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Classes Explanation
A Practical use of classes:
The Problem:
Living forms (plants, animals and humans) haven’t been fully understood by man. There are many things in life that are very complex and the most complex of all is the human being. A simpler life form would be the bacteria. Nowadays, researchers are trying to mimic or imitate the behaviour of living organisms (like bacterial movement). Consider the problem of optimization. Optimization means finding the best point in a given area. How do we decide which point is best? As a simple example, consider an equation with a single variable ‘x’. Let us assume that you have to find the lowest value for that equation (the equation is also called as a function). In this case the best value means the lowest value. Let the equation be:
cost = 10*x + 100 …given that x is greater than 0 and can only take positive integer values between 1 and 100.
What value of x do you think gives the least cost?
If x=1 then cost=110; if x=2 then cost=120 and so on…
It is quite clear that x=1 will give the least cost. Hence the best point (or optimum value) is x=1.
That was very easy. The equation was very simple, involving just one variable. In real application you would have equations involving more than one variable and you will also have many conditions (the only condition we used was that x should lie between 1 and 100).
You can extend the problem to the case of two variables.
Let’s assume that we have to find the least value for a given equation that has two variables (x and y). Let the limits of x and y be between 1 to 100 (and assume that they can take only integer values). The solution space (or the area in which you have to find the best value) will be as below:
It is clear from the above graph that there will be 100*100 (=10,000) possible coordinates (in other words 10,000 possible combined values for x and y). The simplest way to find the minimum value for a function involving x and y would be to substitute all the 10,000 combinations, find the cost for each combination and then choose the minimum value. This seems to be a straightforward and simple method, but you can imagine the computational time needed when the solution space increases in size. And if you include floating point values, you are going to end up with a major problem. This is why we need techniques for optimization. We need to use algorithms that can determine the best point in a given solution space as quickly as possible (i.e. the algorithm should not calculate values for each and every combination).
This is where bacterial movement comes into the picture. You might have heard of the E.Coli bacteria. This bacterium lives in the human body and has a particular method of movement. It keeps searching for areas where there is more food (or nutrition). Assume that the bacteria is initially at some place in your intestine. It will take a small step in one particular direction. If the new place has more nutrition than the previous position, then it will continue to move in the same direction. Otherwise, it will take a different direction and proceed. Suppose the new position has more nutrition, it will take another step in the same direction. Again if the next position is favourable, it will proceed in the same direction. This movement is a very simplified version of the actual movement (there are more complicated steps involved in bacterial movement but we will not deal with them here). Now you can apply the same bacterial movement algorithm to the problem of optimization. The advantage is that you won’t be computing values for each and every point in the solution space (i.e. we will program in such a way that after a particular number of iterations the program will terminate). There are many other organisms in nature which have different ways of movement and this could also be applied to the problem of optimization (in fact algorithms relating to ants, bees and snakes have been developed already).
So, where do we make use of classes and objects?
Bacteria is a general category and it forms the class. All types of bacteria will have some stepsize (stepsize refers to the length of the stride taken by the bacteria in moving once) and some particular direction as well (bacteria are in three-dimensional space and they are free to take any direction). Thus, these features are common to all bacteria. We can then model bacteria as a class. This class could contain the following functions:
• To initialize the starting position of the bacteria. Every bacterium when it is created will have a particular position in 3 dimension. This has to be specified by three variables: x,y and z.
• To generate the direction in which the bacteria should move (in mathematics this is called the unit vector).
• To find the new position after the bacteria moves.
• An algorithm to decide whether to move to the new position or not (move only if new place is better than the present one).
Every instance created from the class ‘bacteria’ is an object. Each object will have a different starting position (decided randomly) and each bacteria will move in different directions (i.e. each one will try to find the best point).
Thus, we can create 10 objects of type ‘bacteria’ and make each bacteria search the solution space. After some particular number of iterations, we can stop the bacterial movement and find out the positions of each of the objects.
Continue with this explanations
or go back to Contents page. |
# Finding the Equation of a Plane from 3 Coplanar Points
• DiamondV
In summary, the method for determining the equation of a plane involves finding the cross product of two vectors on the plane to get the normal vector, and then substituting a point into the Cartesian equation to solve for d. The coefficients of the normal vector represent the x, y, and z components that make up the equation of the plane. This is because the dot product of any vector in the plane with the normal vector must be equal to 0, resulting in the equation ax+by+cz=(ad+be+cf).
DiamondV
## Homework Statement
The method that we are taught on how to determine the equation of a plane is as follows when given 3 coplanar points:
1.
Determine the vectors
2.
Find the cross product of the two vectors.
3.
Substitute one point into the Cartesian equation to solve for d.
## The Attempt at a Solution
I know how to do this but my issue is with the intuition behind it, by getting the cross product of two vectors on the plane we are essentially getting the normal vector of the entire plane. we then take the coefficents of this vector and put it into sort of an equation like this x+y+z=d, then sub a point into this to find d and that's how you get the equation of the plane. I mean what exactly is happening here? How does the coefficents of the x, y and z components(or magnitude of x,y,z, basically whatever is front of the x, y,z) give us the equation of the plane?
The cross product of two vectors in the plane will be a vector ##\vec v=(a,b,c)## that is normal to the plane. Let ##D=(d,e,f)## be any point in the plane. Then for any other point ##P=(x,y,z)## in the plane, the vector ##\vec u## from D to P lies in the plane and hence must be perpendicular to ##\vec v##.
So we have
$$0=\vec u\cdot \vec v =(x-d,y-e,z-f)\cdot(a,b,c)=a(x-d)+b(y-e)+c(z-f)=ax+by+cz-(ad+be+cf)$$
So the equation of the plane is
$$ax+by+cz=(ad+be+cf)$$
## 1. How do you find the equation of a plane from 3 coplanar points?
To find the equation of a plane from 3 coplanar points, you can use the method of cross products. First, determine the vectors between each pair of points and take the cross product of two of these vectors. This will give you the normal vector of the plane. Then, use one of the given points and the normal vector to form the equation of the plane in the form of ax + by + cz = d.
## 2. What are coplanar points?
Coplanar points are points that lie on the same plane. This means that they can be connected by a straight line and do not require any bending or stretching to be placed on the same plane. In other words, the points are in the same two-dimensional space.
## 3. Can you find the equation of a plane with only 2 coplanar points?
No, to find the equation of a plane, you need at least 3 coplanar points. This is because 2 points can only form a straight line, which does not uniquely define a plane.
## 4. What is the significance of the normal vector in finding the equation of a plane?
The normal vector is perpendicular to the plane and is used to define the orientation of the plane. It is also used in the equation of the plane to determine the coefficients of x, y, and z.
## 5. Are there other methods to find the equation of a plane from 3 coplanar points?
Yes, there are other methods such as using the formula for the distance between a point and a plane or using the components of the normal vector. However, the method of cross products is the most commonly used and straightforward method.
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# Egyptian Fractions: The Answer Sheet
Remember the Math Adventurer’s Rule: Figure it out for yourself! Whenever I give a problem in an Alexandria Jones story, I will try to post the answer (relatively) soon afterward. But don’t peek! If I tell you the answer, you miss out on the fun of solving the puzzle. So if you haven’t worked these problems yet, go back to the original post. Figure them out for yourself — and then check the answers just to prove that you got them right.
## The Secret of Egyptian Fractions
Alex made a poster of Egyptian-style fractions, from 1/2 to 9/10. Many of the fractions were easy. She knew that…
$\frac{5}{10} = \frac{4}{8} = \frac{3}{6} = \frac{2}{4} = \frac{1}{2}$
Therefore, as soon as she figured out one fraction, she had the answer to all of its equivalents.
She had the most trouble with the 7ths and 9ths. She tried converting these to other fractions that were easier to work with. For example, 28 has more factors than 7, making 28ths easier to break up into other fractions with one in the numerator.
Alex changed 2/7 into 28ths:
$\frac{2}{7} = \frac{8}{28} = \frac{7}{28} + \frac{1}{28} = \frac{1}{4} + \frac{1}{28}$
The number 36 also has plenty of factors, so Alex tried:
$\frac{7}{9} = \frac{28}{36} = \frac{18}{36} + \frac{9}{36} + \frac{1}{36} = \frac{1}{2} + \frac{1}{4} + \frac{1}{36}$
But then she realized that:
$\frac{7}{9} = \frac{6}{9} + \frac{1}{9} = \frac{2}{3} + \frac{1}{9}$
…which was so much easier that she felt dumb for not noticing it at once.
Alex found Egyptian equivalents for all of the fractions on her chart. How well did you do? Download the answer sheet and compare:
Remember that many other correct answers are possible. For example, Alex chose…
$\frac{8}{9} = \frac{2}{3} + \frac{1}{6} + \frac{1}{18}$
But this answer would be equally valid:
$\frac{8}{9} = \frac{1}{2} + \frac{1}{3} + \frac{1}{18}$
## To Be Continued…
Read all the posts from the January/February 1999 issue of my Mathematical Adventures of Alexandria Jones newsletter.
This site uses Akismet to reduce spam. Learn how your comment data is processed. |
Back to the class
Section 3.1 #2: Express $\sqrt{-21}$ in terms of $i$.
Solution: Compute $$\sqrt{-21} = \sqrt{21}i.$$ Section 3.1 #12: Simplify. Write the answer in the form $a+bi$, where $a$ and $b$ are real numbers:$$(-6-5i)+(9+2i).$$
Solution: Compute $$(-6-5i)+(9+2i)=(-6+9)+(-5i+2i)=3-3i.$$
Section 3.1 #39: Simplify. Write the answer in the form $a+bi$, where $a$ and $b$ are real numbers: $$(1+3i)(1-4i).$$ Solution: Compute $$(1+3i)(1-4i)=1-4i+3i-12i^2=1-i+12=13-i$$
Section 3.1 #74: Simplify. Write the answer in the form $a+bi$, where $a$ and $b$ are real numbers: $$\dfrac{\sqrt{5}+3i}{1-i}.$$ Solution: Compute $$\begin{array}{ll} \dfrac{\sqrt{5}+3i}{1-i} &= \left( \dfrac{\sqrt{5}+3i}{1-i} \right) \left( \dfrac{1+i}{1+i} \right) \\ &= \dfrac{\sqrt{5}+\sqrt{5}i+3i+3i^2}{1-i^2} \\ &= \dfrac{\sqrt{5} + (\sqrt{5}+3)i-3}{1-(-1)} \\ &= \dfrac{(\sqrt{5}-3)+(\sqrt{5}+3)i}{2} \\ &= \dfrac{\sqrt{5}-3}{2} + \dfrac{\sqrt{5}+3}{2} i. \end{array}$$
Section 3.1 #85: Simplify $$(-i)^{71}.$$ Solution: We want to exploit the fact that $i^2=-1$. First rewrite this as $$(-i)^{71}=(-1)^{71} i^{71} = (-1)i i^{70}.$$ To finish, we compute $$(-i)^{71} = -i i^{70} = -i \left( i^2 \right)^{35} = -i (-1)^{35} = -i (-1) = i.$$
Section 3.2 #34: Solve the quadratic equation $$x^2+6x+13=0.$$ Solution: Use the quadratic formula with $a=1$, $b=6$, and $c=13$ to compute $$\begin{array}{ll} x&=\dfrac{-6 \pm \sqrt{6^2-4(1)(13)}}{2}\\ &=\dfrac{-6 \pm \sqrt{36-52}}{2}\\ &=\dfrac{-6 \pm \sqrt{-16}}{2} \\ &= \dfrac{-6 \pm 4i}{2} \\ &= -3 \pm 2i. \end{array}$$
Section 3.2 #42: Solve the quadratic equation $$3t^2+8t+3=0.$$ Solution: Use the quadratic formula iwht $a=3$, $b=8$, and $c=3$ to compute $$\begin{array}{ll} x &= \dfrac{-8 \pm \sqrt{8^2 - 4(3)(3)}}{2(3)} \\ &= \dfrac{-8 \pm \sqrt{64 - 36}}{6} \\ &= \dfrac{-8 \pm \sqrt{28}}{6} \\ &= \dfrac{-8 \pm 2\sqrt{7}}{6} \\ &= -\dfrac{4}{3} \pm \dfrac{\sqrt{7}}{3}. \end{array}$$
Section 3.3 #4: Find the vertex, find the axis of symmetry, determine whether there is a maximum or a minimum value (and find it), and graph the function $$g(x)=x^2+7x-8.$$ Solution: The coefficient of $x^2$ is already $1$ so to complete the square, add and subtract $\left( \dfrac{7}{2} \right)^2$ and then factor to get $$\begin{array}{ll} x^2+7x-8&=x^2+7x-8 + \left( \dfrac{7}{2} \right)^2 - \left( \dfrac{7}{2} \right)^2 \\ &=\left( x+ \dfrac{7}{2} \right)^2 - 8 - \left( \dfrac{7}{2} \right)^2 \\ &=\left( x + \dfrac{7}{2} \right)^2 - \dfrac{32}{4} - \dfrac{49}{4} \\ &=\left( x+ \dfrac{7}{2} \right)^2 - \dfrac{81}{4}. \end{array}$$ From this we see that the vertex is $\left(-\dfrac{7}{2}, -\dfrac{81}{4} \right)$. The axis of symmetry is $x=-\dfrac{7}{2}$. There is no maximum. The minimum value is $-\dfrac{81}{4}$ and it occurs as $x=-\dfrac{7}{2}$.
Section 3.3 #15: Find the vertex, find the axis of symmetry, determine whether there is a maximum or a minimum value (and find it), and graph the function $$g(x)=-2x^2+2x+1.$$ Solution: First factor out the $-2$ to get $$-2\left(x^2-x-\dfrac{1}{2} \right).$$ Inside, we will complete the square by adding and subtracting $\left( - \dfrac{1}{2} \right)^2$: $$\begin{array}{ll} -2\left( x^2-x - \dfrac{1}{2} \right) &= -2 \left( x^2 - x - \dfrac{1}{2} + \left( - \dfrac{1}{2} \right)^2 - \left( -\dfrac{1}{2} \right)^2 \right) \\ &= -2 \left( \left( x-\dfrac{1}{2} \right)^2 - \dfrac{1}{2} - \dfrac{1}{4} \right) \\ &= -2 \left( \left( x -\dfrac{1}{2} \right)^2 - \dfrac{3}{4} \right) \\ &= -2 \left( x-\dfrac{1}{2} \right)^2 + \dfrac{6}{4} \\ &= -2 \left( x - \dfrac{1}{2} \right)^2 + \dfrac{3}{2}. \end{array}$$ From this we see the vertex is $\left( \dfrac{1}{2}, \dfrac{3}{2} \right)$. The axis of symmetry is $x= \dfrac{1}{2}$, there is no minimum, and the maximum is $\dfrac{3}{2}$ which occurs at $x=\dfrac{1}{2}$. |
To solve a rational inequality, we first write the inequality in the format of a rational expression on one side and zero on the other. We will then find the boundaries or endpoints by setting the numerator and denominator each equal to zero and finding solutions for the equations. We can use our boundaries or endpoints to set up intervals on a number line and use test values to determine which intervals satisfy the inequality.
Test Objectives
• Demonstrate the ability to solve a rational inequality
• Demonstrate the ability to write an inequality solution in interval notation
• Demonstrate the ability to graph an interval on the number line
Solving Rational Inequalities Practice Test:
#1:
Instructions: solve each inequality, write in interval notation, graph.
$$a)\hspace{.2em}\frac{x - 7}{x - 1}+ 5 > 0$$
$$b)\hspace{.2em}\frac{x + 5}{x - 4}- 1 < 0$$
#2:
Instructions: solve each inequality, write in interval notation, graph.
$$a)\hspace{.2em}\frac{x + 6}{x - 5}> 1$$
$$b)\hspace{.2em}\frac{5}{x - 1}> \frac{8}{2x + 7}$$
#3:
Instructions: solve each inequality, write in interval notation, graph.
$$a)\hspace{.2em}\frac{3}{x + 2}≤ \frac{4}{x - 9}$$
$$b)\hspace{.2em}\frac{x^2 + 4x - 21}{x^2 - 6x + 9}> 0$$
#4:
Instructions: solve each inequality, write in interval notation, graph.
$$a)\hspace{.2em}\frac{x^2 + 5x + 4}{x^2 + 5x + 6}< 0$$
$$b)\hspace{.2em}\frac{6}{x}> 6x - 5$$
#5:
Instructions: solve each inequality, write in interval notation, graph.
$$a)\hspace{.2em}\frac{x^2 - 4x + 4}{x - 2}≥ 1$$
$$b)\hspace{.2em}\frac{x + 6}{x^2 - 5x - 24}≥ 0$$
Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}x < 1 \hspace{.2em}\text{or}\hspace{.2em}x > 2$$ $$(-\infty, 1) ∪ (2, \infty)$$
$$b)\hspace{.2em}x < 4$$ $$(-\infty, 4)$$
#2:
Solutions:
$$a)\hspace{.2em}x > 5$$ $$(5, \infty)$$
$$b)\hspace{.2em}{-}\frac{43}{2}< x < -\frac{7}{2}\hspace{.2em}\text{or}\hspace{.2em}x > 1$$ $$\left(-\frac{43}{2}, -\frac{7}{2}\right) ∪ (1, \infty)$$
#3:
Solutions:
$$a)\hspace{.2em}{-}35 ≤ x < -2 \hspace{.2em}\text{or}\hspace{.2em}x > 9$$ $$[-35, -2) ∪ (9, \infty)$$
$$b)\hspace{.2em}x < -7 \hspace{.2em}\text{or}\hspace{.2em}x > 3$$ $$(-\infty, -7) ∪ (3, \infty)$$
#4:
Solutions:
$$a)\hspace{.2em}{-}4 < x < -3 \hspace{.2em}\text{or}\hspace{.2em}{-}2 < x < -1$$ $$(-4, -3) ∪ (-2, -1)$$
$$b)\hspace{.2em}x < -\frac{2}{3}\hspace{.2em}\text{or}\hspace{.2em}0 < x < \frac{3}{2}$$ $$\left(-\infty, -\frac{2}{3}\right) ∪ \left(0, \frac{3}{2}\right)$$
#5:
Solutions:
$$a)\hspace{.2em}x ≥ 3$$ $$[3, \infty)$$
$$b)\hspace{.2em}{-}6 ≤ x < -3 \hspace{.2em}\text{or}\hspace{.2em}x > 8$$ $$[-6, -3) ∪ (8, \infty)$$ |
1. ## Circles
Given the equation of a circle is $\displaystyle x^2+(y-6)^2=8$.
And the equation of a second circle is $\displaystyle x^2+y^2=6y+d$, where $\displaystyle d$ is an integer, is the reflection of the circle $\displaystyle x^2+(y-6)^2=8$ about the line $\displaystyle y=k$, find the values of $\displaystyle k$ and of $\displaystyle d$
2. Move the 6y on the same side as y^2 and complete the square for y, to get it in the same form as your first given equation.
Since the second one is a reflection of the first one, their radius is the same, you can find d.
Then, find the midpoint of the centres of the two circles. This will give you the value of k.
Post what you get!
3. Originally Posted by Punch
Given the equation of a circle is $\displaystyle x^2+(y-6)^2=8$.
And the equation of a second circle is $\displaystyle x^2+y^2=6y+d$, where $\displaystyle d$ is an integer, is the reflection of the circle $\displaystyle x^2+(y-6)^2=8$ about the line $\displaystyle y=k$, find the values of $\displaystyle k$ and of $\displaystyle d$
$\displaystyle x^2+y^2=6y+d\Rightarrow x^2+(y-3)^2=d+9$.
Since transformation is reflection, then both circles must have the same radius, hence $\displaystyle 8=d+9\Rightarrow d=-1.$
Both circles have its center on the y-axis $\displaystyle C_1(0,6)$ and $\displaystyle C_2(0,3)$ so $\displaystyle k=\frac{6+3}{2}=\frac{9}{2}.$ |
### Exponents
Because multiplication can be thought of as repeated addition, you can think of exponents as repeated multiplication. This means that 43 is the same as 4 × 4 × 4 or 64.
4means multiplying 4 three times. The result is 64. Here 4 is the base and superscript 3 is the exponent.
Positive and Negative Bases / Even and Odd Exponents
• A positive number taken to an even or odd power remains positive.
• A negative number taken to an odd power remains negative.
• A negative number taken to an even power becomes positive.
### Multiplying and Dividing Exponents
To multiply terms with exponents and the same bases, add the exponents. If the expression contains coefficients, multiply the coefficients as you normally would.
a3 × a2 = a5
When you divide terms with exponents and the same bases, just subtract the exponents. Any coefficients are also divided as usual.
a5 ÷ a3 = a2
To multiply exponential terms with different bases, first make sure the exponents are the same. If they are, multiply the bases and maintain the same exponent. Follow the same procedure when you divide terms with different bases but the same exponents.
a5 × b5 = (ab)5
When you raise a power to another power, multiply the exponents. If your expression includes a coefficient, take it to the same power.
(a3)5 = a15
The value of a base with an exponent of 0 is always 1.
a0 = 1
The value of a base with an exponent of 1 is the same value as the base.
a1 = a
Fractional Exponents
If you see a problem with an exponent in fraction form, consider the top number of the fraction (the numerator) as your actual exponent and the bottom number (the denominator) as the root.
Negative Exponents
A negative exponent works like a positive exponent with a twist. A negative exponent takes the positive exponent and then flips it around so the exponent becomes its reciprocal.
3-3 = 1/33 = 1/27
### Roots
Roots are also known as radicals. Roots are sort of the opposite of exponents. You square 3 to get 9, the square root of 9 is 3. |
# Helping a Struggling Maths Student: Quadrilaterals – Part Two
Discovery goals this week:
• Revise the quadrilaterals of last week
• Go over the nets of a cube ensuring each of the 11 nets are found
• Discover the formulae for the perimeter of all the quadrilaterals which are parallelograms and be able to explain why the formulae work
• Discover the formulae for the area of the quadrilaterals that are parallelograms and be able to explain why they work with these shapes but not the deltoid or trapezium
In order to practice the vocab we learnt last week and introduce a few more words I played a game whereby I gave them each a card with the name of each quadrilateral and I described one to them using mathematical language. Then I asked who I was. They had to choose the card which best fitted my description. I wanted to introduce the word congruent to their list of vocal, used to describe two lines or shapes that are exactly the same in size and shape.
Next, the girls held an unknown quadrilateral name up on their head and had to ask questions to find out who they were:
Each day I had them do a five minute exercise on the computer to make up the different quadrilaterals they had learnt. This was easy, fun and great for revising the properties of each.
Last week the children had played about with magnetix to discover the nets for a cube. They had found 6. As there are 11 I gave them this exercise to do. I was interested to see whether they would have sufficient understanding to visualise the nets they had missed. I asked them first to try without any manipulatives but if they were struggling they were free to use anything that would be helpful. I suggested marshmallows and sticks might be fun!
I brought out the big play blocks and had them build two squares of different sizes. From here we discussed how to find the lengths, perimeter and area of the shape. They have done this before and I knew they would find this fairly easy. We discussed units and the types of units we could use to describe the measurements I had asked for. In the absence of a ruler, they chose blocks and blocks squared as their units. I then asked them to work out the formula. This was interesting because they have done this many, many times before and I was quite surprised they had completely forgotten the formula. They did know how to do it, however and working their way back from their result they came up with the following formulae for finding the perimeter and area of any square:
We did the same exercise with rectangles and then I asked them to figure out why it works. They thought it had something to do with the straight lines until I pointed out that triangles had straight lines but that wasn’t how the perimeter and area of a triangle was found. Then one of them shouted out that it was the four right angles. Yay!
Next we investigating the perimeter and area of a parallelogram and rhombus (a type of parallelogram). I gave them two cardboard cut outs each of a rhombus and parallelogram and two identical ones cut out of squared paper. I asked them to give me some methods to work out the perimeter and area of the two shapes. For the perimeter they said they could measure using a ruler or a piece of string and then measure the piece of string. The area wasn’t so simple, although L10 did say they could count the squares which was accurate if not a little laborious. I asked them whether the formula for the rectangle and square would or wouldn’t work and why or why not. They thought not and after some prodding they mentioned the lack of square corners.
It was at this point I gave them a pair of scissors each and told them to figure out the correct formula! They looked at me with stunned suspicion to which I had to laugh! T11 was now with us and I suggested they look at what they knew to be true from the day before (formula for area of square/rectangle = length x width) and try to find a way to use that formula with the rhombus and the parallelogram. L10 was away! This is the girl who cried everyday over her maths and she is now simply getting it because she can touch and therefore visualise it:
It was T11 who figured out that you placed the two triangles together to make a second rectangle:
Probably a bit unnecessarily, I asked them to demonstrate that the cut out rectangle had the same area as the parallelogram, which they did easily:
Now for the tricky part. They understood that the same formula could be used, but when I asked where the width was on a parallelogram they all thought it was the slanted line. We went back to their cut out versions and found that the width of the cut out rectangle was actually the height of the parallelogram i.e. the distance between the top and bottom line. I wrote out on the board their narration of what was going on:
They were perplexed, because obviously the formula for a parallelogram is slightly different even though the maths is the same: area of a parallelogram = height x length I simply reminded them that letters were only representational of numerical values and not to worry too much. The important thing was to understand the dimensional maths of each shape rather than just learning a set of formulae.
I placed examples of all the quadrilaterals and asked them to put into one pile those for which they had a formula for the area and perimeter and in another pile those they had no formula for. We then discussed what the difference was between the two piles. The children were able to state that the pile with the formulae were all parallelograms whilst the other pile were not. I told them we would be revisiting the second pile after we had investigated triangles as there is a link between the areas of quadrilaterals which are not parallelograms, and triangles.
I had some questions from a year 7 paper (grade 6) which I gave the girls to check their understanding. T11 flew through them, whilst L10 and C10 plodded through. They both got stuck on one when the wording threw them. In this question they were given the area and perimeter of a rectangle and had to work out the length and width. Once I had demonstrated how to work it out they were away! I wrote out a few more to check they understood, which they did. What struck me most about the girls working through these problems was their obvious confidence. Even when they struggled there were no tears just a willingness to listen and understand and then just get on with it. They no longer have maths anxiety.
Next week we will be investigating square numbers, and continuing with area and perimeter. |
# How do solve the following linear system?: 3x - y = -6, y = 5x - 7?
Aug 27, 2017
See a solution process below: $\left(\frac{13}{2} , \frac{51}{2}\right)$
#### Explanation:
Step 1) Because the second equation is already solved for $y$ we can substitute $\left(5 x - 7\right)$ for $y$ in the first equation and solve for $x$:
$3 x - y = - 6$ becomes:
$3 x - \left(5 x - 7\right) = - 6$
$3 x - 5 x + 7 = - 6$
$\left(3 - 5\right) x + 7 = - 6$
$- 2 x + 7 = - 6$
$- 2 x + 7 - \textcolor{red}{7} = - 6 - \textcolor{red}{7}$
$- 2 x + 0 = - 13$
$- 2 x = - 13$
$\frac{- 2 x}{\textcolor{red}{- 2}} = \frac{- 13}{\textcolor{red}{- 2}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 2}}} x}{\cancel{\textcolor{red}{- 2}}} = \frac{13}{2}$
$x = \frac{13}{2}$
Step 2) Substitute $\frac{13}{2}$ for $x$ in the second equation and calculate $y$:
$y = 5 x - 7$ becomes:
$y = \left(5 \cdot \frac{13}{2}\right) - 7$
$y = \frac{65}{2} - \left(\frac{2}{2} \times 7\right)$
$y = \frac{65}{2} - \frac{14}{2}$
$y = \frac{51}{2}$
The Solution Is: $x = \frac{13}{2}$ and $y = \frac{51}{2}$ or $\left(\frac{13}{2} , \frac{51}{2}\right)$ |
News
# Teaching Highest Common Factors (HCF) – The Two Essential Stages Worth of Consideration
Highest Common Factors HCF
This blog looks at Highest Common Factors (or Greatest Common Divisors) of two or more numbers. There are two stages worthy of consideration:
1. Common factors and Highest Common Factors by inspection of 2 or more numbers
2. Calculating by using prime factors
### Common factors of 2 or more numbers
If we examine all the factors of say, 12 and 30 we can compare results to see which factors are common to both numbers.
12 = {1, 2, 3, 4, 6, 12}
30 = {1, 2, 3, 5, 6,10, 15, 30}
We can observe that factors common to both 12 and 30 are 1, 2, 3 and 6. The highest valued factor of both of them is therefore 6. We can do this with more than just two numbers.
The same procedure can be used to find all the factors and by definition, the HCF of all three numbers.
For example look at finding all the factors of 18, 42 and 56.
18 = {1,2, 3, 6, 9, 18}
42 = {1, 2, 3, 6, 7, 14, 21,42}
56 = {1,2, 4, 7, 8, 14, 28, 56}
Factors common to all of them are 1 and 2 only, as such we can see that the highest common factor of all three numbers is 2. In other words, there is no greater number than 2 which will divide all three numbers exactly.
This method, however, can be time consuming, no matter that it will always throw out the HCF of any amount of numbers you care to use.
### Calculating by using prime factors
If we break the numbers down into prime factor form we can get to the solution more efficiently. We will again look at finding the HCF of 12 and 60 to see how prime factors work in the process.
By expressing the number 12 as 22 x 3 and 30 as 2 x 3 x 5 in columns of prime values we can inspect each column to see where common primes exist, note them, then multiply them finally to see the HCF.
In the prime column for ‘2’ we see that 12 has 22 and 30 has 2.
We now need to be clear that the highest common factor of these two numbers is 2 as it is the greatest possible divisor of both 2 and 22 (4). 3 is common to both 12 and 30 and is recorded in the HFC primes column.
Later all values in the HFC primes row will be multiplied to give the HFC of the original numbers. We are left with 2 and 3 in this case which when multiplied give the product 6. We can say that no number greater than 6 divides both 12 and 30 exactly. The same approach can be used with 3 or more numbers.
We will take the earlier example of 18, 42 and 56.
By inspection we see the only prime column that is common to all 3 numbers is 2 as 3 is only a factor of 42 and 7 is a factor of 42 and 56 but not 18.
Again there is a higher power of 2 which is a factor of 56 (23 in this case). As before we choose the lowest power in the column (2) as there is no higher power of 2 which will divide exactly all three.
### Recap
When finding the HCF of 2 or more numbers practice the following stages in order.
1. Convert all numbers to products of their primes
2. Record the primes which are common to all numbers in the problem
3. Record the lowest power that occurs in any prime column (see tables above)
### HCF..or Greatest Common Divisor
This blog assumes that factors of whole numbers have been defined and understood.
When finding the HCF of two or more numbers it makes sense to write the numbers in question as products of their respective primes.
For example as a product of their primes 45 is written 32 x 5 and 60 is written 22 x 3 x 5.
Before looking at an efficient way to calculate HCFs we should look at these two numbers and factor each one out so that we can see the largest factor that is common to both of them.
45 = {1, 3, 5, 9, 15, 45}
60 = {1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60}
We see that there are more than one common factors to both but the largest that they have in common is 15. Armed with this clear evidence we take it back to the prime factor layout to see how and where 15 can be observed there.
15 = 32 x 5
60 = 22 x 3 x 5 |
# How do you implicitly differentiate y= (x-y^2) e^(xy) ?
Jan 25, 2018
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x y} - \left(x - {y}^{2}\right) {e}^{x y} y}{1 + 2 y - \left(x - {y}^{2}\right) {e}^{x y} x}$
#### Explanation:
$y = \left(x - {y}^{2}\right) {e}^{x y}$
We could proceed using the product rule. In implicit differentiation, we must also remember to use the chain rule when differentiating a $y$ since $y$ is a function of $x$. So:
$y = \left(x - {y}^{2}\right) {e}^{x y}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(x - {y}^{2}\right) \cdot {e}^{x y} + \left(x - {y}^{2}\right) \frac{d}{\mathrm{dx}} {e}^{x y}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(1 - 2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) {e}^{x y} + \left(x - {y}^{2}\right) \cdot {e}^{x y} \cdot \frac{d}{\mathrm{dx}} \left\{x y\right\}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(1 - 2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) {e}^{x y} + \left(x - {y}^{2}\right) \cdot {e}^{x y} \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)$
Now gather all terms with $\frac{\mathrm{dy}}{\mathrm{dx}}$ on the left side:
$\frac{\mathrm{dy}}{\mathrm{dx}} + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} - \left(x - {y}^{2}\right) {e}^{x y} x \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x y} + \left(x - {y}^{2}\right) {e}^{x y} y$
$\frac{\mathrm{dy}}{\mathrm{dx}} \left(1 + 2 y - \left(x - {y}^{2}\right) {e}^{x y} x\right) = {e}^{x y} + \left(x - {y}^{2}\right) {e}^{x y} y$
$\to \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x y} - \left(x - {y}^{2}\right) {e}^{x y} y}{1 + 2 y - \left(x - {y}^{2}\right) {e}^{x y} x}$
Jan 25, 2018
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x y {e}^{x y} - {y}^{3} {e}^{x y} + {e}^{x y}}{1 - {x}^{2} {e}^{x y} + x {y}^{2} {e}^{x y} + 2 y {e}^{x y}}$
#### Explanation:
$\text{differentiate using the "color(blue)"product/chain rules}$
$\text{given "y=f(x)g(x)" then}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = f \left(x\right) g ' \left(x\right) + g \left(x\right) f ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{product rule}}$
$\text{given "y=f(g(x))" then}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$
$f \left(x\right) = x - {y}^{2} \rightarrow f ' \left(x\right) = 1 - 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$
$g \left(x\right) = {e}^{x y} \Rightarrow g ' \left(x\right) = {e}^{x y} \times \frac{d}{\mathrm{dx}} \left(x y\right)$
$\textcolor{w h i t e}{\times \times \times \times \times \times x} = {e}^{x y} \left(x \frac{\mathrm{dy}}{\mathrm{dx}} + y\right)$
$\textcolor{w h i t e}{\times \times \times \times \times \times x} = x {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + y {e}^{x y}$
(x-y^2)(xe^(xy)dy/dx+ye^(xy)))+e^(xy)(1-2ydy/dx)
$= {x}^{2} {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + x y {e}^{x y} - x {y}^{2} {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} - {y}^{3} {e}^{x y} + {e}^{x y} - 2 y {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}}$
$y = \left(x - {y}^{2}\right) {e}^{x y} \leftarrow \textcolor{b l u e}{\text{original question}}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} - {x}^{2} {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + x {y}^{2} {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y {e}^{x y} \frac{\mathrm{dy}}{\mathrm{dx}} = x y {e}^{x y} - {y}^{3} {e}^{x y} + {e}^{x y}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} \left(1 - {x}^{2} {e}^{x y} + x {y}^{2} {e}^{x y} + 2 y {e}^{x y}\right) = x y {e}^{x y} - {y}^{3} {e}^{x y} + {e}^{x y}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x y {e}^{x y} - {y}^{3} {e}^{x y} + {e}^{x y}}{1 - {x}^{2} {e}^{x y} + x {y}^{2} {e}^{x y} + 2 y {e}^{x y}}$ |
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# EX.13.3 Q1 Exponents-and-Powers Solutions- NCERT Maths class 7
Go back to 'Ex.13.3'
## Question
Write the following numbers in the expanded forms:
$$279404,3006194,2806196,\\\qquad120719,20068$$
Video Solution
Exponents And Powers
Ex 13.3 | Question 1
## Text Solution
Reasoning:
Expanded form of number means expressing a number using powers of $$10$$ as exponents.
(i) $$279404$$
\begin{align}&=\!\left[\begin{array} =2 \times 100000 + 7 \times 10000 +\\ 9 \times 1000 + 4 \times 100 +\\ 0 \times 10 + 4 \times 1\end{array} \right]\\&=\!\left[\begin{array}=2 \times {{10}^5} + 7 \times {{10}^4} + \\9 \times {{10}^3} + 4 \times {{10}^2} +\\ 0 \times {{10}^1} + 4 \times {{10}^0}\end{array} \right]\end{align}
(ii) $$3006194$$
\begin{align} &=\! \left[\begin{array} = 3 \times 1000000 + \\ 0 \times 10000 + 0 \times 10000 \\ + 6 \times 1000 + 1 \times 100 \\ + 9 \times 10 + 4 \times 1\end{array} \right]\\&= \left[\begin{array}= 3 \times {{10}^6}+0\times{{10}^5} +\\ 0\times {{10}^4} + 6\times {{10}^3} +\\1 \times{{10}^2} + 9 \times {{10}^1} +\\ 4 \times {{10}^0}\end{array} \right]\end{align}
(iii) $$2806196$$
\begin{align}&=\left[\begin{array} = 2 \times 10000 + 0 \times 1000 + \\ 0 \times 100 + 6 \times10 + \\ 8 \times 1\end{array} \right]\\&=\left[\begin{array}= 2 \times {{10}^4} + 0 \times {{10}^3} +\\ 0 \times {{10}^2} + 6 \times {{10}^1} + \\ 8 \times {{10}^0}\end{array} \right]\end{align}
(iv) $$120719$$
\begin{align}&=\! \left[\begin{array} = 1 \times 100000 + 2 \times 10000 + \\ 0 \times 1000 + 7 \times 100 + \\ 1 \times 10 + 9 \times 1\end{array} \right]\\&=\! \left[\begin{array} = 1 \times {10}^5 + 2 \times {10}^4 + \\ 0 \times {10}^3 + 7 \times {10}^2 + \\ 1 \times {10}^1 + 9 \times {10}^0\end{array} \right]\end{align}
(v) $$20068$$
\begin{align}&=\left[\begin{array} = 2 \times 10000 + 0 \times 1000 +\\ 0 \times 100 + 6 \times 10 + 8 \times 1\end{array} \right]\\&=\left[\begin{array}=2 \times {{10}^4} + 0 \times {{10}^3} + 0 \times\\ {{10}^2} + 6 \times {{10}^1} + 8 \times {{10}^0}\end{array}\right]\end{align}
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# Antiderivative of a Fraction: Complete Explanation and Examples
The antiderivative, also called the integral of a function, is the inverse process of taking the derivative of a function.
When we have a function $\dfrac{p}{q}$ where $q \neq 0$, then such an expression is called a fraction, and if we take the antiderivative of such a function, then it will be called the antiderivative of that fraction.
In this topic, we will discuss how to take the antiderivative or integral of a fraction, and we will discuss in detail solving fraction problems using the partial fraction technique of integration.
## What Is the Antiderivative of a Fraction?
The antiderivative, also called the integral of a function, is the inverse process of taking the derivative of a function; if we take the antiderivative of an algebraic function that is written as a fraction, we call it the antidifferentiation of a fraction. We know that a fraction is given in $\dfrac{p}{q}$ with $q \neq 0$. The antiderivative of a fraction can be divided into two types.
To solve antiderivative problems, some basic antiderivative relations must be memorized. For example, the antiderivative of a constant fraction is $\int \dfrac{1}{k} = \dfrac{1}{k} x +c$; the antiderivative of $\frac{1}{x}$ is $ln|x| +c$. Similarly, the antiderivative of $\dfrac{1}{x^{2}}$ is $-\dfrac{1}{x} + c$.
## How To Find the Antiderivative of Fractions
The simple answer to finding the antiderivative of an algebraic expression having multiple or complicated fractions is by using the fraction decomposition or separation of the fraction into smaller parts and then taking the antiderivative of those smaller fractions. Most rational fractions are solved by using partial fractions, while irrational fractions are solved by using the substitution method.
We will now discuss different examples related to fractions and how we can take the antiderivative of fractions with different types of quotients algebraic expressions.
### Antiderivative of a Rational Fraction
A rational fraction is a fraction wherein both the numerator and denominator consist of polynomials. For example, $\dfrac{x + 7}{x}$ is a rational fraction.
We can easily calculate the antiderivative for the above given rational fraction by dividing it into parts. We can write $\dfrac{x + 7}{x}$ as $( \dfrac{x}{x} + \dfrac{7}{x})$. Let us now calculate the antiderivative of the given rational function.
$\int \dfrac{x + 7}{x} = \int(\dfrac{x}{x} + \dfrac{7}{x})$
$\int \dfrac{x + 7}{x} = \int ( 1 + \dfrac{7}{x})$
$\int \dfrac{x + 7}{x} = \int 1 + \int \dfrac{7}{x}$
$\int \dfrac{x + 7}{x} = x – \dfrac{7}{x^{2}}$
It is not necessary that all the rational numbers can easily be divided into parts to find their antiderivative. The denominator can consist of multiple linear factors or repeated linear factors; in such cases, it is advisable to solve the problem using the partial fraction technique.
### Fractions with Two Linear Factors
When we are given a fraction function such that the power/degree of the numerator is less than that of the denominator while the denominator has two distinct linear factors, then we can use a partial fraction to separate the fraction into smaller parts and then find out the antiderivative of the function.
For example, we are given an integral function $\int \dfrac{x}{(x + 3) (4 – x)}$, we will use partial fraction decomposition to separate the given fraction.
$\dfrac{x}{(x + 3) (4 – x)} = \dfrac{A}{(x + 3)} + \dfrac{B} {(4 – x)}$
$\dfrac{x}{(x + 3) (4 – x)} = \dfrac{A}{(x + 3)} + \dfrac{B} {(4 – x)}$
$\dfrac{x}{(x + 3) (4 – x)} = \dfrac{A (4 – x) + B (x-3)}{(x + 3) (4 – x)}$
$x = A (4 – x) + B (x – 3)$
Now we will choose the value of “x” in such a manner that it makes an algebraic expression with “A” or “B” zero. So let us take $x = 3$ and put it in the above equation:
At $x = 3$
$3 = A ( 4 – 3) + B ( 3 – 3)$
$A = 3$
At $x = 4$
$4 = A (4 – 4) + B ( 4 – 3)$
$B = 4$
$\dfrac{x}{(x + 3) (4 – x)} = \dfrac{3}{(x + 3)} + \dfrac{4} {(4 – x)}$
$\int \dfrac{x}{(x + 3) (4 – x)} = \int (\dfrac{3}{x + 3} + \dfrac{4} {4 – x})$
$\int \dfrac{x}{(x + 3) (4 – x)} = \int \dfrac{3}{x + 3} + \int \dfrac{4} {4 – x})$
$\int \dfrac{x}{(x + 3) (4 – x)} = 3 \int \dfrac{1}{x + 3} – 4 \int \dfrac{-1} {4 – x})$
$\int \dfrac{x}{(x + 3) (4 – x)} = 3 ln (x +3) – 4 ln (4 – x) + c$
The examples we have studied so far used definite integrals but with no upper and lower limits. Let us now solve an example with upper and lower limits using the partial fraction decomposition method.
Example 1: Evaluate the given antiderivative function.
$\int_{2}^{4} \dfrac{4}{x (x + 2)}$
Solution:
$\int_{2}^{4} \dfrac{4}{x (x + 2)}$
By using the partial fraction decomposition method, we can write the above equation as:
$\dfrac{4}{x (x + 2)} = \dfrac{A}{x} + \dfrac{B} {(x + 2)}$
$\dfrac{4}{ x (x + 2)} = \dfrac{A}{x} + \dfrac{B} {(x + 2)}$
$\dfrac{4}{x (x + 2)} = \dfrac{A (x + 2) + Bx }{x (x + 2)}$
$4 = A (x + 2) + Bx$
Now we will choose the value of “x” in such a manner that it makes an algebraic expression with “A” or “B” zero. So let us take x = 0 and put it in the above equation:
At $x = 0$
$3 = A ( 0 + 2) + B (0)$
$3 = 2A$
$A = \dfrac{3}{2}$
At $x = -2$
$4 = A (2 – 2) – 2B$
$4 = -2B$
$B = -2$
$\dfrac{x}{(x + 3) (4 – x)} = \dfrac{3}{(x + 3)} + \dfrac{4} {(4 – x)}$
$\int_{2}^{4} \dfrac{x}{(x + 3) (4 – x)} = \int_{2}^{4} (\dfrac{3}{x + 3} + \dfrac{4} {4 – x})$
$\int_{2}^{4} \dfrac{x}{(x + 3) (4 – x)} = \int_{2}^{4} \dfrac{3}{x + 3} + \int_{2}^{4} \dfrac{4} {4 – x})$
$\int_{2}^{4} \dfrac{x}{(x + 3) (4 – x)} = 3 \int_{2}^{4} \dfrac{1}{x + 3} – 4 \int_{2}^{4} \dfrac{-1} {4 – x})$
$\int_{2}^{4} \dfrac{x}{(x + 3) (4 – x)} = [3 ln (x +3) – 4 ln (4 – x) ]_{2}^{4}$
$\int_{2}^{4} \dfrac{x}{(x + 3) (4 – x)} = [3 ln (4 +3) – 4 ln (4 – 4) – 3 ln (2 + 3) + 4 ln (4 – 2) ]$
$\int_{2}^{4} \dfrac{x}{(x + 3) (4 – x)} = ( 5.8377 – 4 – 4.828 + 2.772) = -0.22$
### Fractions With Repeated Factors
When we are given a fraction function such that the power/degree of the numerator is less than that of the denominator while the denominator has repeated linear factors, we have to use a partial fraction to separate the fraction into smaller parts and then find out the antiderivative of the function.
For example, if we are given an integral function $\int \dfrac{x}{(x + 3) (4 – x)}$, we will use partial fraction to separate the given fraction.
$\dfrac{4}{(x – 4)^{2} (x + 4)} = \dfrac{A}{(x – 4)} + \dfrac{B} {(x – 4)^{2}} + \dfrac{C} {(x + 4)}$
$\dfrac{4}{(x – 4)^{2} (x + 4)} = \dfrac{A (x – 4) (x+4) + B (x + 4) + C (x-4)^{2}}{(x – 4)^{2} ( x +4)}$
$4 = A (x – 4) (x + 4) + B (x + 4) + C (x – 4)^{2}$
At $x = 4$
$4 = 0 + B ( 4 + 4) + 0 = B = \dfrac{1}{2}$
At $x = – 4$
$4 = 0 + 0 + C (-4 – 4)^{2}$
$4 = 64 C$
$C = \dfrac{1}{16}$
We know the value of B and C, now let us put x = 0:
At $x = 0$ |
# Find a relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the points $(3, 6)$ and $(-3, 4)$.
Given:
The points $(3, 6)$ and $(-3, 4)$ are equidistant from the point $(x, y)$.
To do:
We have to find a relation between $x$ and $y$.
Solution:
We know that,
The distance between two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is $\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$.
Therefore,
The distance between $(3,6)$ and $(x, y)$ $=\sqrt{(x-3)^{2}+(y-6)^{2}}$
The distance between $(-3,4)$ and $(x, y)$ $=\sqrt{(x+3)^{2}+(y-4)^{2}}$
The points $(3,6)$ and $(-3,4)$ are equidistant from $(x, y)$.
$\therefore \sqrt{(x-3)^{2}+(y-6)^{2}}=\sqrt{(x+3)^{2}+(y-4)^{2}}$
Squaring on both sides, we get,
$\Rightarrow(x-3)^{2}+(y-6)^{2}=(x+3)^{2}+(y-4)^{2}$
$\Rightarrow x^{2}-6 x+9+y^{2}-12 y+36=x^{2}+6 x+9+y^{2}-8y+16$
$\Rightarrow x^{2}-6 x+y^{2}-12 y+36-x^{2}-6 x-y^{2}+8 y-16=0$
$\Rightarrow-12 x-4 y+20=0$
$\Rightarrow -4(3x+ y-5)=0$
$\Rightarrow 3x+y-5=0$
The required relation is $3x+y-5=0$.
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# Computer Lab #2: Integration
```Computer Lab #2: Integration
Introduction: In this laboratory we will be investigating some aspects of the process of finding antiderivatives, using
W∣α as an aid to simplify otherwise tedious hand calculations.
Commands Reviewed:
integrate
but we can trick it by introducing variables:
Symbolic and Numerical Integration
We begin by looking at the integrate command in
partial fractions 1/(x^2-a^2)
=
greater detail. If there are multiple variables involved, it
The situation gets more complicated if there are differperforms integration with respect to the alphabetically first
ent
roots:
variable, considering the others as constant:
integrate cos(a*b)
solve x^2+2*x-10
=
shows that x 2 + 2x − 10 = (x + 1 +
One can specify a variable for integration as a second
argument:
integrate(cos(a*b),b)
=
partial fractions
1/((x+1-sqrt(11))*(x+1+sqrt(11)))
=
A good part of the integration techniques taught in this
course involved suitable substitutions. In fact there is often
no single “correct” substitution, but for a particular problem, there are many substitutions that will work. However
often many of these are very surprising, while the ones
taught in class are somewhat more systematical. In general, examining these nontraditional substitutions is rather
tedious, but with the computer’s help we can do this.
As illustration lets investigate the case of trigonometric substitutions. Often the resulting antiderivative has
no traces of the trigonometric functions left. Knowing
the antiderivative, we can often guess a suitable (nontrigonometric) u-substitution that does the job. (However,
as it requires already knowing an antiderivative, it is of limited practical use. This is the reason why we learn trigonometric substitution in class.)
Let us look as example at ∫ √ d 2x . In class we learned
x −4
that the appropriate substitution would be sec(θ) = x2 , and
this is indeed what
=
=
(You are welcome to use these functions to check your
homework solutions. However we ask that you still do the
computer during exams.)
Partial Fractions
Beyond its use in integration, partial fractions are a useful
tool in other areas of mathematics. (For example many of
you will encounter it again in MATH340 when doing the
Laplace transform.) W∣α can do partial fraction decompositions easily:
partial fractions 1/((x-2)*(x+5)^2)
integrate 1/(sqrt(x^2-4))
=
=
does if we ask it to show the
√ steps. However if we look at
the antiderivative ln (x + x 2 − 4), we might get the idea
√
to try the (non-obvious) substition u = x + x 2 − 4. Let’s
try this:
Note however that W∣α will not introduce irrational
roots on its own:
partial fractions 1/(x^2-3)
=
Substitution
If the integral does not converge, W∣α tells us so (at
least in cases it can verify convergence):
integrate(1/x^5,x=0..1)
=
just returns a single term, but if we explicitly give the factorization, it works as desired.
