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How do you simplify 2 ^ 2 + (13 - 8/4) * 2- (6b - 4) ^ 2?
Jun 6, 2017
$- {\left(6 b - 4\right)}^{2} + 26$ or
$2 \left(- 18 {b}^{2} + 24 b + 5\right)$
Explanation:
Whether you use BODMAS, BIDMAS OR PEDMAS, brackets and parentheses always come first in an expression, followed by exponents and roots, after which comes division and multiplication, and finally addition and subtraction.
For both answers, the first thing to do is to simplify the question. The first thing I do is rearrange the $\cdot 2$ to be in front of the first brackets, $\left(13 - \frac{8}{4}\right)$, so that the expression now looks like this:
${2}^{2} + 2 \left(13 - \frac{8}{4}\right) - {\left(6 b - 4\right)}^{2}$.
The second thing I do is simplify $\frac{8}{4} = 2$, and then solve the first brackets, which becomes $\left(13 - 2\right) = 11$. From this, I can solve the first bit of the expression, like so:
${2}^{2} + 2 \left(11\right) - {\left(6 b - 4\right)}^{2}$
$= 4 + 22 - {\left(6 b - 4\right)}^{2}$
$= 26 - {\left(6 b - 4\right)}^{2}$
For the first answer, I simply rearrange the equation just so:
$- {\left(6 b - 4\right)}^{2} + 26$
and I am done.
Now for the second answer. We have to be careful here because there is a minus sign in front of the second brackets. $- {\left(6 b - 4\right)}^{2}$ does mean to take away what is in the brackets, but when variables are involved and it is not as simple as that, $-$ also means to multiply everything inside the brackets by $- 1$. With that in mind, I expand the brackets like so
$26 - \left(6 b - 4\right) \left(6 b - 4\right) \text{ }$ (since ${\left(a - b\right)}^{2} = \left(a - b\right) \left(a - b\right)$)
and from there I multiply each term to get $26 - \left(36 {b}^{2} - 48 b + 16\right)$.
This is where I multiply each term inside the brackets by $- 1$, so that the signs of each term changes. Doing this I get
$26 - 36 {b}^{2} + 48 b - 16$
From here I simplify by adding $26 + \left(- 16\right)$ to get
$- 36 {b}^{2} + 48 b + 10$.
Although you could stop here, a last step that can be done involves taking out the common factor of 2 from this whole expression, to end up with the final answer of
$2 \left(- 18 {b}^{2} + 24 b + 5\right)$
I hope that helped!
Jun 6, 2017
$- 36 {b}^{2} + 48 b + 10$
Explanation:
In an expression with different operations there is a well-defined order in which they have to be done.
First count how many TERMS there are.
Each arithmetic term will simplify to one value while an algebraic term might simplify to several unlike terms.
Within each term:
do BRACKETS first,
then POWERS AND ROOTS,
then MULTIPLY AND DIVIDE.
The simplified terms are only ADDED ANDSUBTRACTED in the LAST step.
We are working with $3$ terms and will simplify each separately:
Take note of what happens within each term from one line to the next.
$\text{ "color(blue)(2^2)color(green)( +(13-8/4) xx 2)" } \textcolor{red}{- {\left(6 b - 4\right)}^{2}}$
=color(blue)(4)color(green)(" + "(13-2) xx 2)" "color(red)( -(6b-4)(6b-4))
$= \textcolor{b l u e}{4} \textcolor{g r e e n}{\text{ + "(11) xx 2)color(red)(" } - \left(36 {b}^{2} - 24 b - 24 b + 16\right)}$
$= \textcolor{b l u e}{4} \textcolor{g r e e n}{\text{ + "22color(red)(" } - \left(36 {b}^{2} - 48 b + 16\right)}$
$= \textcolor{b l u e}{4} \textcolor{g r e e n}{+ 22 \textcolor{red}{- 36 {b}^{2} + 48 b - 16}} \text{ } \leftarrow$ note that the signs change
Re-arrange the terms with the algebraic terms first, and the subtraction at the end:
$= \textcolor{red}{- 36 {b}^{2} + 48 b} \textcolor{b l u e}{\text{ + "4)color(green)(" + "22)color(red)(" - } 16}$
$= - 36 {b}^{2} + 48 b + 26 - 16$
$= - 36 {b}^{2} + 48 b + 10$
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# What are the critical points of f(x) = sqrt(e^(sqrtx)-sqrtx)?
Feb 25, 2017
$x = 0$
#### Explanation:
$f \left(x\right) = {\left({e}^{{x}^{\frac{1}{2}}} - {x}^{\frac{1}{2}}\right)}^{\frac{1}{2}}$
Through the chain rule:
$f ' \left(x\right) = \frac{1}{2} {\left({e}^{{x}^{\frac{1}{2}}} - {x}^{\frac{1}{2}}\right)}^{- \frac{1}{2}} \frac{d}{\mathrm{dx}} \left({e}^{{x}^{\frac{1}{2}}} - {x}^{\frac{1}{2}}\right)$
Then:
$f ' \left(x\right) = \frac{1}{2} {\left({e}^{{x}^{\frac{1}{2}}} - {x}^{\frac{1}{2}}\right)}^{- \frac{1}{2}} \left({e}^{{x}^{\frac{1}{2}}} \left(\frac{1}{2} {x}^{- \frac{1}{2}}\right) - \frac{1}{2} {x}^{- \frac{1}{2}}\right)$
Factoring from the final parentheses:
$f ' \left(x\right) = \frac{1}{2} {\left({e}^{{x}^{\frac{1}{2}}} - {x}^{\frac{1}{2}}\right)}^{- \frac{1}{2}} \left(\frac{1}{2} {x}^{- \frac{1}{2}}\right) \left({e}^{{x}^{\frac{1}{2}}} - 1\right)$
Rewriting:
$f ' \left(x\right) = \frac{1}{2 {\left({e}^{{x}^{\frac{1}{2}}} - {x}^{\frac{1}{2}}\right)}^{\frac{1}{2}} \left(2 {x}^{\frac{1}{2}}\right)} \left({e}^{{x}^{\frac{1}{2}}} - 1\right)$
$f ' \left(x\right) = \frac{{e}^{\sqrt{x}} - 1}{4 \sqrt{x} \sqrt{{e}^{\sqrt{x}} - \sqrt{x}}}$
If we want to find critical point, we need to find when $f ' \left(x\right) = 0$ or when $f '$ is undefined but $f$ is defined.
Setting $f ' \left(x\right) = 0$ gives ${e}^{\sqrt{x}} - 1 = 0 \implies {e}^{\sqrt{x}} = 1 \implies x = 0$.
This is also when $f '$ is undefined, since $\sqrt{x}$ is in the denominator. We also see that $\sqrt{{e}^{\sqrt{x}} - \sqrt{x}}$ is never undefined, as ${e}^{\sqrt{x}} > \sqrt{x}$ for all $x \ge 0$.
Thus the only critical point is $x = 0$. It's also one of the function's endpoints.
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Home / The science / Mathematics / How fast to count in the mind
# How fast to count in the mind
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Many schoolchildren are appalled at the mere mention of the solution of mathematical examples.
Sometimes the calculations seem so complicated that you can not do without a calculator.
But mathematics is a science, although complex, but natural, and with the help of some mathematical techniques one can learn to produce quite complex mathematical actions in the mind.
Instructions
1
Multiplication of two-digit numbers by 11.
Anyone who knows the multiplication table, for sureRemembers that it is easiest to multiply the number by 10, because no matter how complex the original number was, only zero will be added to its record at the end. However, multiplying by 11 is also very easy! To do this, you need to add both numbers that make up the given number, and assign the first digit to the left, and the second digit to the right.
Example:
31 is the original number.
3 (3 + 1) 1
It turns out 31 * 11 = 341
Do not worry, if you add two digits, you get a two-digit number - just add one to the left digit.
Example:
39 is the original number.
3 (3 + 9) 9
3 + 1 2 9
It turns out that 39 * 11 = 429
2
Multiplication of any number by 4.
One of the most obvious mathematical techniquesIs the multiplication of numbers by 4. To facilitate the work, without multiplying the numbers in the mind, you can multiply the number first by 2 twice in a row, and then add the results.
Example:
745 is the original number.
745 * 2 + 745 * 2 = 2980
Thus, 745 * 4 = 2980
3
Multiplication of any number by 5.
Some people face difficulties when multiplying large numbers by the number 5. In order to quickly multiply the number by 5, you need to split it in half and evaluate the result.
If, as a result of division, an integer is obtained, then it is necessary to assign the number 0 to it.
Example:
1348 is the original number.
1348: 2 = 674 is an integer.
Hence, 1348 * 5 = 6740
If the result is a fractional number, then discard all the digits after the decimal point and assign the number 5.
Example:
5749 is the original number.
5749: 2 = 2874.5 is a fractional number.
Hence, 5749 * 5 = 28745
4
The squaring of a two-digit number ending in the figure 5.
When constructing such a number in a square, it is necessary to square up only its first digit, adding one to it, and add 25 at the end of the number.
Example:
75 is the original number.
7 * (7 + 1) = 56 Assign 25, and get the result: 75 in the square will be equal to 5625.
5
The regrouping method, if one of the numbers is even.
If you need to multiply 2 large numbers and one of them is even, you can simply regroup them.
Example:
32 must be multiplied by 125
32 * 125 = 16 * 250 = 4 * 1000 = 4000
That is, it turns out that 32 * 125 = 4000
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Home | | Maths 8th Std | Rational numbers in Standard form
# Rational numbers in Standard form
Observe the following rational numbers: 4/5, -3/7, 1/6, -4/13, -50/51.
Rational numbers in Standard form
Observe the following rational numbers: . Here, we see that
i) the denominators of these rational numbers are all positive integers
ii) 1 is the only common factor between the numerator and the denominator of each of them and
iii) the negative sign occurs only in the numerator.
Such rational numbers are said to be in standard form.
A rational number is said to be in standard form, if its denominator is a positive integer and both the numerator and denominator have no common factor other than 1.
If a rational number is not in the standard form, then it can be simplified to arrive at the standard form.
The collection of rational numbers is denoted by the letter because it is formed by considering all quotients, except those inv olving division by 0. Decimal numbers can be put in quotient form and hence they are also rational numbers.
Example 1.2
Reduce to the standard form:
Solution:
(i) Method 1:
(dividing by –2,2 and 3 successively)
Method 2:
The HCF of 48 and 84 is 12 (Find it!). Thus, we can get its standard form by dividing it by –12.
(ii) Method 1:
(dividing by –2 and 3 successively)
Method 2:
The HCF of 18 and 42 is 6 (Find it!). Thus, we can get its standard form by dividing it by 6.
Try these
1. Which of the following pairs represents equivalent rational numbers?
(i) −6/4, 18/−12
−6/4 = [−6 × 3] / [4 × 3] = −18 / 12
−6 / 4 = equivalent to −18 / 12
(ii) −4/−20, 1/−5
−4/−20 = [−4 ÷ (−4)] / [−20 ÷ (−4)] = 1/5 ≠ −1/5
−4 / −20 not equivalent to 1 / −5
(iii) −12/−17, 60/85
−12 / −17 = [−12 × −5] / [−17 × −5] = 60 / 85
−12 / −17 equivalent to 60 / 85
2. Find the standard form of:
(i) 36/−96 = [−36 ÷ 12] / [96 ÷ 12] = −3 / 8
(ii) −56/−72 = [−56 ÷ (−8)] / [−72 ÷ (−8)] = 7 / 9
(iii) 27/18 = 1 (9/18) = 1 (1/2)
3. Mark the following rational numbers on a number line.
Solution:
(i) −2 / 3
−2/3 lies between 0 and −1.
The unit part between 0 and −1 is divided into 3 equal parts and second part taken.
(ii) −8 / −5
−8/−5 = 1 (3/5)
1 (3/5) lies between 1 and 2. The unit part between 1 and 2 is divided into 5 equal parts and the third part is taken.
(iii) 5 / −4
5 / −4 = − 5/4 = − 1 ¼
−1 (1/4) lies between −1 and −2. The unit part between −1 and −2 is divided into four equal parts and the first part is taken.
Tags : Numbers | Chapter 1 | 8th Maths , 8th Maths : Chapter 1 : Numbers
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
8th Maths : Chapter 1 : Numbers : Rational numbers in Standard form | Numbers | Chapter 1 | 8th Maths
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# The supergiant star Betelgeuse has a measured angular diameter of 0.044 arcsecond. Its distance has been measured to be 427 light-years. What is the actual diameter of Betelgeuse?
Dec 13, 2015
$861 , 000 , 000 \text{ km}$
#### Explanation:
This is a pretty straight forward trigonometry problem. We can set up a diagram showing that the distance to Betelgeuse and the radius of Betelgeuse make a right angle.
Therefore, we can use the $\sin$ function to find the radius of Betelgeuse. Since $\theta$ is very small, we can use the small angle approximation, $\sin \left(\theta\right) \approx \theta$ if we convert $\theta$ to radians.
$.044 \text{ arc seconds" = 2.13 xx 10^-7 " radians}$
Since $\theta$ is the total diameter of Betelgeuse, we want to use $\sin \left(\theta \text{/} 2\right)$ to calculate the radius.
$r = d \sin \left(\frac{\theta}{2}\right) \approx d \frac{\theta}{2}$
But the radius is $1 \text{/} 2$ of the diameter, $D$, so we have;
$\frac{D}{2} = d \frac{\theta}{2}$
Canceling the $2$s leaves;
$D = d \theta$
Now we have an expression for the diameter, we can plug in what we know.
$D = \left(427 \text{ light years}\right) \left(2.13 \times {10}^{-} 7\right)$
$D = 9.10 \times {10}^{-} 5 \text{ light years}$
Light years are not the most practical units for measuring the diameter of a star, however, so lets convert to $\text{km}$ instead.
$D = \left(9.10 \times {10}^{-} 5 \text{ light years")(9.46 xx 10^12 " km / light year}\right)$
$D = 8.61 \times {10}^{8} \text{ km}$
This is about 600 times the diameter of the sun!
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# How do you solve 3x - 2/7 = (3x)/4 + 4?
Dec 20, 2015
$x = \frac{40}{21}$
#### Explanation:
$3 x - \frac{2}{7} = \frac{3 x}{4} + 4$
Make each term have a denominator of $7$ on the left side and $4$ on the right side.
$\frac{7 \left(3 x\right)}{7} - \frac{2}{7} = \frac{3 x}{4} + \frac{4 \left(4\right)}{4}$
$\frac{21 x}{7} - \frac{2}{7} = \frac{3 x}{4} + \frac{16}{4}$
Subtract the fractions on the left side and add on the right side.
$\frac{21 x - 2}{7} = \frac{3 x + 16}{4}$
Cross multiply.
$7 \left(3 x + 16\right) = 4 \left(21 x - 2\right)$
Multiply.
$21 x + 112 = 84 x - 8$
Isolate for $x$ by bringing all terms with $x$ to the left and all without to the right.
$21 x - 84 x = - 8 - 112$
Solve.
$- 63 x = - 120$
$x = \frac{- 120}{-} 63$
$x = \frac{40}{21}$
Dec 20, 2015
$\textcolor{g r e e n}{x = \frac{40}{21}}$
The solution looks a bit long. This is because I have explained the principles behind the shortcuts usually adopted.
#### Explanation:
$\textcolor{b l u e}{\text{Objective}}$
To have a single $x$ on the left of the equals sign and everything else on the others side.
$\textcolor{b l u e}{\text{Principles used}}$
To end up with something on its own you have to 'remove' from that side the things you do not wish to be there.
For conditions of add or subtract you change what you do not need into the value of 0. This because adding or subtracting 0 has no effect.
