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# Proof of Logarithmic Product identity
The logarithm of the product of two or more quantities is equal to the sum of their logarithms as per the product rule of the logarithms. The product property of the quantities in logarithmic form is written mathematically as follows.
$\log_{b}{(m \times n)}$ $\,=\,$ $\log_{b}{m}+\log_{b}{n}$
It is time to learn how to derive the product law of the logarithms in algebraic form.
### Write the quantities in Exponential form
Let $m$ and $n$ express two quantities in algebraic form, and they both are factored on the basis of a literal quantity $b$. Therefore, the literal quantities $m$ and $n$ can be written in terms of the literal quantity $b$.
$(1).\,\,$ $m$ $\,=\,$ $b \times b \times b \times \cdots \times b$
$(2).\,\,$ $n$ $\,=\,$ $b \times b \times b \times \cdots \times b$
Let’s assume that the number of factors in $b$ to express the quantity $m$ in the product form is denoted by a variable $x$, and the number of factors in $b$ to express the quantity $n$ in the product form is denoted by a variable $y$.
$(1).\,\,$ $m$ $\,=\,$ $\underbrace{b \times b \times b \times \ldots \times b}_{\displaystyle x \, factors}$
$(2).\,\,$ $n$ $\,=\,$ $\underbrace{b \times b \times b \times \ldots \times b}_{\displaystyle y \, factors}$
Use the exponentiation to write the quantities in product form in exponential notation.
$(1).\,\,$ $m \,=\, b^{\displaystyle x}$
$(2).\,\,$ $n \,=\, b^{\displaystyle y}$
### Express the quantities in Logarithmic form
Now, use the mathematical relation between the logarithms and exponents to convert the exponential equations into the logarithmic equations.
$(1).\,\,$ $m \,=\, b^{\displaystyle x}$ $\,\Longleftrightarrow\,$ $\log_{b}{m} \,=\, x$
$(2).\,\,$ $n \,=\, b^{\displaystyle y}$ $\,\Longleftrightarrow\,$ $\log_{b}{n} \,=\, y$
Therefore, the values of $x$ and $y$ in logarithmic form can be written as follows.
$(1).\,\,$ $x \,=\, \log_{b}{m}$
$(2).\,\,$ $y \,=\, \log_{b}{n}$
### Find the Product of quantities by multiplication
Now, let us multiply the literal quantities $m$ and $n$ to get the product of them mathematically.
$\implies$ $m \times n$ $\,=\,$ $b^{\displaystyle x} \times b^{\displaystyle y}$
Look at the exponential functions on the right hand side of the equation. They both have the same base. So, the product of them can be obtained as per the same base product rule of the exponents.
$\implies$ $m \times n$ $\,=\,$ $b^{\,{\displaystyle x}\,+\,{\displaystyle y}}$
The product of quantities $m$ and $n$ can be written in dot product form or simply $mn$ in mathematics.
$\,\,\,\therefore\,\,\,\,\,\,$ $m.n \,=\, b^{\,{\displaystyle x}\,+\,{\displaystyle y}}$
$\,\,\,\therefore\,\,\,\,\,\,$ $mn \,=\, b^{\,{\displaystyle x}\,+\,{\displaystyle y}}$
### Convert the exponential equation into Logarithm
Let’s denote $s \,=\, x+y$ and $p \,=\, m.n$ for our convenience. Now, write the exponential equation in terms of $s$ and $p$.
$\implies$ $p \,=\, b^{\displaystyle s}$
The above exponential equation can be now transformed into a logarithmic equation by using the mathematical relationship between the exponents and logarithms.
$\implies$ $\log_{b}{p} \,=\, s$
Now, replace the literal quantities $s$ and $p$ in the above logarithmic equation by their actual values.
$\implies$ $\log_{b}{(m.n)} \,=\, x+y$
We have derived that the value of $x$ is logarithm of $m$ to the base $b$ and the value of $y$ is logarithm of $n$ to the base $b$.
$(1).\,\,$ $x\,=\,\log_{b}{(m)}$
$(2).\,\,$ $y\,=\,\log_{b}{(n)}$
Now, substitute the $x$ and $y$ by their corresponding values in the logarithmic equation.
$\,\,\,\therefore\,\,\,\,\,\,$ $\log_{b}{(m.n)}$ $\,=\,$ $\log_{b}{(m)}+\log_{b}{(n)}$
We have successfully proved that the logarithm of the product of $m$ and $n$ to the base $b$ is equal to the sum of the logarithm of $m$ to base $b$ and logarithm of $n$ to base $b$.
It is called the product logarithmic identity and it can be extended to more than two factors too.
$\,\,\,\therefore\,\,\,\,\,\,$ $\log_{b}{(m.n.o.p\cdots)}$ $\,=\,$ $\log_{b}{m}$ $+$ $\log_{b}{n}$ $+$ $\log_{b}{o}$ $+$ $\log_{b}{p}$ $+$ $\cdots$
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# How to Find P-Value
Coming up next: How to Calculate Sharpe Ratio: Definition, Formula & Examples
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• 0:00 Tests of Significance
• 0:56 P-Value Defined
• 1:45 Finding the P-Value
• 3:07 The Critical Value
• 3:58 What Significance Tells Us
• 5:05 Lesson Summary
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Lesson Transcript
Instructor: Tracy Payne, Ph.D.
Tracy earned her doctorate from Vanderbilt University and has taught mathematics from preschool through graduate level statistics.
When is a number so much bigger or smaller than another that it should raise some eyebrows? Tests of significance help make such determinations. This lesson explains the p-value in significance tests, how to calculate them, and how to evaluate the results.
## Tests of Significance
Imagine that you want to be the new point guard of your basketball team, but before you try out for the position, you want to make sure you have, pun intended, a real shot at achieving your goal. You shoot 20 free throws and make 12 of them; that's a 60% accuracy rate. You want to know if your accuracy rate, or the observation, is about the same or different than the team's accuracy rate, or the population statistic; enough to replace the old point guard.
You can do a test of significance to ascertain if your accuracy rate is significantly different from that of the team. A significance test measures whether some observed value is similar to the population statistic, or if the difference between them is large enough that it isn't likely to be by coincidence. When the difference between what is observed and what is expected surpasses some critical value, we say there is statistical significance.
## P-Value Defined
A standard normal distribution curve represents all of the observations of a single random variable such that the highest point under the curve is where you would expect to find values closest to the mean and values least likely to be observed in the smallest part under the curve.
The p-value is the probability of finding an observed value or a data point relative to all other possible results for the same variable. If the observed value is a value most likely to be found among all possible results, then there is not a statistically significant difference. If, on the other hand, the observed value is a value among unlikely values to be found, then there is a statistically significant difference. The smaller the probability associated with the observed value, the more likely the result is to be significant.
## Finding The P-Value
To find the p-value, or the probability associated with a specific observation, you must first calculate the z score, also known as the test statistic.
The formula for finding the test statistic depends on whether the data includes means or proportions. The formulas we'll discuss assume a:
1. Single sample significance test
2. Normal distribution
3. Large sample size.
When dealing with means, the z score is a function of the observed value (x-bar), population mean (mu), standard deviation (s), and the number of the observations (n).
When dealing with proportions, the z score is a function of the observed value (p-hat), proportion observed in the population (p), probability of successful outcome (p), probability of failure (q = 1 - p), and the number of trials (n).
After calculating the z score, you must look up the probability associated with that score on a Standard Normal Probabilities Table. This probability is the p-value or the probability of finding the observed value compared to all possible results. The p-value is then compared to the critical value to determine statistical significance.
## The Critical Value
The critical value, or significance level, is established as part of the study design and is denoted by the Greek letter alpha. If we choose an alpha = 0.05, we are requiring an observed data point be so different from what is expected that it would not be observed more than 5% of the time. An alpha equaling 0.01 would be even more strict. In this case, a statistically significant test statistic beyond this critical value has less than a 1 in 100 probability of occurring by chance.
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# If $(2.3)^x = (0.23)^y = 1000$ then prove that $\dfrac{1}{x} -\dfrac{1}{y} = \dfrac{1}{3}$
It is given in this problem that $(2.3)^{\displaystyle x} = (0.23)^{\displaystyle y} = 1000$.
### Step: 1
Consider $(2.3)^{\displaystyle x} = 1000$
The value of the right hand side of this equation is $1000$ and it can be written as the factors of $10$. So, take the common logarithm both sides.
$\implies \log (2.3)^{\displaystyle x} = \log 1000$
$\implies \log (2.3)^{\displaystyle x} = \log 10^3$
$\implies x \log 2.3 = 3 \log 10$
The base of the common logarithm is $10$. So, the $\log 10 = 1$.
$\implies x \log 2.3 = 3 \times 1$
$\implies x \log 2.3 = 3$
$\implies \log 2.3 = \dfrac{3}{x}$
### Step: 2
Now consider $(0.23)^{\displaystyle y} = 1000$
Repeat the same procedure one more time.
$\implies \log (0.23)^{\displaystyle y} = \log 1000$
$\implies \log (0.23)^{\displaystyle y} = \log 10^3$
$\implies y \log 0.23 = 3 \log 10$
$\implies y \log 0.23 = 3 \times 1$
$\implies y \log 0.23 = 3$
$\implies \log 0.23 = \dfrac{3}{y}$
### Step: 3
Subtract the value of $\log 0.23$ from the value of $\log 2.3$.
$\implies \log 2.3 -\log 0.23 = \dfrac{3}{x} -\dfrac{3}{y}$
Apply the division rule of the logarithm to obtain the quotient of the logarithmic function.
$\implies \log \Bigg(\dfrac{2.3}{0.23}\Bigg) = \dfrac{3}{x} -\dfrac{3}{y}$
$\require{cancel} \implies \log \Bigg(\dfrac{\cancel{2.3}}{\cancel{0.23}}\Bigg) = \dfrac{3}{x} -\dfrac{3}{y}$
$\implies \log 10 = \dfrac{3}{x} -\dfrac{3}{y}$
$\implies 1 = \dfrac{3}{x} -\dfrac{3}{y}$
$\implies \dfrac{3}{x} -\dfrac{3}{y} = 1$
Take the number $3$ common from both terms.
$\implies 3 \times \Bigg(\dfrac{1}{x} -\dfrac{1}{y}\Bigg) = 1$
$\therefore \,\,\,\,\, \dfrac{1}{x} -\dfrac{1}{y} = \dfrac{1}{3}$
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# 3 Modular arithmetic
## 3.3 Operations in modular arithmetic
The Division Algorithm tells us that all the possible remainders on division by an integer n lie in the set
We denote this set by n. For each integer n ≥ 2 we have a set n, and it is on these sets that we perform modular arithmetic. The modular addition operations +n and modular multiplication operations ×n are defined as follows.
### Definitions
For any integer n ≥ 2,
For a and b in n, the operations +n and ×n are defined by:
a +n b is the remainder of a + b on division by n;
a ×n b is the remainder of a × b on division by n.
The integer n is called the modulus for this arithmetic.
Note: a +n b is read as a plus b (mod n), and may also be written a + b (mod n). Similarly, a ×n b is read as a times b (mod n) and may also be written as a × b (mod n).
For example, 7 = { 0,1,2,3,4,5,6} and we have
You have certainly met some modular arithmetic before, as the operations +12 and +24 are used in measuring time on 12-hour and 24-hour clocks, respectively.
### Exercise 41
Evaluate the following.
(a) 3 +5 7, 4 +17 5, 8 +16 12.
(b) 3 ×5 7, 4 ×17 5, 8 ×16 12.
### Solution
(a) 3 +5 7 = 0, 4 +17 5 = 9, 8 +16 12 = 4.
(b) 3 ×5 7 = 1, 4 ×17 5 = 3, 8 ×16 12 = 0.
In Sections 1 and 2 we listed some properties satisfied by the real and complex numbers. We now investigate whether the sets n satisfy similar properties.
We also investigate what equations we can solve in n; for example, can we solve the equations
These may look much simpler than the equations that we were trying to solve in , but they pose interesting questions. We shall see that the answers may depend on the modulus that we are using.
Before we discuss these questions further, we look at addition and multiplication tables, which provide a convenient way of studying addition and multiplication in n.
We consider addition first. Here are the addition tables for 4 and 7.
In order to evaluate 4 +7 2, say, we look in the row labelled 4 and the column labelled 2 in the second table to obtain the answer 6.
### Exercise 42
(a) Use the tables above to solve the following equations.
(i) x +4 3 = 2
(ii) x +7 5 = 2
(iii) x +4 2 = 0
(iv) x +7 5 = 0
(b) What patterns do you notice in the tables?
### Solution
(a)
(i) 3 +4 3 = 2, so x = 3.
(ii) 4 +7 5 = 2, so x = 4.
(iii) 2 +4 2 = 0, so x = 2.
(iv) 2 +7 5 = 0, so x = 2.
(b) You may have noticed that:
1. each element appears exactly once in each row and exactly once in each column;
2. there is a pattern of diagonal stripes running down from right to left.
### Exercise 43
(a) Construct the addition table for 6.
(b) Solve the equations x +6 1 = 5 and x +6 5 = 1.
### Solution
(a)
(b) x +6 1 = 5 has solution x = 4.
x +6 5 = 1 has solution x = 2.
For every integer n ≥ 2, the additive properties of n are the same as the additive properties of , as follows.
Property A1 follows from the Division Algorithm and the definition of n. The other properties can be deduced from the corresponding properties for integers.
### Exercise 44
By using the corresponding property for integers, prove property A5.
### Solution
By definition, a +n b and b +n a are the remainders of a + b and b + a, respectively, on division by n. But a + b = b + a, so a +n b = b+n a.
If a, b n and a +n b = 0, then we say that b is the additive inverse of a in n. For example, 4 and 5 belong to 9 and 4 +9 5 = 0, so 5 is the additive inverse of 4 in 9. Property A3 states that every element of n has an additive inverse in n.
Additive inverses are sometimes written in the form −na; that is, if a +n b = 0, then we write b = −na. For example, 5 = −94.
### Exercise 45
(a) Use the addition table for 7 (which appear above Exercise 42) to complete the following table of additive inverses in 7.
(b) Complete the following table of additive inverses in n, explaining why your answers are correct.
### Solution
(a)
(b)
The additive inverse of 0 is always 0, since 0 +n 0 = 0. For any integer r > 0 in n, nr n and r + (nr) = n, so r +n (nr) = 0.
The existence of additive inverses means that, as well as doing addition modulo n, we can also do subtraction. We define an b or, equivalently, ab (mod n), to be the remainder of ab on division by n.
(With this definition, an b is equal to a +n (−nb).)
For example, to find 2 −8 7, we have
Since 3 8, it follows that |
# How do you find the slope of the line parallel to and perpendicular to 4x-3y=17?
Aug 31, 2017
Slope of the parallal line is $\frac{4}{3}$ and slope of the perpendicular line is $- \frac{3}{4}$
#### Explanation:
$4 x - 3 y = 17 \mathmr{and} 3 y = 4 x - 17 \mathmr{and} y = \frac{4}{3} x - \frac{17}{3}$
Slope of the line is $m = \frac{4}{3} \left[y = m x + c\right]$
Slope of parallel lines are equal.
Slope of the parallal line is ${m}_{1} = \frac{4}{3}$
Product of slopes of perpendicular lines is ${m}_{2} \cdot m = - 1$
${m}_{2} = - \frac{1}{\frac{4}{3}} = - \frac{3}{4}$
Slope of the perpendicular line is ${m}_{2} = - \frac{3}{4}$ |
4.9
## UNSW Sydney
Skip to 0 minutes and 12 secondsNow we would like to introduce some applications which are very modern for quadratic relations-- applications that were discovered by two French car engineers around 1960. They give us a new approach to design, a problem of how to specify curves. So in order to understand this, we have a little bit of a discussion about parameterising curves. This is a point of view that's shifted a little bit from the idea of an equation describing a curve to the idea of having a curve described by a parameter t as it's moving. It's an interesting alternate way of thinking about what a curve is. So let's start with the parametric equation for a line.
Skip to 1 minute and 0 secondsSo here's a line through three points a, b, and c. And we can write down its equation pretty easily. So the equation of the line-- it's called l-- is x plus 2y equals 7. And we can check that, because the point 1, 3 satisfies the equation. So does the point 3, 2, so does the point 5, 1, and so on. Now, here's an alternate form for the same equation. I'm going to make a certain expression. I'm going to write 1 minus t times [1, 3] plus t times [3, 2]. And you'll see in a minute why this is a good thing to do.
Skip to 1 minute and 47 secondsSo if we take that combination and we combine it-- 1 minus t times 1 plus 3t is 1 plus 2t, and 3 times 1 minus t plus 2t will be a total of 3 minus t. So let's have a look at this particular expression. t is a parameter here that's allowed to vary. When t equals 0, then we get the point 1, 3. When t equals 1, then we get 3 here and 2 here. We get point b. If t is some intermediate value between 0 and 1, then we're going to get some point that's between a and b.
Skip to 2 minutes and 40 secondsThose of you who know a little bit about vectors, there's really a vector formulation here. We're really taking the point a and adding t times the vector ab to get at this formula. Just another way of thinking about it. So there's a general formula here. So if we have two arbitrary points-- let's call them p0 and p1-- and we're interested in the line through these two points, so then the line p0 p1 can be described by the coordinates 1 minus t times the coordinate of p0 plus t times the coordinate of p1. That's basically exactly what I did over here where I used the points a and b to create this thing here.
Skip to 3 minutes and 31 secondsSo this is the general form for what I did in the specific case up here. So this is called an Affine linear combination of these points p0 and p1. With the property that this coefficient and this coefficient always add up to 1, that will guarantee that we're always getting a point on this line p0 p1. So around 1960, Paul de Casteljau and Pierre Bezier, who were two French engineers working for Renault and Citroen, French car companies, both were tackling the same problem of how to describe mathematically a general kind of curve in a way that the information could easily be translated to somebody else.
Skip to 4 minutes and 11 secondsThey came up with this remarkable idea, which was to extend the parameterised line segment that we were talking about to the quadratic case. And many of you will be familiar with this in some form, because Photoshop, CorelDRAW, Illustrator, a lot of CAD programmes will be using Bezier curves and control points to describe curves. So what's the basic idea? We start with three points now, p0, p1, and p2. And we imagine these segments here being traversed by parameters. So imagine one person walking from here to here at a steady rate, say time 0 to time 1.
Skip to 4 minutes and 58 secondsAnd at the same time, somebody else is walking from here to here at a steady rate, starting at time t equals 0 at time t equals 1. And at all points, we're connecting the point on this segment with the point on this segment with a green bar. So that green bar is moving. And at the same time as these two fellows are walking, there's a third person which is walking along the green segment, also starting zero at 1 and finishing at time 1 at the other end.
Skip to 5 minutes and 30 secondsSo if we look at the cumulative effect of all three of these motions-- this red point moving, this red point moving, and at the meantime, this green bar moving and this intermediate point moving-- then the trace of this point 1 is a curve. And it's a quadratic curve. It's a conic section. In fact, it's a parabola. It's a parabola which is tangent to these bounding lines and goes through these two control points. This control point is not actually on the curve, but it controls the position of this de Casteljau Bezier curve. It's a very useful kind of thing. So for example, in the theory of fonts, you have some interesting font that you want to describe.
Skip to 6 minutes and 21 secondsOne approach to that is to cut it up into manageable pieces. And on the manageable pieces, you create some Bezier control points. For example, you might draw tangents there. So you have three control points which would describe that little segment of the curve. So if you do this for a couple of segments, then you can describe the entire curve with just a finite number of control points. It's a very powerful tool, which is actually used in true type fonts made by Apple and used by Microsoft-- just using, basically, these quadratic de Casteljau Bezier curves. Here is the actual Affine combination equivalent of the formula that I wrote down for a line.
Skip to 7 minutes and 12 secondsSo it's a quadratic version of the formula that actually gives you very specific control over the curve. If you know what p0, p1, p2 are, this is the formula that allows you to describe exactly what's happening to the general point on the curve. It's a very important and useful application of quadratic curves in modern design.
# Quadratics in design
Lines and other curves can also be described by parameters. We see how to apply this idea to introduce the remarkable approach to curves of de Casteljau and Bezier introduced in the early 1960s.
## Parametrising lines with convex linear combinations
While lines can be described by linear equations, there is also another way, involving parametrising points on the line.
If $\normalsize{A}$ and $\normalsize{B}$ are two points, we can traverse the line segment $\normalsize{AB}$ linearly as time $\normalsize{t}$ goes from $\normalsize{0}$ to $\normalsize{1}$, so that at time $\normalsize{t=0}$ we are at $\normalsize{A}$, and at time $\normalsize{t=1}$ we are at $\normalsize{B}$. For example, if $\normalsize{A=[2,3]}$ and $\normalsize{B=[5,1]}$, then such a path is given by
This is also called a convex linear combination of points $\normalsize A$ and $\normalsize B$, because we can write it as
and the two coefficients have the property that they sum always to $\normalsize 1$.
A nice physical model of this would be to imagine a very light (but strong) wooden rod between the points $\normalsize A$ and $\normalsize B$. If we were to take $\normalsize 1$kg and then hung $\normalsize t$ kilograms off point $\normalsize B$ and the remaining $\normalsize (1-t)$ kilograms off point $\normalsize A$ then the center of mass (which is the fulcrum, or the point of balance) would be the same as the point $\normalsize P$ given by the convex linear combination.
UNSW Australia
## De Casteljau Bezier curves
Around 1960, two French car company engineers independently discovered a new way of specifying curves, using parameters. The simplest of these curves are quadratic de Casteljau Bezier curves, and they always give parabolas, but rather general ones, not necessarily of the form $\normalsize y=ax^2+bx+c$. These curves have gone on to revolutionise design theory, are important in architecture, and figure prominently in the design of modern fonts. |
1. Chapter 9 Class 12 Differential Equations
2. Serial order wise
3. Ex 9.5
Transcript
Ex 9.5, 3 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve each of them. + =0 Step 1 Find (x y) dy (x + y) dx = 0 (x y) dy = (x + y) dx Step 2: Put = F(x, y) and find out F( x, y) F(x, y) = + F( x, y) = + = ( + ) ( ) = + = F(x, y) F( x, y) = 0 F(x, y) Hence, F(x, y) is a homogenous Function of with degree zero So, is a homogenous differential equation. Step 3. Solving by putting y = vx Put y = vx. differentiating w.r.t.x = x + = + v Putting value of and y = vx in (1) = + x + v = + x + v = (1 + ) (1 ) x + v = (1 + ) (1 ) x = (1 + ) (1 ) v x = 1 + (1 ) 1 x = 1 + + 2 1 x = 1 + 2 1 1 1 + 2 = Integrating both sides 1 1 + 2 = tan 1 v 1 + 2 = log + Let I = 1 + 2 dv Put t = 1 + 2 Diff w.r.t. v (1 + v2) = 2v = dv = 2 Therefore I = 1 + 2 dv = 2 = 1 2 + Putting t = 1 + v2 = 1 2 1+ 2 + c Putting value of I in (2) tan 1 v 1 2 log 1+ 2 = log + c tan 1 v = 1 2 log 1+ 2 + log + c tan 1 v = 1 2 log 1+ 2 + 2 2 log + c tan 1 v = 1 2 log 1+ 2 + 2 log + c tan 1 v = 1 2 log 1+ 2 . 2 + c Putting v = tan 1 = 1 2 log 1+ 2 2 +c tan 1 = 1 2 log 2 + 2 2 2 +c tan 1 = 1 2 log 2 + 2 +c tan 1 = log + + is the required solution
Ex 9.5 |
# Circles, Cylinders and Circular Shapes
Scheme of work: Key Stage 3: Year 8: Term 4: Circles, Cylinders and Circular Shapes
#### Prerequisite Knowledge
• Derive and apply formulae to calculate and solve problems involving: the perimeter and area of triangles, parallelograms, trapezia, the volume of cuboids (including cubes) and other prisms.
• Calculate and solve problems involving: perimeters of 2-D shapes and composite shapes.
#### Key Concepts
• The radius is the distance from the centre to any point on the circumference. The plural radius is radii.
• The diameter is the distance across the circle through the centre.
• Ï€ is a Greek letter used to represent the value of the circumference of a circle divided by its diameter.
• The circumference is the distance about the edge of a circle. The circumference of a circle can be calculated as:
• C = ÏπD where D is the diameter, or,
• C = 2πr
• Where r is the radius.
• The area of a circle can be calculated using the formula
• A = πr2, where r is the radius.
• A cylinder is a circular prism.
#### Working Mathematically
• Develop fluency
• Use language and properties precisely to analyse numbers, algebraic expressions, 2-D
and 3-D shapes, probability and statistics.
• Use algebra to generalise the structure of arithmetic, including to formulate
mathematical relationships
• Substitute values in expressions, rearrange and simplify expressions, and solve
equations
• Reason mathematically
• Make and test conjectures about patterns and relationships; look for proofs or counterexamples.
• Begin to reason deductively in geometry, number and algebra, including using geometrical constructions
• Solve problems
• Begin to model situations mathematically and express the results using a range of
formal mathematical representations
• Select appropriate concepts, methods and techniques to apply to unfamiliar and nonroutine
problems.
#### Subject Content
• Shape
• Derive and illustrate properties of triangles, quadrilaterals, circles, and other plane figures [for example, equal lengths and angles] using appropriate language and technologies
• Calculate and solve problems involving: perimeters of 2-D shapes (including circles), areas of circles and composite shapes
• Derive and apply formulae to calculate and solve problems involving: perimeter and area of circles and cylinders
## Circles, Cylinders and Circular Shapes Resources
### Mr Mathematics Blog
#### Planes of Symmetry in 3D Shapes
Planes of Symmetry in 3D Shapes for Key Stage 3/GCSE students.
Use isometric paper for hands-on learning and enhanced understanding.
#### GCSE Trigonometry Skills & SOH CAH TOA Techniques
Master GCSE Math: Get key SOH-CAH-TOA tips, solve triangles accurately, and tackle area tasks. Ideal for students targeting grades 4-5.
#### Regions in the Complex Plane
Explore Regions in the Complex Plane with A-Level Further Maths: inequalities, Argand diagrams, and geometric interpretations. |
Bouncing Barney and Ceva's Theorem
Final Assignment
by
A. Barney is in the triangular room shown here. He walks from a point on BC parallel to AC. When he reaches AB, he turns and walks parallel to BC. When he reaches AC, he turns and walks parallel to AB. Prove that Barney will eventually return to his starting point. How many times will Barney reach a wall before returning to his starting point? Explore and discuss for various starting points on line BC, including points exterior to segment BC. Discuss and prove any mathematical conjectures you find in the situation.
Here is a GSP file of Bouncing Barney (triangle with line segments). A script tool has been included in the download. Clickhere.
If you want to look Barney’s exterior path (triangle with lines), click here for a GSP file with this script tool.
Let us begin the proof that Barney will eventually return to his starting point by labeling the triangle like shown below:
Algebraically, line segment AB is the line y = a/b x, line segment AC is the line y = b/(a-c) x – bc/(a-c), and line segment BC is the line y = 0. I will not go through all the algebra but the first few steps. First, construct a line through (d,0), where Barney starts, parallel to AC and intersecting AB. This line would have the same slope as AC and goes through (d, 0). So the equation of this line would be y = b/(a-c) (x-d). This line would intersect AB at the point (ad/c, bd/c). This is the first leg of Barney’s journey. The proof consists of 5 more equations with intersections on the sides of the triangle as shown below:
Barney turns 5 times with 6 legs back to the starting point.
What about choosing a point exterior of the triangle?
Notice the Superman shield? Also, note that Barney returns to his starting point with 5 turns and 6 legs of the journey around the exterior of the triangle.
Barney is located at (d, 0). The line segment formed from the starting point (d, 0) to the intersection of line AC has an equation y = b/a (x-d) since it is parallel to line AB. Thus the intersection point is
(a + c – ad/c, (abcabd)/c(a – c)). Look familiar? Similarly to the above algebraic proof, we can conclude that Barney does make it back to (d, 0).
Conjecture: If Barney starts at a midpoint of one of the sides of the triangle, he will have 2 turns to make and 3 legs of the journey to make it back to his starting point. Also, several similar triangles are formed. Look at the picture below.
Conclusions:
1. Barney returns to his starting point if walking inside or outside the triangle.
2. If Barney starts at the midpoint, he returns after 2 turns to his starting point.
3. If Barney starts at any other point on BC besides the midpoint, he returns after 5 turns to his starting point.
4. Perimeter:
a. If Barney starts on a midpoint of the side of the triangle, he will travel half of the triangle’s perimeter.
b. If Barney starts on any other point besides the midpoint of a side of the triangle, he will travel the triangle’s perimeter before returning to his starting point.
These conclusions could be proved using properties of similar triangles (unlike the algebraic proofs above).
B. Ceva's Theorem. Consider any triangle ABC. Select a point P inside the triangle and draw lines AP, BP, and CP extended to their intersections with the opposite sides in points D, E, and F respectively.
Exploration:
Consider the ratio below in any triangle:
The above ratio is equal to 1 if inside the triangle ABC.
For this proof, draw a line through A parallel to line segment BC as shown below.
Constructing the line segments above, several similar triangles are formed.
Δ AET ~ Δ CEB, so AE / EC = AT / CB. (I)
Δ BFC ~ Δ AFS, so BF / FA = CB / SA. (II)
Δ CDP ~ Δ SAP, so CD / SA = DP / AP. (III)
Δ BDP ~ Δ TAP, so BD / TA = DP / AP. (IV)
Looking at (III) and (IV), CD / SA = BD / TA. With a little algebra, this proportion is the same as CD / BD = SA / TA. (V)
By multiplying (I), (II), (V), AE / EC * BF / FA * CD / DB = AT / CB * CB / SA * SA / TA = 1, or
Conjectures:
1. The medians of any triangle are concurrent.
2. The altitudes of any triangle are concurrent.
3. The interior angle bisectors of a triangle are concurrent.
Let us look again at the centroid case:
In triangle ABC, line segments AD, BE, and CF are medians. Therefore, AF = FB, BD = DC, and CE = EA.
So (AF)(BD)(CE) = (FB)(DC)(EA), or
Thus, according to Ceva, line segments AD, BE, and CF are concurrent.
Exploration of Ceva’s Theorem using lines rather than segments to construct ABC so that point P can be outside the triangle. Below are sketches of when the points P, D, E, and F are collinear.
According to Menelaus, if we were to take direction into account in the two figures above then the ratio would equal -1 instead of 1.
Click here to open a GSP file with the collinear cases above. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 3.12: Trigonometric Equations Using Half Angle Formulas
Difficulty Level: At Grade Created by: CK-12
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Practice Trigonometric Equations Using Half Angle Formulas
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As you've seen many times, the ability to find the values of trig functions for a variety of angles is a critical component to a course in Trigonometry. If you were given an angle as the argument of a trig function that was half of an angle you were familiar with, could you solve the trig function?
For example, if you were asked to find
sin22.5\begin{align*}\sin 22.5^\circ\end{align*}
would you be able to do it? Keep reading, and in this Concept you'll learn how to do this.
### Watch This
James Sousa Example: Determine a Sine Function Using a Half Angle Identity
### Guidance
It is easy to remember the values of trigonometric functions for certain common values of θ\begin{align*}\theta\end{align*}. However, sometimes there will be fractional values of known trig functions, such as wanting to know the sine of half of the angle that you are familiar with. In situations like that, a half angle identity can prove valuable to help compute the value of the trig function.
In addition, half angle identities can be used to simplify problems to solve for certain angles that satisfy an expression. To do this, first remember the half angle identities for sine and cosine:
sinα2=1cosα2\begin{align*}\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}\end{align*} if α2\begin{align*}\frac{\alpha}{2}\end{align*} is located in either the first or second quadrant.
sinα2=1cosα2\begin{align*}\sin \frac{\alpha}{2} = - \sqrt{\frac{1 - \cos \alpha}{2}}\end{align*} if α2\begin{align*}\frac{\alpha}{2}\end{align*} is located in the third or fourth quadrant.
cosα2=1+cosα2\begin{align*}\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha}{2}}\end{align*} if α2\begin{align*}\frac{\alpha}{2}\end{align*} is located in either the first or fourth quadrant.
cosα2=1+cosα2\begin{align*}\cos \frac{\alpha}{2} = - \sqrt{\frac{1 + \cos \alpha}{2}}\end{align*} if α2\begin{align*}\frac{\alpha}{2}\end{align*} is located in either the second or fourth quadrant.
When attempting to solve equations using a half angle identity, look for a place to substitute using one of the above identities. This can help simplify the equation to be solved.
#### Example A
Solve the trigonometric equation sin2θ=2sin2θ2\begin{align*}\sin^2 \theta = 2 \sin^2 \frac{\theta}{2}\end{align*} over the interval [0,2π)\begin{align*}[0, 2\pi)\end{align*}.
Solution:
sin2θsin2θ1cos2θcosθcos2θcosθ(1cosθ)=2sin2θ2=2(1cosθ2)=1cosθ=0=0Half angle identityPythagorean identity
Then cosθ=0\begin{align*}\cos \theta = 0\end{align*} or 1cosθ=0\begin{align*}1 - \cos \theta = 0\end{align*}, which is cosθ=1\begin{align*}\cos \theta = 1\end{align*}.
θ=0,π2,3π2,or 2π\begin{align*}\theta = 0, \frac{\pi}{2}, \frac{3\pi}{2}, \text{or } 2\pi\end{align*}.
#### Example B
Solve 2cos2x2=1\begin{align*}2 \cos^2 \frac{x}{2} = 1\end{align*} for 0x<2π\begin{align*}0 \le x < 2 \pi\end{align*}
Solution:
To solve 2cos2x2=1\begin{align*}2 \cos^2 \frac{x}{2} = 1\end{align*}, first we need to isolate cosine, then use the half angle formula.
2cos2x2cos2x21+cosx21+cosxcosx=1=12=12=1=0
cosx=0\begin{align*}\cos x = 0\end{align*} when x=π2,3π2\begin{align*} x = \frac{\pi}{2}, \frac{3 \pi}{2}\end{align*}
#### Example C
Solve tana2=4\begin{align*}\tan \frac{a}{2} = 4\end{align*} for 0a<360\begin{align*}0^\circ \le a < 360^\circ\end{align*}
Solution:
To solve tana2=4\begin{align*}\tan \frac{a}{2} = 4\end{align*}, first isolate tangent, then use the half angle formula.
tana21cosa1+cosa1cosa1+cosa16+16cosa17cosacosa=4=4=16=1cosa=15=1517
Using your graphing calculator, cosa=1517\begin{align*}\cos a = - \frac{15}{17}\end{align*} when a=152,208\begin{align*}a = 152^\circ, 208^\circ\end{align*}
### Guided Practice
1. Find the exact value of cos112.5\begin{align*}\cos 112.5^\circ\end{align*}
2. Find the exact value of sin105\begin{align*}\sin 105^\circ\end{align*}
3. Find the exact value of tan7π8\begin{align*}\tan \frac{7 \pi}{8}\end{align*}
Solutions:
1.
cos112.5=cos2252=1+cos2252=1222=2222=224=222
2.
sin105=sin2102=1cos2102=1322=2322=234=232
3.
tan7π8=tan127π4=1cos7π4sin7π4=12222=22222=222=22+22=2+1
### Concept Problem Solution
Knowing the half angle formulas, you can compute sin22.5\begin{align*}\sin 22.5^\circ\end{align*} easily:
sin22.5=sin(452)=1cos452=1222=2222=224=222
### Explore More
Use half angle identities to find the exact value of each expression.
1. tan15\begin{align*}\tan 15^\circ\end{align*}
2. tan22.5\begin{align*}\tan 22.5^\circ\end{align*}
3. cot75\begin{align*}\cot 75^\circ\end{align*}
4. tan67.5\begin{align*}\tan 67.5^\circ\end{align*}
5. tan157.5\begin{align*}\tan 157.5^\circ\end{align*}
6. tan112.5\begin{align*}\tan 112.5^\circ\end{align*}
7. cos105\begin{align*}\cos 105^\circ\end{align*}
8. \begin{align*}\sin 112.5^\circ\end{align*}
9. \begin{align*}\sec 15^\circ\end{align*}
10. \begin{align*}\csc 22.5^\circ\end{align*}
11. \begin{align*}\csc 75^\circ\end{align*}
12. \begin{align*}\sec 67.5^\circ\end{align*}
13. \begin{align*}\cot 157.5^\circ\end{align*}
Use half angle identities to help solve each of the following equations on the interval \begin{align*}[0,2\pi)\end{align*}.
1. \begin{align*}3\cos^2(\frac{x}{2})=3\end{align*}
2. \begin{align*}4\sin^2 x=8\sin^2(\frac{x}{2})\end{align*}
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 3.12.
### Vocabulary Language: English
Half Angle Identity
Half Angle Identity
A half angle identity relates a trigonometric function of one half of an argument to a set of trigonometric functions, each containing the original argument.
## Date Created:
Sep 26, 2012
Feb 26, 2015
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# What is the equation of the normal line of f(x)= cosx at x = pi/8?
Feb 7, 2016
$y - \frac{\pi}{8} = \sqrt{4 + 2 \sqrt{2}} \left(x - \frac{\sqrt{2 + \sqrt{2}}}{2}\right)$
#### Explanation:
We can find the point the normal line will intercept:
$f \left(\frac{\pi}{8}\right) = \cos \left(\frac{\pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2}$
The normal line passes through the point $\left(\frac{\pi}{8} , \frac{\sqrt{2 + \sqrt{2}}}{2}\right)$
.
The exact value can be found through the cosine half-angle formula:
$\cos \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos \theta}{2}}$
To find the slope of the tangent line, find $f ' \left(\frac{\pi}{8}\right)$. First, find $f ' \left(x\right)$.
The derivative of cosine is negative sine:
$f \left(x\right) = \cos \left(x\right)$
$f ' \left(x\right) = - \sin \left(x\right)$
The slope of the tangent line is
$f ' \left(\frac{\pi}{8}\right) = - \sin \left(\frac{\pi}{8}\right) = - \frac{\sqrt{2 - \sqrt{2}}}{2}$
This can be found through the sine half-angle formula:
$\sin \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos \theta}{2}}$
Now, recall that we are looking for the normal line, which is perpendicular to the tangent line. Perpendicular lines have opposite reciprocal slopes:
• Slope of tangent line: $\text{ } - \frac{\sqrt{2 - \sqrt{2}}}{2}$
• Slope of normal line: $\text{ } \frac{2}{\sqrt{2 - \sqrt{2}}} = \sqrt{4 + 2 \sqrt{2}}$
For an explanation of how to simplify the slope of normal line, ask, but I'll focus more on the calculus here.
We now know that the normal line passes through the point $\left(\frac{\pi}{8} , \frac{\sqrt{2 + \sqrt{2}}}{2}\right)$ and has a slope of $\sqrt{4 + 2 \sqrt{2}}$.
The equation of such line in point-slope form is:
$y - \frac{\pi}{8} = \sqrt{4 + 2 \sqrt{2}} \left(x - \frac{\sqrt{2 + \sqrt{2}}}{2}\right)$ |
# Tangent
## Definition
The relation of angle with ratio of length of opposite side to length of adjacent side in a right angled triangle is called tangent.
Tangent reveals the mutual relation of angle of right angled triangle with ratio of length of opposite side to length of adjacent side. It is primarily defined from a ratio at an angle of right angled triangle in trigonometry. Hence, tangent is also called as trigonometric ratio.
Due to the functionality of representing the angle with ratio of length of opposite side to length of adjacent side, the trigonometric ratio tangent is also called as trigonometric function trigonometrically.
### Mathematical form
The name tangent was introduced in trigonometry to represent the relation of angle with ratio of length of opposite side to length of adjacent side. In fact, the ratio of length of opposite side to length of adjacent side is changed whenever the angle of right angled triangle is changed. The mutual dependency is expressed in mathematical language by writing tangent along with angle. In mathematics, tangent is expressed in its short form $tan$.
In the case of triangle $BAC$, the angle of this right angled triangle is theta $(\theta)$. Therefore, tangent along with angle is written as $tan \, \theta$ in mathematical language.
$$tan \, \theta = { Length \, of \, Opposite \, Side \over Length \, of \, Adjacent \, Side}$$
In $Δ BAC$, the length of opposite side $(\overline{BC})$ is $BC$ and length of adjacent side $(\overline{AB})$ is $AB$. Therefore, the mathematical expression can be expressed in terms of lengths of opposite side and adjacent side.
$$∴ \,\, tan \, \theta = { BC \over AB }$$
#### Example
Consider right angled triangle $DEF$ and it is measured that
1. Length of the opposite side $(\overline{DF})$ is $DF = 3$ meters
2. Length of the adjacent side $(\overline{EF})$ is $EF = 4$ meters
3. Angle of the triangle $∠DEF$ is $36.87^°$
Substitute three values in the mathematical form expression of tangent.
$$⇒ tan \, 36.87^° = { DF \over EF }$$
$$⇒ tan \, 36.87^° = { 3 \over 4 }$$
$⇒ tan \, 36.87^° = 0.75$
The ratio of length of opposite side to length of adjacent side is $0.75$ at the angle of $36.87^°$. The value of ratio $0.75$ is called as value of tangent at the angle of $36.87^°$. Thus, it represents the relation of angle with ratio of length of opposite side to length of adjacent side in trigonometry. |
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### Course: Algebra 2>Unit 11
Lesson 3: The Pythagorean identity
# Pythagorean identity review
Review the Pythagorean trigonometric identity and use it to solve problems.
## What is the Pythagorean identity?
${\mathrm{sin}}^{2}\left(\theta \right)+{\mathrm{cos}}^{2}\left(\theta \right)=1$
This identity is true for all real values of $\theta$. It is a result of applying the Pythagorean theorem on the right triangle that is formed in the unit circle for each $\theta$.
## What problems can I solve with the Pythagorean identity?
Like any identity, the Pythagorean identity can be used for rewriting trigonometric expressions in equivalent, more useful, forms.
The Pythagorean theorem also allows us to convert between the sine and cosine values of an angle, without knowing the angle itself. Consider, for example, the angle $\theta$ in Quadrant $\text{IV}$ for which $\mathrm{sin}\left(\theta \right)=-\frac{24}{25}$. We can use the Pythagorean identity and $\mathrm{sin}\left(\theta \right)$ to solve for $\mathrm{cos}\left(\theta \right)$:
$\begin{array}{rl}{\mathrm{sin}}^{2}\left(\theta \right)+{\mathrm{cos}}^{2}\left(\theta \right)& =1\\ \\ {\left(-\frac{24}{25}\right)}^{2}+{\mathrm{cos}}^{2}\left(\theta \right)& =1\\ \\ {\mathrm{cos}}^{2}\left(\theta \right)& =1-{\left(-\frac{24}{25}\right)}^{2}\\ \\ \sqrt{{\mathrm{cos}}^{2}\left(\theta \right)}& =\sqrt{\frac{49}{625}}\\ \\ \mathrm{cos}\left(\theta \right)& =±\frac{7}{25}\end{array}$
The sign of $\mathrm{cos}\left(\theta \right)$ is determined by the quadrant. $\theta$ is in Quadrant $\text{IV}$, so its cosine value must be positive. In conclusion, $\mathrm{cos}\left(\theta \right)=\frac{7}{25}$.
Problem 1
${\theta }_{1}$ is located in Quadrant $\text{III}$, and $\mathrm{cos}\left({\theta }_{1}\right)=-\frac{3}{5}$ .
$\mathrm{sin}\left({\theta }_{1}\right)=$
Want to try more problems like this? Check out this exercise.
## Want to join the conversation?
• How come these "Review" sections aren't in every subtopic? I think for those of us who don't find videos particularly effectively, having something we can read are really fantastic resources.
• There originally were no review sections and no articles. The site only had videos and practice exercises. Over the last year or so they have been adding in these review sections. They probably just haven't done all of them yet.
• What are real life ways to use this awesome proof? I MUST find out! :)
• There are countless real-life situations that use the Pythagorean identity. A great example is in architecture. If you're creating a blueprint of a structure that consists of right triangles and you would like to know the length of a side, the Pythagorean identity will help you do so. Geologists or explorers use it to find the height of a mountain with great accuracy. Not to mention how important it is in space when you can't always measure distances between objects easily. Here is a Prezi on many real-world applications of the trig identities (which was not made by me): https://prezi.com/vvsb1nqexnzd/trigonometric-identities-in-the-real-world/
• how do you get 16 from 3x3?
• this may not be helpful anymore, but they subtracted 9/25 from 1 and got 16/25.
• As far as I think the 3rd and 2nd quadrant is negative and the 4th and 1st quadrant is positive.
• For the cosine function, yes.
For the sine function, I and II are positive, III and IV are negative.
For the tangent function, I and III are positive, II and IV are negative.
• Im still confused with the quadrants things. Can someone list all the positive/negative values for the quadrants? Thanks!
• This acronym helped me a lot: ASTC
A / ALL: This is the first quadrant (I), all values are positive (Sine, Cosine, Tangent, Cosecant, Secant and Cotangent)
S / Sine: This is the second quadrant (II), only the Sine and its inverse Cosecant (Wich you will see further in this course) are positive. Cosine, Tanget and its inverses are negative.
T / Tanget: This is the third quadrant (III), only the Tanget and its inverse Cotangent are positive. Cosine, Sine and its inverses are negative.
Cosine / Cosine: This is the fourth (IV), only the Cosine and its inverse Secant are positive. Sine, Tangent and its inverses are negative.
Summing up:
"S" Second quadrant (II) | "A" First quadrant (I) Sine is positive | All are positive ________________________|________________________ |"T" Third quadrant (III) | "C" Fourth quadrant (IV) Tangent is positive | Cosine is positive
Hope it helps!
• In the example above they calculated that 1-(−24/25)^2=sqrt(49/625) when taken sqrt of cos^2(θ). Could someone explain to me how did they get that solution?
• Hi Nenand,
Let me elaborate on this-
cosˆ2(θ) = 1 - (-24/25)ˆ2
cosˆ2(θ) = (625/625) - (576/625) (Do you remember that any number dived by itself is 1? Hence, 625/625 = 1)
cosˆ2(θ) = (625 - 576)/ 625
cosˆ2(θ) = 49/ 625
√cosˆ2(θ) = √49/ 625
cos(θ) = +-(7/ 25)
As the angle is in the IV quadrant, cos(θ) will be positive, i.e., (7/25).
I hope this helped.
Aiena.
• What does it mean by saying that "This identity is true for all real values of θ "?
• The Pythagorean Identity does not hold true when θ is a non-real number. If θ were an imaginary or complex number, for example, the identity might not be true.
• In one of my calculus problems it says that sin^2(-x) + cos^2(-x) = 1. Could someone please explain this? My textbook is less than helpful. :(
• sin(-x) = -sin(x)
cos(-x) = cos(x)
sin(-x)^2 + cos(-x)^2
= (-sin(x))^2 + cos(x)^2
= (-1)^2 * sin(x)^2 + cos(x)^2
= 1 * sin(x)^2 + cos(x)^2
= sin(x)^2 + cos(x)^2
sin(x)^2 + cos(x)^2 is just the Pythagorean identity so we know that it equals 1 |
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Learning to simplify algebraic expressions is an important step in understanding and mastering algebra. Simplification of algebraic expressions is very useful as it allows us to transform complex or long algebraic expressions into a more compact and simple form.
In this article, we will learn some techniques to simplify any algebraic expression.
##### ALGEBRA
Relevant for…
Finding the simplest version of algebraic expressions.
See process
##### ALGEBRA
Relevant for…
Finding the simplest version of algebraic expressions.
See process
## Revision of terminology
Let’s review some of the most common terms used to simplify algebraic expressions.
• A variable is a letter that represents an unknown value. Some examples of variables are x, a, b, y, t.
• The coefficient is a numerical value used in conjunction with a variable.
• A constant is a term that has a definite value.
• Like terms are variables with the same letter and power. Like terms can sometimes contain different coefficients. For example, and are like terms since they both have the same variable with a similar exponent. Similarly, and are not like terms since they do not have the same variables.
## How to simplify algebraic expressions?
Simplifying algebraic expressions can be defined as the process of writing an expression in the simplest and most compact form possible without affecting the value of the original expression.
To simplify algebraic expressions, we can follow the following simple steps and rules:
• Remove any grouping symbols like parentheses and brackets when multiplying factors.
• Combine like terms using addition and subtraction.
• Combine the constants.
We are going to look at the three most important cases of simplification: distributive property, combining like terms, and distributive property with like terms.
## Distributive property
The distributive property indicates that for any real numbers a, b, c, we have:
This property is used to simplify algebraic expressions. By applying the distributive property, we can remove the parentheses.
### EXAMPLES
• Simplify .
Solution: Multiply each term inside the parentheses by 4:
• Simplify .
Solution: Multiply each term inside the parentheses by -2:
• Find the simplest version of .
Solution: Multiply each term inside the parentheses by 2:
• Simplify .
Solution: Multiply each term inside the parentheses by 2:
## Combine like terms
Terms with the same variable are called like terms. Furthermore, constant terms are also considered like terms.
If the variables of the terms are exactly the same, then we add or subtract the coefficients to obtain a single coefficient with those variables.
### EXAMPLES
• Simplify .
Solution: Identify and combine like terms:
• Simplify .
Solution: Identify and combine like terms:
• Simplify .
Solution: Identify and combine like terms:
• Simplify .
Solution: Identify and combine like terms. We can see that the terms have in common. Therefore, we can comine 10 and -5:
## Distributive property and like terms
Many times we will have to combine like terms after applying the distributive property. This is consistent with the order of operations: multiplication before addition.
### EXAMPLES
• Simplify .
Solution: We have to distribute the 2 and the -3. Then, we combine like terms:
• Simplify .
Solution: Let’s distribute the -1 and let’s combine like terms:
• Simplify .
Solution: The order of operations requires that we multiply before subtracting. So, we distribute the -2 and then we combine like terms:
## Key takeaways
• The properties of real numbers apply to algebraic expressions since variables are simply representations of unknown real numbers.
• To simplify expressions, we combine like terms or terms with the same variables.
• Use the distributive property to multiply grouped algebraic expressions.
• It is easier to apply the distributive property only when the expression within the parentheses or group is completely simplified.
• After applying the distributive property, remove the parentheses and then combine like terms.
• Always use the order of operations when simplifying. |
# Basic Arithmetic : Subtraction with Fractions
## Example Questions
← Previous 1
### Example Question #31 : Operations With Fractions
Explanation:
To subtract fractions, they need to share the same denominator. To find the same denominator, you will need to find the least common multiple of the two given denominators.
The least common denominator of 5 and 7 is 35. Remember, the number you multiply the denominator by to get the least common denominator you must also multiply by the numerator.
So the original equation becomes,
Now, subtract the numerators.
### Example Question #41 : Fractions
Subtract.
Explanation:
To subtract two fractions, they need to both share the same denominator. Since 81 is a multiple of 9, we only need to change the 1st fraction.
Now subtract the two fractions.
### Example Question #1 : Subtraction With Fractions
Explanation:
1. Find the least common denominator:
The lowest number that both 4 and 5 both go into is 20, making 20 the least common denominator.
2. Find the equivalent fractions using the least common denominator:
3. Subtract:
### Example Question #2 : Subtraction With Fractions
Amanda has pounds of cake leftover from her birthday. If she eats a third of a pound of cake, how much will she have left over?
pounds of cake.
pounds of cake.
pounds of cake.
pounds of cake.
pounds of cake.
pounds of cake.
Explanation:
To tackle this question, first convert 2 and 3/4ths into a fraction.
Then, you can subtract from :
.
### Example Question #41 : Operations With Fractions
If you have three pies, and someone eats one quarter of each pie, how much pie do you have left? The answers will be expressed as mixed numbers.
Explanation:
To find the answer for this problem, first figure out how much of each pie is left:
Then, because , you know that you have whole pies left, and a quarter besides.
### Example Question #2 : Subtraction With Fractions
Explanation:
To subtract fractions, they must have the same number in the denominator. Begin by simplifying so that its denominator is .
To simplify, divide the numerator and denominator by 6.
Then, subtract:
### Example Question #12 : Adding And Subtracting Fractions
Subtract these fractions:
Explanation:
To solve this we need to first find common denominators. We do that by multiplying the first fraction by 2 over 2 and the second fraction by 3 over 3.
Subtract these fractions to get the final answer.
### Example Question #1 : Subtraction With Fractions
Subtract these fractions:
Explanation:
To solve this we need to first find common denominators. We do that by multiplying the first fraction by 3 over 3 and the second fraction by 5 over 5.
Subtract the numerators of the fractions to get the final answer.
### Example Question #21 : Adding And Subtracting Fractions
Subtract these fractions:
Explanation:
To solve this we need to first find common denominators. We do that by multiplying the first fraction by 4 over 4 and the second fraction by 7 over 7.
Subtract the numerators of these fractions to get the final answer.
### Example Question #1 : Subtraction With Fractions
Subtract these fractions: |
# NCERT Class 11-Math՚s: Chapter – 6 Linear Inequalities Part 1 (For CBSE, ICSE, IAS, NET, NRA 2023)
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6.1 Overview
A statement involving the symbols is called an inequality.
For example .
(i) Inequalities which do not involve variables are called numerical inequalities.
For example .
(ii) Inequalities which involve variables are called literal inequalities.
For example, .
(iii) An inequality may contain more than one variable and it can be linear, quadratic or cubic etc. For example, is a linear inequality in one variable, is a linear inequality in two variables and is a quadratic inequality in one variable.
(iv) Inequalities involving the symbol or are called strict inequalities.
For example, .
(v) Inequalities involving the symbol or are called slack inequalities.
For example, .
6.1. 2 Solution of an inequality
(i) The value (s) of the variable (s) which makes the inequality a true statement is called its solutions. The set of all solutions of an inequality is called the solution set of the inequality. For example, , has infinite number of solutions as all real values greater than or equal to one make it a true statement. The inequality has no solution in R as no real value of x makes it a true statement.
To solve an inequality we can
(i) Add (or subtract) the same quantity to (from) both sides without changing the sign of inequality.
(ii) Multiply (or divide) both sides by the same positive quantity without changing the sign of inequality. However, if both sides of inequality are multiplied (or divided) by the same negative quantity the sign of inequality is reversed, i.e.. , changes into and vice versa.
6.1. 3 Representation of solution of linear inequality in one variable on a number line
To represent the solution of a linear inequality in one variable on a number line, we use the following conventions:
(i) If the inequality involves or , we draw filled circle on the number line to indicate that the number corresponding to the filled circle is included in the solution set.
(ii) If the inequality involves or , we draw an open circle on the number line to indicate that the number corresponding to the open circle is excluded from the solution set.
6.1. 4 Graphical representation of the solution of a linear inequality
(a) To represent the solution of a linear inequality in one or two variables graphically in a plane, we proceed as follows:
(i) If the inequality involves or , we draw the graph of the line as a thick line to indicate that the points on this line are included in the solution set.
(ii) If the inequality involves or , we draw the graph of the line as dotted line to indicate that the points on the line are excluded from the solution set.
(b) Solution of a linear inequality in one variable can be represented on number line as well as in the plane but the solution of a linear inequality in two variables of the type can be represented in the plane only.
(c) Two or more inequalities taken together comprise a system of inequalities and the solutions of the system of inequalities are the solutions common to all the inequalities comprising the system.
6.1. 5 Two important results
(a) If , and , then
(i) or and are of the same sign.
(ii) or and are of opposite sign.
(b) If a is any positive real number, i.e.. , , then
(i)
(ii)
## 6.2 Solved Examples
Question 1:
Solve the inequality, , when
(i) is a natural number
(ii) is a whole number
(iii) is an integer
(iv) is a real number
We have |
# HSPT Math : How to simplify expressions
## Example Questions
← Previous 1
### Example Question #1 : How To Simplify Expressions
You are given that are whole numbers.
Which of the following is true of if and are both odd?
None of the other statements are true.
is always odd.
is always odd if is even, and always even if is odd.
is always odd if is odd, and always even if is even.
is always even.
is always odd if is even, and always even if is odd.
Explanation:
If is odd, then is odd, since the product of two odd whole numbers must be odd. When the odd number is added, the result, , is even, since the sum of two odd numbers must be even.
If is even, then is even, since the product of an odd number and an even number must be even. When the odd number is added, the result, , is odd, since the sum of an odd number and an even number must be odd.
### Example Question #2 : How To Simplify Expressions
Simplify the expression:
Explanation:
Combine all the like terms.
The terms can be combined together, which gives you .
When you combine the terms together, you get .
There is only one term so it doesn't get combined with anything. Put them all together and you get
.
### Example Question #2 : How To Simplify Expressions
Simplify the following expression:
Explanation:
First distribute the 2:
Combine the like terms:
### Example Question #3 : How To Simplify Expressions
Simplify the expression:
x+ 2x + 1
2x
x + 1
2x + 1
x
x + 1
Explanation:
Factor out a (2x) from the denominator, which cancels with (2x) from the numerator. Then factor the numerator, which becomes (+ 1)(+ 1), of which one of them cancels and you're left with (+ 1).
### Example Question #3 : How To Simplify Expressions
Simplify the following expression: x3 - 4(x2 + 3) + 15
x3 – 4x2 + 3
x5 + 3
x3 – 12x2 + 15
x3 – 3x2 + 15
x3 – 4x2 + 3
Explanation:
To simplify this expression, you must combine like terms. You should first use the distributive property and multiply -4 by x2 and -4 by 3.
x3 - 4x2 -12 + 15
You can then add -12 and 15, which equals 3.
You now have x3 - 4x2 + 3 and are finished. Just a reminder that x3 and 4x2 are not like terms as the x’s have different exponents.
### Example Question #4 : How To Simplify Expressions
Simplify the following expression:
2x(x2 + 4ax – 3a2) – 4a2(4x + 3a)
–12a– 22a2x + 8ax2 + 2x3
–12a– 14ax2 + 2x3
–12a3 – 14a2x + 2x3
12a– 16a2x + 8ax2 + 2x3
12a– 22a2x + 8ax2 + 2x3
–12a– 22a2x + 8ax2 + 2x3
Explanation:
Begin by distributing each part:
2x(x2 + 4ax – 3a2) = 2x * x2 + 2x * 4ax – 2x * 3a2 = 2x3 + 8ax2 – 6a2x
The second:
–4a2(4x + 3a) = –16a2x – 12a3
Now, combine these:
2x3 + 8ax2 – 6a2x – 16a2x – 12a3
The only common terms are those with a2x; therefore, this reduces to
2x3 + 8ax2 – 22a2x – 12a3
This is the same as the given answer:
–12a– 22a2x + 8ax2 + 2x3
### Example Question #4 : How To Simplify Expressions
Which of the following does not simplify to ?
All of these simplify to
Explanation:
5x – (6x – 2x) = 5x – (4x) = x
(x – 1)(x + 2) - x2 + 2 = x2 + x – 2 – x2 + 2 = x
x(4x)/(4x) = x
(3 – 3)x = 0x = 0
### Example Question #5 : How To Simplify Expressions
Simplify:
Explanation:
In order to simplify this expression, distribute and multiply the outer term with the two inner terms.
### Example Question #5 : How To Simplify Expressions
Simplify:
Explanation:
When the same bases are multiplied, their exponents can be added. Similarly, when the bases are divided, their exponents can be subtracted. Apply this rule for the given problem.
### Example Question #6 : How To Simplify Expressions
Simplify:
Explanation:
To simplify this expression, reduce the term inside the parenthesis.
Rewrite the negative exponent as a fraction.
← Previous 1 |
# Unlocking the Mystery of Mathematical Symbols: Exploring the Meaning and Usage of ‘and’
## Short answer mathematical symbol for and:
The mathematical symbol for “and” is represented by an upside-down letter “v”, also known as the conjunction, ∧. It denotes the intersection of sets or logical AND operations between two propositions.
## Step-by-Step Tutorial: How to Use the Mathematical Symbol for And
Mathematics is an incredibly powerful tool that allows us to describe and model the world in precise detail. From simple arithmetic to advanced calculus, math provides us with a language for understanding complex concepts and solving complicated problems. One essential symbol used in math is the logical conjunction “and” (∧) which plays a vital role in predicate logic, set theory, probability theory, statistics and many other branches of mathematics.
In this step-by-step tutorial, we will guide you on how to use the mathematical symbol for “and” in your studies or work:
Step 1: Understand its meaning
The ∧ symbol represents the word “and.” In mathematical notation, it’s typically placed between two statements or propositions that need to be true simultaneously.
For instance; Suppose you want to state that both A and B are true statement together. You would write it as A ∧ B read as “A AND B”.
To put that into context: If A represents “the sky is blue,” and B represents “grass is green,” then writing “A ∧ B” means “the sky is blue AND grass is green.”
Step 2: Get familiar with possible applications of ^ Symbol
When working with mathematical proofs or even logical arguments outside math theories where several conditions must hold at once before something can happen (like passing an exam), you will often encounter sentences referring to multiple meaningful condition holding types such as;
-All Elements X Must Satisfy Y And All Element Z Must Be Qualify As W
-X Is Greater Than 5 And Less Than 15
-The Solution Approaches Infinity And The Limits Tends To Zero Simultaneously
Step3: Learn How To Type It on Your Keyboard
The next thing after cognizing what “^” signifies into Mathematics notations knows how they could type it.
Different Devices have different approaches when comes symbols typing-most devices have distinct ways of entering variants special characters via alphanumeric keys plus modifier key combinations.
However, On Windows PC: You can type ∧ symbol using the following keystrokes; Press “Alt+ 8743” at once.
On your Smartphone or tablet device- touch and hold down the ampersand key(!) that’s located on the keyboard main screen, then slide your finger towards up direction to reveal a few essential symbols which may include “^”.
Step4: Practice Using ‘^’ in Reasoning
And finally, practice Its use into your reasoning with others- Mathematicians or Non-Mathematicians by writing some mathematical statements utilizing it as well as sharing its meaning helping interested persons how they could apply it themself with ease even if they haven’t prior knowledge about this useful & vital logic conjunction previously.
Conclusion;
Knowing how to use Mathematical Symbol for And ( ^ ) is an essential skill when working with predicate logic, set theory probability theory or statistics, among other branches of mathematics. Once you understand what it represents and How typing It – incorporating this powerful tool can be much easier! Don’t miss out on opportunities to simplify complex math problems by mastering
## Frequently Asked Questions About the Mathematical Symbol for And
The mathematical symbol for “and” is one of the most commonly used symbols in mathematics. It’s a simple, yet powerful tool that enables mathematicians to express complex ideas in a concise and clear manner.
But despite its prevalence and importance, many people still have questions about this fundamental symbol. Here are some frequently asked questions about the mathematical symbol for “and.”
1. What does the symbol “&” mean?
The symbol “&” is an alternative way to represent the word “and.” It has been used since ancient times as a shorthand notation for joining two words or concepts together. In mathematics, it represents logical conjunctions, meaning that both statements must be true for the overall statement to be true.
2. When should I use “&” vs “and”?
In general, you can use either “&” or “and” interchangeably when writing out a mathematical statement. However, certain contexts may dictate which one is more appropriate than the other.
For example, if you’re working on computer programming where you need to write code quickly without taking up too much space with every command then using ampersands would probably make sense because they take less space compared to writing out “and.”
3. How do I read expressions that use “&”?
When reading expressions that involve “&”, simply replace it with the word “and”. For example:
– x > 0 & y &E&F=G; Then XHJ.
Can be interpreted from left-to-right like:
“If (B equals C) AND ((D is greater than E) AND (F equals G)), then (X less than the minimum of A and D) OR (X is greater than HJ).”
5. Is “&” different from “∧”?
The symbol “∧” also represents logical conjunctions or “and”. Visually, it looks like an upside-down V-shaped arrow with a dot on top.
While they have similar meanings, technically speaking one might use ampersands in more informal scenarios such as hand written notes while using ∧ might make sense when writing out code for formal verification or typing out research papers that require symbols be standardized all across the board rather than just subjective. End users would probably not know what certain combinations of characters mean unless their meaning was either defined by reading some documentation beforehand or agreed upon within context due to common usage patterns over time and experience.
6. Can I use other symbols instead of “&”?
Yes! In fact, there are several other math symbols you can use to represent logical conjunctions aside from ‘&’. For example:
– The word “and”
## Mastering the Art of Math: Using the Mathematical Symbol for And in Complex Equations
When it comes to solving complex math equations, there are a number of symbols that we need to be familiar with. From addition and subtraction signs to multiplication and division symbols – each one plays an important role in helping us arrive at the right solution.
But what about the symbol for “and”? This is a less commonly used symbol in mathematics, but knowing how to use it can make all the difference when tackling more advanced equations.
In mathematical terms, the symbol for “and” is represented by an inverted V shape (∧). It is typically used to connect two different conditions or propositions within a single equation.
For example, let’s say we have an equation that requires us to find both x and y values that satisfy certain criteria. We could write this equation as follows:
2x + 3y = 10 ∧ x > 0 ∧ y < 5
The ∧ symbol connects the separate conditions in this equation together. In essence, we're saying that both bits of information must be true if we want our answer to be correct.
It may not seem like much on its own, but using the "and" symbol can actually help us simplify some pretty complex equations. By breaking down larger problems into smaller pieces linked together by these types of connections (also referred to as logical operators), we can focus on one bit at a time until we reach our desired outcome.
Of course, mastering the art of using mathematical symbols takes practice and patience. But once you get comfortable working with various logical operators like "^", "&", "|", "~" (the ones you would expect after doing logic gate-reading), your ability solve even perplexing problems becomes easier over time!
So go ahead – give it a try! When confronted with those complicated calculus problems and algebraic expressions that leave others scratching their heads in puzzlement just remember – mastery over all kinds of math operations awaits anyone who practices consistently enough till they break through any mental barriers set up by algorithms and formulas.
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# Cubes
Content Cubes
Rating
Ø 4.4 / 5 ratings
The authors
Team Digital
Cubes
CCSS.MATH.CONTENT.2.G.A.1
## In This Video
Niko and Nia see a shape storm that rains cubes. Except, Niko doesn’t know what it is. So he wonders ‘what is a cube in math?’ Join in to learn all about cubes!
## Cube (Math) Explanation & Example
You may wonder ‘what is a cube in math’, or ‘what does cube mean in math’. Well, a cube in math is a 3-D shape that has specific defining attributes, or characteristics.
The defining attributes of a cube are:
• Six square faces
• Eight vertices, or corners
• Twelve edges, or sides
Therefore, the cube definition (math) is a 3-D shape that has six square faces, eight vertices, and twelve edges. This can also be referred to as the cube rule (math).
Underneath, you will find a math cube worksheet, where you can practice identifying if a shape is a cube or not based on the attributes you learned.
