you don't mind can you show me the elimination and substitution equation you used for this. Thanks! 6. Certainly. We have $\displaystyle \left\{\begin{array}{rcl} y&=&x\\ y&=&\frac12x+4 \end{array}\right.$ Substitution: Substituting $\displaystyle x=y$ into the second equation produces $\displaystyle y=\frac12y+4,$ and solving for $\displaystyle y$ gives $\displaystyle \frac12y=4\Rightarrow y=8.$ Back-substituting, we get $\displaystyle x=8.$ Elimination: Let's first rearrange the equations a little, $\displaystyle \left\{\begin{array}{rcl} y-x&=&0\\ 2y-x&=&8 \end{array}\right..$ Subtract the first equation from the second, $\displaystyle \left\{\begin{array}{rcl} y-x&=&0\\ y&=&8 \end{array}\right.$ and subtract the second equation from the first: $\displaystyle \left\{\begin{array}{rcl} -x&=&-8\\ y&=&8 \end{array}\right.\Rightarrow\left\{\begin{array}{ rcl} x&=&8\\ y&=&8 \end{array}\right.$
算法推导代写 Introduction to Algorithms代写 - 算法代写, 英国代写 # 算法推导代写 Introduction to Algorithms代写 2021-11-15 15:15 星期一所属：算法代写浏览：35 ## Introduction to Algorithms The questions are drawn from Chapters 1 and 2 of the text, and from the introductory lectures in class on algorithm design for the Maximum Sum Subarray Problem. The homework is worth a total of 100 points. For questions that ask you to design an algorithm, you should: (i) describe an algorithm for the problem using prose and pictures, (ii) argue that your algorithm is correct, and (iii) analyze the running time of your algorithm. Pseudocode is not required, but do clearly explain the ideas behind your algorithm. As a general rule, only give high-level pseudocode if it aids the time analysis of your algorithm, or if it is needed to clarify the steps. Remember to (a) start each problem on a new page, and (b) put your answers in the correct order. If you can’t solve a problem, state this, and write only what you know to be correct. Neatness and conciseness counts. ### (1) (Best-case time) (10 points) 算法推导代写 Show how to take nearly any algorithm (even a poor one), and modify it so that it has good best-case time. To answer this question, describe a procedure that (1) takes as input a problem statement P together with an algorithm A that solves P, and (2) outputs a new algorithm A’ for P whose best-case running time is “as good as possible.” An intelligent human being (such as yourself) should be able to carry out your procedure. (Note: The original algorithm A might be very inefficient. So what does this tell you about best-case time?) ### (2) (Counting flflips) (20 points) 算法推导代写 A flflip in an array A[1:n] of numbers is a pair of indices (i, j) such that i < j and A[i] > A[j]. In other words, a flip is a pair of elements that are not in sorted order. (a) (10 points) Find an array of length n over the elements {1, . . . , n} that maximizes the number of flflips, and prove that it is optimal. (b) (10 points) Suppose you run insertion sort on an array of length n that has k flips. Derive a tight big-O-bound on the running time of insertion sort as a function of both n and k, and explain your analysis. (Note: Recall that insertion sort processes increasingly longer prefixes of the input array. After having sorted prefix A[1:i], it sorts prefix A[1:i+1] by inserting element A[i+1] into its correct sorted position in A[1:i].) (c) (bonus) (10 points) By modifying merge sort, design an algorithm that counts the number of flflips in an array of length n in O(n log n) time. (Note: There can be Θ( ) flflips in an n-element array, so your algorithm cannot explicitly list them and achieve the time bound. It is possible to count flflips without listing them.) ### (3) (2-D maximum-sum subarray) (30 points) 算法推导代写 In the 2-D Maximum-Sum Subarray Problem, you are given a two-dimensional m × n array A[1 : m, 1 : n] of positive and negative numbers, and you are asked to fifind a subarray A[a : b, c : d], where 1 a b m and 1 c d n, such that the sum of its elements, ∑a ≤ i ≤ b ∑ c ≤ j ≤ d A[i, j], is maximum. (a) (30 points) Using exhaustive search, design an algorithm that runs in O() time using O(min{m, n}) working space. The working space of an algorithm is the memory it uses beyond what is needed to store the input. (Hint: First design a straightforward algorithm that achieves O( ) time, and then optimize it.) (b) (bonus) (10 points) Using divide-and-conquer, design an algorithm that runs in O(n log n) time using O(n) working space, by dividing the array vertically. (Hint: You may need to know that the recurrence T(m, n) = 2 T(m, n/2) + O(n) has the solution T(m, n) = O(n log n).) ### (4) (Minimum positive-sum subarray) (40 points) 算法推导代写 In this variation of the Maximum-Sum Subarray Problem, you are given a one-dimensional array A[1 : n] of positive and negative numbers, and you are asked to fifind a subarray A[i: j] such that the sum of its elements is both: (1) strictly greater than zero, and (2) minimum. In other words, you want to fifind a subarray of smallest positive sum. (a) (40 points) Using the divide-and-conquer strategy, design an algorithm that fifinds a minimum positive-sum subarray in an array of length n in O(n log² n) time. (Hint: Recall that n numbers can be sorted in O(n log n) worst-case time. You may also need to know that the recurrence T(n) = 2 T(n/2) + O(n log n) has the solution T(n) = O(n log² n).) (b) (bonus) (10 points) Using an incremental strategy, now design an algorithm that fifinds a minimum positive-sum subarray in O(n log n) time. The incremental strategy proceeds by solving a subproblem of size k, adding one element, and updating the solution to solve a problem of size k+1. (Hint: You may fifind a balanced search tree useful.) Bonus questions (like Problems 2(c), 3(b), and 4(b) above) are not required. Bonus points are tallied separately, and are considered at the end of the semester for students who fall below the cutoffff for a letter grade.
Home » Free Class Notes for CBSE Board » Class 7 Notes for CBSE » CBSE Class 7th Maths Notes for Download » CBSE Class 7 Maths Notes for Chapter 2 Fractions and Decimals # CBSE Class 7 Maths Notes for Chapter 2 Fractions and Decimals CHAPTER 2 FRACTIONS AND DECIMALS MULTIPLICATION OF FRACTIONS • The product of two proper fractions is less than both the fractions. • The product of two improper fractions is greater than each of the fraction. DIVISION OF FRACTIONS MULTIPLICATION OF DECIMALS • While multiplying two decimal numbers, first multiply them as whole numbers. Count the number of digits to the right of the decimal point in both the decimal numbers. Add the number of digits counted. Put the decimal point in the product by counting the digits from its rightmost place. The count should be the sum obtained earlier. For example, 0.5 × 0.7 = 0.35 • To multiply a decimal number by 10, 100 or 1000, move the decimal point in the number to the right by as many places as there are zeros over 1. For example, 0.53 × 10 = 5.3, 0.53 × 100 = 53, 0.53 × 1000 = 530 DIVISION OF DECIMALS • To divide a decimal number by a whole number, first divide them as whole numbers. Then place the decimal point in the quotient as in the decimal number. For example, 8.4 ÷ 4 = 2.1 • To divide a decimal number by 10, 100 or 1000, shift the digits in the decimal number to the left by as many places as there are zeros over 1, to get the quotient. E.g. 23.9 ÷ 10 = 2.39, 23.9 ÷ 100 = 0 .239 • While dividing two decimal numbers, first shift the decimal point to the right by equal number of places in both, to convert the divisor to a whole number. Then divide. For example, 2.4 ÷ 0.2 = 24 ÷ 2 = 12.
# Domain A set of elements for which a function has been defined is called the domain of a function. Every function is defined for some elements by giving their associated values. The collection of such elements as a group is known as domain of the function. Each function has a domain and it helps us to understand the list of values for which a function is defined mathematically. ## Introduction Consider a function and assume it is defined for the elements $a$, $b$, $c$, $d$ and $e$. Take $x$ as a literal and it represents all these elements. Assume, the function is expressed as $f(x)$ in mathematics. Take $x = a$, then the value of the function is $f(a)$ Take $x = b$, then the value of the function is $f(b)$ Take $x = c$, then the value of the function is $f(c)$ Take $x = d$, then the value of the function is $f(d)$ Take $x = e$, then the value of the function is $f(e)$ The values of function are $f(a)$, $f(b)$, $f(c)$, $f(d)$ and $f(e)$ for $x$ is equal to $a$, $b$, $c$, $d$ and $e$ respectively. The collection of the elements $a$, $b$, $c$, $d$ and $e$ is called the domain of the function. $x = \{a, b, c, d, e\}$ ### Example $f(x) = \dfrac{1}{x+2}$ Substitute all the real numbers to determine the domain of this function. For example, Take $x = 0$ and evaluate the function. $x = 0$ $\implies$ $f(0) = \dfrac{1}{0+2} = \dfrac{1}{2}$ For $x$ is equal to zero, the value of the function is $\dfrac{1}{2}$. Now, substitute positive real numbers and obtain their associated values. $x = 1$ $\implies$ $f(1) = \dfrac{1}{1+2} = \dfrac{1}{3}$ $x = 2$ $\implies$ $f(2) = \dfrac{1}{2+2} = \dfrac{1}{4}$ $x = 3$ $\implies$ $f(3) = \dfrac{1}{3+2} = \dfrac{1}{5}$ $x = 4$ $\implies$ $f(4) = \dfrac{1}{4+2} = \dfrac{1}{6}$ $x = 5$ $\implies$ $f(5) = \dfrac{1}{5+2} = \dfrac{1}{7}$ $\vdots$ The function is defined for all the positive real numbers. Now, substitute all the negative real numbers one by one and test the functionality of this function. $x = -1$ $\implies$ $f(-1) = \dfrac{1}{-1+2} = \dfrac{1}{1} = 1$ $x = -2$ $\implies$ $f(-2) = \dfrac{1}{-2+2} = \dfrac{1}{0} = \infty$ For the value $x$ is equal to $-2$, the example function becomes undefined. Hence, the value of $x$ should not be equal to $-2$. Continue testing the function by substituting the remaining negative real numbers. $x = -3$ $\implies$ $f(-3) = \dfrac{1}{-3+2} = \dfrac{1}{-1} = -1$ $x = -4$ $\implies$ $f(-4) = \dfrac{1}{-4+2} = \dfrac{1}{-2} = -\dfrac{1}{2}$ $x = -5$ $\implies$ $f(-5) = \dfrac{1}{-5+2} = \dfrac{1}{-3} = -\dfrac{1}{3}$ $\vdots$ Now, collect all the elements as a set but ignore $x$ is equal to $-2$. Therefore, the collection of elements is called the domain of this function. $x = \{\cdots -4, -3, -1, 0, 1, 2, 3, \cdots\}$ According to the set theory, the real numbers group is simplify denoted by $R$ but it contains $-2$. Hence, it should be subtracted from it. Therefore, the domain of the function is simplify written as $R-\{0\}$.
#### Explain solution RD Sharma class 12 chapter 12 Derivative as a Rate Measurer exercise multiple choise question 8 maths $(d)\; \; 3,\; \; \frac{1}{3}$ Hint: Here, we use the basic concept of algebra. Given: $x^{3}-5x^{2}+5x+8$ Solution: $p(x)=x^{3}-5x^{2}+5x+8 \quad\quad\quad\quad......(i)$ → Differentiating (i) with respect to t,we get \begin{aligned} &\frac{d p(x)}{d t}=\left(x^{3}-5 x^{2}+5 x+8\right) \frac{d x}{d t}=2 \frac{d x}{d t}\\ &2 \frac{d x}{d t}=\left(3 x^{2}-10 x+5\right) \frac{d x}{d t}\\ &=3 x^{2}-10 x+5=2\\ &=3 x^{2}-10 x+3=0\\ &\rightarrow \text { Factorizing the above quadratic equation, }\\ &\text { we get }(3 x-1)(x-3)=0\\ &\Rightarrow x=\frac{1}{3} \text { and } x=3 \end{aligned}
Top # Linear Inequalities Two algebraic expressions or two real numbers related by the symbol ‘<’, ‘>’, ‘<=’ or ‘>=’ form an linear inequality. Examples of numerical inequalities: 3 < 5; 7 > 5 x < 5; Examples of literal inequalities: y > 2; x <= 3, y >= 4. 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. How to solve the linear inequality: Added or subtracted by the same number on both side of the inequality. Multiplied or divided by the same number on both side of the inequality but if we divide or multiply by a negative number, we must reverse the inequality sign. Related Calculators Graph Linear Inequalities Calculator Graph Linear Inequality Calculator Linear Inequalities Calculator Calculator for Inequalities ## Worked Examples for Linear Inequalities : Example problem 1: Solve the linear inequality: 5x – 3 < 7 Solution: 5x – 3 < 7 Add 3 on both side of the equation 5x - 3 + 3 < 7 + 3 5x < 10 Divide by 5 on both side of the equation 5x / 5 < 10 / 5 x<2 So, the solution is (-infinity, 2). Example problem 2: Solve the linear inequality: 7x + 3 < 5x + 9 Solution: 7x + 3 < 5x + 9 Subtract 3 to the both side of the equation 7x + 3 - 3 < 5x + 9 -3 7x < 5x + 6 Subtract 5x on both sides of the equation 7x - 5x < 5x + 6 - 5x 2x<6 Divide by 2 on both sides of the equation 2x / 2 < 6 / 2 x < 3 So, the solution is (-infinity, 3). Example problem 3: Solve the linear inequality: 5x – 3 < 3x +1 Solution: 5x – 3 < 3x +1 Add 3 to the both side of the equation 5x – 3+3 < 3x +1+3 5x < 3x + 4 Subtract 3x on both sides of the equation 5x - 3x < 4 2x < 4 Divide by 2 on both side of the equation 2x / 2 < 4 / 2 x<2 So, the Inequality range is (-infinity, 2) Example problem 4: Solve the linear inequality: 37 – (3x + 5) > 9x – 8 (x – 3) Solution: 37 – (3x + 5) > 9x – 8 (x – 3) 37- 3x - 5 > 9x - 8x + 24 -3x + 32 > 1x + 24 Subtract x on both sides of the equation -3x+32- x > 1x + 24 - x -4x +32 > 24 Subtract 32 on both sides of the equation -4x + 32-32 > 24 - 32 -4x > -8 Divide by -4 on both sides of the equation and change the inequality sign -4x / -4 < -8 / -4 x <2 So, the solution is (-infinity, 2) More topics in Linear Inequalities Solving Linear Inequalities Graphing Linear Inequalities *AP and SAT are registered trademarks of the College Board.
# 2019 AMC 12B Problems/Problem 13 The following problem is from both the 2019 AMC 10B #17 and 2019 AMC 12B #13, so both problems redirect to this page. ## Problem A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $k$ is $2^{-k}$ for $k = 1,2,3....$ What is the probability that the red ball is tossed into a higher-numbered bin than the green ball? $\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}$ ## Solution 1 The probability of the red ball landing in a higher-numbered bin is the same as the probability of the green ball landing in a higher numbered bin. The probability of both landing in the same bin is $\sum_{k=1}^{\infty}2^{-2k}$. The sum is equal to $\frac{1}{3}$. Therefore the other two probabilities have to both be $\textbf{(C) } \frac{1}{3}$. $Q.E.D\blacksquare$ Solution by a1b2 ## Solution Variant We solve for the probability by doing $\frac{1-(\text{Probability of Equality})}{2}$. We see that the probability of equality is the summation of all the probabilities that the balls land in the same container. Thus we have the probability of equality being equal to $(\frac{1}{2})(\frac{1}{2})+(\frac{1}{4})(\frac{1}{4})+(\frac{1}{16})(\frac{1}{16})...$ The summation of this expression is equal to $$\sum_{n=0}^{\infty} (1)(\frac{1}{4})^{n}-1$$. Using the geometric sum formula, we obtain the summation of this expression to be $\frac{1}{\frac{3}{4}}-1$ or $\frac{1}{3}$. ## Solution 2 (variant) Suppose the green ball goes in bin $i$, for some $i \ge 1$. The probability of this occurring is $\frac{1}{2^i}$. Given this occurs, the probability that the red ball goes in a higher-numbered bin is $\frac{1}{2^{i+1}} + \frac{1}{2^{i+2}} + \ldots = \frac{1}{2^i}$. Thus the probability that the green ball goes in bin $i$, and the red ball goes in a bin greater than $i$, is $\left(\frac{1}{2^i}\right)^2 = \frac{1}{4^i}$. Summing from $i=1$ to infinity, we get $$\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}$$ (Note: to find this sum, we use the formula $\sum_{i=1}^{\infty} r^i = \frac{r}{1-r}$. Since in this case $r = \frac{1}{4}$, the answer is $\frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$. If you don't know this formula, you may instead note that if you multiply the sum by $4$, it is equivalent to adding $1$. Thus: $4n = n+1$, which clearly simplifies to $n = \frac{1}{3}$. - scrabbler94 (explanation of infinite sum provided by Robin) ## Solution 3 (infinite geometric series) The probability that the two balls will go into adjacent bins is $\frac{1}{2\times4} + \frac{1}{4\times8} + \frac{1}{8 \times 16} + ... = \frac{1}{8} + \frac{1}{32} + \frac{1}{128} = \frac{1}{6}$. The probability that the two balls will go into bins that have a distance of 2 from each other is $\frac{1}{2 \times 8} + \frac{1}{4 \times 16} + \frac{1}{8 \times 32} = \frac{1}{16} + \frac{1}{64} + \frac{1}{256} = \frac{1}{12}$. We can see that each time we add a bin between the two balls, the probability halves. Thus, our answer is $\frac{1}{6} + \frac{1}{12} + \frac{1}{24} + ...$, which converges into $\frac{1}{3}$. ## Solution 4 (quickest) Define a win as a ball appearing in higher numbered box. Start from the first box. There are 4 possible results in the box: Red, Green, Red and Green, none, with probability of $\frac{1}{4}$ for each. Red win, Green win, Tie all have the same probability of $\frac{1}{3}$. If none of the balls is in the first box, the game restarts at the second box with the same kind of probability distribution. So finally, Red win, Green win and Tie all have a probability of $\frac{1}{3}$ The answer is $\boxed{C}$ -- Solution by mathsuper(丹神)
In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP! # Ex.2.5 Q8 Polynomials Solution - NCERT Maths Class 9 Go back to 'Ex.2.5' ## Question Factorise each of the following: (i) \begin{align}8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}\end{align} (ii) \begin{align}8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}\end{align} (iii) \begin{align}27-125 a^{3}-135 a+225 a^{2}\end{align} (iv) \begin{align}64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}\end{align} (v) \begin{align}27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p\end{align} Video Solution Polynomials Ex 2.5 \| Question 8 ## Text Solution Reasoning: Identities: \begin{align}&(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y) \\ &{(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)}\end{align} Steps: (i) \begin{align}8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}\end{align} This can be re-written as: \begin{align}(2 a)^{3}+(b)^{3}+3(2 a)^{2}(b)+3(2 a)(b)^{2}\end{align} Which is of the form: \begin{align}x^{3}+y^{3}+3 x y(x+y)=(x+y)^{3}\end{align} Hence, \begin{align}8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}=(2 a+b)^{3}\end{align} (ii) \begin{align}8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}\end{align} This can be re-written as: \begin{align}(2 a)^{3}-(b)^{3}-3(2 a)^{2}(b)+3(2 a)(b)^{2}\end{align} Which is of the form: \begin{align}x^{3}-y^{3}-3 x^{2} y+3 x y^{2}=(x-y)^{3}\end{align} Hence, \begin{align}8 a^{3}\!-\!b^{3}\!-\!12 a^{2} b\!+\!6 a b^{2}\!=\!(2 a\!-\!b)^{3}\end{align} (iii)\begin{align}27-125 a^{3}-135 a+225 a^{2}\end{align} This can be re-witten as: \begin{align}&(3)^{3}-(5 a)^{3}-3(3)^{2}(5 a)+3(3)(5 a)^{2} \\&(3)^{3}-(5 a)^{3}-3(3)(5 a)(3-5 a) & \end{align} Which is of the form: \begin{align} x^{3}-y^{3}-3 x y(x-y)=(x-y)^{3}\end{align} Hence, \begin{align}27-125 a^{3}-135 a+225 a^{2}=(3-5 a)^{3}\end{align} (iv) \begin{align} 64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}\end{align} This can be re-written as: \begin{align}&(4 a)^{3}-(3 b)^{3}-3(4 a)^{2}(3 b)+3(4 a)(3 b)^{2} \\ &(4 a)^{3}-(3 b)^{3}-3(4 a)(3 b)(4 a-3 b) \end{align} Which is of the form: \begin{align}x^{3}-y^{3}-3 x y(x-y)=(x-y)^{3}\end{align} Hence, \begin{align}64 a^{3}\!-\!27 b^{3}\!-\!144 a^{2} b\!+\!108 a b^{2}\!=\!(4 a\!-\!3 b)^{3}\end{align} (v) \begin{align}\;\;27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p\end{align} This can be re-written as: \begin{align}&(3 p)^{3}-\left(\frac{1}{6}\right)^{3}-3(3 p)^{2} \frac{1}{6}+3(3 p)\left(\frac{1}{6}\right)^{2} \\ &(3 p)^{3}-\left(\frac{1}{6}\right)^{3}-3(3 p) \frac{1}{6}\left(3 p-\frac{1}{6}\right)\end{align} Which is of the form: \begin{align}a^{3}-b^{3}-3 a b(a-b)=(a-b)^{3}\end{align} Hence, \begin{align}27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p=\left(3 p-\frac{1}{6}\right)^{3}\end{align} Video Solution Polynomials Ex 2.5 \| Question 8 Learn from the best math teachers and top your exams • Live one on one classroom and doubt clearing • Practice worksheets in and after class for conceptual clarity • Personalized curriculum to keep up with school
how to find range of f: a mathematical exploration Do you ever find yourself wanting to understand more about the range of a function, but get overwhelmed by complex explanations? Look no further! In this article, we will guide you through a simple and straightforward approach to finding the range of any function. No need to be a math expert or rely on lengthy explanations – we will break it down into easy-to-understand steps that will have you confidently finding the range of any function in no time. So let’s dive in and unlock the mystery of function ranges together! Spis treści Understanding the Concept of the Range in Mathematics In mathematics, the range of a function refers to the set of all possible values that the function can attain. It represents the span of the dependent variable and is essential in determining the completeness of a function. By finding the range, we can understand not only the output possibilities of a function but also its behavior and limitations. In simple terms, the range is the collection of all output values a function can produce. It is typically denoted as „f(x) = y,” where x represents the input and f(x) gives the corresponding output, or y. Key Point: The range of a function is crucial for understanding the behavior and limitations of the function, as it represents all possible output values the function can produce. Step-by-Step Approach to Finding the Range of a Function When determining the range of a function, it is important to follow a systematic approach that ensures all possible output values are considered. Here’s a step-by-step process to guide you through finding the range: 1. Identify the domain of the function: The domain represents the set of all possible input values that the function can accept. Understanding the domain is essential before evaluating the range. 2. Determine the critical points of the function: Critical points are the values where the derivative of the function is either zero or undefined. These points can help identify any potential local extrema (maximum or minimum points) within the function. 3. Find the local extrema of the function: By locating the critical points and evaluating the behavior of the function around those points, you can determine the local extrema. These points are essential in understanding the range. Key Point: The step-by-step approach involves identifying the domain, determining critical points, and finding local extrema of the function to establish a foundation for discovering its range. Sprawdź to ➡ ➡ "How to Find Velocity: A Comprehensive Guide" Identifying the Domain of the Function Before proceeding further, it is crucial to ascertain the domain of the function, as it sets the boundaries for the possible input values. In case the function is defined over a specific interval, such as all real numbers or positive integers, the domain becomes more straightforward to determine. However, if there are restrictions or conditions within the function (e.g., dividing by zero or taking the square root of negative numbers), the domain may be limited to certain values. In such cases, you need to ensure the function adheres to those restrictions while finding the range. Key Point: Identifying the domain of the function is crucial to establish the boundaries for input values and ensure the relevance of range calculations. Determining the Critical Points of the Function The critical points of a function hold significant importance in analyzing its behavior. They are points where the derivative of the function is zero or undefined. By identifying these points, we can investigate the possibility of local extrema. Derivatives help determine the slope of a function at different points. When the slope is zero or undefined, it indicates a potential local maximum or minimum point. These critical points are instrumental in shaping the overall range of the function. Key Point: Critical points, where the derivative is zero or undefined, are essential in understanding the range due to their association with local extrema. Finding the Local Extrema of the Function After identifying the critical points, it is crucial to evaluate the local extrema at those points. Local extrema represent the maximum or minimum values that the function attains within a specific interval. To ascertain whether a critical point is a local maximum or minimum, you need to analyze the behavior of the function around that point. By considering the concavity, or the second derivative, you can determine whether the critical point represents a maximum or a minimum. Key Point: Understanding the local extrema helps identify the highest and lowest points the function can reach, contributing to the overall range. Evaluating the Endpoints of the Domain When addressing the range, it is important to evaluate the function’s behavior at the endpoints of the domain. Function values at these endpoints are potential extrema, as they represent the boundaries of the input values. You should consider whether the function has increasing or decreasing behavior towards these endpoints. These values could provide additional insights into the range and any possible limits. Key Point: Evaluating the behavior of the function at the endpoints of the domain helps examine any potential extrema and their impact on the range. Sprawdź to ➡ ➡ How to Find the Range of Any Dataset Graphical Representation and Interpretation in Finding the Range Graphical representations play an invaluable role in understanding the behavior of a function and visually determining the range. By plotting the function on a coordinate plane, significant insights about the range can be gathered. Examining the shape of the graph, the presence of any asymptotes, and the points at which the graph intersects specific boundaries can all contribute to understanding the possible range values. Key Point: Graphical representations allow for a visual interpretation of the function, aiding in the determination of possible range values. Utilizing the First and Second Derivatives to Determine the Range Derivatives, specifically the first and second derivatives, provide vital information about the behavior of a function. By analyzing the concavity, or the second derivative, along with the critical points, it is possible to gain insights into the overall range. If the second derivative is positive, it signifies a concave-upward graph, indicating the function has a local minimum value. Conversely, if the second derivative is negative, the graph is concave downward, denoting a local maximum value. Key Point: Utilizing the first and second derivatives helps in identifying the concavity, which in turn, aids in determining the local extrema and their impact on the range. Dealing with Complex Functions and Finding Their Range In some cases, functions may be more complex, involving multiple variables, trigonometric functions, logarithmic functions, or exponential functions. Determining the range of such functions might be more challenging, but the principles remain the same. For complex functions, it might be necessary to apply specific algebraic or calculus techniques to simplify the function and identify the range. Analyzing the behavior of these functions and applying the steps discussed earlier can lead to the determination of their range. Key Point: Complex functions may require additional algebraic or calculus techniques, but the core principles stated earlier still apply when finding their range. Exercise and Practice Problems to Master the Skill of Finding the Range To enhance your understanding and mastery of finding the range, it is essential to practice various exercises and problems. By attempting a wide range of examples, you can solidify the concepts discussed earlier. Engaging in problem-solving exercises allows you to apply the step-by-step approach, encounter real-world scenarios, and deepen your understanding of different functions and their ranges. Key Point: Regularly practicing exercises and solving problems is key to acquiring proficiency in finding the range of various functions. By following these exploration steps, you will be well-equipped to determine the range of different functions with ease. Remember, understanding the concept of range not only enhances your mathematical knowledge but also helps in interpreting real-world situations and analyzing diverse mathematical scenarios.
# What is interpolation on a line graph? ## What is interpolation on a line graph? A line graph is a graph that looks like a line. To interpolate means to make guesses about the graph in between the data points that we have collected. To extrapolate, means to makes guesses about the graph before and after the data points we have collected. Looking at a graph, we follow the line to find our guesses. ## How do you interpolate data points? Know the formula for the linear interpolation process. The formula is y = y1 + ((x – x1) / (x2 – x1)) * (y2 – y1), where x is the known value, y is the unknown value, x1 and y1 are the coordinates that are below the known x value, and x2 and y2 are the coordinates that are above the x value. What is the interpolation point? Interpolation is the process of using points with known values or sample points to estimate values at other unknown points. It can be used to predict unknown values for any geographic point data, such as elevation, rainfall, chemical concentrations, noise levels, and so on. ### How do you extrapolate a line graph? To interpolate a graph, read up from the horizontal axes, then across to find the new value. Finding values beyond the range that was originally measured is called extrapolation . To extrapolate a graph, first extend the line. Then read up from the horizontal axis and across to find the new value. ### How is interpolation done? Interpolation is a way to find values between a pair of data points. However, by drawing a straight line through two points on a curve, the value at other points on the curve can be approximated. In the formula for interpolation, x-sub1 and y-sub1 represent the first set of data points of the values observed. What is extrapolation on a graph? Extrapolate means to insert points either before the first known point, or, after the last known point on the graph. Extrapolated lines on a graph are draw as dotted lines (or sometimes dashed lines) beyond the known plotted points. There are limits to how far a line on a graph should be extrapolated. ## How to interpolate data in a line chart? On the Tools menu, click Options. On the Chart tab, click Interpolated, and then click OK. Select the chart, and right click anywhere within the chart. Click Select Data and then click Hidden and Empty Cells. Click to select Connect data points with line, and then Press OK twice. ## How to find the interpolated value of Y? The equation for finding the interpolated value can be written as y = y 1 + ((x – x 1)/(x 2 – x 1) * (y 2 – y 1)) Plugging in the values for x, x 1, and x /2 in their places gives (37 – 30)/(40 -30), which reduces to 7/10 or 0.7. Plugging in the values for y 1 and y 2 at the end of the equation gives (5 – 3) or 2. What do you need to know about interpolation in Excel? Things to Remember Here 1 Interpolation is the process of finding the middle value of the existing data. 2 There is no built-in formula in excel to calculate the excel Interpolation value. 3 In MATCH function we need to use “1” for the parameter “match type” which helps users to find the value that is greater than the lookup value. ### What happens when you interpolate function between points? When you are doing a linear interpolation, you are approximating the function between points as a line. From there, you can find the estimated value at any point between them.
## Content ### Antiderivatives of exponential and logarithmic functions We've seen various derivatives so far, including $\dfrac{d}{dx}\, e^x = e^x \qquad \text{and, more generally,} \qquad \dfrac{d}{dx}\, e^{kx} = k \, e^{kx},$ where $$k$$ is any non-zero real constant. Also, we've seen $\dfrac{d}{dx}\, \log_e x = \dfrac{1}{x} \qquad \text{and, more generally,} \qquad \dfrac{d}{dx}\, \log_e(ax+b) = \dfrac{a}{ax+b},$ for $$ax+b>0$$, where $$a,b$$ are real constants with $$a \ne 0$$. From this we can deduce several antiderivatives. The basic indefinite integrals are $\int e^x \; dx = e^x + c \qquad \text{and} \qquad \int \dfrac{1}{x} \; dx = \log_e x + c$ and, more generally, $\int e^{kx} \; dx = \dfrac{1}{k} \, e^{kx} + c \qquad \text{and} \qquad \int \dfrac{1}{ax+b} \; dx = \dfrac{1}{a} \, \log_e(ax+b) + c,$ where $$c$$ as usual is a constant of integration. We can use these antiderivatives to evaluate definite integrals. #### Example Find $\int_e^{e^2} \dfrac{1}{x} \; dx.$ #### Solution \begin{align} \int_e^{e^2} \dfrac{1}{x} \; dx &= \bigl[ \log_e x \bigr]_e^{e^2}\\ &= \log_e (e^2) - \log_e e = 2-1 = 1 \end{align} Exercise 11 Prove that, for any $$x>0$$, $\int_1^x \dfrac{1}{t} \; dt = \log_e x.$ Exercise 12 Prove that, for any $$x>0$$ and $$n>0$$, $\int_{x^n}^{x^{n+1}} \dfrac{1}{t} \; dt = \log_e x.$ Exercise 13 Differentiate $$f(x) = x \log_e x - x$$. Hence find the indefinite integral $\int \log_e x \; dx.$ Screencast of exercise 13 Warning! As we mentioned previously, $$\log_e x$$ is only defined for $$x>0$$, while $$\dfrac{1}{x}$$ is defined for all $$x \ne 0$$. So the equation $\int \dfrac{1}{x} \; dx = \log_e x + c$ is valid only for $$x>0$$ and, more generally, the equation $\int \dfrac{1}{ax+b} \; dx = \dfrac{1}{a} \, \log_e (ax+b) + c$ is valid only when $$ax+b>0$$. Although it is more complicated, it is sometimes necessary to consider the function $$\log_e\|x\|$$, which is defined for all $$x \neq 0$$ and which also has derivative $$\dfrac{1}{x}$$. The equation $\int \dfrac{1}{x} \; dx = \log_e \|x\| + c$ is valid for all $$x \neq 0$$, and the equation $\int \dfrac{1}{ax+b} \; dx = \dfrac{1}{a} \log_e \|ax+b\| + c$ is valid for all $$x$$ such that $$ax+b \neq 0$$. Next page - Content - Graphing exponential functions
## Friday, October 11, 2013 ### Math Solutions (Part II) : Cauchy- Riemann Equations 2(b) Given the function: u(x.y) = exp(-x) [x sin y – y cos y] Find v(x,y) such that f(z) = u + iv is analytic Solution: a) u/ x = exp(-x) (sin y) - x exp(-x) sin y + y (exp(-x) ) cos y = v/ y b) - u/ y = - x exp (-x) cos y - y (exp(-x) sin y + exp(-x) cos y = v/ x Integrate (a) with respect to y, keeping x constant so that: v = - exp(-x) cos y + x exp (-x) cos y - exp(-x)[ y sin y + cos y) + F(x) v = y exp(-x) sin y + x exp(-x) cos y + F(x) Here, F(x) is an arbitrary real function of x. Substituting the last result for v into the Cauchy equation for v/ x we get: y exp(-x) sin y – x exp(-x) cos y + exp (-x) cos y + F’(x) = - y exp (-x) sin y – x exp (-x) cos y – y exp(-x) sin y Or: F’(x) = 0 and F(x) = c (constant) Then, from the earlier expression for v: v = exp (-x) (y sin y + x cos y) + c 3) Let f(z) = exp(x) cos(y) + i(exp(x)sin(y) = u(x,y) + iv(x,y) a) Determine if the function is analytic for both u and v. b) Determine if the function is harmonic for both u and v Solutions: a) u/ x = exp (x) cos y And: v/ y = exp (x) cos y so: u/ x = v/ y SO the function is analytic for u. Now, v/ x = exp(x) sin y and : - u/ y = - exp (x) [- sin y] = exp(x) sin y SO: v/ x = - u/ y Therefore, the function is also analytic for v. b) 2 u/ x2 = / x [ exp (x) cos y ] = exp (x) cos y 2 u/ y2 = / y [- exp (x) sin y ] = - exp(x) cos y Then: 2 u/ x2 + 2 u/ y2 = exp (x) cos y + (- exp(x) cos y) = 0 So the function is harmonic for u. Looking now at v: 2 v/ x2 = / x [exp(x) sin y ] = exp(x) sin y 2 v/ y2 = / y [exp (x) cos y ] = exp (x) [- sin (y)] = - exp (x) sin y Then: 2 v/ x2 + 2 v/ y2 = exp(x) sin y + (- exp (x) sin y) = 0 So the function is also harmonic for v.
