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### Overview
The differential rate law relates the change in concentration of a reactant over a specified time. In other words, it tells us about the overall rate of a reaction from the beginning to the end of the reaction and the rate law also tells us how the speed of a reaction is impacted by the change in concentration. The differential rate law can be integrated with time to describe the change in concentration of reactants with respect to time. Using the integrated rate law expressions, we can find the concentration of a reactant or product present at a particular moment in time after the reaction has started. In this section, we will look at the integration of 1st, 2nd and 0th order reactions and some interesting graphs that the integration produces.
## First order reactions
For a first order reaction, we know that the rate of reaction is dependent on one 1st order reactant.
Order of Reactions Rate Integrated Rate Law 1st rate = k•[A]
### Click to show integration
See how the integrated rate law is derived using calculus.
First order differential rate law:
Integrating both sides by time between t=0 and t = t:
## 2nd order reactions
For a 2nd order reaction, we know that the rate of reaction is dependent on the square of the reactant’s concentration.
Order of reactions Rate Integrated Rate Law 2nd rate = k•[A]²
### Click to show integration
See how the integrated rate law is derived using calculus.
First order differential rate law:
Integrating both sides by time between t=0 and t = t:
## 0th order reactions
A 0th order reaction rate is defined by the decrease in concentration of reactants over time. A zeroth order reaction rate does not depend on the concentration of the reactant, it means that the rate of reaction is constant over time even though the reactants are decreasing as the reaction progresses.
The differential rate law:
### Click to show integration
See how the integrated rate law is derived using calculus.
0th order differential rate law:
Integrating both sides by time between t=0 and t = t:
Rearranging the equation:
## Graphs
The most useful aspect of the integration is to arrive at an equation y = mx + c, as this can be easily graphed. And by distinguishing the patterns of the graphs, we can quickly determine if the reaction is 0th, 1st, or 2nd order.
For a first order reaction, the graph ln[A] vs time is a straight line with a slope -k.
For a second order reaction, a plot of the inverse of a concentration vs time produces a straight line graph with a slope = k.
For a zeroth order reaction, the plot of the concentration of the reactant vs time is a straight line with a negative slope.
## Example
Given the experimental data (see table below), determine the rate law for the reaction with respect to reactant A.
You can determine the order from integrated rate law. Calculate the values of ln[A] and 1/[A] .
Plot the graphs.
[A] vs Time
ln[A] vs time
1/[A] vs time
Since graph ln[A] vs time gives a straight line and the others do not, we know that the reaction is a first order reaction with respect to A.
## Determine time
How long will it take for the reaction to reach 0.0200 M which has a rate constant of k = 3.5 x 10⁻³ s⁻¹. The initial concentration is 0.100 M/
To determine t, we need to know:
[A]₀ – the initial concentration
[A]t – the final concentration
The order of the reacton
The rate constant k is given 3.5 x 10⁻³ s⁻¹ |
# Trick to find multiples mentally
We all know how to recognize numbers that are multiple of $2, 3, 4, 5$ (and other). Some other divisors are a bit more difficult to spot. I am thinking about $7$.
A few months ago, I heard a simple and elegant way to find multiples of $7$:
Cut the digits into pairs from the end, multiply the last group by $1$, the previous by $2$, the previous by $4$, then $8$, $16$ and so on. Add all the parts. If the resulting number is multiple of $7$, then the first one was too.
Example:
$21553$
Cut digits into pairs:
$2, 15, 53$
Multiply $53$ by $1, 15$ by $2, 2$ by $4$:
$8, 30, 53$
$8+30+53=91$
As $91$ is a multiple of $7$ ($13 \cdot 7$), then $21553$ is too.
This works because $100-2$ is a multiple of 7. Each hundreds, the last two digits are 2 less than a multiple of $7 (105 → 05 = 7 - 2, 112 → 12 = 14 - 2, \cdots)$
I figured out that if it works like that, maybe it would work if we consider $7=10-3$ and multiplying by $3$ each digit instead of $2$ each pair of digits.
Exemple with $91$:
$91$
$9, 1$
$9\cdot3, 1\cdot1$
$27, 1$
$28$
My question is: can you find a rule that works with any divisor? I can find one with divisors from $1$ to $19$ ($10-9$ to $10+9$), but I have problem with bigger numbers. For example, how can we find multiples of $23$?
• You'll want to see this KCd blurb. Dec 8, 2011 at 15:49
• I've seen the following method for 7: Take the last digit, double it, and subtract it from the first digits. So $21553$ leads to $2155-3*2=2149$. Next step is $214-2*9=196$. Next $19-2*6=7$. Dec 8, 2011 at 16:04
• How can we find divisors of 23? Very easily: they are just $\pm 1$ and $\pm 23$. Presumably, you want to recognize multiples of 23? Dec 8, 2011 at 16:11
• For $23$, here is something fairly simple, for numbers that have $3$ or $4$ digits, say $abcd$. Take the number $ab$, add to it $3$ times the number $cd$. Call the result $t$. Then $abcd$ is a multiple of $23$ iff $t$ is. One could probably tweak this to be even simpler to execute. Dec 8, 2011 at 16:16
• @ArturoMagidin Yes, I meant multiples. It's corrected. Dec 9, 2011 at 8:46
One needn't memorize motley exotic divisibility tests. There is a universal test that is simpler and much easier recalled, viz. evaluate a radix polynomial in nested Horner form, using modular arithmetic. For example, consider evaluating a $3$ digit radix $10$ number modulo $7$. In Horner form $\rm\ d_2\ d_1\ d_0 \$ is $\rm\: (d_2\cdot 10 + d_1)\ 10 + d_0\ \equiv\ (d_2\cdot 3 + d_1)\ 3 + d_0\ (mod\ 7)\$ since $\rm\ 10\equiv 3\ (mod\ 7)\:.\:$ So we compute the remainder $\rm\ (mod\ 7)\$ as follows. Start with the leading digit then repeatedly apply the operation: multiply by $3$ then add the next digit, doing all of the arithmetic $\rm\:(mod\ 7)\:.\:$
For example, let's use this algorithm to reduce $\rm\ 43211\ \:(mod\ 7)\:.\:$ The algorithm consists of repeatedly replacing the first two leading digits $\rm\ d_n\ d_{n-1}\$ by $\rm\ d_n\cdot 3 + d_{n-1}\:\ (mod\ 7),\:$ namely
$\rm\qquad\phantom{\equiv} \color{red}{4\ 3}\ 2\ 1\ 1$
$\rm\qquad\equiv\phantom{4} \color{green}{1\ 2}\ 1\ 1\quad$ by $\rm\quad \color{red}4\cdot 3 + \color{red}3\ \equiv\ \color{green}1$
$\rm\qquad\equiv\phantom{4\ 3} \color{royalblue}{5\ 1}\ 1\quad$ by $\rm\quad \color{green}1\cdot 3 + \color{green}2\ \equiv\ \color{royalblue}5$
$\rm\qquad\equiv\phantom{4\ 3\ 5} \color{brown}{2\ 1}\quad$ by $\rm\quad \color{royalblue}5\cdot 3 + \color{royalblue}1\ \equiv\ \color{brown}2$
$\rm\qquad\equiv\phantom{4\ 3\ 5\ 2} 0\quad$ by $\rm\quad \color{brown}2\cdot 3 + \color{brown}1\ \equiv\ 0$
Hence $\rm\ 43211\equiv 0\:\ (mod\ 7)\:,\:$ indeed $\rm\ 43211 = 7\cdot 6173\:.\:$ Generally the modular arithmetic is simpler if one uses a balanced system of representatives, e.g. $\rm\: \pm\{0,1,2,3\}\ \:(mod\ 7)\:.$ Notice that for modulus $11$ or $9\:$ the above method reduces to the well-known divisibility tests by $11$ or $9\:$ (a.k.a. "casting out nines" for modulus $9\:$).
• Cool trick! Thanks Dec 3, 2020 at 22:02
One approach is to find some higher multiple that makes it easy. For your example of $23$, note that $3*23+1=70$, so $a(70-1)$ will be a multiple of $23$. Now you have a single digit multiply followed by a subtraction. If you pick $a=301, a(70-1)=301(70-1)=301*70-301=2107-301=1806$, which is a multiple of $23$
• I generally find this method much easier than the standard modular divisibility test. It does require that you memorize a starting list of multiples, but it's not so bad because you can just multiply them by powers of $10$. Dec 8, 2011 at 18:28
• @QiaochuYuan: It seemed the OP's request was to find multiples, not do a divisibility test. But this generalizes. In the spirit of the test for $7$, to test divisibility by $23$, you can take off the last digit, multiply the rest by $7$ and add the last digit. Keep going until you have only two digits left. You just have to find a multiple that ends in $1$ or $9$ to do this. Dec 8, 2011 at 18:40
In general, if you are doing things by dividing the number into groups of $k$ digits, you can think of it as looking at the number in base $10^k.$
More generally, if $B$ is a base, and $n$ is a number with no common factors with $B$, then you can always find an $m$ such that $Bm\equiv 1\pmod n$. Then $X\equiv 0\pmod n$ if and only if $Xm\equiv 0\pmod n$. But if $X=Bu+v$, then $Xm\equiv u+mv\pmod n$. So if we take the last digit, base B, multiply it by $m$ and add it to the other digits, the result is divisible by $n$ if and only if the original number was divisible by $n$.
In the case of $n=23$ base $B=10$, you get $m=7$, so you can take the last digit, multiply it by seven, and add it to the rest.
Or, if you use $B=100$, you get $m=3$, and you can take the lsat two digits, multiply by 3, and add to the other digits.
In general, you can always find a $k$ such that $10^k=1\pmod n$ if $n$ is not even or divisible by 5. Then if you take base $B=10^k$, you get $m=1$, and you can separate $m$ into groups of $k$ digits and add them. That's hardly useful when $k$ is large. For example, the smallest $k$ for $n=23$ is $k=22$, so this part only helps if your starting number was more than 23 digits long.
Usually, you want to find a relatively small pair $(m,k)$ so that $m10^k\equiv \pm 1\pmod n$. Then you take the last $k$ digits, multiplied by $m$, and add to or subtract from the other digits, depending on whether $+1$ or $-1$.
The following is a simple method to check divisibility by $7$ or $13$:
Cut the digits in pairs of 3 and calculate their alternating sum. If this is a multiple of 7 or 13, the original number is.
For example
$12345631241$ leads to
$$241-639+345-12=-65 \,.$$
Thus our number is a multiple of $13$, but not a multiple of $7$.
This works because $1001=7*11*13$ meaning that any number of the form $abcabc$ is a multiple of $7, 11$ and $13$...The trick also works for 11, but there is another simple trick.
Method 2 If the number is relatively prime to 10 (if it is not, you can make it), look for a multiple on $n$ which ends in 1.
Let say that this multiple is $a_1..a_k1$.
Then you simply pick the large number and subtract $a_1...a_k*$last digit from the remaining digits.
The trick works because $a_1...ak1$*last digit is always a multiple of $n$, and subtracting this from the original number you get a multiple of 10..Since $n$ is relatively prime to 10, you can erase the 0 at the end...
A simple such example, for $7$ the smalest such desires multiple is ... 21, which leads to the criteria posted by Thomas Andrews .
Also, for small numbers the following
I will try to explain a general rule using modular congruence. We can show that any integer in base $10$ can be written as $$z = a_0 + a_1 \times 10 + a_2 \times 10^2 + a_3 \times10^3 + \cdots + a_n \times 10^n$$
Lets say we have to find a divisibility rule of $7$,Hence for congruence modulo $7$ have,
$$10 \equiv 3, 10^2 \equiv 2, 10^3 \equiv -1, 10^4 \equiv -3, 10^5 \equiv -2, 10^6 \equiv 1,$$
The successive remainder then repeat. Thus our integer $z$ is divisible by $7$ iff if the remainder expression $$r= a_0 + 3a_1 +2a_2 -a_3-3a_4-2a_5+a_6+3a_7+\cdots$$ is divisible by $7$
To understand why the divisibility of $r$ indicate the divisibility of $z$, find $z-t$ which is given by :$$z-t = a_1 \times (10-3) + a_2 \times (10^2-2) + a_3 \times (10^3+1) + \cdots + a_6 \times (10^6-1) + \cdots$$ Since all this numbers $(10-3),(10^2-2),(10^3+1),\cdots$ are congruent to 0 modulo $7,z-t$ is also, and therefore $z$ leaves the same remainder on division by $7$ as $r$ does.
Using this approach we can derive divisibility of any integer.
This can be explained using Horner's method as below. z=a0 + a1 ×10 + a2 ×102 + a3 × 10^3+⋯+ an × 10^n
i.e. Z = ((((...(an × 10 + an-1)*10 + an-2 )*10 + an-3).....)*10 + a0
( Shift-add representation of number)
Z % 7 = ((((...(an × 3 + an-1)*3 + an-2 )*3 + an-3).....)*3 + a0 (Mod 7)
Now evaluate from left to right using modulo algebra.
To know more about shift- add representation of number and its application in interbase conversion and divisibility refer tinyurl.com/mlxk8pw .
Regard, |
### Home > A2C > Chapter 4 > Lesson 4.1.3 > Problem4-40
4-40.
Find the vertex of each of the following parabolas by averaging the $x$-intercepts. Then write each equation in graphing form.
1. $y=(x−3)(x−11)$
What are the $x$-intercepts of this equation? After finding them, average the $x$-coordinates to find the $x$-value, and then substitute this value into the equation to find the corresponding $y$-value.
$x$-intercepts: $(3,0)$ and $(11,0)$
vertex: $(7,-16)$
equation: $y=(x-7)^2-16$
1. $y=(x+2)(x−6)$
Refer to part (a).
$x$-intercepts: $(-2,0)$ and $(6,0)$
vertex: $(2,-16)$
equation: $y=(x-2)^2-16$
1. $y=x^2−14x+40$
Refer to part (a).
$x$-intercepts: $(10,0)$ and $(4,0)$
vertex: $(7,-9)$
equation: $y=(x-7)^2-9$
1. $y=(x−2)^2−1$
First square $(x-2)$ and combine like terms.
Refer to part (a). |
Advices
# What are math facts up to 20?
Table of Contents
## What are math facts up to 20?
The addition facts to 20 are simply the sums from 0+0 up to 10+10. They’re the building blocks of arithmetic, and usually the first math facts that children master.
What are basic facts in math examples?
A List of Basic Math Facts
• Addition. The numbers used in addition are called addends; the answer is called the sum.
• Subtraction. The top, or higher, number in subtraction is called the minuend, the lower number is the subtrahend and the answer is the difference.
• Multiplication.
• Division.
### What is a basic math fact?
Basic math facts are defined as computations involving the four basic math operations: addition, subtraction, multiplication, and division; using the single-digit numbers, 0 – 9. These basic facts are often referred to in. current literature as “basic number combinations”.
What are some cool math facts?
14 Interesting Math Facts
• I) The numbers on opposite sides of a die always add up to seven.
• II) Zero is an even number.
• III) A useful trick for percentages.
• IV) Every odd number, when written in English, contains an “e”.
## What are 4th grade math facts?
Seven Math Skills your Child Will Learn in 4th Grade
• Multiply bigger numbers.
• Divide with remainders.
• Find factors of numbers 1-100.
• Solve real-world word problems.
• Understand large numbers in various forms.
• Work with fractions.
• Compare decimals.
Why are basic math facts important?
Math facts are important because they form the building blocks for higher-level math concepts. When a child masters his/her math facts, these concepts will be significantly easier and the student will be better equipped to solve them faster.
### What is an interesting fact?
While four babies are born on Earth every second, it’s estimated that around two people pass away at the same time. That means that 105 people die each minute, 6,316 people die each hour, 151,600 people die each day, and 55.3 million people die each year.
How many math facts are there?
Math facts are the “phonics of math.” Except instead of only 44 phonemes from 26 letters, there are 121 facts for addition, another 121 for subtraction, etc. A total of 580 facts if you go up to 144 ÷ 12! No wonder it takes so much practice!
## What should a 5th grader know in math?
In fifth grade, students focus on adding, subtracting, multiplying, and dividing whole numbers, fractions, and decimals. Your kid will become fluent with computing these types of numbers and understanding the relationship between them. Students should also be able to use these numbers in real-world scenarios.
Did you know these 20 cool facts about math?
To prove this, we’ve compiled a list of 20 cool facts about math which we encourage you to share with the children in your life. 1. The word “hundred” comes from the old Norse term, “hundrath”, which actually means 120 and not 100. 2. In a room of 23 people there’s a 50% chance that two people have the same birthday. 3.
### What are 5 interesting facts about math symbols?
Most mathematical symbols weren’t invented until the 16th century. Before that, equations were written in words. 4. “Forty” is the only number that is spelt with letters arranged in alphabetical order. 5. Conversely, “one” is the only number that is spelt with letters arranged in descending order.
What grade level is addition facts to 20?
Addition facts to 20 worksheets. Basic addition facts printables and worksheets for grade 1 math. Addition facts to 20 should be mastered by the second grade. Grade 1 addition worksheets. Great activities for distance learning and independent learning.
## What is the most mind-blowing fact about numbers?
‘Four’ is the only number in the English language that is spelt with the same number of letters as the number itself. 8. Every odd number has an “e” in it. 9. Over 2000 years ago, Eratosthenes estimated the Earth’s circumference using math, without ever leaving Egypt, and he was accurate to within 2% 10. |
10th cbse maths solution for exercise 3.6 part 2
This page 10th cbse maths solution for exercise 3.6 part 2 is going to provide you solution for every problems that you find in the exercise no 3.6
10th CBSE maths solution for Exercise 3.6 part 2
(iv) 5/(x – 1) + 1/(y – 2) = 2
6/(x – 1) – 3/(y – 2) = 1
Solution:
Let 1/(x – 1) = a and 1/(y – 2) = b
5 a + b = 2 --------(1)
6 a – 3 b = 1 --------(2)
(1) x 3 => 15 a + 3 b = 6
6 a – 3 b = 1
-----------------
21 a = 7
a = 7/21
a = 1/3
1/(x-1) = 1/3
3 = x -1
x – 1 = 3
x = 3 + 1
x = 4
Substitute a = 1/3 in the first equation
5(1/3) + b = 2
(5/3 )+ b = 2
b = 2 – (5/3)
b = 1/3
1/(y – 2) = 1/3
3 = y – 2
y – 2 = 3
y = 3 + 2
y = 5
(v) (7 x - 2 y)/xy = 5
(8 x + 7 y)/xy = 15
Solution:
(1) x 7 = > - 14 a + 49 b = 35
(2) x 2 = > 14 a + 16 b = 30
-----------------------
65 b = 65
b = 1
Substitute b = 1 in the first equation
-2 a + 7 (1) = 5
-2a + 7 = 5
- 2a = 5 -7
-2a = -2
a = 1
1/x = a 1/ y = b
1/x = 1 1/y = 1
x = 1 y = 1
(vi) 6 x + 3 y = 6 xy
2 x + 4 y = 5 xy
Solution:
Dividing the whole equation by xy
(6 x/xy) + (3y/xy) = (6xy/xy)
(2x/xy) + (4y/xy) = (5xy/xy)
(6/y) + (3/x) = 6
(2/y) + (4/x) = 5
1/x = a and 1/y = b
3 a + 6 b = 6 ---------(1)
4 a + 2 b = 5 ---------(2)
3 a + 6 b = 6
(2) x 3 =>12 a + 6 b = 15
(-) (-) (-)
---------------------
-9a = - 9
a = 1
Substitute a = 1 in the first equation
3(1) + 6 b = 6
3 + 6 b = 6
6 b = 6 – 3
6 b = 3
b = 3/6
b = 1/2
1/x = a 1/y = b
1/x = 1 1/y = 1/2
x = 1 y = 2
Related pages |
# The integral of 1/sqrt(x^2-1) dx
Up a level : Integrals
Previous page : The integral of 1/sqrt(1-x^2) dx
Next page : The integral of one over sqrt(x^2+1) dx
Now to
$\int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}}$
This can be done using various substitutions. We will look at a hyperbolic substitution at the end of this page. It can also be done using the substitution
$w = x + \sqrt {{x^2} - 1}$
But that is begging the question because it feels like we have to know the answer to come up with that idea. We could also do the substitution
$x = \sec \theta$
But that will require us to know or figure out, that
$\sqrt {{{\sec }^2}\theta - 1} = {\tan } \theta$
which is quite easy to find. You can get it from
$\sin^2 \theta + \cos^2 \theta =1$
in a couple of steps. Worse is that after the substitution we end up with the necessity to find the integral of sec θ, and that is not all too straightforward.
The above can be found in various sources, so we will not go through that here, but instead, we will find the answer to our integral using (simple) complex analysis.
Using complex analysis
We will use complex analysis, or rather just complex numbers and a bit of calculus in a rather straightforward way. Let us first pull out a factor of -1 from the square root, and then the rule from the previous page, to get
$\begin{gathered} I = \int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}} = \int {\frac{{dx}}{{\sqrt {( - 1)(1 - {x^2})} }}} = \hfill \\ \quad = \int {\frac{{dx}}{{i\sqrt {(1 - {x^2})} }}} = \frac{1}{i}{\sin ^{ - 1}}x + C \hfill \\ \end{gathered}$
Technically we are done, but it is in a not all too useful form. To fix this we need to look at the arcsine part. We want it to be able to “absorb” the i.
$I = \frac{1}{i}{\sin ^{ - 1}}x$
then
$\sin (iI) = x$
(We put the constant of integration back later on.) From a previous section, we have that
$\sin \theta = \frac{{{e^{i\theta }} - {e^{ - i\theta }}}}{{2i}}$
So
$x = \sin iI = \frac{{{e^{ - I}} - {e^I}}}{{2i}}$
or
$2ix = {e^{ - I}} - {e^I} = \frac{1}{w} - w$
where
$w = {e^I}$
this gives us
$2ixw = 1 - {w^2}$
or
${w^2} + 2ixw - 1 = 0$
This we can solve using the quadratic formula to get
$\begin{gathered} w = \frac{{ - 2ix \pm \sqrt { - 4{x^2} - 4 \cdot 1 \cdot ( - 1)} }}{{2 \cdot 1}} \hfill \\ \quad = \frac{{ - 2ix \pm \sqrt { - 4{x^2} + 4} }}{2} \hfill \\ \quad = \frac{{ - 2ix \pm \sqrt { - 4} \sqrt {{x^2} - 1} }}{2} \hfill \\ \quad = \frac{{ - 2ix \pm 2i\sqrt {{x^2} - 1} }}{2} \hfill \\ \quad = i\left( {x \pm \sqrt {{x^2} - 1} } \right) \hfill \\ \end{gathered}$
Substituting back eI and adding back the constant of integration will give us
$\begin{gathered} I = \ln \left( {i\left( {x \pm \sqrt {{x^2} - 1} } \right)} \right) + {C_1} \hfill \\ \quad = \ln (i) + \ln \left( {x \pm \sqrt {{x^2} - 1} } \right) + {C_1} \hfill \\ \quad = \ln \left( {x \pm \sqrt {{x^2} - 1} } \right) + C \hfill \\ \end{gathered}$
Here we let the constant absorb the ln(i). Next comes the question of which sign to use. We may start by noting that
$\left( {x + \sqrt {{x^2} - 1} } \right) = {\left( {x - \sqrt {{x^2} - 1} } \right)^{ - 1}}$
and thus that
$\ln \left( {x \pm \sqrt {{x^2} - 1} } \right) = \pm \ln \left( {x + \sqrt {{x^2} - 1} } \right)$
If z is larger than one the logarithm is not negative, since
$x \geqslant 1 \Rightarrow x + \sqrt {{x^2} - 1} \geqslant 1$
and we can also see that this function is increasing, but less and less fast. The function
$\frac{1}{{\sqrt {{x^2} - 1} }}$
should be the derivative of the above logarithmic expression. We can see that is s positive, and decreasing as x>0, and the only option is thus that the integral is the positive variant of the logarithm. So we get
$\int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}} = \ln \left( {x + \sqrt {{x^2} - 1} } \right) + C,\quad x \geqslant 1$
For symmetry reasons we also have
$\int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}} = \ln \left| {x + \sqrt {{x^2} - 1} } \right| + C,\quad \left| x \right| \geqslant 1$
as shown in the graph below.
Let us look at the inverse function
$y = \ln \left( {x + \sqrt {{x^2} - 1} } \right)$
We get
$x = \ln \left( {y + \sqrt {{y^2} - 1} } \right)$
and thus
$\begin{gathered} {e^x} = y + \sqrt {{y^2} - 1} \quad \quad \quad \quad // - y \hfill \\ {e^x} - y = \sqrt {{y^2} - 1} \quad \quad \quad \quad /{/^2} \hfill \\ {e^{2x}} - 2y{e^x} + {y^2} = {y^2} - 1\quad // - {y^2} \hfill \\ {e^{2x}} - 2y{e^x} = - 1\quad \quad \quad \quad \;// \cdot {e^{ - x}} \hfill \\ {e^x} - 2y = - {e^{ - x}}\quad \quad \quad \quad \;\;// + 2y + {e^{ - x}} \hfill \\ {e^x} + {e^{ - x}} = 2y\quad \quad \quad \quad \;\;\;\;// \div 2 \hfill \\ y = \frac{{{e^x} + {e^{ - x}}}}{2} = \cosh x \hfill \\ \end{gathered}$
This is one of the hyperbolic functions. You can find more on those here. So this means that
${\cosh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} - 1} } \right)$
so
$\int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}} = {\cosh ^{ - 1}}x + C,\quad x \geqslant 1$
Using hyperbolic functions directly
If we happen to know our hyperbolic functions we can do the substitution
$\left| \begin{gathered} x = \cosh y \hfill \\ dx = \sinh y\;dy \hfill \\ \end{gathered} \right.$
We also have that
${\cosh ^2}\theta - {\sinh ^2}\theta = 1 \Rightarrow {\cosh ^2}\theta - 1 = {\sinh ^2}\theta$
This gives us
$\begin{gathered} \int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}} = \int {\frac{{\sinh y\;dy}}{{\sqrt {{{\cosh }^2}y - 1} }}} = \int {\frac{{\sinh y\;dy}}{{\sinh y}}} \hfill \\ \quad = \int {dy} = y + C = {\cosh ^{ - 1}}x + C \hfill \\ \end{gathered}$
The above generalized a bit
Now, we get
$\begin{gathered} \int {\frac{{dx}}{{\sqrt {{x^2} - {a^2}} }}} \hfill \\ \quad = \int {\frac{{dx}}{{a\sqrt {{{\left( {\frac{x}{a}} \right)}^2} - 1} }}} \hfill \\ \quad = \frac{{\;\frac{1}{a}\;}}{{\;\frac{1}{a}\;}}{\cosh ^{ - 1}}\frac{x}{a} + C = {\cosh ^{ - 1}}\frac{x}{a} + C \hfill \\ \end{gathered}$
or
$\begin{gathered} \int {\frac{{dx}}{{\sqrt {{x^2} - {a^2}} }}} = {\cosh ^{ - 1}}\frac{x}{a} + {C_1} \hfill \\ \quad = \ln \left( {\frac{x}{a} + \sqrt {\frac{{{x^2}}}{{{a^2}}} - 1} } \right) + {C_1} \hfill \\ \quad = \ln \left( {\frac{x}{a} + \frac{1}{a}\sqrt {{x^2} - {a^2}} } \right) + {C_1} \hfill \\ \quad = \ln \left( {\frac{x}{a} + \frac{1}{a}\sqrt {{x^2} - {a^2}} } \right) + {C_1} \hfill \\ \quad = \ln \left( {x + \sqrt {{x^2} - {a^2}} } \right) - \ln a + {C_1} \hfill \\ \quad = \ln \left( {x + \sqrt {{x^2} - {a^2}} } \right) + C \hfill \\ \end{gathered}$
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# Permutations and Combinations Basics
Friends, in this post we shall discuss about another important topic of Aptitude Section, Permutations and Combinations. Lets start with definitions. Ofcourse all of you might have already tried to learn the standard definitions of Permutations and Combinations and almost fed up with them. So now we are not going to repeat those technical definitions again. Lets try to understand in simple words, what is permutation and what is combination.
In my school days I have a BIG confusion on which one is permutation and which one is combination. Then my uncle told me a simple technique to remember them without confusion.
He said, permutation is always complicated than combination
In simple words
• If the order is important, it is a Permutation.
• If the order is not important, it is a Combination.
Still confused ? lets discuss with examples.
Permutation :
Assume that I have 4 letters (A, B, C and D). Now if anybody asks me to write down all the permutations of 3 of these letters.....
ABC BAC CAB DAB
ACB BCA CBA DBA
ACD BCD CBD DBC
So, there are 24 permutations in total. Here the order is important. In other words ACB is different, BCA is different, CBA is different and ABC is different (even-though they all are formed with same group letters).
Combination :
As we have already discussed, the collection of letters is important here, not the order. That means, if you have ABC in your set that's enough. So you cant claim ABC, ACB, BAC, BCA, CAB, CBA... for combinations. These all are 1 combination of letters A, B and C.
So, from the given 4 letters (A, B, C and D), You can write the combination of 3 of those letters
ABC ABD ACD BCD
hope you have got the basic concept now.
Now lets have a look at the technical side, before going to calculate Permutations and Combinations, you should know the word Factorial.
Factorial : The factorial of a number, represented by n!, is the product of the natural numbers up to and including n
In simple words, the Factorial of the number n is the number of ways that the n elements of a group can be ordered.
So, if somebody ask you a question, how many different ways six people can sit at a table with six chairs,
you should say them its 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.
Note : We treat 0! as 1
How to Calculate Permutations ?
There is a simple formula for calculating permutations.
Number of all permutations of n items, taken r at a time, is given by:
nPr = n(n - 1)(n - 2) ... (n - r + 1)
n above case n is 4 and r is 3
So, nPr = P(n,r) = 24
How to Calculate Combinations ?
Important Note : To calculate combinations,
• First you should calculate all the equivalent permutations.
• Later you should correct this list by cutting out duplicates / repetitions.
Now lets have a look at the mathematical formula for Combinations..
nCr = n! = n(n - 1)(n - 2) ... ... r . (r!)(n - r!) r!
n above case n is 4 and r is 3, so
nCr = C(n,r) = 4
That's all for now friends. In our next post we shall discuss some practice problems on Permutations and Combinations.
Home Work :
Before going to leave, you have a small Home Work kind of Stuff here. The below pdf file consists of some basic shortcut techniques on Permutations and Combinations with some simple examples. Just download this pdf file and prepare well. It will help you getting good idea on the concepts and approach.
1. hello....
After solving the Combination ans is C(n,r) is 8 in the above example... how it came 4....???
if n-4,r=3
n! / (r!)(n-r!)
4*3*2*1 / (3)*(4-3) = 8
1. You just took r instead of r! in the denominator Rashmi...
the solution will be like
(4X3X2X1)/(3X2X1)(1) = 4
hope you got it now...
2. In Denominator formula is factorial r multiple by factorial (n-r). So in denominator fact. of 3 * fact. (4-3). Hope u will understand.
3. Its not "3" in denominator
Its 3!
So (4*3*2*1)/(3*2*1)*(4-3)= 4
4. your way of calculation is wrong
C(n,r)= n!/(r!)*(n-r)
Where n=4, r=3
then combination is
= 4!/3!*(4-3)
4*3*2*1/3*2*1*1
=24/6
Ans is 4.
2. Dear mam,u r really giving vaulable information to us bt am requesting u that dnt delete these links plz bcoz its useful 4us in future hope u understand...thnks
1. We don't delete any links friend. If you encounter any deleted link, then please let us know. we will update :)
3. Very nice
4. so nice of u mam :)
5. so nice of u mam thank u for ur valuable information
6. having problem in understanding this topic. :(
7. 1/1*2 + 1/2*3 + 1/3*4 + ..... + 1/9*10
What may be d right Answer.
Optios are 1) 11/10, 2) 7/10,3) 9/10, 4)1. How can solve this ??
8. Dear admin, i have been refering your site for a long time but one thing tht bothers me is tht u always refer your material for the banking guys y nt engineering and folks related to other folks........nd i also wnt to express my gratitude becuz ur site ws of great help in qualifying various PSUs.........
9. thank u for ur nice work |
# What range imply in math? What’s a function in math? What is a function in math?
For this one particular, remember that the function called the square root is an object that has a name, and when it really is getting expressed, you are able to place the name inside the square brackets. It is known as a function due to its formula. If you would like to express the square root with regards to a product, the formula is then X=R3.
What do you consider would be the equation for the square root? R3. In this case, the square root is named the derivative. So the square root might be described by two formulas.
The initial formula is often represented by the quantity (R3). Then the derivative is represented by the word the. For those who replace the word the together with the word differentiation, you can get the equation for the square root. This would involve a product, a value in addition to a time. This could be a time that is certainly unique in the original.
Here is what’s the equation for the square root? Now it is an ordinary solution between the product plus the worth.
How would you describe what exactly is the square root in terms of a time? Well, it truly is the derivative in the time with respect to the value, and it’s the item of the two values.
So the time may be the derivative with the http://en.wikipedia.com/wiki/Gender_studies#Other_people_whose_work_is_associated worth, which is generally equal towards the item from the time and also the value. As a result, if you add these points up, you get the worth. In the event you take the derivative of your value, you get the time. Hence, you have to take the time into account any time you are acquiring the derivative. So the square root is definitely an expression for time. But what’s a time?
Time is often a price at which we are able to repeat an action or a method. If the action or course of action is continuous, then the time will always be precisely the same.
You can evaluate the time for you to a point on the approach. A value is often believed of as a point within a course of action.
In this case, the factor which has an impact on the time could be the value. Whenever you measure time, that you are measuring how the process is going. So to have the worth, you’ve to add the time to the worth.
### פרטי התקשרות
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# Slope of a Line
Hi. This very small lesson will talk about how to find the slope of a line.
We’ll consider two cases
(i) Slope of a line whose equation is known
(ii) Slope of a line joining two points (we could find the equation first and then use the first method, but that’s really not necessary)
Lets consider the first case.
Let the equation of the line be ax + by + c =0 (general form). Then how do we find its slope?
We know that if the equation of the line looks like y = mx + c (slope-intercept form), then m is the slope of the line.
So.. if we transform in the given equation into the slope-intercept form, then the coefficient of x will give us the slope.
Let’s do this !
The equation ax + by + c =0 can be transformed as y = (-a/b)x + (-c/b). This gives the slope of the line as -a/b. And we’re done!
Therefore the slope of a line (when expressed in the general form) will be -a/b or more generally -(coefficient of x/coefficient of y)
Remember always to transform the equation in the general form, before using this expression.
For example, the slope of the line 2x – y + 10 = 0 will be -(2)/(-1) or 2.
And, the slope of the line 4x = y + 5 will not be (-4)/(1) ! Because it isn’t expressed in the general form.
The equation should first written in that form as 4x – y – 5 = 0, and now the correct slope would be -(4)/(-1), which is 4.
That’s all there to it.
Moving on to the next case.
I’ve already discussed this earlier. Nevertheless, it makes sense to discuss it here once again.
Let A(x1,y1) and B(x2,y2) be two points. We are to calculate the slope of the line joining A and B, or simply the slope of AB.
Here is the figure which I used earlier..
Recall that the slope of a line is the tangent of the angle made by the line with the X-axis (measured anticlockwise).
The line joining A and B is extended to meet the X-axis, as shown. In triangle ACB, tanθ = BC / AC, which equals (y2-y1)/(x2-x1). (details of that here)
Well, tanθ is nothing but the slope. So, we’re done here. The slope of AB equals (y2-y1)/(x2-x1). (i.e. the ratio of the difference of the y coordinates to the difference of the x-coordinates)
## Lesson Summary
1. The slope of the line ax + by + c = 0, is equal to -a/b
2. The slope of the line joining the points (x1,y1) and (x2,y2) is equal to (y2-y1)/(x2-x1)
Remember these two, and you’ll be much comfortable later. In the next lesson, we’ll derive an expression to find the angle between two lines, whose slopes are known. |
# Lesson 1Getting Ready for a Pool PartyDevelop Understanding
## Learning Focus
Graph a function to model a situation.
Interpret and identify key features of the graph.
How can we create a graph without equations or points being given?
How does a graph tell a story?
How do I describe key features of a graph?
## Open Up the Math: Launch, Explore, Discuss
Sylvia has a small pool full of water that needs to be emptied and cleaned, then refilled for a pool party. During the process of getting the pool ready, Sylvia did all of the following activities, each during a different time interval.
Removed water with a single bucket Filled the pool with a hose(same rate as emptying pool) Drained water with a hose(same rate as filling pool) Cleaned the empty pool Sylvia and her two friends removed water with her three buckets Took a break
### 1.
Create a story of Sylvia’s process for emptying, cleaning, and filling the pool. Number the activities given in the table 1–6 to indicate the order in which they occurred in your story.
### 2.
Sketch a possible graph showing the height of the water level in the pool over time. Be sure to include all of the activities Sylvia did to prepare the pool for the party. Remember that only one activity happened at a time. Think carefully about how each section of your graph will look, labeling where each activity occurs.
### 3.
Does your graph represent a function? Why or why not? Would all graphs created for this situation represent a function?
Write a story context that has one interval of increase, one interval of decrease, and one interval when the rate of change is , in any order. Then, graph your story.
## Takeaways
A relationship is a function if and only if:
## Adding Notation, Vocabulary, and Conventions
Key features of functions:
## Lesson Summary
In this lesson, we created graphs to model a story context. We learned the mathematical words to describe the key features of functions. These key features are used to analyze and compare functions.
## Retrieval
### 1.
#### a.
Graph each of the functions on the same coordinate grid.
#### b.
What is similar and what is different about the graphs for these two functions?
### 2.
#### a.
Graph each of the functions on the same coordinate grid.
#### b.
What is similar and what is different about the graphs for these two functions?
### 3.
Solve each of the equations for . |
## how to find the altitude of a triangle
In a right triangle, the altitude for two of the vertices are the sides of the triangle. Two heights are easy to find, as the legs are perpendicular: if the shorter leg is a base, then the longer leg is the altitude (and the other way round). Imagine that you have a cardboard triangle standing straight up on a table. The third altitude of a triangle … Cite. Multiply the result by the length of the remaining side to get the length of the altitude. But the red line segment is also the height of the triangle, since it is perpendicular to the hypotenuse, which can also act as a base. A triangle gets its name from its three interior angles. Right: The altitude perpendicular to the hypotenuse is inside the triangle; the other two altitudes are the legs of the triangle (remember this when figuring the area of a right triangle). So the area of 45 45 90 triangles is: area = a² / 2 To calculate the perimeter, simply add all 45 45 90 triangle sides: Use the Pythagorean Theorem for finding all altitudes of all equilateral and isosceles triangles. Two congruent triangles are formed, when the altitude is drawn in an isosceles triangle. You can classify triangles either by their sides or their angles. Since the two opposite sides on an isosceles triangle are equal, you can use trigonometry to figure out the height. The length of the altitude is the distance between the base and the vertex. c 2 = a 2 + b 2 5 2 = a 2 + 3 2 a 2 = 25 - 9 a 2 = 16 a = 4. An isosceles triangle is a triangle with 2 sides of equal length and 2 equal internal angles adjacent to each equal sides. We can rewrite the above equation as the following: Simplify. How to Find the Equation of Altitude of a Triangle - Questions. Two congruent triangles are formed, when the altitude is drawn in an isosceles triangle. Drag A. In the above triangle the line AD is perpendicular to the side BC, the line BE is perpendicular to the side AC and the side CF is perpendicular to the side AB. The altitude from ∠G drops down and is perpendicular to UD, but what about the altitude for ∠U? In an obtuse triangle, the altitude from the largest angle is outside of the triangle. Equation of the altitude passing through the vertex A : (y - y1) = (-1/m) (x - x1) A (-3, 0) and m = 5/2. The task is to find the area (A) and the altitude (h). Find the area of the triangle [Take \sqrt{3} = 1.732] View solution Find the area of the equilateral triangle which has the height is equal to 2 3 . Find … 1-to-1 tailored lessons, flexible scheduling. Local and online. Learn faster with a math tutor. What about an equilateral triangle, with three congruent sides and three congruent angles, as with △EQU below? The following figure shows the same triangle from the above figure standing up on a table in the other two possible positions: with segment CB as the base and with segment BA as the base. Every triangle has three altitudes. In the animation at the top of the page: 1. In a right triangle, we can use the legs to calculate this, so 0.5 (8) (6) = 24. Now, recall the Pythagorean theorem: Because we are working with a triangle, the base and the height have the same length. Altitude of Triangle. Drag B and C so that BC is roughly vertical. We can use this knowledge to solve some things. How big a rectangular box would you need? In this figure, a-Measure of the equal sides of an isosceles triangle. A right triangle is a triangle with one angle equal to 90°. The altitude to the base of an isosceles triangle … Isosceles: Two altitudes have the same length. The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. Use the below online Base Length of an Isosceles Triangle Calculator to calculate the base of altitude 'b'. In geometry, an altitude of a triangle is a line segment through a vertex and perpendicular to (i.e., forming a right angle with) a line containing the base (the side opposite the vertex). If we take the square root, and plug in the respective values for p and q, then we can find the length of the altitude of a triangle, as the altitude is the line from an opposite vertex that forms a right angle when drawn to the side opposite the angle. Go to Constructing the altitude of a triangle and practice constructing the altitude of a triangle with compass and ruler. First we find the slope of side A B: 4 – 2 5 – ( – 3) = 2 5 + 3 = 1 4. Find the height of an equilateral triangle with side lengths of 8 cm. Step 1. You now can locate the three altitudes of every type of triangle if they are already drawn for you, or you can construct altitudes for every type of triangle. Divide the length of the shortest side of the main triangle by the hypotenuse of the main triangle. Drag it far to the left and right and notice how the altitude can lie outside the triangle. An altitude of a triangle is the line segment drawn from a vertex of a triangle, perpendicular to the line containing the opposite side. Altitude of an Equilateral Triangle Formula. The altitude of a triangle is a line from a vertex to the opposite side, that is perpendicular to that side, as shown in the animation above. To find the height of a scalene triangle, the three sides must be given, so that the area can also be found. The length of the altitude is the distance between the base and the vertex. Imagine you ran a business making and sending out triangles, and each had to be put in a rectangular cardboard shipping carton. For △GUD, no two sides are equal and one angle is greater than 90°, so you know you have a scalene, obtuse (oblique) triangle. Share. The altitude of a triangle: We need to understand a few basic concepts: 1) The slope of a line (m) through two points (a,b) and (x,y): {eq}m = \cfrac{y-b}{x-a} {/eq} Altitude of a Triangle is a line through a vertex which is perpendicular to a base line. Learn how to find all the altitudes of all the different types of triangles, and solve for altitudes of some triangles. The altitude passing through the vertex A intersect the side BC at D. AD is perpendicular to BC. This is identical to the constructionA perpendicular to a line through an external point. In each triangle, there are three triangle altitudes, one from each vertex. How to Find the Altitude? Vertex is a point of a triangle where two line segments meet. In a right triangle, the altitude for two of the vertices are the sides of the triangle. How to find the height of an equilateral triangle An equilateral triangle is a triangle with all three sides equal and all three angles equal to 60°. After working your way through this lesson and video, you will be able to: To find the altitude, we first need to know what kind of triangle we are dealing with. The height or altitude of a triangle depends on which base you use for a measurement. You can find it by having a known angle and using SohCahToa. To get that altitude, you need to project a line from side DG out very far past the left of the triangle itself. The correct answer is A. The above figure shows you an example of an altitude. AE, BF and CD are the 3 altitudes of the triangle ABC. This is done because, this being an obtuse triangle, the altitude will be outside the triangle, where it intersects the extended side PQ.After that, we draw the perpendicular from the opposite vertex to the line. Let AB be 5 cm and AC be 3 cm. The altitude C D is perpendicular to side A B. By their interior angles, triangles have other classifications: Oblique triangles break down into two types: An altitude is a line drawn from a triangle's vertex down to the opposite base, so that the constructed line is perpendicular to the base. Think of building and packing triangles again. To find the height, we can draw an altitude to one of the sides in order to split the triangle into two equal 30-60-90 triangles. The pyramid shown above has altitude h and a square base of side m. The four edges that meet at V, the vertex of the pyramid, each have length e. ... 30 triangle rule but ended up with $\frac{m\sqrt3}{2}$. It will have three congruent altitudes, so no matter which direction you put that in a shipping box, it will fit. We know that the legs of the right triangle are 6 and 8, so we can use inverse tan to find the base angle. Find the area of the triangle (use the geometric mean). In an obtuse triangle, the altitude from the largest angle is outside of the triangle. In each of the diagrams above, the triangle ABC is the same. The altitude is the shortest distance from the vertex to its opposite side. Slope of BC = (y 2 - y 1 )/ (x 2 - x 1) = (3 - (-2))/ (12 - 10) = (3 + 2)/2. Every triangle has 3 altitudes, one from each vertex. The other leg of the right triangle is the altitude of the equilateral triangle, so … Finding an Equilateral Triangle's Height Recall the properties of an equilateral triangle. This line containing the opposite side is called the extended base of the altitude. I searched google and couldn't find anything. An altitude of a triangle is a line segment that starts from the vertex and meets the opposite side at right angles. You only need to know its altitude. Where to look for altitudes depends on the classification of triangle. The above figure shows you an example of an altitude. The side is an upright or sloping surface of a structure or object that is not the top or bottom and generally not the front or back. The area of a triangle having sides a,b,c and S as semi-perimeter is given by. Lesson Summary. For right triangles, two of the altitudes of a right triangle are the legs themselves. Did you ever stop to think that you have something in common with a triangle? The height is the measure of the tallest point on a triangle. The other leg of the right triangle is the altitude of the equilateral triangle, so solve using the Pythagorean Theorem: Anytime you can construct an altitude that cuts your original triangle into two right triangles, Pythagoras will do the trick! This height goes down to the base of the triangle that’s flat on the table. Try this: find the incenter of a triangle using a compass and straightedge at: Inscribe a Circle in a Triangle. For an obtuse triangle, the altitude is shown in the triangle below. The intersection of the extended base and the altitude is called the foot of the altitude. Altitude of a Triangle is a line through a vertex which is perpendicular to a base line. On standardized tests like the SAT they expect the exact answer. Altitude (triangle) In geometry , an altitude of a triangle is a line segment through a vertex and perpendicular to i. To find the area of such triangle, use the basic triangle area formula is area = base * height / 2. For example, the points A, B and C in the below figure. Hence, Altitude of an equilateral triangle formula= h = √(3⁄2) × s (Solved examples will be updated soon) Quiz Time: Find the altitude for the equilateral triangle when its equal sides are given as 10cm. In this triangle 6 is the hypotenuse and the red line is the opposite side from the angle we found. Find the equation of the altitude through A and B. Apply medians to the coordinate plane. Classifying Triangles By their sides, you can break them down like this: Most mathematicians agree that the classic equilateral triangle can also be considered an isosceles triangle, because an equilateral triangle has two congruent sides. Triangles have a lot of parts, including altitudes, or heights. An equilateral … Obtuse: The altitude connected to the obtuse vertex is inside the triangle, and the two altitudes connected to the acute vertices are outside the triangle. Use Pythagoras again! b-Base of the isosceles triangle. Altitude of an Equilateral Triangle. = 5/2. Isosceles triangle properties are used in many proofs and problems where the student must realize that, for example, an altitude is also a median or an angle bisector to find a missing side or angle. How to Find the Altitude of a Triangle Altitude in Triangles. AE, BF and CD are the 3 altitudes of the triangle ABC. Equilateral: All three altitudes have the same length. How to find the altitude of a right triangle. The 3 altitudes always meet at a single point, no matter what the shape of the triangle is. An equilateral triangle is a special case of a triangle where all 3 sides have equal length and all 3 angles are equal to 60 degrees. Today we are going to look at Heron’s formula. Constructing an altitude from any base divides the equilateral triangle into two right triangles, each one of which has a hypotenuse equal to the original equilateral triangle's side, and a leg ½ that length. Because the 30-60-90 triange is a special triangle, we know that the sides are x, x, and 2x, respectively. Notice how the altitude can be in any orientation, not just vertical. First get AC with the Pythagorean Theorem or by noticing that you have a triangle in the 3 : 4 : 5 family — namely a 9-12-15 triangle. It seems almost logical that something along the same lines could be used to find the area if you know the three altitudes. Find the midpoint between (9, -1) and (1, 15). Two heights are easy to find, as the legs are perpendicular: if the shorter leg is a base, then the longer leg is the altitude (and the other way round). As there are three sides and three angles to any triangle, in the same way, there are three altitudes to any triangle. The altitude, also known as the height, of a triangle is determined by drawing a line from the vertex, or corner, of the triangle to the base, or bottom, of the triangle.All triangles have three altitudes. You would naturally pick the altitude or height that allowed you to ship your triangle in the smallest rectangular carton, so you could stack a lot on a shelf. The altitude or height of an equilateral triangle is the line segment from a vertex that is perpendicular to the opposite side. Since every triangle can be classified by its sides or angles, try focusing on the angles: Now that you have worked through this lesson, you are able to recognize and name the different types of triangles based on their sides and angles. The next problem illustrates this tip: Use the following figure to find h, the altitude of triangle ABC. This is a formula to find the area of a triangle when you don’t know the altitude but you do know the three sides. Question 1 : A(-3, 0) B(10, -2) and C(12, 3) are the vertices of triangle ABC . How do you find the altitude of an isosceles triangle? Here we are going to see, how to find the equation of altitude of a triangle. In terms of our triangle, this theorem simply states what we have already shown: since AD is the altitude drawn from the right angle of our right triangle to its hypotenuse, and CD and DB are the two segments of the hypotenuse. Try it yourself: cut a right angled triangle from a piece of paper, then cut it through the altitude and see if the pieces are really similar. In triangle ADB, sin 60° = h/AB We know, AB = BC = AC = s (since all sides are equal) ∴ sin 60° = h/s √3/2 = h/s h = (√3/2)s ⇒ Altitude of an equilateral triangle = h = √(3⁄2) × s. Click now to check all equilateral triangle formulas here. A = S (S − a) (S − b) (S − c) S = 2 a + b + c = 2 1 1 + 6 0 + 6 1 = 7 1 3 2 = 6 6 c m. We need to find the altitude … Calculate the orthocenter of a triangle with the entered values of coordinates. The orthocenter of a triangle is described as a point where the altitudes of triangle meet and altitude of a triangle is a line which passes through a vertex of the triangle and is perpendicular to the opposite side, therefore three altitudes possible, one from each vertex. All three heights have the same length that may be calculated from: h△ = a * √3 / 2, where a is a side of the triangle In the above right triangle, BC is the altitude (height). Given the side (a) of the isosceles triangle. The length of its longest altitude (a) 1675 cm (b) 1o75 cm (c) 2475 cm Hence, Altitude of an equilateral triangle formula= h = √(3⁄2) × s (Solved examples will be updated soon) Quiz Time: Find the altitude for the equilateral triangle when its equal sides are given as 10cm. Drag the point A and note the location of the altitude line. Find the altitude and area of an isosceles triangle. We can construct three different altitudes, one from each vertex. The altitude of the triangle tells you exactly what you’d expect — the triangle’s height (h) measured from its peak straight down to the table. The decimal answer is … If you insisted on using side GU (∠D) for the altitude, you would need a box 9.37 cm tall, and if you rotated the triangle to use side DG (∠U), your altitude there is 7.56 cm tall. Find the altitude of a triangle if its area is 120sqcm and base is 6 cm. Solution : Equation of altitude through A [insert scalene △GUD with ∠G = 154° ∠U = 14.8° ∠D = 11.8°; side GU = 17 cm, UD = 37 cm, DG = 21 cm]. Get better grades with tutoring from top-rated private tutors. The base is one side of the triangle. Here we are going to see how to find slope of altitude of a triangle. h^2 = pq. Consider the points of the sides to be x1,y1 and x2,y2 respectively. Now, the side of the original equilateral triangle (lets call it "a") is the hypotenuse of the 30-60-90 triangle. Properties of Rhombuses, Rectangles, and Squares, Interior and Exterior Angles of a Polygon, Identifying the 45 – 45 – 90 Degree Triangle, The altitude of a triangle is a segment from a vertex of the triangle to the opposite side (or to the extension of the opposite side if necessary) that’s perpendicular to the opposite side; the opposite side is called the base. Kindly note that the slope is represented by the letter 'm'. The isosceles triangle is an important triangle within the classification of triangles, so we will see the most used properties that apply in this geometric figure. In our case, one leg is a base and the other is the height, as there is a right angle between them. That can be calculated using the mentioned formula if the lengths of the other two sides are known. The intersection of the extended base and the altitude is called the foot of the altitude. METHOD 1: The area of a triangle is 0.5 (b) (h). This height goes down to the base of the triangle that’s flat on the table. … The answer with the square root is an exact answer. The altitude is the mean proportional between the … Base angle = arctan(8/6). (Definition & Properties), Interior and Exterior Angles of Triangles, Recognize and name the different types of triangles based on their sides and angles, Locate the three altitudes for every type of triangle, Construct altitudes for every type of triangle, Use the Pythagorean Theorem to calculate altitudes for equilateral, isosceles, and right triangles. For an equilateral triangle, all angles are equal to 60°. But what about the third altitude of a right triangle? Where all three lines intersect is the "orthocenter": An equilateral … (You use the definition of altitude in some triangle proofs.). The task is to find the area (A) and the altitude (h). Heron's Formula to Find Height of a Triangle. For example, say you had an angle connecting a side and a base that was 30 degrees and the sides of the triangle are 3 inches long and 5.196 for the base side. Lets find with the points A(4,3), B(0,5) and C(3,-6). Every triangle has three altitudes, one for each side. Every triangle has three altitudes. In each triangle, there are three triangle altitudes, one from each vertex. Construct the altitude of a triangle and find their point of concurrency in a triangle. Let us find the height (BC). Here is scalene △GUD. The side of an equilateral triangle is 3 3 cm. Altitude of an equilateral triangle is the perpendicular drawn from the vertex of the triangle to the opposite side and is represented as h= (sqrt (3)*s)/2 or Altitude= (sqrt (3)*Side)/2. Intersection of the triangle parts, including the altitude or height of a.. A single point, no matter which direction you put that in a right triangle formed by the red is! Base you use the Pythagorean Theorem: because we are going to,. 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how to find the altitude of a triangle 2021 |
# Scientific Notation, Exponents and Significant Figures by kzp12233
VIEWS: 119 PAGES: 28
• pg 1
``` Scientific Notation,
Exponents and Significant
Figures
Scientific Notation
Scientific Notation – It is notation used to
express very large or very small numbers
using powers of 10.
It is written as a number multiplied by 10x
Example: 1000 = 1 x 103 (10 x 10 x 10)
Here the 1000 is in standard notation
1 x 103 is scientific notation
Scientific Notation
How do we express these terms?
For Scientific Notation:
The number is expressed as >1 and <10 and
then multiplied by a power of 10.
Example: 523,000 = 5.23 x 105
Scientific Notation
Prefix Symbol Standard notation Exponent
giga- G 1 000 000 000 109
mega- M 1 000 000 106
kilo- k 1000 103
deci- d 0.1 10-1
centi- c 0.01 10-2
milli- m 0.001 10-3
micro- µ 0.000 001 10-6
nano- n 0.000 000 001 10-9
pico- p 0.000 000 000 001 10-12
Scientific Notation
So, how do we change from standard
notation to scientific notation?
Move the decimal point to create number
that is between 1 and 10
Example: 7,231,967 = 7.231967 x 106
0.00003433 = 3.433 x 10-5
Scientific Notation
Rules:
The decimal place should end up at to the
right of the first nonzero digit.
The total number of spaces moved becomes
the exponent of 10 in the scientific notation.
If the given number is greater than 1, the
exponent is positive.
If the given number is less than 1 (but >0),
the exponent is negative.
Scientific Notation
Practice writing these in scientific notation.
17 mL
153 kg
24883.5 km
2000 miles
0.4502 g
0.00063401 m
Scientific Notation
You can also use this information to write
a number in standard notation.
Example: 2.3445 x 103 g= 2344.5 g
2.21 x 10-7 m = 0.000000221 m
Scientific Notation
Practice: Write the follow ing in standard notation:
6.423 x 103 g 6423 g
1.002 x 10-6 m 0.000001002 m
5.0023 x 1010 m 50,023,000,000 m
3.3 x 10-9 sec 0.0000000033 sec
Rules of Exponents
Remember that exponents, especially with
powers of 10, help count zeros.
It is easier to see keep track of zeros with
and exponent like 106 than with the
standard notation of 1,000,000.
Using the rules of exponents, you can
multiply and divide exponents easily.
Exponents
Rules of Exponents Example
(10m)(10n) = 10m+n (102)(103 100,000
100*1000)== 105
(10m)n = 10m*n (103)2 = 106
(1000)2 = 1,000,000
10m/10n = 10m-n 106/102 = 10410,000
1,000,000 =
100
10-m = 1/10m 10-8 = 1/108
1 = 0.00000001
100,000,000
100 = 1 = 1 x 10-8
Scientific Notation & Exponents
Prefix Symbol Standard notation Exponent
giga- G 1 000 000 000 109
mega- M 1 000 000 106
kilo- k 1000 103
deci- d 0.1 10-1
centi- c 0.01 10-2
milli- m 0.001 10-3
micro- µ 0.000 001 10-6
nano- n 0.000 000 001 10-9
pico- p 0.000 000 000 001 10-12
Scientific Notation & Exponents
Practice:
Convert numbers or exponents to prefix.
1000 g 1 x 103 g or 1 kg
5.3 x 103 m 5.3 km
4.5 x 10-6 m 4.5 µm
1.7 x 10-3g 1.7 mg
22000 seconds 22 kiloseconds
Scientific Notation & Exponents
2.4 mg 2.4 x 10-3 g
2 km 2 x 103 m
1.6 Mm (megameter) 1.6 x 106 m
15 msec (milliseconds) (1.5 x101) x 10-3 sec
or 1.5 x 10-2 sec
253 km (2.53 x 102) x 103 m
Or 2.53 x 105 m
Exponents
How many milligrams are in a kilogram?
1 kg = 1000 g = 103 g x 1 mg =
10-3g
= 106 mg
Exponents
How many picograms in a microgram?
1 µg = (1 x 10-6 g)(1 pg )=
1x10-12 g
1/(1 x 10-6) pg =
106 pg = 1,000,000 pg
Significant Figures
With scientific measurements, you want to
know accuracy, precision and certainty.
Accuracy – How close a measurement is to an
accepted value
Precision – How close a measurement is to
other measurements of the same thing.
Certainty – Degree of confidence of a
measurement. The last digit to the right is
usually an uncertain digit.
Significant Figures
So for any measured value, we’ll record all
of the certain digits plus an uncertain
digit.
All together, they are the significant
figures of the measurement.
Significant Figure RULES
1. All non zero digits (1,2,3,4,5,6,7,8, and 9) are
significant.
2. Final zeros to the right of the decimal point are
significant.
3. Zeros between two significant digits are
significant.
4. Zeros used for spacing the decimal point are
not significant.
5. For numbers in scientific notation, all of the
digits before the “x 10x” are significant.
How Many Significant Figures?
Measurement # of Sig Figs
135.3 4 sig figs
4.6025 5 sig figs
200,035 6 sig figs
0.0000300 3 sig figs
2.0000300 8 sig figs
0.002 1 sig fig
4.44 x 103 3 sig figs
2.0 x 10-2 2 sig figs
10.00 4 sig figs
10 1 sig fig
102,000 3 sig figs
Significant Figures
Multiplying and Dividing with Sig Figs
When multiplying or dividing measurements, the
answer must have the same number of sig figs
as the measurement with the fewest sig figs.
Example: 22 feet x 9 feet = 198 square
feet…but
Since 9 feet only has 1 sig fig the correct answer
is
200 ft2
Calculate the Area in square blocks
3
2
1
0 1 2 3 4 5
Significant Figures
Calculation Calc’d Answer Ans w/sig figs
2.86 m x
1.824 m 5.21664 m2 5.22 m2
460 miles/
8 hours 57.5 mi/hr 60 mi/hr
98.50 in x
1.82 in 179.27 in2 179 in2
Significant Figures
Calculation Calc’d Answer Ans w/sig figs
2.100 m x
0.0030 m 0.0063 m2 0.0063 m2
10.00 g /
5.000 L 2 g/L 2.000 g/L
4.610 ft x
1.7 ft 7.837 ft2 7.8 ft2
Significant Figures
Defined numbers – part of a definition and
is not measured. So, defined numbers
(unit conversion factors) do not limit the
Also, counting numbers do not limit sig
figs.
Example: You cut a 24 ft piece of wood into 4
pieces. Each is 24 ft/4 = 6.0 ft/piece.
Significant Figures
The sig figs with addition and subtraction are
handled differently than with x and /.
The answer cannot have more certainty than
the least certain measurement.
This means the answer must have the same
number of sig figs to the right of the decimal
as the measurement with the fewest sig figs
to the right of the decimal place.
Significant Figures
Example:
4.271 g (3 sig figs to right of decimal)
2 g (0 sig figs to right of decimal)
10.0 g (1 sig fig to right of decimal)
16.271 g is calculated answer but…
since 2 g has no sig figs to right of
decimal the final answer is 16 g.
```
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# Math Snap
## Suppose $f$ is a linear function over the interval $[a, b]$. ThenSelect one: a. $\operatorname{av}(f)=f\left(\frac{a+b}{2}\right)$ b. $a v(f)=f(a+b)$ c. $\operatorname{av}(f)=f(b-a)$ d. $a v(f)=f\left(\frac{b-a}{2}\right)$
#### STEP 1
Assumptions
1. The function $f$ is linear over the interval $[a, b]$.
2. We are looking for the average value of the function $f$ over the interval $[a, b]$, denoted as $\operatorname{av}(f)$.
#### STEP 2
The average value of a function $f$ over an interval $[a, b]$ is given by the formula:
$\operatorname{av}(f) = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx$
#### STEP 3
Since $f$ is a linear function, it can be expressed in the form $f(x) = mx + c$, where $m$ is the slope and $c$ is the y-intercept.
#### STEP 4
To find the average value of $f$, we need to integrate $f(x)$ over the interval $[a, b]$ and then divide by the length of the interval, which is $b-a$.
$\operatorname{av}(f) = \frac{1}{b-a} \int_{a}^{b} (mx + c) \, dx$
#### STEP 5
Perform the integration of the linear function.
$\int_{a}^{b} (mx + c) \, dx = \left[\frac{m}{2}x^2 + cx\right]_{a}^{b}$
#### STEP 6
Apply the Fundamental Theorem of Calculus to evaluate the integral from $a$ to $b$.