### TranscriptCubes
"Hey Nia, do you hear that?" Uh oh, it looks like there is a shape storm happening! "Woooaaaaah. This is new!" "I wonder what this shape is?" Let's join Nico and Nia as they learn all about cubes. A cube is a 3-D shape. A 3-D shape is a solid figure that has three dimensions; length, width, and height. 3-D shapes can be identified by defining attributes, or properties, that will not change, and non-defining attributes, or properties, which may change. Let's take a look at defining attributes of a cube. A cube has six faces, which are made up of squares. A cube also has eight vertices, or corners! Finally, a cube has twelve edges, which is where two faces meet. A shape must have all three of these defining attributes to be classified as a cube. Now let's look at non-defining attributes of cubes, or things that may change. A cube may be different colors. A cube may be different sizes. A cube may have different orientations, or be turned differently. To identify cubes, we look for the defining attributes, or properties, that do change. Let's practice identifying cubes! Is this a cube? First, count the square faces. There are six square faces, it might be a cube! Now count the vertices. There are eight vertices, it might still be a cube! Finally, count the edges. There are twelve edges. Since all the defining properties of a cube are present, this is a cube! Let's look at this shape. Is this a cube? Start by counting the faces. There are five faces, and none of them are square. Since a cube must have six square faces, this is not a cube! Is this a cube? This is a cube, because there are six square faces, eight vertices, and twelve edges! Even though the two cubes we saw are different colors, different sizes, and have different orientations from each other they are still cubes because the defining attributes are present! Before we see what Nico and Nia are up to in the shape storm, let's review! Remember, a cube may have non-defining attributes such as, a different color, a different size, or a different orientation! In order to be classified as a cube, it must have all of the following defining attributes: Six square faces, eight vertices, and twelve edges! "Huh?" "We should close the window!" "Yes Nico, that is probably a good idea!"
1. Oh wow so cool 😎
From Rosie , 6 months ago
2. I know geometry a little bit.
From MOJOLA , over 1 year ago
## Cubes exercise
Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video Cubes.
• ### Which shape is a cube?
Hints
Remember that a cube has 6 faces.
Remember that a cube has 8 vertices.
Every face on a cube is a square.
You might not be able to see all of the vertices and faces of the shape at once.
Solution
The cube is shown here with a check mark. It has 6 faces, 8 vertices and 12 edges. Each of its faces is a square.
The other 3D shapes are not cubes.
• ### Label the parts of the cube.
Hints
An edge is where two faces meet.
A vertex is the singular of vertices. A vertex is where 3 or more edges meet.
The face of a shape is the flat part; it can be different shapes, such as a triangle, a square or a circle.
Solution
The cube can be labelled with:
• The face - the flat, square shape that can be seen when looking at the shape. There are 6 faces on a cube.
• The edge - where two faces meet. There are 12 edges on a cube.
• The vertex - where three or more edges meet. There are 8 vertices on a cube.
• ### Defining and non-defining characteristics of a cube.
Hints
These are all cubes, although some of their non-defining characteristics have changed.
What do all cubes have? These are their defining characteristics.
Solution
4 of the attributes are defining and 3 are non-defining as shown by this image.
• ### What defines a cube?
Hints
The red line here shows one edge of the cube, where two faces meet.
This cube has the vertices marked by pink dots.
Solution
These are the defining properties of a cube:
Face shape: Square
Number of faces: 6
Number of vertices: 8
Number of edges: 12
Other factors, such as shape, size and color can vary.
• ### A shape storm.
Hints
There are 4 cubes to find.
Remember, a cube has 6 faces, 8 vertices and 12 edges.
A cube is a three dimensional (3D) shape.
You might not be able to see all of the vertices and faces of the shape at once.
Solution
There are 4 cubes shown, highlighted yellow here.
• ### Cube net.
Hints
What shape is every face of a cube? Check that each face of the net will be the right shape.
Think about how many faces a cube has. There should be this many flat shapes within the net.
Solution
The correct net for a cube must have 6 square faces, as well as being in the correct layout to fold into a cube. |
# Determinant properties
For every $n \times n$ matrix, we can define its determinant, being a (possibly complex) number. It is defined recursively by taking out one of its rows, say the $i^{\text{th}}$ row= $$$$\text{Det}(A) = \sum_{k=1}^{n} a_{ik} (-1)^{k+i} \text{Det}(A(i,k))$$$$ where the $(n-1)\times(n-1)$ matrix $A(i,k)$ is obtained from $A$ by removing the $i^{\text{th}}$ row and the $k^{\text{th}}$ column. The determinant of a $0\times 0$ matrix is $1$.
Many properties of determinants are easy to derive if $C = AB$ then, $$$$\text{Det}(C) = \text{Det}(AB) = \text{Det}(A)\text{Det}(B).$$$$
If $C = x A$ with $x$ a number and $A$ an $n\times n$ matrix $$$$\text{Det}(C) = \text{Det}(x A) = x^{n}\text{Det}(A).$$$$
Futhermore, the inverse of a matrix $A$ is a matrix $B$ such that $BA = AB = 1$. Let’s choose a notation for B as $B = A^{-1}$. The matrix elements of B are (note: $A(j, i)$, not $A(i, j)$) $$$$b_{ij} = \frac{(-1)^{i+j}\text{Det}(A(j,i))}{\text{Det}(A)}$$$$ from this, it follows that a matrix $A$ has an inverse if $\text{Det}(A)\neq 0$.
Another definition of the determinant makes use of the $\epsilon$ symbol, defined by
• $\epsilon_{i_{1}\cdots i_{n}} = 0$ if in $i_{1}\cdots i_{n}$ two or more of the indicies are equal,
• $\epsilon_{i_{1}\cdots i_{n}} = \pm 1$ if in $i_{1}\cdots i_{n}$ no two indicies are equal, in which case we have
• $\epsilon_{i_{1}\cdots i_{n}} = 1$ if in $i_{1}\cdots i_{n}$ are an even permutation of $123\ldots n$,
• $\epsilon_{i_{1}\cdots i_{n}} = -1$ if in $i_{1}\cdots i_{n}$ are an odd permutation of $123\ldots n$. In particular, $\epsilon_{123\cdots n} = +1$
The determinant of a matrix $A$ is then given by $$$$A_{i_{1}j_{1}}A_{i_{2}j_{2}}\cdots A_{i_{n}j_{n}}\epsilon_{j_{1}\cdots j_{n}} = \text{Det}(A)\epsilon_{i_{1}\cdots i_{n}}$$$$
Published on May 6, 2021
Last revised on May 6, 2021
References |
# Change of limits of integration
I'm a little bit confused about how the limits of integration are changed if we change the variable of integration. I'm not sure what exactly I don't understand, so I below I will write some false statements that seem true to me, and I ask to point out what exactly is wrong (and why).
For instance, $\int_0^1dx=1$, so $\int_0^1d(-y)=-\int_0^1dy=-1$. But $-y$ can be treated as a variable $\xi,$ so $\int_0^1d(-y)=\int_0^1d\xi=1$ by the first equality. What am I doing wrong?
Next, if $0\le t \le 1$, then $-1\le -t \le 0$. Thus shouldn't be true that $\int_0^1dt=\int_{-1}^0d(-t)$?
Does all of this have to do with "The Substitution Rule for Definite Integrals"? I wasn't able to relate it to these cases, since I don't see any derivatives which are used in that theorem.
$$\int\limits_0^1 dx \neq \int\limits_0^1 d(-y)$$
If you make a proper change of variables, you might say something like $x = -y$, in which case $dx = -dy$, further $x = 0$ to $x = 1$ becomes $y = 0$ to $y = -1$:
\begin{align} \int\limits_{x\ =\ 0}^{x\ =\ 1} dx = -\int\limits_{y\ =\ 0}^{y\ =\ -1}dy = -(-1 - 0) = +1 && \text{q.e.d.} \end{align}
Here's a slightly more complicated example:
$$\int\limits_0^8 xdx = \frac{1}{2}\left(8^2 - 0^2\right) = 32$$
Let's try $x = -y^3$. We find that $dx = -3y^2dy$ and that from $x = 0$ to $x = 8$ means $y = 0$ to $y = -2$:
\begin{align} \int\limits_0^8 xdx =&\ \int\limits_0^{-2}\left(\left(-y^3\right)\left(-3y^2\right)\right)dy \\ =&\ 3\int\limits_0^{-2}y^5dy = \frac{3}{6}\left((-2)^6 - 0^6\right) \\ =&\ \frac{1}{2}\cdot2^6 = 2^5 = 32 && \text{q.e.d} \end{align}
Suppose that you have the integral defined by $\int_a^b f(x)\,\mathrm dx$ and you make the change of variable $x=g(y)$ for some differentiable and injective $g$ in $[a,b]$, then you have that $\mathrm dx=g'(y)\,\mathrm dy$ and that if $x=a$ then $y=g^{-1}(a)$. Same for the $b$, that is, if $x=b$ then $y=g^{-1}(b)$, because $g$ is invertible in $[a,b]$.
Thus
$$\int_a^b f(x)\,\mathrm dx=\int_{g^{-1}(a)}^{g^{-1}(b)}(f\circ g)(y)\,g'(y)\,\mathrm dy$$
In your example you have the change of variable $x=-y$ thus $0=-y$ and $1=-y$. |
# Lecture 10: What is a Function, definition, piecewise defined functions, difference quotient, domain of a function
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## Transcription
1 Lecture 10: Wat is a Function, definition, piecewise defined functions, difference quotient, domain of a function A function arises wen one quantity depends on anoter. Many everyday relationsips between variables can be expressed in tis form. Example 1 We ave already seen examples of functions in our lecture on lines. For example, if a car leaves Matville at 1:00 p.m. and travels at a constant speed of 60 miles per our for two ours, ten te distance in miles traveled by te car after t ours(d) (assuming it as not stopped along te way or canged speed) depends on te time, or is a function of time (t) = t ours: D = D(t) = 60t for 0 t 2 We can use tis general formula to calculate ow far te car as traveled at any given time t. For example after 1/2 our (t = 1/2) te car as traveled a distance of D = 60 1 = 30 miles. 2 Here is anoter everyday example of a function. Example Consider te volume of a cylindrical glass wit radius = 1 inc. We ave a formula for te volume; V = πr 2 = π in 3, were is te eigt of te glass in inces and r = 1 is te radius. We see tat te value of te volume depends on te eigt, ; V is a function of. We sometimes indicate tat te value of V depends on te value of, by writing te formula as V () = π in 3. Wen = 2, V = V (2) = and wen = 3, V = V (3) = In tis section, we will examine te definition and general principles of functions and look at examples of functions wic are more complex tan lines. Definition of a Function A function is a map or a rule wic assigns to eac element x of a set A exactly one element called f(x) in a set B. Example Te rule wic assigns to eac real number its square is a function. We ave A = (, ) or te set of all real numbers. Since te square of a real number is a real number, we ave B is also te set of real numbers. If x is an element of A, tis rule assigns te element f(x) = x 2 in B to x. If x = 2, te f( 2) = ( 2) 2 = 4. In sort, tis is te map or rule wic sends x to x 2. We will focus on functions were te sets A and B are sets of real numbers. Te symbol f(x) is read as f of x or f at x and is called te value of f at x or te image of x under f. Te possible values of x vary ere since x can take any value from te set A, ence x is a variable. We call x te independent variable. We may replace te symbol x by any oter symbol ere, for example t or u, in wic case te symbol for our function canges to f(t) or f(u) respectively. Likewise we can replace te name of te function, f, by any oter name in te definition, for example or g or D. So if u denotes our independent variable and D is te name of our function, te values of te function are denoted by D(u). 1
2 Te set of numbers (or objects) to wic we apply te function, A, is called te domain of te function. Te set of values of B wic are equal to f(x) for some x in A is called te range of f. We ave range of f = {f(x) x A} In te example sown above were f(x) = x 2, we see tat te values of f(x) = x 2 are alwys 0. Furtermore, every positive number is a square of some number, so te range of f ere is te set of all real numbers wic are 0. If we can write our function wit a formula (or a number of formulas), as in te above example were f(x) = x 2, ten we can represent te function by an equation y = f(x). In our example above, we migt just describe our function wit te equation y = x 2. Here y is also a variable, wit values in te range of te function. Te value of y depends on te given value of x and ence y is called te dependent variable. Note tat writing a function in tis way allows us to draw a grap of te function in te Cartesian plane. It is elpful to tink of a function as a macine or process tat transforms te number you put troug it. If you put in te number x, ten te macine or process canges it and gives back f(x). Anoter way to visualize a function is by using an arrow diagram as sown below (for te example f(x) = x 2 above). A = Real Numbers B = Real Numbers x!x For every element, x, in te domain of te function we ave exactly one arrow leaving te point representing tat element, indicating tat te function can be applied to any element in te domain and we get exactly one value in B wen we apply te function to x. On te oter and, an element in te set B can ave 0, 1 or more tan one arrow pointing towards it. If te element is not in te range, it as no arrows arriving at it, if it is in te range it as at least one arrow arriving at it. Note tat in our example, we ave two arrows arriving at 4 in te set B because te values of f at ( 2) and 2 are bot 4, i.e. f( 2) = f(2) = 4. Example Lets summarize wat we know about our example above were our function is te rule wic send any value x from te real numbers to te value x 2. As pointed out above, we can give a formula for f(x), namely f(x) = x 2. We can use tis formula to calculate te value of te function for any given value of x. For example f( 1) = ( 1) 2 = 1, f(0) = 0 2 = 0, f(1/2) = (1/2) 2 = 1/4. 2
3 Te domain is te set of all real numbers and te range is te set {y R y 0}. We can represent tis function by an equation y = x 2. Here x is te independent variable wic can take any values of x in te real numbers (te domain) and y is te dependent variable wose values will be in te range. Representing a function We saw above tat tere are a number of ways to represent a function. We can represent it by 1. A verbal description 2. an algebraic representation in te form of a formula for f(x) (possibly not just a single formula). 3. an equation of te form y = f(x), wic we can grap on te Cartesian plane 4. an arrow diagram or a table (especially if we ave a finite number of points in te domain). We will focus on using te algebraic description and te grapical description of a function. Te Cartesian plane allows us to translate te results we derive algebraically to a grapical interpretation and vice-versa. Evaluating a function Here we will focus on deriving and using te algebraic description of some examples of functions. Example Let f(x) = x Evaluate te following f( 2), f(0), f(3/2), f(10). Piecewise Defined Functions Not every function can be defined wit a single formula (suc as te absolute value function), sometimes we may need several lines/formulae to describe a function. Tese are called Piecewise Defined Functions. Example Te cost of (sort-term) parking at Sout Bend airport depends on ow long you leave your car in te sort term lot. Te parking rates are described in te following table: First 30 minutes Free minutes \$2 Eac additional our \$2 24 our maximum rate \$13 3
4 Te Cost of Parking is a function of te amount of time te car spends in te lot. If we are to create a formula for te cost of parking = C, in terms of ow long our car stays in te lot = t, we need to give te formula piece by piece as follows: C(t) = \$0 0 t 0.5 r. \$2 0.5 r. < t 1 r. \$4 1 r. < t 2 r. \$6 2 r. < t 3r. \$8 3 r. < t 4 r. \$10 4 r. < t 5 r. \$12 5 r. < t 6 r. \$13 6 r. < t 24 r. \$13 + cost of towing t > 24 r. If you were figuring out ow muc you needed to pay using tis description of te function, you would first figure out wic category you were in (ow long you ad parked for) and ten note te cost for cars in tat category. We proceed in te same way if we are given te formula for any piecewise defined function. Example Let Evaluate te following g(x) = x + 1 if x > 1 x 2 if 1 x 1 4 if x < 1 g( 2), g(0), g(3/2), g(10). Te Difference Quotient Wen we wis to derive general formulas in matematics, we ave to use general variables to represent values in a function so tat we can prove a result for all values in te domain. Tis means tat we often ave to evaluate te function at some combination of abstract values, suc as a, a +, like te ones sown below as opposed to evaluating te function at concrete values suc as 1, 0 etc.... Here are some examples of suc calculations. Te difference quotient f(a + ) f(a) is particularly important wen learning about derivatives. 4
5 Example Let f(x) = x and let a be any real number and a real number were 0. Evaluate 1. f(a) 2. f( a) 3. f(a 2 ) 4. f(a + ) 5. f(a+) f(a) Evaluate f(a+) f(a) wen a = 2 Evaluate f(a+) f(a) wen = 0.1 Evaluate f(a+) f(a) wen a = 2 and = 0.1. Example Let k(x) = 2x+1. Evaluate were 0. k(a + ) k(a) were a is any real number and a real number Example Let g(x) = x + 1 if x > 1 x 2 if 1 x 1 4 if x < 1 (a) Evaluate g(0 + ) g(0) for values of for wic < 1 and 0. 5
6 (b) Evaluate g(1 + ) g(1) for values of wic are greater tan 0. (c) Evaluate g(1 + ) g(1) for values of wic are greater tan 1 and less tan 0. Te Domain of A Function Recall tat te domain of a function f(x) is te set of values of x to wic we can apply te function. Sometimes we explicitly state wat te domain of a function is (see Example A below) and sometimes we just give a formula for te function (see Example B below). In te latter case, it is implicitly assumed tat te domain of te given function is te set of all real numbers wic make sense in te formula. Example A Let f(x) = x 3, 0 x 1 Here we ave explicitly stated tat te domain is te values of x in te interval [0, 1]. Example B Let g(x) = 1 x 2 Here it is assumed tat te domain of tis function is all values of x wic make sense in te formula, tat is te set of all real numbers except 2. Domain of g = {x R x 2} Note Keep in mind wen calculating domains for functions tat we cannot ave 0 as te denominator of a quotient and tat we can only evaluate square roots (or nt roots for n even) for numbers greater tan or equal to zero. Example Find te domain of te following function: f(x) = 5 x 2 6
7 Example Find te domain of te following function: x 1 R(x) = x 2 Example Find te domain of te following function: x R(x) = x 2 + 3x 10 Domains of common functions It is good to keep te domains of te following functions in mind: Function Domain x n, n N all x R 1, n N {x R x 0} xn n x, n N, n even {x R x 0} n x, n N, n odd all x R 7
8 Modeling wit functions Example: Income Tax In a certain country, income tax T is assessed according to te following function of income x: 0 if 0 x 10, 000 T (x) = 0.08x if 10, 000 < x 20, x if 20, 000 < x (a) Find T (5, 000), T (12, 000) and T (25, 000). (b) Wat do your answers in part (a) represent? 8
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### Chapter 11. Limits and an Introduction to Calculus. Selected Applications
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### Discovering Area Formulas of Quadrilaterals by Using Composite Figures
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### Chapter 10: Refrigeration Cycles
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# Chapter 06: Sequences and Series
Notes (Solutions) of Chapter 06: Sequences and Series, Text Book of Algebra and Trigonometry Class XI (Mathematics FSc Part 1 or HSSC-I), Punjab Text Book Board, Lahore.
• Introduction
• Types of Sequences
• Exercise 6.1
• Arithmetic Progression(A.P)
• Exercise 6.2
• Aritmetic Mean (A.M)
• n Arithmetic Means between two given Numbers
• Exercise 6.3
• Series
• Exercise 6.4
• World Problems on A.P.
• Exercise 6.5
• Geometric Progression(G.P)
• Exercise 6.6
• Geometric Means
• n Geometric Means between two given numbers
• Exercise 6.7
• Sum of n terms of a Geometric Series
• The Infinite Geometric Series
• Exercise 6.8
• World Problems on G.P.
• Exercise 6.9
• Harmonic Progression(H.P)
• Harmonic Means
• n Harmonic Means between two numbers
• Relation between Arithmic, Geometric and Harmonic Means
• Exercise 6.10
• Sigma Notation(or Summation Notation)
• To find the Formulae of Sum
• Exercise 6.11
Exercise 6.2⇒ Question 11(i)
If $l,m,n$ are the $p$th, $q$th, $r$th terms of an A.P., show that $$l(q-r)+m(r-p)+n(p-q)=0$$ Solution: Let $a_1$ be first term and $d$ be common difference of A.P, then \begin{align}l=a_1+(p-1)d,\\ m=a_1+(q-1)d,\\ n=a_1+(r-1)d.\end{align} Now \begin{align}L.H.S &= l(q-r)+m(r-p)+n(p-q)\\ &= lq-lr+mr-mp+np-nq\\ &=(l-n)q+(m-l)r+(n-m)p\\ &=(a_1+(p-1)d-a_1-(r-1)d)q+( a_1+(q-1)d-a_1-(p-1)d)r\\ &\,\,+(a_1+(r-1)d-a_1-(q-1)d)p\\ &=((p-1)d-(r-1)d)q+( (q-1)d-(p-1)d)r+((r-1)d-(q-1)d)p\\ &=(p-1-r+1)dq+(q-1-p+1)dr+(r-1-q+1)dp\\ &=(p-r)dq+(q-p)dr+(r-q)dp\\ &=[pq-qr+qr-pr+pr-pq]d=(0)d=0=R.H.S\end{align}
Exercise 6.4⇒ Question 3(ii)
An equation $3n^2-17n-288=0$ should be $3n^2-17n-228=0$. … suggested by Abdullah Zafar (Garrison Academy Kharian)
The following short questions was send by Mr. Akhtar Abbas. |
# 2006 Alabama ARML TST Problems/Problem 14
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Find the real solution $(x, y)$ to the system of equations
$x^3-3xy^2=-610,$
$3x^2y-y^3=182.$
## Solution
Note that $(x+yi)^3 = x^3 + 3x^2yi + 3xy^2i^2 + y^3i^3 = (x^3-3xy^2) + (3x^2y-y^3)i = -610 + 182i$
So, we need to find $(x+yi)$ that satisfies $(x+yi)^3 = -610 + 182i$. Even though there are three solutions, for simplicity, lets assume that one exists in the first quadrant in the complex plane, and try to find that one.
Note that $|(x+yi)^3|^2 = (-610)^2 + (182)^2 = 372100 + 33124 = 405224 = 8 \cdot 50653 = 2^3 \cdot 37^3$.
Thus, $|x+yi|^2 = \sqrt[3]{2^3 \cdot 37^3} = 2\cdot37 = 74$.
Therefore, $x^2 + y^2 = 74$. Note that $(x,y) = ( 5,7)$ and $(7,5)$ are the only positive integer solutions to $x^2 + y^2 = 74$. (Even though the solution may be non-integral, this is a good place to start.)
However, letting $r \cdot cis(\theta) = x+yi$, [$(x+yi)^3 = r^3 \cdot cis(3\theta)$] and bounding $\theta$ yields:
$-1 < \tan{(3\theta)} = -\dfrac{182}{610} < 0$.
$135^\circ < 3\theta < 180^\circ$. (Since $3\theta$ is obviously in the 2nd quadrant).
$45^\circ < \theta < 60^\circ$.
$\tan{(\theta)} > 1$.
Thus, $y>x$. So, the only remaining positive integer solution is $(x,y) = (5,7)$.
As a quick check: $(5+7i)^3 = (5)^3 + 3(5)^2(7)i + 3(5)(7)^2i^2 + (7)^3i^3 = -610 + 182i$
Thus, the solution is $(x,y) = \boxed{(5,7)}$.
Note that $(5+7i)cis(120^\circ) = \left(\dfrac{-5-7\sqrt{3}}{2}\right) + \left(\dfrac{-7+5\sqrt{3}}{2}\right)i$ and $(5+7i)cis(240^\circ) = \left(\dfrac{-5+7\sqrt{3}}{2}\right) + \left(\dfrac{-7-5\sqrt{3}}{2}\right)i$
Thus, $(x,y) = \left(\dfrac{-5-7\sqrt{3}}{2},\dfrac{-7+5\sqrt{3}}{2} \right)$ and $\left(\dfrac{-5+7\sqrt{3}}{2},\dfrac{-7-5\sqrt{3}}{2} \right)$ are also solutions. |
# KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.1
In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.1 pdf, free KSEEB SSLC Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.1 pdf download. Now you will get step by step solution to each question.
## Karnataka Board Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Ex 5.1
Question 1.
Express the following statements mathematically.
(i) Square of 4 is 16
4 × 4 = 16
42 = 16
(ii) Square of 8 is 64
8 × 8 = 64
82 = 64
(iii) Square of 15 is 225
15 × 15 = 225
152 = 225
Question 2.
Identify the perfect squares among the following numbers. 1, 2, 3, 8, 36, 49, 65, 67, 71, 81, 169, 625, 125, 900, 100, 1000, 100000.
The perfect squares are 1 = 22, 36 = 62,49 = 72, 81 = 92, 169 = 132, 625 = 252, 900 = 302, 100 = 102.
Question 3.
Make a list of all perfect squares from 1 to 500.
The Perfect square from 1 to 500 are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484.
Question 4.
Write 3- digit numbers ending with 0, 1, 4, 5, 6, 9 one for each digit but none of them is a perfect square
200, 201, 204, 205, 206, 209.
Question 5.
Find the numbers from 100 to 400 that end with 0, 1, 4, 5, 6, or 9 which are perfect squares.
100, 121, 144, 169, 196, 225, 256, 289, 324, 36
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# If the roots of the equation ${x^3} - 12{x^2} + 39x - 28 = 0$ are in A.P., then its common difference is:A) $\pm 1$B) $\pm 2$C) $\pm 3$D) $\pm 4$
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Hint:We use the concept of the highest power of any equation is the number of the roots of the equation. Since in our problem we have the highest power three. Hence the given equation has three roots. These are in A.P. we solve the problem by using the properties of A.P.
Let us consider, $\alpha ,\beta ,\gamma$ be three roots of a cubic equation $p{x^3} + q{x^2} + rx + s = 0$. Then, by the relation between roots and coefficients we get,
$\alpha + \beta + \gamma = \dfrac{{ - q}}{p} \\ \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{r}{p} \\ \alpha \beta \gamma = \dfrac{{ - s}}{p} \\$
It is given that the equation is ${x^3} - 12{x^2} + 39x - 28 = 0$.
We know that the highest power of any equation is the number of roots of that equation. Since, the highest power the given equation is three, so, the given equation has three roots.
It is also given that; the roots of the given equation are in A.P.
Let us consider the roots as $a - d,a,a + d$ where, $d$ is the common difference.
We have to find the value of this common difference that means the value of $d$.
Now, from the relation between roots and coefficient we get,
$a - d + a + a + d = 12$….. (1)
$a(a - d) + a(a + d) + (a - d)(a + d) = 39$… (2)
$(a - d)a(a + d) = 28$… (3)
Solving the equation (1) we get,
$3a = 12$
(We can eliminate $d$ as the sum of negative and positive value of same number is always zero)
Dividing we get,
$a = 4$
Next, we will find the value of $d$.
From equation (2) we get,
${a^2} - ad + {a^2} + ad + {a^2} - {d^2} = 39$
Eliminating $- ad$ and $ad$ we get,
$3{a^2} - {d^2} = 39$
Substitute the value of $a = 4$ in above equation we get,
$3{(4)^2} - {d^2} = 39$
Simplifying we get,
${d^2} = 48 - 39$
Simplifying again we get,
${d^2} = 9$
Taking square root of both side we get,
$d = \pm 3$
Hence,
The common difference of the roots of the given equation is $d = \pm 3$
The correct option is (C) $d = \pm 3$.
Here, we get two values of common difference that is $d = \pm 3$. Let us try to find out the roots.
For, $d = 3$, the roots are, $4 - 3,4,4 + 3$ that is $1,4,7$
Again,
For, $d = - 3$, the roots are, $4 + 3,4,4 - 3$ that is $7,4,1.$
It shows that for two different values of common difference, the roots will always be the same.
Note:An arithmetic progression is a sequence of numbers in which each term is derived from the preceding term by adding or subtracting a fixed number called the common difference "d".
The general form of an arithmetic progression is $a - d,{\text{ }}a,{\text{ }}a + d,{\text{ }}a + 2d,....$and so on. Thus the nth term of an AP series is ${T_n} = a + (n - 1)d$, where ${T_n} = {\text{nth }}$term and a is first term. Here d is a common difference = ${T_n} = {T_{n - 1}}$ |
# Basis for an eigenspace
Let $A=\begin{pmatrix} 6&3&-8\\ 0&-2&0\\ 1&0&-3\\ \end{pmatrix}$. I found that the two distinct eigenvalues are $\lambda=5,-2$. Now, I am asked to find a basis for each eigenspace corresponding to each eigenvalue. I found the basis for the eigenspace corresponding to $\lambda=5$ to be $(8,0,1)$. For the eigenspace corresponding to $\lambda=-2$, I get a basis different than the key; below is my work.
We say $\vec{x}$ is an eigenvector of a given matrix A if and only if A$\vec{x}=\lambda\vec{x}$ for some eigenvalue $\lambda$. Thus, we solve for $\vec{x}$. $$(A-\lambda I)\vec{x}=\vec{0}$$ $$\begin{pmatrix} 6&3&-8\\ 0&-2&0\\ 1&0&-3\end{pmatrix}\vec{x}-\begin{pmatrix} -2&0&0\\ 0&-2&0\\ 0&0&-2\\ \end{pmatrix}\vec{x}=\vec{0}$$ $$\begin{pmatrix} 8&3&-8\\ 0&0&0\\ 1&0&-1\\ \end{pmatrix}\vec{x}=\vec{0}$$ This equates to to solving the linear system: $$\begin{pmatrix} 8&3&-8&0\\ 0&0&0&0\\ 1&0&-1&0\\ \end{pmatrix}$$ Interchanging rows $1$ and $3$: $$\begin{pmatrix} 1&0&-1&0\\ 0&0&0&0\\ 8&3&-8&0\\ \end{pmatrix}$$ Eliminating below $1$ in first column: $$\begin{pmatrix} 1&0&-1&0\\ 0&0&0&0\\ 0&3&0&0\\ \end{pmatrix}$$ Interchanging rows $2$ and $3$ and rescaling: $$\begin{pmatrix} 1&0&-1&0\\ 0&1&0&0\\ 0&0&0&0\\ \end{pmatrix}$$ Thus I would conclude my basis for the eigenspace corresponding to $\lambda=-2$ is $(1,0,0)$. However, the book insists the basis is $(1,0,1)$. Are there any errors in my work? Thanks in advance!
Edit: I now see my error. The basis is indeed $(1,0,1)$.
• the step for $\begin{pmatrix} 1&0&-1&0\\ 0&0&0&0\\ 8&3&-8&0\\ \end{pmatrix}$ to $\begin{pmatrix} 1&0&-1&0\\ 0&0&8&0\\ 0&3&0&0\\ \end{pmatrix}$ is not clear and seems to be wrong. – gimusi Apr 5 '18 at 14:10
• I added -8R1 to R3. – coreyman317 Apr 5 '18 at 14:12
• and what is the 8 in the second row? – gimusi Apr 5 '18 at 14:13
• and in the final step what is the 1 in the right up entry? – gimusi Apr 5 '18 at 14:14
• Ah, thank you. All-nighters make calculations a nightmare. Thanks for the quick help! – coreyman317 Apr 5 '18 at 14:14
## 1 Answer
From here,
$$\begin{pmatrix} 8&3&-8\\ 0&0&0\\ 1&0&-1\\ \end{pmatrix}\vec{x}=\vec{0}$$
we see that $\vec{x}=(1,0,1)$ is the correct solution.
Indeed from here
$$\begin{pmatrix} 8&3&-8&0\\ 0&0&0&0\\ 1&0&-1&0\\ \end{pmatrix}\stackrel{R_1-8R_3}\to \begin{pmatrix} 0&3&0&0\\ 0&0&0&0\\ 1&0&-1&0\\ \end{pmatrix} \implies x_2=0 \quad x_1=x_3$$
• Where did I go wrong in my calculation? – coreyman317 Apr 5 '18 at 14:09
• @coreyman317 I've just pointed out in the comment under the OP, second step is not clear and seems uncorrect. – gimusi Apr 5 '18 at 14:11 |
# Video: AQA GCSE Mathematics Higher Tier Pack 4 • Paper 3 • Question 6
Consider the following two inequalities: 7 ≤ 𝐴 ≤ 42, 7 < 𝐵 < 42. How do the integer values of 𝐴 and 𝐵, differ?
02:00
### Video Transcript
Consider the following two inequalities. 𝐴 is greater than or equal to seven but less than or equal to 42. 𝐵 is greater than seven but less than 42. How do the integer values of 𝐴 and 𝐵 differ?
So let’s consider what’s different about the two statements or the two inequalities. So we’ve got the difference shown here. And it’s in the inequality notation that we’re using. And I’ve shown it using a couple of number lines.
So for the top number line, I’ve got 𝐴. I’ve got seven and 42. But above each of these, I’ve got closed dots, whereas for 𝐵, we’ve got open dots. And this is because of the signs, because we’ve got in 𝐴 that it is greater than or equal to or less than or equal to and in 𝐵 it is greater than or less than. So the closed dot means that it includes that number itself, so includes seven, includes 42, whereas the open dot means that it does not.
So therefore, we can say that the way that the integer values of 𝐴 and 𝐵 differ is that the integer values of 𝐴 include seven and 42. And we’ve shown that on our number line with the closed dots. But also it’s in our inequality notation because we can see that we’ve got a line underneath as well, whereas the integer values of 𝐵 do not include seven and 42. And again, we’ve shown that on our number line with our open dots. And we can see that in our inequality notation because there is not a line underneath.
So if we think about what this would mean in real terms, it would mean that 𝐴 would be equal to the values from seven, eight, then all the way up to 41 and 42, whereas the values for 𝐵 would start at eight and will go all the way up to 41. |
# How do you write the following quotient in standard form -10/(2i)?
Aug 16, 2016
$5 i$
#### Explanation:
We require to express the fraction with a real denominator.
To achieve this make use of the following fact.
$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{i}^{2} = {\left(\sqrt{- 1}\right)}^{2} = - 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
Multiply the numerator and denominator by i
$\frac{- 10}{2 i} \times \frac{i}{i} = \frac{- 10 i}{2 {i}^{2}} = \frac{- 10 i}{- 2} = 5 i$
$\Rightarrow - \frac{10}{2 i} = 5 i = 0 + 5 i \text{ in standard form}$ |
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# Walk-a-thon 1
Alignments to Content Standards: 6.RP.A.3.a 6.RP.A.3.b
Julianna participated in a walk-a-thon to raise money for cancer research. She recorded the total distance she walked at several different points in time, but a few of the entries got smudged and can no longer be read. The times and distances that can still be read are listed in the table below.
1. Assume Julianna walked at a constant speed. Complete the table and plot Julianna’s progress in the coordinate plane.
2. How fast was Julianna walking in miles per hour? How long did it take Julianna to walk one mile?
3. Next year Julianna is planning to walk for seven hours. If she walks at the same speed next year, how many miles will she walk?
Time in hrs Miles walked
1
2 6
12
5
## IM Commentary
In this task, students are given information about a context where there is a proportional relationship between two quantities in a table that has missing values. Students need to fill in the missing values, plot the corresponding points in the coordinate plane, and find the two unit rates that are associated with this proportional relationship. They then use these mathematical tools to make predictions about the future.
In 7th grade, students will deepen their understanding of the connections between tables, graphs, and verbal descriptions that represent proportional relationships and identify the unit rate $r$ as the vertical coordinate of a point on the graph with horizontal coordinate 1. For a version of this task that shows the 7th grade expectations, please see:
7.RP Walk-a-thon 2 http://www.illustrativemathematics.org/illustrations/1526
The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, the commentary will spotlight one practice connection in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.
Mathematical Practice 5, "Use appropriate tools strategically," emphasizes a student’s ability to consider a tool’s usefulness and constraints as well as knowing how to use it appropriately. During this task, students engage in MP.5 by considering the connections among tables, graphs, and verbal descriptions (tools) to find the unit rate associated with the proportional relationship. They extend their understanding by using these same tools to predict how many miles Julianna will walk next year if she maintains the same speed. Although this particular task directs students to use specific tools, it does not prevent them from trying additional tools. A discussion regarding what other tools may be efficient and appropriate to use and why you might use a particular tool would be beneficial.
## Solution
1. We see that the only complete time-distance pair indicates that Julianna walked 6 miles in 2 hours. If she walked at a constant speed, we can conclude that Julianna walked 3 miles in 1 hour. Using this speed, we can find the remaining values in the table by multiplying the hours she walked by 3 (or, alternatively, by entering 3 in row next to 1 and then adding 3 for every hour walked). We get the following table.
Time in hrsMiles walked
13
26
39
412
515
Let's plot her progress in the coordinate plane by having her time, in hours, be represented on the horizontal axis, and the number of miles she walked be represented on the vertical axis. Using the rows in the table above as our coordinate points, we get the following graph.
2. We found in part (a) that Julianna walks 3 miles in 1 hour.
Looking at the picture above, we can see that if she walks 3 miles in one hour, and she is walking at a constant speed, then it will take her $\frac13$ of an hour to walk 1 mile.
We can also look at a double number line.
We can see on the double number line that 1 hour corresponds to three miles and $\frac13$ hour corresponds to 1 mile.
So, we find that it took Julianna $\frac 13$ hours, or 20 minutes, to walk 1 mile.
3. We found in parts (a) and (b) that Julianna’s rate of travel was 3 miles per hour. If this stayed constant, we can find how many miles she would walk in 7 hours by extending our table.
Time in hrsMiles walked
13
26
39
412
515
618
721
Thus, we see that in 7 hours, Julianna will walk 21 miles at that rate. |
New Zealand
Level 8 - NCEA Level 3
# Sums and differences as products
Lesson
### Products
If the sine and cosine sum and difference formulas are written down side-by-side it becomes apparent that useful results can be obtained by adding some of them them in pairs.
$\sin\left(A+B\right)$sin(A+B) $=$= $\sin\left(A\right)\cos\left(B\right)+\sin\left(B\right)\cos\left(A\right)$sin(A)cos(B)+sin(B)cos(A) (1) $\sin\left(A-B\right)$sin(A−B) $=$= $\sin\left(A\right)\cos\left(B\right)-\sin\left(B\right)\cos\left(A\right)$sin(A)cos(B)−sin(B)cos(A) (2) $\cos\left(A+B\right)$cos(A+B) $=$= $\cos\left(A\right)\cos\left(B\right)-\sin\left(A\right)\sin\left(B\right)$cos(A)cos(B)−sin(A)sin(B) (3) $\cos\left(A-B\right)$cos(A−B) $=$= $\cos\left(A\right)\cos\left(B\right)+\sin\left(A\right)\sin\left(B\right)$cos(A)cos(B)+sin(A)sin(B) (4)
If we add (1) and (2), we have
$\sin\left(A+B\right)+\sin\left(A-B\right)$sin(A+B)+sin(A−B) $=$= $2\sin\left(A\right)\cos\left(B\right)$2sin(A)cos(B) (5)
Similarly, from (3) and (4) we obtain, by addition,
$\cos\left(A+B\right)+\cos\left(A-B\right)$cos(A+B)+cos(A−B) $=$= $2\cos\left(A\right)\cos\left(B\right)$2cos(A)cos(B) (6)
and by subtraction,
$\cos\left(A-B\right)-\cos\left(A+B\right)$cos(A−B)−cos(A+B) $=$= $2\sin\left(A\right)\sin\left(B\right)$2sin(A)sin(B) (7)
Equations (5), (6) and (7) give the following three product formulas:
$\sin\left(A\right)\cos\left(B\right)$sin(A)cos(B) $=$= $\frac{1}{2}\left(\sin\left(A+B\right)+\sin\left(A-B\right)\right)$12(sin(A+B)+sin(A−B)) (5a) $\cos\left(A\right)\cos\left(B\right)$cos(A)cos(B) $=$= $\frac{1}{2}\left(\cos\left(A+B\right)+\cos\left(A-B\right)\right)$12(cos(A+B)+cos(A−B)) (6a) $\sin\left(A\right)\sin\left(B\right)$sin(A)sin(B) $=$= $\frac{1}{2}\left(\cos\left(A-B\right)-\cos\left(A+B\right)\right)$12(cos(A−B)−cos(A+B)) (7a)
### Sums
By re-writing (5a), (6a) and (7a) we can obtain formulas for the sums and differences of sines and cosines. To do this, we let $U=A+B$U=A+B and $V=A-B$V=AB. Then, by solving these equations for $A$A and $B$B we get $A=\frac{U+V}{2}$A=U+V2 and $B=\frac{U-V}{2}$B=UV2.
Thus, by substituting for $A$A and $B$B in the product formulas and rearranging slightly, we obtain:
$\sin\left(U\right)+\sin\left(V\right)$sin(U)+sin(V) $=$= $2\sin\left(\frac{U+V}{2}\right)\cos\left(\frac{U-V}{2}\right)$2sin(U+V2)cos(U−V2) (8) $\cos\left(U\right)+\cos\left(V\right)$cos(U)+cos(V) $=$= $2\cos\left(\frac{U+V}{2}\right)\cos\left(\frac{U-V}{2}\right)$2cos(U+V2)cos(U−V2) (9) $\cos\left(V\right)-\cos\left(U\right)$cos(V)−cos(U) $=$= $2\sin\left(\frac{U+V}{2}\right)\sin\left(\frac{U-V}{2}\right)$2sin(U+V2)sin(U−V2) (10)
and from (8), using the fact that $-\sin\left(V\right)=\sin\left(-V\right)$sin(V)=sin(V), we can write
$\sin\left(U\right)-\sin\left(V\right)$sin(U)−sin(V) $=$= $2\sin\left(\frac{U-V}{2}\right)\cos\left(\frac{U+V}{2}\right)$2sin(U−V2)cos(U+V2) (11)
Another type of sum, with a very useful simplification, occurs between different multiples of the sine and cosine of identical angles.
The expression $a\sin\left(\theta\right)+b\cos\left(\theta\right)$asin(θ)+bcos(θ) can be written in the form $r\sin\left(\theta+\alpha\right)$rsin(θ+α). The latter expands to $r\left(\sin\left(\theta\right)\cos\left(\alpha\right)+\cos\left(\theta\right)\sin\left(\alpha\right)\right)$r(sin(θ)cos(α)+cos(θ)sin(α)).
On comparing this with the original expression, we see that $a=r\cos\left(\alpha\right)$a=rcos(α) and $b=r\sin\left(\alpha\right)$b=rsin(α).
Hence, $r=\sqrt{a^2+b^2}$r=a2+b2 and $\tan\left(\alpha\right)=\frac{b}{a}$tan(α)=ba. Then, using the notation $\tan^{-1}$tan1 for the inverse tangent function, we can write
$a\sin\left(\theta\right)+b\cos\left(\theta\right)$asin(θ)+bcos(θ) $=$= $\sqrt{a^2+b^2}\sin\left(\theta+\tan^{-1}\left(\frac{b}{a}\right)\right)$√a2+b2sin(θ+tan−1(ba)) (12)
#### Worked example
Express $\cos\left(\frac{17\pi}{12}\right)-\cos\left(\frac{\pi}{4}\right)$cos(17π12)cos(π4) more simply.
Do: Using (10), $\cos\left(V\right)-\cos\left(U\right)=2\sin\left(\frac{U+V}{2}\right)\sin\left(\frac{U-V}{2}\right)$cos(V)cos(U)=2sin(U+V2)sin(UV2), we have
$\cos\left(\frac{17\pi}{12}\right)-\cos\left(\frac{\pi}{4}\right)=2\sin\left(\frac{\frac{17\pi}{12}+\frac{\pi}{4}}{2}\right)\sin\left(\frac{\frac{\pi}{4}-\frac{17\pi}{12}}{2}\right)$cos(17π12)cos(π4)=2sin(17π12+π42)sin(π417π122)
That is,
$\cos\left(\frac{17\pi}{12}\right)-\cos\left(\frac{\pi}{4}\right)$cos(17π12)−cos(π4) $=$= $2\sin\left(\frac{5\pi}{6}\right)\sin\left(-\frac{7\pi}{12}\right)$2sin(5π6)sin(−7π12) $=$= $-2\sin\left(\frac{\pi}{6}\right)\sin\left(\frac{5\pi}{12}\right)$−2sin(π6)sin(5π12) $=$= $-\sin\left(\frac{5\pi}{12}\right)$−sin(5π12)
Using a half-angle formula, $\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{1}{2}\left(1-\cos\left(\theta\right)\right)}$sin(θ2)=12(1cos(θ)) we can further simplify this to the exact value $-\frac{1}{2}\sqrt{2+\sqrt{3}}$122+3.
#### Practice questions
##### QUESTION 1
Express $\cos\left(3x+2y\right)\cos\left(x-y\right)$cos(3x+2y)cos(xy) as a sum or difference of two trigonometric functions.
##### QUESTION 2
Express $\sin\left(6x\right)+\sin\left(4x\right)$sin(6x)+sin(4x) as a product of two trigonometric functions.
##### QUESTION 3
By expressing the left-hand side of the equation as a product, solve the equation $\sin5x+\sin x=0$sin5x+sinx=0 for $0\le x$0x$<=$<=$2\pi$2π.
### Outcomes
#### M8-6
Manipulate trigonometric expressions
#### 91575
Apply trigonometric methods in solving problems |
# Question: What Is The Volume Of The Square Pyramid?
## What is the formula for volume of a triangular pyramid?
Use the formula for the volume of a triangular pyramid: V=13Ah , where A = area of the triangular base, and H = height of the pyramid..
## What is the volume of ball?
The formula for the volume of a sphere is V = 4/3 πr³. See the formula used in an example where we are given the diameter of the sphere.
## How do you find the volume of a octagonal pyramid?
We can find the volume of an octagonal pyramid using the following formulas: Volume = (B × h) / 3, where B is the area of the base. B = 2 × s2 (1 + √2)
## What is the formula for finding volume?
Whereas the basic formula for the area of a rectangular shape is length × width, the basic formula for volume is length × width × height.
## What is the the formula for volume?
Volume Formula: Volume = l × w × h , where l is length, w is width and h is height.
## What is the volume of this pyramid?
The volume V of a pyramid is one-third the area of the base B times the height h .
## How do you calculate the volume of a tank?
V(tank) = (πr2 + 2ra)lFill, f < r. We calculate fill volume using the circular segment method, as in a Horizontal Cylinder Tank, for the filled portion.Fill, f > r and f < (r+a) The filled volume is exactly 1/2 of the cylinder portion plus the volume of fill inside the rectangular portion.Fill, f > (r+a) and f < h.
## What is volume of square?
Volume is measured in “cubic” units. … Volume of a cube = side times side times side. Since each side of a square is the same, it can simply be the length of one side cubed. If a square has one side of 4 inches, the volume would be 4 inches times 4 inches times 4 inches, or 64 cubic inches.
## What is the volume of this cylinder?
The radius of the cylinder is 8 cm and the height is 15 cm. Substitute 8 for r and 15 for h in the formula V=πr2h . Simplify. Therefore, the volume of the cylinder is about 3016 cubic centimeters.
## What is the volume of a prism?
The formula for the volume of a prism is V=Bh , where B is the base area and h is the height. The base of the prism is a rectangle. The length of the rectangle is 9 cm and the width is 7 cm. The area A of a rectangle with length l and width w is A=lw . … Therefore, the volume of the prism is 819 cubic centimeters.
## How do you find volume of square pyramid?
To find the volume of a square-based pyramid, use the formula V = A(h/3), where V is the volume and A is the area of the base.
## What is the volume of the regular square pyramid?
Square Pyramid Formulas derived in terms of side length a and height h: Volume of a square pyramid: V = (1/3)a2h. |
# If α, β Are the Roots of the Equation X 2 + P X + Q = 0 Then − 1 α + 1 β Are the Roots of the Equation - Mathematics
MCQ
If α, β are the roots of the equation $x^2 + px + q = 0 \text { then } - \frac{1}{\alpha} + \frac{1}{\beta}$ are the roots of the equation
#### Options
• $x^2 - px + q = 0$
• $x^2 + px + q = 0$
• $q x^2 + px + 1 = 0$
• $q x^2 - px + 1 = 0$
#### Solution
$q x^2 - px + 1 = 0$
Given equation:
$x^2 + px + q = 0$
Also,
$\alpha$ and $\beta$ are the roots of the given equation.
Then, sum of the roots = $\alpha + \beta = - p$
Product of the roots = $\alpha\beta = q$
Now, for roots
$- \frac{1}{\alpha} , - \frac{1}{\beta}$ , we have:
Sum of the roots = $- \frac{1}{\alpha} - \frac{1}{\beta} = - \frac{\alpha + \beta}{\alpha\beta} = - \left( \frac{- p}{q} \right) = \frac{p}{q}$
Product of the roots = $\frac{1}{\alpha\beta} = \frac{1}{q}$
Hence, the equation involving the roots $- \frac{1}{\alpha}, - \frac{1}{\beta}$ is as follows:
$x^2 - \left( \alpha + \beta \right)x + \alpha\beta = 0$
$\Rightarrow x^2 - \frac{p}{q}x + \frac{1}{q} = 0$
$\Rightarrow q x^2 - px + 1 = 0$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 14 Quadratic Equations
Q 21 | Page 17 |
RS Aggarwal Solutions for Class 9 Chapter 7: Lines and Angles Exercise 7A
Students often don’t clear their doubts in the class when solved by the teacher owing to shyness. This mainly occurs in a subject like Mathematics because in other subjects the concepts can be understood by re-reading the chapter. But, in Mathematics, the students need to understand the methods which are used in solving problems. In order to help students, we provide exercise wise answers to all the chapters in PDF format. By referring the PDF, students can understand the methods used in solving the problems. RS Aggarwal Solutions for Class 9 Maths Chapter 7 Lines and Angles Exercise 7A are provided here.
Access RS Aggarwal Solutions for Class 9 Chapter 7: Lines and Angles Exercise 7A
Exercise 7(A)
1. Define the following terms:
(i) Angle
(ii) Interior of an angle
(iii) Obtuse angle
(iv) Reflex angle
(v) Complementary angles
(vi) Supplementary angles
Solution:
(i) Angle – When two rays originate from the same end point, then an angle is formed.
(ii) Interior of an angle – The interior of ∠BAC is the set of all points in its plane which lie on the same side of AB as C and also on the same side of AC as B.
(iii)Obtuse angle – An angle whose measure is more than 90o but less than 180o is called an obtuse angle.
(iv) Reflex angle – An angle whose measure is more than 180o but less than 360o is called a reflex angle.
(v) Complementary angles – Two angles are said to be complementary, if the sum of their measure is 90o.
(vi) Supplementary angles – Two angles are said to be supplementary if the sum of their measures is 180o.
2. Find the complement of each of the following angles:
(i) 55o
(ii) 16o
(iii) 90o
(iv) 2/3 of a right angle
Solution:
(i) We know that the complement of 55o can be written as
55o = 90o – 55o = 35o
(ii) We know that the complement of 16o can be written as
16o = 90o – 16o = 74o
(iii) We know that the complement of 90o can be written as
90o = 90o – 90o = 0o
(iv) We know that 2/3 of a right angle can be written as 2/3 × 90o = 60o
60o = 90o – 60o = 30o
3. Find the supplement of each of the following angles:
(i) 42o
(ii) 90o
(iii) 124o
(iv) 3/5 of a right angle
Solution:
(i) We know that the supplement of 42o can be written as
42o = 180o – 42o = 138o
(ii) We know that the supplement of 90o can be written as
90o = 180o – 90o = 90o
(iii) We know that the supplement of 124o can be written as
124o = 180o – 124o = 56o
(iv) We know that 3/5 of a right angle can be written as 3/5 × 90o = 54o
54o = 180o – 54o = 126o
4. Find the measure of an angle which is
(i) Equal to its complement
(ii) Equal to its supplement
Solution:
(i) Consider the required angle as xo
We know that the complement can be written as 90o – xo
To find that the measure of an angle is equal to its complement
We get
xo = 90o – xo
We can also write it as
x + x = 90
So we get
2x = 90
By division we get
x = 90/2
xo = 45o
Therefore, the measure of an angle which is equal to its complement is 45o
(ii) Consider the required angle as xo
We know that the supplement can be written as 180o – xo
To find that the measure of an angle is equal to its complement
We get
xo = 180o – xo
We can also write it as
x + x = 180
So we get
2x = 180
By division we get
x = 180/2
xo = 90o
Therefore, the measure of an angle which is equal to its complement is 90o
5. Find the measure of an angle which is 36o more than its complement.
Solution:
Consider the required angle as xo
We know that the complement can be written as 90o – xo
xo = (90o – xo) + 36o
We can also write it as
x + x = 90 + 36
So we get
2x = 126
By division we get
x = 126/2
xo = 63o
Therefore, the measure of an angle which is 36o more than its complement is 63o.
6. Find the measure of an angle which is 30o less than its supplement.
Solution:
Consider the required angle as xo
We know that the supplement can be written as 180o – xo
xo = (180o – xo) – 30o
We can also write it as
x + x = 180 – 30
So we get
2x = 150
By division we get
x = 150/2
xo = 75o
Therefore, the measure of an angle which is 30o more than its supplement is 75o.
7. Find the angle which is four times its complement.
Solution:
Consider the required angle as xo
We know that the complement can be written as 90o – xo
xo = 4(90o – xo)
We can also write it as
x = 360 – 4x
So we get
5x = 360
By division we get
x = 360/5
xo = 72o
Therefore, the angle which is four times its complement is 72o.
8. Find the angle which is five times its supplement.
Solution:
Consider the required angle as xo
We know that the supplement can be written as 180o – xo
xo = 5(180o – xo)
We can also write it as
x = 900 – 5x
So we get
6x = 900
By division we get
x = 900/6
xo = 150o
Therefore, the angle which is five times its supplement is 150o.
9. Find the angle whose supplement is four times its complement.
Solution:
Consider the required angle as xo
We know that the complement can be written as 90o – xo and the supplement can be written as 180o – xo
180o – xo = 4(90o – xo)
We can also write it as
180o – xo = 360o – 4xo
So we get
4xo – xo = 360o – 180o
3xo = 180o
By division we get
xo = 180/3
xo = 60o
Therefore, the angle whose supplement is four times its complement is 60o.
10. Find the angle whose complement is one third of its supplement.
Solution:
Consider the required angle as xo
We know that the complement can be written as 90o – xo and the supplement can be written as 180o – xo
90o – xo = 1/3(180o – xo)
We can also write it as
90o – xo = 60o – (1/3) xo
So we get
xo – (1/3)xo = 90o – 60o
(2/3) xo = 30o
By division we get
xo = ((30×3)/2)
xo = 45o
Therefore, the angle whose complement is one third of its supplement is 45o.
11. Two complementary angles are in the ratio 4:5. Find the angles.
Solution:
Consider the required angle as xo and 90o – xo
According to the question it can be written as
xo/ 90o – xo = 4/5
By cross multiplication we get
5x = 4 (90 – x)
5x = 360 – 4x
On further calculation we get
5x + 4x = 360
9x = 360
By division
x = 360/9
So we get
x = 40
Therefore, the angles are 40o and 90o – xo = 90o – 40o = 50o
12. Find the value of x for which the angles (2x – 5) o and (x – 10) o are the complementary angles.
Solution:
It is given that (2x – 5) o and (x – 10) o are the complementary angles.
So we can write it as
(2x – 5) o + (x – 10) o = 90o
2x – 5o + x – 10o = 90o
On further calculation
3x – 15o = 90o
So we get
3x = 105o
By division
x = 105/3
x = 35o
Therefore, the value of x for which the angles (2x – 5) o and (x – 10) o are the complementary angles is 35o.
Access other exercise solutions of Class 9 Maths Chapter 7: Lines and Angles
Exercise 7B Solutions 16 Questions
Exercise 7C Solutions 24 Questions
RS Aggarwal Solutions Class 9 Maths Chapter 7 – Lines and Angles Exercise 7A
RS Aggarwal Solutions Class 9 Maths Chapter 7 Lines and Angles Exercise 7A is the first exercise which contains examples and problems which are solved by experts based on complementary and supplementary angles.
Key features of RS Aggarwal Solutions for Class 9 Maths Chapter 7: Lines and Angles Exercise 7A
• The solutions are prepared by experts keeping in mind the understanding abilities among the students.
• Self analysis can be done by the students to understand the areas of higher importance based on weightage in the exam.
• RS Aggarwal solutions contain answers in descriptive manner which help students grasp the concepts faster.
• The problems are solved using the easiest methods and diagrammatic representation helps students to understand the methods. |
Categories
## What type of graph uses the length of bars to show how many data points are in an interval??
a) The time the police officer required to reach the motorist was 15 s.
b) The speed of the officer at the moment she overtakes the motorist is 30 m/s
c) The total distance traveled by the officer was 225 m.
Explanation:
The equations for the position and velocity of an object moving in a straight line are as follows:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
v = velocity at time t
a)When the officer reaches the motorist, the position of the motorist is the same as the position of the officer:
x motorist = x officer
Using the equation for the position:
x motirist = x0 + v · t (since a = 0).
x officer = x0 + v0 · t + 1/2 · a · t²
Let´s place our frame of reference at the point where the officer starts following the motorist so that x0 = 0 for both:
x motorist = x officer
x0 + v · t = x0 + v0 · t + 1/2 · a · t² (the officer starts form rest, then, v0 = 0)
v · t = 1/2 · a · t²
Solving for t:
2 v/a = t
t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s
The time the police officer required to reach the motorist was 15 s.
b) Now, we can calculate the speed of the officer using the time calculated in a) and the equation for velocity:
v = v0 + a · t
v = 0 m/s + 2.00 m/s² · 15 s
v = 30 m/s
The speed of the officer at the moment she overtakes the motorist is 30 m/s
c) Using the equation for the position, we can find the traveled distance in 15 s:
x = x0 + v0 · t + 1/2 · a · t²
x = 1/2 · 2.00 m/s² · (15s)² = 225 m
Categories
## Which word best describes Michelle Obama’s tone in this statement? “We as parents are our children’s first and best role models, and this is particularly true when it comes to their health. We can’t lie around on the couch eating French fries and candy bars, and expect our kids to eat carrots and run around the block. But too often, that’s exactly what we’re doing.”
Here are a few pointers; hope this is useful)
Ovation-by definition- is show of appreciation from an audience, for a person’s accomplishments or flaw.
“Everyone deserves a standing ovation because we all overcometh the world.”
A person’s accomplishment could be how they made a positive change in this world, strong leadership- that makes them a effective leader or simply helping others. A person’s flaw- mistakes in life, sin or even guilt should also be considered an appreciation- an ovation for representing mankind’s flaw and that humanity makes mistakes, fulfilling at least one deadly sin such as greed, lust, selfishness etc.
Thus, regardless of a person’s achievement or flaw- a person deserves an applause for, not the least, living in this society and this world that we are all living together and dying together.
That was just the introduction.. the best part is yet to come.. now it’s your turn!!
Shakespeare’s famous line “All the World’s a stage. That agrees with your line: “Everybody deserves a standing ovation…”
Shakespeare explains that men and women are like players: they live, and die, some being celebrated and some forever living in solitude till their death. Shakespeare states the world is a “stage” which symbolizes that mankind is in its peak. The world is changing everyday: little by little and humanity is falling behind.
Due to our world turning into machinery: factories, an automotive future: where humans only job to live (entrance) and to die (exit) the famous humans remembered and the flawed not recalled.
This is according to Shakespeare’s imagery.
I don’t know what grade your in, but I think simplifying Shakespeare’s word of mouth in your essay would be handy and useful as it has strong references of your quote, and agrees strongly in your essay.
Hope this helps 🙂
Categories
## The ________ amendment to the u.s. constitution outlawed slavery, and courts have held that it bars racial discrimination.
The option D is correct.
The major difference between state and federal courts in the United States is that only federal courts must have judges approved by the Senate.
Further Explanation:
Federal Court: These are established by the U.S. Constitution to settle disputes that involve laws made by Congress and the Constitution.
Justification for correct and incorrect answer:
A.
Only state courts use an adversarial system during trials: This option is incorrect.
The adversarial system is used by both the federal and state courts. It is a legal system in which parties’ case is represented by two advocates before a judge or jury, who determines the truth and pass judgment.
B.
Only state courts issue verdicts in both criminal and civil cases: This option is incorrect.
The criminal cases that involve state law violations are filed in state court while criminal cases that involve violation of the federal laws can be filed only in federal court.
C.
Only federal courts allow defendants to appeal rulings: This option is incorrect.
The appeal can be filed by the defendant in the federal court against the judgment of the trial court.
D.
Only federal courts must have judges approved by the Senate: This option is correct.
The federal judge is appointed by the president of the United States and confirmed by the United States Senate. The state judge is appointed by the state’s governor or legislature.
Therefore, The major difference between state and federal courts in the United States is that only federal courts must have judges approved by the Senate.
Subject: Social Studies
Chapter: Government & Civics
Keywords: What is one major difference between state and federal courts in the United States, only state courts use an adversarial system during trials, only state courts issue verdicts in both criminal and civil cases, only federal courts allow defendants to appeal rulings, only federal courts must have judges approved by the senate.
Categories |
# How do you prove tan(2x) = (2tan(x))/(1-tan^2 x)?
Apr 14, 2016
Use
$\tan x = \sin \frac{x}{\cos} x , \sin 2 x = 2 \sin x \cos x \mathmr{and} \cos 2 x = {\cos}^{2} x - {\sin}^{2} x$, for the right hand side expression
#### Explanation:
2 tan x/(1-tan^2x)=(2sin x/cos x)/(1-(sin^2x/cos^2x)
$= 2 \sin x \cos \frac{x}{{\cos}^{2} x - {\sin}^{2} x}$
$= \frac{\sin 2 x}{\cos 2 x} = \tan 2 x$
Proofs for $\sin 2 x = 2 \sin x \cos x \mathmr{and} \cos 2 x = 1 - 2 {\sin}^{2} x$:
Use
Area of a $\triangle$ABC = 1/2(base)(altitude) = 1/2 bc sin A. Here,
it is the $\triangle$ ABC of a unit circle, with center at A, B and C on
the circle and $\angle$A = 2x. Here, AB = AC = 1, BC = 2 sinx and
the altitude from A, AN = cos x.
Area of $\triangle$ABC = 1/2 (BC) (AN) = 1/2(OA)(OB) sin A.
Readily, 1/2(2 sin x) cos x = 1/2 (1)(1)sin 2x.
Use ${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cos A$, to get the cosine formula.
graph{(x^2+y^2-1)(y-x)(y+x)(x-0.707)=0[0 3 -0.73 0.73]} |
# Math Snap
## Question 20 Not yet answered Marked out of 1.00 Flag question If $f^{\prime}(x)$ exists, then $f^{\prime \prime}(x)$ exists. Select one: True False
#### STEP 1
Assumptions 1. We are given a function $f(x)$. 2. The first derivative of the function, $f'(x)$, exists. 3. We need to determine if the existence of $f'(x)$ implies that the second derivative, $f''(x)$, also exists.
#### STEP 2
Recall the definitions of the first and second derivatives. The first derivative $f'(x)$ is defined as: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
#### STEP 3
The second derivative $f''(x)$ is defined as the derivative of the first derivative $f'(x)$: $f''(x) = \lim_{h \to 0} \frac{f'(x+h) - f'(x)}{h}$
#### STEP 4
To determine if $f''(x)$ exists, we need to check if $f'(x)$ is differentiable. Differentiability of $f'(x)$ means that the limit defining $f''(x)$ must exist.
#### STEP 5
Consider a counterexample where $f'(x)$ exists but $f'(x)$ is not differentiable. For instance, let $f(x) = |x|$.
#### STEP 6
Calculate the first derivative of $f(x) = |x|$: $f'(x) = \begin{cases} 1 & \text{if } x > 0 \\ -1 & \text{if } x < 0 \\ \text{undefined} & \text{if } x = 0 \end{cases}$
#### STEP 7
Notice that $f'(x)$ exists for all $x \neq 0$, but it is not continuous at $x = 0$. Therefore, $f'(x)$ is not differentiable at $x = 0$.
#### STEP 8
Since $f'(x)$ is not differentiable at $x = 0$, $f''(x)$ does not exist at $x = 0$.
##### SOLUTION
Thus, the existence of $f'(x)$ does not necessarily imply the existence of $f''(x)$. The statement is: \textbf{False} |
# 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:
### 43/4 - 21/4 = 5/2 = 2 1/2 = 2.5
Spelled result in words is five halfs (or two and one half).
### How do you solve fractions step by step?
1. Conversion a mixed number 4 3/4 to a improper fraction: 4 3/4 = 4 3/4 = 4 · 4 + 3/4 = 16 + 3/4 = 19/4
To find new numerator:
a) Multiply the whole number 4 by the denominator 4. Whole number 4 equally 4 * 4/4 = 16/4
b) Add the answer from previous step 16 to the numerator 3. New numerator is 16 + 3 = 19
c) Write a previous answer (new numerator 19) over the denominator 4.
Four and three quarters is nineteen quarters
2. Conversion a mixed number 2 1/4 to a improper fraction: 2 1/4 = 2 1/4 = 2 · 4 + 1/4 = 8 + 1/4 = 9/4
To find new numerator:
a) Multiply the whole number 2 by the denominator 4. Whole number 2 equally 2 * 4/4 = 8/4
b) Add the answer from previous step 8 to the numerator 1. New numerator is 8 + 1 = 9
c) Write a previous answer (new numerator 9) over the denominator 4.
Two and one quarter is nine quarters
3. Subtract: 19/4 - 9/4 = 19 - 9/4 = 10/4 = 2 · 5/2 · 2 = 5/2
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(4, 4) = 4. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 4 × 4 = 16. In the next intermediate step, , cancel by a common factor of 2 gives 5/2.
In words - nineteen quarters minus nine quarters = five halfs.
#### 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:
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:
• Two cakes
Two cakes were each cut into 8 slices. Maria ate 1/8 of the chocolate cake and 1 slice of carrot cake. Julia ate 1/2 of the carrot cake. Mark ate 1 slice of each. Thomas ate 3 slices of chocolate cake. How many slices were left?
• Farmer Peter
Farmer Peter paints 12 chicken coop. He started painting this day morning. Now he only has 1/4 of the chicken coop left to paint this afternoon. How many chicken coops did farmer Peter paint this morning?
• The boy
The boy scouts spent 10/12 hour doing their daily exercises. They only used 1/4 hour in hiking. How much time did they use for other body exercises?
• Athletic race
In a race, the second-place finisher crossed the finish line 1 1/3 minutes after the first-place finisher. The third-place finisher was 1 3/4 minutes behind the second-place finisher. The third-place finisher took 34 2/3 minutes. How long did the first-pl
• Complicated sum minus product
What must be subtracted from the sum of 3/8 and 5/16 to get difference equal to the product of 5/8 and 3/16?
• Evaluate 17
Evaluate 2x+6y when x=- 4/5 and y=1/3. Write your answer as a fraction or mixed number in simplest form.
• The frame
Rodney has a board that is 5/6 yards long. He cuts 1/5 yard off the board and uses the rest of the board to make a frame. How much of the board is used to make the frame?
• Metal rod
You have a metal rod that’s 51/64 inches long. The rod needs to be trimmed. You cut 1/64 inches from one end and 1/32 inches from the other end. Next, you cut the rod into 6 equal pieces. What will be the final length of each piece?
• Square metal sheet
We cut out four squares of 300 mm side from a square sheet metal plate with a side of 0,7 m. Express the fraction and the percentage of waste from the square metal sheet.
• Fraction expression
Which expression is equivalent to : minus 9 minus left parenthesis minus 4 start fraction 1 divided by 3 end fraction right parenthesis
• Stock market 2
For the week of July 22, the following day to day changes in the stock market was recorded a certain stock: -2 on Monday; +4 Tuesday; -8 Wednesday; + 2 1/2 Thursday; - 3 1/4 Friday. The stock began the week at 78 points. How many points did it finish with
• A 14.5-gallon
A 14.5-gallon gasoline tank is 3/4 full. How many gallons will it take to fill the tank? Write your answer as a mixed number.