For conditions of multiply or divide you change what you do not want into 1. Multiplying or dividing by 1 has now effect.
What you do to one side of an equation you do to the other to maintain the truth od the = sign.
$\textcolor{b l u e}{\text{Application of principle}}$
Given: $\textcolor{w h i t e}{\text{...}} \textcolor{b r o w n}{3 x - \frac{2}{7} = \frac{3 x}{4} + 4. \ldots \ldots \ldots \left(1\right)}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Moving $\frac{3 x}{4}$ from the right to the left: subtract $\textcolor{b l u e}{\frac{3 x}{4}}$ from both sides of equation (1)
$\textcolor{b r o w n}{\left(3 x - \frac{2}{7}\right) \textcolor{b l u e}{- \frac{3 x}{4}} = \left(\frac{3 x}{4} + 4\right) \textcolor{b l u e}{- \frac{3 x}{4}}}$
$\textcolor{w h i t e}{. .}$
color(brown)(3xcolor(blue)(-(3x)/4)-2/7=4+(3x)/4color(blue)(-(3x)/4)$\textcolor{w h i t e}{. .}$
color(brown)(3xcolor(blue)(-(3x)/4)-2/7=4+0..................(2)
$\textcolor{w h i t e}{. .}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
But $\textcolor{b r o w n}{3 x} = \frac{12 x}{4} \textcolor{w h i t e}{. .}$ so $\textcolor{w h i t e}{. .} \textcolor{b r o w n}{3 x} \textcolor{b l u e}{- \frac{3 x}{4}} = \frac{12 x - 3 x}{4} = \textcolor{p u r p \le}{\frac{9 x}{4}}$
So equation (2) becomes:
color(brown)(color(purple)((9x)/4)-2/7=4.....................(3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
By applying the same process to equation (3) we have
$\frac{9 x}{4} = \frac{28 + 2}{7}$
Multiply both sides by $\textcolor{b l u e}{4}$ giving:
color(brown)(color(blue)(4xx)((9x)/4)=color(blue)(4xx)(30/7)
color(brown)(color(blue)(4)/4xx9x=(color(blue)(4)xx30)/7
$9 x = \frac{120}{7}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Divide both sides by $\textcolor{b l u e}{9}$ ( same as multiply by $\textcolor{b l u e}{\frac{1}{9}}$ )
$\textcolor{b r o w n}{\textcolor{b l u e}{\frac{1}{9} \times} 9 x = \textcolor{b l u e}{\frac{1}{9} \times} \frac{120}{7}}$
color(brown)(9/(color(blue)(9)) xx x=120/(color(blue)(9xx)7)
$x = \frac{120}{63}$
But $\frac{120}{63} = \frac{40}{21}$
$\textcolor{g r e e n}{x = \frac{40}{21}}$
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# 9.2 Mole-Mass and Mass-Mass Calculations
## Learning Objectives
By the end of this section, you will be able to:
• From a given number of moles of a substance, calculate the mass of another substance involved using the balanced chemical equation
• From a given mass of a substance, calculate the moles of another substance involved using the balanced chemical equation
• From a given mass of a substance, calculate the mass of another substance involved using the balanced chemical equation
Mole-mole calculations are not the only type of calculations that can be performed using balanced chemical equations. Recall that the molar mass can be determined from a chemical formula and used as a conversion factor. We can add that conversion factor as another step in a calculation to make a mole-mass calculation, where we start with a given number of moles of a substance and calculate the mass of another substance involved in the chemical equation, or vice versa.
For example, suppose we have the balanced chemical equation:
2 Al + 3 Cl2 → 2 AlCl3
Suppose we know we have 123.2 g of Cl2. How can we determine how many moles of AlCl3 we will get when the reaction is complete? First and foremost, chemical equations are not balanced in terms of grams; they are balanced in terms of moles. So to use the balanced chemical equation to relate an amount of Cl2 to an amount of AlCl3, we need to convert the given amount of Cl2 into moles. We know how to do this by simply using the molar mass of Cl2 as a conversion factor. The molar mass of Cl2 (which we get from the atomic mass of Cl from the periodic table) is 70.90 g/mol. We must invert this fraction so that the units cancel properly:
$123.2 \cancel{\text{ g }\ce{Cl2}}\times \dfrac{1\text{ mol }\ce{Cl2}}{70.90\cancel{\text{ g }\ce{Cl2}}}=1.738\text{ mol }\ce{Cl2}$
Now that we have the quantity in moles, we can use the balanced chemical equation to construct a conversion factor that relates the number of moles of Cl2 to the number of moles of AlCl3. The numbers in the conversion factor come from the coefficients in the balanced chemical equation:
$\dfrac{2\text{ mol }\ce{AlCl3}}{3\text{ mol }\ce{Cl2}}$
Using this conversion factor with the molar quantity we calculated above, we get:
$1.738\cancel{\text{ mol }\ce{Cl2}}\times \dfrac{2\text{ mol }\ce{AlCl3}}{3\cancel{\text{ mol }\ce{Cl2}}}=1.159\text{ mol }\ce{AlCl3}$
So, we will get 1.159 mol of AlCl3 if we react 123.2 g of Cl2.
In this last example, we did the calculation in two steps. However, it is mathematically equivalent to perform the two calculations sequentially on one line:
$123.2\cancel{\text{ g }\ce{Cl2}}\times \dfrac{1\cancel{\text{ mol }\ce{Cl2}}}{70.90\cancel{\text{ g }\ce{Cl2}}}\times \dfrac{2\text{ mol }\ce{AlCl3}}{3\cancel{\text{ mol }\ce{Cl2}}}=1.159\text{ mol }\ce{AlCl3}$
The units still cancel appropriately, and we get the same numerical answer in the end. Sometimes the answer may be slightly different from doing it one step at a time because of rounding of the intermediate answers, but the final answers should be effectively the same.
### Example 9.2a
#### Problem
How many moles of HCl will be produced when 249 g of AlCl3 are reacted according to this chemical equation?
2 AlCl3 + 3 H2O(ℓ) → Al2O3 + 6 HCl(g)
#### Solution
We will do this in two steps: convert the mass of AlCl3 to moles and then use the balanced chemical equation to find the number of moles of HCl formed. The molar mass of AlCl3 is 133.33 g/mol, which we have to invert to get the appropriate conversion factor:
$1.87\cancel{\text{ mol }\ce{AlCl3}}\times \dfrac{6\text{ mol }\ce{HCl}}{2\cancel{\text{ mol }\ce{AlCl3}}}=5.61\text{ mol }\ce{HCl}$
Now we can use this quantity to determine the number of moles of HCl that will form. From the balanced chemical equation, we construct a conversion factor between the number of moles of AlCl3 and the number of moles of HCl:
$\dfrac{6\text{ mol }\ce{HCl}}{2\text{ mol }\ce{AlCl3}}$
Applying this conversion factor to the quantity of AlCl3, we get:
$1.87\cancel{\text{ mol }\ce{AlCl3}}\times \dfrac{6\text{ mol }\ce{HCl}}{2\cancel{\text{ mol }\ce{AlCl3}}}=5.61\text{ mol }\ce{HCl}$
Alternatively, we could have done this in one line:
$249\cancel{\text{ g }\ce{AlCl3}}\times \dfrac{1\cancel{\text{ mol }\ce{AlCl3}}}{133.33\cancel{\text{ g }\ce{AlCl3}}} \times \dfrac{6\text{ mol }\ce{HCl}}{2\cancel{\text{ mol }\ce{AlCl3}}} = 5.60\text{ mol }\ce{HCl}$
The last digit in our final answer is slightly different because of rounding differences, but the answer is essentially the same.
### Exercise 9.2a
How many moles of Al2O3 will be produced when 23.9 g of H2O are reacted according to this chemical equation?
2 AlCl3 + 3 H2O(ℓ) → Al2O3 + 6 HCl(g)
#### Check Your Answer[1]
A variation of the mole-mass calculation is to start with an amount in moles and then determine an amount of another substance in grams. The steps are the same but are performed in reverse order.
### Example 9.2b
#### Problem
How many grams of NH3 will be produced when 33.9 mol of H2 are reacted according to this chemical equation?
N2(g) + 3 H2(g) → 2 NH3(g)
#### Solution
The conversions are the same, but they are applied in a different order. Start by using the balanced chemical equation to convert to moles of another substance and then use its molar mass to determine the mass of the final substance. In two steps, we have:
$33.9\cancel{\text{ mol }\ce{H2}}\times \dfrac{2\text{ mol }\ce{NH3}}{3\cancel{\text{ mol }\ce{H2}}}=22.6\text{ mol }\ce{NH3}$
Now, using the molar mass of NH3, which is 17.03 g/mol, we get:
$22.6\cancel{\text{ mol }\ce{NH3}}\times \dfrac{17.03\text{ g }\ce{NH3}}{1\cancel{\text{ mol }\ce{NH3}}}=385\text{ g }\ce{NH3}$
### Exercise 9.2b
How many grams of N2 are needed to produce 2.17 mol of NH3 when reacted according to this chemical equation?
N2(g) + 3 H2(g) → 2 NH3(g)
#### Check Your Answer[2]
It should be a trivial task now to extend the calculations to mass-mass calculations, in which we start with a mass of some substance and end with the mass of another substance in the chemical reaction. For this type of calculation, the molar masses of two different substances must be used – be sure to keep track of which is which. Again, however, it is important to emphasize that before the balanced chemical reaction is used, the mass quantity must first be converted to moles. Then the coefficients of the balanced chemical reaction can be used to convert to moles of another substance, which can then be converted to a mass.
For example, let us determine the number of grams of SO3 that can be produced by the reaction of 45.3 g of SO2 and O2:
2 SO2(g) + O2(g) → 2 SO3(g)
First, we convert the given amount, 45.3 g of SO2, to moles of SO2 using its molar mass (64.06 g/mol):
$45.3\cancel{\text{ g }\ce{SO2}}\times \dfrac{1\text{ mol }\ce{SO2}}{64.06\cancel{\text{ g }\ce{SO2}}}=0.707\text{ mol }\ce{SO2}$
Second, we use the balanced chemical reaction to convert from moles of SO2 to moles of SO3:
$0.707\cancel{\text{ mol }\ce{SO2}}\times \dfrac{2\text{ mol }\ce{SO3}}{2\cancel{\text{ mol }\ce{SO2}}}=0.707\text{ mol }\ce{SO3}$
Finally, we use the molar mass of SO3 (80.06 g/mol) to convert to the mass of SO3:
$0.707\cancel{\text{ mol }\ce{SO3}}\times \dfrac{80.06\text{ g }\ce{SO3}}{1\cancel{\text{ mol }\ce{SO3}}}=56.6\text{ g }\ce{SO3}$
We can also perform all three steps sequentially, writing them on one line as:
$45.3\cancel{\text{ g }\ce{SO2}}\times \dfrac{1\cancel{\text{ mol }\ce{SO2}}}{64.06\cancel{\text{ g }\ce{SO2}}}\times \dfrac{2\cancel{\text{ mol }\ce{SO3}}}{2\cancel{\text{ mol }\ce{SO2}}}\times \dfrac{80.06\text{ g }\ce{SO3}}{1\cancel{\text{ mol }\ce{SO3}}}=56.6\text{ g }\ce{SO3}$
We get the same answer. Note how the initial and all the intermediate units cancel, leaving grams of SO3, which is what we are looking for, as our final answer.
### Example 9.2c
#### Problem
What mass of Mg will be produced when 86.4 g of K are reacted?
MgCl2(s) + 2 K(s) → Mg(s) + 2 KCl(s)
#### Solution
We will simply follow the steps:
mass K → mol K → mol Mg → mass Mg
In addition to the balanced chemical equation, we need the molar masses of K (39.09 g/mol) and Mg (24.31 g/mol). In one line,
$86.4\cancel{\text{ g K}}\times \dfrac{1\cancel{\text{ mol K}}}{39.09\cancel{\text{ g K}}}\times \dfrac{1\cancel{\text{ mol Mg}}}{2\cancel{\text{ mol K}}}\times \dfrac{24.31\text{ g Mg}}{1\cancel{\text{ mol Mg}}}=26.87 \text{ Mg}$
### Exercise 9.2c
What mass of H2 will be produced when 122 g of Zn are reacted?
Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)
### Example 9.2d
#### Relating Masses of Reactants and Products
What mass of sodium hydroxide, NaOH, would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, Mg(OH)2] by the following reaction?
$\text{MgCl}_2(aq) + 2\text{NaOH}(aq) \longrightarrow \text{Mg(OH)}_2(s) + \text{NaCl}(aq)$
#### Solution
The approach used previously in Example 9.2a and Example 9.2b is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps of the sort learned in the previous chapter are required. The calculations required are outlined in this flowchart:
$16 \;\rule[0.5ex]{5em}{0.1ex}\hspace{-5em}\text{g Mg(OH)}_2 \times \frac{1 \;\rule[0.5ex]{6.0em}{0.1ex}\hspace{-6.0em}\text{mol Mg(OH)}_2}{58.3 \;\rule[0.5ex]{5.0em}{0.1ex}\hspace{-5.0em}\text{g Mg(OH)}_2} \times \frac{2 \;\rule[0.5ex]{5em}{0.1ex}\hspace{-5em}\text{mol NaOH}}{1 \;\rule[0.5ex]{6em}{0.1ex}\hspace{-6em}\text{mol Mg(OH)}_2} \times \frac{40.0 \;\text{g NaOH}}{1 \;\rule[0.5ex]{4.5em}{0.1ex}\hspace{-4.5em}\text{mol NaOH}} = \\22 \;\text{g NaOH}$
### Exercise 9.2d
What mass of gallium oxide, Ga2O3, can be prepared from 29.0 g of gallium metal? The equation for the reaction is:
$4 \text{Ga} + 3\text{O}_2 \longrightarrow 2\text{Ga}_2 \text{O}_3$
### Example 9.2e
#### Relating Masses of Reactants
What mass of oxygen gas, O2, from the air is consumed in the combustion of 702 g of octane, C8H18, one of the principal components of gasoline?
$2\text{C}_8 \text{H}_{18} + 25\text{O}_2 \longrightarrow 16\text{CO}_2 + 18\text{H}_2 \text{O}$
#### Solution
The approach required here is the same as for the Example 3, differing only in that the provided and requested masses are both for reactant species.
$702 \;\rule[0.5ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{g C}_8 \text{H}_{18} \times \frac{1 \;\rule[0.25ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{mol C}_8 \text{H}_{18}}{114.23 \;\rule[0.25ex]{2.75em}{0.1ex}\hspace{-2.75em}\text{g C}_8 \text{H}_{18}} \times \frac{25 \;\rule[0.25ex]{2.5em}{0.1ex}\hspace{-2.5em}\text{mol O}_2}{2 \;\rule[0.25ex]{3.5em}{0.1ex}\hspace{-3.5em}\text{mol C}_8 \text{H}_{18}} \times \frac{32.00 \;\text{g O}_2}{\rule[0.25ex]{2.5em}{0.1ex}\hspace{-2.5em}\text{mol O}_2} = \\2.46 \times 10^3 \;\text{g O}_2$
### Exercise 9.2e
What mass of CO is required to react with 25.13 g of Fe2O3 according to the equation:
$\text{Fe}_2 \text{O}_3 + 3\text{CO} \longrightarrow 2\text{Fe} + 3\text{CO}_2$
#### Check Your Answer[5]
These examples illustrate just a few instances of reaction stoichiometry calculations. Numerous variations on the beginning and ending computational steps are possible depending upon what particular quantities are provided and sought (volumes, solution concentrations, and so forth). Regardless of the details, all these calculations share a common essential component: the use of stoichiometric factors derived from balanced chemical equations. Figure 9.2a provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.