$\left[\frac{m}{2}x^2 + cx\right]_{a}^{b} = \left(\frac{m}{2}b^2 + cb\right) - \left(\frac{m}{2}a^2 + ca\right)$
#### STEP 7
Simplify the expression.
$\left(\frac{m}{2}b^2 + cb\right) - \left(\frac{m}{2}a^2 + ca\right) = \frac{m}{2}(b^2 - a^2) + c(b - a)$
#### STEP 8
Recognize that $b^2 - a^2$ is a difference of squares and can be factored as $(b+a)(b-a)$.
$\frac{m}{2}(b^2 - a^2) + c(b - a) = \frac{m}{2}(b+a)(b-a) + c(b - a)$
#### STEP 9
Factor out $(b-a)$ from the expression.
$\frac{m}{2}(b+a)(b-a) + c(b - a) = (b-a)\left(\frac{m}{2}(b+a) + c\right)$
#### STEP 10
Now divide by $(b-a)$ to find the average value of $f$.
$\operatorname{av}(f) = \frac{(b-a)\left(\frac{m}{2}(b+a) + c\right)}{b-a}$
#### STEP 11
Simplify the expression by canceling out $(b-a)$.
$\operatorname{av}(f) = \frac{m}{2}(b+a) + c$
#### STEP 12
Recognize that $\frac{m}{2}(b+a)$ is equivalent to $m\left(\frac{b+a}{2}\right)$.
$\operatorname{av}(f) = m\left(\frac{b+a}{2}\right) + c$
#### STEP 13
Since $f(x) = mx + c$, we can substitute $x$ with $\frac{b+a}{2}$ to find $f\left(\frac{b+a}{2}\right)$.
$\operatorname{av}(f) = f\left(\frac{b+a}{2}\right)$
##### SOLUTION
The correct answer matches option a:
$\operatorname{av}(f)=f\left(\frac{a+b}{2}\right)$ |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 8.11: Quotient Rule and Higher Derivatives
Difficulty Level: At Grade Created by: CK-12
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Practice Quotient Rule and Higher Derivatives
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You may recall hearing about Becky and her Track and Field competition in a prior lesson. Her boyfriend had taken a picture of her just as she started to pull away from the others on the track. We learned how she might learn to identify her instantaneous speed at just the split second the picture was taken by using Calculus to find a derivative.
What if, instead of just finding her speed at that split second, she wanted to find her acceleration?
Embedded Video:
### Guidance
The Quotient Rule
Theorem: (The Quotient Rule) If f and g are differentiable functions at x and g(x) ≠ 0, then
ddx[f(x)g(x)]=g(x)ddx[f(x)]f(x)ddx[g(x)][g(x)]2\begin{align*}\frac {d}{dx}\left [ \frac{f(x)}{g(x)} \right ]= \frac {g(x) \frac {d}{dx}\left [{f(x)} \right ] - f(x) \frac{d}{dx} \left [{g(x)} \right ]}{\left [{g(x)} \right ]^2}\end{align*}
In simpler notation
\begin{align*}\left ( \frac{f}{g} \right )'=\frac {g\cdot f'-f\cdot g'}{g^2}\end{align*}
Keep in mind that the order of operations is important (because of the minus sign in the numerator) and \begin{align*}\left ( \frac{f}{g} \right )'\neq \frac {f'}{g'}\end{align*}.
Higher Derivatives
If the derivative \begin{align*}\,\! f'\end{align*} of the function \begin{align*}\,\! f\end{align*} is differentiable, then the derivative of \begin{align*}f'\end{align*}, denoted by \begin{align*}\,\! f''\end{align*} is called the second derivative of \begin{align*}\,\! f\end{align*}. We can continue the process of differentiating derivatives and obtain third, fourth, fifth and higher derivatives of \begin{align*}\,\! f\end{align*}. They are denoted by \begin{align*}\,\! f'\end{align*}, \begin{align*}\,\! f''\end{align*}, \begin{align*}\,\! f'''\end{align*}, \begin{align*}\,\! f^{(4)}\end{align*}, \begin{align*}\,\! f^{(5)}\end{align*}, . . . ,
#### Example A
Find \begin{align*}= \frac {dy}{dx}\end{align*} for \begin{align*}y = \frac {x^2-5}{x^3+2}\end{align*}
Solution
\begin{align*}\frac {dy}{dx}\end{align*} \begin{align*}= \frac {d}{dx}\left [ \frac{x^2-5}{x^3+2} \right ]\end{align*}
\begin{align*}= \frac {(x^3+2).(x^2-5)'-(x^2-5).(x^3+2)'}{(x^3+2)^2}\end{align*}
\begin{align*}= \frac {(x^3+2)(2x)-(x^2-5)(3x^2)}{(x^3+2)^2}\end{align*}
\begin{align*}= \frac {2x^4+4x-3x^4+15x^2}{(x^3+2)^2}\end{align*}
\begin{align*}= \frac {-x^4+15x^2+4x}{(x^3+2)^2}\end{align*}
\begin{align*}= \frac {x(-x^3+15x+4)}{(x^3+2)^2}\end{align*}
#### Example B
At which point(s) does the graph of \begin{align*}y = \frac {x} {x^2+9}\end{align*} have a horizontal tangent line?
Solution
Since the slope of a horizontal line is zero, and since the derivative of a function signifies the slope of the tangent line, then taking the derivative and equating it to zero will enable us to find the points at which the slope of the tangent line is equal to zero, i.e., the locations of the horizontal tangents. Notice that we will need to use the quotient rule here:
\begin{align*}y\end{align*} \begin{align*}= \frac{x} {x^2 + 9}\end{align*}
\begin{align*}y'\end{align*} \begin{align*}= \frac{(x^2 + 9) \cdot f'(x) - x \cdot g' (x^2 + 9)} {(x^2 + 9)^2} = 0\end{align*} \begin{align*}= \frac{(x^2 + 9) (1) - x (2x)} {(x^2 + 9)^2} = 0\end{align*}
Multiply both sides by \begin{align*}(x^2 +9)^2\end{align*},
\begin{align*}x^2 + 9 - 2x^2\end{align*} \begin{align*}= 0\end{align*}
\begin{align*}x^2\end{align*} \begin{align*}= 9\end{align*}
\begin{align*}x\end{align*} \begin{align*}= \pm 3\end{align*}
Therefore, at \begin{align*}x = -3\end{align*} and \begin{align*}x = 3\end{align*}, the tangent line is horizontal.
#### Example C
Find the fifth derivative of \begin{align*}f(x) = 2x^4 - 3x^3 + 5x^2 - x - 1\end{align*}
Solution
To find the fifth derivative, we must first find the first, second, third, and fourth derivatives.
\begin{align*}f '(x)\end{align*} = \begin{align*}8x^3 - 9x^2 + 5x - x\end{align*}
\begin{align*}f ' ' (x)\end{align*} = \begin{align*}24x^2 - 18x + 5\end{align*}
\begin{align*}f ' ' ' (x)\end{align*} = \begin{align*}48x - 18\end{align*}
\begin{align*}f^{(4)} (x)\end{align*} = \begin{align*}48\end{align*}
\begin{align*}f^{(5)} (x)\end{align*} =
Concept question wrap-up
Once Becky has calculated her instantaneous speed at a given point on the track by finding the derivative, she could then take the derivative of that function to find her instantaneous acceleration at the same point in the race.
By finding her instantaneous speed and acceleration at different points in the race, she can learn a lot about what points made a difference in her overall success, and also what points she needs to work on.
Better racing through Calculus!
### Vocabulary
Instantaneous velocity is speed in a given direction at a single "snapshot" moment in time.
Instantaneous acceleration is change in velocity calculated at a single instant.
### Guided Practice
Questions
1) Suppose y'(2) = 0 and (y/q)(2) = 0. Find q(2) assuming y(2) = 0
2) Find the derivative of \begin{align*}k(x) = \frac{-2x - 4}{e^x}\end{align*}
3) Given \begin{align*}f(x) = -2x^2 - 4x - 1\end{align*} What is \begin{align*}f''(x)\end{align*}?
4) Given \begin{align*}f(x) = (-x^4 -4x^3 - 5x^2 +3)\end{align*} Find \begin{align*}f''(x)\end{align*} when \begin{align*}x=3\end{align*}
Solutions
1) Begin with the quotient rule:
\begin{align*}\left(\frac{y}{q}\right)' (2) = \left(\frac{y'(2)q(2) - y(2)q'(2)}{q(2)^2}\right)\end{align*} ..... Substitute
\begin{align*}(0) = \left(\frac{(0)q(2) - (0)q'(2)}{q(2)^2}\right)\end{align*} ..... Substituting again with given values
\begin{align*}0 = \left(\frac{(0)q(2)}{q(2)^2}\right)\end{align*} ..... Simplify with:\begin{align*}(0)q'(2) = 0\end{align*}
\begin{align*}0 = \frac{0}{q(2)}\end{align*}
\begin{align*}q(2) = 0\end{align*}
2) Use the quotient rule: Note: \begin{align*}(-2x - 4)' = -2\end{align*} and \begin{align*}(e^x)' = e^x\end{align*}
\begin{align*}\left(\frac{-2x-4}{e^x}\right)' = \frac{(-2)(e^x) - (-2x - 4)(e^x)}{e^{2x}}\end{align*} ..... Substitute
\begin{align*}\frac{2x + 2}{e^x}\end{align*} ..... Simplify
3) Recall that \begin{align*}f''(x)\end{align*} means "The derivative of the derivative of x"
\begin{align*}f'(x) = f'(-2x^2 - 4x -1) \to f'(x) = -4x -4\end{align*} ..... By the power rule
\begin{align*}f''(x) = f'(-4x -4) = -4\end{align*}
\begin{align*}\therefore f''(x) = -4\end{align*}
4) Recall that \begin{align*}f''(x)\end{align*} means "The derivative of the derivative of x"
\begin{align*}f'(x) = -4x^3 -12x^2 - 10x\end{align*} ..... Use the power rule on f(x)
\begin{align*}f''(x) = -12x^2 - 24x - 10\end{align*} ..... Use the power rule on f'(x)
\begin{align*}f''(3) = -12(3)^2 -24(3) - 10 \to -108 -72 - 10 = -190\end{align*} ..... Substitute 3
\begin{align*}\therefore f''(3) = -190\end{align*}
### Practice
Use the Quotient Rule to Solve:
1. Suppose \begin{align*}u'(0) = 98\end{align*} and \begin{align*}(\frac{u}{q})'(0) = 7\end{align*}. Find \begin{align*}q(0)\end{align*} assuming \begin{align*}u(0) = 0\end{align*}
2. Given: \begin{align*} b(x) = \frac{x^2 - 5x + 4}{-5x + 2}\end{align*} What is: \begin{align*} b'(2)\end{align*}?
3. Given: \begin{align*}m(x) = \frac {e^x}{3x + 4}\end{align*} what is \begin{align*} \frac{dm}{dx}\end{align*}?
4. What is \begin{align*}\frac{d}{dx} \cdot \frac{sin (x)}{x - 4}\end{align*}?
5. \begin{align*}q(x) = \frac{x}{sin (x)}\end{align*}
Solve these Higher Order Derivatives:
1. Given: \begin{align*}v(x) = -4x^3 + 3x^2 + 2x + 3\end{align*} What is \begin{align*}v''(x)\end{align*}?
2. Given: \begin{align*}m(x) = x^2 + 5x \end{align*} What is \begin{align*}m''(x)\end{align*}?
3. Given: \begin{align*}d(x) = 3x^4e^x\end{align*} What is \begin{align*}d''(x)\end{align*}?
4. Given: \begin{align*}t(x) = -2x^5sin (x)\end{align*} What is \begin{align*}\frac{d^2t}{dx^2}\end{align*}?
5. What is \begin{align*}\frac{d^2} {dx^2}3x^5e^x\end{align*}?
Solve:
1. \begin{align*}y = \frac{3} {\sqrt{x} + 3}\end{align*}
2. \begin{align*}y = \frac{4x + 1} {x^2 - 9}\end{align*}
3. Newton’s Law of Universal Gravitation states that the gravitational force between two masses (say, the earth and the moon), m and M is equal two their product divided by the squared of the distance r between them. Mathematically, \begin{align*}F = G\frac{mM} {r^2}\end{align*} where G is the Universal Gravitational Constant (1.602 × 10-11 Nm2/kg2). If the distance r between the two masses is changing, find a formula for the instantaneous rate of change of F with respect to the separation distance r.
4. Find \begin{align*}\frac{d} {d \psi} \left [\frac{\psi \psi_0 + \psi^3} {3 - \psi_0}\right ]\end{align*} : where \begin{align*}\psi_0\end{align*} is a constant.
5. Find \begin{align*}\frac{d^3y} {dx^3} |_{x = 1},\end{align*} where \begin{align*}y = \frac{2} {x^3}\end{align*}.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
differentiable
A differentiable function is a function that has a derivative that can be calculated.
Instantaneous acceleration
The instantaneous acceleration of an object is the change in velocity of the object calculated at a specific point in time.
Instantaneous velocity
The instantaneous velocity of an object is the velocity of the object at a specific point in time.
quotient rule
In calculus, the quotient rule states that if $f$ and $g$ are differentiable functions at $x$ and $g(x) \ne 0$, then $\frac {d}{dx}\left [ \frac{f(x)}{g(x)} \right ]= \frac {g(x) \frac {d}{dx}\left [{f(x)} \right ] - f(x) \frac{d}{dx} \left [{g(x)} \right ]}{\left [{g(x)} \right ]^2}$.
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# Math Snap
## Find $x$ if; $3 / 4(5-x)=8-(2 x-1 / 9)$
#### STEP 1
Assumptions 1. We are given the equation $\frac{3}{4}(5-x) = 8 - (2x - \frac{1}{9})$. 2. We need to solve for the variable $x$.
#### STEP 2
First, we need to distribute the $\frac{3}{4}$ across the terms inside the parentheses on the left side of the equation. $\frac{3}{4}(5) - \frac{3}{4}(x) = 8 - (2x - \frac{1}{9})$
#### STEP 3
Perform the multiplication. $\frac{15}{4} - \frac{3x}{4} = 8 - 2x + \frac{1}{9}$
#### STEP 4
Next, we want to combine like terms and simplify the equation. To do this, we need to get all the $x$ terms on one side and the constants on the other side. We can start by adding $2x$ to both sides of the equation. $\frac{15}{4} - \frac{3x}{4} + 2x = 8 + \frac{1}{9} + 2x$
#### STEP 5
Now, we combine the $x$ terms on the left side of the equation. To add $-\frac{3x}{4}$ and $2x$, we need a common denominator. The common denominator is 4, so we convert $2x$ to $\frac{8x}{4}$. $\frac{15}{4} - \frac{3x}{4} + \frac{8x}{4} = 8 + \frac{1}{9}$
#### STEP 6
Combine the $x$ terms. $\frac{15}{4} + \frac{5x}{4} = 8 + \frac{1}{9}$
#### STEP 7
Now, we want to isolate the $x$ term on one side of the equation. We can do this by subtracting $\frac{15}{4}$ from both sides. $\frac{5x}{4} = 8 + \frac{1}{9} - \frac{15}{4}$
#### STEP 8
To combine the constants on the right side, we need a common denominator. The common denominator for 4 and 9 is 36. Convert $8$ to $\frac{288}{36}$ and $\frac{15}{4}$ to $\frac{135}{36}$. $\frac{5x}{4} = \frac{288}{36} + \frac{1}{9} - \frac{135}{36}$
#### STEP 9
Convert $\frac{1}{9}$ to $\frac{4}{36}$ to have the same denominator. $\frac{5x}{4} = \frac{288}{36} + \frac{4}{36} - \frac{135}{36}$
#### STEP 10
Combine the constants on the right side. $\frac{5x}{4} = \frac{288 + 4 - 135}{36}$
#### STEP 11
Perform the addition and subtraction. $\frac{5x}{4} = \frac{157}{36}$
#### STEP 12
To solve for $x$, we need to multiply both sides of the equation by the reciprocal of $\frac{5}{4}$, which is $\frac{4}{5}$. $x = \frac{157}{36} \cdot \frac{4}{5}$
#### STEP 13
Multiply the numerators and denominators. $x = \frac{157 \cdot 4}{36 \cdot 5}$
#### STEP 14
Perform the multiplication. $x = \frac{628}{180}$
#### STEP 15
Simplify the fraction by dividing both the numerator and denominator by their greatest common divisor, which is 4. $x = \frac{628 \div 4}{180 \div 4}$
##### SOLUTION
Perform the division. $x = \frac{157}{45}$ The solution for $x$ is $\frac{157}{45}$. |
# Exponents Lessons
Before we dive into simplifying exponents, let’s take some time to learn exactly what an exponent is. An exponent is a superscript,
or small number written at the top right corner of a number, variable, or set of parentheses. An example of one is shown
below.
23
This tells you to multiply 1 by the number as many times as the exponent says.
The example above is 2 raised to the third power (raised to the third power
means the exponent is 3). This is equivalent to the multiplication problem below, because there is a 1 multiplied by 2 three
times.
1 * 2 * 2 * 2
8
As you can see, the 1 * 2 * 2 * 2 can be simplified to 8 which is the answer to the problem.
## Exponents
Examine the next problem:
-36
This problem is changed to the multiplication problem below. Because of the order of operations (explained in a later lesson)
the exponent is simplified first and then, the negative sign is added to the answer.
1 * 3 * 3 * 3 * 3 * 3 * 3
-729
As you can see, the multiplication simplifies to the number -729. You can do the work for this in your head, on the margin
of your paper, or using a calculator if allowed.
## Exponents
The next problem is shown below:
(-3)6
This time the -3 is inside parentheses. Instead of carrying down the negative sign, each 3 is made negative.
1 * -3 * -3 * -3 * -3 * -3 * -3
729
As you can see, the multiplication simplifies to 729. Note that aside from
the result of the previous example being negative, the result
here is the same. Pay careful attention to finding exponents when negative
signs are involved, as this is a common source of error.
## The zero exponent
Each one of these problems is solved by using a 1 multiplied by the number,
the amount of times the exponent indicates. If the exponent is 0, the 1 isn’t
multiplied by the number at all. Therefore, the answer is 1.
This is an important rule to remember.
510
1
## Zero with an exponent:
In most cases zero with an exponent can be calculated like any other number
and exponent.
04
1 * 0 * 0 * 0 * 0
0
Note that as long as 1 is being multiplied by at least one 0, the end result
is 0. Therefore we can conclude that 0 to any positive exponent is always zero.
Another special case occurs with 00. Zero with an exponent of
zero is undefined, and cannot be calculated.
Be careful not to the rules for zero exponents! Zero to any positive power
is always zero, because no matter how many times you multiply the 1 by
zero the answer will always be zero. But 00 is undefined.
## The 1 exponent:
Consider this example in which rasies a number is raised to the first power.
511
1 * 51
51
If you try any similar example such as 101 or 1001, you
will find that the result is always the original number or base. This is
because 1 times any other number is always equal to the second number.
So to simplify the case where a number is raised to the first power, we can
simply remove the exponent.
## Ten to Any Power
This tip can save you a lot of time: Ten to any power is simply the number 1
followed by a number of zeros indicated by the exponent.
An example is shown below.
105
100,000
Note that the result is a one followed by five zeros because the exponent on
10 was 5.
In general, a problem such as
510
is read as “Five to the tenth power.”
## Special Cases for Reading Exponents
Certain exponents have special ways of being pronounced. These make it a
little easier to say, but it is not necessary that you use them.
• The Second Power: 32 – may be read as either “Three to the second power,” or “Three squared.”
• The Third Power: 103 – may be read as either “Ten to the third power,” or “10 cubed.”
## Exponents of Numbers Resources
Practice Problems / WorksheetPractice all of the methods you learned in this lesson here. Next Lesson: Exponents of VariablesLearn how to simplify a variable inside a parenthesis with an exponent.
## Tutoring
Looking for someone to help you with algebra? At Wyzant, connect with algebra tutors and math tutors nearby. Prefer to meet online? Find online algebra tutors or online math tutors in a couple of clicks.
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# Into Math Grade 2 Module 6 Answer Key Use Place Value
We included HMH Into Math Grade 2 Answer Key PDF Module 6 Use Place Value to make students experts in learning maths.
## HMH Into Math Grade 2 Module 6 Answer Key Use Place Value
Break the Code
Detective Dina is on the lookout for three numbers. Read the clues to help her find the numbers.
The digit in the tens place is 4.
The digit in the ones place is 3 less than 8.
What is my number? ____
The number is 45.
Explanation:
Given that the digit in the tens place is 4 and the digit in the ones place is 3 less than 8 which is 8-3 = 5. So the number is 45.
The digit in the tens place is 2 greater than 5.
The digit in the ones place is 1 less than 3.
What is my number? _____
The number is 72.
Explanation:
Given that the digit in the tens place is 2 greater than 5 which is 5+2 = 7 and the digit in the ones place is 1 less than 3 which is 3-1 =2. So the number is 72.
The digit in the hundreds place is 6.
The digit in the tens place is equal to 2 + 3.
The digit in the ones place is equal to 4 – 1.
What is my number? _______
The number is 653.
Explanation:
Given that the digit in the hundreds place is 6 and the digit in the tens place is equal to 2 + 3 which is 5 and the digit in the ones place is equal to 4 – 1 is 3. So the number is 653.
Turn and Talk
Make up a clue for a two-digit or three-digit number. Trade clues with a partner and solve.
Complete these problems to review prior concepts and skills you will need for this module.
Count by Tens
Question 1.
Count by tens to 100. Circle the numbers as you count.
Compare Numbers to 50
Question 2.
Compare the numbers to find which is greater. Compare 18 and 29.
____ is greater than ____
29 is greater than 18.
Explanation:
In the given numbers 29 is greater than 18.
Question 3.
Compare the numbers to find which is less. Compare 35 and 47.
____ is less than ____
35 is less than 47.
Explanation:
In the given numbers 35 is less than 47.
Compare 2-Digit Numbers Using Symbols
Compare. Write >, <, or =.
Question 4.
46 46 |
# Download LESSON #32: MULTIPLYING POLYNOMIALS
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```Multiplying Polynomials
Lesson 32
In lesson 32, our warm-up has students multiplying binomials. They are going to
review multiplying binomials because they will be multiplying polynomials in this
lesson. Our solve problem, the length of a rectangle is represented by the trinomial
3x2+2x-4. The width of the rectangle is represented by the trinomial x2+5x-1. What
is the area of the rectangle? When we study the problem, or “S” the problem, we will
first underline the question. What is the area of the rectangle? We will then answer
the question, what is this problem asking me to find? This problem is asking me to
find the area of the rectangle. We are going to use the box method to multiply
polynomials. We will represent our first term, x+2, vertically. And we will represent
our second term horizontally. The box method will help keep your students organized
so they will not loose any terms in this process. x(x2)=x3; x(+3x)=3x2; and x(+1)=+x;
+2(x2)=+2x2; +2(+3x)=+6x; +2(+1)=+2. From here to get our final answer, we have
to combine like terms. Our diagonals show us that there are like terms. We only
have one x3; we have 2x2+3x2=5x2; we have 6x+x=7x; +2. This is our final answer.
In problem two, we will multiply the binomial times the trinomial. We will represent
our first binomial vertically. And our trinomial horizontally. We will use the box
method to keep us organized. 2x(x2)=2x3; 2x(-x)=-2x2; 2x(+4)=+8x; +1(x2)=+x2; +1(x)=-x;+1(+4)=+4. We see that once again we have like terms on the diagonal. So,
we have 2x3; +x2-2x2=-x2; -x+8x=+7x; +4. In problem 1, we are going to multiply two
trinomials. We will represent the first trinomial vertically. And the second trinomial
horizontally. Because we have 3 terms times 3 terms, we will end up with 9 terms
after we multiply. So the box method will help us stay organized. x2(x2)=x4; we will
then multiply our x2(3x), so you get +3x3; x2(+2)=+2x2; +2x(x2)= +2x3;
+2x(+3x)=+6x2; +2x(+2)=+4x; +1(x2)= +x2; +1(+3x)=+3x; and +1(+2)=+2. From here
we are going to combine our like terms, which once again are our diagonals. We only
have one x4; +2x3+3x3=5x3; +x2+6x2+2x2=9x2; +3x+4x= 7x; +2.
To multiply our first trinomial, we will represent our trinomial vertically; and the
second trinomial horizontally. 3x2(x2)=3x4; 3x2(+2x)=+6x3; 3x2(-1)=-3x2; -x(x2)=-x3;
-x(+2x)=-2x2; -x(-1), negative times a negative… is a +x. +4(x2)=+4x2; +4(+2x)=+8x;
+4(-1)=-4. Our diagonals are our like terms. So we have 3x4; -x3+6x3=5x3; +4x2-2x2
-3x2=-x2; +8x+x=9x; -4. In problem 3, we are going to multiply a trinomial by a
binomial. We will represent our trinomial vertically and our binomial horizontally.
x2(2x)=2x3; x2(-3)=-3x2; +5x(+2x)=+10x2; +5x(-3)=-15x; +4(+2x)=+8x; +4(-3)=-12.
To find our final answer, we are going to combine our like terms. 2x3; +10x23x2=+7x2; +8x-15x=-7x; -12.
In problem 4, we have two products added together. So we have 2x (x+1) and this is
added to or plus, 3x(x2-5x+1). 2x(x)=2x2; 2x(+1)=2x; 3x(x2)=3x3; 3x(-5x)=-15x2;
3x(+1)=3x. We have a 2x2+2x+3x3-15x2+3x. Our largest number of x’s is 3x3 so we
can write that first. We then have two x2, one is a +2 and one is a -15, so we can
combine those two and get -13x2. And then we have a +2x and +3x, which we can
combine to get 5x. We can use the distributive property to multiply polynomials. We
would have to multiply the x (x2)…to get x3; the x (-x)….to get –x2; and the x(-2)…to get
-2x. We would then have to multiply the +1(x2)…to get +x2; +1(-x)…-x; and +1(-2)…to
get -2. We can combine like terms. We will begin with our x3. There one x2 plus
another x2, Negative and positive, so they actually cancel and leave us with 0x2. -2xx=-3x; -2. To complete problem 2, we will use the distributive property. Simplify the
first product. 4x(x)=4x2; 4x(-2)=-8x. We will add this to the product (x+2) (x+3). We
can use FOIL to multiply these two binomials. x (x)=x2 ; our outer, x (-3)= -3x ; our
inner, +2(x) = 2x ; and our last +2 (-3)=-6. We now need to combine like terms. We
have one 4x2 and another x2 , which gives us 5x2; we have a -8x, -3x, +2x which gives
us -9x; and we have a -6 for a minus 6.
To complete our solve problem, we have already “S” the problem and know that this
problem is asking us to find the area of the rectangle. In “O” organize the facts, we
must identify our facts. The length of a rectangle is represented by the trinomial
3x2+2x-4. This is a fact. The width of the rectangle is represented by the trinomial
x2+5x-1. This is also a fact. Both of these facts are necessary so we will list them. In
“L” lining up our plan, we must choose an operation and because we are finding the
area, we are going to use multiplication. We will write our plan of action in words.
We will multiply the length times the width. In “V” verify your plan with action, we are
going to estimate. Our estimate is going to be a polynomial because our length is a
polynomial and so is our width; so the product will also be a polynomial. We can
carry out our plan. I’m going to multiply the two trinomials using the box method.
Some students may choose to use the distributive property in order to multiply these
polynomials. Because I have a trinomial times a trinomial, I need three rows and
three columns. (3x2+2x-4) and (x2+5x-1). 3x2(x2)=4x4; 3x2(5x)=+15x2; 3x2(-1)=-3x2;
+2x(x2)=2x3; +2x(+5x)=10x2; +2x(-1)=-2x; -4(x2)=-4x2; -4(+5x)=-20x; -4(-1)=+4.
When we look at our diagonals, they are like terms. So we have 3x4;
+2x3+15x3=17x3; -4x2+10x2-3x2=3x2; -20x-2x= -22x; +4. In “E” we will examine our
results. Does your answer make sense? Our question asked, what is the area of the
rectangle? Because our length and width were polynomials, it does make sense to
have a polynomial for an answer. Is your answer reasonable? Our estimate was a
polynomial, so yes our answer is reasonable. And is your answer accurate? You
could have your students check their work using a different method. We will now
write our answer as a complete sentence. The area of the rectangle is
3x4+17x3+3x2-22x+4. To close the lesson, we will look at the essential questions.
Number 1, when multiplying polynomials, is it possible to determine the number of
terms you should have before combining like terms? YES, you can multiply the
number of terms in each factor to determine the total number of terms in the product
before combining like terms. In other words, if you have a trinomial times a trinomial,
there are 3 terms in the first factor and 3 terms in the second factor…so you should
have 9 terms before combining like terms. Question 2, is it possible to have more
than two like terms when multiplying polynomials? YES, depending on the types of
polynomials you are multiplying, you could have more than two like terms.
```
Related documents |
# Fractions
1. UNDERSTANDING A FRACTION
A fraction is a number representing part of a whole. The whole may be a single object or a group of objects and each to be equal.
Definition
The numbers of the form $\frac{p}{q}$, where ‘p’ and ‘q’ are whole numbers and q $\ne$ 0 are called fractions.
Example:
$\frac{1}{2},\frac{2}{3},\frac{1}{4},\frac{3}{5},\frac{8}{12}$etc., are fractions, here $\frac{1}{2}$ means the whole thing can be divided into two equal parts and we taken out 1 part similarly $\frac{3}{3}$ means the whole thing can be divided into three equal parts and we take out two parts.
In the above figure the whole rectangle can be divided into 5 equal parts among which 2 parts are shaded. The shaded portion represents two-fifths and is denoted by $\frac{2}{5}$. Here two-fifths is a fractional number and $\frac{2}{5}$ is a fraction.
$\frac{1}{2}$ read as one-half; $\frac{2}{3}$ as two-thirds; $\frac{1}{4}$ as one-quarter; $\frac{3}{5}$ as three-fifth and $\frac{8}{12}$ read as eight-twelths and so an.
Note:
For a given fractions $\frac{\mathrm{p}}{\mathrm{q}}$, p is called the numerator and q is called denominator. Where p, q may have common factors
Example:
$\frac{5}{12},\frac{1}{2},\frac{2}{3},\frac{6}{8},\frac{10}{12}$, etc, 5 is numerator and 8 is denominator
Fractions can also be written in its lowest terms.
Example:
(i) $\frac{5}{12},\frac{1}{2},\frac{2}{3}$ fractions does not have any common factors in both numerator and denominator.
They are written in lowest terms.
(ii) Fractions $\frac{6}{8},\frac{10}{12}$ have common factors in both numerator and denominator
We know that the numbers of the form $\frac{\mathrm{p}}{\mathrm{q}}$ where p,q $\in$ w (whole numbers), q $\ne$ 0 and p, q does not have any common factors are called rational numbers.
Example:
$\frac{1}{2},\frac{2}{3},\frac{4}{5},\frac{10}{11}$ , etc are rational numbers. Therefore ‘every rational number is a fraction’
2. SIMPLEST FORM OF A FRACTION
If the numerator and denominator of a fraction have no common factor except 1, then the fraction is said to be in its simplest form or in lowest terms. Or if a fraction is said to be in simplest form if the H.C.F of its numerator and denominator is 1.
I. Irreducible Fraction
If a fraction is in simplest form then it is called irreducible fraction.
II. Reducible Fraction
If a fraction is not irreducible fraction then it is called reducible fraction.
III. Method to find Simplest Form of a Fraction:
We can reduce given fraction into its simplest form using any of the following two methods.
Method 1
Divide numerator and denominator of given fraction by their H.C.F
Example:
To find simplest form of $\frac{9}{15}$; the H.C.F of 9 and 15 is 3. Now divide both numerator and denominator of $\frac{9}{15}$ by 3 to get reduce it into simplest form. Hence the simplest form of $\frac{9}{15}$ is $\frac{3}{5}$.
Method 2
In this method, we can divide both numerator and denominator of the given fraction by common factor till we are left with common 1 only.
Example:
We will find simplest form of $\frac{72}{90}$ using this method or follows, $\frac{72}{90}=\frac{4}{5}$. Therefore the simplest form of $\frac{72}{90}$ is $\frac{4}{5}$.
3. REPRESENTING FRACTIONS ON NUMBER LINE
To represent given fraction $\frac{1}{2}$ on a number line, we can divide the length between 0 and 1 into two equal parts and we can take 1 part as a fraction $\frac{1}{2}$.
To represent $\frac{2}{3}$ on a number line, we can divide the length between 0 and 1 into three equal parts and we can take into three equal parts and we can take 2 parts out of 3 as a fraction $\frac{2}{3}$.
4. CLASSIFICATION OF FRACTIONS
I. Like Fractions
Fractions with the same denominator are called like fractions.
Example:
$\frac{1}{9},\frac{2}{9},\frac{4}{9},\frac{8}{9},\frac{16}{9}$ are like fractions.
II. Unlike Fractions
Fractions with different denominators are called unlike fractions.
Example:
$\frac{1}{2},\frac{2}{3},\frac{3}{5},\frac{5}{8}$ are unlike fractions.
III Unit Fractions
Fractions with 1 as numerator are known as unit fractions.
Example:
$\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{12}$, etc, are all unit fractions.
IV. Proper Fractions
A fraction in which the numerator is less than its denominator is called proper fraction.
Example:
$\frac{1}{2},\frac{2}{3},\frac{4}{5},\frac{8}{12}$, etc, are proper fractions lie between the 0 and 1 on a number line.
V. Improper Fractions
A fraction in which the numerator is greater than or equal to its denominator.
Note:
The numerical value of all the improper fraction does not lie between 0 and 1 on a number line.
VI. Mixed Fraction
A combination of a whole number and a proper fraction is called mixed fraction.
Example:
$2\frac{1}{3},3\frac{1}{4},16\frac{2}{3}$, etc, are all mixed fraction.
we can write mixed fraction = whole number part + fractional parts = whole number + of fractional part
In a mixed fraction is whole number and $\frac{1}{3}$ is a proper fraction. $2\frac{1}{3}=2+\frac{1}{3}$
VII. Equivalent Fraction
Two or more fractions representing the same point of a whole are called equivalent fractions.
Eg: $\frac{1}{2}=\frac{2}{4}=\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{6}{12}–$…..are equivalent fractions
In above figures, the shaded regions of each figure are equal i.e., $\frac{1}{2}=\frac{2}{4}=\frac{3}{6}$ are equivalent fractions.
Finding Equivalent Fractions
To find an equivalent fraction of a given fraction, we can multiply or divide both numerator and denominator of the given fraction by the same non-zero number.
Examples:
$\frac{1}{2}=\frac{1×2}{2×2}=\frac{2×3}{4×3}=\frac{1×3}{2×3}=\frac{1×5}{2×5}=$….. are equivalent fractions
$\frac{12}{18}=\frac{12÷2}{18÷2}=\frac{12÷3}{18÷3}=\frac{12÷6}{18÷6}$ are also equivalent fractions
Note:
If two fractions are said to be equivalent fractions, the product of the numerator of the first and denominator of the second is equal to the product of denominator of the first and the numerator of the second. These products are called cross products. If $\frac{\mathrm{p}}{\mathrm{q}},\frac{\mathrm{r}}{\mathrm{s}}$ are two fractions then their cross product is denoted as
and if ps = rq then we say that $\frac{\mathrm{p}}{\mathrm{q}},\frac{\mathrm{r}}{\mathrm{s}}$ are equivalent fractions.
Example: $\frac{1}{2},\frac{2}{4}$ are equivalent fractions since 1 × 4 = 2 × 2 $⇒$4 = 4
5. INTER-CONVERSION OF FRACTIONS
I. Conversion of Unlike Fractions into Like Fractions
To convert unlike fractions we can make denominators of all given fractions equal to their L.C.M (i.e., we convert each of the given fractions into an equivalent fractions having a denominator equal to the L.C.M of all the denominators of the given fraction.)
Example:
To convert the unlike fractions $\frac{3}{4},\frac{5}{6},\frac{4}{9},\frac{1}{8},\frac{7}{12}$ into like fractions first we find L.C.M of denominator of given unlike fractions.
L.C.M of 4, 6, 9, 8, 12 = L.C.M of 8, 9,12 = 72
Now we are equating denominator of all fractions to 72 by multiplying numerators and denominators of all given fractions with a suitable number.
Therefore ; $\frac{1}{8}–\frac{1×9}{8×9}–\frac{9}{72}$ and $\frac{7}{12}=\frac{7×6}{12×6}–\frac{42}{72}$
Therefore $\frac{54}{72},\frac{60}{72},\frac{32}{72},\frac{9}{72},\frac{42}{72}$ are the required like fractions.
I Conversion of mixed Fraction into Improper Fraction
We know that mixed fraction has two parts; one is whole number part and the other is fractional part. To convert mixed fraction into improper fraction, multiply the whole number part with the denominator of the fractional part and add the product to the numerator of the fraction part. This gives the numerator of the improper fraction and its denominator is same as denominator of fraction part.
i.e., if mixed fraction = whole number , then the improper fraction can be expressed as
Improper Fraction =
Example:
Convert mixed fraction $5\frac{2}{3}$ into improper fraction as follows:
$5\frac{2}{3}=5–\frac{2}{3}\frac{\left(5×3\right)–2}{3}=\frac{15+2}{3}=\frac{17}{3}$
III. Convert an Improper Fraction into a Mixed Fraction
We know that improper fraction has greater numerical value of numerator than denominator. To convert an improper fraction into a mixed fraction, divide the numerator by denominator then the quotient so obtained forms the whole number part and the remainder forms numerator of fractional part.
Here denominator of fractional part is same as denominator of given improper fraction. So that mixed fraction can be written as follows
Mixed Fraction Quotient
Procedure to convert $\frac{23}{7}$ into mixed fraction as follows divide 23 by 7 as follows
We have quotient 3 and remainder 2 when 23 divided by 7.
Therefore required fraction = Quotient
Convert $\frac{58}{7}$ into mixed fraction
Required mixed fraction = $8\frac{2}{7}$
6. COMPARISON OF FRACTIONS
To compare two are more fractions, their denominators should be equal. If the denominator of given fractions are not equal, then change each one of the given fractions into an equivalent fractions with denominator equal to the L.C.M of the denominators of the given fractions. Now the new fractions are like fractions so that the comparison between two or more fractions is possible when they are like fractions.
Conditions for comparison of fractions
To compare given fraction we can use following conditions.
(i) Fractions with same denominator
Among the fractions with same denominator, the fraction with greater numerator is greater than the other.
Example:
For the fractions $\frac{2}{9},\frac{14}{9};\frac{14}{9}>\frac{2}{9}$ since both have same denominator and 14 > 2
(ii) Fractions with same numerator
Among two fractions with same numerator. The fraction with smaller denominator is greater than the other.
Eg: For the fractions $\frac{2}{9}$ and $\frac{2}{6};\frac{2}{6}>\frac{2}{9}$ since both have same numerator and 6<9
(iii) Fractions with different numerator and denominator
(A) When compare fractions with different numerator and different denominators, we change them into fractions by equating their denominators to L.C.M of denominators of given fractions and then compare numerators of like fraction by applying first condition.
Example:
For $\frac{4}{5},\frac{6}{7}$; L.C.M of 5, 7 is 35
Now
Now are like fractions (since they have equal denominators) and 30 > 28
$⇒\frac{30}{35}>\frac{28}{35}⇒\frac{6}{7}>\frac{4}{5}$
(B) To compare fractions with different numerators and different denominators we can also use the following method for the fractions $\frac{a}{b},\frac{c}{d}$ find cross products ad and bc.
If ad > bc, then $\frac{a}{b}>\frac{c}{d};$ If ad < bc, then $\frac{a}{b}<\frac{c}{d}$; If ad = bc, then $\frac{a}{b}=\frac{c}{d}$
Example:
Compare the fractions $\frac{3}{5}$ and $\frac{8}{11}$ by considering cross multiplication
The products are 33, 40 and 40 > 33
$⇒5×8>3×11⇒\frac{8}{11}>\frac{3}{5}$
Therefore $\frac{8}{11}>\frac{3}{5}$
7. ARRANGING FRACTIONS IN ASCENDING AND DESCENDING ORDER
Ascending order of fractions means arranging fractions from smaller value to greater value and descending order of fractions means arranging fractions from greater value to smaller value. This can be done by comparison of fractions.
Examples:
Rearrange the following fractions in ascending order and descending order $\frac{2}{3},\frac{4}{5},\frac{7}{15},\frac{11}{20},\frac{23}{30}$
Here the given fractions have different numerators and different denominators.
L.C.M. of 3, 5, 15, 20, 30 = 2 × 3 × 10 = 60
Therefore $\frac{2}{3}=\frac{2×20}{3×20}=\frac{40}{60};$ $\frac{4}{5}=\frac{4×12}{5×12}=\frac{48}{60};$ $\frac{7}{15}–\frac{7×4}{15×4}–\frac{28}{60}$; $\frac{11}{20}–\frac{11×3}{20×3}–\frac{33}{60}$ and $\frac{23}{30}=\frac{23×2}{30×2}=\frac{46}{60}$.
Since 28 < 33 < 40 < 46 < 48
$⇒\frac{28}{60}<\frac{33}{60}<\frac{40}{60}<\frac{46}{60}<\frac{48}{60}$ $⇒\frac{7}{15}<\frac{11}{20}<\frac{2}{3}<\frac{23}{30}<\frac{4}{5}$
Therefore the ascending order of given fractions is $\frac{7}{15},\frac{11}{20},\frac{2}{3},\frac{23}{30},\frac{4}{5}$ and the descending order of given fraction is $\frac{4}{5},\frac{23}{30},\frac{2}{3},\frac{11}{20},\frac{7}{15}$.
8. FUNDAMENTAL OPERATIONS (+, -, ×, ÷) ON FRACTIONS
Addition of two or more fractions is possible, when they are like fractions.
If the given set of fractions have same denominator (i.e., like fraction), then the numerator of sum of all fractions is the sum of numerators of given fractions and denominator is their common denominator.
Therefore sum of all like fractions =
Example: $\frac{5}{6}+\frac{8}{6}+\frac{13}{6}+\frac{22}{6}=\frac{2–8–13–22}{6}=\frac{48}{6}=8$
Step1: We find L.C.M of denominators of all fractions.
Step2: Now convert each of the given fractions into equivalent like fractions by equating their denominators to L.C.M of denominators.
Step3: Now we can add all like fractions which are so obtained in Step2
Step4: Reduce the fraction obtained in Step3 into its lowest terms and convert it into mixed fractions if it is a improper fraction.
Examples:
Step1: Find the sum of $\frac{5}{9},\frac{8}{12},\frac{15}{16}$
Step2: Now $\frac{5}{9}=\frac{5×16}{9×16}=\frac{80}{144}$; $\frac{8}{12}=\frac{8×12}{12×12}=\frac{96}{144}$; $\frac{15}{16}=\frac{15×9}{16×9}=\frac{135}{144}$;
Step3: Now sum = $\frac{5}{9}+\frac{8}{12}+\frac{15}{16}=\frac{80+96+135}{144}=\frac{311}{144}$ is a improper fraction
Step4: Therefore $\frac{5}{9}–\frac{8}{12}+\frac{15}{16}=\frac{311}{144}=2\frac{23}{144}$
If given set of fractions are mixed fractions;
Step1: Convert each of the mixed fractions into an improper fraction.
Example:
Find the sum of $1\frac{2}{5}+2\frac{3}{5}+3\frac{4}{5}$
Method 1:
In above the addends are mixed fractions
Now, $1\frac{2}{5}+2\frac{3}{5}–3\frac{4}{5}=\frac{7}{5}–\frac{13}{5}+\frac{19}{5}=\frac{39}{5}=7\frac{4}{5}$
Method 2:
$1\frac{2}{5}+2\frac{3}{5}–3\frac{4}{5}=\left(1+2+3\right)+\left(\frac{2+3+4}{5}\right)$
$=6–\frac{5}{5}+\frac{4}{5}$
$=6+1+\frac{4}{5}$
$=–7\frac{4}{5}$
Example:
Find sum of $1\frac{2}{3}+2\frac{3}{4}+3\frac{1}{5}$
Method 1:
In above addends are mixed fractions
$1\frac{2}{3}=\frac{5}{3}$; $2\frac{3}{4}=\frac{11}{4}$; $3\frac{1}{5}=\frac{16}{5}$
Now, $1\frac{2}{3}+2\frac{3}{4}+3\frac{1}{5}=\frac{5}{3}+\frac{11}{4}+\frac{16}{5}$
$=\frac{\left(5×20\right)×\left(11×15\right)×\left(16×12\right)}{60}$ = $\frac{100+165+192}{60}=\frac{457}{60}=7\frac{37}{60}$
Method 2:
$1\frac{2}{3}+2\frac{3}{4}+3\frac{1}{5}=\left(1+2+3\right)+\left(\frac{2}{3}–\frac{3}{4}+\frac{1}{5}\right)$
$=6+\frac{\left(2×20\right)+\left(3×15\right)+\left(1×12\right)}{60}$
$=6+\frac{\left(40+45–12\right)}{60}$ = $6+\frac{97}{60}=6+1\frac{37}{60}=7\frac{37}{60}$
II. Subtraction
Subtraction of Like Fractions
Difference of like fractions is a fraction which is having numerator as difference of numerator and denominators is common denominator.
Therefore difference of like fraction =
Examples:
$\frac{9}{13}–\frac{4}{13}=\frac{9–4}{13}=\frac{5}{13}$
$\frac{2}{3}–\frac{1}{5}$
Subtraction of Unlike Fractions
First we find L.C.M of denominators of given fractions.
Convert each of the given fraction into equivalent like fraction by equating their denominators to L.C.M of denominators.
Now we can find difference between like fractions so obtained.
Method 1 :
$\frac{2}{3}–\frac{1}{5}$
L.C.M of 3,5 = 15
$\frac{2}{3}=\frac{2×5}{3×5}=\frac{10}{15}$
$\frac{1}{5}=\frac{1×3}{5×3}=\frac{3}{15}$
$\frac{2}{3}–\frac{1}{5}=\frac{10}{15}–\frac{3}{15}=\frac{10–3}{15}=\frac{7}{15}$
Method 2 :
$\frac{2}{3}–\frac{1}{5}=\frac{\left(2×5\right)–\left(1×3\right)}{15}=\frac{10–3}{15}=\frac{7}{15}$
Subtraction of Mixed Fractions
While subtracting two or more mixed fractions, convert each of mixed fractions in to an improper fraction and then find the difference of fractions so obtained using any of the above methods.
Examples:
Find $3\frac{1}{6}–2\frac{1}{4}+4\frac{1}{3}–2\frac{1}{5}$
Since
Now $3\frac{1}{6}–2\frac{1}{4}+4\frac{1}{3}–2\frac{1}{5}=\frac{19}{6}–\frac{9}{4}–\frac{13}{3}–\frac{11}{5}$
(Therefore L.C.M of 6, 4, 3, 5 = 60)
$=\frac{190–135+260–132}{60}$
(Since 60$÷$6 = 10; 60$÷$3 = 20; 60$÷$4 = 15; 60$÷$5 = 2)
Therefore
III. Multiplication
Multiplication of two fractions is a fraction whose numerator is products of numerators of given fractions and denominators is product of denominators of given fractions, we can define it as follows:
Product of Fractions =
i.e., for any two fractions
Example:
Multiplication of $\frac{2}{5}$ and $\frac{3}{8}$ is $\frac{2}{5}×\frac{3}{8}=\frac{6}{40}=\frac{3}{20}$
For the multiplication of mixed fractions convert them into improper fractions and then consider multiplication.
Example:
$2\frac{3}{4}×1\frac{5}{6}=\frac{11}{4}×\frac{11}{6}=\frac{121}{24}$
Reciprocal or Multiplicative Inverse of a Fraction
If the product of any two fractions is 1, then each one of them is called reciprocal or multiplicative inverse of other.
For the fraction $\frac{\mathrm{p}}{\mathrm{q}},\frac{\mathrm{q}}{\mathrm{p}},\frac{\mathrm{p}}{\mathrm{q}}×\frac{\mathrm{q}}{\mathrm{p}}=1$
Therefore reciprocal of $\frac{\mathrm{p}}{\mathrm{q}}$ is $\frac{\mathrm{q}}{\mathrm{p}}$ and reciprocal of $\frac{\mathrm{q}}{\mathrm{p}}$ is $\frac{\mathrm{p}}{\mathrm{q}}$
Examples:
The reciprocal of $\frac{5}{6}$ is $\frac{6}{5}$
The reciprocal of $1\frac{2}{3}$ is $\frac{3}{5}$
The reciprocal of 2 is $\frac{1}{2}$
IV. Division
To divide one fraction by another fraction, we multiply the dividend fraction by the reciprocal of the divisor.
Examples:
$\frac{3}{5}÷\frac{2}{3}=\frac{3}{5}×\frac{3}{2}=\frac{9}{10}$
$1\frac{2}{3}÷2\frac{2}{5}=\frac{5}{3}÷\frac{12}{5}=\frac{5}{3}×\frac{5}{12}=\frac{25}{36}$
Note:
For any fraction $\frac{a}{b},1×\frac{a}{b}=\frac{a}{b}×1=\frac{a}{b}$ and
Every non zero fraction or rational number has multiplicative inverse.
Zero does not have multiplicative inverse. |
# Often asked: How can you tell if a graph is linear?
Every linear graph is nothing more than a straight line so if there is any curvies in it, it’s not linear. The other way to tell is look at its equation. If the equation can be shaped into Y = MX + B where M and B are numbers, then it’s going to be a linear equation.
• There are two ways to tell if a graph is linear. First of all the shape of the graph is a straight line. Every linear graph is nothing more than a straight line so if there is any curvies in it, it’s not linear. The other way to tell is look at its equation. If the equation can be shaped into Y = MX + B where M and B are numbers, then it’s going to be a linear equation.
## How do you know if a graph is linear or not?
How Can You Tell if a Function is Linear or Nonlinear From a Table? To see if a table of values represents a linear function, check to see if there’s a constant rate of change. If there is, you’re looking at a linear function!
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## What makes a graph linear or nonlinear?
Linear functions make graphs that are perfectly straight lines. Nonlinear functions have graphs that are curved.
## What does a linear graph look like?
Linear functions are those whose graph is a straight line. A linear function has one independent variable and one dependent variable. The independent variable is x and the dependent variable is y. It is also known as the slope and gives the rate of change of the dependent variable.
## How do you identify a linear equation?
Simplify the equation as closely as possible to the form of y = mx + b. Check to see if your equation has exponents. If it has exponents, it is nonlinear. If your equation has no exponents, it is linear.
## What is difference between linear and nonlinear equation?
Linear means something related to a line. All the linear equations are used to construct a line. A non-linear equation is such which does not form a straight line. It looks like a curve in a graph and has a variable slope value.
## What is a linear or nonlinear function?
Linear FunctionA linear function is a relation between two variables that produces a straight line when graphed. Non-Linear FunctionA non-linear function is a function that does not form a line when graphed.
## How do you find out if a function is linear or nonlinear?
To see if a table of values represents a linear function, check to see if there’s a constant rate of change. If there is, you’re looking at a linear function!
## What is linear and nonlinear in English?
Linear text refers to traditional text that needs to be read from beginning to the end while nonlinear text refers to text that does not need to be read from beginning to the end. As their names imply, linear texts are linear and sequential while non-linear and non-sequential.
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## What is the graph of a linear equation?
The graph of a linear equation in two variables is a line (that’s why they call it linear ). If you know an equation is linear, you can graph it by finding any two solutions. (x1,y1) and (x2,y2), plotting these two points, and drawing the line connecting them.
## What is a linear graph used for?
A line graph is commonly used to display change over time as a series of data points connected by straight line segments on two axes. The line graph therefore helps to determine the relationship between two sets of values, with one data set always being dependent on the other set.
## Why is a graph linear?
It always goes up in steps of the same size, so it’s a straight line. This is fine as far as it goes. It identifies the defining property of a linear function—that it has a constant rate of change—and relates that property to a geometric feature of the graph.
## What is called linear equation?
An equation for a straight line is called a linear equation. The general representation of the straight-line equation is y=mx+b, where m is the slope of the line and b is the y-intercept. Linear equations are those equations that are of the first order.
## How can you recognize a linear equation in one variable?
A linear equation in one variable is an equation that can be written in the form ax b c + =, where a, b, and c are real numbers and. Linear equations are also first-degree equations because the exponent on the variable is understood to be 1.
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## How do you determine if an equation is linear in two variables?
If a, b, and r are real numbers (and if a and b are not both equal to 0) then ax+by = r is called a linear equation in two variables. (The “two variables” are the x and the y.) The numbers a and b are called the coefficients of the equation ax+by = r. |
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The arithmetic mean of the first n natural numbers is$A)\dfrac{{\left( {n + 1} \right)}}{2} \\ B)\dfrac{{\left( {n - 1} \right)}}{2} \\ C)\dfrac{n}{2} \\ D){\text{none of the above}} \\$
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Hint: To proceed with a solution we need the sum of n natural numbers which will be helpful to solve the arithmetic mean of the first n natural numbers where the formula of arithmetic mean of the first n natural numbers includes the sum of n natural numbers.
We know that
First n natural numbers are 1, 2, 3, 4……………………..n
We also know that sum of n natural numbers =$\dfrac{{n\left( {n + 1} \right)}}{2}$
Hence we know that
Arithmetic mean of n natural numbers = $\dfrac{{{\text{Sum of natural numbers}}}}{{{\text{Total natural numbers}}}}$
Arithmetic mean = $\dfrac{{\dfrac{{n\left( {n + 1} \right)}}{2}}}{n}$
Arithmetic mean = $\dfrac{{n + 1}}{2}$
Therefore arithmetic mean is $\dfrac{{n + 1}}{2}$
Option A is the correct answer.
NOTE: In this problem to get the arithmetic mean of n natural numbers we need to know the sum of n natural numbers where the formula of arithmetic mean includes .After getting the values we have substituted in formula and proceeded on calculation part. Here the calculation is a simple cancellation of n terms where we get the answer as $\dfrac{{n + 1}}{2}$. |
Thrust and Pressure
Thrust and Pressure
Have you ever wondered why a camel can run in a desert easily? Why an army tank weighing more than a thousand tonne rests upon a continuous chain? Why a truck or a motorbus has much wider tyres? Why cutting tools have sharp edges? In order to address these questions and understand the phenomena involved, it helps to introduce the concepts of the net force in a particular direction (thrust) and the force per unit area (pressure) acting on the object concerned.
Let us try to understand the meanings of thrust and pressure by considering the following situations:
Situation 1 :
You stand on loose sand. Your feet go deep into the sand. Now, lie down on the sand. You will find that your body will not go that deep in the sand. In both cases the force exerted on the sand is the weight of your body.
You have learnt that weight is the force acting vertically downwards. Here the force is acting perpendicular to the surface of the sand. The force acting on an object perpendicular to the surface is called thrust.
When you stand on loose sand, the force, that is, the weight of your body is acting on an area equal to area of your feet. When you lie down, the same force acts on an area equal to the contact area of your whole body, which is larger than the area of your feet. Thus, the effects of forces of the same magnitude on different areas are different. In the above cases, thrust is the same. But effects are different. Therefore the effect of thrust depends on the area on which it acts.
The effect of thrust on sand is larger while standing than while lying. The thrust on unit area is called pressure. Thus,
Pressure =thrust/area
Substituting the SI unit of thrust and area in the above formulae, we get the SI unit of pressure as
N/m2 or N m–2.
In honour of scientist , Blaise Pascal, the SI unit of pressure is called pascal, denoted as Pa.
Let us consider a numerical example to understand the effects of thrust acting on different areas.
Example-1:
A block of wood is kept on a tabletop. The mass of wooden block is 5 kg and its dimensions are 40 cm × 20 cm × 10 cm. Find the pressure exerted by the wooden block on the table top if it is made to lie on the table top with its sides of dimensions (a) 20 cm × 10 cm and (b) 40 cm × 20 cm.
Solution :
The mass of the wooden block = 5 kg
The dimensions = 40 cm × 20 cm × 10 cm
Here, the weight of the wooden block applies a thrust on the table top.
That is,
Thrust = F = m × g
= 5 kg × 9.8 m s–2
= 49 N
Area of a side = length × breadth
= 20 cm × 10 cm
= 200 cm2 = 0.02 m2
From Eq. (20),
Pressure= 49 N ⁄ 0.02m2
= 2450 N m-2.
When the block lies on its side of dimensions 40 cm × 20 cm, it exerts the same thrust.
= 40 cm × 20 cm
= 800 cm2 = 0.08 m2
From Eq. (20),
Pressure= 49 N ⁄ 0.08m2
= 612.5 N m–2
The pressure exerted by the side 20 cm× 10 cm is 2450 N m–2 and by the side 40 cm × 20 cm is 612.5N m–2.
Thus, the same force acting on a smaller area exerts a larger pressure, and a smaller pressure on a larger area. This is the reason why a nail has a pointed tip, knives have sharp edges and buildings have wide foundations.
Pressure in Fluids
All liquids and gases are fluids. A solid exerts pressure on a surface due to its weight.. Similarly, fluids have weight, and they also exert pressure on the base and walls of the container in which they are enclosed. Pressure exerted in any confined mass of fluid is transmitted undiminished in all directions
Summary :
• The total force acting perpendicular on a given surface is called thrust.
• Thrust acting upon a unit area is called pressure.Its S.I. unit is Pascal denoted by ‘P’
• Pressure =Thrust / Area of contact
Cite this Simulator: |
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A 6 in Fraction Form
# 6 in Fraction Form
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I’m going to show you a new way of thinking about science that might help you understand the world in a whole new way.
### Math
Shopping taxes and discounts are two calculations consumers need to understand. Learn the connection between math and shopping, examine examples and calculations for taxes and discounts, and review examples of tax as well as tax and discount calculations.
Learn about mathematical operations with integers using the operations of addition, subtraction, multiplication, and division. Discover the properties of integers and how those properties affect the solution to different types of math problems. (Source: study.com)
### Step
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(12, 6) = 12. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 12 × 6 = 72. In the following intermediate step, cancel by a common factor of 2 gives 7/6.
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. (Source: www.hackmath.net)
An alternative method for finding a common denominator is to determine the least common multiple (LCM) for the denominators, then add or subtract the numerators as one would an integer. Using the least common multiple can be more efficient and is more likely to result in a fraction in simplified form. In the example above, the denominators were 4, 6, and 2. The least common multiple is the first shared multiple of these three numbers.
Unlike adding and subtracting integers such as 2 and 8, fractions require a common denominator to undergo these operations. One method for finding a common denominator involves multiplying the numerators and denominators of all of the fractions involved by the product of the denominators of each fraction. Multiplying all of the denominators ensures that the new denominator is certain to be a multiple of each individual denominator. The numerators also need to be multiplied by the appropriate factors to preserve the value of the fraction as a whole. This is arguably the simplest way to ensure that the fractions have a common denominator. However, in most cases, the solutions to these equations will not appear in simplified form (the provided calculator computes the simplification automatically). Below is an example using this method. (Source: www.calculator.net)
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# The base of an isosceles right triangle is 30 cm. Its area is - Mathematics
MCQ
The base of an isosceles right triangle is 30 cm. Its area is
#### Options
• 225 cm2
• 225 $\sqrt{3}$ cm
• 225 $\sqrt{2}$ cm
• 450 cm2
#### Solution
$\text{Let ABC be the right triangle in which} \angle B = 90° .$
$\text{ Now, base = BC; perpendicular = AB; Hypotenuse = AC }$
$\text{ Now, BC = 30 cm } \left( \text{ given } \right)$
$\text{ Now, ∆ ABC is an isosceles right angled ∆ and we know that hypotenuse is the longest side of the right ∆ }m.$
$\text{ So, AB = BC = 30 cm }$
$\text{ area of ∆ ABC } = \frac{1}{2} \times\text{ base } \times \text{ height }$
$= \frac{1}{2} \times BC \times AB$
$= \frac{1}{2} \times 30 \times 30$
$= 450 {cm}^2$
Concept: Application of Heron’s Formula in Finding Areas of Quadrilaterals
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RD Sharma Mathematics for Class 9
Chapter 17 Heron’s Formula
Exercise 17.4 | Q 2 | Page 24
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# Reviewing Quantitative Relationships
## Overview
In this lesson, students represent quantitative relationships involving rates using tables, graphs, double number lines, and formulas. Students will understand how to create one such representation when given another representation.
# Key Concepts
Quantitative relationships involving rates can be represented using tables, graphs, double number lines, and formulas. One such representation can be used to create another representation. Two rates can describe each situation: the rate and its inverse. For the water pump situation, there are two related formulas: a formula for finding the quantity of water pumped for any amount of time, and a formula for finding the amount of time for any quantity of water.
# Goals and Learning Objectives
• Understand that tables, graphs, double number lines, and formulas can be used to represent the same situation.
• Compare the different representations within a situation and the same representation across similar situations.
• Understand each representation and how to find the rate in each one.
# Lesson Guide
Review the graph and make sure students know how to read the coordinates and interpret the graph.
• The point (10,200) tells you 200 gallons were pumped in 10 seconds.
• The rate is 20 gallons/second.
# Graph of Pumping Water
The graph shows the amount of water pumped over time.
• What does the point (10, 200) tell you?
• At what rate is the water being pumped?
# Lesson Guide
Discuss the Math Mission. Students will represent quantitative relationships in different ways.
SWD: Students with disabilities may struggle to independently find patterns and to observe similarities between formulas. Use your understanding of your students' needs to inform any pre-teaching and/or interventions you may provide.
## Opening
Represent rates in different ways.
# Lesson Guide
As students work, look for these types of responses:
• Students that find the rate by solving for gallons per minute.
• Students that find the rate by solving for minutes per gallon.
• Students who use different types of formulas for the situations, because the inconsistency will cause problems when graphing.
SWD: Color coding of ideas, relationships, and formulas can be helpful for students with special needs. Consider providing these color-coded formulas to students once they have learned about the concepts.
# Mathematical Practices
Mathematical Practice 6: Attend to precision; and Mathematical Practice 1: Make sense of problems and persevere in solving them.
• To find each rate and formula, students must first make sense of the problem situation and attend to precision. In some situations, the rate is given, and students need to write the formula. In other situations, students need to calculate the rate. In some situations, the time is given first and then the number of gallons. In other situations, the number of gallons is given first and then the time.
• Later, students will compare the different situations on the same graph. While graphing, students may have difficulty selecting axes value ranges that will work for all four situations. Identify student work that has different values for the axes and that attends to different levels of precision.
Mathematical Practice 4: Model with mathematics.
• Students will be working from one model in order to create another model of the same situation (for example, from a table to a formula to a graph). Watch for students who are able to model the mathematical relationships correctly.
• Also look for students who are able to use their graph to draw conclusions about the four situations.
# Interventions
Student is having difficulty getting started.