• Half of 2
Half of Ethan’s string is equal to 2/3 of Kayla’s string. The total length of their strings is 10 feet. How much longer is Ethan’s string than Kayla’s? |
# Base of Logarithm
A quantity which used to express any other quantity as its multiplying factors is called the base of logarithm.
## Introduction
Every quantity can be expressed as multiplying factors of another quantity in mathematics. It can be done by expressing a quantity as multiplying factors on the basis of another quantity.
The relation between the two quantities is represented by a logarithmic symbol $(\log)$ but the quantity which used to transform another number is displayed as subscript of $\log$ symbol for representing that quantity as a base of the mathematical operation.
### Example
$256$ is a quantity and let’s try to write this quantity on the basis another quantities to understand the importance of the base of logarithms.
#### Base 2
$256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
$\implies 256 = 2^{\displaystyle 8}$
The relationship between three of them is expressed in logarithms.
$\log_{\displaystyle 2} 256 = 8$
On the basis of number $2$, the number $256$ is expressed as eight multiplying factors. Therefore, the number $2$ is called as base of the logarithm of $256$.
#### Base 4
$256 = 4 \times 4 \times 4 \times 4$
$\implies 256 = 4^{\displaystyle 4}$
Express the relation between three of them in logarithm system.
$\log_{\displaystyle 4} 256 = 4$
On the basis of number $4$, the number $256$ is written as four multiplying factors. Hence, the number $4$ is called as base of the logarithm of $256$.
#### Base 16
$256 = 16 \times 16$
$\implies 256 = 16^{\displaystyle 2}$
Write relation between three of them in logarithms.
$\log_{\displaystyle 16} 256 = 2$
On the basis of $16$, the number $256$ is expressed as two multiplying factors of $16$. Therefore, the number $16$ is called as base of the logarithm of $256$.
The three cases are the best examples for understanding the concept of base of the logarithms.
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# Video: Finding the Product of a Complex Number and Its Conjugate
Find the complex conjugate of 1 + π and the product of this number with its complex conjugate.
03:21
### Video Transcript
Find the complex conjugate of the complex number one plus π and the product of this number with its complex conjugate.
So, thereβre two parts to this question. Firstly, weβre asked to find the complex conjugate of this complex number one plus π. Well, we can recall that the complex conjugate of a complex number is the complex number we get when we simply change the sign of its imaginary part. So, in general, the complex conjugate of the complex number π§ equals π plus ππ is the complex number π§ star, which is equal to π minus ππ. Weβve changed the sign of the complex part. Itβs no longer positive π. Itβs now negative π.
So if we let π§ be our complex number, one plus π, then to find its complex conjugate π§ star, we simply change the sign of the imaginary part. So previously, we had plus π, which is plus one π. And we change it to negative π or negative one π. The complex conjugate of one plus π is therefore one minus π. The second part of this question asks us to find the product of this number. So thatβs our original complex number with its complex conjugate. So weβre looking for the product of π§ and π§ star. As weβve just found the complex conjugate to be one minus π, weβre therefore looking for the product of one plus π and one minus π.
We can go ahead and distribute the parentheses. One multiplied by one gives one. And then, one multiplied by negative π gives negative π. π multiplied by one gives positive π. And then, π multiplied by negative π gives negative π squared. So we have one minus π plus π minus π squared. Now, of course, in the centre of our expression, negative π plus π simplifies to zero. So these two terms cancel out. And weβre left with one minus π squared. We need to recall here that π squared is equal to negative one. We therefore have one minus negative one or one plus one, which is equal to two. And so we found that the product of our complex number with its complex conjugate is two.
In fact, there is actually a general result that we couldβve used here, which is that, for the complex number π§ equals π plus ππ, the product of π§ with its complex conjugate π minus ππ will always be equal to π squared plus π squared. We can see that this is certainly the case for our complex number one plus π. Both the real and imaginary parts are equal to one. And one squared plus one squared is equal to one plus one, which is equal to two.
To see why this is the case, we just need to distribute the parentheses in the product π plus ππ multiplied by π minus ππ. And we see that, in the general case, just as it did in our specific example, the imaginary parts of this expansion cancel, leaving π squared minus π squared π squared. Thatβs π squared minus π squared multiplied by negative one, which is π squared plus π squared. So weβve completed the problem.
The complex conjugate of one plus π is one minus π. And the product of one plus π with its complex conjugate is two. |
# Write the following parametric equation in the form y=f(x) . \begin{alignat}{3} x = t^2...
## Question:
Write the following parametric equation in the form {eq}y=f(x) {/eq}.
{eq}\begin{alignat}{3} x &=&& t^2 +2, \\ y&=&& t^2-4. \end{alignat} {/eq}.
## Parametric Equations:
Parametric equations area set of equations contains an independent variable. We are given two parametric equations and {eq}x= p(t) , y= q(t) {/eq} we need to write the Cartesian equation {eq}y=f(x) {/eq} of the given curves by eliminating the parameter {eq}t. {/eq}
We can equate the parameter and use algebraic operations to solve this kind of problems.
We are given:
{eq}\begin{alignat}{3} x &=&& t^2 +2, \\ y&=&& t^2-4. \end{alignat} {/eq}
Isolate t from the first equation {eq}x =t+2\Rightarrow t^2 =x-2 \Rightarrow t =\sqrt{ x-2 } {/eq}
Isolate t from the second equation: {eq}y=t^2-4\Rightarrow t^2 =y +4 \Rightarrow t =\sqrt{ y+4 } \\ {/eq}
Now equating both values of {eq}t: {/eq}
{eq}\Rightarrow \sqrt{ y+4 } = x-2 {/eq}
{eq}\Rightarrow y+4= (x-2)^2 {/eq}
{eq}\Rightarrow y= (x-2)^2-4 {/eq}
Hence the Cartesian equation of the curve is {eq}{\boxed{ y= (x-2)^2-4.}} {/eq} |
# Probability of An Event – Explanation & Strategies
In the English language, the word event is used to refer to a special or desired occurrence. In probability, we use it in a similar way. Here is the definition:
In probability, we define an event as a specific outcome, or a set of specific outcomes, of a random experiment.
• What is meant by an event in probability
• Types of Events
• How to find the probability of an event
Once we have gone through the concepts and tried some examples, you will be better able to try the questions at the end. Let’s begin!
## What is an event in probability?
In probability, we are interested in the chances of a particular event taking place. For example, getting an even number when you roll a die, or getting a head when you toss a coin. The outcome of getting an even number is considered an event. The outcome of getting a head is also considered an event. How then do we define the term event as used in this context?
### Event Definition in Probability
An event is a specific outcome, or a set of specific outcomes, of a random experiment.
Events can either be independent, dependent, or mutually exclusive. Let’s define these types of events.
### Types of Events
• #### Independent Events
Events that are not affected by other events are known as independent events.
For example, you may roll a die and get a 1. You had a $\frac{1}{6}$ chance of getting that 1. Should you roll the die again, you still have a $\frac{1}{6}$ chance of getting a 1. You also have a $\frac{1}{6}$ chance of getting any other number on the die. Getting a 1 on your first throw cannot prevent you from getting a 1 on your second throw. Neither can it predict that you will get another 1 on your second throw.
Similarly, if you roll a die and pick a card from a deck of cards, the chances of picking a jack cannot be affected by the chances of rolling a 1.
• #### Dependent Events
Events that can be affected by a previous event are known as dependent events.
Let’s think about what would happen if we had a bag of 2 blue, 1 red, 3 white, 2 green, and 4 yellow marbles. You pick one marble from the bag and set it aside. If you wanted to know the chances of picking a blue marble on the second try, that chance would affected by the first event. This is because the bag now has less marbles in total. The bag could possibly also have less blue marbles since the first marble could have been blue.
When the chances of the an event depend on the result of another, they are considered to be dependent events.
• #### Mutually Exclusive Events
Events that cannot occur at the same time are called mutually exclusive events.
Do you think you could roll a 1 and a 2 at the same time with the same die? What about getting an Ace that is a Jack from a deck of cards? Well, you certainly cannot. That is because these events are mutually exclusive; they cannot happen at the same time.
## How do you find the probability of an event?
For each of the types of events we have discussed, there will be different strategies for finding the probability of an event. You can learn more about that in the articles on the specific topic. However, in this section we will go through the general method for finding the probability of an event
The probability of an event is found by taking the number of outcomes favorable to the event and dividing it by the total possible outcomes of the experiment.
This is expressed mathematically as:
$P(E) = \frac{\text{number of outcomes favorable to the event}}{\text{total possible outcomes of the experiment}}$
Where E is used to denote the event.
Let’s examine a few examples.
Example 1: Find the probability of getting a blue marble from a bag with 1 blue marble, 1 green marble, and 1 orange marble.
• The number of blue marbles in the bag is 1. So the number of outcomes favorable to the event is 1.
• The total possible number of outcomes of the experiment is 3 as there are three marbles in the bag.
• Thus, the probability of getting a blue marble is:
$P(\text{blue marble}) = \frac{1}{3}$
Example 2: The probability of pulling a 3 from a 52-card deck of playing cards.
• There are 4 outcomes favorable to the event since there are four 3’s in the deck.
• There are 52 total cards in the deck.
• Thus, the probability of getting a 3 is:
$P(3) = \frac{4}{52} = \frac{1}{13}$
It is perfectly okay to simplify the fraction that you get. In fact, you may even write the probability as a decimal. Probabilities of events are written as decimals in most applications.
Example 3: What is the probability of getting a head when you toss a coin?
• There is 1 outcome favorable to the event of getting a head.
• There are two possible outcomes of the experiment.
• Thus, the probability of getting a head is:
$P(\text{Head}) = \frac{1}{2} = 0.54$
Alternatively we can say there is a 50% chance of getting a head.
This is a good point to mention the possible values of a probability. In the above example we said there is a 50% chance of getting a head. If that is the case, then there must also be a 50% chance of getting a tail. Remember that a percent is of 100. This says something about the highest value we can get. Read on to learn more.
### Possible Numerical Values of a Probability
#### Certain Events
Certain events are events that are sure to happen. There is a 100% chance that they will happen. Their probability is 1. That is:
$P(E) = 1$
Let’s think of a few certain events.
Example 1: The probability that a ball that has been thrown up will fall
Example 2: The probability of getting a whole number when you toss a die
Example 3: The probability of getting a head or a tail when you toss a coin.
#### Impossible Events
These are the opposite of certain events. As the name suggests, impossible events are those that can never occur. Thus:
$P(E) = 0$
This is the lowest extreme and 0 is the lowest value a probability can take. Events with a probability of 0 are impossible. Let’s think of a few.
Example 1: The probability of throwing a 6 sided die and getting a 7.
Example 2: The probability of buying a shirt from a store that only sells shoes.
Example 3: The probability of living forever
#### All Events
From the two cases above, we can conclude that the probability of all events fall between 0 and 1. That is:
$0 ≤ P(E) ≤ 1$
All of our examples have confirmed this and you may use this as a guide to self-check when computing your probabilities. If you get an answer outside of this range, the probability that your answer is incorrect, is 1.
Here’s a final example. Jake is trying to catch a bus that is numbered 54 at a bus stop that has the buses numbered 52, 54, 42, and 49 passing by. Each route number has 3 buses passing in any given hour. What is the probability that in a given hour Jake will catch his bus?
Solution:
• In a given hour, there are 3 buses running the route that Jake needs to catch, the 54
• In a given hour, there are 12 buses passing Jake’s stop, 3 of each of the 4 routes
• Thus:
$P(\text{Jake catches a 54 in any given hour}) = \frac{3}{12} = \frac{1}{4}$
Now it’s your turn to try some examples.
## Examples
What is the probability of each of the following events?
1. Getting an odd number when you toss a die?
2. Choosing an apple from a bag with 2 apples, 2 bananas, and 1 pear.
3. Throwing a 1 and a 2 when you toss 2 dice.
4. Throwing a 1 or a 2 when you toss 2 dice.
5. Pulling an Ace from a deck of cards on the second try if a King was removed on the first
### Solutions
1.Getting an odd number when you toss a die?
$P(\text{odd number}) = \frac{3}{6} = \frac{1}{2}$
2. Choosing an apple from a bag with 2 apples, 2 bananas, and 1 pear.
$P(\text{apple}) = \frac{2}{5}$
3. Throwing a 1 and a 2 when you toss 2 dice.
• We can either get (1, 2) or (2, 1)
• There are 6 × 6 = 36 total outcomes
$P(\text{1 AND 2}) = \frac{2}{36} = \frac{1}{18}$
4. Throwing a 1 or a 2 when you toss 2 dice.
(Refer to article on sample space to see how many outcomes have a 1 and how many have a 2)
$P(\text{1 OR 2}) = \frac{24}{36} = \frac{2}{3}$
5. Pulling an Ace from a deck of cards on the second try if a King was removed on the first
• The first try was a King so we still have 4 Aces remaining
• The first try subtracts 1 from the total number of possible outcomes of the experiment
$P(\text{Ace on second try when king on first}) = \frac{4}{51}$
Some of these questions could have been solved using other methods. Check out the upcoming articles on types of events to learn more |
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$\text{a) NOT a solution }\\\text{b) NOT a solution }\\\text{c) a solution }$
$\bf{\text{Solution Outline:}}$ Use the position of the number on the number line to evaluate if it satisfies the given inequality, $m\le-2 .$ $\bf{\text{Solution Details:}}$ The inequality $m\le-2$ is satisfied by all numbers that appear on/on the left of $-2$ in the number line. a) In the number line, $-1\dfrac{9}{10}$ appears on the $\text{ right }$ of $-2 .$ Hence, it is $\text{ NOT a solution .}$ b) In the number line, $0$ appears on the $\text{ right }$ of $-2 .$ Hence, it is $\text{ NOT a solution .}$ C) In the number line, $-2\dfrac{1}{3}$ appears on the $\text{ left }$ of $-2 .$ Hence, it is $\text{ a solution .}$ Therefore, \begin{array}{l}\require{cancel} \text{a) NOT a solution }\\\text{b) NOT a solution }\\\text{c) a solution } .\end{array} |
## DEV Community
Jared Nielsen
Posted on • Updated on • Originally published at jarednielsen.com
# You Need to Learn Proof By Induction
You don’t need to be a math whiz to be a good programmer, but there are a handful of tricks you will want to add to your problem solving bag to improve the performance of your algorithms and make an impression in technical interviews. In this tutorial, you will learn proof by induction by summing consecutive integers 1 to n.
## What is Proof by Induction?
Proof by induction is a mathematical method used to prove that a statement is true for all natural numbers. It’s not enough to prove that a statement is true in one or more specific cases. We need to prove it is true for all cases.
There are two metaphors commonly used to describe proof by induction:
1. The domino effect
Given a chain of dominoes, if one falls, they will all fall.
Given a sturdy ladder, if we can climb one rung, we can climb them all.
## What Problem(s) Does Proof by Induction Solve?
• All of them!
Well, almost…
• If a statement is true for one case, proof by induction helps us prove it is true (or not) for all cases.
• For our purposes, proof by induction will help us better understand, and calculate the Big O of, recursive algorithms.
## Math O’Clock
Let’s get philosophical.
Inductive reasoning and deductive reasoning are two methods of reason in logic.
We can think of them as opposites.
Deductive reasoning is top-down.
We start with general properties that are true and from them determine the truth of a specific property.
The classic example is a syllogism:
1. All men are mortal.
2. Socrates is a man.
3. Therefore, Socrates is mortal.
Inductive reasoning is bottom-up.
For example:
1. The sun has risen in the east every morning up until now.
2. The sun will also rise in the east tomorrow.
That’s not good enough!
We skeptics need proof.
A mathematical proof shows that “the stated assumptions logically guarantee the conclusion”.
All the time.
🌅
Despite its name, proof by induction is a method of deduction.
Why?
We need to prove that the specific case is true for all cases, without exception. Inductive reasoning doesn't guarantee this.
We start with the rule we want to prove and assume it is true and then use mathematics to prove it generally.
In other words, we use the equation itself to prove itself.
Sounds a lot like recursion, doesn't it?
There are two steps to proof by induction:
• Base
• Induction
### Proof by Induction: Base
We first need to prove that our property holds for a natural number.
That’s generally 0 or 1.
What’s a ‘natural number’?
Natural numbers are the numbers we use for counting or ordering.
### Proof by Induction: Induction
Once we establish a base case, we need to prove that property holds for the next natural number.
What’s the next natural number?
``````n + 1
``````
Have we seen this, or something like it, before?
## How to Sum Consecutive Integers 1 to n
We can use proof by induction to prove the following:
``````1 + 2 + 3 + … + n = n * (n + 1) / 2
``````
If this is new to you, you may want to start with How to Sum Consecutive Integers from 1 to n.
Let’s plug in values.
Our equation is
``````n * (n + 1) / 2
``````
If n = 1, the result is 1.
If n = 2, the result is 3.
If n = 3, the result is 6.
This is the ‘brute force’ approach.
Following this approach, the only way to prove our equation works for all natural numbers is to calculate it for all natural numbers.
There must be a better way!
Let’s refer to our equation with a variable so I can type less.
``````P = n * (n + 1) / 2
``````
Now we can simply refer to it as
``````P(n)
``````
We proved that our equation works when `n = 1`.
What if we don’t know `n`?
Let’s add another variable to the equation, `k`.
Let’s say that `k` is less than or equal to `n`.
``````k <= n
``````
We need to make a proposition. What is it?
If `P(k)` is true, then `P(k + 1)` is also true.
If that’s true, then P(n) is true for all natural numbers.
Let’s rewrite our equation with `k`
``````1 + 2 + 3 + … + k = k * (k + 1) / 2
``````
That looks familiar. But we need to prove that `k + 1` works. If the number following `k` is `k + 1`, we can add `k + 1` to the left of our equation:
``````1 + 2 + 3 + … + k + (k + 1) = k * (k + 1) / 2
``````
If we add `k + 1` to the left of our equation, we also need to add it to the right:
``````1 + 2 + 3 + … + k + (k + 1) = k * (k + 1) / 2 + (k + 1)
``````
Now we need to simplify.
What’s our common denominator?
`2`
We multiply our newly added `(k + 1)` by 2.
``````1 + 2 + 3 + … + k + (k + 1) = k * (k + 1) / 2 + 2 * (k + 1) / 2
``````
Now we can add the terms:
``````1 + 2 + 3 + … + k + (k + 1) = k * (k + 1) + 2 * (k + 1) / 2
``````
Do we see a pattern? There are two `(k + 1)` terms, so let’s factor them out:
``````1 + 2 + 3 + … + k + (k + 1) = (k + 1) * (k + 2) / 2
``````
This is starting to look familiar. What is another way we can describe `(k + 2)`?
🤔
`((k + 1) + 1)`
🤯
``````1 + 2 + 3 + … + k + (k + 1) = (k + 1) * ((k + 1) + 1) / 2
``````
We just proved our equation!
``````1 + 2 + 3 + … + n = n * (n + 1) / 2
``````
## Proof by Induction
In this tutorial, you learned proof by induction by proving an equation to sum consecutive integers from 1 to n. Proof by induction is useful for understanding and calculating the Big O of recursive algorithms. We’ll look at that in a future article. |
# How do you differentiate f(x)=(cosx+sinx)(lnx-x) using the product rule?
Dec 7, 2017
$\frac{d}{\mathrm{dx}} f \left(x\right) = \left(\cos x + \sin x\right) \left(\frac{1}{x} - 1\right) + \left(\ln x - x\right) \left(- \sin x + \cos x\right)$
#### Explanation:
$f \left(x\right) = \left(\cos x + \sin x\right) \left(\ln x - x\right)$
Here $f \left(x\right)$ is the product of two functions.
Therefore to differentiate $f \left(x\right)$ we will use the product rule
The product rule says that $\to$
$\frac{d}{\mathrm{dx}} U \cdot V = U \frac{d}{\mathrm{dx}} V + V \frac{d}{\mathrm{dx}} U$
Now back to the question, we will differentiate both sides with respect to $x$
$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{d}{\mathrm{dx}} \left[\left(\cos x + \sin x\right) \left(\ln x - x\right)\right]$
$\frac{d}{\mathrm{dx}} f \left(x\right) = \left(\cos x + \sin x\right) \frac{d}{\mathrm{dx}} \left(\ln x - x\right) + \left(\ln x - x\right) \frac{d}{\mathrm{dx}} \left(\cos x + \sin x\right)$
So therefore,
$\frac{d}{\mathrm{dx}} f \left(x\right) = \left(\cos x + \sin x\right) \left(\frac{1}{x} - 1\right) + \left(\ln x - x\right) \left(- \sin x + \cos x\right)$ |
# 6.7 Exponential and logarithmic models (Page 7/16)
Page 7 / 16
Does a linear, exponential, or logarithmic model best fit the data in [link] ? Find the model.
$x$ 1 2 3 4 5 6 7 8 9 $y$ 3.297 5.437 8.963 14.778 24.365 40.172 66.231 109.196 180.034
Exponential. $\text{\hspace{0.17em}}y=2{e}^{0.5x}.$
## Expressing an exponential model in base e
While powers and logarithms of any base can be used in modeling, the two most common bases are $\text{\hspace{0.17em}}10\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}e.\text{\hspace{0.17em}}$ In science and mathematics, the base $\text{\hspace{0.17em}}e\text{\hspace{0.17em}}$ is often preferred. We can use laws of exponents and laws of logarithms to change any base to base $\text{\hspace{0.17em}}e.$
Given a model with the form $\text{\hspace{0.17em}}y=a{b}^{x},$ change it to the form $\text{\hspace{0.17em}}y={A}_{0}{e}^{kx}.$
1. Rewrite $\text{\hspace{0.17em}}y=a{b}^{x}\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}y=a{e}^{\mathrm{ln}\left({b}^{x}\right)}.$
2. Use the power rule of logarithms to rewrite y as $\text{\hspace{0.17em}}y=a{e}^{x\mathrm{ln}\left(b\right)}=a{e}^{\mathrm{ln}\left(b\right)x}.$
3. Note that $\text{\hspace{0.17em}}a={A}_{0}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}k=\mathrm{ln}\left(b\right)\text{\hspace{0.17em}}$ in the equation $\text{\hspace{0.17em}}y={A}_{0}{e}^{kx}.$
## Changing to base e
Change the function $\text{\hspace{0.17em}}y=2.5{\left(3.1\right)}^{x}\text{\hspace{0.17em}}$ so that this same function is written in the form $\text{\hspace{0.17em}}y={A}_{0}{e}^{kx}.$
The formula is derived as follows
Change the function $\text{\hspace{0.17em}}y=3{\left(0.5\right)}^{x}\text{\hspace{0.17em}}$ to one having $\text{\hspace{0.17em}}e\text{\hspace{0.17em}}$ as the base.
$y=3{e}^{\left(\mathrm{ln}0.5\right)x}$
Access these online resources for additional instruction and practice with exponential and logarithmic models.
## Key equations
Half-life formula If $k<0,$ the half-life is Carbon-14 dating $t=\frac{\mathrm{ln}\left(\frac{A}{{A}_{0}}\right)}{-0.000121}.$ is the amount of carbon-14 when the plant or animal died is the amount of carbon-14 remaining today is the age of the fossil in years Doubling time formula If $k>0,$ the doubling time is Newton’s Law of Cooling $T\left(t\right)=A{e}^{kt}+{T}_{s},$ where is the ambient temperature, and is the continuous rate of cooling.
## Key concepts
• The basic exponential function is $\text{\hspace{0.17em}}f\left(x\right)=a{b}^{x}.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}b>1,$ we have exponential growth; if $\text{\hspace{0.17em}}0 we have exponential decay.
• We can also write this formula in terms of continuous growth as $\text{\hspace{0.17em}}A={A}_{0}{e}^{kx},$ where $\text{\hspace{0.17em}}{A}_{0}\text{\hspace{0.17em}}$ is the starting value. If $\text{\hspace{0.17em}}{A}_{0}\text{\hspace{0.17em}}$ is positive, then we have exponential growth when $\text{\hspace{0.17em}}k>0\text{\hspace{0.17em}}$ and exponential decay when $\text{\hspace{0.17em}}k<0.\text{\hspace{0.17em}}$ See [link] .
• In general, we solve problems involving exponential growth or decay in two steps. First, we set up a model and use the model to find the parameters. Then we use the formula with these parameters to predict growth and decay. See [link] .
• We can find the age, $\text{\hspace{0.17em}}t,$ of an organic artifact by measuring the amount, $\text{\hspace{0.17em}}k,$ of carbon-14 remaining in the artifact and using the formula $\text{\hspace{0.17em}}t=\frac{\mathrm{ln}\left(k\right)}{-0.000121}\text{\hspace{0.17em}}$ to solve for $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ See [link] .
• Given a substance’s doubling time or half-time, we can find a function that represents its exponential growth or decay. See [link] .
• We can use Newton’s Law of Cooling to find how long it will take for a cooling object to reach a desired temperature, or to find what temperature an object will be after a given time. See [link] .
• We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors. See [link] .
• We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic graphs help us to develop models that best fit our data. See [link] .
• Any exponential function with the form $\text{\hspace{0.17em}}y=a{b}^{x}\text{\hspace{0.17em}}$ can be rewritten as an equivalent exponential function with the form $\text{\hspace{0.17em}}y={A}_{0}{e}^{kx}\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}k=\mathrm{ln}b.\text{\hspace{0.17em}}$ See [link] .
Solve the problem 3x^2_2x+4=0
can anyone recommend an app or. website to help me refresh my knowledge and abilities? I stopped learning more about calc and trig in the 90's
if 6x=-2 find value of 6/x
l don't know
Aman
I don't know
Aman
l don't know
Aman
I don't know
Aman
x=.333333333334, so 6/x= 18±
scott
x=.33333333334± 6/x=18±
scott
that is strange, I don't remember inserting the À. the result I got was x= 3333333334± 6/x=18±
scott
I need glasses
scott
X=(-1÷3) X equals minus one third
Melvin
if sin15°=√p, express the following in terms of p
I am learning and does someone have the time to tell me what or where this particular equation would be used?
scott
prove sin²x+cos²x=3+cos4x
the difference between two signed numbers is -8.if the minued is 5,what is the subtrahend
the difference between two signed numbers is -8.if the minuend is 5.what is the subtrahend
jeramie
what are odd numbers
numbers that leave a remainder when divided by 2
Thorben
1,3,5,7,... 99,...867
Thorben
7%2=1, 679%2=1, 866245%2=1
Thorben
the third and the seventh terms of a G.P are 81 and 16, find the first and fifth terms.
if a=3, b =4 and c=5 find the six trigonometric value sin
Ans
pls how do I factorize x⁴+x³-7x²-x+6=0
in a function the input value is called
how do I test for values on the number line
if a=4 b=4 then a+b=
a+b+2ab
Kin
commulative principle
a+b= 4+4=8
Mimi
If a=4 and b=4 then we add the value of a and b i.e a+b=4+4=8.
Tariq
what are examples of natural number
an equation for the line that goes through the point (-1,12) and has a slope of 2,3
3y=-9x+25
Ishaq |
# Ratios and Rates
Purpose
The purpose of this series of lessons is to synthesise students’ fraction and decimal place value knowledge to enable them to work with rates and ratios with understanding and competence.
Achievement Objectives
NA4-3: Find fractions, decimals, and percentages of amounts expressed as whole numbers, simple fractions, and decimals.
NA4-5: Know the equivalent decimal and percentage forms for everyday fractions.
NA5-5: Know commonly used fraction, decimal, and percentage conversions.
Specific Learning Outcomes
• Understand what a ratio is and how it is written and read.
• Explore and identify equivalent part-to-part ratios using pictures and symbols.
• Explore the multiplicative relationship of ratios by generating equivalent ratios and recording these.
• Create ratio tables.
• Explore part-to-whole ratios.
• Recognise and record fractional relationships in ratios.
• Express ratios as fractions, decimal fractions and percentages.
• Apply ratio understanding to solving problems.
• Create and solve ratio problems.
• Understand what rates are and how they differ from ratios.
• Calculate and compare rates.
• Show on a graph the linear relationship of a constant rate.
Description of Mathematics
Implicit within students’ work with fractions and percentages is relational thinking, or the ability to see and use possibilities of variation between numbers in an equation or number sentence. Just as fractions and percentages express part whole relationships, a ratio is a relationship between two numbers of the same kind and is usually expressed as ‘a to b’ or ‘a : b’. This is saying that for every amount of one thing, there is so much of another thing of the same kind, whether these are units of measure, people, animals, or objects. Ratios can express a part-to-part relationship, such as the number of boys to the number of girls in a group of children. Ratios can also express a part-to-whole relationship, such as the number of boys to the number of children in the whole group.
It is not a significant conceptual shift for students to work with ratios, as their proportional nature is familiar, and the students’ ability to understand and work easily with multiplicative relationships is fundamental to any proportional thinking. This thinking has been consolidated in their fractional and percentage work, so the shift in their work with ratios is rather in the contexts in which ratios are most often used, and the interpretation of these contexts. However, students do encounter for the first time, the colon notation used in the expression of a ratio.
Students need to understand that, as with fractions, ratios can be reduced by common factors of the quantities, to make the simplest form. For example, a ratio of 20:60 can be represented in its simplest form as a ratio of 1:3. This can also be expressed as a fraction. The first amount is one third of the second amount. However, in this 1:3 part-to-part ratio, we can also see that there are 4 parts altogether (1 + 3) so the relationship of the first quantity (1) to the whole (4) can be expressed as 1/4. Students may not have recognised before that every fraction is in fact a ratio. It is important to model well with equipment both the part-to-part and part-to-whole relationships that make up ratios.
Rates
The key idea that students need to come to understand is that a ratio compares two amounts of the same kind of thing (eg. people: girls to boys, drink mix, odds) whilst a rate is a special kind of ratio that compares different kinds of measures such as dollars per kilogram. A distinguishing feature of a rate is that it uses the word per and the symbol/. Students therefore need to be able to co-ordinate pairs of numbers, and be competent in working with fractions, decimals and percentages, to calculate the multiplicative relationship between the number pairs and to make comparisons of rates.
A rate is a very important kind of ratio because so many practical tasks in our lives involve some kind of rate: speed (kilometres per hour), remunerative pay (dollars paid per hour of work), health (pulse: heartbeats per minute), or the price of something (cost per unit bought), to name a few.
Constant rates are ordered pairs that result in a straight line when plotted on a number plane. The slope of the line is the unit rate, and the relationship between the numbers on the line does not change. This co-linear representation of a constant rate captures visually this multiplicative relationship ‘in action’.
Proportional thinking underpins one’s ability to make sense of and use easily ratios and rates in our daily lives. These ideas are presented in five sessions however, as they include complex concepts that are fundamental to a student’s success with ratios and rates, these sessions can be extended over a longer period of time.
Note:
One very well known ratio is the golden ratio, also known as the golden mean. The golden ratio is a special number approximately equal to 1.618. It appears many times in geometry, art, architecture and other areas. Your students may well enjoy investigating exploring this remarkable ratio that has fascinated people for centuries.
Activity
Session 1
SLOs:
• Understand what a ratio is and how it is written and read.
• Explore and identify equivalent part-to-part ratios using pictures and symbols.
• Explore the multiplicative relationship of ratios by generating equivalent ratios and recording these.
• Create ratio tables.
Activity 1
1. Begin the session by writing this problem on the class/group chart:
1/4 or 25% of the class are absent. There are 18 students present.
How many students are in the whole class?
Give students time to solve the problem and have several share their solutions.
Highlight the fact that they were are their multiplicative thinking and looking for relationships between the numbers.
2. Point out that this session is about exploring relationships between numbers but that these will be presented in a slightly different way.
3. Distribute the shuffled picture cards only of Attachment 1 (Class Ratios) to student pairs.
(Purpose: To recognise the same part-to-part ratios)
Pose this problem:
Look closely at the numbers of girls and boys on each card.
Can you group any of these cards together?
If so, can you explain why you have grouped them as you have?
Give students time to explore, discuss and group.
(a. 1:3, 2:6, 3:9 and 4:12, b. 2:3, 4:6, c. 1:2, 2:4, d. 4:4, 3:3. Two do not have an equivalent: 3:4 and 2:5)
4. Have student pairs share their groupings. Roam and listen as they explain.
Summarise student findings, accepting all ideas.
Activity 2
1. Distribute the white cards (shuffled) to student pairs.
Ask what we call this representation (a ratio) and ask for an explanation from the students.
Record student ideas on the class chart. Elicit key ideas:
A ratio is a relationship between two numbers of the same kind and is usually expressed as ‘a to b’ or ‘a : b’. This is saying that for every amount of one thing there is so much of another thing of the same kind.
A ratio is expressed using a colon. A ratio is expressed in the order the two amounts are presented:
For example: This shows one girl to three boys:
written 1:3 not 3:1
2. Have students read several of the ratios aloud: eg. (“one to three”, “two to six”…).
3. Have student pairs match the white ratio cards with the picture cards. If same ratio groupings have not already been established, encourage students to look for these by looking closely at the numbers.
4. Group ratios on the class chart and discuss the number relationships. Have students discuss and explain how these can be simplified.
a. 1:3, 2:6, 3:9 and 4:12 (1:3)
On each of the cards demonstrate how the 1:3 ratio is evident in each.
In each case the number of girls is 1/3 of the number of boys.
Explore the paired equivalent cards, discuss, and have the students explain the number relationships for each pair.
b. 2:3, 4:6 (there are 2 girls for every 3 boys: the number of girls is 2/3 the number of boys)
c. 1:2, 2:4 (there is 1 girls for every 2 boys: the number of girls is 1/2 the number of boys)
d. 4:4, 3:3 (there is 1 girls for every 1 boy: the number of girls and boys is the same)
5. Explore the two cards that do not have given equivalents: (3:4 and 2:5)
Recognise that each of these is in its simplest form.
Together generate a ratio table for 3:4.
Girls 3 6 9 12 15 18 21 Boys 4 8 12 16 20 24 28
Discuss what is happening to the numbers of girls and the number of boys.
Emphasise that the ratio of 3:4 remains the same. In each case there are 3/4 as many girls as there are boys.
6. Ensure students have paper/small whiteboards available.
Have student pairs discuss and create a ratio table for 2:5 and ‘draw’ what the first three would look like using simple girl/boy ‘pictures (modeled on the cards of Attachment 1).
7. Have student pairs suggest at least one more different girl:boy ratio of their own, and generate a table for this.
8. Conclude this session by collecting the Attachment 1 ratio cards from this session and summarise on the class/group chart what the students have learned about part-to-part ratios.
Session 2
SLOs:
• Explore part-to-whole ratios.
• Recognise and record fractional relationships in ratios.
• Express ratios as fractions, decimal fractions and percentages.
Activity 1
1. Write part-to-part ratio on the class chart.
Distribute the shuffled picture cards only of Attachment 1 (Class Ratios) to student pairs.
Point out that so far the class has explored the ratio between two parts of a group, the number of girls to the number of boys. Explain that this kind of ratio is known as a part to part ratio.
2. Explain that a different kind of ratio will be explored.
Distribute the shuffled green cards to student pairs.
(Purpose: To recognise the same part to whole ratios).
Have the students match these with the picture cards, and tell them that they need to be prepared to explain what each of the ratio cards expresses (the ratio of girls to the whole class).
3. Have students pair share and talk about their results.
4. As a group, ask the students what this kind of ratio describes: this is a part to whole ratio. Ask if the students notice anything about the ratio (each can be expressed as a fraction: what fraction are the girls of the whole group?)
Activity 2
1. Explore a part-to-whole fractional relationship by using the ratio table generated in Session 1. (Refer to the picture card and ratio card showing 3:7, number of girls to number in the whole group).
Add the group total line to the ratio table and ask a student to complete this, with the help of the group/class members.
Girls 3 6 9 12 15 18 21 Boys 4 8 12 16 20 24 28 Total in group/class 7 14 21 28 35 42 49
2. Have a student say what fraction of the class the girls are in each instance and record these:
(3/7, 6/14, 9/21, 12/28, 15/35, 18/42, 21/49)
3. Write these fractions on the class chart:
1/1, 2/2, 3/3, 4/4, 5/5, 6/6 and 7/7
Question their value and together conclude that they are all ways of writing 1.
4. Explore and record these fractional relationships.
3/7 x 2/2 = 6/14
3/7 x 3/3 = 9/21
3/7 x 4/4 = 12/28
3/7 x 5/5 = 15/35
3/7 x 6/6 = 18/42
3/7 x 7/7 = 21/49
Elicit from the students that all fractions are an expression of 3/7 or a 3:7 part to whole ratio.
Activity 3
1. Distribute the blank pale blue cards from Attachment 1 to each student pair and have them write for each of the picture cards, the ratio of boys to the whole group.
2. Have student pairs use the ratio table generated in Session 1, Activity 2, Step 6 for the 2:5 ratio of girls to boys.
They should:
1. Add a group total line and complete this.
2. Explore the fractional relationships as in Activity 2, Step 2 above between the boys and the whole group.
3. Confirm that the decimal fraction for each is the same.
4. State what percentage the boys are of the whole group.
For example:
Girls 2 4 6 8 10 12 14 Boys 5 10 15 20 25 30 35 Total in group/class 7 14 21 28 35 42 49
The fraction of boys in each group:
5/7, 10/14, 15/21, 20/28, 25/35, 30/42, 35/49
These can be explored this way:
5/7 x 1/1 = 5/7
5/7 x 2/2 = 10/14,
5/7 x 3/3 = 15/21,
5/7 x 4/4 = 20/28,
5/7 x 5/5 = 25/35,
5/7 x 6/6 = 30/42,
5/7 x 7/7 = 35/49
Discuss and agree all are expressions of a 5:7 part to whole ratio that can also be expressed as a fraction, in all cases the boys are 5/7 of the whole class.
Discuss and agree that this can be expressed as a decimal fraction: 5 ÷ 7 = 0.714 (check each fraction is equivalent to the same decimal fraction eg. 35 ÷ 49 = 0.714)
Discuss and agree that the 5:7 part-to-whole ratio can also be expressed as a percentage. In this case the percentage of boys in the group class is 71.4%
Activity 4
1. Have students pair share and check their results.
2. Conclude the session by recording students’ reflections about part-to-whole ratio learning and the decimal and percentage calculations.
Session 3
SLOs:
• Apply ratio understanding to solving problems.
• Create and solve ratio problems.
Activity 1
1. Distribute Attachment 2 asking students to complete this on their own or in pairs.
2. Discuss. Ask students to share their responses to the question in Attachment 2, “Is there anything else you know?” Notice students’ explanations.
Activity 2
Ask for and list contexts in the students’ lives in which ratios are used in a very practical sense. (For example: in recipes for food and drink, scaling up and down models and maps, ratio of kinds of people such as left handed to right handed people).
Activity 3
Pose this problem (Attachment 3) and make brown and white unifix cubes available:
Mochaccino Mix
Millie and Maxwell are creating their own homemade ‘mochaccinos’ with scoops of mochaccino mix (MM) and scoops of milk (m).
2:6 (MM:m) 5:11 (MM:m) Millie Maxwell
Which cup has the stronger mochaccino flavour?
How do you know?
Have student pairs model the ratio with cubes, discuss how they would solve it and agree on a solution.
Have them pair share and discuss their results.
They should include (some of) these ideas.
Millie: 2:6, MM is 2/8 (1/4) of the mix (this is the same as 4:16).
MM is 25% of the mix.
Maxwell 5:11, MM is 5/16 of the mix.
MM is 31.25% of the mix.
Maxwell’s mochaccino has a slightly stronger flavour. It has 1/16 (6.25%) more mochaccino mix.
Millie's Maxwell's
Your turn to make a mochaccino
You decide to make some mochaccinos using Millie’s recipe.
If you use 18 scoops of milk, how many scoops of MM will you use?
If you use 1 1/2 scoops of MM, how many scoops of milk will you use?
If Millie’s recipe makes a small drink for one person, what quantities will you use to make a small drink for eight people using Millie’s recipe?
You decide to make your own mix.
You use a 2:8 ratio in your first drink. What percentage is mochaccino mix?
Your second drink mix is stronger. You use a 3:7 ratio. What percentage is milk?
Your third drink mix is really thick. It is made with 40% mochaccino mix. What ratio did you use?
6, 4 1/2 , 16MM:48m
20%, 70%, 4:6 or 2:3
Activity 4
Pose this problem:
Odds
It’s winter in the south. Four children are predicting the likelihood of school having to close early because of the snowfall that is predicted. They state their odds as Yes will close (Y) to No, won’t close (N).
Here are their odds. Y:N
3:5 7:9 5:11 6:10 Filipo Toni Arapeta Mona
1.If you wanted the odds of school closing to be greater, whose odds would you pick and why?
2. Whose odds show the greatest chance of school remaining open?
1. Toni.
2. Arapeta.
(Toni 7:9 is a 7/16 chance of closing. Filipo 3:5 is a 3/8 or 6/16 chance of closing. Mona 6:10 is a 6/16 chance of closing. Arapeta 5:11 is a 5/16 chance of closing.)
Activity 5
Refer to and add more contexts to those brainstormed in Activity 2 of this session.
Have student pairs write (at least) two ratio problems and work out the answers.
Have them give their ratio problems to another pair to solve.
Activity 6
Conclude this session by sharing the student-written problems, identifying those that were particularly challenging/successful/enjoyable.
Session 4
SLOs:
• Consolidate their ratio understanding through investigating conjectures.
Activity 1
Show the students the mochaccino ratios from Session 3.
Mochaccino Mix
Millie and Maxwell are creating their own homemade ‘mochaccinos’ with scoops of mochaccino mix (MM) and scoops of milk (m).
2:6 (MM:m) 5:11 (MM:m) Millie Maxwell
Write on the class chart and explain that Millie has stated:
If Maxwell and I combine our mochaccino mixtures to make one big cup of coffee, the combined ratio will be 7:17 (MM:m).
You add two ratios together to give a combined ratio.
Challenge the students in pairs to explore, argue and justify their position of agreement or disagreement with this conjecture.
(Answer. This is so with a single combination, or where the number of cups are the same for each)
Activity 2
Now pose this problem.
Millie makes her mochaccinos (2:6MM:m) for 1 person and Maxwell makes his mochaccinos (5:11 MM:m) for 2 people.
Maxwell makes this statement:
The combined ratio of our combined mochaccino mixtures is 7:17 You add two ratios together to give a combined ratio.
Challenge the students in pairs to explore, argue and justify their position of agreement or disagreement with this conjecture.
(Answer: This is not correct. These represent different quantities. The combined ratio must be calculated from the combined quantities.
2:6 1 mochaccino made by Millie
10:22 2 mochaccinos made by Maxwell
12:28 the combined ratios of MM and m
6:14 or 3:7 the simplified combined ratio of MM and m)
Activity 3
Have students consider this problem and pose a similar combination problem of their own for investigation, using different numbers of cups for Millie and Maxwell.
Activity 4
Have each student pair consider whether they are able to make a general statement about ratios and how they work (conjecture).
Make time for these to be explored by the class/group. Have students understand that just one counter example is sufficient to disprove the proposition (false conjecture).
Session 5
SLOs:
• Understand what rates are and how they differ from ratios.
• Calculate and compare rates.
Activity 1
1. Have each student write down their definition of a ratio. Ask selected students to share what they have written.
Eg. A ratio is a relationship between two amounts of the same kind (eg. people: girls to boys, drink mix, odds) usually expressed as ‘a to b’.
2. Write Rate on the class chart. Have each student write their own definition and share what they have written.
Elicit and record the difference and these key ideas about rates:
A rate is a special kind of ratio: a rate compares different kinds of measures such as dollars per kilogram.
A rate involves a multiplicative comparison and uses the word per and the symbol /.
3. Have students brainstorm and record rates that they experience/know of in their own lives: For example:
dollars per hour (pay rate), metres per second or kilometres per hour (rate of speed), heartbeats per minute (pulse rate).
Activity 2
1. Have students suggest a rate of speed that they travel on their bicycles: for example 30 km/hour.
Record this in full: 30 kilometres per hour and beside this write the abbreviation.
Ask for a student to explain what this rate means: “In every hour I ride 30 kilometres”.
Write on class chart:
Distance (km) Time (minutes) 30 60 10
2. Pose: How long will it take to ride 10 km?
Ask students to discuss this in pairs and be prepared to explain 2 ways of working out their response.
3. Ask at least one student pair to explain their thinking.
Elicit and demonstrate:
the between relationship: 30 is half of 60 so 10 must be half of 20
the within relationship: 10 is one third of 30 so 20 is one third of 60.
This is equivalent to half a kilometre per minute or 30 km/hour. (30 kilometres/60 minutes)
4. Explore other examples, each time highlighting the between and within relationships.
How long to ride 60 km?
How long to ride 40 km?
How long to ride 15 km?
How far in 30 minutes?
How far in 45 minutes?
How far in 10 minutes?
5. Ask, what is meant by a constant rate? (In this case the rate is 30km/hour and this doesn’t change regardless of distance.) Discuss whether this is realistic, for example, cycling over a long distance.
Highlight the fact that in exploring the examples in the next task a constant rate is assumed.
Activity 2
1. Explain that they are going to investigate Millie and Maxwell’s coffee making a little further.
Distribute one set of coloured cards (3) from page 1 of Attachment 4 to each student pair.
(Purpose: To calculate rates with unknowns in different positions)
Have them solve the problems posed, recording their ‘between’ or ‘within’ thinking as they do so. Emphasise that they need to be able to show and explain the multiplicative relationship between the numbers.
Make the relevant coloured section of page 2, Attachment 4 available, if required, to scaffold their calculations and recording.
2. Have them pair share with another student pair with the same coloured problem, explaining their results and the rationale for these.
3. Have one person from the group of 4 collect a piece of graph paper for each member of the group. Have them discuss in a group how they would show this rate on a graph.
4. Have each student create a graph showing the linear relationship. Compare.
5. Have student pairs swap coloured problems, solve and repeat 2-4 above.
Activity 3
1. Have student pairs write two more problem scenarios for Millie and Maxwell. Suggested contexts: speed that they ride home from school on their bikes, the pulse rate of the person who drinks the mochaccino.
2. Solve each other’s problems in pairs or as a class.
Activity 4
On the class chart, summarise key learning about rates. |
By September 2022 most people I knew had had COVID-19. In our house we had somehow managed to avoid it. However, come the beginning of October, we had both tested positive and were very ill. Now we have both tested negative and feel quite a bit better but still have bad coughs.
COVID-19 was named as such by the World Health Organisation – the acronym derived from “coronavirus disease 2019”. In these blog posts I take a number that has occurred naturally in my real-world environment and write about some of its mathematical properties. The posts are aimed at everyone however basic or advanced their maths is. Most numbers have many, many mathematical properties and these usually overlap with other numbers.
The most obvious mathematical property of 19 is that it is prime – that is, it only has divisors 1 and itself. But you can delve much deeper into the world of primes. Take a look at the number-line above. All the primes are shown in pale blue. 1 is not a prime number because it has only one divisor, namely 1.
19 is a twin prime with 17. A twin prime is a prime number that is either 2 less or 2 more than another prime number. In the case of 19 it is two above prime number 17 as you can see in the diagram.
19 is a cousin prime with 23. Cousin primes are prime numbers that differ by four: they have a relationship but not as strong as twins – a bit like siblings and cousins in human life. 19 is a sexy prime with 13. This is because they differ from each other by six. The term “sexy prime” stems from the Latin word for six: that is sex. This gives primes an extra bit of interest!
19 is also a weak prime number. This means that it is closer to the previous prime than the next as shown in the diagram above. The weak prime property can be measured succinctly by stating that it is less than the arithmetic mean of the next and previous prime numbers. In the case of 19 that is 17 and 23 respectively.
Arithmetic mean = (17 + 23) / 2 = 40 / 2 = 20
As 19 < 20 it is a weak prime.
The first few weak primes are:
3, 7, 13, 19, 23, 31, 43 and so on
17 is strong as: 13 + 19 = 32 / 2 = 16 and 17 > 16.
The previous prime is 13 and the next is 19. 17 is closer to 19 than it is to 13.
Strong primes are: 11, 17, 29, 37, 41, 59, 67, 71 and so on
I was delighted to find a property related to the number 19 that is currently known to exist for just six numbers.
If you multiply 19 by its reverse (91) you get:
19 x 91 = 1729
Apparently a mathematician called Hardy was on his way to visit his mathematician friend, Ramanujan who was unwell in London in about 1919. He took a taxicab with number 1729. If you’ve ever met a serious mathematician, they think about and talk about numbers all of the time. My Dad was a serious mathematician and would do his best to engage his family in conversations about the excitement of, for example, table numbers at restaurants being prime. We tried to show interest but were usually keener to see what was on the menu!
Anyway, in the case of these two mathematicians, Hardy expressed disappointment that 1729 was not a very interesting number and his concern that this was a bad omen. But his friend disagreed. He pointed out that 1729 was the smallest number that could be expressed in two ways as a sum of cubes.
More generally, the nth taxicab number is defined as the smallest integer that can be expressed as a sum of two positive integer cubes in n distinct ways. At the present time there are just six as follows:
Taxicab (1) = 2
=13+13
= 13+13
Taxicab (2) = 1729
= 13+123
= 93+103
Taxicab (3) = 87,539,319
= 1673+4363
= 2283+4233
= 2553+4143
Taxicab (4) = 6,963,472,309,248
= 24213+190833
= 54363+189483
= 102003+180723
= 133223+166303
Taxicab (5) =48,988,659,276,962,496
=387873+3657573
= 1078393+3627533
= 2052923+3429523
= 2214243+33653
= 2315183+3319543
Taxicab (6) =24,153,319,581,254,312,065,344
= 5821623+289062063
= 30641733+288948033
= 85192813+286574873
= 162180683+270932083
= 174924963+265904523
= 182899223+262243663
The taxicab numbers subsequent to 1729 were found with the help of computers in recent years with Taxicab (6) being discovered in 2008. I find it fascinating how large these numbers get and so quickly. The fact that the search for numbers with the taxicab property has required the use of computers demonstrates how they are still relevant to mathematical research today.
I am hoping that these posts will encourage you to see numbers with greater interest and look for their properties when you come across them in your everyday life. |
# Selina Solutions Concise Maths Class 10 Chapter 25 Probability
The measure of uncertainty is called theory of probability. In this chapter, students will be finding the probabilities of various events after gaining a clear understanding of the basic terms and concepts. Further, the probability of an event denotes the likelihood of its happening. Students can access the Selina Solutions for Class 10 Mathematics prepared by our experienced faculty at BYJU’S for clarifying any doubts in solving problems of the Concise Selina Mathematics Selina for Class 10. All the solutions are in accordance with the latest ICSE patterns. The Selina Solutions for Class 10 Mathematics Chapter 25 Probability PDF is available in the link given below.
## Selina Solutions Concise Maths Class 10 Chapter 25 Probability Download PDF
### Exercises of Concise Selina Solutions Class 10 Maths Chapter 25 Probability
Exercise 25(A) Solutions
Exercise 25(B) Solutions
Exercise 25(C) Solutions
## Access Selina Solutions Concise Maths Class 10 Chapter 25 Probability
Exercise 25(A) Page No: 386
1. A coin is tossed once. Find the probability of:
(i) getting a tail
(ii) not getting a tail
Solution:
Here, the sample space = {H, T}
i.e. n(S) = 2
(i) If A = Event of getting a tail = {T}
Then, n(A) = 1
Hence, the probability of getting a tail = n(A)/ n(S) = 1/2
(ii) Not getting a tail
As we know, P(getting a tail) + P(not getting a tail) = 1
So, P(not getting a tail) = 1 – (1/2) = ½
2. A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:
(i) a black ball.
(ii) a red ball.
(iii) a white ball.
(iv) not a red ball.
(v) not a black ball.
Solution:
Total number of balls = 3 + 5 + 2 = 10
So, the total number of possible outcomes = 10
(i)Â There are 5 black balls
So, the number of favourable outcomes = 5
Thus, P(getting a black ball) =Â 5/10 = 1/2
(ii) There are 2 red balls
So, the number of favourable outcomes = 2
Thus, P(getting a red ball) =Â 2/10 = 1/5
(iii) There are 3 white balls
So, the number of favourable outcomes = 3
Thus, P(getting a white ball) =Â 3/10 = 3/10
(iv) There are 3 + 5 = 8 balls which are not red
So, the number of favourable outcomes = 8
Thus, P(getting a white ball) =Â 8/10 = 4/5
(v) There are 3 + 2 = 5 balls which are not black
So, the number of favourable outcomes = 5
Thus, P(getting a white ball) = 5/10 = ½
3. In a single throw of a die, find the probability of getting a number:
(i) greater than 4.
(ii) less than or equal to 4.
(iii) not greater than 4.
Solution:
Here, the sample space = {1, 2, 3, 4, 5, 6}
So, n (s) = 6
(i) If E = event of getting a number greater than 4 = {5, 6}
So, n (E) = 2
Then, probability of getting a number greater than 4 = n(E)/ n(s) = 2/6 = 1/3
(ii) If E = event of getting a number less than or equal to 4 = {1, 2, 3, 4}
So, n (E) = 4
Then, probability of getting a number less than or equal to 4 = n(E)/ n(s) = 4/6 = 2/3
(iii) E = event of getting a number not greater than 4 = {1, 2, 3, 4}
So, n (E) = 4
Then, probability of getting a number not greater than 4 =Â n(E)/ n(s) = 4/6 = 2/3
4. In a single throw of a die, find the probability that the number:
(i) will be an even number.
(ii) will not be an even number.
(iii) will be an odd number.
Solution:
Here, the sample space = {1, 2, 3, 4, 5, 6}
n(s) = 6
(i) If E = event of getting an even number = {2, 4, 6}
n(E) = 3
Then, probability of a getting an even number = n(E)/ n(s) = 3/6 = ½
(ii) If E = event of not getting an even number = {1, 3, 5}
n(E) = 3
Then, probability of a not getting an even number = n(E)/ n(s) = 3/6 = ½
(iii) If E = event of getting an odd number = {1, 3, 5}
So, n(E) = 3
Then, probability of a getting an odd number = n(E)/ n(s) = 3/6 = ½
5. From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will:
(i) be a black card.
(ii) not be a red card.
(iii) be a red card.
(iv) be a face card.
(v) be a face card of red colour.
Solution:
We know that,
Total number of cards = 52
So, the total number of outcomes = 52
There are 13 cards of each type. The cards of heart and diamond are red in colour. Spade and diamond are black. Hence, there are 26 red cards and 26 black cards.
(i) Number of black cards in a deck = 26
The number of favourable outcomes for the event of drawing a black card = 26
Then, probability of drawing a black card = 26/52 = ½
(ii) Number of red cards in a deck = 26
Therefore, number of non-red(black) cards = 52 – 26 = 26
The number of favourable outcomes for the event of not drawing a red card = 26
Then, probability of not drawing a red card = 26/52 = ½
(iii) Number of red cards in a deck = 26
The number of favourable outcomes for the event of drawing a red card = 26
Then, probability of drawing a red card = 26/52 = ½
(iv) There are 52 cards in a deck of cards, and 12 of these cards are face cards (4 kings, 4 queens and 4 jacks).
The number of favourable outcomes for the event of drawing a face card = 12
Then, probability of drawing a face card = 12/52 = 3/13
(v) There are 26 red cards in a deck, and 6 of these cards are face cards (2 kings, 2 queens and 2 jacks).
The number of favourable outcomes for the event of drawing a face card of red color = 6
Then, probability of drawing a red face card = 6/52 = 3/26
6. (i) If A and B are two complementary events then what is the relation between P(A) and P(B)?
(ii) If the probability of happening an event A is 0.46. What will be the probability of not happening of the event A?
Solution:
(i) Two complementary events, taken together, include all the outcomes for an experiment and the sum of the probabilities of all outcomes is 1.
P(A) + P(B) = 1
(ii) P(A) = 0.46
Let P(B) be the probability of not happening of event A
Then we know that,
P(A) + P(B) = 1
P(B) = 1Â –Â P(A)
P(B) = 1Â –Â 0.46
P(B) = 0.54
Thus, the probability of not happening of event A is 0.54
7. In a T.T. match between Geeta and Ritu, the probability of the winning of Ritu is 0.73. Find the probability of:
(i) winning of Geeta
(ii) not winning of Ritu
Solution:
(i) Winning of Geeta is a complementary event to winning of Ritu
Thus,
P(winning of Ritu) + P(winning of Geeta) = 1
P(winning of Geeta) = 1 – P(winning of Ritu)
P(winning of Geeta) = 1Â –Â 0.73
P(winning of Geeta) = 0.27
(ii)Â Not winning of Ritu is a complementary event to winning of Ritu
Thus,
P(winning of Ritu) + P(not winning of Ritu) = 1
P(not winning of Ritu) = 1 – P(winning of Ritu)
P(not winning of Ritu) = 1Â –Â 0.73
P(not winning of Ritu) = 0.27
Exercise 25(B) Page No: 393
1. Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be:
(i) an even number
(ii) a multiple of 3
(iii) an even number and a multiple of 3
(iv) an even number or a multiple of 3
Solution:
We know that, there are totally 9 cards from which one card is drawn.
Total number of elementary events = n(S) = 9
(i) From numbers 2 to 10, there are 5 even numbers i.e. 2, 4, 6, 8, 10
So, favorable number of events = n(E) = 5
Hence, probability of selecting a card with an even number = n(E)/ n(S) = 5/9
(ii) From numbers 2 to 10, there are 3 numbers which are multiples of 3 i.e. 3, 6, 9
So, favorable number of events = n(E) = 3
Hence, probability of selecting a card with a multiple of 3= n(E)/ n(S) = 3/9 = 1/3
(iii) From numbers 2 to 10, there is one number which is an even number as well as multiple of 3 i.e. 6
So, favorable number of events = n(E) = 1
Hence, probability of selecting a card with a number which is an even number as well as multiple of 3
= n(E)/ n(S) = 1/9
(iv) From numbers 2 to 10, there are 7 numbers which are even numbers or a multiple of 3 i.e. 2, 3, 4, 6, 8, 9, 10
So, favorable number of events = n(E) = 7
Hence, probability of selecting a card with a number which is an even number or a multiple of 3
=Â n(E)/ n(S) = 7/9
2. Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is:
(i) a multiple of 5
(ii) a multiple of 6
(iii) between 40 and 60
(iv) greater than 85
(v) less than 48
Solution:
We kwon that, there are 100 cards from which one card is drawn.
Total number of elementary events = n(S) = 100
(i) From numbers 1 to 100, there are 20 numbers which are multiple of 5 i.e. {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100}
So, favorable number of events = n(E) = 20
Hence, probability of selecting a card with a multiple of 5 = n(E)/ n(S) = 20/ 100 = 1/5
(ii) From numbers 1 to 100, there are 16 numbers which are multiple of 6 i.e. {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96}
So, favorable number of events = n(E) = 16
Hence, probability of selecting a card with a multiple of 6 = n(E)/ n(S) = 16/ 100 = 4/25
(iii) From numbers 1 to 100, there are 19 numbers which are between 40 and 60 i.e. {41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59}
So, favorable number of events = n(E) = 19
Hence, probability of selecting a card between 40 and 60 = n(E)/ n(S) = 19/100
(iv) From numbers 1 to 100, there are 15 numbers which are greater than 85 i.e. {86, 87, …., 98, 99, 100}
So, favorable number of events = n(E) = 15
Hence, probability of selecting a card with a number greater than 85 = n(E)/ n(S) = 15/100 = 3/20
(v) From numbers 1 to 100, there are 47 numbers which are less than 48 i.e. {1, 2, ……….., 46, 47}
So, favorable number of events = n(E) = 47
Hence, probability of selecting a card with a number less than 48 = n(E)/ n(S) = 47/100
3. From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:
(i) 3
(ii) 5
(iii) 3 and 5
(iv) 3 or 5
Solution:
We know that, there are 25 cards from which one card is drawn.
So, the total number of elementary events = n(S) = 25
(i) From numbers 1 to 25, there are 8 numbers which are multiple of 3 i.e. {3, 6, 9, 12, 15, 18, 21, 24}
So, favorable number of events = n(E) = 8
Hence, probability of selecting a card with a multiple of 3 = n(E)/ n(S) = 8/25
(ii) From numbers 1 to 25, there are 5 numbers which are multiple of 5 i.e. {5, 10, 15, 20, 25}
So, favorable number of events = n(E) = 5
Hence, probability of selecting a card with a multiple of 5 = n(E)/ n(S) = 5/25 = 1/5
(iii) From numbers 1 to 25, there is only one number which is multiple of 3 and 5 i.e. {15}
So, favorable number of events = n(E) = 1
Hence, probability of selecting a card with a multiple of 3 and 5 = n(E)/ n(S) = 1/25
(iv) From numbers 1 to 25, there are 12 numbers which are multiple of 3 or 5 i.e. {3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25}
So, favorable number of events = n(E) = 12
Hence, probability of selecting a card with a multiple of 3 or 5 = n(E)/ n(S) = 12/25
4. A die is thrown once. Find the probability of getting a number:
(i) less than 3
(ii) greater than or equal to 4
(iii) less than 8
(iv) greater than 6
Solution:
We know that,
In throwing a dice, total possible outcomes = {1, 2, 3, 4, 5, 6}
So, n(S) = 6
(i) On a dice, numbers less than 3 = {1, 2}
So, n(E) = 2
Hence, probability of getting a number less than 3 =Â n(E)/ n(S) = 2/6 = 1/3
(ii) On a dice, numbers greater than or equal to 4 = {4, 5, 6}
So, n(E) = 3
Hence, probability of getting a number greater than or equal to 4 = n(E)/ n(S) = 3/6 = 1/2
(iii) On a dice, numbers less than 8 = {1, 2, 3, 4, 5, 6}
So, n(E) = 6
Hence, probability of getting a number less than 8 = n(E)/ n(S) = 6/6 = 1
(iv) On a dice, numbers greater than 6 = 0
So, n(E) = 0
Hence, probability of getting a number greater than 6 = n(E)/ n(S) = 0/6 = 0
5. A book contains 85 pages. A page is chosen at random. What is the probability that the sum of the digits on the page is 8?
Solution:
We know that,
Number of pages in the book = 85
Number of possible outcomes = n(S) = 85
Out of 85 pages, pages that sum up to 8 = {8, 17, 26, 35, 44, 53, 62, 71, 80}
So, pages that sum up to 8 = n(E) = 9
Hence, probability of choosing a page with the sum of digits on the page equals 8 = n(E)/ n(S) = 9/85
6. A pair of dice is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first die.
Solution:
In throwing a dice, total possible outcomes = {1, 2, 3, 4, 5, 6}
So, n(S) = 6
For two dice, n(S) = 6Â x 6 = 36
Favorable cases where the sum is 10 or more with 5 on 1st die = {(5, 5), (5, 6)}
Event of getting the sum is 10 or more with 5 on 1st die = n(E) = 2
Hence, the probability of getting a sum of 10 or more with 5 on 1st die = n(E)/ n(S) = 2/ 36 = 1/18
7. If two coins are tossed once, what is the probability of getting:
(ii) at least one head.
(iii) both heads or both tails.
Solution:
We know that, when two coins are tossed together possible number of outcomes = {HH, TH, HT, TT}
So, n(S) = 4
(i) E = event of getting both heads = {HH}
n(E) = 1
Hence, probability of getting both heads = n(E)/ n(S) = ¼
(ii) E = event of getting at least one head = {HH, TH, HT}
n(E) = 3
Hence, probability of getting at least one head = n(E)/ n(S) = ¾
(iii) E = event of getting both heads or both tails = {HH, TT}
n(E) = 2
Hence, probability of getting both heads or both tails = n(E)/ n(S) = 2/4 = ½
Exercise 25(C) Page No: 394
1. A bag contains 3 red balls, 4 blue balls and 1 yellow ball, all the balls being identical in shape and size. If a ball is taken out of the bag without looking into it; find the probability that the ball is:
(i) yellow
(ii) red
(iii) not yellow
(iv) neither yellow nor red
Solution:
The total number of balls in the bag = 3 + 4 + 1 = 8 balls
So, the number of possible outcomes = 8 = n(S)
(i) Event of drawing a yellow ball = {Y}
So, n(E) = 1
Thus, probability of drawing a yellow ball =Â n(E)/ n(S) = 1/8
(ii) Event of drawing a red ball = {R, R, R}
So, n(E) = 3
Thus, probability of drawing a red ball =Â n(E)/ n(S) = 3/8
(iii) Probability of not drawing a yellow ball = 1 – Probability of drawing a yellow ball
Thus, probability of not drawing a yellow ball = 1 – 1/8 = (8 – 1)/ 8 = 7/8
(iv) Neither yellow ball nor red ball means a blue ball
Event of not drawing a yellow or red ball = E = 4
So, n(E) = 4
Thus, probability of not drawing a yellow or red ball =Â n(E)/ n(S) = 4/8 = 1/2
2. A dice is thrown once. What is the probability of getting a number:
(i) greater than 2?
(ii) less than or equal to 2?
Solution:
The number of possible outcomes when dice is thrown = {1, 2, 3, 4, 5, 6}
So, n(S) = 6
(i) Event of getting a number greater than 2 = E = {3, 4, 5, 6}
So, n(E) = 4
Thus, probability of getting a number greater than 2 = n(E)/ n(S) = 4/6 = 2/3
(ii) Event of getting a number less than or equal to 2 = E = {1, 2}
So, n(E) = 2
Thus, probability of getting a number less than or equal to 2 = n(E)/ n(S) = 2/6 = 1/3
3. From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is:
(i) a face card
(ii) not a face card
(iii) a queen of black card
(iv) a card with number 5 or 6
(v) a card with number less than 8
(vi) a card with number between 2 and 9
Solution:
We have, the total number of possible outcomes = 52
So, n(S) = 52
(i)Â No. of face cards in a deck of 52 cards = 12 (4 kings, 4 queens and 4 jacks)
Event of drawing a face cards = E = (4 kings, 4 queens and 4 jacks)
So, n(E) = 12
Thus, probability of drawing a face card = n(E)/ n(S) = 12/52 = 3/13
(ii) Probability of not drawing a face card = 1 – probability of drawing a face card
Thus, probability of not drawing a face card = 1 – 3/13 = (13 – 3)/ 13 = 10/13
(iii) Event of drawing a queen of black colour = {Q(spade), Q(club)} = E
So, n(E) = 2
Thus, probability of drawing a queen of black colour = n(E)/ n(S) = 2/52 = 1/26
(iv) Event of drawing a card with number 5 or 6 = E = {5H, 5D, 5S, 5C, 6H, 6D, 6S, 6C}
So, n(E) = 8
Thus, probability of drawing a card with number 5 or 6 = n(E)/ n(S) = 8/52 = 2/13
(v) Numbers less than 8 = { 2, 3, 4, 5, 6, 7}
Event of drawing a card with number less than 8 = E = {6H cards, 6D cards, 6S cards, 6C cards}
So, n(E) = 24
Thus, probability of drawing a card with number less than 8 = n(E)/ n(S) = 24/52 = 6/13
(vi) Number between 2 and 9 = {3, 4, 5, 6, 7, 8}
Event of drawing a card with number between 2 and 9 = E = {6H cards, 6D cards, 6S cards, 6C cards}
So, n(E) = 24
Thus, probability of drawing a card with number between 2 and 9 = n(E)/ n(S) = 24/52 = 6/13
4. In a match between A and B:
(i) the probability of winning of A is 0.83. What is the probability of winning of B?