### Airbags
Airbags (Figure 9.2b) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN3. When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN3 to initiate its decomposition:
$2 \text{NaN}_3(s) \longrightarrow 3\text{N}_2(g) + 2\text{Na}(s)$
This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03–0.1 s). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass (~100 g) of NaN3 will generate approximately 50 L of N2.
## Attribution & References
Except where otherwise noted, this page is adapted by Adrienne Richards from:
1. 0.442 mol
2. 30.4 g (Note: here we go from a product to a reactant, showing that mole-mass problems can begin and end with any substance in the chemical equation.)
3. 3.77 g
4. 39.0 g Ga2O3
5. 13
definition
## License
9.2 Mole-Mass and Mass-Mass Calculations Copyright © 2023 by Gregory Anderson; Caryn Fahey; Jackie MacDonald; Adrienne Richards; Samantha Sullivan Sauer; J.R. van Haarlem; and David Wegman is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.
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# Operations with Rational Exponents
## Simplify expressions with negative and fractional exponents.
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Applying the Laws of Exponents to Rational Exponents
The period (in seconds) of a pendulum with a length of L (in meters) is given by the formula \begin{align*}P = 2\pi{(\frac{L}{9.8})}^{\frac{1}{2}}\end{align*}. If the length of a pendulum is \begin{align*}9.8^{\frac{8}{3}}\end{align*}, what is its period?
### Applying the Laws of Exponents to Rational Exponents
When simplifying expressions with rational exponents, all the laws of exponents that were learned previously are still valid. On top of that, the rules of fractions apply as well.
Let's simplify the following.
1. \begin{align*}x^{\frac{1}{2}} \cdot x^{\frac{3}{4}}\end{align*}
Recall from the Product Property of Exponents, that when two numbers with the same base are multiplied we add the exponents. Here, the exponents do not have the same base, so we need to find a common denominator and then add the numerators.
\begin{align*}x^{\frac{1}{2}}\cdot x^{\frac{3}{4}}=x^{\frac{2}{4}}\cdot x^{\frac{3}{4}}=x^{\frac{5}{4}}\end{align*}
This rational exponent does not reduce, so we are done.
1. \begin{align*}\frac{4x^{\frac{2}{3}}y^4}{16x^3y^{\frac{5}{6}}}\end{align*}
This problem utilizes the Quotient Property of Exponents. Subtract the exponents with the same base and reduce \begin{align*}\frac{4}{16}\end{align*}.
\begin{align*}\frac{4x^{\frac{2}{3}}y^4}{16x^3y^{\frac{5}{6}}}=\frac{1}{4}x^{{\left(\frac{2}{3}\right)}-3}y^{\frac{4-5}{6}}=\frac{1}{4}x^{\frac{-7}{3}}y^{\frac{19}{6}}\end{align*}
If you are writing your answer in terms of positive exponents, your answer would be \begin{align*}\frac{y^{\frac{19}{6}}}{4x^{\frac{7}{3}}}\end{align*}. Notice, that when a rational exponent is improper we do not change it to a mixed number.
If we were to write the answer using roots, then we would take out the whole numbers. For example, \begin{align*}y= \frac{19}{6}\end{align*} can be written as \begin{align*}y^{\frac{19}{6}}=y^3y^{\frac{1}{6}}=y^3\sqrt[6]{y}\end{align*} because 6 goes into 19, 3 times with a remainder of 1.
1. \begin{align*}\frac{\left(2x^{\frac{1}{2}}y^6\right)^{\frac{2}{3}}}{4x^{\frac{5}{4}}y^{\frac{9}{4}}}\end{align*}
On the numerator, the entire expression is raised to the \begin{align*}\frac{2}{3}\end{align*} power. Distribute this power to everything inside the parenthesis. Then, use the Powers Property of Exponents and rewrite 4 as \begin{align*}2^2\end{align*}.
\begin{align*}\frac{\left(2x^{\frac{1}{2}}y^6\right)^{\frac{2}{3}}}{4x^{\frac{5}{4}}y^{\frac{9}{4}}}=\frac{2^{\frac{2}{3}}x^{\frac{1}{3}}y^4}{2^2x^{\frac{5}{4}}y^{\frac{9}{4}}}\end{align*}
Combine like terms by subtracting the exponents.
\begin{align*}\frac{2^{\frac{2}{3}}x^{\frac{1}{3}}y^4}{2^2x^{\frac{5}{4}}y^{\frac{9}{4}}} = 2^{\left(\frac{2}{3}\right)-2}x^{\left(\frac{1}{3}\right)-\left(\frac{5}{4}\right)}y^{4-\left(\frac{9}{4}\right)}=2^{\frac{-4}{3}}x^{\frac{-11}{12}}y^{\frac{7}{4}}\end{align*}
Finally, rewrite the answer with positive exponents by moving the 2 and \begin{align*}x\end{align*} into the denominator. \begin{align*}\frac{y^{\frac{7}{4}}}{2^{\frac{4}{3}}x^{\frac{11}{12}}}\end{align*}
### Examples
#### Example 1
Earlier, you were asked to find the period of the pendulum.
Substitute \begin{align*}9.8^{\frac{8}{3}}\end{align*} for L and solve.
\begin{align*}P = 2\pi{(\frac{L}{9.8})}^{\frac{1}{2}}\\ P = 2\pi{(\frac{9.8^{\frac{8}{3}}}{9.8})}^{\frac{1}{2}}\\ P = 2\pi{(\frac{9.8^{\frac{8}{3}}}{9.8^{\frac{3}{3}}})^{\frac{1}{2}}}\\ P = 2\pi{(9.8^{\frac{5}{3}})^{\frac{1}{2}}}\\ P = 2\pi{(9.8)^{\frac{5}{6}}}\end{align*}
Therefore, the period of the pendulum is \begin{align*} P = 2\pi{(9.8)^{\frac{5}{6}}}\end{align*}.
Simplify each expression below. Reduce all rational exponents and write final answers using positive exponents.
#### Example 2
\begin{align*}4d^{\frac{3}{5}} \cdot 8^{\frac{1}{3}}d^{\frac{2}{5}}\end{align*}
Change 4 and 8 so that they are powers of 2 and then add exponents with the same base.
\begin{align*}4d^{\frac{3}{5}} \cdot 8^{\frac{1}{3}}d^{\frac{2}{5}}=2^2 d^{\frac{3}{5}} \cdot \left(2^3\right)^{\frac{1}{3}}d^{\frac{2}{5}}=2^3 d^{\frac{5}{5}}=8d\end{align*}
#### Example 3
\begin{align*}\frac{w^{\frac{7}{4}}}{w^{\frac{1}{2}}}\end{align*}
Subtract the exponents. Change the \begin{align*}\frac{1}{2}\end{align*} power to \begin{align*}\frac{2}{4}\end{align*}.
\begin{align*}\frac{w^{\frac{7}{4}}}{w^\frac{1}{2}}= \frac{w^{\frac{7}{4}}}{w^{\frac{2}{4}}}=w^{\frac{5}{4}}\end{align*}
#### Example 4
\begin{align*}\left(3^{\frac{3}{2}}x^4 y^{\frac{6}{5}}\right)^{\frac{4}{3}}\end{align*}
Distribute the \begin{align*}\frac{4}{3}\end{align*} power to everything inside the parenthesis and reduce.
\begin{align*}\left(3^{\frac{3}{2}}x^4 y^{\frac{6}{5}}\right)^{\frac{4}{3}}=3^{\frac{12}{6}}x^{\frac{16}{3}}y^{\frac{24}{15}}=3^2 x^{\frac{16}{3}}y^{\frac{8}{5}}=9x^{\frac{16}{3}}y^{\frac{8}{5}}\end{align*}
### Review
Simplify each expression. Reduce all rational exponents and write final answer using positive exponents.
1. \begin{align*}\frac{1}{5}a^{\frac{4}{5}}25^{\frac{3}{2}}a^{\frac{3}{5}}\end{align*}
2. \begin{align*}7b^{\frac{4}{3}}49^{\frac{1}{2}}b^{-\frac{2}{3}}\end{align*}
3. \begin{align*}\frac{m^{\frac{8}{9}}}{m^{\frac{2}{3}}}\end{align*}
4. \begin{align*}\frac{x^{\frac{4}{7}}y^{\frac{11}{6}}}{x^{\frac{1}{14}}y^{\frac{5}{3}}}\end{align*}
5. \begin{align*}\frac{8^{\frac{5}{3}}r^5 s^{\frac{3}{4}}t^{\frac{1}{3}}}{2^4 r^{\frac{21}{5}}s^2 t^{\frac{7}{9}}}\end{align*}
6. \begin{align*}\left(a^{\frac{3}{2}}b^{\frac{4}{5}}\right)^{\frac{10}{3}}\end{align*}
7. \begin{align*}\left(5x^{\frac{5}{7}}y^4\right)^{\frac{3}{2}}\end{align*}
8. \begin{align*}\left(\frac{4x^{\frac{2}{5}}}{9y^{\frac{4}{5}}}\right)^{\frac{5}{2}}\end{align*}
9. \begin{align*}\left(\frac{75d^{\frac{18}{5}}}{3d^{\frac{3}{5}}}\right)^{\frac{5}{2}}\end{align*}
10. \begin{align*}\left(\frac{81^{\frac{3}{2}}a^3}{8a^{\frac{9}{2}}}\right)^{\frac{1}{3}}\end{align*}
11. \begin{align*}27^{\frac{2}{3}}m^{\frac{4}{5}}n^{-\frac{3}{2}}4^{\frac{1}{2}}m^{-\frac{2}{3}}n^{\frac{8}{5}}\end{align*}
12. \begin{align*}\left(\frac{3x^{\frac{3}{8}}y^{\frac{2}{5}}}{5x^{\frac{1}{4}}y^{-\frac{3}{10}}}\right)^2\end{align*}
To see the Review answers, open this PDF file and look for section 7.3.
### Notes/Highlights Having trouble? Report an issue.
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Due to system maintenance, CK-12 will be unavailable on 8/19/2016 from 6:00p.m to 10:00p.m. PT.
# 11.16: Coordinate Locations on a Map
Difficulty Level: At Grade Created by: CK-12
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Practice Coordinate Locations on a Map
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Have you ever created a map?
Kevin and his pen pal Charlotte are both creating maps of their neighborhoods to show each other what it looks like where they live. Kevin has decided to name the most important things on his map. He has decided to include his house, his school, the skate park and the library. Since Kevin lives close to each of these things, he is sure that he can draw them on a map.
Kevin has decided to use a coordinate grid to show each location. He wants to send Charlotte a key that will match each location with its accurate coordinates.
Here is Kevin's grid.
Given this map, which coordinates should Kevin use to name each location?
Pay close attention to this Concept and you will learn how to write coordinates to name locations.
### Guidance
When we graphed geometric figures, we used integer coordinates to find the location of each point. Then we graphed each point according to its location. Maps also use integer coordinates to identify different locations. If you look at a map, you will see some numbers and sometimes letters around the border of the map. This can assist you in figuring out the location of cities or even different locations.
Some maps use integers to identify different locations. Let’s look at a map that does this. Here we have used a coordinate grid to identify where different places are in a town. Let’s look at this map.
We can say that Kara’s house is blue, Mark’s house is pink and Chase’s house is green. Each house has coordinates. We can say that the center of each house marks its coordinates on the map.
Kara’s house is at (-3, 1)
Mark’s house is at (3, -2)
Chase’s house is at (3, 4)
Local maps use letters and numbers to identify locations. World maps use degrees written in latitude and longitude. Let’s learn about this real life use of coordinates.
Longitude is the measure of lines vertically on a map.
Latitude is the measure of lines horizontally on a map.
We can measure longitude and latitude using degrees. These degrees are written as ordered pairs.
Here you can see degrees of latitude as horizontal measures. The degrees of longitude are the vertical measures.
We can identify different locations on a map if we have the coordinates of the location. Notice that the degrees of latitude are written first, those are the horizontal degrees, and the degrees of longitude are written second. Those are the vertical degrees.
Practice working in degrees. Identify the states according to their locations in latitude and longitude.
#### Example A
30,83\begin{align*}30^\circ, 83^\circ\end{align*}
Solution:Texas
#### Example B
42,100\begin{align*}42^\circ, 100^\circ\end{align*}
Solution: South Dakota
#### Example C
30,100\begin{align*}30^\circ, 100^\circ\end{align*}
Solution: Texas
Now back to the map.
Here is the original problem once again. Reread the problem and then use what you have learned to write the coordinates to match Kevin’s map.
Kevin and his pen pal Charlotte are both creating maps of their neighborhoods to show each other what it looks like where they live. Kevin has decided to name the most important things on his map. He has decided to include his house, his school, the skate park and the library. Since Kevin lives close to each of these things, he is sure that he can draw them on a map.
Kevin has decided to use a coordinate grid to show each location. He wants to send Charlotte a key that will match each location with its accurate coordinates.
Here is the grid that Kevin starts off with.
Given this map, which coordinates should Kevin use to name each location?
Now that you have finished this Concept, let’s work on writing coordinates to match Kevin’s map.
First, let’s start with his home. His house is located at (4, 5).
His school is located close to his home at (4, 2)
The library is located at (-1, 3).
Finally, the skate park is the farthest away from his home at (0, -3).
Kevin is ready to send his map and coordinates to Charlotte. He can’t wait to see her map.
### Vocabulary
Here are the vocabulary words in this Concept.
the four sections of a coordinate grid
Origin
the place where the x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} axis’ meet at (0, 0)
Ordered Pair
the x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} values used to locate points on a coordinate grid (x,y)\begin{align*}(x,y)\end{align*}
x\begin{align*}x\end{align*} axis
the horizontal axis on the coordinate grid
y\begin{align*}y\end{align*} axis
the vertical axis on the coordinate grid
Coordinates
the x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} values of an ordered pair
Longitude
vertical measure of degrees on a map
Latitude
horizontal measure of degrees on a map
### Guided Practice
Which state is at 45,70\begin{align*}45^\circ, 70^\circ\end{align*}?
To answer this question, we start with the horizontal degrees, the latitude. That says 45\begin{align*}45^\circ\end{align*}. We start at 45 and then move to 70 degrees.
You can see that we are at the state of Maine.
As long as you have values on a map, you can use coordinates to identify any location.
### Video Review
Here is a video for review.
http://video.about.com/geography/Latitude-and-Longitude.htm - This is a video from about.com on latitude and longitude.
### Practice
Directions: Identify each place on the map according to latitude and longitude.
1. What is at 35,70\begin{align*}35^\circ, 70^\circ\end{align*}?
2. What is at 30,90\begin{align*}30^\circ, 90^\circ\end{align*}?
3. What is at 85,70\begin{align*}85^\circ, 70^\circ\end{align*}?
4. What is at 55,90\begin{align*}55^\circ, 90^\circ\end{align*}?
5. What is at 20,40\begin{align*}20^\circ, 40^\circ\end{align*}?
6. What is at 40,80\begin{align*}40^\circ, 80^\circ\end{align*}?
7. What is at 85,100\begin{align*}85^\circ, 100^\circ\end{align*}?
8. What is at 95,80\begin{align*}95^\circ, 80^\circ\end{align*}?
9. What is at 95,100\begin{align*}95^\circ, 100^\circ\end{align*}?
10. What is at 60,30\begin{align*}60^\circ, 30^\circ\end{align*}?
11. What is at 75,30\begin{align*}75^\circ, 30^\circ\end{align*}?
12. What is at 45,40\begin{align*}45^\circ, 40^\circ\end{align*}?
13. What is at 45,45\begin{align*}45^\circ, 45^\circ\end{align*}?
14. What is at 50,70\begin{align*}50^\circ, 70^\circ\end{align*}?
15. What is at 80,50\begin{align*}80^\circ, 50^\circ\end{align*}?
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
$x-$axis
The $x-$axis is the horizontal axis in the coordinate plane, commonly representing the value of the input or independent variable.