• Describe what you need to do for each situation to your partner.
• What information do you know?
• What is the first thing you need to find?
• Once you identify the rate, what will you do?
• How can you use your formula to make a graph of the situation?
Student doesn't understand how to write a formula.
• How do you know your formula is correct?
• What do the variables in your formula represent?
• What formula matches this line on the graph? Explain how you know.
• How are they the same?
• How are they different?
• Can you use both formulas to make the same graph?
Make sure your students are correctly representing the units.
• What does the vertical axis of your graph show? What is the scale? What size intervals did you use?
• What does the horizontal axis of your graph show? What is the scale? What size intervals did you use?
• Does the range of your scales cover all problem situations?
• Students should come up with the same formulas, but they might use different variables for gallons, g, or time, t. Solving for gallons:
• Pump 1: g = 125t
• Pump 2: g = 52t
• Pump 3: g = 260t
• Pump 4: g = 52t
• Solving for minutes:
• Pump 1: t = (1,125)g
• Pump 2: t = (152)g
• Pump 3: t = (1,260)g
• Pump 4: t = (152)g
# Graph the Relationships
Below are four situations describing the rate of water flowing through a pump.
Situation 1: Pump 1 pumps water at a rate of 125 gallons per minute.
Situation 2: Pump 2 pumps water at a constant rate. Pump 2 pumps 364 gallons in 7 minutes.
Situation 3: Pump 3 pumps at a constant rate, represented by the table.
Situation 4: Pump 4 pumps at a constant rate, represented by the double number line.
For each situation, do the following.
1. Write a formula that represents the rate relationship.
2. Graph all four relationships on a single graph.
Situation 3:
Situation 4:
## Hint:
• What rate will you find gallons per minute or minutes per gallon?
• How will you represent this rate as a formula?
• Why should you represent time on the x -axis?
# Preparing for Ways of Thinking
• Look for students who wrote equations that solve for gallons, in particular students whose methods may differ from others.
• Look for students who wrote equations that solve for minutes, in particular students whose methods may differ from the others.
• Look for students who had issues with graphing due to the formula they wrote.
• Look for students who have a clear understanding of the relationship between rate and the slope of the graph.
# Challenge Problem
• On the graph, the line for pump 3 is the steepest because it is pumping water at the fastest rate.
• Pump 1: g = 125t or $t=\left(\frac{1}{125}\right)g$
• Pump 2: g = 52t or $t=\left(\frac{1}{52}\right)g$
• Pump 3: g = 260t or $t=\left(\frac{1}{260}\right)g$
• Pump 4: g = 52t or $t=\left(\frac{1}{52}\right)g$
• Possible answer: We are interested in how many gallons of water there will be after a given time, so the number of gallons depends on the time. Thus, time is the independent variable and gallons is the dependent variable. The formula g = kt indicates that time is the independent variable and g is the dependent variable. The formula t = kg indicates that volume is the independent variable and time is the dependent variable.
# Prepare a Presentation
Prepare a presentation about the four situations.
• Compare the situations. What is similar, and what is different?
• What did you notice about two of the situations?
# Challenge Problem
• Which line on your graph of the rate relationships is the steepest? Explain why.
• In this problem, the time is considered the independent variable and the volume the dependent variable. Explain why.
# Lesson Guide
By the end of the discussion, students should understand each others' methods and make connections between methods.
If any students work on the challenge problems, have them share their thinking about independent and dependent variables.
ELL: When discussing how to find the formulas, make it a point to write, in words, what each of the formulas represents. It is important to provide students with the opportunity to practice a combination of written and oral skills.
# Mathematics
Make note of any mathematical practices that you observe students using.
• How does one student's graph compare with another student's graph?
• What is similar?
• What is different?
• Which graph is easiest to read?
• Did anyone use different strategies? Explain what you did.
• Where do you see the rate in each representation (table, double number line, graph, formula)?
• Which representation of the rate situation is easiest for you to understand? Why?
• What conclusions can you draw from interpreting the graph?
• Which pump is the fastest? How do you know?
• Which pump is the slowest? How do you know?
• How does the graph show that each pump works at a constant rate?
# Ways of Thinking: Make Connections
Take notes as you listen to your classmates’ solutions.
## Hint:
• What strategies did you use to make sense of the four situations?
• Whose method makes most sense to you? Why?
• Where do you see the rate in each representation (table, double number line, graph, formula)?
• Which representation of the rate situation is easiest for you to understand? Why?
• What conclusions can you draw from interpreting the graph?
• Which pump is the fastest? How do you know?
• Which pump is the slowest? How do you know?
• How does the graph show that each pump works at a constant rate?
# A Possible Summary
You can represent rate situations using tables, double number lines, graphs, or formulas. You can use one representation to create another representation.
• Quantitative relationships involving rates can be represented using tables, graphs, double number lines, and formulas. One such representation can be used to create another representation.
• Two rates can describe each situation: each rate is the inverse of the other.
• For the water pump situation, there are two related formulas: a formula for finding the quantity of water pumped for any amount of time and a formula for finding the amount of time for any quantity of water.
# Summary of the Math: Summarize Your Learning
Summarize the mathematics you learned about graphing rates.
## Hint:
• Does your summary explain what each point on your graph represents?
• Does your summary include the term constant rate ?
• Does your summary describe the two rates for each situation and their related formulas?
# Lesson Guide
This task allows you to assess students’ work and determine what difficulties they are having. The results of the Self Check will help you determine which students should work on the Gallery problems and which students would benefit from review before the assessment. Have students work on the Self Check individually.
# Assessment
Have students submit their work to you. Make notes about what their work reveals about their current levels of understanding and their different problem-solving approaches. Do not score students’ work. Share with each student the most appropriate Interventions to guide their thought process. Also note students with a particular issue so that you can work with them in the Putting It Together lesson that follows.
ELL: It might be helpful for ELLs and other students to have a sample or model for the concepts, strategies, and applications that will be addressed in the quiz and the format you want them to follow. Be prepared to address and explicitly re-teach or review vocabulary, concepts, strategies, and applications.
SWD: Students with disabilities may benefit from having the Self Check presented in a variety of ways (as auditory and visual information). Provide students the option of listening to the content for this task. Some students with disabilities may need additional time to complete the Self Check. Be sure to make provisions for the additional time or consider reducing the number of tasks required for students to demonstrate mastery of skill.
# Interventions
Student has trouble getting started
• The scale of the graph is in inches; which animal is more likely to travel just one inch at a time?
Student does not know how to write a formula.
• Look at the formulas you wrote in the previous lessons. How do those formulas relate to this situation?
• What is the general formula to find distance? How can you use this to help you write the formula for this situation?
• What is the general formula to find time? How can you use this to help you write the formula for this situation?
1. The graph most likely shows the speed of a garden snail.
2. In 8 minutes, the creature traveled 0.4 inch.
3. In 3 minutes, the creature traveled 0.15 inch.
4. The creature's rate of speed is 0.05 inch per minute.
5. $d=12\text{\hspace{0.17em}}\mathrm{min}\cdot 0.05\frac{\mathrm{in.}}{\mathrm{min}}=0.6\text{\hspace{0.17em}}\mathrm{in.}$ In 12 minutes, the creature travels 0.6 inch.
6. $t=2\text{\hspace{0.17em}}\mathrm{in.}÷0.05\frac{\mathrm{in.}}{\mathrm{min}}=40\text{\hspace{0.17em}}\mathrm{min}$. It takes the creature 40 minutes to travel 2 inches.
## Formative Assessment
Use the graph to answer the questions.
1. Does this graph most likely show the speed of a garden snail or a cheetah?
2. How far did this creature travel in 8 minutes?
3. How far did this creature travel in 3 minutes?
4. What is this creature’s rate of speed in inches per minute?
5. Solve an equation to find the distance the creature travels in 12 minutes.
6. Solve an equation to find the time it takes the creature to travel 2 inches. |
# NCERT Class 9 Solutions: Number Systems (Chapter 1) Exercise 1.2
Q.1 State whether the following statements are true or false. Justify your answer
1. Every irrational number is a real number.
2. Every point on the number line is of the form , where m is a natural number.
3. Every real number is an irrational number.
Solution:
1. 1st statement is true. Irrational numbers are those which cannot be expressed accurately in the form , i.e. as a fraction. Irrational numbers, when written as decimal numbers, do not terminate, nor do they repeat. Rational numbers on the other hand can be expressed as a fraction. Note that is a repeating decimal, therefore it can be accurately and exactly expressed as a fraction and hence is a rational number. All the rational and irrational numbers are contained in the set of real numbers.
2. 2nd statement is wrong. Every point on the number line is not a square root of a natural number. For example is a point on the number line. It is not square root of a natural number. It is square root of . Similarly 0.5 is the square root of 0.25 and so on.
3. As we explained in 1, 3rd statement is wrong. Even all rational numbers are real numbers.
Q.2 Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.
Solution:
• Square roots of all positive integers are not irrational.
• Example 4, 9, 16, 25, etc. are a positive integers
• And their square roots are 2, 3, 4, 5 which are rational numbers
Q.3 Show how can be represented on the number line.
Solution:
The idea of this construction is to use the Pythagoras theorem which states that in a right triangle:
Another way to understand this is that the area of square drawn at hypotenuse would be the sum of areas of squares drawn on the two sides.
In the following construction we construct a right triangle with two sides whose square add to be 5. Now,.
Step: 1 – Let AB be a line of length 2 inch on number line
Step: 2 – B draw a perpendicular line BC of length 1 inch, then join CA
Step: 3 – in Triangle ABC is a right angled triangle, Applying Pythagoras theorem
Step: 4 – Taking CA as a radius and A as a center draw an arc touching the number line. Since any point on the circumference of a circle is ‘radius’ distance away from the center. Therefore, the point at which number line will get intersected by arc is at distance from 0 because it is a radius of the circle whose center was A.
Explore Solutions for Mathematics |
# 3.6.1: Cardinality
In counting, as it is learned in childhood, the set {1, 2, 3, . . . , n} is used as a typical set that contains elements. In mathematics and computer science, it has become more common to start counting with zero instead of with one, so we define the following sets to use as our basis for counting:
$$N_{0}=\varnothing$$, a set with 0 elements
$$N_{1}=\{0\}$$, a set with 1 element
$$N_{2}=\{0,1\}$$, a set with 2 elements
$$N_{3}=\{0,1,2\}$$, a set with 3 elements
$$N_{4}=\{0,1,2,3\}$$, a set with 4 elements
and so on. In general, $$N_{n}=\{0,1,2, \ldots, n-1\}$$ for each $$n \in \mathbb{N} .$$ For each natural number $$n, N_{n}$$ is a set with $$n$$ elements. Note that if $$n \neq m,$$ then there is no one-to-one correspondence between $$N_{n}$$ and $$N_{m} .$$ This is obvious, but like many obvious things is not all that easy to prove rigorously, and we omit the argument here.
Theorem 4.6.
For each $$n \in \mathbb{N},$$ let $$N_{n}$$ be the set $$N_{n}=\{0,1, \ldots, n-1\} .$$ If $$n \neq m,$$ then there is no bijective function from $$N_{m}$$ to $$N_{n}$$.
We can now make the following definitions:
Definition4.3.
A set A is said to be finite if there is a one-to-one correspondence between $$A$$ and $$N_{n}$$ for some natural number $$n .$$ We then say that $$n$$ is the cardinality of $$A .$$ The notation |A| is used to indicate the cardinality of A. That is, if A is a finite set, then|A| is the natural number n such that there is a one-to-one correspondence between A and $$N_{n} .$$ In layman's terms: $$|A|$$ is the number items in $$A .$$ A set that is not finite is said to be infinite. That is, a set B is infinite if for every n ∈ N, there is no one-to-one correspondence between $$B$$ and $$N_{n}$$.
Fortunately, we don’t always have to count every element in a set individually to determine its cardinality. Consider, for example, the set × B, where and are finite sets. If we already know |A| and |B|, then we can determine |× B| by computation, without explicit counting of elements. In fact, |× B| = |A| · |B|. The cardinality of the cross product × can be computed by multiplying the cardinality of by the cardinality of B. To see why this is true, think of how you might count the elements of A × B. You could put the elements into piles, where all the ordered pairs in a pile have the same first coordinate. There are as many piles as there are elements of A, and each pile contains as many ordered pairs as there are elements of B. That is, there are |A| piles, with |B| items in each. By the definition of multiplication, the total number of items in all the piles is |A| · |B|. A similar result holds for the cross product of more that two finite sets. For example, |× × C| = |A| · |B| · |C|.
It’s also easy to compute |∪ B| in the case where and are disjoint finite sets. (Recall that two sets and are said to be disjoint if they have no members in common, that is, if ∩ = ∅.) Suppose |A| = and |B| = m. If we wanted to count the elements of ∪ B, we could use the numbers from 0 to − 1 to count the elements of and then use the numbers from to n+m−1 to count the elements of B. This amounts to a one-to-one correspondence between $$A \cup B$$ and the set $$N_{n+m} .$$ We see that $$|A \cup B|=n+m$$ That is, for disjoint finite sets and B, |∪ B| = |A| + |B|.
What about ∪ B, where and are not disjoint? We have to be careful not to count the elements of ∩ twice. After counting the elements of A, there are only |B| − |∩ B| new elements in that still need to be counted. So we see that for any two finite sets and B, |∪ B| = |A| + |B| − |∩ B|.
What about the number of subsets of a finite set A? What is the relationship between|A| and |P(A)|? The answer is provided by the following theorem.
Theorem 4.7.
A finite set with cardinality $$n$$ has $$2^{n}$$ subsets.
Proof. Let $$P(n)$$ be the statement "Any set with cardinality $$n$$ has $$2^{n}$$ subsets". We will use induction to show that P(n) is true for all ∈ N.
Base case: For $$n=0, P(n)$$ is the statement that a set with cardinality 0 has $$2^{0}$$ subsets. The only set with 0 elements is the empty set. The empty set has exactly 1 subset, namely itself. Since $$2^{0}=1, P(0)$$ is true.
Inductive case: Let be an arbitrary element of N, and assume that P(k) is true. That is, assume that any set with cardinality $$k$$ has $$2^{k}$$ elements. (This is the induction hypothesis.) We must show that P(+ 1) follows from this assumption. That is, using the assumption that any set with cardinality has 2subsets, we must show that any set with cardinality $$k+1$$ has $$2^{k+1}$$ subsets.
Let $$A$$ be an arbitrary set with cardinality $$k+1 .$$ We must show that $$|\mathcal{P}(A)|=2^{k+1}$$ \. Since |A| > 0, contains at least one element. Let be some element of A, and letB ∖ {x}. The cardinality of is k, so we have by the induction hypothesis that $$|\mathcal{P}(B)|=2^{k} .$$ Now, we can divide the subsets of $$A$$ into two classes: subsets of $$A$$ that do not contain and subsets of that do contain x. Let be the collection of subsets ofA that do not contain x, and let be the collection of subsets of that do contain xand are disjoint, since it is impossible for a given subset of both to contain and to not contain x. It follows that |P(A)| = |∪ Y| = |X| + |Y|.
Now, a member of is a subset of that does not contain x. But that is exactly the same as saying that a member of is a subset of B. So = P(B), which we know contains $$2^{k}$$ members. As for $$X,$$ there is a one-to-one correspondence between $$\mathcal{P}(B)$$ and X. Namely, the function : P(B) → defined by (C) = ∪ {x} is a bijective function. (The proof of this is left as an exercise.) From this, it follows that $$|X|=|\mathcal{P}(B)|=2^{k}$$ Putting these facts together, we see that $$|\mathcal{P}(A)|=|X|+|Y|=2^{k}+2^{k}=2 \cdot 2^{k}=2^{k+1}$$ This completes the proof that P(k) → P(+ 1).
We have seen that the notation $$A^{B}$$ represents the set of all functions from $$B$$ to $$A$$ Suppose $$A$$ and $$B$$ are finite, and that $$|A|=n$$ and $$|B|=m .$$ Then $$\left|A^{B}\right|=n^{m}=|A|^{B |}$$. (This fact is one of the reasons why the notation $$A^{B}$$ is reasonable.) One way to see this is to note that there is a one-to-one correspondence between $$A^{B}$$ and a cross product A×A×···A, where the number of terms in the cross product is m. (This will be shown in one of the exercises at the end of this section.) It follows that ����$$\left|A^{B}\right|=|A| \cdot|A| \cdots|A|=$$ n · · · · n, where the factor occurs times in the product. This product is, by definition, $$n^{m}$$
This discussion about computing cardinalities is summarized in the following theorem:
Theorem 4.8.
Let and be finite sets. Then
• |A×B|=|A|·|B|.
• |AB| = |A|+|B|−|AB|.
• If and are disjoint then|AB|=|A|+|B|.
$$\bullet\left|A^{B}\right|=|A|^{B |}$$
$$\bullet|\mathcal{P}(A)|=2^{|A|}$$
When it comes to counting and computing cardinalities, this theorem is only the be- ginning of the story. There is an entire large and deep branch of mathematics known as combinatorics that is devoted mostly to the problem of counting. But the theorem is already enough to answer many questions about cardinalities.
For example, suppose that |A| = and |B| = m. We can form the set P(× B), which
consists of all subsets of × B. Using the theorem, we can compute that |P(× B)| =$$2^{|A \times B|}=2^{|A| \cdot|B|}=2^{n m}$$. If we assume that and are disjoint, then we can compute that $$\left|A^{A \cup B}\right|=|A|^{|A \cup B|}=n^{n+m}$$.
To be more concrete, let = {a,b,c,d,e} and let = {c,d,ef} where a,bcde, and are distinct. Then |X×Y| = 5·4 = 20 while |XY| = $$5+4-|\{c, d, e\}|=6$$ and $$\left|X^{Y}\right|=5^{4}=625$$
We can also answer some simple practical questions. Suppose that in a restaurant you can
choose one appetizer and one main course. What is the number of possible meals? If Ais the set of possible appetizers and is the set of possible main courses, then your meal is an ordered pair belonging to the set × C. The number of possible meals is |× C|, which is the product of the number of appetizers and the number of main courses.
Or suppose that four different prizes are to be awarded, and that the set of people who are eligible for the prizes is A. Suppose that |A| = n. How many different ways are there to award the prizes? One way to answer this question is to view a way of awarding the prizes as a function from the set of prizes to the set of people. Then, if is the set of prizes, the number of different ways of awarding the prizes is $$\left|A^{P}\right| .$$ since $$|P|=4$$ and $$|A|=n,$$ this is $$n^{4} .$$ Another way to look at it is to note that the people who win the prizes form an ordered tuple (abcd), which is an element of × × × A. So the number of different ways of awarding the prizes is |× × × A|, which is |A| · |A| · |A| · |A|. This is$$|A|^{4},$$ or $$n^{4}$$, the same answer we got before. |
# What is the vertex of f(x)=x^2+4x-5?
Sep 4, 2014
There are several ways to find the vertex of a parabola:
1. Complete the square
2. Use $\frac{- b}{2 a}$
3. Find the average of the two x-intercepts, and substitute that x-value in to find the y-value. (f(x))
4. Use a graphing calculator and the "analyze" feature
1. Here is how I would complete the square:
Replace f(x) with "y" and make sure that the lead coefficient = 1
$y = {x}^{2} + 4 x - 5$
$y + 5 = {x}^{2} + 4 x$
y + 5 + ???_ = ${x}^{2}$ + 4x + _???__
(take half of the coefficient of the x-term, then square it)
so y + 5 + 4 = ${x}^{2}$ + 4x + 4
y + 9 = ${\left(x + 2\right)}^{2}$ (the right side is a perfect square trinomial!)
Then set y + 9 = 0 and x + 2 = 0 to get y = -9 and x = -2 which are the coordinates of the vertex (-2, -9).
Method 2 involves calculating $\frac{- b}{2 a}$ from your standard form equation as given. That is, a = 1, b = 4 and c = -5.
x = $\frac{- 4}{\left(2 \cdot 1\right)}$ = $\frac{- 4}{2}$ = -2.
Now that you have the x-value for the vertex, substitute it into the function and find f(-2) = ${\left(- 2\right)}^{2} + 4 \left(- 2\right) - 5$ = 4 + (-8) - 5 = -9.
Confirmed, (-2,-9) is the vertex.
Method 3. This quadratic is factorable: ${x}^{2} + 4 x - 5 = \left(x + 5\right) \left(x - 1\right)$
So the zeros are x = -5 and 1. Average the zeros: $\frac{- 5 + 1}{2}$ = $\frac{- 4}{2}$ = -2. Repeat finding f(-2) to get -9.
Method 4 involves using technology and not showing any work to your instructor. Is this really how you want to do it? Here is what my calculator would show: |
The Relation Between Integration and Differentiation
An interesting article: Calculus for Dummies by John Gabriel
The derivative of an indefinite integral. The first fundamental theorem of calculus
We corne now to the remarkable connection that exists between integration and differentiation. The relationship between these two processes is somewhat analogous to that which holds between “squaring” and “taking the square root.” If we square a positive number and then take the positive square root of the result, we get the original number back again. Similarly, if we operate on a continuous function f by integration, we get a new function (an indefinite integral off) which, when differentiated, leads back to the original function f. For example, if f(x) = x2, then an indefinite integral A off may be defined by the equation
$A(x)=\int_c^x f(t) \ dt = \int_c^x t^2 \ dt = \frac{x^3}{3} - \frac{c^3}{3},$
where c is a constant. Differentiating, we find A'(x) = x2 = f(x). This example illustrates a general result, called the first fundamental theorem of calculus, which may be stated as follows:
First fundamental theorem of calculus
Let f be a function that is integrable on [a, x] for each x in [a, b]. Let c be such that a ≤ c ≤ b and define a new function A as follows:
$A(x)=\int_c^x f(t) \ dt, \qquad \qquad a \leq x \leq b$
Then the derivative A'(x) exists at eachpoint x in the open interval (a, b) where f is continuous, andfor such x we have
(5.1) A'(x) = f(x).
First we give a geometric argument which suggests why the theorem ought to be true; then we give an analytic proof.
Geometric motivation. Figure 5.1 shows the graph of a function f over an interval [a, b]. In the figure, h is positive and
$\int_x^{x+h} f(t) \ dt = \int_c^{x+h} f(t) \ dt - \int_c^x f(t) \ dt = A(x+h) - A(x)$
The example shown is continuous throughout the interval [x, x + h]. Therefore, by the mean-value theorem for integrals, we have
A(x + h) - A(x) = hf(Z), where x ≤ z ≤ x + h.
Hence we have
(5.2) [A(x + h) - A(x)]/h = f(z),
and, since x ≤ z ≤ x + h, we find that f(z) → f(x) as h → 0 through positive values. A similar argument is valid if h → 0 through negative values. Therefore, A'(x) exists and is equal to f (x).
This argument assumes that the function f is continuous in some neighbourhood of the point x. However, the hypothesis of the theorem refers only to continuity off at a single point x. Therefore, we use a different method to prove the theorem under this weaker hypothesis.
Analytic Proof. Let x be a point of continuity off, keep x fixed, and form the quotient
[A(x + h) - A(x)]/h
To prove the theorem we must show that this quotient approaches the limit f(x) as h → 0. The numerator is
$A(x+h) - A(x) = \int_c^{x+h} f(t) \ dt - \int_c^x f(t) \ dt = \int_x^{x+h} f(t) \ dt.$
If we write f(t) =f(x) + [f(t) -f(x)] in the last integral, we obtain
from which we find
(5.3) $\frac{A(x+h) - A(x)}{h} = f(x) + \frac{1}{h} \int_x^{x+h} [f(t) - f(x)] \ dt$
Therefore, to complete the proof of (5.1), all we need to do is show that
$\lim_{h\rightarrow 0} \ \frac{1}{h} \int_x^{x+h} [f(t) - f(x)] \ dt = 0$
It is this part of the proof that makes use of the continuity off at x.
Let us denote the second term on the right of (5.3) by G(h). We are to prove that G(h) -f 0 as h --f 0. Using the definition of limit, we must show that for every ε > 0 there is a δ > 0 such that
(5.4) |G(h)| < ε whenever 0 < |h| < δ.
Continuity of f at x tells us that, if ε is given, there is a positive δ such that
(5.5) |f(t) -f(x)| < ε/2
whenever(5.6) x - δ < t < x + δ.
If we choose h so that 0 < h < δ, then every t in the interval [x, x + h] satisfies (5.6) and hence (5.5) holds for every such t. Using the property $|\int _x^{x+h} g(t) \ dt| \ \leq \ \int _x^{x+h} |g(t)| \ dt$ with g(t) =f(t) -f(x), we see that the inequality in (5.5) leads to the relation
$|\int _x^{x+h} [f(t) - f(x)] \ dt | \leq \int _x^{x+h} |f(t) - f(x)| \ dt \leq \int _x^{x+h} \frac{1}{2} \epsilon \ dt = \frac{1}{2} h \epsilon < h \epsilon$
If we divide by h, we see that (5.4) holds for 0 < h < δ. If h < 0, a similar argument proves that (5.4) holds whenever 0 < Ihl < 6, and this completes the proof.
The zero-derivative theorem
If a function f is constant on an open interval (a, b), its derivative is zero everywhere on (a, b). We proved this fact earlier as an immediate consequence of the definition of derivative. We also proved, as part (c) of Theorem 4.7, the converse of this statement which we restate here as a separate theorem.
THEOREM 5.2. ZERO-DERIVATIVE THEOREM. If f'(x) = 0 for each x in an open interval I, then f is constant on I.
This theorem, when used in combination with the first fundamental theorem of calculus, leads to the second fundamental theorem which is described in the next section.
5.3 Primitive functions and the second fundamental theorem of calculus
Definition of primitive function
A function P is called a primitive (or an antiderivative) of a function f on an open interval I if the derivative of P is f, that is, if P'(x) = f (x) for all x in I.
For example, the sine function is a primitive of the cosine on every interval because the derivative of the sine is the cosine. We speak of a primitive, rather than the primitive, because if P is a primitive offthen so is P + k for every constant k. Conversely, any two primitives P and Q of the same function f cari differ only by a constant because their difference P - Q has the derivative
P'(x) - Q'(x) = f(x) - f(x) = 0
for every x in I and hence, by Theorem 5.2, P - Q is constant on Z.
The first fundamental theorem of calculus tells us that we cari always construct a primitive of a continuous function by integration. When we combine this with the fact that two primitives of the same function can differ only by a constant, we obtain the second fundamental theorem of calculus.
Second fundamental theorem of calculus
Assume f is continuous on an open interval I, and let P be any primitive off on I. Then, for each c and each x in I, we have
(5.7) $P(x) = P(c) + \int _c^x f(t) \ dt$
Proof. Let $A(x) = \int _c^x f(t) \ dt$ Since f is continuous at each x in I, the first fundamental theorem tells us that A'(x) = f(x) for all x in Z. In other words, A is a primitive off on Z. Since two primitives off can differ only by a constant, we must have A(x) - P(x) = k for some constant k. When x = c, this formula implies -P(c) = k, since A(c) = 0. Therefore, A(x) - P(x) = -P(c), from which we obtain (5.7).
Theorem 5.3 tells us how to find every primitive P of a continuous function f. We simply integrate f from a fixed point c to an arbitrary point x and add the constant P(c) to get P(x). But the real power of the theorem becomes apparent when we write Equation (5.7) in the following form:
(5.8) $\int _c^x f(t) \ dt = P(x) - P(c)$
In this form it tells us that we cari compute the value of an integral by a mere subtraction if we know a primitive P. The problem of evaluating an integral is transferred to another problem-that of finding a primitive P off. In actual practice, the second problem is a great deal casier to deal with than the first. Every differentiation formula, when read in reverse, gives us an example of a primitive of some functionfand this, in turn, leads to an integration formula for this function.
From the differentiation formulas worked out thus far we cari derive the following integration formulas as consequences of the second fundamental theorem.
EXAMPLE 1. Integration of rational powers. The integration formula
(5.9) $\int _a^b x^n \ dx = \frac{b^{n+1}-a^{n+1}}{n+1}$ (n = 0, 1, 2, . . .)
was proved in Section 1.23 directly from the definition of the integral. The result may be rederived and generalized to rational exponents by using the second fundamental theorem. First of all, we observe that the function P defined by the equation
(5.10) P(x) = (xn + 2)/(n + 1)
has the derivative P'(x) = xn if n is any non-negative integer. Since this is valid for all real x, we may use (5.8) to write
$\int _a^b x^n \ dx = P(b) - P(a) = \frac{b^{n+1}-a^{n+1}}{n+1}$
for all intervals [a, b]. This formula, proved for all integers n ≥ 0, also holds for all negative integers except n = -1, which is excluded because n + 1 appears in the denominator. To prove (5.9) for negative n, it suffices to show that (5.10) implies P'(x) = xn when n is negative and ≠ - 1, a fact which is easily verified by differentiating P as a rational function. Of course, when n is negative, neither P(x) nor P'(x) is defined for x = 0, and when we use (5.9) for negative n, it is important to exclude those intervals [a, b] that contain the point x = 0.
The results of Example 3 in Section 4.5 enable us to extend (5.9) to all rational exponents (except -l), provided the integrand is defined everywhere on the interval [a, b] under consideration. For example, if 0 < a < b and n = -1/2, we find
$\int _a^b \frac{1}{\sqrt{x}} \ dx = \int _a^b x^{-\frac{1}{2}} \ dx = \frac{x^{\frac{1}{2}}}{\frac{1}{2}}|_a^b = 2(\sqrt{b} - \sqrt{a})$
This result was proved earlier, using the area axioms. The present proof makes no use of these axioms.
In the next chapter we shall define a general power function f such that f(x) = xc for every real exponent c. We shall find that this function has the derivative f'(x) = cxc - 1 and the primitive P(x) = xc + 1/(C + 1) if c ≠ - 1. This Will enable us to extend (5.9) to all real exponents except - 1.
Note that we cannot get P'(x) = 1/x by differentiation of any function of the form P(x) = xn. Nevertheless, there exists a function P whose derivative is P'(x) = 1/x. To exhibit such a function all we need to do is write a suitable indefinite integral; for example,
$P(x) = \int _x^c \frac{1}{t} \ dt \qquad \qquad if \ x > 0$
This integral exists because the integrand is monotonie. The function SO defined is called the Zogarithm (more specifically, the naturaf logarithm). Its properties are developed systematically in Chapter 6.
EXAMPLE 2. Integration of the sine and cosine. Since the derivative of the sine is the cosine and the derivative of the cosine is minus the sine, the second fundamental theorem also gives us the following formulas:
$\int _a^b \cos x \ dx = \sin x |_a^b = \sin b - \sin a$
$\int _a^b \sin x \ dx = (-\cos x) |_a^b = \cos a - \cos b$
Further examples of integration formulas cari be obtained from Examples 1 and 2 by taking finite sums of terms of the form Ax'“, B sin x, C COS x, where A, B, C are constants.
5.4 Properties of a function deduced. from properties of its derivative
If a function f has a continuous derivative f' on an open interval Z, the second fundamental theorem states that
(5.11) $f(x) = f(c) + \int _c^x f'(t) \ dt$
for every choice of points x and c in Z. This formula, which expresses f in terms of its derivative f ‘, enables us to deduce prolperties of a function from properties of its derivative. Although the following properties have already been discussed in Chapter 4, it may be of interest to see how they cari also be deduced as simple consequences of Equation (5.11).
Suppose f' is continuous and non-negative on I. If x > c, then $\int _c^x f'(t) \ dt \geq 0$, and hence f(x) ≥ f(c). In other words, if the Iderivative is continuous and nonnegative on Z, the function is increasing on Z.
In Theorem 2.9 we proved that the indefinite integral of an increasing function is convex. Therefore, if f' is continuous and increasing on 1, Equation (5.11) shows that f is convex on Z. Similarly, f is concave on those intervals where f' is continuous and decreasing.
Exercises
In each of Exercises 1 through 10, find a primitive of f; that is, find a function P such that P'(x) = f(x) and use the second fundamental theorem to evaluate $\int _a^b f'(x) \ dx$
1. f(x) = 5x3. 6. f(x) = √2x + √x/2, x > 0.
2. f(x) = 4x4 - 12x. 7. f(x) = [2x2 - 6x + 7]/2√2x, x > 0.
3. f(x) = (x + 1)(x3 - 2). 8. f(x) = 2x1/3 - x-1/3, x > 0.
4. f(x) =[x4 +x - 3]/x3 , x ≠ 0. 9. f(x) = 3sinx + 2x5.
5. f(x) = (1 + √x)2, x > 0. 10. f(x) = x4/3 - 5cosx.
11. Prove that there is no polynomial f whose derivative is given by the formula f'(x) = 1/x.
12. Show that $\int _0^x |t| \ dt = \frac{1}{2} x |x|$ for all real x.
13. Show that
$\int _0^x (t + |t|)^2 \ dt = \frac{2x^2}{3} (x + |x|)$ for all real x .
14. A function f is continuous everywhere and satisfies the equation
$\int _0^x f(t) \ dt = -\frac{1}{2} + x^2 + x\sin 2x + \frac{1}{2} \cos 2x$
for all x. Compute f(π/4) and f'(π/4).
15. Find a function f and a value of the constant c such that
$\int _c^x f(t) \ dt = \cos x - \frac{1}{2}$ for all real x .
16. Find a function f and a value of the constant c such that
$\int _c^x tf(t) \ dt = \sin x - x\cos x - \frac{1}{2}x^2$ for all real x .
17. There is a function J defined and continuous for all real x, which satisfies an equation of the form
$\int _0^x f(t) \ dt = \int _x^1 t^2f(t) \ dt + \frac{x^{16}}{8} + \frac{x^{18}}{9} +c$
where c is a constant. Find an explicit formula for f (x) and find the value of the constant c.
18. A function f is defined for all real x by the formula
$f(x) = 3 + \int _0^1 \frac{1 + \sin t}{2 + t^2} dt$
Without attempting to evaluate this integral, find a quadratic polynomial $p(x) = a + bx + cx^2$ such that p(0) = f(0), p'(0) =f'(0), and p''(0) =f''(0).
19. Given a function g, continuous everywhere, such that g( 1) = 5 and $\int _0^1 g(t) \ dt = 2$. Let f(x) = $\frac{1}{2} \int _0^x (x-t)^2 g(t) \ dt$. Prove that
$f'(x) = x \int _0^x g(t) \ dt - \int _0^x tg(t) \ dt$
then compute f''( 1) and f'''( 1).
20. Without attempting to evaluate the following indefinite integrals, find the derivative f'(x) when f(x) is equal to
$\int _0^{x^2} (1+t^2)^{-3} \ dt$.
21. Without attempting to evaluate the integral, compute f'(x) if f is defined by the formula
$f(X) = \int _{x^3}^{x^2} \frac{t^6}{1+t^4} dt$
22. Compute f(2) if f is continuous and satisfies the given formula for all x ≥ 0.
$\int _0^{f(x)} t^2 \ dt = x^2 (1+x)$
23. The base of a solid is the ordinate set of a nonnegative functionf'over the interval [0, a]. All cross sections perpendicular to this interval are squares. The volume of the solid is
a3 - 2acosa + (2 - a2)sina
for every a ≥ 0. Assume f is continuous on [0, a] and calculate f(a).
24. A mechanism propels a particle along a straight line. It is designed so that the displacement of the particle at time t from an initial point 0 on the line is given by the formula f(t) = t2/2 + 2tsint. The mechanism works perfectly until time t = π when an unexpected malfunction occurs. From then on the particle moves with constant velocity (the velocity it acquires at time t = π). Compute the following: (a) its velocity at time t = π; (b) its acceleration at time t = π/2 ; (c) its acceleration at time t = 3π/2; (d) its displacement from 0 at time t = 5π/2.(e) Find a time t > π when the particle returns to the initial point 0, or else prove that it never returns to 0.
25. A particle moves along a straight line. Its position at time t is,f(t). When 0 ≤ t ≤ 1, the position is given by the integral
$f(t) = \int _0^t \frac{1+2\sin XX \cos \pi x}{1+x^2} dx$
(Do not attempt to evaluate this integral.) For t ≤ 1, the particle moves with constant acceleration (the acceleration it acquires at time t = 1). Compute the following: (a) its acceleration at time t = 2; (b) its velocity when t = 1; (c) its velocity when t > 1; (d) the difference f(t) -f(l) when t > 1.
26. In each case, find a function f with a continuous second derivative f'' which satisfies all the given conditions or else explain why auch an example cannot exist.
(a) f''(x) > 0 for every x, f'(0) = 1, f'(1) = 0.
(b) f''(x) > 0 for every x, f'(0) = 1, f'( 1) = 3.
(c) f''(x) > 0 for every x, f'(0) = 1, f(x) ≤ 100 for all x > 0.
(d) f''(x) > 0 for every x, f'(0) = 1, f(x) ≤ 100 for all x < 0.
27. A particle moves along a straight line, its position at time t being f(t). It starts with an initial velocity f(0) = 0 and has a continuous acceleration f''(t) ≥ 6 for all t in the interval 0 ≤ t ≤ 1.Prove that the velocity f'(t) ≥ 3 for all t in some interval [a, b], where 0 ≤ a < b ≤ 1, with b - a = + .
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## How do you describe arithmetic and geometric sequence?
An arithmetic sequence has a constant difference between each consecutive pair of terms. This is similar to the linear functions that have the form y=mx+b. A geometric sequence has a constant ratio between each pair of consecutive terms.
## What is arithmetic sequence summary?
An arithmetic sequence is a sequence where the difference between any two consecutive terms is a constant. The constant between two consecutive terms is called the common difference. The common difference is the number added to any one term of an arithmetic sequence that generates the subsequent term.
What are arithmetic and geometric sequences used for?
Difference between an arithmetic sequence and a geometric sequence
S.No. Arithmetic sequence Geometric sequence
3 Subtraction or addition are used to get terms. Division or Multiplication are used to get terms.
4 Example: 5, 11, 17, 23, 29, 35,… Example: 2, 6, 18, 54, 162,…
### Why is it important to learn arithmetic and geometric sequence?
The arithmetic sequence is important in real life because this enables us to understand things with the use of patterns. An arithmetic sequence is a great foundation in describing several things like time which has a common difference of 1 hour. An arithmetic sequence is also important in simulating systematic events.
### What is arithmetic and geometric?
Meaning. Arithmetic Sequence is described as a list of numbers, in which each new term differs from a preceding term by a constant quantity. Geometric Sequence is a set of numbers wherein each element after the first is obtained by multiplying the preceding number by a constant factor.
What is the meaning of geometric sequence?
A geometric sequence goes from one term to the next by always multiplying or dividing by the same value. The number multiplied (or divided) at each stage of a geometric sequence is called the common ratio. Examples of geometric sequences are the frequencies of musical notes and interest paid by a bank.
#### What is the importance of sequence in real life?
As we discussed earlier, Sequences and Series play an important role in various aspects of our lives. They help us predict, evaluate and monitor the outcome of a situation or event and help us a lot in decision making.
#### What is the importance of arithmetic in our daily life?
Preparing food. Figuring out distance, time and cost for travel. Understanding loans for cars, trucks, homes, schooling or other purposes. Understanding sports (being a player and team statistics)
What is the similarities of arithmetic sequence and geometric sequence?
The arithmetic sequences and geometric sequences are similar because they follow a pattern. In arithmetic sequences, the same number is either added or subtracted to obtain the next number. Similarly, in geometric sequences, the same number is either multiplied or divided by to obtain the next number. |
# 7th Grade NYSE Math FREE Sample Practice Questions
Practicing is undoubtedly the most important factor for success in the 7th Grade NYSE Math test! If your 7th Grade student needs sample practice questions to succeed in the 7th Grade NYSE Math exam, you have found the best article! In this article, we want to acquaint your students with the best 7th Grade NYSE Math sample practice questions. Also, the solutions to all the questions are given in the article so that the students become familiar with how to solve the questions.
Start preparing your student for the 2021 NYSE Math test with our free sample practice questions. Also, make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions students need to practice.
## 10 Sample 7th Grade NYSE Math Practice Questions
1- A chemical solution contains $$4\%$$ alcohol. If there is 24 ml of alcohol, what is the volume of the solution?
☐A. 240 ml
☐B. 480 ml
☐C. 600 ml
☐D. 1200 ml
2- The price of a laptop is decreased by $$10\%$$ to $360. What is its original price? ☐A. 320 ☐B. 380 ☐C. 400 ☐D. 450 3- What is the median of these numbers? $$4, 9, 13, 8, 15, 18, 5$$ ☐A. 8 ☐B. 9 ☐C. 13 ☐D. 15 4- Three times the price of a laptop is equal to five times the price of a computer. If the price of laptop is$200 more than the computer, what is the price of the computer?
☐A. 300
☐B. 500
☐C. 800
☐D. 1500
5- What is the perimeter of a square that has an area of 595.36 feet?
6- Jason is 9 miles ahead of Joe running at 5.5 miles per hour and Joe is running at the speed of 7 miles per hour. How long does it take Joe to catch Jason?
☐A. 3 hours
☐B. 4 hours
☐C. 6 hours
☐D. 8 hours
7- 55 students took an exam and 11 of them failed. What percent of the students passed the exam?
☐A. $$20 \%$$
☐B. $$40 \%$$
☐C. $$60 \%$$
☐D. $$80 \%$$
8- Jason needs an $$75\%$$ average in his writing class to pass. On his first 4 exams, he earned scores of $$68\%, 72\%, 85\%,$$ and $$90\%$$. What is the minimum score Jason can earn on his fifth and final test to pass?
9- A bank is offering $$3.5\%$$ simple interest on a savings account. If you deposit $12,000, how much interest will you earn in two years? ☐A.$420
☐B. $840 ☐C.$4200
☐D. $8400 10- 5 less than twice a positive integer is 83. What is the integer? ☐A. 39 ☐B. 41 ☐C. 42 ☐D. 44 ## Best 7th Grade NYSE Math Workbook Resource for 2021 ## Answers: 1- C $$4\%$$ of the volume of the solution is alcohol. Let $$x$$ be the volume of the solution. Then: $$4\% \space$$ of $$\space x = 24 \space$$ ml $$⇒ 0.04 x = 24 ⇒ x = 24 ÷ 0.04 = 600$$ 2- C Let $$x$$ be the original price. If the price of a laptop is decreased by $$10\%$$ to$360, then:
$$90 \%$$of $$x=360 ⇒ 0.90x=360 ⇒ x=360÷0.90=400$$
3- B
Write the numbers in order:
$$4, 5, 8, 9, 13, 15, 18$$
Since we have 7 numbers (7 is odd), then the median is the number in the middle, which is 9.
4- A
Let L be the price of laptop and C be the price of computer.
$$3$$(L) $$=5$$(C) $$\space$$ and $$\space$$ L $$= 200 +$$ C
Therefore, $$3(200 +$$ C$$)=5$$C $$⇒ 600 + 3$$C $$= 5$$C $$⇒$$ C$$=300$$
5- 97.6
Use the area of square formula.
$$S = a^2 ⇒ 595.36 = a^2 ⇒ a = 24.4$$
Use the perimeter of square formula.
$$P = 4a ⇒ P=4(24.4) ⇒ P = 97.6$$
6- C
The distance between Jason and Joe is 9 miles. Jason running at 5.5 miles per hour and Joe is running at the speed of 7 miles per hour. Therefore, every hour the distance is 1.5 miles less.
$$9 ÷ 1.5 = 6$$
7- D
The failing rate is 11 out of $$55 = \frac{11}{55}$$.
Change the fraction to percent:
$$\frac{11}{55} ×100\%=20\%$$
20 percent of students failed. Therefore, 80 percent of students passed the exam.
8- 60
Jason needs an $$75\%$$ average to pass for five exams. Therefore, the sum of 5 exams must be at lease $$5 × 75 = 375$$
The sum of 4 exams is:
$$68 + 72 + 85 + 90 = 315$$
The minimum score Jason can earn on his fifth and final test to pass is:
$$375 – 315 = 60$$
9- B
Use simple interest formula:
I=prt
(I = interest, p = principal, r = rate, t = time)
I$$=(12000)(0.035)(2)=840$$
10- D
Let $$x$$ be the integer. Then:
$$2x – 5 = 83$$
Add 5 both sides: $$2x = 88$$
Divide both sides by $$2: x = 44$$
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A taxi driver charges for a ride at a rate of 2 dollars for the first half mile and 75 cents for each additional half-mile. Which of the following expressions represents the total charge, in cents for $p$ miles, where $p$ is a positive integer?A. $25\left( {6p + 8} \right)$B. $25\left( {6p + 5} \right)$C. $25\left( {3p + 5} \right)$D. $1.25 + 1.5p$E. $2 + 0.75\left( {p - 1} \right)$
Last updated date: 10th Sep 2024
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Hint: First of all, calculate the charge for the first half-mile in cents. Then find out the remaining number of miles by subtracting a half-mile from the total number of miles i.e., $p$ miles. Further, calculate the charge for these remaining numbers of miles. Add up both the charges to get the total fare for $p$ miles. So, use this concept to reach the solution to the given problem.
Complete step-by-step solution:
Given that the charge for a ride in a certain city is 2 dollars for the first half of a mile and 75 cents for each additional half of a mile.
We have to calculate the charge for $p$ miles.
The charge for first half mile = 2 dollars = 200 cents
The charge for the next half miles = 75 cents
Or
The charge for next one mile = $2 \times 75 = 150$ cents
The remaining number of miles in $p$ miles of distance after the first half mile = $\left( {p - \dfrac{1}{2}} \right)$ miles.
The charge for the remaining $\left( {p - \dfrac{1}{2}} \right)$ miles = $\left( {p - \dfrac{1}{2}} \right) \times 150$ cents
Therefore, the total charge for the $p$ miles = charge for first half mile + charge for $\left( {p - \dfrac{1}{2}} \right)$ miles
$= 200 + \left( {p - \dfrac{1}{2}} \right) \times 150 \\ = 200 + 150p - \dfrac{{150}}{2} \\ = 200 + 150p - 75 \\ = 125 + 150p \\ = 25\left( {6p + 5} \right){\text{ cents}} \\$
Thus, the correct option is B. $25\left( {6p + 5} \right)$
Note: One dollar is equal to hundred cents. In this problem we have converted the charge for the next half miles in to charge for the next one mile. We can also solve this by converting the remaining $\left( {p - \dfrac{1}{2}} \right)$ miles in to number of half miles i.e., $\dfrac{{\left( {p - \dfrac{1}{2}} \right)}}{2}$ miles. Both the methods give the same answer. |
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Consider the function of the from. We shall the check the of the given function at the point x=4 . To check the continuity of the given function we follow the three steps. (i) Value of Function at given Point: We have given value of function at x=4 is equal to 0 .
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It nice seeing this,pls can u treat calculus
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In this tutorial we shall discuss an example to evaluating limits involving cubic expression, in most of the cases if limit involves cubic expression and we can factorize by using method of synthetic division. Let us consider an example which involve cubic expression. limit _ {x->1}(x^(3)+x ...
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In this tutorial we shall discuss an example of limit which involves quadratic function, and to find the value of limit we shall factorize the quadratic first and then solve it for the existence of limit. Let us consider an example which involve quadratic expression. limit _ {x->-1}(x^(2)+ ...
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Let us consider the relation. limit _{x->0} (a^(x)-1. Let y=a^(x)-1 , then 1+y=a^(x) , we have. Consider the relation. 1+y=a^(x). Taking logarithm on both sides, we have. ln (1+y)= ln a^(x) ln (1+y)=x ln a x=( ln (1+y))/( ln a). Also limit _{x- >0}y= limit _{x-> . This shows that y- >0 as x->0 .
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Let us consider the function f defined by the equation f(x)=(2x^(2))/(x^ . Let x take on the values 0, -1, -2, -3, -4, -5, -10, -100, -1000 etc, allowing to decreasing through negative values without bound. In this case, we say that x is approaching negative infinity and write x->- infinity ...
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A physical object that is moving cannot immdeiately disappear at some point and again appear from some where else to continue its motion. Thus, we perceive the path of a moving object as a single, unbroken curve without gaps, jumps or holes. Such curves can be described as continuous.
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In this tutorial we shall discuss an example to evaluating limits involving function with nth power of variable, in most of the cases if limit involves nth power variable expression and to solve using binomial theorem. Let us consider an example which is. limit _{x->a}(x^(n)-a ...
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Let us consider the relation. (1+(1)/(x))^(x). We shall prove this formula with the help of binomial series expansion, we have. (1+(1)/(x))^(x)= Taking limit as x-> infinity both sides, we get. limit _{x-> infinity }(1+(1)/(. Applying limits we have. limit _{x-> infinity }(1+(1)/( ...
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Let us consider the function f defined by the equation f(x)=(2x^(2))/(x^ . Let x take on the values 0, 1, 2, 3, 4, 5, 10, 100, 1000 etc, allowing increasing without bound. The corresponding function values are given in the given table below.
1 |
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# Unit Circle
## Determine exact values of trig ratios for common radian measures
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Trigonometric Ratios on the Unit Circle
What are the exact values of the following trigonmetric functions?
a. $\sin 495^\circ$
b. $\tan \frac{5\pi}{3}$
### Guidance
Recall special right triangles from Geometry. In a $(30^\circ - 60^\circ - 90^\circ)$ triangle, the sides are in the ratio $1:\sqrt{3}:2$ .
In an isosceles triangle $(45^\circ - 45^\circ - 90^\circ)$ , the congruent sides and the hypotenuse are in the ratio $1:1:\sqrt{2}$ .
In a $(30^\circ - 60^\circ - 90^\circ)$ triangle, the sides are in the ratio $1:\sqrt{3}:2$ .
Now let’s make the hypotenuse equal to 1 in each of the triangles so we’ll be able to put them inside the unit circle. Using the appropriate ratios, the new side lengths are:
Using these triangles, we can evaluate sine, cosine and tangent for each of the angle measures.
$& \sin 45^\circ = \frac{\sqrt{2}}{2} \qquad \quad \sin 60^\circ = \frac{\sqrt{3}}{2} \qquad \qquad \quad \ \ \sin 30^\circ = \frac{1}{2} \\& \cos 45^\circ = \frac{\sqrt{2}}{2} \qquad \ \ \cos 60^\circ = \frac{1}{2} \qquad \qquad \qquad \ \ \cos 30^\circ = \frac{\sqrt{3}}{2} \\& \tan 45^\circ = 1 \qquad \quad \ \ \tan 60^\circ = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3} \qquad \quad \tan 30^\circ = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{\sqrt{3}}{2}$
These triangles can now fit inside the unit circle.
Putting together the trigonometric ratios and the coordinates of the points on the circle, which represent the lengths of the legs of the triangles, $(\Delta x, \Delta y)$ , we can see that each point is actually $(\cos \theta, \sin \theta)$ , where $\theta$ is the reference angle. For example, $\sin 60^\circ=\frac{\sqrt{3}}{2}$ is the $y$ – coordinate of the point on the unit circle in the triangle with reference angle $60^\circ$ . By reflecting these triangles across the axes and finding the points on the axes, we can find the trigonometric ratios of all multiples of $0^\circ, 30^\circ$ and $45^\circ$ (or $0, \frac{\pi}{6}, \frac{\pi}{4}$ radians).
#### Example A
Find $\sin \frac{3 \pi}{2}$ .
Solution: Find $\frac{3 \pi}{2}$ on the unit circle and the corresponding point is $(0, -1)$ . Since each point on the unit circle is $(\cos \theta, \sin \theta), \sin \frac{3 \pi}{2}=-1$ .
#### Example B
Find $\tan \frac{7 \pi}{6}$ .
Solution: This time we need to look at the ratio $\frac{\sin \theta}{\cos \theta}$ . We can use the unit circle to find $\sin \frac{7 \pi}{6}=-\frac{1}{2}$ and $\cos \frac{7 \pi}{6}=-\frac{\sqrt{3}}{2}$ . Now, $\tan \frac{7 \pi}{6}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$ .
### More Guidance
Another way to approach these exact value problems is to use the reference angles and the special right triangles. The benefit of this method is that there is no need to memorize the entire unit circle. If you memorize the special right triangles, can determine reference angles and know where the ratios are positive and negative you can put the pieces together to get the ratios. Looking at the unit circle above, we see that all of the ratios are positive in Quadrant I, sine is the only positive ratio in Quadrant II, tangent is the only positive ratio in Quadrant III and cosine is the only positive ratio in Quadrant IV.
Keeping this diagram in mind will help you remember where cosine, sine and tangent are positive and negative. You can also use the pneumonic device - A ll S tudents T ake C alculus, or ASTC , to recall which is positive (all the others would be negative) in which quadrant.
The coordinates on the vertices will help you determine the ratios for the multiples of $90^\circ$ or $\frac{\pi}{2}$ .
#### Example C
Find the exact values for the following trigonometric functions using the alternative method.
a. $\cos 120^\circ$
b. $\sin \frac{5 \pi}{3}$
c. $\tan \frac{7 \pi}{2}$
Solution:
a. First, we need to determine in which quadrant the angles lies. Since $120^\circ$ is between $90^\circ$ and $180^\circ$ it will lie in Quadrant II. Next, find the reference angle. Since we are in QII, we will subtract from $180^\circ$ to get $60^\circ$ . We can use the reference angle to find the ratio, $\cos 60^\circ=\frac{1}{2}$ . Since we are in QII where only sine is positive, $\cos 120^\circ=-\frac{1}{2}$ .
b. This time we will need to work in terms of radians but the process is the same. The angle $\frac{5 \pi}{3}$ lies in QIV and the reference angle is $\frac{\pi}{3}$ . This means that our ratio will be negative. Since $\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}, \sin \frac{5 \pi}{3}=-\frac{\sqrt{3}}{2}$ .
c. The angle $\frac{7 \pi}{2}$ represents more than one entire revolution and it is equivalent to $2 \pi + \frac{3 \pi}{2}$ . Since our angle is a multiple of $\frac{\pi}{2}$ we are looking at an angle on an axis. In this case, the point is $(0, -1)$ . Because $\tan \theta=\frac{\sin \theta}{\cos \theta}, \tan \frac{7 \pi}{2}=\frac{-1}{0}$ , which is undefined. Thus, $\tan \frac{7 \pi}{2}$ is undefined.
Concept Problem Revisit
a. First, we need to determine in which quadrant the angle lies. Since $495^\circ - 360^\circ = 135^\circ$ is between $90^\circ$ and $180^\circ$ it will lie in Quadrant II. Next, find the reference angle. Since we are in QII, we will subtract from $180^\circ$ to get $45^\circ$ . We can use the reference angle to find the ratio, $\cos 45^\circ=\frac{\sqrt {2}}{2}$ . Since we are in QII where only sine is positive, $\cos 495^\circ=-\frac{\sqrt{2}}{2}$ .
b. In the previous example we established that the angle $\frac{5 \pi}{3}$ lies in QIV and the reference angle is $\frac{\pi}{3}$ . This means that the tangent ratio will be negative. Since $\tan \frac{\pi}{3}=\sqrt{3}, \tan \frac{5 \pi}{3}=-\sqrt{3}$ .
### Guided Practice
Find the exact trigonometric ratios. You may use either method.
1. $\cos \frac{7 \pi}{3}$
2. $\tan \frac{9 \pi}{2}$
3. $\sin 405^\circ$
4. $\tan \frac{11 \pi}{6}$
5. $\cos \frac{2 \pi}{3}$
1. $\frac{7 \pi}{3}$ has a reference angle of $\frac{\pi}{3}$ in QI. $\cos \frac{\pi}{3}=\frac{1}{2}$ and since cosine is positive in QI, $\cos \frac{7 \pi}{3}=\frac{1}{2}$ .
2. $\frac{9 \pi}{2}$ is coterminal to $\frac{\pi}{2}$ which has coordinates (0, 1). So $\tan \frac{9 \pi}{2}=\frac{\sin \frac{9 \pi}{2}}{\cos \frac{9 \pi}{2}}=\frac{1}{0}$ which is undefined.
3. $405^\circ$ has a reference angle of $45^\circ$ in QI. $\sin 45^\circ=\frac{\sqrt{2}}{2}$ and since sine is positive in QI, $\sin 405^\circ=\frac{\sqrt{2}}{2}$ .
4. $\frac{11 \pi}{6}$ is coterminal to $\frac{\pi}{6}$ in QIV. $\tan \frac{\pi}{6}=\frac{\sqrt{3}}{3}$ and since tangent is negative in QIV, $\tan \frac{11 \pi}{6}=-\frac{\sqrt{3}}{3}$ .
5. $\frac{2 \pi}{3}$ is coterminal to $\frac{\pi}{3}$ in QII. $\cos \frac{\pi}{3}=\frac{1}{2}$ and since cosine is negative in QII, $\cos \frac{2 \pi}{3}=\frac{1}{2}$ .
### Practice
Find the exact values for the following trigonometric functions.
1. $\sin \frac{3 \pi}{4}$
2. $\cos \frac{3 \pi}{2}$
3. $\tan 300^\circ$
4. $\sin 150^\circ$
5. $\cos \frac{4 \pi}{3}$
6. $\tan \pi$
7. $\cos \left(-\frac{15 \pi}{4}\right)$
8. $\sin 225^\circ$
9. $\tan \frac{7 \pi}{6}$
10. $\sin 315^\circ$
11. $\cos 450^\circ$
12. $\sin \left(-\frac{7 \pi}{2}\right)$
13. $\cos \frac{17 \pi}{6}$
14. $\tan 270^\circ$
15. $\sin(-210^\circ)$
### Vocabulary Language: English
Coterminal
Coterminal
Two angles are coterminal if they are drawn in the standard position and both have terminal sides that are at the same location.
Coterminal Angles
Coterminal Angles
A set of coterminal angles are angles with the same terminal side but expressed differently, such as a different number of complete rotations around the unit circle or angles being expressed as positive versus negative angle measurements.
A quadrant is one-fourth of the coordinate plane. The four quadrants are numbered using Roman Numerals I, II, III, and IV, starting in the top-right, and increasing counter-clockwise.
unit circle
unit circle
The unit circle is a circle of radius one, centered at the origin. |
Ex 8.2
Serial order wise
### Transcript
Ex 8.2, 2 ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. Given: ABCD is rhombus where P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively To prove: PQRS is a rectangle Construction: Join A & C Proof: A rectangle is a parallelogram with one angle 90° First we will prove PQRS is a parallelogram, and prove one angle 90 ° From (1) & (2) PQ ∥ RS and PQ = RS In PQRS, one pair of opposite side is parallel and equal. Hence, PQRS is a parallelogram. Now we prove have to prove PQRS is a rectangle Since AB = BC 1/2AB = 1/2BC So, PB = BQ Now, in Δ BPQ PB = BQ ∴ ∠ 2 = ∠ 1 In ∆ APS & ∆ CQR AP = CQ AS = CR PS = QR ∴ ∆ APS ≅ ∆ CQR ∠ 3 = ∠ 4 Now, AB is a line So, ∠ 3 + ∠ SPQ + ∠ 1 = 180° Similarly, for line BC ∠ 2 + ∠ PQR + ∠ 4 = 180° ∠ 1 + ∠ PQR + ∠ 3 = 180° From (5) & (6) ∠ 1 + ∠ SPQ + ∠ 3 = ∠ 1 + ∠ PQR + ∠ 3 ∴ ∠ SPQ = ∠ PQR Now, PS ∥ QR & PQ is a transversal So, ∠ SPQ + ∠ PQR = 180° ∠ SPQ + ∠ SPQ = 180° 2∠ SPQ = 180° ∠ SPQ = 180"°" /2 = 90° So, PQRS is a parallelogram with one angle 90° ∴ PQRS is a rectangle Hence proved |
Theory:
In the last topic, we have learned how to write a number in expanded form and exponential form.
Now, we shall learn how to find the unit digit of a number which is in exponential form.
Let us consider $$6^3 = 6 \times 6 \times 6$$ $$= 216$$. Here, the unit digit is $$6$$.
Similarly, let us consider $$5^5 = 5 \times 5 \times 5 \times 5 \times 5$$ $$= 3125$$. Here, the unit digit is $$5$$.
It is easier to determine the unit digit of a number whose exponential is small.
Is it possible to find the unit digit of $$26^{34}$$, $$45^{20}$$, $$20^{33}$$?
It isn't easy to find the unit digit by expanding the exponential form. But, there are specific patterns to find the unit digit.
We know that the numbers ends with either of the following digits:
$$1$$ or $$2$$ or $$3$$ or $$4$$ or $$5$$ or $$6$$ or $$7$$ or $$8$$ or $$9$$ or $$0$$.
There are some groups of numbers which following certain patterns when we find the exponential value of the numbers. Now we will learn those groups in detail by observing its pattern.
Let us observe the pattern for the numbers $$0$$, $$1$$, $$5$$ and $$6$$.
Pattern for the number $$0$$:
$$0^2 = 0 \times 0 = \underline{0}$$
$$0^3 = 0 \times 0 \times 0 = \underline{0}$$
$$0^4 = 0 \times 0 \times 0 \times 0 = \underline{0}$$
Pattern for the number $$1$$:
$$1^2 = 1 \times 1 = \underline{1}$$
$$1^3 = 1 \times 1 \times 1 = \underline{1}$$
$$1^4 = 1 \times 1 \times 1 \times 1 = \underline{1}$$
Pattern for the number $$5$$:
$$5^2 = 5 \times 5 = 2\underline{5}$$
$$5^3 = 5 \times 5 \times 5 = 12\underline{5}$$
$$5^4 = 5 \times 5 \times 5 \times 5 = 62\underline{5}$$
Pattern for the number $$6$$:
$$6^2 = 6 \times 6$$ $$= 3\underline{6}$$
$$6^3 = 6 \times 6 \times 6$$ $$= 21\underline{6}$$
$$6^4 = 6 \times 6 \times 6 \times 6$$ $$= 129\underline{6}$$
Here, in the above patterns, we can see that, any number in the exponential form of $$0^x$$, $$1^x$$, $$5^x$$ and $$6^x$$, the unit digit will be the same number where $$x$$ is positive number.
Example:
$$66^2 =$$ $$(11 \times 6)^2 =$$ $$11^2 \times 6^2$$ $$= 121 \times 36$$ $$= 435\underline{6}$$
Here, the unit digit is $$6$$.
Let us observe the pattern for the numbers $$4$$ and $$9$$.
Pattern for the number $$4$$:
$$4^1 = \underline{4}$$ (Odd pattern)
$$4^2 = 4 \times 4$$ $$= 1\underline{6}$$ (Even pattern)
$$4^3 = 4 \times 4 \times 4$$ $$= 6\underline{4}$$ (Odd pattern)
$$4^4 = 4 \times 4 \times 4 \times 4$$ $$= 25\underline{6}$$ (Even pattern)
$$4^5 = 4 \times 4 \times 4 \times 4 \times 4$$ $$= 102\underline{4}$$ (Odd pattern)
Here, we can see that if the power is odd, then the unit digit of the resultant is $$4$$. If the power is even, then the unit digit of the resultant is $$6$$.