We can also use infinity as limit to get the values of
improper integrals:
integrate(1/x^5,x=1..infinity)
=
√
11)(x + 1 − 11), but
partial fractions 1/(x^2+2*x-10)
or use this second argument to specify limits for definite
integration:
integrate(sin(x)^2,x=0..2*Pi)
√
=
1
tells us that x =
u 2 +4
,
2u
4. Find a value for the constant a such that
=
solve u=x+sqrt(x^2-4),x
∫
and
=
5. Find a non-trigonometric substitution that would
solve the integral ∫ (1−x12 )3/2 dx. (Hint: first integrate
this problem and then try substitution by hand for
parts of the result.)
tells us that dx = 21 − u22 du.
Let’s look at the substituted integrand and simplify:
simplify 1/sqrt(((u^2+4)/(2*u))^2-4)
*(1/2-2/u^2)
6. Consider the integral
=
∫ ((x
we get (under the assumption that u > 2, so u 2 − 4 > 0)
the integrand 1/u. For this we can clearly find an antiderivative by hand. Substituting u back, we get the desired result.
Remember that your lab write up should contain not just a
session transcript but text and (if applicable) explanations.
Note that some problems require a work that cannot be
done alone with W∣α, but a mixture of computer and hand
work. In that case you are welcome to write the manual
part by hand.
− 1)(x + 1))−2/3 dx
(a) u = 1/(x + 1)
(b) u = atan(x)
(c) u = acos(x)
(d) u = atan((x − 1)/2)
1. Determine a partial fraction decomposition of
7. Let f (x)
1
(−105 − 48x + 80x 2 + 22x 3 − 18x 4 − 2x 5 + x 6 )
=
−
10
d
and g(x)
dx ln(ln(x))
=
d
ln(ln(ln(x))).
dx
a) Without further calculations, explain which integrals of
Note that the denominator has 6 roots, so you
should get 6 summands!
2
sin(x)
dx?
x
1
Can you explain why a function Si is introduced?
∫
2
Now perform, using W∣α as a tool for differentiation and simplification, the following substitutions.
In each case write down what the result of the substitution is, and with which method one could solve
the resulting integral. (You may use W∣α also as a
tool to find a suitable integration approach.) You do
not need to execute these actual integrations.
Homework
∫
∞
100
3. Does the following W∣α calculation return a correct
result? Think about whether this solution holds for
all values of a. Explain!
integrate x^a
2
a ⋅ e −x dx = 2.
−∞
differentiate (u^2+4)/(2*u)
2. What is the result of the integrating
∞
=
2
f (x)dx,
∫
∞
100
g(x)dx
converge. (The choice of 100 is just as ln(ln(100)) >
0.)
b) Plot f and g on a suitable range. What does this
tell you about convergence of the improper integrals
in a)?
c) Now consider h(x) = g(x)/ f (x). Plot h(x).
What is limx→∞ h(x)? Does the limit comparison
test say anything about comparing the integrals?
``` |
# Verbal Reasoning for Circular Seating Arrangement for SBI Clerk 2020, IBPS Pre, LIC Assistant, IBPS Clerk, SBI PO Pre at Smartkeeda
Directions: Study the following information carefully and answer the questions given beside:
Some persons are sitting around a circular table facing towards the centre and at an equal distance from each other. Each of these persons has a different income greater than 11 Lakhs but less than 20 Lakhs.
Only one seat was vacant.
Anuj’s income is in multiple of 5 but it is not the highest.
There are two seats between Jaya and Reet.
Anuj sits opposite to Faiz but neither of them are sitting adjacent to Jaya.
None of the seats adjacent to Anuj is vacant.
The one, whose income is 19 Lakhs, is sitting on the seat which is to the immediate right of Pari’s seat.
Reet’s income is in perfect square number.
Jaya’s income is in even number.
Pari sits on the seat which is second to the right of one, whose income is 13 Lakhs.
The income of the persons sitting adjacent to vacant seat is the just next number of income the persons sitting adjacent to them.
The number of persons between Reet and dev are same as the number of persons between Anuj and Mala.
Important for :
1
How many persons are sitting around the circular table?
» Explain it
B
Following the final solution we can say that seven persons are sitting around the circular table.
Hence, the correct answer is option B.
Final Solution:
Common Explanation:
Reference:
Some persons are sitting around a circular table facing towards the centre and at an equal distance from each other. Each of these persons has a different income greater than 11 Lakhs but less than 20 Lakhs.
Inference:
Here, we can say that there were at most 8 seats in the arrangement and we keep the rest of the information in mind while solving the puzzle.
Reference:
Anuj sits opposite to Faiz but neither of them are sitting adjacent to Jaya.
Inference:
Using the above information we can say that there were even numbers of seats around the circular table as it is given that Anuj is sitting opposite to Faiz and we know that all these persons are equidistant from each other which cannot be possible if the number of seats is odd.
Here, we will make a mental note that neither Jaya nor Reet are sitting adjacent to Jaya.
Reference:
There are two seats between Jaya and Reet.
Inference:
At this point we can say that were exactly 8 seats in the arrangement because of the number of seats is less than 8 the above hints will not follow.
At this point we can fix the position of Jaya, Reet, Anuj and Faiz as:
Case 1:
Case 2:
Reference:
Anuj’s income is in multiple of 5 but it is not the highest.
Reet’s income is in perfect square number.
Jaya’s income is in even number.
Inference:
After using the above hints, we have:
Anuj’ income = 15 Lakhs
Reet’s income = 16 Lakhs
Here, we will make mental note that income of Jaya was an even number.
Reference:
Pari sits on the seat which is second to the right of one, whose income is 13 Lakhs.
The one, whose income is 19 Lakhs, is sitting on the seat which is to the immediate right of Pari’s seat.
Inference:
Here, there are three possible scenarios in which we can use the above hints in case 1.
Case 1-A:
Case 1-B:
Case 1-C:
Case 2:
Reference:
Only one seat was vacant.
None of the seats adjacent to Anuj is vacant.
Inference:
At this point our Case 1-A, Case 1-B and Case 2 are contradicting with the above hints. So we can say that Case 1-A, Case 1-B and Case 2 are invalid cases.
Reference:
The income of the persons sitting adjacent to vacant seat is the just next number of income the persons sitting adjacent to them.
The number of persons between Reet and Dev are same as the number of persons between Anuj and Mala.
Inference:
After using the above hints, we have:
Case 1-C:
2
Who among the following is sitting opposite to Mala?
» Explain it
D
Following the final solution we can say that Reet is sitting opposite to Mala.
Hence, the correct answer is option D.
Final Solution:
Common Explanation:
Reference:
Some persons are sitting around a circular table facing towards the centre and at an equal distance from each other. Each of these persons has a different income greater than 11 Lakhs but less than 20 Lakhs.
Inference:
Here, we can say that there were at most 8 seats in the arrangement and we keep the rest of the information in mind while solving the puzzle.
Reference:
Anuj sits opposite to Faiz but neither of them are sitting adjacent to Jaya.
Inference:
Using the above information we can say that there were even numbers of seats around the circular table as it is given that Anuj is sitting opposite to Faiz and we know that all these persons are equidistant from each other which cannot be possible if the number of seats is odd.
Here, we will make a mental note that neither Jaya nor Reet are sitting adjacent to Jaya.
Reference:
There are two seats between Jaya and Reet.
Inference:
At this point we can say that were exactly 8 seats in the arrangement because of the number of seats is less than 8 the above hints will not follow.
At this point we can fix the position of Jaya, Reet, Anuj and Faiz as:
Case 1:
Case 2:
Reference:
Anuj’s income is in multiple of 5 but it is not the highest.
Reet’s income is in perfect square number.
Jaya’s income is in even number.
Inference:
After using the above hints, we have:
Anuj’ income = 15 Lakhs
Reet’s income = 16 Lakhs
Here, we will make mental note that income of Jaya was an even number.
Reference:
Pari sits on the seat which is second to the right of one, whose income is 13 Lakhs.
The one, whose income is 19 Lakhs, is sitting on the seat which is to the immediate right of Pari’s seat.
Inference:
Here, there are three possible scenarios in which we can use the above hints in case 1.
Case 1-A:
Case 1-B:
Case 1-C:
Case 2:
Reference:
Only one seat was vacant.
None of the seats adjacent to Anuj is vacant.
Inference:
At this point our Case 1-A, Case 1-B and Case 2 are contradicting with the above hints. So we can say that Case 1-A, Case 1-B and Case 2 are invalid cases.
Reference:
The income of the persons sitting adjacent to vacant seat is the just next number of income the persons sitting adjacent to them.
The number of persons between Reet and Dev are same as the number of persons between Anuj and Mala.
Inference:
After using the above hints, we have:
Case 1-C:
3
What is the income of Pari?
» Explain it
D
Following the final solution we can say that we have no information about the income of Pari but we have only two option left with us that must be lie in between 11 Lakhs and 20 Lakhs i.e 12 Lakhs and 18 Lakhs so we can say that Pari income must be either 12 Lakhs or 18 Lakhs.
Hence, the correct answer is option D.
Final Solution:
Common Explanation:
Reference:
Some persons are sitting around a circular table facing towards the centre and at an equal distance from each other. Each of these persons has a different income greater than 11 Lakhs but less than 20 Lakhs.
Inference:
Here, we can say that there were at most 8 seats in the arrangement and we keep the rest of the information in mind while solving the puzzle.
Reference:
Anuj sits opposite to Faiz but neither of them are sitting adjacent to Jaya.
Inference:
Using the above information we can say that there were even numbers of seats around the circular table as it is given that Anuj is sitting opposite to Faiz and we know that all these persons are equidistant from each other which cannot be possible if the number of seats is odd.
Here, we will make a mental note that neither Jaya nor Reet are sitting adjacent to Jaya.
Reference:
There are two seats between Jaya and Reet.
Inference:
At this point we can say that were exactly 8 seats in the arrangement because of the number of seats is less than 8 the above hints will not follow.
At this point we can fix the position of Jaya, Reet, Anuj and Faiz as:
Case 1:
Case 2:
Reference:
Anuj’s income is in multiple of 5 but it is not the highest.
Reet’s income is in perfect square number.
Jaya’s income is in even number.
Inference:
After using the above hints, we have:
Anuj’ income = 15 Lakhs
Reet’s income = 16 Lakhs
Here, we will make mental note that income of Jaya was an even number.
Reference:
Pari sits on the seat which is second to the right of one, whose income is 13 Lakhs.
The one, whose income is 19 Lakhs, is sitting on the seat which is to the immediate right of Pari’s seat.
Inference:
Here, there are three possible scenarios in which we can use the above hints in case 1.
Case 1-A:
Case 1-B:
Case 1-C:
Case 2:
Reference:
Only one seat was vacant.
None of the seats adjacent to Anuj is vacant.
Inference:
At this point our Case 1-A, Case 1-B and Case 2 are contradicting with the above hints. So we can say that Case 1-A, Case 1-B and Case 2 are invalid cases.
Reference:
The income of the persons sitting adjacent to vacant seat is the just next number of income the persons sitting adjacent to them.
The number of persons between Reet and Dev are same as the number of persons between Anuj and Mala.
Inference:
After using the above hints, we have:
Case 1-C:
4
What is the position of Vacant seat with respect to Dev?
» Explain it
A
Following the final solution we can say that the vacant seat third to the right of Dev.
Hence, the correct answer is option A.
Final Solution:
Common Explanation:
Reference:
Some persons are sitting around a circular table facing towards the centre and at an equal distance from each other. Each of these persons has a different income greater than 11 Lakhs but less than 20 Lakhs.
Inference:
Here, we can say that there were at most 8 seats in the arrangement and we keep the rest of the information in mind while solving the puzzle.
Reference:
Anuj sits opposite to Faiz but neither of them are sitting adjacent to Jaya.
Inference:
Using the above information we can say that there were even numbers of seats around the circular table as it is given that Anuj is sitting opposite to Faiz and we know that all these persons are equidistant from each other which cannot be possible if the number of seats is odd.
Here, we will make a mental note that neither Jaya nor Reet are sitting adjacent to Jaya.
Reference:
There are two seats between Jaya and Reet.
Inference:
At this point we can say that were exactly 8 seats in the arrangement because of the number of seats is less than 8 the above hints will not follow.
At this point we can fix the position of Jaya, Reet, Anuj and Faiz as:
Case 1:
Case 2:
Reference:
Anuj’s income is in multiple of 5 but it is not the highest.
Reet’s income is in perfect square number.
Jaya’s income is in even number.
Inference:
After using the above hints, we have:
Anuj’ income = 15 Lakhs
Reet’s income = 16 Lakhs
Here, we will make mental note that income of Jaya was an even number.
Reference:
Pari sits on the seat which is second to the right of one, whose income is 13 Lakhs.
The one, whose income is 19 Lakhs, is sitting on the seat which is to the immediate right of Pari’s seat.
Inference:
Here, there are three possible scenarios in which we can use the above hints in case 1.
Case 1-A:
Case 1-B:
Case 1-C:
Case 2:
Reference:
Only one seat was vacant.
None of the seats adjacent to Anuj is vacant.
Inference:
At this point our Case 1-A, Case 1-B and Case 2 are contradicting with the above hints. So we can say that Case 1-A, Case 1-B and Case 2 are invalid cases.
Reference:
The income of the persons sitting adjacent to vacant seat is the just next number of income the persons sitting adjacent to them.
The number of persons between Reet and Dev are same as the number of persons between Anuj and Mala.
Inference:
After using the above hints, we have:
Case 1-C:
5
Four of the following five are alike in some way and hence form a group. Which of the following is the one that does not belong to the group?
» Explain it
C
Following the final solution we can say that Jaya is the one that does not belong to the group because of all the persons given in the option Jaya is the only persons whose income is in even number.
Hence, the correct answer is option C.
Final Solution:
Common Explanation:
Reference:
Some persons are sitting around a circular table facing towards the centre and at an equal distance from each other. Each of these persons has a different income greater than 11 Lakhs but less than 20 Lakhs.
Inference:
Here, we can say that there were at most 8 seats in the arrangement and we keep the rest of the information in mind while solving the puzzle.
Reference:
Anuj sits opposite to Faiz but neither of them are sitting adjacent to Jaya.
Inference:
Using the above information we can say that there were even numbers of seats around the circular table as it is given that Anuj is sitting opposite to Faiz and we know that all these persons are equidistant from each other which cannot be possible if the number of seats is odd.
Here, we will make a mental note that neither Jaya nor Reet are sitting adjacent to Jaya.
Reference:
There are two seats between Jaya and Reet.
Inference:
At this point we can say that were exactly 8 seats in the arrangement because of the number of seats is less than 8 the above hints will not follow.
At this point we can fix the position of Jaya, Reet, Anuj and Faiz as:
Case 1:
Case 2:
Reference:
Anuj’s income is in multiple of 5 but it is not the highest.
Reet’s income is in perfect square number.
Jaya’s income is in even number.
Inference:
After using the above hints, we have:
Anuj’ income = 15 Lakhs
Reet’s income = 16 Lakhs
Here, we will make mental note that income of Jaya was an even number.
Reference:
Pari sits on the seat which is second to the right of one, whose income is 13 Lakhs.
The one, whose income is 19 Lakhs, is sitting on the seat which is to the immediate right of Pari’s seat.
Inference:
Here, there are three possible scenarios in which we can use the above hints in case 1.
Case 1-A:
Case 1-B:
Case 1-C:
Case 2:
Reference:
Only one seat was vacant.
None of the seats adjacent to Anuj is vacant.
Inference:
At this point our Case 1-A, Case 1-B and Case 2 are contradicting with the above hints. So we can say that Case 1-A, Case 1-B and Case 2 are invalid cases.
Reference:
The income of the persons sitting adjacent to vacant seat is the just next number of income the persons sitting adjacent to them.
The number of persons between Reet and Dev are same as the number of persons between Anuj and Mala.
Inference:
After using the above hints, we have:
Case 1-C:
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# How do you write the vertex form equation of each parabola given Vertex (8, -1), y-intercept: -17?
Mar 23, 2018
Equation of parabola is y = -1/4(x-8)^2-1 ;
#### Explanation:
Vertex form of equation of parabola is y = a(x-h)^2+k ; (h,k)
being vertex. Here $h = 8 , k = - 1$ Hence equation of parabola is
y = a(x-8)^2-1 ; , y intercept is $- 17 \therefore \left(0 , - 17\right)$ is a point
through which parabola passes , so the point will satisfy the
equation of parabola $- 17 = a {\left(0 - 8\right)}^{2} - 1 \mathmr{and} - 17 = 64 a - 1$
or $64 a = - 16 \mathmr{and} a = - \frac{16}{64} = - \frac{1}{4}$. So equation of parabola
is y = -1/4(x-8)^2-1 ;
graph{-1/4(x-8)^2-1 [-40, 40, -20, 20]} [Ans]
Mar 24, 2018
$y = - \frac{1}{4} {\left(x - 8\right)}^{2} - 1$
#### Explanation:
$\text{the equation of a parabola in "color(blue)"vertex form}$
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$
$\text{here } \left(h , k\right) = \left(8 , - 1\right)$
$\Rightarrow y = a {\left(x - 8\right)}^{2} - 1$
$\text{to find a substitute "(0,-17)" into the equation}$
$- 17 = 64 a - 1$
$\Rightarrow 64 a = - 16 \Rightarrow a = - \frac{1}{4}$
$\Rightarrow y = - \frac{1}{4} {\left(x - 8\right)}^{2} - 1 \leftarrow \textcolor{red}{\text{in vertex form}}$ |
# In a GP, the ratio of the sum of the first three terms is to first six terms is 125
Question:
In a GP, the ratio of the sum of the first three terms is to first six terms is 125: 152. Find the common ratio.
Solution:
Sum of a G.P. series is represented by the formula, $\mathrm{S}_{\mathrm{n}}=\mathrm{a} \frac{\mathrm{r}^{\mathrm{n}}-1}{\mathrm{r}-1}$
hen r≠1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.
Sum of first 3 terms $={ }^{\mathrm{a}} \times \frac{\mathrm{r}^{3}-1}{\mathrm{r}-1}$
Sum of first 6 terms $=a \times \frac{r^{6}-1}{r-1}$
$\therefore \frac{a \times \frac{r^{3}-1}{r-1}}{a \times \frac{r^{6}-1}{r-1}}=\frac{125}{152}$
$\Rightarrow \frac{\left(r^{3}-1\right)}{\left(r^{6}-1\right)}=\frac{125}{152}$
$\Rightarrow 152 r^{3}-152=125 r^{6}-125$
$\Rightarrow 125 r^{6}-152 r^{3}-125+152=0$
$\Rightarrow 125 r^{6}-152 r^{3}+27=0$
$\Rightarrow 125 r^{6}-125 r^{3}-27 r^{3}+27=0$
$\Rightarrow\left(125 r^{3}-27\right)\left(r^{3}-1\right)=0$
Either $125 r^{3}-27=0$ or $r^{3}-1=0$
Either $125 r^{3}=27$ or $r^{3}=1$
Either $r^{3}=\frac{27}{125}$ or $r=1$
Either $r=\frac{3}{5}$ or $r=1$
Since r ≠ 1 [ if r is 1, all the terms will be equal which destroys the purpose ]
$\therefore r=\frac{3}{5}$ |
## 4. Median of Two Sorted Arrays
There are two sorted arrays nums1 and nums2 of size m and n respectively.
Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
Example 1:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
Example 2:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
b'
\n
\n\n
\n
## Solution
\n
\n
#### Approach #1 Recursive Approach [Accepted]
\n
To solve this problem, we need to understand "What is the use of median". In statistics, the median is used for:
\n
\n
Dividing a set into two equal length subsets, that one subset is always greater than the other.
\n
\n
If we understand the use of median for dividing, we are very close to the answer.
\n
First let\'s cut into two parts at a random position :
\n
left_A | right_A\n A[0], A[1], ..., A[i-1] | A[i], A[i+1], ..., A[m-1]\n
\n
Since has elements, so there are kinds of cutting ().
\n
And we know:
\n
\n
\n.
\n
Note: when , is empty, and when , is empty.
\n
\n
With the same way, cut into two parts at a random position :
\n
left_B | right_B\n B[0], B[1], ..., B[j-1] | B[j], B[j+1], ..., B[n-1]\n
\n
Put and into one set, and put and into another set. Let\'s name them and :
\n
left_part | right_part\n A[0], A[1], ..., A[i-1] | A[i], A[i+1], ..., A[m-1]\n B[0], B[1], ..., B[j-1] | B[j], B[j+1], ..., B[n-1]\n
\n
If we can ensure:
\n
\n
\n
1. \n\n
2. \n
3. \n\n
4. \n
\n
\n
then we divide all elements in into two parts with equal length, and one part is always greater than the other. Then
\n
\n\n
\n
To ensure these two conditions, we just need to ensure:
\n
\n
\n
1. \n
\n (or: )
\n if , we just need to set: \n
\n
2. \n
3. \n
\n and \n
\n
4. \n
\n
\n
ps.1 For simplicity, I presume are always valid even if , , , or .\nI will talk about how to deal with these edge values at last.
\n
ps.2 Why ? Because I have to make sure is non-negative since and . If , then may be negative, that will lead to wrong result.
\n
So, all we need to do is:
\n
\n
Searching in , to find an object such that:
\n
\n and where \n
\n
\n
And we can do a binary search following steps described below:
\n
\n
1. Set , , then start searching in \n
2. \n
3. Set , \n
4. \n
5. \n
Now we have . And there are only 3 situations that we may encounter:
\n
\n
• \n
\n and \n
\n Means we have found the object , so stop searching.
\n
• \n
• \n
\n\n
\n Means is too small. We must adjust to get .
\n Can we increase ?
\n \xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0Yes. Because when is increased, will be decreased.
\n \xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0So is decreased and is increased, and may
\n \xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0be satisfied.
\n Can we decrease ?
\n \xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0No! Because when is decreased, will be increased.
\n \xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0So is increased and is decreased, and will
\n \xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0\xc2\xa0be never satisfied.
\n So we must increase . That is, we must adjust the searching range to .
\n So, set , and goto 2.
\n
• \n
• \n
\n:
\n Means is too big. And we must decrease to get .
\n That is, we must adjust the searching range to .
\n So, set , and goto 2.
\n
• \n
\n
6. \n
\n
When the object is found, the median is:
\n
\n
\n when is odd
\n
\n when is even
\n
\n
Now let\'s consider the edges values where may not exist.\nActually this situation is easier than you think.
\n
What we need to do is ensuring that . So, if and are not edges values (means all exist), then we must check both and .\nBut if some of don\'t exist, then we don\'t need to check one (or both) of these two conditions.\nFor example, if , then doesn\'t exist, then we don\'t need to check .\nSo, what we need to do is:
\n
\n
Searching in , to find an object such that:
\n
\n or or and
\n or or where \n
\n
\n
And in a searching loop, we will encounter only three situations:
\n
\n
\n
1. \n or or and
\n or or \n
\n Means is perfect, we can stop searching.
2. \n
3. \n and and \n
\n Means is too small, we must increase it.
4. \n
5. \n and and \n
\n Means is too big, we must decrease it.
6. \n
\n
\n
Thanks to @Quentin.chen for pointing out that: and . Because:
\n
\n
\n\n
\n
\n\n
\n
\n
So in situation 2. and 3. , we don\'t need to check whether and whether .
\n\n\n
Complexity Analysis
\n
\n
• \n
Time complexity: .
\nAt first, the searching range is .\nAnd the length of this searching range will be reduced by half after each loop.\nSo, we only need loops. Since we do constant operations in each loop, so the time complexity is .\nSince , so the time complexity is .
\n
• \n
• \n
Space complexity: .
\nWe only need constant memory to store local variables, so the space complexity is .
\n
• \n
\n
\n
Analysis written by: @MissMary
\n
' |
# Help with Math WordProblem
• Feb 10th 2011, 11:51 AM
thelensboss
Help with Math WordProblem
A 1000L Tank contains 50L of a 25% brine solution. You add x liters of a 75% brine solution to the ank.
a)Show that the concentration C, the proportion of brine to total solution, in the final mixture is:
C= (3x+50)/(4(x+50))
b)Determine the domain of the function based on the physical constraints of the problem. Explain your answer.
c) Graph the concentration function. As the tank is filled, what happens to the rate at which the concentration of brine is increasing? What percent does the concentration of brine appear to approach?
• Feb 10th 2011, 12:02 PM
Quote:
Originally Posted by thelensboss
A 1000L Tank contains 50L of a 25% brine solution. You add x liters of a 75% brine solution to the ank.
a)Show that the concentration C, the proportion of brine to total solution, in the final mixture is:
C= (3x+50)/(4(x+50))
b)Determine the domain of the function based on the physical constraints of the problem. Explain your answer.
c) Graph the concentration function. As the tank is filled, what happens to the rate at which the concentration of brine is increasing? What percent does the concentration of brine appear to approach?
1000 litres is simply the capacity of the tank.
The total solution is 50 litres plus x litres
$\displaystyle 25\%=\frac{25}{100}=\frac{1}{4}$
$\displaystyle 75\%=\frac{3}{4}$
Therefore, the amount of brine in the solution, in litres, is
$\displaystyle \frac{1}{4}(50)+\frac{3}{4}(x)$
Try writing the fraction of brine divided by total amount of solution.
• Feb 10th 2011, 12:34 PM
thelensboss
alright how about the domain based on physical constraints of the problem
• Feb 10th 2011, 12:37 PM
The physical constraint relates on the fact that the solution cannot exceed 1000 litres.
That creates an upper bound for x.
The maximum value of the solution reveals the domain.
• Feb 10th 2011, 12:49 PM
thelensboss
So is the domain X cant equal -50 cuz that doesnt make sense
• Feb 10th 2011, 12:52 PM
No, x can't be negative at all, since a measurement in litres cannot drop below zero.
The solution of 50+x litres cannot exceed 1000 litres.
• Feb 10th 2011, 12:58 PM
thelensboss
Yea so it is basically x must equal 950 or less??
And also can you answer the last question part c because when I graphed it on my calc it did not show
• Feb 10th 2011, 01:13 PM
As x increases towards 950 (and beyond if it was physically possible), the graph approaches $\displaystyle \frac{3x}{4x}=\frac{3}{4}$ |
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# If in a triangle ${\rm{ABC}}$ side $a = (\sqrt 3 + 1){\rm{cm}}$ and $\angle B = {30^\circ }$,$\angle C = {45^\circ }$ then the area of the triangle isA. $\frac{{\sqrt 3 + 1}}{3}\;{\rm{c}}{{\rm{m}}^2}$B. $\frac{{\sqrt 3 + 1}}{2}\;{\rm{c}}{{\rm{m}}^2}$C. $\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}\;{\rm{c}}{{\rm{m}}^2}$D. $\frac{{\sqrt 3 + 1}}{{3\sqrt 2 }}\;{\rm{c}}{{\rm{m}}^2}$
Verified
89.7k+ views
Hint:
In this case, we have been given that in a triangle ABC, the side is $a = (\sqrt 3 + 1){\rm{cm}}$ and two angles $\angle B = {30^\circ }$,$\angle C = {45^\circ }$and here we see that one of the angles is $\angle C = {45^\circ }$and the other is $\angle B = {30^\circ }$then it may be isosceles triangle. We know that all angle’s sum is ${180^\circ }$ in a triangle. So that we have to apply sine rule and also to use area of triangle formula in order to get the required result.
Formula used:
Sine rule:
$\frac{{\rm{a}}}{{\sin {\rm{A}}}} = \frac{{\rm{b}}}{{\sin {\rm{B}}}} = \frac{{\rm{c}}}{{\sin {\rm{C}}}}$
Triangle’s area:
$\frac{1}{2}{\rm{ab}}\sin {\rm{C}}$
Complete step-by-step solution:
We have been provided in the question that in a triangle $ABC$ we have
$a = (\sqrt 3 + 1);\angle B = {30^\circ },\angle C = {45^\circ }$
As we know by sum of angle property that triangle’s sum of angle is ${180^0}$ we can write as
$\Rightarrow \angle {\rm{A}} = {180^\circ } - {75^\circ } = {105^\circ }$
Now, we have to use sine rule
$\frac{{\rm{a}}}{{\sin {\rm{A}}}} = \frac{{\rm{b}}}{{\sin {\rm{B}}}} = \frac{{\rm{c}}}{{\sin {\rm{C}}}}$
Now, we have to substitute the values provided in the data to the above equation, we get
$\Rightarrow \frac{{\sqrt 3 + 1}}{{\sin {{105}^\circ }}} = \frac{{\rm{b}}}{{\sin {{30}^\circ }}} = \frac{{\rm{c}}}{{\sin {{45}^\circ }}}$
From the above expression, we have to determine the value of ${\rm{b}}$ we obtain
$\Rightarrow {\rm{b}} = \frac{{(\sqrt 3 + 1)\sin {{30}^\circ }}}{{\sin {{105}^\circ }}}$
Now, let’s solve numerator and denominator by using trigonometry identity, we get
$\frac{{(\sqrt 3 + 1)}}{{2\sin \left( {{{45}^\circ } + {{60}^\circ }} \right)}}$
On further simplification using trigonometry identity, we obtain
$= \sqrt 2$
We have been already known that the area of triangle is $\frac{1}{2}{\rm{ab}}\sin {\rm{C}}$
By using the above formula we have ti substitute the values obtain previously we get
$= \frac{1}{2}(\sqrt 3 + 1)(\sqrt 2 )\frac{1}{{\sqrt 2 }}$
Now, we have to simplify further we obtain
$= \frac{{\sqrt 3 + 1}}{2}\;{\rm{c}}{{\rm{m}}^2}$
Therefore, if in a triangle ${\rm{ABC}}$ side $a = (\sqrt 3 + 1){\rm{cm}}$ and $\angle B = {30^\circ },\angle C = {45^\circ }$ then the area of the triangle is $\frac{{\sqrt 3 + 1}}{2}\;{\rm{c}}{{\rm{m}}^2}$
Hence, the option B is correct
Note:
It is important to remember that the sum of all the angles must equal 180. Also, keep in mind that the length of a triangle's sides is a positive number. To relate the length and angles of a triangle, the sine and cosine formulas must be used.
Last updated date: 21st Sep 2023
Total views: 89.7k
Views today: 1.89k |
Till now, we discussed about the one sample z test,t test and p test.
We are going to discuss about the two sample z test, p test, t test and paired t test.
Before talking on two sample z test, let us understand the difference between one sample z test and two sample z test.
One sample z test vs Two sample z test :
One sample z test- Let’s say there is a population p which has µ (mean) and standard deviation(σ). We select the sample (q) from p ,which has x(mean) and standard deviation(s). Using one sample z test ,we can conclude from sample whether population mean is changed or not.
Two sample z test- Lets say there are two population and find out whether two population has same mean or different using samples. Using two sample z test, we derive that there is a significant difference of mean for two population or not.
Example 1- There are two cities and if want to compare the mean salaries of two population in the city from the samples whether there is some significant difference between the population mean in the salary or not.
Example-2-Two machines producing some X product and want to compare whether both of the machine has same mean to produce the product or there is significant difference of means between the machines from the samples.
Two sample z test-
One may be confused , why two formula for two sample z test. It is even valid and good question.
Answer is that we can derive the first formula from second formula. Let’s see how
delta= µ1 – µ2
Here, delta is the difference of population means and it is 0 in case need to find whether there is difference of population means or not. So we put delta as 0, it forms the first equation.
Question- There are two machines A and B in the factory and 100 samples are drawn from each machine . Sample mean and standard deviation are 150 and 2 respectively for Machine A and sample mean and standard deviation are 152 and 2.5 respectively for Machine B. We need to find whether there is a significant difference between means of Machine A and B or not with 95% of confidence level ?
Alternate Hypothesis: µ1 ≠µ2
Here delta is 0 so we will use first formula of two sample z test. Numerator will be (152-150) and n1=100,n2=100,σ1=2 and σ2=2.5 , when we put the these values in the formula , denominator will be approx 0.3
z = (152-150)/0.3 = 6.6
z crtical value for 95% confidence level= 1.96
z > z critical so we can reject the null hypothesis ,means alternate hypothesis is correct so there is a change in the population means of two machines.
$${}$$ |
Rationalize the denominator and simplify:
Question:
Rationalize the denominator and simplify:
(i) $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
(ii) $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}$
(iii) $\frac{1+\sqrt{2}}{3-2 \sqrt{2}}$
(iv) $\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}$
(v) $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$
(vi) $\frac{2 \sqrt{3}-\sqrt{5}}{2 \sqrt{2}+3 \sqrt{3}}$
Solution:
(i) $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor
$\sqrt{3}-\sqrt{2}$
$=\frac{(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(\sqrt{3}-\sqrt{2})^{2}}{3-2}$
As we know, $(a-b)^{2}=\left(a^{2}-2 \times a \times b+b^{2}\right)$
$=\frac{3-2 \sqrt{3} \sqrt{2}+2}{1}=5-2 \sqrt{6}$
(ii) $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor
$7-4 \sqrt{3}$
$=\frac{(5+2 \sqrt{3})(7-4 \sqrt{3})}{(7+4 \sqrt{3})(7-4 \sqrt{3})}$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(5+2 \sqrt{3})(7-4 \sqrt{3})}{49-48}$
$=35-20 \sqrt{3}+14 \sqrt{3}-24=11-6 \sqrt{3}$
(iii) $\frac{1+\sqrt{2}}{3-2 \sqrt{2}}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor
$3+2 \sqrt{2}$
$=\frac{(1+\sqrt{2})(3+2 \sqrt{2})}{(3-2 \sqrt{2})(3+2 \sqrt{2})}$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(1+\sqrt{2})(3+2 \sqrt{2})}{9-8}$
$=3+2 \sqrt{2}+3 \sqrt{2}+4=7+5 \sqrt{2}$
(iv) $\frac{2 \sqrt{6}-\sqrt{5}}{3 \sqrt{5}-2 \sqrt{6}}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor
$3 \sqrt{5}+2 \sqrt{6}$
$=\frac{(2 \sqrt{6}-\sqrt{5})(3 \sqrt{5}+2 \sqrt{6})}{(3 \sqrt{5}-2 \sqrt{6})(3 \sqrt{5}+2 \sqrt{6})}$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(2 \sqrt{6}-\sqrt{5})(3 \sqrt{5}+2 \sqrt{6})}{45-24}$
$=\frac{(2 \sqrt{6}-\sqrt{5})(3 \sqrt{5}+2 \sqrt{6})}{21}$
$=\frac{(2 \sqrt{6}-\sqrt{5})(3 \sqrt{5}+2 \sqrt{6})}{21}$
$=\frac{6 \sqrt{30}+24-15-2 \sqrt{30}}{21}$
$=\frac{4 \sqrt{30}+9}{21}$
(v) $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor
$\sqrt{48}-\sqrt{18}$
$=\frac{(4 \sqrt{3}+5 \sqrt{2})(\sqrt{48}-\sqrt{18})}{(\sqrt{48}+\sqrt{18})(\sqrt{48}-\sqrt{18})}$
As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$=\frac{(4 \sqrt{3}+5 \sqrt{2})(\sqrt{48}-\sqrt{18})}{48-18}$
$=\frac{48-12 \sqrt{6}+20 \sqrt{6}-30}{30}$
$=\frac{18+8 \sqrt{6}}{30}$
$=\frac{9+4 \sqrt{6}}{15}$
(vi) $\frac{2 \sqrt{3}-\sqrt{5}}{2 \sqrt{2}+3 \sqrt{3}}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor
$2 \sqrt{2}-3 \sqrt{3}$
$=\frac{(2 \sqrt{3}-\sqrt{5})(2 \sqrt{2}-3 \sqrt{3})}{(2 \sqrt{2}+3 \sqrt{3})(2 \sqrt{2}-3 \sqrt{3})}$
$=\frac{(2 \sqrt{3}-\sqrt{5})(2 \sqrt{2}-3 \sqrt{3})}{8-27}$
$=\frac{(4 \sqrt{6}-2 \sqrt{10})-18+3 \sqrt{15})}{-19}$
$=\frac{(18-4 \sqrt{6}+2 \sqrt{10}-3 \sqrt{15})}{19}$ |
# Vedic Maths Tutorial (interactive
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Vedic Maths Tutorial
Vedic Maths is based on sixteen sutras or principles. These principles are general in nature and can be applied in many ways. In practice many applications of the sutras may be learned and combined to solve actual problems. These tutorials will give examples of simple applications of the sutras, to give a feel for how the Vedic Maths system works. These tutorials do not attempt to teach the systematic use of the sutras. For more advanced applications and a more complete coverage of the basic uses of the sutras, we recommend you study one of the texts available. N.B. The following tutorials are based on examples and exercises given in the book 'Fun with figures' by Kenneth Williams, which is a fun introduction some of the applications of the sutras for children. If you are having problems using the tutorials then you could always read the instructions. Tutorial 1 Tutorial 2 Tutorial 3 Tutorial 4 Tutorial 5 Tutorial 6 Tutorial 7
Tutorial 1
Use the formula ALL FROM 9 AND THE LAST FROM 10 to perform instant subtractions.
q
For example 1000 - 357 = 643 We simply take each figure in 357 from 9 and the last figure from 10.
So the answer is 1000 - 357 = 643 And thats all there is to it! This always works for subtractions from numbers consisting of a 1 followed by noughts: 100; 1000; 10,000 etc.
q
Similarly 10,000 - 1049 = 8951
q
For 1000 - 83, in which we have more zeros than figures in the numbers being subtracted, we simply suppose 83 is 083. So 1000 - 83 becomes 1000 - 083 = 917
http://www.vedicmaths.org/Group%20Files/tutorial/tutorial.asp (1 of 12)2/10/2004 8:36:53 PM
000 .org/Group%20Files/tutorial/tutorial. And you multiply vertically: 2 x 3 to get 6.777 2) 1000 . Think of it like this: The answer is 56.asp (2 of 12)2/10/2004 8:36:53 PM . the first figure of the answer. q Suppose you need 8 x 7 8 is 2 below 10 and 7 is 3 below 10.1101 7) 100 .9876 6) 10.57 8) 1000 .38 Total Correct = = = = = = = = = = = Reset Test Return to Index Tutorial 2 Using VERTICALLY AND CROSSWISE you do not need to the multiplication tables beyond 5 X 5.283 3) 1000 .2345 5) 10000 .321 10) 10.57 9) 10. the last figure of the answer.000 . The diagram below shows how you get it. That's all you do: http://www.505 4) 10.000 .vedicmaths.000 .2 to get 5. You subtract crosswise 8-3 or 7 .Vedic Maths Tutorial (interactive) Try some yourself: 1) 1000 .
using the same method as above.org/Group%20Files/tutorial/tutorial. and multiply the deficiencies together.12 = 86: you can subtract either way. Multply These: 1) 8 8x 2) 9 7x 3) 8 9x 4) 7 7x 5) 9 9x 6) 6 6x Total Correct = Reset Test Here's how to use VERTICALLY AND CROSSWISE for multiplying numbers close to 100. But with VERTICALLY AND CROSSWISE you can give the answer immediately. Not easy. q Suppose you want to multiply 88 by 98. Both 88 and 98 are close to 100. q 7 x 6 = 42 Here there is a carry: the 1 in the 12 goes over to make 3 into 4. you will always get the same answer). 88 is 12 below 100 and 98 is 2 below 100. Try some: http://www.you might think. So 88 x 98 = 8624 This is so easy it is just mental arithmetic.Vedic Maths Tutorial (interactive) See how far the numbers are below 10. subtract one number's deficiency from the other number.asp (3 of 12)2/10/2004 8:36:53 PM . You can imagine the sum set out like this: As before the 86 comes from subtracting crosswise: 88 .2 = 86 (or 98 . And the 24 in the answer is just 12 x 2: you multiply vertically.vedicmaths.
Use VERTICALLY AND CROSSWISE to write the answer straight down! q Multiply crosswise and add to get the top of the answer: 2 x 5 = 10 and 1 x 3 = 3. Then 10 + 3 = 13. The bottom of the fraction is just 3 x 5 = 15.org/Group%20Files/tutorial/tutorial.asp (4 of 12)2/10/2004 8:36:53 PM .vedicmaths. 107 is just 103 + 4 (or 104 + 3). and 12 is just 3 x 4. http://www. just for mental arithmetic Try a few: 1) 102 x 107 = 1) 106 x 103 = 1) 104 x 104 = 4) 109 x 108 = 5) 101 x123 = 6) 103 x102 = Total Correct = Return to Index Reset Test Tutorial 3 The easy way to add and subtract fractions. q Similarly 107 x 106 = 11342 107 + 6 = 113 and 7 x 6 = 42 Again.Vedic Maths Tutorial (interactive) 1) 87 98 x 2) 88 97 x 3) 77 98 x 4) 93 96 x 5) 94 92 x 6) 64 99 7) 98 97 x Total Correct = Reset Test Multiplying numbers just over 100. q 103 x 104 = 10712 The answer is in two parts: 107 and 12.
vedicmaths. The answer is in two parts: 56 and 25.asp (5 of 12)2/10/2004 8:36:53 PM . The last part is always 25. which is 8: so 7 x 8 = 56 http://www. multiplied by the number "one more". The first part is the first number. 7.org/Group%20Files/tutorial/tutorial. q 752 = 5625 752 means 75 x 75.Vedic Maths Tutorial (interactive) You multiply the bottom number together. but the subtract: q Try a few: Total Correct = Return to Index Reset Test Tutorial 4 A quick way to square numbers that end in 5 using the formula BY ONE MORE THAN THE ONE BEFORE. So: q Subtracting is just as easy: multiply crosswise as before.
So we just multiply 3 by 4 (the next number up) to get 12 for the first part of the answer.Vedic Maths Tutorial (interactive) q Similarly 852 = 7225 because 8 x 9 = 72. Diagrammatically: q And 81 x 89 = 7209 We put 09 since we need two figures as in all the other examples. q 32 x 38 = 1216 Both numbers here start with 3 and the last figures (2 and 8) add up to 10.org/Group%20Files/tutorial/tutorial.asp (6 of 12)2/10/2004 8:36:54 PM . Practise some: 1) 43 x 47 = 2) 24 x 26 = 3) 62 x 68 = http://www. Try these: 1) 452 = 2) 652 = 3) 952 = 4) 352 = 5) 152 = Total Correct = Reset Test Method for multiplying numbers where the first figures are the same and the last figures add up to 10.vedicmaths. And we multiply the last figures: 2 x 8 = 16 to get the last part of the answer.
vedicmaths.Vedic Maths Tutorial (interactive) 4) 17 x 13 = 5) 59 x 51 = 6) 77 x 73 = Total Correct = Return to Index Reset Test Tutorial 5 An elegant way of multiplying numbers using a simple pattern. just write down the answer: http://www. 23 below 21: There are 3 steps: a) Multiply vertically on the left: 2 x 2 = 4. This gives the first figure of the answer. We first put. q 21 x 23 = 483 This is normally called long multiplication but actually the answer can be written straight down using the VERTICALLY AND CROSSWISE formula. 6 x 1 + 1 x 3 = 9. q Similarly 61 x 31 = 1891 q 6 x 3 = 18. And thats all there is to it.asp (7 of 12)2/10/2004 8:36:54 PM .org/Group%20Files/tutorial/tutorial. c) Multiply vertically on the right: 1 x 3 = 3 This gives the last figure of the answer. or imagine. b) Multiply crosswise and add: 2 x 3 + 1 x 2 = 8 This gives the middle figure. 1 x 1 = 1 Try these.
However. 14. q 21 x 26 = 546 The method is the same as above except that we get a 2-figure number.Vedic Maths Tutorial (interactive) 1) 14 21 x 2) 22 31 x 3) 21 31 x 4) 21 22 x 5) 32 21 x Total Correct = Reset Test Multiply any 2-figure numbers together by mere mental arithmetic! If you want 21 stamps at 26 pence each you can easily find the total price in your head.46. There were no carries in the method given above. there only involve one small extra step. Practise a few: 1) 21 47 x 2) 23 43 x 3) 32 53 x 4) 42 32 x 5) 71 72 x Total Correct = q Reset Test 33 x 44 = 1452 There may be more than one carry in a sum: Vertically on the left we get 12. so we carry 2 to the left and mentally get 144. so the 1 is carried over to the left (4 becomes 5). Crosswise gives us 24.asp (8 of 12)2/10/2004 8:36:54 PM . in the middle step. http://www. So 21 stamps cost £5.vedicmaths. Then vertically on the right we get 12 and the 1 here is carried over to the 144 to make 1452.org/Group%20Files/tutorial/tutorial.
Return to Index Tutorial 6 Multiplying a number by 11. Multiply by 11: 1) 88 = http://www. no matter how big. To multiply any 2-figure number by 11 we just put the total of the two figures between the 2 figures. q So 72 x 11 = 792 Multiply by 11: 1) 43 = 2) 81 = 3) 15 = 4) 44 = 5) 11 = Total Correct = q Reset Test 77 x 11 = 847 This involves a carry figure because 7 + 7 = 14 we get 77 x 11 = 7147 = 847.Vedic Maths Tutorial (interactive) 6) 32 56 x 7) 32 54 x 8) 31 72 x 9) 44 53 x 10) 54 64 x Total Correct = Reset Test Any two numbers. q 26 x 11 = 286 Notice that the outer figures in 286 are the 26 being multiplied. And the middle figure is just 2 and 6 added up.vedicmaths.org/Group%20Files/tutorial/tutorial.asp (9 of 12)2/10/2004 8:36:54 PM . can be multiplied in one line by this method.
could it be easier? Divide by 9: 1) 61 = remainder http://www. The remainder is just 2 and 3 added up! q 43 / 9 = 4 remainder 7 The first figure 4 is the answer and 4 + 3 = 7 is the remainder . and we add the last pair: 3 + 4 = 7.org/Group%20Files/tutorial/tutorial.Vedic Maths Tutorial (interactive) 2) 84 = 3) 48 = 4) 73 = 5) 56 = Total Correct = q Reset Test 234 x 11 = 2574 We put the 2 and the 4 at the ends.vedicmaths. Multiply by 11: 1) 151 = 2) 527 = 3) 333 = 4) 714 = 5) 909 = Total Correct = Return to Index Reset Test Tutorial 7 Method for diving by 9. and this is the answer. q 23 / 9 = 2 remainder 5 The first figure of 23 is 2.asp (10 of 12)2/10/2004 8:36:54 PM . We add the first pair 2 + 3 = 5.
Vedic Maths Tutorial (interactive) 2) 33 = 3) 44 = 4) 53 = 5) 80 = Total Correct = q remainder remainder remainder remainder Reset Test 134 / 9 = 14 remainder 8 The answer consists of 1.vedicmaths. 14 has one more 9 with 5 left over the final answer will be 93 remainder 5 Divide these by 9: 1) 771 = 2) 942 = 3) 565 = 4) 555 = 5) 777 = remainder remainder remainder remainder remainder http://www.4 and 8. Divide by 9: 6) 232 = 7) 151 = 8) 303 = 9) 212 = 10) 2121 = Total Correct = q remainder remainder remainder remainder remainder Reset Test 842 / 9 = 812 remainder 14 = 92 remainder 14 Actually a remainder of 9 or more is not usually permitted because we are trying to find how many 9's there are in 842. 4 is the total of the first two figures 1+ 3 = 4. and 8 is the total of all three figures 1+ 3 + 4 = 8. 1 is just the first figure of 134. Since the remainder.asp (11 of 12)2/10/2004 8:36:54 PM .org/Group%20Files/tutorial/tutorial.
asp (12 of 12)2/10/2004 8:36:54 PM . N. the following will be displayed depending on how you answered the question :Correct Wrong Answer has more than one part (such as fractions and those answers with remainders). will cause the answer you entered to be checked. If you cannot get this to work then try the text/picture based version of this tutorial. Next to each question is a box (field) into which you can enter the answer to the question. will determine whether you answered the question correctly or not. Moving to the next question. press the 'TAB' key.vedicmaths. The Reset Test button will clear all answers from the test and set the count of correct answers back to zero. JavaScript is used to obtain the interactive nature of these tutorials. others do not. Return to Index http://www. Enter the answer for the question using the numeric keys on the keyboard. Answering remaining parts of the question. Some browsers will update the answer on 'RETURN' being pressed.org/Group%20Files/tutorial/tutorial. Select the first question in each test with the mouse to start a test. To move to the answer field of the next question in the test. Any problems stick to the 'TAB' key. Pressing 'SHIFT TAB' will move the cursor back to the answer field for the previous question.B.Vedic Maths Tutorial (interactive) 6) 2382 = 7) 7070 = Total Correct = Return to Index remainder remainder Reset Test Instructions for using the tutorials Each tutorial has test sections comprising of several questions each.
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# Difference between revisions of "1951 AHSME Problems/Problem 49"
## Problem
The medians of a right triangle which are drawn from the vertices of the acute angles are $5$ and $\sqrt{40}$. The value of the hypotenuse is:
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 2\sqrt{40}\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 2\sqrt{13}\qquad\textbf{(E)}\ \text{none of these}$
## Solution
We will proceed by coordinate bashing.
Call the first leg $2a$, and the second leg $2b$ (We are using the double of a variable to avoid any fractions)
Notice that we want to find $\sqrt{(2a)^2+(2b)^2}$
Two equations can be written for the two medians: $a^2 + 4b^2 = 40$ and $4a^2+b^2= 25$.
Add them together and we get $5a^2+5b^2=65$,
Dividing by 5 gives $x^2+y^2=13$
Multiplying them by 4 gives $4x^2+4y^2=52\implies (2x)^2+(2y)^2=52$, just what we need to find the hypotenuse. Recall that he hypotenuse is $\sqrt{(2a)^2+(2b)^2}$. The value inside the radical is equal to $52$, so the hypotenuse is equal to $\sqrt{52}=\boxed{\textbf{(D)}\ 2\sqrt{13}}$ |
# 4.7.2: Write an Equation Given the Slope and a Point
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## Write an Equation Given the Slope and a Point
Suppose that you sent out a text message to all of your friends, asking them what information was needed to write the equation of a line. One of your friends responded that all you need is the slope of the line and a point on the line. Do you think that your friend was correct? If so, does it matter what point you have, and how could you use this information to come up with the equation?