$y$ axis
The $y$-axis is the vertical number line of the Cartesian plane.
Latitude
Latitude is a coordinate that specifies the north-south location of a point on the Earth's surface.
Longitude
Longitude is a coordinate that specifies the east-west location of a point on the Earth's surface.
Ordered Pair
An ordered pair, $(x, y)$, describes the location of a point on a coordinate grid.
Origin
The origin is the point of intersection of the $x$ and $y$ axes on the Cartesian plane. The coordinates of the origin are (0, 0).
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.
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# Storybook Guide Based on Robie H. Harris’s “Crash! Boom! A Math Tale”
Topic:
## Activity Summary
Use this storybook guide with the book “Crash! Boom! A Math Tale.”
Many public libraries have this book.
Elephant is playing with blocks and decides to build a tower as tall as itself. Elephant makes a stack of one, two, three, four blocks with their longer side facing up. It’s the same height as Elephant until CRASH! BOOM! it all comes down! Elephant tries again and places each block so that its shorter side faces up. Now Elephant needs eight blocks to make a tower “as tall as me.” But CRASH! BOOM! it all comes down again! Find out how Elephant uses different sized blocks to make two more stacks as tall as it is.
#### Words to Learn
NUMBER SYMBOLS AND WORDS
1, 2, 3, 4, 5, 6, 7, 8, more
SPATIAL WORDS
up, down, flat
MEASUREMENT WORDS
tall, short, as tall as, shorter than
Crash! Boom! has ideas about counting, measurement, and spatial relations. While reading the story, your child can learn how:
• To count the number of blocks as we add more to find out how many we have in total. For example, Elephant starts with two blocks and then adds two more. Elephant counts one, two, three, four to find out that now there are four blocks.
• To think about measurement. To make a tall tower, Elephant first arranges the blocks so they are stacked with their taller side facing up, and then makes a tower with their shorter side facing up. But both towers are the same height.
• Spatial and measurement words can mean the opposite of one another: For example, Elephant builds a tower up, but when it crashes it comes down.
• GUESS HOW MANY BLOCKS ELEPHANT WILL USE
If Elephant wants to build a tower as tall as itself, how many blocks like these do you think Elephant will use? Let’s keep reading to see if our guess is right.
• NOTICE WHAT HAPPENS TO THE TOTAL NUMBER OF BLOCKS WHEN WE ADD MORE
Elephant started with two blocks, then added two more. Let’s count to find out how many blocks Elephant has now.
• ON THE LAST TA-DAH PAGE SPREAD
How can one longer block be the same height as eight shorter blocks?
• DISCUSS SPATIAL AND MEASUREMENT WORDS THAT ARE OPPOSITE
Up is the opposite of down and tall is the opposite of short. Can you think of two more words that are opposites?
Try to come up with some of your own questions and comments, too!
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# The maxima for the function is
## Introduction
In mathematics, maxima and minima (the plural of maximum and minimum) are points of a function at which the function takes its largest value (maximum) or smallest value (minimum), respectively. In the case of a local maximum or minimum, this value is also the largest value or smallest value in a neighboring interval. Maxima and minima are therefore local extrema of functions.
## What is a function?
A function is a mathematical relation between two sets, usually denoted by an equation. The first set is called the domain and the second set is called the range. The function assigns a unique output to every input. In other words, for every element in the domain, there is a corresponding element in the range.
## What is a maxima?
In mathematics, the maxima (the plural of maximum) are the highest points of a function. The maxima of a function can be found by taking the derivative and setting it equal to zero. The points where the derivative is zero are called the critical points. The critical points can be maxima, minima, or inflection points. To determine which type ofcritical point it is, you must take the second derivative and evaluating it at the critical point. If the second derivative is positive, it is a maximum; if negative, it is a minimum; and if zero, it is an inflection point.
## How to find the maxima for a function
To find the maxima for a function, you need to take the derivative of the function and set it equal to zero. The maxima will be the points where the derivative is equal to zero.
## Examples
To find the maxima for the function
-Take the derivative of the function.
-Set the derivative equal to zero and solve for x.
-Plug in your x value into the original function to see if it is a maximum or minimum point. If it is a maximum point, it will be positive and if it is a minimum point, it will be negative.
For example, let’s say you have the function f(x)=-2x^2+5x+3 and you want to find its maxima.
To do this, you would take the derivative of the function which would give you f'(x)=-4x+5. Then, you would set this equal to zero and solve for x which would give you x=1/4. You would then plug this value into the original function to get f(1/4)=11/8 which is a maximum point.
## Conclusion
To find the maxima for the function, we take the derivative and set it equal to zero. This gives us the equation:
0 = -2x^3 + 12x^2 – 20x + 8
We can use the Quadratic Formula to solve this equation and find that the maxima for the function is at x = 2.
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# Decoding Mathematical Symbols: Understanding the Meaning Behind the Signs
**Short answer in math what does this symbol mean:**
In mathematics, symbols are used to represent numerical quantities, mathematical operations or relations. The meaning of a specific symbol depends on the context it is used. Common mathematical symbols include + (addition), – (subtraction), × (multiplication) and ÷(division). Other commonly used symbols are ≤(less than or equal to), ≥(greater than or equal to), ≠(not equal to) and ≈(approximately equal to).
## How to Decode Mathematical Symbols: A Guide
Mathematics is a beautiful, yet complex subject that can sometimes feel daunting and confusing. But fear not! If you learn how to decode the various mathematical symbols used in equations and formulas, you’ll soon find yourself falling in love with this subject.
In this guide, we’ll walk you through some of the most common mathematical symbols and what they mean. So sit back, grab your calculator (if needed), and let’s get started!
1. ‘=’
The equal sign (=) is perhaps the most basic symbol in all of mathematics. It simply means “is equal to” or “equals.” For example, 2 + 2 = 4 means that two plus two equals four.
2. ‘+’, ‘-‘, ‘x’, ‘÷’
These are the four basic arithmetic operations: addition (+), subtraction (-), multiplication (x), and division (÷). They are used to manipulate numbers in calculations. For example, if we have two numbers x and y, then x + y means adding x and y together; x – y means subtracting y from x; x ÷ y means dividing x by y; while x multiplied by y (*y can be written as xy) denotes their product.
3. ”
The less than () sign indicates larger ones. These signs help us compare two values quickly without having to write out which number is lesser or greater explicitly everytime.
For instance:
*10 7 — twenty is greater than seven.
4. ‘^’
The caret (^) symbol signifies an exponential function where it raises a number (‘a’) by another number(‘n’). This expression has an equivalent meaning like continuously multiplying of a for n times at once such as pow(a,n).
For example:
*5^4 — reads as “five to the power of four” and is equivalent to five multiplied by itself four times, which equals 625.
5. ‘√’
The square root (√) symbol denotes the radical function and it calculates the inverse operation of exponentiation or finding a number whose square yields given value n.
Take this example:
* √16 means what number multiplied by itself gives you sixteen: that would be four because 4×4=16
6. ‘!’
The exclamation mark (!), known colloquially as factorial sign, represents how many ways we can order distinct objects in a given set while maintaining relative placement within each ordered arrangement.
For instance,
*3! (read three factorial) equates to six, as there are six possible arrangements for three items orderly arranged in different ways i.e.,
{1,2,3}, {1,3,2}, {2,1,3}, {2 ,3 ,1} ,{3 , 1 ,2} and lastly {3 ,
## Step-by-Step Explanation of Common Math Symbols and Their Meaning
Mathematics is a language that transcends geographical boundaries, cultural differences and linguistic barriers. It is often said that math communicates directly with the universe, but to do so effectively, it requires an intricate vocabulary made up of a multitude of symbols. From arithmetic operations like addition (+), subtraction (-), multiplication (×) and division (/) to more advanced concepts such as integrals (∫) or sigma notation (Σ), these strange shapes carry a deep meaning within them.
As you start delving into the world of mathematics, these symbols might seem alien at first; however, once deciphered they become invaluable tools for communication through equations. To help get to grips with some common mathematical symbols and their meanings, we’ll go through each step-by-step in this blog post.
The ubiquitous plus sign may look simple in its design by combining two diagonal lines together but don’t let its simplicity fool you. This symbol represents the operation of adding numbers or quantities together – whether they are positive or negative values.
Subtraction (-)
When subtracting one number from another also known as ‘minus’, use the horizontal dash – appropriately called “the minus” – which divides two distinct numerical entities when placed between them like 7 – 3 = 4. The symbol also doubles up for other forms such as derivative calculus.
Multiplication (×)
To multiply various numbers simply use either a letter x (“times”), asterisk (*) or dot (.). As expected multiplication tells us how many times larger we have scaled our initial number(s).
Division (/)
Just like multiplication “division,” involves multiple different ways to denote value ratios: ÷ / : \$div\$ so any will suffice between two variable sets A/B(or C/D). Division never ends though since fractions represent decimals too except with differing levels of precision depending on rounding rules!
Exponents (^)
This important function largely determines values raised to certain powers denoted next to the symbol ‘x’ for example 2^3 although its suggestive of “squared”, it actually denotes raising a number to any power (e.g., x^n) and thus taking x multiplied by itself n times, helping us check error margins in our calculations.
Equals (=)
The equal sign indicates that two mathematical expressions are equivalent or interchangeable with one another. Used throughout almost all else areas like geometry to calculus and beyond.
Inequality Symbols
> Greater than, < smaller than or = is not greater/smaller.
This trio of operators helps compare quantities based on their values within certain boundaries.
Function Notation (f(x))
Functions expresses how changing variables results in different outputs – denoted as f(x), this operator integrates initial parameters entered into an equation to reveal output when applied under specific circumstances marked by parentheses inside brackets [ ]. Effectively demonstrating complex outcomes via simple input data
Derivatives(d/dt)
Calculus par excellence! Derivatives demonstrate function movement over time – d/dx: means ‘the derivative with respect to x’, referring specifically what happens when
Mathematics has always been an enigma for many students, particularly when it comes to understanding the symbols used in this subject. While some of these symbols are relatively self-explanatory and easy to grasp, others might require a bit more effort before they can be fully comprehended.
To help clarify the air around this topic, we’ve come up with this article that presents several frequently asked questions about understanding symbols in math.
1. What is the ‘equal’ sign?
The equal sign (=) is one of the most basic symbols often used in mathematics. It serves as an indication that two sides or expressions have equivalent values even if they appear differently from each other.
2. What does ‘greater than’ & ‘less than’ mean?
The greater than symbol (>) indicates a value larger than another number while the less than symbol (<) represents a value smaller compared to another number. An essential rule-of-thumb approach towards distinguishing between both of them would be by their orientation on numerical lines: The arrowhead usually points towards small numbers whilst pointing away from higher numbers.
3. What is meant by ‘Pi’?
Pi(π) is yet another symbol peculiar to mathematicians out there featuring prominently within geometry and trigonometry circles involving circle ratios containing radii dimensions along its circumference divided into Pi whose result manifests itself as 3.14/22 over seven or infinite decimal places like 0x4f9…..
4.What does "set" and "subset" mean?
Set (∈), depicts selecting varying arrays/lines of figures organized together through brackets example {7,6,,8}. In contrast, Subset ⊆ portrays picking particular groups possessing one or multiple items having similar inscriptions examples thereof include (-4,-5-7)- {-5}{+∞ ≤ – five}
Therefore,/Subset utilizes fewer elements compared to Set such that everything could belong uniquely inside corresponding sets as well subsets
5.What is the ‘integral’ symbol?
The integral ∫ similar to Pi is another notable icon of mathematics which plays a significant role in calculus/concepts. It represents the addition or displacement on differential rates commonly depicted as Sigma( + ). Symbolization here can often include adding up infinitesimal widths' over various lengths that imply curve insights within applicable sections.
6.How do brackets work?
Brackets conform to different formulas containing numerous orders involving calculations, helping match equations and pinning down order of operations such exemplified by BOMDAS/PEMDAS shorthand representing bracket breaks/posts/power/multiplication/ division/addition/subtraction applied effectively throughout an impending equation consisting of multiple digits towards determining accurate outcomes ergo minimizing errors continually.
In summary understanding symbols in math requires attention, focus, practice repeated efforts for comprehending what each symbol implies during mathematical expressions solving otherwise it could lead individuals astray affecting their academic scores adversely if left unchecked.
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# Probability Problems
Probability is an important branch of mathematics which basically deals with the phenomena of chances or randomness of events. Probability of an event measures how likely it is expected to happen. An event is said to be "most likely" if it has higher probability and "least likely" if it has lower probability. The probability of an event varies between 0 and 1. An event with probability 0 is an impossible event, while event having probability 1 is said to be a certain event.
The formula for probability of an event P(E) is given below:
Few sample problems based on probability theory are given below:
Problem 1: A coin is tossed in air. What is the probability of getting 2 tails.
Solution: When a coin is tossed, the possible outcomes are:
• Tail - Tail
Here, 2 tails appear only once. Therefore number of favorable outcomes = 1
Total number of outcomes = 4
We have,
$P(E)=$$\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes} Probability of getting 2 tails = \frac{1}{4} = 0.25 Problem 2: A pair of dice is rolled. What is the probability of obtaining a sum of 6. Solution: When a pair of dice is rolled, the total 36 outcomes are possible which are listed below: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) Out of which (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) have total of 6. We have, P(E)=$$\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$
Probability of getting a sum of 6 = $\frac{5}{36}$ = 0.14
Problem 3: A card is drawn from the deck of well shuffled cards at random. Find the probability of drawing an ace.