Pattern for the number $$9$$:
$$9^1 = \underline{9}$$ (Odd pattern)
$$9^2 = 9 \times 9$$ $$= 8\underline{1}$$ (Even pattern)
$$9^3 = 9 \times 9 \times 9$$ $$= 72\underline{9}$$ (Odd pattern)
$$9^4 = 9 \times 9 \times 9 \times 9$$ $$= 656\underline{1}$$ (Even pattern)
$$9^5 = 9 \times 9 \times 9 \times 9 \times 9$$ $$= 5904\underline{9}$$ (Odd pattern)
Here, we can see that if the power is odd, then the unit digit is $$9$$. If the power is even, then the unit digit is $$1$$.
Example:
Find the unit digit of the number $$19^{16}$$.
Solution:
Here, the unit digit of base $$19$$ is $$9$$ and power is $$16$$ (even power).
For base ending with $$9$$ having an even power, the unit digit is $$1$$.
Therefore, the unit digit is $$1$$. |
# ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.4
## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.4
These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.4
More Exercises
Question 1.
Find the discriminant of the following equations and hence find the nature of roots:
(i) 3x² – 5x – 2 = 0
(ii) 2x² – 3x + 5 = 0
(iii) 7x² + 8x + 2 = 0
(iv) 3x² + 2x – 1 = 0
(v) 16x² – 40x + 25 = 0
(vi) 2x² + 15x + 30 = 0.
Solution:
(i) 3x² – 5x – 2 = 0
Here a = 3, b = -5, c = -2
Question 2.
Discuss the nature of the roots of the following quadratic equations :
(i) x² – 4x – 1 = 0
(ii) 3x² – 2x + $$\\ \frac { 1 }{ 3 }$$ = 0
(iii) 3x² – 4√3x + 4 = 0
(iv) x² – $$\\ \frac { 1 }{ 2 } x$$ + 4 = 0
(v) – 2x² + x + 1 = 0
(vi) 2√3x² – 5x + √3 = 0
Solution:
(i) x² – 4x – 1 = 0
Here a = 1, b = -4, c = -1
Question 3.
Find the nature of the roots of the following quadratic equations:
(i) x² – $$\\ \frac { 1 }{ 2 } x$$ – $$\\ \frac { 1 }{ 2 }$$ = 0
(ii) x² – 2√3x – 1 = 0 If real roots exist, find them.
Solution:
(i) x² – $$\\ \frac { 1 }{ 2 } x$$ – $$\\ \frac { 1 }{ 2 }$$ = 0
Here a = 1, b = $$– \frac { 1 }{ 2 }$$, c = $$– \frac { 1 }{ 2 }$$
Question 4.
Without solving the following quadratic equation, find the value of ‘p’ for which the given equations have real and equal roots:
(i) px² – 4x + 3 = 0
(ii) x² + (p – 2)x + p = 0.
Solution:
(i) px² – 4x + 3 = 0
Here a = p, b = -4, c = 3
Question 5.
Find the value (s) of k for which each of the following quadratic equation has equal roots :
(i) kx² – 4x – 5 = 0
(ii) (k – 4) x² + 2(k – 4) x + 4 = 0
Solution:
(i) kx² – 4x – 5 = 0
Here a = k, b = -4, c = 5
Question 6.
Find the value(s) of m for which each of the following quadratic equation has real and equal roots:
(i) (3m + 1)x² + 2(m + 1)x + m = 0
(ii) x² + 2(m – 1) x + (m + 5) = 0
Solution:
(i) (3m + 1)x² + 2(m + 1)x + m = 0
Here a = 3m + 1, b = 2(m + 1), c = m
Question 7.
Find the values of k for which each of the following quadratic equation has equal roots:
(i) 9x² + kx + 1 = 0
(ii) x² – 2kx + 7k – 12 = 0
Also, find the roots for those values of k in each case.
Solution:
(i) 9x² + kx + 1 = 0
Here a = 9, b = k, c = 1
Question 8.
Find the value(s) of p for which the quadratic equation (2p + 1)x² – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots.
Solution:
The quadratic equation given is (2p + 1)x² – (7p + 2)x + (7p – 3) = 0
Comparing with ax² + bx + c = 0, we have
Question 9.
If – 5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k.
Solution:
-5 is a root of the quadratic equation
2x² + px – 15 = 0, then
⇒ 2(5)² – p( -5) – 15 = 0
⇒ 50 – 5p – 15 = 0
⇒ 35 – 5p = 0
Question 10.
Find the value(s) of p for which the equation 2x² + 3x + p = 0 has real roots.
Solution:
2x² + 3x + p = 0
Here, a = 2, b = 3, c = p
Question 11.
Find the least positive value of k for which the equation x² + kx + 4 = 0 has real roots.
Solution:
x² + kx + 4 = 0
Here, a = 1, b = k, c = 4
Question 12.
Find the values of p for which the equation 3x² – px + 5 = 0 has real roots.
Solution:
3x² – px + 5 = 0
Here, a = 3, b = -p, c = 5
Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Ex 5.4 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. |
# A triangle has sides with lengths: 2, 9, 10. How do you find the area of the triangle using Heron's formula?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
2
Jim G. Share
Jun 27, 2016
≈ 8.182 square units
#### Explanation:
This is a 2-step process.
Step 1: Calculate half the perimeter (s ) of the triangle.
$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{s = \frac{a + b + c}{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where a , b and c are the 3 sides of the triangle.
let a = 2 , b = 9 and c = 10
$\Rightarrow s = \frac{2 + 9 + 10}{2} = \frac{21}{2} = 10.5$
Step 2: Calculate the area (A ) using.
$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
$A = \sqrt{10.5 \left(10.5 - 2\right) \left(10.5 - 9\right) \left(10.5 - 10\right)}$
=sqrt(10.5xx8.5xx1.5xx0.5)≈8.182" (3 dec. places) "
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Find t if d1 is parallel with d2. The equation of d1: 2x-ty+1=0 The equation of d2: -tx+8y-3=0
hala718 | Certified Educator
d1= 2x-ty +1 =0
Rewrite in standard form:
==> y= (2x+1)/t = (2/t)x + 1/t
==> the slope (m1) = 2/t
d2= -tx+8y-3=0
Rewrite in standard form:
==> y= (tx+3)/8 = (t/8)x + 3/8)
==> the slope (m2)= t/8
Since d1 and d2 are parallel, then m1=m2
==> 2/t = t/8
Cross multiply:
==> t^2 = 16
==> t = +-4
krishna-agrawala | Student
Equation of di1 is:
2x - ty + 1 = 0
We change this equation in the form y = mx + c as follows:
-ty = -2x - 1
y = (-2/-t) x - 1/-t
y =(2/t) + 1/t
Similarly we change t the equation of d2 in the form y = mx + c as follows:
-tx + 8y -3 = 0
8y = tx + 3
y = (t/8)x + 3/8
In a equation of a straight line of the form y = mx + c, the slope of the line is given by m.
Therefore:
Slope of d1 = 2/t
Slope of d2 = t/8
As d1 and d2 are parallel their slopes are equal.
Therefore:
2/t = t/8
multiplying both side of above equation by 8t we get:
16 = t^2
Therefore:
t = 16^(1/2) = 4 and -4
t = 4 and -4
giorgiana1976 | Student
If d1 to be parallel to d2, the slope of d1 has to be equal to the slope of d2:
m1 = m2
Let's find out the slope of d1:
2x-ty+1=0
We'll re-write the equation into the form: y=mx+n, where m is the slope of the line.
For this reason, we'll isolate -ty to the left side:
-ty = -2x - 1
We'll multiply by -1:
ty = 2x+1
We'll divide both sides by t:
y = (2/t)*x + (1/t)
So, the slope of d1 is m1 = 2/t
Now, we'll find out the slope of d2:
-tx+8y-3=0
We'll isolate 8y to the left side:
8y = tx + 3
We'll divide by 8, both sides:
y = tx/8 - 3/8
The slope of d2 is m2 = t/8
But m1=m2, so we'll get:
2/t=t/8
We'll cross multiply:
t^2=16
t1=sqrt16
t1=4
t2=-4
neela | Student
If a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 are the two lines and if they are || then
a1/a2 = b1/b2.
So here the lines are 2x+ty+1 = 0 and -tx+8y-3 = 0.
Therefore a1/a2 = b1/b2 gives: 2/-t = -t/8. Or
8*2 = (-t)(-t) =t^2.
t^2 = 16 . Or t = -4 Or t = 4 are the values of t for which d1 and d2 are ||/ |
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# Parallelograms with the Same Base and Two Parallels
A parallelogram is a flat shape with four sides where both pairs of opposite sides are parallel. In other words, if you were to draw a line through the middle of a parallelogram, both halves would be mirror images of each other.
There are many different types of parallelograms, but in this post we'll focus on those that have the same base and two parallels. Keep reading to learn more about these shapes and how to identify them.
## What is a Parallelogram with the Same Base and Two Parallels?
A parallelogram with the same base and two parallels is a four-sided flat shape where both pairs of opposite sides are parallel AND the distance between the two parallel sides is equal. In other words, if you were to draw a line through the middle of this type of parallelogram, both halves would be mirror images of each other AND the distance between the lines would be equal on both halves.
## How to Identify a Parallelogram with the Same Base and Two Parallels
There are a few different ways that you can identify a parallelogram with the same base and two parallels. The first is by looking at the shape itself. As we mentioned before, this type of parallelogram will have both pairs of opposite sides parallel AND the distance between the two parallel sides will be equal.
Another way to identify this type of parallelogram is by using algebra. If you know that two sides are parallel, you can set up equations to solve for the unknowns. For example, let's say we have a parallelogram with side lengths of 3x+2, x+5, 3x+2, and x+5. We know that the first pair of opposite sides are parallel (3x+2 and 3x+2) as are the second pair of opposite sides (x+5 and x+5). This means that we can set up our equation as 3x+2=x+5. Once we solve for x, we get x=3. This tells us that all four side lengths in our parallelogram are equal to 3(3)+2=11 or 3(5)+5=16. So our final dimensions would either be 3x3 or 5x16.
## Conclusion:
Parallelograms are four-sided shapes where both pairs of opposite sides are parallel. There are many different types of parallelograms, but in this post we focused on those with the same base and two parallels. These shapes can be identified by their mirror image halves and equal distance between their two parallel lines.
## FAQ
### What do you say about area of parallelograms on the same base and between the same parallels not equal about equal none of these?
There is no definitive answer to this question since it depends on the specific dimensions of the parallelograms in question. However, in general, the area of a parallelogram with the same base and two parallels will be equal to the product of the length of the base and the height (distance between the parallel sides). So if two parallelograms have the same base and height, then their areas will be equal. However, if the parallelograms have different base or height dimensions, then their areas will be unequal.
### How do you prove parallelogram are on same base and between same parallels are equal in area?
There is no definitive answer to this question since it depends on the specific dimensions of the parallelograms in question. However, in general, the area of a parallelogram with the same base and two parallels will be equal to the product of the length of the base and the height (distance between the parallel sides). So if two parallelograms have the same base and height, then their areas will be equal. However, if the parallelograms have different base or height dimensions, then their areas will be unequal.
### What do you say about area of triangles on the same base and between same parallels?
There is no definitive answer to this question since it depends on the specific dimensions of the triangles in question. However, in general, the area of a triangle with the same base and two parallels will be equal to one-half the product of the length of the base and the height (distance between the parallel sides). So if two triangles have the same base and height, then their areas will be equal. However, if the triangles have different base or height dimensions, then their areas will be unequal.
### How do you prove two parallelograms are equal?
There is no definitive answer to this question since it depends on the specific dimensions of the parallelograms in question. However, in general, the area of a parallelogram with the same base and two parallels will be equal to the product of the length of the base and the height (distance between the parallel sides). So if two parallelograms have the same base and height, then their areas will be equal. However, if the parallelograms have different base or height dimensions, then their areas will be unequal. |
# 5.1.1 Direct Variation Part 1
(A) Determining whether a quantity varies directly as another quantity
1. If a quantity varies directly as a quantity x, the
(a) y increases when x increases
(b) y decreases when x decreases
2. A quantity varies directly as a quantity x if and only if $\frac{y}{x}=k$ where k is called the constant of variation.
3. y varies directly as x is written as $y\propto x$ .
4. When $y\propto x$ , the graph of against x is a straight line passing through the origin.
(B) Expressing a direct variation in the form of an equation involving two variables
Example 1
Given that y varies directly as x and y = 20 when x = 36 . Write the direct variation in the form of equation.
(C) Finding the value of a variable in a direct variation
1. When y varies directly as x and sufficient information is given, the value of y or x can be determined by using:
$\begin{array}{l}\left(a\right)\text{}y=kx,\text{or}\\ \left(b\right)\text{}\frac{{y}_{1}}{{x}_{1}}=\frac{{y}_{2}}{{x}_{2}}\end{array}$
Example 2
Given that varies directly as x and y = 24 when x = 8, find
(a) The equation relating to x
(b) The value of when = 6
(c) The value of when = 36
Solution:
Method 1: Using y = kx
$\left(a\right)\text{y}\propto x$
y = kx
when y = 24, x = 8
24 = k (8)
k = 3
y = 3x
(b) when x = 6,
y = 3 (6)
y = 18
(c) when y = 36
36 = 3x
x =12
Method 2: $\text{Using}\frac{{y}_{1}}{{x}_{1}}=\frac{{y}_{2}}{{x}_{2}}$
(a) Let x1 = 8 and y1 = 24
$\begin{array}{l}\frac{{y}_{1}}{{x}_{1}}=\frac{{y}_{2}}{{x}_{2}}\to \frac{24}{8}=\frac{{y}_{2}}{{x}_{2}}\\ \frac{3}{1}=\frac{{y}_{2}}{{x}_{2}}\to {y}_{2}=3{x}_{2}\\ \therefore y=3x\end{array}$
(b) Let x1 = 8 and y1 = 24 and x2= 6; find y2.
$\begin{array}{l}\frac{{y}_{1}}{{x}_{1}}=\frac{{y}_{2}}{{x}_{2}}\to \frac{24}{8}=\frac{{y}_{2}}{6}\\ {y}_{2}=\frac{24}{8}\left(6\right)\\ {y}_{2}=18\end{array}$
(c) Let x1 = 8 and y1 = 24 and y2= 36; find x2.
$\begin{array}{l}\frac{{y}_{1}}{{x}_{1}}=\frac{{y}_{2}}{{x}_{2}}\to \frac{24}{8}=\frac{36}{{x}_{2}}\\ 24{x}_{2}=36×8\\ {x}_{2}=12\end{array}$ |
# Math site that shows steps
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## The Best Math site that shows steps
Math site that shows steps can be found online or in math books. One of the most important things that students need to learn is basic math. This includes things like addition, subtraction, multiplication, and division. Basic math is the foundation that students will need in order to succeed in school. Without a strong foundation in math, students will struggle in other subjects as well.
To solve an inverse function, you first need to determine what the function is and what its inverse would be. To do this, you need to know what a function is and what inverse functions are. A function is a set of ordered pairs (x, y) where each x corresponds to a unique y. An inverse function is a function that "undoes" another function. So, if the function is f(x) = 2x + 3, then the inverse function would be
Algebra math problems can be tricky to solve, but there are some tips and tricks that can help. First, make sure you understand the problem and what is being asked. Next, use algebraic methods to solve the problem. Finally, check your work to make sure it is correct.
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An exponential equation solver is a tool that can be used to solve exponential equations. Exponential equations are equations that involve exponents, and can be tricky to solve. An exponential equation solver can be used to find the value of x in an exponential equation.
If you are given a perfect square, then you can just take the square root of that number. For example, the square root of 9 is 3, because 3 squared is 9. However, if you are not given a perfect square, then you will need to use a different method. One method is to use estimation. To do this, you look at the number and find
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# Lesson 8
Relating Area to Circumference
### Lesson Narrative
In the previous lesson, students found that it takes a little more than 3 squares with side lengths equal to the circle’s radius to completely cover a circle. Students may have predicted that the area of a circle can be found by multiplying $$\pi r^2$$. In this lesson students derive that relationship through informal dissection arguments. In the main activity they cut and rearrange a circle into a shape that approximates a parallelogram (MP 3). In an optional activity, they consider a different way to cut and rearrange a circle into a shape that approximates a triangle. In both arguments, one side of the polygon comes from the circumference of the circle, leading to the presence of $$\pi$$ in the formula for the area of a circle.
### Learning Goals
Teacher Facing
• Generalize a process for finding the area of a circle, and justify (orally) why this can be abstracted as $\pi r^2$.
• Show how a circle can be decomposed and rearranged to approximate a polygon, and justify (orally and in writing) that the area of this polygon is equal to half of the circle’s circumference multiplied by its radius.
### Student Facing
Let’s rearrange circles to calculate their areas.
### Required Preparation
You will need one cylindrical household item (like a can of soup) for each group of 2 students. The activity works best if the diameter of the item is between 3 and 5 inches.
If possible, it would be best to give each group 2 different colors of blank paper.
### Student Facing
• I can explain how the area of a circle and its circumference are related to each other.
• I know the formula for area of a circle.
### Glossary Entries
• squared
We use the word squared to mean “to the second power.” This is because a square with side length $$s$$ has an area of $$s \boldcdot s$$, or $$s^2$$.
### Print Formatted Materials
For access, consult one of our IM Certified Partners. |
# What is indirect reasoning? Who uses indirect reasoning?
## Presentation on theme: "What is indirect reasoning? Who uses indirect reasoning?"— Presentation transcript:
What is indirect reasoning? Who uses indirect reasoning?
5-4 Indirect Proof What is indirect reasoning? Who uses indirect reasoning?
You wrote paragraph, two-column, and flow proofs.
Write indirect algebraic proofs. Write indirect geometric proofs.
Direct Reasoning In direct reasoning, you assume that the hypothesis is true and show that the conclusion must also be true. If it is 3pm on a school day, then academic classes at Marian High School are finished for the day.
Indirect Reasoning Indirect reasoning shows that a statement is true by proving that it cannot be false. Assume the opposite—contradict it.
Indirect Reasoning Mark’s car won’t start. He knows that there are three likely reasons for this. His battery is dead His starter doesn’t work. He is out of gas. When a car’s starter needs to be replaced, the car is silent when you try to start it. If the battery is dead, the engine “turns over” slowly, if at all. When Mark tries to start the car, it sounds normal. What do you think is wrong with his car? Out of gas!
Three Key Steps in Indirect Reasoning.
Assume that the statement you are trying to prove is false. Show that this assumption leads to a contradiction of something you know is true. Conclude that your assumption was incorrect, so that the statement you originally wanted to prove must be true.
What would you assume for indirect reasoning?
If it rains, then I will wash my car. It rains and I do not wash my car.
State the Assumption for Starting an Indirect Proof
A. State the assumption you would make to start an indirect proof for the statement is not a perpendicular bisector. Answer: is a perpendicular bisector.
A. B. C. D.
Write an indirect proof to show that if –2x + 11 < 7, then x > 2.
Given: –2x + 11 < 7 Prove: x > 2 Step 1 Indirect Proof: The negation of x > 2 is x ≤ 2. So, assume that x < 2 or x = 2 is true. Step 2 Make a table with several possibilities for x assuming x < 2 or x = 2.
Step 2. Make a table with several possibilities for x
Step 2 Make a table with several possibilities for x assuming x < 2 or x = 2. When x < 2, –2x + 11 > 7 and when x = 2, –2x + 11 = 7. Step 3 In both cases, the assumption leads to a contradiction of the given information that –2x + 11 < 7. Therefore, the assumption that x ≤ 2 must be false, so the original conclusion that x > 2 must be true.
Which is the correct order of steps for the following indirect proof?
Given: x + 5 > 18 Prove: x > 13 I. In both cases, the assumption leads to a contradiction. Therefore, the assumption x ≤ 13 is false, so the original conclusion that x > 13 is true. II. Assume x ≤ 13. III. When x < 13, x + 5 = 18 and when x < 13, x + 5 < 18. A. I, II, III B. I, III, II C. II, III, I D. III, II, I
SHOPPING David bought four new sweaters for a little under \$135
SHOPPING David bought four new sweaters for a little under \$135. The tax was \$7, but the sweater costs varied. Can David show that at least one of the sweaters cost less than \$32? A. Yes, he can show by indirect proof that assuming that every sweater costs \$32 or more leads to a contradiction. B. No, assuming every sweater costs \$32 or more does not lead to a contradiction.
Given: ΔJKL with side lengths
5, 7, and 8 as shown. Prove: mK < mL Write an indirect proof. Indirect Proof: Step 1 Assume that Step 2 By angle-side relationships, By substitution, This inequality is a false statement. Step 3 Since the assumption leads to a contradiction, the assumption must be false. Therefore, mK < mL.
Which statement shows that the assumption leads to a contradiction for this indirect proof?
Given: ΔABC with side lengths 8, 10, and 12 as shown. Prove: mC > mA A. Assume mC ≥ mA + mB. By angle-side relationships, AB > BC + AC. Substituting, ≥ or 12 ≥ 18. This is a false statement. B. Assume mC ≤ mA. By angle-side relationships, AB ≤ BC. Substituting, 12 ≤ 8. This is a false statement.
Who uses Indirect Reasoning?
Auto mechanics Physicians diagnosing diseases CSI Lawyers Eliminating possibilities that contradict a know fact can lead to the actual cause of a problem.
What is indirect reasoning?
In direct reasoning, you assume that the hypothesis is true and show that the conclusion must also be true. Who uses indirect reasoning? Auto mechanics, doctors, police, lawyers…
5-4 Assignment Page 358, 11-20 |
# Find a matrix transformation mapping $\{(1,1,1),(0,1,0),(1,0,2)\}$ to $\{(1,1,1),(0,1,0),(1,0,1)\}$
Find a matrix transformation mapping $\{(1,1,1),(0,1,0),(1,0,2)\}$ to $\{(1,1,1),(0,1,0),(1,0,1)\}$.
$$\begin{bmatrix}1& 0& -1\\0& 1& 1\\0& 0& 1\end{bmatrix}?$$
I understand the concept of Matrix Transformation, I don’t think I’m doing it right.
#### Solutions Collecting From Web of "Find a matrix transformation mapping $\{(1,1,1),(0,1,0),(1,0,2)\}$ to $\{(1,1,1),(0,1,0),(1,0,1)\}$"
We wish to find a $3\times 3$ matrix $T$ such that $TA=B$ where
\begin{align*}
A &=\begin{bmatrix}1 & 0 & 1\\ 1 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}
&
B &= \begin{bmatrix} 1 & 0 & 1\\ 1 & 1 & 0\\ 1 & 0 & 1\end{bmatrix}
\end{align*}
Perhaps the quickest way to find $T$ is to multiply the equation $TA=B$ on the right by $A^{-1}$ to obtain
$$T=BA^{-1}$$
Can you compute $A^{-1}$ and carry out the matrix multiplication?
The columns of the matrix tell you where it sends the standard basis vectors. For instance if I am interested in the third column then I need to determine what the action of our linear operator is on the column vector ,
$$\left[ \begin{array}{c}0 \\ 0 \\ 1 \end{array}\right].$$
This vector can be written as a linear combination of the vectors used to define the linear operator,
$$\left[ \begin{array}{c}0 \\ 0 \\ 1 \end{array}\right] = \left[ \begin{array}{c}1 \\ 0 \\ 2 \end{array}\right] -\left[ \begin{array}{c}0 \\ 1 \\ 0 \end{array}\right] -\left[ \begin{array}{c}1 \\ 1 \\ 1 \end{array}\right].$$
Multiplying both sides by our linear operator $M$ we get,
$$M\left[ \begin{array}{c}0 \\ 0 \\ 1 \end{array}\right] = M\left[ \begin{array}{c}1 \\ 0 \\ 2 \end{array}\right] -M\left[ \begin{array}{c}0 \\ 1 \\ 0 \end{array}\right] -M\left[ \begin{array}{c}1 \\ 1 \\ 1 \end{array}\right].$$
Note that we know what $M$ does to the vectors on the right so we can just substitute those values in and add,
$$M\left[ \begin{array}{c}0 \\ 0 \\ 1 \end{array}\right] = \left[ \begin{array}{c}1 \\ 0 \\ 1 \end{array}\right] -\left[ \begin{array}{c}0 \\ 1 \\ 0 \end{array}\right] -\left[ \begin{array}{c}1 \\ 1 \\ 1 \end{array}\right] =\left[ \begin{array}{c}0 \\ -2 \\ 0 \end{array}\right].$$
The resulting vector is the third column of our matrix,
$$M = \left[\begin{array}{ccc} \ & \ & 0 \\ \ & \ & -2 \\ \ & \ & 0 \end{array}\right].$$
A similar process will yield the other collumns. |
# Which Products Give the Answers in the Times Tables?
Being able to list a bunch of numbers or knowing the answers to all the multiplications isn’t enough in and of itself. You also need to be able to know which multiplications give you a specific answers. For that reason, you need to learn which multiplications are connected to each of the different answers in the times tables.
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Example 1
Which factors in the times tables make up the product 24?
The product 24 can be written as the following multiplications from the times tables: $\begin{array}{llll}\hfill 24& =4\cdot 6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =6\cdot 4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3\cdot 8\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =8\cdot 3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Example 2
Which factors in the times tables give you the product 36?
The product 36 can be written as the following multiplications from the times tables: $\begin{array}{llll}\hfill 36& =4\cdot 9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =9\cdot 4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =6\cdot 6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Example 3
Which factors in the times tables give you the product 30?
The product 30 can be written as a multiplication of the following numbers: $\begin{array}{llll}\hfill 30& =5\cdot 6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =6\cdot 5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3\cdot 10\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =10\cdot 3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Example 4
Which factors in the times tables give you the product 64?
The product 64 can be written as the multiplication of the following numbers from the times tables:
$64=8\cdot 8$
Example 5
Which factors in the times tables give you the product 18?
The product 18 can be written as the multiplication of the following numbers from the times tables: $\begin{array}{llll}\hfill 18& =2\cdot 9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =9\cdot 2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3\cdot 6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =6\cdot 3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ |
When you learn mathematics, you may be taught about the commutative property (or commutative law) and the associative property (or associative law). The commutative property and the associative property are taught in elementary or junior high school.
The commutative property and the associative property are formed by addition and multiplication. The commutative property and the associative property cannot be established by subtraction and division. However, subtraction can be converted to addition and division can be converted to multiplication. This means that we can use these rules in all calculations.
It is important to understand these rules in mathematics. Understanding the commutative property and the associative property will allow us to perform all kinds of calculations. Although we don’t use division anymore in math in junior high school and above, the commutative property and the associative property can help us understand why this is the case.
We will explain how to think about and apply the commutative law and the associative law.
## An Overview of the Commutative Law and the Associative Law
What is the content of the commutative property and the associative property that is taught in elementary or secondary school? The meanings of these are as follows.
• Commutative property: the law that gives the same answer even if the numbers are interchanged
• Associative property: the law that gives the same answer even if you change the place of parentheses.
Regarding the commutative property and the associative property, both of which are used in so many situations, they are essential knowledge when solving math problems. These laws are used in addition and multiplication. But in practice, they can also be applied to subtraction and division.
In other words, we can use the commutative property and the associative property in all calculations. This makes it convenient, and these laws are useful for everyone, including those who do mathematical calculations in high school, college, and even in the business world.
## Under the Commutative Property, Replacing the Numbers Gives the Same Answer
In the commutative property, if you swap numbers, the answer will be the same. The commutative law is satisfied by addition.
For example, we have the following.
It doesn’t matter how you rearrange the order; this is called the commutative property of addition.
The same is true for multiplication. The commutative property is satisfied not only in addition but also in multiplication. As shown below, the answer is the same in multiplication, even if the numbers are interchanged.
It doesn’t matter how you multiply, and this is called the commutative property of multiplication.
### Isn’t the Commutative Property Established by Subtraction and Division?
However, it is said that the commutative law is not valid for subtraction and division. Let’s try to calculate by actually swapping numbers.
For example, if we interchange the numbers for subtraction, we get the following.
Thus, the answers are different. In one equation, the answer is 2, and in the other equation, the answer is -2. Since we subtract different numbers, the answers will naturally be different. This is the reason why it is said that the commutative property is not valid in subtraction.
The same is true for division. In the case of division, it is as follows.
Using the commutative property with division changes the answer to the question. Therefore, the commutative law is not valid for division.
### Correcting the Signs and Making Them Negative Numbers, the Commutative Property Is Established
Since the commutative property is only valid for addition and multiplication, its usage is limited and seems to be meaningless. However, the commutative law that we learn in elementary school or junior high school mathematics is very important.
Why is the commutative property so important even though it only applies to addition and multiplication? It’s because by changing subtraction to addition, the commutative property becomes valid. For division, the commutative property is also available by changing it to multiplication.
Addition and subtraction are the same. Subtraction can be converted to addition. For example, the following are all the same equations and the same answer.
• $4-2=2$
• $4+(-2)=2$
Once you learn positive and negative numbers, you can understand that addition and subtraction are the same.
As mentioned above, the commutative property is valid for addition. So, by changing subtraction to addition of negative numbers, the commutative law is established.
What is important is the fact that the commutative property is valid even for negative numbers. We have explained above that the commutative property is not established for subtraction. In fact, however, we can use the commutative property of addition even for subtraction, because the commutative law is established by changing subtraction to addition.
-Use Fractions in Division and Use the Commutative Property
The same is true for division. By converting division to multiplication, we can use the commutative property of multiplication.
How do we convert division into multiplication? This is done by using fractions. By using reciprocal numbers, we can convert division into multiplication of fractions. Divisions are also multiplication of fractions, as shown below.
As mentioned above, the commutative property is valid in multiplication. As shown below, the answer is the same in multiplication, even if it is a fraction.
If you want to correct subtraction to addition, you just need to change the sign of the plus and minus. But in the case of division, you have to convert it into fractions. This requires a bit of work. In any case, you can use the commutative property even in division.
## The Associative Property That Can Change the Place of Parentheses
When we study the commutative property, there is the associative property that we learn at the same time. What kind of law is the associative law?
The associative property allows us to change the place of the parentheses freely. The same as the commutative property, the associative property is only valid for addition and multiplication. As shown below, the answer is the same even if we change the position of the parentheses in addition.
In math, the rule is to calculate the inside of the parentheses first. It doesn’t matter where the parentheses are placed in addition.
In the same way, the associative property is valid for multiplication.
Why is the associative law valid? It is because the meaning is the same with or without the parentheses. The answer is the same even if you remove the parentheses as follows.
The answer is the same no matter where you put the parenthesis, because it doesn’t matter if there are parentheses or not. The associative property is just a law of the obvious.
### The Associative Law That Can Be Calculated From Anywhere
The associative property is, in essence, a law that you can calculate from anywhere. For example, we have all done multiplication calculations by changing the order of the numbers. For example, how would you calculate the following?
• $2×6×13$
From left to right, the calculations are as follows.
• $\textcolor{red}{2×6}×13=\textcolor{red}{12}×13=156$
You need to calculate $12×13$, which makes the calculation more difficult. So let’s use the associative property and change the order of the numbers. For example, instead of $2×6×13$, we change the equation to $6×13×2$. In this case, we get the following.
• $\textcolor{red}{6×13}×2=\textcolor{red}{78}×2=156$
Thus, the calculation is easier; multiplying a number with fewer digits is less likely to result in calculation errors than multiplying two numbers by two digits.
### The Formula of Subtraction and Division Is Formed by Changing the Sign
The associative property is valid for addition and multiplication formulas. It is the same as the commutative property that cannot be applied to subtraction and division. For example, in subtraction, changing the parentheses will change the answer as follows.
If there are parentheses in the equation, we must first calculate the inside of the parentheses. So when there is subtraction, the answer changes by using the associative property.
The same thing happens in division. The answer changes when the parentheses change, as shown below.
For subtraction and division, the associative property is not valid.
However, in practice, it doesn’t matter if there is subtraction or division. Just like the commutative property, subtraction can be corrected to addition and division can be corrected to multiplication. It is as follows.
It can be modified into an equation for addition or multiplication, so that the associative property is established. The answer is the same no matter where you put the parenthesis. In other words, you can do the calculation from any place first.
In general, the associative property is not available for subtraction and division. However, by correcting it to addition or multiplication equations, the associative law becomes valid. Therefore, the associative property is a rule that can be used in all calculations.
## Use Multiplication of Fractions by Reciprocal Numbers Instead of Using Division in Math
Why do we need to understand the commutative property and the associative property, including elementary school and junior high school math? This is because they are essential for easier calculations. Understanding the commutative property and the associative property is especially important for division calculations.
When you convert subtraction to addition, the method is simple. Just make the addition and then make it a negative number. It is as follows.
• $2\textcolor{red}{-3}=2\textcolor{red}{+(-3)}$
For division, on the other hand, you can’t calculate it as it is. In order to change division to multiplication, you have to modify it to multiply fractions. By using reciprocal, we must change the shape of the numbers.
• $2\textcolor{red}{÷3}=2\textcolor{red}{×\displaystyle\frac{1}{3}}$
You can calculate from anywhere by using only multiplication equations.
We don’t use division in math above secondary school because the commutative property and the associative property is not valid. We don’t use division in math, including junior high school and high school math. Instead, we calculate after converting division into fractions.
And in the case of division, there are many cases where the numbers are indivisible. With fractions, on the other hand, you can get an answer even if you can’t divide. It is important to understand that division is less useful and has almost no use in mathematics.
The reason why we all convert division to multiplication in secondary school and above in math is so that the commutative property and associative property can be used. This makes it much easier to calculate and less likely to make calculation errors.
## Exercises: the Commutative Property and the Associative Property for Positive and Negative Numbers
Q1: Do the following calculation.
1. $(6 ÷ 15) × 5$
2. $-4 ÷ 3 ÷ 6 × 15$
For addition and subtraction, you can solve problems without trouble when you use the commutative law and associative law. On the other hand, if division is involved, you need to correct it to multiply by a fraction. So use the reciprocal number and make it a fraction.
(a)
$(6\textcolor{red}{÷15})×5$
$＝(6\textcolor{red}{×\displaystyle\frac{1}{15}})×5$
$=6×\displaystyle\frac{1}{3}=2$
In the division, we always correct it to multiply of fractions. Then, in this calculation, we calculate $\displaystyle\frac{1}{15}×5=\displaystyle\frac{1}{3}$ first. It’s easier to calculate than $6×\displaystyle\frac{1}{15}$.
(b)
$-4÷3÷6×15$
$=-4×\displaystyle\frac{1}{3}×\displaystyle\frac{1}{6}×15$
$=\textcolor{red}{-4×\displaystyle\frac{1}{6}}×\textcolor{blue}{\displaystyle\frac{1}{3}×15}$
$=\textcolor{red}{-\displaystyle\frac{2}{3}}×\textcolor{blue}{5}$
$=-\displaystyle\frac{10}{3}$
First, we convert division into multiplication of fractions. Then, by using the commutative property and the associative property, we replace the numbers to make them easier to calculate.
Calculating with $-4×\displaystyle\frac{1}{3}$ and $\displaystyle\frac{1}{6}×15$ is difficult. It is also difficult to multiply fractions by each other.
On the other hand, $-4 × \displaystyle\frac{1}{6}$ and $\displaystyle\frac{1}{3} × 15$ allow you to reduce the number, making the calculation easier. In a multiplication-only equation, you can move the numbers freely. In this way, we use the commutative law and the associative law.
## Learn the Definition of the Law and Use it in Formulas
There are several laws in mathematics. One of those laws is the commutative property and the associative property. Why do we need to understand the rule? It is because the laws make it easier to calculate the equations and reduce calculation errors.
Two of the most important laws in mathematics that we all use are the commutative property and the commutative property. Even in elementary school math, we all use the commutative property and the commutative property.
However, you have to understand the rules and definitions of the laws in order to use them in mathematical formulas. The commutative property and the commutative property are only valid for equations with addition or multiplication. So if there is subtraction or division, correct it to addition or multiplication. Note that it is easy to correct subtraction, but with division, you must change it to a fraction.
Also, if you learn the commutative law and the associative law, you can understand why no one uses division in math above junior high school. Division is not useful. Be sure to correct the multiplication of fractions by reciprocal numbers before you calculate equations. |
# Section 1.1 Linear Equations
Section Objectives
1. Recognize a single-variable equation as linear and solve it (0, 1, or inf. many soln's).
2. Translate a problem situation into an equation and solve.
A linear equation is an equation in which each variable appears only with an exponent of 1 and not in the denominator of a fraction nor in a radical.
For now, we will focus on linear equations of one variable. Every linear equation of one variable, $x$, can be written in the form $ax+b=0$.
### Solving Linear Equations
To solve an equation means to find ALL replacements for the variable that make the equation true. We typically solve linear equations by constructing a sequence of simpler, equivalent equations. At some point in this process, the solution becomes obvious.
#### For example...
Solve for $x$: $3(x-1)+x=-x+7$
$3x-3+x=-x+7$
$4x-3=-x+7$
$5x-3=7$
$5x=10$
$x=2$
#### General Guidelines for Solving Linear Equations
• Remove parentheses (if necessary) by using the distributive property.
• Combine like terms on each side of the equation.
• Use addition or subtraction to move all variables to one side of the equation and all constants to the other.
• If possible, use multiplication or division to get an equation of the form $x=$ constant.
• If the step above is not possible, your equation is either an identity (every number is a sol'n) or a contradiction (no number is a sol'n).
Important note: At any step in the process above, it could be helpful to clear fractions by multiplying both sides of the equation by the LCM of all denominators.
### Application Problems
In mathematics, there is a well-known, very broad, four-step problem solving process:
1. Understand the problem.
• What is the problem asking for?
• What information is given?
• Assign variables to unspecified quantities.
• Sketch any helpful tables or diagrams.
2. Devise a plan.
• Translate the problem situation into an equation.
3. Carry out the plan.
4. Look back.
• Check your solution in the original wording of the problem.
Steps 1 & 2 involve defining variables and translating words to equations. We'll do examples in class, but this sheet may help you with your translation skills.
#### Example
When walking, Jose burns 96 calories per mile and Sara burns 64 calories per mile. One day the two of them walk a total of 7 miles. Let $x$ represent the number of miles walked by Sara.
a. Write an algebraic expression for the total number of calories burned by the two of them.
b. Together they burn a total of 505.6 calories. How far did each person walk? |
One of the most practical uses of differentiation is finding the maximum or minimum value of a real-world function. In the following example, you calculate the maximum volume of a box that has no top and that is to be manufactured from a 30-inch-by-30-inch piece of cardboard by cutting and folding it as shown in the figure.
What dimensions produce a box that has the maximum volume? Mathematics often seems abstract and impractical, but here’s an honest-to-goodness practical problem. If a manufacturer can sell bigger boxes for more money, and he or she is making a million boxes, you better believe he or she will want the exact answer to this question:
1. Express the thing you want maximized, the volume, as a function of the unknown, the height of the box (which is the same as the length of the cut).
2. Determine the domain of your function.
The height can’t be negative or greater than 15 inches (the cardboard is only 30 inches wide, so half of that is the maximum height). Thus, sensible values for h are 0 ≤ h ≤ 15.
3. Find the critical numbers of V(h) in the open interval (0, 15) by setting its derivative equal to zero and solving. And don’t forget to check for numbers where the derivative is undefined.
Because 15 isn't in the open interval (0, 15), it doesn’t qualify as a critical number. And because this derivative is defined for all input values, there are no additional critical numbers. So, 5 is the only critical number.
4. Evaluate the function at the critical number, 5, and at the endpoints of the interval, 0 and 15, to locate the function’s max.
The extremum (dig that fancy word for maximum or minimum) you’re looking for doesn’t often occur at an endpoint, but it can — so don’t fail to evaluate the function at the interval’s two endpoints.
You’ve got your answer: a height of 5 inches produces the box with maximum volume (2000 cubic inches). Because the length and width equal 30 – 2h, a height of 5 inches gives a length and width of 30 – 2 · 5, or 20 inches. Thus, the dimensions of the desired box are 5 inches by 20 inches by 20 inches. |
# How do you use implicit differentiation to find (dy)/(dx) given 2xy-y^2=3?
Oct 9, 2016
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{y - 1}$
#### Explanation:
Given
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{2 x y} - \textcolor{b l u e}{{y}^{2}} = \textcolor{g r e e n}{3}$
Note 1:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{\frac{d \left(2 x y\right)}{\mathrm{dx}}} = 2 x \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + y \cdot \frac{d \left(2 x\right)}{\mathrm{dx}} = \textcolor{red}{2 \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y}$
Note 2:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{\frac{d \left({y}^{2}\right)}{\mathrm{dx}}} = \textcolor{b l u e}{2 y \frac{\mathrm{dy}}{\mathrm{dx}}}$
Note 3:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{\frac{d \left(3\right)}{\mathrm{dx}}} = \textcolor{g r e e n}{0}$
Therefore $2 x y - {y}^{2} = 3$
$\Rightarrow \textcolor{red}{2 \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y} - \textcolor{b l u e}{2 y \frac{\mathrm{dy}}{\mathrm{dx}}} = \textcolor{g r e e n}{0}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} - y \frac{\mathrm{dy}}{\mathrm{dx}} = - y$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} \left(1 - y\right) = - y$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{y - 1}$ |
UPSKILL MATH PLUS
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In this section, let us look at the general formula to find the $$n^{th}$$ term of a geometric progression.
We know that the general form of a geometric progression is $$a$$, $$ar$$, $$ar^2$$,$$...ar^{n - 1}$$ with the common ratio $$r$$.
$$\text{Term } 1 = t_1 = a \times r^0 = a \times 1 = a$$
$$\text{Term } 2 = t_2 = a \times r^1 = ar$$
$$\text{Term } 3 = t_3 = a \times r^2 = ar^2$$
$$.$$ $$.$$
$$.$$ $$.$$
$$.$$ $$.$$
$$.$$ $$.$$
$$.$$ $$.$$
$$\text{Term } n = t_{n} = a \times r^{n - 1} = ar^{n - 1}$$
Therefore, the general form of G.P. is $$ar^{n - 1}$$.
Ratio between any two consecutive terms in a G.P.:
$$\frac{t_2}{t_1} = \frac{ar}{a} = r$$
$$\frac{t_3}{t_2} = \frac{ar^2}{ar} = r$$
$$\frac{t_n}{t_{n - 1}} = \frac{ar^n}{ar^{n - 1}} = \frac{ar^n}{a \times r^n \times r^{-1}} = \frac{1}{r{-1}} = r$$
Thus, the ratio between any two consecutive terms of a G.P. is $$r$$.
Condition for three numbers to be in G.P.:
Let the three numbers $$a$$, $$b$$, $$c$$ in G.P.
$$\text{First term }= a$$
$$\text{Second term }= b = ar$$
$$\text{Third term }= c = ar^2$$
$$ac = a \times ar^2 = a^2r^2 = (ar)^2 = b^2$$
Thus, three non-zeroes numbers are in G.P. only if $$b^2 = ac$$.
Important!
1. When the product of the three consecutive terms of a G.P. is given, then we can consider the terms as $$\frac{a}{r}$$, $$a$$, $$ar$$.
2. When the product of the first four terms of a G.P. are given, then we can consider the terms as $$\frac{a}{r^3}$$, $$\frac{a}{r}$$, $$ar$$, $$ar^3$$.
3. When every term of a G.P. is divided or multiplied by the same non-zero constant, the newly-formed sequence is also in G.P. |
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Sine & Cosine ( Math | Algebra | Trig | Sine&Cosine)
Trig Functions: Sine and Cosine Definition
Definition: An algebraic approach
From defining a few general properties of the sine and cosine functions, we can algebraically derive the sine and cosine functions themselves.
First, define the sine and cosine functions to have these properties:
sin x = cos x
cos x = - sin x
sin 0 = 0
cos 0 = 1
Now, write the sine function as an arbitrary power series, in that let sin x = a0x0 + a1x1 + a2x2 + a3x3 + a4x4 + ...
sin 0 = 0 = a0
a0 = 0
By differentiating the power series and equating it with the cosine by the original properties, sin x = cos x = 0 + (1)a1x0 + (2)a2x1 + (3)a3x2 + (4)a4x3 + ...
cos 0 = 1 = (1)a1
a1 = 1
Continuing, cos x = -sin x = -(0 + 0 + (2)(1)a2x0 + (3)(2)a3x1 + (4)(3)a4x2 + ...)
-sin 0 = 0 = -(2)a2
a2 = 0
-sin x = cos x = -(0 + 0 + 0 + (3)(2)(1)a3x0 + (4)(3)(2)a4x1 + ...)
cos 0 = -1 = (3)(2)(1)a3
a3 = -1/3!
cos x = sin x = 0 + 0 + 0 + 0 + (4)(3)(2)(1)a4x0 + ...
sin 0 = 0 = (4)(3)(2)(1)a4
a4 = 0
Continue on and you get values of all an
if n is even then an = 0
if n is odd then an = 1/n! alternating positive and negative.
Or stated as:
a2n = 0
a2n+1 = (-1)n/(2n+1)!
Plugging these values into the equation for sine:
sin x = x1/1! - x3/3! + x5/5! - x7/7! + x9/9! - ... + (-1)nx2n+1/(2n+1)! + ...
= (-1)nx2n+1 / (2n+1)!
and for cosine:
cos x = 1 - x2/2! + x4/4! - x6/6! + x8/8! - ... + (-1)nx2n/(2n)! + ...
= (-1)nx2n / (2n)!
These series converge for all x ÎÂ.
Source: Jeff Yates. |
# How to describe the intersection of two sets?
The following is a homework problem:
Let \begin{align*} W_1 &= \{(a_1, a_2, a_3) \in\mathbb{R}^3 \mid a_1 = 3 a_2\text{ and }a_3 = -a_2\}\\ W_2 &= \{(a_1, a_2, a_3) \in \mathbb{R}^3 \mid a_1 - 4 a_2 - a_3 = 0\} \end{align*} Describe the intersection of W1 and W2 and observe that it is a subspace.
I realize that the intersection is $$\{ (a_1, a_2, a_3) \in \mathbb{R}^3 \mid a_1 = 3 a_2\text{ and }a_3 = -a_2\text{ and }a_1 - 4 a_2 - a_3 = 0 \}.$$ Further, I've observed that this seems to just be the set, $W_1$, based on the conditions. I just kind of haphazardly messed with the formulas to try to figure stuff out with no real methodology.
What is the method to solving these kinds of problems? How do I solve this and similar problems?
-
Assume that $(a_1,a_2,a_3)\in W_1\cap W_2$. As you note, you must have $a_1=3a_2$ and $a_3=-a_2$, in order for the point to be in $W_1$. And you must have $a_1-4a_2-a_3=0$ in order to be in $W_2$.
That means that the point must satisfy the following three equations: \begin{align*} a_1 -3a_2 &= 0\\ a_2 + a_3 &= 0\\ a_1 - 4a_2 - a_3 &= 0 \end{align*} So the intersection consists precisely of all solutions to this system of linear equations.
Here, we obtain that if $a_1 = 3a_2$ and $a_2=-a_3$, then necessarily $a_1-4a_2 - a_3 = 0$, so that the third equation is a consequence of the first two, which is what you observed: it turns out that $W_1\subseteq W_2$, so $W_1\cap W_2 = W_1$.
In general, if your sets are given as solution sets of equations, then finding the intersection is equivalent to solving a system of equations; whether this is easy or not will depend on the equations. And for even more generality, it may be extremely hard to say what the intersection is explicitly; for instance, trying to figure out exactly what integers are in $$\{ p\in\mathbb{Z}\mid p\text{ is prime}\}\cap \{n\in\mathbb{Z}\mid n=p+2\text{ for some prime }p\}$$ is the same as trying to figure out exactly what the twin primes are, and it is still unknown whether there are infinitely many of them or not.... |
# Perpendicular Line Formula
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## Perpendicular Line
Perpendicular lines are the two distinct lines that intersect at each other at 90°. Have you noticed anything common between the joining corners of your walls, or the letter “L”? They are the straight lines known as perpendicular lines that meet each other at a specific angle – the right angle.
We say that a line is perpendicular to another line if the two lines meet at an angle of 90°. Let us understand the concept of perpendicular lines in detail.
## What is Perpendicular?
A perpendicular is a straight line that makes an angle of 90° with another line. 90° is also called a right angle and is marked by a little square between two perpendicular lines as shown in the figure. Here, the two lines intersect at a right angle, and hence, are said to be perpendicular to each other.
Now, look at the examples of lines that are not perpendicular. These lines are either not meeting at all, or intersecting at an angle that is not 90°. Hence, they are not perpendicular.
Perpendicular lines, in math, are two lines that intersect each other and
Some shapes which have perpendicular lines are:
• Square
• Right-angled triangle
• Rectangle
## Properties of Perpendicular Line
We have already seen how the perpendicular lines look like. If there is an “L” shape in a figure, the corresponding angle at the vertex is a right angle. Perpendicular lines always intersect each other, however, all intersecting lines are not always perpendicular to each other. The two main properties of perpendicular lines are:
• Perpendicular lines always meet or intersect each other.
• The angle between any two perpendicular lines is always equal to 90°
## How to Draw Perpendicular Lines?
To draw a perpendicular line, all we need is a scale (ruler), a compass or a protractor. We will discuss how to draw the perpendicular lines step by step using a compass and a protractor. So, we can draw perpendicular lines for a given line in two ways.
• Using a protractor
• Using a compass
## Drawing a Perpendicular Line Using Protractor
A protractor, in math, is considered an important measuring instrument in the geometry box. This tool not only helps us measure an angle in degrees, but also helps in drawing perpendicular lines. To draw a perpendicular line at point P on the given line, follow the steps given below.
• Step 1: Place the baseline of the protractor along the line such that its center is at P.
• Step 2: Mark a point B at 90° of the protractor.
Step 3: Remove the protractor and join P and B. So, BP is a line that is perpendicular to the given line.
## Drawing a Perpendicular Line Using Compass
We can also draw a perpendicular line using a compass. To draw a perpendicular line at a point P on a line, follow the steps given below.
• Step 2: Placing the pointer of the compass at P, construct a semi-circle that cuts the line at A and B.
Step 3: Without disturbing the radius of the compass, draw two arcs that cut the semi-circle at C and D by placing the pointer of the compass at A and B respectively.
Step 4: Keeping the same radius, draw two intersecting arcs that intersect at Q by placing the pointer of the compass at C and D.
Step 5: Join P and Q, which forms a perpendicular on the given line. Now, PQ is perpendicular to AB.
## Perpendicular Lines and Parallel Lines
Two straight lines are said to be parallel if they are equidistant from each other and never meet, no matter how much they may be extended in either directions. Observe the lines shown below to see the difference between perpendicular lines and parallel lines. In the figure, AB is perpendicular to CD and PQ is parallel to RS.
The symbol used to show that the lines are parallel is ||, and we express it as: PQ || RS. You can find parallel lines all around you. For example, zebra crossings on the road or the opposite sides of your ruler, and many others too! The table shown below differentiates between parallel lines and perpendicular lines.
## Difference Between Parallel and Perpendicular Lines
.A perpendicular line is a straight line through a point. It makes an angle of 90 degrees with a particular point through which the line passes. Coordinates and line equation is the prerequisite to finding out the perpendicular line.
Consider the equation of the line is ax + by + c = 0 and coordinates are (x1, y1), the slope should be − a/b. If one line is perpendicular to this line, the product of slopes should be -1. Let m1 and m2 be the slopes of two lines, and if they are perpendicular to each other, then their product will be -1.
### Solved Example
Question: Check whether 2x + 3y + 5 = 0 and 3x – 2y + 1 = 0 are perpendicular or not.
Solution:
The given equations of lines are:
2x + 3y + 5 = 0 and 3x – 2y + 1 = 0
To check whether they are perpendicular to each other, find out the slopes of both lines. If the product of their slopes is -1, these lines are perpendicular to each other.
Since the product of slope is -1, the given lines are perpendicular to each other.
## Equation of a Line Perpendicular to a Line
We will learn how to find the equation of a line perpendicular to a line.
Prove that the equation of a line perpendicular to a given line ax + by + c = 0 is bx – ay + λ = 0, where λ is a constant.
Let m1 be the slope of the given line ax + by + c = 0 and m2
be the slope of a line perpendicular to the given line.
Then,
Algorithm for directly writing the equation of a straight line perpendicular to a given straight line:
To write a straight line perpendicular to a given straight line we proceed as follows:
Step I: Interchange the coefficients of x and y in equation ax + by + c = 0.
Step II: Alter the sign between the terms in x and y of equation i.e., If the coefficient of x and y in the given equation are of the same signs make them of opposite signs and if the coefficient of x and y in the given equation are of the opposite signs make them of the same sign.
Step III: Replace the given constant of equation ax + by + c = 0 by an arbitrary constant.
For example, the equation of a line perpendicular to the line 7x + 2y + 5 = 0 is 2x – 7y + c = 0; again, the equation of a line, perpendicular to the line 9x – 3y = 1 is 3x + 9y + k = 0.
Note:
Assigning different values to k in bx – ay + k = 0 we shall get different straight lines each of which is perpendicular to the line ax + by + c = 0. Thus we can have a family of straight lines perpendicular to a given straight line.
Solved examples to find the equations of straight lines perpendicular to a given straight line
1. Find the equation of a straight line that passes through the point (-2, 3) and perpendicular to the straight line 2x + 4y + 7 = 0.
Solution:
The equation of a line perpendicular to 2x + 4y + 7 = 0 is
4x – 2y + k = 0 …………………… (i) Where k is an arbitrary constant.
According to the problem equation of the perpendicular line 4x – 2y + k = 0 passes through the point (-2, 3)
Then,
4 ∙ (-2) – 2 ∙ (3) + k = 0
⇒ -8 – 6 + k = 0
⇒ – 14 + k = 0
⇒ k = 14
Now putting the value of k = 14in (i) we get, 4x – 2y + 14 = 0
Therefore the required equation is 4x – 2y + 14 = 0.
2. Find the equation of the straight line which passes through the point of intersection of the straight lines x + y + 9 = 0 and 3x – 2y + 2 = 0 and is perpendicular to the line 4x + 5y + 1 = 0.
Solution:
The given two equations are x + y + 9 = 0 …………………… (i) and 3x – 2y + 2 = 0 …………………… (ii)
Multiplying equation (i) by 2 and equation (ii) by 1 we get
2x + 2y + 18 = 0
3x – 2y + 2 = 0
Adding the above two equations we get, 5x = – 20
⇒ x = – 4
Putting x = -4 in (i) we get, y = -5
Therefore, the co-ordinates of the point of intersection of the lines (i) and (ii) are (- 4, – 5).
Since the required straight line is perpendicular to the line 4x + 5y + 1 = 0, hence we assume the equation of the required line as
5x – 4y + λ = 0 …………………… (iii)
Where λ is an arbitrary constant.
By problem, the line (iii) passes through the point (- 4, – 5); hence we must have,
⇒ 5 ∙ (- 4) – 4 ∙ (- 5) + λ = 0
⇒ -20 + 20 + λ = 0
⇒ λ = 0.
Therefore, the equation of the required straight line is 5x – 4y = 0.
### Examples on Perpendicular Lines
Solution
Since the two lines AB and CD intersect each other at right angles, there are 4 right angles at the intersecting point. Hence, Angle AOD = Angle DOB = Angle BOC = Angle AOC = 90°.
Example 2:
In the following figure, AB is perpendicular to CD. If Angle BOC = 90°, then find the value of x.
Solution
Since Angle BOC = 90°, we can write: x + 63° = 90°. Therefore, x = 27°
## FAQs on Perpendicular Line
### What is Perpendicular?
Perpendicular means standing at right angles to the plane of the horizon.
### What are Perpendicular Angles?
The perpendicular angle can be referred to as the right angle (90°) that is formed when two perpendicular lines intersect each other.
### What is the Perpendicular Symbol?
The symbol used in between the two lines to say that they are perpendicular is ⊥.
### What is Perpendicular Lines Formula?
To find the slope of two lines we use the perpendicular line’s formula. The perpendicular line formula is defined as the product of two slopes m1 and m2 is -1. It is mathematically expressed as (m1×m2 = -1) or m1 = -1/m2.
### What is the Difference Between Perpendicular and Parallel Lines?
Perpendicular lines are those lines that intersect each other at right angles (90°). Parallel lines are those lines that do not intersect anywhere and are always the same distance apart.
### Do Perpendicular Lines Touch Each Other?
Yes, one of the properties of perpendicular lines is that they intersect and touch each other.
### How do You Know if Two Lines Are Perpendicular?
If two lines intersect each other making an angle of 90°, then those two lines are perpendicular to each other.
### Are Diagonals of a Rhombus Perpendicular to Each Other?
The diagonals of a rhombus intersect each other at an angle of 90°, so we can say that the diagonals of a rhombus are perpendicular to each other.
### Slope of Perpendicular Lines
Slope of perpendicular lines are such that the slope of one line is the negative reciprocal of the slope of another line. If the slopes of the two perpendicular lines are m1, m2, then we can represent the relationship between the slope of perpendicular lines with the formula m1.m2 = -1. The product of the slope of perpendicular lines is -1.
Let us learn more about the slope of perpendicular lines, their derivation, with the help of examples, FAQs.
## What Is the Slope of Perpendicular Lines?
Slope of a perpendicular line can be computed from the slope of a given line. The product of the slope of a given line and the slope of the perpendicular line is equal to -1. If the slope of a line is m1 and the slope of the perpendicular line is m2, then we have m1.m2 = -1.
The equations of two perpendicular lines are such that the coefficients of x and y are interchanged. For the equation of a line ax + by + c1 = 0, the equation of the perpendicular line is bx – ay + c2 = 0.
## Formula for Slope of Perpendicular Lines
The formula for the slope of two perpendicular lines is that the product of the slopes of individual lines is equal to -1. If the slope of the individual lines is m1 and m2 respectively, then the formula to represent the slope of two perpendicular lines is m1.m2 = -1.
### Formula of Slope of Perpendicular Lines: m1.m2 = -1
Further, the slope of each of the perpendicular lines can be found from the equations of the lines, or from the points on the line. The slope of a line having slope-intercept form of the equation of a line – y = mx + c is m, and the slope of a line having a general equation of a line ax + by + c = 0 is -a/b. Also, the slope of the line passing through any two points (x1, y1), and (x2, y2) is m = (y2 – y1)/(x2 – x1).
## Derivation of Slope of Perpendicular Lines
The slope of the perpendicular line can be derived from the formula of the angle between two lines. For two lines having slopes m1 and m2, the angle between the two lines is obtained using Tanθ.
Tanθ = (m1 – m2)/(1 + m1.m2)
The angle between two perpendicular lines is 90º, and we have Tan90º= ∞
Tan90º = (m1 – m2)/(1 + m1.m2)
∞ = (m1 – m2)/(1 + m1.m2)
n/20 = (m1 – m2)/(1 + m1.m2)
Here the denominator of the right hand side of the expression can be equalized to zero.
1 + m1.m2 = 0
m1.m2 = -1
m2 = -1/m1
Thus the slope of the perpendicular line is equal to the negative inverse of the slope of the given line.
## How to Find Slope of Perpendicular Lines?
The slope of perpendicular lines can be calculated by knowing the slope of one of the two perpendicular lines. Here we take the equation of one of the perpendicular lines as the general form of the equation of a line. The general form of equation of a line is as follows.
ax + by + c = 0
Let us convert this above equation into the slope-intercept form of the equation of a line.
y = -ax/b – c/b
The slope of this line is m1 = -a/b, and we have the slope of perpendicular lines formula as m1.m2 = -1. Thus we can find the slope of the other perpendicular lines as follows.
(-a/b).m2 = -1
m2= b/a
Thus the required equation of slope of the perpendicular line is b/a.
Let us understand this with the help of a simple numeric example. The given equation of a line is 5x + 3y + 7 = 0. Now let us try to find the slope of the perpendicular line.
Comparing this equation 5x + 3y + 7 = 0, with ax + by + c = 0, we have a = 5, b = 3. The slopes of perpendicular lines is m1 = -a/b = -5/3, and m2 = b/a = 3/5.
## Examples on Slope of Perpendicular Lines
Example 2: Find the equation of a line passing through (4, -3), and has the slope of perpendicular line as 2/3.
Solution:
The given point is
= (4, -3), and the slope of the perpendicular line is
.
We know that the product of the slopes of two perpendicular lines is m1.m2 = -1.
2/3 . m2 = -1
m2 = -1 × 3/2
m2 = -3/2
The required equation of the line can be found using the formula of point slope form.
y – (-3) = -3/2(x – 4)
2(y + 3) = -3(x – 4)
2y + 6 = -3x + 12
3x + 2y + 6 – 12 = 0
3x + 2y – 6 = 0
Thus the required equation of the line is 3x + 2y – 6 = 0.
## FAQs on Slope of Perpendicular Lines
### What Is Slope of Perpendicular Lines in Coordinate Geometry?
The slope of perpendicular lines in coordinate geometry is such that the slope of one line is the negative reciprocal of the slope of another line. If the slopes of the lines is m1 and m2 respectively, then we have m1.m2 = -1. The product of the slopes of two perpendicular lines is -1.
### What Is the Formula To Find Slope of Perpendicular Line?
The formula for the slope of perpendicular lines is m1.m2 = -1. The product of the slopes of perpendicular lines is equal to -1. Alternatively, we can say that m2 = -1/m1, that is the slope of one line is equal to the negative reciprocal of another line.
### How To Find Equation Of Line From Slope of Perpendicular Line?
The equation of a line from the slope of a perpendicular line is obtained using the point-slope form or the slope-intercept form of the equation of a line. From the slope of the perpendicular line, we can find the slope of the required line by taking its negative reciprocal. If the slope of the perpendicular line is m1 and the slope of the required line is m2 then we have m2 = -1/m1. Further by using the slope of the line we can find the equation of the line from
, or y = mx + c.
### How To Find the Slope of A Perpendicular Line?
The slope of a perpendicular line from the slope of a given line is obtained by taking the negative reciprocal of the slope of the given line. If the slope of the given line is m1 and the slope of a perpendicular line is m2 then we have m2 = -1/m1.