### Equations Given the Slope and a Point
#### Writing an Equation Given the y−Intercept
Previously, you learned how to graph solutions to two-variable equations in slope-intercept form. This Concept focuses on how to write an equation for a graphed line when given the slope and a point. There are two things you will need from the graph to write the equation in slope-intercept form:
1. The y−intercept of the graph
2. The slope of the line
Having these two pieces of information will allow you to make the appropriate substitutions in the slope-intercept formula. Recall the following:
Slope-intercept form: y=(slope)x+(y−intercept) or y=mx+b
#### Let's write an equation in slope-intercept form for each of the following situations:
1. A line with a slope of 4 and a y−intercept of (0, –3).
Slope-intercept form requires two things: the slope and y−intercept. To write the equation, you substitute the values into the formula.
y=(slope)x+(y−intercept)
y=4x+(−3)
y=4x−3
You can also use a graphed line to determine the slope and y−intercept.
1. The line shown in:
The y−intercept is (0, 2). Using the slope triangle, you can determine the slope is riserun=−3/−1=3/1. Substituting the value 2 for b and the value 3 for m, the equation for this line is y=3x+2.
#### Writing an Equation Given an Ordered Pair
You will not always be given the y−intercept, but sometimes you will be given any point on the line. When asked to write the equation given a graph, it may be difficult to determine the y−intercept. Perhaps the y−intercept is rational instead of an integer. Maybe all you have is the slope and an ordered pair. You can use this information to write the equation in slope-intercept form. To do so, you will need to follow several steps:
Step 1: Begin by writing the formula for slope-intercept form: y=mx+b.
Step 2: Substitute the given slope for m.
Step 3: Use the ordered pair you are given (x,y) and substitute these values for the variables x and y in the equation.
Step 4: Solve for b (the y−intercept of the graph).
Step 5: Rewrite the original equation in Step 1, substituting the slope for m and the y−intercept for b.
#### Let's write an equation in slope-intercept form for the following line:
A line with a slope of 4 that contains the ordered pair (–1, 5).
Step 1: Begin by writing the formula for slope-intercept form.
y=mx+b
Step 2: Substitute the given slope for m.
y=4x+b
Step 3: Use the ordered pair you are given, (–1, 5), and substitute these values for the variables x and y in the equation.
5=(4)(−1)+b
Step 4: Solve for b (the y−intercept of the graph).
5=−4+b
5+4=−4+4+b
9=b
Step 5: Rewrite y=mx+b, substituting the slope for m and the y−intercept for b.
y=4x+9
### Examples
Example 4.7.2.1
Earlier, you were asked if you only need the slope of a line and a point on the line to write the equation of a line.
Solution
You can write the equation of a line with just the slope and a point on the line. As shown in this concept, if you have the slope and the y−intercept, you can easily write the equation in slope-intercept form y=mx+b. If you do not have the y−intercept and instead just have a random point on the line, you can plug the ordered pair and the slope into the general equation and solve for b (the y−intercept of the graph).
Example 4.7.2.2
Write the equation for a line with a slope of –3 containing the point (3, –5).
Solution
Using the five-steps from above:
y=(slope)x+(y−intercept)
y=−3x+b
−5=−3(3)+b
−5=−9+b
4=b
y=−3x+4
### Review
1. What is the formula for slope-intercept form? What do the variables m and b represent?
2. What are the five steps needed to determine the equation of a line given the slope and a point on the graph (not the y−intercept)?
In 3–13, find the equation of the line in slope–intercept form.
1. The line has a slope of 7 and a y−intercept of –2.
2. The line has a slope of –5 and a y−intercept of 6.
3. The line has a slope of -2 and a y−intercept of 7.
4. The line has a slope of 2/3 and a y−intercept of 4/5.
5. The line has a slope of −1/4 and contains the point (4, –1).
6. The line has a slope of 2/3 and contains the point (12,1).
7. The line has a slope of –1 and contains the point (45,0).
8. The slope of the line is −2/3, and the line contains the point (2, –2).
9. The slope of the line is –3, and the line contains the point (3, –5).
### Vocabulary
Term Definition
undefined slope An undefined slope cannot be computed. Vertical lines have undefined slopes.
zero slope A line with zero slope is a line without any steepness, or a horizontal line.
Intercept The intercepts of a curve are the locations where the curve intersects the x and y axes. An x intercept is a point at which the curve intersects the x-axis. A y intercept is a point at which the curve intersects the y-axis.
Video: Writing Linear Equations in Point-Slope Form Given Information
Activity: Write an Equation Given the Slope and a Point Discussion Questions
Study Aid: Determining the Equation of a Line Study Guide
Practice: Write an Equation Given the Slope and a Point |
# 2002 AMC 10A Problems/Problem 25
## Problem
In trapezoid $ABCD$ with bases $AB$ and $CD$, we have $AB = 52$, $BC = 12$, $CD = 39$, and $DA = 5$. The area of $ABCD$ is
$[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("A",A,S); label("B",B,S); label("C",C,N); label("D",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W); [/asy]$
$\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$
## Solution
### Solution 1
It shouldn't be hard to use trigonometry to bash this and find the height, but there is a much easier way. Extend $\overline{AD}$ and $\overline{BC}$ to meet at point $E$:
$[asy] size(250); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), F=(100/13,240/13); draw(A--B--C--D--cycle); draw(D--F--C,dashed); label("$$A$$",A,S); label("$$B$$",B,S); label("$$C$$",C,NE); label("$$D$$",D,W); label("$$E$$",F,N); label("39",(C+D)/2,N); label("52",(A+B)/2,S); label("5",(A+D)/2,E); label("12",(B+C)/2,WSW); [/asy]$
Since $\overline{AB} || \overline{CD}$ we have $\triangle AEB \sim \triangle DEC$, with the ratio of proportionality being $\frac {39}{52} = \frac {3}{4}$. Thus \begin{align*} \frac {CE}{CE + 12} = \frac {3}{4} & \Longrightarrow CE = 36 \\ \frac {DE}{DE + 5} = \frac {3}{4} & \Longrightarrow DE = 15 \end{align*} So the sides of $\triangle CDE$ are $15,36,39$, which we recognize to be a $5 - 12 - 13$ right triangle. Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared), $$[ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{\mathrm{(C)}\ 210}$$
### Solution 2
Draw altitudes from points $C$ and $D$:
$[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label("$$A$$",A,SW); label("$$B$$",B,S); label("$$C$$",C,NE); label("$$D$$",D,N); label("$$D'$$",F,SSE); label("$$C'$$",E,S); label("39",(C+D)/2,N); label("52",(A+B)/2,S); label("5",(A+D)/2,W); label("12",(B+C)/2,ENE); [/asy]$
Translate the triangle $ADD'$ so that $DD'$ coincides with $CC'$. We get the following triangle:
$[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0); draw(A--B--C--cycle); draw(C--F,dashed); label("$$A'$$",A,SW); label("$$B$$",B,S); label("$$C$$",C,N); label("$$C'$$",F,SE); label("5",(A+C)/2,W); label("12",(B+C)/2,ENE); [/asy]$
The length of $A'B$ in this triangle is equal to the length of the original $AB$, minus the length of $CD$. Thus $A'B = 52 - 39 = 13$.
Therefore $A'BC$ is a well-known $(5,12,13)$ right triangle. Its area is $[A'BC]=\frac{A'C\cdot BC}2 = \frac{5\cdot 12}2 = 30$, and therefore its altitude $CC'$ is $\frac{[A'BC]}{A'B} = \frac{60}{13}$.
Now the area of the original trapezoid is $\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{\mathrm{(C)}\ 210}$
### Solution 3
Draw altitudes from points $C$ and $D$:
$[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label("$$A$$",A,SW); label("$$B$$",B,S); label("$$C$$",C,NE); label("$$D$$",D,N); label("$$D'$$",F,SSE); label("$$C'$$",E,S); label("39",(C+D)/2,N); label("52",(A+B)/2,S); label("5",(A+D)/2,W); label("12",(B+C)/2,ENE); [/asy]$
Call the length of $AD'$ to be $y$, the length of $BC'$ to be $z$, and the height of the trapezoid to be $x$. By the Pythagorean Theorem, we have: $$z^2 + x^2 = 144$$ $$y^2 + x^2 = 25$$
Subtracting these two equation yields: $$z^2-y^2=119 \implies (z+y)(z-y)=119$$
We also have: $z+y=52-39=13$.
We can substitute the value of $z+y$ into the equation we just obtained, so we now have:
$$(13) (z-y)=119 \implies z-y=\frac{119}{13}$$.
We can add the $z+y$ and the $z-y$ equation to find the value of $z$, which simplifies down to be $\frac{144}{13}$. Finally, we can plug in $z$ and use the Pythagorean theorem to find the height of the trapezoid.
$$\frac{12^4}{13^2} + x^2 = 12^2 \implies x^2 = \frac{(12^2)(13^2)}{13^2} -\frac{12^4}{13^2} \implies x^2 = \frac{(12 \cdot 13)^2 - (144)^2}{13^2} \implies x^2 = \frac{(156+144)(156-144)}{13^2} \implies x = \sqrt{\frac{3600}{169}} = \frac{60}{13}$$
Now that we have the height of the trapezoid, we can multiply this by the median to find our answer.
The median of the trapezoid is $\frac{39+52}{2} = \frac{91}{2}$, and multiplying this and the height of the trapezoid gets us:
$$\frac{60 \cdot 91}{13 \cdot 2} = \boxed{\mathrm{(C)}\ 210}$$
### Solution 4
We construct a line segment parallel to $\overline{AD}$ from point $C$ to line $\overline{AB},$ and label the intersection of this segment with line $\overline{AB}$ as point $E.$ Then quadrilateral $AECD$ is a parallelogram, so $CE=5, AE=39,$ and $EB=13.$ Triangle $EBC$ is therefore a right triangle, with area $\frac12 \cdot 5 \cdot 12 = 30.$
By continuing to split $\overline{AB}$ and $\overline{CD}$ into segments of length $13,$ we can connect these vertices in a "zig-zag," creating seven congruent right triangles, each with sides $5,12,$ and $13,$ and each with area $30.$ The total area is therefore $7 \cdot 30 = \boxed{\textbf{(C)} 210}.$
Alternative: Instead of creating seven congruent right triangles, we can find the height of parallelogram $AECD$ by drawing an altitude from $D$ to side $AE$, creating the new point $F$. By recognizing that triangle $DAF$ is similar to triangle $BEC$, we can use properties of similar triangles and find that $DE = 12 \cdot \frac{5}{13} = \frac{60}{13}$. Thus, the area of parallelogram $AECD$ is $\frac{60}{13} \cdot 39 = 180$. Finally, we add the areas of the parallelogram $AECD$ and the right triangle $BEC$ together and we get $180 + 30 = \boxed{\textbf{(C)} 210}$. ~scarletsyc
### Solution 2 but quicker
From Solution $2$ we know that the the altitude of the trapezoid is $\frac{60}{13}$ and the triangle's area is $30$. Note that once we remove the triangle we get a rectangle with length $39$ and height $\frac{60}{13}$. The numbers multiply nicely to get $180+30=\boxed{(C) 210}$ -harsha12345
### Quick Time Trouble Solution 5
First note how the answer choices are all integers. The area of the trapezoid is $\frac{39+52}{2} \cdot h = \frac{91}{2} h$. So h divides 2. Let $x$ be $2h$. The area is now $91x$. Trying $x=1$ and $x=2$ can easily be seen to not work. Those make the only integers possible so now you know x is a fraction. Since the area is an integer the denominator of x must divide either 13 or 7 since $91 = 13\cdot7$. Seeing how $39 = 3\cdot13$ and $52 = 4\cdot13$ assume that the denominator divides 13. Letting $y = \frac{x}{13}$ the area is now $7y$. Note that (A) and (C) are the only multiples of 7. We know that A doesn't work because that would mean $h$ is $4$ which we ruled out. So the answer is $\boxed{\textbf{(C)} 210}$. - megateleportingrobots
## See also
2002 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 24 Followed by-(Last question) 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. |
INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith
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• ## Related Books
1 answerLast reply by: Dr Carleen EatonSat Jan 16, 2016 10:41 PMPost by Jessie Carrillo on January 8, 2016On example 1V I also got a not true statement. But, I have a question about the step you took when subtracting 3x from both sides. Can you also add 8 to both sides in which it will create 3x >3x + 32. Afterwords, can you divided 3x in both sides in which you will have 1 > 32?
### Techniques for Multistep Inequalities
• To solve a multi-step inequality, use the same techniques that were discussed in the section on multi-step equations. For review, see the material given for that section.
• If the inequality contains grouping symbols, use the distributive property to remove these symbols and simplify the inequality.
• If the solution of an inequality leads to an inequality that is always true, the solution set of the original inequality is the set of all real numbers.
• If the solution leads to an inequality that is never true, the solution set is the empty set.
### Techniques for Multistep Inequalities
3x + 12 ≥ − 24
• 3x ≥ − 36
x ≥ − 12
5x − 20 < 115
• 5x < 135
x < 27
− 4y − 16 > 48
• − 4y > 64
y <− 16
− 2u + 14 ≤ 56
• − 2u ≤ 42
u ≥ − 21
6r − 3 > 21
• 6r > 24
r > 4
− 8p − 10 < 54
• − 8p < 64
p >− 8
− 3(c − 5) − 6 ≥ 2c + 8
• − 3c + 15 − 6 ≥ 2c + 8
• − 3c + 9 ≥ 2c + 8
• − 3c ≥ 2c − 1
• − 5c ≥ − 1
c ≥ [1/5]
− (4h + 2) − 6 < 8h + 10
• − 4h − 8 − 6 < 8h + 10
• − 4h − 14 < 8h + 10
• − 4h < 8h + 24
• − 12h < 24
h >− 2
2(3f − 5) + 4 ≥ 3(f + 4) − 10
• 6f − 10 + 4 ≥ 3f + 12 − 10
• 6f − 6 ≥ 3f + 2
• 6f ≥ 3f + 8
• 3f ≥ 8
• f[8/3]
f ≥ 3[2/3]
4(i − 4) + 1 < 6(i + 2) − 2i
• 4i − 16 + 1 < 6i + 12 − 2i
• 4i − 15 < 4i + 12
− 15 < 12
Not True
Empty Set
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
### Techniques for Multistep Inequalities
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Similarity to Multistep Equations 0:16
• Negative Numbers
• Example
• Inequalities Containing Grouping Symbols 1:24
• Example
• Special Cases 2:45
• Example: All Real Numbers
• Example: Empty Set
• Example 1: Solve the Inequality 6:05
• Example 2: Solve the Inequality 7:39
• Example 3: Solve the Inequality 9:57
• Example 4: Solve the Inequality 13:56 |
# What is the probability of getting 1 and 5 If a dice is thrown once?
Contents
## What is the probability of getting 5 when a dice is thrown?
Two (6-sided) dice roll probability table
Roll a… Probability
4 6/36 (16.667%)
5 10/36 (27.778%)
6 15/36 (41.667%)
7 21/36 (58.333%)
## What is the probability of getting 5 in a single throw of a die?
Hence, the probability of getting #5 in a single row of dice is 1/6.
## What is the probability of getting a number 5 in cards?
d) The probability of 5 non-ace cards is: (485)(525)=1,712,3042,598,960=0.6588, so the probability of getting 5 card at least one ace is: 1−0.6588=0.34. There are (525) equally likely ways to choose 5 cards. For solving all but the last problem, we count the number of “favourables” and divide by (525).
## When the dice is thrown what is the probability of getting 1 and 5?
So they are mutually exclusive events, therefore their probabilities add to 1. By symmetry we expect that each face is equally likely to appear and so each has probability = 1/6. The outcome of a 5 is one of those events and so has probability = 1/6 of appearing.
## What is the probability of not getting 3 or 5 in a single throw of a dice?
Answer: Under the standard assumptions (an unbiased 6 sided die with sides numbered 1-6), there are two qualifying numbers (3 and 5) out of 6 possibilities, so the probability is 2/6 or 1/3.
THIS IS IMPORTANT: What happens if you bet on a game and it gets Cancelled FanDuel?
## When a die is thrown probability of getting a number less than 5?
Hence, P(values less than 5) = 4/6 = 2/3. To find the complement of rolling a number less than 5, we use the formula P’ = 1 – P, where P’ is the complement of P. = 1/3. Hence, the probability of the complement of rolling a number less than 5 by using a six-sided die is 1/3.
## What is the probability of getting a number 8 in a single throw of a die?
We know that there are only six possible outcomes in a single throw of a die. These outcomes are 1, 2, 3, 4, 5 and 6. Since no face of the die is marked 8, so there is no outcome favourable to 8, i.e., the number of such outcomes is zero. In other words, getting 8 in a single throw of a die, is impossible. |
# How do you find 5% of a number?
5 percent is one half of 10 percent. To calculate 5 percent of a number, simply divide 10 percent of the number by 2. For example, 5 percent of 230 is 23 divided by 2, or 11.5.
## How do you work out 10% of a number?
As finding 10% of a number means to divide by 10, it is common to think that to find 20% of a number you should divide by 20 etc. Remember, to find 10% of a number means dividing by 10 because 10 goes into 100 ten times. Therefore, to find 20% of a number, divide by 5 because 20 goes into 100 five times.
## What percent of your day is 1 hour?
Therefore a day consists of 1440 minutes in total. Therefore, 1 hour and 45 minutes is 7.291% of the total number of hours in a day. Hence the correct option is C.
## What is 10% off?
Now that we know the formula, let’s practice using it to find a 10 percent discount when the original price for an item is \$14. So, a 10 percent discount off of \$14 gives you a discount of \$1.40.
## How much is 10 percent off?
Figuring the Discount While 10 percent of any amount is the amount multiplied by 0.1, an easier way to calculate 10 percent is to divide the amount by 10. So, 10 percent of \$18.40, divided by 10, equates to \$1.84.
## How do you do discounts?
Multiply the original price by the decimal Take the original price of the item and multiply it by the decimal determined in step one. Example: Winter boots originally sold for \$147. Multiply \$147 by 0.25 to find the amount of the discount. \$145 x 0.25 = \$36.75, so the boots are discounted by \$36.75.
## How do you calculate .75 of an hour?
1. Converting Seconds. Convert any seconds to a percentage of a minute by dividing by 60. For example, 14 minutes and 45 seconds equals 14.75 minutes, because 45/60 equals 0.75.
## What percent of 134 is one third of 1206?
1/3 * 1206 = 402. Now 300 percent of 134 is 402 . as this is becuase 402/132 = 3 and multiplying 3 with 100 we 300.
## How do I work out 5 VAT on a gross amount?
Divide gross sale price by 1 + VAT rate For example, if the applicable standard VAT rate is 20%, you’ll divide the gross sales price by 1.2. If the applicable VAT rate is 5%, you’ll divide the gross sales price by 1.05.
## How do I calculate VAT payable?
VAT Payable: VAT Payable = Output VAT – Input VAT = INR ( 25 – 12.50) = INR 12.50 VAT is therefore calculated by deducting tax credit from tax collected during the payment period.
## What is the sale price?
A sale price is the discounted price at which goods or services are being sold. This price is usually offered for a limited period of time, typically to spur sales during a slow period or to sell off excess inventory. Another interpretation of the term is that it is simply the price at which something sells.
## What is net price?
A net price is an estimate of the actual cost you and your family need to pay in a year to cover education expenses for you to attend a particular college or career school. It is the institution’s cost of attendance minus any grants and scholarships for which you may be eligible.
## What percent is 5 out of 22?
Now we can see that our fraction is 22.727272727273/100, which means that 5/22 as a percentage is 22.7273%.
## What is the percentage of 5 of 30?
Now we can see that our fraction is 16.666666666667/100, which means that 5/30 as a percentage is 16.6667%.
## What is 40 as a percentage of 8?
Now we can see that our fraction is 20/100, which means that 8/40 as a percentage is 20%. And there you have it! Two different ways to convert 8/40 to a percentage.
## Why are there 60 minutes in an hour instead of 100?
The Babylonians made astronomical calculations in the sexagesimal (base 60) system they inherited from the Sumerians, who developed it around 2000 B.C. Although it is unknown why 60 was chosen, it is notably convenient for expressing fractions, since 60 is the smallest number divisible by the first six counting numbers …
## How many fourths are there in 3?
In order to find out the total number of fourths which will be there in any number, we will divide this number with the one by fourth of that number. Therefore, there are a total 12 fourths present in 3. |
## NCERT Solutions for Class 6 Maths Chapter 2 Exercise 2.2
Here at MentorAtHome, we provide NCERT Solution for class 6 maths chapter 2 Whole Numbers we study numbers, number lines, addition, multiplication and subtraction using suitable properties like distributive, commutative etc. NCERT solution for class 6 maths chapter 2 help to make your calculation fast.
### NCERT solution for class 6 maths chapter 2 Ex 2.2 : whole numbers
In NCERT Solutions for Class 6 Maths Chapter 2 Exercise 2.2, we study about Properties of whole numbers are:- Addition and multiplication of any 2 whole numbers give a whole number and Subtraction and division of any 2 whole numbers may or may not give a whole number. NCERT Class 6 maths solutions ex 2.2 are designed in a simple way that’s students easily understand.
If you want to Check NCERT Solutions of class 6 for another chapter please select it here.
• Chapter 1 Knowing our Numbers
• Chapter 2 Whole Numbers
• Chapter 3 Playing With Numbers
• Chapter 4 Basic Geometrical Ideas
• Chapter 5 Understanding Elementary Shapes
• Chapter 6 Integers
• Chapter 7 Fractions
• Chapter 8 Decimals
• Chapter 9 Data Handling
• Chapter 10 Mensuration
• Chapter 11 Algebra
• Chapter 12 Ratio And Proportion
• Chapter 13 Introduction To Symmetry
• Chapter 14 Practical Geometry
Here we provide NCERT solutions of class 6 maths chapter 2 exercise 2.2. If you want to see solutions of other chapters you can check them by the above links.
### exercise 2.2
NCERT Solutions for Class 6 Maths Chapter 2 Exercise 2.2
Class 6 Maths exercise 2.2 solutions Question 1
Q.1. Find the sum by suitable rearrangement:
(a) 837 + 208 + 363
(b) 1962 + 453 + 1538 + 647
Ans:-
(a) 837 + 208 + 363 = (837 + 363) + 208
= 1200 + 208 = 1408
(b) 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100 = 4600
Class 6 exercise 2.2 solutions question 2
Q.2. Find the product by suitable rearrangement:
(a) 2 × 1768 × 50
Ans:-
2 x 1768 x 50 = (2 x 50) x 1768 = 176800
(b) 4 × 166 × 25
Ans:-
4 x 166 x 25 = 166 x (25 x 4) = 166 x 100 = 16600
(c) 8 × 291 × 125
Ans:-
8 x 291 x 125 = (8 x 125) x 291 = 1000 x 291 = 291000
(d) 625 × 279 × 16
Ans:-
625 x 279 x 16 = (625 x 16) x 279 = 10000 x 279 = 2790000
(e) 285 × 5 × 60
Ans:-
285 x 5 x 60 = 285 x (5 x 60) = 285 x 300 = (300 – 15)x 300 = 300 x 300 – 15 x 300 = 90000 – 4500 = 85500
(f) 125 × 40 × 8 × 25
Ans:-
125 x 40 x 8 x 25 = (125 x 8) x (40 x 25) = 1000 X 1000 = 1000000
Class 6 exercise 2.2 solutions question 3
Q.3. Find the value of the following:
(a) 297 × 17 + 297 × 3
Ans:-
297 x 17 x 297 x 3 = 297 x (17 + 3)
= 297 x 20 = 297 x 2 x 10
= 594 x 10 = 5940
(b) 54279 × 92 + 8 × 54279
Ans:-
54279 x 92 + 8 x 54279 = 54279 x (92 + 8)
= 54279 x 100 = 5427900
(c) 81265 × 169 – 81265 × 69
Ans:-
81265 x 169 – 81265 x 69
= 81265 x (169 – 69)
= 81265 x 100 = 8126500
(d) 3845 × 5 × 782 + 769 × 25 × 218
Ans:-
3845 x 5 x 782 + 769 x 25 x 218 = 3845 x 5 x 782 + 769 x 5 x 5 x 218
= 3845 x 5 x 782 + (769 x 5) x 5 x 218
= 3845 x 5 x 782 + 3845 x 5 x 218
= 3845 x 5 x 782 + 3845 x 5 x 218
= 3845 x 5 x (782 + 218)
= 3845 x 5 x 1000
= 19225 x 1000
= 19225000
NCERT solution for class 6 maths chapter 2 Ex 2.2 question 4
Q.4. Find the product using suitable properties.
(a) 738 × 103
Ans:-
738 x 103 = 738 x (100 + 3)
= 738 x 100 + 738 x 3 [Using distributive property]
= 73800 + 2214 = 76014
(b) 854 × 102
Ans:-
854 x 102 = 854 x (100 + 2)
= 854 x 100 + 854 x 2 [Using distributive property]
= 85400 + 1708 = 87108
(c) 258 × 1008
Ans:-
258 x 1008 = 258 x (1000 + 8)
= 258 x 1000 + 258 x 8 [Using distributive property]
= 258000 + 2064 = 260064
(d) 1005 × 168
Ans:-
1005 x 168 = (1000 + 5) x 168
= 1000 x 168 + 5 x 168 [Using distributive property]
= 168000 + 840 = 168840
NCERT Solutions for Class 6 Maths Chapter 2 exercise 2.2 Question 5
5. A taxi driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If petrol costs ₹ 44 per litre, how much did he spend in all on petrol?
Ans:-
Petrol filled on Monday = 40 litres
Cost of petrol = ₹44 per litre
Petrol filled on Tuesday = 50 litre
Cost of petrol = ₹44 pet litre
∴ Total money spent in all
= ₹(40 x 44 + 50 x 44)
= ₹(40 + 50) x 44 = ₹90 x 44 = ₹3960
Exercise 2.2 class 6 Question 6
6. A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs ₹ 45 per litre, how much money is due to the vendor per day?
Ans:-
Milk supplied in the morning = 32 litres
Cost of milk = ₹15 per litre
Milk supplied in the evening = 68 litres
Cost of milk = ₹15 per litre
Class 6 Exercise 2.2 Question 7
7. Match the following:
(i) 425 × 136 = 425 × (6 + 30 + 100) (a) Commutativity under multiplication.
(ii) 2 × 49 × 50 = 2 × 50 × 49 (b) Commutativity under addition.
(iii) 80 + 2005 + 20 = 80 + 20 + 2005 (c) Distributivity of multiplication over addition.
Ans:-
(i) ↔ (c), (ii) ↔ (a) and (iii) ↔ (b)
## Key Points
Some important key points in NCERT Solutions for Class 6 Maths Chapter 2 Exercise 2.2
• All natural numbers are whole numbers.
• All whole numbers, except 0, are natural numbers.
• Properties of Whole numbers:
• Whole numbers are closed under addition and also under multiplication.
• Whole numbers are not closed under subtraction and under division.
• Division of a whole number by 0 is not defined.
• Addition and multiplication are commutative for whole numbers.
### Which is the smallest whole numbers
Zero is the smallest whole number.
### Which is the biggest whole number?
We can’t find the greatest whole number. so we can say that infinity is the biggest whole number.
### Is zero a whole number?
The whole numbers are the numbers 0, 1, 2, 3, 4, and so on. Negative numbers are not considered whole numbers. All natural numbers are whole numbers, but all whole numbers are not natural numbers since zero is a whole number but not a natural number.
### Is Pi a real number?
Pi is an irrational number, which is not a real number that cannot be expressed by a simple fraction, pi value is 3.14 or 22/7.
Students are advised to read class 6 NCERT books. You can also download the NCERT book pdf only for Class 6 maths chapter 2.
You may also like to read |
# Algebra math problems
We'll provide some tips to help you select the best Algebra math problems for your needs. We will also look at some example problems and how to approach them.
## The Best Algebra math problems
This Algebra math problems provides step-by-step instructions for solving all math problems. Absolute value equations are two different types of equations. Absolute value is the difference between two numbers. For example, if a number is subtracted from another number, then the absolute value of the second number is what’s being subtracted. Another type of equation is an absolute value equation, which compares two numbers and checks to see whether they’re equal. In absolute value equations, the sentence “The total weight of the boxes is 60 pounds” means that both the total weight and the box weights are 60 pounds. Absolute values are also called positive or real values. To solve absolute value equations, you need to know how to subtract numbers. You can subtract a negative number from a positive one, as long as you remember to use parentheses. For example: (3 -5) ÷ 2 = 1 To solve absolute value equations, you need to know how to subtract numbers. You can subtract a negative number from a positive one, as long as you remember to use parentheses. For example:
Some examples of common types of math problems include addition and subtraction problems, multiplication and division problems, fractions and decimals questions, ratio and proportion questions, geometry questions, probability questions, and graph problem questions. In order to solve a math problem, students must first understand the goal of the question they are being asked to answer. Next, they must identify the variables in the problem. Variables are any values that are being changed or are unknown in the equation being solved. Once these two steps have been completed, students should start working backward through the equation to determine what value must be substituted into each variable in order to reach their desired answer. While all math problems require some form of memorization or calculation, some types of questions will require more advanced skills than others. For this reason, it is important for students to know which type of mathematics problem they are facing before
The HCF can also be used to simplify a problem by eliminating one or more smaller factors from the numerator or denominator. . . . The HCF can also be used to simplify a problem by eliminating one or more smaller factors from the numerator or denominator.
First, you have to use correct capitalization (e.g., the word “the” should be capitalized). Second, you need to use correct punctuation (e.g., an apostrophe to show possession or a question mark or exclamation point to show if something is a statement or a question). Third, you need to spell words and proper names correctly (e.g., “New York” not “New Yrk”). Fourth, you need to use the right number of spaces between words and sentences. Fifth, you need to avoid run-on sentences and grammatical errors. Sixth, you need to avoid using wordy and overused phrases. Finally, you need to write clearly so that your meaning is clear.
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Scan and solve Simplify complex fractions solver Websites about math Algebra math problems Solve for circumference Math site that solves problems |
FreeAlgebra Tutorials!
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
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# Solving Quadratic and Polynomial Equations
When you solve a quadratic equation when you have been given a y-value and need to find all of the corresponding x-values. For example, if you had been given the quadratic equation:
y = x2 + 8 · x +10,
and the y-value,
y = 30,
then solving the quadratic equation would mean finding all of the numerical values of x that work when you plug them into the equation:
x2 + 8 · x +10 = 30.
Note that solving this quadratic equation is the same as solving the quadratic equation:
x2 + 8 · x +10 - 30 = 30 - 30 (Subtract 30 from each side) x2 + 8 · x - 20 = 0 (Simplify)
Solving the quadratic equation x2 + 8 · x - 20 = 0 will give exactly the same values for x that solving the original quadratic equation, x2 + 8 · x +10 = 30, will give.
The advantage of manipulating the quadratic equation to reduce one side of the equation to zero before attempting to find any values of x is that this manipulation creates a new quadratic equation that can be solved using some fairly standard techniques and formulas.
Solving a polynomial equation is exactly the same kind of process as solving a quadratic equation, except that the quadratic might be replaced by a different kind of polynomial (such as a cubic or a quartic).
## The Number of Solutions of a Polynomial Equation
A quadratic is a degree 2 polynomial. This means that the highest power of x that shows up in a quadratic’s formula is x2. The maximum number of solutions that a quadratic function can possibly have is 2.
The maximum number of solutions that a polynomial equation can have is equal to the degree of the polynomial.
It is possible for a polynomial equation to have fewer solutions (or none at all). The degree of the polynomial gives you the maximum number of solutions that are theoretically possible, not the actual number of solutions that will occur.
Example: Solving a Polynomial Equation Graphically
The graph given below shows the graph of the polynomial function:
Use the graph to find all solutions of the polynomial equation:
Solution
Graphically, as the polynomial equation is equal to zero the solutions of the polynomial equation,
will be the x-coordinates of the points where the graph of the polynomial:
touches or crosses the x-axis. If you look carefully at the graph supplied above, the graph of the polynomial touches or cuts the graph at the following points:
x = -2, x = -1, x = 2.
The solutions of the polynomial equation
are x = -2, x = -1 and x = 2. |
# 1.04 Square and triangular numbers
Lesson
## Ideas
We've seen that prime and composite numbers are two types of numbers that can help us with calculations.
### Examples
#### Example 1
The only factors a prime number has is 1 and itself.
Which of the following numbers are prime?
A
10
B
21
C
19
Worked Solution
Create a strategy
Write the factors of each option.
Apply the idea
Option A: 10 has the following factors:1,2,5,10
Option B: 21 has the following factors:1,3,7,21
Option C: 19 has the following factors:1,19
Among the choices, 19 is a prime number as its factors are 1 and itself. So, the correct answer is Option C.
Idea summary
• Every whole number greater than 1 is either a prime number or a composite number
• All even numbers greater than 2 are composite numbers
• To be a prime number, a number can only have itself and 1 as factors
• 0 and 1 are not prime or composite numbers
## Square numbers
### Examples
#### Example 2
Write down the next square number after 16.
Worked Solution
Create a strategy
Find the 5th square number.
Apply the idea
16 is the 4th square number since 4\times 4=16.
To find the square number after 16, we need to find the 5th square number which is:5\times 5=25
25 is the next square number after 16.
Idea summary
If we multiply a number by itself, we make a square number. If we use dots to picture this as an array it will make a square shape.
## Triangular numbers
### Examples
#### Example 3
Write down the next 3 triangular numbers.1,\,3,\,6,\,10,\,15,\,⬚,\,⬚,\,⬚
Worked Solution
Create a strategy
Add the the position of the number we are finding to the previous triangular number.
Apply the idea
To find the 6th triangular number we add 6 to the previous triangular number of 15:
To find the 7th triangular number we add 7 to the previous triangular number of 21:
To find the 8th triangular number we add 8 to the previous triangular number of 28:
Here is the complete list of triangular numbers:1,\,3,\,6,\,10,\,15,\,21,\,28,\,36
Idea summary
With triangular numbers, we add 1 more dot to each new row and create a triangle shape with dots.
### Outcomes
#### MA3-4NA
orders, reads and represents integers of any size and describes properties of whole numbers |
# Areas related to circles – Class 10
CBSE CLASS 10 MATHEMATICS
Areas Related to Circles – Chapter 12 /MCQ
1. Find the area of a sector of angle 90, given the radius of circle is 7 cm?
A. 77 sq cm
B. 77/2 sq cm
C. 76 sq cm
2. If the area and circumference of a circle are numerically equal, then the radius of the circle is —————
A. 2 units
B. 4 units
C. 6 units
3. The radii of two circles are 8 cm and 6 cm respectively. The radius of the circle having area equal to the sum of the areas of the two circles is ——–
A. 8 cm
B. 10 cm
C. 12 cm
4. Find the width of a circular path whose inner circumference is 44 cm and outer circumference is 176 cm?
A. 7 cm
B. 21 cm
C. 28 cm
5. The sum of circumference of two circles is 176 cm. If the radius of one circle is 21 cm, the radius of the second circle is —————
A. 6 cm
B. 7cm
C. 8 cm
6. Find the perimeter of the quadrant of a circle of radius 21 cm?
A. 33 cm
B. 21 cm
C. 75 cm
7. The area of a circle whose circumference is 44 cm ————–
A. 77 sq cm
B. 154 sq cm
C. 231 sq cm
8. In a circle of radius 14 cm an arc subtends an angle of 90 degree at the centre. Then the length of the arc is ————
A. 11 cm
B. 22 cm
C. 33 cm
9. If the perimeter of a protractor is 36 cm, its radius is ——–
A. 6cm
B. 7 cm
C. 8 cm
10. A wheel has diameter 56 cm. How many times will the wheel of a vehicle rotate to cover 880m?
A. 5
B. 7
C. 8
11. If the difference between the circumference and diameter of a circle is
225 cm. Then the radius of the circle is —————–
A. 50.5 cm
B. 52.5 cm
C. 54.5 cm
12. The circumference of a circle exceeds its diameter by 30 cm, and then the radius of the circle is —————
A. 7 cm
B. 9 cm
C. 11 cm
1. 77/2 sq cm
2. 2 units
3. 10 cm
4. 21 cm
5. 7 cm
6. 75 cm
7. 154 sq cm
8. 22 cm
9. 7 cm
10. 5
11. 52.5 cm
12. 7 cm |
# How do you prove that the sum of two sides is greater than the third?
## How do you prove that the sum of two sides is greater than the third?
∠ABD>∠CDB. Hence we have AD>AB( because the side opposite to a larger angle is longer). AD = AC+BC. Hence the sum of two sides of a triangle is larger than the third side.
### Why must the sum of two sides of a triangle be greater than the third side?
In order to draw a triangle, the sum of the lengths of any of the two segments must be greater than the length of the third segment. that a + b is not greater than c. The sum of the lengths of any two sides of a triangle must be greater than the third side.
READ: How many editions of The Little Prince are there?
#### Is the difference of two sides of a triangle is less than the third side?
We know that the definition of a triangle as the polygon having three sides such that the sum of any two sides is greater than the third side. Therefore, we can say that the difference between two sides is less than the third side.
What is the sum of the squares of two sides of a triangle?
When the sum of the squares on two sides of a triangle is equal to the square on the third side, then the triangle is right angled .
Which side of triangle is greater?
The longest side in a triangle is opposite the largest angle, and the shortest side is opposite the smallest angle. Triangle Inequality: In any triangle, the sum of the lengths of any two sides is greater than the length of the third side.
## When the sum of the squares of the length of two sides of a triangle is equal to the square of the length of the third side it is called triangle?
The converse of the Pythagorean Theorem is: If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.
READ: Is there a DDR3 Ryzen?
### What is the statement of the triangle inequality theorem?
triangle inequality, in Euclidean geometry, theorem that the sum of any two sides of a triangle is greater than or equal to the third side; in symbols, a + b ≥ c. In essence, the theorem states that the shortest distance between two points is a straight line.
#### What is the triangle sum theorem?
Theorem: The sum of the measures of the interior angles of a triangle is 180°.
Is the sum of two sides of a triangle twice the median?
Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side. Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side. Hence the sum of any two sides of a triangle is greater than twice the median with respect to the third side.
Is the sum of two sides of a triangle always greater?
READ: What is the Getty art challenge?
In a triangle, Sum of two sides is always greater than the third side.
## What is the sum of two sides > the third side?
Sum of two sides > Third side. Sum of two sides is always greater than the third side. Since sum of any two sides is greater than the third side. Since sum of any two sides is greater than the third side. Since sum of any two sides is greater than the third side.
### How do you prove the perimeter of a triangle is greater?
Prove that the perimeter of a triangle is greater than the sum of its three medians. Prove that if two sides of a triangle are unequal, the longer side has greater angle opposite it. In the figure PQ > PR, QS and RS are the bisectors of Q and R respectively. |
# Science:Math Exam Resources/Courses/MATH100/December 2015/Question 10 (c)/Solution 1
As ${\displaystyle x\to +\infty }$, the rate at which the exponential function ${\displaystyle e^{-x}}$ decays to zero is much faster than the rate at which the two linear functions ${\displaystyle x+2}$ and ${\displaystyle x+1}$ grow; therefore
${\displaystyle \lim _{x\to +\infty }(x+2)(x+1)e^{-x}=0.}$
More rigorously, we can apply l'Hospital's rule twice to compute that
${\displaystyle \lim _{x\to +\infty }(x+2)(x+1)e^{-x}=\lim _{x\to +\infty }{\frac {x^{2}+3x+2}{e^{x}}}=\lim _{x\to +\infty }{\frac {2x+3}{e^{x}}}=\lim _{x\to +\infty }{\frac {2}{e^{x}}}=0.}$
On the other hand, as ${\displaystyle x\to -\infty }$, we have ${\displaystyle \lim _{x\to -\infty }x+2=-\infty }$, ${\displaystyle \lim _{x\to -\infty }x+1=-\infty }$, and ${\displaystyle \lim _{x\to -\infty }e^{-x}=+\infty }$, so
${\displaystyle \lim _{x\to -\infty }(x+2)(x+1)e^{-x}=-\infty \cdot -\infty \cdot +\infty =+\infty .}$
Therefore the only asymptote of the function is the line ${\displaystyle {\color {blue}y=0}}$. |
Home » LETReviewersUPCAT » How To Find the Derivative of a Function: Review of Basic Differentiation
# How To Find the Derivative of a Function: Review of Basic Differentiation
At the heart of calculus are derivatives.
The process of taking the derivative of a function is called differentiation. With differentiation, various insights and information were extracted and became part of our current understanding of the study of physics, computer science, economics, statistics, business analytics, and other related fields.
In this review, you will learn what it means to take the derivative of a function and how to derive it. Furthermore, you will understand the intuition behind derivatives using your prior knowledge of slopes and limits already discussed in the previous reviewers.
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## An Overview of the Derivative of a Function
The derivative of a function f(x) is its rate of change with respect to x. In other words, the derivative of the function tells how sensitive or fast the dependent variable f(x) or y changes with respect to the changes to the value of its independent variable or x
Consider the linear function f(x) = 2x + 4. The derivative of this function is 2 (we will discuss later how to get the derivative). This value of the derivative (2) means that f(x) is increasing two times as fast as x
To show this, let x = 1. With this value, the value of f(x) will be:
f(1) = 2(1) + 4 = 6
Suppose we change the value of x = 1 to x = 2. Take a look at what will happen to the value of f(x):
f(2) = 2(2) + 4
f(2) = 4 + 4
f(2) = 8
From x = 1 to x = 2, the “change in value of x” is 2 – 1 = 1. Meanwhile, from f(1) = 6 to f(2) = 8, “the change in value of f(x)” is 8 – 6 = 2. Notice that the change in the value of f(x) is twice as fast as the change in the value of x. This tells us that f(x) changes its value twice as fast as the changes in the value of x. For this reason, the derivative is 2.
For linear functions, the value of the derivative is a constant. This means that for any values of x, you will always get the same rate of change of f(x) with respect to x. Going back to f(x) = 2x + 4, the rate of change of f(x) with respect to x is always the same regardless of the value of x you will use.
Earlier, we tried to test how fast f(x) changes if we start from x = 1 and change its value to x = 2. We found out that the value of f(x) changes its value twice as fast as x (the value of f(x) changes from 6 to 8).
Let us try to use x = 3. Substituting this value to f(x) = 2x + 4. We have:
f(3) = 2(3) + 4
f(3) = 10
Now, if we change the value of x = 3 to x = 20, we have:
f(20) = 2(20) + 4
f(20) = 44
The change in the value of x, in this case, is from 3 to 20, which is equal to 20 – 3 = 17
The change in f(x) in this case is from 10 to 44, which is equal to 44 – 10 = 34
Surprisingly, the change in the value of f(x) is still twice as fast as the change in the value of x. Hence, the rate of change of f(x) = 2x + 4 with respect to x using the above is still 2. This is what we mean that the derivative of a linear function is a constant value.
However, for those functions that are not linear, the derivative is also a function. For instance, the derivative of f(x) = x2 is 2x. 2x is an example of a derivative function. Note that 2x is not a constant. This means that the rate of change of f(x) = x2 is not a constant value but depends on what value of x is used.
Anyway, we will not go deep with the concept of derivatives as it is too complicated. Our goal here is to provide you with an overview so that you have an idea of what the derivative of a function tells us.
Again, the derivative tells us how fast the value of f(x) changes with respect to x. The derivative of a linear function is a constant, but for nonlinear functions, it depends on the specific value of x used for the derivative function.
Another way to explain derivatives is through geometric representation. The derivative of a function is also the slope of the line that is tangent to the curve or the graph of the given function. In the image below, the curve is the geometric representation of the function f(x), and the slope of line l1, which is tangent or intersects the function at a point, is the function f(x) derivative.
We will not deal too much with the geometric interpretation of the derivative since it is more complicated (and given that we can calculate the derivative without knowing it). If you are interested in how derivatives can be explained using this method, refer to the BONUS section of this article.
In the meantime, you can proceed to the next section of this review, which deals with the notations used for the derivative of a function.
## Derivative Notations
There are different ways to write the derivative of a function. However, in this section, we are going to discuss only two notations that are commonly used.
The first notation is the Lagrange notation, attributed to the Italian mathematician Joseph-Louis Lagrange. This notation uses the symbol f’ (read as “f prime”) or f’(x) (read as “f prime of x”) to refer to the derivative of the function.
For example, the derivative of the function f(x) = x2 is the function 2x. Hence, we can write this derivative function using the Lagrange notation as f’(x) = 2x
The second notation is the Leibniz notation, attributed to the German mathematician Gottfried Wilhelm von Leibniz. It uses the symbol dy/dx to indicate the function f(x) derivative. Where did the “y” come from? Recall that y = f(x), as discussed in our review of functions. Hence, when we say dy/dx, we assume y refers to a function.
We usually use the Lagrange notation when the given function is expressed with f(x). For instance, the derivative of f(x) = x2 is written as f’(x) = 2x. On the other hand, we usually use the Leibniz notation when the given function is expressed with y. For instance, the derivative of y = x2 is written as dy/dx = 2x.
Now that you know the notation used for derivatives, let us discuss the techniques we can use to identify them.
## Differentiation Rules
To find the derivative of a function, we apply the following differentiation rules:
### 1. Derivative of a Constant
The derivative of a constant is zero.”
Recall that every real number is a constant. Hence, if we take the derivative of a real number or a constant, it is always zero (0).
Sample Problem 1: What is the derivative of -5?
Solution: -5 is a constant, so its derivative is simply 0.
Sample Problem 2: What is the derivative of π?
Solution: π is also a constant, so its derivative must be 0.
### 2. General Rule for Differentiation (Power Rule)
“The derivative of a function f(x) = xn is nxn – 1 such that x is variable and n is constant.”
The Power Rule, also called General Rule for Differentiation, provides the general technique to find the derivative of a variable raised to any real number.
According to this rule, the derivative of xn, where x is a variable while n is a constant, can be obtained by these steps:
1. Multiply the variable x by the constant n to get nx.
2. Raise x to n – 1, where n is a constant
Here’s an example:
Sample Problem 1: Use the power rule to determine the derivative of f(x) = x6
Solution: Using the power rule, we can perform these steps:
Step 1: Multiply the variable x by the constant n to obtain nx.
We have n = 6 here, so multiply x by 6 to obtain 6x.
Step 2: Raise x to n – 1, where n is a constant.
We already have 6x and need to raise x to n – 1. Recall that n = 6. So, n – 1 = 5.
Thus, we have 6x5.
Therefore, the derivative of f(x) = x6 is f’(x) = 6x5.
Sample Problem 2: Identify the derivative of f(x) = x2
Solution:
Step 1: Multiply the variable x by the constant n to obtain nx.
We have n = 2 here, so multiply x by 2 to obtain 2x.
Step 2: Raise x to n – 1, where n is a constant.
We already have 2x and need to raise x to n – 1. Recall that n = 2. So, 2 – 1 = 1.
Thus, we have 2x1 or simply 2x.
Therefore, the derivative of f(x) = x2 is f’(x) = 2x.
Sample Problem 3: Determine the derivative of f(x) = x
Solution
Note that we can write the function x as x1. Let us use the steps for applying the power rule:
Step 1: Multiply the variable x by the constant n to obtain nx.
We have n = 1 here, so we multiply x by 1 to obtain 1x or simply x.
Step 2: Raise x to n – 1, where n is a constant.
We have n = 1. Thus, 1 – 1 = 0. This means that we should raise x to the power of 0.
This means that we have x0
However, remember that any quantity raised to 0 equals one according to the zero-exponent rule.
Hence, x0 = 1. This means that the derivative of f(x) = x is 1.
### 3. Derivative of Constant Multiplied by a Function Rule (Multiplication by Constant Rule)
If k is a constant and f(x) is a function, then the derivative of k * f(x) is k * f’(x).”
This property tells us that to identify the derivative of the product of a constant and a function, we take the function’s derivative and then multiply the resulting derivative by the given constant.
Sample Problem 1: What is the derivative of f(x) = 2x5?
Solution: Note that 2x5 is the product of a constant (which is 2) and a function (x5). Hence, we have to get the derivative of the function x5 first, then multiply the result by 2:
The derivative of x5 can be obtained using the power rule.
Step 1: Multiply the variable x by the constant n to get nx.
We have n = 5 here, so multiply x by 5 to get 5x.
Step 2: Raise x to n – 1, where n is a constant.
We already have 5x and need to raise x to n – 1. Recall that n = 5. So, 5 – 1 = 4.
Thus, we have 5x4.
Hence, we have 5x4.
We multiply 5x4 by the constant of 2x5 (which is 2): 2 * 5x4 = 10x4.
Hence, the derivative of f(x) = 2x5 is f’(x) = 10x4.
Sample Problem 2: Identify the derivative of f(x) = 7x4.
Solution:
This time, let us solve the derivative as briefly as possible. Don’t be confused by the brevity of the solution; we only apply the derivative of the constant multiplied by a function rule.
f(x) = 7x4
f’(x) = 7 * 4x3 By power rule, the derivative of x4 is 4x3
f’(x) = 28x3
Sample Problem 3: Compute the derivative of the function f(x) = 11x9.
Solution:
f(x) = 11x9
f’(x) = 11 * 9x8 By power rule, the derivative of x9 is 9x8
f’(x) = 99x8
Sample Problem 4: Solve for dy/dx if y = 9x7.
Solution: Recall that the dy/dx notation or the Leibniz notation only implies that we are looking for the derivative of y = 9x7. We can apply the derivative of a constant multiplied by a function rule:
y = 9x7
dy/dx = 9 * 7x6 By power rule, the derivative of x7 is 7x6
dy/dx = 63x6
### 4. Sum Rule
“The derivative of the sum of two functions is equal to the sum of the respective derivatives of the functions.”