Solution: Total number of cards in a deck = 52
Total number of aces = 4
We have,
$P(E)=$$\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes} Probability of drawing an ace = \frac{4}{52} = 0.08. Problem 4: A bag contains 3 black, 5 red, 4 blue and 6 green marbles. One marble is drawn at random. What is the probability of NOT drawing a green ball. Solution: Let us find the probability of drawing a green ball. Total number of green balls = 6 Total balls = 3 + 5 + 4 + 6 = 18 We have, P(E)=$$\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$
Probability of drawing a green ball = $\frac{6}{18}$ = $\frac{1}{3}$
We know that $P(E)=1-P(Not\ E)$
Probability of NOT drawing a green ball = 1- $\frac{1}{3}$
= $\frac{2}{3}$
Related Calculators Calculation of Probability Binomial Distribution Probability Calculator Binomial Probability Calculator Coin Toss Probability Calculator
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1 DERIVATIVES: RULES Part 1: Derivatives of Polynomial Functions We can use the definition of the derivative in order to generalize solutions and develop ... Read More
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DERIVATIVES: RULESPart 1: Derivatives of Polynomial FunctionsWe can use the definition of the derivative in order to generalize solutions and develop rules to find derivatives. Thesimplest derivatives to find are those of polynomial functions.Example 1: Find the derivative of the constant function f(x) = c using the definition of derivative.Solution: lim f (x + ∆x) − f (x) = lim c−c = 0 ∆x ∆x ∆x→0 ∆x→0So, the derivative of a constant is 0. This corresponds to the graphing of derivatives we did earlier. The graph of aconstant function is a horizontal line and the slope of a horizontal line is 0. f(x) = c f’(x) = 0Symbolically, we write Constant Rule: If f(x) = c, then f ’(x) = 0.Example 2: Find the derivative of each of the following functions based on their functions. The function and itsderivative are pictured in each diagram. A table of values for f and f ’ is also shown. Study the values carefully.The function is given at the right.Complete the list of derivatives f f’ f(x) = x f ’ (x) = ________ f f(fx(x) )==xx2 f’ (x) = ________ f’ 1
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f’ f(x) = x3 f’ (x) = ________ff f(x) = x 4 f’ (x) = ________ f’A pattern is emerging when we take the derivative of a power. The exponent becomes the coefficient of thederivative and the power of the derivative is one less than the power of the function. This is called the power ruleand symbolically it is written as follows. Power Rule: If f(x) = xn, then f ’ (x) = nx n-1.We noted above that the derivative of f (x) = x2 is 2x. What is the derivative of 5x2 ? 16x2 ? 1023x2 ? Isthere a rule that governs multiplying a function by a constant? It is possible to answer this question by once againgoing back to the definition of the derivative. The question we are asking is “what is the derivative of kx2 ”? lim f (x + ∆x) − f (x) = lim k ( x + ∆x)2 − kx2 ∆x ∆x ∆x→0 ∆x→0 = lim kx2 + 2kx∆x + k ∆x 2 − kx2 ∆x ∆x→0 = lim (2kx − k∆x) ∆x→0 = 2kx = k(2x)So, the derivative of a function multiplied by a constant is the constant multiplied by the derivative. Or, Constant Multiple Rule: The derivative of kf(x) is kf ’(x).What is the derivative of a sum or difference of several powers; i.e., what is the derivative of a polynomial? Sum Rule: The derivative of a sum f(x) + g(x) is the sum of the derivatives, f ’ (x) + g ’ (x).Likewise, the derivative of a difference is the difference of the derivatives. Difference Rule: The derivative of a difference f(x) - g(x) is the difference of the derivatives, f ’ (x) - g ’ (x).With these few simple rules, we can now find the derivative of any polynomial. 2
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Example 2: Find the derivative of f (x) = 5x6 − 3x4 + 2x3 − 8x2 +17 Solution: Find the derivative of each term of the polynomial using the constant multiple rule and power rules. Then, add or subtract the derivative of each term, as appropriate. The derivative of 5x6 is 5(6x5 ) = 30x5 . The derivative of −3x4 is −3(4x3 ) = −12x3 . The derivative of 2x3 is 2(3x2 ) = 6x2 . The derivative of −8x2 is −8(2x) = 16x . The derivative of 17 is 0. So, the derivative of f (x) = 5x6 − 3x4 + 2x3 − 8x2 +17 is f '(x) = 30x5 −12x3 + 6x2 −16x .Example 3: Find the slope of the tangent line to the curve f (x) = 3x2 − 4x + 3 at x = 1. Solution: Using the derivative rules, f '(x) = 6x − 4 . At x = 1, the derivative is 6(1) - 4 = 2. This can also be written as f ’ (1) = 2.Homework Exercises Part 1: Derivatives of Polynomial FunctionsFind the derivative of each of the following. 5. f (x) = x6 − 8x1. f (x) = 4x5 − 3x2 6. f (x) = 1 x4 − 1 x2 + 2x − 4 2 42. f (x) = x3 + x2 − x +1 7. f (x) = x6 + x5 + x4 + x3 + x2 + x +13. f (x) = 0.1x10 − 0.5x5 − 0.3x3 − 0.25 8. f (x) = (x + 7)(2x − 3)4. f (x) = (3x − 2)2Find the indicated derivative.9. f '(1) if f (x) = 3x4 − 5x + 2 11. f '(2) if f (x) = 3x5 − 2x2 + 4 12. f '(−3) if f (x) = 2x4 + 4x2 + 610. f ' 1 if f (x) = 5x3 − 3x2 + 2x − 6 2Write the equation of the tangent line at the indicated point.13. f (x) = x2 at (2, 4) 15. f (x) = 4 − x2 at (-1, 3)14. f (x) = (x −1)2 +1 at (1, 1) 16. f (x) = 2 − (x −1)2 at (-2, -7)17. Use the definition of derivative to prove that the derivative of a linear function is a constant; i.e., prove that if f (x) = ax + b then f '(x) = a , using lim f (x + ∆x) − f (x) . ∆x ∆x→0 3
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18. Use the definition of derivative to prove that the derivative of a quadratic function is a linear function; i.e., prove that if f (x) = ax2 + bx + c , then f '(x) = 2ax + b , using lim f (x + ∆x) − f (x) . ∆x ∆x→019. Find all points on the graph of y = x3 − x2 where the tangent line is horizontal.20. Find all points on the graph of y = 1 x3 + x2 − 3x + 1 where the tangent line has slope 1. 321. The height x in feet of a ball above the ground at t seconds is given by the equation x = −16t2 + 40t +100 a. What is the instantaneous velocity at t = 2? b. When is the instantaneous velocity equal to 0?22. There are two tangent lines to the curve y = 4x − x2 that go through the point (2, 5). Find the equations of both of them.23. Suppose f '(0) = 2 , and g '(0) = −1, find a. The derivative of f (0) + g(0) b. The derivative of 2 f (0) − 3g(0)Part 2: Derivatives of Negative and Fractional PowersIn the last set of exercises, you proved that the derivative of a linear function is a constant function; the derivative ofa quadratic function is a linear function; the derivative of a cubic function is a quadratic function. It is possible tocontinue these proofs for any n. We generalized these derivatives by using the power rule and constant, sum anddifference rules. Thus far, we have used the power rule for positive integer powers only. What of negative andfractional powers?Example 1: Find the derivative of f (x) = x−2 = 1 x2 Solution: Substituting into the definition, we get 1 1 f (x + ∆x) − f (x) (x + ∆x)2 − x2 ∆x lim = lim ∆x ∆x→0 ∆x→0 (x2 + 1 ∆x2 ) − 1 2x∆x + x2 = lim ∆x ∆x→0 x2(x2 + x2 + ∆x2 ) − x2 + 2x∆x + ∆x2 ) 2x∆x x2 (x2 + 2x∆x + ∆x2 = lim ∆x ∆x→0 = lim x2 − (x2 + 2x∆x + ∆x2 ) ∆x x2 (x2 + 2x∆x + ∆x2 ) ∆x→0 = lim ∆x −2x∆x − ∆x2 ∆x2 ) x2 (x2 + 2x∆x + ∆x→0 = lim −2x − ∆x ∆x2 ) = −2x = −2 = −2 x −3 x2 (x2 + 2x∆x + x4 x3 ∆x→0 4
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So, the derivative of f (x) = x−2 is f '(x) = −2x−3 . Therefore, the power rule applies to negative exponents.Study the graphs of y = x−2 and y = −2x−3 below. Is x−2 differentiable at every point?From the graphs, we see that x−2 is not differentiable at x = 0.Is it possible to apply the power rule to fractional exponents? Consider the function f (x) = x . 1Example 2: Find the derivative of f (x) = x = x 2 Solution: Substituting into the definition, we get lim f (x + ∆x) − f (x) = lim x + ∆x − x= ∆x ∆x ∆x→0 ∆x→0 = lim x + ∆x − x x + ∆x + x ∆x→0 ∆x x + ∆x + x = lim ∆x( x + ∆x − x x) x + ∆x + ∆x→0 = lim ∆x( ∆x x) x + ∆x + ∆x→0 = lim 1 = 1 = 1 x− 1 ∆x→0 x + ∆x + 2x 2 2 xSo, the derivative of f (x) = x or 1 is f '( x) = 1 x − 1 . Again, the power rule applies. 2 2 f (x) = x2Homework Exercises Part 2: Derivatives of Negative and Fractional PowersFind the derivatives of each of the following functions.1. f (x) = 1 4. f (x) = 3 x x3 12. f (x) = x2 − x2 5. f (x) = 3x2 − 4 x − 43. f (x) = 5 − 2 + x− 1 6. f (x) = 3x5 − 4x3 − 2x2 − x −1 x3 x2 x x 5
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Find the indicated derivative.7. f '(1) if f (x) = 3 − 5 +π 9. f '(32) if f (x) = 35 x − 2 x + 4 x4 x 10. f '(16) if f (x) = 2 4 x + 4x2 + 68. f '12 if f (x) = 5 − 3 + 2 −6 x3 x2 xWrite the equation of the tangent line at the indicated point.11. f (x) = x at (4, 2) 12. f ( x) = x + 2 at (1, 3) x13. If f (x) = 13 − 8x + 2x2 and f '(c) = 4, find c.Part 3: Derivatives of the Sine and Cosine FunctionsLook at the graph of f (x) = sin x and its derivative. What is the derivative of sin x ? If f(x) = sin x, then f ’ (x) = cos x.Do the same for f (x) = cos x . What is the derivative of cos x ?Be careful! The temptation is to say that the derivative of f (x) = cos x is sin x , but note that it is f '(x) = − sin x . If f(x) = cos x, then f ’ (x) = -sin x.Example 1: Find the derivative of f (x) = 2 sin x − 4 cos x . Solution: The derivative of 2 sin x is 2 cos x The derivative of −4 cos x is −4(− sin x) = 4 sin x . So, the derivative of the function is f '(x) = 2 cos x + 4 sin x .Example 2: Find the equation of the tangent line to the graph of y = 3sin x at x = π 3 6
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Solution: The derivative of f (x) = 3sin x is f '(x) = 3cos x . The value of the derivative at x = π is 3 cos π =3 1 = 3 3 3 2 2 The y value at x = π is 3sin π =3 3 = 33 3 3 2 2 So, the tangent line is y − 3 3 = 3 x − π 2 2 3Homework Exercises Part 3: Derivatives of the Sine and Cosine FunctionsFind the derivative of each of the following functions.1. f (x) = 3sin x + 6 cos x 4. f (x) = 2x − 2 cos x 5. f (x) = cos x − x2. f (x) = 1 − sin x x 6. f (x) = sin2 x + cos2 x 1 23. f (x) = 2 sin x − x37. Find the equation of the tangent line to y = sin x at x = 1.8. At time t seconds, the center of a bobbing buoy is 1 sin t meters above or below water level. What is the 2 velocity of the cork at t = 0, /2 and ?9. A weight is hanging from a spring. It is compressed 5 cm above its rest position (x = 0) and released at time t = 0 seconds to bob up and down. Its position at any later time t is x = 5 cos t . What is its velocity at time t? You can simulate the up and down motion of the spring y using parametric equations. Graph simultaneously <<π Use the TRACE key on your calculator to explore the position and velocity functions. If you change to , you can see the up and down motion of the spring.10. A normal line is perpendicular to the tangent line. Find the equations for the lines that are tangent and normal to the curve y = 2 cos x at the point π ,1 . 4 7
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Part 4: Derivatives of the Natural Exponential and Logarithmic FunctionsLook at the graph of f (x) = ex and its derivative. What is the derivative of ex ? If f(x) = ex, then f ’ (x) = ex. The natural exponential function is the only function that has itself as its derivative.Do the same for f (x) = ln x . What is the derivative of ln x ?If f(x) = ln x, then f ’ (x) = 1/x, x > 0.It is possible to use the laws of logarithms to aid in finding derivatives of natural logarithmic functions.Example 1: Find the derivative of f (x) = ln x2 .Solution: We know that log ab = b log a . So, ln x2 = 2 ln x (when x > 0 ). Now, we can apply the constant multiple rule and find the derivative of f (x) : f '(x) = 2 1 = 2 . x xExample 2: ( )Find the derivative of f (x) = ln x2 exSolution: Again, apply a log rule: ln(ab) = ln a + ln b. ( )So, ln x2 ex = ln x2 + ln ex = 2 ln x + x ln e. What is ln e ? Recall that the natural logarithmic function and natural exponential functions are inverses of each other. Therefore, ln e = 1 . So, 2 ln x + x ln e = 2 ln x + x . Now, that we have simplified the function, finding the derivative is simple. f '(x) = 2 1 +1= 2 +1. x x 8
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Homework Exercises Part 4: Derivatives of the Natural Exponential & Logarithmic Fns.Find the derivatives of each of the following.1. f (x) = 5x2 + 4ex 6. f (x) = x3 − ln x2. f (x) = (ln 2)x2 − (ln 3)ex 7. f (x) = π ex + x 8. f (x) = 1 − 3ln x − 3ex3. f (x) = ln x − 1 x 5 x24. f (x) = ln x3 ( )9. f (x) = ln 3x4ex5. f (x) = ln 2x 3ex10. Find the equation of the tangent line to the graph of the function at the point (1, 0). 3 b. f (x) = ln x2 a. f (x) = ln x 211. Consider the function f (x) = 1− ex . a. Find the slope of f(x) at the point where it crosses the x-axis. b. Find the equation of the tangent line to the curve at this point. c. Find the equation of the normal line to the curve at this point.12. Find the equation of the tangent line to the function f (x) = 3x2 − ln x at the point (1, 3).Part 5: DifferentiabilityWhen we examined functions with negative exponents as well as the natural exponential and logarithmic functions,we found that there were certain values where we could not find the derivative of a function. There are actuallyseveral situations that destroy differentiability at a point. In order to determine whether a function is differentiableat a given point, we will again return to the definition of the derivative.Example 1: Consider the function f (x) = x −1 . Find the derivative of this function at x = 1. Solution: Substituting into the definition, we get lim f (x + ∆x) − f (x) = lim 1+ ∆x −1 − 1−1 = ∆x ∆x→0 ∆x ∆x→0 lim ∆x ∆x ∆x→0 The limit does not exist, because the limit from the left and from the right differ: lim ∆x =1 but lim ∆x = −1 ∆x ∆x ∆x→0+ ∆x→0− Therefore, the derivative does not exist at x = 1. Or, in other words, the function f (x) = x −1 is not differentiable at x = 1. 9
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Examine the graph of the function at x = 1. The graph comes to a sharp point at x = 1. Now, study the graph of the derivative as given by the calculator. It supports the results found by using the definition of the derivative. Now, use the command on your calculator to find the derivative of the function at x = 1. Obviously, the calculator is giving an incorrect answer. Why is that? Recall that your calculator uses the symmetric difference quotient when computing the derivative. Essentially, it is averaging the right and left hand limits as it approaches the given value. In this case, the left and right hand limits are -1 and 1, and when averaged they give 0. No matter how small a ∆x you choose, you will always get left and right hand limits of -1 and 1 and you will get an averaged value of 0. The calculator will always give an incorrect answer when the graph comes to a sharp point, so it is important to examine the graph of a function when using the calculator to find the numeric derivative. Note, the derivative of f(x) does exist at other points. You can find f '(0) . It is -1. And, f '(10) = 1.Example 2: Consider the function f (x) = 3 x −1. Find the derivative of this function at x = 0.Solution: Using the power rule, we know f '(x) = 1 x − 2 = 1 = 1 3 3 33 x2 2 3x 3 Obviously, when we try to evaluate this at x = 0, we get an undefined value. What does this mean? 10
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Study the graph of the function. At x = 0, the graph becomes very steep. When you examine the graph of the derivative you find an asymptote at x = 0: What does your calculator tell you? There are two values displayed below, one for a ∆x of 0.0000001 and one for a ∆x of 0.00000000001. Note that the values are getting very large; they are approaching infinity. This means that the function has a vertical tangent line at the point x = 0, a tangent line with an undefined slope. Again, you cannot trust your calculator to find the correct derivative for you at this particular value. It is possible to find the derivative of this function at any other value on your calculator or by using the power rule.Example 3: Consider the piecewise function f (x) = x2 − 2x +1 if x < 1 and find the derivative of the function at x = 1. 2x − x2 if x ≥1Solution: f '(x) = 2x − 2 if x < 1 and f '(x) = 2 − 2x if x 1. For both of these, if we substitute x = 1, we get a value of 0. This implies that there is a horizontal tangent line for the function at x = 1. But does this make sense? Again, examine the graph of the function. There is a discontinuity at x = 1. It doesn’t make sense that there is a tangent line at a point of discontinuity. 11