### How Do You Derive the Relationship Between the Slopes of Perpendicular Lines?
The slope of perpendicular line can be calculated from the trigonometric ratio of Tan. The formula for finding the slope of
## ACT Math : How to find the equation of a perpendicular line
Example Questions
GIA SƯ TOÁN BẰNG TIẾNG ANH
GIA SƯ DẠY SAT
Math Formulas |
# 10th Maths Chapter 1 Exercise 1.5 Solutions
10th Standard Maths Chapter 1 Exercise 1.5 Relations and Functions Guide Book Back Answers Solutions. TN 10th SSLC Samacheer Kalvi Guide. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here
## 10th Maths Chapter 1 Exercise 1.5 Solutions
### 1. Using the functions f and g given below, find fog and gof. Check whether fog = gof.
• (i) f(x) = x – 6, g(x) = x2
• (ii) f(x) = 2x, g(x) = 2×2 – 1
• (iii) f(x) = x+63g(x) = 3 – x
• (iv) f(x) = 3 + x, g(x) = x – 4
• (v) f(x) = 4×2– 1,g(x) = 1 + x
Solu.:
(i) f(x) = x – 6, g(x) = x2
fog(x) = f(g(x)) = f(x2) = x2– 6 …………….. (1)
gof(x) = g(f(x)) = g(x – 6) = (x – 6)2
= x2 + 36 – 12x = x2 – 12x + 36 ……………… (2)
(1) ≠ (2)
∴ fog(x) ≠ gof(x)
(iii) f(x) = x+63 g(x) = 3 – x
(iv) f(x) = 3 + x, g(x) = x – 4
fog(x) = f(g(x)) = f(x – 4) = 3 + x – 4
= x – 1 ………… (1)
gof(x) = g(f(x)) = g(3 + x) = 3 + x – 4
= x – 1 ……………… (2)
Here fog(x) = gof(x)
(v) f(x) = 4×2 – 1, g(x) = 1 + x
fog(x) = f(g(x)) = f(1 + x) = 4(1 + x)2 – 1
= 4(1 + x2 + 2x) – 1 = 4 + 4×2 + 8x – 1
= 4×2 + 8x + 3 ……………. (1)
gof(x) = g(f(x)) = g(4×2 – 1)
= 1 + 4×2 – 1 = 4×2 …………….. (2)
(1) ≠ (2)
∴ fog(x) ≠ gof(x)
### 2. Find the value of k, such that f o g = g o f
(i) f(x) = 3x + 2, g(x) = 6x – k
f(x) = 3x + 2 ;g(x) = 6x – k
fog = f[g(x)]
= f (6x – k)
= 3(6x – k) + 2
= 18x – 3K + 2
g0f= g [f(x)]
= g (3x + 2)
= 6(3x + 2) – k
= 18x + 12 – k
But given fog = gof.
18x – 3x + 2 = 18x + 12 – k
-3k + 2 = 12 – k
-3 k + k = 12-2
-2k = 10
k = −102 = -5
The value of k = -5
(ii) f(x) = 2x – k, g(x) = 4x + 5
f(x) = 2x – k ; g(x) = 4x + 5
fog = f[g(x)]
= f(4x + 5)
= 2(4x + 5) – k
= 8x + 10 – k
gof = g [f(x)]
= g(2x – k)
= 4(2x – k) + 5
= 8x – 4k + 5
But fog = gof
8x + 10 – k = 8x – 4k + 5
-k + 4k = 5 – 10
3k = -5
k = −53
The value of k = −53
Solu.:
### 4. (i) If f (x) = x2 – 1, g(x) = x – 2 find a, if g o f(a) = 1.
(a) Find k, if f(k) = 2k -1 and
fof (k) = 5.
Solu:
(i) f(x) = x2 – 1 ; g(x) = x – 2 .
gof = g [f(x)]
= g(x2 – 1)
= x2 – 1 – 2
= x2 – 3
given gof (a) = 1
a2 – 3 = 1 [But go f(x) = x2 – 3]
a2 = 4
a = 4–√ = ± 2
The value of a = ± 2
(ii) f(k) = 2k – 1 ; fof(k) = 5
fof = f[f(k)]
= f(2k – 1)
= 2(2k – 1) – 1
= 4k – 2 – 1
= 4k – 3
fof (k) = 5
4k – 3 = 5
4k = 5 + 3
4k = 8
k = 84 = 2
The value of k = 2
### 5. Let A,B,C ⊂ N and a function f: A → B be defined by f(x) = 2x + 1 and g : B → C be defined by g(x) = x2. Find the range of fog and gof
Solu.:
f(x) = 2x + 1
g(x) = x2
fog(x) = fg(x)) = f(x2) = 2×2 + 1
gof(x) = g(f(x)) = g(2x + 1) = (2x + 1)2
= 4×2 + 4x + 1
Range of fog is
{y/y = 2×2 + 1, x ∈ N}
Range of gof is
{y/y = (2x + 1)2, x ∈ N}.
### 6. Let f(x) = x2 – 1. Find (i) fof (ii) fofof
Solu.:
f(x) = x2 – 1
(i) fof = f[f{x)]
= f(x2 – 1)
= (x2 – 1)2 – 1
= x4 – 2×2 + 1 – 1
= x4 – 2×2
(ii) fofof = fof[f(x)]
= fof (x2 – 1)
= f(x2 – 1)2 – 1
= f(x4 – 2×2 + 1 – 1)
= f (x4 – 2×2)
fofof = (x4 – 2×2)2 – 1
### 7. If f: R → R and g : R → R are defined by f(x) = x5 and g(x) = x4 then check if f,g are one-one and fog is one-one?
Solu.:
f(x) = x5
g(x) = x4
fog = fog(x) = f(g(x)) = f(x4)
= (x4)5 = x20
f is one-one, g is not one-one.
∵ g(1) = 14 = 1
g(-1) = ( -1)4 = 1
Different elements have same images
fog is not one-one. [∵ fog (1) = fog (-1) = 1]
### 8. Consider the functions f(x), g(x), h(x) as given below. Show that
(f o g) o h = f o(g o h) in each case.
(i) f(x) = x – 1, g(x) = 3x + 1 and h(x) = x2
(ii) f(x) = x2, g(x) = 2x and h(x) = x + 4
(iii) f(x) = x – 4, g(x) = x2 and h(x) = 3x – 5
Solu.:
(i) f(x) = x – 1, g (x) = 3x + 1, h(x) = x2
fog (x) = f[g(x)]
= f(3x + 1)
= 3x + 1 – 1
fog = 3x
(fog) o h(x) = fog [h(x)] ,
= fog (x2)
= 3(x2)
(fog) oh = 3×2 …..(1)
goh (x) = g[h(x)]
= g(x2)
= 3(x2) + 1
= 3×2 +1
fo(goh) x = f [goh(x)]
= f[3×2 + 1]
= 3×2 + 1 – 1
= 3×2 ….(2)
From (1) and (2) we get
(fog) oh = fo (goh)
Hence it is verified
(ii) f(x) = x2 ; g (x) = 2x and h(x) = x + 4
(fog) x = f[g(x)]
= f (2x)
= (2x)2
= 4×2
(fog) oh (x) = fog [h(x)]
= fog (x + 4)
= 4(x + 4)2
= 4[x2 + 8x + 16]
= 4×2 + 32x + 64 …. (1)
goh (x) = g[h(x)]
= g(x + 4)
= 2(x + 4)
= 2x + 8
fo(goh) x = fo [goh(x)]
= f[2x + 8]
= (2x + 8)2
= 4×2 + 32x + 64 …. (2)
From (1) and (2) we get
(fog) oh = fo(goh)
(iii) f(x) = x – 4 ; g (x) = x2; h(x) = 3x – 5
fog (x) = f[g(x)]
= f(x2)
= x2 – 4
(fog) oh (x) = fog [h(x)]
= fog (3x – 5)
= (3x – 5)2 – 4
= 9×2 – 30x + 25 – 4
= 9×2 – 30x + 21 ….(1)
goh (x) = g[h(x)]
= g(3x – 5)
= (3x – 5)2
= 9×2 + 25 – 30x
fo(goh)x = f[goh(x)]
= f[9×2 – 30x + 25]
= 9×2 – 30x + 25 – 4
= 9×2 – 30x + 21 ….(2)
From (1) and (2) we get
(fog) oh = fo(goh)
### 9. Let f ={(-1, 3),(0, -1),(2, -9)} be a linear function from Z into Z . Find f(x).
Solu.:
f ={(-1, 3), (0, -1), 2, -9)
f(x) = (ax) + b ………… (1)
is the equation of all linear functions.
∴ f(-1) = 3
f(0) = -1
f(2) = -9
f(x) = ax + b
f(-1) = -a + b = 3 …………… (2)
f(0) = b = -1
-a – 1 = 3 [∵ substituting b = – 1 in (2)]
-a = 4
a = -4
The linear function is -4x – 1. [From (1)]
### 10.In electrical circuit theory, a circuit C(t) is called a linear circuit if it satisfies the superposition principle given by C(at1 + bt2) = aC(t1) + bC(t2), where a, b are constants. Show that the circuit C(t) = 31 is linear.
Solu.:
Given C(t) = 3t
C(at1) = 3at1 …. (1)
C(bt2) = 3 bt2 …. (2)
Add (1) and (2)
C(at1) + C(bt2) = 3at1 + 3bt2
C(at1 + bt2) = 3at1 + 3bt2
= Cat1 + Cbt2 [from (1) and (2)]
∴ C(at1 + bt2) = C(at1 + bt2)
The superposition principle is satisfied.
∴ C(t) = 3t is a linear function. |
# Alternate Definition of a Group
A set $G$ with binary composition denoted multiplicatively is a group if
(i) The composition is associative.
(ii) For every pair of elements $a,b \in G$, the equations $ax = b$ and $ya = b$ have unique solutions in $G$.
Proof: Binary operation implies that the set $G$, under consideration is closed under the operation. Now to prove that the set $G$ is a group we have to show the left identity exists and each element of $G$ possesses left inverse with respect to the operation under consideration.
It is given that for every pair of elements $a,b \in G$ the equation $ya = b$ has a solution in $G$. Therefore, say $e \in G$ such that $e \cdot a = a$.
Now, let us suppose that $b$ is any arbitrary element of $G$. Therefore, there exist $e \in G$ such that $ax = b$.
Thus
$b = ax$
$\Rightarrow eb = e\left( {ax} \right)$
$\Rightarrow eb = \left( {ea} \right)x$
$\, \Rightarrow eb = ax$
$\Rightarrow eb = b$
Therefore there exist $e \in G$ such that $eb = b, \forall b \in G$
$\therefore e$ is the left identity.
Now, let $b = e \in G$ be an element. $ya = b \Rightarrow ya = e$ $\Rightarrow y$ is inverse of $a$ in $G$.
Let $y = {a^{ - 1}}$ such that ${a^{ - 1}}a = e$, then ${a^{ - 1}} \in G$ as $ya = e$ has got solution in $G$.
Thus ${a^{ - 1}}$ is the left inverse of $a$ in $G$. Therefore each element of possesses left inverse. Hence $G$ is a group for the given composition if the postulate and (ii) are satisfied. |
# What are Factors and Multiples
Let's see what are factors and multiples.
When we talked about divisibility of integers, we said that number a is divisible b if c=a/b is integer number.
If a is divisible by b, then b is a factor of a and a is multiple of b.
For example, since 12/3=4 then 3 is a factor of 12 and 12 is multiple of 3.
Also, since 12/4=3 then 4 is a factor of 12 and 12 is multiple of 4.
It is known that division is inverse of multiplication, so we can write that 12=3*4. Here, we clearly see that 3 and 4 are factors of 12.
But 12 has other factors as well. Since 12=1*12 and 12=2*6 then 1, 2, 6, 12 are also factors of 12.
Nice facts.
• Any number a (except 1) always has at least two factors: 1 and a.
• a is multiple of a and 1.
Indeed, since a=1*a then a has two factors: 1 and a.
Also, a is multiple of a and 1.
Example 1 . Find all positive factors of 8.
We need to check all numbers that are less or equal 8.
1 and 8 are factors of 8.
2 is factor of 8 because 8/2=4 (so 8 is divisible by 2).
4 is also factor of 8 because 8=2*4.
3 is not factor of 8 because 8 is not divisible by 3.
5,6,7 are also not factors of 8.
So, positive factors of 8 are 1,2,4,8.
Next example.
Example 2. Find all factors of 10.
We need to check all numbers that are less or equal 10.
1 and 10 are factors of 10.
Other factors are 2 and 5.
So, positive factors are 1,2,5,10.
Now, we need to find negative factors. For this just write minus in front of positive factors: -1,-2,-5,-10.
So, factors of 10 are -10,-5,-2,-1,1,2,5,10.
Now, let's see how to find multiples.
Example 3. Find positive multiples of 5.
We start from 5, because 5 is multiple of 5.
Now we add 5: 5+5=10. This is another multiple.
Now add 5: 10+5=15. This is another multiple.
In a similar fashion we can find that positive multiples are 5,10,15,20,...
Ellipsis means that there are infinitely many multiples.
Next example.
Example 4. Find all multiples of 8.
First we find positive multiples.
8 is first multiple, 8+8=16 is second, 16+8=24 is third etc.
To find non-positive multiples we subtract 8.
8-8=0, 0-8=-8, -8-8=-16, -16-8=-24 etc.
So, multiples of 8 are ...,-24,-16,-8,0,8,16,24,...
To find multiple of number a we need multiple it by all integers number.
For example, 8*(-3)=-24, 8*(-2)=-16, 8*0=0, 8*1=8, 8*2=16 etc.
Exercise 1. Find all factors of 1.
Next one.
Exercise 2. Find all factors of 24.
Factors of negative numbers are done in same way.
Exercise 3. Find all factors of -25. |
Welcome to my math blog! The purpose of this blog is to help you stay informed about our learning and experiences that have taken place during our math class. I have also included links your child (and you) may want to use in order to supplement math learning in 5th grade.
## Tuesday, November 18, 2014
### Minuend and Subtrahend?!?!
The focus today was adding and subtracting decimal numbers using an input/output table. This was a little more involved than it might sound. To begin, we took a few notes in our journal to help us remember important ideas and new vocabulary. To create a page for your journal, please watch: Add and Subtract Decimals.
Basically, we reviewed that you must ALWAYS LINE UP DECIMAL POINTS when adding or subtracting. We discussed what to do when adding or subtracting whole numbers with decimal numbers.... just turn the whole number into a decimal number (see illustration above). Then we discussed some new vocabulary: minuend and subtrahend ....
The minuend is the number that is being subtracted FROM. The subtrahend is the number that is being SUBTRACTED. The kids had no trouble at all with difference, they have known and used that vocabulary word for years!
Once we finished with these notes we began the assignment, "Under the Weather," from The Mailbox resource Teacher's Helper Aug/Sept 2009. On this assignment, students were working with input/output tables.... a number goes in and a new number comes out using a rule. The rule was given to the students at the top of each table, but the confusion came when working backwards! For example, if the "in" number was given, the student followed the given rule to determine the "out" number. HOWEVER, if the "out" number was given, the student had to work backwards (do the opposite of the rule given). Once students understood the pattern, there was no longer any confusion! To view the instructions for the assignment, please watch: Under the Weather.
Finally, I asked the students to explain their understanding of adding and subtracting decimals by answering the following prompt:
I can conclude from today's activity that if I do not
line up the decimal points when adding or
subtracting decimal numbers then _____________ because _____________. |
# Eureka Math Grade 6 Module 2 Lesson 17 Answer Key
## Engage NY Eureka Math 6th Grade Module 2 Lesson 17 Answer Key
### Eureka Math Grade 6 Module 2 Lesson 17 Opening Exercise Answer Key
Below is a list of 10 numbers. Place each number in the circle(s) that is a factor of the number. Some numbers can be placed in more than one circle. For example, if 32 were on the list, it would be placed in the circles with 2, 4, and 8 because they are all factors of 32.
24; 36; 80; 115; 214; 360; 975; 4,678; 29,785; 414, 940
Discussion:
Divisibility rule for 2: If and only if its unit digit is 0, 2, 4, 6, or 8
Divisibility rule for 4: If and only if its last two digits are a number divisible by 4
Divisibility rule for 5: If and only if its unit digit is 0 or 5
Divisibility rule for 8: If and only if its last three digits are a number divisible by 8
Divisibility rule for 10: If and only if its unit digit is 0
Decimal numbers with fraction parts do not follow the divisibility tests:
Students learn two new divisibility rules today. The rules are used to determine if numbers are divisible by 3 or 9. Start with students who already know the factors of 3 and 9, so they can see that the rule works.
→ What do the numbers 12, 18, 30, 66, and 93 all have in common?
Each is divisible by 3.
→ Calculate the sum of the digits for each given number. For example, the sum of the digits in the number 12 is 3 because 1 + 2 = 3.
Give students time to find the sums. Record the sums on the board.
→ What do these sums have in common?
They are divisible by 3.
→ When the sum of a number’s digits is divisible by 3, the entire number is divisible by 3. Now let’s examine a different set of numbers: 27, 36, 54, 72, and 99. What do these numbers have in common?
They are divisible by 9.
→ Calculate the sum of the digits for each given number.
Provide time for students to find the sums. Record the sums on the board.
→ What do all the sums have in common?
They are divisible by 9.
→ When the sum of the digits is divisible by 3 and 9, the entire number is divisible by 9. Let’s use this knowledge to determine if a large number is divisible by 3, 9, or both. The number 765 is divisible by both 3 and 9.
→ We can use what we know about the distributive property to prove that 765 is divisible by 3 and by 9.
→ Let’s begin by expanding 765.
Display the following progression.
→ We can represent 765 as:
7 × 100 + 6 × 10 + 5 × 1
→ We can further decompose the numbers that are easily seen as divisible by 3 and 9.
→ Let’s decompose 100 to 99 + 1. Why would we do this?
Because we already know that 99 is divisible by 3 and 9.
7(99 + 1) + 6 × 10 + 5
→ We can also decompose 10 to 9 + 1. Why would we do this?
Because we already know that 9 is divisible by 3 and 9.
7(99 + 1) + 6(9+ 1) + 5
→ Let’s use the distributive property further to distribute the factor 7 in the expression. We can represent 7
times the quantity (99 + 1) as 7(99) + 7 × 1, or 7(99) + 7.
7(99) + 7+ 6(9 + 1) + 5
→ We can distribute the factor 6 in the same fashion. How can we distribute the factor 6 in the expression?
We can represent 6 times the quantity (9 + 1) as 6(9) + 6 × 1, or 6(9) + 6.
7(99) + 7 + 6(9) + 6 + 5
→ Since we know that 9 is divisible by both 3 and 9, let’s factor the 9 out of the expression. We can use the commutative and associative properties to easily see this.
7(99) + 7 + 6(9) + 6 + 5
7(9 × 11) + 6(9 × 1) + 7 + 6 + 5
9(7 × 11 + 6) + 7 + 6 + 5
→ Let’s investigate our current expression. Obviously the product of 9(7 × 11 + 6) is divisible by 9 since the 9 is already factored out.
→ What about the sum of 7 + 6 + 5? What is the sum? Is this sum divisible by 3 and 9?
7 + 6 + 5 = 18.
This sum is divisible by 3 and 9.
→ Are 3 and 9 both factors of 18?
Yes
→ What do you notice about the addends 7 + 6 + 5?
They are the digits of our original number 765.
Let’s look once more at the expression:
→ Since the first term is divisible by 3 and 9, the number 765 is divisible by 3 and 9 if and only if 7 + 6 + 5 is also divisible by 3 and 9.
→ This process can be used for any decimal whole number!
Introduce the divisibility rules for 3 and 9. Have students record the rules in their student materials.
Divisibility rule for 3: If the sum of the digits is divisible by 3, then the number is divisible by 3.
Divisibility rule for 9: If the sum of the digits is divisible by 9, then the number is divisible by 9.
### Eureka Math Grade 6 Module 2 Lesson 17 Example Answer Key
Example 1.
This example shows how to apply the two new divisibility rules we just discussed.
Explain why 378 is divisible by 3 and 9.
a. Expand 378.
300 + 70 + 8
3 × 100 + 7 × 1o + 8
b. Decompose the expression to factor by 9.
3(99 + 1) + 7(9 + 1) + 8
3(99) + 3 + 7(9) + 7 + 8
c. Factor the 9.
3(9 × 11) + 3 + 7(9 × 1) + 7 + 8
9(3 × 11 + 7) + 3 + 7 + 8
d. What is the sum of the three digits?
3 + 7 + 8 = 18; the sum of the three digits is 18.
e. Is 18 divisible by 9?
Yes
f. Is the entire number 378 divisible by 9? Why or why not?
The number 378 is divisible by 9 because the sum of the digits is divisible by 9.
g. Is the number 378 divisible by 3? Why or why not?
Three is a factor of 378 because if 9 is a factor of 378, then 3 will also be a factor. OR
The number 378 is divisible by 3 because the sum of the digits is divisible by 3.
Example 2.
Is 3,822 divisible by 3 or 9? Why or why not?
The number 3,822 is divisible by 3 but not by 9 because the sum of the digits is 3 + 8 + 2 + 2 = 15, and 15 is divisible by 3 but not by 9.
### Eureka Math Grade 6 Module 2 Lesson 17 Exercise Answer Key
Circle ALL the numbers that are factors of the given number. Complete any necessary work in the space provided.
Exercise 1.
2,838 is divisible by
The number 2,838 is divisible by 3 because 3 is a factor of 2,838. I know this because the sum of the digits is 21, which is divisible by 3. The number 2,838 is not divisible by 9 because 21 is not divisible by 9, and 2,838 is not divisible by 4 because the last two digits (38) are not divisible by 4.
Exercise 2.
34,515 is divisible by
Explain your reasoning for your choice (s).
The number 34,515 is divisible by 3 and 9 because both 3 and 9 are factors of 34,515. I know this because the sum of the digits is 18, and 18 is divisible by both 3 and 9. The number 34,515 is also divisible by 5 because the unit digit is a 5.
Exercise 3.
10,534,341 is divisible by
The number 10,534,341 is divisible by 3 but not 9 because 3 is afactor of 10,534,341, but 9 is not. I know this because the sum of the digits is 21, which is divisible by 3 but not 9. The number 10, 534, 341 is not divisible by 2 because it does not end with 0, 2, 4, 6, or 8.
Exercise 4.
4,320 is divisible by
The number 4,320 is divisible by 3 and 9 because 3 and 9 are factors of 4,320. I know this because the sum of the digits is 9, which is divisible by 3 and 9. The number 4,320 is also divisible by 10 because 10 is a factor of 4,320. I know this because the unit digit is 0.
Exercise 5.
6,240 is divisible by
The number 6,240 is divisible by 3 but not divisible by 9 because 3 is a factor of 6,240, but 9 is not. I know this because the sum of the digits is 12, which is divisible by 3 but not divisible by 9. The number 6,240 is divisible by 8 because the last three digits (240) is divisible by 8.
### Eureka Math Grade 6 Module 2 Lesson 17 Problem Set Answer Key
Question 1.
Is 32, 643 divisible by both 3 and 9? Why or why not?
The number 32,643 is divisible by both 3 and 9 because the sum of the digits is 18, which is divisible by 3 and 9.
Question 2.
Circle all the factors of 424,380 from the list below.
Question 3.
Circle all the factors of 322,875 from the list below.
Question 4.
Write a 3-digit number that ¡s divisible by both 3 and 4. Explain how you know this number is divisible by 3 and 4.
Answers will vary. Possible student response: The sum of the digits is divisible by 3, and that’s how I know the number is divisible by 3. The last 2 digits are divisible by 4, so the entire number is divisible by 4.
Question 5.
Write a 4-digit number that Is divisible by both 5 and 9. Explain how you know this number Is divisible by 5 and 9.
Answers will vary. Possible student response: The number ends with a 5 or 0, so the entire number is divisible by 5. The sum of the digits is divisible by 9, so the entire number is divisible by 9.
### Eureka Math Grade 6 Module 2 Lesson 17 Exit Ticket Answer Key
Question 1.
Is 26, 341 divisible by 3? If it is, write the number as the product of 3 and another factor. If not, explain. |
# 2.4 Complex numbers
Page 1 / 8
In this section you will:
• Add and subtract complex numbers.
• Multiply and divide complex numbers.
• Simplify powers of $i$ .
Discovered by Benoit Mandelbrot around 1980, the Mandelbrot Set is one of the most recognizable fractal images. The image is built on the theory of self-similarity and the operation of iteration. Zooming in on a fractal image brings many surprises, particularly in the high level of repetition of detail that appears as magnification increases. The equation that generates this image turns out to be rather simple.
In order to better understand it, we need to become familiar with a new set of numbers. Keep in mind that the study of mathematics continuously builds upon itself. Negative integers, for example, fill a void left by the set of positive integers. The set of rational numbers, in turn, fills a void left by the set of integers. The set of real numbers fills a void left by the set of rational numbers. Not surprisingly, the set of real numbers has voids as well. In this section, we will explore a set of numbers that fills voids in the set of real numbers and find out how to work within it.
## Expressing square roots of negative numbers as multiples of $\text{\hspace{0.17em}}i$
We know how to find the square root of any positive real number. In a similar way, we can find the square root of any negative number. The difference is that the root is not real. If the value in the radicand is negative, the root is said to be an imaginary number . The imaginary number $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ is defined as the square root of $\text{\hspace{0.17em}}-1.$
$\sqrt{-1}=i$
${i}^{2}={\left(\sqrt{-1}\right)}^{2}=-1$
We can write the square root of any negative number as a multiple of $\text{\hspace{0.17em}}i.\text{\hspace{0.17em}}$ Consider the square root of $\text{\hspace{0.17em}}-49.$
$\begin{array}{ccc}\hfill \sqrt{-49}& =& \hfill \sqrt{49\cdot \left(-1\right)}\\ & =& \sqrt{49}\sqrt{-1}\hfill \\ & =& 7i\hfill \end{array}$
We use $\text{\hspace{0.17em}}7i\text{\hspace{0.17em}}$ and not $\text{\hspace{0.17em}}-7i\text{\hspace{0.17em}}$ because the principal root of $\text{\hspace{0.17em}}49\text{\hspace{0.17em}}$ is the positive root.
A complex number is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written $\text{\hspace{0.17em}}a+bi\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is the real part and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is the imaginary part. For example, $\text{\hspace{0.17em}}5+2i\text{\hspace{0.17em}}$ is a complex number. So, too, is $\text{\hspace{0.17em}}3+4i\sqrt{3}.$
Imaginary numbers differ from real numbers in that a squared imaginary number produces a negative real number. Recall that when a positive real number is squared, the result is a positive real number and when a negative real number is squared, the result is also a positive real number. Complex numbers consist of real and imaginary numbers.
## Imaginary and complex numbers
A complex number is a number of the form $\text{\hspace{0.17em}}a+bi\text{\hspace{0.17em}}$ where
• $a\text{\hspace{0.17em}}$ is the real part of the complex number.
• $b\text{\hspace{0.17em}}$ is the imaginary part of the complex number.
If $\text{\hspace{0.17em}}b=0,$ then $\text{\hspace{0.17em}}a+bi\text{\hspace{0.17em}}$ is a real number. If $\text{\hspace{0.17em}}a=0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is not equal to 0, the complex number is called a pure imaginary number. An imaginary number is an even root of a negative number.
Given an imaginary number, express it in the standard form of a complex number.
1. Write $\text{\hspace{0.17em}}\sqrt{-a}\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}\sqrt{a}\sqrt{-1}.$
2. Express $\text{\hspace{0.17em}}\sqrt{-1}\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}i.\text{\hspace{0.17em}}$
3. Write $\text{\hspace{0.17em}}\sqrt{a}\cdot i\text{\hspace{0.17em}}$ in simplest form.
## Expressing an imaginary number in standard form
Express $\text{\hspace{0.17em}}\sqrt{-9}\text{\hspace{0.17em}}$ in standard form.
$\begin{array}{ccc}\hfill \sqrt{-9}& =& \sqrt{9}\sqrt{-1}\hfill \\ & =& 3i\hfill \end{array}$
In standard form, this is $\text{\hspace{0.17em}}0+3i.$
Express $\text{\hspace{0.17em}}\sqrt{-24}\text{\hspace{0.17em}}$ in standard form.
$\sqrt{-24}=0+2i\sqrt{6}$
## Plotting a complex number on the complex plane
We cannot plot complex numbers on a number line as we might real numbers. However, we can still represent them graphically. To represent a complex number, we need to address the two components of the number. We use the complex plane , which is a coordinate system in which the horizontal axis represents the real component and the vertical axis represents the imaginary component. Complex numbers are the points on the plane, expressed as ordered pairs $\text{\hspace{0.17em}}\left(a,b\right),$ where $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ represents the coordinate for the horizontal axis and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ represents the coordinate for the vertical axis.
By the definition, is such that 0!=1.why?
(1+cosA+IsinA)(1+cosB+isinB)/(cos@+isin@)(cos$+isin$)
hatdog
Mark
how we can draw three triangles of distinctly different shapes. All the angles will be cutt off each triangle and placed side by side with vertices touching
bsc F. y algebra and trigonometry pepper 2
given that x= 3/5 find sin 3x
4
DB
remove any signs and collect terms of -2(8a-3b-c)
-16a+6b+2c
Will
Joeval
(x2-2x+8)-4(x2-3x+5)
sorry
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
(X2-2X+8)-4(X2-3X+5)=0 ?
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
Y
master
master
Soo sorry (5±Root11* i)/3
master
Mukhtar
2x²-6x+1=0
Ife
explain and give four example of hyperbolic function
What is the correct rational algebraic expression of the given "a fraction whose denominator is 10 more than the numerator y?
y/y+10
Mr
Find nth derivative of eax sin (bx + c).
Find area common to the parabola y2 = 4ax and x2 = 4ay.
Anurag
y2=4ax= y=4ax/2. y=2ax
akash
A rectangular garden is 25ft wide. if its area is 1125ft, what is the length of the garden
to find the length I divide the area by the wide wich means 1125ft/25ft=45
Miranda
thanks
Jhovie
What do you call a relation where each element in the domain is related to only one value in the range by some rules?
A banana.
Yaona
a function
Daniel
a function
emmanuel
given 4cot thither +3=0and 0°<thither <180° use a sketch to determine the value of the following a)cos thither
what are you up to?
nothing up todat yet
Miranda
hi
jai
hello
jai
Miranda Drice
jai
aap konsi country se ho
jai
which language is that
Miranda
I am living in india
jai
good
Miranda
what is the formula for calculating algebraic
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda |
# How Much Does a Pumpkin Cost? {FREE Algebra Lesson!}
For many students, the transition from math computations in the upper elementary grades to Algebra is a challenge. Suddenly, they are having to write mathematical equations and use symbols and letters rather than just numbers. (Although hopefully they have at least been introduced to some of these things prior to formal Algebra). One way to help students ease the transition and make sense of it all is by looking at real and meaningful examples. Today I’m going to share a 3 part Algebra lesson based on finding the cost of a pumpkin!
*Please Note: This post contains affiliate links which help support the work of this site. Read our full disclosure here.*
I recently took my kids to the pumpkin patch to pick out a variety of pumpkins because it’s a really fun Fall activity, but also because there are so many learning activities you can do! But picking out pumpkins isn’t just fun for elementary students! 😉
## Materials Needed For Finding the Cost of a Pumpkin Lesson:
• AÂ scale to measure the weights
• Student lesson handouts (3 pages)
## Teaching and Exploring with this Pumpkin Lesson:
Ideally, you want to have a variety of real pumpkins. If you are a classroom teacher, you could have each student that’s able bring one from home, or you could let your colleagues know and see if a handful of teachers can each bring one pumpkin.
You don’t need tons, 6-8 would probably be sufficient. Just make sure they are a good variety of sizes from very small to relatively large.
Once you’ve gathered your materials, you can begin the first lesson. Here’s basic outline of the lessons:
1. Writing an equation for the cost and then evaluating that equation using the different weights of pumpkins.
2. Writing different equations for the cost given certain restraints, and then re-evaluating for the costs.
3. Given an equation for the total cost, determine the largest pumpkin that can be purchased given a certain amount of money (solving linear equations).
Each lesson builds on the last, and while you do not need to do all three lessons in one day, you do need to keep the data and lesson pages together so students can refer back to the weights and equations they have used.
While I think this will work best and be the most fun if you have actual pumpkins to weigh and consider, (maybe you could even go to a farm to do this!), I know this is not going to be possible for everyone. Whether you don’t have time to gather materials, or don’t have time to weigh all the pumpkins, I understand that it might not be feasible.
To help you, I’ve also included in this lesson 24 data cards that you can cut out and hand out to students telling them the weight of their imaginary pumpkin. They can then share their data with their group or classmates and complete the lessons that way.
### Included in Finding the Cost of a Pumpkin lesson packet:
• Teaching tips and directions
• Student handouts
• 24 data cards with pumpkin weights
*Psst! Guess what? This 3 part lesson is included in my HUGE Algebra Essentials Resource Bundle! Everything you need to teach in a way that makes sense and build a solid foundation for students! Click the graphic below to learn more!
Have younger ones who want to join the pumpkin math fun? Grab this FREE “Runaway Pumpkins” board Game! |
Alternating Series of Real Numbers
Table of Contents
# Alternating Series of Real Numbers
Definition: Let $(a_n)_{n=1}^{\infty}$ be a strictly positive sequence of real numbers ($a_n > 0$ for all $n \in \mathbb{N}$). Then the corresponding alternating series for this sequence is a series of the form $\displaystyle{\sum_{n=1}^{\infty} (-1)^{n+1} a_n}$.
Alternatively one can define an alternating series such that for the sequence of terms $(a_n)_{n=1}^{\infty}$ we have that $a_na_{n+1} < 0$ for all $n \in \mathbb{N}$ which implies that every alternating term in the sum is negative while the remaining terms are positive.
Suppose that we have an alternating sequence:
Then the corresponding alternating series can be thought of as summing all of the positive terms and then adding the sum of all of the negative terms:
For example, the following series are alternating series:
(1)
\begin{align} \quad \sum_{n=1}^{\infty} (-1)^{n+1} \left ( \frac{1}{n} \right ) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + ... \end{align}
(2)
\begin{align} \quad \sum_{n=1}^{\infty} (-1)^{n+1} \left ( \frac{1}{n^2} \right ) = 1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \frac{1}{25} + ... \end{align}
Note that $(-1)^n = \cos ((n+1)\pi)$ for all $n \in \mathbb{N}$ so the two alternating series could be written as $\displaystyle{\sum_{n=1}^{\infty} \frac{\cos ((n+1)\pi)}{n}}$ and $\displaystyle{\sum_{n=1}^{\infty} \frac{\cos ((n+1)\pi)}{n^2}}$. In fact, there are many identities for $(-1)^n$ so the reader should always be aware of such identities that may pop up in numerical series.
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License |
# In quadrilateral ABCD with diagonal BD, measure of angle A=93, measure of angle ADB=43, measure of angle C=3x+5, measure of angle BDC=x+19, and measure of angle DBC=2x+6. Determine if AB is...
In quadrilateral ABCD with diagonal BD, measure of angle A=93, measure of angle ADB=43, measure of angle C=3x+5, measure of angle BDC=x+19, and measure of angle DBC=2x+6. Determine if AB is parallel to DC. Explain your reasoning.
Posted on
The diagonal BD separates the quadrilateral into triangles: `Delta` ABD and `Delta` BCD. The sum of the angles of a triangle is equal to 180 degrees.
Two of the angles of `Delta` ABD are 93 degrees and 43 degrees. Subtract from 180 degrees to find the measure of the third angle: 180 - 93 - 43 = 44. The measure of `/_` ABD = 44 degrees.
The angles of `Delta` BCD are (x + 19), (2x + 6), and (3x + 5). Their sum is 180 degrees. Write this as an equation and solve for x.
(x + 19) + (2x + 6) + (3x + 5) = 180
6x + 30 = 180
6x = 150
x = 25
Substitute 25 in for x to find the measures of the angles of `Delta` BCD.
`/_` BDC = x + 19 = 25 + 19 = 44 degrees
`/_` DBC = 2x + 6 = 2 * 25 + 6 = 56 degrees
`/_` BCD = 3x + 5 = 3 * 25 + 5 = 80 degrees
The diagonal BD acts as a transversal, causing `/_` ABD and `` BDC to be alternate interior angles.
`/_` ABD = 44 degrees
`/_` BDC = 44 degrees
Therefore `/_` ABD = `/_` BDC.
When alternate interior angles are congruent, the opposite sides are parallel. Therefore, side AB is parallel to side CD.
However, the shape is not a parallelogram because in a parallelogram, opposite angles are congruent.
`/_` A = 93 degrees and `/_` C = 80 degrees. `/_` A `!=` `/_` C
`` B = 100 degrees and `/_` D = 87 degrees. `/_` B `` `` D
Quadrilateral ABCD is not a parallelogram (2 pairs of parallel sides). It is a trapezoid (1 pair of parallel sides).
Answer: AB is parallel to DC.
Posted on
Sum of all angles in triangle 180.
so 93+43+angle(dba)=180
angle(DBA)=44
angle(BDC)+angle (DCB)+angle(DBC)=180
x+19+3x+5+2x+6=180
6x+30=180
6x=150
x=25
angle(BDC)= x+19=25+19=44
But it is alternate angle to angle DBA
therefore CD is parallel to BA |
# How do you simplify (2x^-2y^5 )/ (5y^5)?
May 29, 2018
See a solution process below:
#### Explanation:
First, rewrite the expression as:
$\frac{2}{5} \cdot {x}^{-} 2 \cdot {y}^{5} / {y}^{5} \implies$
$\frac{2}{5} \cdot {x}^{-} 2 \cdot 1 \implies$
$\frac{2}{5} \cdot {x}^{-} 2$
Next, use this rule for exponents to eliminate the negative exponent:
${x}^{\textcolor{red}{a}} = \frac{1}{x} ^ \textcolor{red}{- a}$
$\frac{2}{5} \cdot {x}^{\textcolor{red}{- 2}} \implies$
$\frac{2}{5} \cdot \frac{1}{x} ^ \textcolor{red}{- - 2} \implies$
$\frac{2}{5} \cdot {x}^{\textcolor{red}{- 2}} \implies$
$\frac{2}{5} \cdot \frac{1}{x} ^ \textcolor{red}{2} \implies$
$\frac{2}{5 {x}^{2}}$ |
# What are the zeros in x^3+3x^2-3x-5?
Jan 7, 2016
$\left\{- 1 , - 1 + \sqrt{6} , - 1 - \sqrt{6}\right\}$
#### Explanation:
Applying the rational roots theorem, we find that any rational roots of the given expression will be of the form $\frac{p}{q}$ where $p$ is a divisor of $- 5$ and $q$ is a positive divisor of $1$. Then, the possible rational roots are $\pm 1$ and $\pm 5$.
Trying these out, we find that ${\left(- 1\right)}^{3} + 3 {\left(- 1\right)}^{2} - 3 \left(- 1\right) - 5 = 0$ Thus $x - \left(- 1\right) = x + 1$ is a factor of ${x}^{3} + 3 {x}^{2} - 3 x - 5$
Dividing, we get
${x}^{3} + {3}^{2} - 3 x - 5 = \left(x + 1\right) \left({x}^{2} + 2 x - 5\right)$
To find the remaining roots, we can simply apply the quadratic formula
$a {x}^{2} + b x + c = 0 \implies x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
to obtain the roots of ${x}^{2} + 2 x - 5$ as
$\frac{- 2 \pm \sqrt{{2}^{2} - 4 \left(1\right) \left(- 5\right)}}{2 \left(1\right)} = - 1 \pm \sqrt{6}$
The the final set of roots (zeros) of the expression is
$\left\{- 1 , - 1 + \sqrt{6} , - 1 - \sqrt{6}\right\}$ |
Tenebre.xyz | Math | Linear Algebra | Solving Linear Systems for Multiple Variables
Linear algebra is especially suited for solving linear systems with multiple variables.
## Solving Linear Systems with Multiple Variables
Let's say for example, that a restaurant held a special special breakfast to help raise money for the local food bank. Each meal cost $8.50 for adults and$4.50 for children 10 and under. The restaurant wants to donate $1.00 per adult meal and$0.50 for each child's meal sold, but somehow the numbers got mixed up. All they know is that they sold a total of 519 meals, for a total amount of \$2743.50. How much money should the restaurant donate?
Lucky for us, this problem isn't really that difficult to solve using linear algebra, as long as there is a unique solution. Not all such problems have a unique solution, so we must test for solvency first, using a determinant (see Determinants of Matrices). We start by setting up a matrix to perform the determinant:
$$A= \begin{bmatrix} 1&1\\ 8.5&4.5\\ \end{bmatrix}$$
The determinant of the matrix is then:
$$|A|= \begin{vmatrix} 1&1\\ 8.5&4.5\\ \end{vmatrix} = 4.5-8.5 = -4$$
Since the determinant is non-zero, then there is a unique solution available. Now we must decide which method to use to solve the actual problem. I will show two methods, one fast and simple, the other complex and tedious if done by hand (but more useful for computers).
### Elimination Method
The elimination method is a fairly straight-forward algebraic approach that is quick and easy for a person to do by hand. We start by setting up the equation with what is known. We'll let $$x$$ be the number of adult meals sold and $$y$$ be the number of child meals sold, so our setup looks like this:
$$x+y=519$$
$$8.5x+4.5y=2743.5$$
We must select one of the constant coeficients in the bottom row to eliminate from the equation (we'll pick 4.5). We then multiply all terms of the top row by that value. Now subtract the bottom row from the top:
$$\small\begin{array}{cccccc} & 4.5x & + & 4.5y & = & 2335.5\\ - & 8.5x & + & 4.5y & = & 2743.5\\\hline & -4x & + & 0 & = & -408\\ \end{array}\normalsize$$
Now we can solve for $$x$$:
$$-4x=-408$$ → $$x=102$$
Since a total of 519 meals were sold, $$y$$ must be 417. The restaurant therefore sold 102 adult meals and 417 child meals. The total amount that they will donate is then $$102\times 1+417\times 0.5= 310.50$$.
### Inverse Matrix Method
Due to the complexity of this method, calculation is a chore best left to computers. The advantage is that it works well for solving problems with many variables. Since we already tested for solvency, we simply calculate an inverse matrix and then multiply:
We start with the same matrix we used for the determinant above:
$$A= \begin{bmatrix} 1&1\\ 8.5&4.5\\ \end{bmatrix}$$
Now to generate the inverse matrix, we'll use the upward-downward elimination method (the upward-downward elimination method is fairly complex and will be explained elsewhere.)
$$\left[\begin{array}{r|r} A & I_2\\ \end{array}\right] =$$ $$\left[\begin{array}{rr|rr} 1 & 1 & 1 & 0 \\ 8.5 & 4.5 & 0 & 1 \\ \end{array}\right]$$
$$\scriptsize(R_2-8.5R_1)→R_1\small$$ $$\left[\begin{array}{rr|rr} 1 & 1 & 1 & 0 \\ 0 & -4 & -8.5 & 1 \\ \end{array}\right]\normalsize$$
$$\scriptsize(R_2/-4)→R_2\small$$ $$\left[\begin{array}{rr|rr} 1 & 1 & 1 & 0 \\ 0 & 1 & 2.125 & -0.25 \\ \end{array}\right]\normalsize$$
$$\scriptsize(R_1-R_2)→R_1\small$$ $$\left[\begin{array}{rr|rr} 1 & 0 & -1.125 & 0.25 \\ 0 & 1 & 2.125 & -0.25 \\ \end{array}\right]\normalsize$$
Our inverse matrix is now contained on the right-hand side of the augmented matrix:
$$A^{-1}= \begin{bmatrix} -1.125&0.25\\ 2.125&-0.25\\ \end{bmatrix}$$
Now we perform a matrix multiply in order to obtain or result:
$$\small\left[\begin{array}{r} x\\ y\\ \end{array}\right] =\normalsize$$ $$\small\left[\begin{array}{rr} -1.125 & 0.25\\ 2.125 & -0.25\\ \end{array}\right]\normalsize$$ $$\small\left[\begin{array}{r} 519\\ 2743.5\\ \end{array}\right] =\normalsize$$ $$\small\left[\begin{array}{r} (-1.125)(519)+(0.25)(2743.5)\\ (2.125)(519)+(-0.125)(2743.5)\\ \end{array}\right] =\normalsize$$ $$\small\left[\begin{array}{r} 102\\ 417\\ \end{array}\right]\normalsize$$
And viola! We get the answer $$x=102$$ and $$y=417$$.
Don't get too hung up on the complexity of the inverse matrix process. Unless you're going for a Master's degree in mathematics, it's unlikely that anyone will ever expect you to do it manually. Graphing calculators or computers are generally used when dealing with matrices. |
Working with Rational Expressions - THE SAT MATH: ADVANCED MATHEMATICS - SAT 2016
THE SAT MATH: ADVANCED MATHEMATICS
Skill 4: Working with Rational Expressions
Lesson 12: Interpreting and computing with rational expressions
Which of the following is equivalent to for all x greater than 0?
A)
B)
C)
D)
When adding, subtracting, multiplying, or dividing rational expressions, just follow the rules for working with fractions.
• When adding or subtracting fractions, first get a common denominator, then combine numerators.
• When multiplying fractions, just multiply straight across.
• To divide by a fraction, just multiply by its reciprocal.
(Medium) To simplify this difference of fractions, we must find a common denominator.
So the correct answer is (C).
If is equivalent to for all x, which of the following is equivalent to B?
A) 3x – 1
B) 3x + 1
C) 9x2
D) 9x2 – 1
(Hard) It helps to notice that the given rational expression is “improper,” but that the transformed expression is not. Recall that an “improper fraction,” like 5/3, is one in which the numerator is larger than the denominator. Such fractions can also be expressed as “mixed numbers,” which include an integer and a “proper fraction:” 5/3 = 1 ⅔. Similarly, an “improper rational expression” is one in which the degree of the numerator is greater than the degree of the denominator. In the expression , the numerator has a degree of 2 and the denominator has a degree of 1. Just as with improper fractions, we can convert this to a “mixed” expression by just doing the division:
which means that equals . Therefore, the correct answer is (A).
Let x represent the time, in hours, it takes pump A to fill a standard tank, and let y represent the time, in hours, it takes pump A and pump B, working together, to fill the same standard tank. If the equation above represents this situation, then b must represent
A) the time, in hours, it takes pump B, working alone, to fill the standard tank
B) the portion of the standard tank that pump B fills when the pumps work together to fill the entire standard tank
C) the rate, in standard tanks per hour, of pump B
D) the difference between the rates, in standard tanks per hour, of pump B and pump A
Rational expressions are often used to express rates. (Remember: raterational, and ratio all derive from the same Latin root.) When working with rational expressions that represent real quantities, it often helps to think in terms of the rate-units that they represent.
For instance, if t represents the amount of time, in hours, it takes someone to paint n rooms, then t/n represents the number of “hours per room” and n/t represents the number of “rooms per hour.”
(Medium-hard) You may find it helpful to review Chapter 8, Lesson 5, “Rates and unit rates” before tackling this problem. We are told that x represents the number of “hours per tank” for pump A, that is, the number of hours it takes pump A to fill one standard tank. Therefore, its reciprocal, 1/x, must represent the number of “tanks per hour” for pump A, that is, the number of tanks (or fraction of a tank) that pump A can fill in one hour. Likewise, since y represents the number of “hours per tank” when the two pumps work together, 1/y must represent the number of “tanks per hour” that the two pumps can fill when working together.
The essential fact in this situation is that “the rate (in tanks per hour) at which the two pumps work together must equal the sum of the rates (in tanks per hour) of the two pumps working separately.” (For instance, if pump A can fill 2 tanks per hour and pump B can fill 3 tanks per hour, then working together they can fill 5 tanks per hour.)
Since the given equation essentially says, “the rate of pump A plus b = the rate of pump A and pump B working together,” b must represent the rate (in tanks per hour) of pump B. Therefore, the correct answer is (C).
Lesson 13: Simplifying rational expressions
If x = 3a and a ≠ 2, which of the following is equivalent to
A)
B)
C)
D)
Since rational expressions are just fractions (although perhaps complicated ones), we simplify them exactly the same way we simplify any fraction, that is, by cancelling common factors in the numerator and denominator (which is equivalent to dividing by 1), or multiplying numerator and denominator by a convenient factor (which is equivalent to multiplying by 1).
• Factoring and cancelling common factors:
• Multiplying by a common factor:
(Medium) This question is asking us to translate an expression in x into an expression in a, which requires making a substitution. However, it is a bit simpler if we don’t substitute right away, but instead simplify the given expression:
Factor:
Cancel common factor:
Substitute x = 3a:
Divide numerator and denominator by 3:
Therefore the answer is (A). Bonus question: Why did the question have to mention that a ≠ 2?
If for all x > 3, where a and b are constants, what is the value of ab?
A)
B)
C)
D)
(Medium-hard) The expression on the left side of the equation is obnoxious and in desperate need of simplification:
Factor:
Cancel common factor (okay since x > 3):
Divide numerator and denominator by 5:
This last step, which may seem strange, is important because it shows us how the two sides of the equation “match up.” If this equation is to be true for “all x > 3” then a must equal 2/5 and b must equal 3. Therefore, ab = (2/5)(3) = 6/5, and the correct answer is (C). Bonus question: Why did the question mention that x > 3?
Lesson 14: Solving rational equations
If x > 0 and , what is the value of x? [No calculator]
A)
B)
C)
D)
When solving an equation that includes fractions or rational expressions, you may find it helpful to simplify the equation by multiplying both sides by the “common denominator” (that is, the common multiple of the denominators).
Multiply by 5x:
Distribute:
Simplify:
x2 + 5 = 10x
Notice that, in this case, the equation simplifies to a quadratic, which is relatively easy to work with.
(Hard) Let’s apply this strategy to our equation:
Multiply by
Distribute:
Simplify:
(x + 1) – (x – 1) = 2x2 – 2
Simplify:
2 = 2x2 – 2
4 = 2x2
Divide by 2:
2 = x2
Take the square root:
But since the equation states that x > 0, the correct answer is (A).
The function f is defined by the equation f(x) = x2 – 3x – 18 and the function h is defined by the equation . For what value of x does h(x) = 6?
A) –6
B) –3
C) 0
D) 9
(Hard) The first thing we should try to do is simplify the expression for h(x).
Substitute f(x) = x2 – 3x – 18:
Factor using Product-Sum Method:
Cancel common factor:
Solve for x if h(x) = 6:
Multiply by 2:
12 = x + 3
Subtract 3:
9 = x
Therefore, the correct answer is (D).
Exercise Set 5 (No Calculator)
1
If , what is the value of y?
2
If and x > 0, what is the value of x?
3
If , what is the value of x2?
4
If , what is the value of z?
5
Let g(x) = x2 – 9x + 18 and , where a is a constant. If h(4) = , what is the value of a?
6
If for all values of x greater than 1, what is the value of a + b?
7
Which of the following is equivalent to for all x greater than 1?
A)
B)
C)
D)
8
For how many distinct integer values of n is
A) Zero
B) One
C) Two
D) Three
9
If and a > 1, which of the following is equivalent to
A)
B)
C)
D)
Exercise Set 5 (Calculator)
10
If , what is the value of x2 – 10x?
11
For how many positive integer values of k is
12
If g(x) = x2 – 9x + 18 and , what is the value of h(9)?
13
If , what is the value of
14
If , what is the value of c2?
15
If for all values of x, what is the value of a?
16
Which of the following is equivalent to for all positive values of b?
A)
B)
C)
D)
17
Given the system above, what is the value of a + b?
A)
B)
C)
D)
18
If one proofreader takes n hours to edit 30 pages and another takes m hours to edit 50 pages, and together they can edit x pages per hour, which of the following equations must be true?
A)
B)
C)
D)
EXERCISE SET 5 ANSWER KEY
No Calculator
1. 6/5 or 1.2
Multiply by 45:
15 – 9 = 5y
(45 is the least common multiple of the denominators.)
Simplify:
6 = 5y
Divide by 5:
6/5 = y
2. 7
Multiply by 24(x + 1)(x – 1):
24x(x – 1)+ 24(x + 1) = 25(x + 1)(x – 1)
Distribute:
24x2 – 24x + 24x + 24 = 25x2 – 25
Gather like terms:
0 = x2 – 49
49 = x2
Take square root:
±7 = x
Since x must be positive, x = 7.
3. 13/2 or 6.5
Multiply by 5(x – 2)(x + 2):
5(x + 2) – 5(x – 2) = 8(x – 2)(x + 2)
Distribute:
5x + 10 – 5x + 10 = 8x2– 32
Subtract 20 and simplify:
0 = 8x2– 52
52 = 8x2
Divide by 8:
52/8 = 13/2 = x2
Remember, the question asks for the value of x2, not x, so don’t worry about taking the square root.
4. 6/17 or .353
Multiply by 6z:
12z – 6 = –5z
Add 5z and 6:
17z = 6
Divide by 17:
z = 6/17
5. 28
Use definition of g:
Simplify:
Cross-multiply:
4 –a = –24
Add 24 and a:
28 = a
6. 5
Combine fractions:
Simplify:
Since must equal for all values of xa = 3 and b = 2, so a + b = 5.
7. D
Since (1 - x) = -(x - 1):
8. C
Recall from Chapter 7, Lesson 9, on solving inequalities, that we need to consider two conditions. First, if n + 2 is positive (that is, n > –2), we can multiply on both sides without “flipping” the inequality:
n + 5 > 2n + 4
Subtract n and 4:
1 > n
So n must be between –2 and 1, and the integer values of –1 and 0 are both solutions. Next, we consider the possibility n + 2 is negative (that is, n < –2), and therefore multiplying both sides by n + 2 requires “flipping” the inequality:
n + 5 < 2n + 4
Subtract n and 4:
1 < n
But there are no numbers that are both less than –2 and greater than 1, so this yields no new solutions.
9. C
Factor:
Cancel common factors:
Substitute x = 4a:
Cancel common factor:
Calculator
10. 15
Multiply by 5x:
x2 – 15 = 10x
Add 15, subtract 10x:
x2 – 10x = 15
Notice that you should not worry about solving for x!
11. 2
Use common base:
10–k > 10–3
Exponential Law #10:
k > –3
Multiply by –1:
k < 3
Therefore, the two positive integer solutions are 1 and 2.
12. 3/14 or .214
Use definition of g:
Simplify:
13. 81/2 or 40.5
Combine fractions:
Simplify:
Multiply by 9/2:
14. 5
Convert to ×:
Multiply:
Multiply by c2 – 1:
2c2 = 10
Divide by 2:
c2 = 5
15. 2 Notice that the right-hand side of the equation is the “proper” form of the “improper” fraction on the left, and that a is the remainder when the division of the polynomials is completed:
16. D
Common denominator:
Combine:
17. C
Multiply by a:
Divide by 10:
2 = 10a
1/5 = a
Subtract equations:
Multiply by –b:
2 = 6b
Divide by 6:
1/3 = b
Therefore, a + b = 1/5 + 1/3 = 8/15.
18. A The number of pages they can edit together in an hour must equal the sum of the number of pages they can edit separately. The number of pages the first proofreader can edit per hour is 30/n, and the number of pages the second proofreader can edit per hour is 50/m. Since they can edit x pages per hour together, .
NOTE: You can avoid the most common mistakes with this problem by paying attention to the units of each term. The units of two sides, as well as the unit of each term in a sum or difference, must “match.” Notice that the unit for all of the terms is pages/hour.
|
### There's a Limit
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
### Not Continued Fractions
Which rational numbers cannot be written in the form x + 1/(y + 1/z) where x, y and z are integers?
### Comparing Continued Fractions
Which of these continued fractions is bigger and why?
# And So on - and on -and On
##### Stage: 5 Challenge Level:
A well presented solution from Richard of The Royal Hospital School reflected those of a number of other solvers including Kevin of Langley Grammar, Jeff from New Zealand and Andrei of Tudor Vianu School. Well done to all of you.
We are given that:
\begin{eqnarray} F_0(x) &=& 1/(1 - x) \\ F_n(x) &=& F_0 (F_{n-1}(x)) \end{eqnarray}
This implies that: $$F_n(x) = 1 / (1 - F_{n-1}(x))$$ and can be extended to: $$F_n(x) = 1 / (1 - (1 / (1 - F_{n-2}(x))))$$
Creating functions of $x$ when $n = 1, 2$ and $3$ gives:
\begin{eqnarray} F_1(x) &=&1 / (1 - F_0(x))\\ &=& 1 / (1 - (1 / (1 - x)))\\ &=& 1 / ((1 - x - 1) / (1 - x))\\ &=& 1(1 - x) / -x\\ &=& (1 - x) / -x\\ F_2(x) &=& 1 / (1 - F_1(x))\\ &=& 1 / (1 - ((1 - x) / -x))\\ &=& 1 / ((-x - 1 + x) / -x)\\ &=& -x / -1\\ &=& x\\ F_3(x) &=& 1 / (1 - F_2(x))\\ &=& 1 / (1 - x)\\ &=& F_0(x) \end{eqnarray} Therefore the function of x will repeat itself every three times. \begin{eqnarray} F_0(x) &=& F_3(x)\\ F_1(x) &=& F_4(x) \end{eqnarray} Etc. etc. So, to find $F_{2000}(x)$, we must find the remainder given when 2000 is divided by three.$$Mod_3 2000 = 2$$ Thus, $F_{2000}(x) = F_2(x)$, and $F_2(x) = x$.
Therefore: $$F_{2}2000 = 2000$$ |
# Playing With Numbers Class 6 Notes
These are of chapter 3 Playing with Numbers class 6 notes. Hope you like these Playing with Numbers notes and share the page among your friends.
Topics of Divisibility Rules based on factors and Multiple, co -prime is deleted as per Rationalized Syllabus
Table of Content
## What are Factors?
A factor of a number is an exact divisor of that number
Example
1. Factor of 6
1-> Since 1 exactly divides 6
2 -> Since it exactly divides 6
3 -> Since it exactly divides 6
6-> Since it exactly divides 6
### Properties of factors
1. 1 is a factor of every number
2. every number is a factor of itself.
3. every factor of a number is an exact divisor of that number
4. every factor is less than or equal to the given number
5. number of factors of a given number are finite.
## Multiple
Multiple of a number is the numbers obtained by multiplying that numbers with various Natural numbers
Example
Number is 6
Multiple will be
$6 \times 1= 6$
$6 \times 2=12$
$6 \times 3=18$
### Properties of Multiple
1. Every multiple of a number is greater than or equal to that number.
2. number of multiples of a given number is infinite
3. every number is a multiple of itself
## Perfect Number
A number for which sum of all its factors is equal to twice the number is called a perfect number
Example
1. 6
The factors of 6 are 1, 2, 3 and 6.
Now, $1+2+3+6 = 12 = 2 \times 6$.
2. 28
All the factors of 28 are 1, 2, 4, 7, 14 and 28.
Now, $1 + 2 + 4 + 7 + 14 + 28 = 56 = 2 \times 28$.
## Prime Numbers
The numbers other than 1 whose only factors are 1 and the number itself are called Prime numbers.
Example:
2, 3, 5, 7, 11 ,13
We can find list of prime numbers till 100 using Sieve of Eratosthenes method
Step 1: Cross out 1 because it is not a prime number.
Step 2: Encircle 2, cross out all the multiples of 2, other than 2 itself, i.e. 4, 6, 8 and so on.
Step 3: You will find that the next uncrossed number is 3. Encircle 3 and cross out all the multiples of 3, other than 3 itself.
Step 4: The next uncrossed number is 5. Encircle 5 and cross out all the multiples of 5 other than 5 itself.
Step 5: Continue this process till all the numbers in the list are either encircled or crossed out.
All the encircled numbers are prime numbers. All the crossed-out numbers, other than 1 are composite numbers
## Composite Numbers
Numbers having more than two factors (1 and itself) are called Composite numbers
Example:
4, 6, 8 ,9….
## Even Numbers
The numbers which are multiple of 2 are called even numbers
Example
2,4,6,8,10,12,14
Even numbers have 0,2,4,6,8 in it one’s place
## Odd Numbers
The numbers which are not multiple of 2 are called odd numbers
Example
1,3, 5,7,9,11......
Important points about prime numbers based on definition of odd and even numbers
1. 2 is the smallest prime number which is even.
2. every prime number except 2 is odd.
## Common Factors and Multiple
We already studied the factors and Multiple in previous section. We can find common factors and multiple between 2 or more numbers
Example
1) Common factors of 4,12, and 16
Factors of 4 are 1, 2 and 4.
Factors of 12 are 1, 2, 3, 4, 6 and 12.
Factors of 16 are 1, 2, 4, 8 and 16.
Clearly, 1, 2 and 4 are the common factors of 4, 12, and 16.
Important Note
Two numbers having only 1 as a common factor are called co-prime number
Example 3 and 4 are co-prime
## Divisibility Rules based on factors and Multiple, co -prime
1) if a number is divisible by another number then it is divisible by each of the factors of that number
Example
36, is divisible by 18
Now if we find factors of 18 i.e., 1,2 ,3,4,9,18
So, 36 is also divisible by 1,2 ,3,4,9,18
2) If a number is divisible by two co-prime numbers then it is divisible by their product also
Example
45
It is divisible by 3 (4+5=9)
It is divisible by 5
Since 3,5 are coprime. Now the product is 3 X5=15. Now it is divisible by 15 also
3) If two given numbers are divisible by a number, then their sum is also divisible by that number.
Example
15 is divisible by 3
9 is divisible by 3
Sum = 15+9 =24
We can see that it is also divisible by 3
4) If two given numbers are divisible by a number, then their difference is also divisible by that number
Example
15 is divisible by 3
9 is divisible by 3
Difference = 15 - 9 =6
We can see that it is also divisible by 3
## Factorisation and Prime -Factorisation
Factorisation is expressing the number as a product of its factors
So,
36 = 3×12 = 4 X 9
These form is called Factorisation
Prime Factorisation is expressing the number as a product of its prime factors
$36 = 2 \times 2 \times 3 \times 3$
We can find prime factorisation by dividing the numbers with 2, 3, 5, 7 etc. in this order repeatedly so long as the quotient is divisible by that number
## HCF and LCM
a. The Highest Common Factor (HCF) of two or more given numbers is the highest of their common factors. It is also known as Greatest Common Divisor (GCD).
Steps to find HCF or GCD
a. Find the prime factorisation of the numbers
b. Choose the common factors in them
c. Multiply those common factors to obtain HCF
Example
(a) Find the HCF of 8 and 12
Prime Factorisation of the numbers
$8 = 2 \times 2 \times 2$
$12 = 2 \times 2 \times 3$
Common factors are 2,2
So $HCF = 2 \times 2 =4$
(b) Find the HCF of 20, 28 and 36
Prime Factorisation of the numbers
$20 = 2 \times 2 \times 5$
$28 = 2 \times 2 \times 7$
$36=2 \times 2 \times 3 \times 3$
Common factors are 2,2
So $HCF = 2 \times 2 =4$
(b) The Lowest Common Multiple (LCM) of two or more given numbers is the lowest of their common multiples.