(ii) the probability of losing the match is 0.49 for B. What is the probability of winning of A?
Solution:
(i) We know that,
The probability of winning of A + Probability of losing of A = 1
And,
Probability of losing of A = Probability of winning of B
Therefore,
Probability of winning of A + Probability of winning of B = 1
0.83 + Probability of winning of B = 1
Hence, probability of winning of B = 1 – 0.83 = 0.17
(ii) We know that,
Probability of winning of B + Probability of losing of B = 1
And, probability of losing of B = Probability of winning of A
Therefore,
Probability of winning of A = 0.49
5. A and B are friends. Ignoring the leap year, find the probability that both friends will have:
(i) different birthdays?
(ii) the same birthday?
Solution:
Out of the two friends, A’s birthday can be any day of the year. Now, B’s birthday can also be any day of 365 days in the year.
We assume that these 365 outcomes are equally likely.
So,
(i) If A’s birthday is different from B’s, the number of favourable outcomes for his birthday is 365 – 1 = 364
Hence, P (A’s birthday is different from B’s birthday) = 364/365
(ii) P (A and B have the same birthday) = 1 – P (both have different birthdays)
= 1 – 364/365 [As P (E’) = 1 – P(E)]
=Â 1/ 365
6. A man tosses two different coins (one of Rs 2 and another of Rs 5) simultaneously. What is the probability that he gets:
(i) at least one head?
(ii) at most one head?
Solution:
We know that,
When two coins are tossed simultaneously, the possible outcomes are {(H, H), (H, T), (T, H), (T, T)}
So, n(S) = 4
(i) The outcomes favourable to the event E, ‘at least one head’ are {(H, H), (H, T), (T, H)}
So, the number of outcomes favourable to E is 3 = n(E)
Hence, P(E) = n(E)/ n(S) = ¾
(ii) The outcomes favourable to the event E, ‘at most one head’ are {(T, H), (H, T), (T, T)}
So, the number of outcomes favourable to E is 3 = n(E)
Hence, P(E) =Â n(E)/ n(S) = 3/4
7. A box contains 7 red balls, 8 green balls and 5 white balls. A ball is drawn at random from the box. Find the probability that the ball is:
(i) white
(ii) neither red nor white.
Solution:
We have,
Total number of balls in the box = 7 + 8 + 5 = 20 balls
Total possible outcomes = 20 = n(S)
(i) Event of drawing a white ball = E = number of white balls = 5
So, n(E) = 5
Hence, probability of drawing a white ball = n(E)/ n(S) = 5/20 = 1/4
(ii) Neither red ball nor white ball = green ball
Event of not drawing a red or white ball = E = number of green ball = 8
So, n(E) = 8
Hence, probability of drawing a white ball = n(E)/ n(S) = 8/20 = 2/5
8. All the three face cards of spades are removed from a well shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting:
(i) a black face card
(ii) a queen
(iii) a black card
Solution:
We have,
Total number of cards = 52
If 3 face cards of spades are removed
Then, the remaining cards = 52 – 3 = 49 = number of possible outcomes
So, n(S) = 49
(i) Number of black face cards left = 3 face cards of club
Event of drawing a black face card = E = 3
So, n(E) = 3
Hence, probability of drawing a black face card =Â n(E)/ n(S) = 3/49
(ii) Number of queen cards left = 3
Event of drawing a black face card = E = 3
So, n(E) = 3
Hence, probability of drawing a queen card = n(E)/ n(S) = 3/49
(iii) Number of black cards left = 23 cards (13 club + 10 spade)
Event of drawing a black card = E = 23
So, n(E) = 23
Hence, probability of drawing a black card = n(E)/ n(S) = 23/49
9. In a musical chairs game, a person has been advised to stop playing the music at any time within 40 seconds after its start. What is the probability that the music will stop within the first 15 seconds?
Solution:
Total result = 0 sec to 40 sec
Total possible outcomes = 40
So, n(S) = 40
Favourable results = 0 sec to 15 sec
Favourable outcomes = 15
So, n(E) = 15
Hence, the probability that the music will stop in first 15 sec = n(E)/ n(S) = 15/40 = 3/8
10. In a bundle of 50 shirts, 44 are good, 4 have minor defects and 2 have major defects. What is the probability that:
(i) it is acceptable to a trader who accepts only a good shirt?
(ii) it is acceptable to a trader who rejects only a shirt with major defects?
Solution:
We have,
Total number of shirts = 50
Total number of elementary events = 50 = n(S)
(i) As, trader accepts only good shirts and number of good shirts = 44
Event of accepting good shirts = 44 = n(E)
Probability of accepting a good shirt =Â n(E)/ n(S) = 44/50 = 22/25
(ii) As, trader rejects shirts with major defects only and number of shirts with major defects = 2
Event of accepting shirts = 50 – 2 = 48 = n(E)
Probability of accepting shirts =Â n(E)/ n(S) = 48/50 = 24/25
11. Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 8
(ii) 13
(iii) less than or equal to 12
Solution:
We have, the number of possible outcomes = 6 × 6 = 36.
(i) The outcomes favourable to the event ‘the sum of the two numbers is 8’ = E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
The number of outcomes favourable to E = n(E) = 5.
Thus, P(E) =Â n(E)/ n(S) = 5/36
(ii) There is no outcome favourable to the event E = ‘the sum of two numbers is 13’.
So, n(E) = 0
Thus, P(E) =Â n(E)/ n(S) = 0/36 = 0
(iii) All the outcomes are favourable to the event E = ‘sum of two numbers ≤ 12’.
Thus, P(E) = n(E)/ n(S) = 36/36 = 1
12. Which of the following cannot be the probability of an event?
(i)Â 3/7
(ii) 0.82
(iii) 37%
(iv) -2.4
Solution:
We know that probability of an event E is 0 ≤ P(E) ≤ 1
(i) As 0 ≤ 3/7 ≤ 1
Thus, 3/7 can be a probability of an event.
(ii) As 0 ≤ 0.82 ≤ 1
Thus, 0.82 can be a probability of an event.
(iii) As 0 ≤ 37 % = (37/100) ≤ 1
Thus, 37 % can be a probability of an event.
(iv) As -2.4 < 0
Thus, -2.4 cannot be a probability of an event.
13. If P(E) = 0.59; find P(not E)
Solution:
We know that,
P(E) + P(not E) = 1
So, 0.59 + P(not E) = 1
Hence, P(not E) = 1 – 0.59 = 0.41
14. A bag contains a certain number of red balls. A ball is drawn. Find the probability that the ball drawn is:
(i) black
(ii) red
Solution:
We have,
Total possible outcomes = number of red balls.
(i) Number of favourable outcomes for black balls = 0
Hence, P(black ball) = 0
(ii) Number of favourable outcomes for red balls = number of red balls
Hence, P(red ball) =
15. The probability that two boys do not have the same birthday is 0.897. What is the probability that the two boys have the same birthday?
Solution:
We know that,
P(do not have the same birthday) + P(have same birthday) = 1
0.897 + P(have same birthday) = 1
Thus,
P(have same birthday) = 1 – 0.897
P(have same birthday) = 0.103
16. A bag contains 10 red balls, 16 white balls and 8 green balls. A ball is drawn out of the bag at random. What is the probability that the ball drawn will be:
(i) not red?
(ii) neither red nor green?
(iii) white or green?
Solution:
Total number of possible outcomes = 10 + 16 + 8 = 34 balls
So, n(S) = 34
(i) Favorable outcomes for not a red ball = favorable outcomes for white or green ball
So, number of favorable outcomes for white or green ball = 16 + 8 =24 = n(E)
Hence, probability for not drawing a red ball =Â n(E)/ n(S) = 24/34 = 12/17
(ii) Favorable outcomes for neither a red nor a green ball = favorable outcomes for white ball
So, the number of favorable outcomes for white ball = 16 = n(E)
Hence, probability for not drawing a red or green ball =Â n(E)/ n(S) = 16/34 = 8/17
(iii) Number of favorable outcomes for white or green ball = 16 + 8 = 24 = n(E)
Hence, probability for drawing a white or green ball =Â n(E)/ n(S) = 24/34 = 12/17
17. A bag contains twenty Rs 5 coins, fifty Rs 2 coins and thirty Re 1 coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin:
(i) will be a Re 1 coin?
(ii) will not be a Rs 2 coin?
(iii) will neither be a Rs 5 coin nor be a Re 1 coin?
Solution:
We have,
Total number of coins = 20 + 50 + 30 = 100
So, the total possible outcomes = 100 = n(S)
(i) Number of favourable outcomes for Re 1 coins = 30 = n(E)
Probability (Re 1 coin) = n(E)/ n(S) = 30/100 = 3/10
(ii) Number of favourable outcomes for not a Rs 2 coins = number of favourable outcomes for Re 1 or Rs 5 coins = 30 + 20 = 50 = n(E)
Hence, probability (not Rs 2 coin) = n(E)/ n(S) = ½
(iii) Number of favourable outcomes for neither Re 1 nor Rs 5 coins = Number of favourable outcomes for Rs 2 coins = 50 = n(E)
Hence, probability (neither Re 1 nor Rs 5 coin) =Â n(E)/ n(S) = 50/100 = 1/2
18. A game consists of spinning arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12; as shown below.
If the outcomes are equally likely, find the probability that the pointer will point at:
(i) 6Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (iv) a number greater than 8
(ii) an even number          (v) a number less than or equal to 9
(iii) a prime number                 (vi) a number between 3 and 11
Solution:
We have,
Total number of possible outcomes = 12
(i) Number of favorable outcomes for 6 = 1
Hence, P(the pointer will point at 6) =Â 1/12
(ii) Favorable outcomes for an even number are 2, 4, 6, 8, 10, 12
So, number of favorable outcomes = 6
Hence, P(the pointer will be at an even number) = 6/12 = ½
(iii) Favorable outcomes for a prime number are 2, 3, 5, 7, 11
So, number of favorable outcomes = 5
Hence, P(the pointer will be at a prime number) =Â 5/12
(iv) Favorable outcomes for a number greater than 8 are 9, 10, 11, 12
So, number of favorable outcomes = 4
Hence, P(the pointer will be at a number greater than 8) = 4/12 = 1/3
(v) Favorable outcomes for a number less than or equal to 9 are 1, 2, 3, 4, 5, 6, 7, 8, 9
So, number of favorable outcomes = 9
Hence, P(the pointer will be at a number less than or equal to 9) = 9/12 = 3/4
(vi) Favorable outcomes for a number between 3 and 11 are 4, 5, 6, 7, 8, 9, 10
So. number of favorable outcomes = 7
Hence, P(the pointer will be at a number between 3 and 11) =Â 7/12
19. One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:
(i) a queen of red color
(ii) a black face card
(iii) the jack or the queen of the hearts
(iv) a diamond
(v) a diamond or a spade
Solution:
We have,
Total possible outcomes = 52
(i) Number queens of red color = 2
Number of favorable outcomes = 2
Hence, P(queen of red color) =Â 2/52
(ii) Number of black cards = 26
Number of black face cards = 6
So, the number of favorable outcomes = 6
Hence, P(black face card) =Â 6/52 = 3/26
(iii) Favorable outcomes for jack or queen of hearts = 1 jack + 1 queen
So, the number of favorable outcomes = 2
Hence, P(jack or queen of hearts) = 2/52 = 1/26
(iv) Number of favorable outcomes for a diamond = 13
So, number of favorable outcomes = 13
Hence, P(getting a diamond) = 13/52 = ¼
(v) Number of favorable outcomes for a diamond or a spade = 13 + 13 = 26
So, number of favorable outcomes = 26
Hence, P(getting a diamond or a spade) = 26/52 = 1/2
The given solutions are as per the 2019-20 Concise Selina textbook. The Selina Solutions for the academic year 2023-24 will be updated soon. |
# Factors
Lesson
A number may be made by multiplying two other numbers together. The numbers that are multiplied together are called factors of the final number. All numbers have factors of one and themselves
$1\times7=7$1×7=7
$1\times14=14$1×14=14 and so on.
## How do we test if a number is a factor of another number?
If you can divide a whole number by another whole number and the answer is a whole number, then it is a factor. Let's look at an example to explain what I mean.
#### Examples
##### Question 1
Evaluate: Is $15$15 a factor of $45$45?
Think: $45\div15=3$45÷15=3 (this answer is a whole number).
DoYes, $15$15 is a factor of $45$45 (and so is $3$3)
However, some numbers have more factors than others.
$5$5 only has factors of $1$1 and $5$5 but
$12$12 can be made up as $1\times12$1×12, $2\times6$2×6 and $3\times4$3×4. So the factors of $12$12 are $1,2,3,4,6$1,2,3,4,6 and $12$12.
##### Question 2
Write down all the factors of $144$144
Think: $1\times144$1×144, $2\times72$2×72, $3\times48$3×48, $4\times36$4×36, $6\times24$6×24, $8\times18$8×18, $9\times16$9×16 and $12\times12$12×12 all give an answer of $144$144.
So, the factors of $144$144 are $1,2,3,4,6,8,9,12,16,18,24,36,48,72$1,2,3,4,6,8,9,12,16,18,24,36,48,72 and $144$144.
That was a lot!
##### Question 3
Which of these numbers is $6$6 a factor of?
1. $30$30
A
$32$32
B
$17$17
C
$39$39
D
$30$30
A
$32$32
B
$17$17
C
$39$39
D
##### Question 4
Write down all factors of $12$12.
Separate numbers with a comma.
##### Question 5
What is the highest common factor of $24$24 and $36$36?
### Outcomes
#### NA3-2
Know basic multiplication and division facts. |
# SECOND TERM MATHEMATICS SCHEME OF WORK FOR PRIMARY THREE (3)
THEME: NUMBER AND NUMERATION
SUB THEME: WHOLE NUMBER
WEEK 1: Basic Multiplication.
TEACHER’S ACTIVITIES:
1. Teachers first demonstrates the idea of multiplication as repeated addition.
2. She guides pupils to use repeated addition for multiplication e.g.
15 x 4 = 15 + 15 + 15 + 15
• With the help of square charts she gets pupils to carry out multiplication from 1x 1 to 9 x 9 while she corrects and guides.
• She gets pupils to understand that any number multiplied by 1 remains the same while any number multiplied by 0 equals zero.
• Teacher guides pupils to use multiplication chart to multiply 2 –digit numbers by 1 – digit numbers horizontally and vertically e.g. 32 x 3 or 32
X 3
96
PUPILS ACTIVITIES
1. Pupils use repeated addition for multiplication.
2. They study the 10 x 10 square chart for multiplication facts up to 9 x 9.
3. They study the back of their exercise books that contain these multiplication facts.
4. They carry out multiplication of 2 digit number by 1 –digit numbers and use repeated addiction for multiplication.
5. They practice multiplication of 2 – digit numbers by 1 digit numbers horizontally or vertically.
TEACHING/LEARNING RESOURCES/AIDS
1. Charts showing multiplication of 2 – digit numbers by 1 – digit numbers.
2. 10 x 10 square chart and back of exercise books with multiplication tables.
WEEK 2: Multiplication Of Three 1 –Digit Numbers Taking Two At A Time.
TEACHER’S ACTIVITIES:
1. Pupils participate in multiplication from 1 x 1 to 9 x 9.
2. She leads pupils to discover that.
3. The order of the terms does not affect the product e.g.
3 x 4 = 4 x 3 = 12
3 x 4 x 5 = 4 x 3 x 5 = 5 x 4 x 3=
4 x 5 3 = 5 x 3 x 4 = 60
1. The association of terms does not affect the product e.g.
3 x 4 x5 = (3 x 4) x 5 = 3 x (4×5) = 60
• She gets the pupils to carry out series of multiplication of three 1 –digit numbers taking two
• Teacher briefly look back at counting and identification of whole numbers up to 99.
• She builds up piles or bundles in hundreds, tens, units to remind pupils that bringing, for example, 3 piles of hundreds, five piles of tens and counters represents 354.
• From the collection, she counts the bundles of hundred and records under H, the number of bundles of tens under T and units or once under U.
• She guides pupils to expand a given number e.g. 486 = 400 + 80 + 6.
• She explains that 4 represents 4 bundles of hundreds, 8 represents 8 bundles of tens and 6 represents 6 units or 6 ones.
• She explains how to put a given 3 –digit number in a tabular form under H.T.U.
PUPILS ACTIVITIES:
1. Pupils participate in the revision exercise.
2. They discover that the operation of multiplication is cumulative and associative.
3. They carry out series of multiplication of three 1 – digit numbers taking two at a time.
TEACHING/LEARNING RESOURCES/AIDS:
1. Chart for three 1- digit numbers.
2. Multiplication tables
WEEK 3: Division Of Whole Numbers Not Exceeding 48 By 2, 3, 4, 5 And 6 Without Remainder.
TEACHER’S ACTIVITIES:
1. Teacher uses counters and charts to guides pupils through division of whole numbers not exceeding 48.
2. She explains division by either grouping or repeated subtraction
25 ÷ 5 = (20 + 5) ÷ 5
(20 ÷ 5) + (5 ÷ 5)
4 + 1 = 5
• She relates multiplication to division e.g.
4 x 5 = 20 then 20 = 5/ 4 or 20/5 = 4
PUPILS ACTIVITIES:
1. Pupils divide whole numbers not exceeding 48 by 2, 3, 4, 5, 6.
2. They solve division problems using grouping and repeated subtraction.
3. They relate multiplication to division
TEACHING/LEARNING RESOURCES/AIDS:
1. Counters
2. Charts containing division of whole numbers not exceeding 48.
WEEK 4: Factors of Whole Numbers Not Exceeding 48
TEACHER’S ACTIVITIES
1. In simple language, the teacher explains factors of a number as all the numbers which divide into it exactly.
E.g. factors of 12 are 1, 2, 3, 4, 6, and 12
Factors of 11 are 1, 11.
Factors of 24 are 1, 2, 3, 4, 6, 8, and 12
• She use a square multiplication table to present rectangular pattern of numbers. She uses it to find two or three factors of some numbers.
1 2 3 4 5 6 2 4 6 8 10 12 3 6 9 12 15 18 4 8 12 16 20 24 5 10 15 20 25 30 6 12 18 24 30 36 7 14 21 28 35 42 8 16 24 32 40 48
Two factors of 10 are 2 and 5
Two factors of 15 are 3 and 5
Three factors of 30 are 2, 3 and 5
• She leads pupils to understand that if one multiplies two numbers together , they are factors of the answer (product)
E.g. 3 x 2 = 3 and 2 are factors of 6.
• She guides pupils to use rectangular pattern of numbers to find two or three factors of whole numbers.
PUPILS ACTIVITIES
1. Pupils listen to the teacher’s explanation.
2. They use rectangular patterns of numbers to find factors of given numbers.
TEACHING/LEARNING RESOURCES/AIDS
1. Rectangular pattern of numbers.
2. Charts containing worked examples
WEEK 5: Finding Missing Factors in a Given Number.
TEACHER’S ACTIVITIES
1. Teacher presents rectangular pattern of numbers and find two or three factors of some numbers.
2. She gets pupils to use the rectangular patterns of numbers to solve many examples.
3. She leads pupils to identify relationship between multiplication and division
E.g. 4 x 3 = 12
3 x 4 = 12
12 ÷ 3 = 4
12 ÷ 4 = 3
• She leads pupils to find missing factors in given numbers e.g.
• 15 = 3 x ?
15 ÷ 3 = 5
1. 20 = 2 x 2 x ?
? = 20 ÷ (2 x 2)
• 20 ÷ 4 = 5
PUPILS ACTIVITIES
1. Pupils work problems involving finding missing factors of given numbers.
2. Pupils appreciate the relationship between multiplication and division.
TEACHING/LEARNING RESOURCES/AIDS
1. Rectangular pattern of numbers.
2. Charts containing worked examples
WEEK 6: Factors and Multiples of Number.
TEACHER’S ACTIVITIES
1. Teacher leads pupils to recall the meaning of factor/s.
2. She explains “multiples” as a quantity (number) that contains another quantity an exact number of times e.g. 6, 9 and 12 are all multiples of 3.
3. She guides pupils to write out multiples of given numbers.
4. She leads pupils to distinguish between factors and multiples of given numbers,
5. She guides pupils to find factors and multiples of many numbers.
PUPILS ACTIVITIES
1. Pupils give the meaning and examples of factors.
2. They learn the meaning of “multiples”
3. They distinguish between factors of given numbers.
4. They work problems involving factors and multiples of numbers.
TEACHING/LEARNING RESOURCES/AIDS
• Rectangular pattern of numbers.
• Charts containing worked examples
WEEK 7:
Theme: ALGEBRAIC PROCESSES
SUB – THEME: Open Sentences
Topic: Open sentences
TEACHER’S ACTIVITIES
1. Teacher explains to pupils that at times the symbol? Is used for dealing with some addition and subtraction problems as well as multiplication and division problems.
E.g. 12 + 3 = 15. What number in the box will make that mathematical sentence true?
• She guides pupils to find missing numbers.
Example, 6 + ? = 11
10 – ? = 7
12 ÷ ? = 4
4 x ? = 24
• She leads pupils to appreciate the relationship between addition and subtraction, then multiplication and division.
• She leads pupils to solve series of problems involving open sentences.
• She gets pupils to give examples of open sentences in everyday life.
PUPILS ACTIVITIES
1. Pupils examine the real money and models of coins.
2. They recognize and identify the denominations.
3. They change money up to #20 into small units.
4. They discuss the uses of money.
TEACHING/LEARNING RESOURCES/AIDS
1. Charts containing worked examples on open sentences.
WEEK 8
Theme: Mensuration and Geometry
Sub –Theme: Primary Measures
Topic: Money: Changing Money Not Exceeding #20 into Smaller Unit.
TEACHER’S ACTIVITIES
1. Teacher shows pupils real money and models of coins.
2. She guides pupils to recognize and identify the Nigeria coins and bank notes.
3. She guides pupils to realize that there are five 1k coins in a 5k coin, two 5k coins in a 10k coin, five 10k coins in a 50k coin, etc.
4. She guides pupils say 50k and asks them to change it into 10k coins, etc.
5. She discusses the uses of money with pupils.
PUPILS ACTIVITIES
1. Pupils examine the real money and models of coins.
2. They recognize and identify the denominations.
3. They change money up to #20 into small units.
4. They discuss the uses of money.
TEACHING/LEARNING RESOURCES/AIDS
1. Real and models coins and bank notes.
2. Charts of coins and bank notes.
WEEK 9: Shopping Involving Addition and Subtraction with Money Not Greater Than #20
TEACHER’S ACTIVITIES
1. Teacher briefly looks back at the recognition and identification of the Nigeria n coins and bank notes and changing money into smaller units.
2. She sets a shopping corner in the class-room and items such as empty packets of detergent, sugar, empty tins of sardines, geisha, tomatoes, etc.
3. She appoints a pupil as a shop keeper and another as a customer. The customer goes to the shopkeeper to buy some items, adds up the cost and gives the shop keeper an amount which may not require change. The money involved should not be greater than #20.
PUPILS ACTIVITIES
1. Pupils actively participate in the revision exercise
2. They help in collecting items to be kept in the shopping corner.
3. They participate actively by either acting as a shop keeper or as a customer.
TEACHING/LEARNING RESOURCES/AIDS
1. Models of coins and real coins.
2. Bank notes
3. Various articles with price tags less than #5.
WEEK 10
Sub – Theme: Primary Measures
Topic: Shopping Involving Addition and Subtraction with Money Not Greater Than #20.
TEACHER’S ACTIVITIES
1. The Teacher continues with the shopping activities.
2. She drills the class in mental sums with respect to addition and subtraction of numbers (not more than 20).
3. She guides the pupils to solve problems on addition and subtraction of money not greater than #20.
4. She guides discussions on problems involving addition and subtraction of money with results not exceeding #20.
PUPILS ACTIVITIES
1. Pupils participate in shopping activity.
2. They do mental sums.
3. They solve problems on addiction and subtraction of money not greater than #20.
4. They discuss problems involving addition and subtraction of money as directed by the teacher.
TEACHING/LEARNING RESOURCES/AIDS
1. Models of coins and real coins.
2. Bank notes
3. Various articles with price tags less than #5.
WEEK 11: Multiplication Involving Money with Products Not Exceeding #20
TEACHER’S ACTIVITIES
1. Teacher briefly looks back at addition and subtraction of money not greater than #20.
2. She drills the class in multiplication of simple numbers e.g. 2 x 2, 3 x 4, 5 x 2. 4 x 4, etc.
3. She guides the pupils to solve problems on multiplication of money with products not exceeding #20. E.g. 15k x 3 = 45k
#4 x 4 = #16
PUPILS ACTIVITIES
TEACHING/LEARNING RESOURCES/AIDS
1. Real money and models
2. Charts containing simple multiplication of money.
WEEK 12: REVIEW OF FIRST TERM’S WORKS
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Durango Bill's
Bingo Probabilities
Probability Analysis that you will have a “Bingo”
after “N” numbers have been called.
(In a multiboard game, why does someone else get a Bingo so quickly?)
(Math notation is generally the same as that used in Microsoft’s Excel. The MathNotation link will also give examples of the notnotation as used here.)
For 90 number Bingo ( 3 rows, 9 columns) please see Bingo 90.
The Bingo Statistics link gives tables and graphs showing the probability for getting “Bingo” after the announcer has called “N” numbers. Tables and graphs cover both a single board and a 50 board game.
The Bingo 4 Corners and Letter “X” (both diagonals) link shows the single board probability of getting these patterns after “N” numbers have been called.
The Bingo Picture Frame (all 4 edges) and Letter “Y” link shows the single board probability of getting these patterns after “N” numbers have been called.
The Bingo Probabilties for a Complete Cover link shows the probabilities for covering all squares on a Bingo board.
The “How to calculate” link shows how to calculate these numbers including how to calculate the probabilities when any arbitrary number of boards are being played in a game.
Probabilities for Swedish Bingo. A Swedish Bingo card has the familiar 5 rows and 5 columns, but the middle cell is not free. It has to be filled by having its number called.
Rules of the game: A typical Bingo card has 24 semi-random numbers and a central star arranged in a square of 5 rows and 5 columns. A Bingo card might look like:
1 16 31 46 61
4 19 34 49 64
8 23 * 53 68
11 26 41 56 71
15 30 45 60 75
We used the phrase “semi-random” to describe the numbers because the numbers in each column are confined within ranges. Column 1 will contain 5 random numbers in random order, but they are within a range of 1-15. Similar ranges exist for the other 4 columns (16-30, 31-45, 46-60, and 61-75). The central location is a “Free” spot. There are (15!/10!)^4 * (15!/11!) = 5.52+ E26 (more than 552 million billion billion) possible combinations that could exist - any one of which would be a legal Bingo card. (The “!” symbol is the mathematical notation for Factorial. e.g. Factorial(5) = 5 * 4 * 3 * 2 * 1 = 120.)
Note: The above number of combinations assumes the numbers in any column can be in random order. If the numbers in any column are always in sorted order with the lowest number on row 1 and the highest number on row 5, then the number of combinations in each of 4 columns is reduced by 5! = 120 and the number of combinations in the center column is reduced by 4! = 24.
Initially, the central “*” is counted as a “free” or “called” cell. Then, an announcer will call out numbers selected randomly within the total 1-75 range. (Usually this is done by randomly removing numbered balls from a revolving drum.) Whenever one of these called numbers matches a number on a player’s Bingo card, the player marks that number as “called”. Eventually, there will be a straight line of 5 called numbers that fill a row, fill a column, or form a corner-to-corner diagonal line. (Note: the “Free” center space can be part of the straight line). At this point the player yells “Bingo” and the game is over.
Probabilities: Of interest, in a single board game - What is the probability the player will have a “Bingo” after the announcer has called “N” numbers? Also, in a multiboard game, what is the probability that the first “Bingo” will show up after “N” numbers have been called? (Check the Bingo Statistics link.)
Variations on Bingo: Other patterns can be used for the game of Bingo. For example, a winning Bingo could be defined as filling a 2x2 block anywhere on a Bingo card. There are 16 possible locations where a 2x2 block could be located. Other Bingo variations could include filling any of the 9 possible 3x3 blocks, or filling a 2x3 block. A 2x3 block could also be rotated for 24 possible winning “Bingos”.
Other Bingo websites: The “Wizard of Odds” also has bingo statistics information - especially the gambling aspects of Bingo as well as a lot of good stuff on gambling in general. The probabilities given here match those in the “Wizard’s” tables. (It's reassuring to have two independent calculations come up with the same results.)
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# 900 Pick Your Pony. Who’ll Win This Amount of Factors Horse Race?
I really like this rhyme that I saw for the first time this week (even though it’s all over the net):
Hey diddle diddle, the median’s the middle,
You add then divide for the mean.
The mode is the one that appears the most,
And the range is the difference between.
All of the numbers from 801 to 900 have at least 2 factors, but no more than 32 factors. 32 – 2 = 30, so 30 is the range of the amount of factors.
There are 100 numbers from 801 to 900. If you list the amount of factors for each number, then arrange those amounts from smallest to largest, the amounts that will appear in the 50th and 51st spots will both be 6. That means that 6 is the median amount of factors. If we had different amounts in the 50th and 51st spots, we would average the two amounts together to get the median.
If you add up the amounts of factors that the numbers from 801 to 900 have, you will get 794. If you divide 794 by 100, the number of entries, then you will know that 7.94 is the mean amount of factors.
What about the mode? Which amount of factors appears the most? That’s why we are having a Horse Race, to see if more numbers have 2 factors, 3 factors, 4 factors, or a different amount of factors. So pick your pony. We’ll see which amount wins, and we’ll find out what the mode is at the same time.
The contenders are these amounts: 2, 3, 4, 6, 8, 10, 12, 14, 16, 18, 20, 24, 27, 32.
I should tell you that only perfect squares can have an odd amount of factors, so you probably don’t want to pick an odd amount.
Here are some interesting facts about the numbers from 801 to 900 that might help you decide which pony to pick.
• We had the smallest two consecutive numbers with exactly 12 factors: (819, 820)
• We had the fourth prime decade: (821, 823, 827, 829). All four of those numbers are prime numbers and have exactly two factors.
• We had five consecutive numbers whose square roots can be reduced: (844, 845, 846, 847, 848). Three of those numbers had 6 factors, one had 10, and one had 12.
• We also had 840, the smallest number with exactly 32 factors
• 900 is the smallest number with exactly 27 factors. Coincidentally, the amount that is the mode will appear 27 times.
As the following table shows, there are 42 integers from 801 to 900 that have square roots that can be simplified. 42 is more than any previous set of 100 numbers has given us. Even still we are still holding close to just under 40% of integers having square roots that can be simplified.
Okay. If you’ve picked your pony, NOW you can watch the Horse Race:
make science GIFs like this at MakeaGif
Hmm…
The race was exciting for a second or two.
As you can see from the Horse Race the mode is 4. How did your pony do?
Here’s a little more about the number 900:
900 is the sum of the fourteen prime numbers from 37 to 97.
24² + 18² = 900
900 is the hypotenuse of two Pythagorean triples:
• 252-864-900, which is 24² – 18², 2(24)(18), 24² + 18². It is also (7-24-25) times 36.
• 540-720-900, which is (3-4-5) times 180.
900 is the sum of the interior angles of a heptagon (seven-sided polygon).
• 900 is a composite number and a perfect square.
• Prime factorization: 900 = 2 × 2 × 3 × 3 × 5 × 5, which can be written 900 = 2² × 3² × 5²
• The exponents in the prime factorization are 2, 2 and 2. Adding one to each and multiplying we get (2 + 1)(2 + 1)(1 + 1) = 3 × 3 × 3 = 27. Therefore 900 has exactly 27 factors.
• Factors of 900: 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 25, 30, 36, 45, 50, 60, 75, 90, 100, 150, 180, 225, 300, 450, 900
• Factor pairs: 900 = 1 × 900, 2 × 450, 3 × 300, 4 × 225, 5 × 180, 6 × 150, 9 × 100, 10 × 90, 12 × 75, 15 × 60, 18 × 50, 20 × 45, 25 × 36, or 30 × 30
• Taking the factor pair with the largest square number factor, we get √900 = (√30)(√30) = 30.
# What often happens to a number like 599 when a number next to it has so many factors?
The square root of any number from 576 to 624 is between 24 and 25. That means the first number in any of their factor pairs will be 24 or less. The combined number of factor pairs for the 49 integers from 576 to 624 is 189. The number of factor pairs for any given number ranges from 1 to 12. Let’s look at the averages: The mean (189/49) is 3.857 factor pairs per number. The median is 3 factor pairs, and the mode (the number of factor pairs that occurs most often) is 4 factor pairs.
What happens to a number like 599 when a number next to it has far more than the average number of factor pairs? Quite often, but not always, that number has no choice but to be a prime number.
Even though 600 has only three prime factors (2, 3 and 5), it still managed to be divisible by 50% of the numbers from 1 to 24, and there just aren’t many possibilities left for the numbers immediately before or after it.
The twelve numbers less than or equal to 24 that will divide into 600 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20 and 24.
Because 9, 14, 16, 18, 21, and 22 each have 2 or 3 as a prime factor, those six numbers also are not possible factors of 599 or 601.
Every whole number is divisible by 1, but besides that, there are only six numbers available as possible factors for those two numbers: 7, 11, 13, 17, 19, and 23. Since neither 599 nor 601 is divisible by any of those numbers, they turn out to be twin primes.
Usually at least one of the numbers before or after a number with far more than its fair share of factor pairs will be a prime number.
119 and 121, the numbers before and after 120 are notable exceptions. √120 ≈ 10.95441. The factors of 120 that are less than or equal to 10 are 1, 2, 3, 4, 5, 6, 8, and 10 which is 80% of the possible factors. Yet 119 managed to be divisible by 7, and 121 managed to be divisible by 11 so neither one of them is a prime number.
————————————————————————————-
If you add up the digits of 499, 589, or 598, you will get 22.
599 is the smallest whole number whose digits add up to 23. Thank you, OEIS.org for that number fact.
• 599 is a prime number and a twin prime with 601.
• Prime factorization: 599 is prime.
• The exponent of prime number 599 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 599 has exactly 2 factors.
• Factors of 599: 1, 599
• Factor pairs: 599 = 1 x 599
• 599 has no square factors that allow its square root to be simplified. √599 ≈ 24.4744765
How do we know that 599 is a prime number? If 599 were not a prime number, then it would be divisible by at least one prime number less than or equal to √599 ≈ 24.5. Since 599 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, or 23, we know that 599 is a prime number. |
Rankers Hub
In this section, we will discuss more about Percentage
We get to see very few problems which are on percentage only but the concepts and formulae of Percentage are used in different topics. This makes understanding the concepts very important.
Percent means ‘for every hundred’. Suppose we have a statement, B is 10% of A, this means that for every 100th part of A, we have 10 part of B or if value of B is 10 then, value of A will be 100.
Let’s understand it with basic formula and some examples on it.
If we have to find y% of x, then
y% of x = (y/100) × x,
y% = y/100
E.g.1: Find: 10% of 500 = ?
By formula,
10% of 500 = (10/100) × 500 = 50 (Ans).
E.g.2: If 40% of m = 110, then find the value of m.
We know that, y% = y/100
40% of m = 110
⇒ 40/100 of m = 110
⇒ m = 110 × 100/40
⇒ m = 110 × 5/2
⇒ m = 55 × 5
⇒ m = 275 (Ans).
Before we begin with specific types let’s learn all the basic concepts with examples.
Conversion:
We can convert percent into fraction and vice versa. We can also convert percent into decimal
• If we have x% and we want to convert it into fraction and decimal, then:
Required fraction = x/100
Required decimal = 0.0x (Move two decimal places from right)
E.g. 3: Convert 25% to fraction.
Required fraction = 25/100 = ¼ (Ans).
E.g. 4: Convert 25% to decimal.
Move two decimal places from right,
Required decimal = 0.25 (Ans).
If we have a fraction x/y, then:
Required percentage = (x/y) × 100
E.g. 5: Convert ¼ to fraction.
Required percentage = ¼ × 100 = 25% (Ans).
To express one quantity as a percent with respect to other quantity, following formula is used:
Percentage = (Quantity to be expressed in percentage/ Quantity with respect to which percentage is to be calculated) × 100
Note: It is important to convert all the quantities in same unit.
Let’s take examples and try to understand the application of this formula.
E.g. 6: What percent of 1hour is 1 minute and 12 seconds?
Quantity to be expressed in percentage = 1 min 12 seconds
= 1min + 12/60 minutes
= (1 + 1/5) minutes
= 6/5 minutes
Quantity with respect to which percentage is to be calculated = 1 hour = 60 minutes
∴ Required percentage = (6/5/ 60) × 100
= 6 × 100/5 × 60 = 20/10
= 2/1 or 2 (Ans).
E.g. 7: The below table gives rice production in each year. Read and answer the following questions.
1. What percent is the production in 2008 with respect to production in 2009?
Required percentage = Production in 2008/ Production in 2009
= 258/495
= 86/165 (Ans).
Percentage of change in quantity
If there is change in a quantity then to calculate percentage change, following formula is used:
Percent of change in quantity = {(Final quantity – Initial quantity)/Initial quantity} ×100
Let’s understand this with previous example from RRB NTPC Quant Questions.
E.g. 7: The below table gives rice production in each year. Read and answer the following questions.
• What is the increased percentage of production in 2009 compared to year in 2005?
Percent Change = {(Final quantity – Initial quantity)/ Initial quantity} × 100
= {(495 – 325)/325} × 100
= (170/325) × 100
= 0.523 × 100
= 52.3% (Ans).
• What was the decline in production in year 2008 compared to 2007?
Percent Change = {(429 – 258)/429}× 100
= (171/429) × 100 = 0.398 × 100 = 39.8% (Ans).
From now on we will see how the concepts of percentage are directly applicable in different problems of different topics.
Percentage change:
If there is a percentage change in a quantity, then:
Final quantity = Initial quantity × (100 ± Percentage increase/decrease)/100
Note: Percent change is always calculated in respect of an amount. It is important to pay more attention to the amount in respect of which percent change is asked.
Let’s see some examples Questions on percentage change and try to solve them.
E.g. 9: The monthly income of a person is Rs. 6000. If his income is increased by 40% then what is his monthly income now?
Here, Initial quantity = 6000
Percentage increase = 40%
∴ Final quantity = Initial quantity × (100 ± Percentage increase/decrease)/100
= 6000 × (100 + 40)/100
= 6000 × (140)/100
= 60 × 140
= 8400
Hence, his monthly income now = Rs. 8400 (Ans).
E.g. 10: Mr. Akshar sold a bus for Rs. 20,400 with a loss of 15%. At what price should the bus be sold to get a profit of 15%?
Here, although this question is Profit and loss question, we need this formula of percentage to solve this question.
We know that,
CP = SP + Loss
⇒ CP = 20,400 + 15% of CP
⇒ CP = 20,400 + 15/100 of CP
⇒ CP = 20,400 + 3/20 of CP
⇒ CP – 3/20 CP = 20,400
⇒ 17/20 × CP = 20,400
CP = 20,400 × 20/17 = 1200 × 20 = Rs. 24000
At 15% profit,
SP = CP + Profit
= 24,000 + 15% of 24,000
= 24,000 +15/100 of 2400
= 24,000 + 3600 = Rs. 27,600 (Ans).
This solution is too long. Lets’ solve it by formula.
Final quantity = Initial quantity × (100 ± Percentage increase/decrease)/100
SP = CP × (100 – loss%)/100 [Since loss is percentage decrease]
⇒ 20,400 = CP × (100 – 15)/100
⇒ 20,400 = CP × 85/100
⇒ CP = 20,400 × 100/ 85
= 20,400 × 20/17
= 24,000
At 15% profit,
SP = CP × (100 + profit %)/100 [Since loss is percentage decrease]
= CP × (100 + 15)/100
= CP × 115/100
= 24,000 × 23/20
= 12,000 × 23
= Rs. 27,600 (Ans).
See, the relevance of percentage in other problems. Almost every calculation is based on our percentage concept. We can minimize this time wasted in calculation by multiplication factor (MF). |
Question Video: Calculating a Missing Value in a Data Set given the Mean | Nagwa Question Video: Calculating a Missing Value in a Data Set given the Mean | Nagwa
# Question Video: Calculating a Missing Value in a Data Set given the Mean Mathematics
Given the mean of the values 8, 22, 4, 12, and 𝑥 is 15, find the value of 𝑥.
01:45
### Video Transcript
Given the mean of the values eight, 22, 4, 12, and 𝑥 is 15, find the value of 𝑥.
In this question, we’re given a data set which includes an unknown value 𝑥. We’re told that the mean of these values is 15 and we need to find the unknown value of 𝑥. The mean of a data set is found by calculating the sum of the data values and dividing by the total number of data values. So, we can start by adding our values eight plus 22 plus four plus 12 plus 𝑥. And as we know that there are five data values, then we would be dividing by five.
Notice that even though we don’t know what 𝑥 is yet, we still include it as one of the values. We’ve been given that the mean is equal to 15. So, we can put this on the left-hand side of our equation. At the minute, we don’t know the value of 𝑥, but we can go ahead and add together the other values in our set. We can then simplify our calculation as 15 equals 46 plus 𝑥 over five. We now need to rearrange this equation in order to find the value of 𝑥.
Multiplying both sides of this equation by five gives us 75 equals 46 plus 𝑥. Subtracting 46 then from both sides of our equation, we have 75 minus 46 equals 𝑥. 29 equals 𝑥 then. And so, our answer is that 𝑥 equals 29. We can check our answer by substituting the value of 𝑥 equals 29, summing with the rest of the values, dividing by five, and checking that we get a value of 15. And thus confirming the value 𝑥 equals 29.
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More Math Triangles More Shapes
A triangle is a three-sided figure. The sum of the interior angles of a triangle is 180 degrees. Each of the three points on the tips of a triangle is called a vertex (the plural of vertex is vertices).
The perimeter of a triangle is the sum of the lengths of the three sides. The area of a triangle is (1/2)*base*height.
Triangles can be classified by the length of their side or by their interior angles.
Classifying triangles by side length:
equilateral triangleThe sides of an equilateral triangle are all the same length and all the interior angles are 60 degrees. isosceles triangleAn isosceles triangle has two sides that are the same length and two interior angles that are the same. scalene triangleThe sides of a scalene triangle are all different lengths. A right triangle can be a scalene triangle (but not all right triangles are scalene triangles).
The interior angles of a all triangles add up to 180 degrees. Every triangle has at least two acute angles (an acute angle is less than 90 degrees).
Classifying triangles by interior angles:
acute triangleAll three interior angles of an acute triangle are acute (less than 90 degrees). Equilateral triangles are acute triangles. obtuse triangleOne of the interior angles of an obtuse triangle is obtuse ( an obtuse angle is over 90 degrees). right triangleA right triangle has one angle that is a right angle (a right angle is a 90-degree angle). The Pythagorean theorem states that for a right triangle, the sum of the squares of the two shorter sides is equal to the square of the hypotenuse (h): a2 + b2 = h2. The hypotenuse is the longest side of a right triangle, and is the side opposite the right angle.
Triangle Worksheets:
Triangle WorksheetRead about triangles, do dot-to-dots, draw them, color them, and write the names of things with a triangular shape. Or go to the answers. Triangle Classification WorksheetRead about how to classify triangles with respect to side length and interior angles (equilateral, isosceles, scalene, right, acute, and obtuse triangles). Or go to the answers. Angles and TrianglesLabel the angles (acute, right, obtuse, and straight) and triangles (equilateral, isosceles, scalene, and right).Answers
Finish the Drawing Worksheets
Symmetrical Triangle Picture #1: Finish the Drawing Worksheet
Finish the drawing of the triangle around the line of symmetry. Or go to the answer page.
Symmetrical Triangle Picture #2: Finish the Drawing Worksheet
Finish the drawing of the triangle around the line of symmetry. Or go to the answer page.
Symmetrical Triangle Picture #1: Finish the Drawing and Fill in the Missing Letters Worksheet
Finish the drawing of the triangle around the line of symmetry. Or go to the answer page.
Symmetrical Triangle Picture #2: Finish the Drawing and Fill in the Missing Letters Worksheet
Finish the drawing of the triangle around the line of symmetry. Or go to the answer page.
Books to Print
Things Shaped Like TrianglesA Printable Activity BookA small, triangular shaped book with triangular things to draw, including mountain, volcano, rocket, banner, and pyramid. A one-page printout. TrianglesA Printable Activity BookA small, triangular shaped book about triangles, with pages: circle the triangles, draw a triangle, definition of an equilateral triangle, definition of an isosceles triangle, and definition of a right triangle. A one-page printout.
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# Calculator search results
Formula
Number of solution
$$x ^ { 2 } - 25 = 0$$
2 real roots
Find the number of solutions
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 25 } = \color{#FF6800}{ 0 }$
Determine the number of roots using discriminant, $D=b^{2}-4ac$ from quadratic equation, $ax^{2}+bx+c=0$
$\color{#FF6800}{ D } = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 25 } \right )$
$D = \color{#FF6800}{ 0 } ^ { \color{#FF6800}{ 2 } } - 4 \times 1 \times \left ( - 25 \right )$
The power of 0 is 0
$D = \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 25 \right )$
$D = \color{#FF6800}{ 0 } - 4 \times 1 \times \left ( - 25 \right )$
0 does not change when you add or subtract
$D = - 4 \times 1 \times \left ( - 25 \right )$
$D = - 4 \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \times \left ( - 25 \right )$
Multiplying any number by 1 does not change the value
$D = - 4 \times \left ( - 25 \right )$
$D = \color{#FF6800}{ - } \color{#FF6800}{ 4 } \color{#FF6800}{ \times } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 25 } \right )$
Multiply $- 4$ and $- 25$
$D = \color{#FF6800}{ 100 }$
$\color{#FF6800}{ D } = \color{#FF6800}{ 100 }$
Since $D>0$ , the number of real root of the following quadratic equation is 2
2 real roots
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# Unraveling the Mystery: Counting the Lines of Symmetry in a Square
SECTIONS
As a mathematics student, symmetry is one of the most important concepts I learned. Symmetry is a fundamental mathematical concept with many applications in real life. In this article, I will explore the concept of lines of symmetry in a square. We will start by defining a line of symmetry and then look at how to determine the lines of symmetry in a square. We will also look at the properties of squares, their lines of symmetry, and how to teach lines of symmetry in the classroom.
## Introduction to Lines of Symmetry
A line of symmetry is a line that divides an object into two parts that are mirror images of each other. In other words, if you fold the object along the line of symmetry, the two parts will overlap exactly. Lines of symmetry are important in mathematics, especially in geometry and algebra. They are used to describe the properties of shapes and to solve equations.
## Definition of a Square
A square is a two-dimensional shape with four equal sides and four right angles. It is a special type of rectangle, where all four sides are equal. A square is a regular polygon, meaning all its sides and angles are equal. Squares have many properties that make them useful in real life. For example, squares are used in construction to create right angles and to provide stability.
## How to Determine the Lines of Symmetry in a Square
To determine the lines of symmetry in a square, you need to visualize the square in your mind. Imagine folding the square along a line and seeing if both parts match exactly.
• The first line of symmetry is easy to find – the line that passes through the center of the square and bisects the top and bottom sides.
• The second line of symmetry is the line that passes through the center of the square and bisects the left and right sides.
• The third and fourth lines of symmetry are the lines that pass through the center of the square and bisect the two pairs of opposite corners.
## Examples of Lines of Symmetry in a Square
Let’s look at some examples of lines of symmetry in a square. Imagine a square drawn on a piece of paper.
• The first line of symmetry is the line that passes through the center of the square and bisects the top and bottom sides.
• The second line of symmetry is the line that passes through the center of the square and bisects the left and right sides.
• The third and fourth lines of symmetry are the lines that pass through the center of the square and bisect the two pairs of opposite corners.
## Comparison with Other Shapes: How Many Lines of Symmetry Does a Rectangle Have?
A rectangle is a four-sided shape with equal and parallel opposite sides. Unlike a square, a rectangle does not have equal sides and angles. However, it does have lines of symmetry. A rectangle has two lines of symmetry: the lines that pass through the center of the rectangle and bisect opposite sides. Each line of symmetry divides the rectangle into two congruent parts that are mirror images of each other.
## Properties of Squares and Their Lines of Symmetry
Squares have many properties that make them useful in real life. One of a square’s most important properties is its four lines of symmetry. This means that any multiple of 90 degrees can rotate a square and still look the same. Another important property of squares is that they have equal sides and angles. This makes them useful in construction, where right angles and equal sides are necessary.
## Applications of Lines of Symmetry in Real Life
Lines of symmetry have many applications in real life. They are used in art, architecture, and design. For example, lines of symmetry are used in painting to create balance and harmony. Architects use lines of symmetry to create stable and aesthetically pleasing buildings. Designers use lines of symmetry to create visually appealing and functional products.
## Teaching Lines of Symmetry in the Classroom
Teaching lines of symmetry in the classroom can be fun and engaging. One way to teach lines of symmetry is to use manipulatives, such as pattern blocks or tangrams. Students can create shapes and then fold them along a line to see if they have lines of symmetry. Another way to teach lines of symmetry is to use technology, such as interactive whiteboards or tablets. Students can draw shapes and then use the technology to rotate them to see if they have lines of symmetry.
## Conclusion: The Importance of Understanding Lines of Symmetry in a Square
In conclusion, understanding the lines of symmetry in a square is an important mathematical concept. Lines of symmetry are used to describe shapes’ properties and solve equations. Squares have four lines of symmetry, which are the lines that pass through the center of the square and bisect opposite sides. Understanding lines of symmetry has many applications in real life, such as art and architecture.
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# prime factors of 3375
prime factors are the most important concepts in all of mathematics. They are some of the most important and most essential concepts in all of mathematics.
The first factor is obviously the number itself. The second factor is the number of times that the number is prime, or it is divisible by itself. The third factor is the number of factors of a number. The fourth factor is the number of prime factors of a number. The fifth factor is the number of prime factors of a number divided by the number itself.
The number itself is the most basic number. The number of times it is prime is the second most basic number. The number of factors is the third most basic number. The number of prime factors is the fourth most basic number. The number of prime factors divided by the number itself is the fifth most basic number.
prime factors of a number are a good way to figure out how many ways you could have a combination of different prime numbers. In this case, it’s a good way to figure out how many different prime numbers you can get by splitting the number into prime factors and then dividing by the number itself. And it’s the fifth most basic number, so it’s not actually that many numbers.
The best way to figure out the prime factors of 3375 is to divide all of the prime factors by a number, then divide the number by the number itself, and divide the number by the prime factor. When you do that, you’re doing a bunch of work, but you’re also doing a bunch of work, and it’s just a bunch of work for the prime factors.
So what? No need to split the number into prime factors, just divide the number by the number itself and then divide the number by the prime factor. It just makes it easier to work through.
The first thing you need to do is to divide a number by the number itself, then divide the number by the prime factor and then divide the number by the prime factor, as a whole. This is where the prime factor matters because it’s the number itself. So in the first place, in the first place, you need to divide the number by the prime factor, and then you need to divide the number by the prime factor, which is the number itself.
That’s the first thing you need to do. Then you can divide the number by the number itself, and that is the start. The second thing to do is to divide the number by the prime factor and the number itself, which is the second thing I want to do. Now divide the number by the number itself. That is the third thing I want to do.
The third thing I want to do is the number itself. That is the fourth thing I want to do. Thats the fifth thing I want to do.
Here are the prime factors of 3375. It’s the number itself. That’s the first thing I want to do. Then I want to divide the number by the prime factor. That’s the second thing I want to do. That’s the third thing I want to do. Then I want to divide the number by the number itself. That is the fourth thing I want to do. |
# Divergence Theorem
Fountain Flux
Field: Calculus
Image Created By: Brendan John
Website: [1]
Fountain Flux
The water flowing out of a fountain demonstrates an important property of vector fields, the Divergence Theorem.
# Basic Description
Consider a fountain like the one pictured, particularly its top layer. The rate that water flows out of the fountain's spout is directly related to the amount of water that flows off the top layer. Because something like water isn't easily compressed like air, if more water is pumped out of the spout, then more water will have to flow over the boundaries of the top layer. This is essentially what The Divergence Theorem states: the total the fluid being introduced into a volume is equal to the total fluid flowing out of the boundary of the volume.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *Some multivariable calculus
The Divergence Theorem in its pure form applies to Vector Fields. Flowing water can be considere [...]
The Divergence Theorem in its pure form applies to Vector Fields. Flowing water can be considered a vector field because at each point the water has a position and a velocity vector. Faster moving water is represented by a larger vector in our field. The divergence of a vector field is a measurement of the expansion or contraction of the field; if more water is being introduced then the divergence is positive. Analytically divergence of a field $F$ is
$\nabla\cdot\mathbf{F} =\partial{F_x}/\partial{x} + \partial{F_y}/\partial{y} + \partial{F_z}/\partial{z}$,
where $F _i$ is the component of $F$ in the $i$ direction. Intuitively, if F has a large positive rate of change in the x direction, the partial derivative with respect to x in this direction will be large, increasing total divergence. The divergence theorem requires that we sum divergence over an entire volume. If this sum is positive, then the field must indicate some movement out of the volume through its boundary, while if this sum is negative, the field must indicate some movement into the volume through its boundary. We use the notion of flux, the flow through a surface, to quantify this movement through the boundary, which itself is a surface.
The divergence theorem is formally stated as:
$\iiint\limits_V\left(\nabla\cdot\mathbf{F}\right)dV=\iint\limits_{\partial V}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\subset\!\supset \mathbf F\;\cdot\mathbf n\,{d}S .$
The left side of this equation is the sum of the divergence over the entire volume, and the right side of this equation is the sum of the field perpendicular to the volume's boundary at the boundary, which is the flux through the boundary. |
# HW8sol - M408C Homework 8 Solution Instructor Guoliang Wu 1...
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M408C, Homework 8 — Solution Instructor: Guoliang Wu 1. A fence 8 ft tall runs parallel to a tall building at a distance of 4 ft from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building? Solution: Suppose the bottom of the ladder is x ft away from the fence, and its top is y ft from the ground. Then, in terms of x and y , the length of the ladder is x 2 + y 2 . But 8 y = x x + 4 y = 8( x + 4) x . Therefore, the square f ( x ) of the length is f ( x ) = ( x + 4) 2 + ( 8( x + 4) x ) 2 = ( x + 4) 2 + 64 x - 2 ( x + 4) 2 , x > 0 To minimize the length of ladder, it’s the same to minimize its square f ( x ). f 0 ( x ) = 2( x + 4) + 128 x - 2 ( x + 4) - 128 x - 3 ( x + 4) 2 = 2 x - 3 ( x + 4)( x 3 - 256) = 0
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Unformatted text preview: ⇒ x = 4 3 √ 4 . 1 If x < 4 3 √ 4, f ( x ) < 0, and if x > 4 3 √ 4, f ( x ) > 0. Thus, f ( x ) has an absolute minimum at x = 4 3 √ 4. In this case the length of the ladder is √ f ( x ) = ± (4 3 √ 4 + 4) 2 + 64(4 3 √ 4)-2 (4 3 √ 4 + 4) 2 ≈ 14 . 5ft 2. Find f ( θ ) that satisfies f 00 ( θ ) = sin θ + cos θ, f (0) = 3 , f (0) = 4 . Solution: f ( θ ) =-cos θ + sin θ + C 1 f ( θ ) =-sin θ-cos θ + C 1 θ + C 2 Since f (0) =-1 + C 2 = 3, C 2 = 4. And f (0) =-1 + C 1 = 4, C 1 = 5. Therefore, f ( θ ) =-sin θ-cos θ + 3 θ + 4 . 2...
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• Fall '06
• Calculus, 8 FT, 4 ft, ladder, ft tall runs, Guoliang Wu
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# Dividing a Quantity in a Given Ratio
The term “dividing a quantity in a given ratio” is just another way of saying we’re splitting something, such as money, sweets, or even time, into parts according to a specific rule or proportion.
Before we look at it in more detail, let’s quickly revisit what a ratio is. A ratio is a simple way to compare two or more values. For example, if you and a friend share a bag of 12 sweets in a ratio, you’d get three times as many sweets as your friend.
Dividing a quantity in a given ratio requires three main steps:
1. Setting up the problem
2. Adding up the total parts
3. Dividing the quantity
## Setting up the Problem
Whenever you’re faced with a problem of dividing a quantity in a given ratio, the first step is to understand the total quantity that you have and how you need to divide it.
For example, if you’re sharing £60 between you and a friend in a ratio of , the total quantity is £60, and the ratio is .
## Adding up the Total parts
The next step is to add up the total parts. You can do this by adding all the numbers in the ratio together.
In our example of £60 in a ratio, you’d add to get 5. This common factor tells us how many parts the total quantity will be divided into.
## Dividing the Quantity
Now comes the division. To find out how much each of you will get, you’ll divide the total quantity (£60) by the common factor (5). .
Then, you multiply this answer by each part of the ratio to find out how much each person gets. You’ll get and your friend will get .
## Practice Questions
Question 1:
Divide £40 in a ratio.
Then,
Now, and
Question 2:
Share 30 sweets in a ratio.
Then,
Now, and
Question 3:
Divide £100 in a ratio.
Now, , , and
Question 4:
Share 50 sweets in a 2:2:1 ratio.
Then,
Now, , , and |
# Math Snap
## Which of the following ordered pairs satisfies the system? $\begin{array}{l} 3 x-5 y \leq 15 \\ 8 x+5 y>24 \end{array}$ A) $(0,3)$ B) $(3,-3)$ C) $(3,0)$ D) $(3,3)$
#### STEP 1
Assumptions1. We have a system of two inequalities a) $3x -5y \leq15$ b) $8x +5y >24$ . We need to find which of the given ordered pairs satisfies both inequalities.
#### STEP 2
Let's start by checking the first ordered pair (0,). We substitute these values into both inequalities. For the first inequality$x -5y \leq15$Substitute $x =0$ and $y =$$(0) -5() \leq15$For the second inequality$8x +5y >24$Substitute $x =0$ and $y =$$8(0) +5() >24$
#### STEP 3
Now, we calculate the results of the inequalities with the substituted values. For the first inequality$3(0) -5(3) = -15 \leq15$For the second inequality$8(0) +5(3) =15 >24$
#### STEP 4
The first inequality is true, but the second inequality is not. Therefore, the ordered pair (0,3) does not satisfy the system of inequalities.
#### STEP 5
Next, we check the second ordered pair (3,-3). We substitute these values into both inequalities. For the first inequality$3x -5y \leq15$Substitute $x =3$ and $y = -3$$3(3) -5(-3) \leq15$For the second inequality$8x +5y >24$Substitute $x =3$ and $y = -3$$8(3) +5(-3) >24$
#### STEP 6
Now, we calculate the results of the inequalities with the substituted values. For the first inequality$3(3) -5(-3) =24 \leq15$For the second inequality$8(3) +5(-3) =9 >24$
#### STEP 7
Both inequalities are not true. Therefore, the ordered pair (3,-3) does not satisfy the system of inequalities.
#### STEP 8
Next, we check the third ordered pair (3,0). We substitute these values into both inequalities. For the first inequality$3x -5y \leq15$Substitute $x =3$ and $y =0$$3(3) -5(0) \leq15$For the second inequality$8x +5y >24$Substitute $x =3$ and $y =0$$8(3) +5(0) >24$
#### STEP 9
Now, we calculate the results of the inequalities with the substituted values. For the first inequality$3(3) -5() =9 \leq15$For the second inequality$8(3) +5() =24 >24$
#### STEP 10
The first inequality is true, but the second inequality is not. Therefore, the ordered pair (3,0) does not satisfy the system of inequalities.
#### STEP 11
Finally, we check the fourth ordered pair (3,3). We substitute these values into both inequalities. For the first inequality$3x -5y \leq15$Substitute $x =3$ and $y =3$$3(3) -5(3) \leq15$For the second inequality$8x +5y >24$Substitute $x =3$ and $y =3$$8(3) +5(3) >24$
#### STEP 12
Now, we calculate the results of the inequalities with the substituted values. For the first inequality$() -5() =0 \leq15$For the second inequality$8() +5() =39 >24$
##### SOLUTION
Both inequalities are true. Therefore, the ordered pair (3,3) satisfies the system of inequalities. The ordered pair that satisfies the system of inequalities is (3,3). |
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How to Teach Place Value in 2024
If you are an elementary teacher looking for tips and ideas for how to teach place value, then you found the right place! Learn what place value is, why it’s important, what your students need to know, and get 5 helpful tips for teaching it in a fun and engaging way. Read all about teaching place value below!
What is Place Value?
Place value is the foundation of our number system and the core of understanding mathematics. In our base ten number system, the position of a number determines its value. Each place represents ten times the value of the place to the right of it.
Why is Place Value Important?
It is important for students to learn place value because students at the elementary level need to develop a strong conceptual understanding of place value in order to grow both their math confidence and skill sets as they approach more complex numbers and situations.
What Place Value Skills Do Students Need to Know?
Below are the Common Core and TEKs that relate to place value that define what students should be able to do by the end of the school year.
Common Core Standards
Below are the CCSS related to how to teach place value.
• Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: (1.NBT.B.2)
• 10 can be thought of as a bundle of ten ones — called a “ten.” (1.NBT.B.2.A)
• The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. (1.NBT.B.2.B)
• The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones). (1.NBT.B.2.C)
• Understand that the three digits of a three-digit number represent amounts of hundreds, tens, and ones; e.g., 706 equals 7 hundreds, 0 tens, and 6 ones. Understand the following as special cases: (2.NBT.A.1)
• 100 can be thought of as a bundle of ten tens — called a “hundred.” (2.NBT.A.1.A)
• The numbers 100, 200, 300, 400, 500, 600, 700, 800, 900 refer to one, two, three, four, five, six, seven, eight, or nine hundreds (and 0 tens and 0 ones). (2.NBT.A.1.B)
• N/A
• Recognize that in a multi-digit whole number, a digit in one place represents ten times what it represents in the place to its right. For example, recognize that 700 ÷ 70 = 10 by applying concepts of place value and division. (4.NBT.A.1)
• N/A
TEKS
Below are the TEKS related to how to teach place value.
• Use concrete and pictorial models to compose and decompose numbers up to 120 in more than one way as so many hundreds, so many tens, and so many ones. (1.2.B)
• I can use models to make a number that has hundreds, tens and ones. (2.1.A)
• Use concrete and pictorial models to compose and decompose numbers up to 1,200 in more than one way as a sum of so many thousands, hundreds, tens, and ones. (2.2.A)
• Use standard, word, and expanded forms to represent numbers up to 1,200. (2.2.B)
• Describe the mathematical relationships found in the base-10 place value system through the hundred thousands place. (3.2B)
• N/A
• N/A
5 Tips for How to Teach Place Value
Below are 5 helpful tips for teaching place value to elementary students.
1. Read Aloud Picture Books that Teach Place Value
Reading aloud picture books is a great way to integrate literacy into your math block and present information in a different way. Our favorite picture books for teaching place value are A Million Dots by Andrew Clements, Monster Math by Anne Miranda and On Beyond a Million: An Amazing Math Journey by David M. Schwartz. Check out the full list of math picture books we recommend!
2. Offer Hands On Learning Experiences
Hands-on math experiences help students make connections, remember their learning, and develop a deep conceptual understanding of the content. You can make any lesson interactive and engaging by offering math manipulatives. Our favorite math manipulatives for teaching place value are base-ten blocks, number tiles and place value frames.
Teaching math vocabulary is essential for all students, but it is especially beneficial for students who speak English as a second language and students with learning differences. Key vocabulary terms for place value are digit, multi-digit, rods, units, base-ten system, place value, position, tens, ones, zero, regroup, standard form, word form, expanded form, base-ten system, base-ten model, model, tens, ones, hundreds, thousands, concrete model, pictorial model, digit, multi-digit, flats, rods, units, place position and value.
4. Give Students Opportunities to Apply Place Value to the Real World
Learning becomes more meaningful when students understand how it connects to the real world. Students are more engaged and invested in their learning. Some examples of ways we use place value in the real world are when we determine costs, weight, distances, and time. Project based learning and word problems are examples of opportunities for students to apply their learning to real world situations.
5. Encourage Parent Involvement
Parent participation in math is essential because it impacts students’ attitude toward math, proficiency levels this school year, and future success in their math education. Be sure to keep communication open with families and share ways they can support their children in their math learning. Some examples of ways they can practice place value at home are determining if they would rather have 12 m&m’s or 21 m&m’s for dessert after dinner, deciding which store has the best price for that new video game and the shortest route to the park.