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These three examples illustrate three ways that differentiability can be destroyed: 1. When the graph of a function comes to a sharp point. There is no tangent line at this point. 2. When the derivative of a function is undefined at a particular value and the graph of the function becomes so steep that it appears almost vertical itself. The tangent line is a vertical line with an undefined slope. 3. When the graph of a function is not continuous at a value. There is no tangent line at this point.These situations are illustrated below. a bc dAt points a and b, the function is discontinuous and therefore not differentiable. There are no tangent lines to thecurve at these points. At point c, the function is continuous but not differentiable since the curve comes to a sharppoint. There is no tangent line at this point. At point d, the function is again continuous, but not differentiable.However, there is a vertical tangent line at this point.Homework Exercises Part 5: DifferentiabilityExamine the graphs of each of the following functions. Indicate the points at which the functions are notdifferentiable, and give the reason this occurs.1. f (x) = 1− x 5. f (x) = 2x x −12. f (x) = 5 x4 6. f (x) = 3 ( x − 3)23. f (x) = 1 7. f (x) = 4 x x4. f (x) = x2 − 4 8. f (x) = x2 4 x2 −9. Determine whether the functions below are differentiable at x = 2. a. f (x) = x2 +1 if x ≤ 2 b. g(x) = 1 x + 1 if x<2 4x −3 if x > 2 2 2 x if x ≥ 2 12
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10. Consider the pictured function below: a. Where on the interval −1 < x < 7 does the limit of the function fail to exist? b. Where on the interval −1 < x < 7 does the function fail to be continuous? c. Where on the interval −1 < x < 7 does the function fail to be differentiable? d. Where on the interval −1 < x < 7 is f '(x) = 0 ?Determine if the following statements are true or false.11. If f '(x) = g '(x) for all x, then f (x) = g(x) for all x.12. If f (x) = π 4 , then f '(x) = 4π 3 .13. If f '(c) exists, then f is continuous at c.14. The graph of y = 3 x has a tangent line at x = 0 but the derivative of the function does not exist there.15. The derivative of a polynomial is a polynomial.16. If a function is continuous at a point, then it is differentiable at that point.17. If a function has derivatives from both the right- and the left-sides of a point, then it is differentiable at that point.Part 6: Other Notations and Higher Order DerivativesThe f '(x) notation is one of the most common notations for derivatives, but there are others: dy , Dx f , d ( f ) dx dxThese three notations use “d” or “D” to indicate taking the derivative of a function named y or f: the derivative ofy with respect to x or the derivative of f with respect to x. In these cases, y or f is the dependent variable; x is theindependent variable. d( )a. ( )b. Dx ax2 + bx + cExample 1: Find each of the indicated derivatives: dt 2t3 − sin tSolutions: For part a, you are being asked to find the derivative of the given function with respect to the ( )independent variable t: d dt 2t3 − sin t = 6t2 − cos t 13
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For part b, you are being asked to find the derivative of the given function with respect to the independent variable x. This indicates that a, b, and c are constants. ( )Dx ax2 + bx + c = 2ax + bThere are several applications for higher order derivatives; i.e., derivatives of derivatives. We will examine theseapplications in the next chapter. It is important to learn the notation and practice the process before learning theapplications. The derivative of the first derivative is the second derivative. It can be denoted byf ''(x), Dx2 y or d2y . A summary of notations is given below. dx2 Derivative f notation D notation dy notation first f '(x) Dx y dx f ''(x) Dx2 y second f '''(x) Dx3 y dy third f iv (x) Dx4 y fourth f v (x) Dx5 y dx fifth f (n) (x) Dxn y nth d2y dx2 d3y dx3 d4y dx4 d5y dx5 dny dxnExample 2: Find the first five derivatives of f (x) = sin x Solution: f '(x) = cos x f ''(x) = − sin x f '''(x) = − cos x f iv (x) = sin x f v (x) = cos xExample 3: Find the first four derivatives of y = ln x Solution: dy = 1 d3y = 2 dx x dx3 x3 d2y = − 1 d4y = − 6 dx2 x2 dx4 x4 14
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One type of problem that we have studied throughout the last two units has been position and velocity. We learnedthat the change in position (x) with respect to time (t) yielded velocity; i.e., velocity = dx . The change in velocity dt(v) with respect to time yields acceleration; i.e., acceleration = dv = d2x dt dt 2Example 4: An object moves along a path so that its position x is defined by the functionx(t) = 2t3 − 24t + 8 , where x is measured in centimeters and t in seconds (t > 0). Determine the velocitywhen t = 1 and t = 6. When is the velocity 0? Determine the acceleration at t = 1. Solution: Velocity = dx = 6t 2 − 24 . So, at t = 1, the velocity is -18 cm/sec. At t = 6, the dt velocity is 192 cm/sec. The velocity is 0 when 6t 2 − 24 = 0 . Solving this equation, we get 6t 2 − 24 = 6(t2 − 4) = 6(t − 2)(t + 2) t = ±2 . Since t > 0, the velocity is 0 at t = 2 seconds. Acceleration = dv = 12t . At t = 1, the acceleration is 12 cm/sec2. dtHomework Exercises Part 6: Other Notations and Higher Order DerivativesFind the first three derivatives for each of the following functions.1. y = x3 − 3x2 + 6x − 5 5. f (x) = x2. y = cos x 6. f (x) = 1 x3. y = ex ( )7. y = ln x24. f (x) = 1 8. f (x) = sin x + cos x x29. Without using any formulas, find the following derivatives. Explain your reasoning. ( )a. Dx5 x4 − 3x2 + 5x − 7 b. Dx5 (sin x) ( )c. Dx9 ex10. Suppose f (x) = ax2 + bx + c and f (1) = 5, f '(1) = 3, f ''(1) = −4. Find a, b, and c.11. The position of an object moving along a coordinate line is given by x(t) = 2 − 2 sin t . Find the velocity and acceleration of the object at time t = π . 412. The position of an object thrown upward on the moon is given by the function x(t) = −2.7t 2 + 27t + 6 where x is measured in feet and t is measured in seconds. a. Find expressions for the velocity and acceleration of the object. b. Find the time when the object is at its highest point by finding the time when the velocity is 0. What is the height at this time? 15
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13. The graphs of f(x), f'(x) and f''(x) are shown. Choose the letter (a, b, or c) that represents each. bc aTrue/False14. If y = (x −1)(x + 2)(x + 5)(x − 7) then d5y =0. dx515. The second derivative represents the rate of change of the first derivative.16. If the velocity of an object is constant, then its acceleration is 0.Part 7: Other Differentiation RulesConsider the function, y = x sin x . This is a product of two functions for which we know the derivatives. We alsoknow that the derivative of the sum of two functions is the sum of the derivatives. Is this true of the product of twoderivatives? If it was true that the derivative of the product of two functions is the product of the derivatives of thetwo functions, then the derivative of x sin x should be (1)(cos x) . We can check this graphically with thecalculator. The function and its derivative are pictured below. Y1 Y2It is obvious that the graph of the derivative of x sin x is not cos x . Therefore, the derivative of the product of twofunctions is NOT the product of the derivatives of the two functions.What of the derivative of the quotient of two functions? Consider the function y = ex . If the derivative of the xquotient of two functions is the quotient of the derivatives of the two functions, then the derivative should be orex = ex1 Y1 Y2This is not the graph of ex . Therefore, the derivative of the quotient of two functions is NOT the quotient of thederivatives of the two functions. 16
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What of the derivative of the function of a function (the composite of two functions)? Consider the function( ) ( )y = sin x2 +1 Is the derivative of this function simply cos x2 +1 ? Y1( )Y2 y = cos x2 +1Again, the rule is not so simple. The derivative of the composite of two functions is NOT simply the derivative ofthe principal function composed with the given function. A special rule called the “chain rule” is needed in thiscase.Chain Rule: If f is differentiable at x, g is differentiable at f(x), and h(x) = g(f(x)), then h isdifferentiable at x and h ’ (x) = g ’ (f(x)) . f ’ (x).For the example above, if we let f (x) = x2 +1 and g(x) = sin x , then rule,h(x) = g( f (x)), f '(x) = 2x, and g '(x) = cos x . Therefore, according to the( )h '(x) = cos( f (x)) 2x = cos(x2 +1) 2x = 2x cos x2 +1 .The graph of this function is pictured below.( )Note that it is the same function pictured in Y2 above, the derivative of y = sin x2 +1So, in order to apply the Chain Rule, a composite function must be decomposed into its component parts, thederivatives of each of these component parts must be determined and the Rule applied. 17
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Example 1: Find the derivative of h(x) = 4 − x2 Solution: 1 Let f (x) = 4 − x2 and g(x) = x = x 2 . Thus, f '(x) = −2x and g '( x) = 1 x− 1 . 2 2 Using the rule, we get ( )h 1 − 1 1 − 1 (−2x) = −x '(x) = 2 ( f )( x) 2 (−2x) = 2 4−x 2 4 − x2Example 2: Find the derivative of h(x) = e2x Solution: Let f (x) = 2x and g(x) = ex . Thus, f '(x) = 2 and g '(x) = ex . Using the rule, we get h '(x) = e f (x) (2) = e2x (2) = 2e2xThe two rules yet to be learned are the product and quotient rules; and, the chain rule must be more completelyexplored. In combination with the rules already learned, they will allow you to find the derivative of any function.The product and quotient rules will be studied next year in calculus, as well as combinations of all the rules in muchmore complicated functions.Homework Exercises Part 7: Other Differentiation RulesFor problems 1 - 6, find h '(x). . 4. h(x) = (cos x)21. h(x) = (2x + )1 10 ( )5. h(x) = sin x2( )2. h(x) = ln x3 −1 6. h(x) = 1 x2 + 43. h(x) = esin xFor problems 7 - 16, determine if the derivatives rules from this chapter apply. If they do, find the derivative. Ifthey do not, indicate why.7. y = 3x −1 12. y = sin2 x8. y = 1 −π 13. y = 1 x x2 2x3 −9. y = 2x 14. y = e4 x10. y = xx 15. y = (ln 2)ex11. y = ex4 16. y = π x + ex 18
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True/False17. If y= 1 then dy = −3 . 20. If f (x) = 2x , then f '(x) = 1. 3x dx x2 2x18. If y = π 2 , then dy = 2π . 21. If y = 1+ x , then dy = 2 1 x . dx dx 1+19. If y = x , then dy = 1 . π dx πOptional Exercises:22. Using the function f (x) = x sin x , verify the product rule: ( fg ) '= f 'g + fg '.23. Using the function f (x) = ex , verify the quotient rule: f ' f 'g − fg ' x g g2 =Homework Exercises ReviewFind the derivative of each of the following: 7. f (x) = sin x − 2 cos x − 1 x1. f (x) = 4x3 − 3x22. f (x) = 2x4 −1 8. f (x) = 3ex − 2 ln x x2 13. f (x) = x− x 9. f (x) = π −π 2 x34. f (x) = 2 10. f (x) = (tan x)(cos x) 3x25. f (x) = (x −1)(x2 − 2x + 3) 11. f (x) = x x26. f (x) = ln (2x3 )(2ex ) 12. f (x) = x2 +1 xFind the second derivative for each of the following:13. f (x) = x2 − 2 15. f (x) = 2 sin x − 3cos x x ( )16. f (x) = ln x414. f (x) = (2x −1)(3x + 4)17. Find the equation of the normal line and the tangent line to the curve y = ln x at x = 2.18. Sketch the graph of f (x) = 4 − x − 2 . a. Is f continuous at x = 2? Why or why not? b. Is f differentiable at x = 2? Why or why not? 19
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19. Sketch the graph of f (x) = x2 + 4x + 2 if x < −2 1− 4x − x2 if x ≥ −2 a. Is f continuous at x = -2? Why or why not? b. Is f differentiable at x = -2? Why or why not?20. Let f be the real-valued functioned defined by f (x) = x a. Give the domain and range of f. b. Determine the slope of the line tangent to the graph of f at x = 4. c. Determine the y-intercept of the line tangent to the graph of f at x = 4. d. Give the coordinates of the point on the graph of f where the tangent line is parallel to y = x − 2 .21. AP Calculus Problem: 1978 AB 1 Given f (x) = x3 − x2 − 4x + 4 . The point (a, b) is on f(x) and a tangent line passes through (a, b) and (0, -8), which is not on the graph of f(x). Find a and b.True/False22. If a function is continuous, then it is differentiable.23. If f (x) = g(x) + c , then f '(x) = g '(x) .24. If f(x) is an nth degree polynomial, then f (n) (x) = 0 .25. If f(x) is an nth degree polynomial, then f (n+1) (x) = 026. The acceleration of an object can be negative.27. If f (x) = 1 , then f (4) (x) = 24 . x x528. If f(x) is differentiable at x = c, then f'(x) is differentiable at x = c.29. If a graph of a function has a tangent line at a point, then it is differentiable at that point.30. If y= f (x) g(x) , then dy = f '( x) g '(x) . dx 20
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# Conditional Probability
## P(B|A) = P(A and B)/ P(A)
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Intersection of Compound Events
#### Objective
In this lesson, you will learn about calculating the probability that all of a series of independent events will occur in a single experiment.
#### Concept
There is a classic example of probability studies involving a coin flip. Everyone knows that the probability of getting heads on a single flip is 50%, which means that every time you flip it, there is also a 50% probability of getting tails. The question is, if you have flipped a coin 99 times and got heads every time, what is the probability of getting heads the next time?
After this lesson on the intersection of compound events, we’ll return to this question and see how it does (and doesn’t!) fit with the concept.
#### Watch This
http://youtu.be/xSc4oLA9e8o Khan Academy – Compound Probability of Independent Events
#### Guidance
It should make sense intuitively that the more specific or restricted you make the details of an event, the less probable it becomes for that event to occur. The concept of calculating the total probability of multiple events strung together is the same idea.
If I flip a coin once there are only two possible outcomes:
$H \ (heads) \ \text{or} \ T \ (tails)$
If I flip the coin twice, there are four possibilities:
$H+T \ \text{or} \ H+H \ \text{or} \ T+H \ \text{or} \ T+T$
We know there are a total of four possible outcomes from two coin flips: $HT$ , $HH$ , $TH$ , and $TT$ , and only one of them: $HH$ , results in the outcome we want to calculate. Using the simple probability formula, we get:
$P(HH)=\frac{1 \ outcome}{4 \ possible \ outcomes}=\frac{1}{4} \ or \ 25 \%$
Example A
What is the probability of flipping a coin four times and getting tails all four times?
Solution: Create a table listing all of the possible outcomes:
Now we can look at the bottom row and see that there are a total of 16 possibilities, only one of which is four tails in a row. The probability, therefore, is:
$P(4 \ tails)=\frac{1 \ outcome}{16 \ outcomes}=\frac{1}{16}=6.25\%$
Example B
What is the probability of rolling two even numbers in a row on a standard six-sided die?
Solution: Create a table listing all possible outcomes:
$P(two \ evens)=\frac{9 \ favorable \ outcomes}{36 \ total \ outcomes}=\frac{9}{36}=\frac{1}{4}$
Reducing to:
$P(two \ evens)=\frac{1}{4} \ or \ 25\%$
Example C
What is the probability of spinning two 2’s in a row OR two 4’s in a row on a spinner with the numbers 1-4?
Solution: Create a table listing all possible outcomes, and highlight the favorable ones:
Out of a total of 16 possible outcomes, only 2 fit our description, which gives us:
$P(2's \ or \ 4's )=\frac{2 \ \text{favorable} \ \text{outcomes}}{16 \ \text{possible} \ \text{outcomes}}=\frac{2}{16}=\frac{1}{8} \ or \ 12.5\%$
Concept Problem Revisited
The question is, if you have flipped a coin 99 times and got heads every time, what is the probability of getting heads the next time?