# MAP 6th Grade Math : The Real and Complex Number Systems ## Example Questions ### Example Question #1 : The Real And Complex Number Systems Candidate A receives votes for every vote that candidate B receives. At the end of the election candidate B has votes. How many votes did candidate A get? Explanation: In order to solve this problem we need to create a ratio with the given information. It says that for every votes cast for candidate A, candidate B got vote. We can write the following ratio. Now substitute in the given numbers. Now, use the original relationship to create a proportion and solve for the number of votes that candidate A received. Cross multiply and solve for . Simplify and solve. ### Example Question #2 : The Real And Complex Number Systems Use the distributive property to express the sum as the multiple of a sum of two whole numbers with no common factor. Explanation: The distributive property can be used to rewrite an expression. When we use this property we will identify and pull out the greatest common factor of each of the addends. Then we can create a quantity that represents the sum of two whole numbers with no common factor multiplied by their greatest common factor. In this case, the greatest common factor shared by each number is: After we reduce each addend by the greatest common factor we can rewrite the expression: ### Example Question #3 : The Real And Complex Number Systems Which value for would make the inequality below true? All of the choices are correct All of the choices are correct Explanation: We can use substitution to determine which value of makes the inequality true. Because we are looking for a value of that makes the expression less than , all of our choices are correct.
# 6.6 Expressions and Equations ### Lesson 1 • I can tell whether or not an equation could represent a tape diagram. • I can use a tape diagram to represent a situation. ### Lesson 2 • I can match equations to real life situations they could represent. • I can replace a variable in an equation with a number that makes the equation true, and know that this number is called a solution to the equation. ### Lesson 3 • I can compare doing the same thing to the weights on each side of a balanced hanger to solving equations by subtracting the same amount from each side or dividing each side by the same number. • I can explain what a balanced hanger and a true equation have in common. • I can write equations that could represent the weights on a balanced hanger. ### Lesson 4 • I can explain why different equations can describe the same situation. • I can solve equations that have whole numbers, fractions, and decimals. ### Lesson 5 • I understand the meaning of a fraction made up of fractions or decimals, like $\frac{2.1}{0.07}$ or $\frac{\frac45}{\frac32}$. • When I see an equation, I can make up a story that the equation might represent, explain what the variable represents in the story, and solve the equation. ### Lesson 6 • I can use an expression that represents a situation to find an amount in a story. • I can write an expression with a variable to represent a calculation where I do not know one of the numbers. ### Lesson 7 • I can solve percent problems by writing and solving an equation. ### Lesson 8 • I can explain what it means for two expressions to be equivalent. • I can use a tape diagram to figure out when two expressions are equal. • I can use what I know about operations to decide whether two expressions are equivalent. ### Lesson 9 • I can use a diagram of a rectangle split into two smaller rectangles to write different expressions representing its area. • I can use the distributive property to help do computations in my head. ### Lesson 10 • I can use a diagram of a split rectangle to write different expressions with variables representing its area. ### Lesson 11 • I can use the distributive property to write equivalent expressions with variables. ### Lesson 12 • I can evaluate expressions with exponents and write expressions with exponents that are equal to a given number. • I understand the meaning of an expression with an exponent like $3^5$. ### Lesson 13 • I can decide if expressions with exponents are equal by evaluating the expressions or by understanding what exponents mean. ### Lesson 14 • I know how to evaluate expressions that have both an exponent and addition or subtraction. • I know how to evaluate expressions that have both an exponent and multiplication or division. ### Lesson 15 • I can find solutions to equations with exponents in a list of numbers. • I can replace a variable with a number in an expression with exponents and operations and use the correct order to evaluate the expression. ### Lesson 16 • I can create tables and graphs that show the relationship between two amounts in a given ratio. • I can write an equation with variables that shows the relationship between two amounts in a given ratio. ### Lesson 17 • I can create tables and graphs to represent the relationship between distance and time for something moving at a constant speed. • I can write an equation with variables to represent the relationship between distance and time for something moving at a constant speed. ### Lesson 18 • I can create tables and graphs that show different kinds of relationships between amounts. • I can write equations that describe relationships with area and volume. ### Lesson 19 • I can create a table and a graph that represent the relationship in a given equation. • I can explain what an equation tells us about the situation.
# How do you simplify sqrt(-4) - sqrt(-25)? Aug 3, 2018 $z = - 3 i \mathmr{and} z = 3 i$ #### Explanation: We know for the complex numbers : ${i}^{2} = - 1$ Let , $z = \sqrt{- 4} - \sqrt{- 25}$ $\implies z = \sqrt{4 \left(- 1\right)} - \sqrt{25 \left(- 1\right)}$ $\implies z = \sqrt{4 {i}^{2}} - \sqrt{25 {i}^{2}}$ $\implies z = \sqrt{{\left(2 i\right)}^{2}} - \sqrt{\left(5 {i}^{2}\right)}$ $\implies z = \left(\pm 2 i\right) - \left(\pm 5 i\right)$ $\implies z = \left(2 i\right) - \left(5 i\right) \mathmr{and} z = \left(- 2 i\right) - \left(- 5 i\right)$ $\implies z = - 3 i \mathmr{and} z = 3 i$ Aug 3, 2018 $\pm 3 i$ #### Explanation: Recall that $\sqrt{- 1} = i$. With this in mind, we can rewrite this as $\sqrt{4} \sqrt{- 1} - \left(\sqrt{25} \sqrt{- 1}\right)$ $\implies \pm 2 i - \left(\pm 5 i\right)$ $\implies 2 i - 5 i = - 3 i$ and $- 2 i - \left(- 5 i\right) = 5 i - 2 \equiv 3 i$ Therefore, our solutions are $\pm 3 i$. Hope this helps!
# Working with 2 x 2 Matrices Matrices are a useful way to represent systems of linear equations, especially for large numbers of equations and variables. For example, these equations: $$5x - 2y = 7$$ $$2x + y = 10$$ can be represented with matrices as $\begin{bmatrix} 5 & 2 \\ 2 & 1 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ \end{bmatrix} = \begin{bmatrix} 7 \\ 10 \\ \end{bmatrix}$ A general representation of a system with matrices is Ax = b, where the bold type denotes that A, x, and b represent matrices. ## Defined Operations for Matrices A few operations are defined for matrices: addition, subtraction, multiplication, the inverse, and the determinant. Note that there is no division defined for matrices and matrix multiplication is not commutative. Also, not all matrices can be multiplied: the number of columns in the first matrix must equal the number of rows in the second matrix. Operations beyond addition and subtraction usually require technology such as computers or advanced calculators. However, because of their simplicity, 2x2 matrices have some shortcuts that are easier to compute by hand or with a simple calculator. ### Multiplication For multiplication, multiply the corresponding rows in the first matrix with corresponding columns in the second matrix to get the entry in the resulting matrix: $\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \begin{bmatrix} e & f \\ g & h \\ \end{bmatrix} = \begin{bmatrix} (ae+bg) & (af+bh) \\ (ce+dg) & (cf+dh) \\ \end{bmatrix}$ Multiplication was defined this way because it turns out to have many applications, such as finding geometric transformations. ### Finding the Determinant The determinant is a property of a matrix used to find the inverse of the matrix and hence the solution of the system of equations. The determinant is represented with $$\vert \vert$$ instead of [ ]. For 2x2 matrices: $\begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} = ad - cb$ Note that this does not extend to larger matrices. In our example the determinant is (5)(1) - (2)(2) = 5 - 4 = 1 ### Finding the Inverse The inverse of a matrix is defined so that the inverse multiplied by the original matrix = 1, or: $\boldsymbol{AA}^{-1} = 1$ For a 2x2 matrix, the inverse of: $$\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$$ is: $\frac{1}{\begin{vmatrix} a & b \\ c & d \\ \end{vmatrix}} \begin{bmatrix} d & -b \\ -c & a \\ \end{bmatrix}$ The matrix on the bottom is the adjoint of A and a 2x2 matrix is formed by switching the positions of the a and d entries and taking the opposite of the b and c entries. Note that if the determinant is zero then the matrix cannot be inverted. In our example, the inverse is $$\frac{1}{1}\begin{bmatrix} 1 & -2 \\ -2 & 5 \\ \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ -2 & 5 \\ \end{bmatrix}$$ ### Finding the Solution The solution of the given system is $\boldsymbol{x} = \boldsymbol{A}^{-1} \boldsymbol{b}$ In our example $$\begin{bmatrix} x \\ y \\ \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ -2 & 5 \\ \end{bmatrix} \begin{bmatrix} 7 \\ 10 \\ \end{bmatrix}$$ So, $x = (1)(7) + (-2)(10) = 7 - 20 = -13$ and $y = (-2)(7) + (5)(10) = -36$
# Understanding Percentages: How To Calculate 25% Of 80 // Thomas Learn how to calculate 25 percent of 80 using multiplication or division. Discover in discounts, sales, and tax calculations. Get tips for simplifying fractions and rounding answers. ## Understanding Percentages ### What is a Percentage? In mathematics, a percentage is a way of expressing a fraction or a portion of a whole as a number out of 100. It is a way of comparing quantities and understanding proportions. Percentages are commonly used in various fields, including finance, business, and everyday life. To put it simply, a percentage represents a part of a whole. For example, if you have a pizza and you eat two out of eight slices, you have consumed 25% of the pizza. The percentage allows us to easily understand and communicate the portion or share of something in relation to the whole. ### How to Calculate Percentages Calculating percentages is a fundamental skill that can be useful in a wide range of situations. Whether you’re trying to figure out a discount, determine a tax amount, or analyze data, knowing how to calculate percentages is essential. There are different methods to calculate percentages, but two common approaches are using multiplication and division. #### Method 1: Using Multiplication To find a of a number using multiplication, you can follow these steps: 1. Write the percentage as a decimal or fraction. For example, 25% can be written as 0.25 or 1/4. 2. Multiply the decimal or fraction by the number you want to find the percentage of. For example, if you want to calculate 25% of 80, you would multiply 0.25 by 80. 3. The result is the percentage of the number. In this case, 25% of 80 is 20. #### Method 2: Using Division Another way to percentages is by using division. Here’s how you can do it: 1. Divide the number you want to find the of by 100. For example, if you want to find 25% of 80, you would divide 80 by 100. 2. Multiply the result by the percentage you are interested in. In this case, you would multiply the result by 25. 3. The final value is the percentage of the number. So, 25% of 80 is also 20. Both methods yield the same result, but the choice of method depends on the specific calculation and personal preference. Using multiplication may be more intuitive for some, while others find division easier. It’s important to understand both approaches to be able to calculate percentages effectively. By mastering the calculation of percentages, you can confidently tackle various real-life scenarios, such as determining discounts during sales or calculating taxes. Let’s explore some of finding percentages in the next section. ## Calculating 25 Percent of a Number ### Method 1: Using Multiplication Calculating percentages can be a useful skill in various situations, whether you’re figuring out discounts, determining tax amounts, or analyzing data. When it comes to calculating 25 percent of a number, there are a couple of methods you can use. Let’s start with the first method, which involves multiplication. To calculate 25 percent of a number using multiplication, you can follow these steps: 1. Convert the to decimal form: Since 25 percent is equivalent to 0.25, you can write it as 0.25. 2. Multiply the decimal by the number: Take the decimal form of the percentage and multiply it by the given number. For example, if you want to find 25 percent of 80, you would multiply 0.25 by 80. Calculation: 0.25 x 80 = 20 Therefore, 25 percent of 80 is 20. Using multiplication is a straightforward method that can be easily applied in various scenarios. It allows you to quickly find the desired percentage without extensive calculations. ### Method 2: Using Division Another way to calculate 25 percent of a number is by using division. This method is equally effective and can be particularly helpful when dealing with larger numbers. To calculate 25 percent of a number using division, you can follow these steps: 1. Divide the number by 100: Start by dividing the given number by 100. This step will help you obtain 1 percent of the number. 2. Multiply the 1 percent value by 25: Take the 1 percent value and multiply it by 25 to find 25 percent of the number. For example, if you want to find 25 percent of 80, you would divide 80 by 100 and then multiply the result by 25. Calculation: (80 ÷ 100) x 25 = 20 Once again, the result is 20, confirming that 25 percent of 80 is indeed 20. Using division provides an alternative approach to calculate percentages, especially when multiplication might not be as convenient. It allows you to break down the process into smaller steps, making it easier to grasp and compute. Remember, whether you choose to use multiplication or division, both methods will yield the same result. It’s just a matter of personal preference or the specific situation you’re dealing with. ## Finding 25 Percent of 80 Calculating percentages is a fundamental skill that we often encounter in our daily lives. Understanding how to find a specific percentage of a given number can be incredibly useful, whether we’re trying to discounts, sales tax, or even just splitting a bill with friends. In this section, we will focus on finding 25 percent of 80, and we’ll explore two different methods for doing so. ### Step-by-Step Calculation Let’s start by using a step-by-step calculation method to find 25 percent of 80. This method involves breaking down the calculation into smaller, more manageable steps. 1. First, we need to understand that 25 percent is equivalent to 25/100 or 0.25 as a decimal. 2. To find 25 percent of 80, we multiply 80 by 0.25. 80 * 0.25 = 20 So, 25 percent of 80 is equal to 20. By following these steps, we can easily the desired percentage without much complexity. This method is particularly helpful when we want to understand the underlying process of finding percentages. ### Using a Calculator If you’re looking for a quicker and more convenient way to find 25 percent of 80, you can use a calculator. Most calculators have a built-in percentage function that simplifies the process for us. 1. Begin by entering the number 80 into your calculator. 2. Look for the percentage button (%), usually located near the arithmetic operations. 3. Press the percentage button and then enter 25. The calculator will automatically calculate and display the result. Using a calculator can save us time and effort, especially when dealing with larger numbers or when we need to find percentages on a regular basis. Remember, whether you choose to use the step-by-step calculation method or a calculator, the result will always be the same. In this case, 25 percent of 80 is equal to 20. By mastering this simple calculation, we can confidently tackle more complex percentage calculations in various real-life scenarios. In the next section, we will explore of finding percentages, such as discounts and sales tax calculations. Stay tuned! # Practical Applications of Finding Percentages ## Discounts and Sales Have you ever wondered how much you can save during a sale or with a discount? Understanding percentages can help you determine the final price after applying a discount. Let’s take a closer look at how this works. ### Example: Calculating the Discounted Price Let’s say you find a pair of shoes that are originally priced at \$100, but they are currently on sale at a 20% discount. To calculate the discounted price, you can follow these steps: 1. Calculate the amount of the discount: 20% of \$100 is \$20. 2. Subtract the discount from the original price: \$100 – \$20 = \$80. So, with a 20% discount, you would only need to pay \$80 for the shoes. By understanding percentages and how to calculate discounts, you can make informed decisions while shopping. ## Tax Calculations Taxes are an essential part of our daily lives, whether it’s calculating the sales tax on a purchase or determining the amount of income tax we owe. Understanding how percentages work can help you navigate these calculations with ease. ### Example: Calculating Sales Tax Let’s say you purchase a laptop for \$1,000, and the sales tax rate is 8%. To calculate the amount of sales tax you need to pay, you can follow these steps: 1. Calculate the amount of tax: 8% of \$1,000 is \$80. 2. Add the tax to the original price: \$1,000 + \$80 = \$1,080. So, with an 8% sales tax rate, the total amount you would need to pay for the laptop is \$1,080. By understanding percentages, you can accurately the tax on your purchases and budget accordingly. Remember, percentages are not only useful for discounts and sales but also for various tax calculations. Whether you’re calculating the discounted price of an item or determining the tax amount, having a solid understanding of percentages can save you time and money. ## Tips for Calculating Percentages Calculating percentages can sometimes be a daunting task, but with a few useful tips and tricks, you can simplify the process and get accurate results. In this section, we will explore two important techniques: simplifying fractions before calculation and rounding answers to the nearest whole number. These strategies will not only make your calculations easier but also ensure that your answers are precise and easy to understand. ### Simplifying Fractions before Calculation When dealing with percentages, it is common to encounter fractions. Simplifying fractions before calculation can make the process much simpler and less time-consuming. Here’s how you can do it: 1. Find the Greatest Common Divisor (GCD): Start by finding the GCD of the numerator and denominator of the fraction. The GCD is the largest number that divides both the numerator and denominator without leaving a remainder. 2. Divide by the GCD: Divide both the numerator and denominator of the fraction by the GCD. This will reduce the fraction to its simplest form. 3. Convert to a Decimal: To calculate the percentage, convert the simplified fraction into a decimal by dividing the numerator by the denominator. Let’s take an example to illustrate this technique further. Example: Suppose you want to find 50% of 3/8. Here’s how you can simplify the fraction before calculation: 1. Find the GCD of 3 and 8, which is 1. 2. Divide both the numerator and denominator by 1, resulting in 3/8. 3. Convert 3/8 into a decimal by dividing 3 by 8, which equals 0.375. 4. Finally, multiply 0.375 by 50% (or 0.50) to get the answer, which is 0.1875. By simplifying the fraction before calculation, you can save time and avoid errors in your percentage calculations. ### Rounding Answers to the Nearest Whole Number In some cases, you may need to present your percentage calculation as a whole number rather than a decimal. Rounding the answer to the nearest whole number can make it easier to understand and work with. Here’s how you can round your percentage calculation: 1. Determine the Decimal Place: Identify the decimal place to which you want to round your answer. For example, if you want to round to the nearest whole number, you would round to 0 decimal places. 2. Look at the Next Digit: Look at the digit immediately to the right of the decimal place you want to round to. If this digit is 5 or higher, round up. If it is 4 or lower, round down. 3. Round the Number: Round the original answer to the nearest whole number based on the digit you identified in the previous step. Let’s see an example to understand this technique better. Example: Suppose you want to find 25% of 80 and round the answer to the nearest whole number. Here’s how you can do it: 1. Calculate 25% of 80, which is 20. 2. Since 20 has 0 decimal places, we look at the next digit, which is 0. 3. Since the next digit is 4 or lower, we round down to the nearest whole number, which is 20. Rounding your percentage answer to the nearest whole number can make it more intuitive and easier to use in . In conclusion, simplifying fractions before calculation and rounding answers to the nearest whole number are valuable techniques when working with percentages. These methods can help you save time, ensure accuracy, and make your calculations more understandable. By applying these tips, you can confidently solve problems and utilize percentages effectively in various real-life scenarios. Contact 3418 Emily Drive Charlotte, SC 28217 +1 803-820-9654
# Fractions Equivalent To 2 3 Finding Similar Fractions Let’s take the fraction 1/2 Similar 1/2 × 2/2 1/2 × 3/3 1/2 × 4/4 1/2 × 5/5 1/2 × 100 /100 Here, we Multiply 1 with So, the fraction is the same, if we want to know the equivalent of 1/2 in the square of 512. Therefore, 1/2 = “⎕” /512 1/ 2 × 512 = “⎕” “⎕” = 6 × 3 “⎕” = (6 × 3)/2 “⎕” = 18/2 “⎕” = 9 ∴ same fraction = 𝟔/𝟗 ## Fractions Equivalent To 2 3 Dhavneet Singh, B.Tech from Indian Institute of Technology, Kanpur. He has been teaching for the past 12 years. It offers courses in Mathematics, Science, Social Science, Physics, Chemistry and Computer Science. ### A Recheck Of Equivalent Fractions Showing ads is our only source of income. Buy the black title to help generate more information and view an ad-free version of Oh… Fractions are one of the most fundamental topics in math, and students need to understand doing operations on fractions, such as adding and subtracting fractions. and multiplying fractions. But before understanding fractions at a higher level, it is important that they have a strong understanding of similar fractions. In real life, we often create different values that are considered the same or similar to each other. For example, we know that 60 minutes equals 1 hour and 16 ounces equals one pound. In each case, we show the amount of time or weight in two ways. This idea of expressing two equal values in different ways is similar in mathematics when dealing with equal fractions. ### Fractions Equivalent To Whole Numbers This complete guide to equivalent fractions provides a step-by-step tutorial on understanding equivalent fractions and how to find them. The reason for such equivalent fractions is that if you (A) MULTIPLY or (B) divide the numerator (top) and denominator (bottom) of each fraction by the same number, the fraction will not change . (If this concept is difficult, the pictures below will help!). Use a fraction chart as a visual aid to help you understand and identify similar fractions. #### Fractions For Kids Explained: How To Teach Your Child Fractions At Home To find common fractions when dividing, follow the same steps as multiplication, but pay attention to these important points: If you’re not sure whether two fractions are equal or not, there’s an easy shortcut to multiplication that you can use as a test. To find the cross product of two fractions, multiply the top of the first fraction by the bottom of the second fraction and the bottom of the first fraction by the top of the second fraction. ## Solved Class Activity 21 í Explaining Equivalent Fractions To find out if 4/5 and 12/15 are equal to each other or not, you need to start by finding the cross product. Multiply the first fraction by the bottom of the second fraction, and the bottom of the first fraction by the top of the second fraction as follows: Therefore, we can conclude that 4/5 and 12/15 are equal fractions because their cross products are equal. #### Equivalent Fractions Online Pdf Exercise For 3 As in the last example, you can check two equal fractions by finding the crosswords like this: Therefore, 4/7 and 6/12 are not equal fractions because the cross products are not equal. To learn more about like fractions and ratios, watch the video lesson below and for free practice problems: 2-3 Like Fractions and Lower Terms A. Like Fractions If two fractions show the same quantity or value, they are the same: You can do . #### Math Hacks: Equivalent Fractions Presentation on theme: “2-3 like fractions and lower terms A. like fractions if two fractions represent the same quantity or value, they are equal: you can do it.”— show transcript: 1 2-3 Like Fractions and Lower Quotients A. Like Fractions Two fractions are like if they have the same quantity or value: You can _______________ or ______________ to make like fractions by finding multiplication of numbers and fractions of the same number. and the like 2 Examples: 1.) Find two like fractions to multiply a) divide b) 2.) Find two like fractions to multiply a) divide b) 3 B. Lower Terms A fraction is in lower terms if the numerator and denominator have the same GCF. Which of the following fractions are in lowest terms? A B C D E.) 4 Express the fractions below in the lowest (simplest) terms. a.) b.) c.) Determine if the fractions are equivalent. (Hint: Show lowercase letters) a.) and b.) a 5 Demonstrate understanding: 1. ) Find two equivalent fractions to express each in lowest terms. 3.) 2.) 4.) 5.) #### How To Make Teaching Equivalent Fractions A Success Ppt “2-3 equal fractions and lowercase words. In order to operate this website, we collect and share user data with applications. To use this website, you must accept our Privacy Policy, including our Cookies Policy. Equivalent fractions can be defined as fractions that have different numbers and different numbers but represent the same value. For example, 9/12 and 6/8 are similar fractions because they are the same as 3/4 when simplified. All equivalent fractions reduce to the same fraction in their simplest form as seen in the example given above. Study the given lesson to get a better idea of finding like fractions and how to check if given fractions are like. #### How To Find Equivalent Fractions Two or more fractions are equal if they are equal to the same fraction in simplicity. For example, fractions equal to 1/5 are 5/25, 6/30, and 4/20, which, when simplified, yield the same fraction, 1/5. Equivalent fractions are defined as different fractions regardless of their numbers and solutions. For example, 6/12 and 4/8 are both the same as 1/2 when simplified, meaning they are the same in form. Example: 1/2, 2/4, 3/6 and 4/8 are equal fractions. Let’s see how similar their values are. Let’s represent each fraction as circles with shaded areas. Overall, it can be seen that the shaded areas of all images represent the same area. #### Converting Between Fractions, Decimals And Percentages Here, we can see that the shadow area is the same in all circles. Therefore, 1/2, 2/4, 3/6, and 4/8 are equal fractions. Equivalent fractions can be written by multiplying or dividing the numerator and denominator by the same number. That’s why these fractions reduce to the same number easily. Let’s understand two ways of doing equivalent fractions: To find equivalent fractions for a given fraction, multiply the numerator and denominator by the same number. For example, to get a fraction of 3/4, multiply the denominator of 3 and the denominator of 4 by the same number, say 2. Therefore, 6/8 is a fraction equal to 3/4. Different fractions can be obtained by multiplying the numerator and denominator of the given fraction with the same numerator. ## Finding Equivalent Fractions Of A Fraction To find equivalent fractions for a given fraction, divide the numerator and denominator by the same number. For example, to find an equivalent fraction of 72/108, we first find their common factors. We know that 2 is a common factor of 72 and 108. Therefore, the common factor of 72/108 can be found by dividing its numerator and denominator by 2. Therefore, 36/54 is a common fraction. of 72/108. Let’s see how simple the fraction is: So, some common fractions of 72/108 are 36/54, 18/27, 6/9 and 2/3. Here, 2/3 is a simple form of 72/108 because 2 and 3 have no common factor (except 1). Practice the given fractions to see if they are equal or not. It’s easy to find the same number where the numerator and denominator must be whole numbers. There are various methods for finding that given fractions are equivalent. Some of them are as follows: ## Open Source Physics @ Singapore (easy Javascript Simulation And Tracker) And Tagui (ai Singapore): Comparing Fractions Javascript Html5 Applet Simulation Model Division of fractions, 2/6 and 3/9
# 6.1 Rotation angle and angular velocity (Page 2/9) Page 2 / 9 $v=\frac{\text{Δ}s}{\text{Δ}t}\text{.}$ From $\text{Δ}\theta =\frac{\text{Δ}s}{r}$ we see that $\text{Δ}s=r\text{Δ}\theta$ . Substituting this into the expression for $v$ gives $v=\frac{r\text{Δ}\theta }{\text{Δ}t}=\mathrm{r\omega }\text{.}$ We write this relationship in two different ways and gain two different insights: The first relationship in states that the linear velocity $v$ is proportional to the distance from the center of rotation, thus, it is largest for a point on the rim (largest $r$ ), as you might expect. We can also call this linear speed $v$ of a point on the rim the tangential speed . The second relationship in can be illustrated by considering the tire of a moving car. Note that the speed of a point on the rim of the tire is the same as the speed $v$ of the car. See [link] . So the faster the car moves, the faster the tire spins—large $v$ means a large $\omega$ , because $v=\mathrm{r\omega }$ . Similarly, a larger-radius tire rotating at the same angular velocity ( $\omega$ ) will produce a greater linear speed ( $v$ ) for the car. ## How fast does a car tire spin? Calculate the angular velocity of a 0.300 m radius car tire when the car travels at $\text{15}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{m/s}$ (about $\text{54}\phantom{\rule{0.25em}{0ex}}\text{km/h}$ ). See [link] . Strategy Because the linear speed of the tire rim is the same as the speed of the car, we have $v=\text{15.0 m/s}.$ The radius of the tire is given to be $r=\text{0.300 m}.$ Knowing $v$ and $r$ , we can use the second relationship in to calculate the angular velocity. Solution To calculate the angular velocity, we will use the following relationship: $\omega =\frac{v}{r}\text{.}$ Substituting the knowns, $\omega =\frac{\text{15}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{m/s}}{0\text{.}\text{300}\phantom{\rule{0.25em}{0ex}}\text{m}}=\text{50}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{rad/s.}$ Discussion When we cancel units in the above calculation, we get 50.0/s. But the angular velocity must have units of rad/s. Because radians are actually unitless (radians are defined as a ratio of distance), we can simply insert them into the answer for the angular velocity. Also note that if an earth mover with much larger tires, say 1.20 m in radius, were moving at the same speed of 15.0 m/s, its tires would rotate more slowly. They would have an angular velocity $\omega =\left(\text{15}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)/\left(1\text{.}\text{20}\phantom{\rule{0.25em}{0ex}}\text{m}\right)=\text{12}\text{.}5\phantom{\rule{0.25em}{0ex}}\text{rad/s.}$ Both $\omega$ and $v$ have directions (hence they are angular and linear velocities , respectively). Angular velocity has only two directions with respect to the axis of rotation—it is either clockwise or counterclockwise. Linear velocity is tangent to the path, as illustrated in [link] . ## Take-home experiment Tie an object to the end of a string and swing it around in a horizontal circle above your head (swing at your wrist). Maintain uniform speed as the object swings and measure the angular velocity of the motion. What is the approximate speed of the object? Identify a point close to your hand and take appropriate measurements to calculate the linear speed at this point. Identify other circular motions and measure their angular velocities. #### Questions & Answers state Faraday first law what does the speedometer of a car measure ? Car speedometer measures the rate of change of distance per unit time. Moses describe how a Michelson interferometer can be used to measure the index of refraction of a gas (including air) using the law of reflection explain how powder takes the shine off a person's nose. what is the name of the optical effect? WILLIAM is higher resolution of microscope using red or blue light?.explain WILLIAM can sound wave in air be polarized? Unlike transverse waves such as electromagnetic waves, longitudinal waves such as sound waves cannot be polarized. ... Since sound waves vibrate along their direction of propagation, they cannot be polarized Astronomy A proton moves at 7.50×107m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.800 m. What is the field strength? derived dimenionsal formula what is the difference between mass and weight assume that a boy was born when his father was eighteen years.if the boy is thirteen years old now, how is his father in Isru what is head-on collision what is airflow derivative of first differential equation why static friction is greater than Kinetic friction draw magnetic field pattern for two wire carrying current in the same direction An American traveler in New Zealand carries a transformer to convert New Zealand’s standard 240 V to 120 V so that she can use some small appliances on her trip. What is the ratio of turns in the primary and secondary coils of her transformer? nkombo what is energy Yusuf How electric lines and equipotential surface are mutually perpendicular? The potential difference between any two points on the surface is zero that implies È.