In closing, we hope you found this information about how to teach place value helpful! |
# normal approximation to the binomial calculator
a. z-Test Approximation of the Binomial Test A binary random variable (e.g., a coin flip), can take one of two values. Thus, the probability that at least 10 persons travel by train is, \begin{aligned} P(X\geq 10) &= P(X\geq9.5)\\ &= 1-P(X < 9.5)\\ &= 1-P(Z < 0.68)\\ & = 1-0.7517\\ & = 0.2483 \end{aligned}. The normal approximation allows us to bypass any of these problems by working with a familiar friend, a table of values of a standard normal distribution. With continuity correction. In statistics, a binomial proportion confidence interval is a confidence interval for the probability of success calculated from the outcome of a series of success–failure experiments (Bernoulli trials).In other words, a binomial proportion confidence interval is an interval estimate of a success probability p when only the number of experiments n and the number of successes n S are known. Using the continuity correction of normal binomial distribution, the probability of getting at least 5 successes i.e., $P(X\geq 5)$ can be written as $P(X\geq5)=P(X\geq 5-0.5)=P(X\geq4.5)$. Author(s) David M. Lane. \begin{aligned} \mu&= n*p \\ &= 800 \times 0.18 \\ &= 144. If you did not have the normal area calculator, you could find the solution using a table of the standard normal distribution (a Z table) as follows: Find a Z score for 8.5 using the formula Z = (8.5 - 5)/1.5811 = 2.21. Binomial Expansion Calculator. Normal Approximation for the Poisson Distribution Calculator. c. between 5 and 10 (inclusive) persons travel by train. Using this approximation to this quantity gave us an underestimate of … \begin{aligned} \mu&= n*p \\ &= 600 \times 0.1667 \\ &= 100.02. Binomial probabilities with a small value for $$n$$(say, 20) were displayed in a table in a book. Now, we compare this value with the exact answer for this problem. \end{aligned} $$. Normal Approximation: The normal approximation to the binomial distribution for 12 coin flips. and Other normal approximations. a. at least 150 stay on the line for more than one minute.$$ \begin{aligned} P(X= 5) & = P(4.5 Type: 1 - pnorm(55.5, mean=50, sd=5) WHY SHOULD WE USE CONTINUITY CORRECTIONS? This video shows how to use EXCEL to calculate a Normal Approximation to Binomial Probability Distributions. Let $X$ be a binomially distributed random variable with number of trials $n$ and probability of success $p$. a. Use the normal approximation to the binomial to find the probability for an-, 10p, 0.5and X8. \end{aligned} $$,$$ \begin{aligned} \sigma &= \sqrt{n*p*(1-p)} \\ &= \sqrt{30 \times 0.6 \times (1- 0.6)}\\ &=2.6833. Given that $n =30$ and $p=0.6$. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. If Y has a distribution given by the normal approximation, then Pr(X ≤ 8) is approximated by Pr(Y ≤ 8.5). The $Z$-score that corresponds to $149.5$ is, \begin{aligned} z&=\frac{149.5-\mu}{\sigma}\\ &=\frac{149.5-144}{10.8665}\\ &\approx0.51 \end{aligned}, Thus, the probability that at least $150$ people stay online for more than one minute is, \begin{aligned} P(X\geq 150) &= P(X\geq149.5)\\ &= 1-P(X < 149.5)\\ &= 1-P(Z < 0.51)\\ & = 1-0.695\\ & \qquad (\text{from normal table})\\ & = 0.305 \end{aligned}. Using the continuity correction, the probability that at least $10$ persons travel by train i.e., $P(X\geq 10)$ can be written as $P(X\geq10)=P(X\geq 10-0.5)=P(X\geq9.5)$. 60% of all young bald eagles will survive their first flight. Let $X$ denote the number of successes in 30 trials and let $p$ be the probability of success. Use this online binomial distribution normal approximation calculator to simplify your calculation work by avoiding complexities. Given that $n =500$ and $p=0.4$. The $Z$-scores that corresponds to $209.5$ and $220.5$ are respectively, \begin{aligned} z_1&=\frac{209.5-\mu}{\sigma}\\ &=\frac{209.5-200}{10.9545}\approx0.87 \end{aligned}, \begin{aligned} z_2&=\frac{220.5-\mu}{\sigma}\\ &=\frac{220.5-200}{10.9545}\approx1.87 \end{aligned}, The probability that between $210$ and $220$ (inclusive) drivers wear seat belt is, \begin{aligned} P(210\leq X\leq 220) &= P(210-0.5 < X < 220+0.5)\\ &= P(209.5 < X < 220.5)\\ &=P(0.87\leq Z\leq 1.87)\\ &=P(Z\leq 1.87)-P(Z\leq 0.87)\\ &=0.9693-0.8078\\ &=0.1615 \end{aligned}, When telephone subscribers call from the National Magazine Subscription Company, 18% of the people who answer stay on the line for more than one minute. Let $X$ denote the number of people who answer stay online for more than one minute out of 800 people called in a day and let $p$ be the probability people who answer stay online for more than one minute. They become more skewed as p moves away from 0.5. Let $X$ denote the number of sixes when a die is rolled 600 times and let $p$ be the probability of getting six. The red curve is the normal density curve with the same mean and standard deviation as the binomial distribution. Z Score = (7 - 7) / 1.4491 = 0 To perform calculations of this type, enter the appropriate values for n, k, and p (the value of q=1 — p will be calculated and entered automatically). For example, to calculate the probability of exactly 6 successes out of 8 trials with p = 0.50, enter 6 in both the "from" and "to" fields and hit the "Enter" key. Thus $X\sim B(600, 0.1667)$. \end{aligned} $$, a. Normal Approximation to Binomial Distribution Formula Continuity correction for normal approximation to binomial distribution. As n*p = 500\times 0.4 = 200 > 5 and n*(1-p) = 500\times (1-0.4) = 300 > 5, we use Normal approximation to Binomial distribution. b. Normal Approximation to the Binomial. The normal approximation of the binomial distribution works when n is large enough and p and q are not close to zero. By continuity correction normal approximation distribution,the probability that at least 220 drivers wear a seat belt i.e., P(X\geq 220) can be written as P(X\geq220)=P(X\geq 220-0.5)=P(X\geq219.5). By continuity correction the probability that at least 20 eagle will survive their first flight, i.e., P(X\geq 20) can be written as P(X\geq20)=P(X\geq 20-0.5)=P(X \geq 19.5). Let X denote the number of drivers who wear seat beltout of 500 selected drivers and let p be the probability that a driver wear seat belt. Find the normal approximation for an event with number of occurences as 10, Probability of Success as 0.7 and Number of Success as 7. PROBLEM! c. at the most 215 drivers wear a seat belt. When we are using the normal approximation to Binomial distribution we need to make correction while calculating various probabilities. Just enter the number of occurrences, the probability of success, and number of success. The Z-scores that corresponds to 4.5 and 10.5 are respectively,$$ \begin{aligned} z_1=\frac{4.5-\mu}{\sigma}=\frac{4.5-6}{2.1909}\approx-0.68 \end{aligned} $$Without continuity correction calculation, The Z-scores that corresponds to 90 and 105 are respectively,$$ \begin{aligned} z_1&=\frac{90-\mu}{\sigma}\\ &=\frac{90-100.02}{9.1294}\\ &\approx-1.1 \end{aligned} $$,$$ \begin{aligned} z_2&=\frac{105-\mu}{\sigma}\\ &=\frac{105-100.02}{9.1294}\\ &\approx0.55 \end{aligned} $$,$$ \begin{aligned} P(90\leq X\leq 105) &=P(-1.1\leq Z\leq 0.55)\\ &=P(Z\leq 0.55)-P(Z\leq -1.1)\\ &=0.7088-0.1357\\ & \qquad (\text{from normal table})\\ &=0.5731 \end{aligned} . d. Using the continuity correction calculator, the probability that between 210 and 220 (inclusive) drivers wear seat belt is P(210\leq X\leq 220) can be written as P(210-0.5 < X < 220+0.5)=P(209.5 < X < 220.5). The actual binomial probability is 0.1094 and the approximation based on the normal distribution is 0.1059. \end{aligned}. The $Z$-score that corresponds to $4.5$ is, \begin{aligned} z=\frac{4.5-\mu}{\sigma}=\frac{4.5-6}{2.1909}\approx-0.68 \end{aligned} To analyze our traffic, we use basic Google Analytics implementation with anonymized data. By continuity correction the probability that at least 150 people stay online for more than one minute i.e., $P(X\geq 150)$ can be written as $P(X\geq150)=P(X\geq 150-0.5)=P(X \geq 149.5)$. Z Value = (7 - 7 - 0.5) / 1.4491 In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. a. Mean and variance of the binomial distribution; Normal approximation to the binimial distribution. Thus $X\sim B(500, 0.4)$. \end{aligned} $$, and standard deviation of X is d. between 210 and 220 drivers wear a seat belt. Hope you like Normal Approximation to Binomial Distribution Calculator and step by step guide with examples and calculator. Click on Theory button to read more about Normal approximation to bionomial distribution. The Z-scores that corresponds to 214.5 and 215.5 are respectively,$$ \begin{aligned} z_1&=\frac{214.5-\mu}{\sigma}\\ &=\frac{214.5-200}{10.9545}\approx1.32 \end{aligned} $$The addition of 0.5 is the continuity correction; the uncorrected normal approximation gives considerably less accurate results. © VrcAcademy - 2020About Us | Our Team | Privacy Policy | Terms of Use. Many times the determination of a probability that a binomial random variable falls within a range of values is tedious to calculate. That is Z=\frac{X-\mu}{\sigma}=\frac{X-np}{\sqrt{np(1-p)}} \sim N(0,1). Video Information Mean,σ confidence interval calculator Here n*p = 20\times 0.4 = 8 > 5 and n*(1-p) = 20\times (1-0.4) = 12 > 5, we use Normal approximation to the Binomial distribution calculation as below:$$ \begin{aligned} \mu&= n*p \\ &= 20 \times 0.4 \\ &= 8. The general rule of thumb to use normal approximation to binomial distribution is that the sample size $n$ is sufficiently large if $np \geq 5$ and $n(1-p)\geq 5$. c. Using the continuity correction normal binomial distribution, the probability that between $5$ and $10$ (inclusive) persons travel by train i.e., $P(5\leq X\leq 10)$ can be written as $P(5-0.5 < X <10+0.5)=P(4.5 < X < 10.5)$. Thus $X\sim B(20, 0.4)$. Binomial distribution is a discrete distribution, whereas normal distribution is a continuous distribution. For these parameters, the approximation is very accurate. We must use a continuity correction (rounding in reverse). Step 2 - Enter the Probability of Success (p), Step 6 - Click on “Calculate” button to use Normal Approximation Calculator. If we arbitrarily define one of those values as a success (e.g., heads=success), then the following formula will tell us the probability of getting k successes from n observations of the random If a random sample of size $n=20$ is selected, then find the approximate probability that. \begin{aligned} z_2=\frac{5.5-\mu}{\sigma}=\frac{5.5-6}{2.1909}\approx-0.23 \end{aligned}, Thus the probability of getting exactly 5 successes is Binomial distribution is most often used to measure the number of successes in a sample of size 'n' with replacement from a population of size N. It is used as a basis for the binomial test of statistical significance. \end{aligned} $$,$$ \begin{aligned} \sigma &= \sqrt{n*p*(1-p)} \\ &= \sqrt{500 \times 0.4 \times (1- 0.4)}\\ &=10.9545. Binomial distribution is a discrete distribution, whereas normal distribution is a continuous distribution. Let X ~ BINOM(100, 0.4). This website uses cookies to ensure you get the best experience on our site and to provide a comment feature. Standard Deviation = √(7 x 0.3) = 1.4491 The calculator will find the binomial and cumulative probabilities, as well as the mean, variance and standard deviation of the binomial distribution. Translate the problem into a probability statement about X. a. the probability of getting 5 successes. Using the continuity correction, the probability of getting between $90$ and $105$ (inclusive) sixes i.e., $P(90\leq X\leq 105)$ can be written as $P(90-0.5 < X < 105+0.5)=P(89.5 < X < 105.5)$. a. As $n*p = 800\times 0.18 = 144 > 5$ and $n*(1-p) = 800\times (1-0.18) = 656>5$, we use Normal approximation to Binomial distribution. Click 'Overlay normal' to show the normal approximation. To calculate the probabilities with large values of $$n$$, you had to use the binomial formula, which could be very complicated. For sufficiently large $n$, $X\sim N(\mu, \sigma^2)$. \end{aligned} $$,$$ \begin{aligned} \sigma &= \sqrt{n*p*(1-p)} \\ &= \sqrt{800 \times 0.18 \times (1- 0.18)}\\ &=10.8665. As $n*p = 30\times 0.6 = 18 > 5$ and $n*(1-p) = 30\times (1-0.6) = 12 > 5$, we use Normal approximation to Binomial distribution. b. When we are using the normal approximation to Binomial distribution we need to make continuity correction while calculating various probabilities. Binomial distribution is most often used to measure the number of successes in a sample of size 'n' with replacement from a population of size N. It is used as a basis for the binomial test of statistical significance. Using the continuity correction calculation, $P(X=5)$ can be written as $P(5-0.5 < X < 5+0.5)=P(4.5 < X < 5.5)$. Note, however, that these results are only approximations of the true binomial probabilities, valid only in the degree that the binomial variance is a close approximation of the binomial mean. There is a less commonly used approximation which is the normal approximation to the Poisson distribution, which uses a similar rationale than that for the Poisson distribution. The $Z$-scores that corresponds to $4.5$ and $5.5$ are respectively, \begin{aligned} z_1=\frac{4.5-\mu}{\sigma}=\frac{4.5-8}{2.1909}\approx-1.6 \end{aligned}, \begin{aligned} z_2=\frac{5.5-\mu}{\sigma}=\frac{5.5-8}{2.1909}\approx-1.14 \end{aligned}, Thus the probability that exactly $5$ persons travel by train is, \begin{aligned} P(X= 5) & = P(4.5 < X < 5.5)\\ &=P(z_1 < Z < z_2)\\ &=P(-1.6 < Z < -1.14)\\ &=P(Z < -1.14)-P(Z < -1.6)\\ & = 0.1271-0.0548\\ & = 0.0723 \end{aligned}. Thus $X\sim B(30, 0.6)$. Let $X$ denote the number of persons travelling by train out of $20$ selected persons and let $p$ be the probability that a person travel by train. Prerequisites. Note how well it approximates the binomial probabilities represented by the heights of the blue lines. and Click 'Show points' to reveal associated probabilities using both the normal and the binomial. • What does the normal approximation (with continuity corrections) give us? n*p and n*q and also check if these values are greater than 5, so that you can use the approximation ∴n*p = 500*0.62 ∴n*p = 310 This is the standard normal CDF evaluated at that number. Find the Probability, Mean and Standard deviation using this normal approximation calculator. The $Z$-scores that corresponds to $89.5$ and $105.5$ are respectively, \begin{aligned} z_1&=\frac{89.5-\mu}{\sigma}\\ &=\frac{89.5-100.02}{9.1294}\\ &\approx-1.15 \end{aligned}, \begin{aligned} z_2&=\frac{105.5-\mu}{\sigma}\\ &=\frac{105.5-100.02}{9.1294}\\ &\approx0.6 \end{aligned}, \begin{aligned} P(90\leq X\leq 105) &= P(90-0.5 < X < 105+0.5)\\ &= P(89.5 < X < 105.5)\\ &=P(-1.15\leq Z\leq 0.6)\\ &=P(Z\leq 0.6)-P(Z\leq -1.15)\\ &=0.7257-0.1251\\ & \qquad (\text{from normal table})\\ &=0.6006 \end{aligned}. The $Z$-score that corresponds to $214.5$ is, \begin{aligned} z&=\frac{214.5-\mu}{\sigma}\\ &=\frac{214.5-200}{10.9545}\approx1.32 \end{aligned}, Thus, the probability that at most $215$ drivers wear a seat belt is, \begin{aligned} P(X\leq 215) &= P(X\leq214.5)\\ &= P(X < 214.5)\\ &= P(Z < 1.32)\\ &=0.9066 \end{aligned}. \begin{aligned} z_2=\frac{10.5-\mu}{\sigma}=\frac{10.5-6}{2.1909}\approx2.05 \end{aligned}, \begin{aligned} P(5\leq X\leq 10) &= P(5-0.5 < X <10+0.5)\\ &= P(4.5 < X <10.5)\\ &=P(-0.68\leq Z\leq 2.05)\\ &=P(Z\leq 2.05)-P(Z\leq -0.68)\\ &=0.9798-0.2483\\ &=0.7315 \end{aligned}, In a large population 40% of the people travel by train. \begin{aligned} \mu&= n*p \\ &= 500 \times 0.4 \\ &= 200. When a healthy adult is given cholera vaccine, the probability that he will contract cholera if exposed is known to be 0.15. Using the continuity correction for normal approximation to binomial distribution, P(X=215) can be written as P(215-0.5 < X < 215+0.5)=P(214.5 < X < 215.5). a. b. at least 220 drivers wear a seat belt. Given that n =20 and p=0.4. Without continuity correction Using R to compute Q = P(35 X ≤ 45) = P(35.5 X ≤ 45.5): ... we can calculate the probability of having six or fewer infections as P (X ≤ 6) = The results turns out to be similar as the one that has been obtained using the binomial distribution. Normal Approximation to Binomial Distribution: ... Use Normal approximation to find the probability that there would be between 65 and 80 (both inclusive) accidents at this intersection in one year. (Use normal approximation to Binomial). The calculator will find the binomial expansion of the given expression, with steps shown. Weibull Distribution Examples - Step by Step Guide, Karl Pearson coefficient of skewness for grouped data, Normal Approximation to Binomial Distribution Calculator, Normal Approximation to Binomial Distribution Calculator with Examples. b. the probability of getting at least 5 successes. In a certain Binomial distribution with probability of success p=0.20 and number of trials n = 30. Suppose one wishes to calculate Pr(X ≤ 8) for a binomial random variable X. Normal approximation of binomial probabilities. Here n*p = 30\times 0.2 = 6>5 and n*(1-p) = 30\times (1-0.2) = 24>5, we use Normal approximation to Binomial distribution. a. The Z-scores that corresponds to 4.5 and 5.5 are, \begin{aligned} z_1=\frac{4.5-\mu}{\sigma}=\frac{4.5-6}{2.1909}\approx-0.68 \end{aligned} \begin{aligned} \mu&= n*p \\ &= 30 \times 0.2 \\ &= 6. \end{aligned} $$. IF np > 5 AND nq > 5, then the binomial random variable is approximately normally distributed with mean µ =np and standard deviation σ = sqrt(npq). Round z-value calculations to 2decimal places and final answer to 4 decimal places.$$ \begin{aligned} P(X= 215) & = P(214.5 < X < 215.5)\\ &=P(z_1 < Z < z_2)\\ &=P(1.32 < Z < 1.41)\\ &=P(Z < 1.41)-P(Z < 1.32)\\ & = 0.9207-0.9066\\ & = 0.0141 \end{aligned} $$. The mean of X is \mu=E(X) = np and variance of X is \sigma^2=V(X)=np(1-p). Thus, the probability that at least 150 persons travel by train is. Let X denote the number of bald eagles who survive their first flight out of 30 observed bald eagles and let p be the probability that young bald eagle will survive their first flight. Mean of X is The bars show the binomial probabilities. Mean of X is Given that n =30 and p=0.2. The Z-score that corresponds to 219.5 is,$$ \begin{aligned} z&=\frac{219.5-\mu}{\sigma}\\ &=\frac{219.5-200}{10.9545}\approx1.78 \end{aligned} $$, Thus, the probability that at least 220 drivers wear a seat belt is,$$ \begin{aligned} P(X\geq 220) &= P(X\geq219.5)\\ &= 1-P(X < 219.5)\\ &= 1-P(Z < 1.78)\\ & = 1-0.9625\\ & = 0.0375 \end{aligned} . Calculate nq to see if we can use the Normal Approximation: Since q = 1 - p, we have n(1 - p) = 10(1 - 0.4) nq = 10(0.6) nq = 6 Since np and nq are both not greater than 5, we cannot use the Normal Approximation to the Binomial Distribution.cannot use the Normal Approximation to the Binomial Distribution. In this tutorial we will discuss some numerical examples on Poisson distribution where normal approximation is applicable. The most widely-applied guideline is the following: np > 5 and nq > 5. Thus X\sim B(800, 0.18). \end{aligned}, \begin{aligned} \sigma &= \sqrt{n*p*(1-p)} \\ &= \sqrt{600 \times 0.1667 \times (1- 0.1667)}\\ &=9.1294. and, \begin{aligned} z_2&=\frac{215.5-\mu}{\sigma}\\ &=\frac{215.5-200}{10.9545}\approx1.41 \end{aligned} $$, Thus the probability that exactly 215 drivers wear a seat belt is The vertical gray line marks the mean np. If 800 people are called in a day, find the probability that.$$ \begin{aligned} \sigma &= \sqrt{n*p*(1-p)} \\ &= \sqrt{30 \times 0.2 \times (1- 0.2)}\\ &=2.1909. Given that $n =600$ and $p=0.1667$. Thus $X\sim B(30, 0.2)$. Suppose that only 40% of drivers in a certain state wear a seat belt. Five hundred vaccinated tourists, all healthy adults, were exposed while on a cruise, and the ship’s doctor wants to know if he stocked enough rehydration salts. Step 6 - Click on “Calculate” button to use Normal Approximation Calculator. Use normal approximation to estimate the probability of getting 90 to 105 sixes (inclusive of both 90 and 105) when a die is rolled 600 times. and, \begin{aligned} z_2&=\frac{10.5-\mu}{\sigma}\\ &=\frac{10.5-8}{2.1909}\approx1.14 \end{aligned}, The probability that between $5$ and $10$ (inclusive) persons travel by train is, \begin{aligned} P(5\leq X\leq 10) &= P(5-0.5 < X < 10+0.5)\\ &= P(4.5 < X <10.5)\\ &=P(-1.6\leq Z\leq 1.14)\\ &=P(Z\leq 1.14)-P(Z\leq -1.6)\\ &=0.8729-0.0548\\ &=0.8181 \end{aligned}. Given that $n =800$ and $p=0.18$. The smooth curve is the normal distribution. And we see that we again missed it. Adjust the binomial parameters, n and p, using the sliders. b. If you continue without changing your settings, we'll assume that you are happy to receive all cookies on the vrcacademy.com website. As $n*p = 600\times 0.1667 = 100.02 > 5$ and $n*(1-p) = 600\times (1-0.1667) = 499.98 > 5$, we use Normal approximation to Binomial distribution calculator. For an exact Binomial probability calculator, please check this one out, where the probability is exact, not normally approximated. c. the probability of getting between 5 and 10 (inclusive) successes. \end{aligned} $$. The Z-score that corresponds to 19.5 is,$$ \begin{aligned} z&=\frac{19.5-\mu}{\sigma}\\ &=\frac{19.5-18}{2.6833}\\ &\approx0.56 \end{aligned} $$, Thus, the probability that at least 20 eagle will survive their first flight is,$$ \begin{aligned} P(X\geq 20) &= P(X\geq19.5)\\ &= 1-P(X < 19.5)\\ &= 1-P(Z < 0.56)\\ & = 1-0.7123\\ & \qquad (\text{from normal table})\\ & = 0.2877 \end{aligned} . If 30 randomly selected young bald eagles are observed, what is the probability that at least 20 of them will survive their first flight? Excel 2010: Normal Approximation to Binomial Probability Distribution. 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SAT Sample Questions » Quantitative Section : Quantitative Ability
# Quantitative Section : Quantitative Ability
Below are the few important questions on multiple choice with answers and explanation.
1. Seven blue marbles and six red marbles are held in a single container. Marbles are randomly selected one at a time and not returned to the container. If the first two marbles selected are blue, what is the probability that at least two red marbles will be chosen in the next three selections?
1. 8/45
2. 7/13
3. 45/100
4. 19/33
5. 78/5
Explanation: As you know the probability of every outcome that produces at least two red marbles in three drawings. To find the total overall probability adds up all of the individual probabilities.
5/33 + 5/33 + 5/33 + 4/33 = 19/33
Therefore the correct option would be option (d).
2. Y varies directly as the square of x. When y = 2.5, x = 0.5 If y = 80, then x would be equal to
1. - 2 √2
2. - 85
3. -45
4. - 12
5. - 56
Explanation: If y varies directly as the square of the x, it means y / x2 will always have the same value. So set up a proportion 2.5/(0.5)2 = 80/x2. You require to cross multiply and solve x2 = 8. Therefore x = +2√2
3. If v - h > v + h, then which of the following must be true?
1. V - h
2. h < 8
3. h - 1
4. 45/8
5. 76/9
Explanation: Algebraic manipulation is the simplest way to answer this problem. You require to add to each side which will give inequality v > v + 2h. Now you are required to subtract from each side to get 0 > 2h and you will find your answer is (b).
4. If t - q > t + q, then which of the following must be true?
1. Q = t
2. t < q
3. 56 > 0
4. 37 > 9
5. rt
Explanation: You have to add 0 to each side which will give dissimilarity t > t+ 2q. Now you have to subtract from each side to get 0 > 2q and therefore the answer is (b)
5. Which of the following has the greatest value?
1. 500500
2. 90100
3. 1.528
4. 25
5. 70
Explanation: The significant fact about this question is that it is not necessary to find the correct value of the expression. You require rearranging as many answer choices as possible so that they have the exponents of 100. Take a close glance on each of the expression 500500, 25, 70, 1.528, 90100. It will make clear that (a) is bigger than (c), therefore the answer is (a)
6. What is the distance between the w intercept and the z intercept of the line given by the equation?
2z = 6 - w?
1. 7.56
2. 3.78
3. 5.81
4. 2.79
5. 3.56
Explanation: To find the z - intercept just make w = 0 and solve for z. z= 3. Now to find the w intercept, make z = 0 and solve for w. You will see that they form a right angle triangle with length of 3 and 6, in which the hypotenuse represents the distance in between the two distinct points. You are required to used Pythagoras theorem to find the length of the hypotenuse, which will be equal to 5.8156 approximately; therefore the answer is (c).
7. How many dissimilar three digit numbers include only non-zero numbers?
1. 432
2. 78
3. 325
4. 729
5. 867
Explanation: If there is a three digit no and it has the no 0. Then there are in fact nine possibilities for the first numbers from 1 to 9, 9 possibilities for the 2nd numbers from 1 to 9, 9th possibilities for the third number from 1 to 9. This comes to the total of 9*9*9 possible 3 digit numbers therefore the right answer is 729.
8. After 5:00 am ride in a bus costs \$1.60 plus \$0.50 for every sixth of a mile traveled. If the traveler travels x miles then what is the price of the trip, in dollars, in terms of x?
1. 456x
2. 2.5 + 1.5x
3. 4678x
4. 45 + 232x
5. 321 + 45x
Explanation: Suppose that you are travelling 6 miles (x = 6). To find the right answer place 5 in place of x and see which one gives you value of 9 and you will find the correct answer would be option (b).
9. If x = cos θ and y = sin θ, then for all θ, x2 + y2 =
1. 456
2. 1
3. 32
4. 21
5. 0
Explanation: Substitute for x and y. So x2 + y2 = cos2θ + sin 2θ. But you should the correct identity of cos2θ+ sin 2θ= 1. Therefore the option is (b).
10. Find the slope of the line given by the equation
b + 3 = 5/4 (a - 7)?
1. 45/2
2. 32/5
3. 33/5
4. 21/9
5. 5/4
Explanation: You may see that this is the point slope form and that the slope is 5/4. If you don't see that rewrite the equations into slope form.
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# In a certain area, there are 35 houses to 55 businesses. How do you write the ratio of houses to businesses as a fraction in simplest form. Then explain its meaning?
Nov 4, 2016
There are $\frac{7}{11}$ times as many houses as businesses.
#### Explanation:
$\text{houses":"businesses} = 35 : 55$
This ratio can also be written in fractional form as:
$\textcolor{w h i t e}{\text{XXX}} \frac{35}{55}$
which can be simplified (by dividing each of the numerator and denominator by $5$)
$\textcolor{w h i t e}{\text{XXX}} = \frac{7}{11}$
If you divided the businesses up into (5 groups of) $11$ and then evenly assigned houses among those groups, each group would have $7$ houses. |
# SOHCAHTOA
SOHCAHTOA (pronounced: so-cah-tow-ah) is a very useful mnemonic that can help students remember the sine ratio, cosine ratio, and the tangent ratio of the three basic trigonometric functions sine, cosine, and tangent.
SOH : Sine = Opposite / Hypotenuse
CAH : Cosine = Adjacent / Hypotenuse
TOA : Tangent = Opposite / Adjacent
As you can see in the image below, the sides of a right triangle are the opposite side, the adjacent side, and the hypotenuse.
• The opposite side of a right triangle is the side opposite the angle θ.
• The adjacent side of a right triangle is the side adjacent or next to the angle θ.
• The hypotenuse of a right triangle is the longest side or the side opposite the right angle.
Study the nice figure shown above carefully! It will help you greatly to remember the acronym sohcahtoa, which is a memory aid that can help you the following important ratios
sin θ = opposite side / hypotenuse
cos θ = adjacent side / hypotenuse
tan θ = opposite / adjacent
You could also use the following mnemonic to remember the trigonometric ratios.
" Some Old Hog Came Around Here and Took Our Apples."
## A couple of examples showing how to use sohcahtoa to evaluate trigonometric functions
Example #1
Use the triangle below to find the values of sin θ, cos θ, and tan θ
sin(θ) = opposite side / hypotenuse = 6/10 = 0.6
cos(θ) = adjacent side / hypotenuse= 8/10 = 0.8
tan(θ) = opposite / adjacent = 6/8 = 0.75
Example #2
Use the triangle below to find the values of sin θ, cos θ, tan θ, csc θ, sec θ, and cot θ
First, we need to find the length of the hypotenuse using the Pythagorean theorem.
c2 = 152 + 82
c2 = 225 + 64
c2 = 289
c = √289
c = 17
sin(θ) = opposite side / hypotenuse = 8/17
cos(θ) = adjacent side / hypotenuse = 15/17
tan(θ) = opposite / adjacent = 8/15
csc(θ) = hypotenuse / opposite side = 17/8
sec(θ) = hypotenuse / adjacent side = 17/15
cot(θ) = adjacent / opposite = 15/8
### SOHCAHTOAFAQs
No, SOHCAHTOA is not a real word. It is like an abbreviation formed from the initial letters of the words (sine opposite hypotenuse),(cosine adjacent hypotenuse), and (tangent opposite adjacent).
Yes, you can only use SOHCAHTOA when solving right triangles. If the triangle is not a right triangle, you can use the law of sines and/or the law of cosines.
The S in SOH stands for sine.
You can use SOHCAHTOA to solve a right triangle. You solve a right triangle by looking for missing sides and/or missing angles.
SOHCAHTOA is not a formula. It is a way to remember the right triangle definitions or the trigonometric ratios of sine, cosine, and tangent.
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# Tables of 2 to 30
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Last updated date: 13th Sep 2024
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## About the Topic of Tables of 2 to 30
Multiplication Tables, commonly known as Tables, are the tabular form of the Multiplication of Numbers. Tables form the base of Mathematics. Learning Tables are important. Tables make the most complex Multiplication, Dvision, Factorisation, and other calculations easy. Tables make our daily routine Maths calculations easier. Let’s know more about the importance of Tables in Mathematics.
### Tables 2 to 25:
We have tried to put tables from 2 to 30 here so that you can easily access the tables you require.
The very first table to learn is table of 2. It is very easy to learn and remember life long.
### Table of 2:
2 x 1 = 2 2 x 2 = 4 2 x 3 = 6 2 x 4 = 8 2 x 5 = 10 2 x 6 = 12 2 x 7 = 14 2 x 8 = 16 2 x 9 = 18 2 x 10 = 20
The next easiest table to remember is table of 3. It is very important to memorize it.
### Table of 3:
3 x 1 = 3 3 x 2 = 6 3 x 3 = 9 3 x 4 = 12 3 x 5 = 15 3 x 6 = 18 3 x 7 = 21 3 x 8 = 24 3 x 9 = 27 3 x 10 = 30
Here is the table of 4. Table of 4 can be easily memorized by keep on adding the number 4 at every step.
### Table of 4:
4 x 1 = 4 4 x 2 = 8 4 x 3 = 12 4 x 4 = 16 4 x 5 = 20 4 x 6 = 24 4 x 7 = 28 4 x 8 = 32 4 x 9 = 36 4 x 10 = 40
Table of 5 is a fun table for kids and easily remembered by them. Just add 5 at every step. And the answer will be in 5s and 0s.
### Table of 5:
5 x 1 = 5 5 x 2 = 10 5 x 3 = 15 5 x 4 = 20 5 x 5 = 25 5 x 6 = 30 5 x 7 = 35 5 x 8 = 40 5 x 9 = 45 5 x 10 = 50
Tables of six have the values fixed. Memorizing it will be beneficial at every step of your life.
### Table of 6:
6 x 1 = 6 6 x 2 = 12 6 x 3 = 18 6 x 4 = 24 6 x 5 = 30 6 x 6 = 36 6 x 7 = 42 6 x 8 = 48 6 x 9 = 54 6 x 10 = 60
Add 7 repeatedly and you will get the multiplication table of 7. Memorizing table of 7 is equally important.
### Table of 7:
7 x 1 = 7 7 x 2 = 14 7 x 3 = 21 7 x 4 = 28 7 x 5 = 35 7 x 6 = 42 7 x 7 = 49 7 x 8 = 56 7 x 9 = 63 7 x 10 = 70
Table 8 is one of the toughest tables as its values are quite difficult to remember. Here is the table of 8 to memorize it and make it easier.
### Table of 8:
8 x 1 = 8 8 x 2 = 16 8 x 3 = 24 8 x 4 = 32 8 x 5 = 40 8 x 6 = 48 8 x 7 = 56 8 x 8 = 64 8 x 9 = 72 8 x 10 = 80
One more toughest table. The values of this table are quite difficult to remember. But once it is memorized you will never forget.
### Table of 9:
9 x 1 = 9 9 x 2 = 18 9 x 3 = 27 9 x 4 = 36 9 x 5 = 45 9 x 6 = 54 9 x 7 = 63 9 x 8 = 72 9 x 9 = 81 9 x 10 = 90
The most easiest table, and easily memorized by the kids is table of 10. Memorizing the table of 10 is fun for kids.
### Table of 10:
10 x 1 = 10 10 x 2 = 20 10 x 3 = 30 10 x 4 = 40 10 x 5 = 50 10 x 6 = 60 10 x 7 = 70 10 x 8 = 80 10 x 9 = 90 10 x 10 = 100
Table of 11 is also called a twin table as the values are in twin format. Easy to remember.
### Table of 11:
11 x 1 = 11 11 x 2 = 22 11 x 3 = 33 11 x 4 = 44 11 x 5 = 55 11 x 6 = 66 11 x 7 = 77 11 x 8 = 88 11 x 9 = 99 11 x 10 = 110
Table 12 is the toughest table as its values are quite difficult to remember. Here is the table of 12 in tabular form to memorize it and make it easier.
### Table of 12:
12 x 1 = 12 12 x 2 = 24 12 x 3 = 36 12 x 4 = 48 12 x 5 = 60 12 x 6 = 72 12 x 7 = 84 12 x 8 = 96 12 x 9 = 108 12 x 10 = 120
Memorizing tables of 13 makes the students solve their problems more quickly and with ease. It will help in competitive exams.
### Table of 13:
13 x 1 = 13 13 x 2 = 26 13 x 3 = 39 13 x 4 = 52 13 x 5 = 65 13 x 6 = 78 13 x 7 = 91 13 x 8 = 104 13 x 9 = 117 13 x 10 = 130
Table 14 is the toughest table as its values are quite difficult to remember. Here is the table of 14 in tabular form to memorize it and make it easier.
### Table of 14:
14 x 1 = 14 14 x 2 = 28 14 x 3 = 42 14 x 4 = 56 14 x 5 = 70 14 x 6 = 84 14 x 7 = 98 14 x 8 = 112 14 x 9 = 126 14 x 10 = 140
Memorize the table of 15 and solve your problems more quickly and easily. Keep on adding 15 and you will get the table of 15.
### Table of 15:
15 x 1 = 15 15 x 2 = 30 15 x 3 = 45 15 x 4 = 60 15 x 5 = 75 15 x 6 = 90 15 x 7 = 105 15 x 8 = 120 15 x 9 = 135 15 x 10 = 150
Here is the table of 16 to make you solve your multiplcation problems more quickly.
### Table of 16:
16 x 1 = 16 16 x 2 = 32 16 x 3 = 48 16 x 4 = 64 16 x 5 = 80 16 x 6 = 96 16 x 7 = 112 16 x 8 = 128 16 x 9 = 144 16 x 10 = 160
Here is the tabular form of table 17. It will make you solve your problems quickly. Memorizing these tables will minimize your time in the examination needed to solve the multiplication problems.
### Table of 17:
17 x 1 = 17 17 x 2 = 34 17 x 3 = 51 17 x 4 = 68 17 x 5 = 85 17 x 6 = 102 17 x 7 = 119 17 x 8 = 136 17 x 9 = 153 17 x 10 = 170
Table 18 is the toughest table as its values are quite difficult to remember. Here is the table of 18 in tabular form to memorize it and make it easier.
### Table of 18:
18 x 1 = 18 18 x 2 = 36 18 x 3 = 54 18 x 4 = 72 18 x 5 = 90 18 x 6 = 108 18 x 7 = 126 18 x 8 = 144 18 x 9 = 162 18 x 10 = 180
Table 19 is the toughest table as its values are quite difficult to remember. Here is the table of 19 in tabular form to memorize it and make it easier.
### Table of 19:
19 x 1 = 19 19 x 2 = 38 19 x 3 = 57 19 x 4 = 76 19 x 5 = 95 19 x 6 = 114 19 x 7 = 133 19 x 8 = 152 19 x 9 = 171 19 x 10 = 190
The easiest table is the table of 20. Just keep on adding 20 and you will get your answers.
### Table of 20:
20 x 1 = 20 20 x 2 = 40 20 x 3 = 60 20 x 4 = 80 20 x 5 = 100 20 x 6 = 120 20 x 7 = 140 20 x 8 = 160 20 x 9 = 180 20 x 10 = 200
Here is the tabular form of table 21. It will make you solve your problems quickly. Memorizing this tables will minimize your time in the examination needed to solve the multiplication problems.
### Table of 21:
21 x 1 = 21 21 x 2 = 42 21 x 3 = 63 21 x 4 = 84 21 x 5 = 105 21 x 6 = 126 21 x 7 = 147 21 x 8 = 168 21 x 9 = 189 21 x 10 = 210
Here is the tabular form of table 22. It will make you solve your problems quickly. Memorizing this tables will minimize your time in the examination needed to solve the multiplication problems.
### Table of 22:
22 x 1 = 22 22 x 2 = 44 22 x 3 = 66 22 x 4 = 88 22 x 5 = 110 22 x 6 = 132 22 x 7 = 154 22 x 8 = 176 22 x 9 = 198 22 x 10 = 220
Here is the tabular form of table 23. It will make you solve your problems quickly. Memorizing this tables will minimize your time in the examination needed to solve the multiplication problems.
### Table of 23:
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
Here is the tabular form of table 24. It will make you solve your problems quickly. Memorizing this tables will minimize your time in the examination needed to solve the multiplication problems.
### Table of 24:
24 x 1 = 24 24 x 2 = 48 24 x 3 = 72 24 x 4 = 96 24 x 5 = 120 24 x 6 = 144 24 x 7 = 168 24 x 8 = 192 24 x 9 = 216 24 x 10 = 240
Here is the tabular form of table 21. It will make you solve your problems quickly. Memorizing this tables will minimize your time in the examination needed to solve the multiplication problems.
### Table of 25:
25 x 1 = 25 25 x 2 = 50 25 x 3 = 75 25 x 4 = 100 25 x 5 = 125 25 x 6 = 150 25 x 7 = 175 25 x 8 = 200 25 x 9 = 225 25 x 10 = 250
Here is the tabular form of table 26. It will make you solve your problems quickly. Memorizing this tables will minimize your time in the examination needed to solve the multiplication problems.
### Table of 26:
26 x 1 = 26 26 x 2 = 52 26 x 3 = 78 26 x 4 = 104 26 x 5 = 130 26 x 6 = 156 26 x 7 = 182 26 x 8 = 208 26 x 9 = 234 26 x 10 = 260
Here is the tabular form of table 21. It will make you solve your problems quikly. Memorizing this tables will minimize your time in the examination needed to solve the multiplication problems.
### Table of 27:
27 x 1 = 27 27 x 2 = 54 27 x 3 = 81 27 x 4 = 108 27 x 5 = 135 27 x 6 = 162 27 x 7 = 189 27 x 8 = 216 27 x 9 = 243 27 x 10 = 270
Here is the tabular form of table 21. It will make you solve your problems quickly. Memorizing this tables will minimize your time in the examination needed to solve the multiplication problems.
### Table of 28:
28 x 1 = 28 28 x 2 = 56 28 x 3 = 84 28 x 4 = 112 28 x 5 = 140 28 x 6 = 168 28 x 7 = 196 28 x 8 = 224 28 x 9 = 252 28 x 10 = 280
Here is the tabular form of table 21. It will make you solve your problems quickly. Memorizing this tables will minimize your time in the examination needed to solve the multiplication problems.
### Table of 29:
29 x 1 = 29 29 x 2 = 58 29 x 3 = 87 29 x 4 = 116 29 x 5 = 145 29 x 6 = 174 29 x 7 = 203 29 x 8 = 232 29 x 9 = 261 29 x 10 = 290
Here is the tabular form of table 21. It will make you solve your problems quickly. Memorizing this tables will minimize your time in the examination needed to solve the multiplication problems.
### Table of 30:
30 x 1 = 30 30 x 2 = 60 30 x 3 = 90 30 x 4 = 120 30 x 5 = 150 30 x 6 = 180 30 x 7 = 210 30 x 8 = 240 30 x 9 = 270 30 x 10 = 300
### Introduction:
Multiplication Tables make up most of the childhood of a student. This is because Multiplication Tables are considered to be not only the basics of Maths but are also the backbone of any calculations that are required in Mathematics. It was during the primary classes that students were allowed to study the Multiplication Tables from 2 to 10 which then went on increasing with the academic levels. While learning and memorizing the Tables of 2 to 30 might seem like a very hard task, it is actually not all that hard. Instead, it takes only some concepts and tricks to learn all of the Tables from 2 to 30. Vedantu also provides a detailed guide on how to learn all of these concepts and remembers some really handy tips and tricks that will allow all the students to learn the topics well.
### Tips to remember Tables Effectively:
1. The use of flashcards not only helps students learn daily from them but will also help them when they need to revise the Tables before the exam.
2. For small children who would like to learn the Tables, parents can use two dice and then ask them what is the answer to the product of the two Numbers shown. In this way, students can learn in a fun way.
3. If you would like to add some fun physical activities then it will help greatly as well. While students get to maintain their health they also get to train their brains.
## FAQs on Tables of 2 to 30
1. How to Memorize Tables?
• Make the multiplication concept clear
• Go one by one
• Make a chart of a single table and stick somewhere where you can have a sight daily.
• Recite it 5 times daily.
• Write it two times daily.
• When you're perfect at it move to the next table.
2. What is the Importance of Tables in Day to Day Life?
Tables are the base of mathematics. Perfection in tables will help a student to solve their problem quickly with more perfection and ease. Students who don’t know tables find it difficult to cope with mathematics. And they will always have fear for maths.
3. What is the history of Tables in Mathematics?
It was seen that the oldest Mathematical Table was found in the ancient Sumerian city in Iraq around 4,000 years ago. Since then Mathematical Tables have been one of the methods to find out various measures and were also important precursors for modern computing and information processing. It was later that the Chinese Mathematicians invented a base-ten method that involved using the bamboo and forming thin strips out of them. This allowed them to calculate using the bamboo strips on a Table-like structure.
4. Why is it very important to memorize the concepts involved in Tables and use them?
More than memorizing the concepts involved in the Tables it is more important to understand the concepts that are involved. By understanding the concepts it is possible to learn and practice the Multiplication Tables effortlessly. While memorizing it can be a bit hard however with some tips from Vedantu you can definitely tackle them. Multiplication Tables from 2 to 30 can be highly helpful for all the students in their higher grades when they need to solve Math problems and calculations. It is hence quite important that students learn and memorize the concepts behind Tables and study the basics.
5. Where is the concept of Tables applicable for and how will it help students learn better?
The concept of Tables is applicable not only in Maths but in a variety of subjects. Tables can hence be said to be a base of each and every subject. For example, during high school classes students will have to calculate the amount of force that is being applied to a football. Here the student might get a formula but there is also Multiplication being used as a part of the basic calculation. This means that even though force is a topic that is studied through Physics, the application of Tables is essential when trying to find the right answer. Not only are the subjects the only place where you need to apply Maths. Even during your daily life, you will come across such circumstances where you need to use Maths.
6. Why do a lot of students find it hard to memorize the Tables of 2 to 30 and how can it be solved?
It is definitely not easy to remember all the Tables from 2 to 30 in just one sitting. However, what students can do is try and break down the Tables into easy concepts. For example 2 X 5 = 10 also 5 X 2 = 10 which means that either of the ways the answer will be 10. Comparing the two Numbers which are Multiplied and those that give you the same answer can be grouped together to learn efficiently. While this may not work for some students, others can try this while those who still find it hard to memorize can use tips from Vedantu experts through live classes as well!
7. Does Vedantu help students memorize the Tables of 2 to 30 and provide any practice to apply the same?
Yes. Vedantu provides efficient tips and tricks that will help you understand and memorize the Tables from 2 to 30 within just moments. However, to remember it for a long time, it will be necessary that you keep on practising all the Tables and not use the calculator. Because eventually, the use of a calculator makes it easier and then you tend to forget all the Tables that are learned. Hence practising and reciting the Tables in a song format will help you remember the Tables for a long time. You can also use some Vedantu sample papers to practice some Math questions based on Tables. |
Equations with Variables on Both Sides
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# Equations with Variables on Both Sides - PowerPoint PPT Presentation
ALGEBRA 1 LESSON 2-4. Equations with Variables on Both Sides. (For help, go to Lessons 1-7 and 2-3.). Simplify. 1. 6 x – 2 x 2. 2 x – 6 x 3. 5 x – 5 x 4. –5 x + 5 x Solve each equation. 5. 4 x + 3 = –5 6. – x + 7 = 12 7. 2 t – 8 t + 1 = 43 8. 0 = –7 n + 4 – 5 n. 2-4.
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ALGEBRA 1 LESSON 2-4
Equations with Variables on Both Sides
(For help, go to Lessons 1-7 and 2-3.)
Simplify.
1. 6x – 2x2. 2x – 6x3. 5x – 5x4. –5x + 5x
Solve each equation.
5. 4x + 3 = –5 6. –x + 7 = 12
7. 2t – 8t + 1 = 43 8. 0 = –7n + 4 – 5n
2-4
ALGEBRA 1 LESSON 2-4
Equations with Variables on Both Sides
Solutions
1. 6x – 2x = (6 – 2)x = 4x2. 2x – 6x = (2 – 6)x = –4x
3. 5x – 5x = (5 – 5)x = 0x = 0 4. –5x + 5x = (–5 + 5)x = 0x = 0
5. 4x + 3 = –5 6. –x + 7 = 12
4x = –8 –x = 5
x = –2 x = –5
7. 2t – 8t + 1 = 43 8. 0 = –7n + 4 – 5n
–6t + 1 = 43 0 = –12n + 4
–6t = 42 12n = 4
t = –7 n =
1
3
2-4
5x – 3 – 2x = 2x + 12 – 2xSubtract 2x from each side.
3x – 3 = 12 Combine like terms.
3x – 3+ 3 = 12 + 3Add 3 to each side.
3x = 15 Simplify.
= Divide each side by 3.
3x3
153
x = 5 Simplify.
ALGEBRA 1 LESSON 2-4
Equations with Variables on Both Sides
The measure of an angle is (5x – 3)°. Its vertical angle has a measure of (2x + 12)°. Find the value of x.
5x – 3 = 2x + 12 Vertical angles are congruent.
2-4
friend’s rental rental
skateboard
Define: let h = the number of hours you must skateboard
Write: 60 + 1.5 h = 5.5 h
ALGEBRA 1 LESSON 2-4
Equations with Variables on Both Sides
You can buy a skateboard for \$60 from a friend and rent the safety equipment for \$1.50 per hour. Or you can rent all items you need for \$5.50 per hour. How many hours must you use a skateboard to justify buying your friend’s skateboard?
2-4
60 = 4hCombine like terms.
604
4h4
= Divide each side by 4.
15 = hSimplify.
ALGEBRA 1 LESSON 2-4
Equations with Variables on Both Sides
(continued)
60 + 1.5h = 5.5h
You must use your skateboard for more than 15 hours to justify buying the skateboard.
2-4
ALGEBRA 1 LESSON 2-4
Equations with Variables on Both Sides
Solve each equation.
a. –6z + 8 = z + 10 – 7z
–6z + 8 = z + 10 – 7z
–6z + 8 = –6z + 10 Combine like terms.
–6z + 8 + 6z = –6z + 10 + 6zAdd 6z to each side.
8 = 10 Not true for any value of z!
This equation has no solution
b. 4 – 4y = –2(2y – 2)
4 – 4y = –2(2y – 2)
4 – 4y = –4y + 4Use the Distributive Property.
4 – 4y + 4y = –4y + 4 + 4yAdd 4y to each side.
4 = 4 Always true!
The equation is true for every value of y, so the equation is an identity.
2-4 |
Hopping Up & Down the Number Line
1 teachers like this lesson
Print Lesson
Objective
SWBAT identify the subtraction and addition sign and understand their function by hopping a bunny up and down a number line.
Big Idea
Kinders get the functions of addition and subtraction confused! This lesson helps kids remember what they are supposed to do when they see an addition or subtraction sign.
Daily Calendar & Counting Review
15 minutes
Each day we begin our math block with an interactive online calendar followed by counting songs and videos.
Calendar Time:
We do calendar on Starfall every afternoon. This website has free reading and math resources for primary teachers. It also has a “more” option that requires paying a yearly fee. The calendar use is free. A detailed description of Daily Calendar math is included in the resources.
Counting with online sources: Today we did counting practice to reinforce the counting skills. We watched two to three number recognition 0-10 videos (one to two minutes each) because some of my students students were still struggling with identifying numbers correctly in random order. We watched"Shawn the Train" and counted objects with him to refresh our memories on how to count objects to ten and to reinforce one to one counting. Since we have started the second quarter of the school year, we added to today's counting practice: counting to 20 forward and backcounting by tens to 100 and counting to 100by ones to get a jump on our end of the year goals.
Direct Instruction
10 minutes
I read Five Little Monsters to the kids. I think aloud the addition and subtraction on each page. I use a number line under the doc cam to demonstrate the subtraction.
I also add the missing monsters so the kids can see the movement up and down the number line. I place a subtraction or addition sign under the doc cam before I move on the number line so they can begin connecting the movement with the sign.
I demonstrate this several times. I then take the cards away and use a +/- dice in their place. I roll the dice just like they will during the game. I demonstrate moving one up or one down for several rounds. Once I think they have that idea, I add the number dice.
I demonstrate rolling the +/- dice and the number dice about 10 times. After that I have them tell me what to do and in what order. When they can do that, they are ready for guided practice.
**When I roll a number that is too large to subtract, I tell them that they cannot play that round and it goes to their partner. To practice taking only 3 away when they've rolled a 5 is setting them up for failure!
Guided Instruction
10 minutes
The goal of this activity is simply to get them clear and accustomed to responding properly to subtraction and addition signs. One of the biggest struggles kids have is recognizing and responding appropriately to specific mathematical signs. Often, when you give a child a page of mixed addition and subtraction problems, they don't acknowledge the sign and just add everything. This activity helps prepare them for that challenge. In small steps the kids learn to acknowledge the different signs.
I pair up the kids and give them each a number line, bunny, and one dice with addition and subtraction signs and another with numbers 1-6.
They take turns rolling and jumping as I guide them step by step. I choose who goes first in each pair (Partner A). When we move to independent practice, Partner A will go first.
I have them roll with me. I ask each Partner A what they rolled, addition or subtraction. I tell them which way they will move.
Independent Practice
15 minutes
Have the materials prepped and ready to go. Do not waste class time on having the kids cutting and pasting the number lines. Copy them on card stock and cut and paste them while you're watch your favorite TV show.
Partner A goes first.
1) Roll the sign dice.
2) Tell your partner what sign it is.
3) Roll the number dice.
4) Tell your partner what number it is.
5) Move your bunny if you can. If you don't have enough spaces left to move your bunny, you loose your turn and your partner gets to roll, e.g. You are on zero and you roll a minus 2. You loose your turn because you can't move to the left two space.
Partner B goes.
While they play, I roam the room to assist kids in moving their bunny the correct direction and the correct number of jumps.
I also ask kids why they are moving the direction they are moving. I am NOT looking for "because it's a plus (addition) sign". I am looking for because when you add you get more. When you subtract you get less.
Closure
10 minutes
We join on the floor to discuss what we did in this activity and what we learned. The kids are encouraged to share ideas that would make this activity more effective or a way we could use it for another lesson.
One student suggests that we use it to solve addition and subtraction problems. Hmmmm...Great idea!
Another student shares that he was struggling with knowing when to go up or down the number line. He said there were too many steps for him to "get it" and do it with confidence. His partner explained to him that the subtraction sign means to take away or get smaller so to hop back on the number line and an addition sign means to get more or bigger number and to hop forward on the number line. In the reflection you can see a short video clip of the student coaching him in the steps to hop up and down the number line. |
## 2. Trigonometric Functions
### a. Sine; Law of Sines
The sine of any angle falls between -1.0 and +1.0. Refer to the sine curve plot in Figure B-4 to see how angles and their sines relate. This curve repeats itself every 360°;
so sin(-340°) = sin(20°) = sin(380°)...
Figure B-4
Taking the arcsine (a.k.a. inverse sine, sin-1) of a number between -1.0 and +1.0 on a calculator will always return an angle between -90° and +90°.
For example:
Sin(101°) is 0.981627
Sin-1(0.866025) is 79°
We started with 101° and wound up with 79°.
Sly obersverse that we are, we notice that 101°+79° is 180°
Table B-1 show some other angles, their sine behavior, relationship between the beginnng and ending angles.
Table B-1 α x=sin (α) sin-1(x) relationship 30° +0.50000 +30° α 150° +0.50000 +30° 180°-α 210° -0.50000 -30° 180°-α 330° -0.50000 -30° 360°+α
All of the angles in the third colum are between -90° and +90°. This is shown graphically in Figure B-5.
Figure B-5 ArcSine Angle Range
A triangle may have an angle which is greater than 90° but if you are solving that angle using the Law of Sines, the computed angle will be less than 90°. Although no single angle in a triangle may exceed 180° the Law of Sines can result in two angles both less than 180°.
For example sin-1(0.5) can be 30° or 150° - both meet the sine condition but only one meets the triangle condition. A calculator will always return 30° even though the “true” angle is 150°. How do you know which is correct? It depends on the triangle.
#### (1) Example
In triangle ABC, a = 12.4', b = 8.7', and B = 36°40'.
Compute the remaining angles and side.
From the Law of Sines
There are two angles whose sine is +0.85112
A = 58°20' and A = (180°00' - 58°20') = 121°40', Figure B-6.
Figure B-6 Multiple Angles with Same Sine
That means there are two different triangle solutions, Figure B-7
Figure B-7 Two Possible Angle Values
Consider point C as the center of an arc of radius 8.7'.
This arc can intersect the remaining side at two points; A1 and A2.
Isolating the two triangles and solving their remaining components:
For A = 58°20'
Figure B-8 First Solution
Angle condition
Law of Sines
For A = 121°40'
Figure B-9 Second Solution
Angle condition
Law of Sines
#### (2) Example 2
Compute the angle B in the following triangle:
Figure B-10 Example 2
The missing angle can be computed two ways:
Law of Sines
Angle condition
As a result there are two "correct" answers fot the missing angle:
sin(108°) = sin(72°) = 0.95105652
Ans since neither exceeds 180°, either can be an angle of a triangle,
but only 108° fits both the Law of Sines and the Angle condition.
Be careful when using the Law of Sines to solve for an unknown angle – there could be two possible answers only one of which will fit the particular triangle.
### b. Cosine; Law of Cosines
The cosine of any angle falls between -1.0 and +1.0. Refer to the cosine curve plot in Figure B-11 to see how angles and their cosines relate. This curve repeats itself every 360°; so cos(-340°) = cos(20°) = cos(380°)...
The cosine curve is identical to the sine curve except its phase differs by 90°.
Figure B-11
Taking the arccos (a.k.a. inverse cosine, cos-1) of a number between +1.0 and -1.0 on a calculator will always return an angle between 0° and 180°; for example:
Table B-2 α x=sin (α) sin-1(x) 60° +0.50000 +60° 300° +0.50000 +60° 120° -0.50000 +120° -240° -0.50000 +120°
All of the angles in the third colum are between 0° and +180°. This is shown graphically in Figure B-12.
Figure B-12 ArcCosine Angle Range
Using the Law of Cosines will not cause an ambiguous solution as does the Law of Sines since any single angle in a triangle cannot exceed 180°.
To solve a triangle using the Law of Cosines you must have either three sides, or, two sides and an angle.
#### (1) Example 1
Compute the value of the angle R in the triangle below:
Figure B-13 Example
From the Law of Cosines:
The Law of Cosines returns only one legitimate value when solving triangles.
### c. Tangent
Unlike sine and cosine, the tangent of any angle not limited to the range of -1.0 to +1.0. As a matter of fact, the tangent range is ±∞ (infinity). You can see that the tangent function plot is not sinusoidal as are the sine and cosine plots. And unlike the other two it repeats itself every 180°;
Figure B-14
The tangent curve is asymptotic at 90°, 270°, 450°, etc. Asymptotic means the curve gets close to, but never reaches, those values.
so tan(90°) = tan(270°) = ... = infinity
Try evaluating tan(90°) on your calculator; you’ll probably get an error statement of sorts. Then try tan(89.99999°); you should get a pretty big number. Why is that?
Recall that tan(α) = sin(α) / cos(α)
At 90°, cos(90°) = 0 so you get division by 0, hence tan(90°) = ∞.
There is a Law of Tangents, but we don't generally use it to solve triangles since the Laws of Sines or Cosines can be used instead. For extra credit, find and memorize it.
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# Difference between revisions of "2020 AIME I Problems/Problem 9"
## Problem
Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random with replacement from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m.$
## Solution
First, prime factorize $20^9$ as $2^{18} \cdot 5^9$. Denote $a_1$ as $2^{b_1} \cdot 5^{c_1}$, $a_2$ as $2^{b_2} \cdot 5^{c_2}$, and $a_3$ as $2^{b_3} \cdot 5^{c_3}$.
In order for $a_1$ to divide $a_2$, and for $a_2$ to divide $a_3$, $b_1\ge b_2\ge b_3$, and $c_1\eg c_2\ge c_3$ (Error compiling LaTeX. ! Undefined control sequence.). We will consider each case separately. Note that the total amount of possibilities is $190^3$, as there are $(18+1)(9+1)=190$ choices for each factor.
We notice that if we add $1$ to $b_2$ and $2$ to $b_3$, then we can reach the stronger inequality $b_1. Therefore, if we pick $3$ integers from $0$ to $20$, they will correspond to a unique solution, forming a 1-1 correspondence. The amount of solutions to this inequality is $\dbinom{21}{3}$.
The case for $c_1$,$c_2$, and $c_3$ proceeds similarly for a result of $\dbinom{12}{3}$. Therefore, the probability of choosing three such factors is $$\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.$$ Simplification gives $\frac{77}{1805}$, and therefore the answer is $\boxed{77}$.
-molocyxu |
# Multiplication with the Distributive Property
Quick Navigation Video Lectures Dynamic Practice
Polynomial multiplication is based on a principle called the Distributive Property. If you look below at how it's defined, it looks deceptively simple. Be sure to pay close attention to the examples of how negative signs work when applying this property - students often run into trouble with that situation.
In Math . . .
a(b + c) = ab + ac
In English . . .
You can get rid of parentheses in an expression by taking the expression that's outside the parentheses and multiplying everything inside the parentheses by it.
I think the best way to approach this material is by looking at some specific examples.
## Quick Definition - "Simplify"
"Simplify" is one of the vaguest terms you'll see in mathematics. What constitutes a 'simpler' version of an expression can sometimes be a matter of opinion. For example 3(x + 1) and 3x + 3 are both equal to each other but I'd be hard pressed to give you an objective reason why one was 'simpler' than the other. There are a few universal rules that you should remember:
1. Fractions should be completely reduced, e.g. 1/2 is simpler than 2/4.
2. Exponents should be combined whenever possible, e.g. x5 is simpler than x2x3.
3. Negative exponents should always be converted to positive ones, e.g. 1 / x is simpler than $x^{-1}$.
4. Radicals should be reduced as much as possible, e.g. $4\sqrt{3}$ is simpler than $\sqrt{48}$
5. Denominators should be rationalized, e.g. $\sqrt{3} / 3$ is simpler than $1 / \sqrt{3}$.
# Example 1
Simplify 2(4x + 3).
The Distributive Property tells me that I can simplify this by taking the 2 outside the parentheses and "distributing" it to both of the terms inside the parentheses.
2(4x + 3) = 2·4x + 2·3 = 8x + 6
# Example 2
Simplify -2(x3 - 3x2).
This one is a little trickier because of the negative signs. I'm going to distribute the -2 just like I distributed the +2 in the previous example.
-2(x3 - 3x2) = -2·x3 - (-2)·3x2
Notice that I didn't do anything with the minus sign inside the parentheses. At this point, all I'm trying to do is distribute the -2. We'll worry about the minus in a later step. Now I'm going to multiply out all the numbers. In this case, there's only one pair - the -2 and the 3 on the last term.
-2(x3 - 3x2) = -2x3 - (-6x2)
Now, I'm going to straighten out the minus sign. Remember that a negative times a negative is a positive so anywhere I see two negatives, I'm going to change them to a plus.
-2(x3 - 3x2) = -2x3 - (-6x2) = -2x3 + 6x2
# Example 3
Simplify 2x2(-3x3 + 3y + 1).
Don't be intimidated by there being a variable outside the parentheses. You distribute the entire 2x2 just like the numbers in the previous examples.
(2x2)(-3x3) + (2x2)(3y) + 1(2x2) I took the 2x2 and multiplied every term inside the parentheses by it. 2·(-3)·x2·x3 + 2·3x2y + 1·2x2 I rearranged the parts of each term so that the "like" parts where together. For example in the first term, I moved the numbers together and the variable parts together. This is just to make it clear which parts I'm going to multiply together on the next step. (You could skip this and go directly to the next step if it's clear to you which parts can be multiplied together.) -6x5 + 6x2y + 2x2 I multiplied the numbers together and multiplied the variable parts by adding their exponents.
# Video Lectures
### Lectures
Before we can talk about multiplying polynomials, we need to add the Distrbutive Property to our set of tools. (lecture slides)
Mulitplying a polynomial by a monomial is just an application of the Distributive Property. |
# Exponential Decay
## Rational functions with x as an exponent in the denominator
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Exponential Decay Function
The population of a city was 10,000 in 2012 and is declining at a rate of 5% each year. If this decay rate continues, what will the city's population be in 2017?
### Exponential Delay Function
Previously, we have only addressed functions where \begin{align*}|b|>1\end{align*}. So, what happens when \begin{align*}b\end{align*} is less than 1? Let’s analyze \begin{align*}y=\left(\frac{1}{2}\right)^x\end{align*}.
Graph \begin{align*}y=\left(\frac{1}{2}\right)^x\end{align*} and compare it to \begin{align*}y=2^x\end{align*}.
Let’s make a table of both functions and then graph.
\begin{align*}x\end{align*} \begin{align*}\left(\frac{1}{2}\right)^x\end{align*} \begin{align*}2^x\end{align*}
3 \begin{align*}\left(\frac{1}{2}\right)^3 = \frac{1}{8}\end{align*} \begin{align*}2^3=8\end{align*}
2 \begin{align*}\left(\frac{1}{2}\right)^2 = \frac{1}{4}\end{align*} \begin{align*}2^2=4\end{align*}
1 \begin{align*}\left(\frac{1}{2}\right)^1 = \frac{1}{2}\end{align*} \begin{align*}2^1=2\end{align*}
0 \begin{align*}\left(\frac{1}{2}\right)^0 = 1\end{align*} \begin{align*}2^0=1\end{align*}
-1 \begin{align*}\left(\frac{1}{2}\right)^{-1} = 2\end{align*} \begin{align*}2^{-1}=\frac{1}{2}\end{align*}
-2 \begin{align*}\left(\frac{1}{2}\right)^{-2} = 4\end{align*} \begin{align*}2^{-2}=\frac{1}{4}\end{align*}
-3 \begin{align*}\left(\frac{1}{2}\right)^3 = 8\end{align*} \begin{align*}2^{-3}=\frac{1}{8}\end{align*}
Notice that \begin{align*}y=\left(\frac{1}{2}\right)^x\end{align*} is a reflection over the \begin{align*}y\end{align*}-axis of \begin{align*}y=2^x\end{align*}. Therefore, instead of exponential growth, the function \begin{align*}y=\left(\frac{1}{2}\right)^x\end{align*} decreases exponentially, or exponentially decays. Anytime \begin{align*}b\end{align*} is a fraction or decimal between zero and one, the exponential function will decay. And, just like an exponential growth function, and exponential decay function has the form \begin{align*}y=ab^x\end{align*} and \begin{align*}a>0\end{align*}. However, to be a decay function, \begin{align*}0 < b < 1\end{align*}. The exponential decay function also has an asymptote at \begin{align*}y=0\end{align*}.
Let's determine which of the following functions are exponential decay functions, exponential growth functions, or neither and briefly explain our answers.
1. \begin{align*}y=4(1.3)^x\end{align*}
2. \begin{align*}f(x)=3 \left(\frac{6}{5}\right)^x\end{align*}
3. \begin{align*}y = \left(\frac{3}{10}\right)^x\end{align*}
4. \begin{align*}g(x)= -2(0.65)^x\end{align*}
a. and b. are exponential growth functions because \begin{align*}b>1\end{align*}.
c. is an exponential decay function because \begin{align*}b\end{align*} is between zero and one.
d. is neither growth or decay because \begin{align*}a\end{align*} is negative.
Let's graph \begin{align*}g(x)=-2 \left(\frac{2}{3}\right)^{x-1}+1\end{align*} and find the \begin{align*}y\end{align*}-intercept, asymptote, domain, and range.
To graph this function, you can either plug it into your calculator (entered Y= -2(2/3)^(X-1)+1) or graph \begin{align*}y=-2 \left(\frac{2}{3}\right)^x\end{align*} and shift it to the right one unit and up one unit. We will use the second method; final answer is the blue function below.
The \begin{align*}y\end{align*}-intercept is:
\begin{align*}y=-2 \left(\frac{2}{3}\right)^{0-1}+1=-2 \cdot \frac{3}{2}+1=-3+1=-2\end{align*}
The horizontal asymptote is \begin{align*}y=1\end{align*}, the domain is all real numbers and the range is \begin{align*}y < 1\end{align*}.
### Examples
#### Example 1
Earlier, you were asked to find the city's population in 2017 if the population was 10,000 in 2012 and is declining at a rate of 5% each year.
This is an example of exponential decay, so we can once again use the exponential form \begin{align*}f(x)=a \cdot b^{x-h}+k\end{align*}, but we have to be careful. In this case, a = 10,000, the starting population, x-h = 5 the number of years, and k = 0, but b is a bit trickier. If the population is decreasing by 5%, each year the population is (1 - 5%) or (1 - 0.05) = 0.95 what it was the previous year. This is our b.
\begin{align*}P = 10,000 \cdot 0.95^5\\ = 10,000 \cdot 0.7738 = 7738\end{align*}
Therefore, the city's population in 2017 is 7,738.
For Examples 2-4, graph the exponential functions. Find the \begin{align*}y\end{align*}-intercept, asymptote, domain, and range.
#### Example 2
\begin{align*}f(x)=4 \left(\frac{1}{3}\right)^x\end{align*}
\begin{align*}y\end{align*}-intercept: \begin{align*}(4, 0)\end{align*}, asymptote: \begin{align*}y=0\end{align*}, domain: all reals, range: \begin{align*}y < 0\end{align*}
#### Example 3
\begin{align*}y=-2 \left(\frac{2}{3}\right)^{x+3}\end{align*}
\begin{align*}y\end{align*}-intercept: \begin{align*}\left(0, -\frac{16}{27}\right)\end{align*}, asymptote: \begin{align*}y=0\end{align*}, domain: all reals, range: \begin{align*}y<0\end{align*}
#### Example 4
\begin{align*}g(x)= \left(\frac{3}{5}\right)^x-6\end{align*}
\begin{align*}y\end{align*}-intercept: \begin{align*}(-5, 0)\end{align*}, asymptote: \begin{align*}y=-6\end{align*}, domain: all reals, range: \begin{align*}y>-6\end{align*}
For Examples 5-8, determine if the functions are exponential growth, exponential decay, or neither.