Steps to find LCM
(a) Find the prime factorisation of the numbers
(b) look for the maximum occurrence of all the prime factors in these numbers
(c) The LCM of the numbers will be the product of the prime factors counted the maximum number of times they occur in any of the numbers
Example
(a) Find the LCM of 8 and 12
Prime Factorisation of the numbers
$8 = 2 \times 2 \times 2$
$12 = 2 \times 2 \times 3$
So, $LCM = (2 \times 2 \times 2) \times (3) = 24$
(b) Find the LCM of 20, 28 and 36
Prime Factorisation of the numbers
$20 = 2 \times 2 \times 5$
$28 = 2 \times 2 \times 7$
$36=2 \times 2 \times 3 \times 3$
$LCM = (2 \times 2) \times (3) \times (5) \times (7) =540$
LCM using division method
Here we divide the given numbers by common prime number until the remainder is a prime number or one. LCM will be the product obtained by multiplying all divisors and remaining prime numbers.
Steps are
(1) We place number in the line
(2) We start dividing the number by least prime number which is common among all of them or group of them
(3) Keep dividing by least until we have 1's in the remainder
(4) LCM is the product of the divisors
Example
(1) Find the LCM of 14 and 20
$LCM = 2 \times 2 \times 7 \times 5=140$
(2) Find the LCM of 12 ,18 and 44
$LCM = 2 \times 2 \times 3 \times 3 \times 11=396$
### Practice Question
Question 1 What is $\frac {1}{2} + \frac {3}{4}$ ?
A)$\frac {5}{4}$
B)$\frac {1}{4}$
C)$1$
D)$\frac {4}{5}$
Question 2 Pinhole camera produces an ?
A)An erect and small image
B)an Inverted and small image
C)An inverted and enlarged image
D)None of the above |
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# 6.5: Review Problems
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1. Show that the pair of conditions:
$(1)~~~\left\{\begin{matrix}L(u+v) = L(u) + L(v)\\L(cv) = cL(v)\end{matrix}\right.$
(valid for all vectors $$u,v$$ and any scalar $$c$$) is equivalent to the single condition:
$$(2)~~~L(ru + sv) = rL(u) + sL(v)$$
(for all vectors $$u,v$$ and any scalars $$r$$ and $$s$$).
Your answer should have two parts. Show that (1) $$\Rightarrow$$ (2), and then show that (2) $$\Rightarrow$$ (1),
2. If $$f$$ is a linear function of one variable, then how many points on the graph of the function are needed to specify the function? Give an explicit expression for $$f$$ in terms of these points.
3.
a) If $$p\begin{pmatrix}1\\2\end{pmatrix}=1$$ and $$p\begin{pmatrix}2\\4\end{pmatrix}=3$$ is it possible that $$p$$ is a linear function?
b) If $$Q(x^{2})=x^{3}$$ and $$Q(2x^{2})=x^{4}$$ is it possible that $$Q$$ is a linear function from polynomials to polynomials?
4. If $$f$$ is a linear function such that
$$f\begin{pmatrix}1\\2\end{pmatrix}=0{\rm ,~and~} f\begin{pmatrix}2\\3\end{pmatrix}=1\, ,$$
then what is $$f\begin{pmatrix}x\\y\end{pmatrix}$$?
5. Let $$P_{n}$$ be the space of polynomials of degree $$n$$ or less in the variable $$t$$. Suppose $$L$$ is a linear transformation from $$P_{2} \rightarrow P_{3}$$ such that $$L(1) = 4$$, $$L(t)=t^{3}$$, and $$L(t^{2}) = t-1$$.
a) Find $$L(1+t+2t^{2})$$.
b) Find $$L(a+bt+ct^{2})$$.
c) Find all values $$a,b,c$$ such that $$L(a+bt+ct^{2})=1+3t+2t^{3}$$.
6. Show that the operator $$\cal{I}$$ that maps $$f$$ to the function $$\cal{I}f$$ defined by $$\cal{I}f(x):=\int_{0}^{x}f(t)dt$$ is a linear operator on the space of continuous functions.
7. Let $$z \in \mathbb{C}$$. Recall that we can express $$z = x + iy$$ where $$x,y \in \mathbb{R}$$, and we can form the $$\textit{complex conjugate} of \(z$$ by taking $$\overline{z} = x - iy$$. The function $$c \colon \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$$ which sends $$(x, y) \mapsto (x, -y)$$ agrees with complex conjugation.
a) Show that $$c$$ is a linear map over $$\mathbb{R}$$ ($$\textit{i.e.}$$ scalars in $$\mathbb{R}$$).
b) Show that $$\overline{z}$$ is not linear over $$\mathbb{C}$$ |
Browse Questions
# Prove that the coefficient of $x^n$ in $(1+x)^{2n}$ is twice the coefficient of $x^n$ in $(1+x)^{2n-1}$
Toolbox:
• General term in the expansion of $(1+x)^{2n}$ is $T_{r+1}=C(2n,n)x^n$
General term in the expansion of $(1+x)^{2n}$ is
$T_{r+1}=C(2n,n)x^n$
Putting $r=n$ we have
$T_{n+1}=C(2n,n)x^n$
Coefficient of $x^n=C(2n,n)$
Again general term in the expansion of $(1+x)^{2n-1}$ is
$T_{r+1}=C(2n-1,r)x^r$
Putting $r=n$ we have
$T_{n+1}=C(2n-1,n)x^n$
Coefficient of $x^n$ in the expansion of $x^n$ is $C(2n-1,n)$
According to the problem we have to prove that
$C(2n,n)=2\times C(2n-1,n)$
Or $\large\frac{2n!}{n!(2n-n)!}=$$2\times \large\frac{(2n-1)!}{n!(2n-1-n)!} \large\frac{2n!}{n!n!}=$$2\times \large\frac{(2n-1)!}{n!(n-1)!}$
Multiplying $N^r$ and $D^r$ by n on RHS we have
$\large\frac{2n(2n-1)!}{n!n(n-1)!}$
i.e $\large\frac{2n!}{n!n!}=\frac{2n!}{n!n!}$ which is true.
Hence proved. |
# Class 8 RD Sharma – Chapter 1 Rational Numbers – Exercise 1.5
Last Updated : 05 Nov, 2020
### Question 1. Multiply:
(i) 7/11 by 5/4
Solution:
7/11 × 5/4
Multiplying numerator with numerator of other rational number and denominator with denominator
= (7 × 5)/(11 × 4)
= 35/44
(ii) 5/7 by -3/4
Solution:
(5/7) × (-3/4)
Multiplying numerator with numerator of other rational number and denominator with denominator
= (5 × -3)/(7 × 4)
= -15/28
(iii) -2/9 by 5/11
Solution:
-2/9 × 5/11
Multiplying numerator with numerator of other rational number and denominator with denominator
= (-2 × 5)/(9 × 11)
= -10/99
(iv) -3//17 by -5/-4
Solution:
-3/17 × 5/4
Multiplying numerator with numerator of other rational number and denominator with denominator
= (-3 × 5)/(17 × 4)
= -15/68
(v) 9/-7 by 36/-11
Solution:
(9/-7) × (36/-11)
Multiplying numerator with numerator of other rational number and denominator with denominator
= (-9 × -36)/(7 × 11)
(vi) -11/13 by -21/7
Solution:
Multiplying numerator with numerator of other rational number and denominator with denominator
= (-11 × -21)/(13 × 7)
= 231/91
(vii) -3/5 by -4/7
Solution:
Multiplying numerator with numerator of other rational number and denominator with denominator
= (-3 × -4)/(5 × 7)
= 12/35
(viii) -15/11 by 7
Solution:
-15/11×7/1
Multiplying numerator with numerator of other rational number and denominator with denominator
=(-15×7)/(11×1)
=-105/11
### Question 2. Multiply
(i) -5/17 by 51/-60
Solution:
(-5/17) × (51/-60)
Multiplying numerator with numerator of other rational number and denominator with denominator
= (-5 × -51)/(17 × 60)
Common factor of 5 and 60
= 51/17 × 12
51 and 12 have 3 as common factor
= 17/17 × 4
= 1/4
(ii) -6/11 by -55/36
Solution:
(-6/11) × (-55/36)
Multiplying numerator with numerator of other rational number and denominator with denominator
= (-6 × -55)/(11 × 36)
Common factor of 6 and 36, 55 and 11
= 5/6
(iii) -8/25 by -5/16
Solution:
(-8/25) × (-5/16)
Multiplying numerator with numerator of other rational number and denominator with denominator
= (-8 × -5)/(25 × 16)
Common factor of 8 and 16, 5 and 25
= 1/5 × 2
= 1/10
(iv) 6/7 by -49/36
Solution:
(6/7) × (-49/36)
Multiplying numerator with numerator of other rational number and denominator with denominator
= (6 × -49)/(7 × 36)
Common factor of 6 and 36, 49 and 7
= -7/6
(v) 8/-9 by -7/-16
Solution:
-8/9 × 7/16
Multiplying numerator with numerator of other rational number and denominator with denominator
= (-8 × 7)/(9 × 16)
Common factor of 8 and 16
= -7/9 × 2
= -7/18
(vi) -8/9 by 3/64
Solution:
-8/9 × 3/64
Multiplying numerator with numerator of other rational number and denominator with denominator
= (-8 × 3)/(9 × 64)
Common factor of 8 and 64, 3 and 9
= -1/3 × 8
= -1/24
### Question 3. Simplify each of the following and express the result as arational number in standard form:
(i) (-16/21) × (14/5)
Solution:
Multiplying numerator with numerator of other rational number and denominator with denominator
= (-16 × 14)/(21 × 5)
Common factor of 21 and 14
= (-16 × 2)/(3 × 5)
= -32/15
(ii) (7/6) × (-3/28)
Solution:
Multiplying numerator with numerator of other rational number and denominator with denominator
= (7 × -3)/(6 × 28)
Common factor of 7, 28, 3 and 6
= -1/2 × 4
= -1/8
(iii) (-19/36) × 16
Solution:
Multiplying numerator with numerator of other rational number and denominator with denominator
= (-19 × 16)/(36 × 1)
Common factor of 16 and 36
= (-19×4)/(9 × 1)
=-76/9
(iv) (-13/9) × (27/-26)
Solution:
Multiplying numerator with numerator of other rational number and denominator with denominator
=(-13× -27)/(9×26)
Common factor of 27 and 9 , 13 and 26
=(-1 × -3)/(2)
=3/2
(v) (-9/16) × (-64/-27)
Solution:
Multiplying numerator with numerator of other rational number and denominator with denominator
= (-9 × 64) / (16 × 27)
Common factor of 9 and 27, 64 and 16
= (-1 × 4) / (1 × 3)
= -4/3
(vi) (-50/7) × (14/3)
Solution:
Multiplying numerator with numerator of other rational number and denominator with denominator
= (-50 × 14)/(7 × 3)
Common factor of 14 and 7
= (-50 × 2)/(3)
= -100/3
(vii) (-11/9) × (-81/-88)
Solution:
Multiplying numerator with numerator of other rational number and denominator with denominator
= (-11 × 81)/(9 × 88)
Common factor of 1, 88, 9 and 81
= (-1 × 9)/(1 × 8)
= -9/8
(viii) (-5/9) × (72/-25)
Solution:
Multiplying numerator with numerator of other rational number and denominator with denominator
= (-5 * -72)/(9 × 25)
Common factor of 5 and 25, 9 and 72
= (-1 × -8)/(1 × 5)
= 8/5
### 4. Simplify:
(i) ((25/8) × (2/5)) – ((3/5) × (-10/9))
Solution:
= ((25 × 2)/(8 × 5)) – ((3 × -10)/(5 × 9))
= 50/40 – (-30)/45
= 5/4 + 2/3
LCM of 4 and 3 is 12
= (5 × 3 + 2 × 4)/12
= (15 + 8)/12
= 23/12
(ii) ((1/2) × (1/4)) + ((1/2) × 6)
Solution:
= ((1 × 1)/(2 × 4)) + ((1 × 6)/(1 × 2))
= 1/8 + 3/1
LCM of 8 and 1 is 8
= (1 × 1 + 3 × 8)/8
= (1 + 24)/8
= 25/8
(iii) (-5 × (2/15)) – (-6 × (2/9))
Solution:
= ((-5 × 2)/(15 × 1)) – ((-6 × 2)/(1 × 9))
= (-10/15) – (-12/9)
Common factor of 10 and 15, 12 and 9
= -2/5 + 4/3
LCM of 5 and 3 is 15
= (-2 × 3 + 4 × 4)/15
= (-6 + 16)/15
= 10/15
Common factor of 10 and 15
= 2/3
(iv) ((-9/4) × (5/3)) + ((13/2) × (5/6))
Solution:
= (-9 × 5)/(4 × 3) + (13 × 5)/(2 × 6)
Common factor of 9 and 3
= (-45/12) + (65/12)
As denominators are same
= (-45 + 65)/12
= (20)/12
Common factor of 20 and 12
= 5/3
(v) ((-4/3) × (12/-5)) + ((3/7) × (21/15))
Solution:
= (-4 * -12)/(3 × 5) + ((3 × 21)/(7 × 15))
= 48/15 + 3/5 (Common factor 3 and 15, 21 and 7)
LCM of 15 and 5 is 15
= (48 + 3 × 3)/15
= (48 + 9)/15
= 57/15
Common factor of 57 and 15
= 19/5
(vi) ((13/5) × (8/3)) – ((-5/2) × (11/3))
Solution:
= (13 × 8)/(5 × 3) – ((-5 × 11)/(2 × 3))
= 104/15 – 55/6
LCM of 15 and 6 is 3 × 5 × 2 = 30
= (104 × 2 + 55 × 5)/30
= (208 + 275)/30
= 483/30
(vii) ((13/7) × (11/26)) — ((-4/3) × (5/6))
Solution:
= ((13 × 11)/(7 × 26)) – ((-4 × 5)/(3 × 6))
Common factor of 13 and 26, 4 and 6
= 11/7 × 2 – (-2 × 5/3 × 3)
= 11/14 + 10/9
LCM of 14 and 9 is 126
= (11 × 9 + 10 × 14)/126
= (99 + 140)/126
= 239/126
### Question 5. Simplify:
(i) ((3/2) × (1/6)) + ((5/3) × (7/2) – (13/8) × (4/3))
Solution:
= (3 × 1)/(2 × 6) + (5 × 7)/(3 × 2) – (13 × 4)/(8 × 3)
Common factor of 3 and 6, 4 and 8
= 1/4 + 35/6 – 13/6
LCM of 4 and 6 is 12
= (1 × 3 + 35 × 2 – 13 × 2)/12
= (3 + 70 – 26)/12
= (73 – 26)/12
= 47/12
(ii) ((1/4) × (2/7)) — (5/14) × (-2/3) + (3/7) × (9/2)
Solution:
= (1 × 2)/(4 × 7) – (5 × -2)/(14 × 3) + (3 × 9)/(7 × 2)
Common factor of 2 and 4, 2 and 14
= 1/14 – (-5/21) + 27/14
LCM of 21 and 14 is 7 × 2 × 3 = 42
= 1/14 + 5/21 + 27/14
LCM of 14 and 21 is 2 × 7 × 3 = 42
= (1 × 3 + 5 × 2 + 27 × 3)/42
= (3 + 10 + 81)/42
= (94)/42
(iii) ((13/9) × (-15/2)) + ((7/3) × (8/5) + (3/5) × (1/2))
Solution:
= (13 × -15)/(9 × 2) + ((7 × 8)/(3 × 5) + (3 × 1)/(5 × 2))
Common factor of 9 and 15
= (13 × -5)/(3 × 2) + ((56/15) + 3/10)
= -65/6 + 56/15 + 3/10
6 = 2 × 3
15 = 3 × 5
10 = 2 × 5
LCM is 2 × 3 × 5 = 30
= (-65 × 5 + 56 × 2 + 3 × 3)/30
= (-325 + 112 + 9)/30
= (-325 + 121)/30
= -204/30
(iv) ((3/11) × (5/6)) – (9/12) × (4/3) + (5/13) × (6/15)
Solution:
= (3 × 5)/(11 × 6) – ((9 × 4)/(12 × 3) + (5 × 6)/(13 × 15))
Common factor of 3 and 6, 9 and 12, 5 and 15
= 5/22 – 1/1 + 2/13
= 5/22 – 1/1 + 2/13
LCM of 22,1 and 13 is 286
= (5 × 13 – 286 + 2 × 22)/286
= (65 – 286 + 44)/286
= (65 – 330)/286
= -177/286
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# How do you find the scalar and vector projections of $b$ onto a given $a = < 3,\; - 6,\;2 > ,\;b = < 1,\;1,\;1 > ?$
Last updated date: 12th Aug 2024
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Hint:Scalar projection of $b$ onto $a$ is given as $\dfrac{{\overrightarrow a .\;\overrightarrow b }}{{\left| a \right|}}$ and vector projection of $b$ onto $a$ is given as $\dfrac{{\overrightarrow a .\;\overrightarrow b }}{{{{\left( {\left| a \right|} \right)}^2}}}\overrightarrow a$
Where $\overrightarrow a .\;\overrightarrow b$ represents the dot product between $a\;{\text{and}}\;b$, $\left| a \right|$ represents the magnitude of the vector $a$.
Use the above information to find the respective scalar and vector projections.
In order to find the scalar projection of $b$ onto $a$ we have to first find the dot product of $a\;{\text{and}}\;b$. Dot product of two vectors $x = < a,\;b,\;c > \;{\text{and}}\;y = < e,\;f,\;g >$ is given as follows
$\overrightarrow x .\overrightarrow y = < a,\;b,\;c > . < e,\;f,\;g > = (a \times e + b \times f + c \times g)$
From the above formula of dot product performing the dot product of $a\;{\text{and}}\;b$, we will get,
$\overrightarrow a .\;\overrightarrow b = < 3,\; - 6,\;2 > . < 1,\;1,\;1 > = \left( {3 \times 1 + ( - 6) \times 1 + 2 \times 1} \right) = 3 - 6 + 2 = - 1 \\ \Rightarrow \overrightarrow a .\;\overrightarrow b = - 1 \\$
Now we have to find the magnitude of $a$ in order to find the scalar projection of $b$ onto $a$. Magnitude of a vector $x = < a,\;b,\;c >$ is given as
$\left| x \right| = \left| {\sqrt {{a^2} + {b^2} + {c^2}} } \right|$
Using this to find the magnitude of $a = < 3,\; - 6,\;2 >$, we will get
$\left| a \right| = \left| {\sqrt {{3^2} + {{( - 6)}^2} + {2^2}} } \right| = \left| {\sqrt {9 + 36 + 4} } \right| = \left| {\sqrt {49} } \right| = \left| { \pm 7} \right| = 7$
Scalar projection of a vector $x = < a,\;b,\;c >$ on to the vector $y = < e,\;f,\;g >$ is given as $\dfrac{{\overrightarrow y .\;\overrightarrow x }}{{\left| y \right|}}$
With the use of this formula the scalar projection of $b$ onto $a$ will be given as $\dfrac{{\overrightarrow a .\;\overrightarrow b }}{{\left| a \right|}}$
Putting the values we will get
$\dfrac{{\overrightarrow a .\;\overrightarrow b }}{{\left| a \right|}} = \dfrac{{ - 1}}{7}$
Now we will find the vector projection of $b$ onto $a$
Vector projection of a vector $x = < a,\;b,\;c >$ on to the vector $y = < e,\;f,\;g >$ is given as $\dfrac{{\overrightarrow y .\;\overrightarrow x }}{{{{\left( {\left| y \right|} \right)}^2}}}\overrightarrow y$
Therefore vector projection of $b$ onto $a$ will be given as $\dfrac{{\overrightarrow a .\;\overrightarrow b }}{{{{\left( {\left| a \right|} \right)}^2}}}\overrightarrow a$
Hence we are familiar to these terms in the vector projection of $b$ onto $a$
So directly putting their values we will get
$\therefore\dfrac{{\overrightarrow a .\;\overrightarrow b }}{{{{\left( {\left| a \right|} \right)}^2}}}\overrightarrow a = \dfrac{{ - 1}}{{{7^2}}} < 3,\; - 6,\;2 > = \dfrac{{ - 1}}{{49}} < 3,\; - 6,\;2 >$
Therefore the required scalar and vector projections of the given vectors are $\dfrac{{ - 1}}{7}\;{\text{and}}\;\dfrac{{ - 1}}{{49}} < 3,\; - 6,\;2 >$ respectively.
Note:The given vectors $a = < 3,\; - 6,\;2 > \;{\text{and}}\;b = < 1,\;1,\;1 >$ can also be written as $a = \left( {3\widehat {\text{i}},\; - 6\widehat {\text{j}},\;2\widehat {\text{k}}} \right)\;{\text{and}}\;b = \left( {\widehat {\text{i}},\;\widehat {\text{j}},\;\widehat {\text{k}}} \right)$ where it $3\widehat {\text{i}}$ is read as “3 i-cap” it is also a form vector notation. Also the actual formula for dot product of two vectors $\overrightarrow x \;{\text{and}}\;\overrightarrow y$ is given as follows
$\overrightarrow x .\overrightarrow y = xy\cos \theta ,\;{\text{where}}\;\theta$ is the acute angle between the two vectors. |
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# Generation of Equivalent Expression
Sep 17, 2022
### Key Concepts
• Use properties of operations to write equivalent expressions
• Write equivalent expressions by combining like terms
• Identify equivalent expressions
## 4.1 Generate equivalent expressions
To generate an equivalent expression to another expression, we have to be aware of the parts of an algebraic expression.
We can use properties to combine like terms in an expression.
For example, let us consider the algebraic expression
3x + 2x + 4
You can add/subtract the coefficients of the like terms to combine them.
3x + 2x + 4 = 5x + 4
### 4.2.1 Use properties of operations to write equivalent expressions
Equivalent expressions have the same value regardless of the value that is substituted for the same variable in the expression.
Example 1:
Use properties operation to write equivalent expressions.
– 1/2(x+8)
Solution:
-1/2(x+8) =-1/2x + (-1/2 ) . 8
=-1/2x +(-4)
= -4 + (−1/2x)
### 4.2.2 Write equivalent expressions by combining like terms
Example1:
Write an equivalent expression by combining the like terms.
-6x+2y+4x
Solution:
-6x+5y+4x
-6x + 4x+5y
(-6+4) x+5y
-2x+5y
Example2:
Write an equivalent expression by combining the like terms.
2y+5y−5+8
Solution:
Combine the like terms of the first expression.
Here, the terms 2y and 5y are like terms.
Also, −5 and 8 can be combined to get 3.
### 4.2.3 Identify equivalent expressions
Example1:
Which of the expressions below is equaling?
Solution:
4(2x–1) = 4(2x) – 4(1)
= 8x – 4
Example2:
1. Which of the following expressions are equivalent to 7x+1?[Text Wrapping Break]2(2x−1) + 3(x+1)
1. 5(x+1) + 2(x–2)
1. 7(x–1) +2
Solution:
1. 2(2x−1) +3(x+1) =2(2x) +2(−1) +3(x)+3(1)
=4x−2+3x+3
=4x+3x−2+3
=7x+1
1. 5(x+1) +2(x−2) =5(x)+5(1) +2(x)+2(−2)
=5x+5+2x−4
=5x+2x+5−4
=7x+1
1. 7(x−1) +2 =7(x)+7(−1) +2
=7x−7+2
=7x−5
## Exercise:
1. Write an equivalent expression. (a.) -3(8+5g) (b.) (x+6)+3y
2. Use properties of operation to write an expression equivalent to 5x + +3x-3.
3. Use the associative property to write an expression equivalent to (w + 6) +4.
4. Which of the following expressions is equivalent to 4x-3 for all values of x?
5. 2(2x−3)
6. 2(2x-1)-1
7. 2(2x+1)-4
8. Write an equivalent expression to 5(x-1) +7.
9. Which expressions are equivalent to -6n +(-12) +4n?
10. 4(n-3)-6n
11. 2(2n-6)
12. Andre wrote the expression —2 + 4x ÷3 to represent the relationship shown in the table.
Write two other expressions that also represent the relationship shown in the table.
8. Write equivalent expressions by combining the like terms.
4x + 3x + 5y
9. Write equivalent expressions by combining the like terms.
3x + 2y + 4x + 7 – y.
10. Use properties of operation to write an expression equivalent to 6x + + 4x – 4.
### What have we learned:
• Understand equivalent expressions
• Understand how to use properties of operations to write equivalent expressions
• Identify like terms
• Write equivalent expressions by combining like terms
• Identify equivalent expressions
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […] |
# What makes the standard deviation helpful in the calculation
## Variance
The Variance is one of the most important Dispersion parameters in the statistics. Find out how the Variance defined is what value it describes and what the Difference from the standard deviation is.
With our Video do you understand the topic without any problems - lean back and let ‘explain it to you! What are you waiting for?
You want to understand how exactly the Calculate variance leaves or what the Standard deviation is? Then take a look at our separate article! Also on the topic Empirical variance we have a contribution.
### Variance explained simply
Finding the variance for the distribution of a random variable (population variance) is easier when you understand what it means. Let's first look at how it is defined.
The variance is the mean squared deviation of all values in a random experiment from their expected value. The formula for the variance is:
You practically estimate how far the individual values of the random experiment are from the expected value. Then you square the deviation. The whole thing is best illustrated graphically. In this context is also the Standard deviation important. It is the root of the variance. Another important distinction is whether it is the variance of a random variable (i.e. the corresponding probability distribution is known) or whether it is a sample (the probability distribution is unknown). In the latter case you have to use the Sample variance to calculate.
### Variance example
Is that too abstract for you? Then imagine the following example:
You have two different games of chance: With the first you can either win or lose € 100 with the same probability, with the second you win exactly one euro or lose one euro. Although both games of chance have exactly the same expected value, namely 0, their variance is very different. This is because the possible results are differently far from the expected value.
### Calculate variance
There is a simple procedure to calculate the variance: First you have to determine the expected value, then insert the individual values in the formula and then calculate the variance. In our article Calculate variance let's go into more detail about the procedure and the formula of the variance.
In this example we can easily determine the variance: First we need the expected value. In both cases it is 0. You calculate this by calculating the individual values times their probability of occurrence and adding them together. If you are unsure how to come up with it, have a look at our Video on the expected value at. Then we can insert the values into the formula for the variance and get two different values of the variance for our game of chance:
Bank note:
Coin:
So you see: Although the expected value is the same, the variance can be very different. This is because the possible events, in the case of the bank note, are further away from the expected value than in the case of the coin.
### Variance in statistics
The variance is a measure of statistics and stochastics, which the Spread of the data around the mean indicates. Since there is a difference in the formula, it can only be calculated for cardinally scaled data. You can also tell from the formula that the values are squared, which makes it difficult to interpret. Therefore, the standard deviation is usually used to interpret the spread of the data. If you do not know the probability of occurrence for the events, we will use the sample variance. This weights the individual values equally and forms a distorted or undistorted estimator of the variance. If you want to learn more about it, check out our article Sample variance at!
### Difference variance standard deviation
So if we want to interpret the dispersion around the mean, it is not so easy to do with the variance. Instead, we can do the Standard deviation use. But what is the difference between these two values?
The standard deviation is the square root of the variance
You can see from the formula: The standard deviation is nothing other than the root of the variance. Based on our example, we get a standard deviation of € 100 and € 1 - this is how far the values are on average from the mean.
In order to be able to compare individual random experiments with one another and to be able to interpret the values better, it is therefore often helpful to calculate the standard deviation. |
A parallelogram is a regular polygon with four sides and vertices. It is a quadrilateral formed by two pairs of parallel lines. A parallelogram with right angles is known as a rectangle. We will discuss how to prove that the Area of a Parallelogram is Equal to that of a Rectangle Between the Same Parallel Lines.
## Area of a Parallelogram is Equal to that of a Rectangle Between the Same Parallel Lines – Theorem
Prove that Area of the parallelogram is equal to that of the rectangle on the same base and of the same height that is between the same parallel lines.
Proof:
PS = QR [opposite sides of the parallelogram PQRS]
PN = QM [Opposite sides of the rectangle PQMN]
∠PNS = ∠QMR [Noth are right angles, PQMN being a rectangle]
∆PNS ≅ ∆QMR
ar(∆PNS) = ar(∆QMR)
We know that the rectangle and parallelogram are on the same base and between the same parallel lines and the height of the parallel line at every point is constant.
Let b be the base of the rectangle
Let h be the height of the line is h
The height of the parallel line is nothing but the breadth of the rectangle.
Area of the rectangle = length × breadth
Also, we can write it as base × height
The area of the parallelogram is base × height
Therefore, area of a rectangle = Area of the parallelogram
Hence Proved.
### FAQs on Area of a Parallelogram is Equal to that of a Rectangle Between the Same Parallel Lines
1. What is the relation between the area of parallelogram and rectangle standing on the same base and between the same parallels?
Parallelograms and rectangles on the same base and between the same parallels are equal in area.
2. Are parallel lines equal in a parallelogram?
A quadrilateral with two pairs of parallel lines is known as a parallelogram. The opposite sides of a parallelogram are always equal in length and parallel.
3. Is a parallelogram a rectangle True or false?
Parallelograms are quadrilaterals with two sets of parallel sides. Since squares must be quadrilaterals with two sets of parallel sides, then all squares are parallelograms. A parallelogram is a rectangle. |
Eventually distributed points in any direction can called line.
Eventually distributed points in particular direction can called straight line.
See below fig to understand different type of lines.
angeles
When two lines are intersect then there will be one corner formed this corner is called vertex. In the above fig1-e two line AB and CD are intersecting at the point O. Due to the intersection of two lines there are four angles. Try to name that four angles.
complementary angles
If the sum of two angle are 90º then the are called complementary angles.
Now try to answer following questions.
1. Can two acute angles be complement to each other?
2. can two obtuse angles be complement to each other?
3. Can two right angles be complement to each other?
From the below figure identify which pair of angles are complementary.
Supplementary angles : If the sum of 2 angles are equal to 180°, then that angles are called Supplementary angles.
1. Can two obtuse angles be supplementary ?
2. Can two acute angles be supplementary ?
3. Can two right angles be supplementary ?
Check for some exercises in the test book.
Adjacent angles: If two angles are adjacent angles then they have a common vertex .a common arm and there non -common arm are the different side of the common arm. In the below figure-2 we can see a adjacent angles.
Here the angle ∠ABD and ∠DBC are adjacent angle. But also note that ∠ ABC is not adjacent angle. More over when there is two adjacent angle, then the sum is always equal to the angle formed by two non-common arms
that is ∠ABD+∠CBD=∠ABC
LINEAR PAIR OF ANGLE
If the non-common arm of an adjacent angle are opposite ray the angles are called linear pair of angle. As shown in the fig3, AB and BC are non common arm and the angle ABD and angle CBD are linear pair of angle. Linear pair are supplementary.
TRy to answer the following questions
1. Can two obtuse angles be linear pair ?
2. Can two acute angles be linear pair ?
3. Can two right angles be linear pair ?
Now try to do some exercises from textbook
5 thoughts on “Lines and angles Grade-7”
1. Rufus shaji says:
“When two lines are intersect then there will be one corner formed this corner is called vertex”
Sir, if all lines intersect will there be still one vertex only?
1. Two line are intersecting means, two lines are passing through one point. And this point is called vertex. If many lines are pass through one point there will be only one point of intersection. And there will be only one vertex.
2. Rufus shaji says:
sir the hw you gave to study the:
Find the values of the angles x, y and z
can demomstrate it one time |
The Weighted Digit Sums of Multiples of Any Digit of Five or Greater is that Digit
San José State University
applet-magic.com
Thayer Watkins
Silicon Valley,
& the Gateway
to the Rockies
USA
The Weighted Digit Sums
of Multiples of Any Digit of
Five or Greater is that Digit
## Multiples of Nine
It is well known that the digit sums of all multiples of 9 are 9. Digit sum means that the sum of the digits of any result is computed until the result is a single digit. For example, consider 11*9=99. The sum of the digits of 99 is 18 and the sum of the digits of 18 is 9. Thus the digit sum of 99 is 9.
A far more general relationship prevails; i.e.,
## Illustrations
Define the weight h for a digit m as 10−m. The weighted sum of two digits ab of a number is h*a+b. For example, the weight for 8 is 2. Consider the two digit multiples of 8; 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, and 96. Here are their weighted digit sums WDS
#### WDS(16) = 2*1 + 6 = 8 WDS(24) = 2*2 + 4 = 8 WDS(32) = 2*3 + 2 = 8 WDS(40) = 2*4 + 0 = 8 WDS(48) = 2*4 + 8 = WDG(16) = 8 WDS(56) = 2*5 + 6 = WDS(16) = 8 WDS(64) = 2*6 + 4 = WDS(16) = 8 WDS(72) = 2*7 + 2 = WDS(16) = 8 WDS(80) = 2*8 + 0 = WDS(16) = 8 WDS(88) = 2*8 + 8 = WDS(24) = 8 WDS(96) = 2*9 + 6 = WDS(24) = 8
Here is an example of a three digit multiple of 8
#### WDS(14*8) = WDS(112) = WDS(WDS(11),2) = WDS(32) = 8
The weight for 7 is h=3. For the first few multiples of 7.
#### WDS(14) = 3*1 + 4 = 7 WDS( 21) = 3*2 + 1 = 7 WDS( 28) = 3*2 + 8 = WDS(14) = 7 WDS( 35) = 3*3 + 5 = WDS(14) = 7 WDS( 42) = 3*4 + 2 = WDS(14) = 7 WDS( 49) = 3*4 + 9 = WDS(21) = 7
Here are cases of 6 with its weight of 4.
#### WDS(12) = 1*4 + 2 = 6 WDS(18) = 1*4 + 8 = WDS(12) = 6 WDS(24) = 2*4 + 4 = WDS(12) = 6 WDS( 30) = 3*4 + 0 = WDS(12) = 6 WDS(36) = 3*4 + 6 = WDS(18) = 6 WDS(42) = 4*4 + 2 = WDS(18) = 6 WDS(48) = 4*4 + 8 = WDS(24) = 6 WDS(54) = 5*4 + 4 = WDS(24) = 6
Here are cases of 5 with its weight of 5.
#### WDS(10) = 1*5 + 0 = 5 WDS(15) = 1*5 + 5 = WDS(10) = 5 WDS(20) = 2*5 + 0 = WDS(10) = 5 WDS(25) = 2*5 + 5 = WDS(15) = 5 WDS(30) = 3*5 + 0 = WDS(15) = 5 WDS(35) = 3*5 + 5 = WDS(20) = 5 WDS(40) = 4*5 + 0 = WDS(20) = 5 WDS(45) = 4*5 + 5 = WDS(25) = 5
In contrast here are the results for m=4
#### WDS(8) = 0*6 + 8 = 8 ≠ 4 WDS(12) = 1*6 + 2 = 8 ≠ 4 WDS(16) = 1*6 + 6 = WDS(12) = 8 ≠ 4 WDS(20) = 2*6 + 0 = WDS(12) = 8 ≠ 4 WDS(24) = 2*6 + 4 = WDS(16) = WDS(12) = 8 ≠ 4 WDS(28) = 2*6 + 8 = WDS(20) = WDS(12) = 8 ≠ 4 WDS(32) = 3*6 + 2 = WDS(20) = 8 ≠ 4 WDS(36) = 3*6 + 6 = WDS(24) = WDS(16) = 8 ≠ 4
While the weighted digit sums of multiples of 4 are not equal to 4 they are equal a value equivalent to 4 in terms their remainders upon division by 4.
Now consider the weighted digit sums of multiples of 3. For m=3, h=7.
#### WDS(6) = 0*7 + 6 = 6 ≠ 3 WDS(9) = 0*7 + 9 = 9 ≠ 3 WDS(12) = 1*7 + 2 = 9 ≠ 3 WDS(15) = 1*7 + 5 = WDS(12) = 9 ≠ 3 WDS(18) = 1*7 + 8 = WDS(15) = 9 ≠ 3 WDS(21) = 2*7 + 1 = WDS(15) = 9 ≠ 3 WDS(24) = 2*7 + 4 = WDS(18) = 9 ≠ 3 WDS(27) = 2*7 + 7 = WDS(21) = 9 ≠ 3
As with the case of m=4 the weighted sums of multiples of 3 are multiples of 3. Thus while the proposition under consideration does not hold for m<5 some more general proposition would hold. \
The proof of this particular proposition is given Elsewhere and a more general proposition is dealt with in Weighted Digit Sums as Remainders.
## Numbers Greater than Ten
Consider m=11. Its weight h is equal to 10−11= −1. Example:
WDS(12*11) = WDS(132) = (−1)1 + 3 + (−1)2 = 0
Now consider m-12. Its weight h=10−12 = −2. Examples:
WDS(4*12) = WDS(48) = (−2)*4 + 8 = 0
WDS(12*12) = WDS(144) = WDS(WDS(14),4) = WDS(24) = (−2)*2 + 4 = 0
A Tentative General Rule
#### The Weighted Digit Sum of any Multiple of a Number m>10 is either −m or Zero
A wild example: Let m=19. Then h=−9. Consider 38=2*19. WDS(38)=(−9)*3 + 8 = −27 +8 = −19.
But the general rule is not fully formulated. Take m=20 so h=−10. Consider 40=2*20. WDS(40)=(−10)*4 = −40. WDS(−40)=(−10)(−4)=40.
(To be continued.) |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Independent Events
## Recognize outcomes that do and do not depend on other outcomes.
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Independent Events
Now back to the Talent Show.
The Talent Show has been a huge success and the judges have had a very difficult task. There are three prizes that will be given. After a lot of deliberation, the judges have narrowed it down to the following five finalists.
Given these standings, what is the probability that a seventh grader and an eighth grader will be selected for an award?
This is best solved by thinking about dependent and independent events. Pay close attention to this Concept and you will know how to figure out the probability by the end of it.
### Guidance
Suppose you have two events:
Event A\begin{align*}A\end{align*}: Spin red on spinner A\begin{align*}A\end{align*}
Event B\begin{align*}B\end{align*}: Spin purple on spinner B\begin{align*}B\end{align*}
The probability of these events is easy enough to compute. In general:
P(event)=favorable outcomestotal outcomes\begin{align*}P (\text{event}) = \frac{\text{favorable outcomes}}{\text{total outcomes}}\end{align*}
So:
P(red)P(purple)=14=13\begin{align*}P(\text{red}) & = \frac{1}{4}\\ P (\text{purple}) & = \frac{1}{3}\end{align*}
Now a question arises. Does event A\begin{align*}A\end{align*} affect the probability of event B\begin{align*}B\end{align*} in any way? That is, does the arrow landing on red in the first spinner affect the way the arrow lands in the second spinner? If not, then the two events are said to be independent events.
If the outcome of one event has no effect on the outcome of a second event, then the two events are independent events.
Events A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} above are independent events. No matter how many times you spin spinner A\begin{align*}A\end{align*}, the outcome of spinning spinner A\begin{align*}A\end{align*} does not affect the outcome of spinning spinner B\begin{align*}B\end{align*}.
Look at this scenario.
Jeremy flips a coin two times. Event A\begin{align*}A\end{align*} is the first coin flip. Event B\begin{align*}B\end{align*} is the second coin flip. Are the two coin flips independent events?
Ask yourself, Can the outcome of the first event in any way change the outcome of the second event? If not, then the two events are independent.
Suppose Jeremy’s first flip comes up heads. Does that in any way affect the outcome of the second flip? Is it now more likely to come up heads or tails?
In fact, the first flip does not affect the second flip. The probability of heads in the second flip is 12\begin{align*}\frac{1}{2}\end{align*}, no matter what the first flip was. Similarly, the probability of tails in the second flip is also 12\begin{align*}\frac{1}{2}\end{align*}, no matter what the first flip was. So the two events are independent.
Now that you know about independent events, you can learn about dependent events. If one event does depend on another event, then the events depend on each other.
Look at this scenario.
Mariko pulls a red sock from the laundry bag. Does this change the probability that the next sock Mariko pulls out of the bag will be red?
Here, the act of taking a sock out of the bag changes the situation. For the first sock, the probability of pulling out a red sock was:
P(red)=favorable outcomestotal outcomes=36=12\begin{align*}P(\text{red}) = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{3}{6} = \frac{1}{2}\end{align*}
For the second sock, there are now only 5 socks left in the bag and only 2 of them are red. So the probability of pulling out a red sock now for the second sock is:
P(red)=favorable outcomestotal outcomes=25\begin{align*}P(\text{red}) = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{2}{5}\end{align*}
Clearly, the first event affected the outcome of the second event in this situation. So the two events are NOT independent. In other words, they are dependent events.
If the outcome of one event has an effect on the outcome of a second event, then the two events are dependent events.
Write the difference between dependent events and independent events in your notebook.
Try a few of these on your own. Determine whether the events described are dependent or independent events.
#### Example A
A box contains a penny, a nickel, a dime, and a quarter. What is the probability of pulling a quarter out of the box, putting it in your pocket, then pulling a penny out of the box?
Solution: Dependent events
#### Example B
In a laundry bag with 3 red socks and 3 blue socks, Mariko pulls out a blue sock, sees it’s the wrong sock and returns it to the bag. Now Mariko pulls out a second sock. What is the probability that it will be red?
Solution: Independent events
#### Example C
In a laundry bag with 3 red socks and 3 blue socks, Mariko pulls out a blue sock and keeps it out. Now Mariko pulls out a second sock. What is the probability that it will be blue?
Solution: Dependent events
Here is the original problem once again. Reread it and then begin to work on figuring out the probability question.
The Talent Show has been a huge success and the judges have had a very difficult task. There are three prizes that will be given. After a lot of deliberation, the judges have narrowed it down to the following five finalists.
Given these standings, what is the probability that a seventh grader and an eighth grader will be selected for an award?
Before we even begin to figure out the probability, we first need to decide if these are independent events or dependent events.
Think about it this way, if we select a seventh grader for an award, then the number of possible finalists for the next award changes from 5 to 4. One event depends on the other event. Therefore, these are dependent events.
To figure out the probability of dependent events, we can multiply. We want to figure out the probability of a seventh and an eighth grader being selected for an award.
P(7thand 8th)=2514\begin{align*}P(7^{th} \text{and} \ 8^{th}) = \frac{2}{5} \cdot \frac{1}{4}\end{align*}
We have 2 out of 5 for seventh graders. Then once a seventh grader is selected, we go to four finalists. One is an eighth grader, so we have a 1 out of 4 chance of having an eighth grader for an award.
Next, we multiply.
P(7thand 8th)=2514=320\begin{align*}P(7^{th} \text{and} \ 8^{th}) = \frac{2}{5} \cdot \frac{1}{4} = \frac{3}{20}\end{align*}
What is this as a percent?
320=15100=15%\begin{align*}\frac{3}{20} = \frac{15}{100} = 15\%\end{align*}
There is only a 15% chance of both a seventh grader and an eighth grader as award winners!
### Vocabulary
Independent Events
events where one event does not impact the result of another.
Dependent Events
events where one event does impact the result of another.
### Guided Practice
Here is one for you to try on your own.
Does this situation describe dependent or independent events?
Kelsey has a drawer full of earrings and necklaces. She has four pairs of earrings and six necklaces in the drawer. Kelsey first takes out a necklace. She isn't happy with it, so she puts it back into the drawer. What is the probability of pulling another necklace?
This situation describes independent events because Kelsey put the necklace back in the drawer.
### Practice
Directions: Write whether each pair of events is dependent or independent.
1. A: Mike rolls a number cube. B: Leah spins a red-blue-green spinner.
2. A: In a game of Go Fish, the probability of one player drawing a Queen from the deck. B: On the next player’s turn, the probability of drawing a Queen.
3. A: The probability that a randomly ordered pizza will be large. B: The probability that the same randomly ordered pizza will be deep-dish.
4. A: The probability that a randomly ordered 2-topping pizza will have pepperoni. B: The probability that the same randomly ordered 2-topping pizza will have mushrooms.
5. A: The probability of flipping a coin tails 5 times in a row. B: The probability of the sixth flip turning out to be heads.
6. A: In a 4-team league, the probability of the Rockets finishing in first place. B: In a 4-team league, the probability of the Sharks finishing in first place.
7. A: On a roll of a number cube, the probability of rolling 6. B: On a second roll of a number cube, the probability of rolling 6.
8. A: In a spelling bee, the probability of the first contestant being given the word khaki from a list of 10 words. B: In a spelling bee, the probability of the second contestant getting the word khaki from the same list of words.
9. A: The probability that it will snow on Tuesday. B: The probability that Tuesday will fall on an odd day of the month.
10. A: The probability that it will be below 32 degrees on Tuesday. B: The probability that it will snow on Tuesday.
11. A: The probability that it will snow on Tuesday. B: The probability that school will be cancelled on Tuesday.
12. A: The probability that the first Wednesday in June will fall on an even day of the month. B: The probability that the first Thursday in June will fall on an even day of the month.
13. A: The probability that the first Wednesday in June will fall on an even day of the month. B: The probability that the first Thursday in June will be sunny.
14. A: The probability that a coin will land on heads. B: The probability that a number cube will land on 5.
15. A: The probability that the first spin of a red-blue-green spinner will land on green. B: The probability that the second spin of a red-blue-green spinner will land on green.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Dependent Events
In probability situations, dependent events are events where one outcome impacts the probability of the other.
Disjoint Events
Disjoint or mutually exclusive events cannot both occur in a single trial of a given experiment.
Independent Events
Two events are independent if the occurrence of one event does not impact the probability of the other event.
probability
The chance that something will happen. |
# Chapter-1-Reviewing-linear-equations
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```Chapter 1
1
Reviewing linear equations
Objectives
To solve linear equations in one unknown.
To construct linear equations.
To use linear equations to solve problems.
To solve simultaneous linear equations by substitution and elimination methods.
To solve linear inequalities.
To use and transpose formulas.
Many problems may be solved by first translating them into mathematical equations and then
solving the equations using algebraic techniques. An equation is solved by finding the value
or values of the variables that would make the statement true.
Consider the equation 2x + 11 = 3 − 2x. If x = −2, then
LHS = 2(−2) + 11 = 7
and
RHS = 3 − 2(−2) = 7
The statement is true when x = −2. The solution to the equation is therefore x = −2. In this
case there is no other value of x that would give a true statement.
The equations that we deal with in this chapter are called linear equations since they are
related to the equation of a straight line.
Linear equations have either one solution (as shown in the example above), no solutions or
infinitely many solutions. The equation 2x + 3 = 2x + 4 has no solutions, since no value of x
makes the statement true. The equation 2(x + 3) = 2x + 6 has infinitely many solutions, since
it is true for all values of x.
We note that the equation x2 = 16 has exactly two solutions, x = 4 or x = −4, but this
equation is not linear.
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
Cambridge University Press
ISBN 978-1-107-52013-4
© Evans et al. 2016
Photocopying is restricted under law and this material must not be transferred to another party.
2
Chapter 1: Reviewing linear equations
1A Linear equations
A linear equation (in one unknown) is a particular type of polynomial equation in which the
variable is to the first power. The following are examples of linear equations:
3x − 5 = 11,
z − 3 2z − 5
+
= 11
4
3
7 − 2t = 8t − 11,
In each of these equations, the variable is to the first power.
The following are examples of non-linear polynomial equations:
x2 − x − 12 = 0
(quadratic),
2x3 − x = 0
(cubic),
x4 = 16
(quartic)
In each of these equations, the highest power of the variable is not the first power. You have
met some of these in your previous study, and they are also in later chapters of this book.
Solving linear equations
It is important when setting out the solution to an equation that each step is written under the
previous one with the equals signs aligned. This careful setting out makes the algebra easy to
check. Unsystematic methods, such as guess and check, will generally be of limited use for
more complicated equations.
It is often helpful to look at how the equation has been constructed so that the steps necessary
to ‘undo’ the equation can be identified. It is most important that the steps taken to solve the
equation are done in the correct order.
Linear equations of the form ax + b = c
Many linear equations that arise in applications are of the form ax + b = c.
Example 1
Solve the equation 3x + 4 = 16 for x.
Solution
Explanation
3x + 4 = 16
3x = 12
Subtract 4 from both sides.
x=4
Divide both sides by 3.
Check:
LHS = 3(4) + 4 = 16
RHS = 16
∴ the solution is correct.
Once a solution has been found it may be
checked by substituting the value back into
both sides of the original equation to ensure
that the left-hand side (LHS) equals the
right-hand side (RHS).
The first three equations in the above example are equivalent equations. Obtaining the second
two equations enables us to solve the first equation.
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
Cambridge University Press
ISBN 978-1-107-52013-4
© Evans et al. 2016
Photocopying is restricted under law and this material must not be transferred to another party.
1A Linear equations
3
Given an equation, an equivalent equation can be formed by:
adding or subtracting the same number on both sides of the equation
multiplying or dividing both sides of the equation by the same non-zero number.
Importantly, two equivalent equations have the same solution. By forming suitable equivalent
equations, we solve linear equations.
Equations with the unknown on both sides
Group all the terms containing the variable on one side of the equation and the remaining
terms on the other side.
Example 2
Solve 4x + 3 = 3x − 5.
Solution
Explanation
4x + 3 = 3x − 5
x + 3 = −5
Subtract 3x from both sides and then
subtract 3 from both sides.
x = −8
Check:
LHS = 4(−8) + 3 = −29
The solution can be checked as previously
shown.
RHS = 3(−8) − 5 = −29
∴ the solution is correct.
Equations containing brackets
A frequently used first step is to remove brackets and then to follow the procedure for solving
an equation without brackets.
Example 3
Solve 3(2x + 5) = 27.
Solution
3(2x + 5) = 27
6x + 15 = 27
6x = 12
Explanation
We note that since 27 is divisible by 3, the
following method is also possible:
3(2x + 5) = 27
2x + 5 = 9
x=2
2x = 4
Check:
LHS = 3(2 × 2 + 5) = 27
x=2
RHS = 27
∴ the solution is correct.
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
Cambridge University Press
ISBN 978-1-107-52013-4
© Evans et al. 2016
Photocopying is restricted under law and this material must not be transferred to another party.
4
Chapter 1: Reviewing linear equations
Equations containing fractions
A frequently used first step is to multiply both sides of the equation by the lowest common
multiple of the denominators of the fractions.
Example 4
Solve
x
x
−2= .
5
3
Solution
x
x
−2=
5
3
x
x
× 15 − 2 × 15 = × 15
5
3
3x − 30 = 5x
−2x = 30
x = −15
Check:
LHS =
−15
− 2 = −3 − 2 = −5
5
RHS =
−15
= −5
3
Explanation
The denominators of the fractions are 3
and 5. The lowest common multiple of 3
and 5 is 15.
Multiply both sides of the equation by 15.
This means that each term of the LHS
and the RHS of the equation is multiplied
by 15.
∴ the solution is correct.
Example 5
Solve
x − 3 2x − 4
−
= 5.
2
3
Solution
2x − 4
x−3
×6−
×6=5×6
2
3
3(x − 3) − 2(2x − 4) = 30
3x − 9 − 4x + 8 = 30
−x = 31
Explanation
Remember that the line separating the
numerator and the denominator (the
vinculum) acts as brackets.
Multiply both sides of the equation by 6,
the lowest common multiple of 2 and 3.
x = −31
Check:
−31 − 3 2 × (−31) − 4
−
2
3
−34 −66
−
= −17 + 22 = 5
=
2
3
LHS =
RHS = 5
∴ the solution is correct.
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
Cambridge University Press
ISBN 978-1-107-52013-4
© Evans et al. 2016
Photocopying is restricted under law and this material must not be transferred to another party.
1A Linear equations
5
Using the TI-Nspire
To find the solution to the linear equation,
use a Calculator application.
> Algebra > Solve.
Enter the equation
x − 3 2x − 4
−
=5
2
3
Press enter to obtain the solution.
Select
menu
Note: A template for fractions may be obtained by pressing
ctrl ÷
.
For more details on the use of the calculator refer to the TI-Nspire appendix in the
Interactive Textbook.
Using the Casio ClassPad
Go to the M screen and turn on the
keyboard.
Select the fraction icon N found in
the Math1 keyboard.
Enter the equation
x − 3 2x − 4
−
=5
2
3
Highlight the equation using the
stylus and select Interactive >
Equation/Inequality > solve.
Tap on OK to obtain the solution.
(Note that the default variable is x.)
Note: For more details on the use of the calculator refer to the Casio ClassPad appendix
in the Interactive Textbook.
Literal equations
An equation for the variable x in which all the coefficients of x, including the constants, are
pronumerals is known as a literal equation.
Example 6
Solve ax + b = cx + d for x.
Solution
Explanation
ax + b = cx + d
ax − cx = d − b
Collect terms in x on the left-hand side and
constants on the right-hand side.
(a − c)x = d − b
Factorise the left-hand side.
x=
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
d−b
a−c
Cambridge University Press
ISBN 978-1-107-52013-4
© Evans et al. 2016
Photocopying is restricted under law and this material must not be transferred to another party.
6
Chapter 1: Reviewing linear equations
Using the TI-Nspire
To solve the literal equation ax + b = cx + d,
use a Calculator application.
Select menu > Algebra > Solve.
Enter ax + b = cx + d as shown.
Press enter to obtain the solution.
Note: Ensure a multiplication sign is placed
between the letters of the expression,
otherwise the calculator will read them
as a single variable rather than a product.
That is, enter a × x and not ax.
Using the Casio ClassPad
To solve the literal equation ax + b = cx + d, first go
to the M screen and turn on the keyboard.
Select the Var keyboard. This will bring up the
variables.
Enter the equation ax + b = cx + d and highlight it.
Select Interactive > Equation/Inequality > solve and
ensure that the variable selected is x.
Tap on OK to obtain the solution.
If necessary, the answer may be simplified further
by copying the answer into the next line and then
selecting Interactive > Transformation > simplify.
Section summary
An equation is solved by finding the value or values of the variables that would make
the statement true.
A linear equation is one in which the variable is to the first power.
There are often several different ways to solve a linear equation. The following steps
provide some suggestions:
1 Expand brackets and, if the equation involves fractions, multiply through by the
lowest common denominator of the terms.
2 Group all of the terms containing a variable on one side of the equation and the
terms without the variable on the other side.
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
Cambridge University Press
ISBN 978-1-107-52013-4
© Evans et al. 2016
Photocopying is restricted under law and this material must not be transferred to another party.
1A
1A Linear equations
7
Exercise 1A
1
2
Solve each of the following equations for x:
a x+3=6
b x−3=6
c 3−x=2
d x + 6 = −2
e 2 − x = −3
f 2x = 4
g 3x = 5
h −2x = 7
i −3x = −7
j
3
b x+b=a
c ax = b
d
l
x
=b
a
a 2y − 4 = 6
b 3t + 2 = 17
c 2y + 5 = 2
e 2a − 4 = 7
f 3a + 6 = 14
g
x
+5=9
3
j 3 − 5y = 12
k −3x − 7 = 14
−5x
= −2
7
e
ax
=c
b
d 7x − 9 = 5
y
− 11 = 6
8
h
t 1 1
+ =
3 6 2
l 14 − 3y = 8
4 Solve the following linear equations:
a 6x − 4 = 3x
5
Example 3
b x − 5 = 4x + 10
a 2(y + 6) = 10
b 2y + 6 = 3(y − 4)
d 5(y − 3) = 2(2y + 4)
e x − 6 = 2(x − 3)
3
x x
+ = 10
h x+4= x
2 3
2
2(1 − 2x)
2 4(2 − 3x)
− 2x = − +
j
3
5
3
Example 5
6
e
8
c 2(x + 4) = 7x + 2
y+2
=4
3
7x + 3 9x − 8
i
=
2
4
4y − 5 2y − 1
k
−
=y
2
6
f
Solve the following literal equations for x:
a ax + b = 0
7
c 3x − 2 = 8 − 2x
Solve the following linear equations:
g
Example 4
Example 6
−3x
=2
5
Solve the following linear equations:
i
Example 2
k
Solve each of the following literal equations for x:
a x−b=a
Example 1
3x
=5
4
b cx + d = e
x x
+ =1
a b
f
a b
+ =1
x x
c a(x + b) = c
d ax + b = cx
g ax − b = cx − d
h
Solve each of the following for x:
a 0.2x + 6 = 2.4
b 0.6(2.8 − x) = 48.6
d 0.5x − 4 = 10
e
Solve
9 Solve
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
1
(x − 10) = 6
4
c
ax + c
=d
b
2x + 12
= 6.5
7
f 6.4x + 2 = 3.2 − 4x
b − cx a − cx
+
+ 2 = 0 for x.
a
b
b
a+b
a
+
=
for x.
x+a x−b x+c
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© Evans et al. 2016
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8
Chapter 1: Reviewing linear equations
1B Constructing linear equations
As stated earlier, many problems can be solved by translating them into mathematical
language and using an appropriate mathematical technique to find the solution. By
representing the unknown quantity in a problem with a symbol and constructing an equation
from the information, the value of the unknown can be found by solving the equation.
Before constructing the equation, each symbol and what it stands for (including the units)
should be stated. It is essential to remember that all the elements of the equation must be in
units of the same system.
Example 7
A chef uses the following rule for cooking a turkey:
‘Allow 30 minutes for each kilogram weight of turkey and then add an extra 15 minutes.’
If the chef forgot to weigh a turkey before cooking it, but knew that it had taken 3 hours to
cook, calculate how much it weighed.
Solution
Let the weight of the turkey be x kilograms.
Then the time taken is (30x + 15) minutes.
∴
30x + 15 = 180
30x = 165
x = 5.5
The turkey weighed 5.5 kilograms.
Explanation
Assign a variable to the quantity that is to
be found. In this example, the weight of
the turkey is x kilograms.
Find, in terms of x, the time to cook the
turkey. Then form the equation. Note that
3 hours is 180 minutes.
State the solution to the problem in words.
Example 8
Find the area of a rectangle whose perimeter is 1.08 m, if it is 8 cm longer than it is wide.
Solution
Let length = cm.
Then width = ( − 8) cm.
Perimeter = 2 × length + 2 × width
= 2 + 2( − 8)
= 4 − 16 cm
Perimeter = 108 cm
∴
4 − 16 = 108
Explanation
We know that
Perimeter = 2 × length + 2 × width
and that the width is 8 cm less than the
length. Let cm be the length. Then the
width is ( − 8) cm.
Find the perimeter in terms of . Find the
length and width, and hence find the area.
4 = 124
= 31 cm
The length is 31 cm and the width is 23 cm.
Therefore the area is 31 × 23 = 713 cm2 .
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
Cambridge University Press
ISBN 978-1-107-52013-4
© Evans et al. 2016
Photocopying is restricted under law and this material must not be transferred to another party.
1B
1B Constructing linear equations
9
Example 9
Adam normally takes 5 hours to travel between Higett and Logett. One day he increases
his speed by 4 km/h and finds the journey from Higett to Logett takes half an hour less
than the normal time. Find his normal speed.
Solution
Let x km/h be his normal speed.
The distance from Higett to Logett is
x × 5 = 5x kilometres.
Explanation
In problems such as this, the speed is the
average speed.
We note that
Adam’s new speed is (x + 4) km/h.
9
Hence (x + 4) × = 5x
2
9(x + 4) = 10x
distance = speed × time
9x + 36 = 10x
36 = x
His normal speed is 36 km/h.
Section summary
Steps for solving a word problem with a linear equation:
Read the question carefully and write down the known information clearly.
Identify the unknown quantity that is to be found.
Assign a variable to this quantity.
Form an expression in terms of x (or the variable being used) and use the other relevant
information to form the equation.
Solve the equation.
Write a sentence answering the initial question.
Exercise 1B
1
For each of the following, write an equation using the variable x, then solve the
equation for x:
a
b
c
d
e
A number plus two is equal to six.
A number multiplied by three is equal to ten.
Six is added to a number multiplied by three and the result is twenty-two.
Five is subtracted from a number multiplied by three and the result is fifteen.
Three is added to a number. If the result of this is multiplied by six, then fifty-six is
obtained.
f Five is added to a number and the result divided by four gives twenty-three.
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
Cambridge University Press
ISBN 978-1-107-52013-4
© Evans et al. 2016
Photocopying is restricted under law and this material must not be transferred to another party.
10
Chapter 1: Reviewing linear equations
2
\$48 is divided among three students, A, B and C. If B receives three times as much
as A, and C receives twice as much as A, how much does each receive?
3
The sum of two numbers is 42, and one number is twice the other. Find the two
numbers.
1B
Example 7
4 A chef uses the following rule for cooking food on a spit: ‘Allow 20 minutes for each
kilogram weight and then add an extra 20 minutes.’ If the chef forgot to weigh the food
before cooking it but knew that it had taken 3 hours to cook, calculate how much it
weighed.
Example 8
5
Find the area of a rectangle whose perimeter is 4.8 m, if it is 0.5 m longer than it is
wide.
6
Find three consecutive whole numbers with a sum of 150.
7
Find four consecutive odd numbers with a sum of 80.
8
Two tanks contain equal amounts of water. They are connected by a pipe and
3000 litres of water is pumped from one tank to the other. One tank then contains
6 times as much water as the other. How many litres of water did each tank contain
originally?
9
A 120-page book has p lines to a page. If the number of lines were reduced by three
on each page, the number of pages would need to be increased by 20 to give the same
amount of writing space. How many lines were there on each page originally?
Example 9
Example 9
10 A rower travels upstream at 6 km/h and back to the starting place at 10 km/h. The total
journey takes 48 minutes. How far upstream did the rower go?
11
A shopkeeper buys a crate of eggs at \$1.50 per dozen. He buys another crate,
containing 3 dozen more than the first crate, at \$2.00 per dozen. He sells them all
for \$2.50 a dozen and makes \$15 profit. How many dozens were there in each of the
crates?
12
Jess walked for 45 minutes at 3 km/h and then ran for half an hour at x km/h. At the end
of that time she was 6 km from the starting point. Find the value of x.
13
A man travels from A to B at 4 km/h and from B to A at 6 km/h. The total journey takes
45 minutes. Find the distance travelled.
14
A boy is 24 years younger than his father. In two years time the sum of their ages will
be 40. Find the present ages of father and son.
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11
1C Simultaneous equations
1C Simultaneous equations
A linear equation that contains two unknowns, e.g. 2y + 3x = 10, does not have a single
solution. Such an equation actually expresses a relationship between pairs of numbers,
x and y, that satisfy the equation. If all possible pairs of numbers (x, y) that satisfy
the equation are represented graphically, the result is a straight line; hence the name
linear relation.
y
If the graphs of two such equations are drawn on
the same set of axes, and they are non-parallel,
the lines will intersect at one point only. Hence
there is one pair of numbers that will satisfy both
equations simultaneously.
The intersection point of two straight lines can
be found graphically; however, the accuracy of
the solution will depend on the accuracy of the
graphs.
2x
–y
=4
4
3
2
–3 –2 –1
1
1
2
3
x
0
–1
–2
Alternatively, the intersection point may be found
algebraically by solving the pair of simultaneous
equations. We shall consider two techniques for
solving simultaneous equations.
–3
–4
x+
(1, –2)
2y
=–
3
Example 10
Solve the equations 2x − y = 4 and x + 2y = −3.