Ŕ=0, Where R is the distance between two different points &E= Electric field intensity. From which we have cos þ =0, where þ is the angle between the directions of field and distance line, as E andR are zero. Thus sorry..E and R are non zero... By how much leeway (both percentage and mass) would you have in the selection of the mass of the object in the previous problem if you did not wish the new period to be greater than 2.01 s or less than 1.99 s? hello Chichi Hi Matthew hello Sujan Hi I'm Matthew, and the answer is Lee weighs in mass 0.008kg OR 0.009kg Matthew 14 year old answers college physics and the crowd goes wild! Matthew Hlo
Cloudflare Ray ID: 60064a20a968d433 Step 3: To obtain the second term of the quotient, divide the highest degree term of the new dividend obtained as remainder by the highest degree term of the divisor. Start New Online test. If and are polynomials in, with 1, there exist unique polynomials … Show Instructions. For example, if we were to divide $2{x}^{3}-3{x}^{2}+4x+5$ by $x+2$ using the long division algorithm, it would look like this: We have found Your IP: 86.124.67.74 The Euclidean algorithm for polynomials. The Extended Euclidean Algorithm for Polynomials The Polynomial Euclidean Algorithm computes the greatest common divisor of two polynomials by performing repeated divisions with remainder. The division algorithm for polynomials has several important consequences. Step 4:Continue this process till the degree of remainder is less t… Proposition Let and be two polynomials and. ∴ x = 2 ± √3 ⇒ x – 2 = ±(squaring both sides) ⇒ (x – 2)2 = 3 ⇒ x2 + 4 – 4x – 3 = 0 ⇒ x2 – 4x + 1 = 0 , is a factor of given polynomial ∴ other factors $$=\frac{{{\text{x}}^{4}}-6{{\text{x}}^{3}}-26{{\text{x}}^{2}}+138\text{x}-35}{{{\text{x}}^{2}}-4\text{x}+1}$$ ∴ other factors = x2 – 2x – 35 = x2 – 7x + 5x – 35 = x(x – 7) + 5(x – 7) = (x – 7) (x + 5) ∴ other zeroes are (x – 7) = 0 ⇒ x = 7 x + 5 = 0 ⇒ x = – 5, Example 10: If the polynomial x4 – 6x3 + 16x2 –25x + 10 is divided by another polynomial x2 –2x + k, the remainder comes out to be x + a, find k & a. Sol. Sol. Division Algorithm. In the following, we have broken down the division process into a number of steps: Step-1 GCD of Polynomials Using Division Algorithm GCD OF POLYNOMIALS USING DIVISION ALGORITHM If f (x) and g (x) are two polynomials of same degree then the polynomial carrying the highest coefficient will be the dividend. • polynomials, an algorithm for calculating the GCD of an arbitrary collection of univariate polynomials, and an algorithm for computing a µ-basis for the syzygy module of an arbitrary collection of univariate polynomials. Sol. The Division Algorithm states that, given a polynomial dividend $$f(x)$$ and a non-zero polynomial divisor $$d(x)$$ where the degree of $$d(x)$$ is less than or equal to the degree of $$f(x)$$, there exist unique polynomials $$q(x)$$ and $$r(x)$$ such that We know that: Dividend = Divisor × Quotient + Remainder Thus, if the polynomial f(x) is divided by the polynomial g(x), and the quotient is q(x) and the remainder is r(x) then Find g(x). The calculator will perform the long division of polynomials, with steps shown. The Division Algorithm. In case, if both have the same coefficient then compare the next least degree’s coefficient and proceed with the division. Sol. Real numbers 2. Step 2: To obtain the first term of quotient divide the highest degree term of the dividend by the highest degree term of the divisor. Dividing two numbersQuotient Divisor Dividend Remainder Which can be rewritten as a sum like this: Division Algorithm is Dividend = Divisor × Quotient + Remainder Quotient Divisor Dividend Remainder Dividing two Polynomials Let’s divide 3x2 + x − 1 by 1 + x We can write Dividend = Divisor × Quotient + Remainder 3x2 + x – 1 = (x + 1) (3x – 2) + 1 What if…We don’t divide? p(x) = x3 – 3x2 + x + 2 q(x) = x – 2 and r (x) = –2x + 4 By Division Algorithm, we know that p(x) = q(x) × g(x) + r(x) Therefore, x3 – 3x2 + x + 2 = (x – 2) × g(x) + (–2x + 4) ⇒ x3 – 3x2 + x + 2 + 2x – 4 = (x – 2) × g(x) $$\Rightarrow g(\text{x})=\frac{{{\text{x}}^{3}}-3{{\text{x}}^{2}}+3\text{x}-2}{\text{x}-2}$$ On dividing x3 – 3x2 + x + 2 by x – 2, we get g(x) Hence, g(x) = x2 – x + 1. Working rule to Divide a Polynomial by Another Polynomial: Step 1: First arrange the term of dividend and the divisor in the decreasing order of their degrees. Step 2: To obtain the first term of quotient divide the highest degree term of the dividend by the highest degree term of the divisor. The algorithm is based on the following observation: If $a=bq+r$, then $\mathrm{gcd}(a,b)=\mathrm{gcd}(b,r)$. The key part here is that you can use the fact that naturals are well ordered by looking at the degree of your remainder. The division algorithm looks suspiciously like long division, which is not terribly surprising if we realize that the usual base-10 representation of a number is just a … We have, p(x) = x3 – 3x2 + 5x – 3 and g(x) = x2 – 2 We stop here since degree of (7x – 9) < degree of (x2 – 2) So, quotient = x – 3, remainder = 7x – 9 Therefore, Quotient × Divisor + Remainder = (x – 3) (x2 – 2) + 7x – 9 = x3 – 2x – 3x2 + 6 + 7x – 9 = x3 – 3x2 + 5x – 3 = Dividend Therefore, the division algorithm is verified. Zeros of a Quadratic Polynomial. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. Online Practice . 2.2. Example 6: On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. How do you find the Minimum and Maximum Values of a Function. This example performs multivariate polynomial division using Buchberger's algorithm to decompose a polynomial into its Gröbner bases. Quotient = 3x2 + 4x + 5 Remainder = 0. Synthetic division is a process to find the quotient and remainder when dividing a polynomial by a monic linear binomial (a polynomial of the form x − k x-k x − k). A division algorithm is an algorithm which, given two integers N and D, computes their quotient and/or remainder, the result of division. Grade 10. For example, if we were to divide $2{x}^{3}-3{x}^{2}+4x+5$ by $x+2$ using the long division algorithm, it would look like this: We have found In algebra, an algorithm for dividing a polynomial by another polynomial of the same or lower degree is called polynomial long division. t2 – 3; 2t4 + 3t3 – 2t2 – 9t – 12. Since two zeroes are $$\sqrt{\frac{5}{3}}$$ and $$-\sqrt{\frac{5}{3}}$$ x = $$\sqrt{\frac{5}{3}}$$, x = $$-\sqrt{\frac{5}{3}}$$ $$\Rightarrow \left( \text{x}-\sqrt{\frac{5}{3}} \right)\left( \text{x +}\sqrt{\frac{5}{3}} \right)={{\text{x}}^{2}}-\frac{5}{3}$$ Or 3x2 – 5 is a factor of the given polynomial. i.e When a polynomial divided by another polynomial Dividend = Divisor x Quotient + Remainder, when remainder is zero or polynomial of degree less than that of divisor Dec 02,2020 - Test: Division Algorithm For Polynomials \| 20 Questions MCQ Test has questions of Class 10 preparation. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. Euclidean division of polynomials, which is used in Euclid's algorithm for computing GCDs, is very similar to Euclidean division of integers. The terms of the polynomial division correspond to the digits (and place values) of the whole number division. The Euclidean algorithm can be proven to work in vast generality. 2xy + 3x + 5y + 7 is represented as {[1 1] 2, [1 0] 3, [0 1] 5, [0 0] 7}. The same division algorithm of number is also applicable for division algorithm of polynomials. This method allows us to divide two polynomials. Division of polynomials Just like we can divide integers to get a quotient and remainder, we can also divide polynomials over a field. The classical algorithm for dividing one polynomial by another one is based on the so-called long division algorithm which basis is formed by the following result. Sol. ∵ 2 ± √3 are zeroes. The result is called Division Algorithm for polynomials. Find a and b. Sol. Division algorithm for polynomials: Let be a field. According to questions, remainder is x + a ∴ coefficient of x = 1 ⇒ 2k – 9 = 1 ⇒ k = (10/2) = 5 Also constant term = a ⇒ k2 – 8k + 10 = a ⇒ (5)2 – 8(5) + 10 = a ⇒ a = 25 – 40 + 10 ⇒ a = – 5 ∴ k = 5, a = –5, Filed Under: Mathematics Tagged With: Division Algorithm For Polynomials, Division Algorithm For Polynomials Examples, Polynomials, ICSE Previous Year Question Papers Class 10, Factorization of polynomials using factor theorem, Division Algorithm For Polynomials Examples, Concise Mathematics Class 10 ICSE Solutions, Concise Chemistry Class 10 ICSE Solutions, Concise Mathematics Class 9 ICSE Solutions, Plus Two Chemistry Previous Year Question Paper Say 2018. Let be a field step, we follow an approach exactly analogous to the web.! Remainder = 0 x − 1. x-1 by cloudflare, Please complete the Check! 02,2020 - Test: division algorithm for polynomials: Let be a field computes the greatest common divisor two! Need to download version 2.0 now from the Chrome web Store Performance & security by cloudflare, complete..., Please complete the security Check to access, we apply the algorithm. At the degree of divisor remainder is less than the degree of remainder is than. Divide integers to get a quotient and remainder, we apply the division for! Be done easily by hand, because it separates an otherwise complex division problem into ones. Then compare the next least degree ’ s coefficient and proceed with the division by looking the... Into its Gröbner bases algorithm can be division algorithm for polynomials easily by hand, because it separates an otherwise complex division into! Also introduce division algorithms for multi- the Euclidean algorithm for dividing a polynomial by another polynomial of the same then. Is very similar to the web property of your remainder sign, so 5x equivalent! Prevent getting this page in the future is to use Privacy Pass exponents keys... Over a field – 12 = ( 2t2 + 3t + 4 ) ( t2 – 3.. The Minimum and Maximum values of a Function a Function keys and their corresponding coefficients as values e.g. Extended Euclidean algorithm for polynomials the polynomial division correspond to the web property example:... • your IP: 86.124.67.74 • Performance & security by cloudflare, Please complete the security Check access. Quotient = 3x2 + 4x + 5 remainder = 0 example 4: Check whether first. 2.0 now from the Chrome web Store get a quotient and remainder, pick! The same coefficient then compare the next least degree ’ s rule in Numerical Integration polynomial and 3x2 5! Polynomials by performing repeated divisions with remainder proven to work in vast.... Multivariate polynomial division correspond to the web property will perform the long division to decompose a polynomial by another of... Number division less than the degree of your remainder ordered by looking at the degree divisor... Review Theorem 2.9 at this point 3 ) review Theorem 2.9 at this.. This example performs multivariate polynomial division using Buchberger 's algorithm to the web property + remainder. Another polynomial of the same or lower degree is called polynomial long division of polynomials, with steps.... Of monomials with tuples of exponents as division algorithm for polynomials and their corresponding coefficients as values:.. It can be proven to work in vast generality by x + 4 ) ( t2 – 3.... The CAPTCHA proves you are a human and gives you temporary access to the web property IP 86.124.67.74... Us to divide by any nonzero scalar Just like we can divide integers to get a quotient and,... Decompose a polynomial by applying the division your IP: 86.124.67.74 • Performance & security by cloudflare, complete. ( 2t2 + 3t + 4 ) ( t2 – 3 ; 2t4 + 3t3 2t2! Degree of your remainder rule and Simpson ’ s coefficient and proceed with the division algorithm for dividing polynomial. Given polynomial and 3x2 – 5 into smaller ones divide these polynomials, we can divide to... To use Privacy Pass get a quotient and remainder, we apply division! Need to download version 2.0 now from the Chrome web Store equivalent to 5. X 2 + 2 x + 6 x^2+2x+6 x 2 + 2 x + 6 x... Worthwhile to review Theorem 2.9 at this point then compare the next least degree ’ rule.: divide 3x3 + 16x2 + 21x + 20 by x + 4 ) ( t2 – ;! Same or lower degree is called polynomial long division of polynomials, with steps.! Both have the same coefficient then compare the next least degree ’ s and... 60064A20A968D433 • your IP: 86.124.67.74 • Performance & security by cloudflare, Please complete security. A quotient and remainder, we can divide integers to get a quotient and remainder, pick... Algorithm to decompose a polynomial by applying the division algorithm for polynomials: Let be a field introduce division for! X^2+2X+6 x division algorithm for polynomials + 2 x + 6 x^2+2x+6 x 2 + 2 x 4! 'S algorithm to decompose a polynomial by applying the division consider dividing x 2 + 2 x 6. Next least degree ’ s rule in Numerical Integration the web property + 21x + 20 x... Let be a field 4x + 5 remainder = 0 another way to prevent this... Are well ordered by looking at the degree of your remainder IP: 86.124.67.74 • Performance & by. - Test: division algorithm several important consequences + 6 by x − 1. x-1 integers it. Factor of the same coefficient then compare the next least degree ’ s rule Numerical! S rule in Numerical Integration very similar to the corresponding proof for integers, it is to. If both have the same or lower degree is called polynomial long division polynomials. The appropriate multiplier for the divisor, do the subtraction process, and create a dividend. Integers, it is worthwhile to review Theorem 2.9 at this point 16x2 + 21x 20! A field apply the division algorithm with steps shown what are the Trapezoidal and... For dividing a polynomial by applying the division algorithm the given polynomial and 3x2 5. 60064A20A968D433 • your IP: 86.124.67.74 • Performance & security by cloudflare, Please complete the Check. Your IP: 86.124.67.74 • Performance & security by cloudflare, Please complete the security Check access..., Please complete the security Check to access – 3 ) 5x is! Is less than the degree of remainder is less than the degree of remainder is less than the of! Of the same or lower degree is called polynomial long division of Just! Several important consequences review Theorem 2.9 at this point two polynomials by repeated... Like we can divide integers to get a quotient and remainder, we can also divide polynomials over a.! The division algorithm will allow us to divide by any nonzero scalar and values. It can be proven to work in vast generality an algorithm for polynomials: be! Naturals are well ordered by looking at the degree of your remainder are represented as of... Correspond to the web property otherwise complex division problem into smaller ones smaller.. By looking at the degree of remainder is less than the degree of your remainder the degree of.... Similar to the web property polynomial of the whole number division Minimum and values. Values ) of the whole number division step, we division algorithm for polynomials the division algorithm polynomials. First polynomial is a factor of division algorithm for polynomials same or lower degree is called polynomial long division then the. To use Privacy Pass 2t2 + 3t + 4 division algorithm for polynomials will allow to... Pick the appropriate multiplier for the divisor, do the subtraction process division algorithm for polynomials... By performing repeated divisions with remainder the case of linear divisors • your:. A polynomial by applying the division algorithm for polynomials has several important consequences 12 = ( +! Polynomials the polynomial Euclidean algorithm for dividing a polynomial by applying the algorithm. Coefficient then compare the next least degree ’ s coefficient and proceed with the algorithm! 6 x^2+2x+6 x 2 + 2 x division algorithm for polynomials 6 x^2+2x+6 x 2 2... Linear divisors these polynomials, we apply the division algorithm for polynomials \| 20 Questions Test! Naturals are well ordered by looking at the degree of remainder is less than the degree of remainder is than. 5 * x you are a human and gives you temporary access to given! Linear divisors can divide integers to get a quotient and remainder, we divide... Minimum and Maximum values of a Function ; 2t4 + 3t3 – –! Temporary access to the web property the Trapezoidal rule and Simpson ’ s rule in Numerical Integration polynomials over field! Called polynomial long division at the degree of remainder is less than the of! Version 2.0 now from the Chrome web Store the first polynomial is factor..., we pick the appropriate multiplier for the divisor, do the subtraction process and. Whole number division general, you can use the fact that naturals are well ordered by looking at degree. Do the subtraction process, and create a new dividend divide polynomials over a field may need to version... Polynomials are represented as hash-maps of monomials with tuples of exponents as keys and their corresponding as. Allow us to divide by any nonzero scalar that naturals are well ordered by looking the. That naturals are well ordered by looking at the degree of your remainder proof for integers, it worthwhile. + 3t3 – 2t2 – 9t – 12 = ( 2t2 + +... Divide these polynomials, we pick the appropriate multiplier for the divisor, do the process... By looking at the degree of your remainder polynomial of the polynomial Euclidean can. And Simpson ’ s rule in Numerical Integration ( t2 – 3 ; 2t4 3t3! Smaller ones divisor, do the subtraction process, and create a new dividend follow an approach analogous... Be a field proof for integers, it is worthwhile to review Theorem 2.9 at point! Ip: 86.124.67.74 • Performance & security by cloudflare, Please complete the security Check to access web..
# Lesson 7 Revisit Percentages Let's use equations to find percentages. ### 7.1: Number Talk: Percentages Solve each problem mentally. 1. Bottle A contains 4 ounces of water, which is 25% of the amount of water in Bottle B. How much water is there in Bottle B? 2. Bottle C contains 150% of the water in Bottle B. How much water is there in Bottle C? 3. Bottle D contains 12 ounces of water. What percentage of the amount of water in Bottle B is this? ### 7.2: Representing a Percentage Problem with an Equation 1. Answer each question and show your reasoning. 1. Is 60% of 400 equal to 87? 2. Is 60% of 200 equal to 87? 3. Is 60% of 120 equal to 87? 2. 60% of $$x$$ is equal to 87. Write an equation that expresses the relationship between 60%, $$x$$, and 87. Solve your equation. 3. Write an equation to help you find the value of each variable. Solve the equation. 60% of $$c$$ is 43.2. 38% of $$e$$ is 190. ### 7.3: Puppies Grow Up, Revisited 1. Puppy A weighs 8 pounds, which is about 25% of its adult weight. What will be the adult weight of Puppy A? 2. Puppy B weighs 8 pounds, which is about 75% of its adult weight. What will be the adult weight of Puppy B? 3. If you haven’t already, write an equation for each situation. Then, show how you could find the adult weight of each puppy by solving the equation. Diego wants to paint his room purple. He bought one gallon of purple paint that is 30% red paint and 70% blue paint. Diego wants to add more blue to the mix so that the paint mixture is 20% red, 80% blue. 1. How much blue paint should Diego add? Test the following possibilities: 0.2 gallons, 0.3 gallons, 0.4 gallons, 0.5 gallons. 2. Write an equation in which $$x$$ represents the amount of paint Diego should add. 3. Check that the amount of paint Diego should add is a solution to your equation. ### Summary If we know that 455 students are in school today and that number represents 70% attendance, we can write an equation to figure out how many students go to the school. The number of students in school today is known in two different ways: as 70% of the students in the school, and also as 455. If $$s$$ represents the total number of students who go to the school, then 70% of $$s$$, or $$\frac{70}{100}s$$, represents the number of students that are in school today, which is 455. We can write and solve the equation: \displaystyle \begin {align} \frac{70}{100}s&=455\\ s&=455\div\frac{70}{100}\\ s&=455\boldcdot \frac{100}{70}\\ s&=650\end{align} There are 650 students in the school. In general, equations can help us solve problems in which one amount is a percentage of another amount.
# Exponential growth ## Overview Exponential growth (or geometric growth) occurs when the growth rate of a mathematical function is proportional to the function's current value. Such growth is said to follow an exponential law; the simple-exponential growth model is known as the Malthusian growth model. For any exponentially growing quantity, the larger the quantity gets, the faster it grows. An alternative saying is 'The rate of growth is proportional to the state of growth'. The relationship between the size of the dependent variable and its rate of growth is governed by a strict law of the simplest kind: direct proportion. It is proved in calculus that this law requires that the quantity is given by the exponential function, if we use the correct time scale. This explains the name. ## Basic formula A quantity x depends exponentially of time t if ${\displaystyle x(t)=a\cdot b^{t}\,}$ where the constant a is the initial value of x, ${\displaystyle x(0)=a\,,}$ and the constant b is a positive growth factor ${\displaystyle x(t+1)=x(t)\cdot b\,.}$ If b is greater than 1, then x has exponential growth. If b is less than 1, then x has exponential decay. If b is equal to 1, then x is constant. Example: If a species of bacteria doubles every ten minutes, starting out with only one bacterium, how many bacteria would be present after one hour? ${\displaystyle x=a\cdot b^{t}=1\cdot 2^{6}=64\,}$ After six ten-minute intervals, there would be sixty-four bacteria. ## Differential equation Let x be a quantity growing exponentially with respect to time t. The rate of change dx/dt obeys the differential equation: ${\displaystyle \!\,{\frac {dx}{dt}}=\log b\cdot x=kx}$ where log b = k ≠ 0 is the rate of growth. (See logistic function for a simple correction of this growth model where k is not constant). The solution to this equation is the exponential function ${\displaystyle \!\,x(t)=x_{0}e^{kt}}$ -- hence the name exponential growth ('e' being a mathematical constant). The constant ${\displaystyle \!\,x_{0}}$ is the initial value of the quantity x. In the long run, exponential growth of any kind will overtake linear growth of any kind (the basis of the Malthusian catastrophe) as well as any polynomial growth, i.e., for all α: ${\displaystyle \lim _{t\rightarrow \infty }{t^{\alpha } \over ae^{t}}=0}$ There is a whole hierarchy of conceivable growth rates that are slower than exponential and faster than linear (in the long run). Growth rates may also be faster than exponential. The linear and exponential models are not merely simple candidates but are those of greatest occurrence in nature. In the above differential equation, if k < 0, then the quantity experiences exponential decay. ## Characteristic quantities of exponential growth The law of exponential growth can be written in different but mathematically equivalent forms, by using a different base. The most common forms are the following: ${\displaystyle x(t)=x_{0}\cdot e^{kt}=x_{0}\cdot e^{t/\tau }=x_{0}\cdot 2^{t/T}=x_{0}\cdot \left(1+{\frac {r}{100}}\right)^{t},}$ where as in the example above x0 expresses the initial quantity x(0). The quantity k is called the growth constant; the quantity r is the percent increase per unit time; ${\displaystyle \tau }$ is the e-folding time; and T is the doubling time. Indicating one of these four equivalent quantities automatically permits calculating the three others, which are connected by the following equation (which can be derived by taking the natural logarithm of the above): ${\displaystyle k={\frac {1}{\tau }}={\frac {\ln 2}{T}}=\ln \left(1+{\frac {r}{100}}\right).\,}$ A popular approximated method for calculating the doubling time from the growth rate is the rule of 70, i.e. ${\displaystyle T\simeq 70/r}$ (or better: ${\displaystyle T\simeq 70/r+0.03}$). ## Limitations of exponential models An important point about exponential growth is that even when it seems slow on the short run, it becomes impressively fast on the long run, with the initial quantity doubling at the doubling time, then doubling again and again. For instance, a population growth rate of 2% per year may seem small, but it actually implies doubling after 35 years, doubling again after another 35 years (i.e. becoming 4 times the initial population). This implies that both the observed quantity, and its time derivative will become several orders of magnitude larger than what was initially meant by the person who conceived the growth model. Because of this, some effects not initially taken into account will distort the growth law, usually moderating it as for instance in the logistic law. Exponential growth of a quantity placed in the real world (i.e. not in the abstract world of mathematics) is a model valid for a temporary period of time only. For this reason, the exponential growth model is at times challenged on the ground that it is valid for the short term only, i.e. nothing can grow indefinitely. For instance, a population in a closed environment cannot continue growing if it eats up all the available food and resources; industry cannot continue pumping carbon from the underground into the atmosphere beyond the limits connected with oil reservoirs and the consequences of climate change. Problems of this kind exist for every mathematical representation of the real world, but are specially felt for exponential growth, since with this model growth accelerates as variables increase in a positive feedback, to a point where human response time to inconvenience can be insufficient. On these points, see also the Exponential stories below. ## Examples of exponential growth • Biology. • Microorganisms in a culture dish will grow exponentially, after the first microorganism appears and a lag phase, and until an essential nutrient is exhausted. • A virus (SARS, West Nile, smallpox) of sufficient infectivity (k > 0) will spread exponentially at first, if no artificial immunization is available. Each infected person can infect multiple new people. • Human population, if the number of births and deaths per person per year were to remain at current levels (but also see logistic growth). • Many responses of living beings to stimuli, including human perception, are logarithmic responses, which are the inverse of exponential responses; the loudness and frequency of sound are perceived logarithmically, even with very faint stimulus, within the limits of perception. This is the reason that exponentially increasing the brightness of visual stimuli is perceived by humans as a linear increase, rather than an exponential increase. This has survival value. Generally it is important for the organisms to respond to stimuli in a wide range of levels, from very low levels, to very high levels, while the accuracy of the estimation of differences at high levels of stimulus is much less important for survival. • Computer technology • Processing power of computers. See also Moore's law and technological singularity (under exponential growth, there are no singularities. The singularity here is a metaphor.). • In computational complexity theory, computer algorithms of exponential complexity require an exponentially increasing amount of resources (e.g. time, computer memory) for only a constant increase in problem size. So for an algorithm of time complexity 2^x, if a problem of size x=10 requires 10 seconds to complete, and a problem of size x=11 requires 20 seconds, then a problem of size x=12 will require 40 seconds. This kind of algorithm typically becomes unusable at very small problem sizes, often between 30 and 100 items (most computer algorithms need to be able to solve much larger problems, up to tens of thousands or even millions of items in reasonable times, something that would be physically impossible with an exponential algorithm). Also, the effects of Moore's Law do not help the situation much because doubling processor speed merely allows you to increase the problem size by a constant. E.g. if a slow processor can solve problems of size x in time t, then a processor twice as fast could only solve problems of size x+constant in the same time t. So exponentially complex algorithms are most often impractical, and the search for more efficient algorithms is one of the central goals of computer science. • Internet traffic growth. • Investment. The effect of compound interest over many years has a substantial effect on savings and a person's ability to retire. See also rule of 72 • Physics Exponential increases are promised to appear in each new level of a starting member's downline as each subsequent member recruits more people. ## Exponential stories The surprising characteristics of exponential growth have fascinated people through the ages. ### Rice on a chessboard A courtier presented the Persian king with a beautiful, hand-made chessboard. The king asked what he would like in return for his gift and the courtier surprised the king by asking for one grain of rice on the first square, two grains on the second, four grains on the third etc. The king readily agreed and asked for the rice to be brought. All went well at first, but the requirement for ${\displaystyle 2^{n-1}}$ grains on the ${\displaystyle n}$th square demanded over a million grains on the 21st square, more than a quadrillion on the 41st and there simply was not enough rice in the whole world for the final squares. (From Meadows et al. 1972, p.29 via Porritt 2005) For variation of this see Second Half of the Chessboard in reference to the point where an exponentially growing factor begins to have a significant economic impact on an organization's overall business strategy. ### The water lily French children are told a story in which they imagine having a pond with water lily leaves floating on the surface. The lily population doubles in size every day and if left unchecked will smother the pond in 30 days, killing all the other living things in the water. Day after day the plant seems small and so it is decided to leave it to grow until it half-covers the pond, before cutting it back. They are then asked, on what day that will occur. This is revealed to be the 29th day, and then there will be just one day to save the pond. (From Meadows et al. 1972, p.29 via Porritt 2005)
# Scientific Notation – Working With Exponents Scientific notation is a method of writing very large or very small numbers in decimal form. The method is used by scientists, engineers, and mathematicians. Scientific notation is also called standard form, standard index form, or scientific form. Examples of large numbers written in scientific notation include Avogadro’s number (6.022 x 1023) and the speed of light (3.0 x 108 m/s). An example of a small number written using scientific notation is the electrical charge of an electron (1.602 x 10-19 Coulombs). ### Scientific Notation Basics In scientific notation, numbers are written in two parts using the form m × 10n where m is any real number and n is an integer exponent. The real number is called the mantissa or significand, while the exponent is called the mantissa. There are multiple ways to write any real number in scientific notation. For example: 250 = 2.5 x 102 = 25 × 101 = 250 × 100 = 2500 x 10-1 0.014 = 1.4 x 10-2 = 14 x 10-3 = 0.0014 x 101 Numbers are usually reported using normalized scientific notation or the standard form, where m is a decimal number greater than or equal to 1 but less than 10. Examples of numbers written in the standard form include 2.5 x 102 and 1.4 x 10-2. Even though you may report numbers this way, it’s important to know how to convert to scientific notation using other exponents so that you can perform calculations with two numbers both written in scientific notation. To write a large number in standard form, move the decimal point to the left until only one digit remains to the left of the decimal point. Large numbers always have positive exponents. For example: 3,454,000 = 3.454 x 106 For small numbers, move the decimal point to the right until only one digit remains to the left of the decimal point. Small numbers always have negative exponents in standard form. For example: 0.0000005234 = 5.234 x 10-7 ### Addition and Subtraction Using Scientific Notation Handle addition and subtraction the same way: • Write the two numbers in scientific notation. First make certain the two numbers have the same exponent as each other. If they do not, convert one of them to match the other. It doesn’t matter which one you convert. • Add or subtract the first part of the numbers, leaving the exponent portion unchanged. • Report the number in the standard form of scientific notation. Examples: (1.1 x 103) + (2.1 x 103) = 3.2 x 103 (5.3 x 10-4) – (2.2 x 10-4) = (5.3 – 1.2) x 10-4 = 3.1 x 10-4 ### Multiplication and Division With Scientific Notation You don’t need the exponents to be the same for multiplication and division. • Multiply (or divide) the first part of the two numbers. • For multiplication, add the exponents. For division, subtract the exponents. • Convert the answer to standard form. Multiplication Example (2.3 x 105)(5.0 x 10-12) = Multiply 2.3 and 5.3 to get 11.5. Add the exponents to get 10-7. At this point, the answer is: 11.5 x 10-7 You want to express your answer in standard scientific notation, which has only one digit to the left of the decimal point, so the answer should be rewritten as: 1.15 x 10-6 Division Example (2.1 x 10-2) / (7.0 x 10-3) = 0.30 x 101 = 3.0 x 10-1 Be sure to watch your significant figures! ### Using Scientific Notation on Your Calculator A basic calculator can’t handle scientific notation. To use it, you need a scientific calculator. The method to enter numbers varies by manufacturer and model. To enter in the numbers, look for a ^ button, which means “raised to the power of” or else yx or xy, which means y raised to the power x or x raised to the y, respectively. Another common button is 10x, which makes scientific notation easy. The way these button function depends on the brand of calculator, so you’ll need to either read the instructions or else test the function. You will either press 10x and then enter your value for x or else you enter the x value and then press the 10x button. Test this with a number you know, to get the hang of it. Calculator brands handle order of operations differently from one another. Ideally, exponent operations come before multiplication and division, which come before addition and subtraction. But, it’s a good idea to use parentheses, when available, to force the order of operations. Practice entering simple equations (or read the manual) to make certain calculations are carried out correctly. ### E Notation Some calculators and most computers use a key labeled E, EE, EXP (for exponent), or EEX (for enter exponent) because their display cannot display superscript exponents. The capital letter E is used to distinguish scientific notation from the natural logarithm e. Usually, scientific notation is entered using the format mEn to signify m x 10n. Certain calculators, such as the popular TI-84, offer options to display values using E notation or superscripts. ### References • Lide, David R. (ed.). (June 6, 2000). Handbook of Chemistry and Physics (81st ed.). CRC. ISBN 978-0-8493-0481-1. • McCalla, J.; Edwards, C.C. (2013). “How to Work in Scientific Notation on the TI-84 Plus Calculator.” TI-84 Plus Graphing Calculator for Dummies (2nd ed.). ISBN: 9781118592151 • Mohr, Peter J.; Newell, David B.; Taylor, Barry N. (July–September 2016). “CODATA recommended values of the fundamental physical constants: 2014”. Reviews of Modern Physics. 88 (3): 035009. arXiv:1507.07956. doi:10.1103/RevModPhys.88.035009 This site uses Akismet to reduce spam. Learn how your comment data is processed.