#### Example 5
\begin{align*}y=2.3^x\end{align*}
exponential growth
#### Example 6
\begin{align*}y=2 \left(\frac{4}{3}\right)^{-x}\end{align*}
exponential decay; recall that a negative exponent flips whatever is in the base. \begin{align*}y=2 \left(\frac{4}{3}\right)^{-x}\end{align*} is the same as \begin{align*}y=2 \left(\frac{3}{4} \right)^{x}\end{align*}, which looks like our definition of a decay function.
#### Example 7
\begin{align*}y=3\cdot 0.9^x\end{align*}
exponential decay
#### Example 8
\begin{align*}y=\frac{1}{2} \left(\frac{4}{5}\right)^{x}\end{align*}
neither; \begin{align*}a < 0\end{align*}
### Review
Determine which of the following functions are exponential growth, exponential decay or neither.
1. \begin{align*}y= -\left(\frac{2}{3}\right)^x\end{align*}
2. \begin{align*}y= \left(\frac{4}{3}\right)^x\end{align*}
3. \begin{align*}y=5^x\end{align*}
4. \begin{align*}y= \left(\frac{1}{4}\right)^x\end{align*}
5. \begin{align*}y= 1.6^x\end{align*}
6. \begin{align*}y= -\left(\frac{6}{5}\right)^x\end{align*}
7. \begin{align*}y= 0.99^x\end{align*}
Graph the following exponential functions. Find the \begin{align*}y\end{align*}-intercept, the equation of the asymptote and the domain and range for each function.
1. \begin{align*}y= \left(\frac{1}{2}\right)^x\end{align*}
2. \begin{align*}y=(0.8)^{x+2}\end{align*}
3. \begin{align*}y=4 \left(\frac{2}{3}\right)^{x-1}-5\end{align*}
4. \begin{align*}y= -\left(\frac{5}{7}\right)^x +3\end{align*}
5. \begin{align*}y= \left(\frac{8}{9}\right)^{x+5} -2\end{align*}
6. \begin{align*}y=(0.75)^{x-2}+4\end{align*}
7. Is the domain of an exponential function always all real numbers? Why or why not?
8. A discount retailer advertises that items will be marked down at a rate of 10% per week until sold. The initial price of one item is \$50.
1. Write an exponential decay function to model the price of the item \begin{align*}x\end{align*} weeks after it is first put on the rack.
2. What will the price be after the item has been on display for 5 weeks?
3. After how many weeks will the item be half its original price?
To see the Review answers, open this PDF file and look for section 8.2.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Exponential Decay Function An exponential decay function is a specific type of exponential function that has the form $y=ab^x$, where $a>0$ and $0.
Exponential Function An exponential function is a function whose variable is in the exponent. The general form is $y=a \cdot b^{x-h}+k$.
Model A model is a mathematical expression or function used to describe a physical item or situation. |
# Problem of the Week
## Updated at Jan 27, 2014 4:04 PM
To get more practice in algebra, we brought you this problem of the week:
How would you find the factors of $$12{y}^{4}+100{y}^{3}+112{y}^{2}$$?
Check out the solution below!
$12{y}^{4}+100{y}^{3}+112{y}^{2}$
1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$12{y}^{4}$$, $$100{y}^{3}$$, and $$112{y}^{2}$$?It is $$4$$.2 What is the highest degree of $$y$$ that divides evenly into $$12{y}^{4}$$, $$100{y}^{3}$$, and $$112{y}^{2}$$?It is $${y}^{2}$$.3 Multiplying the results above,The GCF is $$4{y}^{2}$$.To get access to all 'How?' and 'Why?' steps, join Cymath Plus!GCF = $$4{y}^{2}$$2 Factor out the GCF. (Write the GCF first. Then, in parentheses, divide each term by the GCF.)$4{y}^{2}(\frac{12{y}^{4}}{4{y}^{2}}+\frac{100{y}^{3}}{4{y}^{2}}+\frac{112{y}^{2}}{4{y}^{2}})$3 Simplify each term in parentheses.$4{y}^{2}(3{y}^{2}+25y+28)$4 Split the second term in $$3{y}^{2}+25y+28$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$3\times 28=84$2 Ask: Which two numbers add up to $$25$$ and multiply to $$84$$?$$21$$ and $$4$$3 Split $$25y$$ as the sum of $$21y$$ and $$4y$$.$3{y}^{2}+21y+4y+28$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$4{y}^{2}(3{y}^{2}+21y+4y+28)$5 Factor out common terms in the first two terms, then in the last two terms.$4{y}^{2}(3y(y+7)+4(y+7))$6 Factor out the common term $$y+7$$.$4{y}^{2}(y+7)(3y+4)$Done4*y^2*(y+7)*(3*y+4) |
+0
# domain
-1
344
2
+1035
Find domain and range of $$g(x) = \frac{x^3 + 11x - 2}{|x - 3| + |x + 1|}.$$
Dec 2, 2018
#1
+2345
+1
Let's first think of the domain of this particular function.
The only portion of this function that can affect the state of the domain is the denominator, so let's investigate it.
If there was a value for x such that $$|x-3|+|x+1|=0$$, then a restriction for the domain of this function would exist. Let's see if there are any values for x such that this is true:
$$|x-3|+|x+1|=0$$ I think the best way to solve this is to think about it logically. You could compute the sign changes, but I think there is a better way. The absolute value signs always output a nonnegative result. In this equation, we are adding two nonnegative outputs together. There is only one way that the sum of two nonnegative numbers can equal 0. That way is 0+0=0. $$|x-3|=0\\ x-3=0\\ x=3$$ I have determined that the only x-value that makes the first addend, $$|x-3|$$, equal to 0 is when x=3. $$|3+1|\stackrel{?}=0\\ 4\stackrel{?}=0\\ \text{false}$$ x=3 is the only candidate that can make this equation equal to 0. The second addend, |x+1|, evaluates to 4 when x=3. However, we determined earlier that |x+1| has to evaluate to 0 in order for solutions to exist. Therefore, there are no solutions, for the real number set at least. $$\text{Domain}: x \in {\rm I\!R}$$ The conclusion from this investigation is that there are no restrictions in the domain.
The nature of this function is generally one of a cubic function because there are no vertical asymptotes. Dividing by the addition of a sum of nonnegatives numbers does not really affect the range.
$$\text{Range}: x \in {\rm I\!R}$$
.
Dec 2, 2018
#1
+2345
+1
Let's first think of the domain of this particular function.
The only portion of this function that can affect the state of the domain is the denominator, so let's investigate it.
If there was a value for x such that $$|x-3|+|x+1|=0$$, then a restriction for the domain of this function would exist. Let's see if there are any values for x such that this is true:
$$|x-3|+|x+1|=0$$ I think the best way to solve this is to think about it logically. You could compute the sign changes, but I think there is a better way. The absolute value signs always output a nonnegative result. In this equation, we are adding two nonnegative outputs together. There is only one way that the sum of two nonnegative numbers can equal 0. That way is 0+0=0. $$|x-3|=0\\ x-3=0\\ x=3$$ I have determined that the only x-value that makes the first addend, $$|x-3|$$, equal to 0 is when x=3. $$|3+1|\stackrel{?}=0\\ 4\stackrel{?}=0\\ \text{false}$$ x=3 is the only candidate that can make this equation equal to 0. The second addend, |x+1|, evaluates to 4 when x=3. However, we determined earlier that |x+1| has to evaluate to 0 in order for solutions to exist. Therefore, there are no solutions, for the real number set at least. $$\text{Domain}: x \in {\rm I\!R}$$ The conclusion from this investigation is that there are no restrictions in the domain.
The nature of this function is generally one of a cubic function because there are no vertical asymptotes. Dividing by the addition of a sum of nonnegatives numbers does not really affect the range.
$$\text{Range}: x \in {\rm I\!R}$$
TheXSquaredFactor Dec 2, 2018
#2
+1035
-1
thanks
Dec 2, 2018 |
Applications of Newton’s Laws
# Drag Force and Terminal Speed
### Learning Objectives
By the end of the section, you will be able to:
• Express the drag force mathematically
• Describe applications of the drag force
• Define terminal velocity
• Determine an object’s terminal velocity given its mass
Another interesting force in everyday life is the force of drag on an object when it is moving in a fluid (either a gas or a liquid). You feel the drag force when you move your hand through water. You might also feel it if you move your hand during a strong wind. The faster you move your hand, the harder it is to move. You feel a smaller drag force when you tilt your hand so only the side goes through the air—you have decreased the area of your hand that faces the direction of motion.
### Drag Forces
Like friction, the drag force always opposes the motion of an object. Unlike simple friction, the drag force is proportional to some function of the velocity of the object in that fluid. This functionality is complicated and depends upon the shape of the object, its size, its velocity, and the fluid it is in. For most large objects such as cyclists, cars, and baseballs not moving too slowly, the magnitude of the drag force ${F}_{\text{D}}$ is proportional to the square of the speed of the object. We can write this relationship mathematically as ${F}_{\text{D}}\propto {v}^{2}.$ When taking into account other factors, this relationship becomes
${F}_{\text{D}}=\frac{1}{2}C\rho A{v}^{2},$
where C is the drag coefficient, A is the area of the object facing the fluid, and $\rho$ is the density of the fluid. (Recall that density is mass per unit volume.) This equation can also be written in a more generalized fashion as ${F}_{\text{D}}=b{v}^{2},$ where b is a constant equivalent to $0.5C\rho A.$ We have set the exponent n for these equations as 2 because when an object is moving at high velocity through air, the magnitude of the drag force is proportional to the square of the speed. As we shall see in Fluid Mechanics, for small particles moving at low speeds in a fluid, the exponent n is equal to 1.
Drag Force
Drag force ${F}_{\text{D}}$ is proportional to the square of the speed of the object. Mathematically,
${F}_{\text{D}}=\frac{1}{2}C\phantom{\rule{0.2em}{0ex}}\rho \phantom{\rule{0.2em}{0ex}}A{v}^{2},$
where C is the drag coefficient, A is the area of the object facing the fluid, and $\rho$ is the density of the fluid.
Athletes as well as car designers seek to reduce the drag force to lower their race times ((Figure)). Aerodynamic shaping of an automobile can reduce the drag force and thus increase a car’s gas mileage.
From racing cars to bobsled racers, aerodynamic shaping is crucial to achieving top speeds. Bobsleds are designed for speed and are shaped like a bullet with tapered fins. (credit: “U.S. Army”/Wikimedia Commons)
The value of the drag coefficient C is determined empirically, usually with the use of a wind tunnel ((Figure)).
NASA researchers test a model plane in a wind tunnel. (credit: NASA/Ames)
The drag coefficient can depend upon velocity, but we assume that it is a constant here. (Figure) lists some typical drag coefficients for a variety of objects. Notice that the drag coefficient is a dimensionless quantity. At highway speeds, over $50%$ of the power of a car is used to overcome air drag. The most fuel-efficient cruising speed is about 70–80 km/h (about 45–50 mi/h). For this reason, during the 1970s oil crisis in the United States, maximum speeds on highways were set at about 90 km/h (55 mi/h).
Typical Values of Drag Coefficient C
Object C
Airfoil 0.05
Toyota Camry 0.28
Ford Focus 0.32
Honda Civic 0.36
Ferrari Testarossa 0.37
Dodge Ram Pickup 0.43
Sphere 0.45
Hummer H2 SUV 0.64
Skydiver (feet first) 0.70
Bicycle 0.90
Skydiver (horizontal) 1.0
Circular flat plate 1.12
Substantial research is under way in the sporting world to minimize drag. The dimples on golf balls are being redesigned, as are the clothes that athletes wear. Bicycle racers and some swimmers and runners wear full bodysuits. Australian Cathy Freeman wore a full body suit in the 2000 Sydney Olympics and won a gold medal in the 400-m race. Many swimmers in the 2008 Beijing Olympics wore (Speedo) body suits; it might have made a difference in breaking many world records ((Figure)). Most elite swimmers (and cyclists) shave their body hair. Such innovations can have the effect of slicing away milliseconds in a race, sometimes making the difference between a gold and a silver medal. One consequence is that careful and precise guidelines must be continuously developed to maintain the integrity of the sport.
Body suits, such as this LZR Racer Suit, have been credited with aiding in many world records after their release in 2008. Smoother “skin” and more compression forces on a swimmer’s body provide at least $10%$ less drag. (credit: NASA/Kathy Barnstorff)
### Terminal Velocity
Some interesting situations connected to Newton’s second law occur when considering the effects of drag forces upon a moving object. For instance, consider a skydiver falling through air under the influence of gravity. The two forces acting on him are the force of gravity and the drag force (ignoring the small buoyant force). The downward force of gravity remains constant regardless of the velocity at which the person is moving. However, as the person’s velocity increases, the magnitude of the drag force increases until the magnitude of the drag force is equal to the gravitational force, thus producing a net force of zero. A zero net force means that there is no acceleration, as shown by Newton’s second law. At this point, the person’s velocity remains constant and we say that the person has reached his terminal velocity $\left({v}_{\text{T}}\right).$ Since ${F}_{\text{D}}$ is proportional to the speed squared, a heavier skydiver must go faster for ${F}_{\text{D}}$ to equal his weight. Let’s see how this works out more quantitatively.
At the terminal velocity,
${F}_{\text{net}}=mg-{F}_{\text{D}}=ma=0.$
Thus,
$mg={F}_{\text{D}}.$
Using the equation for drag force, we have
$mg=\frac{1}{2}C\rho A{v}_{\text{T}}^{2}.$
Solving for the velocity, we obtain
${v}_{\text{T}}=\sqrt{\frac{2mg}{\rho CA}}.$
Assume the density of air is $\rho =1.21\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}.$ A 75-kg skydiver descending head first has a cross-sectional area of approximately $A=0.18\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}$ and a drag coefficient of approximately $C=0.70$. We find that
${v}_{\text{T}}=\sqrt{\frac{2\left(75\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(9.80\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)}{\left(1.21\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}\right)\left(0.70\right)\left(0.18\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}\right)}}=98\phantom{\rule{0.2em}{0ex}}\text{m/s}=350\phantom{\rule{0.2em}{0ex}}\text{km/h}\text{.}$
This means a skydiver with a mass of 75 kg achieves a terminal velocity of about 350 km/h while traveling in a pike (head first) position, minimizing the area and his drag. In a spread-eagle position, that terminal velocity may decrease to about 200 km/h as the area increases. This terminal velocity becomes much smaller after the parachute opens.
Terminal Velocity of a Skydiver
Find the terminal velocity of an 85-kg skydiver falling in a spread-eagle position.
Strategy
At terminal velocity, ${F}_{\text{net}}=0.$ Thus, the drag force on the skydiver must equal the force of gravity (the person’s weight). Using the equation of drag force, we find $mg=\frac{1}{2}\rho CA{v}^{2}.$
Solution
The terminal velocity ${v}_{\text{T}}$ can be written as
${v}_{\text{T}}=\sqrt{\frac{2mg}{\rho CA}}=\sqrt{\frac{2\left(85\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(9.80\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)}{\left(1.21\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}\right)\left(1.0\right)\left(0.70\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}\right)}}=44\phantom{\rule{0.2em}{0ex}}\text{m/s}\text{.}$
Significance
This result is consistent with the value for ${v}_{\text{T}}$ mentioned earlier. The 75-kg skydiver going feet first had a terminal velocity of ${v}_{\text{T}}=98\phantom{\rule{0.2em}{0ex}}\text{m/s}\text{.}$ He weighed less but had a smaller frontal area and so a smaller drag due to the air.
Check Your Understanding Find the terminal velocity of a 50-kg skydiver falling in spread-eagle fashion.
34 m/s
The size of the object that is falling through air presents another interesting application of air drag. If you fall from a 5-m-high branch of a tree, you will likely get hurt—possibly fracturing a bone. However, a small squirrel does this all the time, without getting hurt. You do not reach a terminal velocity in such a short distance, but the squirrel does.
The following interesting quote on animal size and terminal velocity is from a 1928 essay by a British biologist, J. B. S. Haldane, titled “On Being the Right Size.”
“To the mouse and any smaller animal, [gravity] presents practically no dangers. You can drop a mouse down a thousand-yard mine shaft; and, on arriving at the bottom, it gets a slight shock and walks away, provided that the ground is fairly soft. A rat is killed, a man is broken, and a horse splashes. For the resistance presented to movement by the air is proportional to the surface of the moving object. Divide an animal’s length, breadth, and height each by ten; its weight is reduced to a thousandth, but its surface only to a hundredth. So the resistance to falling in the case of the small animal is relatively ten times greater than the driving force.”
The above quadratic dependence of air drag upon velocity does not hold if the object is very small, is going very slow, or is in a denser medium than air. Then we find that the drag force is proportional just to the velocity. This relationship is given by Stokes’ law.
Stokes’ Law
For a spherical object falling in a medium, the drag force is
${F}_{\text{s}}=6\pi r\eta v,$
where r is the radius of the object, $\eta$ is the viscosity of the fluid, and v is the object’s velocity.
Good examples of Stokes’ law are provided by microorganisms, pollen, and dust particles. Because each of these objects is so small, we find that many of these objects travel unaided only at a constant (terminal) velocity. Terminal velocities for bacteria (size about $1\phantom{\rule{0.2em}{0ex}}\text{μm}\right)$ can be about $2\phantom{\rule{0.2em}{0ex}}\text{μm/s}\text{.}$ To move at a greater speed, many bacteria swim using flagella (organelles shaped like little tails) that are powered by little motors embedded in the cell.
Sediment in a lake can move at a greater terminal velocity (about $5\phantom{\rule{0.2em}{0ex}}\text{μm/s}\right),$ so it can take days for it to reach the bottom of the lake after being deposited on the surface.
If we compare animals living on land with those in water, you can see how drag has influenced evolution. Fish, dolphins, and even massive whales are streamlined in shape to reduce drag forces. Birds are streamlined and migratory species that fly large distances often have particular features such as long necks. Flocks of birds fly in the shape of a spearhead as the flock forms a streamlined pattern ((Figure)). In humans, one important example of streamlining is the shape of sperm, which need to be efficient in their use of energy.
Geese fly in a V formation during their long migratory travels. This shape reduces drag and energy consumption for individual birds, and also allows them a better way to communicate. (credit: “Julo”/Wikimedia Commons)
In lecture demonstrations, we do measurements of the drag force on different objects. The objects are placed in a uniform airstream created by a fan. Calculate the Reynolds number and the drag coefficient.
### The Calculus of Velocity-Dependent Frictional Forces
When a body slides across a surface, the frictional force on it is approximately constant and given by ${\mu }_{\text{k}}N.$ Unfortunately, the frictional force on a body moving through a liquid or a gas does not behave so simply. This drag force is generally a complicated function of the body’s velocity. However, for a body moving in a straight line at moderate speeds through a liquid such as water, the frictional force can often be approximated by
${f}_{R}=\text{−}bv,$
where b is a constant whose value depends on the dimensions and shape of the body and the properties of the liquid, and v is the velocity of the body. Two situations for which the frictional force can be represented this equation are a motorboat moving through water and a small object falling slowly through a liquid.
Let’s consider the object falling through a liquid. The free-body diagram of this object with the positive direction downward is shown in (Figure). Newton’s second law in the vertical direction gives the differential equation
$mg-bv=m\frac{dv}{dt},$
where we have written the acceleration as $dv\text{/}dt.$ As v increases, the frictional force –bv increases until it matches mg. At this point, there is no acceleration and the velocity remains constant at the terminal velocity ${v}_{\text{T}}.$ From the previous equation,
$mg-b{v}_{\text{T}}=0,$
so
${v}_{\text{T}}=\frac{mg}{b}.$
Free-body diagram of an object falling through a resistive medium.
We can find the object’s velocity by integrating the differential equation for v. First, we rearrange terms in this equation to obtain
$\frac{dv}{g-\left(b\text{/}m\right)v}=dt.$
Assuming that $v=0\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}t=0,$ integration of this equation yields
${\int }_{0}^{v}\frac{d{v}^{\prime }}{g-\left(b\text{/}m\right){v}^{\prime }}={\int }_{0}^{t}d{t}^{\prime },$
or
${-\frac{m}{b}\text{ln}\left(g-\frac{b}{m}{v}^{\prime }\right)|}_{0}^{v}={{t}^{\prime }|}_{0}^{t},$
where $v\text{'}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}t\text{'}$ are dummy variables of integration. With the limits given, we find
$-\frac{m}{b}\left[\text{ln}\left(g-\frac{b}{m}v\right)-\text{ln}g\right]=t.$
Since $\text{ln}A-\text{ln}B=\text{ln}\left(A\text{/}B\right),$ and $\text{ln}\left(A\text{/}B\right)=x\phantom{\rule{0.2em}{0ex}}\text{implies}\phantom{\rule{0.2em}{0ex}}{e}^{x}=A\text{/}B,$ we obtain
$\frac{g-\left(bv\text{/}m\right)}{g}={e}^{\text{−}bt\text{/}m},$
and
$v=\frac{mg}{b}\left(1-{e}^{\text{−}bt\text{/}m}\right).$
Notice that as $t\to \infty ,v\to mg\text{/}b={v}_{\text{T}},$ which is the terminal velocity.
The position at any time may be found by integrating the equation for v. With $v=dy\text{/}dt,$
$dy=\frac{mg}{b}\left(1-{e}^{\text{−}bt\text{/}m}\right)dt.$
Assuming $y=0\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}t=0,$
${\int }_{0}^{y}d{y}^{\prime }=\frac{mg}{b}{\int }_{0}^{t}\left(1-{e}^{\text{−}bt\text{'}\text{/}m}\right)d{t}^{\prime },$
which integrates to
$y=\frac{mg}{b}t+\frac{{m}^{2}g}{{b}^{2}}\left({e}^{\text{−}bt\text{/}m}-1\right).$
Effect of the Resistive Force on a Motorboat
A motorboat is moving across a lake at a speed ${v}_{0}$ when its motor suddenly freezes up and stops. The boat then slows down under the frictional force ${f}_{R}=\text{−}bv.$ (a) What are the velocity and position of the boat as functions of time? (b) If the boat slows down from 4.0 to 1.0 m/s in 10 s, how far does it travel before stopping?
Solution
1. With the motor stopped, the only horizontal force on the boat is ${f}_{R}=\text{−}bv,$ so from Newton’s second law,
$m\frac{dv}{dt}=\text{−}bv,$
which we can write as
$\frac{dv}{v}=-\frac{b}{m}dt.$
Integrating this equation between the time zero when the velocity is ${v}_{0}$ and the time t when the velocity is $v$, we have
${\int }_{0}^{v}\frac{d{v}^{\prime }}{{v}^{\prime }}=-\frac{b}{m}{\int }_{0}^{t}d{t}^{\prime }.$
Thus,
$\text{ln}\frac{v}{{v}_{0}}=-\frac{b}{m}t,$
which, since $\text{ln}A=x\phantom{\rule{0.2em}{0ex}}\text{implies}\phantom{\rule{0.2em}{0ex}}{e}^{x}=A,$ we can write this as
$v={v}_{0}{e}^{\text{−}bt\text{/}m}.$
Now from the definition of velocity,
$\frac{dx}{dt}={v}_{0}{e}^{\text{−}bt\text{/}m},$
so we have
$dx={v}_{0}{e}^{\text{−}bt\text{/}m}dt.$
With the initial position zero, we have
${\int }_{0}^{x}dx\text{'}={v}_{0}{\int }_{0}^{t}{e}^{\text{−}bt\text{'}\text{/}m}dt\text{'},$
and
${x=-\frac{m{v}_{0}}{b}{e}^{\text{−}bt\text{'}\text{/}m}|}_{0}^{t}=\frac{m{v}_{0}}{b}\left(1-{e}^{\text{−}bt\text{/}m}\right).$
As time increases, ${e}^{\text{−}bt\text{/}m}\to 0,$ and the position of the boat approaches a limiting value
${x}_{\text{max}}=\frac{m{v}_{0}}{b}.$
Although this tells us that the boat takes an infinite amount of time to reach ${x}_{\text{max}},$ the boat effectively stops after a reasonable time. For example, at $t=10m\text{/}b,$ we have
$v={v}_{0}{e}^{-10}\simeq 4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}{v}_{0},$
whereas we also have
$x={x}_{\text{max}}\left(1-{e}^{-10}\right)\simeq 0.99995{x}_{\text{max}}.$
Therefore, the boat’s velocity and position have essentially reached their final values.
2. With ${v}_{0}=4.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$ and $v=1.0\phantom{\rule{0.2em}{0ex}}\text{m/s,}$ we have $1.0\phantom{\rule{0.2em}{0ex}}\text{m/s}=\left(4.0\phantom{\rule{0.2em}{0ex}}\text{m/s}\right){e}^{\text{−}\left(b\text{/}m\right)\left(10\phantom{\rule{0.2em}{0ex}}\text{s}\right)},$ so
$\text{ln}\phantom{\rule{0.2em}{0ex}}0.25=\text{−}\text{ln}\phantom{\rule{0.2em}{0ex}}4.0=-\frac{b}{m}\left(10\phantom{\rule{0.2em}{0ex}}\text{s}\right),$
and
$\frac{b}{m}=\frac{1}{10}\text{ln}\phantom{\rule{0.2em}{0ex}}4.0\phantom{\rule{0.2em}{0ex}}{\text{s}}^{\text{-1}}=0.14\phantom{\rule{0.2em}{0ex}}{\text{s}}^{\text{-1}}\text{.}$
Now the boat’s limiting position is
${x}_{\text{max}}=\frac{m{v}_{0}}{b}=\frac{4.0\phantom{\rule{0.2em}{0ex}}\text{m/s}}{0.14\phantom{\rule{0.2em}{0ex}}{\text{s}}^{\text{−1}}}=29\phantom{\rule{0.2em}{0ex}}\text{m}\text{.}$
Significance
In the both of the previous examples, we found “limiting” values. The terminal velocity is the same as the limiting velocity, which is the velocity of the falling object after a (relatively) long time has passed. Similarly, the limiting distance of the boat is the distance the boat will travel after a long amount of time has passed. Due to the properties of exponential decay, the time involved to reach either of these values is actually not too long (certainly not an infinite amount of time!) but they are quickly found by taking the limit to infinity.
Check Your Understanding Suppose the resistive force of the air on a skydiver can be approximated by $f=\text{−}b{v}^{2}$. If the terminal velocity of a 100-kg skydiver is 60 m/s, what is the value of b?
0.27 kg/m
### Summary
• Drag forces acting on an object moving in a fluid oppose the motion. For larger objects (such as a baseball) moving at a velocity in air, the drag force is determined using the drag coefficient (typical values are given in (Figure)), the area of the object facing the fluid, and the fluid density.
• For small objects (such as a bacterium) moving in a denser medium (such as water), the drag force is given by Stokes’ law.
### Key Equations
Magnitude of static friction ${f}_{\text{s}}\le {\mu }_{\text{s}}N$ Magnitude of kinetic friction ${f}_{k}={\mu }_{k}N$ Centripetal force ${F}_{\text{c}}=m\frac{{v}^{2}}{r}\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}{F}_{\text{c}}=mr{\omega }^{2}$ Ideal angle of a banked curve $\text{tan}\phantom{\rule{0.2em}{0ex}}\theta =\frac{{v}^{2}}{rg}$ Drag force ${F}_{D}=\frac{1}{2}C\rho A{v}^{2}$ Stokes’ law ${F}_{\text{s}}=6\pi r\eta v$
### Conceptual Questions
Athletes such as swimmers and bicyclists wear body suits in competition. Formulate a list of pros and cons of such suits.
The pros of wearing body suits include: (1) the body suit reduces the drag force on the swimmer and the athlete can move more easily; (2) the tightness of the suit reduces the surface area of the athlete, and even though this is a small amount, it can make a difference in performance time. The cons of wearing body suits are: (1) The tightness of the suits can induce cramping and breathing problems. (2) Heat will be retained and thus the athlete could overheat during a long period of use.
Two expressions were used for the drag force experienced by a moving object in a liquid. One depended upon the speed, while the other was proportional to the square of the speed. In which types of motion would each of these expressions be more applicable than the other one?
As cars travel, oil and gasoline leaks onto the road surface. If a light rain falls, what does this do to the control of the car? Does a heavy rain make any difference?
The oil is less dense than the water and so rises to the top when a light rain falls and collects on the road. This creates a dangerous situation in which friction is greatly lowered, and so a car can lose control. In a heavy rain, the oil is dispersed and does not affect the motion of cars as much.
Why can a squirrel jump from a tree branch to the ground and run away undamaged, while a human could break a bone in such a fall?
### Problems
The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the terminal velocity (in meters per second and kilometers per hour) of an 80.0-kg skydiver falling in a pike (headfirst) position with a surface area of $0.140\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}$.
115 m/s or 414 km/h
A 60.0-kg and a 90.0-kg skydiver jump from an airplane at an altitude of $6.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\text{m}$, both falling in the pike position. Make some assumption on their frontal areas and calculate their terminal velocities. How long will it take for each skydiver to reach the ground (assuming the time to reach terminal velocity is small)? Assume all values are accurate to three significant digits.
A 560-g squirrel with a surface area of $930\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{2}$ falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a drag coefficient for a horizontal skydiver.) What will be the velocity of a 56-kg person hitting the ground, assuming no drag contribution in such a short distance?
${v}_{\text{T}}=25\phantom{\rule{0.2em}{0ex}}\text{m/s;}{\text{v}}_{2}=9.9\phantom{\rule{0.2em}{0ex}}\text{m/s}$
To maintain a constant speed, the force provided by a car’s engine must equal the drag force plus the force of friction of the road (the rolling resistance). (a) What are the drag forces at 70 km/h and 100 km/h for a Toyota Camry? (Drag area is $0.70\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}$) (b) What is the drag force at 70 km/h and 100 km/h for a Hummer H2? (Drag area is $2.44\phantom{\rule{0.2em}{0ex}}{\text{m}}^{2}\right)$ Assume all values are accurate to three significant digits.
By what factor does the drag force on a car increase as it goes from 65 to 110 km/h?
${\left(\frac{110}{65}\right)}^{2}=2.86$ times
Calculate the velocity a spherical rain drop would achieve falling from 5.00 km (a) in the absence of air drag (b) with air drag. Take the size across of the drop to be 4 mm, the density to be $1.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}$, and the surface area to be $\pi {r}^{2}$.
Using Stokes’ law, verify that the units for viscosity are kilograms per meter per second.
Stokes’ law is ${F}_{\text{s}}=6\pi r\eta v.$ Solving for the viscosity, $\eta =\frac{{F}_{\text{s}}}{6\pi rv}.$ Considering only the units, this becomes $\left[\eta \right]=\frac{\text{kg}}{\text{m}·\text{s}}.$
Find the terminal velocity of a spherical bacterium (diameter $2.00\phantom{\rule{0.2em}{0ex}}\text{μm}$) falling in water. You will first need to note that the drag force is equal to the weight at terminal velocity. Take the density of the bacterium to be $1.10\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}$.
Stokes’ law describes sedimentation of particles in liquids and can be used to measure viscosity. Particles in liquids achieve terminal velocity quickly. One can measure the time it takes for a particle to fall a certain distance and then use Stokes’ law to calculate the viscosity of the liquid. Suppose a steel ball bearing (density $7.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}$, diameter 3.0 mm) is dropped in a container of motor oil. It takes 12 s to fall a distance of 0.60 m. Calculate the viscosity of the oil.
$0.76\phantom{\rule{0.2em}{0ex}}\text{kg/m}·\text{s}$
Suppose that the resistive force of the air on a skydiver can be approximated by $f=\text{−}b{v}^{2}.$ If the terminal velocity of a 50.0-kg skydiver is 60.0 m/s, what is the value of b?
A small diamond of mass 10.0 g drops from a swimmer’s earring and falls through the water, reaching a terminal velocity of 2.0 m/s. (a) Assuming the frictional force on the diamond obeys $f=\text{−}bv,$ what is b? (b) How far does the diamond fall before it reaches 90 percent of its terminal speed?
a. 0.049 kg/s; b. 0.57 m
(a) What is the final velocity of a car originally traveling at 50.0 km/h that decelerates at a rate of $0.400\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ for 50.0 s? Assume a coefficient of friction of 1.0. (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent?
A 75.0-kg woman stands on a bathroom scale in an elevator that accelerates from rest to 30.0 m/s in 2.00 s. (a) Calculate the scale reading in newtons and compare it with her weight. (The scale exerts an upward force on her equal to its reading.) (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent?
a. 1860 N, 2.53; b. The value (1860 N) is more force than you expect to experience on an elevator. The force of 1860 N is 418 pounds, compared to the force on a typical elevator of 904 N (which is about 203 pounds); this is calculated for a speed from 0 to 10 miles per hour, which is about 4.5 m/s, in 2.00 s). c. The acceleration $a=1.53\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}g$ is much higher than any standard elevator. The final speed is too large (30.0 m/s is VERY fast)! The time of 2.00 s is not unreasonable for an elevator.
(a) Calculate the minimum coefficient of friction needed for a car to negotiate an unbanked 50.0 m radius curve at 30.0 m/s. (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent?
As shown below, if $M=5.50\phantom{\rule{0.2em}{0ex}}\text{kg,}$ what is the tension in string 1?
189 N
As shown below, if $F=60.0\phantom{\rule{0.2em}{0ex}}\text{N}$ and $M=4.00\phantom{\rule{0.2em}{0ex}}\text{kg,}$ what is the magnitude of the acceleration of the suspended object? All surfaces are frictionless.
As shown below, if $M=6.0\phantom{\rule{0.2em}{0ex}}\text{kg,}$ what is the tension in the connecting string? The pulley and all surfaces are frictionless.
15 N
A small space probe is released from a spaceship. The space probe has mass 20.0 kg and contains 90.0 kg of fuel. It starts from rest in deep space, from the origin of a coordinate system based on the spaceship, and burns fuel at the rate of 3.00 kg/s. The engine provides a constant thrust of 120.0 N. (a) Write an expression for the mass of the space probe as a function of time, between 0 and 30 seconds, assuming that the engine ignites fuel beginning at $t=0.$ (b) What is the velocity after 15.0 s? (c) What is the position of the space probe after 15.0 s, with initial position at the origin? (d) Write an expression for the position as a function of time, for $t>30.0\phantom{\rule{0.2em}{0ex}}\text{s}\text{.}$
A half-full recycling bin has mass 3.0 kg and is pushed up a $40.0\text{°}$ incline with constant speed under the action of a 26-N force acting up and parallel to the incline. The incline has friction. What magnitude force must act up and parallel to the incline for the bin to move down the incline at constant velocity?
12 N
A child has mass 6.0 kg and slides down a $35\text{°}$ incline with constant speed under the action of a 34-N force acting up and parallel to the incline. What is the coefficient of kinetic friction between the child and the surface of the incline?
The two barges shown here are coupled by a cable of negligible mass. The mass of the front barge is $2.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{kg}$ and the mass of the rear barge is $3.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{kg}\text{.}$ A tugboat pulls the front barge with a horizontal force of magnitude $20.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N,}$ and the frictional forces of the water on the front and rear barges are $8.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N}$ and $10.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N,}$ respectively. Find the horizontal acceleration of the barges and the tension in the connecting cable.
${a}_{x}=0.40\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ and $T=11.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N}$
If the order of the barges of the preceding exercise is reversed so that the tugboat pulls the $3.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\text{-kg}$ barge with a force of $20.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N},$ what are the acceleration of the barges and the tension in the coupling cable?
An object with mass m moves along the x-axis. Its position at any time is given by $x\left(t\right)=p{t}^{3}+q{t}^{2}$ where p and q are constants. Find the net force on this object for any time t.
m(6pt + 2q)
A helicopter with mass $2.35\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{kg}$ has a position given by $\stackrel{\to }{r}\left(t\right)=\left(0.020\phantom{\rule{0.2em}{0ex}}{t}^{3}\right)\stackrel{^}{i}+\left(2.2t\right)\stackrel{^}{j}-\left(0.060\phantom{\rule{0.2em}{0ex}}{t}^{2}\right)\stackrel{^}{k}.$ Find the net force on the helicopter at $t=3.0\phantom{\rule{0.2em}{0ex}}\text{s}\text{.}$
Located at the origin, an electric car of mass m is at rest and in equilibrium. A time dependent force of $\stackrel{\to }{F}\left(t\right)$ is applied at time $t=0$, and its components are ${F}_{x}\left(t\right)=p+nt$ and ${F}_{y}\left(t\right)=qt$ where p, q, and n are constants. Find the position $\stackrel{\to }{r}\left(t\right)$ and velocity $\stackrel{\to }{v}\left(t\right)$ as functions of time t.
$\stackrel{\to }{v}\left(t\right)=\left(\frac{pt}{m}+\frac{n{t}^{2}}{2m}\right)\stackrel{^}{i}+\left(\frac{q{t}^{2}}{2}\right)\stackrel{^}{j}$ and $\stackrel{\to }{r}\left(t\right)=\left(\frac{p{t}^{2}}{2m}+\frac{n{t}^{3}}{6m}\right)\stackrel{^}{i}+\left(\frac{q{t}^{3}}{60m}\right)\stackrel{^}{j}$
A particle of mass m is located at the origin. It is at rest and in equilibrium. A time-dependent force of $\stackrel{\to }{F}\left(t\right)$ is applied at time $t=0$, and its components are ${F}_{x}\left(t\right)=pt$ and ${F}_{y}\left(t\right)=n+qt$ where p, q, and n are constants. Find the position $\stackrel{\to }{r}\left(t\right)$ and velocity $\stackrel{\to }{v}\left(t\right)$ as functions of time t.
A 2.0-kg object has a velocity of $4.0\stackrel{^}{i}\phantom{\rule{0.2em}{0ex}}\text{m/s}$ at $t=0.$ A constant resultant force of $\left(2.0\stackrel{^}{i}+4.0\stackrel{^}{j}\right)\phantom{\rule{0.2em}{0ex}}\text{N}$ then acts on the object for 3.0 s. What is the magnitude of the object’s velocity at the end of the 3.0-s interval?
9.2 m/s
A 1.5-kg mass has an acceleration of $\left(4.0\stackrel{^}{i}-3.0\stackrel{^}{j}\right)\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}.$ Only two forces act on the mass. If one of the forces is $\left(2.0\stackrel{^}{i}-1.4\stackrel{^}{j}\right)\phantom{\rule{0.2em}{0ex}}\text{N,}$ what is the magnitude of the other force?
A box is dropped onto a conveyor belt moving at 3.4 m/s. If the coefficient of friction between the box and the belt is 0.27, how long will it take before the box moves without slipping?
1.3 s
Shown below is a 10.0-kg block being pushed by a horizontal force $\stackrel{\to }{F}$ of magnitude 200.0 N. The coefficient of kinetic friction between the two surfaces is 0.50. Find the acceleration of the block.
As shown below, the mass of block 1 is ${m}_{1}=4.0\phantom{\rule{0.2em}{0ex}}\text{kg,}$ while the mass of block 2 is ${m}_{2}=8.0\phantom{\rule{0.2em}{0ex}}\text{kg}\text{.}$ The coefficient of friction between ${m}_{1}$ and the inclined surface is ${\mu }_{\text{k}}=0.40.$ What is the acceleration of the system?
$5.4\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$
A student is attempting to move a 30-kg mini-fridge into her dorm room. During a moment of inattention, the mini-fridge slides down a 35 degree incline at constant speed when she applies a force of 25 N acting up and parallel to the incline. What is the coefficient of kinetic friction between the fridge and the surface of the incline?
A crate of mass 100.0 kg rests on a rough surface inclined at an angle of $37.0\text{°}$ with the horizontal. A massless rope to which a force can be applied parallel to the surface is attached to the crate and leads to the top of the incline. In its present state, the crate is just ready to slip and start to move down the plane. The coefficient of friction is $80%$ of that for the static case. (a) What is the coefficient of static friction? (b) What is the maximum force that can be applied upward along the plane on the rope and not move the block? (c) With a slightly greater applied force, the block will slide up the plane. Once it begins to move, what is its acceleration and what reduced force is necessary to keep it moving upward at constant speed? (d) If the block is given a slight nudge to get it started down the plane, what will be its acceleration in that direction? (e) Once the block begins to slide downward, what upward force on the rope is required to keep the block from accelerating downward?
a. 0.60; b. 1200 N; c. $1.2\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ and 1080 N; d. $-1.2\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2};$ e. 120 N
A car is moving at high speed along a highway when the driver makes an emergency braking. The wheels become locked (stop rolling), and the resulting skid marks are 32.0 meters long. If the coefficient of kinetic friction between tires and road is 0.550, and the acceleration was constant during braking, how fast was the car going when the wheels became locked?
A crate having mass 50.0 kg falls horizontally off the back of the flatbed truck, which is traveling at 100 km/h. Find the value of the coefficient of kinetic friction between the road and crate if the crate slides 50 m on the road in coming to rest. The initial speed of the crate is the same as the truck, 100 km/h.
0.789
A 15-kg sled is pulled across a horizontal, snow-covered surface by a force applied to a rope at 30 degrees with the horizontal. The coefficient of kinetic friction between the sled and the snow is 0.20. (a) If the force is 33 N, what is the horizontal acceleration of the sled? (b) What must the force be in order to pull the sled at constant velocity?
A 30.0-g ball at the end of a string is swung in a vertical circle with a radius of 25.0 cm. The rotational velocity is 200.0 cm/s. Find the tension in the string: (a) at the top of the circle, (b) at the bottom of the circle, and (c) at a distance of 12.5 cm from the center of the circle $\left(r=12.5\phantom{\rule{0.2em}{0ex}}\text{cm}\right).$
a. 0.186 N; b. 774 N; c. 0.48 N
A particle of mass 0.50 kg starts moves through a circular path in the xy-plane with a position given by $\stackrel{\to }{r}\left(t\right)=\left(4.0\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}3t\right)\stackrel{^}{i}+\left(4.0\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}3t\right)\stackrel{^}{j}$ where r is in meters and t is in seconds. (a) Find the velocity and acceleration vectors as functions of time. (b) Show that the acceleration vector always points toward the center of the circle (and thus represents centripetal acceleration). (c) Find the centripetal force vector as a function of time.
A stunt cyclist rides on the interior of a cylinder 12 m in radius. The coefficient of static friction between the tires and the wall is 0.68. Find the value of the minimum speed for the cyclist to perform the stunt.
13 m/s
When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0 cm. The body and spring are placed on a horizontal frictionless surface and rotated about the held end of the spring at 2.0 rev/s. How far is the spring stretched?
Railroad tracks follow a circular curve of radius 500.0 m and are banked at an angle of $5.00\text{°}$. For trains of what speed are these tracks designed?
20.7 m/s
A plumb bob hangs from the roof of a railroad car. The car rounds a circular track of radius 300.0 m at a speed of 90.0 km/h. At what angle relative to the vertical does the plumb bob hang?
An airplane flies at 120.0 m/s and banks at a $30\text{°}$ angle. If its mass is $2.50\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{kg,}$ (a) what is the magnitude of the lift force? (b) what is the radius of the turn?
a. 28,300 N; b. 2540 m
The position of a particle is given by $\stackrel{\to }{r}\left(t\right)=A\left(\text{cos}\phantom{\rule{0.2em}{0ex}}\omega t\stackrel{^}{i}+\text{sin}\phantom{\rule{0.2em}{0ex}}\omega t\stackrel{^}{j}\right),$ where $\omega$ is a constant. (a) Show that the particle moves in a circle of radius A. (b) Calculate $d\stackrel{\to }{r}\text{/}dt$ and then show that the speed of the particle is a constant ${A}_{\omega }.$ (c) Determine ${d}^{2}\stackrel{\to }{r}\text{/}d{t}^{2}$ and show that a is given by${a}_{\text{c}}=r{\omega }^{2}.$ (d) Calculate the centripetal force on the particle. [Hint: For (b) and (c), you will need to use $\left(d\text{/}dt\right)\left(\text{cos}\phantom{\rule{0.2em}{0ex}}\omega t\right)=\text{−}\omega \phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\omega t$ and $\left(d\text{/}dt\right)\left(\text{sin}\phantom{\rule{0.2em}{0ex}}\omega t\right)=\omega \phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\omega t.$
Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown below. The coefficient of kinetic friction between the blocks and the surface is 0.25. If each block has an acceleration of $2.0\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ to the right, what is the magnitude F of the applied force?
25 N
As shown below, the coefficient of kinetic friction between the surface and the larger block is 0.20, and the coefficient of kinetic friction between the surface and the smaller block is 0.30. If $F=10\phantom{\rule{0.2em}{0ex}}\text{N}$ and $M=1.0\phantom{\rule{0.2em}{0ex}}\text{kg}$, what is the tension in the connecting string?
In the figure, the coefficient of kinetic friction between the surface and the blocks is ${\mu }_{\text{k}}.$ If $M=1.0\phantom{\rule{0.2em}{0ex}}\text{kg,}$ find an expression for the magnitude of the acceleration of either block (in terms of F, ${\mu }_{\text{k}},$ and g).
$a=\frac{F}{4}-{\mu }_{k}g$
Two blocks are stacked as shown below, and rest on a frictionless surface. There is friction between the two blocks (coefficient of friction $\mu$). An external force is applied to the top block at an angle $\theta$ with the horizontal. What is the maximum force F that can be applied for the two blocks to move together?
A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it rests is 0.24. What maximum distance can the truck travel (starting from rest and moving horizontally with constant acceleration) in 3.0 s without having the box slide?
14 m
A double-incline plane is shown below. The coefficient of friction on the left surface is 0.30, and on the right surface 0.16. Calculate the acceleration of the system.
### Challenge Problems
In a later chapter, you will find that the weight of a particle varies with altitude such that $w=\frac{mg{r}_{0}{}^{2}}{{r}^{2}}$ where ${r}_{0}{}^{}$ is the radius of Earth and r is the distance from Earth’s center. If the particle is fired vertically with velocity ${v}_{0}{}^{}$ from Earth’s surface, determine its velocity as a function of position r. (Hint: use ${a}^{}dr={v}^{}dv,$ the rearrangement mentioned in the text.)
$v=\sqrt{{v}_{0}{}^{2}-2g{r}_{0}\left(1-\frac{{r}_{0}}{r}\right)}$
A large centrifuge, like the one shown below, is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries. (a) At what angular velocity is the centripetal acceleration 10g if the rider is 15.0 m from the center of rotation? (b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in the bottom accompanying figure. At what angle $\theta$ below the horizontal will the cage hang when the centripetal acceleration is 10g? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free-body diagram of the forces to see what the angle $\theta$ should be.)
A car of mass 1000.0 kg is traveling along a level road at 100.0 km/h when its brakes are applied. Calculate the stopping distance if the coefficient of kinetic friction of the tires is 0.500. Neglect air resistance. (Hint: since the distance traveled is of interest rather than the time, x is the desired independent variable and not t. Use the Chain Rule to change the variable: $\frac{dv}{dt}=\frac{dv}{dx}\phantom{\rule{0.2em}{0ex}}\frac{dx}{dt}=v\frac{dv}{dx}.\right)$
78.7 m
An airplane flying at 200.0 m/s makes a turn that takes 4.0 min. What bank angle is required? What is the percentage increase in the perceived weight of the passengers?
A skydiver is at an altitude of 1520 m. After 10.0 seconds of free fall, he opens his parachute and finds that the air resistance, ${F}_{\text{D}}$, is given by the formula ${F}_{\text{D}}=\text{−}bv,$ where b is a constant and v is the velocity. If $b=0.750,$ and the mass of the skydiver is 82.0 kg, first set up differential equations for the velocity and the position, and then find: (a) the speed of the skydiver when the parachute opens, (b) the distance fallen before the parachute opens, (c) the terminal velocity after the parachute opens (find the limiting velocity), and (d) the time the skydiver is in the air after the parachute opens.
a. 53.9 m/s; b. 328 m; c. 4.58 m/s; d. 257 s
In a television commercial, a small, spherical bead of mass 4.00 g is released from rest at $t=0$ in a bottle of liquid shampoo. The terminal speed is observed to be 2.00 cm/s. Find (a) the value of the constant b in the equation $v=\frac{mg}{b}\left(1-{e}^{\text{−}bt\text{/}m}\right),$ and (b) the value of the resistive force when the bead reaches terminal speed.
A boater and motor boat are at rest on a lake. Together, they have mass 200.0 kg. If the thrust of the motor is a constant force of 40.0 N in the direction of motion, and if the resistive force of the water is numerically equivalent to 2 times the speed v of the boat, set up and solve the differential equation to find: (a) the velocity of the boat at time t; (b) the limiting velocity (the velocity after a long time has passed).
a. $v=20.0\left(1-{e}^{-0.01t}\right);$ b. ${v}_{\text{limiting}}=20\phantom{\rule{0.2em}{0ex}}\text{m/s}$
### Glossary
drag force
force that always opposes the motion of an object in a fluid; unlike simple friction, the drag force is proportional to some function of the velocity of the object in that fluid
terminal velocity
constant velocity achieved by a falling object, which occurs when the weight of the object is balanced by the upward drag force |
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# Calculus II - 5
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Stewart Calculus Section 7.4
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• ### Calculus II - 5
1. 1. 7.4 Integration of Rational FunctionsFirst step: if the rational function isimproper, decompose it to a polynomial plusa proper rational function.Second step: factor the dominator to be aproduct of linear factors and/or irreduciblequadratic factors.Third step: decompose the proper rationalfunction to be a sum of partial fractions.
2. 2. CASE I: the dominator is a product ofdistinct linear factors.CASE II: the dominator is a product oflinear factors, some of which are repeated.CASE III: the dominator contains distinctirreducible quadratic factors.Case IV: the dominator contains repeatedirreducible quadratic factors.
3. 3. Ex: find +
4. 4. Ex: find + I CA S E IIWe recognize + is irreducible.
5. 5. Ex: find + I CA S E IIWe recognize + is irreducible.Complete the square: + =( ) +
6. 6. Ex: find + I CA S E II We recognize + is irreducible. Complete the square: + =( ) += ( + ) − = + +
7. 7. Ex: find + I CA S E II We recognize + is irreducible. Complete the square: + =( ) += ( + ) − = + + = + +
8. 8. Ex: find + I CA S E II We recognize + is irreducible. Complete the square: + =( ) += ( + ) − = + + = + + = ( + ) +
9. 9. Ex: find + I CA S E II We recognize + is irreducible. Complete the square: + =( ) += ( + ) − = + + = + + = ( + ) + = ( + ) +
10. 10. +Ex: find +
11. 11. +Ex: find + + +We assume = + CAS E III + +
12. 12. +Ex: find + + +We assume = + CAS E III + +and solve = , = , = .
13. 13. +Ex: find + + +We assume = + CAS E III + +and solve = , = , = . +so = + + + +
14. 14. +Ex: find + + +We assume = + CAS E III + +and solve = , = , = . +so = + + + + = | |+ ( + ) +
15. 15. + +Ex: find ( + )
16. 16. + +Ex: find ( + ) CAS E IV + + + +We assume = + ( + ) + ( + )
17. 17. + +Ex: find ( + ) CAS E IV + + + +We assume = + ( + ) + ( + )and solve = , = , = , = .
18. 18. + +Ex: find ( + ) CAS E IV + + + +We assume = + ( + ) + ( + )and solve = , = , = , = . + + +so = ( + ) + ( + )
19. 19. + +Ex: find ( + ) CAS E IV + + + +We assume = + ( + ) + ( + )and solve = , = , = , = . + + +so = ( + ) + ( + ) = ( + )+ + + ( + )
20. 20. +Ex: find
21. 21. +Ex: find √Let = +
22. 22. +Ex: find √Let = + zation Rati onali +then =
23. 23. +Ex: find √Let = + zation Rati onali +then = = +
24. 24. +Ex: find √Let = + zation Rati onali +then = = + = + | | | + |+
25. 25. +Ex: find √Let = + zation Rati onali +then = = + = + | | | + |+ + = + + + + + |
# 2.4 Hypothesis testing of single mean and single proportion: examples (Page 3/6)
Page 3 / 6
Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65.He samples ten statistics students and obtains the scores
• 65
• 65
• 70
• 67
• 66
• 63
• 63
• 68
• 72
• 71
. He performs a hypothesis test using a 5% level of significance. The data are from a normal distribution.
Set up the Hypothesis Test:
A 5% level of significance means that $\alpha =0.05$ . This is a test of a single population mean .
${H}_{o}$ : $\mu$ $=65\phantom{\rule{20pt}{0ex}}$ ${H}_{a}$ : $\mu$ $>65$
Since the instructor thinks the average score is higher, use a " $>$ ". The " $>$ " means the test is right-tailed.
Determine the distribution needed:
Random variable: $\overline{X}$ = average score on the first statistics test.
Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given . You are only given $n=10$ sample data values. Notice also that the data come from a normal distribution. This means that thedistribution for the test is a student's-t.
Use ${t}_{\text{df}}$ . Therefore, the distribution for the test is ${t}_{9}$ where $n=10$ and $\text{df}=10-1=9$ .
Calculate the p-value using the Student's-t distribution:
$\text{p-value}=P\left($ $\overline{x}$ $>67$ ) $=0.0396$ where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.
Interpretation of the p-value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 67 or more.
Compare $\alpha$ and the p-value:
Since $\alpha =.05$ and $\text{p-value}=0.0396$ . Therefore, $\alpha >\text{p-value}$ .
Make a decision: Since $\alpha >\text{p-value}$ , reject ${H}_{o}$ .
This means you reject $\mu =65$ . In other words, you believe the average test score is more than 65.
Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.
The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:
Put the data into a list. Press STAT and arrow over to TESTS . Press 2:T-Test . Arrow over to Data and press ENTER . Arrow down and enter 65 for ${\mu }_{0}$ , the name of the list where you put the data, and 1 for Freq: . Arrow down to $\mu :$ and arrow over to $>{\mu }_{0}$ . Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the p-value ( $p=0.0396$ ) but it also calculates the test statistic (t-score) for the sample mean, the sample mean, and the sample standarddeviation. $\mu >65$ is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with $t=1.9781$ (test statistic) and $p=0.0396$ (p-value). Make sure when you use Draw that no other equations are highlighted in $Y=$ and the plots are turned off.
Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentageis the same or different from 50% . Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level ofsignificance.
Set up the Hypothesis Test:
The 1% level of significance means that $\alpha =0.01$ . This is a test of a single population proportion .
${H}_{o}$ : $p$ $=0.50\phantom{\rule{20pt}{0ex}}$ ${H}_{a}$ : $p$ $\ne 0.50$
The words "is the same or different from" tell you this is a two-tailed test.
Calculate the distribution needed:
Random variable: $P\text{'}$ = the percent of of first-time brides who are younger than their grooms.
Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for $P\text{'}$ , the estimated proportion.
$P\text{'}$ ~ $N$ $\left(p,\sqrt{\frac{p\cdot q}{n}}\right)\phantom{\rule{20pt}{0ex}}$ Therefore, $P\text{'}$ ~ $N$ $\left(0.5,\sqrt{\frac{0.5\cdot 0.5}{100}}\right)$ where $p=0.50$ , $q=1-p=0.50$ , and $n=100$ .
Calculate the p-value using the normal distribution for proportions:
$\text{p-value}=P\left(\mathrm{p\text{'}}$ $()$ $0.47$ or $\mathrm{p\text{'}}>0.53$ ) $=0.5485$
where $x=53$ , $p\text{'}=\frac{x}{n}$ $=\frac{53}{100}=0.53$ .
Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion $p\text{'}$ is 0.53 or more OR 0.47 or less (see the graph below).
$\mu =p=0.50$ comes from ${H}_{o}$ , the null hypothesis.
$p\text{'}$ $=0.53$ . Since the curve is symmetrical andthe test is two-tailed, the $p\text{'}$ for the left tail is equal to $0.50-0.03=0.47$ where $\mu =p=0.50$ . (0.03 is the differencebetween 0.53 and 0.50.)
Compare $\alpha$ and the p-value:
Since $\alpha =0.01$ and $\text{p-value}=0.5485$ . Therefore, $\alpha$ $()$ $\text{p-value}$ .
Make a decision: Since $\alpha$ $()$ $\text{p-value}$ , you cannot reject ${H}_{o}$ .
Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides that are younger than their grooms isdifferent from 50%.
The p-value can easily be calculated using the TI-83+ and the TI-84 calculators:
Press STAT and arrow over to TESTS . Press 5:1-PropZTest . Enter .5 for ${p}_{0}$ , 53 for $x$ and 100 for $n$ . Arrow down to Prop and arrow to not equals ${p}_{0}$ . Press ENTER . Arrow down to Calculate and press ENTER . The calculator calculates the p-value ( $p=0.5485$ ) and the test statistic (z-score). Prop not equals .5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with $z=0.6$ (test statistic) and $p=0.5485$ (p-value). Make sure when you use Draw that no other equations are highlighted in $Y=$ and the plots are turned off.
The Type I and Type II errors are as follows:
The Type I error is to conclude that the proportion of first-time brides that are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%.(Reject the null hypothesis when the null hypothesis is true).
The Type II error is there is not enough evidence to conclude that the proportion of first time brides that are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.)
use the y -intercept and slope to sketch the graph of the equation y=6x
how do we prove the quadratic formular
hello, if you have a question about Algebra 2. I may be able to help. I am an Algebra 2 Teacher
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Seidu
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Opoku
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4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years? Kala Reply lim x to infinity e^1-e^-1/log(1+x) given eccentricity and a point find the equiation Moses Reply 12, 17, 22.... 25th term Alexandra Reply 12, 17, 22.... 25th term Akash College algebra is really hard? Shirleen Reply Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table. Carole I'm 13 and I understand it great AJ I am 1 year old but I can do it! 1+1=2 proof very hard for me though. Atone Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily. Vedant hi vedant can u help me with some assignments Solomon find the 15th term of the geometric sequince whose first is 18 and last term of 387 Jerwin Reply I know this work salma The given of f(x=x-2. then what is the value of this f(3) 5f(x+1) virgelyn Reply hmm well what is the answer Abhi If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10 Augustine A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place. Kimberly Reply Jeannette has$5 and \$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
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why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Berger describes sociologists as concerned with
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Radical expressions are written in simplest terms when. The terms are unlike radicals. The terms are like radicals. Another way to do the above simplification would be to remember our squares. The index is as small as possible. Therefore, in every simplifying radical problem, check to see if the given radical itself, can be simplified. If the indices and radicands are the same, then add or subtract the terms in front of each like radical. In this section we will define radical notation and relate radicals to rational exponents. Example 1: Adding and Subtracting Square-Root Expressions Add or subtract. Simplify each of the following. In other words, these are not like radicals. Step 2. Example 1: Add or subtract to simplify radical expression: $2 \sqrt{12} + \sqrt{27}$ Solution: Step 1: Simplify radicals This is because some are the pinyin for the dictionary radical name and some are the pinyin for what the stroke is called. Mathematically, a radical is represented as x n. This expression tells us that a number x is … Multiplying Radicals – Techniques & Examples A radical can be defined as a symbol that indicate the root of a number. We will also give the properties of radicals and some of the common mistakes students often make with radicals. The above expressions are simplified by first transforming the unlike radicals to like radicals and then adding/subtracting When it is not obvious to obtain a common radicand from 2 different radicands, decompose them into prime numbers. Click here to review the steps for Simplifying Radicals. Do not combine. Decompose 12 and 108 into prime factors as follows. Step 2: To add or subtract radicals, the indices and what is inside the radical (called the radicand) must be exactly the same. Use the radical positions table as a reference. To see if they can be combined, we need to simplify each radical separately from each Simplify: $$\sqrt{16} + \sqrt{4}$$ (unlike radicals, so you can’t combine them…..yet) Don’t assume that just because you have unlike radicals that you won’t be able to simplify the expression. Combining Unlike Radicals Example 1: Simplify 32 + 8 As they are, these radicals cannot be combined because they do not have the same radicand. Combine like radicals. Subtraction of radicals follows the same set of rules and approaches as addition—the radicands and the indices must be the same for two (or more) radicals to be subtracted. For example with丨the radical is gǔn and shù is the name of a stroke. We will also define simplified radical form and show how to rationalize the denominator. Simplify each radical. Example 1. You probably already knew that 12 2 = 144, so obviously the square root of 144 must be 12.But my steps above show how you can switch back and forth between the different formats (multiplication inside one radical, versus multiplication of two radicals) to help in the simplification process. Subtract Radicals. To avoid ambiguities amongst the different kinds of “enclosed” radicals, search for these in hiragana. For example, to view all radicals in the “hang down” position, type たれ or “tare” into the search field. If you don't know how to simplify radicals go to Simplifying Radical Expressions. A radical expression is any mathematical expression containing a radical symbol (√). A. (The radicand of the first is 32 and the radicand of the second is 8.) In the three examples that follow, subtraction has been rewritten as addition of the opposite. No radicals appear in the denominator. The steps in adding and subtracting Radical are: Step 1. B. The radicand contains no fractions. The radicand contains no factor (other than 1) which is the nth or greater power of an integer or polynomial. Yes, you are right there is different pinyin for some of the radicals. Square root, cube root, forth root are all radicals. Simplify radicals. 12 and 108 into prime factors as follows this is because some the... Show how to rationalize the denominator every Simplifying radical problem, check see! Addition of the common mistakes students unlike radicals examples make with radicals the given radical,... Which is the name of a stroke ) which is the name of a.. ” radicals, search for these in hiragana factors as follows shù is name! Define simplified radical form and show how to simplify radicals go to Simplifying radical problem, to! And shù is the nth or greater power of an integer or polynomial a.! Radical can be unlike radicals examples if the given radical itself, can be defined as a that... 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# Unit2.pdf
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8 Feb 2013
School
Department
Professor
Math 1225A/B
Unit 2:
Review of Introductory Calculus
Differentiation Rules and Other Calculus Techniques
(text reference: Sections 5.3 and 5.4 – mixed together
custom text pgs. 19 - 36)
c
V. Olds 2012
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16 Unit 2
2 Differentiation of Exponential and Logarithmic Functions
In this section we review all the differentiation rules you learnt in your Introductory Calculus course,
mostly in the context of finding derivatives of functions involving exponentials and logarithms. We
start with the derivatives of the natural exponential and logarithmic functions. Then later we derive
formulas for the derivatives of more general exponential and logarithmic functions, i.e. with any
base b(with b > 0 and b6= 1, of course). Along the way, we will also review implicit differentiation
and logarithmic differentiation, as well as reviewing what derivatives tell us about the shape of the
graph of a function.
Derivatives of the Natural Exponential and Logarithmic Functions
Recall the Definition of Derivative.
Definition 2.1. For any function f, the derivative function f(x) is the function given by
f(x) = lim
h0
f(x+h)f(x)
h
provided this limit exists. (If the limit DNE, we say that fis not differentiable.)
Remember: This is the formula for the instantaneous rate of change in f(x) – which is the slope of
the tangent line to y=f(x). So if we evaluate f(x) at x=a, we get the slope of the tangent line
to y=f(x) at x=a.
You probably remember doing things like the following example.
Example 2.1.Use the definition of derivative to find f(x) if f(x) = x2.
Solution:
f(x) = lim
h0
f(x+h)f(x)
h= lim
h0
(x+h)2x2
h
= lim
h0
x2+ 2xh +h2x2
h= lim
h0
2xh +h2
h
= lim
h0
h(2x+h)
h= lim
h0(2x+h) = 2x
Unfortunately, applying the definition of derivative to find the derivative of f(x) = exor the deriva-
tive of f(x) = ln xrequires mathematics that is beyond the scope of this course, to evaluate the
limits. Never mind that. We will simply accept the following simple rules:
Rule: d
dx (ex) = exand Rule: d
dx (ln x) = 1
x
Notice: We said that ewas the base such that y=excrosses the y-axis with slope 1. For f(x) = ex,
the rule says that f(x) = ex, so when x= 0 we get f(0) = e0= 1. As promised, the slope (of the
tangent line) as the curve crosses the y-axis (i.e. at x= 0) is 1.
Now that we know the 2 facts mentioned above (the Rules), we can use them to recognize a
certain limit as giving the derivative of some particular function, at some value, according to the
definition of derivative. That allows us to find the value of the specified limit without having to
evaluate it, by instead evaluating the derivative function.
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Unit 2 17
That is, if we recognize that a limit has the form lim
h0f(a+h)f(a)
hfor some function f(x)
and some value a, then we know that this limit is giving f(x) when evaluated at x=a, so we can
find the value of the limit simply by evaluating f(a). We find the derivative of the function f(x)
which we see being used, and then evaluate it at the particular avalue we see in the limit.
Example 2.2.Evaluate lim
h0ln(2 + h)ln 2
h.
Solution:
Looking at this limit, we see that we can’t evaluate it in the usual way, because we don’t know how
to get the hout of the logarithm, and we can’t cancel the hin the denominator while the hin the
numerator is inside a logarithm. However, we do see a limit which has a familiar form. First of
all, we are taking the limit as hgoes to 0, and the denominator is just h. Also, in the numerator,
we see something being done first to 2 + hand then to 2. The thing that’s being done to each of
these is “taking the natural logarithm of”. And the numerator is the difference of those two natural
logarithms.
This tells us that something’s being done with the function that takes the natural logarithm of
a number. That is, we’ve got the function f(x) = ln x. Expressed in terms of that function, the
numerator of the limit is f(2 + h)f(2). That is, we’re evaluating f(x+h)f(x) at the particular
value x= 2. And in fact using f(x) = ln x, we can express the whole limit as
lim
h0f(x+h)f(x)
hevaluated at x= 2
But we know that lim
h0f(x+h)f(x)
h=f(x), and so we simply need to evaluate f(2) to find
the value of the given limit. That is, since we know that for f(x) = ln xthe derivative function is
f(x) = 1
x, we have:
lim
h0ln(x+h)ln x
h=f(x) = 1
xand so lim
h0ln(2 + h)ln 2
h=f(2) = 1
2
Of course, we don’t need to use the definition of derivative every time we want to evaluate a
derivative. You learned a variety of differentiation rules, including: the constant multiplier rule, the
sum and difference rules, the power rule, the product rule and the quotient rule. Let’s briefly review
what those rules say and how they are used.
The constant multiplier rule:
For any differentiable function fand any constant c, the derivative of ctimes f(x) is ctimes
f(x). For instance, for f(x) = 2exwe get f(x) = 2 d
dx (ex)= 2ex.
The sum and difference rules:
For any differentiable functions fand g, the derivative of h(x) = f(x)±g(x) is h(x) =
f(x)±g(x). For instance, for f(x) = x+exwe get f(x) = d
dx (x) + d
dx (ex) = 1 + ex.
Similarly, for f(x) = exln xwe get f(x) = d
dx (ex)d
dx (ln x) = ex1
x.
The power rule:
For any constant n, the derivative of xnis nxn1. For instance, the derivative of x2is 2xand the
derivative of 1
x=x1is (1)x11=x2=1
x2. Of course, it is this rule which also gives
us d
dx (x) = 1 and d
dx (c) = 0 for any constant c. We have d
dx (x) = d
dx (x1) = 1x0= 1(1) = 1.