Solution
Explanation
Method 1: Substitution
2x − y = 4
(1)
x + 2y = −3
(2)
Using one of the two equations, express
one variable in terms of the other variable.
From equation (2), we get x = −3 − 2y.
Substitute in equation (1):
2(−3 − 2y) − y = 4
−6 − 4y − y = 4
Then substitute this expression into the
other equation (reducing it to an equation
in one variable, y). Solve the equation for y.
−5y = 10
y = −2
Substitute the value of y into (2):
x + 2(−2) = −3
Substitute this value for y in one of the
equations to find the other variable, x.
x=1
Check in (1): LHS = 2(1) − (−2) = 4
RHS = 4
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A check can be carried out with the other
equation.
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12
Chapter 1: Reviewing linear equations
Method 2: Elimination
2x − y = 4
(1)
x + 2y = −3
(2)
To eliminate x, multiply equation (2) by 2
and subtract the result from equation (1).
When we multiply equation (2) by 2, the
pair of equations becomes:
2x − y = 4
(1)
2x + 4y = −6
(2 )
If one of the variables has the same
coefficient in the two equations, we can
eliminate that variable by subtracting one
equation from the other.
It may be necessary to multiply one of
the equations by a constant to make the
coefficients of x or y the same in the two
equations.
Subtract (2 ) from (1):
−5y = 10
y = −2
Now substitute for y in equation (2) to
find x, and check as in the substitution
method.
Note: This example shows that the point (1, −2) is the point of intersection of the graphs of
the two linear relations.
Using the TI-Nspire
Calculator application
Simultaneous equations can be solved in a
Calculator application.
Use menu > Algebra > Solve System of
Equations > Solve System of Equations.
Complete the pop-up screen.
Enter the equations as shown to give the
solution to the simultaneous equations
2x − y = 4 and x + 2y = −3.
Note: The solution can also be found with
solve(2x − y = 4 and x + 2y = −3, x, y).
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1C Simultaneous equations
13
Graphs application
The simultaneous equations can also be solved graphically in a Graphs application.
Entering the equations:
The equations can be entered directly in
the form a · x + b · y = c using menu >
Graph Entry/Edit > Equation > Line >
a · x + b · y = c.
Enter the equations as shown.
Hint: Use to enter the second equation.
Alternatively:
The equations can be rearranged to make y the subject. The equations in this form are
−3 − x
.
2
Enter these in the default function entry line.
f 1(x) = 2x − 4 and f 2(x) =
Note: If the entry line is not visible, press tab or double click in an open area. Pressing
enter
will hide the entry line.
Finding the intersection point:
Use menu > Geometry > Points & Lines > Intersection Point(s).
Use the touchpad to move the cursor to select each of the two graphs.
The intersection point’s coordinates will
appear on the screen. Press
Intersection Point(s) tool.
esc
to exit the
Note: You can also find the intersection point using menu > Analyze Graph >
Intersection.
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14
Chapter 1: Reviewing linear equations
Using the Casio ClassPad
To solve the simultaneous equations algebraically:
Go to the M screen and turn on the keyboard.
Open the Math1 keyboard and select the simultaneous
equations icon ~.
Enter the two equations
2x − y = 4
x + 2y = −3
into the two lines.
Type x, y in the bottom-right square to indicate the
variables.
Select EXE .
The simultaneous equations can also be solved graphically:
Tap on the graph icon \$ to display the graph screen.
Using your stylus, highlight the first equation 2x − y = 4
and drag it down into the graph screen. Lift the stylus off
the screen for the graph to appear.
Repeat by highlighting the second equation x + 2y = −3
and dragging it down into the graph screen. Lift the stylus
off the screen for the second graph to appear.
To find the solution, tap into the graph screen to select it,
and then select Analysis > G-Solve > Intersection.
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1C Simultaneous equations
15
The geometry of simultaneous equations
Two distinct straight lines are either parallel or meet at a point.
There are three cases for a system of two linear equations with two variables.
Case 1
Example
Solutions
Geometry
2x + y = 5
Unique solution:
x = 3, y = −1
Two lines meeting at a point
No solutions
Distinct parallel lines
Infinitely many solutions
Two copies of the same line
x−y=4
Case 2
2x + y = 5
2x + y = 7
Case 3
2x + y = 5
4x + 2y = 10
This is further discussed in Chapter 2.
Section summary
We have two methods for solving simultaneous linear equations in two variables by hand.
1 Substitution
Make one of the variables the subject in one of the equations and substitute for that
variable in the other equation.
For example, for the equations:
3x + y = 6
(1)
5x + 3y = −2
(2)
From (1), we have y = 6 − 3x.
Substitute this into (2), and solve the resulting equation 5x + 3(6 − 3x) = −2 for x.
Substitute this value for x back into one of the original equations to find the value of
the other variable, y.
Check that your solution satisfies the original equations.
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16
1C
Chapter 1: Reviewing linear equations
2 Elimination
In this method, one of the variables is eliminated. For example, with the same
equations considered above:
3x + y = 6
(1)
5x + 3y = −2
(2)
Decide which variable to eliminate. We will eliminate y.
If necessary, multiply both sides of one or both equations by a number to obtain the
same or opposite coefficient for the variable to be eliminated. With this example,
multiply (1) by 3:
9x + 3y = 18
(1 )
Add or subtract the equations to eliminate the chosen variable. In this example,
subtract (2) from (1 ).
Solve the resulting equation, and then proceed as with the substitution method.
Exercise 1C
1
Solve each of the following pairs of simultaneous equations by the substitution method:
a y = 2x + 1
b y = 5x − 4
y = 3x + 2
y = 3x + 6
d y − 4 = 3x
e y − 4x = 3
y − 5x + 6 = 0
2
2y − 5x + 6 = 0
2y − 3x = 4
c x − 2y = 6
x+y=3
x + 6y = 10
Solve each of the following pairs of simultaneous linear equations by either the
substitution or the elimination method:
a 2x − 3y = 7
y = 5 − 3x
b 2x − 5y = 10
c 2m − 1 = n
4x + 3y = 7
d 7x − 6y = 20
3x + 4y = 2
g 15x − 4y = 6
9x − 2y = 5
4
f y − 4x = 6
b y−x=5
x − y = 10
3
y = 5x + 10
Solve each of the following pairs of simultaneous equations by the elimination method:
a x+y=6
Example 10
c y = 2 − 3x
2n + m = 8
e 3s − 1 = t
f 4x − 3y = 1
5s + 2t = 20
4y − 5x = 2
h 2p + 5q = −3
i 2x − 4y = −12
7p − 2q = 9
2y + 3x − 2 = 0
For each of the following pairs of simultaneous linear equations, state whether there is
one, none or infinitely many solutions:
a 3x + y = 6
b 3x + y = 6
6x + 2y = 7
Cambridge Senior Maths AC/VCE
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6x + 2y = 12
c 3x + y = 6
6x − 2y = 7
d 3x − y = 6
6x + 2y = 7
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1D Constructing simultaneous linear equations
17
1D Constructing simultaneous linear equations
Skillsheet
Problems involving two unknowns can often be solved by using simultaneous equations with
two variables. The following examples show how this may be done.
Example 11
The sum of two numbers is 24 and their difference is 96. Find the two numbers.
Solution
Let x and y be the two numbers. Then
x + y = 24
(1)
x − y = 96
(2)
Explanation
The problem can also be solved by
eliminating x. Subtracting (2) from (1)
gives 2y = −72 and hence y = −36.
Add equations (1) and (2):
The problem can also be solved by
substitution. From (1), we have y = 24 − x.
Substitute in (2).
2x = 120
x = 60
The values found for x and y have to make
each of the equations true. The equation
which has not been used in the final
substitution is the one to use for the check.
Substitute in equation (1):
60 + y = 24
y = −36
The two numbers are 60 and −36.
Check in (2):
60 − (−36) = 96
Example 12
3 kg of jam and 2 kg of butter cost \$29, and 6 kg of jam and 3 kg of butter cost \$54.
Find the cost per kilogram of jam and butter.
Solution
Let the cost of 1 kg of jam be x dollars and
the cost of 1 kg of butter be y dollars.
Then
3x + 2y = 29
(1)
and
6x + 3y = 54
(2)
Multiply (1) by 2:
6x + 4y = 58
(1 )
Subtract (1 ) from (2):
−y = −4
y=4
Substitute in (2):
6x + 3(4) = 54
6x = 42
x=7
Jam costs \$7 per kg and butter \$4 per kg.
Cambridge Senior Maths AC/VCE
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Explanation
The unknowns are the cost per kilogram of
jam and the cost per kilogram of butter.
Three kilograms of jam and two kilograms
of butter cost \$29.
Six kilograms of jam and three kilograms
of butter cost \$54.
Check in the original problem:
3 kg of jam = \$21 and 2 kg of butter = \$8
Total = \$29
6 kg of jam = \$42 and 3 kg of butter = \$12
Total = \$54
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18
Chapter 1: Reviewing linear equations
1D
Section summary
Steps for solving a word problem with simultaneous linear equations:
Read the question carefully and write down the known information clearly.
Identify the two unknown quantities that are to be found.
Assign variables to these two quantities.
Form expressions in terms of x and y (or other suitable variables) and use the other
relevant information to form the two equations.
Solve the system of equations.
Write a sentence answering the initial question.
Exercise 1D
1
Find two numbers whose sum is 138 and whose difference is 88.
2
Find two numbers whose sum is 36 and whose difference is 9.
3
Six stools and four chairs cost \$58, while five stools and two chairs cost \$35.
a How much do ten stools and four chairs cost?
b How much do four stools cost?
c How much does one stool cost?
4
A belt and a wallet cost \$42, while seven belts and four wallets cost \$213.
a How much do four belts and four wallets cost?
b How much do three belts cost?
c How much does one belt cost?
Use simultaneous equations to solve the following.
Example 11
5 Find a pair of numbers whose sum is 45 and whose difference is 11.
6
In four years time a mother will be three times as old as her son. Four years ago she
was five times as old as her son. Find their present ages.
7
A party was organised for thirty people at which they could have either a hamburger or
a pizza. If there were five times as many hamburgers as pizzas, calculate the number
of each.
8
Two children had 110 marbles between them. After one child had lost half her marbles
and the other had lost 20 they had an equal number. How many marbles did each child
start with and how many did they finish with?
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1D
1D Constructing simultaneous linear equations
9
Example 12
19
One hundred and fifty tickets were sold for a basketball match and \$560 was the total
amount collected. Adult tickets were sold at \$4.00 each and child tickets were sold at
\$1.50 each. How many adult tickets and how many child tickets were sold?
10
The sum of the numerator and denominator of a fraction expressed in simplest form
is 17. If 3 is added to the numerator, the value of the fraction will be 1. What is the
fraction?
11
Linda thinks of a two-digit number. The sum of the digits is 8. If she reverses the digits,
the new number is 36 greater than her original number. What was Linda’s original
number?
12
Tickets to a musical cost \$30 for adults and \$12 for children. At one particular
performance 960 people attended and \$19 080 was collected in ticket sales. Find the
number of adults and the number of children who attended the performance.
13
An investor received \$1400 interest per annum from a sum of money, with part of it
invested at 10% and the remainder at 7% simple interest. This investor found that if
she interchanged the amounts she had invested she could increase her return by \$90 per
annum. Calculate the total amount invested.
14 A shopkeeper sold his entire stock of shirts and ties in a sale for \$10 000. The shirts
were priced at 3 for \$100 and the ties \$20 each. If he had sold only half the shirts and
two-thirds of the ties he would have received \$6000. How many of each did he sell in
the sale?
15
A tent manufacturer produces two models, the Outback and the Bush Walker. From
earlier sales records it is known that 20 per cent more of the Outback model is sold than
the Bush Walker. A profit of \$200 is made on each Outback sold, but \$350 is made on
each Bush Walker. If during the next year a profit of \$177 000 is planned, how many of
each model must be sold?
16
Oz Jeans has factories in Mydney and Selbourne. At the Mydney factory, fixed costs
are \$28 000 per month and the cost of producing each pair of jeans is \$30. At the
Selbourne factory, fixed costs are \$35 200 per month and the cost of producing each
pair of jeans is \$24. During the next month Oz Jeans must manufacture 6000 pairs of
jeans. Calculate the production order for each factory, if the total manufacturing costs
for each factory are to be the same.
17
A tea wholesaler blends together three types of tea that normally sell for \$10, \$11 and
\$12 per kilogram so as to obtain 100 kilograms of tea worth \$11.20 per kilogram. If the
same amounts of the two higher priced teas are used, calculate how much of each type
must be used in the blend.
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Chapter 1: Reviewing linear equations
1E Solving linear inequalities
Skillsheet
An inequality is a mathematical statement that contains an inequality symbol rather than an
equals sign: for example, 2x + 1 < 4. When you solve the inequality 2x + 1 < 4, you answer
the question:
‘Which numbers x satisfy the property that 2x + 1 is less than 4?’
You will find that your answers can be described using a number line. This is a good way to
represent the solution, as there are infinitely many numbers that satisfy an inequality such as
2x + 1 < 4. For example:
1
+ 1 = 2 < 4, 2(−1) + 1 = −1 < 4
2(1) + 1 = 3 < 4, 2(0) + 1 = 1 < 4, 2
2
To solve linear inequalities, proceed exactly as for equations with the following exception:
When multiplying or dividing both sides by a negative number, the ‘direction’ of the
inequality symbol is reversed.
Example 13
Solve the inequality 2x + 1 < 4.
Solution
Explanation
2x + 1 < 4
2x < 3
3
x<
2
Subtract 1 from both sides.
Divide both sides by 2.
The solution can be represented on a real number line.
–2
–1
0
1
3
2
2
Note: In a number-line diagram, the ‘endpoint’ of an interval is indicated with a closed circle
if the point is included and with an open circle if it is not.
Example 14
Solve the inequality 3 − 2x ≤ 4.
Solution
Explanation
3 − 2x ≤ 4
−2x ≤ 1
x≥−
–2 –1 – 1 0
2
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
Subtract 3 from both sides.
1
2
Divide both sides by −2. Note that the inequality
symbol is reversed.
1
2
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1E Solving linear inequalities
21
Example 15
Solve the inequality
2x + 3 3 − 4x
>
+ 2.
5
3
Solution
Explanation
2x + 3 3 − 4x
>
+2
5
3
Multiply both sides by 15, the lowest
common denominator of 5 and 3.
3(2x + 3) > 5(3 − 4x) + 30
3(2x + 3) − 5(3 − 4x) > 30
Collect the terms containing x on the
left-hand side of the inequality.
6x + 9 − 15 + 20x > 30
26x − 6 > 30
36
x>
26
∴
x>
18
13
Using the TI-Nspire
The inequality can be solved in a Calculator
application.
Choose solve( ) from the Algebra menu to
give the solution to
2x + 3 3 − 4x
>
+2
5
3
Note: For the inequality signs template, press ctrl = .
Using the Casio ClassPad
To solve the inequality:
Go to the M screen and turn on the keyboard.
Select the fraction icon N found in Math1 .
Enter the inequality
2x + 3 3 − 4x
>
+2
5
3
Note: The inequality sign can be found in
Math3
.
Highlight the inequality using the stylus.
Select Interactive > Equation/Inequality > solve and
ensure that the variable selected is x.
Tap on OK to obtain the solution.
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22
1E
Chapter 1: Reviewing linear equations
Section summary
We can add or subtract the same number on both sides of an inequality, and the
resulting inequality is equivalent to the original.
We can multiply or divide both sides of an inequality by a positive number, and the
resulting inequality is equivalent to the original.
If we multiply or divide both sides of an inequality by a negative number, then we must
reverse the inequality sign so that the resulting inequality is equivalent.
Exercise 1E
Example 13, 14
1 Solve each of the following inequalities for x:
a x+3<4
d
x
≤4
3
g 6 − 2x > 10
Example 14, 15
2
b x−5>8
c 2x ≥ 6
e −x ≥ 6
f −2x < −6
h
−3x
≤6
4
i 4x − 4 ≤ 2
Solve for x in each of the following and show the solutions on a real number line:
1
c (x + 1) − x > 1
a 4x + 3 < 11
b 3x + 5 < x + 3
2
d
1
(x + 3) ≥ 1
6
e
2
(2x − 5) < 2
3
f
3x − 1 2x + 3
−
< −2
4
2
g
4x − 3 3x − 3
−
<3
2
3
h
1 − 7x
≥ 10
−2
i
5x − 2 2 − x
−
> −1
3
3
3
a For which real numbers x is 2x + 1 a positive number?
b For which real numbers x is 100 − 50x a positive number?
c For which real numbers x is 100 + 20x a positive number?
4
In a certain country it costs \$1 to send a letter weighing less than 20 g. A sheet of paper
weighs 3 g. Write a suitable inequality and hence state the maximum number of pages
that can be sent for \$1. (Ignore the weight of the envelope in this question.)
5 A student receives marks of 66 and 72 on two tests. What
is the lowest mark she can obtain on a third test to have an
average for the three tests greater than or equal to 75?
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1F Using and transposing formulas
23
1F Using and transposing formulas
Skillsheet
An equation containing symbols that states a relationship between two or more quantities
is called a formula. An example of a formula is A = w (area = length × width). The value
of A, called the subject of the formula, can be found by substituting in given values of and w.
Example 16
Find the area of a rectangle with length () 10 cm and width (w) 4 cm.
Solution
Explanation
A = w
A = 10 × 4
Substitute = 10 and w = 4.
A = 40 cm
2
Sometimes we wish to rewrite a formula to make a different symbol the subject of the
formula. This process is called transposing the formula. The techniques for transposing
formulas include those used for solving linear equations detailed in Section 1A.
Example 17
Transpose the formula v = u + at to make a the subject.
Solution
Explanation
v = u + at
v − u = at
Subtract u from both sides.
v−u
=a
t
Divide both sides by t.
If we wish to evaluate an unknown that is not the subject of the formula, we can either
substitute the given values for the other variables and then solve the resulting equation, or we
can first transpose the formula and then substitute the given values.
Example 18
Evaluate p if 2(p + q) − r = z, and q = 2, r = −3 and z = 11.
Solution
Explanation
Method 1: Substituting then solving
2(p + 2) − (−3) = 11
2p + 4 + 3 = 11
First substitute q = 2, r = −3 and z = 11.
Then solve for p.
2p = 4
p=2
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24
Chapter 1: Reviewing linear equations
Method 2: Transposing then substituting
2(p + q) − r = z
First solve for p.
2(p + q) = z + r
p+q=
∴
z+r
2
p=
z+r
−q
2
p=
11 + (−3)
−2
2
Substitute q = 2, r = −3 and z = 11.
p=2
Example 19
A path x metres wide surrounds a rectangular lawn. The lawn is metres long and
b metres wide. The total area of the path is A m2 .
a Find A in terms of , b and x.
b Find b in terms of , A and x.
Solution
a
b + 2x m
bm
m
+ 2x m
The area of the path is
A = (b + 2x)( + 2x) − b
= b + 2x + 2xb + 4x2 − b
∴
A = 2x + 2xb + 4x2
b A − (2x + 4x2 ) = 2xb
Therefore
b=
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A − (2x + 4x2 )
2x
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1F Using and transposing formulas
25
Example 20
For each of the following, make c the subject of the formula:
a e=
√
3c − 7a
Solution
e=
a
b
√
3c − 7a
1 1
1
− =
a b c−2
1 1
1
− =
a b c−2
b
Square both sides of the equation:
e2 = 3c − 7a
Therefore
3c = e2 + 7a
e2 + 7a
c=
3
Establish common denominator on the
left-hand side of the equation:
1
b−a
=
ab
c−2
Take the reciprocal of both sides:
ab
=c−2
b−a
ab
Therefore c =
+2
b−a
Using the TI-Nspire
Literal equations can be solved for a given
variable in a Calculator application.
Use solve( ) from the Algebra menu to make
c the subject of the formula
1
1 1
− =
a b c−2
Using the Casio ClassPad
To solve a literal equation for a given variable:
Go to the M screen and turn on the keyboard.
Select the fraction icon N found in Math1 .
Select the Var keyboard.
Enter and highlight the equation
1 1
1
− =
a b c−2
Select Interactive > Equation/Inequality > solve and
ensure that you change the selected variable to c.
Tap on OK to obtain the solution.
If necessary, copy the solution and use Interactive >
Transformation > expand to produce a neater answer.
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
Cambridge University Press
ISBN 978-1-107-52013-4
© Evans et al. 2016
Photocopying is restricted under law and this material must not be transferred to another party.
26
1F
Chapter 1: Reviewing linear equations
Section summary
A formula relates different quantities: for example, the formula A = πr2 relates the
radius r with the area A of the circle.
The variable on the left is called the subject of the formula: for example, in the formula
A = πr2 , the subject is A.
To calculate the value of a variable which is not the subject of a formula:
Method 1 Substitute the values for the known variables, then solve the resulting
equation for the unknown variable.
Method 2 Rearrange to make the required variable the subject, then substitute values.
Exercise 1F
1
For each of the following, find the value of the letter in parentheses:
a c = ab, a = 6, b = 3 (c)
c c = ab, a = 6, c = 18 (b)
√
e c = a, a = 9 (c)
u
g p = , u = 10, v = 2 (p)
v
b r = p + q, p = 12, q = −3 (r)
d r = p + q, p = 15, r = −3 (q)
√
f c = a, c = 9 (a)
u
h p = , p = 10, v = 2 (u)
v
2 For each of the following, construct a formula using the given symbols:
a
b
c
d
e
Example 16
S , the sum of three numbers a, b and c
P, the product of two numbers x and y
the cost, \$C, of five CDs which each cost \$p
the total cost, \$T , of d chairs which cost \$p each and c tables which cost \$q each
the time, T , in minutes, of a train journey that takes a hours and b minutes
3 Find the values of the following:
a E = IR, when I = 5 and R = 3
b C = pd, when p = 3.14 and d = 10
RT
E
c P=
, when R = 60, T = 150 and V = 9 d I = , when E = 240 and R = 20
V
R
e A = πr, when π = 3.14, r = 5 and = 20
f S = 90(2n − 4), when n = 6
Example 17
Example 18
4 For each of the following, make the symbol indicated the subject of the formula:
5
a PV = c; V
b F = ma; a
c I = Prt; P
d w = H + Cr; r
e S = P(1 + rt); t
f V=
2R
; r
R−r
Find the value of the unknown symbol in each of the following:
T +2
, when D = 10, P = 5
a D=
P
b A = 12 bh, when A = 40, h = 10
c V = 13 πr2 h, when π = 3.14, V = 100, r = 5
d A = 12 h(a + b), when A = 50, h = 5, a = 10
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
Cambridge University Press
ISBN 978-1-107-52013-4
© Evans et al. 2016
Photocopying is restricted under law and this material must not be transferred to another party.
1F
27
1F Using and transposing formulas
6
The diagram represents the brick wall of a
dwelling with three windows. Each of the
windows is h m high and w m wide. Other
dimensions are as shown.
a
b
c
d
7
Find the length of the wall.
Find the height of the wall.
Find the total area of the three windows.
Find the total area of brickwork.
bm
am
bm
am
bm
bm
am
bm
am
bm
A lampshade has a metal frame consisting of two circular hoops of
radii p cm and q cm joined by four straight struts of length h cm.
The total length of metal is T cm.
a
i Find an expression for T in terms of p, q and h.
ii Find T when p = 20, q = 24 and h = 28.
b The area of the material covering the frame is A cm2 , where
A = πh(p + q). Find an expression for p in terms of A, h, q and π.
8
Example 19
Find the value of the unknown symbol in each of the following:
T−M
a a
a P=
, P = 6, T = 8, M = 4
b H = + , H = 5 and a = 6
D
3 b
r r
90(2n − 4)
c a=
,a=6
d R = + , a = 2 and R = 4
n
a 3
9 Right-angled triangles XYZ and ABC are
similar.
YZ
ZX
XY
=
=
=k
AB BC CA
B
c cm
A
If AB = c cm and AC = b cm, find:
Y
X
Z
C
b cm
a the area, D cm2 , of the shaded region in terms of c, b and k
b k in terms of D, b and c
c the value of k if D = 2, b = 3 and c = 4.
10
Two rectangles each with dimensions
c cm × b cm are used to form a cross as shown.
The arms of the cross are all of equal length.
b
a Find the perimeter, P cm, of the cross in
c
terms of b and c.
2
b Find the area, A cm , of the cross in terms of b and c.
c Find b in terms of A and c.
Example 20
b cm
c
b cm
c cm
11 For each of the following, make the symbol in brackets the subject of the formula:
√
a+x b−y
a a = a + 2b (b)
b
=
(x)
a−x b+y
x
v2
2
c px = 3q − r (r)
d
= 1 − 2 (v)
y
u
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
Cambridge University Press
ISBN 978-1-107-52013-4
© Evans et al. 2016
Photocopying is restricted under law and this material must not be transferred to another party.
Review
28
Chapter 1: Reviewing linear equations
Chapter summary
A linear equation is one in which the variable is to the first power.
AS
It is often helpful to look at how the equation has been constructed so that the steps
Nrich
necessary to ‘undo’ the equation can be identified. It is most important that the steps taken
to solve the equation are done in the correct order.
An equation for the variable x in which all the coefficients of x, including the constants,
are pronumerals is known as a literal equation: for example, ax + b = c.
The two methods for solving simultaneous linear equations are substitution and
elimination.
An inequality is a mathematical statement that contains an inequality symbol rather than
an equals sign: for example, 2x + 1 < 4.
To solve linear inequalities, proceed exactly as for equations except that, when multiplying
or dividing both sides by a negative number, the ‘direction’ of the inequality symbol is
reversed.
An equation containing symbols that states a relationship between two or more quantities
is called a formula. An example of a formula is A = w (area = length × width). The
subject of this formula is A.
If we wish to evaluate an unknown that is not the subject of the formula, we can either
substitute the given values for the other variables and then solve the resulting equation, or
we can first transpose the formula and then substitute the given values.
Technology-free questions
1
Solve each of the following equations for x:
a 2x + 6 = 8
d
b 3 − 2x = 6
3−x
=6
5
g 3(2x + 1) = 5(1 − 2x)
c 2x + 5 = 3 − x
e
x
=4
3
h
3x + 2 3 − x
+
=5
5
2
f
2 Solve each of the following for t:
at + b
b
=d
a a−t =b
c
at + b
a−t
d
=c
e
=1
b−t
ct − b
3
13x
− 1 = 10
4
c a(t − c) = d
f
1
=d
at + c
Solve each of the following inequalities for x:
a 2 − 3x > 0
3 − 2x
≥ 60
5
3 − 2x x − 7
d
−
≤2
5
6
b
c 3(58x − 24) + 10 < 70
4 Make x the subject of the formula z = 12 x − 3t. Find x when z = 4 and t = −3.
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
Cambridge University Press
ISBN 978-1-107-52013-4
© Evans et al. 2016
Photocopying is restricted under law and this material must not be transferred to another party.
Chapter 1 review
A number d is equal to the square of a number e plus twice a number f .
a Find a formula for d in terms of e and f .
b Make f the subject of the formula.
c Find f when d = 10 and e = 3.
6
The surface area of a sphere of radius r is given by the formula A = 4πr2 . Calculate the
surface area of a sphere of radius 10 cm. Give your answer in terms of π.
7
The volume of metal in a tube is given by the formula V = π[r2 − (r − t)2 ], where is
the length of the tube, r is the radius of the outside surface and t is the thickness of the
material. Find V when:
a = 100, r = 5 and t = 0.2
8
9
b = 50, r = 10 and t = 0.5
For each of the following, make the variable in brackets the subject of the formula:
a A = πrs (r)
b T = P(1 + rw) (w)
n− p
c v=
(r)
d ac = b2 + bx (x)
r
u + v
t.
Let s =
2
a Find the value of s if u = 10, v = 20 and t = 5.
b Find the value of t if u = 10, v = 20 and s = 120.
10
The volume, V cm3 , of a cylinder is given by V = πr2 h, where r cm is the radius and
h cm is the height. Find the radius of the cylinder if the volume of the cylinder is
500π cm3 and the height is 10 cm.
11
A rope of length 205 m is cut into 10 pieces of one length and 5 pieces of another
length. The total length of three of the first 10 lengths exceeds that of two the second
length by 2 m. Find the lengths of the pieces.
12
1
m
If I add one to the numerator of a fraction it simplifies to . If I subtract one from
n
5
m
1
the denominator it simplifies to . Find the fraction .
7
n
13
Mr Adonis earns \$7200 more than Mr Apollo, and Ms Aphrodite earns \$4000 less
than Mr Apollo. If the total of the three incomes is \$303 200, find the income of each
person.
14
Solve each of the following pairs of simultaneous equations for a and b:
a 4a − b = 11
3a + 2b = 6
15
Review
5
29
b a = 2b + 11
4a − 3b = 11
A motorist travelled a total distance of 424 km, and had an average speed of 80 km/h on
highways and 24 km/h while passing through towns. If the journey took six hours, find
how long the motorist spent travelling on highways.
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
Cambridge University Press
ISBN 978-1-107-52013-4
© Evans et al. 2016
Photocopying is restricted under law and this material must not be transferred to another party.
Review
30
Chapter 1: Reviewing linear equations
Multiple-choice questions
1 The solution of the linear equation 3x − 7 = 11 is
11
−3
4
A
B
C
3
3
4
x 1
+ = 2 then x =
3 3
2
1
A
B
3
3
D 6
E −6
D 5
E 7
D x=2
E x = −2
2 If
C
7
3
3 The solution of the equation x − 8 = 3x − 16 is
11
−8
A x=
B x=
C x=4
3
3
4 The statement that 7 is 11 times the result of subtracting 2 from x can be written as
A 7 = 11(x − 2)
D
B 11x − 2 = 7
11
=7
x−2
E
C 7 = 11(x + 2)
7
−2= x
11
5 The solution of the simultaneous equations 2x − y = 10 and x + 2y = 0 is
A x = −2 and y = 3
D x = 6 and y = 2
B x = 2 and y = −3
E x = 1 and y = −8
C x = 4 and y = −2
6
I bought x CDs for \$a and y DVDs for \$b. The average price paid, in dollars, is
a+b
xa + yb
y+x
a b
x+y
A
B
C
D
E
+
a+b
y+x
y+x
xa + yb
x y
7
The solution of the equation
A x=
8
8
5
10
5
13
C x=1
D x=−
1
5
72 + 15z
> 4 are
3
C z = −4
D z<4
E x = −1
The values of z that satisfy the inequality
A z>4
9
B x=
x + 1 2x − 1
−
= x is
4
6
B z > −4
hw + k
then
w
k
A w=
A−h
3Ah
D w=
−k
2
E z < −4
If A =
ht + k
A
2
E w = h(A + k)
3
B w=
C w=
A − 2k
2h
Bronwyn walks one lap of an oval at 2.5 km/h and then jogs another eight laps of the
oval at 5 km/h. If it takes her 30 minutes in total, how long in metres is each lap?
A 200 m
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
B 250 m
C 300 m
D 350 m
E 400 m
Cambridge University Press
ISBN 978-1-107-52013-4
© Evans et al. 2016
Photocopying is restricted under law and this material must not be transferred to another party.
Chapter 1 review
Review
11
31
Which of the following equations has no solutions?
A 3x + 6 = 2x + 4
D 3x + 6 = 2x + 6
B 2(x + 3) = 2x + 6
E 2x + 4 = 2x + 6
C 3x − 4 = 5 − x
12 Which of the following equations is true for all values of x?
A 4x − 6 = 2x + 4
D 7x + 6 = −2x + 6
B 5(x + 3) = 5x + 15
E 2x + 4 = 2x + 6
C 3x − 4 = 10 − x
Extended-response questions
1
2
3
The formula for converting degrees Celsius, C, to degrees Fahrenheit, F, is
9
F = C + 32.
5
a Convert 30◦ F to Celsius.
b Convert 30◦ C to Fahrenheit.
c If x◦ C = x◦ F find x.
d If (x + 10)◦ C = x◦ F find x.
e If 2x◦ C = x◦ F find the value of x.
f If k◦ F = (−3k)◦ C find k.
1 1 2
For a spherical mirror of radius r cm, + = , where u cm is the distance from
v u r
the mirror to the object and v cm is the distance from the mirror to the image. The
v−r
.
magnification is given by m =
r−u
a Find r in terms of v and u from the first formula.
b Find m in terms of v and u only.
The diagram shows a section of wire
mesh w metres in width and metres
in length.
wm
a Find an expression in terms of w
and for the total length of wire
required for the mesh.
b
m
i If w = 3, find an expression in terms of w for the total length of wire required.
ii If the length of wire used is 100 m, find the value of w and the value of .
c The total length of wire, L m, required for another type of rectangular mesh of
dimensions x m by y m, is given by the formula L = 6x + 8y.
i Find y in terms of x and L.
ii Find y if L = 200 and x = 4.
d A third type of mesh can also be used to cover a rectangular region of dimensions
x m by y m. In this case, the type of mesh introduced in part c requires 100 m of
wire and so 6x + 8y = 100. This third type of mesh requires 80 m and this gives the
equation 3x + 2y = 40. Find the values of x and y.
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
Cambridge University Press
ISBN 978-1-107-52013-4
© Evans et al. 2016
Photocopying is restricted under law and this material must not be transferred to another party.
Review
32
Chapter 1: Reviewing linear equations
4
Tom leaves town A and travels towards town B at a constant speed of u km/h. At the
same time, Julie leaves town B and travels towards town A at a constant speed of
v km/h. Town B is d km from town A.
a How far has each travelled after t hours?
b By considering that the sum of their distances travelled must be d km when they
meet, find:
i the time it takes for them to meet
ii their distance from town A when they meet.
c If u = 30, v = 50 and d = 100, find the time it takes for them to meet and their
distance from town A.
5
Xiu travels from town A to town B at u km/h and then returns at v km/h. Town A is
d km from town B.
a Find the average speed at which Xiu travels for the complete journey, in terms of u
and v. Remember that
total distance travelled
average speed =
total time taken
b If it takes T hours to travel from A to B, find the time taken:
i for the return trip from B to A, in terms of T , u and v
ii for the entire trip, in terms of T , u and v.
6 A man on a bicycle rides one-third of the way from town A to town B at a speed a km/h
and the remainder of the way at 2b km/h.
a If the distance between the two towns is 9 km, find the time taken to ride from
A to B.
If the man had travelled at a uniform rate of 3c km/h, he could have ridden from A to B
and back again in the same time.
2 1 1
b Show that = + .
c a b
c i Make c the subject of this formula.
ii Find c, when a = 10 and b = 20.
7
A man walks 70 km. He walks x km at 8 km/h and y km at 10 km/h.
a Find the length of time he was walking at 8 km/h in terms of x, and the length of
time he was walking at 10 km/h in terms of y.
b Find his average speed in terms of x and y.
c If the man walks at 10 km/h for the time he was walking at 8 km/h and at 8 km/h for
the time he was walking at 10 km/h, he walks 72 km. Find x and y.
8 Prove that the lines with equations 2y − x = 2, y + x = 7 and y − 2x = −5 meet at the
one point.
Cambridge Senior Maths AC/VCE
Mathematical Methods 1&2
Cambridge University Press
ISBN 978-1-107-52013-4
© Evans et al. 2016
Photocopying is restricted under law and this material must not be transferred to another party.
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Transcript
```Quadratic Equations
P.7
• A quadratic equation in x is an equation
that can be written in the standard form
• ax2 + bx + c = 0
• where a, b, and c are real numbers with a
not equal to 0. A quadratic equation in x is
also called a second-degree polynomial
equation in x.
The Zero-Product Principle
If the product of two algebraic expressions is
zero, then at least one of the factors is equal
to zero.
If AB = 0, then A = 0 or B = 0.
Factoring
1. If necessary, rewrite the equation in the form
ax2 + bx + c = 0, moving all terms to one side,
thereby obtaining zero on the other side.
2. Factor.
3. Apply the zero-product principle, setting each
factor equal to zero.
4. Solve the equations in step 3.
5. Check the solutions in the original equation.
Text Example
• Solve 2x2 + 7x = 4 by factoring and then using the
zero-product principle.
Step 1 Move all terms to one side and obtain
zero on the other side. Subtract 4 from both sides
and write the equation in standard form.
2x2 + 7x - 4 = 4 - 4
2x2 + 7x - 4 = 0
Step 2 Factor.
2x2 + 7x - 4 = 0
(2x - 1)(x + 4) = 0
Solution cont.
• Solve 2x2 + 7x = 4 by factoring and then
using the zero-product principle.
Steps 3 and 4 Set each factor equal to
zero and solve each resulting equation.
2x-1=0
or x + 4 = 0
2x=1
x = -4
x = 1/2
Example
(2x + -3)(2x + 1) = 5
4x2 - 4x - 3 = 5
4x2 - 4x - 8 = 0
4(x2-x-2)=0
4(x - 2)*(x + 1) = 0
x - 2 = 0, and x + 1 = 0
So x = 2, or -1
The Square Root Method
If u is an algebraic expression and d is a
positive real number, then u2 = d has exactly
two solutions.
If u2 = d, then u = d or u = -d
Equivalently,
If u2 = d then u = d
Completing the Square
If x2 + bx is a binomial then by adding (b/2) 2,
which is the square of half the coefficient of
x, a perfect square trinomial will result.
That is,
x2 + bx + (b/2)2 = (x + b/2)2
Text Example
What term should be added to the binomial x2
+ 8x so that it becomes a perfect square
trinomial? Then write and factor the
trinomial.
The term that should be added is the square of
half the coefficient of x. The coefficient of x
is 8. Thus, (8/2)2 = 42. A perfect square
trinomial is the result.
x2 + 8x + 42 = x2 + 8x + 16 = (x + 4)2
x2 + 6x = 7
B divided
by 2,
squared
x2 + 6x + 9 = 7 + 9
x + 3
2
= 16
x + 3 = 4
x = 4 - 3
x = 1, or - 7
Don’t forget
sides
- b b - 4ac
x=
2a
2
x 2 - 8x + 5 = 0
x 2 - 8x + 5 = 0
- (-8) ( -8) 2 - 4(1)(5)
x=
2(1)
8 64 - 20
x=
2
8 44
x=
2
8 2 11
x=
2
2( 4 11)
x=
2
x = 4 11
The Discriminant and the Kinds of Solutions
to ax2 + bx +c = 0
Discriminant
b2 – 4ac
Kinds of solutions
to ax2 + bx + c = 0
b2 – 4ac > 0
Two unequal real solutions
Graph of
y = ax2 + bx + c
Two x-intercepts
b2 – 4ac = 0
One real solution
(a repeated solution)
One x-intercept
b2 – 4ac < 0
No real solution;
two complex imaginary
solutions
No x-intercepts
```
Related documents |
Wiki
Bài 28, 29, 30 trang 22 SGK Toán 9 tập 2 – Giải bài toán bằng cách
H2 Heading: Solving Problems 28, 29, and 30 on Page 22 of the 9th-Grade Mathematics textbook
Problem 28: Finding Two Natural Numbers with Specific Properties
Question:
Find two natural numbers such that their sum is 1006, and if the larger number is divided by the smaller number, the quotient is 2 and the remainder is 124.
Solution:
Let’s assume the larger number as (x) and the smaller number as (y). (Condition: x > y; x, y ∈ ℕ*)
According to the given condition that the sum of the two numbers is 1006, we have: (x + y = 1006).
Since the larger number divided by the smaller number gives a quotient of 2 and a remainder of 124, we can write: (x = 2y + 124) (with y > 124).
We have the following system of equations:
``````⇔ {x + y = 1006
x - 2y = 124}``````
Simplifying further:
``````⇔ {x + y = 1006
3y = 882}``````
Solving for the variables:
``````⇔ {x = 1006 - y
y = 294}``````
``````⇔ {x = 1006 - 294
y = 294}``````
``````⇔ {x = 712
y = 294} (satisfies the condition)``````
Therefore, the two natural numbers we are looking for are 712 and 294.
Problem 29: Solving a Classic Problem
Question:
Solve the following classic problem:
Seventeen fresh fruits, consisting of oranges and tangerines, are divided amongst a hundred people. If each orange is divided into three equal parts and each tangerine is divided into ten equal parts, how many oranges and tangerines are there in total?
Solution:
Let’s assume the number of tangerines as (x) and the number of oranges as (y). The condition is that x and y should be positive integers.
Since there are seventeen fruits in total, we can write the equation: (x + y = 17) (1)
Since each orange is divided into three equal parts and each tangerine is divided into ten equal parts, we can write the equations:
For oranges: (3y parts)
For tangerines: (10x parts)
Also, given that there are a hundred parts in total, we can write the equation: (10x + 3y = 100) (2)
From equations (1) and (2), we have the following system of equations:
``````⇔ {x + y = 17
10x + 3y = 100}``````
Simplifying further:
``````⇔ {3x + 3y = 51
10x + 3y = 100}``````
``````⇔ {3x + 3y - (10x + 3y) = 51 - 100
10x + 3y = 100}``````
``````⇔ {-7x = -49
10x + 3y = 100}``````
Solving for x and y:
``````⇔ {x = 7
3y = 100 - 10x}``````
``````⇔ {x = 7
y = 10} (satisfies the condition)``````
Therefore, there are 10 tangerines and 7 oranges in total.
Problem 30: Calculating Distance and Departure Time of a Car Journey
Question:
A car travels from point A to point B, intending to arrive at B at 12:00 PM. If the car travels at a speed of 35 km/h, it arrives 2 hours late. If it travels at a speed of 50 km/h, it arrives 1 hour early. Find the distance of the journey (AB) and the departure time of the car from A.
Solution:
Let’s assume the distance of the journey as (x) km and the estimated travel time from A to B as (y) hours. The condition is that x > 0 and y > 1 (as the car arrives 1 hour early).
+) Case 1:
When the car travels at a speed of 35 km/h:
Since the car arrives 2 hours late, the total travel time is (y + 2) hours.
The distance traveled can be calculated using the formula (S = v * t), where S is distance (in km), v is speed (in km/h), and t is time (in hours). So, we have: (35(y + 2)) km.
Since the distance remains constant, we can set up the equation: (x = 35(y + 2)) (1).
+) Case 2:
When the car travels at a speed of 50 km/h:
Since the car arrives 1 hour early, the total travel time is (y – 1) hours.
The distance traveled can be calculated using the formula (S = v * t), where S is distance (in km), v is speed (in km/h), and t is time (in hours). So, we have: (50(y – 1)) km.
Since the distance remains constant, we can set up the equation: (x = 50(y – 1)) (2).
From equations (1) and (2), we have the following system of equations:
``````⇔ {x = 35(y + 2)
x = 50(y - 1)}``````
Simplifying further:
``````⇔ {x = 35y + 70
x = 50y - 50}``````
``````⇔ {x - 35y = 70 (1)
x - 50y = -50 (2)}``````
Subtracting equation (2) from equation (1), we get:
``````⇔ {15y = 120
x - 50y = -50}``````
Solving for y:
``````⇔ {y = 8
x = -50 + 50y}``````
``````⇔ {y = 8
x = -50 + 50 * 8}``````
``````⇔ {y = 8
x = 350} (satisfies the condition)``````
Therefore, the distance of the journey (AB) is 350 km, and the car departs from A at 4:00 AM.
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## Section10.2Sieve of Eratosthenes
The Sieve of Eratosthenes is a method for finding all primes up to (and possibly including) a given natural $$n\text{.}$$ This method works well when $$n$$ is relatively small, allowing us to determine whether any natural number less than or equal to $$n$$ is prime or composite.
We now explain how the Sieve of Eratosthenes can be used to find all prime numbers up to a given natural number. Recall that $$a$$ is a multiple of $$b$$ means that $$b$$ divides $$a\text{,}$$ see Definition 4.1.1.
### Strategy10.2.1.Sieve of Eratosthenes.
To find all prime numbers up to a given integer $$n$$ we proceed as follows.
#### (a)
List all integers from $$2$$ to $$n$$
#### (b)
The first integer on the list is $$2\text{,}$$ and it is prime. Mark out all multiples of $$2$$ that are bigger than $$2$$ because they are composite.
We do not have to compute anything, as we can simply mark out every second number starting at $$2\text{.}$$
#### (c)
The next integer on the list that is not marked out is 3, and it is prime. Mark out all multiples of 3 that are bigger than 3 because they are composite. (Note that some of these, such as 6, will already be marked out).
We do not have to compute anything, as we can simply mark out every third number starting at $$3\text{.}$$
#### (d)
The next integer on the list that is not marked out is 5, and it is prime. Mark out all multiples of 5 that are bigger than 5 because they are composite.
We do not have to compute anything, as we can simply mark out every fifth number starting at $$5\text{.}$$
#### (e)
Continue in this way until there is no next integer on the list that is not marked out.
The integers that are not marked out are all of the primes up to (and possibly including) $$n\text{.}$$
The advantage of sieving is that any computing effort is replaced by counting.
In the video in Figure 10.2.1 we apply the sieve of Eratosthenes to the natural numbers up to $$100\text{.}$$
In Figure 10.2.2 we show the initial steps as well as the end result of the Sieve of Eratosthenes up to $$100\text{.}$$ Namely:
1. All integers from $$2$$ to $$100$$ in Figure 10.2.(a)
2. All multiples of $$2$$ marked out in Figure 10.2.(b)
3. All multiples of $$2$$ and $$3$$ marked out in Figure 10.2.(c)
The unmarked numbers are all primes numbers up to $$100\text{,}$$ Counting them we find that there are 25 primes up to $$100\text{.}$$
Explore the sieving process in Figure 10.2.3. Click on a number to have all its multiples marked by changing the field color to red and crossing them out. Numbers that you have clicked appear on green background. When there are no white fields left, the numbers in green fields are all the prime numbers up to 192.
It is most efficient to follow the strategy presented in Strategy 10.2.1. In particular that means first clicking on $$2\text{,}$$ then clicking on the first number that remained white, which is $$3$$ and so on. To illustrate the sieving better multiples are marked with a delay and different shades of red are used for multiples of different numbers.
The Sieve of Eratosthenes can be used to produce tables of primes such as in Table 10.2.4 without the need of factoring by simply counting.
In Checkpoint 10.2.5 count the number of prime numbers between two given numbers.
The number of primes from 0 to 10 is
The number of primes from 10 to 20 is
The number of primes from 20 to 30 is
The number of primes from 30 to 40 is
The number of primes from 40 to 50 is
The number of primes from 50 to 60 is
The number of primes from 60 to 70 is
The number of primes from 70 to 80 is
The number of primes from 80 to 90 is
The number of primes from 90 to 100 is
$$4$$
$$4$$
$$2$$
$$2$$
$$3$$
$$2$$
$$2$$
$$3$$
$$2$$
$$1$$
Checkpoint 10.2.5 illustrates that the number of prime numbers in sets of consecutive numbers of the same cardinality decreases as the numbers increase. |
Algorithm and Flowchart to represent a number as sum of two prime numbers
[3891 views]
What are prime numbers?
A number is said to be a prime number when it has only two factors, that is, when the factors of the number are 1 and the number itself. Example: 2, 5, 17, 19, etc.
Sum of Prime Numbers:
There are many theories which state that any number greater than 2 can be expressed as the sum of two prime numbers. We must also keep in mind that the sum of two prime numbers does not necessarily have to be a prime number. In this article, we will check whether a given number can be displayed as the sum of two prime numbers.
For example:
15 = 2 + 13
25 = 2 + 23
Now let’s take a look at the algorithm and flowchart to check whether two given numbers are twin prime or not, for better understanding.
Algorithm to represent a given number as the sum of two prime numbers:
To avoid redundancy, we will use the concept of functions. Here, we will use a function to check whether a number is prime or not. While writing the program, we can call this function any number of times we want.
CheckPrime(num):
Assumption: For this algorithm, we assume that the function takes a number ‘num’, to be checked and returns true if the number is prime, otherwise, returns false.
Step 1: Start Step 2: Initialize number of factors of num, f = 0 Step 3. Initialize i = 1 Step 4. Repeat until i<=num: 4.1: If num % i == 0: 4.2: Increment f by 1 4.3: Increment i by 1 Step 5: If f = 2, then: 5.1: Return True Step 6: Else: 6.1 Return False Step 7. Stop
Algorithm to display a number as sum of two prime numbers:
Step 1: Start Step 2: Read the number from the user, say n Step 3: Initialize loop variable i = 2 Step 4: Initialize flag = false Step 5: Repeat While i <= n/2: 5.1: If CheckPrime(i) == true, then: 5.2: If CheckPrime(n - i) == true, then: 5.2.a: Display i, n-i 5.2.b: flag = true 5.3: Increment i by 1 Step 6: If flag = false, then: 6.1: Display “Number cannot be displayed as sum of two prime numbers” Step 7: Stop
Explanation:
In this problem, we need to check whether a number can be displayed as the sum of two prime numbers. To check this, we need to check whether a number is prime or not, multiple times. Hence, we have written a separate function for doing the prime number check. This function will return true if the number is prime, else returns false.
To do the prime number checking, we must find out the number of factors of the number. We first initialize the number of factors, ‘f’ as 0. Now, we start a loop from 1 to num, where the loop variable is ‘i’. If num% i=0, f is incremented by 1. In this line, we are checking whether the number is divisible by ‘i’ or not. If yes, f is incremented. If the number of factors is equal to 2, the function returns 0, else it returns false.
Now, coming to the main algorithm, we start off by taking the number to be checked as user input and store it in a variable, say ‘n’. We initialize a flag as false and the loop variable I as 2. Now, we check whether i is a prime number. If yes, we check whether ‘n-i’ is a prime number. If both the numbers are prime, we display the numbers and the flag is set as true. Therefore, the number ‘n’ can be displayed as the sum of i and n-i. In the same manner, we check for other pairs. Once the loop is terminated, if the flag is still false, this means the number cannot be displayed as the sum of two prime numbers. Hence, an appropriate message is displayed and the algorithm is terminated.
Note: Here ‘%’ is the modulus operator which returns the remainder value after division.
Flowchart to represent a number as sum of two prime numbers:
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Important Questions For Class 12 Maths Chapter 1 - Relations and Functions - GMS - Learning Simply
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# Important Questions For Class 12 Maths Chapter 1 - Relations and Functions
Important Questions for Class 12 Maths Chapter 1 – Relations and functions are provided here for the students to score good marks in the class 12 board Maths examination. Here, all the important problems are covered as per the NCERT book. By practising and going through the important questions, students can get the confidence to solve the problems easily and efficiently.
Here, we are going to discuss the important questions in class 12 chapter 1 – Relations and functions with solutions. Before that, let us learn the basic concepts covered in class 12 maths chapter 1. In relations and functions chapter 1, it covers the concept of relations, functions and binary operations.
What is the Relation?
Let us consider two sets A and B. Then a relation R from Set A into Set B is defined as the subset of A × B. The relation can be classified into three different types, such as:
• Reflexive Relation
• Symmetric Relation
• Transitive Relation
What is a Function?
A function is defined as the relation between two sets, in which every element in the set 1 associate to the elements in set 2. In other words, if “f” is a function from A to B, then every element in the set B is the image of an element in set A. The different types of functions are:
• One to One function
• One to Many functions
• Onto function
• One to one Correspondence
Also, check:
## Class 12 Chapter 1 – Relations and Functions Important Questions with Solutions
Go through the solved problems given below. Here, all the important questions for class 12 chapter 1 maths problems are given. The problems given here are frequently asked in the previous board examinations. Practice these problems well and solve the practice problems provided here.
Q.1: Show that the Signum Function f: R → R, given by
Solution:
Check for one to one function:
For example:
f(0) = 0
f(-1) = -1
f(1) = 1
f(2) = 1
f(3) = 1
Since, for the different elements say f(1), f(2) and f(3), it shows the same image, then the function is not one to one function.
Check for Onto Function:
For the function,f: R →R
In this case, the value of f(x) is defined only if x is 1, 0, -1
For any other real numbers(for example y = 2, y = 100) there is no corresponding element x.
Thus, the function “f” is not onto function.
Hence, the given function “f” is neither one-one nor onto.
Q.2: If f: R → R is defined by f(x) = x2 − 3x + 2, find f(f(x)).
Solution:
Given function:
f(x) = x2 − 3x + 2.
To find f(f(x))
f(f(x)) = f(x)2 − 3f(x) + 2.
= (x2 – 3x + 2)2 – 3(x2 – 3x + 2) + 2
By using the formula (a-b+c)2 = a2+ b2+ c2-2ab +2ac-2ab, we get
= (x2)2 + (3x)2 + 22– 2x2 (3x) + 2x2(2) – 2x2(3x) – 3(x2 – 3x + 2) + 2
Now, substitute the values
= x4 + 9x2 + 4 – 6x3 – 12x + 4x2 – 3x2 + 9x – 6 + 2
= x4 – 6x3 + 9x2 + 4x2 – 3x2 – 12x + 9x – 6 + 2 + 4
Simplify the expression, we get,
f(f(x)) = x4 – 6x3 + 10x2 – 3x
Q.3: Let A = N × N and * be the binary operation on A defined by (a, b) * (c, d) = (a + c, b + d). Show that * is commutative and associative. Find the identity element for * on A, if any.
Solution:
Check the binary operation * is commutative :
We know that, * is commutative if (a, b) * (c, d) = (c, d) * (a, b) ∀ a, b, c, d ∈ R
L.H.S =(a, b) * (c, d)
=(a + c, b + d)
R. H. S = (c, d) * (a, b)
=(a + c, b + d)
Hence, L.H.S = R. H. S
Since (a, b) * (c, d) = (c, d) * (a, b) ∀ a, b, c, d ∈ R
* is commutative (a, b) * (c, d) = (a + c, b + d)
Check the binary operation * is associative :
We know that * is associative if (a, b) * ( (c, d) * (x, y) ) = ((a, b) * (c, d)) * (x, y) ∀ a, b, c, d, x, y ∈ R
L.H.S = (a, b) * ( (c, d) * (x, y) ) = (a+c+x, b+d+y)
R.H.S = ((a, b) * (c, d)) * (x, y) = (a+c+x, b+d+y)
Thus, L.H.S = R.H.S
Since (a, b) * ( (c, d) * (x, y) ) = ((a, b) * (c, d)) * (x, y) ∀ a, b, c, d, x, y ∈ R
Thus, the binary operation * is associative
Checking for Identity Element:
e is identity of * if (a, b) * e = e * (a, b) = (a, b)
where e = (x, y)
Thus, (a, b) * (x, y) = (x, y) * (a, b) = (a, b) (a + x, b + y)
= (x + a , b + y) = (a, b)
Now, (a + x, b + y) = (a, b)
Now comparing these, we get:
a+x = a
x = a -a = 0
Next compare: b +y = b
y = b-b = 0
Since A = N x N, where x and y are the natural numbers. But in this case, x and y is not a natural number. Thus, the identity element does not exist.
Therefore, the operation * does not have any identity element.
Q.4: Let f : N → Y be a function defined as f (x) = 4x + 3, where, Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse.
Solution:
Checking for Inverse:
f(x) = 4x + 3
Let f(x) = y
y = 4x + 3
y – 3 = 4x
4x = y – 3
x = (𝑦 − 3)/4
Let g(y) = (𝑦 − 3)/4
where g: Y → N
Now find gof:
gof= g(f(x))
= g(4x + 3) = [(4𝑥 + 3) − 3]/4
= [4𝑥 + 3 − 3]/4
=4x/4
= x = IN
Now find fog:
fog= f(g(y))
= f [(𝑦 − 3)/4]
=4[(𝑦 − 3)/4] +3
= y – 3 + 3
= y + 0
= y = Iy
Thus, gof = INand fog = Iy,
Hence, f is invertible
Also, the Inverse of f = g(y) = [𝒚 – 3]/ 4
Q. 5: Let A = R {3} and B = R – {1}. Consider the function f: A →B defined by f (x) = (x- 2)/(x -3). Is f one-one and onto? Justify your answer.
Solution:
Given function:
f (x) = (x- 2)/(x -3)
Checking for one-one function:
f (x1) = (x1– 2)/ (x1– 3)
f (x2) = (x2-2)/ (x2-3)
Putting f (x1) = f (x2)
(x1-2)/(x1-3)= (x2-2 )/(x2 -3)
(x1-2) (x2– 3) = (x1– 3) (x2-2)
x1 (x2– 3)- 2 (x2 -3) = x1 (x2– 2) – 3 (x2– 2)
x1 x2 -3x1 -2x2 + 6 = x1 x2 – 2x1 -3x2 + 6
-3x1– 2x2 =- 2x1 -3x2
3x2 -2x2 = – 2x1 + 3x1
x1= x2
Hence, if f (x1) = f (x2), then x1 = x2
Thus, the function f is one-one function.
Checking for onto function:
f (x) = (x-2)/(x-3)
Let f(x) = y such that y B i.e. y ∈ R – {1}
So, y = (x -2)/(x- 3)
y(x -3) = x- 2
xy -3y = x -2
xy – x = 3y-2
x (y -1) = 3y- 2
x = (3y -2) /(y-1)
For y = 1, x is not defined But it is given that. y ∈ R – {1}
Hence, x = (3y- 2)/(y- 1) ∈ R -{3} Hence, f is onto.
### Practice Problems for Class 12 Maths Chapter 1
1. Show that the function f: R → R is given by f(x) = x3 is injective.
2. Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
3. Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
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# Lesson 2: Statistical Questions
Let's look more closely at data and the questions they can help to answer.
## 2.1: Pencils on A Plot
1. Measure your pencil to the nearest $\frac14$-inch. Then, plot your measurement on the class dot plot.
2. What is the difference between the longest and shortest pencil lengths in the class?
3. What is the most common pencil length?
4. Find the difference in lengths between the most common length and the shortest pencil.
## 2.2: What’s in the Data?
Ten sixth-grade students at a school were each asked five survey questions. Their answers to each question are shown here.
data set A data set B data set C data set D data set E 0 1 1 3 0 0 0 2 1 1 12 12 12 12 12 12 12 12 12 12 6 5 7 6 4 5 3 4 6 8 6 6 6 6 6 6 6 6 6 6 3 7 9 11 6 4 2 16 6 10
1. Here are the five survey questions. Match each question to a data set that could represent the students’ answers. Explain your reasoning.
• Question 1: Flip a coin 10 times. How many heads did you get?
Data set ______ Reason:
• Question 2: How many books did you read in the last year?
Data set ______ Reason:
• Question 3: What grade are you in?
Data set ______ Reason:
• Question 4: How many dogs and cats do you have?
Data set ______ Reason:
• Question 5: How many inches are in 1 foot?
Data set ______ Reason:
2. Discuss with a partner: How are Question 3 and Question 5 different from the other questions?
## 2.3: What Makes a Statistical Question?
These three questions are examples of statistical questions:
• What is the most common color of the cars in the school parking lot?
• What percentage of students in the school have a cell phone?
• Which kind of literature—fiction or nonfiction—is more popular among students in the school?
These three questions are not examples of statistical questions:
• What color is the principal’s car?
• Does Elena have a cell phone?
• What kind of literature—fiction or nonfiction—does Diego prefer?
1. Study the examples and non-examples. Discuss with your partner:
• How are the three statistical questions alike? What do they have in common?
• How are the three non-statistical questions alike? What do they have in common?
• What makes a question a statistical question?
Pause here for a class discussion.
2. Read each question. Think about the data you might collect to answer it and whether you expect to see variability in the data. Complete each blank with “Yes” or “No.”
1. How many cups of water do my classmates drink each day?
• Is variability expected in the data? _______
Is the question statistical? _______
1. Where in town does our math teacher live?
• Is variability expected in the data? _______
Is the question statistical? _______
1. How many minutes does it take students in my class to get ready for school in the morning?
• Is variability expected in the data? _______
Is the question statistical? _______
1. How many minutes of recess do sixth-grade students have each day?
• Is variability expected in the data? _______
Is the question statistical? _______
1. Do all students in my class know what month it is?
• Is variability expected in the data? _______
Is the question statistical? _______
## 2.4: Sifting for Statistical Questions
1. Your teacher will give you and your partner a set of cards with questions. Sort them into three piles: Statistical Questions, Not Statistical Questions, and Unsure.
2. Compare your sorting decisions with another group of students. Start by discussing the two piles that your group sorted into the Statistical Questions and Not Statistical Questions piles. Then, review the cards in the Unsure pile. Discuss the questions until both groups reach an agreement and have no cards left in the Unsure pile. If you get stuck, think about whether the question could be answered by collecting data and if there would be variability in that data.
3. Record the letter names of the questions in each pile.
• Statistical questions:
• Non-statistical questions:
## Summary
We often collect data to answer questions about something. The data we collect may show variability, which means the data values are not all the same.
Some data sets have more variability than others. Here are two sets of figures.
Set A has more figures with the same shape, color, or size. Set B shows more figures with different shapes, colors, or sizes, so set B has greater variability than set A.
Both numerical and categorical data can show variability. Numerical sets can contain different numbers, and categorical sets can contain different categories or types.
When a question can only be answered by using data and we expect that data to have variability, we call it a statistical question. Here are some examples.
• Who is the most popular musical artist at your school?
• When do students in your class typically eat dinner?
• Which classroom in your school has the most books?
To answer the question about books, we may need to count all of the books in each classroom of a school. The data we collect would likely show variability because we would expect each classroom to have a different number of books.
In contrast, the question “How many books are in your classroom?” is not a statistical question. If we collect data to answer the question (for example, by asking everyone in the class to count books), the data can be expected to show the same value. Likewise, if we ask all of the students at a school where they go to school, that question is not a statistical question because the responses will all be the same.
## Glossary
variability
#### variability
Variability is the tendency of a data set to have different data values.
statistical question
#### statistical question
A statistical question is a question that can only be answered by using data and where we expect the data to have variability. |
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# How to Solve Exponential Equations
Exponential equations may look intimidating, but solving them requires only basic algebra skills. Equations with exponents that have the same base can be solved quickly. In other instances, it is necessary to use logs to solve. Even this method, however, is simple with the aid of a scientific calculator.
### Method 1 Equating Two Exponents with the Same Base
1. 1
Determine whether the two exponents have the same base. The base is the big number in an exponential expression.[1] You can only use this method when you are presented with an equation that has an exponent on either side, and each exponent has the same base.
• For example, ${\displaystyle 6^{5+y}=6^{3}}$ has an exponent on either side of the equation, and each exponent has the same base (6).
2. 2
Ignore the base. Since the exponents are equal and have the same base, their exponents must be equal. As such, you can ignore the base and write an equation for the exponents only.[2]
• For example, in the equation ${\displaystyle 6^{5+y}=6^{3}}$, since both exponents have the same base, you would write an equation for the exponents: ${\displaystyle 5+y=3}$.
3. 3
Solve the equation. To do this, you need to isolate the variable. Remember that whatever you do to one side of an equation, you must do to the other side of the equation.