In symbols, the derivative of f(x) + g(x) is f’(x) + g’(x)
This rule tells us that if two functions are added, the derivative of their sum is just the sum of the respective derivative of the addends.
Consider f(x) = x2 + x3. This function is composed of x2 and x3. The derivative of f(x) can be obtained by taking the respective derivatives of x2 and x3 using the power rule and then adding these derivatives:
Hence, the derivative of f(x) = x2 + x3 is f’(x) = 2x + 3x2
Sample Problem: Identify the derivative of f(x) = x5 + 2x4 + 3x2
Solution: Since the derivative of f(x) can be obtained by getting the respective derivatives of x5, 2x4, and 3x2 and then adding them, we have:
Hence, the answer is f’(x) = 5x4 + 8x3 + 6x.
### 5. Difference Rule
“The derivative of the difference of two functions is equal to the difference of the respective derivatives of the functions.”
In symbols, the derivative of f(x) – g(x) = f’(x) – g’(x)
The difference rule is just the subtraction counterpart of the Sum Rule for Derivatives. If we want to identify the derivative of the difference between two functions, we need to subtract the derivative of the functions.
Sample Problem 1: Identify the derivative of f(x) = x4 – x3
Solution: The derivative of x4, determined using the power rule for derivatives, is 4x3.
Meanwhile, the derivative of x3, determined using the power rule for derivatives, is 3x2.
Hence, the derivative of f(x) = x4 – x3 is the difference between the respective derivatives of x4 and x3 (4x3 and 3x2, respectively).
Therefore, the answer is f’(x) = 4x3 – 3x2.
Sample Problem 2: Identify the derivative of f(x) = 2x3 – x
Solution: The derivative of f(x) = 2x3 – x can be obtained by getting the derivative of 2x3 and x using the power rule and then subtracting them:
Hence, the derivative of f(x) = 2x3 – x is f’(x) = 6x2 – 1.
Sample Problem 3: Compute the derivative of f(x) = x5 + 2x3 – 7x2.
Solution: Notice that this problem’s given function uses addition and subtraction. We can identify the derivative of this function by determining the derivative of each term and then retaining the operations.
Therefore, the derivative of f(x) = x5 + 2x3 – 7x2 is f’(x) = 5x4 + 6x2 – 14x.
### 6. Product Rule
“The derivative of the product of two functions is equal to the sum of the product of the derivative of the first function and the second function, and the product of the derivative of the second function and the first function.”
In symbols: The derivative of f(x) g(x) = [f’(x) * g(x)] + [g’(x) * f(x)]
The product rule allows us to identify the derivative of the product of two functions.
I know that the product rule is not that intuitive. For this reason, we will perform the product rule using the steps below:
1. Obtain the derivative of the first function.
2. Multiply the result you obtained in Step 1 by the second function.
3. Obtain the derivative of the second function.
4. Multiply the result you obtained in Step 3 by the first function.
5. Add the products you have obtained in steps 2 and 4. The result is the derivative of the product of the functions.
Let us answer some examples to help us understand this differentiation rule:
Sample Problem: What is the derivative of f(x) = (x + 2)(x2 – 1)?
Solution:
Step 1: Obtain the derivative of the first function.
The first function is x + 2. Obtaining its derivative:
Hence, the derivative of x + 2 is simply 1.
Step 2: Multiply the result obtained in Step 1 by the second function.
We multiply one by the second function, which is x2 – 1
1 * (x2 – 1) = x2 – 1
So, for this step, we have x2 – 1.
Step 3: Obtain the derivative of the second function.
The second function is x2 – 1. Obtaining its derivative:
Hence, we have 2x as the derivative of x2 – 1.
Step 4: Multiply the result you obtained in Step 3 by the first function.
We multiply 2x by the first function, which is x + 2. Therefore, we have 2x * (x + 2) or just 2x(x + 2).
Step 5: Add the products you have obtained in steps 2 and 4. The result is the derivative of the product of the functions.
Note that in step 2, we got x2 – 1; in step 4, we got 2x(x + 2).
Thus, adding these values will result in x2 – 1 + 2x(x + 2).
We can simplify the expression above as
x2 – 1 + 2x(x + 2)
x2 – 1 + 2x2 + 4x Distributive Property
3x2 + 4x – 1 Combining similar terms
Hence, the derivative of f(x) = (x + 2)(x2 – 1) is f’(x) = 3x2 + 4x – 1
### 7. Quotient Rule
“The derivative of the quotient of two functions is equal to the difference between the product of the denominator and the derivative of the numerator and the product of the numerator and the derivative of the denominator, divided by the square of the denominator.”
In symbols: the derivative of
A powerful tool that helped me understand how to apply the quotient rule is the following short poem:
“Low dee high minus high dee low square the bottom and way we go.”
In this poem, low means the denominator, dee means “derivative,” and high means the numerator.
Low dee high means denominator (low) times the derivative (dee) of the numerator (high).
High dee low means numerator (high) times the derivative (dee) of the denominator (low).
Next, you subtract the results you have obtained from “low dee high” and “high dee low.”
Meanwhile, square the bottom means square the denominator.
Lastly, we divide the difference between the “low dee high” and “high dee low” by the “square of the bottom.”
You can also remember the “shorter” version of the poem:
Isn’t it easier to deal with math if we remember the formulas in a fun way?
Let us try to apply this technique through an example.
Sample Problem: Determine the derivative of the following
Solution:
• The numerator is 6x, so this will be the “high.”
• Meanwhile, the denominator is x2 + 1, which will be the “low.”
• The derivative of the numerator, 6x, is just 6. So, our “dee high” is 6.
• The derivative of the denominator, x2 + 1, is just 2x. So our “dee low” is 2x.
Let’s get “low dee high” first. “Low” is x2 + 1 while “dee high” is 6. So, “low dee high” is 6*(x2 + 1).
Now, let us obtain “high dee low.” The “high” is 6x while the “dee low” is 2x. So, “high dee low” is 6x*(2x).
Meanwhile, the square of the bottom is just the square of (x2 + 1), which is (x2 + 1)2.
Let us now complete the “shorter” version of the poem:
Let’s simplify the expression above:
Thus, the answer to this problem is:
The expanded form of this expression is:
## Differentiating a Composition of Functions
Do you still remember the composition of functions? Simply put, this function is “inside” or contained by a function. For example, in f(x) = x2 and g(x) = x + 1, the composite function f(g(x)) means that we put g(x) inside f(x):
Thus, the composite function f(g(x)) is equal to (x + 1)2.
In a composite function, you must identify the outer and inner functions. The outer function is where you plug in or insert the inner function.
In f(g(x)) = (x + 1)2, the outer function is x2 since it is the one where you insert x + 1 (which is the inner function).
Sample Problem 1: Identify the outer and inner functions of the composition f(x) = (x – 3)3
Solution: (x – 3)3 can be broken down into x3 as the outer function and x – 3 as the inner function. Note that as you plug in x – 3 to x3, you will obtain (x – 3)3.
Sample Problem 2: Identify the outer and inner functions of the composition f(x) = (3 + x2)2
Solution: The given function can be broken down into x2 as the outer function and 3 + x2 as the inner function. You will obtain the given function as you plug in 3 + x2 to x2.
### Chain Rule
The chain rule is a special differentiation rule that allows us to identify the derivative of a composite function.
The chain rule states that “the derivative of the composition of a function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.”
In symbols: D[f(g(x))] = f’(g(x)) g’(x)
To make it easier for us to apply the chain rule, we will perform it using the steps below:
1. Identify the outer and inner functions.
2. Take the derivative of the outer function.
3. Evaluate the derivative of the outer function at the inner function.
4. Take the derivative of the inner function.
5. Multiply the result in Step 3 by the result in Step 4.
Sample Problem 1: Use the chain rule to differentiate f(x) = (x + 7)2
Solution: Let us apply the chain rule to differentiate (x + 7)2.
Step 1: Identify the outer and inner functions.
The outer function is x2, and the inner function is x + 7.
Step 2: Take the derivative of the outer function.
By applying the power rule, the derivative of the outer function x2 is simply 2x.
Step 3: Evaluate the derivative of the outer function at the inner function.
We perform this by simply plugging the inner function into the derivative of the outer function. Again, the inner function is x + 7, while the derivative of the outer function is 2x. Hence, we have the following:
2x
2(x + 7) Plug-in x + 7
Hence, we have 2(x + 7)
Step 4: Take the derivative of the inner function.
The inner function is x + 7, and by applying the differentiation rules we have learned, its derivative is 1.
Hence, we have 1.
Step 5: Multiply the result in Step 3 by the result in Step 4.
The result in step 3 is 2(x + 7), while for step 4, the result is 1.
Multiplying them: 2(x + 7) * 1 = 2(x + 7)
The answer is 2(x + 7).
Sample Problem 2: Use the chain rule to identify the derivative of (x2 – 9)5
Solution:
Step 1: Identify the outer and inner functions.
The outer function is x5, and the inner function is x2 – 9.
Step 2: Take the derivative of the outer function.
By applying the power rule, the derivative of the outer function x5 is simply 5x4.
Step 3: Evaluate the derivative of the outer function at the inner function.
We perform this by simply plugging the inner function into the derivative of the outer function. The inner function is x2 – 9, so we plug in this function to the computed derivative of the outer function, which is 5x4
5x4
5(x2 – 9)4 Plug-in x2 – 9
Hence, we have 5(x2 – 9)4
Step 4: Take the derivative of the inner function.
The inner function is x2 – 9. Applying the differentiation rules we learned in the previous section, its derivative must be 2x.
Step 5: Multiply the result in Step 3 by the result in Step 4.
The result in step 3 is 5(x2 – 9)4, while step 4’s result is 2x.
Multiplying them: 5(x2 – 9)4 * 2x = 10x(x2 – 9)4
The answer is 10x(x2 – 9)4.
## Derivatives of Other Functions (Optional)
In this section, you’ll get an overview of the derivatives of other functions, such as the exponential and trigonometric functions.
### 1. Derivative of an Exponential Function
An exponential function is a function whose value is a constant raised to a specific variable.
For instance, f(x) = 2x, g(x) = 3x, h(x) = 90.10x, r(x) = 0.9x4 are all examples of exponential function.
Generally, an exponential function is in the form f(x) = ax, where a is a constant and x is a variable.
What is the derivative of an exponential function?
The derivative of an exponential function f(x) = ax is f’(x) = ax ln x, where ln pertains to the “natural logarithm,” a special function used in algebra and calculus. Technically, this is the logarithm of a number to the base e (the concept of logarithms is beyond the scope of our reviewer, so we take this as it is).
Sample Problem: Find the derivative of f(x) = 3
Solution: The derivative of an exponential function f(x) = ax is just f’(x) = ax ln x. Hence, the derivative of f(x) = 3x must be f’(x) = 3x ln x.
### 2. Derivative of Trigonometric Functions
Shown below are the derivatives of the trigonometric functions which we discussed in our trigonometry review. We will not discuss how their derivatives were derived because the “math” behind them is complicated.
## Geometric Interpretation of the Derivative of a Function
The derivative of a function refers to the rate of change of a function with respect to its variable. Geometrically, this is the slope of the line that is tangent to the function’s graph.
Consider the graph of the function below.
Obtaining its slope is not straightforward since the function above is a curve. In other words, we cannot use the slope formula or the techniques we have discussed in identifying the slope of a line.
What we are going to do is to take a point on this curve. Let us label this point as (x, y).
Suppose we add some arbitrary constant to x, say ∆x, and y, say ∆y so that the resulting point will be on the curve. This resulting point will be the point (x + ∆x, y + ∆y).
Since we are dealing with a function, let us use f(x) instead of y.
Hence, the point (x, y) can be written as (x, f(x)), and the point (x + ∆x, y + ∆y) can be written as (x + ∆x, f(x + ∆x)).
Connect these two points in a straight line. Let us call this line l1
Using the points (x, y) and(x + x, f(x +x)), we can calculate the slope of line l1: Recall that the formula for slope is
where (x1, y1) and (x2, y2) are coordinates of the line.
Hence the slope of line l1 can be expressed as
However, we are not interested in the slope of line l1. Remember that the derivative of a function (or a curve) is the slope of the tangent line. The tangent line of the given curve would look like what’s shown below. The tangent line is the one colored in red.
Now from the slope of line l1, how can we derive the slope of the red-colored line or the tangent line, which is also the derivative of this function or curve?
Notice that if we let the value of x be smaller and smaller, we will achieve the slope of the red line from the slope of l1. In other words, we must make the value of x closer and closer to 0.
Hence, the slope of the tangent line of a curve and its derivative is defined by:
Sample Problem: Find the derivative of f(x) = x2 using the definition of the derivative.
Solution:
We have f(x) = x2. This means that by adding an arbitrary constant, x, we have f(x + x) = (x + x)2
By square of binomial: f(x + x) = (x + x)2 = x2 + 2xx + x2
Using the definition of the derivative:
From our calculation, the derivative of f(x) = x2 is 2x.
Fortunately, we don’t have to do the long and intimidating method above in taking the derivative of a function. The differentiation rules we have learned in this review are already enough for you to obtain the derivative of a function.
Next topic: Basic Integration
Previous topic: Limits
## Test Yourself!
### 3. Math Mock Exam + Answer Key
Written by Jewel Kyle Fabula
### Jewel Kyle Fabula
Jewel Kyle Fabula is a Bachelor of Science in Economics student at the University of the Philippines Diliman. His passion for learning mathematics developed as he competed in some mathematics competitions during his Junior High School years. He loves cats, playing video games, and listening to music. |
# Table of 23
Table of 23 gives the repeated addition of the number 23 for the “n” number of times. We are learning maths basic principles and fundamentals, since our childhood in primary and junior classes. Maths tables are also one of the fundamentals of Mathematics. Along with the table of 23, tables of 24, 25, 26, 27, up to 30, should be memorized by students to do calculations quickly.
For example, there are 4 rooms in a house each has 23 cupboards in it. How to calculate the total number of cupboards in all four rooms? We need to add 23 for 4 times repeatedly to get the answer, i.e. 23+23+23+23 = 92. Therefore, there is a total of 92 cupboards. Or we can also calculate it like 4 times of 23, 4 x 23 = 92.
## Table of 23 Chart
Table of 23 is useful not only for students of schools or colleges but also for the candidates appearing for competitive exams conducted at the state level, national level, or international level. These exams have maths aptitude questions which consist of multiplication calculations. Aspiring candidates should learn the Maths tables to solve the calculations easily and quickly.
## Multiplication Table of 23
Let us have a look at the table of number 23 for 1 to 20 times.
23 x 1 = 23 23 x 2 = 46 23 x 3 = 69 23 x 4 = 92 23 x 5 = 115 23 x 6 = 138 23 x 7 = 161 23 x 8 = 184 23 x 9 = 207 23 x 10 = 230 23 x 11 = 253 23 x 12 = 276 23 x 13 = 299 23 x 14 = 322 23 x 15 = 345 23 x 16 = 368 23 x 17 = 391 23 x 18 = 414 23 x 19 = 437 23 x 20 = 460
In the table of 23, you can see, 23 is the multiplicand and the numbers from 1 to 20 are multipliers. So, in the table of 23, the number 23 is multiplied by the other numbers to get the results.
### 23 Times Table Examples
Example1:
Evaluate 3 plus 2 times 23.
Solution:
3 plus 2 times 23 = 3+2(23)
= 3+46
=49
Hence, 3 plus 2 times 23 is 49.
Example 2:
What would be the answer if 23 is added to 5 times 23.
Solution:
23 added to 5 times 23 means 23+5(23)
= 23 + 115
= 138
Hence, 23 added to 5 times 23 is 138.
More Maths Table Table Of 20 Table Of 21 Table Of 22 Table Of 24 Table Of 25 Table Of 26 Table Of 27 Table Of 28 Table Of 29 Table Of 30
Stay tuned with BYJU’S – The Learning App and download the app to learn all Maths-related concepts quickly by watching more exciting videos.
## Frequently Asked Questions on Tables of 23
Q1
### What is 23 times table?
The multiplication table of 23 or 23 times table is the multiplication of the number 23 with other natural numbers to get the result.
Q2
### Write the table of 23 from 1 to 10.
The table of 23 from 1 to 10 are:
23 × 1 = 23
23 × 2 = 46
23 × 3 = 69
23 × 4 = 92
23 × 5 = 115
23 × 6 = 138
23 × 7 = 161
23 × 8 = 184
23 × 9 = 207
23 × 10 = 230
Q3
### Write the 23 times table from 11 to 20.
The table of 23 from 11 to 20 are:
23 × 11 = 253
23 × 12 = 276
23 × 13 = 299
23 × 14 = 322
23 × 15 = 345
23 × 16 = 368
23 × 17 = 391
23 × 18 = 414
23 × 19 = 437
23 × 20 = 460
Test your knowledge on Table Of 23 |
Keyah Math Module 13, Level 4
Mathematical Content : Circumference of a circle, geometry, volume of a sphere
# Impact Processes: Meteor Crater, Arizona
What was the size of the meteorite that formed Meteor Crater?
Warm-up Questions
1. Have you ever seen a crater that was formed by a meteor impact?
2. After looking a pictures of Meteor Crater, how wide and how deep do you think it is?
3. How big would the meteor have been to form this size crater?
4. How fast would it have been travelling?
Understand the Problem:
A meteor falling toward the Earth is propelled by gravitational attraction. Because it is moving, the meteorite has an energy of movement or kinetic energy (KE), which is described by the equation:
where M is the mass of the meteorite and s is its velocity.
If the meteorite is accelerating downward, its KE must be increasing as the square of the velocity! If the meteorite is big enough, it will pass through the atmosphere without burning up completely. When it strikes the surface of the Earth, its velocity and KE go to zero in an instant, but the law of conservation of energy holds that the energy is not simply lost; it is transferred to the surroundings as heat, light, and work, sending out shock waves and excavating a crater far larger than the meteorite itself.
Consider the factors that would determine how “large” a crater is formed. In part, this would depend on geological conditions specific to the impact site, such as the mechanical properties of soils and rocks. However, one might also assume that KE of the meteorite is a more important factor: the more energy delivered upon impact, the “bigger” the crater that is excavated. (We will use diameter to represent crater “size,” because as a crater is eroded away through time, its diameter changes far less than its depth.)
If KE is the most important (controlling) factor, and we can find a mathematical relationship between KE and crater diameter, we can take the dimensions of Meteor Crater and calculate the KE, and then the mass, of the offending meteorite. From the mass, we can then calculate the "size" (more precisely the volume) of the meteorite, because volume and mass are related by density, and we have actual fragments of the meteorite on which density has been measured.
Gather Data:
What data are available to help us solve this problem?
The impact that formed Meteor Crater is beyond history; evidence suggests that it occurred about 50,000 years ago. An impact of this size has not been observed on Earth in recorded time (and most would likely consider that a good thing!). But in the last century, for better or for worse, human beings have devised and experimented with a process of comparable destructive power: nuclear explosions. Until the advent of treaties restricting the practice, nations tested nuclear weapons by detonating them at or just beneath the surface.
Sedan, NV nuclear testing explosion and resulting crater. Image source http://rst.gsfc.nasa.gov/
In the United States, most nuclear weapons testing took place at the Nevada Test Site, in the Mojave and Great Basin Deserts of south-central Nevada, on the homelands of the Newe (Western Shoshone) people. The Newe had no say in the testing, and continue to work for restoration of these lands.
Nuclear explosions often excavated craters identical to those attributed to meteorite impacts. TheKE released in these blasts was known to the weapons designers, so here was a relationship between energy and crater size. This information was eventually made public, and planetary geoscientists made use of this relationship to estimate the KE needed to excavate impact craters of various sizes and ages, on Earth and other planets.
Some of these data, for what are thought to be actual meteorite impact craters are tabulated here. Crater diameter is reported in meters (m) and KE in joules (J).
Crater Diameter(m) Kinetic Energy of Impact (J) x 1018 Brent 3800 2.461 Deep Bay 12000 15.85 Boltysh 23000 310 Clearwater Lake West 32000 1000 Manicouagan 70000 14500 Sudbury 140000 205000
Source: Roddy: Dence et al., 1977.
We can use mathematical regression on these data to derive an equation relating these two variables:
Notation KE = kinetic energy (in J) released by the impact of the meteorite D = diameter (in meters) of the resulting crater
This will enable us to calculate KE of impact for any Earth crater of known diameter, such as Meteor Crater. However, our ultimate target (so to speak!) is the volume of the meteorite, and for that we will need two more equations:
Information:
This equation models kinetic energy,
where M = mass of the meteorite in kilograms (kg)
#### s= velocity of the meteorite in meters per second (meters/sec)
How fast do meteorites typically travel? The average is about 20,000 meters/sec, so let’s use that value for s.
Information This equation models density, ρ = density of the meteorite in kilograms per cubic meter (kg/meters3) V = volume of the meteorite in cubic meters (meters3)
Solve the Problem:
Use the online applets or your calculators to answer the questions below.
Problem 1: Make points from the data provided in the table: the first coordinate should be the diameter of the crater and the second coordinate should be the kinetic energy. Write the second coordinate as shown in the table, just remember that the real value for KE must be multiplied by 1018. Use the power regression applet.
Problem 2: Use power regression to find the equation of best fit. Round constants to 3 decimal places.
Information and Notation: The equation as given by the power regression applet has the general form In this problem represents KE, the kinetic energy in Joules (J), and represents D, the diameter of the crater in meters So the form is. '' and '' are constants determined by the regression of the data. The coefficient '' should be in scientific notation. In this case, with c in the form x.xxx). We can express this equation as:
We will use this equation for craters with diameters between 500 and 140,000 meters, 500 < D < 140,000.
Information: Meteor Crater is about 1,200 meters in diameter.
Problem 3: Now you are ready to determine the size of the meteorite that formed Meteor Crater. Use the regression equation you developed to find the kinetic energy (KE) of impact. You can do this calculation in the Math Pad in the Math Tool Chest. [Use Math Pad] Use the kinetic-energy equation to find the mass of the meteorite. As noted above, use s = 20,000 meters/sec. Record your answer. [Use Math Pad] Use the density equation to find the volume of the meteorite. The iron-nickel fragments found at the site have a density of about 7,800 kg/meters3. Record your answer. [Use Math Pad] Compare the size of the meteorite to the size of something else familiar to you. Assume The meteorite was approximately spherical as it plunged to Earth.
Information You can find its radius using the equation for the volume of a sphere, where V = volume of the meteorite in cubic meters (meters3) r = radius of the meteorite in meters π ≈ 3.14159, In Math Pad type 'pi' for π
Follow-up Questions
Review the concepts from geology that were used in this study. Review the math you used to answer the questions above. Do you think the equations you used give accurate estimates of the size of the meteor? How do scientists find equations like this?
This material is based upon work supported by the National Science Foundation under Grant GEO-0355224. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of the National Science Foundation. |
1. Chapter 4 Class 11 Mathematical Induction
2. Serial order wise
Transcript
Ex 4.1, 7: Prove the following by using the principle of mathematical induction for all n โ N: 1.3 + 3.5 + 5.7 + โฆ + (2n โ 1) (2n + 1) = (๐(4๐2 + 6๐ โ 1))/3 Let P(n) : 1.3 + 3.5 + 5.7 + โฆ + (2n โ 1) (2n + 1) = (๐(4๐2 + 6๐ โ 1))/3 For n = 1, L.H.S = 1.3 = 3 R.H.S = (1(4.12 + 6.1 โ 1))/3 = (4 + 6 โ 1)/3 = 9/3 = 3 L.H.S. = R.H.S โด P(n) is true for n = 1 Assume P(k) is true 1.3 + 3.5 + 5.7 + โฆ + (2k โ 1) (2k + 1) = (๐(4๐2 + 6๐ โ 1))/3 We will prove that P(k + 1) is true. 1.3 + 3.5 + 5.7 + โฆ + (2(k + 1) โ 1).(2(k + 1) + 1) = (๐ + 1)(4(๐ + 1)^2 + 6(๐ + 1) โ 1 )/3 1.3 + 3.5 + 5.7 + โฆ + (2k + 2 โ 1).(2k + 2 + 1) = (๐ + 1)(4(๐^2 + 1 + 2๐)+ 6๐ + 6 โ 1)/3 1.3 + 3.5 + 5.7 + โฆ + (2k + 1).(2k + 3) = (๐ + 1)(4๐^2 +4(1) +4(2๐) + 6๐ + 6 โ 1)/3 1.3 + 3.5 + 5.7 + โฆ + (2k โ 1) (2k + 1) + (2k + 1).(2k + 3) = (๐ + 1)(4๐^2 + 4 + 8๐ + 6๐ + 6 โ 1)/3 = (๐ + 1)(4๐^2 +14๐ + 9)/3 = ((๐(4๐^2 +14๐ + 9)+ 1(4๐^2 +14๐ + 9)))/3 = ((4๐^3 +18๐^2 + 23๐ + 9))/3 Thus, P(k +1) :1.3 + 3.5 + 5.7 + โฆ + (2k โ 1) (2k + 1) + (2k + 1).(2k + 3) = ((4๐^3 +18๐^2 + 23๐ + 9))/3 We have to prove P(k+1) from P(k) i.e. (2) from (1) From (1) 1.3 + 3.5 + 5.7 + โฆ + (2k โ 1) (2k + 1) = (๐(4๐2 + 6๐ โ 1))/3 Adding (2k+1).(2k+3) both sides 1.3 + 3.5 + 5.7 + โฆ + (2k โ 1) (2k + 1) + (2k + 1).(2k + 3) = (๐(4๐2 + 6๐ โ 1))/3 + (2k + 1).(2k + 3) = (๐(4๐2 + 6๐ โ 1) + 3(2๐ + 1)(2๐ + 3))/3 = (๐(4๐2 + 6๐ โ 1) + 3(2๐(2๐ + 3) + 1(2๐ + 3)))/3 = (๐(4๐2 + 6๐ โ 1) + 3(2๐(2๐) +2๐(3) + 2๐ + 3))/3 = (๐(4๐2 + 6๐ โ 1) + 3(4๐^2+ 6๐ + 2๐ + 3))/3 = (๐(4๐2 + 6๐ โ 1) + 3(4๐^2+8๐ + 3))/3 = (๐(4๐2 + 6๐ โ 1) + (3(4๐^2 ) +3(8๐) + 3(3)))/3 = (๐(4๐2 + 6๐ โ 1) + (12๐^2 + 24๐ + 9))/3 = (4๐3 + 6๐^2 โ ๐ + (12๐^2 + 24๐ + 9))/3 = (4๐3 + 6๐^2 + 12๐^2 โ ๐ + 24๐ + 9)/3 = ((4๐^3 +18๐^2 + 23๐ + 9))/3 Thus, 1.3 + 3.5 + 5.7 + โฆ + (2k โ 1) (2k + 1) + (2k + 1).(2k + 3) = ((4๐^3 +18๐^2 + 23๐ + 9))/3 which is the same as P(k +1) โด P(k + 1) is true whenever P(k) is true. โด By the principle of mathematical induction, P(n) is true for n, where n is a natural number
Serial order wise |
# Angles and the Cauchy-Schwarz inequality
In Rn, where vectors are no longer represented by arrows, its not clear how we should define the angle between two vectors. We could try to define angles by using the formula
which we derived for geometric vectors by using the law of cosines for triangles, but there's a problem: if the right-hand side has to equal the cosine of some angle, how do we know it's between –1 and +1?
It turns out that you can prove it's between –1 and +1, as a consequence of the Cauchy-Schwarz inequality.
Cauchy-Schwarz Inequality
If u and v are vectors in Rn, then
(uv)2 ≤ (uu)(vv).
Equality holds if and only if u and v are parallel.
Proof: If either vector is a scalar multiple of the other, say u = av, then
(u•u)(v•v) = {(av)•(av)}(v•v) = a2(v•v)2 = {a(v•v)}2 = (av•v)2 = (u•v)2 .
If neither vector is a scalar multiple of the other, define the vector w = au + bv for scalars a = –uv and b = uu. Calculate ww and simplify:
w•w = a2(u•u) + 2ab(u•v) + b2(v•v) = (u•v)2(u•u)– 2(u•v)(u•u)(u•v) + (u•u)2(v•v) = –(u•v)2(u•u) + (u•u)2(v•v).
Since neither of u and v is a scalar multiple of the other, w can't be 0. Then ww is positive, so we have
(uv)2(uu) < (uu)2(vv).
Divide both sides by uu (which is positive) to get
(uv)2 < (uu)(vv).
Now, take positive square roots of both sides of this inequality: |uv| ≤ (uu)1/2(vv)1/2, i.e.
|uv| ≤ |u||v|.
which is equivalent to
|u||v|uv|u||v|.
Divide both sides by |u||v| to get
.
This last inequality says that its middle can indeed be the cosine of some angle. Our usual formula for the angle between two vectors does make sense in Rn, so we can use it as a definition.
For two non-zero vectors u = (u1, u2, ..., un) and v = (v1, v2, ..., vn) in Rn, The angle θ between u and v is defined by the formula . u and v are said to be orthogonal whenever this angle is π/2, i.e. whenever u•v = 0.
The Cauchy-Schwartz inequality also lets us prove one of the essential properties of norms in Rn.
Four essential rules for norms of vectors in Rn
For any vectors u and v in Rn and any scalar k,
If u0, then |u| > 0
|0| = 0
|ku| = |k||u|
|u + v||u| + |v| (triangle inequality)
The first three of these rules are easy to prove. To explain the triangle inequality for geometric vectors, we interpreted them as sides of a triangle and use elementary geometry. In Rn, we don't have geometric triangles, but we can still prove this inequality as a consequence of the Cauchy-Schwartz inequality.
Proof of the triangle inequality: Calculate:
|u + v|2 = (u + v)•(u + v) = u•u + 2u•v + v•v = |u|2 + 2u•v + |v|2 ≤ |u|2 + 2|u•v| + |v|2 (any number is less than or equal to its absolute value).
But (from the Cauchy-Schwarz inequality)
|uv| ≤ |u||v|.
Then
|u + v|2|u|2 + 2|u||v| + |v|2
i.e.
|u + v|2 ≤ {|u| + |v|}2.
Take positive square roots:
|u + v||u| + |v|. |
Math resources Geometry
Transformations
Rotations
# Rotations
Here you will learn about rotations, including how to rotate a shape around a fixed point, and how to describe clockwise rotations and counterclockwise rotations.
Students will first learn about rotations as part of geometry in 8 th grade.
## What are rotations?
Rotations are transformations that turn a shape around a fixed point by a certain angle measure. This movement changes the shape’s orientation but not its shape or size.
To rotate a shape, you need:
• a center of rotation
• an angle of rotation (given in degrees)
• a direction of rotation – either clockwise or counterclockwise
(Counterclockwise direction is sometimes known as anticlockwise direction.)
For example,
Rotate shape A \; 90^{\circ} clockwise around a fixed point.
Shape A has been rotated a quarter turn clockwise to give shape B.
For example,
Rotate shape A \; 180^{\circ} around a fixed point.
Shape A has been rotated a half turn to give shape B.
Whether the direction is clockwise or counterclockwise is irrelevant.
Using tracing paper can be very useful when using rotations.
The original figure is called the object or the preimage, and the rotated figure is called the image or rotated image.
For rotations, the object shape and the image shape are congruent because they are the same shape and the same size. As the lengths of the shapes have been kept the same, the shapes are said to have isometry.
## Common Core State Standards
How does this relate to 8 th grade math and high school math?
• Grade 8 – Geometry (8.G.A.3)
Describe the effect of dilations, translations, rotations, and reflections on two-dimensional figures using coordinates.
• High School – Geometry – Congruence (HS.G.CO.A.5)
Given a geometric figure and a rotation, reflection, or translation, draw the transformed figure using, for example, graph paper, tracing paper, or geometry software. Specify a sequence of transformations that will carry a given figure onto another.
• High School – Geometry – Congruence (HS.G.CO.B.6)
Use geometric descriptions of rigid motions to transform figures and to predict the effect of a given rigid motion on a given figure; given two figures, use the definition of congruence in terms of rigid motions to decide if they are congruent.
## How to rotate a shape around a fixed point
In order to rotate a shape around a fixed point:
1. Trace the shape.
2. Rotate the tracing paper around the center of rotation.
3. Draw the rotated shape onto the grid.
## Rotations examples
### Example 1: rotate a shape around a fixed point
Rotate the shaded shape 90^{\circ} clockwise around the fixed point:
1. Trace the shape.
Use a pencil and trace the shape onto a piece of tracing paper.
2Rotate the tracing paper around the center of rotation.
Use the pencil and put the tip onto the fixed point. Pivot the tracing paper a quarter turn clockwise.
3Draw the rotated shape onto the grid.
Carefully lift the tracing paper and draw the rotated shape in the correct position.
Note – one of the vertices of the triangle has not moved. So the original triangle and the rotated triangle share a point. This is also known as an invariant point of the shape.
### Example 2: rotate a shape around a fixed point
Rotate the shaded shape 180^{\circ} around the fixed point:
Trace the shape.
Rotate the tracing paper around the center of rotation.
Draw the rotated shape onto the grid.
### Example 3: rotate a shape around a center of rotation
Rotate the shaded shape 90^{\circ} counterclockwise around (3,3)\text{:}
Before we can start, we need to mark the center of rotation on the diagram.
Trace the shape.
Rotate the tracing paper around the center of rotation.
Draw the rotated shape onto the grid.
### Example 4: rotate a shape around a center of rotation
Rotate the shaded shape 180^{\circ} around O\text{:}
Before we can start, we need to mark the center of rotation on the diagram.
O stands for the Origin of the coordinate grid and has the coordinates (0,0).
Trace the shape.
Rotate the tracing paper around the center of rotation.
Draw the rotated shape onto the grid.
## How to describe a rotation
In order to describe a rotation:
1. Trace the shape.
2. Rotate the tracing paper.
3. Write down the description.
## Describing rotations examples
### Example 5: describe a rotation
Describe the rotation of shape A to shape B .
Trace the shape.
Rotate the tracing paper.
Write down the description.
### Example 6: describe a rotation
Describe the rotation of shape A to shape B
Trace the shape.
Rotate the tracing paper.
Write down the description.
### Teaching tips for rotations
• Provide students with transparent paper or tracing paper to manually rotate shapes. This helps them visualize how each rotated point moves.
• Use different colors to show how vertices of polygons move between quadrants after rotation.
### Easy mistakes to make
• Rotating by an incorrect angle
Adding a small arrow onto your tracing paper can help to rotate the shape by the correct angle.
• Confusing clockwise and counterclockwise
You may need to look at a clock to remind yourself of the difference between clockwise and counterclockwise.
• The origin
The origin of a coordinate grid has the coordinates (0,0). It is commonly denoted as O. It is used often as the center of rotation.
• Misunderstanding the position of the center of rotation
The center of rotation can be within the object shape.
For example,
• Not understanding alternative angles and directions
A rotation of 270 degrees in a clockwise direction is a correct alternative to a 90 degree rotation in a counterclockwise direction.
A rotation of 270 degrees in a counterclockwise direction is a correct alternative to a 90 degree rotation in a clockwise direction.
### Practice rotation questions
1. Rotate the shaded shape 180^{\circ} around the center of rotation:
The object shape has to be rotated a half-turn. It needs to have been rotated around the center of rotation. It can not have been reflected.
2. Rotate the shaded shape 90^{\circ} counterclockwise around the center of rotation:
The object shape has to be rotated 90^{\circ} counterclockwise. The center of rotation should be used. The additional extra dotted lines may help to make this rotation clearer.
3. Rotate the shaded shape 90^{\circ} clockwise around (0,0)\text{:}
The object shape has to be rotated 90^{\circ} clockwise. The center of rotation should be the origin. As you can see, the shape starts in the 1 st quadrant but the rotated image is in the 4 th quadrant (having rotated across the x -axis.) The additional dotted lines help to make this rotation clearer.
4. Rotate the shaded shape 180^{\circ} around (- \, 1,0)\text{:}
The center of rotation should be the (- \, 1,0). The additional dotted lines help to make this rotation clearer.
5. Describe the rotation of shape A to shape B.
Rotation of 90^{\circ} counterclockwise around the origin
Rotation 90^{\circ} clockwise around (1,1)
Rotation of 90^{\circ} clockwise around (1,1)
Rotation of 90^{\circ} clockwise around the origin
Make sure you know which is the original object shape and which is the image shape. The additional dotted lines help to make this rotation clearer. The center of rotation is (0,0), the origin.
6. Describe the rotation of shape A to shape B.
Rotation of 180^{\circ} around (- \, 1,- \, 1)
Rotation of 180^{\circ} around (1, 1)
Rotation of 90^{\circ} counterclockwise around the origin
Reflection of 180^{\circ} around (0,0)
Make sure you know which is the original object shape and which is the image shape. Since this is a half-turn the direction of the 180^{\circ} is not needed. The shape rotates across the y -axis. The additional dotted lines help to make this rotation clearer. The center of rotation is (- \, 1,- \, 1).
## Rotations FAQs
What is a rotation in math?
A rotation is a transformation that turns a figure around a fixed point, known as the center of rotation or point of rotation, by a specified angle.
Can rotations be combined with other types of transformations?
Yes, rotations can be combined with translations, reflections, and other rotations to form more complex transformations.
Can a rotation be clockwise or counterclockwise?
Yes, rotations can be in either direction. Positive angles typically represent counterclockwise rotations, while negative angles represent clockwise rotations.
How is a rotation in math similar to the rotation of earth?
A rotation in math is similar to Earth’s rotation because both involve turning around a fixed point or axis while maintaining the shape and size of the object. In math, a figure rotates around a point on the coordinate plane, while Earth rotates around its axis.
Is a rotation a rigid transformation?
Yes, a rotation is a rigid transformation because it turns a figure around a fixed point without changing its shape or size.
## Still stuck?
At Third Space Learning, we specialize in helping teachers and school leaders to provide personalized math support for more of their students through high-quality, online one-on-one math tutoring delivered by subject experts.
Each week, our tutors support thousands of students who are at risk of not meeting their grade-level expectations, and help accelerate their progress and boost their confidence. |
# Amazing simple Math problems II
We will start again like in the first part with a simple problem: we draw a circle and we split that circle in 10 equal regions, then we make the horizontal diameter more thick so that is more clear which one is the upper half and which one is the lower half, then we put each digit from 0 to 9, so in total 10 digits, in each section of the circle like in the image below:
How can you arrange the digits so that if you add the digits in the upper half and the digits in the lower half you get consecutive numbers? (Example: 7+1+8+9+4=29, 3+2+0+6+5=16, so the numbers 16, 29 are not consecutive, you need to arrange the digits so that you get consecutive numbers)
--> Consecutive numbers means numbers that follow each other in order and they have a difference of 1 between every two numbers.
We will start again with a simple problem: we draw a circle and this time we split that circle in 9 equal regions like in the image below:
Then we split that circle again in 3 equal parts like in the image below, look at the grey color sides:
Now we have the digits from 1 to 9 so in total 9 digits. We can put each digit in each section of the circle like this:
This is just an example of how digits are arranged in the sections of the circle and the condition is that the digits can't be repeated so the rule is: each section of the circle has a different digit from those 9. Your task is to answer the following questions:
a) How can you arrange the digits so that if you add the digits which are placed in each of those 3 grey sections will give you the same sum? (Example: 1+2+7=10, 6+4+8=18, 9+3+5=17 so the sums are not equal, you need to arrange the digits so that you get equal sums)
b) a) How can you arrange the digits so that if you add the digits which are placed in each of those 3 grey sections will give you consecutive numbers? (Example: 1+2+7=10, 6+4+8=18, 9+3+5=17 so the numbers 10, 18, 17 are not consecutive, you need to arrange the digits so that you get consecutive numbers)
c) How can you arrange the digits so that if you multiply the digits which are placed in each of those 3 grey sections will give you the same product? (Example: 1*2*7=14, 6*4*8=192, 9*3*5=135 so the products are not equal, you need to arrange the digits so that you get equal products) |
# Least Squares
## Contents
Carl Heiles has written the definitive lab document on least squares fitting, and you should definitely read at least the lite version in the reference materials above. That being said, all the examples in the document are in IDL, and so this page will focus on performing those examples with Python. Note that all figures and equations referenced are those within Carl Heiles’ document.
This is also the first lab where you will really be working with error and the propagation of uncertainty, so it is highly recommended that you look over chapter 3 of Taylor’s error analysis text (the basics covered there will be up for game for quizzes).
## 1 A Numerical Example and Its Solution in IDL
### 1.1 Generation of the numerical example
Suppose that we make four measurements of the angle y and we want to fit to a parabolic function in time t. In the notation of Equation 1.1, s would be unity, t the time, and u the time squared, so the number of unknowns is three (N = 3). Because there are four independent measurements (M = 4) the subscripts run from m = 0 →3. Suppose that the four values of time are 5, 7, 9, 11.
First, we create the matrix X in Python,
X = np.zeros(N, M, dtype = float) = np.zeros(3, 4, dtype = float)
and then we populate it with numbers. In your own work, you would normally do this by reading a data file and transferring the numbers to the matrix using IDL commands; to work through this example, you might manually type them in. After populating the matrix, in direct correspondence with Equation 2.1a we have sm = 1, tm = timem , um = time2 :
\begin{align} {X} = \left[ \begin{array}{ccc} 1 & 5 & 25 \\ 1 & 7 & 49 \\ 1 & 9 & 81 \\ 1 & 11 & 121 \\ \end{array} \; \right] \; \end{align}\,\!
Suppose that the four measured values of y are (Equation 2.1c),
\begin{align} { Y} = \left[ \begin{array}{c} 142 \\ 168 \\ 211 \\ 251 \\ \end{array} \; \right] \; \end{align}\,\!
Figure 4.1 shows the datapoints, together with the fitted curve. In Python, you can generate a column vector like Y by creating an array,
Y = np.array([142, 168, 211, 251]),
and a matrix like X by using the np.matrix function, providing each of the columns in a list, as,
X = np.matrix([[1,1,1,1],[5,7,9,11],[25,49,81,121]])
## 2 Solution of the Numerical Example in Python
In Python we calculate the normal equation matrices and denote the [α] in Equation 2.4a by XX:
XX = np.dot(X,np.transpose(X))
and we denote the [β] in Equation 2.4b by XY:
XY = np.dot(Y,np.transpose(X))
We can take the inverse of [α] (XX) in Python by:
XXI = np.linalg.inv(XX)
Finally, we can find the least-squares fitted quantities a by multiplying these two matrices,
a = np.dot(XY,XXI)
In Python we denote the matrix of predicted values $\bar{y_ m}$ by YBAR, which is,
YBAR = np.dot(a,X),
and we can also define the residuals in Y as,
DELY = Y - YBAR
In Python we can also denote s2 in Equations 3.1 and 3.6 by s_sq and write
s_sq = np.dot(DELY/(M-N),np.transpose(DELY)),
or
s_sq = np.sum(DELY**2)/(M-N)
It is always a good idea to plot all three quantities (the measured values Y, the fitted values YBAR, and the residuals DELY to make sure your fit looks reasonable and to check for bad datapoints.
To get the error in the derived coefficients, we need a way to select the diagonal elements of a matrix. Obviously, any N by N matrix has N diagonal elements; a convenient way to get them is,
diag_elems = np.dot(NbyNmatrix,np.identity(N))
In Python, we define the variances of the N derived coefficients by vardc (think of "variances of derived coefficients"). You can find this as in Equation 3.7 from,
vardc = s_sq*diag_elems
## 3 Discussion of the numerical example
For this numerical example, the solution (the array of derived coefficients) is,
\begin{align} { a} = \left[ \begin{array}{c} 96.6250 \\ 4.5000 \\ 0.8750 \\ \end{array} \; \right] \; \end{align}\,\!
and the errors in the derived coefficients (the square root of the σ2’s of the derived coefficients, i.e. $[\sigma ^2_ n]^{1/2}$, or in Python, np.sqrt(vardc)) are:
\begin{align} { \sigma _ a} = \left[ \begin{array}{c} 34.012 \\ 9.000 \\ .5590 \\ \end{array} \; \right] \; \end{align}\,\!
These results look horrible: the uncertainties are large fractions of the derived coefficients. The reason: we have specifically chosen an example with terrible covariance. And the great thing is that this can be fixed easily (at least in this case–certainly not always), without taking more data! To address this issue, look to section 5 of Least-Squares Lite. |
# Calculate Volume of Cylindrical Shells using Integration
Contributed by:
In this section, we will learn:
How to apply the method of cylindrical shells to find out the volume of a solid.
1. 6
APPLICATIONS OF INTEGRATION
2. APPLICATIONS OF INTEGRATION
6.3
Volumes by
Cylindrical Shells
In this section, we will learn:
How to apply the method of cylindrical shells
to find out the volume of a solid.
3. VOLUMES BY CYLINDRICAL SHELLS
Some volume problems are very
difficult to handle by the methods
discussed in Section 6.2
4. VOLUMES BY CYLINDRICAL SHELLS
Let’s consider the problem of finding the
volume of the solid obtained by rotating about
the y-axis the region bounded by y = 2x2 - x3
and y = 0.
5. VOLUMES BY CYLINDRICAL SHELLS
If we slice perpendicular to the y-axis,
we get a washer.
However, to compute the inner radius and the outer
we would have to
solve the cubic
equation y = 2x2 - x3
for x in terms of y.
That’s not easy.
6. VOLUMES BY CYLINDRICAL SHELLS
Fortunately, there is a method—the
method of cylindrical shells—that is
easier to use in such a case.
7. CYLINDRICAL SHELLS METHOD
The figure shows a cylindrical shell
and height h.
8. CYLINDRICAL SHELLS METHOD
Its volume V is calculated by subtracting
the volume V1 of the inner cylinder from
the volume of the outer cylinder V2 .
9. CYLINDRICAL SHELLS METHOD
Thus, we have:
V V2 V1
2 2
r2 h r h 1
2 2
(r2 r )h 1
(r2 r1 )(r2 r1 )h
r2 r1
2 h(r2 r1 )
2
10. CYLINDRICAL SHELLS METHOD Formula 1
Let ∆r = r2 – r1 (thickness of the shell) and
r 12 r2 r1
Then, this formula for the volume of a
cylindrical shell becomes:
V 2 rhr
11. CYLINDRICAL SHELLS METHOD
V 2 rhr
The equation can be remembered as:
V = [circumference] [height] [thickness]
12. CYLINDRICAL SHELLS METHOD
Now, let S be the solid
obtained by rotating
region bounded by
y = f(x) [where f(x) ≥ 0],
y = 0, x = a and x = b,
where b > a ≥ 0.
13. CYLINDRICAL SHELLS METHOD
Divide the interval [a, b] into n subintervals
[xi - 1, xi ] of equal width xi and let be x
the midpoint of the i th subinterval.
14. CYLINDRICAL SHELLS METHOD
The rectangle with
base [xi - 1, xi ] and
height f ( xi ) is rotated
The result is a
cylindrical shell with
height f ( xi ) , and
thickness ∆x.
15. CYLINDRICAL SHELLS METHOD
Thus, by Formula 1, its volume is
calculated as follows:
Vi (2 xi )[ f ( xi )]x
16. CYLINDRICAL SHELLS METHOD
So, an approximation to the volume V of S
is given by the sum of the volumes of
these shells:
n n
V Vi 2 xi f ( xi )x
i 1 i 1
17. CYLINDRICAL SHELLS METHOD
The approximation appears to become better
as n →∞.
However, from the definition of an integral,
we know that:
n b
lim 2 xi f ( xi )x 2 x f ( x)dx
n a
i 1
18. CYLINDRICAL SHELLS METHOD Formula 2
Thus, the following appears plausible.
The volume of the solid obtained by rotating
about the y-axis the region under the curve
y = f(x) from a to b, is:
b
V 2 xf ( x)dx
a
where 0 ≤ a < b
19. CYLINDRICAL SHELLS METHOD
The argument using cylindrical shells
makes Formula 2 seem reasonable,
but later we will be able to prove it.
20. CYLINDRICAL SHELLS METHOD
Here’s the best way to remember
the formula.
Think of a typical shell,
cut and flattened,
circumference 2πx,
height f(x), and
thickness ∆x or dx:
b
2 x f( x)
a dx
circumference height thickness
21. CYLINDRICAL SHELLS METHOD
This type of reasoning will be helpful
in other situations—such as when we
rotate about lines other than the y-axis.
22. CYLINDRICAL SHELLS METHOD Example 1
Find the volume of the solid obtained by
rotating about the y-axis the region
bounded by y = 2x2 - x3 and y = 0.
23. CYLINDRICAL SHELLS METHOD Example 1
We see that a typical shell has
height f(x) = 2x2 - x3.
24. CYLINDRICAL SHELLS METHOD Example 1
So, by the shell method,
the volume is: 2
V 2 x 2 x x dx
2 3
0
2
3 4
2 x (2 x x )dx
0
4 5 2
2 x x
1
2
1
5 0
2 8 325 165
25. CYLINDRICAL SHELLS METHOD Example 1
It can be verified that the shell method
gives the same answer as slicing.
The figure shows
a computer-generated
picture of the solid
whose volume we
computed in the
example.
26. Comparing the solution of Example 1 with
the remarks at the beginning of the section,
we see that the cylindrical shells method
is much easier than the washer method
for the problem.
We did not have to find the coordinates of the local
maximum.
We did not have to solve the equation of the curve
for x in terms of y.
27. However, in other examples,
the methods learned in Section 6.2
may be easier.
28. CYLINDRICAL SHELLS METHOD Example 2
Find the volume of the solid obtained
by rotating about the y-axis the region
between y = x and y = x2.
29. CYLINDRICAL SHELLS METHOD Example 2
The region and a typical shell
are shown here.
We see that the shell has radius x, circumference 2πx,
and height x - x2.
30. CYLINDRICAL SHELLS METHOD Example 2
Thus, the volume of the solid is:
1
V 2 x x x 2 dx
0
1
2 x x dx
2 3
0
3 4 1
x x
2
3 4 0 6
31. CYLINDRICAL SHELLS METHOD
As the following example shows,
the shell method works just as well
if we rotate about the x-axis.