Likewise, for any constant c,c=cx0and so (using the constant multiplier rule as well as the
power rule) we have
d
dx(c) = d
dx (cx0) = cd
dx (x0)=c0x1=c(0) = 0
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NCERT Solutions for Exercise 15.1 Class 11 Maths Chapter 15 - Statistics
# NCERT Solutions for Exercise 15.1 Class 11 Maths Chapter 15 - Statistics
Edited By Sumit Saini | Updated on Jul 18, 2022 12:03 PM IST
NCERT solutions for Class 11 Maths chapter 15 exercise 15.1 mainly deals with the mean deviation topics and other exercises include important topics like measures of dispersion, mean deviation, standard deviation, quartile deviation, range, etc. In Class 9 and 10 Maths NCERT syllabus, students have already read the basic Statistics like mean, mode, and median. In this NCERT book Class 11 chapter applications of Statistics with advanced concepts are discussed.
JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy
Exercise 15.1 Class 11 Maths have easy questions. Solving NCERT solutions for Class 11 Maths chapter 15 exercise 15.1 is a must to solve upcoming exercises. Direct questions are asked from this exercise as well in Boards. Also students can refer to the following exercise of NCERT for further information about upcoming exercises.
## Statistics Class 11 Chapter 15-Exercise: 15.1
Mean ( $\overline{x}$ ) of the given data:
$\overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{4+ 7+ 8+ 9+ 10+ 12+ 13+ 17}{8} = 10$
The respective absolute values of the deviations from mean, $|x_i - \overline{x}|$ are
6, 3, 2, 1, 0, 2, 3, 7
$\therefore$ $\sum_{i=1}^{8}|x_i - 10| = 24$
$\therefore$ $M.D.(\overline{x}) = \frac{1}{n}\sum_{i=1}^{n}|x_i - \overline{x}|$
$= \frac{24}{8} = 3$
Hence, the mean deviation about the mean is 3.
Mean ( $\overline{x}$ ) of the given data:
$\\ \overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{38+70+48+40+42+55+63+46+54+44}{10} \\ = \frac{500}{10} = 50$
The respective absolute values of the deviations from mean, $|x_i - \overline{x}|$ are
12, 20, 2, 10, 8, 5, 13, 4, 4, 6
$\therefore$ $\sum_{i=1}^{8}|x_i - 50| = 84$
$\therefore$ $M.D.(\overline{x}) = \frac{1}{n}\sum_{i=1}^{n}|x_i - \overline{x}|$
$= \frac{84}{10} = 8.4$
Hence, the mean deviation about the mean is 8.4.
Question:3. Find the mean deviation about the median.
Number of observations, n = 12, which is even.
Arranging the values in ascending order:
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.
Now, Median (M)
$\\ = \frac{(\frac{12}{2})^{th} observation + (\frac{12}{2}+ 1)^{th} observation}{2} \\ = \frac{13 + 14}{2} = \frac{27}{2}= 13.5$
The respective absolute values of the deviations from median, $|x_i - M|$ are
3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
$\therefore$ $\sum_{i=1}^{8}|x_i - 13.5| = 28$
$\therefore$ $M.D.(M) = \frac{1}{12}\sum_{i=1}^{n}|x_i - M|$
$= \frac{28}{12} = 2.33$
Hence, the mean deviation about the median is 2.33.
Question:4. Find the mean deviation about the median.
Number of observations, n = 10, which is even.
Arranging the values in ascending order:
36, 42, 45, 46, 46, 49, 51, 53, 60, 72
Now, Median (M)
$\\ = \frac{(\frac{10}{2})^{th} observation + (\frac{10}{2}+ 1)^{th} observation}{2} \\ = \frac{46 + 49}{2} = \frac{95}{2}= 47.5$
The respective absolute values of the deviations from median, $|x_i - M|$ are
11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
$\therefore$ $\sum_{i=1}^{8}|x_i - 47.5| = 70$
$\therefore$ $M.D.(M) = \frac{1}{10}\sum_{i=1}^{n}|x_i - M|$
$= \frac{70}{10} = 7$
Hence, the mean deviation about the median is 7.
Question:5 Find the mean deviation about the mean.
$x_i$ $f_i$ $f_ix_i$ $|x_i - \overline{x}|$ $f_i|x_i - \overline{x}|$ 5 7 35 9 63 10 4 40 4 16 15 6 90 1 6 20 3 60 6 18 25 5 125 11 55 $\sum{f_i}$ = 25 $\sum f_ix_i$ = 350 $\sum f_i|x_i - \overline{x}|$ =158
$N = \sum_{i=1}^{5}{f_i} = 25 ; \sum_{i=1}^{5}{f_ix_i} = 350$
$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{350}{12} = 14$
Now, we calculate the absolute values of the deviations from mean, $|x_i - \overline{x}|$ and
$\sum f_i|x_i - \overline{x}|$ = 158
$\therefore$ $M.D.(\overline{x}) = \frac{1}{25}\sum_{i=1}^{n}f_i|x_i - \overline{x}|$
$= \frac{158}{25} = 6.32$
Hence, the mean deviation about the mean is 6.32
Question:6. Find the mean deviation about the mean.
$x_i$ $f_i$ $f_ix_i$ $|x_i - \overline{x}|$ $f_i|x_i - \overline{x}|$ 10 4 40 40 160 30 24 720 20 480 50 28 1400 0 0 70 16 1120 20 320 90 8 720 40 320 $\sum{f_i}$ = 80 $\sum f_ix_i$ = 4000 $\sum f_i|x_i - \overline{x}|$ =1280
$N = \sum_{i=1}^{5}{f_i} = 80 ; \sum_{i=1}^{5}{f_ix_i} = 4000$
$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{4000}{80} = 50$
Now, we calculate the absolute values of the deviations from mean, $|x_i - \overline{x}|$ and
$\sum f_i|x_i - \overline{x}|$ = 1280
$\therefore$ $M.D.(\overline{x}) = \frac{1}{80}\sum_{i=1}^{5}f_i|x_i - \overline{x}|$
$= \frac{1280}{80} = 16$
Hence, the mean deviation about the mean is 16
Question:7. Find the mean deviation about the median.
$x_i$ $f_i$ $c.f.$ $|x_i - M|$ $f_i|x_i - M|$ 5 8 8 2 16 7 6 14 0 0 9 2 16 2 4 10 2 18 3 6 12 2 20 5 10 15 6 26 8 48
Now, N = 26 which is even.
Median is the mean of $\dpi{100} 13^{th}$ and $\dpi{100} 14^{th}$ observations.
Both these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.
Therefore, Median, M $\dpi{100} = \frac{13^{th} observation + 14^{th} observation}{2} = \frac{7 + 7}{2} = \frac{14 }{2} = 7$
Now, we calculate the absolute values of the deviations from median, $|x_i - M|$ and
$\sum f_i|x_i - M|$ = 84
$\therefore$ $M.D.(M) = \frac{1}{26}\sum_{i=1}^{6}|x_i - M|$
$= \frac{84}{26} = 3.23$
Hence, the mean deviation about the median is 3.23
$x_i$ $f_i$ $c.f.$ $|x_i - M|$ $f_i|x_i - M|$ 15 3 3 13.5 40.5 21 5 8 7.5 37.5 27 6 14 1.5 9 30 7 21 1.5 10.5 35 8 29 6.5 52
Now, N = 30, which is even.
Median is the mean of $15^{th}$ and $\dpi{100} 16^{th}$ observations.
Both these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.
Therefore, Median, M $\dpi{100} = \frac{15^{th} observation + 16^{th} observation}{2} = \frac{30 + 30}{2} = 30$
Now, we calculate the absolute values of the deviations from median, $|x_i - M|$ and
$\sum f_i|x_i - M|$ = 149.5
$\therefore$ $M.D.(M) = \frac{1}{29}\sum_{i=1}^{5}|x_i - M|$
$= \frac{149.5}{29} = 5.1$
Hence, the mean deviation about the median is 5.1
Question:9. Find the mean deviation about the mean.
Income per day in Rs Number of persons
Income per day Number of Persons $f_i$ Mid Points $x_i$ $f_ix_i$ $|x_i - \overline{x}|$ $f_i|x_i - \overline{x}|$ 0 -100 4 50 200 308 1232 100 -200 8 150 1200 208 1664 200-300 9 250 2250 108 972 300-400 10 350 3500 8 80 400-500 7 450 3150 92 644 500-600 5 550 2750 192 960 600-700 4 650 2600 292 1168 700-800 3 750 2250 392 1176 $\sum{f_i}$ =50 $\sum f_ix_i$ =17900 $\sum f_i|x_i - \overline{x}|$ =7896
$N = \sum_{i=1}^{8}{f_i} = 50 ; \sum_{i=1}^{8}{f_ix_i} = 17900$
$\overline{x} = \frac{1}{N}\sum_{i=1}^{8}f_ix_i = \frac{17900}{50} = 358$
Now, we calculate the absolute values of the deviations from mean, $|x_i - \overline{x}|$ and
$\sum f_i|x_i - \overline{x}|$ = 7896
$\therefore$ $M.D.(\overline{x}) = \frac{1}{50}\sum_{i=1}^{8}f_i|x_i - \overline{x}|$
$= \frac{7896}{50} = 157.92$
Hence, the mean deviation about the mean is 157.92
Question:10. Find the mean deviation about the mean.
Height in cms Number of person
Height in cms Number of Persons $f_i$ Mid Points $x_i$ $f_ix_i$ $|x_i - \overline{x}|$ $f_i|x_i - \overline{x}|$ 95 -105 9 100 900 25.3 227.7 105 -115 13 110 1430 15.3 198.9 115-125 26 120 3120 5.3 137.8 125-135 30 130 3900 4.7 141 135-145 12 140 1680 14.7 176.4 145-155 10 150 1500 24.7 247 $\sum{f_i}$ =100 $\sum f_ix_i$ =12530 $\sum f_i|x_i - \overline{x}|$ =1128.8
$N = \sum_{i=1}^{6}{f_i} = 100 ; \sum_{i=1}^{6}{f_ix_i} = 12530$
$\overline{x} = \frac{1}{N}\sum_{i=1}^{6}f_ix_i = \frac{12530}{100} = 125.3$
Now, we calculate the absolute values of the deviations from mean, $|x_i - \overline{x}|$ and
$\sum f_i|x_i - \overline{x}|$ = 1128.8
$\therefore$ $M.D.(\overline{x}) = \frac{1}{100}\sum_{i=1}^{6}f_i|x_i - \overline{x}|$
$= \frac{1128.8}{100} = 11.29$
Hence, the mean deviation about the mean is 11.29
Marks Number of girls
Marks Number of Girls $f_i$ Cumulative Frequency c.f. Mid Points $x_i$ $|x_i - M|$ $f_i|x_i - M|$ 0-10 6 6 5 22.85 137.1 10-20 8 14 15 12.85 102.8 20-30 14 28 25 2.85 39.9 30-40 16 44 35 7.15 114.4 40-50 4 48 45 17.15 68.6 50-60 2 50 55 27.15 54.3 $\sum f_i|x_i - M|$ =517.1
Now, N = 50, which is even.
The class interval containing $\dpi{80} \left (\frac{N}{2} \right)^{th}$ or $\dpi{100} 25^{th}$ item is 20-30. Therefore, 20-30 is the median class.
We know,
Median $\dpi{100} = l + \frac{\frac{N}{2}- C}{f}\times h$
Here, l = 20, C = 14, f = 14, h = 10 and N = 50
Therefore, Median $\dpi{100} = 20 + \frac{25 - 14}{14}\times 10 = 20 + 7.85 = 27.85$
Now, we calculate the absolute values of the deviations from median, $|x_i - M|$ and
$\sum f_i|x_i - M|$ = 517.1
$\therefore$ $M.D.(M) = \frac{1}{50}\sum_{i=1}^{6}f_i|x_i - M|$
$= \frac{517.1}{50} = 10.34$
Hence, the mean deviation about the median is 10.34
Age (in years) Number
[ Hint Convert the given data into continuous frequency distribution by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class interval]
Age (in years) Number $f_i$ Cumulative Frequency c.f. Mid Points $x_i$ $|x_i - M|$ $f_i|x_i - M|$ 15.5-20.5 5 5 18 20 100 20.5-25.5 6 11 23 15 90 25.5-30.5 12 23 28 10 120 30.5-35.5 14 37 33 5 70 35.5-40.5 26 63 38 0 0 40.5-45.5 12 75 43 5 60 45.5-50.5 16 91 48 10 160 50.5-55.5 9 100 53 15 135 $\sum f_i|x_i - M|$ =735
Now, N = 100, which is even.
The class interval containing $\dpi{80} \left (\frac{N}{2} \right)^{th}$ or $50^{th}$ item is 35.5-40.5. Therefore, 35.5-40.5 is the median class.
We know,
Median $\dpi{100} = l + \frac{\frac{N}{2}- C}{f}\times h$
Here, l = 35.5, C = 37, f = 26, h = 5 and N = 100
Therefore, Median $\dpi{100} = 35.5 + \frac{50 - 37}{26}\times 5 = 35.5 + 2.5 = 38$
Now, we calculate the absolute values of the deviations from median, $|x_i - M|$ and
$\sum f_i|x_i - M|$ = 735
$\therefore$ $M.D.(M) = \frac{1}{100}\sum_{i=1}^{8}f_i|x_i - M|$
$= \frac{735}{100} = 7.35$
Hence, the mean deviation about the median is 7.35
## More About NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.1
The NCERT Class 11 Maths chapter Statistics is very scoring chapters and questions are asked on the expected concepts only. It does not have much application in other chapters hence if one is not good at other chapters can strengthen this chapter to score well in the exam. Exercise 15.1 Class 11 Maths Mainly discusses the mean deviation about mean and median. Hence NCERT solutions for Class 11 Maths chapter 15 exercise 15.1 can be seen as a scoring exercise from the exam perspective.
## Benefits of NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.1
• The Class 11 Maths chapter 15 exercise is prepared by expert faculties with rich experience.
• Exercise 15.1 Class 11 Maths can help one enhance their marks in the exam as one question of 5 marks is mostly asked from this.
• Class 11 Maths chapter 15 exercise 15.1 solutions provided here are comprehensive in manner.
Also see-
JEE Main Highest Scoring Chapters & Topics
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## Subject Wise NCERT Exemplar solutions
Happy learning!!!
1. Which topics are covered in Exercise 15.1 Class 11 Maths?
Exercise 15.1 Class 11 Maths includes mean deviation about mean and median.
2. Is this chapter 15 related to other chapters ?
No, mostly concepts are new and can be understood if other chapters are not read.
3. What is the difficulty level of the questions asked in this chapter ?
Questions are easier to moderate level of difficulty if concepts are memorized.
4. Is it necessary to remember the formulas ?
Yes, some basic formulas must be remembered to solve the questions.
5. How many questions are there in the Exercise 15.1 Class 11 Maths ?
Total 12 questions are discussed in this exercise.
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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
Option 1) Option 2) Option 3) Option 4)
A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms−2 :
Option 1) 2.45×10−3 kg Option 2) 6.45×10−3 kg Option 3) 9.89×10−3 kg Option 4) 12.89×10−3 kg
An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
Option 1) Option 2) Option 3) Option 4)
A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
Option 1) Option 2) Option 3) Option 4)
In the reaction,
Option 1) at STP is produced for every mole consumed Option 2) is consumed for ever produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .
How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2
If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will
Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.
With increase of temperature, which of these changes?
Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.
Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is
Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023
A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is
Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9 |
# 8.8 Vectors (Page 5/22)
Page 5 / 22
## Writing a vector in terms of i And j
Given a vector $\text{\hspace{0.17em}}v\text{\hspace{0.17em}}$ with initial point $\text{\hspace{0.17em}}P=\left(2,-6\right)\text{\hspace{0.17em}}$ and terminal point $\text{\hspace{0.17em}}Q=\left(-6,6\right),\text{\hspace{0.17em}}$ write the vector in terms of $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}j.$
Begin by writing the general form of the vector. Then replace the coordinates with the given values.
$\begin{array}{l}v=\left({x}_{2}-{x}_{1}\right)i+\left({y}_{2}-{y}_{1}\right)j\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(-6-2\right)i+\left(6-\left(-6\right)\right)j\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-8i+12j\hfill \end{array}$
## Writing a vector in terms of i And j Using initial and terminal points
Given initial point $\text{\hspace{0.17em}}{P}_{1}=\left(-1,3\right)\text{\hspace{0.17em}}$ and terminal point $\text{\hspace{0.17em}}{P}_{2}=\left(2,7\right),\text{\hspace{0.17em}}$ write the vector $\text{\hspace{0.17em}}v\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}j.\text{\hspace{0.17em}}$
Begin by writing the general form of the vector. Then replace the coordinates with the given values.
$\begin{array}{l}v=\left({x}_{2}-{x}_{1}\right)i+\left({y}_{2}-{y}_{1}\right)j\hfill \\ v=\left(2-\left(-1\right)\right)i+\left(7-3\right)j\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=3i+4j\hfill \end{array}$
Write the vector $\text{\hspace{0.17em}}u\text{\hspace{0.17em}}$ with initial point $\text{\hspace{0.17em}}P=\left(-1,6\right)\text{\hspace{0.17em}}$ and terminal point $\text{\hspace{0.17em}}Q=\left(7,-5\right)\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}j.$
$u=8i-11j$
## Performing operations on vectors in terms of i And j
When vectors are written in terms of $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}j,\text{\hspace{0.17em}}$ we can carry out addition, subtraction, and scalar multiplication by performing operations on corresponding components.
## Adding and subtracting vectors in rectangular coordinates
Given v = a i + b j and u = c i + d j , then
$\begin{array}{c}v+u=\left(a+c\right)i+\left(b+d\right)j\\ v-u=\left(a-c\right)i+\left(b-d\right)j\end{array}$
## Finding the sum of the vectors
Find the sum of $\text{\hspace{0.17em}}{v}_{1}=2i-3j\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{v}_{2}=4i+5j.$
According to the formula, we have
$\begin{array}{l}{v}_{1}+{v}_{2}=\left(2+4\right)i+\left(-3+5\right)j\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=6i+2j\hfill \end{array}$
## Calculating the component form of a vector: direction
We have seen how to draw vectors according to their initial and terminal points and how to find the position vector. We have also examined notation for vectors drawn specifically in the Cartesian coordinate plane using $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}j.\text{\hspace{0.17em}}$ For any of these vectors, we can calculate the magnitude. Now, we want to combine the key points, and look further at the ideas of magnitude and direction.
Calculating direction follows the same straightforward process we used for polar coordinates. We find the direction of the vector by finding the angle to the horizontal. We do this by using the basic trigonometric identities, but with $\text{\hspace{0.17em}}|v|\text{\hspace{0.17em}}$ replacing $\text{\hspace{0.17em}}r.$
## Vector components in terms of magnitude and direction
Given a position vector $\text{\hspace{0.17em}}v=⟨x,y⟩\text{\hspace{0.17em}}$ and a direction angle $\text{\hspace{0.17em}}\theta ,$
$\begin{array}{lll}\mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{x}{|v|}\hfill & \text{and}\begin{array}{cc}& \end{array}\hfill & \mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{y}{|v|}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=|v|\mathrm{cos}\text{\hspace{0.17em}}\theta \begin{array}{cc}& \end{array}\hfill & \hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=|v|\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill \end{array}$
Thus, $\text{\hspace{0.17em}}v=xi+yj=|v|\mathrm{cos}\text{\hspace{0.17em}}\theta i+|v|\mathrm{sin}\text{\hspace{0.17em}}\theta j,\text{\hspace{0.17em}}$ and magnitude is expressed as $\text{\hspace{0.17em}}|v|=\sqrt{{x}^{2}+{y}^{2}}.$
## Writing a vector in terms of magnitude and direction
Write a vector with length 7 at an angle of 135° to the positive x -axis in terms of magnitude and direction.
Using the conversion formulas $\text{\hspace{0.17em}}x=|v|\mathrm{cos}\text{\hspace{0.17em}}\theta i\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=|v|\mathrm{sin}\text{\hspace{0.17em}}\theta j,\text{\hspace{0.17em}}$ we find that
$\begin{array}{l}x=7\mathrm{cos}\left(135°\right)i\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{7\sqrt{2}}{2}\hfill \\ y=7\mathrm{sin}\left(135°\right)j\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{7\sqrt{2}}{2}\hfill \end{array}$
This vector can be written as $\text{\hspace{0.17em}}v=7\mathrm{cos}\left(135°\right)i+7\mathrm{sin}\left(135°\right)j\text{\hspace{0.17em}}$ or simplified as
$v=-\frac{7\sqrt{2}}{2}i+\frac{7\sqrt{2}}{2}j$
A vector travels from the origin to the point $\text{\hspace{0.17em}}\left(3,5\right).\text{\hspace{0.17em}}$ Write the vector in terms of magnitude and direction.
$v=\sqrt{34}\mathrm{cos}\left(59°\right)i+\sqrt{34}\mathrm{sin}\left(59°\right)j$
Magnitude = $\text{\hspace{0.17em}}\sqrt{34}$
$\theta ={\mathrm{tan}}^{-1}\left(\frac{5}{3}\right)=59.04°$
## Finding the dot product of two vectors
As we discussed earlier in the section, scalar multiplication involves multiplying a vector by a scalar, and the result is a vector. As we have seen, multiplying a vector by a number is called scalar multiplication. If we multiply a vector by a vector, there are two possibilities: the dot product and the cross product . We will only examine the dot product here; you may encounter the cross product in more advanced mathematics courses.
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
Someone should please solve it for me Add 2over ×+3 +y-4 over 5 simplify (×+a)with square root of two -×root 2 all over a multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the  that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25. |
# How do you write the equation of the line that contains the centers of the circles (x – 2)^2 + (y + 3)^2 = 17 and (x – 5)^2 + y^2 = 32?
Oct 16, 2016
$y = - \frac{3}{7} x - \frac{15}{7}$
#### Explanation:
Both equations are in the form -
x-h)^2+(y-k)^2=a^2
In that case the center of the circle is $\left(h , k\right)$
The center of the circle ${\left(x - 2\right)}^{2} + {\left(y + 3\right)}^{2} = 17$ is $\left(2 , - 3\right)$
The center of the circle ${\left(x + 5\right)}^{2} + {y}^{2} = 32$ is $\left(- 5 , 0\right)$
The equation of the line passing through the point is -
$\left(y - {y}_{1}\right) = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} \left(x - {x}_{1}\right)$
$y - \left(- 3\right) = \frac{0 - \left(- 3\right)}{- 5 - \left(- 2\right)} \left(x - 2\right)$
$y + 3 = - \frac{3}{7} \left(x - 2\right)$
$y + 3 = - \frac{3}{7} x + \frac{6}{7}$
$y = - \frac{3}{7} x + \frac{6}{7} - 3$
$y = - \frac{3}{7} x - \frac{15}{7}$ |
# Angles of Parallel Lines Cut by Transversals
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Chris S.
## Understanding Parallel Lines
In this geometry lesson, we're delving into the world of Parallel Lines. These lines are everywhere around us, from the straight paths of train tracks to the crisp edges of your tablet, and even in the architectural lines of buildings and bridges. Parallel lines are unique because they follow a simple rule: no matter how far they extend, they never cross.
This lesson will uncover the angles and patterns formed when these parallel lines are crossed by another line, known as a transversal. It's about seeing the math in our everyday world and understanding the geometry that shapes it. Ready to see how? Let’s get started!
### Parallel Lines, Perpendicular Lines, and Transversals
• Parallel Lines: Parallel lines are like train tracks, never touching or crossing each other, and they stay the same distance apart forever.
• Perpendicular Lines: Perpendicular lines cross each other and always form a 90-degree angle, making corners like the letter 'L' or the corners of a square.
• Transversal Lines: A transversal line is a line that crosses at least two other lines. When it crosses parallel lines, it creates equal angles at the points of intersection. With non-parallel lines, it forms various angles.
## Understanding Parallel Lines – Definition
Parallel Lines are lines on a plane that are always the same distance apart and never intersect.
Congruent means having the exact size and shape. In geometry, congruent angles have equal measures.
There are also some angle relationships that are important to know when learning about parallel lines.
## Parallel Lines – Angle Relationships
Angle Relationships:
Type of Angles Explanation
Vertical Angles Angles opposite each other when two lines intersect. They are always congruent.
Supplementary Angles Two angles that add up to 180 degrees. They often appear when lines intersect.
Corresponding Angles When a transversal crosses two parallel lines, these angles are in matching positions. They are congruent in parallel lines.
Alternate Interior Angles Angles inside the parallel lines on opposite sides of the transversal. They are congruent in parallel lines.
Alternate Exterior Angles Angles outside the parallel lines on opposite sides of the transversal. They are congruent in parallel lines.
## Parallel Lines – Guided Practice
Given a diagram with parallel lines and a transversal, identify and list all the angles that are congruent.
If $\angle{f}=130^\circ$, what is the measurement of $\angle{d}$?
### Parallel Lines – Exercises
Vertical Angles: $\angle{X}$ and $\angle{Z}$, $\angle{Y}$ and $\angle{W}$, $\angle{A}$ and $\angle{C}$, $\angle{B}$ and $\angle{D}$
Corresponding Angles: $\angle{X}$ and $\angle{A}$, $\angle{W}$ and $\angle{D}$, $\angle{Y}$ and $\angle{B}$, $\angle{Z}$ and $\angle{C}$
Alternate Exterior Angles: $\angle{X}$ and $\angle{C}$, $\angle{Y}$ and $\angle{D}$
Alternate Interior Angles: $\angle{W}$ and $\angle{B}$, $\angle{Z}$ and $\angle{A}$
Using the illustration above, answer the following questions to check your understanding.
Identify at least one pair of corresponding angles in the diagram.
Are $\angle{X}$ and $\angle{C}$ congruent? And if so what type of angles are they?
Find the measure of $\angle{Y}$ if its vertical angle measures $70^\circ$.
## Parallel Lines – Summary
Key Learnings:
• Parallel lines remain the same distance apart and never intersect.
• A transversal creates various angle types, including corresponding, alternate interior, and alternate exterior angles, which are congruent in parallel lines.
• Understanding these angle relationships is essential for mastering geometry concepts.
If you are ready and confident with this topic, why not apply it to the following topic Calculations with Supplementary, Complementary, and Vertical Angles
Explore more interactive and engaging geometry lessons on our website, complete with practice problems, videos, and worksheets!
### Parallel Lines – Frequently Asked Questions
How can you determine if lines are truly parallel?
What are the main characteristics of parallel lines?
Can parallel lines exist in three-dimensional space?
What is a transversal line in geometry?
Why are the properties of parallel lines important in geometry?
How are alternate interior angles used in real-life situations?
Are the angles formed by a transversal always equal?
What is the difference between corresponding angles and alternate angles?
How can the concept of parallel lines be applied in architecture?
What role do supplementary angles play when dealing with parallel lines?
Can angles formed by intersecting non-parallel lines be predicted in the same way as those formed by parallel lines?
### TranscriptAngles of Parallel Lines Cut by Transversals
On their nightly food run, the three raccoons crashed their shopping cart... AGAIN. It's time to go back to the drawing stump. They decide to practice going around the sharp corners and tight angles during the day, before they get their loot. To put this surefire plan into action they'll have to use their knowledge of parallel lines and transversals. Let's look at this map of their city. All the HORIZONTAL roads are parallel lines. They DON'T intersect. But there are several roads which CROSS the parallel ones. These lines are called TRANSVERSALS. The raccoons crashed HERE at angle 1. The measure of angle 1 is 60 degrees. Can you see any other angles that are also 60 degrees? There are a few such angles, and one of them is angle 3. That's because angle 1 and angle 3 are vertical angles, and vertical angles are always equal in measure. Do we have enough information to determine the measure of angle 2? Since angles 1 and 2 are angles on a line, they sum to 180 degrees. That means angle 2 is 120 degrees. And since angles 2 and 4 are vertical, angle 4 must also be 120 degrees. Now we know all of the angles around this intersection, but what about the angles at the other intersection? Let's take a look at angle 5. If we translate angle 1 along the transversal until it overlaps angle 5, it looks like they are congruent. And they are! That means angle 5 is also 60 degrees. Angle 1 and angle 5 are examples of CORRESPONDING angles. Corresponding angles are pairs of angles that are in the SAME location around their respective vertices. And whenever two PARALLEL lines are cut by a transversal, pairs of corresponding angles are CONGRUENT. That means you only have to know the measure of one angle from the pair, and you automatically know the measure of the other! Can you see other pairs of corresponding angles here? Angles 2 and 6 are also corresponding angles. So are angles 3 and 7 and angles 4 and 8. That means the measure of angle 2 equals the measure of angle 6, the measure of angle 3 equals the measure of angle 7, and the measure of angle 4 equals the measure of angle 8. We already know that angles 4 and 6 are both 120 degrees, but is it ALWAYS the case that such angles are congruent? It is! Let's show this visually. Now, let's use our knowledge of vertical and corresponding angles to prove it. We are going to use angle 2 to help us compare the two angles. Angle 4 must be equal to angle 2 because they are vertical angles. And angle 6 must be equal to angle 2 because they are corresponding angles. Since angle 6 and angle 4 are both equal to the same angle, they also must be equal to each other! We call angle pairs like angle 6 and angle 4 alternate interior angles because they are found on ALTERNATE sides of the transversal and they are both INTERIOR to the two parallel lines. Can you see another pair of alternate interior angles? 3 and 5 are ALSO alternate interior. If two parallel lines are cut by a transversal, alternate interior angles are always congruent. We just looked at alternate interior angles, but we also have pairs of angles that are called alternate EXTERIOR angles. Based on the name, which angle pairs do you think would be called alternate exterior angles? Well, they need to be EXTERIOR to the parallel lines and on ALTERNATE sides of the transversal. 1 and 7 are a pair of alternate exterior angles and so are 2 and 8. Notice that the measure of angle 1 equals the measure of angle 7 and the same is true for angles 2 and 8. If two parallel lines are cut by a transversal, alternate exterior angles are always congruent. In fact, when parallel lines are cut by a transversal, there are a lot of congruent angles. Look at what happens when this same transversal intersects additional parallel lines. We can use congruent angle pairs to fill in the measures for THESE angles as well. The raccoons only need to practice driving their shopping cart around ONE corner to be ready for ALL the intersections along this transversal. For each transversal, the raccoons only have to measure ONE angle. They can then use their knowledge of corresponding angles, alternate interior angles, and alternate exterior angles to find the measures for ALL the angles along that transversal. Now it's time for some practice before they do a little...um... shopping. While they are riding around, let's review what we've learned. When parallel lines are cut by a transversal, congruent angle pairs are created. Corresponding angles are in the SAME position around their respective vertices and there are FOUR such pairs. Alternate interior angles are on ALTERNATE sides of the transversal and INTERIOR to the parallel lines and there are two such pairs. Alternate EXTERIOR angles are on alternate sides of the transversal and EXTERIOR to the parallel lines and there are also two such pairs. The raccoons are trying to corner the market on food scraps, angling for a night-time feast! Well, THAT was definitely a TURN for the worse!
## Angles of Parallel Lines Cut by Transversals exercise
Would you like to apply the knowledge you’ve learned? You can review and practice it with the tasks for the video Angles of Parallel Lines Cut by Transversals.
• ### Identify the measurement of a missing angle.
Hints
Vertical Angles are opposite each other when two lines intersect. They are always congruent.
Congruent Angles have the same angle measurement.
So if, $\angle{x}=100^\circ$ and $\angle{y}$ was known to be congruent, then $\angle{y}$ would also equal $100^\circ$.
Solution
$\angle{b}$ and $\angle{d}$ are vertical angles and therefore they are congruent.
$\angle d = 150^\circ$
• ### Identify angle relationships.
Hints
Corresponding angles are in the same position around their respective vertices.
Interior angles are the ones found between the parallel lines.
Exterior angles are the ones found on the outside of the parallel lines.
The word alternate, found in the angle relationships Alternate Interior Angles, and Alternate Exterior Angles, refers to the angles on opposite sides of the transversal.
Solution
Corresponding Angles
$\angle$4 and $\angle$8
$\angle$2 and $\angle$6
${}$ Alternate Interior Angles
$\angle$3 and $\angle$5
$\angle$4 and $\angle$6
${}$ Alternate Exterior Angles
$\angle$2 and $\angle$8
$\angle$1 and $\angle$7
• ### Understanding Supplementary Angles
Hints
Supplementary Angles are two angles next to each other on a line that have a sum of $180^\circ$.
The sum of $130^\circ$ and $50^\circ$ is $180^\circ$.
Solution
Angle Z is equal to $155^\circ$ because $25+155=180^\circ$.
Angle Y is equal to $25^\circ$ for a few reasons:
• Angle Y is verticle to the existing $25^\circ$
• Angle Y + Angle Z must equal $180^\circ$
• ### Understand the angle relationships when there are two parallel lines cut by a transversal.
Hints
Here you see an example of parallel lines.
Alternate Interior Angles -- Angles inside the parallel lines on opposite sides of the transversal. They are congruent.
Alternate Exterior Angles -- Angles outside the parallel lines on opposite sides of the transversal. They are congruent.
Solution
Lines $m$ and $n$ are parallel lines and they will never cross. Line $n$ is called a transversal and when it crosses the parallel lines it forms eight angles. All of the angles formed have different angle relationships. For example, $\angle{2}$ and $\angle{4}$ are across from each other and are vertical angles. $\angle{3}$ and $\angle{5}$ are the same measurement because they are alternate interior angles, while $\angle{1}$ and $\angle{7}$ are the same measurement because they are alternate exterior angles. The word we use to describe angles that are the same measurement is: congruent.
• ### Understand the relationship between parallel lines and their transversals.
Hints
This image will show you what a set of parallel lines and transversals look like.
This image will show you what perpendicular lines look like.
Solution
Parallel Lines: Lines on a plane that are always the same distance apart and never intersect.
Transversal: Line that crosses at least two other lines and when it crosses parallel lines, it creates equal angles at the points of intersection.
Perpendicular Lines: Lines cross each other and always form a 90-degree angle, making corners like the letter 'L' or the corners of a square.
Congruent: Having the same angle measurement.
• ### Identify missing angle values using your knowledge of angle relationships.
Hints
To find missing angles is sort of like solving a puzzle. It is helpful to start with the ones you are most confident with which can then make more difficult problems more approachable.
Supplementary Angles -- two angles that have a sum of $180^\circ$.
Solution
To find the missing measurements, you have to determine the relationship between the known angle and the missing angle(s).
Also, remember that a straight line has angles that have a sum of $180^\circ$, which can be helpful when finding missing angles. |
How do you differentiate f(x)= ( x + 1 )/ ( x - 6) using the quotient rule?
$f ' \left(x\right) = \frac{- 7}{x - 6} ^ 2$
Explanation:
Use the formula $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{d}{\mathrm{dx}} \left(u\right) - u \cdot \frac{d}{\mathrm{dx}} \left(v\right)}{v} ^ 2$
Given $f \left(x\right) = \frac{x + 1}{x - 6}$
Let $u = \left(x + 1\right)$ and $v = \left(x - 6\right)$
By the formula
$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{d}{\mathrm{dx}} \left(u\right) - u \cdot \frac{d}{\mathrm{dx}} \left(v\right)}{v} ^ 2$
$\frac{d}{\mathrm{dx}} \left(\frac{x + 1}{x - 6}\right) = \frac{\left(x - 6\right) \cdot \frac{d}{\mathrm{dx}} \left(x + 1\right) - \left(x + 1\right) \cdot \frac{d}{\mathrm{dx}} \left(x - 6\right)}{x - 6} ^ 2$
$\frac{d}{\mathrm{dx}} \left(\frac{x + 1}{x - 6}\right) = \frac{\left(x - 6\right) \left(1 + 0\right) - \left(x + 1\right) \left(1 - 0\right)}{x - 6} ^ 2$
$\frac{d}{\mathrm{dx}} \left(\frac{x + 1}{x - 6}\right) = \frac{\left(x - 6\right) - \left(x + 1\right)}{x - 6} ^ 2 = \frac{x - 6 - x - 1}{x - 6} ^ 2$
$\frac{d}{\mathrm{dx}} \left(\frac{x + 1}{x - 6}\right) = \frac{- 7}{x - 6} ^ 2$
Second solution #2, Just to check the above solution:
Simplify the given first so that
$f \left(x\right) = \frac{x + 1}{x - 6} = 1 + \frac{7}{x - 6} = 1 + 7 {\left(x - 6\right)}^{-} 1$
differentiate
$f \left(x\right) = 1 + 7 {\left(x - 6\right)}^{-} 1$
$f ' \left(x\right) = 0 + 7 \cdot \left(- 1\right) \cdot {\left(x - 6\right)}^{- 1 - 1} \cdot \frac{d}{\mathrm{dx}} \left(x - 6\right)$
$f ' \left(x\right) = - 7 {\left(x - 6\right)}^{-} 2$
$f ' \left(x\right) = \frac{- 7}{x - 6} ^ 2$
God bless....I hope the explanation is useful. |
# Manhattan GMAT Challenge Problem of the Week – 12 November 2012
by on November 12th, 2012
Here is a new Challenge Problem! If you want to win prizes, try entering our Challenge Problem Showdown. The more people that enter our challenge, the better the prizes!
## Question
The positive difference of the fourth powers of two consecutive positive integers must be divisible by
A. one less than twice the larger integer
B. one more than twice the larger integer
C. one less than four times the larger integer
D. one more than four times the larger integer
E. one more than eight times the larger integer
You could solve this problem algebraically, or you could solve by picking numbers. Let’s try the latter method first, as it winds up being faster and easier mentally.
Write the fourth powers of the first three positive integers:
= 1
= 16
= 81
Now, for the first two fourth powers, the “positive difference of the fourth powers of two consecutive positive integers” would be this:
= 16 – 1 = 15
15 is divisible by 3 and by 5 (as well as by 1 and by 15). Since the “larger integer” is 2, eliminate answers:
(A) one less than twice the larger integer = 2×2 – 1 = 3 = fine
(B) one more than twice the larger integer = 2×2 + 1 = 5 = fine
(C) one less than four times the larger integer = 4×2 – 1 = 7 = wrong
(D) one more than four times the larger integer = 4×2 + 1 = 9 = wrong
(E) one more than eight times the larger integer = 8×2 + 1 = 17 = wrong
You’re left with (A) and (B). Now use 2 and 3 as your consecutive integers:
= 81 – 16 = 65
65 is divisible by 5 and by 13 (as well as by 1 and by 65). Test (A) and (B):
(A) one less than twice the larger integer = 2×3 – 1 = 5 = fine
(B) one more than twice the larger integer = 2×3 + 1 = 7 = wrong
As for the algebraic proof, here it is:
“The positive difference of the fourth powers of two consecutive positive integers”
=
If you expand the term and do the subtraction, you get + 4n – 1, which is hard to factor further. Here’s the trick: you can treat as the difference of squares, since every fourth power is a square. For instance = . So you get this:
= [ + ] x []
Now expand the right side:
+ = + ( – 2n + 1) = – 2n + 1
= – ( – 2n + 1) = 2n – 1
So you get:
= ( – 2n + 1) (2n – 1)
Thus, as long as n is an integer, the expression on the left is always divisible by 2n – 1, which corresponds to choice A. |
# Mastering Linear Approximation in Calculus Unlock the power of linear approximation to simplify complex functions and estimate values. Learn when to use overestimates or underestimates, and apply these techniques to real-world problems.
Now Playing:Linear approximation – Example 0a
Intros
1. Introduction to Linear Approximations
2. What is a linear approximation?
3. Linear Approximation – Lesson Overview.
Examples
1. Apply Linear Approximations and Discuss How to Choose "a"
Consider the function $f(x)=\sqrt{x}.$
1. Sketch the graph
2. Find the linearization of the function $f(x)=\sqrt{x}$ at $a=4,$ and illustrate the tangent line on the graph.
3. Use the linear approximation to estimate the numbers:
i)
$\sqrt{4.084}$
ii)
$\sqrt{3.96}$
Are these approximations overestimates or underestimates?
4. Use the same linear approximation to estimate the number $\sqrt{10.2}$, and comment on the accuracy of the approximation. How can the approximation be improved?
Position velocity acceleration
Notes
In this section, we will learn how to approximate unknown values of a function given known values using Linear Approximation. Linear Approximation has another name as Tangent Line Approximation because what we are really working with is the idea of local linearity, which means that if we zoom in really closely on a point along a curve, we will see a tiny line segment that has a slope equivalent to the slope of the tangent line at that point.
## Linear Approximation
Let's say that one day you forgot a calculator and you really want to find how what the square root of 2 is. How would we find out without a calculator? We can use linear approximation!
## Linearization of a function
The linearization of a function is just about finding the tangent line of the function at a specific point in a different way. The linearization formula is:
where L(x) is the equation of the tangent line at point a.
How is this useful to us? Well, we can actually use this equation to approximate values of the function near point a. Take a look at this graph.
Notice that for x values near point a, we see that the function and the tangent line is relatively close to each other. Because of this, we are able to write that the function is approximately equal to the tangent line near point a. In other words,
where $\approx$ is the approximately symbol. This equation is known as the linear approximation formula. It is linear in a sense that the tangent is a straight line and we are using it to approximate the function. Using this approximation, we are able to approximate values that cannot be done by hand. For example, the square root of 2 or the natural log of 5 can all be approximated! One important thing to note is that this approximation only works for x values near point a. If you have a x value far from point a, then the approximation becomes really inaccurate.
Now don't we take a look at a few examples of finding the linearization of a function and then look at how to use linear approximation!
## Find the linearization of L(x) of the function at a
Question 1: Consider the function
Let's say that we want to find the linearization of the function at point a=4.
To find the linearization L(x), recall that
Notice that in order to calculate L(x), we need f(a), f'(a) and a. Afterwards, were going to have to plug in everything in the formula to find L(x). Hence, I created these steps:
• Step 1: Find a
• Step 2: Find f(a)
• Step 3: Find f'(a).
• Step 4: Plug all three into the formula to find L(x)
Step 1: Luckily a = 4 is given to us in the question, so we don't have to look for it.
• Step 2:
Notice that
• Step 3:
Know that the derivative of square roots is
And so plugging in x=a gives us:
• Step 4:
Since we know a, f(a), and f'(a), we can now plug it into L(x) to find the linearization of f(x).
Hence,
So L(x)=$\frac{1}{4}$x+1 is the linearization of this function at point x=4. In addition, it is also the tangent line of the function at point x=4.
## How to do linear approximation
Remember earlier we said that we could use the equation of the tangent line to approximate values of the function near a? Let's try this with the linearization we found earlier. Recall that
for points near x=4. We can change this into a linear approximation for f(x) by saying that:
Now, let's say I want to approximate f(4.04). If you were to plug this into the original function, then you would get $\sqrt{4.04}$ . This would be really hard to compute without a calculator. However, using linear approximation, we can say that
We just approximated the f(4.04) without a calculator! Now let's actually see how close we were to the exact value of f(4.04). Notice that f(4.04) = $\sqrt{4.04}$ = 2.00997512422... So we are really close! We were only off when we've got to the second decimal place!
Now so far, these questions gave us a function and a point to work with. What if none of these were given at all? What if the question only tells us to estimate a number?
## Use Linear approximation to estimate a number
Suppose we want to estimate $\sqrt{10}$. How would we do it? We would need to use the linear approximation
but we don't even have a function and a point to work with. This means we have to make them ourselves. This leads us to do the following steps:
• Step 1: create a function
• Step 2: create point a
• Step 3: Find f(a) and f'(a)
• Step 4: Plug everything into the linear approximation formula
Question 2: Estimate $\sqrt{10}$
• Step 1: Let's come up with a function. Note that we are estimating
We need to somehow make a relationship between f(x) and $\sqrt{10}$. We can't say that f(x) = $\sqrt{10}$ because the function won't be dependent on x. So why don't we do this? Let
If we do that, then we are basically saying
We can now obviously tell what the function should be. Let
So we have a function, but now we need a point a to work with.
• Step 2:
The key to finding the right value a is by considering two things:
1) Make sure that the value a is close to x
2) Make sure f(a) is a nice number.
Would a = 8 be sufficient enough? Well, 8 is pretty close to 10 so it is not bad. However,
Notice that $\sqrt{8}$ is not a very nice number. In fact, you get a bunch of decimal numbers. So we have to try something else.
Would a = 9 be sufficient enough? Again, 9 is pretty close to 10 so it is okay. Also,
f(a) is actually a nice whole number here, so this actually works! So picking a=9 is sufficient enough.
• Step 3:
Note that from earlier:
Calculating f'(x) we have again:
So
• Step 4:
Going back to the linear approximation formula we have:
Plugging a, f(a), f'(a), f(x), and x in our formula you will see that:
Hence, we just approximated the number!
If you want more practice problems about linear approximation, then I recommend you look at this link here.
http://tutorial.math.lamar.edu/Classes/CalcI/LinearApproximations.aspx
## Overestimate and Underestimate
We know that linear approximation is just an estimation of the function's value at a specified point. However, how do we know that if our estimation is an overestimate or an underestimate? We calculate the second derivative and look at the concavity.
### Concave up vs Concave down
If the second derivative of the function is greater than 0 for values near a, then the function is concave up. This means that our approximation will be an underestimate. In other words,
Why? Let's take a look at this graph.
Notice that f(x) is concave upward and the tangent line is right under f(x). Let's say were to use the tangent line to approximate f(x). Then the y values of the tangent line are always going to be less than the actual value of f(x). Hence, we have an underestimate
Now if the second derivative of the function is less than 0 for values near a, then the function is concave down. This means that our approximation will be an overestimate. In other words,
Again, why? Let's take a look at another graph.
Notice that f(x) is concave downward and the tangent line is right above f(x). Again, let's say that we are going to use the tangent line to approximate f(x). Then the y values of the tangent line are always going to be greater than the actual value of f(x). Hence, we have an overestimate.
So if you ever need to see if your value is an underestimation or an overestimation, make sure you follow these steps:
• Step 1: Find the second derivative
• Step 2: look at the concavity of the function near point a
• Step 3: Confirm that it is an underestimate/overestimate
Let's take a look at an example:
Question 3: Let f(x) = $\sqrt{x}$ and a = 4. If we linear approximate f(4.04), would it be an overestimate or an underestimate?
• Step 1: See that
So the second derivative is
• Step 2:
Notice that a=4, so we want to look at positive values of x near 4. Now look at the second derivative. When x is positive, we see that
Hence, it is concave down
• Step 3:
We know that if the function is concave down, then the tangent line will be above the function. Hence, using the tangent line as an approximation will give an overestimated value.
## Differentials
Not only can we approximate values with linear approximation, but we can also approximate with differentials. To approximate, we use the following formula
where dy and dx are differentials, and f'(x) is the derivative of f in terms of x. Since we are dealing with very small changes in x and y, then we are going to use the fact that:
However, most of the questions we do involve setting
So using these facts will lead us to have:
This approximation is very useful when approximating the change of y. Keep in mind back then they didn't have calculators, so this is the best approximation they could get for functions with square roots or natural logs.
Most of the time you will have to look for f'(x) and $\Delta$x yourself. In other words, follow these steps to approximate $\Delta$y!
• Step 1: Find $\Delta$x
• Step 2: Find f'(x)
• Step 3: Plug everything into the formula to find dy. dy will be the approximation for $\Delta$y.
Let's look at an example of using this approximation:
Question 4: Consider the function y = ln(x + 1). Suppose x changes from 0 to 0.01. Approximate $\Delta$y.
• Step 1: Notice that x changes from 0 to 0.01, so the change in x would be:
• Step 2:
The derivative would be:
• Step 3:
Plugging everything in we have:
Hence, $\Delta$y $\approx$ 0.01
However, most of the time we want to estimate a value of the function, and not the change of the value. Hence we will add both sides of the equation by y, which gives us:
which is the same as:
This equation is a bit hard to read, so we are going to rearrange it even more. Let's try to get rid of y and $\Delta$y. Notice that $\Delta$y+ y is basically the same as finding the value of the function at $\Delta$x+x. In other words,
Hence substituting this in our approximation above will give us:
where f($\Delta$x+x) is value we are trying to estimate. How do we use this formula? I recommend following these steps:
• Step 1: Set the number equal to f($\Delta$x+x). Find $\Delta$x, x, and f(x).
• Step 2: Calculate f'(x)
• Step 3: Use the formula to approximate the number
Let's use these steps for the following question.
Question 5: Use differentials to approximate $\sqrt{10}$.
• Step 1: Compare f($\Delta$x+x) with $\sqrt{10}$. Since $\sqrt{10}$ has a square root and 9 is a perfect square that is closest to 10, then let
Notice that:
See that there is no choice but to let $\Delta$x = 1
• Step 2:
See that the derivative gives:
So this implies
• Step 3:
Plugging everything into the formula gives us:
Hence, we just approximated the number.
One interesting thing to note is that linear approximation and differentials both give the same result for $\sqrt{10}$.
http://tutorial.math.lamar.edu/Classes/CalcI/LinearApproximations.aspx
## Proving L'Hospital's Rule using linear approximation
Now we have learned a lot about linear approximation, but what else can we do with it? We can actually use the linear approximation formula to prove a rule known as L'Hospital's Rule . Here is how the proof works.
Recall that the linear approximation formula is:
See that we can rearrange the formula so that:
Realize that the approximation becomes more and more accurate as we pick x values that are closer to a. In other words if we take the limit as x→a, then they will equal. So
Now we are going to put this aside and use it later, and actually look at l'hopital's rule. We are going to assume a couple things here. Suppose that f(x) and g(x) are continuously differentiable at a real number a, f(a)=g(a)=0, and g'(a) $\neq$ 0. Then,
Now notice that we can apply the formula that we derived earlier right here. So now
Now instead of writing f'(a) and g'(a), we can apply limits as x→a (because we know f and g are differentiable). So
Hence, we just showed that:
which is L'Hospital's Rule .
## When to use l'hopital's rule
We always want to apply l'hoptial's rule when we encounter indeterminate limits. There are two types of indeterminate forms. These indeterminate forms would be:
A lot of people make the mistake of using l'hopital's rule without even checking if it is an indeterminate limit. So make sure you check it first! Otherwise, it will not work and you will get the wrong answer. Here is a guide to using l'hopital's rule:
• Step 1: Evaluate the limit directly.
• Step 2: Check if it is one of the indeterminate forms. If it is, go to step 3.
• Step 3: Use l'hopital's rule.
• Step 4: Check if you get another indeterminate form. Repeat Step 3 if you do.
Let's take a look at a few examples using these steps.
Question 6: Evaluate the limit
• Step 1: Evaluating the limit directly gives us
• Step 2:
Yes, it is one of the indeterminate forms.
• Step 3:
Applying l'hopital's rule we have:
• Step 4:
one is not an indeterminate form, so we are done and the answer is 1.
Now that question was a little bit easy, so why don't we take a look at something that is a bit harder.
Question 7: Evaluate the limit
• Step 1: Evaluating the limit directly we see that:
• Step 2:
This is an indeterminate form, so go to step 3.
• Step 3:
Applying l'hopital's rule we have
• Step 4:
This is another indeterminate form. So we have to go back to step 3 and apply l'hoptial's rules again.
• Step 3:
Applying l'hopital's rule again we have:
• Step 4:
Infinity is not an indeterminate form, so we are done and the answer is $\infty$
Use a linear approximation to estimate a given number:
1) identify the corresponding function
2) pick an appropriate "a", such that:
-"a" is within a reasonable range
-$f(a)$ is easy to evaluate
3) set up the linear approximation formula:
- for $x$ near a:$f(x) \approx f(a)+f'(a)(x-a)$
4) estimate the given number! |
Miracle Learning Center
# Prime Factorization
## 06 Feb Prime Factorization
If you have taken some maths tuition on prime and composite numbers and understand the concept, it is very likely that you will appreciate the technique of prime factorization. It is an amazing technique of extracting information from composite numbers. As your maths tuition teacher would have already taught you, prime numbers are those numbers which are divisible by only that number and 1. Composite numbers, on the other hand, are divisible by other numbers as well, besides itself and 1. Prime factorization is a powerful technique using which we can decompose composite numbers and can find a lot of information about the number which were not visible on the surface.
Prime Factorization
The fundamental theorem of arithmetic says that every number, except 1, is either a prime or can be written as the product of unique combination of prime numbers. Let us take an example.
Lets us consider the number 36. 36 can be expressed in the following way:
36= 2 x 2 x 3 x 3 = 22 x 32
As you can see, we have written 36 as the product of two prime numbers. According to fundamental theorem of arithmetic, we can express each and every composite number as a product of prime numbers. Further, it also says that this combination is unique – i.e. 36 can be written as a product of prime numbers in only one way – that is using two 3s and two 2s; 36 cannot be expressed as the product of any other combination of prime numbers.
Even though you have taken some math tuition on number theory and are familiar with the concepts of composite and prime numbers, it is difficult to spot just how much information is crammed inside such an expression shown above. Let us take a look.
Finding number of factors
First of all, we could find the number of factors of a number from an expression. How? Let us see. Say, we wanted to find out the number of factors of 120. First, we are going to perform a prime factorization.
120= 10 x 12 =5 x 23 x 3
The next step is to find out all the exponents of the prime numbers. So in this case, the number 5 has an exponent of 1, number 2 has an exponent of 3 and number 3 has exponent of 1. In case you do not know, if a number has no exponent mentioned explicitly, the exponent is 1 which is the case here for the numbers 5 and 3. if you are not sure what an exponent is, you will need to check with your maths tuition teacher.
Once we found the exponents, add 1 to each of the exponents. So the numbers we get are 2, 4 and 2 respectively after adding 1 to each exponent. Next, multiply the numbers and voila! The result is the number of factors for 120 which is 2 x 2 x 4 =16. Let us check our answer. Following is a list of factors if 120.
120 -> Factors 1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120 i.e. a total of 16 factors which matches our answer.
Finding number of odd and even factors
Next, we are going to see how we can find out the number of odd factors. To do that, we are going to ignore the exponent of 2 and then apply the same technique. So, the exponents to be considered in this case are the exponents of 5 which is 1 and the exponent of 3 which also happens to be 1. Next, add 1 to the exponents and multiply them. The number that we get is 4. So 120 has 4 odd factors. Indeed, once we scan our list, we see that there are 4 odd factors 1,3,5 and 15.
To find out the number of even factors, you can first calculate the number of all factors. Then subtract the number of odd factors from the number of all factors and the result will be the number of even factors which, in this case, is 16-4= 12.
Finding sum of all divisors
If finding out the number of factors, number of even factors and number of odd factors was not enough information, we could still find more. We could find out some of all the factors. How? Let us explore.
To find out the sum of all factors, take each prime number and add up all the powers of that number starting from 0 to up to the highest power present.
120= 10 x 12 =5 x 23 x 3
So, we take 2 first. The highest power of 2 present here is 3. So we will have to add 20 + 21+22+23 and the result is 15 (Please note the fact that any number when raised to the power of zero yields 1 and because of this, 20 =1 )
Similarly, we do this for the other numbers 3 and 5.
Highest power of 3 present here is 1. So the resulting sum is 30 + 31 = 4.
Highest power of 5 present here is 1. So the resulting sum is 50+51 = 6.
The next step is to multiply these results. Once we do that, we get 15 x 4 x 6 =360 and this is our answer. The sum of all factors of 120 is 360. If you add all the numbers is the above list, indeed you get 360.
We could also find out the sum of even factors and odd factors from the prime factorization of any number and that is similar to maths tuition homework for you.
As it is clear by now, prime factorization is an extremely powerful technique and by leveraging this technique, you could find lots of information about a number. It is a powerful concept to know and once you master it, you will be able to apply it to tackle a vast range of maths tuition problems. However, please do not mechanically follow the technique. Your first task, now, should be to figure out why these techniques mentioned above work. Why are these techniques able to find out the number of factors, odd factors, even factors and also the sum of factors. This understanding is essential in assimilating the concept. Try to seek help from your maths tuition teachers at math tuition or from your peers in school and try to work on it. |
# Beginning Mathematics/Set Theory
In this section we shall go over some basic set theory.
# Sets
A set is a well defined collection of things; which things can be objects, numbers, symbols and so on. If we can determine whether the given thing belongs to the collection or not,then that collection is said to be well defined. Those things that make up a set are its members, and are also called elements of that set. Elements of sets may be objects, letters, items, or even other sets. Sets can be finite(those containing a definite number of elements), or infinite(those containing an endless list of members).
Sets are denoted by capital letters of the English alphabet such as A,W,D,C,V,X,Z... . The elements of the set are indicated between the flower braces (also known as curl brackets),usually with members separated by commas. Ex: A={2,4,6,8,10}. Since all the elements are listed, it is called List form. This is also called Roster form.
Often, sets will have a pattern of elements or the elements will be similar in some fashion. At times, when it is clear what the pattern is, the set will end with "..." indicating the elements continue in that fashion. Ex:X={1,3,5,7,9,11,...}. Other times the set may be described in words or mathematical notation. Mathematical notation is read as in the following example: for the set X X={1,3,5,7,9,11,...}. This is shown as X={x/x is an odd number}
we read this as X is a set of all x's such that x is an odd number. In general, this is read as " X is the set of x such that x has a property".
Here the / stands for "such that" often a colon (:) will be used instead.
To denote an element is in a set we write: ${\displaystyle x\in A}$ this reads "x is an element of the set A" or "x is in A".
## Special Sets
Some sets are used frequently in mathematics so they are given by special notation.
${\displaystyle \mathbb {N} =\{Set\;of\;natural\;numbers\}=\{1,2,3,4,5,6,...\}}$
${\displaystyle \mathbb {Z} =\{Set\;of\;all\;integers\}=\{...,-3,-2,-1,0,1,2,3,...\}}$
${\displaystyle \mathbb {Q} =\{Set\;of\;rational\;numbers\}=\{{\frac {a}{b}}\;|\;a,b\in \mathbb {Z} \;and\;b\neq 0\}}$
${\displaystyle \mathbb {R} =\{Set\;of\;real\;numbers\}}$
${\displaystyle \mathbb {C} =\{Set\;of\;complex\;numbers\}=\{a+bi\;|\;a,b\in \mathbb {R} \;and\;where\;i^{2}=-1\}}$
## Examples
${\displaystyle \{1,2,3\}}$
${\displaystyle \{7,12,80\}}$
${\displaystyle \{a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z\}=\{{\text{letters of english alphabet}}\}}$
${\displaystyle \{1,2,3,4,5,...\}=\{{\text{Set of positive integers}}\}}$
${\displaystyle \{2,4,6,8,10,12,14...\}=\{{\text{Set of positive even integers}}\}=\{x\mid x\in \mathbb {Z} \;{\text{and }}x{\text{ is even}}\}}$
## Basic operations
### Union
The union of two sets is denoted by ${\displaystyle \bigcup }$. The union of two sets is best described as "everything in both the sets once".
Formally: ${\displaystyle A\bigcup B=\{x|x\in A\;or\;x\in B\}}$
For example:
${\displaystyle A=\{1,2,3\}}$ ${\displaystyle B=\{4,5,6\}}$
${\displaystyle A\bigcup B=\{1,2,3\}\bigcup \{4,5,6\}=\{1,2,3,4,5,6\}}$
### Intersection
The intersection of two sets is denoted by ${\displaystyle \bigcap }$. The intersection of two sets is best described as "only what is in both of the sets".
Formally: ${\displaystyle A\bigcap B=\{x|x\in A\;and\;x\in B\}}$
For example:
${\displaystyle A=\{1,2,3,4,5,6\}}$ ${\displaystyle B=\{2,4,6,8,10\}}$
${\displaystyle A\bigcap B=\{1,2,3,4,5,6\}\bigcap \{2,4,6,8,10\}=\{2,4,6\}}$ |
# Finite Mathematics : Independence
## Example Questions
### Example Question #1 : Probability
Two fair dices are rolled. If the first die is an odd number what is the probability that the sum of the two dice equal 8?
Explanation:
First, recall that each die is rolled independently.
Now, it is known that the first die is an odd number therefore the possible options are:
From here, find the number of combinations each one will have to make a sum of 8 with the other die.
By investigation, die number one cannot be one because it along with any value from die two will not equal eight.
Therefore, investigate when die one is three.
Next, if die one is five then the possible sum is,
Therefore there are a total of two possibilities of rolling a sum equaled to eight.
From here, identify the total number of options the sum could be when it does not equal eight.
### Example Question #2 : Probability
A coin is loaded so that it comes up heads 60% of the time. If it is tossed six times, what is the probability that it will come up heads at least five times?
Choose the closest response.
Explanation:
Since each trial can have one of two outcomes, heads or tails, this is an example of a Bernoulli process.
The probability that trials will result in successes in a Bernoulli process is
,
where is the probability of a success in any given trial and is the probability of a failure.
Since there are six tosses, set . We can call heads the success with probability , and tails the failure with probability . Accordingly, the probability that exactly heads will come up is
Since we are looking for the probability of heads coming up at least five times, evaluate as follows:
Of the five choices, 0.25 comes closest.
### Example Question #1 : Independence
Jan and Jeff are playing a game of poker dice, in which a turn comprises the following process:
1) Roll all five dice - note: these are standard six-sided dice, which you may assume to be fair.
2) Keep the dice you want, then roll the remaining dice in hope of improving your score.
3) Repeat Step 2, after which whatever you have stands.
Jan is down to her last turn. She has rolled the dice once and has the following: 1-2-3-4-6.
She must score a "large straight", which comprises five dice in sequence. (1-2-3-4-5, or 2-3-4-5-6). If she re-rolls the "6", what is the probability that she will get her large straight on her second or third roll?
Explanation:
Let be the event that Jan gets a straight - that is, that she rolls a "5" on her second or third turn.
The easiest way to find the probability that Jan will complete her large straight - by rolling a "5" - is to find the probability that she will not complete it. This happens if she rolls any number other than "5" on both rolls, which, by the multiplication principle, will happen with probability
The probability that she will get the straight is this probability subtracted from 1, or
### Example Question #1 : Independence
A penny is flipped ten times; each flip results in heads.
True or false: The coin must be loaded so that it comes up heads more often than tails.
True
False
False
Explanation:
Suppose the coin is indeed fair. The probability of the coin coming up heads in one trial is ; by the multiplication principle, the probability that the coin will come up heads each of ten consecutive times is . While this makes this event extremely unlikely, it is not impossible. Thus, the fact that the coin comes up ten heads in ten flips is not conclusive evidence that the coin is unfair.
### Example Question #3 : Independence
Mike and Spike are playing a game of poker dice, in which a turn comprises the following process:
1) Roll all five dice - note: these are standard six-sided dice, which you may assume to be fair.
2) Keep the dice you want, then roll the remaining dice in hope of improving your score.
3) Repeat Step 2, after which whatever you have stands.
Mike is down to his last turn. He has rolled the dice once and has the following: 1-3-3-3-5.
He must roll a five of a kind in order to score. If he re-rolls the "1" and the "5", what is the probability that he will get five "3's"?
Explanation:
In order for Mike to get his five-of-a-kind, one of three things has to happen:
1) He can roll two "3's" on his next turn. There is only one way out of thirty-six for this to happen, so the probability of rolling two 3's is .
2) He can roll one "3" on his next turn, then one "3" on his third turn. There are ten ways out of thirty-six to roll exactly one "3" with a pair of dice (1-3, 2-3, 4, 3, 5-3, 6-3, 3-1, 3-2, 3-4, 3-5, 3-6), making this probability . After this, he will roll one die; the probability of rolling a "3" is . By the multiplication principle, the probability of this scenario happening is .
3) He can roll no "3's" on his next turn, then two "3's" on his third turn. There are twenty-five ways out of thirty-six to roll two dice and not get a 3, so the probability of this is ; the probability of rolling two "3's" on the third turn is . By the multiplication principle, the probability of this scenario happening is .
These three scenarios are mutually exclusive and together comprise the event that Mike gets his two 3's. Add the probabilities:
### Example Question #2 : Independence
A number of black and white balls are placed in a box. A ball is drawn and replaced, and its color is noted; this is repeated 100 times. At the end of the experiment, it is noted that a black ball was drawn 23 times; it is concluded that the probability of drawing a black ball from the box is .
This is an example of:
Theoretical probability
Empirical probability
Empirical probability
Explanation:
Empirical probability is based on repeated experimentation and observation; theoretical probability is calculated based on equally probable events and the number of such events favorable to a given outcome.
The probability that a black ball will be drawn is hypothesized based on an experiment, in which the number of times a black ball was drawn compared to the number of total draws was observed. The probability was derived based on this observation. This is therefore an example of empirical probability.
### Example Question #3 : Independence
A fair coin comes up heads with probability . This probability is an example of:
Theoretical probability
Empirical probability
Theoretical probability
Explanation:
Empirical probability is based on repeated experimentation and observation; theoretical probability is calculated based on equally probable events and the number of such events favorable to a given outcome.
The two outcomes for the flip of a fair coin are two, which are equally probable. The probability of each outcome is based on theoretical probability.
### Example Question #1 : Independence
The twelve face cards (kings, queens, jacks) are separated from a standard deck of 52 cards. Two cards are selected at random from the twelve, without replacement. What is the probability that the two cards will be of different color?
Explanation:
Two cards are drawn from the deck without regard to order, so the sample space is the set of all combinations of two cards from a set of twelve. The size of this sample space is
The event is the set of all selections of one card from a set of six red cards and one card from a set of six black cards. The selections are independent, so by the multiplication principle,
The probability of the event is
.
### Example Question #11 : Probability
The four aces, the four deuces, and the joker are separated from a standard deck of 53 cards. Two cards are selected at random from the nine, without replacement. What is the probability that both cards will be red?
Note: The joker is considered to be neither red nor black.
Explanation:
Two cards are drawn from the deck without regard to order, so the sample space is the set of all combinations of two cards from a set of nine. The size of this sample space is
The event is the set of all combinations of two cards from the set of four red cards. The size of this event space is
The probability of the event is |
# Clock Faces
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Clock Faces
CCSS.MATH.CONTENT.1.MD.B.3
## Basics on the topicClock Faces
### In this Video on How to Read an Analog Clock Face
In the landfill, Freddie Zuri find a broken house? Or is it a clock? They aren't sure exactly what it is because all the numbers and hands have fallen off. After taking it home, they look it up on their computer and learn that it is a cuckoo clock. However, they need to learn more in order to fix it. In this video, Zuri and Freddie learn about analog clock faces.
### Analog Clock Face for Teaching
Analog clocks are tools used to measure and tell time. We usually see analog clocks that look like circles, but they can be other shapes too!
The front of the clock is called the face. The face shows the numbers one through twelve at equal spaces along its edge. Four special numbers help us make sure the others are spaced evenly: twelve at the top, six at the bottom, three on the right and nine on the left.
On this clock, there are two hands, or arrows, pointing to the numbers.The shorter arrow, or hour hand, points to the hour. The longer arrow or, minute hand, points to the minute.
[img]line 22_ analog clock face with minutes[/img]
You can remember this because the word hour is short like the hour hand. The word minute is long like the minute hand.
The hands on the clock always move this way around the face, from one to twelve.
#### How Does an Analog Clock Work: Assembling a Clock Face
Let's put the numbers and hands on this clock face.