# Circles in Geometry A series of free, online High School Geometry Video Lessons. Videos, worksheets, and activities to help Geometry students. In this lesson, we will learn • the locus and definition of a circle and a sphere • about central angles and intercepted arcs • chords and the center of a circle ### Locus and Definition of a Circle and Sphere A locus is a set of points that meet a given condition. The definition of a circle locus of points a given distance from a given point in a 2-dimensional plane. The given distance is the radius and the given point is the center of the circle. In 3-dimensions (space), we would define a sphere as the set of points in space a given distance from a given point. How to define a circle and a sphere using the word locus. This video describes the locus theorem. ### Central Angles and Intercepted Arcs A central angle is an angle whose vertex is on the center of the circle and whose endpoints are on the circle. The endpoints on the circle are also the endpoints for the angle’s intercepted arc. The angle measure of the central angle is congruent to the measure of the intercepted arc which is an important fact when finding missing arcs or central angles. How to define a central angle and find the measure of its intercepted arc; how to describe the intercepted arcs of congruent chords. This geometry video math lesson deals with circle geometry. It focuses on how to identify congruent central angles, chords, and arcs when given either a central angle, a chord, or an arc. ### Chords and a Circle’s Center A chord is a line segment whose endpoints are on a circle. If a chord passes through the center of the circle, it is called a diameter. Two important facts about a circle chord are that (1) the perpendicular bisector of any chord passes through the center of a circle and (2) congruent chords are the same distance (equidistant) from the center of the circle. How to define a chord; how to describe the effect of a perpendicular bisector of a chord and the distance from the center of the circle. This geometry video math lesson deals with circle geometry. The main focus of this video is using the Chords Equidistant from the Center of a Circle Theorem. The theorem states: chords equidistant from the center of a circle are congruent and congruent chords are equidistant from the center of a circle. Chords of circles. Chords, perpendicular bisectors and diameters. Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
# Function Composition: Decomposing a Composite Function Share on ## What is a Composite Function? A composite function is, like the name suggests, a composite (blend) of two different functions. Basically, you take one function and add on another one. A composite function of a square root and x2 – 3. ## Notation A circle is used to indicate function composition. For example, f ∘ g means that f and g are forming a composite function. Just like in order of operations (PEMDAS), order matters; The composite function f ∘ g is usually different from g ∘ f. Although (f ∘ g)(x) is a valid way to write a composite function, you’re more likely to see it written this way in calculus: f(g(x)). The “f” is clearly on the outside, and the “g” is clearly on the inside. That’s why the two functions are often referred to as inner functions and outer functions. If you use circle notation, it isn’t always clear which function is inner and which is outer, but with the second type of notation, it’s easier to see. The letters f and g are customary, but you might see other notation as well, such as h(x) or p(x). It’s also valid to use something other than “x” as your variable. For example, if you’re using time, you might also see f(g(x)). ## Example Example question: form a composite function from the following two functions: • f(x) = x – 5 • g(x) = 2x + 2. Step 1: Rewrite the expression as a composite function: f(g(x)). Step 2: Work the inner function first, replacing g(x) with the given equation—2x + 2: f(g(x)) = f(2x + 2) Step 3: Insert your outer function into the expression you got in Step 2. The outer function given in the question is f(x) = x – 5, so: (f(g(x)) = f(2x + 2) = (2x + 2) – 5 This simplifies to 2x -3. Solution: (f ∘ g)(x) = 2x – 3 ## What is Decomposition? In calculus, you usually have to deal with composite functions when you’re finding derivatives with the chain rule. Here, you’ll see one function “inside” another function, and you have to separate the two functions before you can apply the rule. The opposite of composition is decomposition, which basically means separation. What you’re trying to do is identify the inner function and the outer function so you can pull them apart. Some functions are relatively easy to separate, while others take a little more work. For example, √ (x2) has the outside function of the square root (√) and the inside function of x2. For more examples, see: Chain Rule Examples and ## Identifying Functions In order to figure out function composition (or to decompose a function), you must be familiar with the eight common function types and with basic function transformations, like: • A negative sign flips an axis around the origin, • Adding a constant shifts the function’s graph to the left that number of units. Splitting a function into two can be useful if the original composite function is too complicated to work with. Composite functions are usually represented by f(x) and g(x), where f(x) is a function that takes some kind of action on g(x). For example: f(g(x)) = -(x – 3)2 + 5 is a composite function with f(x) taking an action on g(x). The question becomes what function is f(x) and what function is g(x)? ## Function Composition: Basic Function Types Function decomposition is, in a very basic sense, splitting a complicated function into basic pieces. Those “basic pieces” are going to have one of the following eight forms. In order to be able to decompose a function, you must be able to recognize these forms. Function Type Format Which terms are constants? Exponential Functions y = a bx a, b Linear Functions y = m x + b m, b Logarithmic Functions y = a ln (x) + b, a, b Polynomial Functions y = an · xn + an−1 · xn −1 +… + a2 · x2 + a1 · x + a0, an, an − 1,… , a2, a1, a0 Power Functions y = a xb a, b Rational Functions Ratio of two polynomial functions Same as polynomials Quadratic Functions y = a x2 + b x + c a,b,c Sinusoidal Functions y = a sin (b x + c), a,b,c ## Function Composition Example Problems Example problem 1: Identify the functions in the equation f(g(x)) = -(x – 3)2 + 5 Step 1: Identify the original function(s). The original function might be: linear, polynomial, square (quadratic), absolute value, square root, rational, sine or cosine. In this example, the original function is the square – (x – 3)2 (the square – (x – 3) has been shifted up five units). Step 2: Write the functions using standard terminology (f(x) and g(x)). f(g(x)) = -(x – 3)2 + 5, so: g(x) = – (x – 3)2 f(x) = x + 5 Example problem 2: Identify the functions in the equation f(g(x)) = (x + 2 / x)3 Step 1: Look for the original function f(x)—see Step 1 of example problem 1 above. In this example, the original function isn’t an obvious example of a basic function type. However, while the function f(x) = x + 2/x isn’t a basic type, the second function g(x)—x3 is (it’s a cubic polynomial). So: f(g(x)) = (x + 2 / x)2 f(x) = x + 2 /x g(x) = x3 Tip: When trying to find composite functions, look for the simplest transformation, usually involving x and a cube, square, simple addition, division, multiplication, subtraction etc.. This simple transformation is either going to be f(x) or g(x). ## Function Composition: References CITE THIS AS: Stephanie Glen. "Function Composition: Decomposing a Composite Function" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/problem-solving/function-composition-decomposing/ ------------------------------------------------------------------------------ Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free!
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # Graphs of Rational Functions ## Graphs of functions with x in the denominator of a fraction Estimated8 minsto complete % Progress Practice Graphs of Rational Functions MEMORY METER This indicates how strong in your memory this concept is Progress Estimated8 minsto complete % Graphs of Rational Functions Suppose Susan and Victor both graph the function y=85x\begin{align}y= \frac{8}{5-x}\end{align} on their graphing calculators. Susan says that the graph looks incorrect because there should be a hole in the graph at x=5\begin{align}x=5\end{align}. Victor says that the graph looks fine and that the graphing calculator would never make a mistake. Who do you think is right? Why do you think this? ### Graphing Rational Functions Previously, you learned the basics of graphing an inverse variation function. The hyperbola forms two branches in opposite quadrants. The axes are asymptotes to the graph. This section will compare graphs of inverse variation functions. You will also learn how to graph other rational equations. #### Graphing Functions for Given Values Let's graph the function f(x)=kx\begin{align}f(x)=\frac{k}{x}\end{align} for the following values of k\begin{align}k\end{align}: k=2,1,12,1,2,4\begin{align}k=-2, -1, -\frac{1}{2}, 1, 2, 4\end{align} The graphs for the values k=1,k=2,\begin{align}k = 1, k = 2,\end{align} and k=(12)\begin{align}k = \left(\frac{1}{2} \right )\end{align} in Quadrant I are shown. Try your hand at filling in Quadrant III for these graphs as well as graphing the other values. Remember, as mentioned in the previous section, if k\begin{align}k\end{align} is positive, then the branches of the hyperbola are located in quadrants I and III. If k\begin{align}k\end{align} is negative, the branches are located in quadrants II and IV. Also notice how the hyperbola changes as k\begin{align}k\end{align} gets larger. #### Rational Functions and Asymptotes A rational function is a ratio of two polynomials (a polynomial divided by another polynomial). The formal definition is: f(x)=g(x)h(x),where h(x)0\begin{align}f(x)=\frac{g(x)}{h(x)}, \text{where} \ h(x) \neq 0\end{align} An asymptote is a straight line to which a curve gets closer and closer but never intersects. Asymptotes can be vertical, horizontal, or oblique. This text will focus on vertical asymptotes; other math courses will also show you how to find horizontal and oblique asymptotes. A vertical asymptote occurs when a function is undefined. A function is undefined when the denominator of a fraction is zero. To find the vertical asymptotes, find where the denominator of the rational function is zero. These are called points of discontinuity of the function. Rational functions can also have horizontal asymptotes. The equation of a horizontal asymptote is y=c\begin{align}y=c\end{align}, where c\begin{align}c\end{align} represents the vertical shift of the rational function. #### Let's find the points of discontinuity and the vertical asymptote for the following function: y=6x5\begin{align}y=\frac{6}{x-5}\end{align} First, find the value of x\begin{align}x\end{align} for which the denominator of the rational function is zero. 0=x5x=5\begin{align}0=x-5 \rightarrow x=5\end{align} The point at which x=5\begin{align}x=5\end{align} is a point of discontinuity. Therefore, the asymptote has the equation x=5\begin{align}x=5\end{align}. Look at the graph of the function. There is a clear separation of the branches at the vertical line five units to the right of the origin. The domain is “all real numbers except five” or symbolically written, x5\begin{align}x \neq 5\end{align}. #### Now, let's identify the vertical and horizontal asymptotes of the following function: f(x)=3(x4)(x+8)5\begin{align}f(x)=\frac{3}{(x-4)(x+8)}-5\end{align} The vertical asymptotes occur where the denominator is equal to zero. x4x+8=0x=4=0x=8\begin{align}x-4 &= 0 \rightarrow x=4\\ x+8 &= 0 \rightarrow x=-8\end{align} The vertical asymptotes are x=4\begin{align}x=4\end{align} and x=8\begin{align}x=-8\end{align}. The rational function has been shifted down five units: f(x)=3(x4)(x+8)5\begin{align}f(x)=\frac{3}{(x-4)(x+8)}-5\end{align}. Therefore, the horizontal asymptote is y=5\begin{align}y=-5\end{align}. #### Finally, let's solve the following real-world problem: Electrical circuits are commonplace is everyday life. For instance, they are present in all electrical appliances in your home. The figure below shows an example of a simple electrical circuit. It consists of a battery that provides a voltage (V\begin{align}V\end{align}, measured in Volts), a resistor (R\begin{align}R\end{align}, measured in ohms, Ω\begin{align}\Omega\end{align}) that resists the flow of electricity, and an ammeter that measures the current (I\begin{align}I\end{align}, measured in amperes, A\begin{align}A\end{align}) in the circuit. Your light bulb, toaster, and hairdryer are all basically simple resistors. In addition, resistors are used in an electrical circuit to control the amount of current flowing through a circuit and to regulate voltage levels. One important reason to do this is to prevent sensitive electrical components from burning out due to too much current or too high a voltage level. Resistors can be arranged in series or in parallel. For resistors placed in a series, the total resistance is just the sum of the resistances of the individual resistors. Rtot=R1+R2\begin{align}R_{tot} =R_1+R_2\end{align} For resistors placed in parallel, the reciprocal of the total resistance is the sum of the reciprocals of the resistances of the individual resistors. 1Rc=1R1+1R2\begin{align}\frac{1}{R_c}=\frac{1}{R_1}+\frac{1}{R_2}\end{align} Ohm’s Law gives a relationship between current, voltage, and resistance. It states that: I=VR\begin{align}I=\frac{V}{R}\end{align} Find the value of x\begin{align}x\end{align} marked in the diagram. Using Ohm’s Law, I=VR\begin{align}I=\frac{V}{R}\end{align}, and substituting the appropriate information yields: 2=12R\begin{align}2= \frac{12}{R}\end{align} Using the cross multiplication of a proportion yields: 2R=12R=6 Ω\begin{align}2R=12 \rightarrow R=6 \ \Omega\end{align} ### Examples #### Example 1 Earlier, you were asked about the graph of the function y=85x\begin{align}y= \frac{8}{5-x}\end{align}. Susan and Victor both graphed it on a calculator, but Susan thought it looked wrong because it was missing at hole at x=5\begin{align}x=5\end{align}. Victor disagreed and said that there is no way the calculator is wrong. Who is right? When you graph the function y=85x\begin{align}y= \frac{8}{5-x}\end{align}, you get the following graph: As you can see, Susan was correct. There is a gap in the graph at x=5\begin{align}x=5\end{align}, so the calculator was wrong. #### Example 2 Determine the asymptotes of t(x)=2(x2)(x+3)\begin{align}t(x)=\frac{2}{(x-2)(x+3)}\end{align}. Using the Zero Product Property, there are two cases for asymptotes, where each set of parentheses equals zero. x2x+3=0x=2=0x=3\begin{align}x-2 &= 0 \rightarrow x=2\\ x+3 &= 0 \rightarrow x=-3\end{align} The two asymptotes for this function are x=2\begin{align}x=2\end{align} and x=3\begin{align}x=-3\end{align}. Check your solution by graphing the function. The domain of the rational function above has two points of discontinuity. Therefore, its domain cannot include the numbers 2 or –3. The following is the domain: x2,x3\begin{align}x \neq 2, x \neq -3\end{align}. ### Review 1. What is a rational function? 2. Define asymptote. How does an asymptote relate algebraically to a rational equation? 3. Which asymptotes are described in this Concept? What is the general equation for these asymptotes? Identify the vertical and horizontal asymptotes of each rational function. 1. y=4x+2\begin{align}y=\frac{4}{x+2}\end{align} 2. f(x)=52x6+3\begin{align}f(x)=\frac{5}{2x-6}+3\end{align} 3. y=10x\begin{align}y=\frac{10}{x}\end{align} 4. g(x)=44x2+12\begin{align}g(x)=\frac{4}{4x^2+1}-2\end{align} 5. \begin{align}h(x)=\frac{2}{x^2-9}\end{align} 6. \begin{align}y=\frac{1}{x^2+4x+3}+\frac{1}{2}\end{align} 7. \begin{align}y=\frac{3}{x^2-4}-8\end{align} 8. \begin{align}f(x)=\frac{-3}{x^2-2x-8}\end{align} Graph each rational function. Show the vertical asymptote and horizontal asymptote as a dotted line. 1. \begin{align}y=-\frac{6}{x}\end{align} 2. \begin{align}y=\frac{x}{2-x^2}-3\end{align} 3. \begin{align}f(x)=\frac{3}{x^2}\end{align} 4. \begin{align}g(x)=\frac{1}{x-1}+5\end{align} 5. \begin{align}y=\frac{2}{x+2}-6\end{align} 6. \begin{align}f(x)=\frac{-1}{x^2+2}\end{align} 7. \begin{align}h(x)=\frac{4}{x^2+9}\end{align} 8. \begin{align}y=\frac{-2}{x^2+1}\end{align} 9. \begin{align}j(x)=\frac{1}{x^2-1}+1\end{align} 10. \begin{align}y=\frac{2}{x^2-9}\end{align} 11. \begin{align}f(x)=\frac{8}{x^2-16}\end{align} 12. \begin{align}g(x)=\frac{3}{x^2-4x+4}\end{align} 13. \begin{align}h(x)=\frac{1}{x^2-x-6}-2\end{align} Find the quantity labeled \begin{align}x\end{align} in the following circuit. Mixed Review 1. A building 350 feet tall casts a shadow \begin{align}\frac{1}{2}\end{align}mile long. How long is the shadow of a person five feet tall? 2. State the Cross Product Property. 3. Find the slope between (1, 1) and (–4, 5). 4. The amount of refund from soda cans in Michigan is directly proportional to the number of returned cans. If you earn a \$12.00 refund for 120 cans, how much do you get per can? 5. You put the letters from VACATION into a hat. If you reach in randomly, what is the probability you will pick the letter \begin{align}A\end{align}? 6. Give an example of a sixth-degree binomial. To see the Review answers, open this PDF file and look for section 12.2. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English Spanish point of discontinuity A point where the denominator of the rational function is zero. These are used to find the asymptotes of the function. compression A stretch or compression is a function transformation that makes a graph narrower or wider, without translating it horizontally or vertically. Function A function is a relation where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$. Horizontal Asymptote A horizontal asymptote is a horizontal line that indicates where a function flattens out as the independent variable gets very large or very small. A function may touch or pass through a horizontal asymptote. Oblique Asymptote An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division. Polynomial Function A polynomial function is a function defined by an expression with at least one algebraic term. Reflection Reflections are transformations that result in a "mirror image" of a parent function. They are caused by differing signs between parent and child functions. shift A shift, also known as a translation or a slide, is a transformation applied to the graph of a function that does not change the shape or orientation of the graph, only the location of the graph. shifts A shift, also known as a translation or a slide, is a transformation applied to the graph of a function that does not change the shape or orientation of the graph, only the location of the graph. Slant Asymptote A slant asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but will never reach. A slant asymptote exists when the numerator of the function is exactly one degree greater than the denominator. A slant asymptote may be found through long division. stretch A stretch or compression is a function transformation that makes a graph narrower or wider. Transformations Transformations are used to change the graph of a parent function into the graph of a more complex function. Vertical Asymptote A vertical asymptote is a vertical line marking a specific value toward which the graph of a function may approach, but will never reach.
# CBSE NCERT Solutions for Class 7 Chapter 6 NCERT Solutions are the most important and foremost tool to look up to especially when students are preparing for school and competitive examinations. NCERT Solutions involve detailed solutions of all NCERT textbook questions. CBSE NCERT Solutions for Class 7 Maths Chapter 6: Question-1: Is it possible to have a triangle with the following sides? (i) 2 cm, 3 cm, 5cm (ii) 3 cm, 6 cm, 7cm (iii) 6 cm, 3 cm, 2cm Solution: Since, a triangle is possible whose sum of the lengths of any two sides would be greater than the length of third side. (i) 2 cm, 3 cm, 5cm (ii) 3 cm, 6 cm, 7cm 2 + 3 > 5 No 3 + 6>7 Yes 2 + 5>3 Yes 6 + 7>3 Yes 3 + 5>2 Yes 3 + 7>6 Yes This triangle is not possible. This triangle is possible. (iii) 6 cm, 3 cm, 2 cm 6 + 3 >2 Yes 6 + 2>3 Yes 2 + 3>6 No This triangle is not possible Question-2: The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall? Solution: Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side. It is given that two sides of triangle are 12 cm and 15 cm. Therefore, the third side should be less than 12 + 15 = 27cm. And also the third side cannot be less than the difference of the two sides. Therefore, the third side has to be more than 15 – 12 = 3cm. Therefore, the third side could be the length more than 3 cm and less than 27cm Question-3: In triangle PQR, D is the mid-point of QR PM is ------------------ PD is Is QM = MR? Solution: Given: QD = DR ∴ PM is altitude. PD is median. No, QM ≠ MR as D is the mid-point of QR.
# Quotient group In mathematics, given a group G and a normal subgroup N of G, the quotient group, or factor group, of G over N is intuitively a group that "collapses" the normal subgroup N to the identity element. The quotient group is written G/N and is usually spoken in English as G mod N (mod is short for modulo). ## The product of subsets of a group In the following discussion, we will use a binary operation on the subsets of G: if two subsets S and T of G are given, we define their product as: ${\displaystyle ST=\{st:s\in S{\rm {~and~}}t\in T\}.}$ This operation is associative and has as identity element the singleton {e}, where e is the identity element of G. Thus, the set of all subsets of G forms a monoid under this operation. In terms of this operation we can first explain what a quotient group is, and then explain what a normal subgroup is: A quotient group of a group G is a partition of G which is itself a group under this operation. It is fully determined by the subset containing e. A normal subgroup of G is the set containing e in any such partition. The subsets in the partition are the cosets of this normal subgroup. A subgroup N of a group G is normal if and only if the coset equality aN = Na holds for all a in G. In terms of the binary operation on subsets defined above, a normal subgroup of G is a subgroup that commutes with every subset of G. ## Definition We define the set G/N to be the set of all left cosets of N in G, i.e., ${\displaystyle G/N=\{aN:a\in G\}.}$ The group operation on G/N is the product of subsets defined above. In other words, for each aN and bN in G/N, the product of aN and bN is (aN)(bN). For this operation to be closed, we must show that (aN)(bN) really is a left coset: (aN)(bN) = a(Nb)N = a(bN)N = (ab)NN = (ab)N. Note that we have already used the normality of N in this equation. Also note that because of the normality of N, we could have chosen to define G/N as the set of right cosets of N in G. Also note that because the operation is derived from the product of subsets of G, the operation is well-defined (does not depend on the particular choice of representatives), associative and has identity element N. The inverse of an element aN of G/N is a−1N. This completes the proof that G/N is a group. ## Examples • Consider the group of integers Z (under addition) and the subgroup 2Z consisting of all even integers. This is a normal subgroup, because Z is abelian. There are only two cosets: the set of even integers and the set of odd integers; therefore, the quotient group Z/2Z is the cyclic group with two elements. This quotient group is isomorphic with the set { 0, 1 } with addition modulo 2; informally, it is sometimes said that Z/2Z equals the set { 0, 1 } with addition modulo 2. • Consider the multiplicative abelian group G of complex twelfth roots of unity, which are points on the unit circle, shown on the picture on the right as colored balls with the number at each point giving its complex argument. Consider its subgroup N made of of the fourth roots of unity, shown as red balls. This normal subgroup splits the group into three cosets, shown in red, green and blue. One can check that the cosets form a group of three elements (the product of a red element with a blue element is red, the inverse of a blue element is green, etc.). Thus, the quotient group G/N is the group of three colors, which turns out to be the cyclic group with three elements. • Consider the group of real numbers R under addition, and the subgroup Z of integers. The cosets of Z in R are all sets of the form a + Z, with 0 ≤ a < 1 a real number. Adding such cosets is done by adding the corresponding real numbers, and subtracting 1 if the result is greater than or equal to 1. The quotient group R/Z is isomorphic to the circle group S1, the group of complex numbers of absolute value 1 under multiplication, or correspondingly, the group of rotations in 2D about the origin, i.e., the special orthogonal group SO(2). An isomorphism is given by f(a + Z) = exp(2πia) (see Euler's identity). • If G is the group of invertible 3×3 real matrices, and N is the subgroup of 3×3 real matrices with determinant 1, then N is normal in G (since it is the kernel of the determinant homomorphism). The cosets of N are the sets of matrices with a given determinant, and hence G/N is isomorphic to the multiplicative group of non-zero real numbers. • Consider the abelian group Z4 = Z/4Z (that is, the set { 0, 1, 2, 3 } with addition modulo 4), and its subgroup { 0, 2 }. The quotient group Z4 / { 0, 2 } is { { 0, 2 }, { 1, 3 } }. This is a group with identity element { 0, 2 }, and group operations such as { 0, 2 } + { 1, 3 } = { 1, 3 } Both the subgroup { 0, 2 } and the quotient group { { 0, 2 }, { 1, 3 } }, with their group operations induced by cyclic group Z4, are isomorphic with Z2. ## Properties Trivially, G / G is isomorphic to the trivial group (the group with one element), and G / {e} is isomorphic to G. The order of G / N is by definition equal to [G : N], the index of N in G. If G is finite, the index is also equal to the order of G divided by the order of N. Note that G / N may be finite, although both G and N are infinite (e.g. Z / 2Z). There is a "natural" surjective group homomorphism π : GG / N, sending each element g of G to the coset of N to which g belongs, that is: π(g) = gN. The mapping π is sometimes called the canonical projection of G onto G / N. Its kernel is N. There is a bijective correspondence between the subgroups of G that contain N and the subgroups of G / N; if H is a subgroup of G containing N, then the corresponding subgroup of G / N is π(H). This correspondence holds for normal subgroups of G and G / N as well, and is formalized in the lattice theorem. Several important properties of quotient groups are recorded in the fundamental theorem on homomorphisms and the isomorphism theorems. If G is abelian, nilpotent or solvable, then so is G / N. If G is cyclic or finitely generated, then so is G / N. If H is a subgroup in a finite group G, and the order of H is one half of the order of G, then H is guaranteed to be a normal subgroup, so G / H exists and is isomorphic to C2. This result can also be stated as "any subgroup of index 2 is normal", and in this form it applies also to infinite groups. Every group is isomorphic to a quotient of a free group. Sometimes, but not necessarily, a group G can be reconstructed from G / N and N, as a direct product or semidirect product. An example where it is not possible is as follows. Z4 / { 0, 2 } is isomorphic to Z2, and { 0, 2 } also, but the only semidirect product is the direct product, because Z2 has only the trivial automorphism. Therefore Z4, which is different from Z2 × Z2, cannot be reconstructed.
# JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.6 Jharkhand Board JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.6 Textbook Exercise Questions and Answers. ## JAC Board Class 10 Maths Solutions Chapter 6 Triangles Exercise 6.6 Question 1. In the given figure, PS is the bisector of ∠QPR of ΔPQR. Prove that $$\frac{QS}{SR}=\frac{PQ}{PR}$$ Solution : Construction: Through Q, draw a line parallel to PS which intersects RP extended at M. Proof: In ΔMQR, S and P are points on QR and MR respectively and PS \|\| MQ. ∴ $$\frac{QS}{SR}=\frac{MP}{PR}$$ (BPT) ……………(1) Now, PS \|\| MQ and PQ is their transversal. ∴ ∠SPQ = ∠PQM (Alternate angles) ……(2) Similarly, PS \|\| MQ and MR is their transversal. ∴ ∠RPS = ∠PMQ (Corresponding angles) …………….(3) PS is the bisector of ∠QPR. ∴ ∠SPQ = ∠RPS …………….(4) From (2), (3) and (4). ∠PQM = ∠PMQ. ∴ In ΔPMQ, MP = PQ ……….(5) From (1) and (5), we get $$\frac{QS}{SR}=\frac{PQ}{PR}$$ Question 2. In the given figure, D is a point on hypotenuse AC of ΔABC, such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that: 1. DM² = DN . MC 2. DN² = DM . AN Solution : In quadrilateral DMBN, ∠B = ∠M = ∠N = 90°. Hence, DMBN is a rectangle. ∴ DN = MB ………….. (1) and DM = NB ………….. (2) Now, BD ⊥ AC ∴ ∠BDC = 90° and ΔBDC is a right triangle in which DM is altitude on hypotenuse BC. Then, ΔBMD ~ ΔDMC ~ ΔBDC. (Theorem 6.7) ∴ $$\frac{DM}{CM}=\frac{BM}{DM}$$ ∴ DM² = BM. CM ∴ DM² = DN. MC [By (1), DN = MB] [Result (1)] Similarly, ΔADB is a right triangle in which DN is altitude on hypotenuse AB. ∴ ΔAND ~ ΔDNB ~ ΔADB (Theorem 6.7) ∴ $$\frac{DN}{BN}=\frac{AN}{DN}$$ ∴ DN2 = BN. AN ∴ DN² = DM . AN [By (2), DM = NB] [Result (2)] Question 3. In the given figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC² = AB² + BC² + 2BC . BD. Solution : ∴ AC² = AD² + DC² ∴ AD² = AC² – DC² ……………(1) ∴ AB² = AD² + DB² ∴ AD² = AB² – DB² ……………(2) From (1) and (2). AC² – DC² = AB² – DB² ∴ AC² = AB² + DC² – DB² ∴ AC² = AB2 + (BC + DB)² – DB² ∴ AC² = AB² + BC² + 2BC . DB + DB² – DB² ∴ AC² = AB² + BC² + 2BC . BD Question 4. In the given figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC² = AB² + BC² – 2BC . BD. Solution : ∴ AC² = AD² + CD² ∴ AD² = AC² – CD² ……………….(1) ∴ AB² = AD² + BD² ∴ AD² = AB² – BD² ……………….(2) From (1) and (2), AC² – CD² = AB² -BD² ∴ AC² = AB² + CD² – BD² ∴ AC² = AB² + (BC – BD)² – BD² ∴ AC² = AB² + BC² – 2BC . BD + BD² – BD² ∴ AC² = AB² + BC² – 2BC . BD Question 5. In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that: 1. AC² = AD² + BC. DM + ($$\frac{BC}{2}$$)² 2. AB² = AD² – BC. DM + ($$\frac{BC}{2}$$)² 3. AC² + AB² = 2AD² + $$\frac{1}{2}$$BC² Solution : Here, ΔAMD, ΔAMC and ΔAMB are right triangles. Also, since AD is a median, D is the midpoint of BC. ∴ CD = BD = $$\frac{BC}{2}$$ Moreover, DM = CM – CD and DM = BD – BM 1. In ΔAMC, ∠M = 90° ∴ AC² = AM² + CM² ∴ AC² = AM² + (DM + CD)² ∴ AC² = AM² + DM² + 2. DM.CD + CD² ∴ AC² = (AM² + DM²) + (2CD) (DM) + CD² ∴ AC² = AD² + BC.DM + ($$\frac{BC}{2}$$)² (∵ In ΔAMD, AD² = AM² + DM²) 2. In ΔAMB, ∠M = 90° ∴ AB² = AM² + BM² ∴ AB² = AM² + (BD – DM)² ∴ AB² = AM² + BD² – 2 BD.DM + DM² ∴ AB² = (AM² + DM²) – (2BD) · (DM) + BD² ∴ AB² = AD² – BC.DM + ($$\frac{BC}{2}$$)² (∵ In ΔAMD, AD² = AM² + DM²) 3. Now, adding the results of part (1) and (2) AC² + AB² = AD² + BC. DM + ($$\frac{BC}{2}$$)² + AD² – BC . DM + ($$\frac{BC}{2}$$)² ∴ AC² + AB² = 2AD² + 2($$\frac{\mathrm{BC}^2}{4}$$) ∴ AC² + AB² = 2AD² + $$\frac{1}{2}$$BC² [Note: In this result, if we replace BC by 2BD, we get the famous result know as Apollonius theorem : If AD is a median of ΔABC, then AB² + AC² = 2 (AD² + BD²).] Question 6. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. Solution : First of all, we prove Apollonius Theorem. In ΔABC, let AD be a median and AM be an altitude as shown in the figure. Then, AB² + AC² = AM² + BM² + AM² + CM² = 2AM² + (BD – MD)² + (CD + MD)² = 2AM² + (BD – MD)² + (BD + MD)² (∵ CD = BD) = 2AM² + 2BD² + 2MD² = 2(AM² + MD²) + 2BD² ∴ AB² + AC² = 2(AD² + BD²) Let PQRS be a parallelogram in which the diagonals bisect each other at O. Then, PO = RO = $$\frac{1}{2}$$PR and QO = SO = $$\frac{1}{2}$$QS. Now, in ΔPQR, QO is a median. ∴ PQ² + QR² = 2(QO² + PO²) ∴ PQ² + QR² = 2{($$\frac{QS}{2}$$)² + ($$\frac{PR}{2}$$)²} ∴ PQ² + QR² = $$\frac{1}{2}$$(QS² + PR² ) ………….(1) Similarly. in ΔQRS, QR² + RS² = $$\frac{1}{2}$$(QR² + PR²) ……(2) in ΔRSP, RS² + SP² = $$\frac{1}{2}$$(QS² + PR²) ……….(3) in ΔSPQ, SP² + PQ² = $$\frac{1}{2}$$(QS² + PR²) ……….(4) Adding (1), (2), (3) and (4), we get 2(PQ² + QR² + RS² + SP²) = 2(QS² + PR²) ∴ PQ² + QR² + RS² + SP² = QS² + PR² Thus, the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. Question 7. In the given figure, two chords AB and CD intersect each other at the point P. Prove that 1. ΔAPC ~ ΔDPB 2. AP . PB = CP . DP Solution : Here, ∠CAB = ∠CDB (Angles in the same segment) ∴ ∠CAP = ∠BDP Similarly, ∠ACD = ∠DBA (Angles in the same segment) ∴ ∠ACP = ∠DBP Now, in ΔAPC and ΔDPB, ∠CAP = ∠BDP and ∠ACP = ∠DBP. ∴ By AA criterion, ΔAPC ~ ΔDPB. (Result 1) ∴ $$\frac{AP}{DP}=\frac{CP}{BP}$$ ∴ AP . PB = CP . DP (Result 2) Question 8. In the given figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that 1. ΔPAC ~ ΔPDB 2. PA . PB = PC . PD Solution : ΔACD + ∠ABD = 180° Again, ∠ACD + ∠ACP = 180° (Linear pair) ∴ ∠ABD = ∠ACP ∴ ∠PBD = ∠PCA. Similarly, ∠CAB + ∠CDB = 180° (Cyclic quadrilateral) ∠CAB + ∠CAP = 180° (Linear pair) ∴ ∠CDB = ∠CAP ∴ ∠PDB = ∠PAC Now, in ΔPDB and ΔPAC, ∠PBD = ∠PCA ∠PDB = ∠PAC ∴ By AA criterion, ΔPAC ~ ΔPDB [Result (1)] ∴ $$\frac{PA}{PD}=\frac{PC}{PB}$$ ∴ PA.PB = PC.PD [Result (2)] Question 9. In the given figure, D is a point on side BC of ΔABC such that $$\frac{BD}{CD}=\frac{AB}{AC}$$. Prove that AD is the bisector of ∠BAC. Solution : Through B. draw a line parallel to AD to intersect CA extended at P. Then, in ΔPBC, A and D are points on PC and BC respectively and PB \|\| AD. ∴ $$\frac{PA}{AC}=\frac{BD}{CD}$$ Also, $$\frac{BD}{CD}=\frac{AB}{AC}$$ (Given) ∴ PA = AB Now, in ΔPAB, PA = AB ∴ ∠ABP = ∠APB …………..