We simply have to draw a diagram to identify
the radius and height of a shell.
32. CYLINDRICAL SHELLS METHOD Example 3
Use cylindrical shells to find the volume of
the solid obtained by rotating about the x-axis
the region under the curve y x from 0 to 1.
This problem was solved using disks in Example 2
in Section 6.2
33. CYLINDRICAL SHELLS METHOD Example 3
To use shells, we relabel the curve
y x
as x = y2.
the x-axis, we see that
a typical shell has
2πy, and height 1 - y2.
34. CYLINDRICAL SHELLS METHOD Example 3
So, the volume is: 1
V 2 y 1 y 2 dy
0
1
3
2 ( y y )dy
0
2 4 1
y y
2
2 4 0 2
In this problem, the disk method was simpler.
35. CYLINDRICAL SHELLS METHOD Example 4
Find the volume of the solid obtained by
rotating the region bounded by y = x - x2
and y = 0 about the line x = 2.
36. CYLINDRICAL SHELLS METHOD Example 4
The figures show the region and a cylindrical
shell formed by rotation about the line x = 2,
which has radius 2 - x, circumference
2π(2 - x), and height x - x2.
37. CYLINDRICAL SHELLS METHOD Example 4
So, the volume of the solid is:
0
V 2 2 x x x dx2
1
0
2 x 3 x 2 x dx
3 2
1
4 1
x 3 2
2 x x
4 0 2 |
Difference between revisions of "1986 AIME Problems/Problem 7"
Problem
The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence.
Solution
Solution 1
Rewrite all of the terms in base 3. Since the numbers are sums of distinct powers of 3, in base 3 each number is a sequence of 1s and 0s (if there is a 2, then it is no longer the sum of distinct powers of 3). Therefore, we can recast this into base 2 (binary) in order to determine the 100th number. $100$ is equal to $64 + 32 + 4$, so in binary form we get $1100100$. However, we must change it back to base 3 for the answer, which is $3^6 + 3^5 + 3^2 = 729 + 243 + 9 = 981$.
Solution 2
Notice that the first term of the sequence is $1$, the second is $3$, the fourth is $9$, and so on. Thus the $64th$ term of the sequence is $729$. Now out of $64$ terms which are of the form $729$ + $'''S'''$, $32$ of them include $243$ and $32$ do not. The smallest term that includes $243$, i.e. $972$, is greater than the largest term which does not, or $854$. So the $95$th term will be $972$, then $973$, then $975$, then $976$, and finally $981$ |
# Your question: What does it mean to share equally?
Contents
On dividing the whole or a group of objects is into equal parts, we get equal shares. We need to divide an object or a whole number into equal parts to distribute it equally. These equal parts have to be the same in measurements like weight, volume, dimensions, numbers etc.
## Why is equal sharing important?
There are two ways you can approach teaching division: equal sharing and equal grouping. Children should experience both concepts, but here’s the interesting bit, research suggests that teaching equal grouping before equal sharing will help your pupils develop a deeper understanding of division.
## What is an equal sharing problem?
Equal sharing problems are partitive division problems where the amount in each group is unknown.
## Is partitioning the same as equal sharing?
By splitting a large table into smaller, individual tables, queries that access only a fraction of the data can run faster because there is less data to scan. So Partitioning is not same as equally sharing.
## How do you work out unequal sharing?
Unequal Sharing
1. Subtract the difference from the total. 210 – 10 = 200.
2. Divide your answer by the number of people sharing. In our case, we divide by 2. …
3. The answer you got in step 2 is the amount that Jack got. To find Jill’s share, take the answer in step 2 and add 10 (because she got 10 mangoes MORE than Jack).
IMPORTANT: Is IQ option legit?
## Does the shape of an equal share change its size?
One of the shares can be described as a third of the whole shape. The whole shape can be described as three thirds. If a shape is partitioned into four equal shares, the shares can be described as fourths.
Common Core Standards Partitioned Circles & Rectangles
Standard: MA.2.FR.1.1
## What is sharing Year 1?
Sharing equally is a kind of division. When something is shared equally between people, you are working out how much each person gets and how big each share is. Watch this video explaining how to share equally rather than unequally. It goes on to show some examples of sharing numbers in twos, fives and tens.
## What is fair share for kids?
You can explain that when you “share” a group of items, you are “dividing” it into smaller groups. A “fair share” means an “equal” amount for each person. a child pick out a story card. Read the story aloud and have the children act out the skit. |
# How do you factor m^4 - n ^4?
Jun 24, 2018
$\left({m}^{2} - {n}^{2}\right) \left({m}^{2} + {n}^{2}\right)$
#### Explanation:
Remember the ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
Note that ${m}^{4}$ is the same as ${\left({m}^{2}\right)}^{2}$ and the same for $n$. So we may write this as:
$\textcolor{w h i t e}{\text{d")a^2color(white)("dd")-color(white)("d")b^2color(white)("dd") ->color(white)("dd}} \left(a - b\right) \left(a + b\right)$
${\left({m}^{2}\right)}^{2} - {\left({n}^{2}\right)}^{2} \to \left({m}^{2} - {n}^{2}\right) \left({m}^{2} + {n}^{2}\right)$
Jun 24, 2018
$\left({m}^{2} + {n}^{2}\right) \left(m + n\right) \left(m - n\right)$
#### Explanation:
What we have is a difference of squares. Recall that if we have an expression
${\textcolor{s t e e l b l u e}{a}}^{2} - {\textcolor{s t e e l b l u e}{b}}^{2}$
This has an expansion of
$\textcolor{p u r p \le}{\left(a + b\right)} \textcolor{s t e e l b l u e}{\left(a - b\right)}$
Let's rewrite our expression as
${\textcolor{s t e e l b l u e}{\left({m}^{2}\right)}}^{2} - {\textcolor{s t e e l b l u e}{\left({n}^{2}\right)}}^{2}$
Since we have a difference of squares, we can rewrite this as
$\textcolor{p u r p \le}{\left({m}^{2} + {n}^{2}\right)} \textcolor{s t e e l b l u e}{\left({m}^{2} - {n}^{2}\right)}$
What we have in purple is a sum of squares, which can't be factored with real numbers, but we have another difference of squares in blue.
Factoring this, we get
$\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{p u r p \le}{\left({m}^{2} + {n}^{2}\right)} \textcolor{s t e e l b l u e}{\left(m + n\right) \left(m - n\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}$
Hope this helps! |
Lesson Objectives
• Demonstrate the ability to solve a linear inequality in one variable
• Learn how to solve a three-part linear inequality
## Solving a Three-Part Linear Inequality in One Variable
In some cases, we will see what is known as a "three-part" inequality. To solve a three-part inequality, we isolate the variable in the middle. We will perform the same action to each part until we accomplish our goal of:
some number < x < some number
Let's look at a few examples.
Example 1: Solve each inequality, write in interval notation, graph.
-11 ≤ 3x - 5 ≤ -2
Since 5 is being subtracted away from x, we need to add 5 to each part:
-11 + 5 ≤ 3x - 5 + 5 ≤ -2 + 5
-6 ≤ 3x ≤ 3
We will divide each part by 3, the coefficient of x: $$\require{cancel}\frac{-6}{3}≤ \frac{3}{3}x ≤ \frac{3}{3}$$ $$\frac{-2\cancel{6}}{\cancel{3}}≤ \frac{1\cancel{3}}{\cancel{3}}x ≤ \frac{1\cancel{3}}{\cancel{3}}$$ $$-2 ≤ x ≤ 1$$ Interval Notation:
[-2, 1]
Graphing the Interval on the Number Line: Example 2: Solve each inequality, write in interval notation, graph.
-7 ≤ x - 1 ≤ 9
To isolate x in the middle, let's add 1 to each part:
-7 + 1 ≤ x - 1 + 1 ≤ 9 + 1
-6 ≤ x ≤ 10
Interval Notation:
[-6, 10]
Graphing the Interval: Example 3: Solve each inequality, write in interval notation, graph.
-90 < -9x ≤ -27
To isolate x in the middle, let's divide each part by (-9). Remember, this means we have to flip each inequality symbol.
-90/-9 > -9/-9 x ≥ -27/-9
10 > x ≥ 3
Write this in the direction of the number line:
3 ≤ x < 10
Interval Notation:
[3, 10)
Graphing the Interval:
#### Skills Check:
Example #1
Solve each inequality. $$5 < -7 - 2x < 13$$
A
$$-1 < x < 5$$
B
$$-10 < x < -6$$
C
$$-\frac{2}{3}< x < -\frac{1}{3}$$
D
$$-\frac{7}{5}< x < 2$$
E
$$x > -4$$
Example #2
Solve each inequality. $$-68 ≤ 10x - 8 < -58$$
A
$$-6 ≤ x < -5$$
B
$$-7 < x < 3$$
C
$$-1 < x < \frac{12}{5}$$
D
$$-5 ≤ x ≤ 5$$
E
$$-6 < x ≤ -5$$
Example #3
Solve each inequality. $$61 > 7x + 5 > 19$$
A
$$1 < x < 4$$
B
$$\frac{2}{5}< x < 13$$
C
$$-7 < x < 8$$
D
$$-8 < x < 17$$
E
$$2 < x < 8$$ |
# How do you simplify 4/(x-2)-3/(x+1)+2/(x^2-x-2)?
$\frac{x + 12}{\left(x - 2\right) \left(x + 1\right)} = \frac{x + 12}{{x}^{2} - x - 2}$
#### Explanation:
$\frac{4}{x - 2} - \frac{3}{x + 1} + \frac{2}{{x}^{2} - x - 2}$
We need the denominators to be the same. We can do that by multiplying through with various forms of the number 1:
$\frac{4}{x - 2} \left(1\right) - \frac{3}{x + 1} \left(1\right) + \frac{2}{\left(x - 2\right) \left(x + 1\right)}$
$\frac{4}{x - 2} \left(\frac{x + 1}{x + 1}\right) - \frac{3}{x + 1} \left(\frac{x - 2}{x - 2}\right) + \frac{2}{\left(x - 2\right) \left(x + 1\right)}$
$\frac{4 \left(x + 1\right)}{\left(x - 2\right) \left(x + 1\right)} - \frac{3 \left(x - 2\right)}{\left(x + 1\right) \left(x - 2\right)} + \frac{2}{\left(x - 2\right) \left(x + 1\right)}$
$\frac{4 x + 4}{\left(x - 2\right) \left(x + 1\right)} - \frac{3 x - 6}{\left(x + 1\right) \left(x - 2\right)} + \frac{2}{\left(x - 2\right) \left(x + 1\right)}$
$\frac{\left(4 x + 4\right) - \left(3 x - 6\right) + 2}{\left(x - 2\right) \left(x + 1\right)}$
$\frac{4 x + 4 - 3 x + 6 + 2}{\left(x - 2\right) \left(x + 1\right)}$
$\frac{x + 12}{\left(x - 2\right) \left(x + 1\right)} = \frac{x + 12}{{x}^{2} - x - 2}$ |
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• This is an assessment test.
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• Kindly take the tests in this series with a pre-defined schedule.
## Geometry and Mensuration: Test 24
Congratulations - you have completed Geometry and Mensuration: Test 24.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%%
Question 1
AB, EF and CD are parallel lines. Given that, EF= 5 cm GC = 10 cm, AB = 15 cm and DC = 18 cm. What is the value of AC?
A 20 cm B 24 cm C 25 cm D 28 cm
Question 1 Explanation:
$\begin{array}{l}\because AB||EF||CD\\\frac{EF}{CD}=\frac{EG}{GC}\\=>\frac{EF}{18}=\frac{5}{10}\\=>EF=9\\In\vartriangle ABC\text{ }and\vartriangle EFC\text{ }\left( similar\text{ }triangles \right),\\\frac{EF}{AB}=\frac{EC}{AC}\\=>9/15=15/AC\\AC=(15\times 15)/9=25cm\end{array}$
Question 2
In the given triangle, AB is parallel to PQ. AP = c, PC = b, PQ = a, AB = x. What is the value of x?
A $\displaystyle a+\frac{ab}{c}$ B $\displaystyle a+\frac{bc}{a}$ C $\displaystyle b+\frac{ca}{b}$ D $\displaystyle a+\frac{ac}{b}$
Question 2 Explanation:
$\begin{array}{l}~ABC\text{ }is\text{ }congruent\text{ }to~PQC,\\\frac{AC}{PC}=\frac{BC}{QC}=\frac{AB}{PQ}\\=>x=\frac{AC\times PQ}{PC}=\frac{(b+c)a}{b}=a+\frac{ac}{b}\end{array}$
Question 3
In the given figure, QR is parallel to AB, DR is parallel to QB. What is the number of distinct pairs of similar triangles?
A 1 B 2 C 3 D 4
Question 3 Explanation:
$\displaystyle \begin{array}{l}The\text{ }triangles\vartriangle PDR\text{ }and\vartriangle PQB,\vartriangle PQR\text{ }and\vartriangle PAB,\vartriangle DRQ\text{ }and\vartriangle QBA\text{ }are\text{ }similar.\\Thus\text{ }there\text{ }are\text{ }three\text{ }possible\text{ }pairs\text{ }of\text{ }triangles.\\Correct\text{ }option\text{ }is\text{ }\left( c \right).\end{array}$
Question 4
If the angels of a triangle are 30°, 60° and 90°, then what is the ratio of the corresponding sides?
A 1: 2: 3 B 1: 1: √2 C 1: √3: 2 D 1: √2: 2
Question 4 Explanation:
$\begin{array}{l}\tan \,60=\sqrt{3}\\Thus\text{ }the\text{ }ratio\text{ }of\text{ }corresponding\text{ }sides\text{ }=1:\text{ }\surd 3:\text{ }2\\\end{array}$
Question 5
I is the incentre of ∆ABC, ABC= 60o and ACB= 50o. The BIC is:
A 55o B 125o C 70o D 65o
Question 5 Explanation:
$\displaystyle \begin{array}{l}\angle BIC\text{ }=~=\text{ }180\text{ }\frac{1}{2}\left( ABC+ACB \right)\\\text{=180 - }30-25\text{ }=\text{ }125.\\Correct\text{ }option\text{ }is\text{ }\left( b \right)\end{array}$
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### Some(?) of the Parts
A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle
### Baby Circle
A small circle fits between two touching circles so that all three circles touch each other and have a common tangent? What is the exact radius of the smallest circle?
### Logosquares
Ten squares form regular rings either with adjacent or opposite vertices touching. Calculate the inner and outer radii of the rings that surround the squares.
# Xtra
##### Age 14 to 18 Challenge Level:
Two solutions to this problem have been forthcoming from different students at the same school - Madras College. Thank you to Mike and Euan who used lots of trigonometry as well as to Thom who likewise resorted to double angles and the cosine rule and reduced the problem to solving a quadratic equation. Thom was also able to show the significance of the two roots.
\eqalign{ \beta &=& \frac{\pi}{6} - \frac{\alpha}{2} \\ \cos\beta &=& \frac{\sqrt{3}}{2}\cdot\frac{3\sqrt{3}}{2\sqrt{7}} + \frac{1}{2}\cdot\frac{1}{2\sqrt{7}} \\ \; &=& \frac{10}{4\sqrt{7}} = \frac{5}{2\sqrt{7}}}
Using the cosine rule on $\triangle ABP$ \eqalign{ 4 &=& x^2 + 7 - 2x\sqrt{7}\cos\beta \\ \; &=& x^2 + 7 - 5x} Therefore $x^2 - 5x + 3 = 0$ $$x = \frac{5\pm\sqrt{13}}{2}$$ Both solutions satisfy the triangle inequality for $\triangle ABP$, namely $\sqrt{7} - 2 < x < \sqrt{7} + 2$. The diagram can be redrawn to show the trapezium $BPQC$ flipped down producing the much smaller equilateral triangle of side $x$ units.
A solution which just needs Pythagoras's Theorem was sent in by Ewan from King Edward VII School, Sheffield. See if you can work it out from the diagram below, then reveal the hidden text to check your answer.
The diagram shows the median AU at point A cutting the triangle in half. The length AU is in two parts, $y$ and $z$. Since the triangle is equilateral, $y+z=\frac{\sqrt 3}{2}x$. (You may like to prove this e.g. by trigonometry) To work out $y$ use Pythagoras's Theorem:
\begin{align}
y^2+\left(\frac{1}{2}\right)^2 &=\left(\sqrt 7\right)^2 \\
y^2 &= \frac{27}{4} \\
y &= \frac{3\sqrt 3}{2}
\end{align}
and use $y+z=\frac{\sqrt 3}{2}x$ to give $z = (x-3)\frac{\sqrt 3}{2}$.
Then more Pythagoras's Theorem and our values found above give
\begin{align}
z^2+\left(\frac{x}{2} - \frac{1}{2}\right)^2 &=2^2 \\
\frac{3(x-3)^2}{4} + \frac{(x-1)^2}{4} = 4
\end{align}
which simplifies to get the same equation as Thom and the others: $x^2 -5x + 3 = 0$.
What a simple solution, Well done! |
Teacher resources and professional development across the curriculum
Teacher professional development and classroom resources across the curriculum
Observing Student Connections
Introduction | Building Viewpoints | Questions and Answers #1 | More Building Viewpoints | Questions and Answers #2 | Observe a Classroom | Classroom Practice | Your Journal
Again, after you have thought about the questions you would ask this student, look at the questions suggested here. Before you select "Show Answer" to read the given answers, try to imagine how Seth would respond.
Teacher: Let's construct this building on a building mat and look at it from each side. Show Answer
Student: (Constructs the building using unit cubes)
Teacher: Get down to eye level so that you are facing the front of the building. Can you describe what you see? Show Answer
Student: I see three columns with two cubes, one cube, and four cubes.
Teacher: Look straight on. Can you tell me what you see in the middle? Show Answer
Student: I see one cube and then three cubes behind it. Oh –– so the middle stack should have three. And then the right column would have four cubes.
Teacher: How would you draw what you just described on the graph paper? Show Answer
Student: I would draw three columns with two cubes, three cubes, and four cubes, from left to right.
Teacher: Now look at the building from the back. Can you describe what you see? Show Answer
Student: I see four cubes on the left, three in the middle, and two on the right.
Teacher: Draw what you just described. Do you notice anything about the front view and the back view? Show Answer
Student: Yes, the back is the reverse of the front.
Teacher: Look at the building from the right view. Is your drawing correct? Show Answer
Student: No, I should have put two cubes on the left, since that is the tallest stack, then two cubes in the middle, and four on the right stack.
Teacher: Look at the building from the left. Can you predict what it will be? Show Answer
Student: I will see four on the left, two in the middle, and two on the right.
Teacher: Draw the left view you just described. How does it compare to the right view? Show Answer
Student: The left is the reverse of the right.
Teaching Math Home | Grades 6-8 | Connections | Site Map | © | |
IDENTIFYING FUNCTIONS USING GRAPHS
Graphs can be used to display relationships between two sets of numbers. Each point on a graph represents an ordered pair. The first coordinate in each ordered pair is the input value. The second coordinate is the output value. The graph represents a function if each input value is paired with only one output value.
Example 1 :
The graph given below shows the relationship between the number of hours students spent studying for an exam and the marks scored in the exam. Determine whether the relationship represented by the graph is a function.
Solution :
From the graph, it is clear that if a students spends 2 hours for studying, he will be able to score 70 marks in the exam. And if he spends 9 hours for studying, he will be able to score 90 marks in the exam.
So, we can consider the number of hours of studying as input values and marks scored in the exam as output values.
The points represent the following ordered pairs in the form (input, output) or (x, y) :
(1, 70), (2, 70), (2, 85), (3, 75), (5, 80), (6, 82), (7, 88), (9, 90), (9, 95) and (12, 98).
In the above order pairs, there is only one output value for each input value. And no input value has more than one output value.
Since, there is only one output value for each input value, the relationship represented by the graph is a function.
Example 2 :
The graph shows the relationship between the heights and weights of the members of a basketball team. Is the relationship represented by the graph a function ? Explain.
Solution :
From the graph, it is clear that if the height of a member is 73 inches, his weight will be 180 lbs.
So, we can consider the height as input value and weight as output value.
The points represent the following ordered pairs in the form (input, output) or (x, y) :
(68, 160), (70, 165), (70, 175), (71, 170), (71, 185), (73, 180) and (74, 190)
Notice that 70 is paired with both 165 and 175, and 71 is paired with both 170 and 185. These input values are paired with more than one output value
Since, there is more than one output value for the input values 70 and 71, the relationship represented by the graph is not a function.
Kindly mail your feedback to v4formath@gmail.com
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Which of the following statements is true?
This question was previously asked in
DSSSB TGT Maths Female Subject Concerned - 18 Nov 2018 Shift 3
View all DSSSB TGT Papers >
1. A number is rational if and only if its square is rational.
2. A number n is odd if and only if n(n + 1) is even
3. An integer n is odd if and only if n2 + 2n is odd.
4. A number is irrational if and only if its square is irrational.
Option 3 : An integer n is odd if and only if n2 + 2n is odd.
Detailed Solution
Concept:
Rational number: It can be defined as any number which can be represented in the form of p/q where q ≠ 0. This means denominator and numerator are integers and the denominator is not equal to zero.
Even and odd: A number which is divisible by 2 and generates a remainder of 0 is called an even number. Number which does not divisible by 2 called odd number.
Explanation:
Statement 1: A number is rational if and only if its square is rational.
We know that, √2 is irrational number and its square i.e. 2 is is an rational number. Hence, above is not true always,
Statement 2: A number n is odd if and only if n(n + 1) is even.
We know that if n is even, n + 1 will be odd and if n is odd, n + 1 will be even. So, n(n + 1) will be even always because product of one even and one odd is always even. Hence, we can not say that if n(n + 1) is even, 'n' will be odd definitely.
Statement 3: An integer n is odd if and only if n2 + 2n is odd.
n2 + 2n = n(n + 2)
We know that if 'n' is even, n + 2 is also even and if 'n' is odd, n + 2 is also odd. Since, product of two even number is even and product of two odd number is odd.
Hence, if n2 + 2n is odd, 'n' will also be is odd.
Statement 4: A number is irrational if and only if its square is irrational.
We know that, √2 is irrational number and its square i.e. 2 is is an rational number. Hence we can not say that any number will be irrational if and only if its square is irrational.
From the above discussion, we can conclude that, option 3 is correct. |
# Lecture 006
## Distance
Find the distance between a point $Q$ and a line $L$: - Let line $L = \overrightarrow{Q} + t \overrightarrow{v}$ - then the distance $\frac{\|\overrightarrow{QP} \times \overrightarrow{v}\|}{\|\overrightarrow{v}\|}$ - this can be proved by constructing parallelogram with $\overrightarrow{QP}$ and $\overrightarrow{v}$ where $\overrightarrow{v}$ is starts from $Q$ by definition. - think about projection for vector $Q$ on line $L$. However, instead of taking the projection (dot product), we take cross product due to taking opposite.
Distance between a plane ($\overrightarrow{n}$ with point $Q$) and a point ($P$) - $d = \|\text{proj}_n{\overrightarrow{QP}}\| = |\text{comp}_n \overrightarrow{QP}| = \frac{|\overrightarrow{QP} \cdot n|}{\|n\|}$
## Conic Section
### 2D
2D conic section: Solution to $Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$ (intersection of a plane with double cone)
### 3D
The first four can be obtained by imagine:
1. The coefficient of a variable change from positive to negative indicate breaking a "wall" on that axis. If no "walls" are broken, that's a ellipse.
2. Breaking 1 "wall" suggest that the axis never hits the surface, forms hyperboloid of one sheet
3. Breaking 2 "wall"s on $x$ and $y$ direction forms "hyperboloid of two sheets"
4. Now, the constant term indicates how "separate" two surfaces are. When it is positive, two surface separate (if all other are positive in normal case). When it is zero, it becomes sort of like "elliptic cone". When it is negative, it inverse other terms, so we can rearrange to make it positive.
Other than the elliptic cone (which is taken to be nonhomogeneous for more subtle reasons) the others are homogeneous if all three variables appear to the second power and nonhomogeneous if at least one of them does not appear to the second power.
3D conic section: $Ax^2 + By^2 + Cz^2 + Dxy + Exz + Fyz + Gx + Hy + Jz + K = 0$
• Ellipsoid (point)
• Hyperboloid of 2 sheets
• Hyperboloid of 1 sheet
• Elliptic Cone
• Elliptic Paraboloid
• Hyperbolic Paraboloid
Homogeneous (definition in polynomial): all variable to highest power, all constant non zero
Classify 3D Conics:
• Constant Zero (Elliptic Cone, Elliptic Paraboloid, Hyperbolic Paraboloid): non-homogeneous
• All Variables 2 Power: Elliptic Cone
• One of Variable Only 1 Power
• Difference Relation between Other Variables: Elliptic Cone
• Add Relation between Other Variables: Elliptic Paraboloid
• Constant Non-Zero (Ellipsoid, all Hyperboloid): homogeneous
• Has No Minus: Ellipsoid
• Has One Minus: Hyperboloid with One Sheet
• Has Two Minus: Hyperboloid with Two Sheets
• Has Three Minus: No Solution
Other Types:
• Cylinders: $y^2 = 4z$
• Circular Cylinders: $y^2 + z^2 = 4$ |
# A Fraction Number Line Lesson
Fractions can really give students fits! How can we develop conceptual understanding with fractions?
I taught this lesson with a group of struggling 5th grade math learners during an after-school tutoring class. Each student stood at a whiteboard on the wall of my classroom. My goal was to help students conceptually understand why we must find a common denominator when adding fractions. To do this, I asked the students to construct their own number lines. All steps beyond #1 are directions I spoke to students as they wrote on their boards. These were the steps in my lesson:
1. I modeled how to make a fraction number line: arrows on both ends showing numbers go infinitely in both directions; small dashes pinpoint exact locations of numbers; begin with zero; space each mark evenly; starting at zero, build your number line and work across the number line rather than starting with a 0 and 1 and subdividing the line into equal size lengths. I wasn’t modeling fraction concepts per se; just creating the number line. (Previous number line work showed students struggling with this.)
2. Create a number line that shows thirds from 0 to 1.
3. Draw a parallel number line below the thirds number line. Put 0 and 1 directly under the 0 and 1 in the top number line. Now subdivide the new number line into sixths.
4. Draw arrows down to the second number line where fractions line up.
5. Solve 1/3 + 1/6 using one of the number lines.
6. Repeat the above steps with a number line showing fourths and then eighths. (See top photo.)
7. Add a third parallel number line showing sixths. (See bottom photo.)
8. Use the proper number line to solve 1/4 + 3/8.
Each time I asked the students to create something, I redirected with small hints and asked students to peer tutor until all students had a perfect model. We constantly discussed the models to help students make sense of their work. In step 7, we discussed how very few fractions aligned with the number line above (sixths and eighths.)
Students made excellent progress as shown in their work on the boards.
Inspiration for this lesson came from my esteemed colleague, Kristian Quiocho. Kristian is one of the most knowledgeable and passionate teachers I know.
### Tim Bedley
Tim Bedley has been teaching elementary school since 1988. He was recognized as the 2013 Riverside County Teacher of the year. Tim is also the founder of America's number one educational rock band, Rockin' the Standards. He also produces two podcast found on iTunes: The Bedley Bros. and The 5-Minute MishMash. Tim and his brother Scott are co-founders of Global School Play Day, a grassroots movement to promote unstructured play with today's youths.
## 5 thoughts on “A Fraction Number Line Lesson”
1. Would it be possible to question the students in the beginning? Maybe – How can you show thirds or sixths (or which ever the desired fraction is) on a number line? instead of modeling it for them?
Do you have cuisenaire rods? As someone who has also struggled with fractions, rods helped me understand fractions a lot more & then connect those to the number line. The number lines are more abstract and the rods provide a concrete model. I think Kristian has some.
I love the use of the parallel number lines. It’s kind of a scaffold for when they move to the problem onto one number line.
Questioning is always where the money is at, and it’s the hardest part. I wonder what questions Ryan and Kristian would ask?
I was thinking about your goal. How do we know whether or not students understand at the end of this why we must find a common denominator? I wonder if you could show them some incorrect answers to these questions? Say Student last year answered these questions this way… Open it up and ask them to prove or disprove why they are correct or incorrect. You may get some good discourse going there. My kids learn a TON from disproving things. Tells them more about what they are doing.
Thanks for sharing! I love learning and thinking about math in the upper grades. I’m itching to get out of first.
1. Tim Bedley says:
Would it be possible to question the students in the beginning? Maybe – How can you show thirds or sixths (or which ever the desired fraction is) on a number line? instead of modeling it for them?
Great point. Actually, I presented this lesson in isolation, but we had already been doing some number line work in previous lessons. I only modeled the proper way to make a number line. I did not discuss how to properly place certain fractions on the number line.
Do you have cuisenaire rods?
I have been using concrete models before this lesson on number lines. I have these fraction cube connector things. Not sure of their exact name. I definitely want to get into using Kristian’s cuisenaire rods.
I wonder if you could show them some incorrect answers to these questions?
LOVE this! I will be stealing that idea.
Thanks again, Jamie.
2. There’s a lot to think about with fractions on the number line, which is also incredibly an important tool for kids to deeply understand fraction equivalence, and hence fraction operations. Most classrooms avoid it and stick with area models, which has much to do with why middle and high school students continue to be lost when dealing with fraction concepts. We recently did a 4th grade lesson study on the topic of fractions on the number line, researched, planned, taught, observed and debriefed. The write up our shared understandings. In short, kids generally see fractions as 2 numbers (b/c of area models) and need a length model. However, they need to understand that fractions are represented as a LENGTH, not a mark. The name of that place on the line is named that (e.g. 3/4) because it is the length of three units of length 1/4 (called 1/4 because it takes four of those lengths to make 1, which is described by ITS (1) distance or length from zero) placed end to end or concatenated – like measuring. Here’s the summary document…
http://www.lcsc.edu/media/4890900/Cuisenaire-Rods-on-the-Number-Line-shared-understandings.pdf
1. Tim Bedley says:
Thanks so much for sharing, Ryan. I am very excited to read about your lesson study! The “2 numbers” insight is huge. Very helpful.
I actually failed to mention a step that I did at the beginning of the lesson with the kids. I first took the fraction connector manipulatives and laid them sideways. I used them to create a number line to show the kids how they related. Also, I circled the segments on the number line showing them that the fractions were actually those segments rather than the dashes along the number line. |
The statement 3+3=6 serves as a/an what to the conjecture that the sum of two odd numbers is an odd number
Question
The statement 3+3=6 serves as a/an what to the conjecture that the sum of two odd numbers is an odd number
0
1. A proposition states a fact.
If we can only find even a single one situation that satisfies all the conditions of the proposition, but which turns out to be false, it immediately invalidates the proposition.
This situation is an example of a “counter example”.
If the proposition is odd+odd=odd, then
3+3=6 is a counter example of the proposition.
The statement 3+3=6 serves as a/an “contradiction” to the conjecture that the sum of two odd numbers is an odd number.
Step-by-step explanation:
Consider the provided statement.
3+3=6
Here, we can observe that if we add two odd numbers we will get an even number. As shown in the above statement.
This is the property which helps us to conclude that sum of two odd numbers is always an even number.
You can take some example:
7+7=14
11+11=22
The conjecture is: The sum of two odd numbers is an odd number.
Which is a false.
Thus, the provided statement is contradictory to the conjecture because the sum of two odd numbers is not an odd number.
So, we can say that 3+3=6 serves as a/an contradiction to the conjecture that the sum of two odd numbers is an odd number. |
Intended for healthcare professionals
Home/4. Statements of probability and confidence intervals
# 4. Statements of probability and confidence intervals
We have seen that when a set of observations have a Normal distribution multiples of the standard deviation mark certain limits on the scatter of the observations. For instance, 1.96 (or approximately 2) standard deviations above and 1.96 standard deviations below the mean (±1.96SD mark the points within which 95% of the observations lie.
## Reference ranges
We noted in Chapter 1 that 140 children had a mean urinary lead concentration of 2.18 µmol24hr, with standard deviation 0.87. The points that include 95% of the observations are 2.18 ± (1.96 × 0.87), giving a range of 0.48 to 3.89. One of the children had a urinary lead concentration of just over 4.0 µmol24hr. This observation is greater than 3.89 and so falls in the 5% beyond the 95% probability limits. We can say that the probability of each of such observations occurring is 5% or less. Another way of looking at this is to see that if one chose one child at random out of the 140, the chance that their urinary lead concentration exceeded 3.89 or was less than 0.48 is 5%. This probability is usually used expressed as a fraction of 1 rather than of 100, and written µmol24hr
Standard deviations thus set limits about which probability statements can be made. Some of these are set out in Table A (Appendix table A.pdf). To use to estimate the probability of finding an observed value, say a urinary lead concentration of 4 µmol24hr, in sampling from the same population of observations as the 140 children provided, we proceed as follows. The distance of the new observation from the mean is 4.8 – 2.18 = 2.62. How many standard deviations does this represent? Dividing the difference by the standard deviation gives 2.62/0.87 = 3.01. This number is greater than 2.576 but less than 3.291 in , so the probability of finding a deviation as large or more extreme than this lies between 0.01 and 0.001, which maybe expressed as 0.001P < 0.01. In fact Table A shows that the probability is very close to 0.0027. This probability is small, so the observation probably did not come from the same population as the 140 other children.
To take another example, the mean diastolic blood pressure of printers was found to be 88 mmHg and the standard deviation 4.5 mmHg. One of the printers had a diastolic blood pressure of 100 mmHg. The mean plus or minus 1.96 times its standard deviation gives the following two figures:
88 + (1.96 x 4.5) = 96.8 mmHg
88 – (1.96 x 4.5) = 79.2 mmHg.
We can say therefore that only 1 in 20 (or 5%) of printers in the population from which the sample is drawn would be expected to have a diastolic blood pressure below 79 or above about 97 mmHg. These are the 95% limits. The 99.73% limits lie three standard deviations below and three above the mean. The blood pressure of 100 mmHg noted in one printer thus lies beyond the 95% limit of 97 but within the 99.73% limit of 101.5 (= 88 + (3 x 4.5)).
The 95% limits are often referred to as a “reference range”. For many biological variables, they define what is regarded as the normal (meaning standard or typical) range. Anything outside the range is regarded as abnormal. Given a sample of disease free subjects, an alternative method of defining a normal range would be simply to define points that exclude 2.5% of subjects at the top end and 2.5% of subjects at the lower end. This would give an empirical normal range. Thus in the 140 children we might choose to exclude the three highest and three lowest values. However, it is much more efficient to use the mean 2 SD, unless the data set is quite large (say >400).
## Confidence intervals
The means and their standard errors can be treated in a similar fashion. If a series of samples are drawn and the mean of each calculated, 95% of the means would be expected to fall within the range of two standard errors above and two below the mean of these means. This common mean would be expected to lie very close to the mean of the population. So the standard error of a mean provides a statement of probability about the difference between the mean of the population and the mean of the sample.
In our sample of 72 printers, the standard error of the mean was 0.53 mmHg. The sample mean plus or minus 196 times its standard error gives the following two figures:
88 + (1.96 x 0.53) = 89.04 mmHg
88 – (1.96 x 0.53) = 86.96 mmHg.
This is called the 95% confidence interval , and we can say that there is only a 5% chance that the range 86.96 to 89.04 mmHg excludes the mean of the population. If we take the mean plus or minus three times its standard error, the range would be 86.41 to 89.59. This is the 99.73% confidence interval, and the chance of this range excluding the population mean is 1 in 370. Confidence intervals provide the key to a useful device for arguing from a sample back to the population from which it came.
The standard error for the percentage of male patients with appendicitis, described in Chapter 3, was 4.46. This is also the standard error of the percentage of female patients with appendicitis, since the formula remains the same if p is replaced by 100 – p. With this standard error we can get 95% confidence intervals on the two percentages:
60.8 (1.96 x 4.46) = 52.1 and 69.5
39.2 (1.96 x 4.46) = 30.5 and 47.9.
These confidence intervals exclude 50%. Can we conclude that males are more likely to get appendicitis? This is the subject of the rest of the book, namely inference .
With small samples – say under 30 observations – larger multiples of the standard error are needed to set confidence limits. This subject is discussed under the tdistribution (Chapter 7).
There is much confusion over the interpretation of the probability attached to confidence intervals. To understand it we have to resort to the concept of repeated sampling. Imagine taking repeated samples of the same size from the same population. For each sample calculate a 95% confidence interval. Since the samples are different, so are the confidence intervals. We know that 95% of these intervals will include the population parameter. However, without any additional information we cannot say which ones! Thus with only one sample, and no other information about the population parameter, we can say there is a 95% chance of including the parameter in our interval. Note that this does not mean that we would expect with 95% probability that the mean from another sample is in this interval. In this case we are considering differences between two sample means, which is the subject of the next chapter.
## Common questions
What is the difference between a reference range and a confidence interval?
There is precisely the same relationship between a reference range and a confidence interval as between the standard deviation and the standard error. The reference range refers to individuals and the confidence intervals to estimates . It is important to realise that samples are not unique. Different investigators taking samples from the same population will obtain different estimates, and have different 95% confidence intervals. However, we know that for 95 of every 100 investigators the confidence interval will include the population mean interval.
## When should one quote a confidence interval?
There is now a great emphasis on confidence intervals in the literature, and some authors attach them to every estimate they make. In general, unless the main purpose of a study is to actually estimate a mean or a percentage, confidence intervals are best restricted to the main outcome of a study, which is usually a contrast (that is, a difference) between means or percentages. This is the topic for the next two chapters.
## Exercises
4.1 A count of malaria parasites in 100 fields with a 2 mm oil immersion lens gave a mean of 35 parasites per field, standard deviation 11.6 (note that, although the counts are quantitative discrete, the counts can be assumed to follow a Normal distribution because the average is large). On counting one more field the pathologist found 52 parasites. Does this number lie outside the 95% reference range? What is the reference range?
4.2 What is the 95% confidence interval for the mean of the population from which this sample count of parasites was drawn? |
# How to Find the Percentage Difference Between Two Numbers?
0
242
Whether you’re a student trying to figure out your grade on a test or a business owner trying to calculate your company’s profit margins, understanding how to find the percentage difference between two numbers is a valuable skill. While there are a few different methods, you can use to calculate this figure; some are more accurate than others. This blog post will show you how to find the percentage difference between two numbers and explain when you should use each method.
## Method 1: The Basic Formula
The most basic way to find the percentage difference between two numbers is to use this formula:
Percentage Difference = (|Number 1 – Number 2| / ((Number 1 + Number 2)/2)) * 100
Let’s say you want to find the percentage difference between 20 and 30. Plugging those numbers into the formula, we get:
Percentage Difference = (|20 – 30| / ((20 + 30)/2)) * 100 = (10 / 25) * 100 = 40%
## Method 2: The Relative Change Formula
Another way to find the percentage difference between two numbers is to use the relative change formula. This formula is often used when one of the numbers is very small relative to the other. The relative change formula is as follows:
Percentage Difference = ((Number 1 – Number 2)/Number 2) * 100
For example, let’s say we want to find the percentage change in sales from one month to the next. Here, we would use the relative change formula since sales figures are often relatively small numbers. If our sales went from \$1,000 last month to \$1,200 this month, we would plug those numbers into the formula like so:
Percentage Difference = ((\$1,200 – \$1,000)/\$1,000) * 100 = 20%
## Method 3: The Percentage Change Point Formula
Another way to calculate percentage change is using the “percentage change point” method. To use this method, subtract one number from another and divide that result by one of the original numbers. Then, multiply that result by 100. The mathematical equation looks like this:
Percentage Difference Point = ((Number 1 – Number 2)/Number 1) * 100
For example, if we wanted to calculate what happened to our sales from last month to this month using this method, it would look like this: Percentage Difference Point = ((\$1,200 – \$1,000)/\$1,000) * 100 = 0.2 * 100 = 20%
## Method 4: The Average Absolute Deviation Approach
Finally, another method you can use to find percentage change is called the average absolute deviation approach. To Calculate percentage change using this method, first, take the absolute value of each number’s deviation from their mean average. Next, divide their sum by two times n-1, where n equals the total data pairs in your sample set (or, in our case, full months). Finally, multiply that result by 100, and you have your answer!
Thus we see that no matter which method you choose for finding a percentage change between two sets of data, there will always be some level of error involved due simply to rounding errors in some instances; however, these methods will give you a reasonable estimate for what has occurred over a given period.
Conclusion:
No matter which method you choose for finding a percentage change between two sets of data, there will always be some level of error involved due to simply rounding errors in some instances; however, these methods will give you a reasonable estimate of what has occurred over a given period. |
## Aptitude Chapter-10
Bodhak competitive bank exams percentages in this tutorial we are going to learn Races and Games, Percentages, Profit and Loss, Permutations and Combinations
Races and Games:
If two or more persons running in a competition then it is called the race. In races and games you have to remember some points:
1)In a race of 100 meters, A gives a start of 20 meters to B. That means A is at the starting point and B is at 20meters.
#### competitive bank exams percentages
2) In a race of 100 meters A beats B by 20 meters that mean A reached the 100m and
B at 80 meters. In a 100 m race, A covers the distance in 36 sec and B in 45 sec. In this race A beats B by?
1. a) 25 m b) 20 m c) 30 m d) 80 m
Solution: race is 100 m
A = 100 m = 36 sec
B = 100 m = 45 sec
IN 36 SEC A reaches to 100 m. In 36 sec B reaches to ?
B = 45 = 100, 36=?
(36*100)/45=80
So A = 36 = 100, B = 36 = 80
A Beats B By 20 meters.
Percentages :
Percentage means per 100
In percentages, every thing considered as 100.
Percentage concept is used to compare persons on a single platform.
Note: In percentages unknown things considered as 100.
### Profit and Loss:
We can find calculate the Profit and Loss using the following:
Profit percentage or loss percentage is calculated only on cost price.
Discount is allowed only on marked price.
Marked price means the extra price on an item than cost price.
In profit and loss, we can do problems by using one shortcut that always takes cost price (C.P) = 100.
If you want profit then add.
If you want loss then subtract.
S.P = Selling Price
C.P= Cost price
Profit = S.P – C.P (S.P > C.P)
Loss = C.P – S.P (C.P > S.P)
#### Permutations and Combinations:
Permutation means arrangements and we denote it by P.
Combinations mean selections and we denote it by C.
Permutations are NPR that means we can arrange n observations in r ways.
Combinations are NCR that means from n observations r observations are selected. |
# Split a set with two kind of elements and preserving the proportion of elements
I have to solve the following problem:
Given a set of + and - elements, I have to split this set into n sets. These subsets must meet the following requirements:
1. All will have similar number of elements (similar means that will differ in 0 or 1 elements).
2. All will have the same proportion of + and - elements than in the original set.
I have a set with 11 elements, (- - - - - - - + + + +), with 7 - and 4 +, and I have to split it into 3 subsets.
I know how to do the first point. I will have three subsets with 4 elements, 4 elements and 3 elements. I have test my formulas with many sizes and it works perfectly.
My problem comes with keeping the same proportion. This is what I did:
1. I have divided the number of elements in the set by the number of - elements and + elements.
7 / 11 = 0.64. 64% of the elements are -. 4 / 11 = 0.36. 36% of the elements are +.
1. I have multiplied this proportion with the number of elements on each subset.
4 * 0.64 = 2.56 - elements. 4 * 0.36 = 1.44 + elements.
1. I can't add 2.56 elements, so I have decided to round up the bigger number, and to floor the smallest one.
2.56 => 3 - elements. 1.44 => 1 + elements.
The next subset, with 4 elements, will have:
3 - elements. 1 + elements.
And the latest one, with 3 elements, will have:
3 * 0.64 = 1.92 => 2 - elements. 3 * 0.36 = 1.08 => 1 + elements.
But if you add all of the elements on each subset, you will get:
3 + 3 + 2 = 8 - elements. 1 + 1 + 1 = 3 - elements.
Which is wrong.
How can I fix this error?
• It's not possible to satisfy condition (2) unless the number of $+$'s and number of $-$'s are both multiples of the number $n$ of subsets into which we're dividing (or if the number of $+$'s or number of $-$'s is zero). – Travis Nov 8 at 19:20 |
## Special Series
In this page special series we are going to see formulas to find the sum of the special series.Here you can find four formulas.The first formula is the sum of natural numbers,second is sum of squares, third formula is for sum of cubes and fourth is for sum of odd numbers.
(i) Sum of first n natural numbers
n (n + 1) / 2
(ii) Sum of squares
n (n + 1) (2n + 1) / 6
(iii) Sum of cubes
[ n (n + 1) / 2 ]2
Example 1:
Find the sum of first 75 natural numbers.
Solution:
1 + 2 + 3 + 4 + ................. + 75
= 75 ( 75 + 1 )/2
= (75 x 76) / 2
= 75 x 38
= 2850
Therefore the sum of first 75 natural number is 2850.
Example 2:
Find the sum of 15 + 16 + 17 + .............. + 80
Solution:
= ( 1 + 2 + 3 +........+ 80 ) - ( 1 + 2 + 3 + ..........+ 14)
= [80 (80+1)/2] - [14 (14+1)]/2]
= [80x 81]/2 - [14 x 15]/2
= (40 x 81) - (7x15)
= 3240 - 105
= 3135
Example 3:
Find the sum of 1 + 4 + 9 + ............. + 1600
Solution:
The given series can be written in the form
12 + 22 + 32+............ +402
= 40 (40+1) (2x40 + 1) / 6
= (40 x 41 x 81) / 6
= 20 x 41 x 27
= 22140
Example 4:
Find the sum of the following series
113 + 12 3 + ............... +k3 where k = 50
= (13 + 23 + ..........+50 3) - (13 + 23 + ..........+ 10 3)
= [(50 x 51) / 2]2 - [(10 x 11)/2] 2
= 1625625 - 3025
= 1622600
Quote on Mathematics
“Mathematics, without this we can do nothing in our life. Each and everything around us is math.
Math is not only solving problems and finding solutions and it is also doing many things in our day to day life. They are: |
Solving Quadratic Equations by Quadratic Formula Worksheet - Page 3 | Problems & Solutions
• Page 3
21.
A tennis player hits a ball when it is 8 feet off the ground. The ball is hit with an upward velocity of 8 feet per second. After the ball is hit, its height $h$ (in feet) is modeled by $h$ = -16$t$2 + 8$t$ + 8, where $t$ is the time in seconds. How long will it take the ball to reach the ground?
a. 3 b. 4 c. 1 d. 2
#### Solution:
h = -16t2 + 8t + 8
[Original equation.]
0 = -16t2 + 8t + 8
[Substitute 0 for h, since at ground level the height h = 0.]
t = {-8 ± √[(8)2 - 4(-16)(8)]}/[2(-16)]
[Substitute the values in the quadratic formula: a = -16, b = 8 and c = 8.]
= -8±√(64+512) / -32
[Simplify.]
= -8±√576 / -32
= -8±24 / -32
= -32 / -32 = 1
[Since t represents time, discard negative value.]
22.
Chris drops a ball from a height of 100 feet above the ground. Calculate the time taken by the ball to hit the ground, if its height is given by the equation $h$ = -16$t$2 + 100, where $t$ is the time in seconds.
a. 3 b. 2.25 c. 2.5 d. 2.75
#### Solution:
h = -16t2 + 100.
[Original equation.]
0 = -16t2 + 100
[Substituting 0 for h, since at ground level h = 0.]
t = {-0 ± √[(-0)2 - 4(-16)(100)]}/[2(-16)]
[Substitute the values in the quadratic formula: a = -16, b = 0 and c = 100.]
= 0±√(0+6400) / -32
[Simplify.]
= 0±√6400 / -32
[Simplify inside the grouping.]
= 0±80 / -32
= -80 / -32 = 2.5
[Since t represents time, rounding it to a positive value.]
So, the ball takes 2.5 sec to reach the ground.
23.
Andy throws a pencil from a building with an initial downward velocity of - 8 feet per second. How long will the pencil take to reach the ground, if the height of the pencil from the ground is modeled by the equation $h$ = - 16$t$2 - 8$t$ + 48, where $t$ is the time in seconds?
a. 1.5 b. 2 c. 1.75 d. 1.25
#### Solution:
h = - 16t2 - 8t + 48
[Original equation.]
0 = - 16t2 - 8t + 48
[Substitute values and write in the standard form.]
t = {- (- 8) ± √[(- 8)2 - 4(- 16)(48)]}/2(- 16)
[Substitute the values in the quadratic formula.]
= 8±√(64+3072) / -32
[Simplify.]
= 8±√3136 / -32
= 8±56 / -32
[Simplify.]
= -48 / -32 = 1.5
[Evaluating the radical and rounding the solution to a positive value as t represents time.]
24.
What are the values of $a$, $b$ and $c$ in the equation 4$f$2 + 3$f$ - 38 = 0, which is in the standard form?
a. $a$ = -4, $b$ = 3 and $c$ = 38 b. $a$ = 4, $b$ = 3 and $c$ = -38 c. $a$ = 4, $b$ = -3 and $c$ = 38 d. None of the above
#### Solution:
The standard equation is ax2 + bx + c = 0 when a ≠ 0.
4f2 + 3f - 38 = 0
[Original equation.]
a = 4, b = 3 and c = -38
[Compare the original equation with the standard equation.]
25.
Write the equation $\frac{1}{3}$$g$2 - 3 = - $\frac{14}{15}$$g$ in the standard form.
a. 5$g$2 + 14$g$ - 45 = 0 b. -5$g$2 - 14$g$ - 45 = 0 c. 5$g$2 - 14$g$ - 45 = 0 d. None of the above
#### Solution:
13g2 - 3 = - 1415g
[Original equation.]
13g2 - 3 + 1415g = 0
[Add 14 / 15g to each side.]
5g2 - 45 + 14g = 0
[Multiply the equation by LCM, 15.]