This is a very common example of something called the gambler’s fallacy. It is not a good example of calculating the intersection of compound events because of the way it is worded . The question as written is essentially asking about a single flip of the coin, which is always $\frac{50}{50}$ , because a coin has no memory.
From the standpoint of an example of what we have been studying in this chapter, the more useful, and dramatically more difficult question would be:
What is the probability of flipping a coin 100 times and getting heads every time?
If you want to know the probability of flipping 100 heads in a row, you could either draw a really long chart of all of the possibilities (like the one in Example A, but much longer), or you could use the multiplication rule that we will be learning in the next lesson. Check it out!
#### Vocabulary
An independent event is an event whose outcome is not directly affected by another event. (a coin flip, for example)
A favorable outcome is an outcome of an event that meets a set of initial specifications.
Two mutually exclusive events cannot both occur at the same time, (e.g. both heads and tails on the same coin flip).
#### Guided Practice
1. What is the probability of pulling 1 red marble, replacing it, then pulling another red marble out of a bag containing 4 red and 2 white marbles?
2. What is the probability of a spinner landing on “2” and then a “3”, or “6” if there are 6 equally spaced points on the spinner?
3. What is the probability of pulling a red and then a black card at random from a standard deck (replacing the first card after drawing)?
4. What probability of picking a red and then a green marble from a bag with 5 red and 1 green marbles in it (replacing the first marble after the draw)?
5. What is the probability of shaking the hand of a student wearing red and then a student wearing blue if you randomly shake the hands of two people in a row in a room containing 3 students in blue and 2 in red?
Solutions:
1. Make a chart: $& \underline{\text{first pull:} \qquad \quad r \qquad \qquad r \qquad \quad \ r \qquad \quad \ \ r \qquad \quad \ w \qquad \quad w \quad \ \ }\\& \text{second pull:} \ {\color{red}r r r r}w w \ \ {\color{red}r r r r} w w \ \ {\color{red}r r r r}w w \ \ {\color{red}r r r r}w w \ \ r r r r w w \ \ r r r r w w$
The four sets of four red “ $r$ ’s” represent the favorable outcomes out of the total of 36, therefore $P(2 \ red)=\frac{16}{36}=\frac{4}{9} \ or \ 44.4\overline 4 \%$
2. Make a chart: $& \underline{\text{first spin:} \qquad \qquad 1 \qquad \qquad \quad 2 \qquad\qquad \ 3 \qquad \qquad \ 4 \qquad \qquad \quad 5 \qquad \qquad \quad 6 \qquad }\\& \text{second spin:} \ 1 \ 2 \ 3 \ 4 \ 5 \ 6 \ \ 1 \ 2 \ {\color{red}3} \ 4 \ 5 \ {\color{red}6} \ \ 1 \ 2 \ 3 \ 4 \ 5 \ 6 \ \ 1 \ 2 \ 3 \ 4 \ 5 \ 6 \ \ 1 \ 2 \ 3 \ 4 \ 5 \ 6 \ \ 1 \ 2 \ 3 \ 4 \ 5 \ 6$
The red numbers 3 and 6 represent the two favorable outcomes out of 36 total, therefore $P(2 \ and \ (3 \ or \ 6))=\frac{2}{36} \ or \ \frac{1}{18} \ or \ 5.5\overline{5} \%$
3. There are 26 black and 26 red cards in the deck, so the probability on the first pull is $P(red)=\frac{26 \ \text{red cards}}{52 \ \text{total cards}}=\frac{26}{52}=\frac{1}{2} \ or \ 50 \%$ On the second pull, we again have a 50% chance of favorable outcome, but that 50% only applies to the half of the first pulls that were favorable. Therefore: $P(\text{red} \ then \ \text{black})=50\% \ of \ 50\%=25\%$
4. Make a chart: $& \underline{\text{first pull:} \qquad \quad r \quad \qquad \ r \qquad \quad r \qquad \quad r \qquad \quad \ r \qquad \quad g \quad \ }\\& \text{second pull:} \ \ r r r r r {\color{red}g} \ \ r r r r r {\color{red}g} \ \ r r r r r {\color{red}g} \ \ r r r r r {\color{red}g} \ \ \ r r r r r {\color{red}g} \ \ r r r r r g$ Of the 36 possible outcomes, only 5 fit the description of red the first time, and green the second time (noted by the red “ $g$ ’s”. Therefore $P(\text{red} \ then \ \text{green})=\frac{5}{36}$
5. Make a chart:
So, out of the 25 possible handshake possibilities, 6 of them fit the requirements of red first, then blue:
$P(\text{red} \ then \ \text{blue})=\frac{6}{25}$
#### Practice
Questions 1-6: Suppose you have an opaque bag filled with 4 red and 3 green balls. Assume that each time a ball is pulled from the bag, it is random, and the ball is replaced before another pull.
1. Create a chart of all possible outcomes of an experiment consisting of pulling one ball from the bag at random, noting the color and replacing it, then pulling another.
2. How many possible outcomes are there?
3. What is the probability of randomly pulling a red ball from the bag, returning it, and pulling agreen ball on your second pull?
4. What is the probability of randomly pulling a red ball both times?
5. What is the probability of pulling a green ball both times?
6. Is the probability of pulling a red followed by a green different than pulling a green followed by a red?
Questions 7 – 12: Suppose you have two standard dice, one red and one blue.
7. Construct a probability distribution table or diagram for an experiment consisting of one roll of the red die followed by one roll of the blue one.
8. How many possible outcomes are there?
9. Is there an apparent mathematical relationship between the number of sides on the dice and the number of possible outcomes?
10. What is the probability of rolling a 2 on the red die and a 1, 3, or 5 on the blue one?
11. What is the probability of rolling an even number on the red die and an odd on the blue one?
12. Do the probabilities of a particular outcome change based on which die is rolled first? Why or why not?
Questions 13 – 16: Suppose you have a spinner with 5 equally-spaced color sections: red, blue, green, yellow, and orange.
13. Construct a probability distribution detailing the possible outcomes of three consecutive spins. You may wish to use only the first letter, or a single color-coded hash mark, to represent each possibility, as there will be many of them.
14. How many possible outcomes are there?
15. Is there an apparent mathematical relationship between the number of sections on the spinner, the number of spins, and the number of possible outcomes? If so, what is the relationship?
16. What is the probability of spinning red, then green, and then orange?
### Vocabulary Language: English
conditional probability
conditional probability
The probability of a particular dependent event given the outcome of the event on which it occurs.
conditional probability formula
conditional probability formula
The conditional probability formula is P(A/B) = P(AUB)/P(B)
Dependent Events
Dependent Events
In probability situations, dependent events are events where one outcome impacts the probability of the other.
Favorable Outcome
Favorable Outcome
A favorable outcome is the outcome that you are looking for in an experiment.
Independent Events
Independent Events
Two events are independent if the occurrence of one event does not impact the probability of the other event.
Multiplication Rule
Multiplication Rule
States that for 2 events (A and B), the probability of A and B is given by: P(A and B) = P(A) x P(B).
Mutually Exclusive Events
Mutually Exclusive Events
Mutually exclusive events have no common outcomes.
Sample Space
Sample Space
In a probability experiment, the sample space is the set of all the possible outcomes of the experiment.
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# RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment
In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 18 Areas of Circle, Sector and Segment Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 18 Areas of Circle, Sector and Segment Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Maths book pdf download. Now you will get step by step solution to each question.
### RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment
Question 1:
Circumference of circle = 2πr = (2×227×352)cm = 110 cm
∴ Area of circle = πr2 = (227×352×352) cm2
= 962.5 cm2
Question 2:
Circumference of circle = 2πr = 39.6 cm
Question 3:
Area of circle = πr2 = 301.84
Circumference of circle = 2πr = (2×227×9.8) = 61.6 cm
Question 4:
Let radius of circle be r
Then, diameter = 2 r
circumference – Diameter = 16.8
Circumference of circle = 2πr = (2×227×3.92) cm = 24.64 cm
Question 5:
Let the radius of circle be r cm
Then, circumference – radius = 37 cm
Question 6:
Area of square = (side)2 = 484 cm2
⇒ side = 484−−−√cm = 22 cm
Perimeter of square = 4 × side = 4 × 22 = 88 cm
Circumference of circle = Perimeter of square
Question 7:
Area of equilateral = 3√4a2 = 121√3
Perimeter of equilateral triangle = 3a = (3 × 22) cm
= 66 cm
Circumference of circle = Perimeter of circle
2πr = 66
⇒ (2×227×r) cm = 66
⇒ r = 10.5 cm
Area of circle = πr2 = (227×10.5×10.5) cm2
= 346.5 cm2
Question 8:
Let the radius of park be r meter
Question 9:
Let the radii of circles be x cm and (7 – x) cm
Circumference of the circles are 26 cm and 18 cm
Question 10:
Area of first circle = πr2 = 962.5 cm2
Area of second circle = πR2 = 1386 cm2
Width of ring R – r = (21 – 17.5) cm = 3.5 cm
Question 11:
Area of outer circle = πr21 = (227×23×23) cm2
= 1662.5
Area of inner circle = πr22 = (227×12×12) cm2
= 452.2 cm2
Area of ring = Outer area – inner area
= (1662.5 – 452.5) cm2 = 1210 cm2
Question 12:
Inner radius of the circular park = 17 m
Width of the path = 8 m
Outer radius of the circular park = (17 + 8)m = 25 m
Area of path = π[(25)2-(17)2] = cm2
Area = 1056 m2
Question 13:
Let the inner and outer radii of the circular tacks be r meter and R meter respectively. Then
Inner circumference = 440 meter
Since the track is 14 m wide every where.
Therefore,
Outer radius R = r + 14m = (70 + 14) m = 84 m
Outer circumference = 2πR
= (2×227×84)m = 528 m
Rate of fencing = Rs. 5 per meter
Total cost of fencing = Rs. (528 × 5) = Rs. 2640
Area of circular ring = πR2 – πr2
Cost of levelling = Rs 0.25 per m2
Cost of levelling the track = Rs(6776 × 0.25) = Rs. 1694
Question 14:
Let r m and R m be the radii of inner circle and outer boundaries respectively.
Then, 2r = 352 and 2R = 396
Width of the track = (R – r) m
Area the track = π(R2 – r) = π (R+r)(R-r)
Question 15:
Area of rectangle = (120 × 90)
= 10800 m2
Area of circular lawn = [Area of rectangle – Area of park excluding circular lawn]
= [10800 – 2950] m2 = 7850 m2
Area of circular lawn = 7850 m2
⇒ πr2 = 7850 m2
Hence, radius of the circular lawn = 50 m
Question 16:
Area of the shaded region = (area of circle with OA as diameter) + (area of semicircle ∆DBC) – (area of ∆BCD)
Area of circle with OA as diameter = πr2
OB = 7 cm, CD = AB = 14 cm
Area of semicircle ∆DBC =
= 72
Question 17:
Diameter of bigger circle = AC = 54 cm
Radius of bigger circle = AC2
= (542) cm = 27 cm
Diameter AB of smaller circle = AC – BC = 54-10 = 44 cm
Radius of smaller circle = 442 cm = 22 cm
Area of bigger circle = πR2 = (227×27×27) cm2
= 2291. 14 cm2
Area of smaller circle = πr2 = (227×22×22) cm2
= 1521. 11 cm2
Area of shaded region = area of bigger circle – area of smaller circle
= (2291. 14 – 1521. 11) cm2 = 770 cm2
Question 18:
PS = 12 cm
PQ = QR = RS = 4 cm, QS = 8 cm
Perimeter = arc PTS + arc PBQ + arc QES
Area of shaded region = (area of the semicircle PBQ) + (area of semicircle PTS)-(Area of semicircle QES)
Question 19:
Length of the inner curved portion
= (400 – 2 × 90) m
= 220 m
Let the radius of each inner curved part be r
Inner radius = 35 m, outer radius = (35 + 14) = 49 m
Area of the track = (area of 2 rectangles each 90 m × 14 m) + (area of circular ring with R = 49 m, r = 35 m
Length of outer boundary of the track
Question 20:
OP = OR = OQ = r
Let OQ and PR intersect at S
We know the diagonals of a rhombus bisect each other at right angle.