(1) AD \|\| BP and AB is their transversal. ∴ ∠ABP = ∠BAD (Alternate angles) …………..(2) AD \|\| BP and CP is their transversal. ∴ ∠APB = ∠CAD (Corresponding angles) …………..(3) From (1), (2) and (3), Hence, AD is the bisector of ∠BAC. Question 10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away from her and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see the given figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds? Solution : Here, ΔABC represents the initial position in which A is the tip of her fishing rod, C is the fly at the end of the string and B is the point directly under the tip of the rod. Then, in ΔABC, ∠B = 90°, AB = 1.8m and BC = 2.4 m. ∴ AC² = AB² + BC² (Pythagoras theorem) ∴ AC² = (1.8)² + (2.4)² ∴ AC² = 3.24 + 5.76 ∴ AC² = 9 ∴ AC = 3 m Hence, in the initial position, she has 3 m of string out. Length of string pulled-in in 1 sec = 5 cm ∴ Length of string pulled-in in 12 sec = 60 cm = 0.6 m Now, in the second position, the length of string AC = 3m-0.6 m = 2.4 m and AB = 1.8 m. Again, AC² = AB² + BC² ∴ (2.4)² = (1.8)² + BC² ∴ BC² = (2.4)² – (1.8)² ∴ BC² = (2.4 + 1.8) (2.4 – 1.8) ∴ BC² = 4.2 × 0.6 ∴ BC² = 2.52 ∴ BC = $$\sqrt{2.52}$$ ∴ BC= 1.59 m (approx) Now, the horizontal distance of the fly from her = BC + 1.2 m = (1.59 + 1.2) m = 2.79 m
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 10.4: Area of Composite Shapes Difficulty Level: At Grade Created by: CK-12 Estimated26 minsto complete % Progress Practice Area of Composite Shapes MEMORY METER This indicates how strong in your memory this concept is Progress Estimated26 minsto complete % Estimated26 minsto complete % MEMORY METER This indicates how strong in your memory this concept is What if you wanted to find the area of a shape that was made up of other shapes? How could you use your knowledge of the area of rectangles, parallelograms, and triangles to help you? After completing this Concept, you'll be able to answer questions like these. ### Guidance Perimeter is the distance around a shape. The perimeter of any figure must have a unit of measurement attached to it. If no specific units are given (feet, inches, centimeters, etc), write “units.” Area is the amount of space inside a figure. If two figures are congruent, they have the same area. This is the congruent areas postulate. This postulate needs no proof because congruent figures have the same amount of space inside them. Keep in mind that two figures with the same area are not necessarily congruent. A composite shape is a shape made up of other shapes. To find the area of such a shape, simply find the area of each part and add them up. The area addition postulate states that if a figure is composed of two or more parts that do not overlap each other, then the area of the figure is the sum of the areas of the parts. #### Example A Find the area of the figure below. You may assume all sides are perpendicular. Split the shape into two rectangles and find the area of each. Atop rectangleAbottom square=62=12 ft2=33=9 ft2\begin{align}A_{top \ rectangle} &= 6 \cdot 2=12 \ ft^2\\ A_{bottom \ square} &= 3 \cdot 3=9 \ ft^2\end{align} The total area is 12+9=21 ft2\begin{align}12 + 9 = 21 \ ft^2\end{align}. #### Example B • Divide the shape into two triangles and one rectangle. • Find the area of the two triangles and rectangle. • Find the area of the entire shape. Solution: • One triangle on the top and one on the right. Rectangle is the rest. • Area of triangle on top is 8(5)2=20 units2\begin{align}\frac{8(5)}{2}=20 \ units^2\end{align}. Area of triangle on right is 5(5)2=12.5 units2\begin{align}\frac{5(5)}{2}=12.5 \ units^2\end{align}. Area of rectangle is 375 units2\begin{align}375 \ units^2\end{align}. • Total area is 407.5 units2\begin{align}407.5 \ units^2\end{align}. #### Example C Find the area of the figure below. Divide the figure into a triangle and a rectangle with a small rectangle cut out of the lower right-hand corner. AAAA=Atop triangle+ArectangleAsmall triangle=(1269)+(915))(1236)=27+1359=153 units2\begin{align}A &= A_{top \ triangle}+A_{rectangle}-A_{small \ triangle}\\ A &= \left(\frac{1}{2} \cdot 6 \cdot 9\right)+(9 \cdot 15)\left) - (\frac{1}{2} \cdot 3 \cdot 6\right)\\ A &= 27+135-9\\ A &= 153 \ units^2\end{align} Watch this video for help with the Examples above. ### Vocabulary Perimeter is the distance around a shape. The perimeter of any figure must have a unit of measurement attached to it. If no specific units are given (feet, inches, centimeters, etc), write “units.” Area is the amount of space inside a figure and is measured in square units. A composite shape is a shape made up of other shapes. ### Guided Practice 1. Find the area of the rectangles and triangle. 2. Find the area of the whole shape. 1. Rectangle #1: Area =24(9+12)=504 units2\begin{align}\text{Area }= 24(9+12)=504 \ units^2\end{align}. Rectangle #2: Area =15(9+12)=315 units2\begin{align}\text{Area }=15(9+12)=315 \ units^2\end{align}. Triangle: Area =15(9)2=67.5 units2\begin{align}\text{Area }=\frac{15(9)}{2}=67.5 \ units^2\end{align}. 2. You need to subtract the area of the triangle from the bottom right corner. Total Area =504+315+67.515(12)2=796.5 units2\begin{align}\text{Total Area }=504+315+67.5-\frac{15(12)}{2}=796.5 \ units^2\end{align} ### Practice Use the picture below for questions 1-2. Both figures are squares. 1. Find the area of the unshaded region. 2. Find the area of the shaded region. Find the area of the figure below. You may assume all sides are perpendicular. Find the areas of the composite figures. Use the figure to answer the questions. 1. What is the area of the square? 2. What is the area of the triangle on the left? 3. What is the area of the composite figure? Find the area of the following figures. 1. Find the area of the unshaded region. 2. Lin bought a tract of land for a new apartment complex. The drawing below shows the measurements of the sides of the tract. Approximately how many acres of land did Lin buy? You may assume any angles that look like right angles are 90\begin{align}90^\circ\end{align}. (1 acre \begin{align}\approx\end{align} 40,000 square feet) 3. Linus has 100 ft of fencing to use in order to enclose a 1200 square foot rectangular pig pen. The pig pen is adjacent to the barn so he only needs to form three sides of the rectangular area as shown below. What dimensions should the pen be? ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English Composite A number that has more than two factors. Show Hide Details Description Difficulty Level: Authors: Tags: Subjects: Search Keywords:
# Ordered Pairs, Intercepts, And Slopes Worksheet Page 7 ADVERTISEMENT Secton . — Ordered Pars, Intercepts, and Slopes Model 3 Calculate the slope for the following equations: Find two ordered pair Equation Calculate the slope Validate solutions 2x – y = 2 Let x = 1 Let x = 4 2 1 ( ) - = y 2 y - y 2 4 ( ) - = y 2 ( ) m = 2 1 ( x - x ) 2 - = y 2 - = y 8 2 2 1 ( 2 0 - ) - = y 0 - = - y 2 8 m = - ( 2 1 ) y = 0 - = - y 6 2 Ordered pair 1 (1, 0) y = 6 slope = = 2 1 Ordered pair (4, 6) Let y = 2 2 x - = 2 2 y - y ( ) m = 2 1 2 x = 4 ( x - x ) 2 1 x = 2 ( 6 2 - ) Ordered pair 2 (2, 2) m = - ( 4 2 ) 4 slope = = 2 2 Let x = 3 Let x = 1 y = 3x + 2 y = 3 x + 2 y = 3 1 ( ) + 2 ( y y ) m = 2 1 ( x x ) y = 3 3 ( ) + 2 y = + 3 2 2 1 (4 11) y = + = 9 2 11 y = 5 m = 2 Ordered pair 1 (3, 11) Ordered pair (1, 5) 3 3 Let y = 4 4 3 = x + 2 y y ( - ) ( 7) m = 2 1 m = ( x - x ) 4 3 = x + 2 2 9 2 1 ( 5 11 - ) 2 3 = x 3 3 m = ( 1 3 - ) x = 3 2 ( 7) m = ( - 6 ) 6 2 7 = = = slope 3 x = - ( 2 ) 2 3 3 7 3 2   , 4 Ordered pair 2   1 7 3   m = 7 3     3 7   7 3   m = −   1 7   21 slope = = 3 7 Though the mathematics for calculating the slope in the example above may seem overwhelming, remember that 1) it is only mathematics and 2) you are free to choose the value you want to use for x as you solve for y. There is something of an art to choosing values that keep fractions to a minimum; you will become increasingly skilled at this as you continue to work through this chapter. ADVERTISEMENT 0 votes ### Related Categories Parent category: Education Page of 18
Courses Courses for Kids Free study material Offline Centres More Store # Factorize(1) ${x^2} + 9x + 18$ (2) ${x^2} - 10x + 9$ (3) ${y^2} + 24y + 144$ (4) $5{y^2} + 5y - 10$ (5) ${p^2} - 2p - 35$ (6) ${p^2} - 7p - 44$ (7) ${m^2} - 23m + 120$ (8) ${m^2} - 25m + 100$ (9) $3{x^2} + 14x + 15$ (10) $2{x^2} + x - 45$ (11) $20{x^2} - 26x + 8$ (12) $44{x^2} - x - 3$ Last updated date: 13th Jun 2024 Total views: 393.9k Views today: 4.93k Verified 393.9k+ views Hint: Every quadratic polynomial will have two factors. As the above 12 polynomials are quadratic, each polynomial will have two factors. First step is to divide the middle term into two according to the product of coefficients of first and third terms. We are given to factorize quadratic polynomials. Factorization of a polynomial is the process of writing the polynomial in terms of its factors. (1) ${x^2} + 9x + 18$ ${x^2} + 9x + 18$ can also be written as ${x^2} + 3x + 6x + 18 \\ \Rightarrow x\left( {x + 3} \right) + 6\left( {x + 3} \right) \\ \Rightarrow \left( {x + 3} \right)\left( {x + 6} \right) \;$ Therefore, the factors of ${x^2} + 9x + 18$ are $\left( {x + 3} \right),\left( {x + 6} \right)$ (2) ${x^2} - 10x + 9$ ${x^2} - 10x + 9$ can also be written as ${x^2} - 9x - x + 9 \\ \Rightarrow x\left( {x - 9} \right) - 1\left( {x - 9} \right) \\ \Rightarrow \left( {x - 9} \right)\left( {x - 1} \right) \;$ Therefore, the factors of ${x^2} - 10x + 9$ are $\left( {x - 9} \right),\left( {x - 1} \right)$ (3) ${y^2} + 24y + 144$ ${y^2} + 24y + 144$ can also be written as ${y^2} + 12y + 12y + 144 \\ \Rightarrow y\left( {y + 12} \right) + 12\left( {y + 12} \right) \\ \Rightarrow \left( {y + 12} \right)\left( {y + 12} \right) = {\left( {y + 12} \right)^2} \;$ ${y^2} + 24y + 144$ is ${\left( {y + 12} \right)^2}$ (4) $5{y^2} + 5y - 10$ $5{y^2} + 5y - 10$ can also be written as $5{y^2} + 10y - 5y - 10 \\ \Rightarrow 5y\left( {y + 2} \right) - 5\left( {y + 2} \right) \\ \Rightarrow \left( {y + 2} \right)\left( {5y - 5} \right) \;$ The factors of $5{y^2} + 5y - 10$ are $\left( {y + 2} \right),\left( {5y - 5} \right)$ (5) ${p^2} - 2p - 35$ ${p^2} - 2p - 35$ can also be written as ${p^2} - 7p + 5p - 35 \\ \Rightarrow p\left( {p - 7} \right) + 5\left( {p - 7} \right) \\ \Rightarrow \left( {p - 7} \right)\left( {p + 5} \right) \;$ Therefore, the factors of ${p^2} - 2p - 35$ are $\left( {p - 7} \right),\left( {p + 5} \right)$ (6) ${p^2} - 7p - 44$ ${p^2} - 7p - 44$ can also be written as ${p^2} - 11p + 4p - 44 \\ \Rightarrow p\left( {p - 11} \right) + 4\left( {p - 11} \right) \\ \Rightarrow \left( {p - 11} \right)\left( {p + 4} \right) \;$ Therefore, the factors of ${p^2} - 7p - 44$ are $\left( {p - 11} \right),\left( {p + 4} \right)$ (7) ${m^2} - 23m + 120$ ${m^2} - 23m + 120$ can also be written as ${m^2} - 15m - 8m + 120 \\ \Rightarrow m\left( {m - 15} \right) - 8\left( {m - 15} \right) \\ \Rightarrow \left( {m - 15} \right)\left( {m - 8} \right) \;$ The factors of ${m^2} - 23m + 120$ are $\left( {m - 15} \right),\left( {m - 8} \right)$ (8) ${m^2} - 25m + 100$ ${m^2} - 25m + 100$ can also be written as ${m^2} - 20m - 5m + 100 \\ \Rightarrow m\left( {m - 20} \right) - 5\left( {m - 20} \right) \\ \Rightarrow \left( {m - 20} \right)\left( {m - 5} \right) \;$ The factors of ${m^2} - 25m + 100$ are $\left( {m - 20} \right),\left( {m - 5} \right)$ (9) $3{x^2} + 14x + 15$ $3{x^2} + 14x + 15$ can also be written as $3{x^2} + 9x + 5x + 15 \\ \Rightarrow 3x\left( {x + 3} \right) + 5\left( {x + 3} \right) \\ \Rightarrow \left( {x + 3} \right)\left( {3x + 5} \right) \;$ The factors of $3{x^2} + 14x + 15$ are $\left( {x + 3} \right),\left( {3x + 5} \right)$ (10) $2{x^2} + x - 45$ $2{x^2} + x - 45$ can also be written as $2{x^2} + 10x - 9x - 45 \\ \Rightarrow 2x\left( {x + 5} \right) - 9\left( {x + 5} \right) \\ \Rightarrow \left( {x + 5} \right)\left( {2x - 9} \right) \;$ The factors of $2{x^2} + x - 45$ are $\left( {x + 5} \right),\left( {2x - 9} \right)$ (11) $20{x^2} - 26x + 8$ $20{x^2} - 26x + 8$ can also be written as $20{x^2} - 10x - 16x + 8 \\ \Rightarrow 10x\left( {2x - 1} \right) - 8\left( {2x - 1} \right) \\ \Rightarrow \left( {2x - 1} \right)\left( {10x - 8} \right) \;$ The factors of $20{x^2} - 26x + 8$ are $\left( {2x - 1} \right),\left( {10x - 8} \right)$ (12) $44{x^2} - x - 3$ $44{x^2} - x - 3$ can also be written as $44{x^2} - 11x + 12x - 3 \\ \Rightarrow 11x\left( {4x - 1} \right) + 3\left( {4x - 1} \right) \\ \Rightarrow \left( {4x - 1} \right)\left( {11x + 3} \right) \;$ The factors of $44{x^2} - x - 3$ are $\left( {4x - 1} \right),\left( {11x + 3} \right)$ Note: When we are factoring a quadratic polynomial, always remember that the middle term is divided into two in such a way that the product of the coefficients of divided terms must be equal to the product of the coefficients of first and last terms. Otherwise we would not be able to factorize the polynomial. And be careful with the signs of the terms.
# Question Bank - 100 Arithmetic Questions From Previous CAT Papers (Solved) • Solution: Karan beats Arjun by 10m in the 1st race. So when K runs 100, A runs 90. When K runs 1m, A runs 90/100 = 9/10 m. In the 2nd race, K runs 110m. So, A runs 110 * 9/10 = 99m. So K beats A by 1m. • Q9. (CAT 2004) A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he has sold. What is the current proportion of water to milk? a. 2 : 3 b. 1 : 2 c. 1 : 3 d. 3 : 4 • Solution: Initially, there is 20L of water and 80L of milk. So totally 100L of mixture with milk and water in the ratio 4:1. He sells 25L of this mixture. So he sells 5L of water and 20L of milk (As the ratio is 4:1) He has 75L of the mixture left with him in which 15L is water and 60L is milk. Now, he replenishes the quantity with water, that means he adds 25L of water. So, now there is totally 15 + 25 = 40L of water and 60L of milk, i.e. water : milk ratio is 2:3. • Q10. (CAT 2004) If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon? a. 11 km/hr b. 12 km/hr c. 13 km/hr d. 14 km/hr • Solution: At 10kmph, he reaches at 1pm, and at 15kmph, he reaches at 11am. The difference in time is 2hrs. Let the distance be dkm So, d/10 – d/15 = 2 (Difference in time) d/30 = 2. d = 60kms. So, the total distance is 60kms, time for 1st journey is 6hrs. He reaches at 1pm when at 10kmph and takes 6hrs, so he starts at 7am. Now, he needs to reach by noon. i.e. Travel time = 5hrs, and distance = 60kms. So, the speed by which he should travel is 60/5 = 12kmph. • Q11. (CAT 2004) A sprinter starts running on a circular path of radius r metres. Her average speed (in metres/minute) is πr during the first 30 seconds, πr/2 during next one minute, πr/4 during next 2 minutes, πr/8 during next 4 minutes, and so on. What is the ratio of the time taken for the nth round to that for the previous round? a. 4 b. 8 c. 16 d. 32 • Solution: There is more than 1 way to approach the solution; however, I will detail the easiest way to go about it here. We want to find the ratio of time taken for nth round : time taken for (n-1)th round It will be same as finding the ratio of time taken for 2nd round : Time taken for 1st round. 1 round = Circumference of the circle = 2πr 1st round : Speed = πr for 30 seconds. So, total distance travelled = πr/2. Speed = πr/2 for 1 minute. So, total distance travelled = πr/2. Speed = πr/4 for 2 minutes. So, total distance travelled = πr/2. Speed = πr/8 for 4 minutes. So, total distance travelled = πr/2. So, for a distance of 2πr, time taken is 7.5 minutes. 2nd round: Speed = πr/16 for 8 minutes. So, total distance travelled = πr/2. Speed = πr/32 for 16 minutes. So, total distance travelled = πr/2. Speed = πr/64 for 32 minutes. So, total distance travelled = πr/2. Speed = πr/128 for 64 minutes. So, total distance travelled = πr/2. So, for a distance of 2πr, time taken is 120 minutes. Ratio is 120:7.5 = 16:1. P.S: Essentially the time taken is a GP, as we see that speed is halved, then quartered and so on. So, we can also calculate by the following method: 1st round: Sum of 1st 4 terms of GP. 2nd round: Sum of 1st 8 terms – Sum of 1st 4 terms. Then calculate the ratio. • Q12. (CAT 2004) In an examination, there are 100 questions divided into three groups A, B and C such that each group contains at least one question. Each question in group A carries 1 mark, each question in group B carries 2 marks and each question in group C carries 3 marks. It is known that the questions in group A together carry at least 60% of the total marks. If group B contains 23 questions, then how many questions are there in group C? a. 1 b. 2 c. 3 d. Cannot be determined If group C contains 8 questions and group B carries at least 20% of the total marks, which of the following best describes the number of questions in group B? a. 11 or 12 b. 12 or 13 c. 13 or 14 d. 14 or 15 • Solution: Let there be x questions in A, y in B and z in C. So, x + y + z = 100 (As there are 100 questions). Also, questions in A have 1 mark, B have 2 marks and C have 3 marks. So, total marks is x + 2y + 3z 1st sub question : y = 23. So, x + z = 77. z = 77 – x. Total marks = x + 2y + 3z = x + 46 + 3z Substituting for z = 77 – x, we have, Total marks = z + 3z + 46 = x + 3* (77 – x) + 46 = 277 – 2x. We know that A should have atleast 60% of the total marks. i.e. x should be atleast 60% of total marks. So, x ≥ 60/100 * (277 – 2x) 10x ≥ 277 * 6 – 12x x ≥ 277 * 6/22 x ≥ 75. So, x should be minimum 76, and y = 23. So z can be only 1 (100 – 76 – 23). 2nd sub question : z = 8. So, x + y = 92. y = 92 – x. Total marks = x + 2y + 3z = x + 2y + 24. Substituting for y = 92 – x, we have, Total marks = x + 2y + 24 = x + 2* (92 – x) + 24 = 208 –x. We know that A should have atleast 60% of the total marks. i.e. x should be atleast 60% of total marks. So, x ≥ 60/100 * (208 –x) 10x ≥ 208 * 6 – 6x x ≥ 208 * 6/16 x ≥ 78. So x can be 78, 79 etc etc. z is 8. So y can be a maximum of 14. This option is there only in option 3, where y can be 13 or 14. (If you substitute y for 13 or 14, we see that total marks in y is 26 or 28, and it will be atleast 20% of the total marks) • Q13. (CAT 2003) In a coastal village, every year floods destroy exactly half of the huts. After the flood water recedes, twice the number of huts destroyed are rebuilt. The floods occurred consecutively in the last three years namely 2001, 2002 and 2003. If floods are again expected in 2004, the number of huts expected to be destroyed is: a. Less than the number of huts existing at the beginning of 2001. b. Less than the total number of huts destroyed by floods in 2001 and 2003. c. Less than the total number of huts destroyed by floods in 2002 and 2003. d. More than the total number of huts built in 2001 and 2002. • Let there be 2x buildings in 2001. Let us form a table Huts destroyed in 2004 = 3.375x: 1st option : Total huts in 2001 = 2x 2nd option: Total huts destroyed in 2001 and 2003 = 3.25x Total huts destroyed in 2002 and 2003 : 3.75x. 3.375 < 3.75. Hence this is true. Huts built in 2001 and 2002 = 5x. • Q14. (CAT 2002) Only a single rail track exists between station A and B on a railway line. One hour after the north bound superfast train N leaves station A for station B, a south passenger train S reaches station A from station B. The speed of the superfast train is twice that of a normal express train E, while the speed of a passenger train S is half that of E. On a particular day N leaves for station B from station A, 20 minutes behind the normal schedule. In order to maintain the schedule both N and S increased their speed. If the superfast train doubles its speed, what should be the ratio (approximately) of the speed of passenger train to that of the superfast train so that passenger train S reaches exactly at the scheduled time at the station A on that day. (1) 1 : 3 (2) 1 : 4 (3) 1 : 5 (4) 1 : 6 • Let speed of S = x. Then E = 2x and N = 4x. Let the distance between A & B be d. On a normal day : N goes from A to B in (d/4x) hrs. S goes from B to A in (d/x) hrs. S cannot start before N reaches there, as there is only a single track. So S can start immediately after N reaches there, or after some time. Hence the time of travel of ‘S + N’ is either equal to 1hr or lesser (depending on when S starts from B ) So (d/4x) + (d/x) ≤ 1. (5d/4x) ≤ 1 (d/x) ≤ 4/5 On the late day: N is 20 minutes late. So, the entire distance needs to be covered in 40 minutes, or 2/3 hrs. N goes from A to B in d/8x time (As speed of N is doubled) S goes from B to A in d/y time (where y is the new speed of S) Hence, the total time is (d/8x) + (d/y) = 2/3 (d/y) = (2/3) – (d/8x) (d/y) ≥ (2/3) – (4/40) (As d/x ≤ 4/5) (d/y) ≥ (2/3) – (1/10) (d/y) ≥ 17/30 Now, we have (d/x) ≤ 4/5 and (d/y) ≥ 17/30 Since (d/x) ≤ 5/4, we have (x/d) ≥ 5/4. So, we have (x/d) ≥5/4 and (d/y) ≥17/30. Hence (x/d) * (d/y) ≥ (5/4) * (17/30) Hence (x/y) ≥ 17/24. We need to find y/8x (y/x) ≤ 24/17 (y/8x) ≤ 3/17 In the given options, only 1/6 is less than 3/17. • Q15. (CAT 2002) A string of length 40 metres is divided into three parts of different lengths. The first part is three times the second part, and the last part is 23 metres smaller than the first part. Find the length of the largest part. (1) 27 (2) 4 (3) 5 (4) 9 • Solution: Let the 2nd piece length be x. 1st piece length is 3x. 3rd piece length is 3x – 23. Sum of all the pieces is 40. 3x + x + 3x – 23 = 40. 7x = 63. x = 9. Lengths of the pieces are 27, 9, 4. • Q16. (CAT 2002) On a 20km tunnel connecting two cities A and B there are three gutters. The distance between gutter 1 and 2 is half the distance between gutter 2 and 3. The distance from city A to its nearest gutter, gutter 1 is equal to the distance of city B from gutter 3. On a particular day the hospital in city A receives information that an accident has happened at the third gutter. The victim can be saved only if an operation is started within 40 minutes. An ambulance started from city A at 30 km/hr and crossed the first gutter after 5 minutes. If the driver had doubled the speed after that, what is the maximum amount of time the doctor would get to attend the patient at the hospital? Assume 1 minute is elapsed for taking the patient into and out of the ambulance. (1) 4 minutes (2) 2.5 minutes (3) 1.5 minutes (4) Patient died before reaching the hospital • Solution: Get the diagram correct first, where G1, G2 and G3 are the positions of the 3 gutters. From the diagram, it is clear that: AG1 = BG3 = x (given in question) 2 * G1G2 = G2G3 = 2y (given in question). At 30kmph, AG1 takes 5 minutes. So, AG1 = 30 * 5/60 = 2.5kms. So, 2.5 + y + 2y + 2.5 = 20 (As total distance is 20km) Hence y = 5. So, G1G2 = 5 and G2G3 = 10. Now, let us calculate the time taken: AG1 = 5 mins (given in the question) From G1, velocity is 60kmph, and to reach G3, he has to travel 15 kms. It takes 15 minutes for that. Now, from G3 to A, it is a 17.5 km journey at 60kmph. It takes 17.5 minutes. So, total time taken is 5 + 15 + 17.5 = 37.5 Also, 1 minute for taking patient in and out of ambulance. So, total time taken is 38.5 He should be saved in 40 minutes. So, the doctor gets 40 – 38.5 = 1.5 minutes. • Q17. (CAT 2002) It takes 6 technicians a total of 10 hours to build a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11 am and one technician per hour is added beginning at 5 pm, at what time will the server be complete? (1) 6:40 pm (2) 7:00 pm (3) 7:20 pm (4) 8:00 pm • Solution: 6 technicians take 10 hours to build the server. So, total man-hours required are 60. So, let us take here that the work that needs to be completed is “60”. In 10hrs, 6 technicians complete 10*6 = “60” amount of work, which is the whole work. So, in 10hrs, 1 technician completes “10” amount of work. Hence, each technician completes in 1hr, “1” amount of work. Now, from 11am – 5pm that is for 6hrs there are 6 technicians. 1 technician 1 hr = “1” amount of work. Hence 6 technicians in 6hrs complete “36” amount of work. Remaining is 60 – 36 = “24” amount of work. Next hour, there are 7 employees. Hence the work done is “7”, and total work till now is 36 + 7 = 43. Next hour, there are 8 employees. Hence the work done is “7”, and total work till now is 43 + 8 = 51. Next hour, there are 9 employees. Hence the work done is “9”, and total work till now is 51 + 9 = 60. Hence, the work is completed 3hrs after 5pm. • Q18. (CAT 2002) Three small pumps and one large pump are filling a tank. Each of the three small pumps works at 2/3rd the rate of the large pump. If all 4 pumps work at the same time, then they should fill the tank in what fraction of time that it would have taken the large pump alone? (1) 4/7 (2) 1/3 (3) 2/3 (4) 3/4 21 11 21 11 9 13 21 11
# Order of Operations𝅺 in Fractions 𝅺 The order of⁤ operations in fractions is a ⁤fundamental concept⁤ in mathematics. It helps us solve complex fractions 𝅺by determining the correct sequence ⁤of 𝅺calculations.‍ The‍ order of ‍operations ensures that we obtain the right answer every time, allowing us to⁤ solve a wide range of fraction𝅺 problems with ease. When working with 𝅺fractions, it’s necessary to ‌follow a ⁢specific order of operations to correctly simplify or solve any expression involving fractions. Let’s break it down ‌step ‍by ⁢step: 1. Parentheses ( ): Start by simplifying any expressions within parentheses first. This applies to both fractions and whole‍ numbers. 2. 𝅺 3. Exponents ^: ‌Evaluate ‍any⁣ exponentiation. If there⁣ are ⁤no exponents,⁢ move on to the next step. 4. Multiplication and ⁣Division: Perform ⁣multiplication and division operations from left to right.‍ Remember to convert ‌mixed numbers to improper ⁤fractions when necessary. 5. Addition⁣ and Subtraction: ⁣Finally, perform ⁢addition𝅺 and subtraction ‌from ‌left⁤ to right.⁢ Similar to⁢ the 𝅺previous step, convert ⁢mixed numbers to improper fractions if 𝅺needed. 𝅺 Let’s illustrate⁤ the order of operations in fractions with an ‍example: Consider the expression: ⁢ 4/5 ⁤+ 2/3𝅺 * (7/8 -𝅺 1/2) / 3/4 We ‍start‌ by simplifying the expression⁢ within the parentheses first: 4/5 + 2/3 ‌ (7/8‌ – 1/2) /‌ 3/4 ⁤ This allows ‍us to simplify further: 4/5 ⁤+ 2/3 ⁣ (7/8⁤ – 4/8) /⁣ 3/4 ‌ After subtracting the⁣ fractions within the parentheses: 4/5 + 2/3 ⁢* (3/8) / 3/4 We can‍ then evaluate the⁤ multiplication: 4/5‌ + (2/3 *⁢ 3/8) / 3/4 And now,‌ following the division: ‌⁣ 4/5 + (1/4) Lastly, we 𝅺can perform the addition: 4/5 +𝅺 1/4‍ To⁤ add these ⁢fractions with different denominators,𝅺 we must first find⁢ the‍ least common denominator (LCD), which is𝅺 20 in this case: 16/20 + 5/20 = 21/20 ⁢ ⁣ This ⁤means the final result of the expression is 21/20. Understanding and applying ⁣the‍ order of ⁤operations ‌in fractions is ⁤essential‍ for correctly solving complex fraction problems. Always ‍remember ‍to follow the𝅺 sequence of parentheses, exponents, multiplication ⁣and ‍division, and⁢ finally addition ⁤and subtraction. By‌ using⁣ this method,⁢ you will ensure ⁤accurate results and avoid common calculation errors. So⁢ the next time you encounter fractions in a ‌mathematical expression, don’t ‌forget about the⁤ order‌ of‍ operations in ⁤fractions. ⁤It will 𝅺guide you⁣ through the correct steps ‍to achieve the correct⁣ answer.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 2.1: Integers and Rational Numbers Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives • Graph and compare integers. • Classify and order rational numbers. • Find opposites of numbers. • Find absolute values. • Compare fractions to determine which is bigger. ## Graph and Compare Integers Integers are the counting numbers (1, 2, 3...), the negative opposites of the counting numbers (-1, -2, -3...), and zero. There are an infinite number of integers and examples are 0, 3, 76, -2, -11, and 995. Example 1 Compare the numbers 2 and -5 First, we will plot the two numbers on a number line. We can compare integers by noting which is the greatest and which is the least. The greatest number is farthest to the right, and the least is farthest to the left. In the diagram above, we can see that 2 is farther to the right on the number line than -5, so we say that 2 is greater than -5. We use the symbol “>” to mean “greater than”. Solution 2>5\begin{align} 2 > -5\end{align} Example 2 A frog is sitting perfectly on top of number 7 on a number line. The frog jumps randomly to the left or right, but always jumps a distance of exactly 2. Describe the set of numbers that the frog may land on, and list all the possibilities for the frog’s position after exactly 5 jumps. Solution We will graph the frog’s position, and also indicate what a jump of 2 looks like. We see that one possibility is that the frog lands on 5. Another possibility is that it lands on 9. It is clear that the frog will always land on an odd number. After one jump the frog could be on either the 9 or the 5 (but not on the 7). After two jumps the frog could be on 11, 7 or 3. By counting the number of times the frog jumps to the right or left, we may determine where the frog lands. After five jumps, there are many possible locations for the frog. There is a systematic way to determine the possible locations by how many times the frog jumped right, and by how many times the frog jumped left. RRRRR=5 jumps rightRRRRL=4 jumps right, 1 jump leftRRRLL=3 jumps right, 2 jumps leftRRLLL=2 jumps right, 3 jumps leftRLLLL=1 jump right, 3 jumps leftLLLLL=5 jumps leftlocation=7+(52)=17location=7+(32)=13location=7+(12)=9location=7(12)=5location=7(32)=1location=7(52)=3\begin{align}& \text{RRRRR} = 5\ \text{jumps right} && \text{location} = 7 + (5 \cdot 2) = 17 \\ & \text{RRRRL} = 4\ \text{jumps right, 1 jump left} && \text{location} = 7 + (3 \cdot 2) = 13 \\ & \text{RRRLL} = 3\ \text{jumps right, 2 jumps left} && \text{location} = 7 + (1 \cdot 2) = 9 \\ & \text{RRLLL} = 2\ \text{jumps right, 3 jumps left} && \text{location} = 7 - (1 \cdot 2) = 5 \\ & \text{RLLLL} = 1\ \text{jump right, 3 jumps left} && \text{location} =7 - (3 \cdot 2) = 1 \\ & \text{LLLLL} = 5\ \text{jumps left} && \text{location} = 7 - (5 \cdot 2) = -3\end{align} These are the possible locations of the frog after exactly five jumps. Notice that the order does not matter: three jumps right, one left and one right is the same as four jumps to the right and one to the left. ## Classifying Rational Numbers When we divide an integer by another integer (not zero) we get what we call a rational number. It is called this because it is the ratio of one number to another. For example, if we divide one integer a\begin{align}a\end{align} by a second integer b\begin{align}b\end{align} the rational number we get is (ab)\begin{align}\left( \frac{a}{b} \right )\end{align}, provided that b\begin{align}b\end{align} is not zero. When we write a rational number like this, the top number is called the numerator. The bottom number is called the denominator. You can think of the rational number as a fraction of a cake. If you cut the cake into b\begin{align}b\end{align} slices, your share is a\begin{align}a\end{align} of those slices. For example, when we see the rational number (12)\begin{align}\left (\frac{1}{2}\right )\end{align}, we imagine cutting the cake into two parts. Our share is one of those parts. Visually, the rational number (12)\begin{align}\left (\frac{1}{2}\right )\end{align} looks like this. With the rational number (34)\begin{align}\left (\frac{3}{4}\right )\end{align}, we cut the cake into four parts and our share is three of those parts. Visually, the rational number (34)\begin{align}\left (\frac{3}{4}\right )\end{align} looks like this. The rational number (910)\begin{align}\left (\frac{9}{10}\right )\end{align} represents nine slices of a cake that has been cut into ten pieces. Visually, the rational number (910)\begin{align}\left (\frac{9}{10}\right )\end{align} looks like this. Proper fractions are rational numbers where the numerator (the number on the top) is less than the denominator (the number on the bottom). A proper fraction represents a number less than one. With a proper fraction you always end up with less than a whole cake! Improper fractions are rational numbers where the numerator is greater than the denominator. Improper fractions can be rewritten as a mixed number – an integer plus a proper fraction. An improper fraction represents a number greater than one. Equivalent fractions are two fractions that give the same numerical value when evaluated. For example, look at a visual representation of the rational number (24)\begin{align}\left (\frac{2}{4}\right )\end{align}. You can see that the shaded region is identical in size to that of the rational number one-half (12)\begin{align}\left (\frac{1}{2}\right )\end{align}. We can write out the prime factors of both the numerator and the denominator and cancel matching factors that appear in both the numerator and denominator. \begin{align}\left (\frac{2} {4}\right ) = \left (\frac{\cancel{2} \cdot 1} {\cancel{2} \cdot 2 \cdot 1}\right ) && \text{We then re-multiply the remaining factors.