• For example:
${\displaystyle 5+y=3}$
${\displaystyle 5+y-5=3-5}$
${\displaystyle y=-2}$
4. 4
Check your work. To make sure that your answer is correct, plug the value you found for the variable back into the original equation, and simplify the expression. The two sides should be equal.
• For example, if you found that ${\displaystyle y=-2}$, you would substitute ${\displaystyle -2}$ for ${\displaystyle y}$ in the original equation:
${\displaystyle 6^{5+y}=6^{3}}$
${\displaystyle 6^{5-2}=6^{3}}$
${\displaystyle 6^{3}=6^{3}}$
### Method 2 Equating an Exponent and a Whole Number
1. 1
Isolate the exponential expression. Make sure that there is an exponential expression on one side of the equation, and a whole number on the other side. If not, you need to rework the equation so that the exponent is alone on one side.[3]
• For example, if your are trying to solve ${\displaystyle 3^{x-5}-2=79}$, you first need to isolate ${\displaystyle 3^{x-5}}$ by adding 2 to each side of the equation:
${\displaystyle 3^{x-5}-2=79}$
${\displaystyle 3^{x-5}-2+2=79+2}$
${\displaystyle 3^{x-5}=81}$
2. 2
Rewrite the equation. You need to determine whether the whole number can be converted to an exponent with the same base as the other exponent.[4] If you can’t convert the whole number in this way, you cannot use this method.
• For example, look at the equation ${\displaystyle 3^{x-5}=81}$. You need to change 81 to an exponent with a base of 3, so that it matches the other exponential expression in the equation. By factoring out 3, you should see that ${\displaystyle 3\times 3\times 3\times 3=81}$, so ${\displaystyle 3^{4}=81}$. The new equation then becomes ${\displaystyle 3^{x-5}=3^{4}}$.
3. 3
Write the equation for the exponents only. Since you converted the whole number, you now have two exponential expressions with the same base. Since the bases are the same, you can ignore them and focus on the exponents.
• For example, since ${\displaystyle 3^{x-5}=3^{4}}$ has two exponents with a base of 3, you can ignore the base and simply look at the equation ${\displaystyle x-5=4}$.
4. 4
Solve for the variable. To do this, you need to isolate the variable on one side of the equation. Make sure that whatever you do to one side, you also do to the other side.
• For example:
${\displaystyle x-5=4}$
${\displaystyle x-5+5=4+5}$
${\displaystyle x=9}$
5. 5
Check your work. You can see if your answer is correct by plugging the solution you found back into the original equation. After simplifying each expression, both sides of the equation should be equal. If they aren’t, you miscalculated and need to try again.
• For example, if you found that ${\displaystyle x=9}$, you would plug in ${\displaystyle 9}$ for ${\displaystyle x}$ in the original equation and simplify:
${\displaystyle 3^{x-5}=81}$
${\displaystyle 3^{9-5}=81}$
${\displaystyle 3^{4}=81}$
${\displaystyle 81=81}$
### Method 3 Using Logs for Terms without the Same Base
1. 1
Make sure that the exponential expression is isolated. One side of the equation should be the exponent, the other should be the whole number. If not, modify the equation so the exponent is alone on one side.
• For example, you need to isolate the expression ${\displaystyle 4^{3+x}}$ in the equation ${\displaystyle 4^{3+x}-8=17}$ by adding 8 to both sides:
${\displaystyle 4^{3+x}-8=17}$
${\displaystyle 4^{3+x}-8+8=17+8}$
${\displaystyle 4^{3+x}=25}$
2. 2
Rewrite the equation. Set up the equation so that you are taking the log of both sides. A log is the inverse of an exponent.[5]. You can find a base-10 log using most scientific calculators. For now, you are just rewriting the equation, indicating you are taking the log of each side.
• For example, if you take the base-10 log of both sides of ${\displaystyle 4^{3+x}=25}$, you would rewrite the equation like this: ${\displaystyle {\text{log}}4^{3+x}={\text{log}}25}$.
3. 3
Rewrite the log of the exponent. Rewrite it using the rule ${\displaystyle {\text{log}}a^{r}=r{\text{log}}a}$. Rewriting the exponential expression this way will allow you to simplify and solve the equation. Do not calculate the logs yet.[6]
• For example, ${\displaystyle {\text{log}}4^{3+x}={\text{log}}25}$ can be rewritten as ${\displaystyle (3+x){\text{log}}4={\text{log}}25}$
4. 4
Isolate the variable. To solve, you need to rewrite the equation so that one side contains the variable, and the other side contains all of the numbers. You will need to divide each side of the equation by the log of the exponential expression. You will also need to add or subtract any constants to both sides, and perform any other necessary operations.
• For example, to isolate the ${\displaystyle x}$ in ${\displaystyle (3+x){\text{log}}4={\text{log}}25}$, you first need to divide each side of the equation by ${\displaystyle {\text{log}}4}$, then subtract 3 from both sides:
${\displaystyle (3+x){\text{log}}4={\text{log}}25}$
${\displaystyle (3+x){\frac {{\text{log}}4}{{\text{log}}4}}={\frac {{\text{log}}25}{{\text{log}}4}}}$
${\displaystyle 3+x={\frac {{\text{log}}25}{{\text{log}}4}}}$
${\displaystyle 3+x-3={\frac {{\text{log}}25}{{\text{log}}4}}-3}$
${\displaystyle x={\frac {{\text{log}}25}{{\text{log}}4}}-3}$
5. 5
Find the logs in the equation. You can do this using a scientific calculator. Type the number you are finding the log of, then hit the ${\displaystyle {\text{LOG}}}$ button. Rewrite the equation using these new values for the logs.
• For example, to find ${\displaystyle {\text{log}}25}$, hit ${\displaystyle 25}$, then ${\displaystyle {\text{LOG}}}$ on your calculator, to get about 1.3979. To find ${\displaystyle {\text{log}}4}$, hit ${\displaystyle 4}$, then ${\displaystyle {\text{LOG}}}$ on your calculator, to get about 0.602. You new equation will now be ${\displaystyle x={\frac {1.3979}{0.602}}-3}$.
6. 6
Complete the calculations. This will give you the value of the variable. Your answer will be approximate since you rounded when finding the logs. Remember to use the order of operations when making your calculations. For more instructions on how to calculate using the order of operations, read Evaluate an Expression Using PEMDAS.
• For example, in ${\displaystyle x={\frac {1.3979}{0.602}}-3}$ you should divide first, then subtract:
${\displaystyle x={\frac {1.3979}{0.602}}-3}$
${\displaystyle x=2.322-3}$
${\displaystyle x=-0.678}$.
## Article Info
Categories: Exponents and Logarithms
In other languages:
Italiano: Risolvere le Equazioni Esponenziali, Español: resolver ecuaciones exponenciales, Português: Resolver Equações Exponenciais, Deutsch: Eine Exponentialgleichung lösen, 中文: 解指数方程, Русский: решать показательные уравнения, Français: résoudre des équations avec exposant(s) inconnu(s), Bahasa Indonesia: Menyelesaikan Persamaan Eksponensial
Thanks to all authors for creating a page that has been read 31,719 times. |
+0
# PLS HELP ME GUYS
0
36
5
+184
A square is drawn such that one of its sides coincides with the line y = 7, and so that the endpoints of this side lie on the parabola y = 2x^2 + 8x + 4. What is the area of the square? Please help, I'm terrible at quadratics!
Jun 14, 2023
#1
0
First, we need to find the two points on the parabola where y = 7. We can do this by setting y equal to 7 and solving for x. This gives us the following two equations:
2x^2 + 8x + 4 = 7 2x^2 + 8x - 3 = 0
We can solve this equation using the quadratic formula. This gives us the following two solutions:
x = -4 or x = -1/2
These are the two points where the side of the square intersects the parabola.
Now, we need to find the length of the side of the square. This is simply the distance between the two points we just found. This distance is equal to:
Code snippet
|(-4) - (-1/2)| = 9/2
Therefore, the area of the square is equal to:
(9/2)^2 = 81/4 = 20.25
Jun 14, 2023
#3
+118623
+3
You need to find out how long one side is.
The the ends of the side are the intersection of y=7 and y = 2x^2 + 8x + 4
Solve simultaneously
$$7=2x^2+8x+4\\ 2x^2+8x-3=0\\ x=\frac{-8\pm\sqrt{64+24}}{4}\\ x=\frac{-8\pm\sqrt{88}}{4}\\ x=\frac{-8\pm2\sqrt{22}}{4}\\ x=\frac{-8\pm2\sqrt{22}}{4}\\ x=-2\pm\frac{\sqrt{22}}{2}\\ \text{Subtract the little from the big and you get the side length of }\;\;\\\ \frac{2\sqrt{22}}{2}=\sqrt{22}\\ \text{So area is 22 }units^2$$
Jun 14, 2023 |
# Multiplication of rational numbers (Math 0)
https://arbital.com/p/multiplication_of_rational_numbers_math_0
by Patrick Stevens Jul 10 2016 updated Aug 1 2016
"Multiplication" is the idea of "now do the same as you just did, but instead of doing it to one apple, do it to some other number".
[summary: Multiplication is making a number by "doing to some quantity what would usually be done to the quantity $1$".]
We've seen how to add and subtract pairs of rational numbers. But the natural numbers have another operation on them: multiplication.
Remember, a given rational number represents what we get when we cut an apple into pieces all of the same size, then take some number %%note:Possibly more than we actually made, and possibly negative!%% of the little pieces. The product of $\frac{a}{m}$ and $\frac{b}{n}$ %%note: Recall that $\frac{a}{m}$ is "$a$ copies of the little-piece we get when we cut an apple into $m$ equal pieces.%% is what we call "$\frac{a}{m}$ multiplied by $\frac{b}{n}$", and it answers the question "What happens if we do the procedure that would make $\frac{b}{n}$, but instead of starting by cutting one apple into $n$ pieces, we started by cutting $\frac{a}{m}$ apples into $n$ pieces?".
We write the product of $\frac{a}{m}$ and $\frac{b}{n}$ as $\frac{a}{m} \times \frac{b}{n}$.
# Example
It's hopefully easy to see that $1 \times \frac{b}{n} = \frac{b}{n}$. Indeed, the definition is "what do we get if we would make $\frac{b}{n}$, but instead of starting by cutting one apple, we started by cutting $1$ apple?"; but that's just the same! It's like saying "What if, instead of putting bread around my sandwich filling, I tried putting bread?" - I haven't actually changed anything, and I'll still get the same old sandwich %%note:or $\frac{b}{n}$%% out at the end.
How about $2 \times \frac{3}{5}$? (Strictly speaking, I should probably be writing $\frac{2}{1}$ instead of $2$, but this way saves a bit of writing. $\frac{2}{1}$ means "two copies of the thing I get when I cut an apple into one piece"; but an apple cut into one piece is just that apple, so $\frac{2}{1}$ just means two apples.) Well, that says "instead of cutting one apple, we cut two apples" into $\frac{3}{5}$-sized pieces.
From now on, my pictures of apples will get even worse: rather than being circles, they'll now be squares. It just makes the diagrams easier to understand.
In the picture, we have two apples (squares) which I've drawn next to each other, separated by a dashed line. Then I've taken $\frac{3}{5}$ of the whole shape (shaded in red): that is, to the group of two apples I have done the procedure that would create $\frac{3}{5}$ if it were done to one apple alone.
Notice, though, that this divides neatly into $\frac{3}{5}$ of the left-hand apple, and $\frac{3}{5}$ of the right-hand apple. So the red-shaded area comes to $\frac{3}{5} + \frac{3}{5}$, which you already know how to calculate: it is $\frac{6}{5}$.
# General integer times fraction
Can you work out, from the case of $2 \times \frac{3}{5}$ above, what $m \times \frac{a}{n}$ is, where $m$ is an integer?
%hidden(Show solution): It is $\frac{a \times m}{n}$.
Indeed, the procedure to get $\frac{a}{n}$ is: we split $1$ into $n$ equal pieces, and then take $a$ of them. So the procedure to get $m \times \frac{a}{n}$ is: we split $m$ into $n$ equal pieces, and then take $a$ of them.
But each of the pieces we've just made by splitting $m$—that is, those demarcated by the longer solid lines in the $2 \times \frac{3}{5}$ diagram above—can be viewed as being $m$ copies of what we get by splitting $1$. (In the diagram above, we have $2$ copies of that which we get by splitting $1$: namely the two copies indicated by the dashed line.) So we can view the second procedure as: we split $1$ into $n$ equal pieces %%note:In the diagram above, there are $5$ such equal pieces, and right now we're looking only at one square, not at both squares joined together.%%, and then take $a$ of them %%note:In the diagram above, $a$ is $3$: this has given us the red shaded bit of one of the squares.%%, and then do this $m$ times. %%note: In the diagram above, $m$ is $2$: we're finally looking at the two squares joined together into a rectangle.%%
This produces $a \times m$ pieces, each of size $\frac{1}{n}$, and hence the rational number $\frac{a \times m}{n}$. %
You should check that you get the right answer for a different example: $-5 \times \frac{2}{3}$. %%hidden(Show solution): This is "do the procedure that makes $\frac{2}{3}$, but instead of starting with $1$, start with $-5$".
So we take five anti-apples, and divide them into thirds (obtaining $15$ anti-chunks of size $\frac{1}{3}$ each, grouped as five groups of three); and then we take two chunks out of each group of three, obtaining $10$ anti-chunks of apple in total.
So $-5 \times \frac{2}{3} = \frac{-10}{3}$, in accordance with the rule of $n \times \frac{a}{n} = \frac{a \times m}{n}$. %%
# General fraction times fraction
[todo: examples and pictures] [todo: instant rule]
# Order doesn't matter
Notice that while it was fairly obvious that order doesn't matter during addition (that is, $\frac{a}{m} + \frac{b}{n} = \frac{b}{n} + \frac{a}{m}$), because it's simply "putting two things next to each other and counting up what you've got", it's not all that obvious that the product of two fractions should be independent of the order we multiplied in. However, you should check, from the general expression above, that it actually is independent of the order.
Why is this? Why should it be that "do the procedure that made $\frac{b}{n}$, but starting from $\frac{a}{m}$ instead of $1$" and "do the procedure that made $\frac{a}{m}$, but starting from $\frac{b}{n}$ instead of $1$" give the same answer?
Well, remember the diagram we had for $2 \times \frac{3}{5}$ (remembering that that is "do the procedure that would make $\frac{3}{5}$, but instead of doing it to $1$, we do it to $2$):
What would we get if we rotated this diagram by a quarter-turn?
But wait! The shaded bit is just what we get when we do the procedure that makes $2$ (namely "put two copies of the shape next to each other"), but instead of doing it on the single (upper-most) square, we do it to the version of the number $\frac{3}{5}$ that is represented by the shaded bit of the upper-most square! And that is exactly what we would do to get $\frac{3}{5} \times 2$.
In general, $\frac{a}{m} \times \frac{b}{n}$ is the same as $\frac{b}{n} \times \frac{a}{m}$, because the two just "come from the same diagram, rotated by a quarter-turn". They are measuring the same amount of stuff, because the amount of stuff in a diagram doesn't change simply because we rotated it.
## Another example
We'll do $\frac{-5}{7} \times \frac{2}{3}$.
[todo: this example]
# Meditation: why the notation makes sense
At this point, a digression is in order. We have already seen the notation $\frac{a}{n}$ for "take an apple; divide it into $n$ pieces, each $\frac{1}{n}$-sized; and then take $a$ of the chunks". In the language of multiplication that we've now seen, that is "do what we would do to make $a$, but do it starting from a $\frac{1}{n}$-chunk instead of $1$". That is, $\frac{a}{n}$ is just $\frac{1}{n} \times a$.
And we can do that in a different way: we can take $a$ apples, divide each into $n$ chunks, and then just draw one of the chunks from each apple. In the language of multiplication, that is just "do what we would do to make a $\frac{1}{n}$-chunk, but do it to $a$ instead of $1$". That is, $\frac{a}{n} = a \times \frac{1}{n}$.
Recalling that $a$ is just $\frac{a}{1}$, our notation $\frac{a}{n}$ is simply the same as $\frac{a}{1} \times \frac{1}{n}$, as an instance of the "instant rule" $\frac{a}{1} \times \frac{1}{n} = \frac{a \times 1}{1 \times n} = \frac{a}{n}$.
# Inverses: putting things in reverse
Remember that we had "anti-apples" as a way of making nothing ($0$) by adding to some quantity of apples. In a similar vein, we can "invert" multiplication.
Whenever $a$ is not $0$, we can find a rational number $\frac{c}{d}$ such that $\frac{a}{b} \times \frac{c}{d} = 1$. (Notice that we've got $1$ as our "base point" now, rather than the $0$ that addition had.)
Indeed, using the instant rule, we see that $\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}$, so to make $1$ we want $a \times c$ to be the same as $b \times d$.
But we can do that: if we let $c = b$ and $d = a$, we get the right thing, namely $\frac{a \times b}{b \times a} = \frac{a \times b}{a \times b} = \frac{1}{1} = 1$.
So $\frac{b}{a}$ works as an inverse to $\frac{a}{b}$. And this is why we needed $a$ not to be $0$: because $\frac{b}{a}$ isn't actually a rational number unless $a$ is nonzero.
## Intuition
We've seen how this definition follows from the instant rule. Where does it actually come from, though?
[todo: intuition] |
### Revising Division 278502 ÷ 399 = 698
This is a good revision example because in one instance we revise the first number of the quotient on rod I down to 7 only to find that we must revise it down a second time to 6 in order to compete the first part of the division problem. (See steps 3 - 6 below for clarification.)
Step 1: Set the dividend on rods J through O and the divisor on rods CDE (Fig 1)
Fig. 1 `Step 1` ```A B C D E F G H I J K L M N O . . . . . 0 0 3 9 9 0 0 0 0 2 7 8 5 0 2``` ` `
Step 2:
In looking at the problem it's evident that 27÷ 3 = 9 but in order to continue we need a remainder so we'll go lower and choose 8 as the first number in our answer.
Step 2a: Set 8 on rod I. Multiply 3 x 8 and subtract the product 24 from rods JK (Fig.2)
Fig. 2 `Step 2` ```A B C D E F G H I J K L M N O . . . . . 0 0 3 9 9 0 0 0 0 2 7 8 5 0 2 (8) Step 2 - 2 4 Step 2a 0 0 3 9 9 0 0 0 8 0 3 8 5 0 2```
We have an error. With only 38 remaining on rods KL we don't have enough to subtract 9 x 8 = 72, in other words we've subtracted 3 from our dividend one too many times. Solution: we must revise both the quotient and the dividend.
Step 3: Reduce the quotient by 1 making it 7.
Step 3a: Continue the revision by adding 3 to the dividend on rod K (Fig 3)
Fig. 3 `Step 3` ```A B C D E F G H I J K L M N O . . . . . 0 0 3 9 9 0 0 0 8 0 3 8 5 0 2 - 1 Step 3 + 3 Step 3a 0 0 3 9 9 0 0 0 7 0 6 8 5 0 2```
Step 4:
Having completed the revision in the above step we have 68 on rods KL which this is enough to continue. Multiply 9 x 7 and subtract the product 63 from rods JK (Fig.4)
Fig. 4 `Step 4` ```A B C D E F G H I J K L M N O . . . . . 0 0 3 9 9 0 0 0 7 0 6 8 5 0 2 (7) - 6 3 Step 4 0 0 3 9 9 0 0 0 7 0 0 5 5 0 2```
Once again we have an error. In order to continue we need to be able to subtract a further 9 x 7 = 63, this time from rods LM. With only 55 available, we've subtracted 39 one too many times from the dividend. Once again, we must revise both the quotient and the dividend.
Step 5: Reduce the quotient by 1 making it 6.
Step 5a: Continue to revise by adding 39 to the dividend on rods KL (Fig.5)
Fig. 5 `Step 5` ```A B C D E F G H I J K L M N O . . . . . 0 0 3 9 9 0 0 0 7 0 0 5 5 0 2 - 1 Step 5 + 3 9 Step 5a 0 0 3 9 9 0 0 0 6 0 4 4 5 0 2```
Step 6:
Having completed the revision in the above step we now have 391 on rods KLM which is enough to continue. Multiply 9 x 6 and subtract the product 54 from rods LM (Fig. 6)
Fig. 6 `Step 6` ```A B C D E F G H I J K L M N O . . . . . 0 0 3 9 9 0 0 0 6 0 4 4 5 0 2 (6) - 5 4 Step 6 0 0 3 9 9 0 0 0 6 0 3 9 1 0 2```
Step 7:
For the next number in the quotient it seems reasonable to choose 9. Therefore set 9 on rod J and multiply 9 times each of the numbers in the divisor. Subtract each product from the dividend leaving the remainder on rods LMNO (Fig. 7)
Fig. 7 `Step 7` ```A B C D E F G H I J K L M N O . . . . . 0 0 3 9 9 0 0 0 6 9 3 9 1 0 2 (9) - 2 7 Step 7 - 8 1 Step 7a - 8 1 Step 7b 0 0 3 9 9 0 0 0 6 9 0 3 1 9 2```
Step 8:
Once again it looks like 9 might be a good choice for the next answer in the quotient. Set 9 on rod K, multiply 3 x 9 and subtract the product 27 from rods LM (Fig. 8)
Fig. 8 `Step 8` ```A B C D E F G H I J K L M N O . . . . . 0 0 3 9 9 0 0 0 6 9 9 3 1 9 2 (9) - 2 7 Step 8 0 0 3 9 9 0 0 0 6 9 9 0 4 9 2```
Another error has occurred. With only 49 on rods MN we don't have enough to subtract 9 x 9 = 81.
Step 9: Revise by subtracting 1 from the quotient.
Step: 9a: Continue the revision by adding 3 to the dividend.
Fig. 9 `Step 9` ```A B C D E F G H I J K L M N O . . . . . 0 0 3 9 9 0 0 0 6 9 9 0 4 9 2 - 1 Step 9 + 3 Step 9a 0 0 3 9 9 0 0 0 6 9 8 0 7 9 2```
Step 10:
Multiply the revised quotient 9 x 8 and subtract the product 72 from rod MN
Step 10a: Multiply 9 x 8 again and subtract the product 72 from rods NO (Fig.10)
Fig. 10 `Step 10` ```A B C D E F G H I J K L M N O . . . . . 0 0 3 9 9 0 0 0 6 9 8 0 7 9 2 (8) - 7 2 Step 10 - 7 2 Step 10a 0 0 3 9 9 0 0 0 6 9 9 0 0 0 0``` |
# Geometry Chapter 6 Resource Book Answers
Geometry Chapter 6 Resource Book Answers: A Comprehensive Guide
Geometry is an intriguing branch of mathematics that deals with the properties, measurements, and relationships of points, lines, angles, and shapes. Chapter 6 of the Geometry Resource Book delves into the world of polygons, exploring their properties, classifications, and calculations. In this article, we will provide answers to commonly asked questions related to the geometry Chapter 6 Resource Book, along with some unique facts about polygons.
Polygon Facts:
1. A polygon is a closed figure made up of straight line segments called sides. It must have at least three sides and three vertices (corner points).
2. The word “polygon” originates from the Greek words “poly” meaning many and “gonia” meaning angles.
3. Polygons are classified based on the number of sides they possess. For example, a triangle has three sides, a quadrilateral has four sides, and a pentagon has five sides.
4. The sum of the interior angles of any polygon can be calculated using the formula (n – 2) x 180 degrees, where n represents the number of sides.
5. A regular polygon is one in which all sides and angles are equal. Regular polygons have a special property – the sum of their exterior angles is always 360 degrees, regardless of the number of sides.
1. What is the difference between a convex and a concave polygon?
A convex polygon has all interior angles less than 180 degrees, and all sides lie on the same side of the polygon. In contrast, a concave polygon has at least one interior angle greater than 180 degrees, and some sides cross the interior of the polygon.
2. How can I determine the number of diagonals in a polygon?
The number of diagonals in a polygon can be found using the formula n(n-3)/2, where n is the number of sides. For example, a hexagon (n=6) has 9 diagonals.
3. What is the formula to find the measure of each interior angle of a regular polygon?
The measure of each interior angle of a regular polygon can be calculated using the formula [(n-2) x 180]/n, where n is the number of sides.
4. How can I identify if a polygon is regular or irregular?
A polygon is regular if all of its sides and angles are equal. To determine if a polygon is regular, measure the lengths of its sides and the size of its angles. If they are all equal, the polygon is regular; otherwise, it is irregular.
5. How do I calculate the area of a regular polygon?
The area of a regular polygon can be computed using the formula A = (1/2) x apothem x perimeter, where the apothem is the distance from the center of the polygon to the midpoint of any side.
6. What is the difference between a regular and an irregular polygon?
A regular polygon has all sides and angles equal, while an irregular polygon has sides and angles of varying lengths and measures.
7. Can a polygon have more than one obtuse angle?
Yes, a polygon can have more than one obtuse angle. For example, a hexagon can have two obtuse angles.
8. How can I determine the number of sides in a polygon if I know the sum of its interior angles?
The number of sides in a polygon can be found using the formula n = (180 x (s-2))/s, where n is the number of sides and s is the sum of the interior angles.
9. What is the difference between a regular and a non-regular polygon?
A regular polygon has all sides and angles equal, while a non-regular polygon has sides and angles of varying lengths and measures.
10. Are all rectangles considered squares?
No, all rectangles are not considered squares. A square is a special type of rectangle in which all sides are equal.
11. How can I determine if a polygon is concave or convex?
A polygon is concave if at least one of its interior angles is greater than 180 degrees. If all interior angles are less than 180 degrees, the polygon is convex.
12. How do I calculate the perimeter of a polygon?
To calculate the perimeter of a polygon, add up the lengths of all its sides.
13. Is it possible to have a polygon with zero sides?
No, it is not possible to have a polygon with zero sides. A polygon must have at least three sides to be considered a polygon.
In conclusion, Chapter 6 of the Geometry Resource Book delves into the fascinating world of polygons. Understanding the properties and calculations associated with these figures is essential for mastering geometry. This article has provided answers to frequently asked questions, along with some intriguing facts about polygons. So, dive into the world of geometry and explore the wonders of polygons! |
# Arithmetic sequences
#### Everything You Need in One Place
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0/6
##### Examples
###### Lessons
1. Arithmetic sequence formula
Consider the arithmetic sequence: 5, 9, 13, 17, … .
1. Identify the common difference.
2. Determine the seventh term of the sequence.
3. Which term in the sequence has a value of 85?
2. Determine $t_1,d,t_n$ for the sequences in which two terms are given
1. $t_4=14$, $t_{10}=32$
2. $t_3=-14$, $t_{12}=-59$
3. Three consecutive terms of an arithmetic sequence are written in the form:
$1+2x,7x,3+4x$
Solve for the value of x.
###### Topic Notes
An arithmetic sequence (arithmetic progression) is a number sequence with a common difference between successive terms. By using the arithmetic sequence formula, we can easily find the value of a term and the common difference in the sequence.
• arithmetic sequence: a sequence with a common difference between successive terms
• The nth term, ${t_n}$ ,of an arithmetic sequence:
${t_n} = {t_1} + \left( {n - 1} \right)d$
where, ${t_n}$: nth term
${t_1}$: first term
$d$ : common difference |
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# A Semi-Detailed Lesson Plan
in Grade 10 Mathematics
I- Objectives
## a. Identify the Probability of Compound Events
b. Illustrate events, and Union and Intersection of events
c. Illustrate the probability of a union of two events and intersection events
d. Illustrate and find probabilities of Mutually Exclusive Events
e. Appreciate the value of compound events
f. Solve the problems of probability of compound events
## II- Subject Matter
Topic: Probability of Compound Events
Reference: Mathematics
Pages: 320-338
Time Frame: 3 session
III- Materials
Chalk, Manila Paper, Cartolina, Pentel pen
IV- Learning Strategies
1. Preparation
a. Prayer
b. Checking of Attendance
2. Lesson Proper
a. Motivation
The teacher will let the students stand and sing an action song called “lean forward”
b. Discussion
Probability- the chances of the given event will occur
Experiments- any process of observation
Outcomes- results of an experiments
Sample Space- set of all outcomes
## Consider rolling a die.
a. “Getting a number 5” is called a simple event
b. “ Getting a 6” is also called a simple event
Probability of simple events- if each of the outcomes in a sample space is equally likely to occur
## P(E) = Number of way the event will occur
Number of possible outcomes
Or P(E) = Number of outcomes in the event
Number of outcomes in the sample space
a. a 5
b. a 6
c. An odd number
## Compound Events- events which consists of more than one outcomes.
Example: “Getting a 6 and a 1”, when two dice are rolled is an event consisting of {(1,6), (6,1)} as
outcomes. This is a compound event.
“Intersection and Union of Events”
The teacher will let the students analyze and solve the questions in activity 3 which are shown
in the Venn diagram.
## Extra-Curricular activities participated by Senior students
Questions:
1. How many students are in the senior class?
2. How many students participated in athletics?
3. If a student is randomly chosen, what is the probability that the student participated only in
drama and band?
(The teacher will discuss the probabilities of grade 10 students joining either soccer (S) or basketball (B).
)
## Use the Venn diagram to find the probabilities
a. P(B)
b. P(S)
c. P(B Π S)
d. P(B U S)
e. P(B’ Π S’)
Complement of an Event- is the set of all outcomes that are Not in the event. This means that if the
probability that the event would not occur. Thus,
P(A’) = 1- P(A)
## “Mutually Exclusive and Not Mutually Exclusive Events”
Mutually Exclusive Events- If two events, A and B cannot occur at the same time. Then the probability
that either A or B occurs is the sum of their probabilities. In symbols,
P(A or B)= P(A) + P(B)
Example: The event of getting a 5 in the event of getting a number divisible y 3 from the set
{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
A B
P(A or B)
Not Mutually Exclusive Events- If two events, A and B, are not mutually exclusive, then the probability
that either A or B occur is the sum of their probabilities decreased by the probability of both occurring.
In symbols,
## P(A or B)= P(A) + P(B) – P(A and B)
Example: The event of getting a number divisible by 3 or the event of getting a number divisible by 4 in
the set
{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
P(A or B)
c. Drill/Exercise
Consider the situation below and answer the questions that follow.
1. Out of 5200 households surveyed, 2107 had a dog, 807 had a cat, and 303 had both a
dog and a cat. What is the probability that a randomly selected household has a dog or a
cat?
2. Rhian likes to wear colored shirts. She has 15 shirts in the closet. Five of these are blue,
four are in different shades of red, and the rest are of different colors. What is the
probability that she will wear a blue or a red shirt?
IV-Evaluation
Direction: Consider each problem below. Draw a Venn diagram for each. Determine whether the events
are mutually exclusive or not mutually exclusive. Then, find the probability.
1. Mario has 45 red chips, 12 blue chips, and 24 white chips. What is the probability that Mario
randomly selects a red chip or a white chip?
2. Of 240 students, 176 are on the honor roll, 48 are members of the varsity team, and 36 are in
the honor roll and are also members of the varsity team. What is the probability that a
randomly selected student is on the honor roll or is a member of the varsity team?
3. Ruby’s dog has 8 puppies. The puppies include white females, 3 mixed-color females, 1 white
male, and 2 mixed-color males. Ruby wants to keep one puppy. What is the probability that she
randomly chooses a puppy that is female and white?
4. Carl’s basketball shooting records indicate that for any frame, the probability that he will score
in a two-point shoot is 30%, a three-point shoot, 45%, and neither, 25%. What is the probability
that Cindy will score either in a two-point shoot or in a three-point shoot?
V-Agreement
a. Assignment
(Study and advance the next topic which is Independent and Dependent Events.)
Prepared by:
Jesiryl V. Piedad
Student-teacher
Checked by:
Mr. Ryan Y. Gelsano
Critic-teacher |
Check out illinois-ipass.com, another site I earn revenue from ads on
# What is a 30 60 90 Triangle?
A 30-60-90 triangle is a special type of right triangle that has some unique properties and is named after the measures of its angles. In a 30-60-90 triangle, the three angles are measured at 30 degrees, 60 degrees, and 90 degrees, hence the name.
One of the most interesting and useful properties of 30-60-90 triangles is their relationship between the side lengths. In a 30-60-90 triangle, the hypotenuse is always twice as long as the shorter side, and the longer side is always sqrt(3) times longer than the shorter side. This makes it easy to calculate the lengths of the sides given the length of one side, and it also makes these triangles useful for solving real-world problems in fields such as construction, engineering, and trigonometry.
Another interesting property of 30-60-90 triangles is that they are isosceles triangles. This means that the two smaller angles are equal, and the two smaller sides are equal in length. This property can be useful in solving problems and making deductions about the triangle.
The 30-60-90 triangle is a fundamental concept in trigonometry and is often used in geometry, engineering, and other technical fields. It is also a common topic in high school and college math classes. Understanding the properties of 30-60-90 triangles and how to use them can be a valuable skill for anyone interested in pursuing a career in science, technology, engineering, or mathematics.
## What are the rules for 30 60 90 triangles?
The rules for 30-60-90 triangles are based on the angles and side lengths of the triangle. Here are some important rules to remember when working with 30-60-90 triangles:
1. Angle Measures: In a 30-60-90 triangle, one angle measures 30 degrees, one angle measures 60 degrees, and the final angle measures 90 degrees.
2. Side Lengths: The side lengths in a 30-60-90 triangle are related by a ratio. The hypotenuse, which is the side opposite the 90-degree angle, is always twice as long as the shortest side of the triangle, known as the “30-degree side”. The middle side, known as the “60-degree side”, is always sqrt(3) times longer than the 30-degree side.
3. Isosceles Triangle: A 30-60-90 triangle is an isosceles triangle, meaning that the two smaller angles are equal and the two smaller sides are equal in length.
4. Right Triangle: A 30-60-90 triangle is a special type of right triangle, meaning that it has one 90-degree angle.
5. Pythagorean Theorem: The Pythagorean theorem, which states that the sum of the squares of the two shorter sides of a right triangle is equal to the square of the hypotenuse, can be used to solve problems involving 30-60-90 triangles.
6. Trigonometry: The ratios of the side lengths in a 30-60-90 triangle can be used in trigonometry to solve problems and make deductions about the triangle.
By understanding and applying these rules, you can effectively use 30-60-90 triangles to solve real-world problems and tackle challenging mathematical concepts. In conclusion, 30-60-90 triangles are special right triangles that have unique properties and relationships between their side lengths and angles. Understanding and utilizing these properties can be a valuable tool in solving problems and making deductions in a variety of fields. This right triangle calculator is a good tool.
A 30-60-90 triangle is a special type of right triangle with several unique properties and relationships between its angles and side lengths.
You can check out my post on the AP Calc AB class. |
Fractions Multiplication: Mixed and Improper Fractions
Here's what we do if to multiply mixed fractions such as 2 1/4 × 1 2/3:
1) Change any mixed fraction to an improper fraction.
2) Multiply the numerators and multiply the denominators.
1) 2 × 4 = 8. 8 + 1 = 9. So, 2 1/4 = 9/4.
Also, 1 × 3 = 3. 3 + 2 = 5. So, 1 2/3 = 5/3.
If you don't understand this then please review Fractions Introduction.
2) We rewrite 2 1/4 × 1 2/3 as 9/4 × 5/3.
Multiply the numerators: 9 × 5 = 45.
Multiply the denominators: 4 × 3 = 12.
3) 45/12 is simplified to 3 9/45: 12 goes into 45 3 times with a remainder of 9. In turn, 3 9/45 is simplified to 3 1/5:
Factors of 45: 1, 5, 9, 45
Factors of 9: 1, 3, 9
9 ÷ 9 = 1
45 ÷ 9 = 5
If you don't understand this then please review Fractions Introduction.
Here's another example: 4 1/5 × 3 1/2:
Once again, we:
1) Change any mixed fraction to an improper fraction.
2) Multiply the numerators and multiply the denominators.
1) 4 × 5 = 20. 20 + 1 = 21. So, 4 1/5 = 21/5.
Also, 3 × 2 = 6. 6 + 1 = 7. So, 3 1/2 = 7/2.
If you don't understand this then please review Fractions Introduction.
2) We rewrite 4 1/5 × 3 1/2 as 21/5 × 7/2.
Multiply the numerators: 21 × 7 = 147.
Multiply the denominators: 5 × 2 = 10. |
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Shvoong Home>Science>Mathematics>Handout Kelas Xi Summary
# Handout Kelas Xi
DIFFERENSIAL
Standard Competence
6. Use the concept of Function limit and the derived function in problem solving.
Basic Competence
6.3Use the concept and the order of differential in calculation of the derived function.
Indicators
1. Count the simple derived function by using differential definition
2. Explaining the physic meaning of differential at any point
3. Explaining the geometry meaning of differential at any point
4. Determine the change velocity of function value to independent variable
5. Use the order of differential to determine the derived algebra function and trigonometry function
6. Determine the derived composition by chain rule
7. Determine the tangent line equation at any curve
1st meeting (3 x 45 minutes)
Pretest
1. Determine the function h if we have the composition function where and
2. Determine ,for
A. Algebraic Function Derivative
1. The Definition of Derivative
Average velocity
In this case, s is the function of time (t), for example f (t). In the previous chapter, you have learned that speed in , is declared with the following formula.
The formula is called the change rate of distance toward time.
Now, observe the figure on the side. The gradient of the line passing through the point A and B, for example .
if the point B is approaching then the gradient of line becomes the gradient of curve tangent in point A as point B will be coinciding with point A. therefore, if point B A then
For example the function is given. The derivative of this function toward , written as or can be determined use the following formula.
If the limit exist.
In compliance with the concept of the change rate of distance toward, a definition of derivative of a function is made as follows.
The function is given, the function derivative is the function of which value in point c is if the limit value exist.
Further to determine we perceive as a constant, but random so that we would obtain the function derivative of is as follow.
Aside from the notation above, the derivative of function f(x) can be written by the notation this notation is called the Leibniz notation as it was initially suggested by a German mathematician named Gottfried Wilhelm Leibniz (1646-1716).
Example:
Determine the function derivative of
Solution:
1. Using the derivative formula , determine the following function derivative.
a.
b.
c.
d.
e.
f.
2. Using the derivative formula , determine the following function derivative function in .
a.
b.
c.
d.
e.
f.
3. We have the function
a. Determine the function derivative.
b. Determine the value of x so that
2nd meeting (2x45 minutes)
The formula of Function Derivative
Using the formula of function derivative we can determine the derivative of constant function, the derivative of identity function, the derivative of exponent function , and the formula of function derivative with real constant.
a. The constant function derivative , where c is the real constant.
Published: June 17, 2011
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We received lots of good solutions to this problem - well done everyone! Many of you spotted that Mr McGregor should put $7$ plants in his potting shed at the beginning, and put $8$ plants in each garden. Well done to Henry from Finton House School, Ruth from Manchester High School for Girls, Liam from Wilbarston School, Mel from Christ Church Grammar School, Rachel from Beecroft Public School in Australia, Yanqing from Devenport High School for Girls and Daniel from Junction City High School for their detailed explanations of how they arrived at the answer.
Henry from Finton House School wrote:
"$1 \times2 \times2 \times2 = 8$.
8 therefore seems likely to be the number in the garden. Let's try it.
The number in the shed at the end must be the number in the garden.
Now what number do we double to get to $8$? It must be $4$.
$4 + 8 = 12$. $12$ divided by $2 = 6$.
$6 + 8 = 14$. $14$ divided by $2 = 7$."
Liam used similar logic:
"Just work backwards from the last garden. Imagine there to be $8$ plants in each garden. (You can't have odd numbers in a garden as the last garden must be double the whole number of plants left after the 2nd garden was planted. I chose $8$ because it's a conveniently sized power of 2.) There must have been $4$ plants left after the 2nd garden was planted so before it was planted there must have been $12$ which is double $6$. $6+8=14$. So Mr McGregor needs to put $14/2$ or $7$ plants in his magic potting shed at the beginning!"
Yanqing and Rachel used algebra. Here is Yanqing's solution:
"First, we make the number of plants put in the shed $n$, and the number planted each night $x$. So by the first morning, the number has doubled to $2n$ in the shed. We plant $x$ of them, leaving $2n-x$ in the shed overnight. By the second morning, we have $2(2n-x)=4n-2x$ in the shed. Planting $x$ of them, we are left with $4n-2x-x=4n-3x$ in the shed. By the third morning, there should be $2(4n-3x)=8n-6x$ plants in the shed. There need to be $x$ plants in the shed, as we need to plant all of them, so $8n-6x=x$ and $8n=7x$.
We can now say that the ratio of $n$ to $x$ is $7:8$, so the smallest values for $n$ and $x$, where they are both positive whole numbers, are obviously $7$ and $8$.
Other numbers which will work are all multiples of $7$."
Rachel also found that $8n= 7x$ and concluded that:
"Now you can see that $8n$ or $7x$ could equal $56$, which makes $n = 7$ and $x = 8$.
This works when you try it out, and if you multiply both numbers by another number, those new numbers work too."
Daniel concluded that:
"If you want to have the same amount of plants in each garden you must start with a multiple of $7$ plants in the shed and each day plant the same multiple of $8$ plants in the garden."
James from C.G.S.B found such a solution:
"Start with $35$ ... then put $40$ in each garden"
And so did Mel from Christ Church Grammar School!
"You start off with $301$ plants in the shed. You put $344$ in each garden." |
# Basic Descriptive Statistics Healey, Chapter 2
## Presentation on theme: "Basic Descriptive Statistics Healey, Chapter 2"— Presentation transcript:
Basic Descriptive Statistics Healey, Chapter 2
Percentages, Ratios and Rates, Frequency Distributions, Charts and Graphs
Outline: Percentages and Proportions Ratios, Rates, and % Change
Frequency Distributions Charts and Graphs
Percentages and Proportions Formulae:
Percentages and Proportions (cont.)
Report relative size. Compare the number of cases in a specific category to the number of cases in all categories. Compare a part (specific category) to a whole (all categories). The part is the numerator (f ). The whole is the denominator (N).
Percentages and Proportions (cont.)
Suppose you have a group of 229 sociology majors, of which 97 are female and 132 are male. What percentage of this group is female? The whole is the number of people in the group. The part is the number of females.
Percentages and Proportions (cont.)
To identify the whole and the part, use the keywords of and is. of identifies the whole (N) is identifies the part (f)
Percentages and Proportions: Example
What % of social science majors is female? of (whole) = all sociology majors = 229 is (part) = female sociology majors 97 (97/229) * 100 = (.4236) * 100 = 42.36% 42.36% of sociology majors are female
Ratio Compares the relative sizes of categories.
Compares parts to parts. Ratio = f1 / f2 f1 - number of cases in first category f2 number of cases in second category
Ratio (cont.) In a class of 23 females and 19 males, the ratio of males to females is: 19/23 = 0.83 For every female, there are 0.83 males. In the same class, the ratio of females to males is: 23/19 = 1.21 For every male, there are 1.21 females.
For Practice: With a partner, try Healey, #2.1 in 2/3e (a-e)
a. % married in A ____ ? in ____ ? b. Ratio single to married in A ___ ? in B ___ ? c. Proportion widowed in A ____? in B ____? d. % single living in B ____? e. Ratio unmarried (living together) to married in A ____ ? in B ____?
Rate (cont.) Expresses the number of actual occurrences of an event (births, deaths, homicides) vs. the number of possible occurrences per some unit of time.
Rate (cont.) Birth rate is the number of births divided by the population size times 1000 per year. If a town of 2300 had 17 births last year, the birth rate is: (17/2300) * 1000 = (.00739) * 1000 = 7.39 The town had 7.39 births for every 1000 residents.
For Practice: With a partner, try Healey #2.3
Rate for bank robberies? ______ Rate for murders? ______ Rate for auto thefts? ______
Percentage Change Measures the relative increase or decrease in a variable over time. Formula:
Percentage Change (cont.)
f1 is the first (or earlier) frequency. f2 is the second (or later) frequency. Change can also be calculated with proportions, rates, or other values.
Percentage Change: Example
In 1990, a city had a murder rate of 7.3. By 2000, the rate had increased to 10.7. What was the relative change? (10.7 – 7.3 / 7.3) * 100 = (3.4 / 7.3) * 100 = 46.58% The rate increased by 46.58%.
Frequency Distribution
This is a report in the form of a table of the number of times each score of a variable occurred. The categories of the frequency distribution must be stated in a way that permits each case to be counted in one and only one category. Categories must not overlap (they should be “mutually exclusive.”) Table should have a title and clearly labeled categories and columns.
Example: Frequency Distribution for Age (Interval width = 2 years)
Class Interval Age Frequency % 18-19 11 55 20-21 5 25 22-23 2 10 24-25 1 26-27 Total 20 100
Stated Class Limits and Real Class Limits
In the previous table, the limits of the intervals appear to have a “gap” between categories: the scores of the variable are organized into discrete intervals. These are the stated class limits. Some calculations require that the “gap” be eliminated so the intervals appear continuous. To do this, you have to find the real class limits by adding half the distance to the upper limit, and subtracting half the distance from the lower limit. In the Age table, the gap is equal to 1, so you would add and subtract half that distance, or .5, to either end of the interval.
Midpoints You will also need to find the midpoints of the intervals for some statistical calculations and for graphing purposes. The midpoints can be found by adding the upper and lower limits together an then dividing the total by 2. See Age example below for the real class limits and midpoints.
Example: Frequency Distribution for Age with Real Class Limits and Midpoints.
Class Interval Stated Limits Real Class Limits Midpoints 18-19 17.5 – 19.5 18.5 20-21 19.5 – 21.5 20.5 22-23 21.5 – 23.5 22.5 24-25 23.5 – 25.5 24.5 26-27 25.5 – 27.5 26.5
Cumulative Frequency and Percentage
Class Interval Age Frequency Cumulative Frequency % Cumulative % 18-19 11 55 20-21 5 16 25 80 22-23 2 18 10 90 24-25 1 19 95 26-27 20 100 Total
Grouping Data (Interval-Ratio)
When you have interval-ratio data, you may have many scores to put into a frequency distribution, so the data may have to be grouped into intervals with widths of 5, 10, or sometimes more, depending on the range of the scores. All intervals should be equal in size and should not overlap. Do not use more than 15 intervals (10 intervals is a good “rule of thumb” to follow.) Once you have decided on your interval width and number of intervals, construct the table in the same way as you would for nominal and ordinal data.
For Practice: Try Healey, #2.5. Construct a frequency distribution for the variables Sex and Age, including a column for %. Complete this question as part of your homework.
Graphs And Charts Histograms, pie and bar graphs and line charts (also called “frequency polygons”) can be constructed to present frequency distributions graphically. Graphs and charts are commonly used ways of presenting “pictures” of research results. Graphs can be constructed by hand or can easily be generated using a software program like Excel or SPSS.
Graphs and Charts (cont.)
Histograms and frequency polygons are commonly used for interval-ratio data. Bar graphs and pie charts are most often used for nominal or ordinal data. Read rest of chapter in Healey for details on how to construct graphs and charts.
Sample Pie Chart: Marital Status (N = 20)
Sample Bar Chart: Marital Status Of Respondents (N = 20)
Histogram for Age Group
Marriage And Divorce Rates Over Time
How would you describe the patterns?
Homework Questions: 1. Complete Healey and Prus #2.5 2. Healey, #2.2
SPSS Exercise: Read SPSS section at end of Ch. 2 In the computer lab, try #2.1 and #2.2 (in SPSS section) for practice |
# Indirect and direct relationship math
### What are the different types of mathematical relationships?
Direct variation describes a simple relationship between two variables. We say y varies directly with x (or as x, in some textbooks) if. Inverse or Indirect Variation refers to relationships of two Variation, like we saw above for a Direct Variation. Mathematical Relationships in Science. measuring The points fall close enough to the straight line to conclude that this is a linear or direct relationship.
If we scale up x by it's a different green color, but it serves the purpose-- we're also scaling up y by 2. To go from 1 to 2, you multiply it by 2. To go from negative 3 to negative 6, you're also multiplying by 2. So we grew by the same scaling factor.
To go from negative 3 to negative 1, we also divide by 3. We also scale down by a factor of 3. So whatever direction you scale x in, you're going to have the same scaling direction as y.
## Direct, Inverse, Joint and Combined Variation
That's what it means to vary directly. Now, it's not always so clear. Sometimes it will be obfuscated. So let's take this example right over here. And I'm saving this real estate for inverse variation in a second. You could write it like this, or you could algebraically manipulate it. Or maybe you divide both sides by x, and then you divide both sides by y. These three statements, these three equations, are all saying the same thing.
So sometimes the direct variation isn't quite in your face. But if you do this, what I did right here with any of these, you will get the exact same result. Or you could just try to manipulate it back to this form over here.
And there's other ways we could do it. We could divide both sides of this equation by negative 3.
## indirect relationship
And now, this is kind of an interesting case here because here, this is x varies directly with y. Or we could say x is equal to some k times y. And in general, that's true. If y varies directly with x, then we can also say that x varies directly with y.
It's not going to be the same constant. It's going to be essentially the inverse of that constant, but they're still directly varying.
### What is indirect relationship? definition and meaning - klokkenluideronline.info
Now with that said, so much said, about direct variation, let's explore inverse variation a little bit. Inverse variation-- the general form, if we use the same variables. And it always doesn't have to be y and x. It could be an a and a b. It could be a m and an n. If I said m varies directly with n, we would say m is equal to some constant times n. Now let's do inverse variation. So let me draw you a bunch of examples. And let's explore this, the inverse variation, the same way that we explored the direct variation.
And let me do that same table over here. In an inverse relationship, an increase in one quantity leads to a corresponding decrease in the other. Faster travel means a shorter journey time.
How Does y Vary with x? Scientists and mathematicians dealing with direct and inverse relationships are answering the general question, how does y vary with x? Here, x and y stand in for two variables that could be basically anything. By convention, x is the independent variable and y is the dependent variable.
So the value of y depends on the value of x, not the other way around, and the mathematician has some control over x for example, she can choose the height from which to drop the ball. When there is a direct or inverse relationship, x and y are proportional to each other in some way. Direct Relationships A direct relationship is proportional in the sense that when one variable increases, so does the other. Using the example from the last section, the higher from which you drop a ball, the higher it bounces back up.
A circle with a bigger diameter will have a bigger circumference. Examples include viscous drag in mechanical systems, resistance in electronic oscillators, and absorption and scattering of light in optical oscillators. Damping not based on energy loss can be important in other oscillating systems such as those that occur in biological systems. Sine Wave Relationship The graphs of the sine and cosine functions are sinusoids of different phases.
The sine wave or sinusoid is a mathematical curve that describes a smooth repetitive oscillation. It is named after the function sine, of which it is the graph.
Lab Activities and Resources What are Mathematical Relationships What is a mathematical relationship and what are the different types of mathematical relationships that apply to the laboratory exercises in the following activities. What is the relationship between how much a spring stretches and the force pulling on the spring?
• Intro to direct & inverse variation
What is the relationship between the mass of a ball and its volume assuming a constant density? What is the relationship between the intensity of a beam of light and the distance from a light source? What is a the relationship between how the distance travels and the time in travel for an accelerating object?
What is the relationship between how much light passes through a Polaroid filter and the angle the filter is rotated? What is the relationship between current, voltage when there is a constant resistance in an electric circuit.
Radioactive Decay- - Problem: What is the relationship between the decay of radioactive material and the time allowed for the decay? Water Pressure - - Problem: What is the relationship between water pressure and depth of water? What is the relationship between the height a ball bounces and the number of times it has bounced? |
## Table 2 To 50
#### Does 50 come in table of 2?
Multiplication Table of 2 |Repeated Addition by 2’s|Read & Write the Table of 2
• Repeated addition by 2’s means the multiplication table of 2.
• (i) When there are 5 bunches of 2 apples in each bunch.
• By repeated addition we can show 2 + 2 + 2 + 2 + 2 = 10
• Then, two 5 times or 5 tows
• 5 × 2 = 10
• Therefore, there are 10 apples.
• (ii) When there are 9 packets of 2 chocolates in each packet.
• By repeated addition we can show 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 18
• Then, two 9 times or 9 tows
• 9 × 2 = 18
• Therefore, there are 18 chocolates.
We will learn how to use the number line for counting the multiplication table of 2. (i) Start at 0. Hop 2, three times. Stop at 6,3 twos are 6 3 × 2 = 6 (ii) Start at 0. Hop 2, four times. Stop at _, Thus, it will be 8 4 twos are 8 4 × 2 = 8 (iii) Start at 0. Hop 2 nine times.
1. Stop at _, Thus, it will be 18
2. 9 twos are 18 9 × 2 = 18
3. How to read and write the table of 2?
The above diagram will help us to read and write the 2 times table.
Read 1 two is 2 2 twos are 4 3 twos are 6 4 twos are 8 5 twos are 10 6 twos are 12 7 twos are 14 8 twos are 16 9 twos are 18 10 twos are 20 11 twos are 21 12 twos are 24 Write 1 × 2 = 2 2 × 2 = 4 3 × 2 = 6 4 × 2 = 8 5 × 2 = 10 6 × 2 = 12 7 × 2 = 14 8 × 2 = 16 9 × 2 = 18 10 × 2 = 20 11 × 2 = 22 12 × 2 = 24
ul>
• Now we will learn how to do forward counting and backward counting by 2’s.
• Forward counting by 2’s: 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50,
• Backward counting by 2’s:, 50, 48, 46, 44, 42, 40, 38, 36, 34, 32, 30, 28, 26, 24, 22, 20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 0.
• Didn’t find what you were looking for? Or want to know more information about, Use this Google Search to find what you need.
Multiplication Table of 2 |Repeated Addition by 2’s|Read & Write the Table of 2
#### What is the table of 2 till 100?
2 Table till 100
Table of 2 till 60 Table 2 till 100
2 x 50 = 100 2 x 90 = 180
2 x 51 = 102 2 x 91 = 182
2 x 52 = 104 2 x 92 = 184
2 x 53 = 106 2 x 93 = 186
#### In which table 50 comes?
Table of 50 is a multiplication table that results in the product of consecutive natural numbers with the number 50. For example, two times 50 = 50 + 50 = 100, three times 50 = 50 + 50 + 50 = 150, and so on. On this page, you can find the table of 50 up to 20.
## Is 2 a multiple of 50?
Video Lesson on Common Multiples –
• The lowest common multiple of 50 with any other given number is the lowest number which both 50 and the given number can divide.
• For example,
• The lowest common multiple of 50 and 80
You might be interested: Tata Ipl Points Table 2023
Multiples of 50 are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500, 550, 600, 650, 700, Multiples of 80 are 80, 160, 240, 320, 400, 480, 560, 640, 720, 800,
Clearly, the lowest common multiple or LCM of 50 and 80 is 400. That is, 400 is the lowest number which both 50 and 80 can divide.
• Every multiple of 50 is an even number since 50 itself is an even number.
• The lowest multiple of 50 is 50 itself (50 × 1 = 50).
• The greatest multiple of 50 cannot be determined as there are infinitely many multiples of 50.
• The n th multiple of 50 can be determined by 50 × n
• 50 is a factor of each multiple of 50.
• All multiples of 50 form a sequence, rather an arithmetic progression in which each term differs by 50.
• No multiple of 50 is a prime number, as every multiple of 50 has at least three multiple 1, 50 and the multiple of 50 itself.
1. Example 1:
2. Find the average of the first 20 multiples of 50.
3. Solution:
4. The first 20 multiples of 50 are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500, 550, 600,
5. 650, 700, 750, 800, 850, 900, 950, 1000
6. Average of the first 20 multiples of 50 = ½ × Sum of first 20 multiples of 50= 10500/2 = 525
7. Example 2:
8. Find the least common multiple of 50, 60 and 70.
9. Solution:
10. By prime factorisation method,
L.C.M of 50, 60 and 70 = 2 × 2 × 5 × 5 × 3 × 7 = 2100 Example 3: In each class of a school, there are 50 students. There are 20 classes in that school.
• How many total students are there in the school?
• Solution:
• The total number of students will be in multiples of 50 since there are 50 students in each
• class.
• Number of classes = 20
• Total number of students = 50 × 20 = 1000
• There are 1000 students.
All those numbers which come in the times table of 50 are multiples of 50. The first ten multiples of 50 are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500. The lowest multiple of 50 is 50 itself but the greatest multiple of 50 can not be determined as there are infinitely many multiples of 50.
No multiple of 50 is a prime number as each multiple of 50 will have at least three factors 1, 50 and the multiple itself. But the prime factors of 50 are 2 and 5. Starting with 50 count the numbers adding 50, or simply multiplying the number with 50 to get that multiple of 50. Common multiples of 50 with any other given number is the multiple of both 50 and the given number(s).
It is a multiple of 50 which both 50 and the given number(s) can divide. : What are the Multiples of 50 | Solved | Multiple of a number
#### What is 2 numbers to make 50?
50 = 1 x 50, 2 x 25, or 5 x 10.
## What is the value of 2 to the power of 50?
What is 2 to the Power of 50? | Thinkster Math The first step is to understand what it means when a number has an exponent. The “power” of a number indicates how many times the base would be multiplied by itself to reach the correct value.
The second step is to write the number in the base-exponent form, and lastly calculate what the final result would be. Consider the example of 2 to the power of 4: in exponent form that would be 2 4 2^4 2 4 2 ⋅ 2 ⋅ 2 ⋅ 2 2\cdot2\cdot2\cdot2 2 ⋅ 2 ⋅ 2 ⋅ 2 2 4 = 1 6 2^4 = 16 2 4 = 1 6 So re-applying these steps to our particular problem, we first convert our word problem to a base-exponent form of: 2 5 0 2^ 2 5 0 To simplify this, all that is needed is to multiply it out:
, To solve this, we need to multiply the base, 2 by itself, 4 times – = 16. So,2 x 2 x 2 x 2 x, (for a total of 50 times) = 1125899906842624 Therefore, 2 to the power of 50 is 1125899906842624. : What is 2 to the Power of 50? | Thinkster Math
## How many times 2 comes from 1 to 100?
Detailed Solution Digit 2 appears 20 times in first 100 natural numbers.
#### Is 50 in the 5 times table?
5 × 7 = 35 (5 + 5 + 5 + 5 + 5 + 5 + 5 = 35) 5 × 8 = 40 (5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 = 40) 5 × 9 = 45 (5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 = 45) 5 × 10 = 50 (5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 = 50)
## Is 2 a multiple of 48?
Get the Multiples of More Numbers Here – Stay tuned with BYJU’S – The Learning App and download the app to learn the multiples of different numbers and all Maths-related concepts easily by exploring more videos. The multiples of 2 are the numbers that are exactly divided by 2 and leave no remainder.
The multiples of 2 are 2, 4, 6, 8, 10 and so on. The first 10 multiples of 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 The multiples of 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, The multiples of 3 are 3, 6, 9, 12, 15, 18, 21,, Hence, some of the common multiples of 2 and 3 are 6, 12, 18, etc. The smallest common multiple of 2 and 3 is 6.
The multiples of 2 between the numbers 50 and 60 are 52, 54, 56, and 58. Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” buttonCheck your score and answers at the end of the quiz Visit BYJU’S for all Maths related queries and study materials
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### Is zero a multiple of 2?
Multiple Question: Children (and adults) are often uncertain whether the multiples of, say, 12 are the numbers one can multiply (like 3 and 4) to make 12, or the numbers that one can make by multiplying 12 times other numbers. The terms multiple and factor are often confused. What are the multiples of a number? By example:
Multiples of 3, like –9, –6, –3, 0, 3, 6, 9, 12, 15 are formed by multiplying 3 by any integer (a “whole” number, negative, zero, or positive, such as–3, –2, –1, 0, 1, 2, 3).Multiples of 12, like –36, –24, –12, 0, 12, 24, 36, 48, 60, are all 12 × n, where n is an integer.Multiples of 2, like –8, –6, –4, –2, 0, 2, 4, 6, 8, 10, 12, are all even, 2 × any integer.
Generally: The multiples of an integer are all the numbers that can be made by multiplying that integer by any integer. Because 21 can be written as 3 × 7, it is a multiple of 3 (and a multiple of 7). Though 21 can also be written as 2 × 10, it is not generally considered a multiple of 2 (or 10), because the word multiple is generally ( always in K–12 mathematics) used only in the context of integers.
Keeping the concept clear: When naming the multiples of a number, children (and adults!) often forget to include the number, itself, and are often unsure whether or not to include 0. The multiples of 3 include 3 times any integer, including 3 × 0 and 3 × 1. So 3 “is a multiple of 3” (though a trivial one) and 5 “is a multiple of 5” (again, trivial). Zero is a multiple of every number so (among other things) it is an even number. When asked for the “smallest” multiple (for example, the least common multiple ), the implication is that only positive multiples are meant. Thus 6 is the “least” common multiple of 3 and 2 even though 0 and –6 (and so on) are also multiples that 3 and 2 have in common, and they are less than 6. Keeping the language clear: It is imprecise to refer to a number as “a multiple” without saying what it is a multiple of, The number 12 is “a multiple of 4” or “a multiple of 6” but not just “a multiple.” (It is not, for example, “a multiple” of 5.) Numbers are multiples of something, not just “multiples.” Also, 6 is a of 12, not a multiple of 12. And 12 is a multiple of 6, not a factor of 6. A fine point: The term multiple —like factor and divisible —is generally used only to refer to results of multiplication by a whole number.
### Is 2 a multiple of 45?
Are the multiples of 45 always odd? Explain. – Odd numbers are those numbers that are not divisible by 2. Even numbers are those numbers that are divisible by 2. Hence, 45 is not divisible by 2 it is an odd number. But, there are infinite odd and even multiples of 45. For example,
• 45×4 = 180
• 45×6 = 270
180 and 270 are even multiples of 45.
• 45×1 = 45
• 45×3 = 135
135 and 225 are odd multiples of 45.
## What percent 50 is 2?
Required Percentage = % = 4%.2 is what percent of 50?
#### How many pairs are in 50?
FAQs on Factors of 50 – List all the factors of 50 1, 2, 5, 10, 25 and 50 are the 50 factors. What are the composite factors of 50? 10, 25 and 50 are the 50 composite factors. Composite factors are those that have more than 2 factors other than 1 and itself.
What are the factors of 50 in pairs? (1, 50), (2, 25) and (5, 10) are the 3 pair of factors. What are the positive factors for 50? 1, 2, 5, 10, 25, and 50 are the positive factors for 50. What are the negative factors of 50? The negative factors of 50 are -1, -2, -5, -10, -25 and -50 We have covered all the necessary information regarding the factors of 50.
We hope that this blog has proven useful and easy to understand.
#### What are the tables of 2?
What is the 2 Times Table Chart?
2 × 1 = 2 2 × 6 = 12
2 × 2 = 4 2 × 7 = 14
2 × 3 = 6 2 × 8 = 16
2 × 4 = 8 2 × 9 = 18
2 × 5 = 10 2 × 10 = 20
### Does 45 come in the table of 2?
45 comes in the table of 1,3,5,8,15 and 45. |
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