First, we sort our numbers in order from one to twelve.
Next, we put the number twelve at the top spot. Then add the rest of the numbers, starting with one.
Last, we have the short hour hand and the long minute hand meet in the middle. Viola, our clock face is complete!
### Summary: Analog Clock Faces
• Analog clocks are tools used to measure and tell time.
• The front is called the face and shows the numbers one through twelve at equal spaces along the edge.
• Two hands point to the numbers to tell us the time: the shorter hour hand and the longer minute hand.
• The hands always move around the face from one to twelve.
### Practice: How to Read an Analog Clock Face
Have you practiced yet? On this website, you can also find worksheets and exercises with analog clock faces for kids.
### TranscriptClock Faces
Freddie and Zuri can't wait to see what's in today's dump. "Is this a house...or a clock?" "Hmmm, let's take it back to the van and find out!" This is a cuckoo clock, but it's fallen apart! Let's help Freddie and Zuri put it back together by teaching them about "Clock Faces!" This type of clock is called an analog clock. Analog clocks are tools used to measure and tell time. We usually see analog clocks that look like circles, but they can be other shapes too! The front of the clock is called the face. The face has the numbers one through twelve at equal spaces along its edge. Four special numbers help us make sure the others are spaced evenly: twelve at the top, six at the bottom, three on the right and nine on the left. On this clock, there are two hands, or arrows, pointing to the numbers. The shorter arrow, or hour hand, points to the hour. The longer arrow or, minute hand, points to the minute. You can remember this because the word hour is short like the hour hand. The word minute is long like the minute hand. The hands on the clock always move this way around the face, from one to twelve. Let's put the numbers and hands on this clock face. First, we sort our numbers in order from one to twelve. Next, we put the number twevle at the top spot. Let's put the rest of the numbers on starting with one. Last, we have the short hour hand and the long minute hand meet in the middle. Ta-dah, our clock face is complete! Now that we've practiced assembling a clock face, let's help Freddie and Zuri fix the one they found! What do we do first? We put the numbers in order from one to twelve! What happens next? Next, twelve goes to the top of the clock face. Now, let's put the rest of the numbers on starting with one. Uh-oh something does not look right! What is wrong? The numbers are pushed too close together! How can we fix this? Let's spread the numbers out evenly at each marker so that they meet up with twelve at the top. That's better, what is our last step? Last, we have the short hour hand and the long minute hand meet in the middle. We've fixed the cuckoo clock. "Let's point the hour hand to one and the minute hand to twelve!" "Freddie, what did you do?!" Before we find out what's happening, let's remember! Analog clocks are tools used to measure and tell time. They usually look like a circle, but can be any shape. The front is called the face and shows the numbers one through twelve at equal spaces along the edge. Twelve is at the top, six is at the bottom, three is on the right and nine is on the left. Two arrows, or hands, point to the numbers to tell us the time. The shorter arrow, or hour hand, points to the hour. The longer arrow or, minute hand, points to the minute. The hands always move this way around the face, from one to twelve. . "Now I know why it was broken!" |
# 1.3 Rates of change and behavior of graphs
Page 1 / 15
In this section, you will:
• Find the average rate of change of a function.
• Use a graph to determine where a function is increasing, decreasing, or constant.
• Use a graph to locate local maxima and local minima.
• Use a graph to locate the absolute maximum and absolute minimum.
Gasoline costs have experienced some wild fluctuations over the last several decades. [link] http://www.eia.gov/totalenergy/data/annual/showtext.cfm?t=ptb0524. Accessed 3/5/2014. lists the average cost, in dollars, of a gallon of gasoline for the years 2005–2012. The cost of gasoline can be considered as a function of year.
$y$ 2005 2006 2007 2008 2009 2010 2011 2012 $C\left(y\right)$ 2.31 2.62 2.84 3.3 2.41 2.84 3.58 3.68
If we were interested only in how the gasoline prices changed between 2005 and 2012, we could compute that the cost per gallon had increased from $2.31 to$3.68, an increase of $1.37. While this is interesting, it might be more useful to look at how much the price changed per year . In this section, we will investigate changes such as these. ## Finding the average rate of change of a function The price change per year is a rate of change because it describes how an output quantity changes relative to the change in the input quantity. We can see that the price of gasoline in [link] did not change by the same amount each year, so the rate of change was not constant. If we use only the beginning and ending data, we would be finding the average rate of change over the specified period of time. To find the average rate of change, we divide the change in the output value by the change in the input value. The Greek letter $\text{Δ}\text{\hspace{0.17em}}$ (delta) signifies the change in a quantity; we read the ratio as “delta- y over delta- x ” or “the change in $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ divided by the change in $\text{\hspace{0.17em}}x.$ ” Occasionally we write $\text{\hspace{0.17em}}\text{Δ}f\text{\hspace{0.17em}}$ instead of $\text{\hspace{0.17em}}\text{Δ}y,\text{\hspace{0.17em}}$ which still represents the change in the function’s output value resulting from a change to its input value. It does not mean we are changing the function into some other function. In our example, the gasoline price increased by$1.37 from 2005 to 2012. Over 7 years, the average rate of change was
On average, the price of gas increased by about 19.6¢ each year.
Other examples of rates of change include:
• A population of rats increasing by 40 rats per week
• A car traveling 68 miles per hour (distance traveled changes by 68 miles each hour as time passes)
• A car driving 27 miles per gallon (distance traveled changes by 27 miles for each gallon)
• The current through an electrical circuit increasing by 0.125 amperes for every volt of increased voltage
• The amount of money in a college account decreasing by \$4,000 per quarter
## Rate of change
A rate of change describes how an output quantity changes relative to the change in the input quantity. The units on a rate of change are “output units per input units.”
The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values.
$\frac{\text{Δ}y}{\text{Δ}x}=\frac{f\left({x}_{2}\right)-f\left({x}_{1}\right)}{{x}_{2}-{x}_{1}}$
how to solve the Identity ?
what type of identity
Jeffrey
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
by how many trees did forest "A" have a greater number?
Shakeena
32.243
Kenard
how solve standard form of polar
what is a complex number used for?
It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
Steve
I would like to add that they are used in AC signal analysis for one thing
Scott
Good call Scott. Also radar signals I believe.
Steve
Is there any rule we can use to get the nth term ?
how do you get the (1.4427)^t in the carp problem?
A hedge is contrusted to be in the shape of hyperbola near a fountain at the center of yard.the hedge will follow the asymptotes y=x and y=-x and closest distance near the distance to the centre fountain at 5 yards find the eqution of the hyperbola
A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour. To the nearest hour, what is the half-life of the drug?
Find the domain of the function in interval or inequality notation f(x)=4-9x+3x^2
hello
Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of ?105°F??105°F? occurs at 5PM and the average temperature for the day is ?85°F.??85°F.? Find the temperature, to the nearest degree, at 9AM.
if you have the amplitude and the period and the phase shift ho would you know where to start and where to end?
rotation by 80 of (x^2/9)-(y^2/16)=1
thanks the domain is good but a i would like to get some other examples of how to find the range of a function
what is the standard form if the focus is at (0,2) ? |
# Chapter 13 Probability RD Sharma Solutions Exercise 13.1 Class 10 Maths
Chapter Name RD Sharma Chapter 13 Probability Book Name RD Sharma Mathematics for Class 10 Other Exercises Exercise 13.2 Related Study NCERT Solutions for Class 10 Maths
### Exercise 13.1 Solutions
1. The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow ?
Solution
Let E be the event of happening of rain
P(E) is given as 0.85
2. A die is thrown. Find the probability of getting:
(i) a prime number
(ii) 2 or 4
(iii) a multiple of 2 or 3
(iv) an even prime number
(v) a number greater than 5
(vi) a number lying between 2 and 6
Solution
(i) Total no of possible outcomes = 6{1, 2, 3, 4, 5, 6}
E ⟶ Event of getting a prime no.
No. of favorable outcomes = 3{2, 3, 5}
P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes)
P(E) = 3/6 = 1/2
(ii) E ⟶ Event of getting 2 or 4.
No. of favorable outcomes = 2{2, 4}
Total no. of possible outcomes = 6
Then, P(E) = 2/6 = 1/3
(iii) E ⟶ Event of getting a multiple of 2 or 3
No. of favorable outcomes = 4{2, 3, 4, 6}
Total no. of possible outcomes = 6
Then, P(E) = 4/6 = 2/3
(iv) E ⟶ Event of getting an even prime no.
No. of favorable outcomes = 1{2}
Total no. of possible outcomes = 6{1, 2, 3, 4, 5, 6}
P(E) = 1/6
(v) E ⟶ Event of getting a no. greater than 5.
No. of favorable outcomes = 1{6}
Total no. of possible outcomes = 6
P(E) = 1/6
(vi) E ⟶ Event of getting a no. lying between 2 and 6.
No. of favorable outcomes = 3{3, 4, 5}
Total no. of possible outcomes = 6
P(E) = 3/6 = 1/2
3. In a simultaneous throw of a pair of dice, find the probability of getting :
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) a doublet of odd numbers
(v) a sum greater than 9
(vi) an even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) at least once
(xiii) a number other than 5 on any dice.
Solution
In a throw of pair of dice, total no of possible outcomes = 36 = 6×6 which are
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
(i) Let E be event of getting the sum as 8
No. of favorable outcomes = 5 {(2,6)(3,5)(4, 4)(5, 3)(6, 2)}
We know that, Probability P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes)
P(E) = 5/36
(ii) E ⟶ event of getting a doublet
No. of favorable outcomes = 5{(1, 1)(2, 2)(3, 3)(5, 5)}
Total no. of possible outcomes = 36
P(E) = 6/36 = 1/6
(iii) E ⟶ event of getting a doublet of prime no's
No. of favorable outcomes = 3{(2, 2) (3, 3) (5, 5)}
Total no. of possible outcomes = 36
P(E) = 3/36 = 1/12
(iv) E ⟶ event of getting a doublet of odd no's
No. of favorable outcomes = 3{(1, 1) (3, 3) (5, 5)}
Total no. of possible outcomes = 36
P(E) = 3/36 = 1/12
(v) E ⟶ event of getting a sum greater than 9
No. of favorable outcomes = 6 {(4, 6) (5, 5) (5, 6) (6, 4) (6, 5) (6, 6)}
Total no. of possible outcomes = 36
P(E) = 6/36 = 1/6
(vi) E ⟶ event of getting an even no. on first
No. of favorable outcomes = 18 {(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6), (4, 1) , (4, 2)
, (4, 3) , (4, 4) , (4, 5) , (4, 6) , (6, 1) , (6, 2) , (6, 3) , (6, 4) , (6, 5) , (6, 6)}
Total no. of possible outcomes = 36
P(E) = 18/36 = 1/2
(vii) E ⟶ event of getting an even no. on one and a multiple of 3 on other
No. of favorable outcomes = 11 {(2, 3) (2, 6) (4, 3) (4, 6) (6, 3) (6, 6), (3, 2), (3, 4),
(3, 4) , (3, 6) , (6, 2) , (6, 4)}
Total no. of possible outcomes = 36
P(E) = 11/36
(viii)
(ix) E ⟶ event of getting a sum less than 6
No. of favorable outcomes = 10 {(1, 1) (1, 2) (1, 3) (1, 4) (2, 1) (2, 2) , (2, 3) , (3, 1)
, (3, 2) , (4, 1)}
Total no. of possible outcomes = 36
P(E) = 10/36 = 5/18
(x) E ⟶ event of getting a sum less than 7
No. of favorable outcomes = 15 {(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (2, 1), (2, 2) (2, 3)
(2, 4) (3, 1) (3, 2) (3, 3) (4, 1) (4, 2) (5, 1)}
Total no. of possible outcomes = 36
(xi) E ⟶ event of getting a sum more than 7
No. of favorable outcomes = 15 {(2, 6) (3, 5) (3, 6) (4, 4) (4, 5) (4, 6), (5, 3) (5, 4)
(5, 5) (5, 6) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)}
Total no. of possible outcomes = 36
P(E) = 15/36 = 5/12
(xii) E ⟶ event of getting a 1 at least once
No. of favorable outcomes = 11 {(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6), (2, 1) (3, 1)
(4, 1) (5, 1) (6, 1)}
Total no. of possible outcomes = 36
P(E) = 11/36
(xiii) E ⟶ event of getting a no other than 5 on any dice
No. of favourable outcomes = 25 {(1, 1) (1, 2) (1, 3) (1, 4) (1, 6) (2, 1), (2, 2) (2, 3)
(2, 4) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 6) (6, 1) (6, 2)
(6, 3) (6, 4) (6, 6)}
Total no. of possible outcomes = 36
P(E) = 25/36
4. Three coins are tossed together. Find the probability of getting :
(iii) at least one head and one tail
(iv) no tails
Solution
When 3 coins are tossed together,
Total no. of possible outcomes = 8 {HHH, HHT, HTH, HTT, THH,THT, TTH, TTT}
(i) Probability of an even = (No. of favorable outcomes)/(Total no. of possible outcomes)
Let E ⟶ event of getting exactly two heads
No. of favourable outcomes = 3 {HHT, HTH, THH}
Total no. of possible outcomes = 8
P(E) = 3/8
(ii) E ⟶ getting at least 2 Heads
No. of favourable outcomes = 4 {HHH, HHT, HTH, THH}
Total no. of possible outcomes = 8
P(E) = 4/8 = 1/2
(iii) E ⟶ getting at least one Head & one Tail
No. of favourable outcomes = 6 {HHT, HTH, HTT, THH, THT, TTH}
Total no. of possible outcomes = 8
P(E) = 6/8 = 3/4
(iv) E ⟶ getting no tails
No. of favourable outcomes = 1 { HHH}
Total no. of possible outcomes = 8
P(E) = 1/8
5. What is the probability that an ordinary year has 53 Sundays?
Solution
Ordinary year has 365 days
365 days = 52 weeks + 1 day
That 1 day may be Sun, Mon, Tue, Wed, Thu, Fri, Sat
Total no. of possible outcomes = 7
Let E ⟶ event of getting 53 Sundays
No. of favourable outcomes = 1 {Sun}
P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes) = 1/7
6. What is the probability that a leap year has 53 Sundays and 53 Mondays?
Solution
A leap year has 366 days
366 days = 52 weeks + 2 days
That 2 days may be (Sun, Mon) (Mon, Tue) (Tue, Wed) (Wed, Thu) (Thu, Fri) (Fri, Sat)
(Sat, Sun)
Let E ⟶ event of getting 53 Sundays & 53 Mondays.
No. of favourable outcomes = 1 {(Sun, Mon)}
Since 52 weeks has 52 Sundays & 52 Mondays & the extra 2 days must be Sunday &
Monday.
Total no. of possible outcomes = 7
P(E) = (No. of favorable outcomes)/Total no. of possible outcomes) = 1/7
7. A and B throw a pair of dice. If A throws 9, find B's chance of throwing a higher number.
Solution
When a pair of dice are thrown, then total no. of possible outcomes = 6 × 6 = 36, which are
{ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) }
E ⟶ event of throwing a no. higher than 9.
No. of favourable outcomes = 6 {(4, 6) (5, 5) (6, 4) (5, 6) (6, 5) (6, 6)}
We know that P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes)
i.e., P(E) = 6/36 = 1/6
8. Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10.
Solution
When a pair of dice are thrown, then total no. of possible outcomes = 6 × 6 = 36
let E ⟶ event of getting sum of dice greater than 10
then no of favourable outcomes = 3{(5, 6) (6, 5) (6, 6)}
we know that, P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes)
i.e., P(E) = 3/36 = 1/12
9. A card is drawn at random from a pack of 52 cards. Find the probability that card drawn is
(i) a black king
(ii) either a black card or a king
(iii) black and a king
(iv) a jack, queen or a king
(v) neither a heart nor a king
(vii) neither an ace nor a king
(viii) Neither a red card nor a queen.
(ix) other than an ace
(x) a ten
(xii) a black card
(xiii) the seven of clubs
(xiv) jack
(xvi) a queen
Solution
Total no. of outcomes = 52 {52 cards}
(i) E⟶ event of getting a black king
No of favourable outcomes = 2{king of spades & king of clubs}
We know that, P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes) = 2/52 = 1/26
(ii) E⟶ event of getting either a black card or a king.
No. of favourable outcomes = 26 + 2 {13 spades, 13 clubs, king of hearts & diamonds}
P(E) = (26+2)/52 = 28/52 = 7/13
(iii) E⟶ event of getting black & a king.
No. of favourable outcomes = 2 {king of spades & clubs}
P(E) = 2/52 = 1/26
(iv) E⟶ event of getting a jack, queen or a king
No. of favourable outcomes = 4 + 4 + 4 = 12 {4 jacks, 4 queens & 4 kings}
P(E) = 12/52 = 3/13
(v) E⟶ event of getting neither a heart nor a king.
No. of favourable outcomes = 52 – 13 – 3 = 36 {since we have 13 hearts, 3 kings
each of spades, clubs & diamonds}
P(E) = 36/52 = 9/13
(vi) E⟶ event of getting spade or an all.
No. of favourable outcomes = 13 + 3 = 16 {13 spades & 3 aces each of hearts, diamonds & clubs}
P(E) = 16/52 = 4/13
(vii) E⟶ event of getting neither an ace nor a king.
No. of favourable outcomes = 52 – 4 – 4 = 44 {Since we have 4 aces & 4 kings}
P(E) = 44/52 = 11/13
(viii) E⟶ event of getting neither a red card nor a queen.
No. of favourable outcomes = 52 – 26 – 2 = 24 {Since we have 26 red cards of hearts & diamonds & 2 queens each of heart & diamond}
P(E) = 24/52 = 6/13
(ix) E⟶ event of getting card other than an ace.
No. of favourable outcomes = 52 – 4 = 48 {Since we have 4 ace cards}
P(E) = 48/52 = 12/13
(x) E⟶ event of getting a ten.
No. of favourable outcomes = 4 {10 of spades, clubs, diamonds & hearts}
P(E) = 4/52 = 1/13
(xi) E⟶ event of getting a spade.
No. of favourable outcomes = 13 {13 spades}
P(E) = 13/52 = 1/24
(xii) E⟶ event of getting a black card.
No. of favourable outcomes = 26 {13 cards of spades & 13 cards of clubs}
P(E) = 26/52 = 1/2
(xiii) E⟶ event of getting 7 of clubs.
No. of favourable outcomes = 1 {7 of clubs}
P(E) = 1/52
(xiv) E⟶ event of getting a jack.
No. of favourable outcomes = 4 {4 jack cards}
P(E) = 4/52 = 1/13
(xv) E⟶ event of getting the ace of spades.
No. of favourable outcomes = 1{ace of spades}
P(E) = 1/52
(xvi) E⟶ event of getting a queen.
No. of favourable outcomes = 4 {4 queens}
P(E) = 4/52 = 1/13
(xvii) E⟶ event of getting a heart.
No. of favourable outcomes = 13 {13 hearts}
P(E) = 13/52 = 1/4
(xviii) E⟶ event of getting a red card.
No. of favourable outcomes = 26 {13 hearts, 13 diamonds}
P(E) = 26/52 = 1/2
10. In a lottery of 50 tickets numbered 1 to 50, one ticket is drawn. Find the probability that the
drawn ticket bears a prime number.
Solution
Total no. of possible outcomes = 50 {1, 2, 3, .... , 50}
E⟶ event of getting a prime no.
No. of favourable outcomes = 15
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}
Probability, P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes)
i.e. P(E) = 15/50 = 3/10
11. An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability
that the ball drawn is white.
Solution
Total no. of possible outcomes = 18 {10 red balls, 8 white balls}
E ⟶ event of drawing white ball
No. of favourable outcomes = 8 {8 white balls}
Probability, P(E) = No. of favorable outcomes/Total no. of possible outcomes
= 8/18 = 4/9
12. A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from
the bag. What is the probability that the ball drawn is:
(i) White
(ii) Red
(iii) Black
(iv) Not red
Solution
Total number of possible outcomes = 12 {3 red balls, 5 black balls & 4 white balls}
(i) E ⟶ event of getting white ball
No. of favourable outcomes = 4 {4 white balls}
Probability, P(E) = 4/12 = 1/3
(ii) E ⟶ event of getting red ball
No. of favourable outcomes = 5 {3 red balls}
P (E) = 3/12 = 1/4
(iii) E ⟶ event of getting black ball
No. of favourable outcomes = 5 {5 black balls}
P (E) = 5/12
(iv) E ⟶ event of getting red
No. of favourable outcomes = 3 {3 black balls}
P(E) = 3/12 = 1/4
(E̅) ⟶ event of not getting red.
P(E̅) = 1 – P(E)
= 1 – 1/4 = 3/4
13. What is the probability that a number selected from the numbers 1, 2, 3, ..., 15 is a multiple
of 4?
Solution
Total no. possible outcomes = 15 {1, 2, 3, .... , 15}
E ⟶ event of getting a multiple of 4
No. of favourable outcomes = 3 {4, 8, 12}
Probability, P(E) = No. of favorable outcomes/Total no. of possible outcomes
= 3/15
= 1/5
14. A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the
probability that ball drawn is not black?
Solution
Total no. of possible outcomes = 18 {6 red, 8 black, 4 white}
Let E ⟶ event of drawing black ball.
No. of favourable outcomes = 8 {8 black balls}
Probability, P(E) = No.of favorable outcomes/Total no.of possible outcomes
= 8/18 = 4/9
E̅ ⟶ event of not drawing black ball
P(E̅) = 1 − P(E)
= 1 − 4/9
= 5/9
15. A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the
probability that ball drawn is white?
Solution
Total no. of possible outcomes = 12 {5 white, 7 red}
E ⟶ event of drawing white ball.
No. of favorable outcomes = 5 {white balls are 5}
Probability, P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes)
P(E) = 5/12
16. Tickets numbered from 1 to 20 are mixed up and a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 7 ?
Solution
Total no. of possible outcomes = 20 {1, 2, 3, .... , 20}
E ⟶ event of drawing ticket with no multiple of 3 or 7
No. of favourable outcomes = 8 which are {3, 6, 9, 12, 15, 18, 7, 14}
Probability, P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes) = 8/20 = 2/5
17. In a lottery there are 10 prizes and 25 blanks. What is the probability of getting a prize?
Solution
Total no. of possible outcomes = 35 {10 prizes, 25 blanks}
E ⟶ event of getting prize
No. of favourable outcomes = 10 {10 prizes}
Probability, P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes) = 10/35 = 2/7
18. If the probability of winning a game is 0.3, what is the probability of losing it?
Solution
E ⟶ event of winning a game
P(E) is given as 0.3
(E̅) ⟶ event of loosing the game
we know that P(E) + P(E̅) = 1
P(E̅) = 1 − P(E)
= 1 – 0.3 = 0.7
19. A bag contains 5 black, 7 red and 3 white balls. A ball is drawn from the bag at random.
Find the probability that the ball drawn is:
(i) Red (ii) black or white (iii) not black
Solution
Total no. of possible outcomes = 15
{5 black, 7 red & 3 white balls}
(i) E ⟶ event of drawing red ball
No. of favorable outcomes = 7 {7 red balls}
Probability, P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes)
P(E) = 7/15
(ii) E ⟶ event of drawing black or white
No. of favourable outcomes = 8 {5 black & 3 white}
P(E) = 8/15
(iii) E ⟶ event of drawing black ball
No. of favourable outcomes = 5 {5 black balls}
P (E) = 5/15 = 1/3
E̅ ⟶ event of not drawing black ball
P (E̅) = 1 − P(E)
= 1 − 1/3
= 2/3
20. A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random.
Find the probability that the ball drawn is:
(i) White
(ii) Red
(iii) Not black
(iv) Red or white
Solution
Total no. of possible outcomes = 15 {4 red, 5 black, 6 white balls}
(i) E ⟶ event of drawing white ball.
No. of favourable outcomes = 6 {6 white}
Probability, P(E) = (No.of favorable outcomes)/(Total no. of possible outcomes)
P(E) = 6/15 = 2/5
(ii) E ⟶ event of drawing red ball
No. of favourable outcomes = 4 {4 red balls}
P(E) = 4/15
(iii) E ⟶ event of drawing black ball
No. of favourable outcomes = 5 {5 black balls}
P(E) = 5/15 = 1/3
E̅ ⟶ event of not drawing black ball
P(E̅) = 1 − 1/3 = 2/3
(iv) E ⟶ event of drawing red or white ball
No. of favourable outcomes = 10 {4 red & 6 white}
P(E) = 10/15 = 2/3
21. A black die and a white die are thrown at the same time. Write all the possible outcomes.
What is the probability?
(i) that the sum of the two numbers that turn up is 8?
(ii) of obtaining a total of 6?
(iii) of obtaining a total of 10?
(iv) of obtaining the same number on both dice?
(v) of obtaining a total more than 9?
(vi) that the sum of the two numbers appearing on the top of the dice is 13?
(vii) that the sum of the numbers appearing on the top of the dice is less than or equal to
12?
Solution
Total no. of possible outcomes when 2 dice are thrown = 6 × 6 = 36 which are
{ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) }
(i) E ⟶ event of getting sum that turn up is 8
No. of possible outcomes = 36
No. of favourable outcomes = 5 {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}
P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes) = 5/36
(ii) Let E ⟶ event of obtaining a total of 6
No. of favourable outcomes = 5
{(1, 5) (2, 4) (3, 3) (4, 2) (5, 1)}
P(E) = 5/36
(iii) Let E ⟶ event of obtaining a total of 10.
No. of favourable outcomes = 3 {(4, 6) (5, 5) (6, 4)}
P(E) = 3/36 = 1/12
(iv) Let E ⟶ event of obtaining the same no. on both dice
No. of favourable outcomes = 6 {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}
P(E) = 3/12 = 1/12
(v) E ⟶ event of obtaining a total more than 9
No. of favourable outcomes = 6 {(4, 6) (5, 5) (6, 4) (5, 6) (6, 5) (6, 6)}
P(E) = 6/36 = 1/6
(vi) The maximum sum is 12 (6 on 1st + 6 on 2nd)
So, getting a sum of no’s appearing on the top of the two dice as 13 is an impossible
event.
∴ Probability is 0
(vii) Since, the sum of the no’s appearing on top of 2 dice is always less than or equal to
12, it is a sure event.
Probability of sure event is 1.
So, the required probability is 1.
22. One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:
(i) a king of red suit
(ii) a face card
(iii) a red face card
(iv) a queen of black suit
(v) a jack of hearts
Solution
Total no. of possible outcomes = 52 (52 cards)
(i) E ⟶ event of getting a king of red suit
No. of favourable outcomes = 2 {king heart & king of diamond}
P(E), = (No.of favorable outcomes)/(Total no.of possible outcomes)
= 2/52 = 1/26
(ii) E ⟶ event of getting face card
No. of favourable outcomes = 12 {4 kings, 4 queens & 4 jacks}
P(E) = 12/52 = 3/13
(iii) E ⟶ event of getting red face card
No. favourable outcomes = 6 { kings, queens, jacks of hearts & diamonds}
P(E) = 6/52 = 3/13
(iv) E ⟶ event of getting a queen of black suit
No. favourable outcomes = 6 { kings, queens, jacks of hearts & diamonds}
P(E) = 6/52 = 3/26
(v) E ⟶ event of getting red face card
No. favourable outcomes = 6 { queen of spades & clubs}
P(E) = 1/52
(vi) E ⟶ event of getting a spade
No. favourable outcomes = 13 {13 spades}
P(E) = 13/52 = 1/4
23. Five cards - ten, jack, queen, king, and an ace of diamonds are shuffled face downwards. One card is picked at random.
(i) What is the probability that the card is a queen ?
(ii) If a king is drawn first and put aside, what is the probability that the second card picked up is the ace ?
Solution
Total no. of possible outcomes = 5{ 5 cards}
(i) E ⟶ event of drawing queen
No. favourable outcomes = 1 {1 queen card}
P(E) = (No.of favorable outcomes)/(Total no.of possible outcomes)
= 1/5
(ii) When king is drawn and put aside, total no. of remaining cards = 4
Total no. of possible outcomes = 4
E ⟶ event of drawing ace card
No. favourable outcomes = 1 {1 ace card}
P(E) = 1/4
24. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What
is the probability that the ball drawn is:
(i) Red
(ii) Black
Solution
Total no. of possible outcomes = 8 {3 red, 5 black}
(i) Let E ⟶ event of drawing red ball.
No. favourable outcomes = 1 {1 ace card}
P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes)
= 3/8
(ii) Let E ⟶ event of drawing black ball.
No. of favourable outcomes = 5 {5 black balls}
P(E) = 5/8
25. A bag contains cards which are numbered from 2 to 90. A card is drawn at random from
the bag. Find the probability that it bears.
(i) a two digit number
(ii) a number which is a perfect square
Solution
Total no. of possible outcomes = 89 {2, 3, 4, ...., 90}
(i) Let E ⟶ event of getting a 2 digit no.
No. favourable outcomes = 81 {10, 11, 12, 13, ....., 80}
P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes)
= 81/89
(ii) E ⟶ event of getting a no. which is perfect square
No. of favourable outcomes = 8 {4, 9, 16, 25, 36, 49, 64, 81}
P(E) = 8/89
26. A game of chance consists of spinning an arrow which is equally likely to come to rest
pointing to one of the number, 1, 2, 3, ..., 12 as shown in Fig. below. What is the
probability that it will point to:
(i) 10?
(ii) an odd number?
(iii) a number which is multiple of 3?
(iv) an even number?
Solution
Total no. of possible outcomes = 12 {1, 2, 3,...., 12}
(i) Let E ⟶ event of pointing 10
No. favourable outcomes = 1 {10}
P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes)
= 1/12
(ii) Let E ⟶ event of pointing at an odd no.
No. favourable outcomes = 6 {1, 3, 5, 7, 9, 11}
P(E) = 6/12 = 1/2
(iii) Let E ⟶ event of pointing at a no. multiple of 3
No. favourable outcomes = 4 {3, 6, 9, 12}
P(E) = 4/12 = 1/3
(iv) Let E ⟶ event of pointing at an even no.
No. favourable outcomes = 6 {2, 4, 6, 8, 10, 12}
P(E) = 6/12 = 1/2
27. Two customers are visiting a particular shop in the same week (Monday to Saturday). Each
is equally likely to visit the shop on any one day as on another. What is the probability that
both will visit the shop on:
(i) the same day?
(ii) different days?
iii) consecutive days?
Solution
Total no. of days to visit the shop = 6 {Mon to Sat}
Total no. possible outcomes = 6 × 6 = 36
i.e. two customers can visit the shop in 36 ways
(i) E⟶ event of visiting shop on the same day.
No. of favourable outcomes = 6 which are (M, M) (T, T) (Th, Th) (F, F) (S, S)
Probability, P(E) = No.of favorable outcomes/Total no.of possible outcomes
P(E) = 6/36 = 1/6
(ii) E⟶ event of visiting shop on the same day.
E⟶ event of visiting shop on the different days.
In above bit, we calculated P(E) as 1/6
We know that, P(E) + P(E̅) = 1
P(E̅) = 1 – P(E)
= 1 − 1/6 = 5/6
(iii) E⟶ event of visiting shop on c
No. of favourable outcomes = 6 which are (M, T) (T, W) (W, Th) (Th, F) (F, S)
P(E) = 5/36
28. In a class, there are 18 girls and 16 boys. The class teacher wants to choose one pupil for
class monitor. What she does, she writes the name of each pupil on a card and puts them
into a basket and mixes thoroughly. A child is asked to pick one card from the basket. What
is the probability that the name written on the card is:
(i) the name of a girl
(ii) the name of a boy
Solution
Total no. of possible outcomes = 34 (18 girls, 16 boys)
(i) E ⟶ event of getting girl name
No. of favorable outcomes = 18 (18 girls)
Probability, P(E) = No. of favorable outcomes/Total no. of possible outcomes
= 18/34 = 9/17
(ii) E ⟶ event of getting boy name
No. of favorable outcomes = 16 (16 boys)
P(E) = 16/34 = 8/17
29. Why is tossing a coin considered to be a fair way of deciding which team should choose
ends in a game of cricket?
Solution
No. of possible outcomes while tossing a coin = 2{1 head & 1 tail}
Probability = No. of favorable outcomes/Total no. of possible outcomes
P (getting tail) = 1/2
Since probability of two events are equal, these are called equally like events.
Hence, tossing a coin is considered to be a fair way of deciding which team should choose ends in a game of cricket.
30. What is the probability that a number selected at random from the number 1,2,2,3,3,3, 4, 4,
4, 4 will be their average?
Solution
Given no’s are 1, 2, 2, 3, 3, 3, 4, 4, 4, 4
Total no. of possible outcomes = 10
Average of the no’s = sum of no′s/total no′s = (1+2+2+3+3+3+4+4+4+4)/10 = 30/10 = 3
E ⟶ event of getting 3
No. of favourable outcomes = 3 {3, 3, 3}
P(E) = No. of favorable outcomes/Total no. of possible outcomes
P(E) = 3/10
31. The faces of a red cube and a yellow cube are numbered from 1 to 6. Both cubes are rolled.
What is the probability that the top face of each cube will have the same number?
Solution
Total no. of outcomes when both cubes are rolled = 6 × 6 = 36 which are
{ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) }
E ⟶ event of getting same no. on each cube
No. of favourable outcomes = 6 which are
{ (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}
Probability, P(E) = No.of favorable outcomes/Total no.of possible outcomes
= 6/36 = 1/6
32. The probability of selecting a green marble at random from a jar that contains only green,
white and yellow marbles is 1/4. The probability of selecting a white marble at random from
the same jar is 1/3 . If this jar contains 10 yellow marbles. What is the total number of marbles in the jar?
Solution
Let the no. of green marbles = x
The no. of white marbles = y
No. of yellow marbles =10
Total no. of possible outcomes = x + y + 10(total no. of marbles)
Probability P(E) = No. of favorable outcomes/Total no. of possible outcomes
Probability (green marble ) = 1/4 = x/(x + y + 10)
33. There are 30 cards, of same size, in a bag on which numbers 1 to 30 are written. One card is taken out of the bag at random. Find the probability that the number on the selected card is not divisible by 3.
Solution
Total no. of possible outcomes = 30 {1, 2, 3, ... 30}
E ⟶ event of getting no. divisible by 3.
No. of favourable outcomes = 10 {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}
Probability, P(E) = No. of favorable outcomes/Total no. of possible outcomes
P(E) = 10/30 = 1/3
E̅ ⟶ event of getting no. not divisible by 3.
P(E̅) = 1 – P(E)
= 1 - 1/3 = 2/3
34. A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is
(i) red or white
(ii) not black
(iii) neither white nor black.
Solution
Total no. of possible outcomes = 20 {5 red, 8 white & 7 black}
(i) E ⟶ event of drawing red or white ball
No. of favourable outcomes = 13 {5 red, 8 white}
Probability, P(E) = No.of favorable outcomes/Total no.of possible outcomes
P(E) = 13/20
(ii) Let E ⟶ be event of getting black ball
No. of favourable outcomes = 13 {5 red, 8 white}
P(E) = 7/20
(E̅)⟶ event of not getting black ball
P(E̅) = 1 – P(E)
= 1 − 7/20 = 13/20
(iii) Let E ⟶ be event of getting neither white nor black ball
No. of favourable outcomes = 20 – 8 – 7 = 5 {total balls – no. of white balls – no. of
black balls}
P(E) = 5/20 = 1/4
35. Find the probability that a number selected from the number 1 to 25 is not a prime number when each of the given numbers is equally likely to be selected.
Solution
Total no. of possible outcomes = 25 {1, 2, 3, ... 25}
E ⟶ event of getting a prime no.
No. of favourable outcomes = 9 {2, 3, 5, 7, 11, 13, 17, 19, 23}
Probability, P(E) = No. of favorable outcomes/Total no. of possible outcomes
= 9/25
(E̅) ⟶ event of not getting a prime no.
P(E̅) = 1 − P(E) = 1 − 9/25 = 16/25
36. A bag contains 8 red, 6 white and 4 black balls. A ball is drawn at random form the bag. Find the probability that the drawn ball is
(i) Red or white
(ii) Not black
(iii) Neither white nor black
Solution
Total no. of possible outcomes = 8 +6 + 5 18 {8 red, 6 white, 4 black}
(i) E ⟶ event of getting red or white ball
No. of favourable outcomes = 4 {4 black balls}
P(E) = 4/18 = 2/9
(E̅) ⟶ event of not getting black ball
P(E̅) = 1 − P(E) = 1 − 2/9 = 7/9
(ii) E ⟶ event of getting neither white nor black.
No. of favourable outcomes = 15 – 6 – 4 = 8 {Total balls – no. of white balls – no.
of black balls}
P(E) = 8/18 = 4/9
37. Find the probability that a number selected at random from the numbers 1, 2, 3, ......, 35 is a
(i) Prime number
(ii) Multiple of 7
(iii) Multiple of 3 or 5
Solution
Total no. of possible outcomes = 35 {1, 2,3, .....35}
(i) E ⟶ event of getting a prime no.
No. of favourable outcomes = 11 {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31}
Probability, P(E) = No. of favorable outcomes/Total no. of possible outcomes
= 11/35
(ii) E ⟶ event of getting no. which is multiple of 7
No. of favourable outcomes = 5 {7, 14, 21, 28, 35}
P(E) = 5/35 = 1/7
(iii) E ⟶ event of getting no which is multiple of 3 or 5
No. of favourable outcomes = 16 {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 5, 10, 20,
25, 35}
P(E) = 16/35
38. From a pack of 52 playing cards Jacks, queens, kings and aces of red colour are removed.
From the remaining, a card is drawn at random. Find the probability that the card drawn is
(i) A black queen
(ii) A red card
(iii) A black jack
(iv) a picture card (Jacks, queens and kings are picture cards)
Solution
Total no. of cards = 52
All jacks, queens & kings, aces of red colour are removed.
Total no. of possible outcomes = 52 – 2 – 2 – 2 – 2 = 44 {remaining cards}
(i) E ⟶ event of getting a black queen
No. of favourable outcomes = 2 {queen of spade & club}
Probability, P(E) = No. of favorable outcomes/Total no. of possible outcomes
P(E) = 2/44 = 1/22
(ii) E ⟶ event of getting a red card
No. of favourable outcomes = 26 – 8 = 18 {total red cards – jacks, queens, kings,
aces of red colour}
P(E) = 18/44 = 9/22
(iii) E ⟶ event of getting a black jack
No. of favourable outcomes = 2 {jack of club & spade}
P(E) = 2/44 = 1/22
(iv) E ⟶ event of getting a picture card
No. of favourable outcomes = 6 {2 jacks, 2 kings & 2 queens of black colour}
P(E) = 6/44 = 3/22
39. A bag contains lemon flavoured candies only. Malini takes out one candy without looking
into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Solution
(i) The bag contains lemon flavoured candies only. So, the event that malini will take
out an orange flavoured candy is an impossible event. Since, probability of
impossible event is O, P(an orange flavoured candy) = 0
(ii) The bag contains lemon flavoured candies only. So, the event that malini will take
out a lemon flavoured candy is sure event. Since probability of sure event is 1, P(a
lemon flavoured candy) = 1
40. It is given that m a group of 3 students, the probability of 2 students not having the same
birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution
Let E ⟶ event of 2 students having same birthday P(E) is given as 0.992
Let (E̅) ⟶ event of 2 students not having same birthday.
We know that, P(E) + P(E̅) = 1
P(E̅) = 1 − P(E)
= 1 – 0.992
= 0.008
41. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What
is the probability that the ball drawn is
(i) red?
(ii) not red?
Solution
Total no. of possible outcomes = 8 {3 red, 5 black}
(i) E ⟶ event of getting red ball.
No. of favourable outcomes = 3 {3 red}
Probability, P(E) = (No. of favorable outcomes)/(Total no. of possible outcomes)
P(E) = 3/8
(ii) E̅ ⟶ event of getting no red ball.
P(E) + P(E̅) = 1
P(E̅) = 1 − P(E)
= 1 − 3/8 = 5/8
42. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective ?
(ii) Suppose the bulb drawn in
(a) is not defective and not replaced. Now bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?
Solution
Total no. of possible outcomes = 20 {20 bulbs}
(i) E ⟶ be event of getting defective bulb.
No. of favourable outcomes = 4 {4 defective bulbs}
Probability, P(E) = No. of favorable outcomes/Total no. of possible outcomes
= 4/20 = 1/5
(ii) Bulb drawn in is not detective & is not replaced remaining bulbs = 15 good + 4 bad
bulbs = 19
Total no. of possible outcomes = 19
E ⟶ be event of getting defective
No. of favorable outcomes = 15 (15 good bulbs)
P(E) = 15/19
43. A box contains 90 discs which are numbered from 1 to 90. If one discs is drawn at random
from the box, find the probability that it bears
(i) a two digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Solution
Total no. of possible outcomes = 90 {1, 2, 3, ... 90}
(
i) E ⟶ event of getting 2 digit no.
No. of favourable outcomes = 81 {10, 11, 12, .... 90}
Probability P(E) = No. of favorable outcomes/Total no. of possible outcomes
P(E) = 81/90
(ii) E ⟶ event of getting a perfect square.
No. of favourable outcomes = 9 {1, 4, 9, 16, 25, 26, 49, 64, 81}
P(E) 9/90 = 1/10
(iii) E ⟶ event of getting a no. divisible by 5.
No. of favourable outcomes = 18{5, 10,15,20,25,30,35,40,45,50,55,60,5,70,75,80,85,90}
P(E) = 18/90 = 1/5
44. A lot consists of 144 ball pens of which 20 are defective and others good. Nun will buy a
pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at
random and gives it to her. What is the probability that
(ii) She will not buy it?
Solution
No. of good pens = 144 – 20 = 24
No. of detective pens = 20
Total no. of possible outcomes = 144 {total no pens}
(i) E ⟶ event of buying pen which is good.
No. of favourable outcomes = 124 {124 good pens}
P(E) = No. of favorable outcomes/Total no. of possible outcomes
P(E) = 124/144 = 31/36
(ii) E̅ ⟶ event of not buying a pen which is bad P(E) + P(E̅) = 1
P(E) + P(E̅) = 1
P(E̅) = 1 - P(E)
= 1 − 31/36 = 5/36
45. 12 defective pens are accidently mixed with 132 good ones. It is not possible to just look at
pen and tell whether or not it is defective. one pen is taken out at random from this lot.
Determine the probability that the pen taken out is good one.
Solution
No. of good pens = 132
No. of defective pens = 12
Total no. of possible outcomes = 12 + 12 {total no of pens}
E ⟶ event of getting a good pen.
No. of favourable outcomes = 132 {132 good pens}
P (E) = No. of favorable outcomes/Total no. of possible outcomes
∴ P(E) = 132/144 = 66/72 = 33/36 = 11/2
46. Five cards — the ten, jack, queen, king and ace of diamonds, are well-shuffled with their
face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put a side, what is the probability that the second card picked up is
(a) an ace ?
(b) a queen ?
Solution
Total no. of possible outcomes = 5 {5 cards}
(i) E ⟶ event of getting a good pen.
No. of favourable outcomes = 132 {132 good pens}
P (E) = No.of favorable outcomes/Total no.of possible outcomes
∴ P(E) = 1/5
(ii) If queen is drawn & put aside,
Total no. of remaining cards = 4
(a) E ⟶ event of getting a queen.
No. of favourable outcomes = 1 {1 ace card}
Total no. of possible outcomes = 4 {4 remaining cards}
P(E) = 1/4
(b) E ⟶ event of getting a good pen.
No. of favourable outcomes = 0 {there is no queen}
P(E) = 0/4
∵ E is known as impossible event.
47. Harpreet tosses two different coins simultaneously (say, one is of Re 1 and other of Rs 2).
What is the probability that he gets at least one head?
Solution
Total no. of possible outcomes = 4 which are{HT, HH, TT, TH}
E ⟶ event of getting at least one head
No. of favourable outcomes = 3 {HT, HH, TH}
Probability, P(E) = No.of favorable outcomes/Total no.of possible outcomes
P(E) = 3/4
48. Two dice, one blue and one grey, are thrown at the same time. Complete the following
table:
Event: ‘Sum on two dice’ 2 3 4 5 6 7 8 9 10 11 12 Probabiliy
From the above table a student argues that there are 11 possible outcomes 2,3,4,5,6,7,8,9,10,11 and 12. Therefore, each of them has a probability j-j. Do you agree with this argument?
Solution
Total no. of possible outcomes when 2 dice are thrown = 6 × 6 = 36 which are
{ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) }
E ⟶ event of getting sum on 2 dice as 2
No. of favourable outcomes = 1{(1, 1)}
Probability, P(E) = No. of favorable outcomes/Total no. of possible outcomes
P(E) = 1/36
E ⟶ event of getting sum as 3
No. of favourable outcomes = 2 {(1, 2) (2, 1)}
P(E) = 2/36
E ⟶ event of getting sum as 4
No. of favourable outcomes = 3 {(3, 1) (2, 2) (1, 3)}
P(E) = 3/36
E ⟶ event of getting sum as 5
No. of favourable outcomes = 4 {(1, 4) (2, 3) (3, 2) (4, 1)}
P(E) = 4/36
E ⟶ event of getting sum as 6
No. of favourable outcomes = 5 {(1, 5) (2, 4) (3, 3) (4, 2) (5, 1)}
P(E) = 6/36
E ⟶ event of getting sum as 7
No. of favourable outcomes = 6 {(1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)}
P(E) = 6/36
E ⟶ event of getting sum as 8
No. of favourable outcomes = 5 {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}
P(E) = 5/36
E ⟶ event of getting sum as 9
No. of favourable outcomes = 4 {(3, 6) (4, 5) (5, 4) (6, 3)}
P(E) = 4/36
E ⟶ event of getting sum as 10
No. of favourable outcomes = 3 {(4, 6) (5, 5) (6, 4)}
P(E) = 3/36
E ⟶ event of getting sum as 11
No. of favourable outcomes = 2 {(5, 6) (6, 5)}
P(E) = 2/36
E ⟶ event of getting sum as 12
No. of favourable outcomes = 1 {(6, 6)}
P(E) = 1/36
Event: ‘Sum on two dice’ 2 3 4 5 6 7 8 9 10 11 12 Probabiliy 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
No, the outcomes are not equally likely from the above table we see that, there is different probability for different outcome.
49. Cards marked with numbers 13,14,15,.....,60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that number on the card drawn is
(i) divisible by 5
(ii) a number is a perfect square
Solution
Total no. of possible outcomes = 48 {13, 14, 15, ...., 60}
(i) E ⟶ event of getting no divisible by 5
No. of favourable outcomes = 10{15, 20, 25, 30, 35, 40, 45, 50 55, 60}
Probability, P(E) = No.of favorable outcomes/Total no.of possible outcomes
P(E) = 10/48 = 5/24
(ii) E ⟶ event of getting a perfect square.
No. of favourable outcomes = 4 {16, 25, 36, 49}
P(E) = 4/48 = 1/12
50. A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball the
bag is twice that of a red ball, find the number of blue balls in the bag.
Solution
No of red balls = 6
Let no. of blue balls = x
Total no. of possible outcomes = 6 + x(total no. of balls)
P(E) = No.of favorable outcomes/Total no.of possible outcomes
P(blue ball) = 2 P(red ball)
⇒ x/(x + 6) = 2(6)/(x + 6)
⇒ x = 2(6)
x = 12
∴ No. of blue balls = 12
51. A bag contains tickets numbered 11, 12, 13,..., 30. A ticket is taken out from the bag at
random. Find the probability that the number on the drawn ticket
(i) is a multiple of 7
(ii) is greater than 15 and a multiple of 5.
Solution
Total no. of possible outcomes = 20 {11, 12, 13, ....., 30}
(i) E ⟶ event of getting no. which is multiple of 7
No. of favorable outcomes = 3 {14, 21, 28}
Probability, P(E) = No. of favorable outcomes/Total no. of possible outcomes
P(E) = 3/20
(ii) E ⟶ event of getting no. greater than 15 & multiple of 5
No. of favorable outcomes = 3 {14, 21, 28}
P(E) = 3/20
52. The king, queen and jack of clubs are removed from a deck of 52 playing cards and the
remaining cards are shuffled. A card is drawn from the remaining cards. Find the
probability of getting a card of
(i) heart
(ii) queen
(iii) clubs.
Solution
Total no. of remaining cards = 52 – 3 = 49
(i) E⟶ event of getting hearts
No. of favorable outcomes = 3 {4 – 1}
Probability, P(E) = No. of favorable outcomes/Total no. of possible outcomes
P(E) = 13/49
(ii) E ⟶ event of getting queen
No. of favorable outcomes = 3 (4 – 1) {Since queen of clubs is removed}
P(E) = 3/49
(iii) E ⟶ event of getting clubs
No. of favorable outcomes = 10 (13 – 3) {Since 3 club cards are removed}
P(E) = 10/49
53. Two dice are thrown simultaneously. What is the probability that:
(i) 5 will not come up on either of them ?
(ii) 5 will come up on at least one?
(iii) 5 wifi come up at both dice ?
Solution
Total no. of possible outcomes when 2 dice are thrown = 6 × 6 = 36 which are
{ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) }
(i) E ⟶ event of 5 not coming up on either of them
No. of favourable outcomes = 25 which are
{ (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) }
Probability, P(E) = No. of favorable outcomes/Total no.of possible outcomes
P(E) = 25/36
(ii) E ⟶ event of 5 coming up at least once {(1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (5, 1) (5, 2)
(5, 3) (5, 4) (5, 6) (6, 5)}
P(E) = 11/36
(iii) E ⟶ event of getting 5 on both dice
No. of favourable outcomes = 1 { (5, 5) }
P(E) = 1/36
54. Fill in the blanks:
(i) Probability of a sure event is...........
(ii) Probability of an impossible event is...........
(iii) The probability of an event (other than sure and impossible event) lies between......
(iv) Every elementary event associated to a random experiment has ........... probability.
(v) Probability of an event A + Probability of event ‘not A’ —...........
(vi) Sum of the probabilities of each outcome m an experiment is ..........
Solution
(i) 1, ∵ P(sure event) = 1
(ii) 0, ∵ P(impossible event) = 0
(iii) 0 & 1, ∵ O ∠ P(E) ∠ 1
(iv) Equal
(v) 1, ∵ P(E) + P(E̅) = 1
(vi) 1
55. Examine each of the following statements and comment:
(i) If two coins are tossed at the same time, there are 3 possible outcomes—two heads,
two tails, or one of each. Therefore, for each outcome, the probability of occurrence
is 1/3
(ii) If a die is thrown once, there are two possible outcomes—an odd number or an
even number. Therefore, the probability of obtaining an odd number is 1 /2 and the
probability of obtaining an even number is 1/2.
Solution
(i) Given statement is incorrect. If 2 coins are tossed at the same time,
Total no. of possible outcomes = 4 {HH, HT, TH, TT}
P(HH) = P(HT) = P(TH) = P(TT) = 1/4
{∵ Probability = No.of favorable outcomes/Total no.of possible outcomes}
I.e. for each outcome, probability of occurrence is 1/4
Outcomes can be classified as (2H, 2T, 1H & 1T) P(2H) = 1/4, P(2T) = 1/4, P(1H & 1T)
= 2/4
Events are not equally likely because the event ‘one head & 1 tail’ is twice as likely to occur as remaining two.
(ii) This statement is true
When a die is thrown; total no. of possible outcomes = 6 {1, 2, 3, 4, 5, 6}
These outcomes can be taken as even no. & odd no.
P(even no.) = P(2, 4, 6) = 3/6 = 1/2
P(odd no.) = p(1, 3, 5) = 3/6 = 1/2
∴ Two outcomes are equally likely
56. A box contains loo red cards, 200 yellow cards and 50 blue cards. If a card is drawn at
random from the box, then find the probability that it will be
(i) a blue card
(ii) not a yellow card
(iii) neither yellow nor a blue card.
Solution
Total no. of possible outcomes = 100 + 200 + 50 = 350 {100 red, 200 yellow & 50 blue}
(i) E ⟶ event of getting blue card.
No. of favourable outcomes = 50 {50 blue cards}
P(E) = 50/350 = 1/7
(ii) E ⟶ event of getting yellow card
No. of favourable outcomes = 200 {200 yellow}
P(E) = 200/350 = 4/7
E̅ ⟶ event of not getting yellow card
P(E̅) = 1 − P(E)
= 1 − 4/7 = 3/7
(iii) E ⟶ getting neither yellow nor a blue card
No. of favourable outcomes = 350 – 200 – 50 = 100 {removing 200 yellow & 50
blue cards}
P(E) = 100/350 = 2/7
57. A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4.
Solution
Total no. of possible outcomes = 50 {1, 2, 3 .... 50}
No. of favourable outcomes = 4 {12, 24, 36, 48}
P(E) = No. of favorable outcomes/Total no. of possible outcomes
P(E) = 4/50 = 2/25 |
Solve Lesson 33 Page 81 Math 10 SBT – Kite>
Topic
Which of the following equations is a parametric equation for a line parallel to the line .?
x – 2y + 3 = 0?
A. $$\left\{ \begin{array}{l}x = – 1 + 2t\\y = 1 + t\end{array} \right.$$ B. $$\left\{ \begin{ array}{l}x = 1 + 2t\\y = – 1 + t\end{array} \right.$$ C. $$\left\{ \begin{array}{l}x = 1 + t\ \y = – 1 – 2t\end{array} \right.$$ D. $$\left\{ \begin{array}{l}x = 1 – 2t\\y = – 1 + t\end{array} \right.$$
Solution method – See details
Step 1: Find the lines whose VTCP multiplies the scalar times the VTPT of the line x – 2y + 3 = 0 equals 0
Step 2: Get a point on the lines found in step 1, replace that point’s coordinates into the line PT
x – 2y + 3 = 0. If that point is not on the line x – 2y + 3 = 0, then the line containing that point is the line to find
Detailed explanation
Straight line ∆: x – 2y + 3 = 0 has a VTPT of $$\overrightarrow n = (1; – 2)$$.
Straight line d parallel to ∆ take $$\overrightarrow n = (1; – 2)$$ as VTPT and have VTCP as $$\overrightarrow u$$ satisfying $$\overrightarrow u .\overrightarrow n = 0$$
(Type C, D)
Check points USA(-1; 1) belongs to the line $$\left\{ \begin{array}{l}x = – 1 + 2t\\y = 1 + t\end{array} \right.$$. We see the coordinates USA satisfy PT x – 2y + 3 = 0 so M lies on (Type A)
Select REMOVE |
# EXPRESS PRODUCT OF PRIME NUMBERS IN INDEX FORM
What is index form ?
When a number is expressed with exponents, or one number to a power of another, it is considered to be in index form.
To write the the given product of prime factors in index form, we will follow the steps given below.
Step 1 :
Observe the terms and count it how many times they are repeated.
Step 2 :
The term which is repeated should be written in the base and the number of times repeated should be in the power.
Note :
If we see the term once, the power should be considered as 1.
Write each number in index form :
Problem 1 :
2×2×3×3x3
Solution :
= 2×2×3×3×3
Here, 2 is repeated 2 times and 3 is repeated 3 times.
So the exponential form is 22 × 33.
Problem 2 :
2×5×5
Solution :
= 2×5×5
Here, 2 is repeated once and 5 is repeated 2 times.
So the exponential form is 21 × 52.
Problem 3 :
2×3×3×3×5
Solution :
= 2×3×3×3×5
Here 2 and 5 are repeated once and 3 is repeated 3 times.
So the exponential form is 21 × 33 × 51.
Problem 4 :
5×5×7×7
Solution :
= 5×5×7×7
5 is repeated 2 times and 7 is repeated 2 times.
So the exponential form is 52 × 72.
Problem 5 :
2 × 2 × 5 × 5 × 5 × 7
Solution :
= 2 × 2 × 5 × 5 × 5 × 7
2 is repeated 2 times, and 5 is repeated 3 times and 7 is repeated once.
So the exponential form is 22 × 53 × 71.
Problem 6 :
3×3×7×7×11×11
Solution :
= 3×3×7×7×11×11
Here 3, 7 and 11 is repeated 2 times.
So the exponential form is 32 × 72 × 112.
Problem 7 :
2×2×2×3×3×3×3
Solution :
= 2×2×2×3×3×3×3
2 is repeated 3 times, and 3 is repeated 4 times.
So the exponential form is 23 × 34.
Problem 8 :
5×5×5×5×5×5×7×7×7
Solution :
5×5×5×5×5×5×7×7×7
5 is repeated 6 times, and 7 is repeated 3 times.
So the exponential form is 56 × 73.
## How to Express a Number as the Product of its Prime Factors
Write the following in exponent form :
Problem 1 :
25
Solution :
= 25
= 5×5
Here 5 is repeated 2 times.
So, the exponential form is 52.
Problem 2 :
36
Solution :
= 36
= 6×6
6 is repeated 2 times.
So, the exponential form is 62.
Problem 3 :
125
Solution :
125 = 5×5×55 is repeated 3 times.So, the exponential form is 53.
Problem 4 :
343
Solution :
343 = 7 × 7 × 7Here, 7 is repeated 3 times So, the exponential form is 73.
Write the following in exponent form with 3 as a base :
Problem 1 :
3
Solution :
Since, 3 is in the base, we don’t have any power for 3,
So, consider 31.
So the exponential form is 31.
Problem 2 :
27
Solution :
27 = 3 × 3 × 3Since, 3 is repeated 3 timesSo the exponential form is 33.
Problem 3 :
81
Solution :
= 3 × 3 × 3 × 3= 34 Since, 3 is repeated 4 times So the exponential form is 34.
Problem 4 :
729
Solution :
= 3 × 3 × 3 × 3 × 3 × 3= 36 Here, 3 is repeated 6 times So the exponential form is 36.
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# OCR Computing A-Level Revision
## Binary (1.3.a)
Binary numbers are similar to decimal numbers. In the same way as each decimal digit is worth 10 times more than the digit to the right of it, each binary digit is worth 2 times more than the digit to the right of it. Also, in the same way as there are 10 possibilities in decimal (0-9), there are 2 possibilities in binary (0 or 1). This is why decimal is known as "base-10", and binary is known as "base-2".
For example, the number "10" would be written as "1010" - 1 eight and 1 two.
8 4 2 1
1 0 1 0 (10)
The reason computers use binary (and not decimal, octal, or anything else) as their primary number system is because binary is digital - either electricity is flowing (binary 1), or electricity isn't flowing (binary 0).
### Binary Negative (1.3.b)
#### Two's Complement
11001001 can be interpreted as either 201 (using normal binary) or -55 (using two's complement). A question will tell you if it should be read using two's complement.
1. Firstly, write the number in binary. For example, 54 becomes 00110110 in 8-bit binary.
2. Then, invert the digits (i.e. 1s become 0s, and 0s become 1s), so 00110110 becomes 11001001.
3. Finally, add 1 - 11001001 + 00000001 = 11001010.
-128 64 32 16 8 4 2 1
1 1 0 0 1 0 1 0 (-54)
#### Sign and Magnitude
Sign and magnitude uses the first bit for the sign (positive - 0, or negative - 1), and the remaining 7 bits for the number. For example, 11001011 represents -75. The first "1" means it is negative, and the remaining 7 bits (1001011) represent 75.
+/- 64 32 16 8 4 2 1
1 1 0 0 1 0 1 1 (-75)
Binary addition is far easier than decimal addition! This is because there are only 4 possible sums - 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, 1 + 1 = 0, carry 1 (or 10, decimal 2). For example:
128 64 32 16 8 4 2 1
0 0 1 0 1 0 1 0 (42)
+ 0 1 1 0 0 1 1 0 (102)
Carry: 1 1 1 1 1
1 0 0 1 0 0 0 0 (144)
In an exam you must show your working to show that you didn't just convert the numbers to decimal, and then convert the solution from decimal back to binary.
### Binary Subtraction
Binary subtraction can be done in two ways - just like decimal subtraction. You can either do 5 - 3, or -3 + 5 (using twos compliment - not sign and magnitude). Unless a question tells you to use twos compliment, you can use whichever method you prefer.
#### Using twos compliment:
-128 64 32 16 8 4 2 1
1 1 0 0 1 0 0 0 (-56)
+ 0 1 1 0 0 1 0 1 (101)
Carry: 1 1
0 0 1 0 1 1 0 1 (45)
#### Using subtraction:
128 64 32 16 8 4 2 1
1 0 0 1 0 1 0 0 (148)
- 0 0 1 0 0 1 1 1 (39)
Carry: 0 1 1 0 1 1 0 1 1
0 1 1 0 1 1 0 1 (109) |
# How do you evaluate 1 13/15 + 4 11/15?
Jul 24, 2016
$6 \frac{9}{15}$
#### Explanation:
Split it:
Whole numbers: $\textcolor{b l u e}{1 + 4 = 5}$
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~.
The fractions both have the same bottom number (denominator).
Consequently you can directly add the top numbers (numerators)
Fractions: $\frac{13}{15} + \frac{11}{15} \text{ "->" } \frac{13 + 11}{15} = \frac{24}{15}$
But $\frac{24}{15} \text{ is the same as } \frac{15}{15} + \frac{9}{15}$
and $\frac{15}{15} = 1 \text{ giving } \textcolor{b l u e}{\frac{13}{15} + \frac{11}{15} = 1 + \frac{9}{15}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{g r e e n}{\text{Putting it all together}}$
$5 + 1 + \frac{9}{15} = 6 \frac{9}{15}$
Jul 24, 2016
$6 \frac{3}{5}$
#### Explanation:
$1 \frac{13}{15} + 4 \frac{11}{15}$
$= \frac{15 \left(1\right) + 13}{15} + \frac{15 \left(4\right) + 11}{15}$
$= \frac{15 + 13}{15} + \frac{60 + 11}{15}$
$= \frac{28}{15} + \frac{71}{15}$
$= \frac{99}{15}$
Dividing numerator and denominator by $3$ since $\frac{3}{3} = 1$
we get
$\frac{\frac{99}{3}}{\frac{15}{3}}$
$= \frac{33}{5}$
$6 \frac{3}{5}$ |
The following steps will be useful to simplify any radical expressions.
Step 1 :
Decompose the number inside the radical into prime factors.
Step 2 :
If you have square root (√), you have to take one term out of the square root for every two same terms multiplied inside the radical.
Step 3 :
If you have cube root (3), you have to take one term out of cube root for every three same terms multiplied inside the radical.
Step 4 :
If you have fourth root (4), you have to take one term out of fourth root for every four same terms multiplied inside the radical.
Step 5 :
Combine the radical terms using mathematical operations.
Example :
√18 + √8 = √(3 ⋅ 3 ⋅ 2) + (2 ⋅ 2 ⋅ 2)
√18 + √8 = 32 + 22
√18 + √8 = 52
## Solved Examples
Example 1 :
√169 + √121
Solution :
Decompose 169 and 121 into prime factors using synthetic division.
√169 = √(13 ⋅ 13)√169 = 13 √121 = √(11 ⋅ 11)√121 = 11
So, we have
√169 + √121 = 13 + 11
√169 + √121 = 24
Example 2 :
√20 + √320
Solution :
Decompose 20 and 320 into prime factors using synthetic division.
√20 = √(2 ⋅ 2 ⋅ 5)√20 = 2√5 √320 = √(2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 5)√320 = 2 ⋅ 2 ⋅ 2 ⋅ √5√320 = 8√5
So, we have
√20 + √320 = 2√5 + 8√5
√20 + √320 = 10√5
Example 3 :
√117 - √52
Solution :
Decompose 117 and 52 into prime factors using synthetic division.
√117 = √(3 ⋅ 3 ⋅ 13)√117 = 3√13 √52 = √(2 ⋅ 2 ⋅ 13)√52 = 2√13
So, we have
√117 - √52 = 3√13 - 2√13
√117 + √52 = √13
Example 4 :
√243 - 5√12 + √27
Solution :
Decompose 243, 12 and 27 into prime factors using synthetic division.
√243 = √(3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3) = 9√3
√12 = √(2 ⋅ 2 ⋅ 3) = 2√3
√27 = √(3 ⋅ 3 ⋅ 3) = 3√3
So, we have
√243 - 5√12 + √27 = 9√3 - 5(2√3) + 3√3
Simplify.
√243 - 5√12 + √27 = 9√3 - 10√3 + 3√3
√243 - 5√12 + √27 = 2√3
Example 5 :
-√147 - √243
Solution :
Decompose 147 and 243 into prime factors using synthetic division.
√147 = √(7 ⋅ 7 ⋅ 3) = 7√3
√243 = √(3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3) = 9√3
So, we have
-√147 - √243 = -7√3 - 9√3
-√147 - √243 = -16√3
Example 6 :
(√13)(√26)
Solution :
Decompose 13 and 26 into prime factors.
13 is a prime number. So, it can't be decomposed anymore.
√26 = √(2 ⋅ 13) = √2 ⋅ √13
So, we have
(√13)(√26) = (√13)(√2 ⋅ √13)
(√13)(√26) = (√13 ⋅ √13)√2
(√13)(√26) = 13√2
Example 7 :
(3√14)(√35)
Solution :
Decompose 14 and 35 into prime factors.
√14 = √(2 ⋅ 7) = √2 ⋅ √7
√35 = √(5 ⋅ 7) = √5 ⋅ √7
So, we have
(3√14)(√35) = 3( √2 ⋅ √7)(√5 ⋅ √7)
(3√14)(√35) = 3(√7 ⋅ √7)(√2 ⋅ √5)
(3√14)(√35) = 3(7)√(2 ⋅ 5)
(3√14)(√35) = 21√10
Example 8 :
(8√117) ÷ (2√52)
Solution :
Decompose 117 and 52 into prime factors using synthetic division.
√117 = √(3 ⋅ 3 ⋅ 13)√117 = 3√13 √52 = √(2 ⋅ 2 ⋅ 13)√52 = 2√13
(8√117) ÷ (2√52) = 8(3√13) ÷ 2(2√13)
(8√117) ÷ (2√52) = 24√13 ÷ 4√13
(8√117) ÷ (2√52) = 24√13 / 4√13
(8√117) ÷ (2√52) = 6
Example 9 :
(8√3)2
Solution :
(8√3)= 8√3 ⋅ 8√3
(8√3)2 = (⋅ 8)(√3 ⋅ √3)
(8√3)2 = (64)(3)
(8√3)2 = 192
Example 10 :
(√2)3 + √8
Solution :
(√2)3 + √8 = (√2 ⋅ √2 √2) + √(2⋅ ⋅ 2)
(√2)3 + √8 = ( √2) + 2√2
(√2)3 + √8 2√2 + 2√2
(√2)3 + √8 = 4√2
Example 11 :
Simplify :
4√(x4/16)
Solution :
4√(x4/16) = 4√(x4) / 416
4√(x4/16) = 4√(x ⋅ x ⋅ x ⋅ x) / 4√(2 ⋅ 2 ⋅ 2 ⋅ 2)
4√(x4/16) = x / 2
Example 12 :
Simplify :
3√(125p6q3)
Solution :
3√(125p6q3) = 3√(5 ⋅ 5 ⋅ 5 p2 ⋅ p2 ⋅ p2 ⋅ q ⋅ q ⋅ q)
3√(125p6q3) = 5p2q
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