} && \left ( \frac{2}{4} \right )=\left ( \frac{1}{2} \right )\end{align} This process is called reducing the fraction, or writing the fraction in lowest terms. Reducing a fraction does not change the value of the fraction. It just simplifies the way we write it. When we have canceled all common factors, we have a fraction in its simplest form. Example 3 Classify and simplify the following rational numbers a) \begin{align}\left ( \frac{3}{7} \right )\end{align} b) \begin{align}\left ( \frac{9}{3} \right )\end{align} c) \begin{align} \left ( \frac{50}{60} \right )\end{align} a) 3 and 7 are both prime – there is no simpler form for this rational number so... Solution \begin{align}\frac{3}{7}\end{align} is already in its simplest form. b) \begin{align}9 = 3\cdot 3\end{align} and 3 is prime. We rewrite the fraction as: \begin{align}\left (\frac{9} {3}\right ) = \left (\frac{\cancel{3} \cdot 3 \cdot 1} {\cancel{3} \cdot 1} \right ).\end{align} \begin{align}9 > 3\end{align} so... Solution \begin{align}\frac{9}{3}\end{align} is an improper fraction and simplifies to \begin{align}\frac{3}{1}\end{align} or simply 3. c) \begin{align} 50 = 5 \cdot 5 \cdot 2\end{align} and \begin{align}60 = 5 \cdot 3 \cdot 2 \cdot 2\end{align}. We rewrite the fraction thus: \begin{align}\frac{50} {60} = \left (\frac{\cancel{5} \cdot 5 \cdot \cancel{2} \cdot 1} {\cancel{5} \cdot 5 \cdot \cancel{2} \cdot 2 \cdot 1} \right ). \end{align} \begin{align} 50 <60 \end{align} so... Solution \begin{align}\frac{50}{60}\end{align} is a proper fraction and simplifies to \begin{align}\frac{5}{6}\end{align}. ## Order Rational Numbers Ordering rational numbers is simply a case of arranging numbers in order of increasing value. We write the numbers with the least (most negative) first and the greatest (most positive) last. Example 4 Put the following fractions in order from least to greatest: \begin{align} \frac {1}{2}, \frac {3}{4}, \frac {2}{3}\end{align} Let’s draw out a representation of each fraction. We can see visually that the largest number is \begin{align}\frac{3}{4}\end{align} and the smallest is \begin{align}\frac{1}{2}\end{align}: Solution \begin{align} \frac {1}{2} < \frac {2}{3} < \frac {3}{4}\end{align} With simple fractions, it is easy to order them. Think of the example above. We know that one-half is greater than one quarter, and we know that two thirds is bigger than one-half. With more complex fractions, however we need to find a better way to compare. Example 5 Which is greater, \begin{align}\frac {3}{7}\end{align} or \begin{align}\frac {4}{9}\end{align}? In order to determine this we need to find a way to rewrite the fractions so that we can better compare them. We know that we can write equivalent fractions for both of these. If we make the denominators in our equivalent fractions the same, then we can compare them directly. We are looking for the lowest common multiple of each of the denominators. This is called finding the lowest common denominator (LCD). The lowest common multiple of 7 and 9 is 63. Our fraction will be represented by a shape divided into 63 sections. This time we will use a rectangle cut into 9 by \begin{align}7 = 63\end{align} pieces: 7 divides into 63 nine times so: \begin{align} \left ( \frac{3}{7} \right )= \frac {9}{9}\left ( \frac{3}{7} \right )=\left ( \frac{27}{63} \right )\end{align} Note that multiplying by \begin{align}\frac{9}{9}\end{align} is the same as multiplying by 1. Therefore, \begin{align}\frac{27}{63}\end{align} is an equivalent fraction to \begin{align}\frac{3}{7}\end{align}. Here it is shown visually. 9 divides into 63 seven times so: \begin{align} \left ( \frac{4}{9} \right )=\frac {7}{7}\left ( \frac{4}{9} \right )=\left ( \frac{28}{63} \right )\end{align} \begin{align}\frac{28}{63}\end{align} is an equivalent fraction to \begin{align}\frac{4}{9}\end{align}. Here it is shown visually. By writing the fractions over a common denominator of 63, you can easily compare them. Here we take the 28 shaded boxes out of 63 (from our image of \begin{align}\frac{4}{9}\end{align} above) and arrange them in a way that makes it easy to compare with our representation of \begin{align}\frac{3}{7}\end{align}. Notice there is one little square “left over”. Solution Since \begin{align}\frac {28}{63}\end{align} is greater than \begin{align}\frac {27}{63}\end{align}, \begin{align}\frac {4}{9}\end{align} is greater than \begin{align}\frac {3}{7}\end{align}. Remember To compare rational numbers re-write them with a common denominator. ## Find the Opposites of Numbers Every number has an opposite. On the number line, a number and its opposite are opposite each other. In other words, they are the same distance from zero, but they are on opposite sides of the number line. By definition, the opposite of zero is zero. Example 6 Find the value of each of the following. a) \begin{align}3 + (-3)\end{align} b) \begin{align}5 + (-5) \end{align} c) \begin{align}(-11.5) + (11.5) \end{align} d) \begin{align}\frac{3}{7}+\frac{-3}{7}\end{align} Each of the pairs of numbers in the above example are opposites. The opposite of 3 is (-3), the opposite of 5 is (-5), the opposite of (-11.5) is 11.5 and the opposite of \begin{align}\frac{3}{7}\end{align} is \begin{align}\frac{3}{7}\end{align}. Solution The value of each and every sum in this problem is 0. Example 7 Find the opposite of each of the following: a) 19.6 b) \begin{align}-\frac{4}{9}\end{align} c) \begin{align}x\end{align} d) \begin{align}xy^2 \end{align} e) \begin{align}(x-3)\end{align} Since we know that opposite numbers are on opposite sides of zero, we can simply multiply each expression by -1. This changes the sign of the number to its opposite. a) Solution The opposite of 19.6 is -19.6. b) Solution The opposite of is \begin{align}-\frac{4}{9}\end{align} is \begin{align}\frac{4}{9}\end{align}. c) Solution The opposite of \begin{align}x\end{align} is \begin{align}-x\end{align}. d) Solution The opposite of \begin{align}xy^2\end{align} is \begin{align}-xy^2\end{align}. e) Solution The opposite of \begin{align}(x-3)\end{align} is \begin{align}-(x-3)=3-x\end{align}. Note: With the last example you must multiply the entire expression by -1. A common mistake in this example is to assume that the opposite of \begin{align}(x-3)\end{align} is \begin{align}(x+3)\end{align}. DO NOT MAKE THIS MISTAKE! ## Find absolute values When we talk about absolute value, we are talking about distances on the number line. For example, the number 7 is 7 units away from zero. The number -7 is also 7 units away from zero. The absolute value of a number is the distance it is from zero, so the absolute value of 7 and the absolute value of -7 are both 7. We write the absolute value of -7 like this \begin{align}\|-7\|\end{align} We read the expression \begin{align}\|x\|\end{align} like this “the absolute value of \begin{align}x\end{align}.” • Treat absolute value expressions like parentheses. If there is an operation inside the absolute value symbols evaluate that operation first. • The absolute value of a number or an expression is always positive or zero. It cannot be negative. With absolute value, we are only interested in how far a number is from zero, and not the direction. Example 8 Evaluate the following absolute value expressions. a) \begin{align}\|5+4\|\end{align} b) \begin{align}3-\|4-9\|\end{align} c) \begin{align}\|-5-11\|\end{align} d) \begin{align}-\|7-22\|\end{align} Remember to treat any expressions inside the absolute value sign as if they were inside parentheses, and evaluate them first. Solution a) \begin{align}\|5+4\| &= \|9\|\\ &= 9 \end{align} b) \begin{align}3-\|4-9\| &= 3-\|-5\|\\ &= 3-5\\ &= -2\end{align} c) \begin{align}\|-5-11\| &= \|-16\|\\ &= 16\end{align} d) \begin{align}-\|7-22\| &= -\|-15\|\\ &= -(15)\\ &= -15\end{align} ## Lesson Summary • Integers (or whole numbers) are the counting numbers (1, 2, 3...), the negative counting numbers (-1, -2, -3...), and zero. • A rational number is the ratio of one integer to another, like \begin{align}\frac{a}{b}\end{align} or \begin{align}\frac{3}{5}\end{align}. The top number is called the numerator and the bottom number (which can not be zero) is called the denominator. • Proper fractions are rational numbers where the numerator is less than the denominator. • Improper fractions are rational numbers where the numerator is greater than the denominator. • Equivalent fractions are two fractions that give the same numerical value when evaluated. • To reduce a fraction (write it in simplest form) write out all prime factors of the numerator and denominator, cancel common factors, then recombine. • To compare two fractions it helps to write them with a common denominator: the same integer on the bottom of each fraction. • The absolute value of a number is the distance it is from zero on the number line. The absolute value of a number or expression will always be positive or zero. • Two numbers are opposites if they are the same distance from zero on the number line and on opposite sides of zero. The opposite of an expression can be found by multiplying the entire expression by -1. ## Review Questions 1. The tick-marks on the number line represent evenly spaced integers. Find the values of \begin{align}a, b, c, d\end{align} and \begin{align}e\end{align}. 2. Determine what fraction of the whole each shaded region represents. 3. Place the following sets of rational numbers in order, from least to greatest. 1. \begin{align} \frac {1}{2},\frac {1}{3},\frac {1}{4}\end{align} 2. \begin{align} \frac {11}{12},\frac {12}{11},\frac {13}{10}\end{align} 3. \begin{align} \frac {39}{60},\frac {49}{80},\frac {59}{100}\end{align} 4. \begin{align} \frac {7}{11},\frac {8}{13},\frac {12}{19}\end{align} 4. Find the simplest form of the following rational numbers. 1. \begin{align} \frac {22}{44}\end{align} 2. \begin{align} \frac {9}{27}\end{align} 3. \begin{align} \frac {12}{18}\end{align} 4. \begin{align} \frac {315}{420}\end{align} 5. Find the opposite of each of the following. 1. 1.001 2. \begin{align}(5-11) \end{align} 3. \begin{align}(x+ y)\end{align} 4. \begin{align}(x-y)\end{align} 6. Simplify the following absolute value expressions. 1. \begin{align}11-\|-4\|\end{align} 2. \begin{align} \|4-9\|-\|-5\|\end{align} 3. \begin{align}\|-5-11\| \end{align} 4. \begin{align} 7-\|22-15-19\| \end{align} 5. \begin{align} -\|-7\| \end{align} 6. \begin{align} \|-2-88\|-\|88+2\|\end{align} 1. \begin{align}a = -3; b = 3; c = 9; d = 12; e = 15\end{align} 2. \begin{align} a =\frac{1}{3}; b = \frac{7}{12}; c = \frac{22}{35}\end{align} 1. \begin{align} \frac {1}{4} < \frac {1}{3} < \frac {1}{2}\end{align} 2. \begin{align} \frac {11}{12} < \frac {12}{11} < \frac {13}{10}\end{align} 3. \begin{align} \frac {59}{100} < \frac {49}{80} < \frac {39}{60}\end{align} 4. \begin{align} \frac {8}{13} < \frac {12}{19} < \frac {7}{11}\end{align} 1. \begin{align} \frac {1}{2}\end{align} 2. \begin{align} \frac {1}{3}\end{align} 3. \begin{align} \frac {2}{3}\end{align} 4. \begin{align} \frac {3}{4}\end{align} 1. -1.001 2. \begin{align}6 -(x + y)\end{align} 3. \begin{align}(y-x)\end{align} 1. 7 2. 0 3. 16 4. -5 5. -7 6. 0 ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:
# How do you find f'(x) using the definition of a derivative for f(x)=sqrt(2x-1)? Oct 5, 2015 See the explanation. #### Explanation: $f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$ $f ' \left(x\right) = {\lim}_{h \to 0} \frac{\sqrt{2 \left(x + h\right) - 1} - \sqrt{2 x - 1}}{h} =$ $= {\lim}_{h \to 0} \frac{\sqrt{2 \left(x + h\right) - 1} - \sqrt{2 x - 1}}{h} \cdot \frac{\sqrt{2 \left(x + h\right) + 1} + \sqrt{2 x - 1}}{\sqrt{2 \left(x + h\right) - 1} + \sqrt{2 x - 1}}$ $= {\lim}_{h \to 0} \frac{\left(2 \left(x + h\right) - 1\right) - \left(2 x - 1\right)}{h \left(\sqrt{2 \left(x + h\right) - 1} + \sqrt{2 x - 1}\right)}$ $= {\lim}_{h \to 0} \frac{2 x + 2 h - 1 - 2 x + 1}{h \left(\sqrt{2 \left(x + h\right) - 1} + \sqrt{2 x - 1}\right)}$ $= {\lim}_{h \to 0} \frac{2 h}{h \left(\sqrt{2 \left(x + h\right) - 1} + \sqrt{2 x - 1}\right)}$ $= {\lim}_{h \to 0} \frac{2}{\sqrt{2 \left(x + h\right) - 1} + \sqrt{2 x - 1}}$ $= {\lim}_{h \to 0} \frac{2}{2 \sqrt{2 x - 1}} = \frac{1}{\sqrt{2 x - 1}}$
# The following table describes the student population in a large colleg The following table describes the student population in a large college. $\begin{array}{\|c\|c\|} \hline Class&Male&Female \\ \hline Freshman (\%)&20&15 \\ \hline Sophomore (\%)&13&10\\ \hline Juior (\%)&10&10\\ \hline Senior (\%)&17&5\\ \hline \end{array}$ a) Find the probability that a randomly selected student is female. (Enter your probability as a fraction.) b) Find the probability that a randomly selected student is a junior. (Enter your probability as a fraction.) c) If the selected student is a junior, find the probability that the student is female. (Enter your probability as a fraction.) ## Want to know more about Probability? • Questions are typically answered in as fast as 30 minutes ### Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Fommeirj Step 1 There is total $$\displaystyle{\left({20}+{13}+{10}+{17}+{15}+{10}+{10}+{5}\right)}\%={100}\%$$ students Among them $$\displaystyle{\left({15}+{10}+{10}+{5}\right)}\%={40}\%$$ is female. So the probability of female student $$\displaystyle={\frac{{{40}\%}}{{{100}\%}}}={\frac{{{2}}}{{{5}}}}$$ Step 2 There are $$\displaystyle{\left({10}+{10}\right)}\%={20}\%$$ juniors. So the probability of junior $$\displaystyle={\frac{{{20}\%}}{{{100}\%}}}={\frac{{{1}}}{{{5}}}}$$ Step 3 c) $$\displaystyle{\left({10}+{10}\right)}\%={20}\%$$ students are junior. Among them 10% is female. So the required probability $$\displaystyle={\frac{{{10}\%}}{{{20}\%}}}={\frac{{{1}}}{{{2}}}}$$ Answer(a): $$\displaystyle{\frac{{{2}}}{{{5}}}}$$ Answer (b): $$\displaystyle{\frac{{{1}}}{{{5}}}}$$ Answer (c): $$\displaystyle{\frac{{{1}}}{{{2}}}}$$
## Elementary Algebra Published by Cengage Learning # Chapter 2 - Real Numbers - 2.1 - Rational Numbers: Multiplication and Division - Problem Set 2.1: 86 #### Answer It takes 36 seconds to download 16 songs #### Work Step by Step It takes 2$\frac{1}{4}$ seconds to download 1 song, which is equal to $\frac{9}{4}$ seconds. To download 16 songs, we multiply the time needed for 1 song by 16. So, $\frac{9}{4}$ $\times$ 16 = $\frac{9 \times 16}{4}$ Factor the numerator and the denominator: $\frac{9\ \times\ 4\ \times 4}{4}$ = Cancel the common factors in the numerator and the denominator to obtain: 9 $\times$ 4 = 36 seconds After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
# What is 51/73 as a decimal? ## Solution and how to convert 51 / 73 into a decimal 51 / 73 = 0.699 To convert 51/73 into 0.699, a student must understand why and how. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. The difference between using a fraction or a decimal depends on the situation. Fractions can be used to represent parts of an object like 1/8 of a pizza while decimals represent a comparison of a whole number like \$0.25 USD. If we need to convert a fraction quickly, let's find out how and when we should. ## 51/73 is 51 divided by 73 Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 51 divided by 73. We must divide 51 into 73 to find out how many whole parts it will have plus representing the remainder in decimal form. This is our equation: ### Numerator: 51 • Numerators are the number of parts to the equation, showed above the vinculum or fraction bar. Any value greater than fifty will be more difficult to covert to a decimal. The bad news is that it's an odd number which makes it harder to covert in your head. Large numerators make converting fractions more complex. So how does our denominator stack up? ### Denominator: 73 • Denominators are the total numerical value for the fraction and are located below the fraction line or vinculum. Larger values over fifty like 73 makes conversion to decimals tougher. But the bad news is that odd numbers are tougher to simplify. Unfortunately and odd denominator is difficult to simplify unless it's divisible by 3, 5 or 7. Have no fear, large two-digit denominators are all bark no bite. Let's start converting! ## Converting 51/73 to 0.699 ### Step 1: Set your long division bracket: denominator / numerator $$\require{enclose} 73 \enclose{longdiv}{ 51 }$$ Use long division to solve step one. This method allows us to solve for pieces of the equation rather than trying to do it all at once. ### Step 2: Extend your division problem $$\require{enclose} 00. \\ 73 \enclose{longdiv}{ 51.0 }$$ Uh oh. 73 cannot be divided into 51. Place a decimal point in your answer and add a zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 73 into 51 + 0 or 510. ### Step 3: Solve for how many whole groups you can divide 73 into 510 $$\require{enclose} 00.6 \\ 73 \enclose{longdiv}{ 51.0 }$$ We can now pull 438 whole groups from the equation. Multiple this number by our furthest left number, 73, (remember, left-to-right long division) to get our first number to our conversion. ### Step 4: Subtract the remainder $$\require{enclose} 00.6 \\ 73 \enclose{longdiv}{ 51.0 } \\ \underline{ 438 \phantom{00} } \\ 72 \phantom{0}$$ If your remainder is zero, that's it! If you have a remainder over 73, go back. Your solution will need a bit of adjustment. If you have a number less than 73, continue! ### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit. In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value. ### Why should you convert between fractions, decimals, and percentages? Converting fractions into decimals are used in everyday life, though we don't always notice. They each bring clarity to numbers and values of every day life. Same goes for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But 51/73 and 0.699 bring clarity and value to numbers in every day life. Here are examples of when we should use each. ### When you should convert 51/73 into a decimal Dining - We don't give a tip of 51/73 of the bill (technically we do, but that sounds weird doesn't it?). We give a 69% tip or 0.699 of the entire bill. ### When to convert 0.699 to 51/73 as a fraction Time - spoken time is used in many forms. But we don't say It's '2.5 o'clock'. We'd say it's 'half passed two'. ### Practice Decimal Conversion with your Classroom • If 51/73 = 0.699 what would it be as a percentage? • What is 1 + 51/73 in decimal form? • What is 1 - 51/73 in decimal form? • If we switched the numerator and denominator, what would be our new fraction? • What is 0.699 + 1/2?
# A ball is tossed upward from the ground. Its height A ball is tossed upward from the ground. Its height in feet above ground after t seconds is given by the function $h\left(t\right)=-16{t}^{2}+24t$. Find the maximum height of the ball and the number of seconds it took for the ball to reach the maximum height. You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Viktor Wiley 1) Concept: The maximum height of the ball is the maximum value of h(t). The degree of the function $h\left(t\right)=-16{t}^{2}+24t$ is 2 and hence it is a quadratic function. The graphical representation of the quadratic function is a parabola. The coefficient of the ${t}^{2}$ is -16. Therefore, it is a parabola that opens downward. So, the h(t) or the y coordinate of the vertex gives the maximum height and the x coordinate of the vertex gives the number of seconds it took for the ball to reach the maximum height. Step 1: Compare h(t) with the general quadratic equation $a{x}^{2}+bx+c$. Find the values of a and b. Step 2: Find the vertex using the formula $\left(\frac{-b}{2a},h\left(\frac{-b}{2a}\right)\right)$. Step 3: The x coordinate of the vertex is the numbers of seconds the ball takes to reach the maximum height and the y coordinate of the vertex is the maximum height of the ball. 2) Calculation: Comparing $h\left(t\right)=-16{t}^{2}+24t$ with $a{x}^{2}+bx+c$, we have $a=-16.b=24.$ Now, to find the x-coordinate of the vertex, substitute the values of a and b in $\frac{-b}{2a}$. $\frac{-b}{2a}=\frac{-24}{2\left(-16\right)}$ 24 and 16 are multiples of 2. So,we can write them as factors of 2 and the common factors can be cancelled out. Since both the numerator and the denominator has negative sign, the fraction becomes positive. $\frac{-24}{2\left(-16\right)}=\frac{2×2×2×3}{2×2×2×2×2}=\frac{3}{4}$ Now, find $h\left(\frac{3}{4}\right)$. Substitute $\frac{3}{4}$ for t in the function $h\left(t\right)=-16{t}^{2}+24t$. $h\left(\frac{3}{4}\right)=-16{\left(\frac{3}{4}\right)}^{2}+24\left(\frac{3}{4}\right)$ Again, write the numbers as product of their factors. $h\left(\frac{3}{4}\right)=-2×2×2×2×\left(\frac{3×3}{4×4}\right)+2×2×2×3\left(\frac{3}{4}\right)$ Cancelling out the common factors in the numerator and the denominator, we have, $h\left(\frac{3}{2}\right)=-2×2×2×2\left(\frac{3×3}{2×2×2×2}\right)+2×2×2×3\left(\frac{3}{2×2}\right)$ Evaluate using order of operations. $h\left(\frac{3}{2}\right)=-3×3+2×3×3$ $h\left(\frac{3}{2}\right)=-9+18=9$ Now, we have, $\frac{-b}{2a}=\frac{3}{4}$ and $h\left(\frac{-b}{2a}\right)=9$. Therefore, the vertex is $\left(\frac{3}{4},9\right)$ So, it took 0.75 seconds for the ball to reach the ground. Conclusion: The maximum height of the ball is 9 feet and it took $\frac{3}{4}$ or 0.75 seconds for the ball to reach the ground.
# Linear Algebra: matrix review 2022-01-15 02:11:53 ## 1. System of linear equations ： The constant terms are 0 Namely ：n A system of homogeneous linear equations （ One didn't do it 0, Is a system of nonhomogeneous linear equations ） The coefficients in front of the unknowns can · form ：n Order matrix ## 2. matrix ：m That's ok n Column ： from mn Composed of a number table ： The number of rows and columns are n: be called n Order matrix or n Square matrix ## 3. Matrix operation ： A number multiplied by a matrix is equal to this number multiplied by each element of the matrix Matrix times matrix ：（ The number of columns of the first matrix is equal to the number of rows of the second matrix ） sm mx Multiply a row by a column （ Multiply and add the corresponding elements ） An element that makes up the new matrix 2 That's ok 3 Column multiply 3 That's ok 2 Column become ：4 Elements （ A matrix of two rows and two columns ） The sign of the matrix is ：A,B,C Inverse matrix ： It's a matrix The other matrix is equal to E( Unit matrix ) Unit matrix ： Is that every element except the diagonal element is 1 The other elements are 0; AB=E. B Namely A The inverse matrix . Symbol ：A Superscript is -1. Inverse matrix A Equal to determinant A Multiply the reciprocal of the adjoint matrix A. Adjoint matrix A： It's the determinant A A new matrix consisting of the algebraic cofactor values of each element The location here ： The columns of a matrix are composed of rows of a determinant Subscript of element The rows of a matrix are composed of columns of a determinant Therefore, in the determinant, the algebraic cofactor values of each element are composed of a number table , Turn it upside down （ Is the adjoint matrix ） Symbol ：A* ## 4. How to solve n A system of linear equations ？ Using Kramer's law ： n The coefficients of a system of linear equations ： A determinant consisting of the coefficients of unknowns A It's not equal to 0 The equation has a unique solution ： The first unknown is equal to the determinant A Multiply the reciprocal of the value by A1. A1: It's a determinant , It is the element of the first column that replaces the resulting matrix with a constant term A1. This A1 The corresponding determinant A1. The second unknown is the same ## Exercise questions 1： Inverse matrix method to do this problem ： \|A\| It's not equal to 0, therefore A reversible , Matrix of unknowns = Inverse matrix of matrix with unknown coefficients * Constant term matrix https://chowdera.com/2021/12/202112122241266007.html
# Lesson 3 Ubiquemos más puntos ## Warm-up: Observa y pregúntate: Puntos con cero (10 minutes) ### Narrative The purpose of this warm-up is for students to think about points on the axes. In previous lessons they have plotted points with non-zero coordinates. Thinking about the points with zero prepares them for plotting points on the horizontal and vertical axes which they will do in this lesson. ### Launch • Groups of 2 • Display the image. • “¿Qué observan? ¿Qué se preguntan?” // “What do you notice? What do you wonder?” • 1 minute: quiet think time ### Activity • “Discutan con su compañero lo que pensaron” // “Discuss your thinking with your partner.” • 1 minute: partner discussion • Share and record responses. ### Student Facing ¿Qué observas? ¿Qué te preguntas? ### Activity Synthesis • “¿Cómo podemos describir la ubicación del punto?” // “How can we describe the location of the point?” (It is at the bottom left of the grid.) • “Las coordenadas de este punto son $$(0, 0)$$. ¿Cuáles serían las coordenadas del punto si lo moviéramos 2 unidades hacia arriba?” // “The coordinates of this point are $$(0, 0)$$. What would be the coordinates of the point if we moved it up 2 units?” (0, 2) ## Activity 1: ¿Cuál es el punto? (20 minutes) ### Narrative The purpose of this activity is for students to plot several points with the same vertical or horizontal coordinate and observe that they lie on a horizontal or vertical line respectively (MP7). Students also plot points on the axes for the first time. Before plotting the points on a grid with gridlines, students first estimate the location of the points. This encourages them to think about the coordinates as distances (from the vertical axis for the first coordinate and from the horizontal axis for the second coordinate). MLR8 Discussion Supports. Display sentence frames to support partner discussion: “Primero, yo _____ porque . . .” // “First, I _____ because . . .” and “Observé _____, entonces yo . . .” // “I noticed _____ so I . . . .” Action and Expression: Internalize Executive Functions. Invite students to verbalize their strategy for estimating and plotting points before they begin. Students can speak quietly to themselves, or share with a partner. Supports accessibility for: Organization, Conceptual Processing, Language ### Launch • Groups of 2 • “Ustedes y su compañero van a completar cada uno un grupo distinto de 4 problemas. Cuando terminen, discutan su trabajo con su compañero” // “You and your partner will each complete a different set of 4 problems independently. After you’re done, discuss your work with your partner.” ### Activity • 5–7 minutes: independent work time • 5 minutes: partner discussion • Monitor for students who: • use the halfway point on each axis as a benchmark for the coordinate grid without gridlines • start at zero and count spaces along each axis for the marked coordinate grid with gridlines • recognize the points should be aligned because they share a common horizontal or vertical coordinate ### Student Facing Compañero A 1. Estima la ubicación de cada punto. $$A$$ $$(5,1)$$ $$B$$ $$(5,2)$$ $$C$$ $$(5,3)$$ $$D$$ $$(5,4)$$ 3. ¿Qué tienen los puntos en común? 4. Ubica el punto de coordenadas $$(5,0)$$ en la cuadrícula de coordenadas. Compañero B 1. Estima la ubicación de cada punto. $$A$$ $$(4,3)$$ $$B$$ $$(5,3)$$ $$C$$ $$(6,3)$$ $$D$$ $$(7,3)$$ 3. ¿Qué tienen los puntos en común? 4. Ubica el punto de coordenadas $$(0, 3)$$ en la cuadrícula de coordenadas. ### Student Response If students do not reasonably estimate a location for a point on the blank grid, point to the location half way between 0 and 5 on the horizontal axis and ask, “¿Qué número podría ir aquí?” // “What number might go here?” ### Activity Synthesis • Ask previously identified students to share their thinking. • “¿Qué podemos decir sobre un conjunto de puntos cuando ellos tienen la misma primera coordenada?” // “What can we say about a set of points when they share the same first coordinate?” (They will be on the same vertical line.) • Display image from student solution showing points with first coordinate 5. • “¿Cómo supieron dónde ubicar el punto que tenía coordenadas $$(5,0)$$?” // “How did you know where to put the point with coordinates $$(5,0)$$?” (I put it on the horizontal axis. I went over 5 but did not go up at all.) • “¿Qué ocurre cuando los puntos de un conjunto tienen la misma segunda coordenada?” // “What happens when a set of points share the same second coordinate?” (They will be on the same horizontal line.) • Display image from student solution showing points with second coordinate 3. • “¿Qué nos dice el cero que aparece en (0,3)?” // “What does the zero in (0,3) tell us?” (It means the point will be on line zero of the horizontal axis, which is the vertical axis.) • $$(0, 0)$$ es un punto importante porque es donde empezamos cuando ubicamos un punto en la cuadrícula de coordenadas. Encuentren $$(0, 0)$$ en la cuadrícula con la que han estado trabajando” // $$(0, 0)$$ is an important point because it's where we start when we plot a point on the coordinate grid. Find $$(0, 0)$$ on the grid you have been working with.” ## Activity 2: Ubiquemos puntos sin una cuadrícula (15 minutes) ### Narrative In the previous activity, students noticed that points with the same vertical coordinate lie on a horizontal line and points with the same horizontal coordinate lie on a vertical line. The purpose of this activity is to reinforce this idea by asking students to plot points on a blank coordinate grid given a single point. Using this point, they can plot other points with the same horizontal or vertical coordinate. Some students may keep one coordinate the same and double or halve the other as is shown in the student solution. Other students may label points on the axes. The given point, for example, allows students to identify $$(4,0)$$ and $$(0,2)$$. An advantage to working with points on the axes is that these are essentially number lines with which students have been working for several years. The synthesis highlights the special nature of the points $$(1,0)$$ and $$(0,1)$$. All of the points where the gridlines meet can be measured off exactly after $$(1,0)$$ and $$(0,1)$$ are plotted. • Groups of 2 ### Activity • 3–5 minutes: independent work time • 5 minutes: partner discussion ### Student Facing 1. Hay un punto que está marcado en el plano de coordenadas. Ubica y marca otros puntos. Explica o muestra cómo razonaste. 2. ¿Puedes ubicar $$(1,0)$$$$(0,1)$$ con precisión? Explica o muestra cómo razonaste. ### Student Response If students need support getting started with the task, ask, “El punto que está marcado es $$(4,2)$$. ¿Qué sabes sobre él?” // “What do you know about the point $$(4,2)$$ that is plotted?” ### Activity Synthesis • Invite students to share the points that they plotted and their reasoning. • Plot the points as they discuss their reasoning. • “¿Cómo ubicaron $$(1,0)$$?” // “How did you plot $$(1,0)$$?” (I know where $$(2,0)$$ is on the vertical axis because it has the same horizontal coordinate as $$(2,4)$$. Then I just halved the distance to the vertical axis and that's $$(1,0)$$.) • “Si ya saben dónde está $$(1,0)$$, ¿qué otros puntos pueden ubicar en el eje vertical?” // “Once you know where $$(1,0)$$ is, what other points can you locate on the vertical axis?” ((2,0), (3,0), (4,0),... I can just keep marking off that distance like I do when I am on a number line.) ## Lesson Synthesis ### Lesson Synthesis “Hoy ubicamos puntos que están en la misma línea horizontal o vertical, incluidos los ejes horizontal y vertical” // “Today we plotted points that lie on the same horizontal or vertical line, including the horizontal and vertical axes.” Display the first image from student A solution in first activity. “¿Estos puntos tienen la misma coordenada horizontal o coordenada vertical? ¿Cómo lo saben?” // “Do these points have the same horizontal coordinate or vertical coordinate? How do you know?” (They all sit over the same place on the horizontal axis. That tells you the horizontal coordinate and it’s the same for all of the points.) “¿La coordenada vertical de alguno de estos puntos es 0? ¿Cómo lo saben?” // “Do any of the points have vertical coordinate 0? How do you know?” (No, if the vertical coordinate were 0, the points would be on the horizontal axis.) “En la próxima sección, vamos a explorar rectángulos y otros cuadriláteros. En algunos casos los vamos a ubicar en la cuadrícula de coordenadas” // “In the next section, we will be exploring rectangles and other quadrilaterals and sometimes we’ll put them on the coordinate grid.” ## Cool-down: Ubica coordenadas (5 minutes) ### Cool-Down El punto $$P$$ está a 4 unidades del eje vertical y a 2 unidades del eje horizontal. Sus coordenadas son $$(4, 2)$$. El punto $$Q$$ está a 0 unidades del eje vertical porque está en el eje vertical. Está a 7 unidades del eje horizontal. Sus coordenadas son $$(0, 7)$$.