5g2 + 14g - 45 = 0
[Rewrite the equation in the standard form.] |
# How do you find the critical numbers of y= 2x-tanx?
Nov 5, 2016
$x = \frac{\pi}{4} + \frac{k \pi}{2} , \frac{\pi}{2} + k \pi , k \in \mathbb{Z}$
#### Explanation:
Note that a critical number of a function $f$ will occur at $x = a$ when $f ' \left(a\right) = 0$ or $f ' \left(a\right)$ is undefined.
So, we first need to find the derivative of the function.
$y = 2 x - \tan x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 - {\sec}^{2} x$
So, critical values will occur when:
$0 = 2 - {\sec}^{2} x$
Or:
${\sec}^{2} x = 2 \text{ "=>" "secx=+-sqrt2" "=>" } \cos x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$
Note that this occurs at $x = \frac{\pi}{4} , \frac{3 \pi}{4} , \frac{5 \pi}{4} , \frac{7 \pi}{4.} . .$ which can be summarized as $x = \frac{\pi}{4} + \frac{k \pi}{2}$ where $k \in \mathbb{Z}$, which means that $k$ is an integer.
Furthermore, note that $2 - {\sec}^{2} x$ is undefined for some values:
$2 - {\sec}^{2} x = 2 - \frac{1}{\cos} ^ 2 x$
This is undefined when $\cos x = 0$, which occurs at $x = \frac{\pi}{2} , \frac{3 \pi}{2} , \frac{5 \pi}{2.} . .$ or $x = \frac{\pi}{2} + k \pi , k \in \mathbb{Z}$.
So, we have critical values at $x = \frac{\pi}{4} + \frac{k \pi}{2} , \frac{\pi}{2} + k \pi , k \in \mathbb{Z}$. |
# How do you solve the system -x-5y-5z=2, 4x-5y+4z=19, and x+5y-z=-20?
Jan 23, 2017
Verify that the determinant of the coefficient matrix is not zero.
If so, then create and augmented matrix by appending of column of constants.
Perform elementary row operations, until solved.
#### Explanation:
Let A equal the coefficient matrix :
A = [ (-1,-5,-5), (4,-5,4), (1,5,-1) ]
Verify that the determinant is not zero:
|A| = |(-1,-5,-5), (4,-5,4), (1,5,-1) | = -140
NOTE: I wrote all of the steps to compute the determinant but it made the write up too long. Please read the reference link. The determinant is not zero, therefore there is a unique solution.
Let B = the column of constants:
$B = \left[\begin{matrix}2 \\ 19 \\ - 20\end{matrix}\right]$
Write augmented matrix by appending B to A:
A|B = [ (-1,-5,-5,|,2), (4,-5,4,|,19), (1,5,-1,|,-20) ]
Perform elementary row operations until the left is an identity matrix :
${R}_{1} + {R}_{3} \to {R}_{3}$
[ (-1,-5,-5,|,2), (4,-5,4,|,19), (0,0,-6,|,-18) ]
${R}_{3} / - 6$
[ (-1,-5,-5,|,2), (4,-5,4,|,19), (0,0,1,|,3) ]
$- 4 {R}_{3} + {R}_{2} \to {R}_{2}$
[ (-1,-5,-5,|,2), (4,-5,0,|,7), (0,0,1,|,3) ]
$5 {R}_{3} + {R}_{1} \to {R}_{1}$
[ (-1,-5,0,|,17), (4,-5,0,|,7), (0,0,1,|,3) ]
$4 {R}_{1} + {R}_{2} \to {R}_{2}$
[ (-1,-5,0,|,17), (0,-25,0,|,75), (0,0,1,|,3) ]
${R}_{2} / - 25$
[ (-1,-5,0,|,17), (0,1,0,|,-3), (0,0,1,|,3) ]
$5 {R}_{2} + {R}_{1} \to {R}_{1}$
[ (-1,0,0,|,2), (0,1,0,|,-3), (0,0,1,|,3) ]
$\left(- 1\right) {R}_{1}$
[ (1,0,0,|,-2), (0,1,0,|,-3), (0,0,1,|,3) ]
We have an identity matrix on the left, therefore, the right contains the values of the variables:
$x = - 2 , y = - 3 , z = 3$
check:
$- x - 5 y - 5 z = 2$
$4 x - 5 y + 4 z = 19$
$x + 5 y - z = - 20$
$- \left(- 2\right) - 5 \left(- 3\right) - 5 \left(3\right) = 2$
$4 \left(- 2\right) - 5 \left(- 3\right) + 4 \left(3\right) = 19$
$\left(- 2\right) + 5 \left(- 3\right) - \left(3\right) = - 20$
$2 = 2$
$19 = 19$
$- 20 = - 20$
This checks. |
#### Need solution for RD Sharma Maths Class 12 Chapter 15 Tangents and Normals Excercise Fill in the blanks Question 18
$\left ( \frac{1}{4} ,\frac{1}{2}\right )$
Hint:
First find the slope of curve $y^{2}=x$, then compare with slope $45^{\circ}$ with x-axis.
i.e. slope $\tan\frac{\pi}{4}=1$
Given:
Given curve, $y^{2}=x$, the tangent at which makes an angle of $45^{\circ}$ with x-axis.
To find:
We have to find the point on the given curve.
Solution:
We have
$y^{2}=x$ … (i)
Differentiate both side with respect to $x$, we get
$\Rightarrow$ $2y\frac{dy}{dx}=1$ $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$
$\Rightarrow$ $\frac{dy}{dx}=\frac{1}{2y}$ … (ii)
Also tangent makes an angle of $45^{\circ}\left ( \frac{\pi}{4} \right )$ with x-axis
$\Rightarrow$ $\frac{dy}{dx}=\tan\left ( \frac{\pi}{4} \right )$ $\left[\because \tan \left(\frac{\pi}{4}\right)=1\right]$
$\Rightarrow$ $\frac{dy}{dx}=1$ … (iii)
From equation (ii) and (iii), we get
\begin{aligned} &\Rightarrow \quad \frac{1}{2 y}=1 \\ &\Rightarrow \quad y=\frac{1}{2} \end{aligned}
Put value of $y$ in equation $y^{2}=x$
$\Rightarrow$ $x=\frac{1}{4}$
Hence required point is $\left ( \frac{1}{4} ,\frac{1}{2}\right )$ |
# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### (2/5) : (1/6) = 12/5 = 2 2/5 = 2.4
Spelled result in words is twelve fifths (or two and two fifths).
### How do you solve fractions step by step?
1. Divide: 2/5 : 1/6 = 2/5 · 6/1 = 2 · 6/5 · 1 = 12/5
Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 1/6 is 6/1) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators. In the next intermediate step, the fraction result cannot be further simplified by canceling.
In words - two fifths divided by one sixth = twelve fifths.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
adding fractions: 2/4 + 3/4
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Unit rate
Find unit rate: 6,840 customers in 45 days
• The third
The one-third rod is blue, one-half of the rod is red, the rest of the rod is white and measures 8 cm. How long is the whole rod?
• Sewing
Beth's mother can sew 235 pairs of short pants in 6 days while Lourdes can sew 187 pairs in 8 days. How many more pairs of short pants can Beth's mother sew?
• Pumps
6 pump fills the tank for 3 and a half days. How long will fill the tank 7 equally powerful pumps?
• Almonds
Rudi has 4 cups of almonds. His trail mix recipe calls for 2/3 cup of almonds. How many batches of trail mix can he make?
• How many 16
How many three-tenths are there in two and one-fourths?
• Golf balls
Of the 28 golf balls, 1/7 are yellow. How many golf balls are yellow? Use the model to help you. Enter your answer in the box.
• Cutting wire
If you cut a 3 ½ ft length wire into pieces that are 2 inches long, how many pieces of wire will you have?
• The balls
You have 108 red and 180 green balls. You have to be grouped into the bags so that the ratio of red to green in each bag was the same. What smallest number of balls may be in one bag?
• Statues
Diana is painting statues. She has 7/8 of a liter of paint remaining. Each statue requires 1/20 of a liter of paint. How many statues can she paint?
• Video game
Nicole is playing a video game where each round lasts 7/12 of an hour. She has scheduled 3 3/4 hours to play the game. How many rounds can Nicole play?
• Jordan
Jordan wants to bring in cookie cakes to share with his 23 classmates for his birthday. His mom is willing to buy no more than 6 cookie cakes and Jordan must share the cakes equally between his classmates (and himself). Answer the questions to help Jordan
• Christmas
Calculate how much of the school year (202 days long) take Christmas holidays 19 days long. Expressed as a decimal number and as a percentage. |
Others
# UNIQUE NUMBERS WHICH CAN BE SUCH IN THE FORM OF ROMAN NUMBERS
Have you ever noticed that the watch you are wearing or the clock of your house has different symbols that depict the natural numbers? Like there signs which are reconsidered as actual numbers but, various signs and symbols are in use to represent those numbers. Not only in the watches or wall clock, but these symbols are also used as codes to send a paritcular message. Even in our class numbers, these symbols are in use. These symbols are in used since Roman times. Romans used these roman numerals instead of using actual numbers. These numerals are still in use as it is a concept of learning these numbers in mathematics. These roman numbers cannot fully replace the natural numbers but, sometimes these can be used to represent the numbers.
## Let’s discuss more roman numerals:
1. Roman numerals use limited symbols to represent the numbers, here:
I represent the number 1.
V represents the number 5.
X represents the number 10.
L represents the number 50.
C represents the number 100.
By using these limited numbers, students can make other numbers too. For example, By joining I and V, gives IV represents the number four.
1. Students have to learn the first line from one to ten of the roman numbers with the symbols that represent ten, fifty, hundred, or thousand. By learning these numbers students will be able to make the other roman numeral by themselves.
2. There are mainly three rules that a person should remember while writing the roman numerals. These rules are related to repetition, addition, and subtraction. These rules are very important while writing the roman numbers, as these rules will not be applied then the roman number will be considered invalid.
3. In the rule of repetition, If numbers are written twice in roman numerals, their value will increase up to 2 or 3 times. The symbols such as I, V, or M have a limit to get repeated. These symbols can be repeated only three times and not more than and Symbols like L, V, and D will never get repeated.
4. In the rule of addition, If a symbol with a smaller value has written on the right side of a symbol with a bigger value, a person can get a new symbol for representing a different number. For example, if we write V on the right side of X it will give us a new symbol which is XV which is equal to 15.
5. In the rule of subtraction, If a symbol of a smaller value is written on the left side of the symbol with a bigger value. Then the symbols will get subtracted. For example, if we right X on the left side of C then we will get XC and this means that the numbers will be subtracted. But the symbols V, L, and D will not be cancelled from any symbol but, only I, X, or C are used in subtracting the symbols, which can be only 6 combinations.
6. If there are long Roman numerals, it can be difficult for students to identify the number. If they know the whole topic and related to it, that will be easier for the students to learn and apply their knowledge while identifying the roman numbers. For example, if the number is 3817, in this we have to expand the numbers and represent their symbols and then calculate it. Here it would be 3000+800+10+6, that will give us MMMDCCCXVI.
From the points mentioned above, we can say that learning about roman numerals is very interesting. If students want to get extra knowledge and in-depth information about these roman digits, They can get this from the website of Cuemath. They also provide math worksheets so that students can practice on their own and improve. Learning about their numbers is unique and fun, people should show their interest in creating and solving these numbers.
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# California CCSS Mathematics Grades 1-3
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3 Number and Operations - Fractions Grade 3 expectations in this domain are limited to fractions with denominators 2, 3, 4, 6, and 8. None None Develop understanding of fractions as numbers. 3.NF.1. Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a/b as the quantity formed by a parts of size 1/b. 3.NF.2. Understand a fraction as a number on the number line; represent fractions on a number line diagram. a. Represent a fraction 1/b on a number line diagram by defining the interval from 0 to 1 as the whole and partitioning it into b equal parts. Recognize that each part has size 1/b and that the endpoint of the part based at 0 locates the number 1/b on the number line. b. Represent a fraction a/b on a number line diagram by marking off a lengths 1/b from 0. Recognize that the resulting interval has size a/b and that its endpoint locates the number a/b on the number line. 3.NF.3. Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size. a. Understand two fractions as equivalent (equal) if they are the same size, or the same point on a number line. b. Recognize and generate simple equivalent fractions, e.g., 1/2 = 2/4, 4/6 = 2/3). Explain why the fractions are equivalent, e.g., by using a visual fraction model. c. Express whole numbers as fractions, and recognize fractions that are equivalent to whole numbers. Examples: Express 3 in the form 3 = 3/1; recognize that 6/1 = 6; locate 4/4 and 1 at the same point of a number line diagram. d. Compare two fractions with the same numerator or the same denominator by reasoning about their size. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model. 3
5 Geometry 2.G.1. Recognize and draw shapes having specified attributes, such as a given number of angles or a given number of equal faces. 5 Identify triangles, quadrilaterals, pentagons, hexagons, and cubes. { 5 Sizes are compared directly or visually, not compared by measuring.} 2.G.2. Partition a rectangle into rows and columns of same-size squares and count to find the total number of them. 2.G.3. Partition circles and rectangles into two, three, or four equal shares, describe the shares using the words halves, thirds, half of, a third of, etc., and describe the whole as two halves, three thirds, four fourths. Recognize that equal shares of identical wholes need not have the same shape. 1.G.1. Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size); build and draw shapes to possess defining attributes. 1.G.2. Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape. 4 { 4 Students do not need to learn formal names such as right rectangular prism. } 1.G.3. Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares. 3.G.1. Understand that shapes in different categories (e.g., rhombuses, rectangles, and others) may share attributes (e.g., having four sides), and that the shared attributes can define a larger category (e.g., quadrilaterals). Recognize rhombuses, rectangles, and squares as examples of quadrilaterals, and draw examples of quadrilaterals that do not belong to any of these subcategories. 3.G.2. Partition shapes into parts with equal areas. Express the area of each part as a unit fraction of the whole. For example, partition a shape into 4 parts with equal area, and describe the area of each part as 1/4 of the area of the shape. Standards for Mathematical Practice (K-12) 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others 4. Model with mathematics 5. Use appropriate tools strategically 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. 5
Grade 3 In grade 3, instructional time should focus on four critical areas: (1) developing understanding of multiplication and division and strategies for multiplication and division within 100; (2) developing
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For the Nebraska Department of Education Comparison of the to the Nebraska Standards for Mathematics, Grades K 12 September 2013 (Final) About McREL Mid-continent Research for Education and Learning (McREL) |
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NCERT solutions for Class 10 Mathematics Textbook chapter 8 - Introduction to Trigonometry [Latest edition]
Course
Textbook page
Chapter 8: Introduction to Trigonometry
Ex. 8.3Ex. 8.4
NCERT solutions for Class 10 Mathematics Textbook Chapter 8 Introduction to Trigonometry Exercise 8.3 [Pages 189 - 190]
Ex. 8.3 | Q 1.1 | Page 189
Evaluate (sin 18^@)/(cos 72^@)
Ex. 8.3 | Q 1.2 | Page 189
Evaluate (tan 26^@)/(cot 64^@)
Ex. 8.3 | Q 1.4 | Page 189
Evaluate cosec 31° − sec 59°
Ex. 8.3 | Q 2.2 | Page 189
Show that cos 38° cos 52° − sin 38° sin 52° = 0
Ex. 8.3 | Q 3 | Page 189
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A
Ex. 8.3 | Q 4 | Page 189
If tan A = cot B, prove that A + B = 90
Ex. 8.3 | Q 5 | Page 189
If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.
Ex. 8.3 | Q 7 | Page 190
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°
NCERT solutions for Class 10 Mathematics Textbook Chapter 8 Introduction to Trigonometry Exercise 8.4 [Pages 193 - 194]
Ex. 8.4 | Q 1 | Page 193
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Ex. 8.4 | Q 2 | Page 193
Write all the other trigonometric ratios of ∠A in terms of sec A.
Ex. 8.4 | Q 3.1 | Page 193
Evaluate
(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)
Ex. 8.4 | Q 3.2 | Page 193
Evaluate sin25° cos65° + cos25° sin65°
Ex. 8.4 | Q 4.1 | Page 193
Choose the correct option. Justify your choice.
9 sec2 A − 9 tan2 A =
• 1
• 9
• 8
• 0
Ex. 8.4 | Q 4.2 | Page 193
Choose the correct option. Justify your choice.
(1 + tan θ + sec θ) (1 + cot θ − cosec θ)
• 0
• 1
• 2
• -1
Ex. 8.4 | Q 4.3 | Page 193
Choose the correct option. Justify your choice.
(secA + tanA) (1 − sinA) =
• secA
• sinA
• cosecA
• cosA
Ex. 8.4 | Q 4.4 | Page 193
Choose the correct option. Justify your choice.
(1+tan^2A)/(1+cot^2A)
• secA
• −1
• cotA
• tanA
Ex. 8.4 | Q 5.01 | Page 193
Prove the following identities, where the angles involved are acute angles for which the expressions are defined
(cosec θ – cot θ)^2 = (1-cos theta)/(1 + cos theta)
Ex. 8.4 | Q 5.02 | Page 193
Prove the following identities, where the angles involved are acute angles for which the expressions are defined
cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A
Ex. 8.4 | Q 5.03 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined
(tantheta)/(1-cottheta) + (cottheta)/(1-tantheta) = 1+secthetacosectheta
Ex. 8.4 | Q 5.04 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(1+ secA)/sec A = (sin^2A)/(1-cosA)
[Hint : Simplify LHS and RHS separately]
Ex. 8.4 | Q 5.05 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA
Ex. 8.4 | Q 5.06 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
sqrt((1+sinA)/(1-sinA)) = secA + tanA
Ex. 8.4 | Q 5.07 | Page 194
Prove the following identities, where the angles involved are acute angles for which the expressions are defined
(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta
Ex. 8.3Ex. 8.4
NCERT solutions for Class 10 Mathematics Textbook chapter 8 - Introduction to Trigonometry
NCERT solutions for Class 10 Mathematics Textbook chapter 8 (Introduction to Trigonometry) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Class 10 Mathematics Textbook solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 10 Mathematics Textbook chapter 8 Introduction to Trigonometry are Introduction to Trigonometry, Introduction to Trigonometry Examples and Solutions, Trigonometric Ratios, Trigonometric Ratios of an Acute Angle of a Right-angled Triangle, Trigonometric Ratios of Some Specific Angles, Trigonometric Ratios of Complementary Angles, Trigonometric Identities, Proof of Existence, Relationships Between the Ratios, Trigonometric Ratios of Complementary Angles, Trigonometric Identities, Trigonometric Ratios of Complementary Angles, Trigonometric Identities.
Using NCERT Class 10 solutions Introduction to Trigonometry exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in NCERT Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 10 prefer NCERT Textbook Solutions to score more in exam.
Get the free view of chapter 8 Introduction to Trigonometry Class 10 extra questions for Class 10 Mathematics Textbook and can use Shaalaa.com to keep it handy for your exam preparation
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# HW6 - Section 6.1 1. F ( x ) is an anti derivative of f ( x...
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Unformatted text preview: Section 6.1 1. F ( x ) is an anti derivative of f ( x ) if F ( x ) = f ( x ). Taking the derivative we find F ( x ) = x 2 + 4 x- 1 = f ( x ) so that F ( x ) is an anti derivative of f ( x ) . 3. F ( x ) is an anti derivative of f ( x ) if F ( x ) = f ( x ). Taking the derivative we find F ( x ) = (1 / 2)(2 x 2- 1)- . 5 (4 x ) = 2 x 2 x 2- 1 = f ( x ) so that F ( x ) is an anti derivative of f ( x ) . 5. G ( x ) is an anti derivative of f ( x ) if F ( x ) = f ( x ). Taking the derivative we find G ( x ) = 2 = f ( x ) so that G ( x ) is an anti derivative of f ( x ) . The set of all anti derivatives of f ( x ) is found by adding an arbitrary constant to G ( x ) . Thus any anti derivative of f ( x ) is of the form 2 x + C . 7. G ( x ) is an anti derivative of f ( x ) if F ( x ) = f ( x ). Taking the derivative we find G ( x ) = x 2 = f ( x ) so that G ( x ) is an anti derivative of f ( x ) . The set of all anti derivatives of f ( x ) is found by adding an arbitrary constant to G ( x ) . Thus any anti derivative of f ( x ) is of the form x 2 + C . 9. R 6 dx = 6 x + C. (Rule 1) 11. R x 3 dx = 1 1+3 x 1+3 + C = 1 4 x 4 + C. (Rule 2) 13. R x- 4 dx = 1 1+(- 4) x 1+(- 4) + C =- 1 3 x- 3 + C. (Rule 2) 15. R x 2 / 3 dx = 1 1+(2 / 3) x 1+(2 / 3) + C = 3 5 x 5 / 3 + C. (Rule 2) 17. R x- 5 / 4 dx = 1 1+(- 5 / 4) x 1+(- 5 / 4) + C =- 4 x- 1 / 4 + C. (Rule 2) 19. R 2 x 2 dx = R 2 x- 2 dx = 2 R x- 2 dx = (2) 1 1+(- 2) x 1+(- 2) + C = (2)(- 1) x- 1 + C =- 2 x- 1 + C. (Rules 2 and 3) 21. R t dt = R t 1 / 2 dt = ( ) 1 1+(1 / 2) t 1+(1 / 2) + C = ( ) 2 3 t 3 / 2 + C = 2 3 t 3 / 2 + C. (Rules 2 and 3) 23. R (3- 2 x ) dx = R 3 dx- 2 R x dx = 3 x- (2) 1 1+1 x 1+1 = 3 x- x 2 + C. (Rules 1, 2, 3, and 4) 25. R ( x 2 + x + x- 3 ) dx = R x 2 dx + R x dx + R x- 3 dx = 1 1+2 x 1+2 + 1 1+1 x 1+1 + 1 1+(- 3) x 1+(- 3) + C = 1 3 x 3 + 1 2 x 2 +- 1 2 x- 2 + C. (Rules 2 and 4) 27. R 4 e x dx = 4 R e x dx = 4 e x + C. (Rules 3 and 5) 29. R 1 + x + e x dx = R 1 dx + R x dx + R e x dx = x + 1 1+1 x 1+1 + e x + C = x + 1 2 x 2 + e x + C. (Rules 1, 2, 4, and 5) 31. R 4 x 3- 2 x 2- 1 dx = 4 R x 3 dx- 2 R x- 2 dx- R 1 dx = (4) 1 1+3 x 1+3- (2) 1 1+(- 2) x 1+(- 2)- x + C = x 4 + 2 x- 1- x + C. (Rules 1, 2, 3, and 4) 33. R x 5 / 2 + 2 x 3 / 2- x dx = R x 5 / 2 dx + 2 R x 3 / 2 dx- R x dx = 1 1+(5 / 2) x 1+(5 / 2) +(2) 1 1+(3 / 2) x 1+(3 / 2)- 1 1+1 x 1+1 + C = (2 / 7) x 7 / 2 +(4 / 5) x 5 / 2- (1 / 2) x 2 + C. (Rules 2, 3, and 4) 35. R x + 3 x dx = R x 1 / 2 dx + 3 R x- 1 / 2 dx = 1 1+(1 / 2) x 1+(1 / 2) + (3) 1 1+(- 1 / 2) x 1+(- 1 / 2) + C = (2 / 3) x 3 / 2 + 6 x 1 / 2 + C. (Rules 2, 3, and 4) 37. R u 3 +2 u 2- u 3 u du = R u 3 3 u + 2 u 2 3 u- u 3 u du = R u 2 3 + 2 u 3- 1 3 du = (1 / 3) R u 2 du +(2 / 3) R u du- R 1 3 du = (1 / 3) 1 1+2 u 1+2 + (2 / 3) 1 1+1 u 1+1- (1 / 3) u + C = (1 / 9) u 3 + (1 / 3) u 2- (1 / 3) u + C....
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HW6 - Section 6.1 1. F ( x ) is an anti derivative of f ( x...
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Ingredient Math: Calculating minimum ingredient proportions
In a previous article here, I showed how to calculate the maximum proportion of a ingredient listed on a food label. I derived the following simple formula:
Maximum percentage of the Nth ingredient = (100 / N)
This time I’ll use similar logic to derive a formula for the minimum amount of an ingredient, based on only its order in the listing and the total number of ingredients.
Let’s start with the simple case of a product with two ingredients: “sugar, cocoa”.
With a little thought, we can see that sugar cannot be any less than 50%. If it was, then cocoa would have to be greater than 50% (since it equals 100% – sugar) and that is impossible based on the ordering rule that the most prominent ingredients, by weight, are listed first.
What about cocoa? Clearly it has no minimum since it could be 0.00001%, leaving the remaining 99.9999% to sugar.
Next let’s try a product with three ingredients: “sugar, cocoa, shredded coconut”.
Using similar logic, we will see the minimum amount of sugar is 33%, since any less than that would mean the other ingredients would have to have a higher proportion than sugar. If we assumed sugar was 20%, then the other two ingredients must add up to 80%, and therefore one of them must be present in at least 40% proportion – but that is greater than sugar which is listed first!
And the second ingredient? Just like the second ingredient in the two-ingredient example, this can be arbitrarily small. For example: 99.9% sugar, 0.09% cocoa, and 0.01% shredded coconut.
To generalize these results:
Minimum percentage of the 1st ingredient = (100 / T), where T is the number of total ingredients
Minimum percentage of the other ingredients = infinitesimally close to zero
As with the calculations of maximum proportion in my previous article, if we have additional information about the other ingredients we can use that to determine the minimum level of the other ingredients. For the same example, if we sifted and then weighed the shredded coconut (third ingredient), we could determine its overall proportion by weight. Lets say this is 10%. Then we could reason that cocoa, the second ingredient, takes up a minimum of 10% total weight since it must be equal or greater to the amount of shredded coconut. From this we could also conclude sugar (the first ingredient) is at most 80% of the total weight.
You can use nutritional values such as protein and fiber to infer how much of certain ingredients are present, especially if those have a large proportion of a certain element.
You may feel that this sort of calculation may seem unlikely to have any practical use, but if you find yourself saying “I eat so-and-so product because it contains a large amount of X, which is healthy”, then you can use it to see the minimum level of that ingredient you are consuming.
Another use is if you are trying to make a homemade version of something and you want to get a feel for the minimum and maximum amount of it in the product.
Ingredient Math – estimating worst (or best) case for ingredient proportions
Have you ever wondered how much of a certain ingredient is really present in a sweet, or any food product?
You probably know that ingredients are listed on food labels in order or prevalence, with the most predominant ingredient first. You may have even known this was determined by weight. But in this article I will discuss a method to get an estimate for the maximum of each ingredient’s percentage of total weight – just by using the ordered ingredient list.
To derive this formula, lets start with a very simple example, a product with just “coffee and sugar”. Since coffee is listed first we know it has higher or equal amount of total weight when compared to sugar.
Is there anything we can do to determine about how much the first ingredient, coffee, is really in the product? The answer is no because coffee could be almost 100% to almost 0% of the total weight, with sugar filling in the remaining space. (Actually, there is a trick to determine the amount here since the second ingredient is sugar, which I’ll discuss later in this article).
Well, if you think about it, there can’t be more than 50% sugar, by weight, since any more of that would mean there was more sugar than coffee, which we know is not the case.
So we’ve learned something important – that there is no more than 50% sugar in the product. This would apply to another second ingredient when there are two total ingredients.
What if there were three or more total ingredients? We would get the same result, because the other ingredients could be in trace amounts (practically 0%), so the “50% maximum for the second ingredient” rule would still apply.
What about the maximum amount of the third ingredient? Using the same logic you will see it cannot be above 33.3%, since any more of that would mean it is in greater proportion than the first and second ingredients. And for the forth ingredient you get a maximum, by weight, of 25%.
Turning this into a simple formula we get the following:
Maximum percentage of the Nth ingredient = (100 / N)
So for the 5th ingredient, you would get (100 / 5) = 20% maximum weight of that ingredient.
If you use formula along with the serving size you can determine the maximum weight of any of the ingredients per serving. Pretty handy if you want to minimize your intake of certain things.
If you want to take this to the next step, you can infer more information when one more more ingredients are a type of sugar. For example, if a product contains “coffee, sugar” and has 3 grams of sugar per 15 gram serving, then you know right away there is 20% sugar and 80% coffee in this product. Keep in mind that the grams of sugar listed includes any type of sugar, so if you have multiple ingredients which contain some type of sugar (even fruits) then the calculation gets a little trickier.
Besides knowing there is a certain percentage of sugar, you can use that to deduce information about other ingredients.
For example, if the imaginary product I just described had a third ingredient, say “coffee, sugar, vanilla”, then you would know that there is 20% or less vanilla because sugar is 20% or less. This assumes that there is no sugar in the vanilla, otherwise it would be harder to make any definitive conclusions.
Similarly, if you know how much protein is in each ingredient, you can figure out even more using the supplied protein in grams.
You can also leverage information about other ingredients to deduce additional information about the other ingredients. For example if a product had “milk, sugar, guar gum, vanilla”, you would know that the proportion of vanilla is much less than 25% since guar gum is typically used in relatively small doses. (I’ve tried overusing guar gum in homemade ice cream – its not pretty!)
I love thinking about food and ingredients from a methodical, logical point of view since it allows me to apply science to my everyday life.
References
http://www.fda.gov/Food/GuidanceRegulation/GuidanceDocumentsRegulatoryInformation/LabelingNutrition/ucm064880.htm |
# 3.7: Counting Objective: To find the counts of various combinations and permutations, as well as their corresponding probabilities CHS Statistics.
## Presentation on theme: "3.7: Counting Objective: To find the counts of various combinations and permutations, as well as their corresponding probabilities CHS Statistics."— Presentation transcript:
3.7: Counting Objective: To find the counts of various combinations and permutations, as well as their corresponding probabilities CHS Statistics
Warm-Up Alfred is trying to find an outfit to wear to take Beatrice on their first date to Burger King. How many different ways can he make an outfit out of this following clothes: Pants: Green, Baby Blue, Black, Grey Shirt: Red, Pink, Plaid, Blue, Lime Green Tie: Polka dot, Stripped
Fundamental Counting Principle For a sequence of two events in which the first event can occur in m ways and the second event can occur in n ways, the events together can occur a total of m n ways. Example: You are purchasing a new car. The possible manufacturers, car sizes, and colors are listed. Manufacturer: Ford, GM, Honda Car Size: compact, midsize Color: White, red, black, and green How many different ways can you select one manufacturer, one car size, and one color?
Fundamental Counting Principle (cont.) Example: The access code for a garage door consists of four digits. How many codes are possible if: Each digit can be used only once and not repeated? Each digit can be repeated? Each digit can be repeated but the first digit cannot be 8 or 9?
Factorial Rule Examples: How many ways can 5 people be seated on a bench? How many ways can a class of 50 be ranked by grades? To answer questions like these, we will use the factorial rule. Factorial Rule A collection of n different items can be arranged in order n! different ways. n! = n x (n – 1) x (n – 2) x (n – 3) x … 5! = 9! = 2! =
Factorial Rule (cont.) Examples: How many ways can 5 people be seated on a bench? How many ways can a class of 50 be ranked by grades?
Permutations Example: Forty-three sprinters race in a 5K. How many ways can they finish first, second, and third? Can we use the factorial rule? Why or why not?
Permutations (When all Items Are Different) Permutations: When r items are selected from n available items (without replacement). Therefore, the order matters. Calculate the following permutations:
Permutations (cont.) Example: Forty-three sprinters race in a 5K. How many ways can they finish first, second, and third?
Distinguishable Permutations
Distinguishable Permutations (cont.) Example: A building contractor is planning to develop a subdivision. The subdivision is to consist of 6 one-story houses, 4 two-story houses, and 2 split-level houses. In how many distinguishable ways can the houses be arranged?
Combinations Example: You are picking 3 different flavors to put on your banana split. You can choose from 25 different flavors. How many ways can this be done? Does the order matter here?
Combinations Combination Rule: When order does not matter, and we want to calculate the number of ways (combinations) r items can be selected from n different items. RECAP: When different orderings of the same items are counted separately, we have a permutation problem, but when different orderings of the same items are not counted separately, we have a combination problem.
Combinations (cont.) Calculate the following combinations: Example: You are picking 3 different flavors to put on your banana split. You can choose from 25 different flavors. How many ways can this be done? Example: You want to buy three different CDs from a selection of 5 CDs. How many ways can you make your selection?
Combinations (cont.) Example: A states department of transportation plans to develop a new section of interstate highway and receives 16 bids. The state plans to hire four of the companies. How many different ways can the companies be selected? Example: The manager of an accounting department want to form a three-person advisory committee from the 20 employees in the department. In how many ways can the manager form this committee?
Probability Using Permutation and Combination A student advisory board consists of 17 members. Three members serve as the boards chair, secretary, and webmaster. What is the probability of selecting at random the three members that will hold these positions? You have 11 letters consisting of one M, four Is, four Ss, and two Ps. If the letters are randomly arranged in order, what is the probability that the arrangement spells the word Mississippi?
Probability Using Permutation and Combination (cont.) Find the probability of being dealt five diamonds from a standard deck of playing cards? A food manufacturer is analyzing a sample of 400 corn kernels for the presence of a toxin. In this sample, three kernels have dangerously high levels of the toxin. If four kernels are randomly selected from the sample, what is the probability that exactly one kernel contains a dangerously high level of the toxin?
Probability Using Permutation and Combination (cont.) A jury consists of five men and seven women. Three are selected at random for an interview. Find the probability that all three are men?
Assignment 3.7 Practice
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# Frank Solutions for Class 9 Maths Chapter 9 Indices
Frank Solutions for Class 9 Maths Chapter 9 Indices is an important study material for students from an exam point of view. These solutions are designed by subject-matter experts at BYJU’S who have vast conceptual knowledge. Students can practise these solutions on a daily basis, which will help them analyse the important questions that will appear in the final examination.
Chapter 9 solutions consist of problems based on Indices. The solutions are curated in such a way that students can grasp the concepts effortlessly. These solutions are available both online and offline modes and can be used as per their requirements. Students who aim to score good marks in the final exams can download Frank Solutions for Class 9 Maths Chapter 9 Indices PDF from the link available below and practise regularly.
## Access Frank Solutions for Class 9 Maths Chapter 9 Indices
1. Evaluate the following:
(i) 60
(ii) (1 / 2)-3
(iii)
(iv)
(v) (0.008)(2 / 3)
(vi) (0.00243)(-3 / 5)
(vii)
(viii)
Solution:
(i) 60 = 1
(ii) (1 / 2)-3 = (2)3
We get,
(1 / 2)-3 = 8
(iii)
= 26
We get,
= 64
(iv)
= 36
We get,
= 729
(v) (0.008)(2 / 3) = (0.23)(2 / 3)
= (0.2)3 × 2 / 3
= (0.2)2
We get,
(0.008)(2 / 3) = 0.04
(vi) (0.00243)(-3 / 5) = {1 / (0.00243)(3 / 5)}
= {1 / (0.35)(3 / 5)}
= {1 / (0.3)3}
We get,
(0.00243)(-3 / 5) = (1 / 0.027)
(vii)
=
=
= 56 × 1 / 6
We get,
= 5
(viii)
= (64 / 27)(2 / 3)
= (4 / 3)3 × 2 / 3
= (4 / 3)2
We get,
= (16 / 9)
2. Evaluate the following:
(a) 94 ÷ 27(-2 / 3)
(b) 7-4 × (343)(2 / 3) ÷ (49)(- 1 / 2)
(c) (64 / 216)(2 / 3) × (16 / 36)(- 3 / 2)
Solution:
(a) 94 ÷ 27(-2 / 3) = {(3)2}4 ÷ {(3)3}(-2 / 3)
On further calculation, we get,
= (3)2 × 4 ÷ (3)3 × (- 2 / 3) [Using (am)n = amn]
= (3)8 ÷ (3)-2
= (3)8 – (-2) [Using am ÷ an = am – n]
= (3)8 + 2
= (3)10
This can be written as,
= (3)2 × 5
= {(3)2}5
= (9)5
We get,
= 59049
(b) 7-4 × (343)(2 / 3) ÷ (49)(- 1 / 2)
On calculating further, we get,
= 7-4 × (73)(2 / 3) ÷ (72)(- 1 / 2)
= 7-4 × 73 × 2 / 3 ÷ 72 × (- 1 / 2)
= 7-4 × 72 ÷ 7-1
= 7-4 + 2 – (-1) [Using am × an = am + n and am ÷ an = am – n]
= 7-4 + 2 + 1
We get,
= 7-1
= (1 / 7) [Using a-m = (1 / am)]
(c) (64 / 216)(2 / 3) × (16 / 36)(- 3 / 2)
On further calculation, we get,
= {(26)2 / 3 / (63)2 / 3} × {(24)-3 / 2 / (62)– 3 / 2}
= {(2)6 × 2 / 3 / (6)3 × 2 / 3} × {(2)4 × (-3 / 2) / (6)2 × (- 3 / 2)}
[Using (am)n = amn]
= {(2)2 × 2 / (6)2} × {(2)2 × (- 3) / (6)– 3}
= {(2)4 / (6)2} × {(2)-6 / (6)-3}
We get,
= {(2)4 / (6)2} × {(6)3 / (2)6}
[Using a-m = (1 / am)
= {(2)4 / (2)6} × {(6)3 / (6)2}
= (2)4 – 6 × (6)3 – 2
[Using am ÷ an = am – n]
= (2)-2 × (6)1
= (1 / 22) × 6
= (1 / 4) × 6
We get,
= (3 / 2)
3. Write each of the following in the simplest form:
(a) (a3)5 × a4
(b) a2 × a3 ÷ a4
(c) a1 / 3 ÷ a-2 / 3
(d) a-3 × a2 × a0
(e) (b-2 – a-2) ÷ (b-1 – a-1)
Solution:
(a) (a3)5 × a4 = (a)3 × 5 × a4
[Using (am)n = amn], we get,
= (a)15 × a4
= a15 + 4
[Using am × an = am + n]
We get,
= a19
(b) a2 × a3 ÷ a4 = a2 + 3 – 4
[Using am × an = am + n and am ÷ an = am – n]
We get,
= a1
= a
(c) a1 / 3 ÷ a-2 / 3 = a(1 / 3) – (- 2 / 3)
[Using am ÷ an = am – n]
We get,
= a(1 / 3) + (2 / 3)
= a(1 + 2) / 3
= a1
= 0
(d) a-3 × a2 × a0 = a-3 + 2 + 0
[Using am × an = am + n]
= a-1
We get,
= (1 / a)
(e) (b-2 – a-2) ÷ (b-1 – a-1)
This can be written as,
= (1/b2 – 1/a2) / (1/b – 1/a)
= {(1/b)2 – (1/a)2} / (1/b – 1/a)
= {(1/b + 1/a) (1/b – 1/a)} / (1/b – 1/a)
We get,
= (1/b + 1/a)
4. Evaluate the following:
(i) (23 × 35 × 242) / (122 × 183 × 27)
(ii) (43 × 37 × 56) / (58 × 27 × 33)
(iii) (122 × 75-2 × 35 × 400) / (482 × 15-3 × 525)
(iv) (26 × 5-4 × 3-3 × 42) / (83 × 15-3 × 25-1)
Solution:
(i) (23 × 35 × 242) / (122 × 183 × 27)
This can be written as,
= {23 × 35 × (23 × 3)2} / (22 × 3)2 × (2 × 32)3 × (33)
= (23 × 35 × 26 × 32) / (24 × 32 × 23 × 36 × 33)
= (29 × 37) / (27 × 311)
On further calculation, we get,
= (29 – 7 / 311 – 7)
= 22 / 34
We get,
= (4 / 81)
(ii) (43 × 37 × 56) / (58 × 27 × 33)
This can be written as,
= (22)3 × 37 – 3} / (58 – 6 × 27)
= (26 × 37 – 3) / (58 – 6 × 27)
On further calculation, we get,
= (26 × 34) / (52 × 27)
= {34 / (52 × 27 – 6)}
= {81 / (52 × 21)}
We get,
= (81 / 50)
(iii) (122 × 75-2 × 35 × 400) / (482 × 15-3 × 525)
This can be written as,
= (122 × 35 × 400 × 153) / (482 × 525 × 752)
= {(22 × 3)2 × (7 × 5) × (24 × 52) × (3 × 5)3} / {(24 × 3)2 × (3 × 52 × 7) × (3 × 52)2}
On calculating further, we get,
= (24 + 4 × 32+ 3 × 51 + 2 + 3 × 7) / (28 × 32 + 1 + 2 × 54 + 2 × 7)
= (28 × 35 × 56 × 7) / (28 × 35 × 56 × 7)
We get,
= 1
(iv) (26 × 5-4 × 3-3 × 42) / (83 × 15-3 × 25-1)
= (26 × 42 × 153 × 251) / (83 × 54 × 33)
This can be written as,
= {(26 × (22)2 × (3 × 5)3 × (52)1} / {(23)3 × 54 × 33}
= (26 + 4 × 33 × 53 + 2) / (29 × 33 × 54)
= (210 – 9 × 55 – 4)
= (2 × 5)
We get,
= 10
5. Simplify the following and express with positive index:
(a) 3p-2q3 ÷ 2p3q-2
(b) {(p-3)2 / 3}1 / 2
Solution:
(a) 3p-2q3 ÷ 2p3q-2
This can be written as,
= (3p-2q3) / (2p3q-2)
= (3 / 2) {(p-2 / p3) × (q3 / q-2)}
= (3 / 2) {(p-2 ÷ p3) × (q3 ÷ q-2)}
= (3 / 2) {(p– 2 – 3) × (q3 – (- 2))} [Using am ÷ an = am – n]
= (3 / 2) {(p-5) × (q5)}
= (3 / 2) {(1 / p5) × (q5)}
We get,
= (3q5 / 2p5)
(b) {(p-3)2 / 3}1 / 2
= p-3 × (2 / 3) × (1 / 2)
[Using (am)n = amn]
= p– 1
We get,
= (1 / p)
6. Evaluate the following:
(i) {1 – (15 / 64)}– 1 / 2
(ii) (8 / 27)– 2 / 3 – (1 / 3)– 2 – 70
(iii) 95 / 2 – 3 × 50 – (1 / 81) – 1 / 2
(iv) (27)2 / 3 × (8)– 1 / 6 ÷ (18)– 1 / 2
(v) (16)3 / 4 + 2 (1 / 2)-1 × 30
(vi)
Solution:
(i) {1 – (15 / 64)}– 1 / 2
On taking LCM, we get,
= {(64 – 15) / 64}– 1 / 2
= (49 / 64) – 1 / 2
= (64 / 49)1 / 2
We get,
= (8 / 7)
(ii) (8 / 27)– 2 / 3 – (1 / 3)– 2 – 70
= (27 / 8)2 / 3 – (1 / 3)-2 – 70
= (27 / 8)2 / 3 – (3)2 – 1
On further calculation, we get,
= (3 / 2)3 × 2 / 3 – 9 – 1
= (3 / 2)2 – 10
= (9 / 4) – 10
On taking LCM, we get,
= {(9 – 40) / 4}
= (- 31 / 4)
(iii) 95 / 2 – 3 × 50 – (1 / 81) – 1 / 2
On further calculation, we get,
= 32 × 5 / 2 – 3 × 1 – (81)1 / 2
= 35 – 3 – 92 × 1 / 2
= 243 – 3 – 9
We get,
= 231
(iv) (27)2 / 3 × (8)– 1 / 6 ÷ (18)– 1 / 2
This can be written as,
= 33 × 2 / 3 × {1 / (23 × 1 / 6)} ÷ (1 / 18)1 / 2
= (32) / (21 / 2) × (2 × 32)1 / 2
= (32 / 21 / 2) × 21 / 2 × 3
We get,
= 32 + 1
= 33
= 27
(v) (16)3 / 4 + 2 (1 / 2)-1 × 30
= 24 × 3 / 4 + 2 × 2 × 1
On further calculation, we get,
= 23 + 4
= 8 + 4
We get,
= 12
(vi)
= (1 / 22)1 / 2 + (0.1)-1 – 32
= (1 / 2) + (0.1)-1 – 32
= (1 / 2) + (1 / 0.1) – 9
= (1 / 2) + (10 / 1) – 9
= (1 / 2) + 1
On taking LCM, we get,
= {(1 + 2) / 2}
= (3 / 2)
7. Simplify the following:
(a) (27x9)2 / 3
(b) (8x6y3)2 / 3
(c) (64a12 / 27b6)– 2 / 3
(d) (36m-4 / 49n-2)– 3 / 2
(e) (a1 / 3 + a– 1 / 3) (a2 / 3 – 1 + a– 2 / 3)
(f) ÷
(g) {(am){m – (1 / m)}}(1 / m + 1)
(h) xm + 2n. x3m – 8n ÷ x5m – 60
(i) (81)3 / 4 – (1 / 32)– 2 / 5 + 81 / 3. (1 / 2)-1. 20
(j) (27/343)2/3 ÷ {1/ (625/1296)1/4} x 536 /
Solution:
(a) (27x9)2 / 3
This can be written as,
= (33x9)2 / 3
= (33)2 / 3(x9)2 / 3 [Using (a × b)n = an × bn]
On calculating further,
We get,
= (3)3 × 2 / 3(x)9 × 2 / 3 [Using (am)n = amn]
= (3)2x3 × 2
We get,
= 9x6
(b) (8x6y3)2 / 3
This can be written as,
= (23x6y3)2 / 3
= (23)2 / 3 (x6)2 / 3(y3)2 / 3 [Using (a × b)n = an × bn]
= (2)3 × 2 / 3(x)6 × 2 / 3(y)3 × 2 / 3 [Using (am)n = amn]
= (2)2(x)2 × 2(y)2
We get,
= 4x4y2
(c) (64a12 / 27b6)– 2 / 3
This can be written as,
= {(26a12) / (33b6)}– 2 / 3
= {26 × (- 2 / 3)a12 × (- 2 / 3)} / {33 × (- 2 / 3)b6 × (- 2 / 3)}
[Using (a × b)n = an × bn and (a / b)n = (an / bn)]
On further calculation, we get,
= (2– 4a– 8) / (3– 2b– 4)
= (32b4) / (24a8) [Using a-n = (1 / an)]
We get,
= (9b4 / 16a8)
(d) (36m-4 / 49n-2)– 3 / 2
This can be written as,
= {(62m-4) / (72n– 2)}– 3 / 2
= {62 × (- 3 / 2) m-4 × (- 3 / 2)} / {72 × (- 3 / 2) n– 2 × (- 3 / 2)}
[Using (a × b)n = an × bn and (a / b)n = (an / bn)]
On further calculation, we get,
= (6-3 m6) / (7– 3n3)
= (73m6) / (63n3)
[Using a– 1 = (1 / an)]
We get,
= (343m6) / (216n3)
(e) (a1 / 3 + a– 1 / 3) (a2 / 3 – 1 + a– 2 / 3)
= a1 / 3(a2 / 3 – 1 + a– 2 / 3) + a– 1 / 3(a2 / 3 – 1 + a– 2 / 3)
On simplification, we get,
= (a1 / 3 × a2 / 3 – a1 / 3 × 1 + a1 / 3 × a– 2 / 3) + (a– 1 / 3 × a2 / 3 – a– 1 / 3 × 1 + a– 1 / 3 × a– 2 / 3)
= {a(1 / 3) + (2 / 3) – a1 / 3 × 1 + a(1 / 3) + (- 2 / 3)} + {a(- 1 / 3) + (2 / 3) – a– 1 / 3 × 1 + a(- 1 / 3) + (- 2 / 3)}
[Using am × an = am + n]
= {a1 – a1 / 3 + a– 1 / 3} + {a1 / 3 – a– 1 / 3 + a– 1}
= {a – a1 / 3 + a– 1 / 3 + a1 / 3 – a– 1 / 3 + (1 / a)}
We get,
= {a + (1 / a)}
(f)
÷
This can be written as,
= (x4y2)1 / 3 ÷ (x5y– 5)1 / 6
On calculating further, we get,
= {x4 × (1 / 3) y2 × (1 / 3)} ÷ {x5 × (1 / 6) y– 5 × (1 / 6)}
[Using (am)n = amn]
= {x(4 / 3) y(2 / 3)} ÷ {x(5 / 6) y(- 5 / 6)}
= {x(4 / 3)y(2 / 3)} / {x(5 / 6) y(- 5 / 6)}
= x(4 / 3) – (5 / 6) y(2 / 3) – (- 5 / 6) [Using am ÷ an = am – n]
= x(1 / 2) y(3 / 2)
= x(1 / 2) (y3)(1 / 2)
[Using (am)n = amn]
= √x √y3
We get,
=
(g) {(am){m – (1 / m)}}(1 / m + 1)
= (a) m × {m – (1 / m)} × {1 / (m + 1)}
[Using (am)n = amn]
Now,
Consider,
m × {m – (1 / m)} × {1 / (m + 1)}
= (m2 – 1) × {1 / (m + 1)}
= m2 × {1 / (m + 1)} – 1 × {1 / (m + 1)}
= {m2 / (m + 1)} – {1 / (m + 1)}
= {(m2 – 1)} / {(m + 1)}
= {(m – 1) (m + 1)} / (m + 1)
We get,
= (m – 1)
Therefore, (a) m × {m – (1 / m)} × {1 / (m + 1)} = am – 1
(h) xm + 2n. x3m – 8n ÷ x5m – 60
= xm + 2n + 3m – 8n – 5m – (- 60)
[Using am × an = am + n and am ÷ an = am – n]
= xm + 2n + 3m – 8n – 5m + 60
We get,
= x– m – 6n + 60
(i) (81)3 / 4 – (1 / 32)– 2 / 5 + 81 / 3. (1 / 2)-1. 20
This can be written as,
= (34)(3 / 4) – (1 / 25)(- 2 / 5) + (23)(1 / 3). (1 / 2)– 1 × 1
[Using a0 = 1]
= 34 × (3 / 4) – {1 / 25 × (- 2 / 5)} + 23 × (1 / 3). (2)1
= 33 – {1 / 2– 2} + 21. (2)1
= 33 – 22 + 2(1 + 1)
[Using am × an = am + n and a– n = (1 / an)]
= 33 – 22 + 22
We get,
= 33
= 27
(j) (27/343)2/3 ÷ {1/ (625/1296)1/4} x 536 /
This can be written as,
= {33 / (73}2/3 ÷ [1/{54 / (24 x 34)}1/4] x (23 x 67) /
= (33/ 73)2/3 ÷ [1/ {54 / (24 x 34)}1/4] x (23 x 67) / (33)1/3
= (33 x 2/3 / 73 x 2/3) ÷ [1/ {54 x ¼ / (24 x ¼ x 34 x ¼)] x (23 x 67) / (33 x 1/ 3)
Using (am)n = am n
= (32 / 72) ÷ [1 / {51 / (21 x 31)}] x (23 x 67) / 31
= (32 / 72) ÷ {(21 x 31) / 51} x (23 x 67) / 31
= (32 / 72) x {51 / (21 x 31)} x (23 x 67) / 31
On further calculation, we get,
= 32 – 1 – 1 x 23 – 1 x 51 x 7-2 x 67
Using a am ÷ an = am – n
= 30 x 22 x 51 x 7-2 x 67
= 1 x 4 x 51x 7-2 x 67
= 1340 / 49
= 27.34
8. Simplify the following:
(i) (5x × 7 – 5x) / (5x + 2 – 5x + 1)
(ii) (3x + 1 + 3x) / (3x + 3 – 3x + 1)
(iii) (2m × 3 – 2m) / (2m + 4 – 2m + 1)
(iv) (5n + 2 – 6.5n + 1) / (13. 5n – 2. 5n + 1)
Solution:
(i) (5x × 7 – 5x) / (5x + 2 – 5x + 1)
On taking common terms, we get,
= {5x (7 – 1)} / {5x + 1 (5 – 1)}
= (5x – x – 1 × 6) / 4 [Using am ÷ an = am – n]
= (5– 1 × 6) / 4
= {6 / (5 × 4)}
We get,
= (3 / 10)
(ii) (3x + 1 + 3x) / (3x + 3 – 3x + 1)
On taking common terms, we get,
= {3x (3 + 1)} / {3x (33 – 3)}
= {4 / (27 – 3)}
= (4 / 24)
We get,
= (1 / 6)
(iii) (2m × 3 – 2m) / (2m + 4 – 2m + 1)
On taking common terms, we get,
= {2m (3 – 1)} / {2m (24 – 2)}
= {2 / (16 – 2)}
= (2 / 14)
We get,
= (1 / 7)
(iv) (5n + 2 – 6.5n + 1) / (13. 5n – 2. 5n + 1)
On taking common terms, we get,
= {5n (52 – 6 × 5)} / {5n (13 – 2 × 5)}
= (25 – 30) / (13 – 10)
We get,
= (- 5 / 3)
9. Solve for x:
(a) 22x + 1 = 8
(b) 3 × 7x = 7 × 3x
(c) 2x + 3 + 2x + 1 = 320
(d) 9 × 3x = (27)2x – 5
(e) 22x + 3 – 9 × 2x + 1 = 0
(f) 1 = px
(g) p3 × p– 2 = px
(h) p– 5 = (1 / px + 1)
(i) 22x + 2x + 2 – 4 × 23 = 0
(j) 9 x 81x = 1/ 27x – 3
(k) 22x – 1 – 9 × 2x – 2 + 1 = 0
(l) : 5x = 25: 1
(m) = (0.6)2 – 3x
(n) = (27– 1) / (125– 1)
(o) 9x + 4 = 32 × (27)x + 1
Solution:
(a) 22x + 1 = 8
This can be written as,
22x + 1 = 23
2x + 1 = 3
2x = 3 – 1
2x = 2
We get,
x = 1
(b) 3 × 7x = 7 × 3x
(7x / 7) = (3x / 3)
7x – 1 = 3x – 1 [Using am ÷ an = am – n]
7x – 1= 3x – 1 × 1
7x – 1 = 3x – 1 × 70
[Using a0 = 1]
x – 1 = 0
We get,
x = 1
(c) 2x + 3 + 2x + 1 = 320
This can be written as,
2x + 3 + 2x + 1 = 26 × 5
2x. 23 + 2x.21 = 26 × 5
On taking common terms, we get,
2x (23 + 21) = 26 × 5
2x (8 + 2) = 26 × 5
2x (10) = 26 × 5
2x (10 / 5) = 26
2x. 2 = 26
(2x.2) / 26 = 1
2x + 1 – 6 = 1 × 20
2x – 5 = 1 × 20
x – 5 = 0
We get,
x = 5
(d) 9 × 3x = (27)2x – 5
This can be written as,
32 × 3x = (33)2x – 5
32 × 3x = 33 × (2x – 5)
On further calculation, we get,
32 + x = 36x – 15
1 = (36x – 15) / (32 + x)
1 = 36x – 15 – 2 – x
30 = 35x – 17
5x – 17 = 0
5x = 17
We get,
x = (17 / 5)
(e) 22x + 3 – 9 × 2x + 1 = 0
This can be written as,
22x.23 – 9 × 2x + 1 = 0
Put 2x = t
Then,
22x = t2
So,
22x.23 – 9 × 2x + 1 = 0 becomes,
8t2 – 9t + 1 = 0
8t2 – 8t – 1t + 1 = 0
On taking common terms, we get,
8t (t – 1) – 1 (t – 1) = 0
(t – 1) = 0 or (8t – 1) = 0
t = 1 or t = (1 / 8)
2x = 1 or 2x = (1 / 23)
2x = 20 or 2x = 2-3
Hence,
x = 0 or x = – 3
(f) 1 = px
p0 = px [Using a0 = 1]
Therefore,
x = 0
(g) p3 × p– 2 = px
p3 + (- 2) = px [Using am × an = am + n]
p3 – 2 = px
p1 = px
Hence,
x = 1
(h) p– 5 = (1 / px + 1)
p-5 × px + 1 = 1
p– 5 + x + 1 = 1 [Using am × an = am + n]
px – 4 = p0
x – 4 = 0
We get,
x = 4
(i) 22x + 2x + 2 – 4 × 23 = 0
This can be written as,
22x + 2x + 2 – 22 × 23 = 0
22x + 2x.22 – 22 + 3 = 0 [Using am × an = am + n]
22x + 2x.22 – 25 = 0
22x + 2x. 4 – 32 = 0
Put 2x = t
Then, 22x = t2
22x + 2x.4 – 32 = 0 becomes,
t2 + 4t – 32 = 0
t2 + 8t – 4t – 32 = 0
On taking common terms, we get,
t (t + 8) – 4 (t + 8) = 0
t + 8 = 0 or t – 4 = 0
t = – 8 or t = 4
2x = – 8 or 2x = 4
2x = – 23 or 2x = 22
Now,
Consider second equation,
2x = 22
We get,
x = 2
(j) 9 x 81x = 1/ 27x – 3
This can be written as,
32 x 34x = 1/ 33 (x – 3)
32 x 34x = 1/ 33x – 9
Using (am)n = amn
32 x 34x x 33x – 9 = 1
32 + 4x + 3x – 9 = 1 x 30
On further calculation, we get,
2 + 4x + 3x – 9 = 0
7x – 7 = 0
7x = 7
x = 1
(k) 22x – 1 – 9 × 2x – 2 + 1 = 0
This can be written as,
22x. 2-1 – 9 × 2x. 2– 2 + 1 = 0
Let 2x = t,
So, 22x = t2
Then,
22x.2– 1 – 9 × 2x.2– 2 + 1 = 0 becomes,
(t2 / 2) – 9 × (t / 22) + 1 = 0
(t2 / 2) – (9t / 4) + 1 = 0
Taking LCM, we get,
2t2 – 9t + 4 = 0
2t2 – 8t – t + 4 = 0
2t (t – 4) – 1 (t – 4) = 0
t – 4 = 0 or 2t – 1 = 0
t = 4 or t = (1 / 2)
Hence,
2x = 4 or 2x = (1 / 2)
2x = 22 or 2x = 2– 1
Therefore,
x = 2 or x = – 1
(l)
: 5x = 25: 1
This can be written as,
(
/ 5x) = (25 / 1)
(
/ 5x) = (52 / 1)
= 52 × 5x
= 52 + x
x2 = 2 + x
x2 – x – 2 = 0
(x – 2) (x + 1) = 0
x – 2 = 0 or x + 1 = 0
Therefore,
x = 2 or x = – 1
(m)
= (0.6)2 – 3x
This can be written as,
{1 + (2 / 3)}1 / 2 = (6 / 10)2 – 3x
On taking LCM, we get,
(5 / 3)1 / 2 = (3 / 5)2 – 3x
(3 / 5)– 1 / 2 = (3 / 5)2 – 3x
(- 1 / 2) = 2 – 3x
-1 = 2 (2 – 3x)
– 1 = 4 – 6x
– 1 – 4 = – 6x
– 5 = – 6x
Hence,
x = (5 / 6)
(n)
= (27– 1) / (125– 1)
This can be written as,
(3 / 5)(x + 3) × (1 / 2) = {(33)-1 / (53)-1}
(3 / 5)(x + 3) / 2 = (3 / 5)– 3
(x + 3) / 2 = – 3
x + 3 = – 3 × 2
x + 3 = – 6
x = – 6 – 3
We get,
x = – 9
(o) 9x + 4 = 32 × (27)x + 1
This can be written as,
9x + 4 = 32 × (33)x + 1
32 (x + 4) = 32 × 33x + 3
32x + 8 = 32 + 3x + 3
Hence,
2x + 8 = 2 + 3x + 3
2x + 8 = 3x + 5
3x – 2x = 8 – 5
We get,
x = 3
10. Find the value of k in each of the following:
(i) = 2k
(ii) = xk
(iii) × = 3k
(iv) (1 / 3)– 4 ÷ 9 (- 1 / 3) = 3k
Solution:
(i)
= 2k
This can be written as,
8(1 / 3) × (- 1 / 2) = 2k
(23)(1 / 3) × (- 1 / 2) = 2k
2(- 1 / 2) = 2k
Therefore,
k = (- 1 / 2)
(ii)
= xk
This can be written as,
{(x2)1 / 3}1 / 4 = xk
On further calculation, we get,
(x2)1 / 12 = xk
x(2 / 12) = xk
x(1 / 6) = xk
Hence,
k = (1 / 6)
(iii)
×
= 3k
This can be written as,
{(32)(1 / 2)}– 7 × {(3)(1 / 2)}– 5 = 3k
3– 7 × 3(- 5 / 2) = 3k
3-7 – 5 / 2 = 3k
3(- 14 – 5) / 2 = 3k
3(- 19 / 2) = 3k
Therefore,
k = (- 19 / 2)
(iv) (1 / 3)– 4 ÷ 9 (- 1 / 3) = 3k
This can be written as,
(3– 1)– 4 ÷ (32)– 1 / 3 = 3k
34 ÷ 3(- 2 / 3) = 3k
34 – (- 2 / 3) = 3k
34 + 2 / 3 = 3k
3(14 / 3) = 3k
We get,
k = (14 / 3)
11. If a = 2(1 / 3) – 2(- 1 / 3), prove that 2a3 + 6a = 3
Solution:
Given
a = 2(1 / 3) – 2(- 1 / 3)
This can be written as,
a = 2(1 / 3) – {1 / 2(1 / 3)}
On taking cube on both sides, we get,
a3 = [2(1 / 3) – {1 / 2(1 / 3)}]3
On further calculation, we get,
a3 = 2 – (1 / 2) – 3 [2(1 / 3) – {1 / 2(1 / 3)}]
a3 = {(4 – 1) / 2} – 3a
a3 = (3 / 2) – 3a
We get,
2a3 + 6a = 3
12. If x = 3(2 / 3) + 3(1 / 3), prove that x3 – 9x – 12 = 0
Solution:
Given
x = 32 / 3 + 31 / 3
x3 = 32 + 3 + 3 × 32 / 3 × 31 / 3 (32 / 3 + 31 / 3)
x3 = 9 + 3 + 3 × 32 / 3 + 1 / 3 (x)
x3 = 12 + 9x
We get,
x3 – 9x – 12 = 0
13. If and abc = 1, prove that x + y + z = 0
Solution:
Let
= k
So,
a1 / x = k, b1 / y = k, c1 / z = k
We get,
a = kx, b = ky , c = kz
Also, given that,
abc = 1
kx × ky × kz = 1
kx + y + z = k0
We get,
x + y + z = 0
14. If ax = by = cz and b2 = ac, prove that y = 2xz / (z + x).
Solution:
Let ax = by = cz = k
So,
a = k1 / x, b = k1 / y, c = k1 / z
Also given that,
b2 = ac
k2 / y = k1 / x × k1 / z
k2 / y = k1 / x + 1 / z
(2 / y) = (1 / x) + (1 / z)
(2 / y) = (z + x) / zx
We get,
y = {2zx / (z + x)}
15. Show that: {1 / (1 + ap – q)} + {1 / (1 + aq – p)} = 1
Solution:
Consider LHS of the equation, i.e,
{1 / (1 + ap – q)} + {1 / (1 + aq – p)}
On taking LCM, we get,
= {(1 + aq – p) + (1 + ap – q)} / (1 + ap – q) (1 + aq – p)
= (2 + a– (p – q) + ap – q) / (1 + ap – q) (1 + a– (p – q))
= 2 + a– (p – q) + ap – q / (1 + a– (p – q) + ap – q + ap – q. a– (p – q))
= 2 + a– (p – q) + ap – q / (1 + a– (p – q) + ap – q + ap – q – p + q
= 2 + a– (p – q) + ap – q / (1 + a– (p – q) + ap – q + a0)
= 2 + a– (p – q) + ap – q / (1 + a– (p – q) + ap – q + 1)
= 2 + a– (p – q) + ap – q / (2 + a– (p – q) + ap – q)
We get,
= 1
= RHS
Therefore,
LHS = RHS
Hence, proved |
## Four ways of looking at a Pentagon, part 1
Suppose there is a pentagon. How do we know the relationship between the edges?
Well, we can imagine the Pentagon is made of Triangles, and look at the Spreads between the lines that make up those triangles.
Pentagon halved and divided into 5 equal triangles
Now if we imagine the Pentagon triangle’s lines form spokes, as though the Pentagon were inside a wheel, we can imagine there is a relationship between those lines orientation. We could think of the Angle between the lines… but what about Rational Trigonometry, and it’s Spread? I’m going to use my amateur reading of Wildberger’s Divine Proportions for inspiration here.
Let’s imagine the lines are all related by the Spread between them. For example we could imagine splitting the pentagon in half, then splitting that into 5 triangles, and each triangle has one point that forms the axle of the “wheel”. Then, each axle point has two lines coming out, and they each have the same spread, s.
The lines of the Pentagon’s triangles have the same spread between them, s
But how does this help us calculate this relationship as a number? It can help us if we imagine that we can group the triangles together and look for patterns in the Spread equations.
We can focus on only a few at a time, for example two triangles together form a new spread, s2
The tricky bit is that Spreads do not add simply like Angles do. For example a 30 degree angle has a Spread of $\dfrac{1}{4}$. Doubling that angle gives us 60 degrees, but the Spread corresponding to a 60 degree angle is $\dfrac{3}{4}$. Clearly $\dfrac{3}{4}$ is not the same as two times $\dfrac{1}{4}$.
What you do instead is look at the Spread Polynomials. Two lines separated by $s$ each form a new spread $s_2$, which is calculated as follows:
$s_2=4s_1(1-s_1)$
This is called a Spread Polynomial. In fact, it is the Spread Polynomial called $S_2$ in Wildberger’s book on page 104, it is basically used for doubling a spread and shows up in all kinds of patterns involving spreads.
So if we had a spread of $\dfrac{1}{4}$ for 30 degrees, and we double it, we get a spread of
$s_2=4\dfrac{1}{4}(1-\dfrac{1}{4})$
$s_2=(1-\dfrac{1}{4})$
$s_2=\dfrac{3}{4}$
That is the Spread corresponding to 60 degrees, which makes sense as 60 is double the angle of 30.
Spread polynomials are not just analogues for doubling an angle… there is a spread polynomial $S_3$ for tripling an angle, $S_4$ for quadrupling, and on and on. But they are not quite as simple as that.
The tricky bit, the interesting bit, is that each Spread actually represents two different angles. The spread of $\frac{1}{4}$ actually represents both the 30 degree angle and the 150 degree angle. In the angle system, these angles are distinct, but in the Spread system, they are the same number. $\frac{1}{4}$. We are just thinking about the lines themselves and their separation, and there is only one number that describes that separation.
So when we are adding Spreads, maybe we can think about it as if we are adding multiple different possibilities of angles, at the same time. For example, if I say I’m “adding” Spread $\frac{1}{4}$ to Spread $\frac{1}{4}$, does that mean I’m adding a 30 degrees angle to another 30 degrees angle to get 60? Or, does it mean I’m adding a 150 degree angle to a 30 degree angle to get 180 degrees?
Let’s take a closer look at the “Equal Spreads” theorem of Divine Proportions on page 94. It shows that when calculating the combination of two equal spreads, $r$, you basically get two answers:
$r=0, r=4s(1-s)$
For our little example above of 150 degrees + 30 degrees, that makes sense, because 180 degrees is represented by a Spread of $0$. So adding spreads of $\frac{1}{4}$ is actually like adding either 30+30 to get 60 (spread $\frac{1}{4}$) or 30+150 to get 180 (spread $0$).
What of 150+150? Yes, that makes 300, which is basically 60, which corresponds to $\frac{3}{4}$, which is the $r=4s(1-s)$ solution as we calculated above.
But what if we go for the tripling of a spread?
As shown on page 101 of Divine Proportions, the calculation of a Spread equation for combining three equal spreads, when derived using the Triple Spread Formula, actually has two solutions, $r$.
$r=s, r=s*(3-4s)^2$
That means that when you are presented with three equal spreads, and try to find the spread of their combo, you can get multiple answers. Imagine again our example of 30 degree angle, or two lines separated by a spread of $\frac{1}{4}$. What would ‘triple’ that angle bring? In angle based trigonometry, it brings you 90 degrees. If you use the Three Equal Spread formula … you get this:
$r=\frac{1}{4}, r=\frac{1}{4}(3-4\frac{1}{4})^2$
$r=\frac{1}{4}, r=\frac{1}{4}(2)^2$
$r=\dfrac{1}{4}, r=1$
Well, $r=1$ makes sense… that is a right angle in Spread land. Perpendicular. 90 degrees. But what about this $r=\frac{1}{4}$ business?
This is where my amateur understanding is going to have to speculate a bit. From what I can gather, $\frac{1}{4}$ is kind of like a special spread solution, a bit like 0 was for doubling the spread in the example above. I get this impression from Spread polynomials, rotations and the butterfly effect, by S. Goh and N.J. Wildberger where they describe The Special Spreads.
$0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1$
If you think about adding 150 degrees to a 30 degree angle to another 150 degree angle… you wind up with 180 degrees plus 150, which is basically 330 degrees. But 330 degrees is basically describing two lines that are at a 30 degree angle from each other. Again, this spread would be $\frac{1}{4}$.
In that sense, maybe you can think of combining three spreads of $\frac{1}{4}$ and getting a result of $\frac{1}{4}$. An addition of two 150 degree angles and a single 30 degree angle gives you 330, which is a spread of $\frac{1}{4}$. An addition of two 30 degree angles and a single 150 degree angle would give you an angle of 210 degrees, which is again a 30 degree separation, which gives us a spread of $\frac{1}{4}$ again.
I could be wrong about all this, but it might be worth thinking about as a geometrical interpretation of “why” trying to use Spread polynomials to “Add angles” doesn’t work exactly the same way as you might think, if you grew up thinking about adding angles together.
It’s almost like spreads are some sort of simple version of Quantum Computing where you calculate multiple solutions to multiple possibilities at the same time.
But what about Pentagons again? There we have 5 spreads! How to combine those?
Well there are multiple ways. Let’s first consider Wildberger’s Divine Proportions, page 166, Exercise 14.3, we can imagine that all five spreads together form a spread of 0. This means solving this equation for $S_5$:
$s(5-20s+16s^2)^2=0$
I’m sure there is a way to do this by hand, but I was never good at factoring polynomials by hand. Fortunately in the modern day there is something called Symbolic Mathematics and computers that can manipulate it. Here is a simple program using the Python language and an add-on for it called SymPy .
>>> from sympy import *
>>> s=symbols(‘s’)
>>> print solve( s*(5-20*s+16*s**2)**2, s )
[0, -sqrt(5)/8 + 5/8, sqrt(5)/8 + 5/8]
Cool! That is the same answer that Wildberger gives in the book, if we assume that 0 here is a “trivial zero” of the equation. Not bad for a three line program. |
NCERT Math Magic Solutions for Class 2 Maths Chapter 12 Give and Take
# NCERT: Math-Magic Solutions for Class-2, Chapter-12:Give and Take
Give and Take is the extension of addition and subtraction concept and its use in real life.
This chapter have exercises on
Adding and subtracting 2-digit numbers Missing numbers in addition and subtraction word problems on addition and subtraction
Adding and subtracting 2-digit numbers Missing numbers in addition and subtraction word problems on addition and subtraction
The solutions for Math-Magic Chapter-12 have been created and verified by our experienced subject matter experts, according to the CBSE syllabus and guidelines of NCERT. For practice, our subject matter experts have created very interactive, activity-based, and Image-based worksheets on these topics to enhance learning.
Chapter 12: Give and Take
Question 1:
Now find how many necklaces and loose beads the other children take.
Razia- 12 = 10 + 2 = 1 ten + 2 ones
Reema- 17 = 10 + 7 = 1 ten + 7 ones
Aarif- 24 = 20 + 4 = 2 tens + 4 ones
Sonu- 35 = 30 + 5 = 3 tens + 5 ones
Simar- 31 = 30 + 1 = 3 tens + 1 one
Tens are the number of necklaces, and ones are the number of loose beads. Therefore, the correct answer is:
Question 2: How many beads are taken by Razia and Reema altogether?
Answer:Since Razia has taken 12 beads and Reema has taken 17 beads, both of them have altogether taken
12 + 17 = 29
Therefore, Razia and Reema have taken 29 beads altogether.
Practice Time
Question 3: How many beads are taken by Razia and Sonu?
___________beads are taken by Razia and Sonu.
Answer:The total number of beads taken by both Razia and Sonu are 12 + 35.
Therefore, 47 beads are taken by Razia and Sonu.
Add by writing and also without writing
Question 4: How many beads do they have together?
A)
B)
A)
Therefore, Sonu has 11 more beads than Aarif.
B)
Therefore, Aarif has 12 more beads than Razia.
Practice Time
Question 6: Tanisha has 17 pencils. Siya has 25 pencils. How many pencils are there in all?
Therefore, there are 42 pencils altogether.
Question 7: In Muneeza's class, there are 13 English story books and 22 Hindi story books. How many story books are there in all?
Answer: To find the total number of story books, add 13 and 22.
Therefore, there are 35 story books altogether.
Question 8: Sakshi had 23 fruits. She ate 15 fruits. How many fruits are left?
Answer: To find the number of fruits left, subtract 15 from 23.
Therefore, 8 fruits are left.
Question 9: Daljeet has 35 marbles. Arvind has 25 marbles. How many marbles do they have in all?
Therefore, there are 60 marbles altogether.
Question 10: Nisha has 32 bangles. Sukhi has 16 bangles. How many more bangles does Nisha have?
Answer: To find how many more bangles Nisha has more than compared to Sukhi, subtract 16 from 32.
Therefore, Nisha has 16 marbles more than Sukhi.
Venkatesha's Canteen
Question 11: Help Venkatesha to make the bills.
Answer: The total bill for Dosa and Uthappam is:
23 + 28 = Rs 51
The total bill for Idli and Coffee is:
15 + 8 = Rs 23
The total bill for Dahi Vada and Chilli Rice is:
25 + 18 = Rs 43
The total bill for Soup and Noodles is:
27 + 15 = Rs 42
Question 12: I have to pay 31 rupees, but I have Rs 40.
I will get back ________ rupees.
Answer: To find the amount to be received back, subtract 31 from 40.
Therefore, you will get back 9 rupees.
Practice Time
Question 13: Shekhar has 32 rupees. He bought a ball for 17 rupees. How much money is left with him?
Answer: To find the amount left with Shekhar, subtract 17 from 32.
Therefore, 15 rupees is left with Sekhar.
Question 14: Soni bought biscuits for 24 rupees and a packet of chips for 16 rupees. How much money will she pay? Try doing it without writing!
Question 15: Fantoosh had 64 rupees. He spent 39 rupees at the fair. How much money is left with him?
Answer: To find the amount left with Fantoosh, subtract 39 from 64.
Therefore, Fantoosh has 25 rupees left with him after spending the money at the fair.
• NCERT Solutions for Class 2 Maths
• - |
# Question Video: Simplifying Algebraic Expressions with Negative Exponents Using Laws of Exponents Mathematics
Simplify π₯β· Γ π₯β»β΅ Γ π₯β΄, where π₯ β 0.
01:24
### Video Transcript
Simplify π₯ to the seventh times π₯ to the negative fifth times π₯ to the fourth, where π₯ is not equal to zero.
We can use the rule where π₯ to the π times π₯ to the π is equal to π₯ to the π plus π. So when we multiply things with like bases, we add their exponents. So we need to take this, keep our like base cause theyβll have a base of π₯, and then add our exponents: seven plus negative five plus four, which gives us π₯ to the sixth.
And we get π₯ to the sixth. Now there is another way to do this. It may take a little bit longer, but letβs give it a try. So in the original problem, we have π₯ to the negative fifth power. When you have a negative exponent and itβs on a numerator, we move it to the denominator and make it positive.
Or vice versa, if it was a negative exponent on the denominator, we can move it up to the numerator. So what we can do, keep π₯ to the seventh and π₯ to the fourth on the numerator but move the π₯ to the negative fifth. So just like we did before, we need to add the seven and four together.
And seven plus four, and we get π₯ to the 11th. Now when we divide, we need to subtract our exponents. And 11 minus 5 gives us π₯ to the sixth. So once again π₯ to the sixth is our final answer. |
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# A boy stands at $78.4m$ from a building and throws a ball which just enters a window $39.2m$ above the ground. Calculate the velocity of projection of the ball.
Last updated date: 13th Sep 2024
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Hint: In the question they have given maximum height and range of projection for the body which is there in projectile motion. By using the given data we will find the angle of projection then substituting in the equation of range of projection we will find the initial velocity of the body.
Formulas used:
Maximum height, ${H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ ……………..$\left( 1 \right)$
Range of projection, $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ ……………..$\left( 2 \right)$
Complete step-by-step solution:
Given:
Range of projection, $R = 78.4m + 78.4m = 156.8m$
Maximum height , ${H_{\max }} = 39.2m$
Take, acceleration due to gravity , $g = 9.8m{s^{ - 2}}$
Using equation $\left( 1 \right)$
That is, ${H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
$39.2 = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2 \times g}}$ …………… $\left( 3 \right)$
Using equation $\left( 2 \right)$
That is, $R = \dfrac{{{u^2}\sin 2\theta }}{g}$
$156.8 = \dfrac{{{u^2}\sin 2\theta }}{g}$ …………………$\left( 4 \right)$
$156.8 = \dfrac{{{u^2}2\sin \theta \cos \theta }}{g}$ ……………. $\left( 5 \right)$ $\left[ {\because \sin 2\theta = 2\sin \theta \cos \theta } \right]$
Divide equation $\left( 3 \right)$ and equation $\left( 5 \right)$
$\dfrac{{39.2}}{{156.8}} = \dfrac{{\dfrac{{{u^2}{{\sin }^2}\theta }}{{2 \times g}}}}{{\dfrac{{{u^2}2\sin \theta \cos \theta }}{g}}}$
$4 = \dfrac{{\dfrac{{\sin \theta }}{2}}}{{\dfrac{{2\cos \theta }}{1}}}$
Therefore, $\tan \theta = \dfrac{4}{4}$
$\tan \theta = 1$
$\theta = {\tan ^{ - 1}}1$
$\theta = {45^ \circ }$
Substituting in equation $\left( 4 \right)$ we get
$156.8 = \dfrac{{{u^2}\sin 90}}{{9.8}}$
$u = \sqrt {1536.64}$
Therefore, $u = 39.2m{s^{ - 1}}$
Note: Projectile motion is the form of motion experienced by a launched body that is motion of a body which is projected or thrown into the air with the angle made by the object with respect to the ground or x-axis. |
# Math Insight
### Approximating a nonlinear function by a linear function
Approximating a nonlinear function by a linear function.
Linear functions are nice and easy to deal with. But, in many cases and applications, one doesn't have a linear function to work with. Instead, one has a nonlinear function. One part of calculus can be viewed as approximating nonlinear functions by linear functions.
#### The secant line
How can we approximate nonlinear functions by linear functions? One way is through the secant line, which is a line through two points on the graph of the function.
Let's denote the function by $f$. To form the secant line, pick two values of $x$, say $x=x_0$ and $x=x_1$. Draw a line through $(x_0,f(x_0)$ and $(x_1,f(x_1))$. This line is the secant line.
The secant line approximates the function near those two points. The slope of the secant line gives the average rate of change. The slope of the secant line is \begin{align*} \text{average rate of change} = \frac{\Delta f}{\Delta x} = \frac{f(x_1)-f(x_0)}{x_1-x_0}. \end{align*}
If $f$ were a linear function, its slope wouldn't depend on the choice of $x_0$ and $x_1$. For nonlinear $f$, one will get different slope as change values of $x$. In the below applet, you can change the values of $x_0$ and $x_1$ and see how the secant line changes. In the applet, we've relabeled $x_1$ as $x_0+\Delta x$, and you change $x_1$ by changing the difference $\Delta x$ between $x_0$ and $x_1$. In terms of $\Delta x$, the slope of the secant line is \begin{align*} \text{average rate of change} = \frac{\Delta f}{\Delta x} = \frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}. \end{align*}
The slope of a secant line. The line (in red) passing through two points (large blue and smaller cyan) on the graph of the function $f$ (in green) is a secant line. The large blue point is at the location $(x_0,f(x_0))$, which you can change by dragging the point or typing in a value for $x_0$. The smaller cyan point is at the location $(x_0+\Delta x,f(x_0+\Delta x))$, which you can change by dragging the point. You can also change $\Delta x$ by typing in a value or changing the length of the purple $\Delta x$ line segment (drag small cyan point). You cannot change $\Delta y$ directly, as it is calculated as $\Delta y = f(x_0+\Delta x)-f(x_0)$. The slope of the secant line is $\frac{\Delta y}{\Delta x}$. As you let $\Delta x$ approach zero, the two points become closer together, and the secant line becomes closer to the tangent line of the graph of $f$. However, if you set $\Delta x=0$, then the secant line is not defined, and the slope $\frac{\Delta y}{\Delta x}=\frac{0}{0}$ is also not defined. However, if $\Delta x$ is very small, but not zero, the secant line becomes very close to the tangent line, which can be thought of as the limit of the secant line as $\Delta x$ approaches zero. If you click the “show limit for $\Delta x=0$” check box, then when you enter $\Delta x=0$, the applet instead shows the limiting tangent line. The slope of the tangent line, denoted by $\frac{dy}{dx}$ is the limit $\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}$ and is also displayed. You can use the buttons at the top to zoom in and out as well as pan the view.
#### The tangent line
The secant line is a good approximation of the average rate that $f$ changes over the period between $x_0$ and $x_1$. However, the secant line isn't a good approximation of $f$ when we zoom in on the point $(x_0,f(x_0))$. As we zoom in, we see that the secant line and the graph of $f$ match at the point $(x_0,f(x_0))$ but then move in different directions. The graph of $f$ and the secant line have different slopes.
To make the secant line a better approximation for the zoomed-in view, we need to shrink the $\Delta x$ used to calculate the secant line. If you make $\Delta x$ smaller and smaller in the above applet, the secant line becomes a better and better approximation for the function right near the value of $x_0$.
What if we want the best possible approximation of $f$ right near the point $x_0$? Unfortunately, the solution cannot be as simple as setting $\Delta x$ equal to zero. If you make $\Delta x$ zero, then the secant line does not exist. When $\Delta x=0$, the two points $(x_0,f(x_0))$ and $(x_0+\Delta x, f(x_0+\Delta x))$ are the same point and therefore do not determine a line. The formula for the slope is also not defined. If we plug in $\Delta x=0$, then the change in $f$, $\Delta f$ is also zero. The slope of the secant line would be $$\frac{\Delta f}{\Delta x}=\frac{0}{0}=\text{undefined}.$$
To keep the slope and the secant line defined, we need to keep $\Delta x$ from being exactly zero. On the other hand, we want to make $\Delta x$ very small, so that we can match the behavior of $f$ right near the point $x_0$. The solution to this problem is to use a limit, one of the fundamental notions underlying calculus.
Right now, we won't make the notion of a limit precise, but just give the general idea. The idea is that, although we can't make $\Delta x$ exactly equal to zero, we can make it smaller and smaller. As we make $\Delta x$ very close to zero, the secant line matches the graph of the function around the point $x_0$ very well.
The limit is the idea of making $\Delta x$ arbitrarily small. As you can explore with the applet, as you start to make $\Delta x$ very small, such as $\Delta x = 0.01$, $\Delta x = 0.001$, or $\Delta x = 0.0001$, the secant line and its slope $\frac{\Delta f}{\Delta x}$ change very little. In fact, the slope is approaching a particular value, and the secant line is approaching a particular line, as $\Delta x$ gets close zero.
The value that the slope $\frac{\Delta f}{\Delta x}$ is approaching as $\Delta x$ gets closer and closer to zero is called the limit of $\frac{\Delta f}{\Delta x}$ as $\Delta x$ approaches zero, which we write as \begin{align*} \lim_{\Delta x \to 0} \frac{\Delta f}{\Delta x} &= \lim_{\Delta x \to 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}. \end{align*} We denote this slope by $\frac{df}{dx}$ and refer to it as the instantaneous rate of change of the function at $x_0$ \begin{align*} \text{instantaneous rate of change} &= \frac{df}{dx} = \lim_{\Delta x \to 0} \frac{\Delta f}{\Delta x}\\ &= \lim_{\Delta x \to 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}. \end{align*}
The line through the point $(x_0,f(x_0))$ with slope $\frac{df}{dx}$ is called a tangent line to the graph of $f$. The slope $\frac{df}{dx}$ is the derivative of $f$ at $x=x_0$. To emphasize that this slope is computed at the point $x=x_0$, we often write the derivative as $$\frac{df}{dx}\bigg|_{x=x_0}.$$ We also denote the derivative by $f'(x_0)$. Both notations refer to the same thing. The derivative is the slope of the tangent line, but we also say that the slope of the function itself is the derivative.
The derivative of $f$ depends on the point $x_0$ where it is evaluated. A nonlinear function will have a different slope at different points. To explore the derivative and tangent line of $f$ in the above applet, check the “show limit for $\Delta x=0$” checkbox. Then, move the point around to see how the tangent line and its slope vary as you change $x_0$. |
PDF chapter test
When it comes to solving the algebraic equation, identities play a vital role to solve them. In this lesson, we are going to explore the geometrical proof of identities.
Let us first see the varies identities which we will discuss here:
1. $\left(x+a\right)\left(x+b\right)\phantom{\rule{0.147em}{0ex}}={x}^{2}+x\left(a+b\right)+\mathit{ab}$
2. $$(a + b^2) = a^2 + 2ab + b^2$$
3. $$(a - b^2) = a^2 - 2ab + b^2$$
4.
Let us take each identity one by one and discuss the proof of that identity.
Identity $$1$$: $\left(x+a\right)\left(x+b\right)\phantom{\rule{0.147em}{0ex}}={x}^{2}+x\left(a+b\right)+\mathit{ab}$.
Let us construct a rectangular diagram of four regions. One region is square-shaped with a dimension of $$4 × 4$$ (Orange). Also, the other three regions are rectangle in shape with dimensions $$4×2$$ (Blue), $$3 × 4$$ (Yellow) and $$3 × 2$$ (Green).
By observing the above rectangle, we can notice that:
$$\text{Area of the bigger rectangle}$$ $$=$$ $$\text{Area of a square (Orange)}$$ $$+$$ $$\text{Area of three rectangles}$$
$$(4 + 3)$$ $$(4 + 2)$$ $$=$$ $$(4 × 4) + (4 × 2) + (3 × 4) + (3 × 2)$$
Now, we simplify the LHS and RHS of the above expression.
LHS $$=$$ $$(4 + 3)$$ $$(4 + 2)$$ $$=$$ $$7×6 = 42$$
RHS $$=$$ $$(4 × 4) + (4 × 2) + (3 × 4) + (3 × 2)$$
RHS $$=$$ $$16+8+12+6 = 42$$
Therefore, LHS $$=$$ RHS
Similarly, if we use the variables in this case instead of number we get:
Let one side of a rectangle be $$(x +a)$$, and the other side be $$(x + b)$$ units.
Then, $$\text{the total area of the rectangle }AEGI$$ $$=$$ $$\text{length } \times \text{breadth}$$ $$=$$ $$(x+a)(x+b)$$………….$$(1)$$
$$\text{The area of the rectangle }AEGI$$ $$=$$ $$\text{The area of the square }ABCD$$ $$+$$ $$\text{The area of the rectangle }BEFD$$ $$+$$ $$\text{The area of the rectangle }DFGH$$ $$+$$ $$\text{The area of the rectangle }CDHI$$.
$$\text{The area of the rectangle AEGI}$$ $$=$$ ${x}^{2}+\mathit{ax}+\mathit{ab}+\mathit{bx}$
$={x}^{2}+x\left(a+b\right)+\mathit{ab}$……………..$$(2)$$
From the equation, $$(3)$$ and $$(4)$$ we get $\left(x+a\right)\left(x+b\right)\phantom{\rule{0.147em}{0ex}}={x}^{2}+x\left(a+b\right)+\mathit{ab}$.
Therefore, $\left(x+a\right)\left(x+b\right)\phantom{\rule{0.147em}{0ex}}={x}^{2}+x\left(a+b\right)+\mathit{ab}$ is a identity.
Example:
Simplify the expression $\left(a+3\right)\left(a+11\right)$ using the identity $\left(x+a\right)\left(x+b\right)\phantom{\rule{0.147em}{0ex}}={x}^{2}+x\left(a+b\right)+\mathit{ab}$.
The expression is $\left(a+3\right)\left(a+11\right)$.
Now write the given expression $\left(a+3\right)\left(a+11\right)$ with respect to the given identity $$(x+a)(x+b)$$ $$=$$ $$x^2+(a+b)x+ab$$.
$\begin{array}{l}\left(x+a\right)\left(x+b\right)=\left(a+3\right)\left(a+11\right)\\ \\ =\left({a}^{2}+a\left(3+11\right)+3×11\right)\end{array}$
Now simplify the expression.
$\left({a}^{2}+14a+33\right)$.
Therefore, $\left(a+3\right)\left(a+11\right)$ $$=$$ $\left({a}^{2}+14a+33\right)$. |
# Wrapping spheres around spheres Mark Behrens (Dept of Mathematics)
## Presentation on theme: "Wrapping spheres around spheres Mark Behrens (Dept of Mathematics)"— Presentation transcript:
Wrapping spheres around spheres Mark Behrens (Dept of Mathematics)
Spheres?? 1-dimensional sphere (Circle) 2-dimensional sphere (Sphere) 3-dimensional sphere and higher… (I’ll explain later!) S1S1 S2S2 S n (n > 2)
Wrapping spheres? Wrapping S 1 around S 1 - Wrapping one circle around another circle - Wrapping rubber band around your finger
Wrapping S 1 around S 2 -Wrapping circle around sphere -Wrapping rubber band around globe …another example…
…and another example. Wrapping S 2 around S 1 - Wrapping sphere around circle - Flatten balloon, stretch around circle
Goal: Understand all of the ways to wrap S k around S n ! n and k are positive numbers Classifying the ways you can wrap is VERY HARD! Turns out that interesting patterns emerge as n and k vary. We’d like to do this for not just spheres, but for other geometric objects – spheres are just the simplest!
Plan of talk Explain what I mean by “higher dimensional spheres” Work out specific low-dimensional examples Present data for what is known Investigate number patterns in this data
n-dimensional space 1-dimensional space: The real line To specify a point, give 1 number x x y 2-dimensional space: The Cartesian plane To specify a point, give 2 numbers 01 (x,y) = (4,3)
3-dimensional space -The world we live in - To specify a point, give 3 numbers (x,y,z). x y z (x,y,z)
Higher dimensional space Points in 4-dimensional space are specified with 4 numbers (x,y,z,w) Points in n-dimensional space are specified with n numbers: (x 1, x 2, x 3, ……, x n )
The circle S 1 is the collection of all points (x,y) in 2-dimensional space of distance 1 from the origin (0,0). Higher dimensional spheres: 1
The sphere S 2 is the collection of all points (x,y,z) in 3-dimensional space of distance 1 from the origin (0,0,0). Higher dimensional spheres: x y z 1
S 3 is the collection of all points (x,y,z,w) in 4-dimensional space of distance 1 from the origin (0,0,0,0). Higher dimensional spheres: S n-1 is the collection of all points (x 1,…,x n ) in n-dimensional space of distance 1 from the origin.
Spheres: another approach (This will help us visualize S 3 ) S 1 is obtained by taking a line segment and gluing the ends together:
Spheres: another approach (This will help us visualize S 3 ) S 2 is obtained by taking a disk and gluing the opposite sides together:
Spheres: another approach (This will help us visualize S 3 ) S 3 is obtained by taking a solid ball and gluing the opposite hemispheres together:
Spheres: another approach (This will help us visualize S 3 ) You can think of S 3 this way: If you are flying around in S 3, and fly through the surface in the northern hemisphere, you reemerge in the southern hemisphere.
Wrapping S 1 around S 1 For each positive integer n, we can wrap the circle around the circle n times
Wrapping S 1 around S 1 We can wrap counterclockwise to get the negative numbers -2 -3
The unwrap A trivial example: just drop the circle onto the circle. The unwrap wraps 0 times around
Equivalent wrappings We say that two wrappings are equivalent if one can be adjusted to give the other This wrapping is equivalent to… …this wrapping. (the “wrap 1”) For example:
Winding number Every wrapping of S 1 by S 1 is equivalent to “wrap n” for some integer n. Which wrap is this equivalent to? Handy trick: 1) Draw a line perpendicular to S 1 2) Mark each intersection point with + or – depending on direction of crossing 3) Add up the numbers – this is the “winding number” +1 +1 1 – 1 + 1 = 1
What have we learned: The winding number gives a correspondence: Ways to wrap S 1 around S 1 The integers: …-2, -1, 0, 1, 2, …
Wrapping S 1 around S 2
What have we learned: Every way of wrapping S 1 around S 2 is equivalent to the “unwrap” FACT: the same is true for wrapping any sphere around a larger dimensional sphere. REASON: there will always be some place of the larger sphere which is uncovered, from which you can “push the wrapping off”.
Wrapping S 2 around S 2 : Wrap 0 Wrap 1 Wrap 2 (Get negative wraps by turning sphere inside out)
“Winding number” +1 Same trick for S 1 works for S 2 for computing the “winding number” Winding number = 1 + 1 = 2
Fact: The winding number gives a correspondence: Ways to wrap S 2 around S 2 The integers: …-2, -1, 0, 1, 2, …
General Fact! The winding number gives a correspondence: Ways to wrap S n around S n The integers: …-2, -1, 0, 1, 2, …
Summary: Ways to wrap S n around S n The integers: …-2, -1, 0, 1, 2, … Ways to wrap S k around S n k < n Only the unwrap Ways to wrap S k around S n k > n ???
Wrapping S 2 around S 1 : Consider the example given earlier: In fact, this wrap is equivalent to The unwrap, because you can “shrink the balloon”
What have we learned: Ways to wrap S 2 around S 1 Only the unwrap This sort of thing always happens, and we have: Turns out that this is just a fluke! There are many interesting ways to wrap S n+k around S n for n > 1, and k > 0.
Wrapping S 3 around S 2 : Recall: we are thinking of S 3 as a solid ball with the northern hemisphere glued to the southern hemisphere. Consider the unwrap: S3S3 S2S2 1) Take two points in S 2 2) Examine all points in S 3 that get sent to these two points. 3) Because the top and bottom are identified, these give two separate circles in S 3.
Hopf fibration: a way to wrap S 3 around S 2 different from the unwrap S3S3 S2S2
S3S3 S2S2 For this wrapping, the points of S 3 which get sent to two points of S 2 are LINKED!
Keyring model of Hopf fibration
Fact: Counting the number of times these circles are linked gives a correspondence: Ways to wrap S 3 around S 2 The integers: …-2, -1, 0, 1, 2, …
n=2n=3n=4n=5n=6n=7n=8n=9n=10n=11 k=1Z222222222 k=22222222222 k=3212Z*1224 k=41222 2000000 k=5222 2Z00000 k=62324*32222222 k=7315 3060120Z*120240 k=8152228*62323 2424 232322 k=922 2323 2323 2323 2424 2525 2424 Z*2 3 2323 k=102 12*240*4* 2*3 2 18*8 24*28 2 *2*3 2 24*212*22 2 *3 k=1112*284*2 2 84*2 5 504*2 2 504*4504*2 504 Number of ways to wrap S n+k around S n Note: “Z” means the integers Some of the numbers are factored to indicate that there are distinct ways of wrapping
n=2n=3n=4n=5n=6n=7n=8n=9n=10n=11 k=1Z222222222 k=22222222222 k=3212Z*1224 k=41222 2000000 k=5222 2Z00000 k=62324*32222222 k=7315 3060120Z*120240 k=8152228*62323 2424 232322 k=922 2323 2323 2323 2424 2525 2424 Z*2 3 2323 k=102 12*240*4* 2*3 2 18*8 24*28 2 *2*3 2 24*212*22 2 *3 k=1112*284*2 2 84*2 5 504*2 2 504*4504*2 504 Number of ways to wrap S n+k around S n The integers form an infinite set – the only copies of the integers are shown in red. This pattern continues. All of the other numbers are finite!
n=2n=3n=4n=5n=6n=7n=8n=9n=10n=11 k=1Z222222222 k=22222222222 k=3212Z*1224 k=41222 2000000 k=5222 2Z00000 k=62324*32222222 k=7315 3060120Z*120240 k=8152228*62323 2424 232322 k=922 2323 2323 2323 2424 2525 2424 Z*2 3 2323 k=102 12*240*4* 2*3 2 18*8 24*28 2 *2*3 2 24*212*22 2 *3 k=1112*284*2 2 84*2 5 504*2 2 504*4504*2 504 Number of ways to wrap S n+k around S n STABLE RANGE: After a certain point, these values become independent of n
Stable values k = 10k = 11k = 12k = 13k = 14k = 15k = 16k = 17k = 18 2*3504032 480*22 2424 8*2 Below is a table of the stable values for various k. k = 1k = 2k = 3k = 4k = 5k = 6k = 7k = 8k = 9 22240022402 2323
Stable values k = 10k = 11k = 12k = 13k = 14k = 15k = 16k = 17k = 18 (2)(3)2 3 *3 2 *703(2)(2)(2 5 *3*5) (2) (2)(2) (2 3 )(2) Below is a table of the stable values for various k. k = 1k = 2k = 3k = 4k = 5k = 6k = 7k = 8k = 9 222 3 *30022 4 *3*5(2)(2)(2)(2)(2) Here are their prime factorizations.
Stable values k = 10k = 11k = 12 = 2 2 *3 k = 13k = 14k = 15k = 16 = 2 4 k = 17k = 18 (2)(3)2 3 *3 2 *703(2)(2)(2 5 *3*5) (2) (2)(2) (2 3 )(2) Below is a table of the stable values for various k. k = 1k = 2k = 3k = 4 = 2 2 k = 5k = 6k = 7k = 8 = 2 3 k = 9 222 3 *30022 4 *3*5(2)(2)(2)(2)(2) Note that there is a factor of 2 i whenever k+1 has a factor of 2 i-1 and is a multiple of 4
Stable values k = 10k = 11k = 12 = 4*3 k = 13k = 14k = 15k = 16 = 4*4 k = 17k = 18 (2)(3)2 3 *3 2 *703(2)(2)(2 5 *3*5) (2) (2)(2) (2 3 )(2) Below is a table of the stable values for various k. k = 1k = 2k = 3k = 4 = 4 k = 5k = 6k = 7k = 8 = 4*2 k = 9 2223*323*30022 4 *3*5(2)(2)(2)(2)(2) There is a factor of 3 i whenever k+1 has a factor of 3 i-1 and is divisible by 4
Stable values k = 10k = 11k = 12k = 13k = 14k = 15k = 16 = 8*2 k = 17k = 18 (2)(3)2 3 *3 2 *703(2)(2)(2 5 *3*5) (2) (2)(2) (2 3 )(2) Below is a table of the stable values for various k. k = 1k = 2k = 3k = 4k = 5k = 6k = 7k = 8 = 8 k = 9 222 3 *30022 4 *3*5(2)(2)(2)(2)(2) There is a factor of 5 i whenever k+1 has a factor of 5 i-1 and is divisible by 8
Stable values k = 10k = 11k = 12 = 12 k = 13k = 14k = 15k = 16k = 17k = 18 (2)(3)2 3 *3 2 *703(2)(2)(2 5 *3*5) (2) (2)(2) (2 3 )(2) Below is a table of the stable values for various k. k = 1k = 2k = 3k = 4k = 5k = 6k = 7k = 8k = 9 222 3 *30022 4 *3*5(2)(2)(2)(2)(2) There is a factor of 7 i whenever k+1 has a factor of 7 i-1 and is divisible by 12
What’s the pattern? Note that: 4 = 2(3-1) 8 = 2(5-1) 12 = 2(7-1) In general, for p a prime number, there is a factor of p i if k+1 has a factor of p i-1 and is divisible by 2(p-1). The prime 2 is a little different… …..2(2-1) does not equal 4!
Beyond… It turns out that all of the stable values fit into patterns like the one I described. The next pattern is so complicated, it takes several pages to even describe. We don’t even know the full patterns after this – we just know they exist! The hope is to relate all of these patterns to patterns in number theory.