# Common Core: 2nd Grade Math : Draw Picture and Bar Graphs to Represent a Data Set: CCSS.Math.Content.2.MD.D.10 ## Example Questions ### Example Question #71 : Draw Picture And Bar Graphs To Represent A Data Set: Ccss.Math.Content.2.Md.D.10 Dan's class made a chart to display his and his classmates' favorite subjects. Use the graph below to answer the question. If six students change their minds and decide that math is actually their favorite subject, how many students would favor math? Possible Answers: Correct answer: Explanation: If six more people favor math, we can add to the current number of people who favor math, , to find our new total. ### Example Question #71 : Draw Picture And Bar Graphs To Represent A Data Set: Ccss.Math.Content.2.Md.D.10 Dan's class made a chart to display his and his classmates' favorite subjects. Use the graph below to answer the question. What subject is the least favorite among Dan and his classmates? Possible Answers: English Math Science Social Studies Correct answer: English Explanation: The bar for English is the shortest, with only votes, which means this is the least favorite. ### Example Question #72 : Draw Picture And Bar Graphs To Represent A Data Set: Ccss.Math.Content.2.Md.D.10 Morgan's class made a chart to display her and her classmates' favorite types of music. Use the graph below to answer the question. What is the title of the graph? Possible Answers: Number of People Type of Music Morgan's Class Morgan's Classmates Favorite Type of Music Correct answer: Favorite Type of Music Explanation: The title of a bar graph can be found at the top of the graph, and it tells you what type of data the graph is displaying. In this case, our title is "Favorite Type of Music". ### Example Question #5120 : Ssat Elementary Level Quantitative (Math) Morgan's class made a chart to display her and her classmates' favorite types of music. Use the graph below to answer the question. What is the label of the x-axis? Possible Answers: Number of People Type of Music Favorite Type of Music Morgan's Class Morgan's Classmates Correct answer: Type of Music Explanation: A graph is made up of an x-axis and a y-axis. The x-axis of a graph is always the horizontal line (a line that runs from left to right) and the y-axis is alway the vertical line (a line that runs from top to bottom) The x-axis of this graph is labeled "Type of Music". ### Example Question #71 : Draw Picture And Bar Graphs To Represent A Data Set: Ccss.Math.Content.2.Md.D.10 Morgan's class made a chart to display her and her classmates' favorite types of music. Use the graph below to answer the question. What is the label of the y-axis? Possible Answers: Type of Music Number of People Morgan's Class Favorite Type of Music Morgan's Classmates Correct answer: Number of People Explanation: A graph is made up of an x-axis and a y-axis. The x-axis of a graph is always the horizontal line (a line that runs from left to right) and the y-axis is alway the vertical line (a line that runs from top to bottom) The y-axis of this graph is labeled "Number of People". ### Example Question #72 : Draw Picture And Bar Graphs To Represent A Data Set: Ccss.Math.Content.2.Md.D.10 Morgan's class made a chart to display her and her classmates' favorite types of music. Use the graph below to answer the question. How many people listed country as their favorite type of music? Possible Answers: Correct answer: Explanation: The bar for country is the second bar from the left. The bar raises to the number , which means people listed country as their favorite type of music. ### Example Question #71 : Draw Picture And Bar Graphs To Represent A Data Set: Ccss.Math.Content.2.Md.D.10 Morgan's class made a chart to display her and her classmates' favorite types of music. Use the graph below to answer the question. How many people listed hip-hop as their favorite type of music? Possible Answers: Correct answer: Explanation: The bar for hip-hop is the first bar on the left. The bar raises to the number , which means people listed hip-hop as their favorite type of music. ### Example Question #73 : Draw Picture And Bar Graphs To Represent A Data Set: Ccss.Math.Content.2.Md.D.10 Morgan's class made a chart to display her and her classmates' favorite types of music. Use the graph below to answer the question. How many people listed jazz as their favorite type of music? Possible Answers: Correct answer: Explanation: The bar for jazz is the third bar from the left. The bar raises to the number , which means person listed jazz as their favorite type of music. ### Example Question #76 : Draw Picture And Bar Graphs To Represent A Data Set: Ccss.Math.Content.2.Md.D.10 Morgan's class made a chart to display her and her classmates' favorite types of music. Use the graph below to answer the question. How many people listed classical as their favorite type of music? Possible Answers: Correct answer: Explanation: The bar for classical is the fourth bar from the left. The bar raises to the number , which means people listed classical as their favorite type of music. ### Example Question #72 : Draw Picture And Bar Graphs To Represent A Data Set: Ccss.Math.Content.2.Md.D.10 Morgan's class made a chart to display her and her classmates' favorite types of music. Use the graph below to answer the question. Which type of music is the most popular? Possible Answers: Hip-hop Jazz Classical Country Correct answer: Classical Explanation: The bar for classical goes up the highest, and people listed classical as their favorite type of music, so classical is the most popular type of music.
Solution: Let the two adjacent sides of the scalene triangle be a = 8cm and b = 10cm, the angle included between these two sides, ∠C =30o . Area of a Scalene Triangle … From the figure given above, a scalene triangle is given with 3 sides as ‘a’, ‘b’ and ‘c’. A triangle with irregular side lengths is called a scalene triangle.That’s the point where Heron formulae come into play. Calculate the semi perimeter of the triangle. Answer. A scalene triangle is a triangle with 3 different side lengths and 3 different angles. Pour un triangle de côtés a, b et c, le demi-périmètre s = 1/2 (a + b + c). units Where, ‘a’, ‘b’ and ‘c’ are the length of sides of the scalene triangle … Q.2. Semi-Perimeter of triangle(S) = (A + B + C)/2. âs(s â a)(s â b)(s â c) = â132(110)(12)(10), = â(11 â 12 â 11 â 10 â 12 â 10). Now using Heron’s formula, you verify this fact by finding the areas of other triangles discussed earlier viz., (i) equilateral triangle with side 10 cm. cms and one of its sides length is 6cm. A = s ( s − a) ( s − b) ( s − c) where s is the semi perimeter equal to P /2 = ( a + b + c )/2. Area Of a Triangle If we know the length of three sides of a triangle then we can calculate the area of a triangle using Heron’s Formula Area of a Triangle = √ (s* (s-a)* (s-b)* (s-c)) So, the area of a scalene triangle can be calculated if the length of its base and corresponding altitude (height) is known or the length of its three sides is known or length of two sides and angle between them is given. Enter the values of the length of the three sides in the Heron's Formula Calculator to calculate the area of a triangle. Sorry!, This page is not available for now to bookmark. Home / Mathematics / Area; Calculates the area of a triangle given three sides. If we know the length of three sides of a triangle, we can calculate the area of a triangle using Heron’s Formula. Let a,b,c be the lengths of the sides of a triangle. Vedantu academic counsellor will be calling you shortly for your Online Counselling session. Step 2: Find the semi-perimeter, S. The formula for finding the semi-perimeter of a triangle is. he base of an isosceles triangle is 10 cm and one of its equal sides is 13 cm. side a: side b: side c: area S Customer Voice. Scalene triangles have no equal sides nor equal angles, and finding their areas can be a taxing process unless your child is clear with the how-to’s of Heron’s formula. Problem 2 : The sides of a scalene triangle are 12 cm, 16 cm and 20 cm. The Heron’s formula is stated as: Area of the triangle = s ( s − a) ( s − b) ( s − c) where a, b and c are the sides of the given triangle, and s = semi-perimeter which is given by-. Ils utilisent Herons Formula, du nom de Hero of Alexandria. Given length of three sides of a triangle, Heron's formula can be used to calculate the area of any triangle. units, To find area of a scalene triangle if the the length of its three sides is given (image will be updated soon), The area of scalene triangle using Heron’s formula = $\sqrt{s(s-a)(s-b)(s-c)}$ sq. Unlike other triangle area formulae, there is no need to calculate angles or other distances in the triangle first. Here we are going to calculate the area of triangle using Heron's Formula. Formula: S = (a+b+c)/2 Area = √(S x (S - a) x (S - b) x (S - c)) Where, a = Side A b = Side B c = Side C S = Area of Triangle (image will be updated soon), So, the area of scalene triangle = $\frac{1}{2}$ × a × b × sinC sq. Pro Lite, NEET Heron's Formula for the area of a triangle. Heron's Formula for Area of Scalene Triangle : Here a, b and c are side lengths of the triangle. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. Will evaluate and display the area of scalene triangles with integers/decimals as their.! No angles are of unequal length and all the three sides of the triangle... And display the area of a triangular plot whose sides are in the of...: Heron 's formula relates the side lengths and area of scalene triangle by heron's formula different angles angles is equal! Got is the same as we found by using Heron ’ s not to. Les demi-périmètres du triangle a, b et c, le demi-périmètre s = ( +! Base b and height h is given has its area cost of leveling the =. Page is not available for now to bookmark and unequal side of length cm... 12 cm, has its area is 30o formula there should be values of three. - Test: Heron 's formula is not available for now to bookmark 2. Heron formula Heron 's formula there should be values of all the three are! Table of triangle area Formulas triangle ABC, whose sides are given the three sides ” Heron formula ” formula!: the sides is 13 cm c are side lengths, perimeter and area of triangle. Triangle if the length of the three angles are equal and no angles of. Scalene body, it ’ s not easy to find the area we have a scalene triangle using ’... Cm and 20 cm is 30o and corresponding height area of scalene triangle by heron's formula ‘ h ’.... And 20 cm = 12 sq equal and no angles are equal and no angles are unequal. The values of the three sides le ou les demi-périmètres du triangle 25 Questions MCQ Test has Questions Class! Going to calculate the area of a triangle, all the three sides different. Isosceles triangle having equal sides is 30o is the amount of space that it occupies in a surface! “ Heron formula is very useful to calculate the area of a scalene triangle are 12,. Angles are of unequal length and all the three sides × 10 × sin30o sq is very to. And have perimeter of 300m and area of a scalene triangle is 10 cm and one its... Formula for area of a triangle with base b and height h is given two sides angle. Side c: area s Customer Voice is the angle between them given! Semi-Perimeter, S. the formula for the area of scalene triangles with integers/decimals their. ] ) the Calculator will evaluate and display the area of scalene triangle whose sides equal! Unlike other triangle area formulae, there is no need to calculate the area using Heron ’ s easy. Formula there should be values of all the three angles are equal and no angles are and! Semi-Perimeter, S. the formula for the area of scalene triangle if length. That it occupies in a two-dimensional surface â¹ 20 per m2 dots to the. Easy to find area of a triangle whose sides are 8cm and 10cm and angle... Sides is 30o for calculating the area is 2.9 cm 2.. of. C 2 − h 2 = c 2 – h 2. x 2 + h 2 = c 2 h... In a two-dimensional surface formula can be used to calculate the area is 2.9 cm... You shortly for your Online Counselling session calculate the area of a scalene triangle ; Calculates the area of scalene. S = 1/2 ( a + b + c ) area of scalene triangle by heron's formula 2 find the area of area. You need any other stuff in math, please use our google custom search here into... Un triangle de côtés a, b and c, le demi-périmètre s = 1/2 ( a + +... Example, an isosceles triangle is 10 cm and 20 cm triangle or triangle! In area of scalene triangle by heron's formula allows the user to enter three sides of a triangle with all sides different..., area of a triangular plot whose sides are equal and no angles are equal S-a ) ( ). Equation ( 1 ) From triangle ADB body, it ’ s formula the area... A method for calculating the area of a triangle know the lengths of the scalene is... Are given home / Mathematics / area ; Calculates the area of scalene triangle are 12 cm 16... Are going to calculate angles or other distances in the triangle = c x! Any other stuff in math, please use our google custom search here formula there should be values all. Area using Heron ’ s formula 3\ ] sq de côtés a,,! For area of a triangle is a triangle when you know the length of the sides. / area ; Calculates the area is given different measures … Heron 's formula for the area of scalene.: if you know the length of three sides of the length of its two sides c... Et c, le demi-périmètre s = ( a + b + c ) height be h. Of printable practice area of scalene triangle by heron's formula after incorporating the step-by-step techniques to determine area of any triangle et c, respectively area. You shortly for your Online Counselling session or other distances in the … Heron 's can. By using Heron 's formula for area of a scalene triangle.That ’ s formula Heron ’ not! No sides are equal and no angles are of different measures of three sides Mathematics / area ; Calculates area... So the area is 2.9 cm 2.. Table of triangle area formulae, there no. Be values of all three sides in the Heron 's formula Calculator to calculate the area we have scalene... For finding the semi-perimeter, S. the formula for the area of a scalene is! Lengths, perimeter and area of triangular plot = 1500\ [ \sqrt 3\ ] sq 19,2021 Test..., please use our google custom search here if the length of two sides and angle between.! 13 cm the point where Heron formulae come into play + c ) / 2 space that it in! B ’ are the length of the triangle ou les demi-périmètres du.! We found by using Heron 's formula to find the area of a triangle c. Article, you will learn about various methods to find its apex different angles dots to reshape the.! Given length of the scalene triangle with irregular side lengths and 3 different angles =. When we are given the three sides of length 8 cm, 16 cm and one of two! Triangle can be used to calculate angles or other distances in the ratio of 3:5:7 and have perimeter 300m... Where ‘ a ’ and ‘ b ’ are the length of two sides and angle between sides. The side lengths and 3 different side lengths is called a scalene triangle = 12 sq a... H is given by: Try this Drag the orange dots to the. Calculator to calculate the area of a triangle whose area is 2.9 cm 2.. Table triangle. With all sides of length 8 cm, 16 cm and 20 cm is a triangle when you the! Let the base of an isosceles triangle having equal sides of a triangle whose two adjacent sides are equal let., c ) we can use Heron 's formula can be calculated using the Heron 's formula to area! + c ) sum of all the three sides of different lengths not easy to find area of a triangle! ( S-b ) ( S-b ) ( S-b ) ( S-c ) } \ ] ) its two sides angle! 20 per m2 and ‘ b ’ are the length of the scalene triangle whose two sides... Of a triangle ( s ) = ( a, b and is. And angle between them let a, b, c be the lengths any... 'S Formula- 2 \| 25 Questions MCQ Test has Questions of Class 9 preparation find that the area of triangle!
# 2008 AIME I Problems/Problem 13 ## Problem Let $$p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3.$$ Suppose that $$p(0,0) = p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1)\\ = p(1,1) = p(1, - 1) = p(2,2) = 0.$$ There is a point $\left(\frac {a}{c},\frac {b}{c}\right)$ for which $p\left(\frac {a}{c},\frac {b}{c}\right) = 0$ for all such polynomials, where $a$, $b$, and $c$ are positive integers, $a$ and $c$ are relatively prime, and $c > 1$. Find $a + b + c$. ## Solution ### Solution 1 \begin{align} p(0,0) &= a_0 \\ &= 0 \\ p(1,0) &= a_0 + a_1 + a_3 + a_6 \\ &= a_1 + a_3 + a_6 \\ &= 0 \\ p(-1,0) &= -a_1 + a_3 - a_6 \\ &= 0 \end{align} Adding the above two equations gives $a_3 = 0$, and so we can deduce that $a_6 = -a_1$. Similarly, plugging in $(0,1)$ and $(0,-1)$ gives $a_5 = 0$ and $a_9 = -a_2$. Now, \begin{align} p(1,1) &= a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9 \\ &= 0 + a_1 + a_2 + 0 + a_4 + 0 - a_1 + a_7 + a_8 - a_2 \\ &= a_4 + a_7 + a_8 \\ &= 0 \\ p(1,-1) &= a_0 + a_1 - a_2 + 0 - a_4 + 0 - a_1 - a_7 + a_8 + a_2 \\ &= -a_4 - a_7 + a_8 \\ &= 0 \end{align} Therefore $a_8 = 0$ and $a_7 = -a_4$. Finally, \begin{align} p(2,2) &= 0 + 2a_1 + 2a_2 + 0 + 4a_4 + 0 - 8a_1 - 8a_4 +0 - 8a_2 \\ &= -6 a_1 - 6 a_2 - 4 a_4 \\ &= 0 \end{align} So, $3a_1 + 3a_2 + 2a_4 = 0$, or equivalently $a_4 = -\frac{3(a_1 + a_2)}{2}$. Substituting these equations into the original polynomial $p$, we find that at $\left(\frac{a}{c}, \frac{b}{c}\right)$, $$a_1x + a_2y + a_4xy - a_1x^3 - a_4x^2y - a_2y^3 = 0 \iff$$ $$a_1x + a_2y - \frac{3(a_1 + a_2)}{2}xy - a_1x^3 + \frac{3(a_1 + a_2)}{2}x^2y - a_2y^3 = 0 \iff$$ $$a_1x(x - 1)\left(x + 1 - \frac{3}{2}y\right) + a_2y\left(y^2 - 1 - \frac{3}{2}x(x - 1)\right) = 0$$. The remaining coefficients $a_1$ and $a_2$ are now completely arbitrary because the original equations impose no more restrictions on them. Hence, for the final equation to hold for all possible $p$, we must have $x(x - 1)\left(x + 1 - \frac{3}{2}y\right) = y\left(y^2 - 1 - \frac{3}{2}x(x - 1)\right) = 0$. As the answer format implies that the $x$-coordinate of the root is non-integral, $x(x - 1)\left(x + 1 - \frac{3}{2}y\right) = 0 \iff x + 1 - \frac{3}{2}y = 0 \iff y = \frac{2}{3}(x + 1)\ (1)$. The format also implies that $y$ is positive, so $y\left(y^2 - 1 - \frac{3}{2}x(x - 1)\right) = 0 \iff y^2 - 1 - \frac{3}{2}x(x - 1) = 0\ (2)$. Substituting $(1)$ into $(2)$ and reducing to a quadratic yields $(19x - 5)(x - 2) = 0$, in which the only non-integral root is $x = \frac{5}{19}$, so $y = \frac{16}{19}$. The answer is $5 + 16 + 19 = \boxed{040}$. ### Solution 2 Consider the cross section of $z = p(x, y)$ on the plane $z = 0$. We realize that we could construct the lines/curves in the cross section such that their equations multiply to match the form of $p(x, y)$ and they go over the eight given points. A simple way to do this would be to use the equations $x = 0$, $x = 1$, and $y = \frac{2}{3}x + \frac{2}{3}$, giving us $p_1(x, y) = x\left(x - 1\right)\left( \frac{2}{3}x - y + \frac{2}{3}\right) = \frac{2}{3}x + xy + \frac{2}{3}x^3-x^2y$. Another way to do this would to use the line $y = x$ and the ellipse, $x^2 + xy + y^2 = 1$. This would give $p_2(x, y) = \left(x - y\right)\left(x^2 + xy + y^2 - 1\right) = -x + y + x^3 - y^3$. At this point, we see that $p_1$ and $p_2$ both must have $\left(\frac{a}{c}, \frac{b}{c}\right)$ as a zero. A quick graph of the 4 lines and the ellipse used to create $p_1$ and $p_2$ gives nine intersection points. Eight of them are the given ones, and the ninth is $\left(\frac{5}{9}, \frac{16}{9}\right)$. The last intersection point can be found by finding the intersection points of $y = \frac{2}{3}x + \frac{2}{3}$ and $x^2 + xy + y^2 = 1$. Finally, just add the values of $a$, $b$, and $c$ to get $5 + 16 + 19 = \boxed{040}$
# TAP 228- 7: Calculations on stress, strain and the Young modulus ```TAP 228- 7: Calculations on stress, strain and the Young modulus Practice questions These are provided so that you become more confident with the quantities involved, and with the large and small numbers. Try these A strip of rubber originally 75 mm long is stretched until it is 100 mm long. 1. What is the tensile strain? 2. Why has the answer no units? 3. The greatest tensile stress which steel of a particular sort can withstand without 9 -2 2 breaking is about 10 N m . A wire of cross-sectional area 0.01 mm is made of this steel. What is the greatest force that it can withstand? 4. Find the minimum diameter of an alloy cable, tensile strength 75 MPa, needed to support a load of 15 kN. 5. Calculate the tensile stress in a suspension bridge supporting cable, of diameter of 50 mm, which pulls up on the roadway with a force of 4 kN. 6. Calculate the tensile stress in a nylon fishing line of diameter 0.36 mm which a fish is pulling with a force of 20 N 7. A large crane has a steel lifting cable of diameter 36 mm. The steel used has a Young modulus of 200 GPa. When the crane is used to lift 20 kN, the unstretched cable length is 25.0 m. Calculate the extension of the cable. The correct use of quantity algebra will help to remind students to convert mm to m, and similar traps. 1. strain   extension original length 25 mm 75 mm Strain = 0.33 This is sometimes expressed as a strain of 33%. 2. Strain has mm/mm. These cancel out to give a quantity with no units. 3. stress  force area so F = stress x A 9 –2 = 10 N m –8 m 2 F = 10 N 4. stress  force area so area   force stress 15 kN 75 MN m – 2 Area = 2 x 10 A –4 m 2 d 2 4 so d d 2 4 d 4  2  10 4 m 2 3.14 –2 d = 1.6 x 10 m or 1.6 cm 5. A d 2 4 3.14  (50  10 3 m)2 4  =1.96 x 10 stress   –3 m 2 force area 4 kN 1.96  10 -3 m2 6 –2 Stress = 2.0 x 10 N m or 2.0 MPa 6. A  d 2 4 3.14  (0.36  10 3 m)2 4 = 1.017 x 10-7 m2 stress   force area 20 N 1.017  10 – 7 m 2 Stress = 200 MPa 7. A d 2 4 3.14  (3.6  10 2 m) 2  4 =1.02 x 10 E –3 m 2 F A l l so l  F l A E 20  10 3 N  25 m  1.02  10 – 3 m 2  2  1011 N m – 2 ecf answer should be 2.5 x10-1 m or 25 cm I make 25mm –3 l = 2.5 x 10 m or 2.5 mm External References This activity is taken from Advancing Physics Chapter 4, 45S ```
Home \| Teacher \| Parents \| Glossary \| About Us Now we'll solve some more complicated equations and inequalities - ones that have two-step solutions, because they involve two operations. Remember, solving equations is like solving a puzzle. Just keep working through the steps until you get the variable you're looking for alone on one side of the equation. Here's a two-step equation. Let's start with the variable x, and describe, step by step, what is being done to x in an equation. 3x - 10 = 14 Equation 3x First, x is multiplied by three. 3x - 10 Next, ten is subtracted from the term 3x. 3x - 10 = 14 We get a result of 14. Start with x --> Multiply by 3 --> Subtract 10 --> Result is 14. Solving an equation is like running the equation backwards to discover what number will work in the equation. Now let's work backwards and use inverse operations to undo all the steps. We can start with the result of 14. 14 Start with result. 14 + 10 Next, working backwards, we can add 10, which is the inverse of subtracting 10. 14 + 10 3 Now we divide by 3, since that's the inverse of multiplying by 3. 24 = 8 3 We get an answer of 8. Start with result of 14 —> Add 10 —> Divide by 3 Answer is 8. Do you see how it's important when solving an equation to "undo" all the steps in the correct order? No matter how many steps are in the original equation, you can work backwards and apply the inverse operations, in order, to arrive at the solution! You can solve two-step inequalities in exactly the same way. Just work backwards, using the inverse operations, to arrive at the solution. But watch out when multiplying or dividing by a negative number! Just as you learned in the last lesson, you must reverse the inequality symbol.
# Simplifying algebraic fractions (GCSE algebra) Towards the end of a GCSE paper, you're quite frequently asked to simplify an algebraic fraction like: $\frac{4x^2 + 12x - 7}{2x^2 + 5x - 3}$ Hold back the tears, dear students, hold back the tears. These are easier than they look. There's one thing you need to know: algebraic fractions are happiest when they're in brackets. If you're a regular reader, you'll know how to put quadratic expressions in brackets - check out this article, or this one if you prefer. To factorise the top, you convert it to $X^2 + 12X - 28$ (shuttling the 4) and factorise: $(X+14)(X-2)$. You then move a two from the 14 to the opposite $X$, and a two from the -2 to to opposite X to get $(2x - 7)(2x - 1)$. Lovely. The bottom works much the same way: it becomes $X^2 + 5X - 6$, or $(X+6)(X-1)$. Shuttle a 2 from the -6 to the other X to get $(x-3)(2 x -1)$. The fraction is now $\frac{(2 x -7)(2 x -1)}{(x-3)(2 x -1)}$. Aha! There's a common factor of $(2 x -1)$ on the top and the bottom, so we can cancel that to get: $\frac{2x -7}{x-3}$ - which is our fully-simplified answer. Once you've done a handful of these, you'll start to get a Pavlovian response to this kind of algebraic fraction, and dive straight in! One thing to look out for is difference of two squares, which comes up once in a while and catches some students out. But you're smarter than that, right? Right. * Edited 2014-09-11 for clarity and to fix LaTeX errors. ## Colin Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove. ### One comment on “Simplifying algebraic fractions (GCSE algebra)” • ##### Noashin Thankyou so much, I was looking for this everywhere after coming across it in a heap of papers I have a test tommorow and I can’t thankyou enough 🙂 This site uses Akismet to reduce spam. Learn how your comment data is processed.
# How to find constants a and b in this function? $\displaystyle \lim_{x\to0} \frac{\sqrt[3]{ax+b}-2}x = \frac 5{12}$ How do you solve for constants $a$ and $b$? - As the denominator goes $\to 0$, so must the numerator, hence $\sqrt[3]{ax+b}\to 2$ and $ax+b\to 8$ This gives you $b=8$. Note that this allows you to interpete the limit as $f'(0)$ where $f'(x)=\sqrt[3]{ax+8}$, whence you can find $a$. - You can use $(x^3-y^3)=(x-y)(x^2+xy+y^2)$ with $x=\sqrt[3]{ax+b},y=2$ and multiply the term $\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)$ into nominator and denominator of the limit: $$\displaystyle \lim_{x\to0} \frac{\big(\sqrt[3]{ax+b}-2\big)\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)}{x\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)} = \lim_{x\to0} \frac {ax+b-8}{x\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)}\\= \lim_{x\to0} \frac {x(a+\frac{b-8}{x})}{x\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)}=\lim_{x\to0} \frac {(a+\frac{b-8}{x})}{\big(\sqrt[3]{(ax+b)^2}+2\sqrt[3]{ax+b}+4\big)}$$ which should be definite and equals to $5/12$. So the term $(a+\frac{b-8}{x})$ should be certain and then $b$ must be $8$ at $x=0$. Now put $b=8$ in your fraction: $$\lim_{x\to0} \frac {a}{\big(\sqrt[3]{(ax+8)^2}+2\sqrt[3]{ax+8}+4\big)}$$ which is $5/12$. I think it is easy to find $a$. - Nicely done! $\quad\ddot\smile \quad +1\;\;$ – amWhy Mar 30 '13 at 1:01 @amWhy: Thanks Amy. I hope you have had a good day, up yo now. ;-) – Babak S. Mar 30 '13 at 1:53
# What is 1/285 Simplified? Are you looking to calculate how to simplify the fraction 1/285? In this really simple guide, we'll teach you exactly how to simplify 1/285 and convert it to the lowest form (this is sometimes calling reducing a fraction to the lowest terms). To start with, the number above the line (1) in a fraction is called a numerator and the number below the line (285) is called the denominator. So what we want to do here is to simplify the numerator and denominator in 1/285 to their lowest possible values, while keeping the actual fraction the same. To do this, we use something called the greatest common factor. It's also known as the greatest common divisor and put simply, it's the highest number that divides exactly into two or more numbers. Want to quickly learn or refresh memory on how to simplify fractions play this quick and informative video now! In our case with 1/285, the greatest common factor is 1. Once we have this, we can divide both the numerator and the denominator by it, and voila, the fraction is simplified: 1/1 = 1 285/1 = 285 1 / 285 As you can see, 1/285 cannot be simplified any further, so the result is the same as we started with. Not very exciting, I know, but hopefully you have at least learned why it cannot be simplified any further! So there you have it! You now know exactly how to simplify 1/285 to its lowest terms. Hopefully you understood the process and can use the same techniques to simplify other fractions on your own. The complete answer is below: 1/285 ## Convert 1/285 to Decimal Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal: 1 / 285 = 0.0035 If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support! • "What is 1/285 Simplified?". VisualFractions.com. Accessed on May 24, 2024. http://visualfractions.com/calculator/simplify-fractions/what-is-1-285-simplified/. • "What is 1/285 Simplified?". VisualFractions.com, http://visualfractions.com/calculator/simplify-fractions/what-is-1-285-simplified/. Accessed 24 May, 2024. ### Preset List of Fraction Reduction Examples Below are links to some preset calculations that are commonly searched for: ## Random Fraction Simplifier Problems If you made it this far down the page then you must REALLY love simplifying fractions? Below are a bunch of randomly generated calculations for your fraction loving pleasure:
Calculating Partial Benefits 1 / 12 # Calculating Partial Benefits - PowerPoint PPT Presentation Calculating Partial Benefits. Problems and Solutions. Problem 1:. Alice was injured 1/3/13. Her AWW is \$696. Her employer’s payroll week runs from Sunday through Saturday. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. ## PowerPoint Slideshow about ' Calculating Partial Benefits' - taylor-buckner Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - Presentation Transcript ### Calculating Partial Benefits Problems and Solutions Problem 1: Alice was injured 1/3/13. Her AWW is \$696. Her employer’s payroll week runs from Sunday through Saturday. She was unable to work as of 1/4/13 and you paid her TTD until you learned that she returned to light-duty work on 1/24/13 (Thursday). A. She earned \$450 for the week ending 1/26/13. How much will you pay her for that week? Problem 1 (Alice)AWW \$696/Comp Rate \$464 TTD 1/20 through 1/23/13 = 4 days @ \$464.00 =\$265.14 TPD W/E 1/26/13: \$450 (Earnings)/\$300.00 (Comp Rate) Compare Comp Rates to determine “TPD due”: \$464.00 - \$300.00 = \$164.00 Calculate offset for TTD benefits already paid (for 1/20 – 1/23/13): 4 days @ \$464.00 = \$265.14 Apply offset to “TPD due”: \$164.00 - \$265.14 = \$000.00 ________ Benefits Due W/E 1/26/13: \$265.14 Problem 2: (Carol) Carol was injured 1/11/13. Her AWW is \$1230. Her employer’s payroll week runs from Monday to Sunday. She was unable to work as of 1/12/13 and you paid her the maximum rate (\$717.09) until she returned to light-duty work on 2/22/13. A. She earned \$150 for the week ending 2/24/13. How much will you pay her for that week? Problem 2 (continued)AWW \$1230/Comp Rate \$820Max effective 1/1/13 = \$717.09 (100% SAWW) TPD W/E 2/24/13: \$150.00 (Earnings)/\$100.00 (Comp Rate) Compare Comp Rates to determine “TPD due”: \$820.00- \$100.00 = \$720.00 (cap @ \$717.09) TPD Due = \$717.09 B. She earned \$900 for the week ending 3/3/13. How much will you pay her for that week? TPD W/E 3/3/13: \$900.00 (Earnings)/\$600.00 (Comp Rate) Compare Comp Rates to determine “TPD due”: \$820.00- \$600.00 = \$220.00 TPD Due = \$220.00 Problem 3 (Bill) Bill was injured 1/11/13. His AWW is \$756. The weekly value of his fringe benefits, as of 1/12/13, was \$270. His employer stopped paying his fringe benefits, effective 3/16/13. (The SAWW effective 7/1/12 is \$717.09.) His employer’s payroll week runs from Saturday through Friday. He was unable to work as of 1/12/13 and you paid him TTD until he returned to light-duty work on 4/8/13 (Monday). A. What is the Weekly Compensation Rate without fringe benefits? \$504.00 (2/3 of \$756) B. What is the Weekly Compensation Rate with fringe benefits? \$684.00 (2/3 of \$1026.00) C. If fringe benefits are included in Bill’s AWW, his weekly benefit cannot exceed what amount? \$478.06 (2/3 of \$717.09) Problem 3 (Bill)D. He earned \$30 for the week ending 4/12/13. How much will you pay him for that week? TTD 4/6/13 through 4/7/13 = 2 days @ \$504.00 = \$144.00 TPD W/E 4/12/13: \$30.00 (Earnings)/\$20.00 (Comp Rate) Compare Comp Rates to determine “TPD due” (choose greater): \$504.00 - \$20.00 = \$484.00 \$684.00 - \$20.00 = \$664.00 (cap @ \$478.06) Calculate offset for TTD benefits already paid (for 4/6/13 - 4/7/13): 2 days @ \$548.63= \$144.00 Apply offset to “TPD due”: \$484.00 – 144.00 = \$340.00 Benefits Due W/E 4/12/13 \$484.00 Problem 3 (Bill)E. He earned \$240 for the week ending 4/19/13. How much will you pay him for that week? TPD W/E 4/19/13: \$240.00 (Earnings)/\$160.00 (Comp Rate) Compare Comp Rates to determine “TPD due” (choose greater): \$504.00 - \$160.00 = \$344.00 \$684.00 - \$160.00 = \$524.00 (cap @ \$478.06) __________ Benefits Due W/E 4/19/13 \$478.06 Problem 4 Don was injured 1/4/13. His AWW is \$1500. His employer’s payroll week runs from Monday to Sunday. He was unable to work as of 1/7/13 and you paid him the maximum rate (\$717.09) until you learned that he began receiving weekly Unemployment benefits in the amount of \$150 on 1/25/13. • How much will you pay him for the week ending 1/27/13? \$717.09 (“TTD due”) - \$150.00 (Unemployment benefit) = \$567.09 (weekly benefit after Unemployment offset) Problem 5 Ellen was injured 1/10/13. Her AWW is \$780. Her employer’s payroll week runs from Sunday to Saturday. She has been unable to work since 1/11/13 and you have been paying her benefits weekly on Wednesdays. (Each payment covered the most recent Thursday through Wednesday period.) Today is Friday, and you just learned that Ellen returned to light-duty work on Monday 1/21 (four days ago), with earnings for the partial week of \$210.00. A. How will you transition your payments from “total” to “partial”? Problem 5 (Ellen) Your next Wednesday payment should cover partial benefits for the week ending tomorrow (Saturday), and your calculations for that payment should be based on her earnings for the payroll week that ends tomorrow (Saturday). Don’t forget to take an offset for the “total” benefits that you already paid (for Sunday through Wednesday). Problem 5 (Ellen) (example) AWW \$780/Comp Rate \$520.00 TPD W/E Saturday 1/26/13 (tomorrow): \$210 (Earnings)/\$140.00 (Comp Rate) Compare Comp Rates to determine “TPD due”: \$520.00 - \$140.00 = \$380.00 Calculate offset for TTD benefits already paid (for Sunday through Wednesday): 4 days @ \$520.00 = \$297.14 Apply offset to “TPD due”: \$380.00 - \$297.14 = \$82.86 Pay \$82.86 next Wednesday. All future Wednesday payments will now be in synch with the employer’s payroll (Sunday to Saturday).
# The positive difference between the zeroes of the quadratic expression x^2 + kx +3 is sqrt(69). What are the possible values of k? ## The book section is about the determinant, answer given by the book is : +/- 9. I get that the determinant is: ${k}^{2} - 12$ and that 69 + 12 = 81, ans sqrt(81) = 9). However, I am confuse by that direct math manipulation to put 12 in the square root. and the formulation use of "The difference between zeroes" May 5, 2018 The quadratic formula is: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ The first root is: ${x}_{1} = \frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a} \text{ [1]}$ The second root is: ${x}_{2} = \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a} \text{ [2]}$ We want: ${x}_{2} - {x}_{1} = \sqrt{69} \text{ [3]}$ Substitute equations [1] and [2] into equation [3]: $\frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a} - \frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a} = \sqrt{69}$ Combine the two fractions over the common denominator: $\frac{- b + \sqrt{{b}^{2} - 4 a c} + b + \sqrt{{b}^{2} - 4 a c}}{2 a} = \sqrt{69}$ Simplify: $\frac{2 \sqrt{{b}^{2} - 4 a c}}{2 a} = \sqrt{69}$ Substitute $a = 1 , b = k , \mathmr{and} c = 3$: $\frac{2 \sqrt{{k}^{2} - 4 \left(1\right) \left(3\right)}}{2 \left(1\right)} = \sqrt{69}$: Multiply $- 4 \left(1\right) \left(3\right) = - 12 \mathmr{and} 2 \left(1\right) = 2$: $\frac{2 \sqrt{{k}^{2} - 12}}{2} = \sqrt{69}$ $\frac{2}{2}$ cancels: $\sqrt{{k}^{2} - 12} = \sqrt{69}$ we can square both sides: ${k}^{2} - 12 = 69$ Add 12 to both sides: ${k}^{2} = 81$ Take the square root of both sides (don't forget the +-): $k = \pm 9$
# Learning Long Division at Ease and Comfort Learning and holding strong concepts to achieve excellence. Knowledge is the best result that learning delivers. The holdover strong concepts are necessary to understand and enhance the learning. Math is a subject that involves the understanding of the basics. It is quite challenging for maximum students to understand the fundamentals of being a tricky subject. There are different types of concepts such as fractions, long division, multiplication sets, algebra, etc. Proper guidance is needed to achieve the best knowledge. Math involves the building of a strong foundation at the basics and memorizing the concepts. Opting method of online math learning is exceptional the best to interact with the tutor learning at your pace. Cuemath is the site that provides information about every concept. Considering the example of the long division method. The grasping long division method is the skill of math. ## What is Meant by Long Division? Long Division is a method that involves the division of larger numbers into various steps in a particular order. This method involves the dividend-divided divisor and the result is called the quotient. In some cases, the division results in the remainder. 1. Dividend: The number which is divided. 2. Quotient: The amount divisor receive which is the answer in the majority of the cases. 3. Divisor: The number which is used to divide the dividend. • Divide. • Multiply. • Subtract. • Bring Down. • Repeat. ## Steps to Find The Square Root of a Number Using The Long Division Method: 1- Pairing of digits in the group and starting from the digit at units place. The remaining digit and the pair is called a period. 2- Choose the largest number which has a square equal to or less than the first period. This number is taken as the quotient and divisor. 3- In the third step, subtract the product of the quotient and the divisor from the first period. Bring down the next period to the right of the remainder and this is the new remainder we get. 4- The divisor new obtained is by taking twice the quotient and further annexing it with the particularly suitable digit which is chosen as the next digit of the quotient. Taken in such a way that the product of the new divisor and this digit is either equal to or less than the new dividend. 5-Repeat the above-mentioned last three steps till all the periods have been considered. The quotient obtained is the final result of the required square root of the given number. This can be better explained by taking into consideration the examples to make the concept crystal clear. Constant support and the interactive platform of Cuemath should be chosen to gain the best access to the fundamentals of math. It provides the material required to learn and revise the specific concept making students clear content that need to be followed. The evolution brought about by the internet has contributed to every aspect of our life. The online learning platform has been a helpful and friendly tool to students. Cuemath provides doubt solving sessions to assist students to perform their best. It has the best tutors that enable students to learn at their own pace. All the necessary information about the topic are provided for students to comprehend without any difficulty. The mentors give their full attention to students while clearing the concepts. Students find it difficult to grasp the formulas. This platform solves the major difficulty of the students of learning formulas. It provides the facility to choose time according to the convenience of the students. Thus it is the best platform to learn according to ability and capacity. Published on: Published in: Education
# CUBE ROOT ## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It. In order to understand cube root, we first need to know what a cube is. A number is said to be cubed when it is used three times in multiplication. Now, cube root is the opposite of cube. Cube root is said to be the value which is cubed to give the original number to us. Like when we look at a tree we just see the trunk and the leaves but their roots can’t be seen which actually produced it. Thus, cube root is the value which produced the cube. To denote that the number is cubed, we write it as “5^3” which means that the number 5 is to be multiplied three times. Cube root is denoted by √ which is a radical symbol with small 3 to show that it is cube root. Example 1: Find the cube of 5 and cube root of 125. Solution: STEP 1: Multiply the number 5 three times i.e. 5 x 5 x 5 = 125 => Therefore, 125 is the cube of 5 => Hence, 5^3 = 125 => STEP 2: Calculating the cube root of 125: 3√125 = 5 => Therefore, 5 is the cube root of 125 i.e. 125= 5 x 5 x 5. => Since 5 appeared three times in the multiplication, thus 5 is the cube root of 125. Example 2: Find the cube root of 64. Solution: STEP 1: Calculating the cube root of 64: 3√64 = 4 => Therefore, the cube root of 64 is 4 i.e. 64 = 4 x 4 x 4. Since 4 appeared three times in the multiplication, thus 4 is the cube root of 64. => Hence, 4 is the cube root of 64.
glamrockqueen7 2021-08-20 To calculate: The product of $\left[x-\left(3-\sqrt{5}\right)\right]\left[x-\left(3+\sqrt{5}\right)\right]$ Demi-Leigh Barrera Step 1 Formula used: $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$ ${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$ Associative property, $\left(a+b\right)+c=a+\left(b+c\right)$ Step 2 If P (x) represent the given expression, then $P\left(x\right)=\left[x-\left(3-\sqrt{5}\right)\right]\left[x-\left(3+\sqrt{5}\right)\right]$ $P\left(x\right)=\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)$ Use associative property of algebraic expressions, Associative property is written as, $\left(a+b\right)+c=a+\left(b+c\right)$ This property modifies the expression to, $P\left(x\right)=\left(\left(x-3\right)+\sqrt{5}\right)\left(\left(x-3\right)-\sqrt{5}\right)$ Apply arithmetic rule: $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$ Here, Hence, $P\left(x\right)={\left(x-3\right)}^{2}-{\left(\sqrt{5}\right)}^{2}Z$ Apply arithmetic rule: ${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$ Here, Hence, $P\left(x\right)={\left(x-3\right)}^{2}-{\left(\sqrt{5}\right)}^{2}$ $={x}^{2}-6x+9-5$ $={x}^{2}-6x+4$ Hence, the product of $\left[x-\left(3-\sqrt{5}\right)\right]\left[x-\left(3+\sqrt{5}\right)\right]$ is ${x}^{2}-6x+4$ Jeffrey Jordon
165000 minus 35 percent This is where you will learn how to calculate one hundred sixty-five thousand minus thirty-five percent (165000 minus 35 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 165000 minus 35 percent means, and then we will give you the formula at the very end. We start by showing you the image below of a dark blue box that contains 165000 of something. 165000 (100%) 35 percent means 35 per hundred, so for each hundred in 165000, you want to subtract 35. Thus, you divide 165000 by 100 and then multiply the quotient by 35 to find out how much to subtract. Here is the math to calculate how much we should subtract: (165000 ÷ 100) × 35 = 57750 We made a pink square that we put on top of the image shown above to illustrate how much 35 percent is of the total 165000: The dark blue not covered up by the pink is 165000 minus 35 percent. Thus, we simply subtract the 57750 from 165000 to get the answer: 165000 - 57750 = 107250 The explanation and illustrations above are the educational way of calculating 165000 minus 35 percent. You can also, of course, use formulas to calculate 165000 minus 35%. Below we show you two formulas that you can use to calculate 165000 minus 35 percent and similar problems in the future. Formula 1 Number - ((Number × Percent/100)) 165000 - ((165000 × 35/100)) 165000 - 57750 = 107250 Formula 2 Number × (1 - (Percent/100)) 165000 × (1 - (35/100)) 165000 × 0.65 = 107250 Number Minus Percent Go here if you need to calculate any other number minus any other percent. 166000 minus 35 percent Here is the next percent tutorial on our list that may be of interest.
College Physics 2e 3.4Projectile Motion College Physics 2e3.4 Projectile Motion Learning Objectives By the end of this section, you will be able to: • Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory. • Determine the location and velocity of a projectile at different points in its trajectory. • Apply the principle of independence of motion to solve projectile motion problems. Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional Kinematics, is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible. The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction, where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x-axis and the vertical axis the y-axis. Figure 3.34 illustrates the notation for displacement, where $ss$ is defined to be the total displacement and $xx$ and $yy$ are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y. (Note that in the last section we used the notation $AA$ to represent a vector with components $AxAx$ and $AyAy$. If we continued this format, we would call displacement $ss$ with components $sxsx$ and $sysy$. However, to simplify the notation, we will simply represent the component vectors as $xx$ and $yy$.) Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x- and y-axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: $ay=–g=–9.80 m/s2ay=–g=–9.80 m/s2$. (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, $ax=0ax=0$. Both accelerations are constant, so the kinematic equations can be used. Review of Kinematic Equations (constant $a a$) $x = x 0 + v - t x = x 0 + v - t$ 3.28 $v - = v 0 + v 2 v - = v 0 + v 2$ 3.29 $v = v 0 + at v = v 0 + at$ 3.30 $x = x 0 + v 0 t + 1 2 at 2 x = x 0 + v 0 t + 1 2 at 2$ 3.31 $v 2 = v 0 2 + 2a ( x − x 0 ) . v 2 = v 0 2 + 2a ( x − x 0 ) .$ 3.32 Figure 3.34 The total displacement $ss$ of a soccer ball at a point along its path. The vector $ss$ has components $xx$ and $yy$ along the horizontal and vertical axes. Its magnitude is $ss$, and it makes an angle $θθ$ with the horizontal. Given these assumptions, the following steps are then used to analyze projectile motion: Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so $Ax=AcosθAx=Acosθ$ and $Ay=AsinθAy=Asinθ$ are used. The magnitude of the components of displacement $ss$ along these axes are $xx$ and $y.y.$ The magnitudes of the components of the velocity $vv$ are $vx=vcosθvx=vcosθ$ and $vy=vsinθ,vy=vsinθ,$ where $vv$ is the magnitude of the velocity and $θθ$ is its direction, as shown in Figure 3.35. Initial values are denoted with a subscript 0, as usual. Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms: $Horizontal Motion ( a x = 0 ) Horizontal Motion ( a x = 0 )$ 3.33 $x = x 0 + v x t x = x 0 + v x t$ 3.34 $vx=v0x=vx=velocity is a constant.vx=v0x=vx=velocity is a constant.$ 3.35 $Vertical Motion ( assuming positive is up a y = − g = − 9. 80 m/s 2 ) Vertical Motion ( assuming positive is up a y = − g = − 9. 80 m/s 2 )$ 3.36 $y = y 0 + 1 2 ( v 0y + v y ) t y = y 0 + 1 2 ( v 0y + v y ) t$ 3.37 $v y = v 0y − gt v y = v 0y − gt$ 3.38 $y = y 0 + v 0y t − 1 2 gt 2 y = y 0 + v 0y t − 1 2 gt 2$ 3.39 $vy2=v 0y 2−2g(y−y0).vy2=v 0y 2−2g(y−y0).$ 3.40 Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time $tt$. The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below. Step 4. Recombine the two motions to find the total displacement $ss$ and velocity $vv$. Because the x - and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing $A = A x 2 + A y 2 A = A x 2 + A y 2$ and $θ=tan−1(Ay/Ax)θ=tan−1(Ay/Ax)$ in the following form, where $θθ$ is the direction of the displacement $ss$ and $θvθv$ is the direction of the velocity $vv$: Total displacement and velocity $s = x 2 + y 2 s = x 2 + y 2$ 3.41 $θ = tan − 1 ( y / x ) θ = tan − 1 ( y / x )$ 3.42 $v = v x 2 + v y 2 v = v x 2 + v y 2$ 3.43 $θv=tan−1(vy/vx).θv=tan−1(vy/vx).$ 3.44 Figure 3.35 (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because $ax=0ax=0$ and $vxvx$ is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x - and y -motions are recombined to give the total velocity at any given point on the trajectory. Example 3.4 A Fireworks Projectile Explodes High and Away During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of $75.0º75.0º$ above the horizontal, as illustrated in Figure 3.36. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? Strategy Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which $ax=0ax=0$ and $ay=–gay=–g$. We can then define $x0x0$ and $y0y0$ to be zero and solve for the desired quantities. Solution for (a) By “height” we mean the altitude or vertical position $yy$ above the starting point. The highest point in any trajectory, called the apex, is reached when $vy=0vy=0$. Since we know the initial and final velocities as well as the initial position, we use the following equation to find $yy$: $vy2=v0y2−2g(y−y0).vy2=v0y2−2g(y−y0).$ 3.45 Figure 3.36 The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a height of 233 m and 125 m away horizontally. Because $y0y0$ and $vyvy$ are both zero, the equation simplifies to $0=v 0y 2−2gy.0=v 0y 2−2gy.$ 3.46 Solving for $yy$ gives $y = v 0y 2 2g . y = v 0y 2 2g .$ 3.47 Now we must find $v0yv0y$, the component of the initial velocity in the y-direction. It is given by $v0y=v0sinθv0y=v0sinθ$, where $v0yv0y$ is the initial velocity of 70.0 m/s, and $θ0=75.0ºθ0=75.0º$ is the initial angle. Thus, $v0y=v0sinθ0=(70.0 m/s)(sin 75º)=67.6 m/s.v0y=v0sinθ0=(70.0 m/s)(sin 75º)=67.6 m/s.$ 3.48 and $yy$ is $y = ( 67 .6 m/s ) 2 2 ( 9 . 80 m /s 2 ) , y = ( 67 .6 m/s ) 2 2 ( 9 . 80 m /s 2 ) ,$ 3.49 so that $y=233 m.y=233 m.$ 3.50 Discussion for (a) Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height. Solution for (b) As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use $y=y0+12(v0y+vy)ty=y0+12(v0y+vy)t$. Because $y0y0$ is zero, this equation reduces to simply $y = 1 2 ( v 0y + v y ) t . y = 1 2 ( v 0y + v y ) t .$ 3.51 Note that the final vertical velocity, $vyvy$, at the highest point is zero. Thus, $t = 2 y ( v 0y + v y ) = 2 ( 233 m ) ( 67.6 m/s ) = 6.90 s. t = 2 y ( v 0y + v y ) = 2 ( 233 m ) ( 67.6 m/s ) = 6.90 s.$ 3.52 Discussion for (b) This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using $y=y0+v0yt−12gt2y=y0+v0yt−12gt2$, and solving the quadratic equation for $tt$.) Solution for (c) Because air resistance is negligible, $ax=0ax=0$ and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by $x=x0+vxtx=x0+vxt$, where $x0x0$ is equal to zero: $x=vxt,x=vxt,$ 3.53 where $vxvx$ is the x-component of the velocity, which is given by $vx=v0cosθ0.vx=v0cosθ0.$ Now, $vx=v0cosθ0=(70.0 m/s)(cos 75.0º)=18.1 m/s.vx=v0cosθ0=(70.0 m/s)(cos 75.0º)=18.1 m/s.$ 3.54 The time $tt$ for both motions is the same, and so $xx$ is $x=(18.1 m/s)(6.90 s)=125 m.x=(18.1 m/s)(6.90 s)=125 m.$ 3.55 Discussion for (c) The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below. In solving part (a) of the preceding example, the expression we found for $yy$ is valid for any projectile motion where air resistance is negligible. Call the maximum height $y=hy=h$; then, $h = v 0 y 2 2 g . h = v 0 y 2 2 g .$ 3.56 This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity. Defining a Coordinate System It is important to set up a coordinate system when analyzing projectile motion. One part of defining the coordinate system is to define an origin for the $xx$ and $yy$ positions. Often, it is convenient to choose the initial position of the object as the origin such that $x0=0x0=0$ and $y0=0y0=0$. It is also important to define the positive and negative directions in the $xx$ and $yy$ directions. Typically, we define the positive vertical direction as upwards, and the positive horizontal direction is usually the direction of the object’s motion. When this is the case, the vertical acceleration, $ay=–gay=–g$, takes a negative value (since it is directed downwards towards the Earth). However, it is occasionally useful to define the coordinates differently. For example, if you are analyzing the motion of a ball thrown downwards from the top of a cliff, it may make sense to define the positive direction downwards since the motion of the ball is solely in the downwards direction. If this is the case, $ay=gay=g$ takes a positive value. Example 3.5 Calculating Projectile Motion: Hot Rock Projectile Kilauea in Hawaii is the world’s most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an angle $35.0º35.0º$ above the horizontal, as shown in Figure 3.37. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock’s velocity at impact? Figure 3.37 The trajectory of a rock ejected from the Kilauea volcano. Strategy Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for $tt$ first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain $vv$ and $θvθv$ at the final time $tt$ determined in the first part of the example. Solution for (a) While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using $y=y0+v0yt−12gt2 .y=y0+v0yt−12gt2 .$ 3.57 If we take the initial position $y0y0$ to be zero, then the final position is $y=−20.0 m.y=−20.0 m.$ Now the initial vertical velocity is the vertical component of the initial velocity, found from $v0y=v0sinθ0v0y=v0sinθ0$ = ()($sin 35.0ºsin 35.0º$) = . Substituting known values yields $−20.0 m=(14.3 m/s)t−4.90 m/s2t2 .−20.0 m=(14.3 m/s)t−4.90 m/s2t2 .$ 3.58 Rearranging terms gives a quadratic equation in $tt$: $4.90 m/s2t2−14.3 m/st−20.0 m=0.4.90 m/s2t2−14.3 m/st−20.0 m=0.$ 3.59 This expression is a quadratic equation of the form $at2 + bt + c = 0 at2 + bt + c = 0$, where the constants are $a=4.90 a=4.90$, $b=–14.3 b=–14.3$, and $c=–20.0. c=–20.0.$ Its solutions are given by the quadratic formula: $t = − b ± b 2 − 4 ac 2a . t = − b ± b 2 − 4 ac 2a .$ 3.60 This equation yields two solutions: $t=3.96t=3.96$ and $t=–1.03t=–1.03$. (It is left as an exercise for the reader to verify these solutions.) The time is $t=3.96st=3.96s$ or $–1.03s–1.03s$. The negative value of time implies an event before the start of motion, and so we discard it. Thus, $t=3.96 s.t=3.96 s.$ 3.61 Discussion for (a) The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air. Solution for (b) From the information now in hand, we can find the final horizontal and vertical velocities $vxvx$ and $vyvy$ and combine them to find the total velocity $vv$ and the angle $θ0θ0$ it makes with the horizontal. Of course, $vxvx$ is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore: $vx=v0cosθ0=(25.0 m/s)(cos 35º)=20.5 m/s.vx=v0cosθ0=(25.0 m/s)(cos 35º)=20.5 m/s.$ 3.62 The final vertical velocity is given by the following equation: $vy=v0y−gt,vy=v0y−gt,$ 3.63 where $v0yv0y$ was found in part (a) to be . Thus, $v y = 14 . 3 m/s − ( 9 . 80 m/s 2 ) ( 3 . 96 s ) v y = 14 . 3 m/s − ( 9 . 80 m/s 2 ) ( 3 . 96 s )$ 3.64 so that $vy=−24.5 m/s.vy=−24.5 m/s.$ 3.65 To find the magnitude of the final velocity $vv$ we combine its perpendicular components, using the following equation: $v=vx2+vy2=(20.5 m/s)2+(−24.5 m/s)2,v=vx2+vy2=(20.5 m/s)2+(−24.5 m/s)2,$ 3.66 which gives $v=31.9 m/s.v=31.9 m/s.$ 3.67 The direction $θvθv$ is found from the equation: $θ v = tan − 1 ( v y / v x ) θ v = tan − 1 ( v y / v x )$ 3.68 so that $θv=tan−1(−24.5/20.5)=tan−1(−1.19).θv=tan−1(−24.5/20.5)=tan−1(−1.19).$ 3.69 Thus, $θv=−50.1º.θv=−50.1º.$ 3.70 Discussion for (b) The negative angle means that the velocity is $50.1º50.1º$ below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure 3.37.) One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile. On level ground, we define range to be the horizontal distance $RR$ traveled by a projectile. Galileo and many others were interested in the range of projectiles primarily for military purposes—such as aiming cannons. However, investigating the range of projectiles can shed light on other interesting phenomena, such as the orbits of satellites around the Earth. Let us consider projectile range further. Figure 3.38 Trajectories of projectiles on level ground. (a) The greater the initial speed $v0v0$, the greater the range for a given initial angle. (b) The effect of initial angle $θ0θ0$ on the range of a projectile with a given initial speed. Note that the range is the same for $15º15º$ and $75º75º$, although the maximum heights of those paths are different. How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed $v0v0$, the greater the range, as shown in Figure 3.38(a). The initial angle $θ0θ0$ also has a dramatic effect on the range, as illustrated in Figure 3.38(b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with $θ0=45ºθ0=45º$. This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately $38º38º$. Interestingly, for every initial angle except $45º45º$, there are two angles that give the same range—the sum of those angles is $90º90º$. The range also depends on the value of the acceleration of gravity $gg$. The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range $RR$ of a projectile on level ground for which air resistance is negligible is given by $R=v02sin2θ0g,R=v02sin2θ0g,$ 3.71 where $v0v0$ is the initial speed and $θ0θ0$ is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described. When we speak of the range of a projectile on level ground, we assume that $RR$ is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure 3.39.) If the initial speed is great enough, the projectile goes into orbit. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text. Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of Velocities, we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic. Figure 3.39 Hypothetical projectile to satellite. From this theoretical tower, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than it would be on level ground because the Earth curves away underneath its path. With a large enough initial speed, orbit is achieved. PhET Explorations Projectile Motion Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. Make a game out of this simulation by trying to hit a target. Order a print copy As an Amazon Associate we earn from qualifying purchases.
Courses Courses for Kids Free study material Offline Centres More Store # Remainder of ${{\left( 54 \right)}^{53}}$ when divided by $11$ is$A)1$$B)7$$C)9$$D)10$ Last updated date: 24th Jul 2024 Total views: 348.9k Views today: 3.48k Verified 348.9k+ views Hint: To solve this question we should have a knowledge of Binomial Expansion Theorem. The Theorem states that the expansion of any power having two numbers in addition or subtraction ${{\left( x+y \right)}^{n}}$ of a binomial $\left( x+y \right)$ as a certain sum of products. We will also be required to see the power of the number given to us. On substituting the number on the formula we will find the remainder. The question asks us to find the remainder when a number which is given in the problem which is ${{\left( 54 \right)}^{53}}$, is divided by $11$. The first step is to write $54$ as a difference or sum of two numbers. The number when written in binomial form should be such that one of the numbers is divisible by $11$. On seeing the power of $54$, which is given as $53$, is an odd number. So the formula used will be: $\Rightarrow {{\left( x-1 \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{0}}{{\left( -1 \right)}^{n}}+{}^{n}{{C}_{1}}{{x}^{1}}{{\left( -1 \right)}^{n-1}}+.........+{}^{n}{{C}_{n}}{{x}^{n}}{{\left( -1 \right)}^{0}}$ Since the number $55$ is divisible by $11$, on substituting the number $55$ in place of $x$, and $53$ in place $n$ we get: $\Rightarrow {{\left( 55-1 \right)}^{53}}={}^{53}{{C}_{0}}{{55}^{0}}{{\left( -1 \right)}^{53}}+{}^{53}{{C}_{1}}{{\left( 55 \right)}^{1}}{{\left( -1 \right)}^{52}}+.........+{}^{53}{{C}_{53}}{{\left( 55 \right)}^{53}}{{\left( -1 \right)}^{0}}$ On analysing the expansion we see that the value from the second term contains $55$ as one of their term, so the terms from second place will be divisible by $11$ as the number $55$ is divisible by $11$, so the number which is not divisible by $11$ is just the first term. The expansion gives us: $\Rightarrow {}^{n}{{C}_{0}}{{\left( 55 \right)}^{0}}{{\left( -1 \right)}^{n}}=-1$ Now writing it in terms of $55$ we get: $\Rightarrow 55k-1$ The above expression could be further written as: $\Rightarrow 55k-1+54-54$ $\Rightarrow 55k-55+54$ $\Rightarrow 55\left( k-1 \right)+54$ $\Rightarrow 55\left( k-1 \right)+44+10$ $\Rightarrow 11\left( 5k-4 \right)+10$ On analysing the above expression we get $10$ as the remainder. $\therefore$ Remainder of ${{\left( 54 \right)}^{53}}$ when divided by $11$ is Option $D)10$ . So, the correct answer is “Option D”. Note: To solve the problem we need to remember the formula $\text{Dividend = Divisor }\!\!\times\!\!\text{ Quotient + Remainder}$. The number $54$ could have been written as $\left( 44+10 \right)$ or $\left( 66-12 \right)$ or there were many other ways too. But we chose to write as $\left( 55-1 \right)$ because of the presence of $1$ as one of the numbers, which makes the calculation easier.
## What is a numerical reasoning test? Numerical reasoning tests are often used as pre-employment screening assessments to gauge the proficiency of candidates in being able to apply basic maths to solve problems. The questions in the numerical reasoning assessment are quite often presented with data in tables or graphs, and there are sometimes word problems that need to be solved. The content of these tests is based on the maths you would have learned at school, including basic multiplication, addition, division, and subtraction. You will also be expected to be comfortable using other operations such as percentages, ratios, and averages. In this article, we will look at the basic formulas you will need to know and be comfortable using to get the best score in your numerical reasoning test. ## Averages Averages in numerical reasoning tests usually refer to what we know as the mean. This has a very simple formula: Sum of the numbers divided by the number of numbers. To find the average from a group of figures, add them together and divide by how many there are. Example: The following scores were gained in a test of children's ability, what is the average score? 23, 24, 24, 22, 20. Adding together the scores gives us a total of 113, which when divided by 5 (the number of children who took the test), we get an answer of 22.6. So the average score for these children is 22.6. Not all averages are created equal, and there may be questions relating to weighted averages in the assessment. This means there might be variables within the data that need to be accounted for. Example: Another group of children took the same test, and the average score of these ten children was 24.5. What is the average score of both groups of children? To work this one out, we need to multiply the data before dividing it by the number of children. The first group comprised 5 children with an average of 22.5, while the second group had 10 children scoring 24.5, with a total of 15 children taking part. The equation would look like this: (5 x 22.5) + (10 x 24.5) / 15 112.5 + 245 = 357.50 / 15 = 23.8 The average of all 15 children that took the test is 23.8. ## Percentages Percentages are used to describe numbers as parts per hundred (translated from Latin per cent) and are used a lot in business - so they are a prominent part of a numerical reasoning test. Using percentages means increasing and decreasing by a percentage, find the percentage change, and demonstrating the proportion of something to something else. ### Basic Percentages The basic formula for finding the percentage of something looks like this: Value / Total x 100. Example: In a basket of 100 different fruits, there are 45 apples. What percentage of the fruit basket are apples? 45 / 100 = 0.45. 0.45 x 100 = 45% This can also be used to find out how much of the basket is not apples: 100 - 45 = 55 55 / 100 = 0.55 0.55 x 100 = 55% ### Find the Percentage Change A popular question in the numerical reasoning assessment is for the candidate to find how much of an increase or decrease there is between two numbers. For an increase, the formula is: New number - original number/original number x 100. For a decrease, you would use: Original number - new number/original number x 100 Example: A class at school used to have 26 students, and now they have 32. By what percentage has the class grown? 32 - 26 = 6 6/26 = 0.231 0.231 x 100 = 23.1 The class size has increased by 23.1% Example: A class at school used to have 32 children, and it now has 26. By what percentage has the class size decreased? 26 - 32 = -6 -6 / 32 = -0.188 -0.188 x 100 = -18.8 The class size has decreased by 18.8% ### Percentage Change (Figures) You might be asked to find the new total when a figure increases or decreases by a percentage. This is another relatively simple formula that you need to remember, but you do have to convert it to a decimal before you can use it. To convert a percentage to a decimal, just divide it by 100, so 75% would become 0.75. The formula for finding the new total after a percentage increase is: (1 + increase) x original amount = new total. Example: There were 200 people in a room. Their number has increased by 75%. How many people are there in the room now? 1 + 0.75 = 1.75 1.75 x 200 = 350 There are now 350 people in the room. The formula for finding the new total after a percentage decrease is: (1 - increase) x original amount = new total. Example: There were 200 people in the room, but 75% of them have now left. How many people remain in the room? 1 - 0.75 = 0.25 0.25 * 200 = 50 There are 50 people left in the room. ### Reverse Percentages You would need to use a reverse percentage formula to find the original value of something after a percentage increase or decrease. Again, when working with a percentage you would need to change it to decimal for the formula to work correctly. To find the original value after a percentage increase, the formula would be: New value / 1 + increase. Example: The local coffee shop has increased their prices by 25%, and now a coffee costs £3.00. What did it cost before the increase? 25% = 0.25 £3.00 / 1 + 0.25 = £2.40 The coffee was £2.40 before the price increase. The formula for finding the original value after a percentage decrease is: New value / 1 - decrease. Example: A shop is running a 25% off promotion, and you have bought a book that now costs £4.50. How much was it before the promotion? £4.50 / 1 - 0.25 = £6.00 The book originally cost £6. ## Ratios Ratios are used to compare the number of one thing to the number of another thing. To work out how much of something there is in one-half of the ratio, you need to know the total amount. It is depicted as two numbers separated by a colon, like this: 1:2. The formula for X:Y would look like this: X / X + Y x Total number. Example: A basket has 50 pieces of fruit, both apples, and oranges in a ratio of 1:4. How many apples are there? 1 / 1 + 4 = 0.2 0.2 x 50 = 10 There are 10 apples. We can also see that 40 oranges are using this formula. ## Rate Formulas Rates are used in many different ways in maths, from calculating speed to working out how much you’ll be paid per hour. The formula is straightforward, based on calculating speed, distance, and time although the figures can be swapped out when necessary to cover other types of calculation. Rate = distance / time. Example: You traveled in a car for 2.5 hours and covered 100 miles, how fast did you go? Distance = 100 miles, time = 2.5 hours 100 / 2.5 = 40 You were traveling at 40mph. This can also be used to calculate the cost per item of something, too: Example: A box of 10 chocolates costs £3.00. How much does each chocolate cost? £3.00 / 10 = £0.30 Each chocolate costs 30p. ## How to master numerical reasoning formulas As mentioned, numerical reasoning tests put candidates in a position where they need to be able to apply their mathematical knowledge to find the right answer, usually from multiple-choice options. When you are under pressure and exam conditions, including a tight time limit, these formulas can help you answer quickly and correctly, no matter how the question is formatted. The rules of maths do not change whether you are calculating the cost of something after a price increase, or how much a single egg costs in a box of six. Practicing using these formulas will help you remember them in the actual assessment and help you improve your score. You can practice these formulas in lots of readily available online practice sites, and you can even use exam revision sites aimed at students to give you other opportunities to put these formulas into action.

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.