Some patterns for the prime 5 |
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# Solving Math Problem
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Sub Topics Solving math problem is a very helpful tool to get answers to calculus, algebra, trigonometry or any other topic you are having trouble with. Mathematics is the abstract study of topics. This page is about solving math problems and solutions which will let you know about this concept with clear understanding. This page gives a detailed solution of some of the topics in mathematics as many students find it very difficult to solve math problems. Students can learn to solve algebra problems, percentage problems, vectors etc.
## Solving Algebra Problems
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Algebra is a branch of mathematics which helps in finding the value of the unknown variables written in combination of terms.
It is the foundation stone of mathematics. Letters and symbols are known as variables.
Examples : x + 7 = 12, xyz + 37xy + z = 0, when an equation contains variables, you will often have to solve for one of those variables.
Example 1: Solve 7x + 120 = 0.
Solution:
Given equation is 7x + 120 = 0
Put all the variables aside and value on the other side.
7x = - 120
Divide both sides by 7, to find the value of 'x'.
$\frac{7x}{7}$ = − $\frac{120}{7}$
x = 17.14
Example 2 : Solve 5x$^{2}$ + 12x + 8 = 0
Solution: Compare given equation with quadratic general equation, we get
a = 5,
b = 12 and
c = 8
Quadratic equation formula :
x = $\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
Substitute the values of a, b and c in the above formula
x = $\frac{-12\pm\sqrt{12^{2}-4*5*8}}{2*5}$
x = $\frac{-12\pm\sqrt{-16}}{10}$
There exist two values for x, let they be $x_{1}$ and x$_{2}$
x$_{1}$ = $\frac{-12+\sqrt{-16}}{10}$
= - $\frac{6}{5}$ + $\frac{2}{5}$ i
x$_{2}$ = $\frac{-12-\sqrt{-16}}{10}$
= - $\frac{6}{5}$ - $\frac{2}{5}$ i
Therefore the values of x$_{1}$ and x$_{2}$ are - $\frac{6}{5}$ + $\frac{2}{5}$ i and - $\frac{6}{5}$ - $\frac{2}{5}$ i.
Example 3: By using completing square method solve 4x$^{2}$ + 3x + 7 = 0.
Solution:
Through out the equation divide by 4 as the given equation is not in standard form.
x$^{2}$ + $\frac{3}{4}$x + $\frac{7}{4}$ = 0
To the right move all the constant terms
x$^{2}$ + $\frac{3}{4}$x = - $\frac{7}{4}$
coefficient of x term = $\frac{3}{4}$
Square half of the coefficient of x term, $\frac{9}{64}$
Add $\frac{9}{64}$ to both sides
x$^{2}$ + $\frac{3}{4}$x + $\frac{9}{64}$ = - $\frac{7}{4}$ + $\frac{9}{64}$
x$^{2}$ + $\frac{3}{4}$x + $\frac{9}{64}$ = - $\frac{103}{64}$
On to the left we write the perfect square
(x + $\frac{3}{8}$)$^{2}$ = - $\frac{103}{64}$
On both sides take the square root
x + $\frac{3}{8}$ = $\pm\sqrt{-\frac{103}{64}}$
Solve for x
x = -$\frac{3}{8}$ $\pm\sqrt{-\frac{103}{64}}$
or x = - $\frac{3}{8}$ $\pm$ $\frac{1}{8}$ $\sqrt{103}$i
## Solving Percentage
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Percentage is a number as a fraction of 100. In a mathematical language percent means out of 100. They are used in media to describe difference in interest rates, VAT, success rate, exam results etc.,
To change a fraction to a percentage, divide the numerator by the denominator and multiply by 100%.
Example 1: In a playground there are 2,000 people. Out of them 825 are players. Find the percentage of players in the class?
Solution:
Number of people in the playground = 2,000
Players in the playground = 825
To find percentage of players
= $\frac{825}{2000}$ * 100
= 41.25%
= (1.07, 1.71) |
# Product to Sum Formulas for Sine and Cosine
## Relation of the product of two trigonometric functions to a sum or difference.
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Product to Sum Formulas for Sine and Cosine
Let's say you are in class one day, working on calculating the values of trig functions, when your instructor gives you an equation like this:
Can you solve this sort of equation? You might want to just calculate each term separately and then compute the result. However, there is another way. You can transform this product of trig functions into a sum of trig functions.
Read on, and by the end of this lesson, you'll know how to solve this problem by changing it into a sum of trig functions.
### Product to Sum Formulas for Sine and Cosine
Here we'll begin by deriving formulas for how to convert the product of two trig functions into a sum or difference of trig functions.
There are two formulas for transforming a product of sine or cosine into a sum or difference. First, let’s look at the product of the sine of two angles. To do this, we need to start with the cosine of the difference of two angles.
The following product to sum formulas can be derived using the same method:
#### Using the Product to Sum Formula
1. Change to a sum.
Use the formula . Set and .
2. Change to a product.
Use the formula . Therefore, and . Solve the second equation for and plug that into the first.
. Again, the sum of and is and the difference is .
3. Solve .
Use the formula .
### Examples
#### Example 1
Earlier, you were asked to solve sin 75°sin15°.
Changing to a product of trig functions can be accomplished using
Substituting in known values gives:
#### Example 2
Express the product as a sum:
Using the product-to-sum formula:
#### Example 3
Express the product as a sum:
Using the product-to-sum formula:
#### Example 4
Express the product as a sum:
Using the product-to-sum formula:
### Review
Express each product as a sum or difference.
Express each sum or difference as a product.
To see the Review answers, open this PDF file and look for section 3.14.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes |
## Mathematics
### Asymptotic:
Asymptotic means convergence in the infinity. For example, the two functions:
f(x) = sin(x)/x
and
g(x) = 1/ex
are asymptotically identical when x goes to infinity.
f(x) = sin(x)/x and g(x) = 1/ex both functions approach zero as x goes to + infinity f(x) = 1/x This hyperbola approaches the x-axis as x +infinity or -infinity and and the y-axis as x goes to zero.
### Radix or Base of a Number System:
Given any integer number b (the base), b >= 2, and any integer number N, N >= 0, there is a unique representation of N as:
N = a0 + a1 b + a2 b2 + ...
where all integer numbers ai satisfy:
0 <= ai < b
and all but a finite number of them are non-zero. Then, it is possible to represent any number N in the base b using those coefficients.
When b is lower than 10, it's conventional to represent N in base b by the decimal representations of ai, with a subindex b to avoid confusion. For example:
1000 = 1*(103) + 0*(102) + 0*(101)+ 0*(100)
or
1000 = 2*(73) + 6*(72) + 2*(71) + 6(70)
So the representation of 1000 in base 7 is:
100010 = 26267
When the base is greater than 10, it's possible to use letters to represent the ai that are greater than or equal to 10 (as in Hexadecimal notation (base 16), used in pre-Bureaucracy computers) or by representing them in the decimal notation, and separating them with symbols (as in the Sexagesimal notation (base 60) for Hours or Angles: 13:10:30)
This representation can be adapted for fractional numbers. Any number x satisfying 0 <= x < 1 can be represented as:
x = a-1 / b + a-2 / b2 + ...
where the a-i, as above, satisfy <4>:
0 <= a-i < b
but they do not satisfy the uniqueness property and an infinite number of them may be non-zero. In fact, their uniqueness can only be violated in the case where they can be replaced by a finite number of non-zero elements.
Combining the two results, there is an (almost) unique representation of all positive numbers in any base.
### Base Two:
Base Two is the simplest system, since it uses only two symbols for the digits. However, the disadvantage is that numbers in base two tend to occupy too much space:
100010 = 11111010002
Base 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Base 2 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111 10000 10001 10010 10011 10100
### Base Six:
The Dura system uses base six.
Base 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Base 6 1 2 3 4 5 10 11 12 13 14 15 20 21 22 23 24 25 30 31 32
### Base Ten:
The system most used by pre-Contact Earthlings. It's widespread use on Earth (with the notable exception of the Mayan Civilization, that used base 20, and the Mesopotamian Civilizations, that used base 60) suggests that this system was brought by Humanity's Patron.
### Torus (Mathematics):
The torus is the two dimensional surface generated by a circle that rotates around a line external to it. A doughnut-like shape.
In topology, this concept is extended: the n-dimensional torus (n >= 2) is the cartesian product of n circles, or:
Tn = S1 x S1 x ... (n times)
Back to Main Index
e3v5r1 |
## § Finite differences and Umbral calculus
Umbral calculus lays out a large collection of "umbral" / "shadowy" coincidences across combinatorics and calculus. Here, I'll lay out some of these that I learnt from Concrete Mathematics. I hope to use this for myself to motivate a bunch of combinatorics. I'll provide an interesting proof of why $\sum{1 \leq k < n} k^2 = k(k-1)/2$ using this umbral calculus.
#### § Discrete Derivative: Forward difference
We begin by trying to build a discrete version of the derivative operator. The derivative of f will be denoted using f'. The forward difference of $f: \mathbb R \rightarrow \mathbb R$ is defined as:
$\delta f: \mathbb R \rightarrow \mathbb R; (\delta f)(x) = f(x + 1) - f(x)$
Immediately, we can see that this operator is linear:
\begin{aligned} &\delta(f + g) &= (f+g)(x+1) - (f+g)(x) \\ &= (f(x+1) - f(x)) + (g(x+1)-g(x)) \\ &= (\delta f) + (\delta g) \\ &\delta(\alpha f)(x) &= (\alpha f)(x+1) - (\alpha f)(x) \\ &= \alpha \cdot (f(x+1) - f(x)) \\ &= \alpha (\delta f) \end{aligned}
it obeys a slightly corrupted version of the chain rule, $(fg)' = f' g + g' f$:
\begin{aligned} &\delta(fg) \\ &= (fg)(x+1) - (fg)(x) \\ &= f(x+1)g(x+1) - f(x)g(x) + 0 \\ &= f(x+1)g(x+1) - f(x)g(x) + [f(x)g(x+1) - f(x)g(x+1)] \\ &= g(x+1)[f(x+1) - f(x)] + f(x)[g(x+1) - g(x)] \\ &= g(x+1)(\delta f)(x) + f(x) (\delta g)(x) \\ &= (S \delta f)(x) + (f \delta g)(x) [(Sh)(x) \equiv h(x+1)] \\ &= (S \delta f + f \delta g)(x) \\ \end{aligned}
We need this new $S$ operator to shift the function's input from $x$ to $x+1$.
#### § Falling factorial as polynomials
cool, now that we seem to have a linear derivative operator that behaves roughly sanely, let's test it out! The first reasonable target is a polynomial, so let's try $x^2$:
$\delta(x^2) = (x+1)^2 - x^2 = 2x + 1$
This is disappointing, it does not behave very well :( However, there is an analogue that does well. This is the falling factorial, defined as:
$x^{(n)} \equiv x(x-1)(x-2)\cdots(x-n+1)$
For example:
$x^{(0)} = 1 \\ x^{(1)} = x \\ x^{(2)} = x(x-1) \\ x^{(3)} = x(x-1)(x-2) \\$
Let's try and apply our discrete difference $\delta$:
\begin{aligned} &\delta(x^{(2)}) \\ & = (x+1)(x-1+1) - x(x-1) \\ & = (x+1)(x) - x(x-1) \\ & = x*2 = 2x(1) \\ &\delta(x^{(3)}) \\ &= (x+1)(x-1+1)(x-2+1) - x(x-1)(x-2) \\ &= (x+1)(x)(x-1) - x(x-1)(x-2) \\ &= x(x-1)((x+1) - (x-2)) = 3x(x-1) = 3x^{(2)} \\ \end{aligned}
These falling factorials look pretty unnatural though, why do we care? Well, once we build some integral calculus, we can handle our problem of $\sum_{1 \leq i < k} i^2$ using this calculus, by rewriting $i^2$ in terms of these falling factorials.
#### § Sums as discrete integrals
We want to think of $\sum_{0 \leq i < n} f(i)$ as the correct variant of $\int_0^n f(i) di$. The most important property of an integral is the fundamental theorem of calculus:
$\int_a^b f'(i) di = f(b) - f(a) \mapsto \sum_{a \leq i < b} (\delta f)(i) =?= f(b) - f(a)$
we can check if the assertion is true:
\begin{aligned} &\sum_{a \leq i < b} (\delta f)(i) \\ &= [f(a+1) - f(a)] + [f(a+2) - f(a+1)] + [f(a+3)-f(a+2)] + \cdots + [f(b) - f(b-1)] \\ &= f(b) - f(a) \quad \text{(The sum telescopes)} \end{aligned}
Sweet, so we just kicked a theory of calculus of the ground. Let's put this to some use:
#### § Gauss' famous formula from discrete calculus
Let's begin by deriving the closed form for $[1\cdot(k-1)]$ naturals:
$\sum_{0 \leq i < n} i = \sum_{0 \leq i < n} i^{(1)} = i^{(2)}/2 \big|_{0}^n = n(n-1)/2$
Let's now derive the closed form form the sum of squares of $[1\cdot(k-1)]$:
\begin{aligned} &\sum_{0 \leq i < n} i^2 \\ &= \sum_{0 \leq i < n} i*(i-1) + i \\ &= \sum_{0 \leq i < n} i^{(2)} + i^{(1)} \\ &= n^{(3)}/2 + n^{(2)}/2 \\ &= n(n-1)(n-2)/3 + n(n-1)/2 \\ &= n(n-1)(n/3 - 2/3 + 1/2) \\ &= n(n-1)(2n - 1)/6 \\ \end{aligned}
Trying to perform this process in general does beg a question: how do we convert from $x^n$ into some combination of rising and falling factorials? It turns out that Stirling Numbers will show up to aid this conversion. But before that, let's see some more connections.
#### § $2^x$ as the combinatorial version of $e^x$
We want to find the analogue of the exponential function $e^x$, which satisfies the equation $f'(x) = f(x)$. Setting this up in the discrete case:
\begin{aligned} &d'f(x) = f(x) | f(0) = 1 \\ &f(x+1) - f(x) = f(x) | f(0) = 1 \\ &f(x+1) = 2f(x) | f(0) = 1 \\ &f(n) = 2^n \end{aligned}
What does this buy us? It gives us a nice proof that $\sum_{k} nCk = 2^n$. It proceeds by taking the taylor of $e^x$, "combinatorializing it", and then simplifying:
\begin{aligned} &e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \\ &2^x = \sum_{n=0}^\infty \frac{x^{(n)}}{n!} \\ & = \sum_{n=0}^\infty \frac{x*(x-1)*(x-2)*\cdots(x-n+1)}{n!} \\ & = \sum_{n=0}^\infty \frac{x*(x-1)*(x-2)*\cdots(x-n+1)}{n!} \end{aligned}
#### § Harmonic series as the combinatorial version of logarithm
In the continuous case, we have that $\int 1/x = \log x$. In the discrete case, we get $\sum_{i=1}^n 1/x \equiv H_n$, the sum of the first $n$ harmonic numbers. This explains why the harmonic numbers keep showing up in different places across discrete math --- We're often trying to integrate some kind of $1/x$ quantity. This also may tell us that $H_n$ and $2^n$ are somewhat "weak inverses" of each other. I haven't thought about this weak inverse thing too much.
#### § Stirling numbers of the first kind for convering between polynomials and falling factorials
We can express $x^{(n)} = \sum_{k=0}^n [n, k] x^k$ where $[n, k]$ are the (yet to be defined) unsigned stirling numbers of the first kind. (aside: I wonder if this can be derived from some kind of mobius inversion). We begin my in fact defining the stirling numbers of the first kind, $[n, k]$ as the coefficients of the rising factorial:
$x^{(n)} = x(x+1)(x+2) \dots (x+n-1) = \sum_{i=0}^n [n, i] x^i$
The question as a combinatorialist is:
what do the unsigned striling numbers of the first kind, $[n, i]$, count?
$[n, i]$ counts the number of permutations of $n$ elements with $i$ disjoint cycles.
There is a more evocative definition, which goes like this:
$[n, i]$ counts the number of ways to seat $n$ people at $i$ circular tables, such that no table is left empty.
For example, in the case of the permutations of the set $\{1, 2, 3\}$, we have the permutations:
\begin{aligned} (1)(2)(3) \text{3 cycles} \\ (12)(3) \text{2 cycles} \\ (1)(23) \text{2 cycles} \\ (2)(13) \text{2 cycles} \\ (132) \text{1 cycle} \\ (123) \text{1 cycle} \\ \end{aligned}
So, this gives the counts:
• $[3, 1] = 2$
• $[3, 2] = 3$
• $[3, 3] = 1$
From the seating people perspective, here's how this goes:
• $[3, 1]$ is the number of ways to seat 3 people at 1 table. The firstperson sits at the table in only one way. The second person can sit to theleft or to the right of the first person, but this is symmetric: they toohave only one way. Now the third person can sit to the left of the firstperson or to the left of the second person. So there are two ways.In total, there are only two ways. Key idea: On a circular table, itonly matters who is to your left, since there is no difference between leftand right.The second person has two choices: they
• $[3, 3]$ is the number of ways to seat 3 people at 3 tables so that no tableis empty. We are forced to place one person per table.
• $[3, 2]$ is the number of ways to seat 3 people at 2 tables so that no tableis empty. So we will need to have two people on the first table,and then the final personal on the second table.Once we pick which two people sit together in the first table, we are done,because for upto two people, it doesn't matter how they sit at a table,the situation is symmetric.
These stirling numbers satisfy a recurrence:
$[n+1, k] = n[n, k] + [n, k-1]$
This can be seen combinatorially. If we want the number permutations of $n+1$ objects with $k$ cycles, we can either:
1. Take the permutations of $n$ objects with $k-1$ cycles, $[n, k-1]$, and theninsert the new $(n+1)$th object into a new cycle all by itself. That is,we create a new permutation where $p(n+1) = (n+1)$.
2. Take the permutation of $n$ objects with $k$ cycles, and insert this new $(n+1)$thobject into any of the $k$ cycles. If the original permutation had x -p-> y,we can insert this new object as x -p-> * -p->y for any x. Thus, there are$n$ choices of locations to inser * --- as the image of all possible xs.This gives us the $n[n, k]$ term.
Another cute fact of the unsigned stirling numbers of the first kind $[n, k]$ is that since permutations are partitioned by their number of cycles, we have:
$\sum_{k=1}^n [n, k] = n!$ |
# Parents guide to Year 5 Maths
Parents’ Guide ‘Must Dos’ by the end of Year 5
Number
Read, write and order numbers to 1,000,000
Count on and back in steps of 100,1,000, 10,000 up to 1,000,000
Round any number to the nearest 10, 100, 1,000, 10,000 or 100,000
Estimate and calculate additions involving 2, 3-digit numbers (to nearest 10)
Estimate and calculate subtractions involving 2, 3-digit numbers (to nearest 10)
Estimate and calculate multiplications involving a 2-digit and a 3-digit numbers (to nearest 10)
Estimate and calculate divisions of a 3-digit number by a 1-digit number (to nearest 10)
Subtract one 5-digit number from another, using column subtraction
Rapid Recall: Mentally add a 2-digit number to a 3-digit number
Rapid Recall: Mentally add a 1000s number to a 5-digit number
Rapid Recall: Mentally subtract a 2-digit number from a 3 or 4-digit number
Rapid Recall: Mentally subtract a 1000s number from a 5-digit number
Know all factors that make up numbers to 100
Recognise all prime numbers to 100
Multiply a 4-digit number by a 2-digit number using formal methods
Divide a 4-digit number by a 1-digit number (with remainders)
Multiply numbers by 10, 100 and 1,000
Parents’ guide ‘Must Dos’ by the end of Year 5
Number
Divide number by 10, 100 and 1,000
Recognise and use square numbers and square roots
Know and use the symbols (.),(3) and (√) accurately
Fractions: Compare and order fractions whose denominators are multiples of the same number
Convert mixed numbers to improper fractions and visa versa
Add and subtract fractions with the same denominator
Multiple proper fractions and mixed fractions by whole numbers
Number
Decimal Fractions: Read and write decimal numbers as fractions (up to hundredths)
Round decimals with two decimal places to nearest whole number and one decimal place
Read, write, order and compare decimal numbers with up to three decimal places
Percentages: Recognise the % symbol
Understand that percentage is measured as part of 100
Solve problems which require knowing percentage and decimal value of ½, ¼, ⅕, ⅖ and ⅘
Parents’ Guide ‘Must Dos’ by the end of Year 5
Shape and measures
Measure angles in degrees
Draw a given angle accurately
Know and use reflex angles
Construct shapes from given dimensions
Identify 3D shapes from 2D representations
Identify, describe and represent the position of a shape following reflection and translation
Add, subtract, multiply and divide units of measure (using decimal notation)
Understand and use basic equivalence between metric and common imperial units
Calculate, estimate and compare areas of squares and rectangles using cm squared (cm.) and metre squared (m.)
Recognise volume in practical contexts
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Wednesday , May 18 2022
# NCERT 5th Class (CBSE) Mathematics: Introduction To Algebra
We know that a mathematical expression uses mathematical symbols instead of words. For example, the words ‘twenty-eight increased by five’, will be shown as the mathematical expression 28 + 5.
Can you give the mathematical expressions for these words?
1. Seventy three minus forty -two
2. Twelve less two
3. Six times fifteen
4. Thirty plus eleven take away nine
Sometimes, in a mathematical expression, we use letters to represent numbers we do not know.
If we say that there were some people at a party and 7 people left, we can show this as an expression, n-7, where n represents the number of people in the party.
If there were 12 people at the party, then n would represent 12. If there were 20 people at the party, then n would represent 20.
These letters are called variable because they can represent any number.
CHECKPOINT: When we write an expression using variables, words and/or operations, it is called an algebraic expression.
Let p be the number of pencils in your pencil box. Study the algebraic expressions for each:
12 pencils more than the number of pencils in your box → p + 12 or 12 + p
4 pencils less than the number of pencils in your box → p – 4
Three times the number of pencils in your box. → 3 × p also written as 3p
The number of pencils in your box divided by 4 → p ÷ 4 also written as p/4
Here, the expressions have been put into words. The unknown number or the variable is w.
18 + → w added to 18
10 – w subtracted from 10
7→ The product of w and 7
6 ÷ → 6 divided by w
If we know that w = 3, Then expressions above can be simplified and a value can be obtained for each expression.
If we substitute 3 in place of w, we get
18 + w = 18 + 3 = 21
10 – w = 10 – 3 = 7
7w = 7 × 3 = 21
6 ÷ w = 6 ÷ 3 = 2
## छोटी-सी हमारी नदी 5 NCERT CBSE Hindi Rimjhim Chapter 17
छोटी-सी हमारी नदी 5th Class NCERT CBSE Hindi Book Rimjhim Chapter 17 प्रश्न: इस कविता के पद … |
Common Core: 7th Grade Math : Finding Volume of a Rectangular Prism
Example Questions
Example Question #1 : Finding Volume Of A Rectangular Prism
An aquarium is shaped like a perfect cube; the perimeter of each glass face is meters. If it is filled to the recommended capacity, then, to the nearest hundred cubic liters, how much water will it contain?
Insufficient information is given to answer the question.
Note:
Explanation:
A perfect cube has square faces; if a face has perimeter meters, then each side of each face measures one fourth of this, or meters. The volume of the tank is the cube of this, or
cubic meters.
Its capacity in liters is liters.
of this is
liters.
This rounds to liters, the correct response.
Example Question #17 : Solve Problems Involving Area, Volume And Surface Area Of Two And Three Dimensional Objects: Ccss.Math.Content.7.G.B.6
Calculate the volume of the provided figure.
Explanation:
In order to solve this problem, we need to recall the volume formula for a rectangular prism:
Now that we have the correct formula, we can substitute in our known values and solve:
Example Question #1 : Finding Volume Of A Rectangular Prism
Calculate the volume of the provided figure.
Explanation:
In order to solve this problem, we need to recall the volume formula for a rectangular prism:
Now that we have the correct formula, we can substitute in our known values and solve:
Example Question #2 : Finding Volume Of A Rectangular Prism
Calculate the volume of the provided figure.
Explanation:
In order to solve this problem, we need to recall the volume formula for a rectangular prism:
Now that we have the correct formula, we can substitute in our known values and solve:
Example Question #3 : Finding Volume Of A Rectangular Prism
A rectangular prism has the following dimensions:
Length:
Width:
Height:
Find the volume.
Explanation:
Given that the dimensions are: , , and and that the volume of a rectangular prism can be given by the equation:
, where is length, is width, and is height, the volume can be simply solved for by substituting in the values.
This final value can be approximated to .
Example Question #981 : Act Math
Solve for the volume of a prism that is 4m by 3m by 8m.
Explanation:
The volume of the rectangle
so we plug in our values and obtain
.
Example Question #21 : Prisms
The dimensions of Treasure Chest A are 39” x 18”. The dimensions of Treasure Chest B are 16” x 45”. Both are 11” high. Which of the following statements is correct?
Treasure Chest A and B can hold the same amount of treasure.
Treasure Chest A has the same surface area as Treasure Chest B.
Treasure Chest B can hold more treasure.
Treasure Chest A can hold more treasure.
There is insufficient data to make a comparison between Treasure Chest A and Treasure Chest B.
Treasure Chest B can hold more treasure.
Explanation:
The volume of B is 7920 in3. The volume of A is 7722 in3. Treasure Chest B can hold more treasure.
Example Question #1 : Finding Volume Of A Rectangular Prism
A rectangular prism has a width of 3 inches, a length of 6 inches, and a height triple its length. Find the volume of the prism.
Explanation:
A rectangular prism has a width of 3 inches, a length of 6 inches, and a height triple its length. Find the volume of the prism.
Find the volume of a rectangular prism via the following:
Where l, w, and h are the length width and height, respectively.
We know our length and width, and we are told that our height is triple the length, so...
Now that we have all our measurements, plug them in and solve:
Example Question #6 : Finding Volume Of A Rectangular Prism
The above diagram shows a rectangular solid. The shaded side is a square. In terms of , give the volume of the box. |
# How to Solve Simultaneous Equations Using Substitution Method
Simultaneous equations are two linear equations with two unknown variables that have the same solution. Solving equations with one unknown variable is a simple matter of isolating the variable; however, this isn’t possible when the equations have two unknown variables. By using the substitution method, you must find the value of one variable in the first equation, and then substitute that variable into the second equation.[1] While it involves several steps, the substitution method for solving simultaneous equations requires only basic algebra skills.
Part 1
Part 1 of 2:
### Finding the Value of y
1. 1
Choose the equation you want to work with first. It doesn’t matter which equation you choose, but you might want to look for one that will give you numbers that are easier to work with.[2]
• For example, if your simultaneous equations are 1) ${\displaystyle x+2y=-4}$ and 2) ${\displaystyle 2x+5y=1}$, you will probably want to begin with the first equation, because the ${\displaystyle x}$ is already by itself.
2. 2
Isolate the variable in the first equation. You could also start by isolating the y variable (or whatever other variable the equation uses).[3]
• For example, if you are beginning with ${\displaystyle x+2y=-4}$, you could solve for ${\displaystyle x}$ by subtracting 2y from each side.
${\displaystyle x+2y=-4}$
${\displaystyle x=-4-2y}$
3. 3
Plug in the value of into the second equation. Place the value in parentheses for clarity.[4]
• For example, if you found ${\displaystyle x=-4-2y}$ in the first equation, plug in ${\displaystyle -4-2y}$ for ${\displaystyle x}$ in the second equation:
${\displaystyle 2x+5y=1}$
${\displaystyle 2(-4-2y)+5y=1}$
4. 4
Find the value of in the second equation. Remember to follow the order of operations.[5]
• For example, to solve for ${\displaystyle y}$ in the equation ${\displaystyle 2(-4-2y)+5=1}$, first use the distributive property to multiply.
${\displaystyle 2(-4-2y)+5y=1}$
${\displaystyle -8-4y+5y=1}$
${\displaystyle -8+y=1}$
${\displaystyle y=9}$
Part 2
Part 2 of 2:
### Finding the Value of x
1. 1
Plug in the value into either equation. It doesn’t matter which equation you use, as long as you use the original equation, or an equation where you’ve isolated the ${\displaystyle x}$ variable. This will allow you to find the value for ${\displaystyle x}$.[6]
• If you plug the ${\displaystyle y}$ value back into the second equation with the ${\displaystyle x}$ substitution, you will not be able to find the value of ${\displaystyle x}$.[7]
• For example, if you found ${\displaystyle y=9}$, plug in ${\displaystyle 9}$ for ${\displaystyle y}$ in the first equation:
${\displaystyle x+2y=-4}$
${\displaystyle x+2(9)=-4}$
2. 2
Find the value of . Remember to follow the order of operations.[8]
• For example, to solve for ${\displaystyle x}$ in the equation ${\displaystyle x+2(9)=-4}$, first multiply, and then subtract 18 from each side to find the value of ${\displaystyle x}$.
${\displaystyle x+2(9)=-4}$
${\displaystyle x+18=-4}$
${\displaystyle x=-22}$.
3. 3
Check your work. To do this, substitute the values you found for ${\displaystyle x}$ and ${\displaystyle y}$ into both equations, and verify that the resulting equation are true.[9]
• For example, if you found ${\displaystyle y=9}$ and ${\displaystyle x=-22}$, substitute these values into both equations.
• So, for the first equation:
${\displaystyle (-22)+2(9)=-4}$
${\displaystyle -22+18=-4}$
${\displaystyle -4=-4}$
• For the second equation:
${\displaystyle 2(-22)+5(9)=1}$
${\displaystyle -44+45=1}$
${\displaystyle 1=1}$
## Community Q&A
Search
• Question
Is there an easier method for simultaneous equations than what is already on this website?
Not really. Here's the simplest example possible: let's say x + y = 3 and x - y = 1. Solve the second equation for x by adding y to both sides: (x - y) + y = 1 + y. So x = 1 + y. Take that value of x, and substitute it into the first equation given above (x + y = 3). With that substitution the first equation becomes (1+y) + y = 3. That means 1 + 2y = 3. Subtract 1 from each side: 2y = 2. So y = 1. Substitute that value of y into either of the two original equations, and you'll get x = 2.
• Question
Is 3x + 4y = 52 and 5x + y = 30 solvable by substitution? I've tried solving twice and I just can't get the last part right.
Yes, it's solvable. Take the second equation, and subtract 5x from both sides: y = (30 - 5x). Plug that value of y back into the other equation: 3x + 4(30 - 5x) = 52. So (3x + 120 - 20x) = 52, and (-17x + 120) = 52. Then (-17x) = -68, and x = 4. Plug that value of x into either of the original equations: 3(4) + 4y = 52, so 12 + 4y = 52, and 4y = 40, so that y = 10. Check your work by plugging the found values of x and y into either of the original equations.
• Question
Which is easiest between the elimination, graphical, and substitution methods of solving simultaneous equations?
It depends on what kind of equation you have. One skill you should develop is knowing when to use what so you can manage your time wisely.
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# 2001 AIME I Problems/Problem 4
## Problem
In triangle $ABC$, angles $A$ and $B$ measure $60$ degrees and $45$ degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$, and $AT=24$. The area of triangle $ABC$ can be written in the form $a+b\sqrt{c}$, where $a$, $b$, and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c$.
## Solution
After chasing angles, $\angle ATC=75^{\circ}$ and $\angle TCA=75^{\circ}$, meaning $\triangle TAC$ is an isosceles triangle and $AC=24$.
Using law of sines on $\triangle ABC$, we can create the following equation:
$\frac{24}{\sin(\angle ABC)}$ $=$ $\frac{BC}{\sin(\angle BAC)}$
$\angle ABC=45^{\circ}$ and $\angle BAC=60^{\circ}$, so $BC = 12\sqrt{6}$.
We can then use the Law of Sines area formula $\frac{1}{2} \cdot BC \cdot AC \cdot \sin(\angle BCA)$ to find the area of the triangle.
$\sin(75)$ can be found through the sin addition formula.
$\sin(75)$ $=$ $\frac{\sqrt{6} + \sqrt{2}}{4}$
Therefore, the area of the triangle is $\frac{\sqrt{6} + \sqrt{2}}{4}$ $\cdot$ $24$ $\cdot$ $12\sqrt{6}$ $\cdot$ $\frac{1}{2}$
$72\sqrt{3} + 216$
$72 + 3 + 216 =$ $\boxed{291}$ |
# Ratios and Proportions
Proportional Relationships (Common Core Standards 7.RP.2, 7.RP.2.a-d) This page describes in detail how to read the graph of proportional quantities.
Directly proportional, or varying directly are other terms that mean proportional, or that y=kx for constant of proportionality k and proportional quantities x and y. Notice that k is the slope of the graph of proportional quantities x and y. The point (0,0) will always be on the graph of proportional quantities x and y. Grade7commoncoremath - Unit 1 Ratios and Proportional Relationships. Unit 2: Ratios and Proportional Relationships (7.RP)Big Ideas:Developing understanding of and applying proportional reasoning:Students extend their understanding of ratios and develop understanding of proportionality to solve single- and multi-step problems.
Students use their understanding of ratios and proportionality to solve a wide variety of percent problems, including those involving discounts, interest, taxes, tips, and percent increase or decrease. Students solve problems about scale drawings by relating corresponding lengths between the objects or by using the fact that relationships of lengths within an object are preserved in similar objects.
Students graph proportional relationships and understand the unit rate informally as a measure of the steepness of the related line, called the slope.
## 7.RP.3
7.RP.2. 7.RP.1. 7.RP - GA. 4mula Fun! :): More Proportion Help. WKU stands for Word Ratio, Known Ratio Unknown Ratio.
Using this method when solving proportion problems will help make sure that you set up equivalent ratios that will make sense as you Cross Multiply an Divide. The word problem states "The Rangers scored 3 runs in the first inning. If they continue at this rate, how many runs will be scored in the entire game without any overtime? " After we read through the problem, we go through and circle the words that we are comparing, in this case runs and innings. Glowing Rectangles. People spend lots of time staring at their screens (glowing rectangles). Kids spend hours each day using television, computers, mobile devices and so on.
Use that interest to motivate your student’s analysis of screen ratios. Getty Images. |
# How do you factor: 2x^3 + 7x^2 - 3x -18?
Apr 5, 2018
${x}^{2} \left(2 x + 7\right) - 3 \left(x + 6\right)$
Here's how I did it:
#### Explanation:
$2 {x}^{3} + 7 {x}^{2} - 3 x - 18$
We will factor this by grouping:
$\left(2 {x}^{3} + 7 {x}^{2}\right) + \left(- 3 x - 18\right)$
To factor, we have to see what everything has in common. For $2 {x}^{3} + 7 {x}^{2}$, both expressions have ${x}^{2}$ in them, so we can take ${x}^{2}$ out:
${x}^{2} \left(2 x + 7\right)$
For $- \left(3 x + 18\right)$, they both have $- 3$ in common, so we take that out:
$- 3 \left(x + 6\right)$
Now we combine these two expressions:
${x}^{2} \left(2 x + 7\right) - 3 \left(x + 6\right)$ |
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# A charge $q = 10\,{\text{mC}}$ is distributed uniformly over the circumference of a ring of radius $3\,{\text{m}}$ placed on x-y plane with its centre at origin. Find the electric potential at a point $P\left( {0,0,4\,{\text{m}}} \right)$A. $18\,{\text{V}}$B. $1.8 \times {10^2}\,{\text{V}}$C. $1.8 \times {10^3}\,{\text{V}}$D. $1.8 \times {10^7}\,{\text{V}}$
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343.2k+ views
Hint: First of all, we will find the shortest distance between the point and the charge. Then we will apply the formula and substitute the required values and manipulate accordingly to find the electric potential.
Complete step by step answer:In the given question, we are supplied with the following data:
The charge present which is uniformly distributed over the circumference of a ring is $10\,{\text{mC}}$ .
The radius of the ring is $3\,{\text{m}}$ .
We are asked to find the electric potential at a point $P\left( {0,0,4\,{\text{m}}} \right)$ .
To proceed the numerical, we will convert the unit of charge to S.I units:
We know,
$1\,{\text{mC}} = 1 \times {10^{ - 3}}\,{\text{C}}$
So, we have:
$10\,{\text{mC}} = 10 \times {10^{ - 3}}\,{\text{C}}$
Now, we need to find the electric potential due to the given charge, at a point which is at a distance of $3\,{\text{m}}$ from the charge.
So, we calculate the shortest distance between them by using the Pythagoras theorem, as given below:
$r = \sqrt {{3^2} + {4^2}} \\ r = \sqrt {9 + 16} \\ r = \sqrt {25} \\ r = 5\,{\text{m}} \\$
Therefore, the shortest distance has come out to be $5\,{\text{m}}$ .
Now, to find the electric potential we apply the formula, as given below:
$V = \dfrac{{Kq}}{r}$ …… (1)
Where,
$V$ indicates electric potential at a point.
$K$ indicates Coulomb’s constant.
$q$ indicates point charge.
$r$ indicates the shortest distance between the circumference and the given point.
Substituting the required values in the equation (1), we get:
$V = \dfrac{{Kq}}{r} \\ V = \dfrac{{9 \times {{10}^9} \times 10 \times {{10}^{ - 3}}}}{5} \\ V = \dfrac{{90 \times {{10}^6}}}{5} \\ V = 1.8 \times {10^7}\,{\text{V}} \\$
Hence, electric potential at a point $P\left( {0,0,4\,{\text{m}}} \right)$ is $1.8 \times {10^7}\,{\text{V}}$ .
The correct option is D.
Note:While solving the numerical, many students tend to make mistake by taking the distance as $3\,{\text{m}}$ into account, however it is $5\,{\text{m}}$ , as it is located in space. So, we need to find the shortest distance between the point charge and the point mentioned in the question. |
# How do you combine (-2t^4+6t^2+5)-(-2t^4+5t^2+1)?
Jul 24, 2015
Group and combine terms with equivalent powers of $t$ to get:
$\textcolor{w h i t e}{\text{XXXX}}$${t}^{2} + 4$
#### Explanation:
(color(red)(−2t^4)color(blue)(+6t^2)color(green)(+5))−(color(red)(−2t^4)color(blue)(+5t^2)color(green)(+1))
$\textcolor{w h i t e}{\text{XXXX}}$Possibly easier (and less error prone) if we change the main subtraction to addition:
=(color(red)(−2t^4)color(blue)(+6t^2)color(green)(+5))+(color(red)(2t^4)color(blue)(-5t^2)color(green)(-1))
$\textcolor{w h i t e}{\text{XXXX}}$Group by powers of $t$
$= \textcolor{red}{- 2 {t}^{4} + 2 {t}^{4}} \textcolor{b l u e}{+ 6 {t}^{2} - 5 {t}^{2}} \textcolor{g r e e n}{+ 5 - 1}$
$\textcolor{w h i t e}{\text{XXXX}}$Combine terms with equal powers of "t"
$= \textcolor{red}{0} \textcolor{b l u e}{+ 1 {t}^{2}} \textcolor{g r e e n}{+ 4}$ |
# SSAT Upper Level Math : Areas and Perimeters of Polygons
## Example Questions
← Previous 1 3 4 5 6 7 8 9 10 11
### Example Question #1 : How To Find The Perimeter Of A Hexagon
A regular hexagon has perimeter 9 meters. Give the length of one side in millimeters.
Explanation:
One meter is equal to 1,000 millimeters, so the perimeter of 9 meters can be expressed as:
9 meters = millimeters.
Since the six sides of a regular hexagon are congruent, divide by 6:
millimeters.
### Example Question #1 : How To Find The Perimeter Of A Hexagon
A hexagon with perimeter 60 has four congruent sides of length . Its other two sides are congruent to each other. Give the length of each of those other sides in terms of .
Explanation:
The perimeter of a polygon is the sum of the lengths of its sides. Let:
Length of one of those other two sides
Now we can set up an equation and solve it for in terms of :
### Example Question #1 : Perimeter Of Polygons
Two sides of a hexagon have a length of , two other sides have the length of , and the rest of the sides have the length of . Give the perimeter of the hexagon.
Explanation:
The perimeter of a polygon is the sum of the lengths of its sides. So we can write:
### Example Question #4 : How To Find The Perimeter Of A Hexagon
A regular hexagon has perimeter 15 feet. Give the length of one side in inches.
Explanation:
As the six sides of a regular hexagon are congruent, we can write:
feet; is the length of each side.
One feet is equal to 12 inches, so we can write:
inches
### Example Question #5 : How To Find The Perimeter Of A Hexagon
Each interior angle of a hexagon is 120 degrees and the perimeter of the hexagon is 120 inches. Find the length of each side of the hexagon.
Explanation:
Since each interior angle of a hexagon is 120 degrees, we have a regular hexagon with identical side lengths. And we know that the perimeter of a polygon is the sum of the lengths of its sides. So we can write:
inches
### Example Question #1 : How To Find The Perimeter Of A Hexagon
A hexagon with perimeter of 48 has three congruent sides of . Its other three sides are congruent to each other with the length of . Find .
Explanation:
The perimeter of a polygon is the sum of the lengths of its sides. Since three sides are congruent with the length of and the rest of the sides have the length of we can write:
Now we should solve the equation for :
### Example Question #7 : How To Find The Perimeter Of A Hexagon
A regular pentagon has sidelength one foot; a regular hexagon has sidelength ten inches. The perimeter of a regular octagon is the sum of the perimeters of the pentagon and the hexagon. What is the measure of one side of the octagon?
Explanation:
A regular polygon has all of its sides the same length. The pentagon has perimeter ; the hexagon has perimeter . The sum of the perimeters is , which is the perimeter of the octagon; each side of the octagon has length .
### Example Question #8 : How To Find The Perimeter Of A Hexagon
Find the perimeter of a hexagon with a side length of .
Explanation:
A hexagon has six sides. The perimeter of a hexagon is:
Substitute the side length.
### Example Question #1 : How To Find The Area Of A Hexagon
Find the area of a regular hexagon that has side lengths of .
Explanation:
Use the following formula to find the area of a regular hexagon:
.
Now, substitute in the length of the side into this equation.
### Example Question #1 : Area Of Polygons
Find the area of a regular hexagon that has a side length of .
Explanation:
Use the following formula to find the area of a regular hexagon:
.
Now, substitute in the length of the side into this equation.
← Previous 1 3 4 5 6 7 8 9 10 11 |
# Ideas in Geometry/Logic
## Lesson 10 logic
Logic In order to understand logic, there are certain key words. These key words consist of: not, and, or, if-then, and if-and-only-if. Suppose I have a statement P= I love tea. The only way this statement can be false, is if P is false. In other words, since P is I love tea, then I must love tea in order for this statement to be true. When we use the symbol ¬, this represents not. In order to deny the truth of the above statement for P, we just put ¬ in front of the P: ¬P=It is not the case that I love tea. The only way ¬P can be true is when P is false. Truth table : File:Untitled.jpg
By using the truth table, we can see that not P is simply the opposite of P. It swaps true for false, and false for true. When we use the symbol ∧, it represents and. Let’s take a look at the statement: I love tea and I’m happy, let P=I love tea, and Q=Im happy. This statement is only true when P and Q are true. For example, I need to love tea and be happy in order for this statement to be true. Truth Table: File:P and Q real.pdf
Looking at the truth table, we see that if P is true, but Q is false, then the statement is false. When P and Q are both false, the statement is still false.
When we look at the statement (¬P)∧(¬Q), it is saying that it is not P and it is not Q. This is the same thing as saying its not P or Q.
We can relate this concept back to venn diagrams. The symbol ∧ can also be the same as the symbol ∩. The symbol ∩ means intersection. In a venn diagram, X ∩ Y are elements that overlap X and Y. When we use the statement P ∧ Q, we are saying both P and Q need to be true in order for the whole statement to be true.
¬(A∧B) and ¬(A ∩ B) both represent “not (A intersects B).. In a venn diagram, this means that this is everything outside of the overlap of A and B.
We can also relate this concept to probability. If two events E1 and E2 are independent, meaning that the occurrence of one has no effect on the occurrence of the other, then the probabiloity that E1 and E2 will happen is the product of the probability that E1 happens with the probability that E2 happens. Symbolically, if P(Ei) is the probability that event Ei happens, then: P(E1 ∧ E2)=P(E1)*P(E2). When we ask “what is the probability that a die will roll a five and a six, we multiply 1/6*1/6 which equal 1/36 chance of rolling a five and a six. Similarly, for the statement Both P ∧ Q, the statement can hold true only when both P and Q are true.
When we use the symbol V, this represents or. Let’s look at the statement: I love tea or I am happy. P=I love tea. Q=I am happy. We can see that the only way this statement can hold true either P or Q is true. For example, if I love tea, but I am not happy, this statement is still true. When I hate tea, but I am happy, the statement is still held true. The only way this statement can be false is when both P and Q are false. It is one or the other, or even both. Truth Table: File:P or q real.pdf
Looking at the truth table, we can see that the statement not (p or Q) has completely the opposite results from (P or Q). The only way the statement (not P) or (not Q) can be false is when both are not P and not Q. In other words, the statement is false when P and Q both hold true. We see that not (P or Q) have the same results as (not P) or (not Q).
When we look at the statement ¬(P V Q), the results for this statement is different from (¬P)V(¬)Q. It is saying different things. (¬P)V(¬)Q means its not P or its not Q. In other words, its either not one or not the other. Whereas, the statement ¬(P V Q), it is saying its not P or Q. We can also say its not P and its not Q.
Here is a link to an example of this concept: Section 2.2 #2 this example goes back to using "not" and relating with statements with "or" and "and"
We can relate this concept to probability. The probability (P) that E2 or E2 will happen is the sum of the probability that E1 happens with the probability that E2 happens. Symbolically, this means that if P(Ei) is the probability that event Ei happens, then: P(E1 V E2)=P(E1)+P(E2). This means that the probability the event 1 will happen or event 2 will happen. We are finding either one or the other that can happen, and we add them all up. Similarly, in the statement PVQ, we know the statement will hold true when either P or Q are true.
When we use the symbol =>, the symbol represents if-then statements. Let’s look at the statement: If I am under 21, then I can’t drink. P=I am under 21, Q=I can’t drink. Both P and Q need to relate. If this, then that. The statement will be false, only when P is true but Q is false. For example, If I am under 21, then I can’t drink, this statement is false because I cannot drink if I am under 21. On the other hand, If I am not under 21, then I can’t drink, this can hold true. It can hold true because even if I am not under 21, I don’t have to drink for different reasons. Maybe one person just cannot drink because they need to drive, or they get sick when they drink. Truth Table:
Looking at the truth table, we see that when both statements are false, it still holds true. For example, If I am not under 21 then I can drink, this statement holds true because since I am not under 21, I can drink. We can simply say that when P=>Q, this means P implies Q.
Here are links for a helpful example: If-Then Truth Tables, Section 2.2 #3
Lastly, if you see if-and-only-if, it is denoted by the symbol<=>. this is the same thing as (p=>Q) ∧ (Q=>P). In other words they are both relational. If anlly only hold true (If P then its Q), and (if Q then its P). Smoo1244 00:17, 7 November 2010 (UTC) |
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# Basic Math Formulas That Are Used in Algebra
Basic math formulas are used so often in real world problems, that it's important to learn how to utilize these formulas early in your Algebra studies.
You've learned how to evaluate algebraic expressions, and now you will use this skill to solve real world problems that involve math formulas.
Let's start with 2 formulas that you are very familiar with. The area and perimeter formulas are used quite often in math. You may be familiar with these formulas, but if not take a look at the following diagram.
## Perimeter Formulas
The perimeter of a figure is the sum of the length of its sides. This is the length around the outside of the figure.
## Area Formulas
The area of a figure is the amount of surface the figure covers.
## Let's Look Specifically at the Area and Perimeter of a Rectangle
Now, that you reviewed the formula, we'll use our area and perimeter formulas for a rectangle in order to solve a problem.
## Example 1 - Finding the Perimeter of a Rectangle
Now, that was pretty easy, huh? Simply substitute the given values for each variable, and evaluate. Let's try the area formula.
## Example 2 - Finding the Area of a Rectangle
There are so many formulas that we could use as examples! There are several formulas for circles and think about all of the formulas that you use in science. This is definitely an algebra skill that is used in everyday life.
Let's take a look at one more formula, just for practice. This is an easy scientific formula - the distance formula.
## Example 3 - The Distance Formula
I"m sure that you've been using formulas for a while in your math studies. I'm hoping that you know have a better sense of the difference between perimeter and area. If not, please make sure that you watch the video above where I am better able to explain these differences. |
The P Series Test Examples 1
# The p-Series Test Examples 1
Recall from The p-Series Test page that:
Theorem (p-Series Test): The special series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ is convergent if $p > 1$ and divergent if $p ≤ 1$.
We will now look at some examples of specifically applying the p-Series test.
## Example 1
Using the p-Series test determine if the series $\sum_{n=1}^{\infty} \frac{n^2}{n^4}$ is convergent or divergent.
Simplifying this series down we get that $\sum_{n=1}^{\infty} \frac{n^2}{n^4} = \sum_{n=1}^{\infty} \frac{1}{n^2}$ and so $p = 2$. Since $p = 2 > 1$, then by the p-series test this series is convergent.
## Example 2
Using the p-Series test determine if the series $\sum_{n=1}^{\infty} \frac{\sec ^2 n - \tan ^2 n}{n}$ is convergent or divergent.
We note the trigonometric identity that $\sec ^2 n - \tan ^2 n = 1$, and so $\sum_{n=1}^{\infty} \frac{\sec ^2 n - \tan ^2 n}{n} = \frac{1}{n}$, and so $p = 1$. Since $p = 1 ≤ 1$ we have that this series is divergent.
## Example 3
Using the p-Series test determine if the series $\sum_{n=1}^{\infty} \frac{\cos (2\pi n)}{n^3}$ is convergent or divergent.
We note that $\cos (2 \pi n) = 1$ for every $n \in \mathbb{N}$ and so $\sum_{n=1}^{\infty} \frac{\cos (2\pi n)}{n^3} = \frac{1}{n^3}$. So $p = 3$ and since $p = 3 ≥ 1$, then by the p-series test this series is convergent. |
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