Therefore we have
Question 21:
Diameter of the inscribed circle = Side of the square = 10 cm
Radius of the inscribed circle = 5 cm
Diameter of the circumscribed circle
= Diagonal of the square
= (√2×10) cm
Radius of circumscribed circle = 5√2 cm
(i) Area of inscribed circle = (227×5×5) = 78.57 cm2
(ii) Area of the circumscribed circle
Question 22:
Let the radius of circle be r cm
Then diagonal of square = diameter of circle = 2r cm
Area of the circle = πr2 cm2
Question 23:
Let the radius of circle be r cm
Let each side of the triangle be a cm
And height be h cm
Question 24:
Radius of the wheel = 42 cm
Circumference of wheel = 2πr = (2×227×42) = 264 cm
Distance travelled = 19.8 km = 1980000 cm
Number of revolutions = 1980000264 = 7500
Question 25:
Radius of wheel = 2.1 m
Circumference of wheel = 2πr = (2×227×2.1) = 13.2 m
Distance covered in one revolution = 13.2 m
Distance covered in 75 revolutions = (13.2 × 75) m = 990 m
= 9901000 km
Distance a covered in 1 minute = 99100 km
Distance covered in 1 hour = 99100×60 km = 59.4 km
Question 26:
Distance covered by the wheel in 1 revolution
The circumference of the wheel = 198 cm
Let the diameter of the wheel be d cm
Hence diameter of the wheel is 63 cm
Question 27:
Radius of the wheel = r = 602 = 30 cm
Circumference of the wheel = 2πr = (2×227×30) = 13207 cm
Distance covered in 140 revolution
Distance covered in one hour = 2641000×60 = 15.84 km
Question 28:
Distance covered by a wheel in 1minute
Circumference of a wheel = 2πr = (2×227×70) = 440 cm
Number of revolution in 1 min = 121000440 = 275
Question 29:
Area of quadrant = 14 πr2
Circumference of circle = 2πr = 22
Question 30:
Area which the horse can graze = Area of the quadrant of radius 21 m
Area ungrazed = [(70×52) – 346.5] m2
= 3293.5 m2
Question 31:
Each angle of equilateral triangle is 60°
Area that the horse cannot graze is 36.68 m2
Question 32:
Each side of the square is 14 cm
Then, area of square = (14 × 14) cm2
= 196 cm2
Thus, radius of each circle 7 cm
Required area = area of square ABCD – 4 (area of sector with r = 7 cm, θ= 90°)
Area of the shaded region = 42 cm2
Question 33:
Area of square = (4 × 4) cm2
= 16 cm2
Radius of inner circle = 2/2 = 1 cm
Area of circle at the center = πr2 = (3.14 × 1 × 1) cm2
= 3.14 cm2
Area of shaded region = [area of square – area of four corner quadrants – area of circle at the centre]
= [16 – 3.14 – 3.14] cm= 9.72 cm2
Question 34:
Area of rectangle = (20 × 15) m= 300 m2
Area of 4 corners as quadrants of circle
Area of remaining part = (area of rectangle – area of four quadrants of circles)
= (300 – 38.5) m2 = 261.5 m2
Question 35:
Ungrazed area
Question 36:
Question 37:
Area of flower bed = (area of quadrant OPQ) – (area of the quadrant ORS)
Question 38:
Let A, B, C be the centres of these circles. Joint AB, BC, CA
Required area=(area of ∆ABC with each side a = 12 cm) – 3(area of sector with r = 6, θ = 60°)
The area enclosed = 5.76 cm2
Question 39:
Let A, B, C be the centers of these circles. Join AB, BC, CA
Required area= (area of ∆ABC with each side 2) – 3[area of sector with r = a cm, θ = 60°]
Question 40:
Let A, B, C, D be the centres of these circles
Join AB, BC, CD and DA
Side of square = 10 cm
Area of square ABCD
= (10 × 10) cm2
= 100 cm2
Area of each sector =
= 19.625 cm2
Required area = [area of sq. ABCD – 4(area of each sector)]
= (100 – 4 × 19.625) cm2
= (100 – 78.5) = 21.5 cm2
Question 41:
Required area = [area of square – areas of quadrants of circles]
Let the side = 2a unit and radius = a units
Area of square = (side × side) = (2a × 2a) sq. units = 4a2 sq.units
Question 42:
Let the side of square = a m
Area of square = (a × a) cm = a2m2
Side of square = 40 m
Therefore, radius of semi circle = 20 m
Area of semi circle =
= 628 m2
Area of four semi circles = (4 × 628) m= 2512 m2
Cost of turfing the plot of of area 1 m2 = Rs. 1.25
Cost of turfing the plot of area 2512 m= Rs. (1.25 × 2512)
= Rs. 3140
Question 43:
Area of rectangular lawn in the middle
= (50 × 35) = 1750 m2
Radius of semi circles = 352 = 17.5 m
Area of lawn = (area of rectangle + area of semi circle)
= (1750 + 962.5) m2 = 2712.5 m2
Question 44:
Area of plot which cow can graze when r = 16 m is πr2
= (227×10.5×10.5)
= 804.5 m2
Area of plot which cow can graze when radius is increased to 23 m
= (227×10.5×10.5)
= 1662.57 m2
Additional ground = Area covered by increased rope – old area
= (1662.57 – 804.5)m2 = 858 m2
Question 45:
Given: ABC is right angled at A with AB = 6 cm and AC = 8 cm
Let us join OA, OB and OC
ar(∆AOC) + ar(∆OAB) + ar(∆BOC) = ar(∆ABC)
Question 46:
Question 47:
Area of region ABCDEFA = area of square ABDE + area of semi circle BCD – area of ∆AFE
Question 48:
Side of the square ABCD = 14 cm
Area of square ABCD = 14 × 14 = 196 cm2
Radius of each circle = 144 = 3.5 cm
Area of the circles = 4 × area of one circle
Area of shaded region = Area of square – area of 4 circles
= 196 – 154 = 42 cm2
Question 49:
Diameter AC = 2.8 + 1.4
= 4.2 cm
Radius r1 = 4.22 = 2.1 cm
Length of semi-circle ADC = πr= π × 2.1 = 2.1 π cm
Diameter AB = 2.8 cm
Length of semi- circle AEB = πr2= π × 1.4 = 1.4 π cm
Diameter BC = 1.4 cm
Radius r3 = 1.42 = 0.7 cm
Length of semi – circle BFC = π × 0.7 = 0.7 π cm
Perimeter of shaded region = 2.1 + 1.4 + 0.7 = 4.2 π cm
= 4.2 × 227 = 13.2 cm
Question 50:
Area of shaded region = Area of ∆ABC + Area of semi-circle APB + Area of semi circle AQC – Area of semicircle BAC
Further in ∆ABC, ∠A = 90°
Adding (1), (2), (3) and subtracting (4)
Question 51:
In ∆PQR, ∠P = 90°, PQ = 24 cm, PR = 7 cm
Area of semicircle
Area of ∆PQR = 12 × 7 × 24 cm2 = 84 cm2
Shaded area = 245.31 – 84 = 161.31 cm2
Question 52:
ABCDEF is a hexagon.
∠AOB = 60°, Radius = 35 cm
Area of sector AOB
Area of ∆AOB =
= 530.425 cm2
Area of segment APB = (641.083 – 530.425) cm= 110.658 cm2
Area of design (shaded area) = 6 × 110.658 cm2 = 663.948 cm2
= 663.95 cm2
Question 53:
In ∆ABC, ∠A = 90°, AB = 6cm, BC = 10 cm
Area of ∆ABC =
Let r be the radius of circle of centre O
Question 54:
Area of equilateral triangle ABC = 49√3 cm2
Let a be its side
Area of sector BDF =
Area of sector BDF = Area of sector CDE = Area of sector AEF
Sum of area of all the sectors
= 773 × 3 cm2 = 77 cm2
Shaded area = Area of ∆ABC – sum of area of all sectors
= 49√3 – 77 = (84.77 – 77.00) cm2
= 77.7 cm2
Question 55:
In ∆ABC, ∠B = 90°, AB = 48 cm, BC = 14 cm
Area of semi-circle APC
Shaded area = Area of ∆ABC + Area of semi-circle APC – Area of quadrant BDC
= ( 336+982.14-154 ) cm2
= ( 1318.14-154 ) cm= 1164.14 cm2
Question 56:
Its area =
Area of ∆ABD = (12×16×16) cm2
= 128 cm2
Area of segment DEB
Area of segment DFB = 5127 cm2
Total area of segments = 2 × 5127 cm2 = 10247 cm2
Shaded area = Area of square ABCD – Total area of segments
Question 57:
Radius of circular table cover = 70 cm
Area of the circular cover =
Shaded area = Area of circle – Area of ∆ABC
= (15400 – 6365.1)
Question 58:
Area of the sector of circle =
r = 14 cm and θ = 45°
Question 59:
Length of the arc
Length of arc = ( 17.5 × 227 ) cm = 55 cm
Area of the sector =
= ( 227 × 183.75 ) cm2 = 577.5 cm2
Question 60:
Length of arc of circle = 44 cm
Radius of circle = 17.5 cm
Area of sector =
= ( 22 × 17.5) cm2 = 385 cm2
Question 61:
Let sector of circle is OAB
Perimeter of a sector of circle =31 cm
OA + OB + length of arc AB = 31 cm
6.5 + 6.5 + arc AB = 31 cm
arc AB = 31 – 13
= 18 cm
Question 62:
Area of the sector of circle =
Question 63:
Length of the pendulum = radius of sector = r cm
Question 64:
Length of arc =
Circumference of circle = 2πr
Area of circle =
= 962.5 cm2
Question 65:
Circumference of circle = 2πr
Question 66:
Angle described by the minute hand in 60 minutes θ = 360°
Angle described by minute hand in 20 minutes
Required area swept by the minute hand in 20 minutes
= Area of the sector(with r = 15 cm and θ = 120°)
Question 67:
θ = 56° and let radius is r cm
Area of sector =
Question 68:
Question 69:
In 2 days, the short hand will complete 4 rounds
∴ Distance travelled by its tip in 2 days
=4(circumference of the circle with r = 4 cm)
= (4 × 2 × 4) cm = 32 cm
In 2 days, the long hand will complete 48 rounds
∴ length moved by its tip
= 48(circumference of the circle with r = 6cm)
= (48 × 2 × 6) cm = 576 cm
∴ Sum of the lengths moved
= (32 + 576) = 608 cm
= (608 × 3.14) cm = 1909.12 cm
Question 70:
∆OAB is equilateral.
So, ∠AOB = 60°
Length of arc BDA = (2π × 12 – arc ACB) cm
= (24π – 4π) cm = (20π) cm
= (20 × 3.14) cm = 62.8 cm
Area of the minor segment ACBA
Question 71:
Let AB be the chord of circle of centre O and radius = 6 cm such that ∠AOB = 90°
Area of sector = OACBO
Area of ∆AOB =
Area of minor segment ACBA
= (area of sector OACBO) – (area of ∆OAB)
= (28.29 – 18) cm2 = 10.29 cm2
Area of major segment BDAB
Question 72:
Let OA = 5√2 cm , OB = 5√2 cm
And AB = 10 cm
Area of ∆AOB =
= 25 cm2
Area of minor segment = (area of sector OACBO) – (area of ∆OAB)
= ( 39.25 – 25 ) cm2 = 14.25 cm2
Question 73:
Area of sector OACBO
Area of minor segment ACBA
Question 74:
Let AB be the chord of circle of centre O and radius = 30 cm such that AOB = 60°
Area of the sector OACBO
Area of ∆OAB =
Area of the minor segment ACBA
= (area of the sector OACBO) – (area of the ∆OAB)
=(471 – 389.25) cm2 = 81.75 cm2
Area of the major segment BADB
= (area of circle) – (area of the minor segment)
= [(3.14 × 30 × 30) – 81.75)] cm2 = 2744.25 cm2
Question 75:
Let the major arc be x cm long
Then, length of the minor arc = 15 x cm
Circumference =
Question 76:
Radius of the front wheel = 40 cm = 25 m
Circumference of the front wheel =
Distance moved by it in 800 revolution
Circumference of rear wheel = (2π × 1)m = (2π) m
Required number of revolutions =
All Chapter RS Aggarwal Solutions For Class 10 Maths
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All Subject NCERT Exemplar Problems Solutions For Class10
All Subject NCERT Solutions For Class 10
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How Do You Reduce Fractions?
# How Do You Reduce Fractions?
To reduce or simplify fractions, look for common factors in the numerator (the term on top) and the denominator (the term on the bottom). Divide both terms by the greatest common factor to reduce the fraction to its lowest terms.
1. Identify the prime factors in both the numerator and denominator
Break each term down into its prime factors, and write those factors in the numerator and denominator of an equivalent fraction. Consider the example 14/20; breaking this fraction into prime factors results in this equation: 14/20 = 7 x 2 / 2 x 2 x 5.
2. Identify the common factors in the numerator and denominator
Look at the factors in the numerator (2 and 7), and look at the factors in the denominator (2 and 5). The common factor between the numerator and denominator is 2, which means you can reduce both numerator and denominator by a factor of 2.
3. Simplify the fraction by removing common factors
Cross the 2 out in the numerator, and cross one of the 2's out in the denominator. Remember that you can only cross out as many 2's (or other factors) as there are in both the numerator and denominator. The example from Step 2 (7 x 2 / 2 x 2 x 5) now reduces to 7 / 2 x 5. Multiply out to find the solution, which is 7/10.
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# Logarithm Formula
Logarithm are the inverse phenomena of exponential. Log function “undo” the exponential function. Consider a number x in the exponent equals to a fixed number b, the logarithmic value of b will be equal to the number x. Such as:
$\large e^{x} = b$
Taking log on both the sides, we have
$\large log _{e}\; e^{x} = log_{e} \; b$
$\large x = log_{e} \; b$
Note: It is to be noted that the logarithmic values of positive numbers is only known, i.e. x>0
Logarithmic value of 0- The logarithmic value of zero is undefined.
log(0) = Undefined
Logarithmic value of Negative number- The logarithmic value of negative numbers are imaginary, i.e. we don’t have a real value for negative numbers.
Below given are some important identities of the logarithmic function.
#### Trivial Identities
$\large \log _{b} (1) = 0; \; because \; b^{0}=1; \; b> 0$
$\large \log _{b} (b) = 1; \; because \; b^{1}=b$
#### Basic Logarithm Formulas
$\large \log _{b} (xy) = \log _{b}(x) + \log _{b}(y)$
$\large \log _{b}\left ( \frac{x}{y} \right ) = \log _{b}(x) – \log _{b}(y)$
$\large \log_{b}(x^{d})= d \log_{b}(x)$
$\large \log_{b}(\sqrt[y]{x})= \frac{\log_{b}(x)}{y}$
$\large c\log_{b}(x)+d\log_{b}(y)= \log_{b}(x^{c}y^{d})$
#### Changing the Base
$\large \log_{b}a = \frac{\log_{d}(a)}{\log_{d}(b)}$
$\large \log_{b} (a+c) = \log_{b}a + \log_{b}\left ( 1 + \frac{c}{a} \right )$
$\large \log_{b} (a-c) = \log_{b}a + \log_{b}\left ( 1 – \frac{c}{a} \right )$
$\large x^{\frac{\log(\log(x))}{\log(x)}} \; = \; \log(x)$
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# How many 13 card hands having exactly 11 diamonds can be dealt?
Here’s information about what you’re looking for in How many 13 card hands having exactly 11 diamonds can be dealt?. While every brand has a “Help Center” (or “Support Center”), when we look for relevant information or links, there are often the answers to some questions can’t be found. This is why this website was created, and hopefully it will help you.
## What is the number of 13 card hands that can be dealt from a deck of 52 cards?
There are 40 possibilities for the thirteenth card (because it can be anything except the ♦7, ♠K, ♣4, or any of the nine other cards already dealt). That means that the number of possible hands is 52×51×50×···×42×41×40 = 3954242643911239680000.
## How many different 13 card hands include the ace and king of spades?
There is only one ace of spades, one king of spades and one queen of spades in the deck of cards. Therefore, we take them out, we have now selected 3 of the 13 cards. We have to select 10 more cards out of the 52-3 = 49 cards.
## How many different 5 card hands can be dealt from a deck of 52 cards if the hand consists of exactly 2 aces?
There are 6 choices for the 2 Aces based on 4 suits in a standard deck: Clubs, Hearts, Diamonds, Spades. For each of these choices there are 4 choices for the 3 Kings (basically one choice for each suit not included). This gives a combination of 6×4=24 possible hands.
## What is a 13 card bridge hand?
A Bridge hand consists of 13 random cards taken from a deck that holds 52 cards. The total number of possible Bridge hands is thus: COMBIN(52, 13) = 635,013,559,600.
## What is the number of 13 card hands that can be dealt from a deck of 52 cards?
There are 40 possibilities for the thirteenth card (because it can be anything except the ♦7, ♠K, ♣4, or any of the nine other cards already dealt). That means that the number of possible hands is 52×51×50×···×42×41×40 = 3954242643911239680000.
## How many 4 card hands can a 13 card deck have?
SOLUTION: As there are 13 diamonds in a standard deck of 52 cards, we are counting the number of combinations of 4 cards chosen from the 13 diamonds in the deck. This gives C =P / 4! = (13Χ12Χ11Χ10) / (4Χ3Χ2Χ1) = 17160 / 24 = 715 hands consisting entirely of diamonds.
## How many different 13 card bridge hands are there that contain all four aces?
to select 9 of the 48 remaining cards. Thus there are 1,677,106,640 ways to select a 13-card bridge hand with all four aces.
## What is a 13 card bridge hand?
A Bridge hand consists of 13 random cards taken from a deck that holds 52 cards. The total number of possible Bridge hands is thus: COMBIN(52, 13) = 635,013,559,600.
## What is the number of 13-card hands that can be dealt from a deck of 52 cards?
There are 40 possibilities for the thirteenth card (because it can be anything except the ♦7, ♠K, ♣4, or any of the nine other cards already dealt). That means that the number of possible hands is 52×51×50×···×42×41×40 = 3954242643911239680000.
## What is 13th hand?
There are special names for specific types of hands. A ten, jack, queen, king, or ace is called an “honor.” Getting the three top cards (ace, king, and queen) of three suits and the ace, king, and queen, and jack of the remaining suit is called 13 top honors. Getting all cards of the same suit is called a 13-card suit.
## How many 4 card hands can a 13-card deck have?
SOLUTION: As there are 13 diamonds in a standard deck of 52 cards, we are counting the number of combinations of 4 cards chosen from the 13 diamonds in the deck. This gives C =P / 4! = (13Χ12Χ11Χ10) / (4Χ3Χ2Χ1) = 17160 / 24 = 715 hands consisting entirely of diamonds.
## How many different 13-card bridge hands are there that contain all four aces?
to select 9 of the 48 remaining cards. Thus there are 1,677,106,640 ways to select a 13-card bridge hand with all four aces.
## How many different 13-card hands are there?
There are 40 possibilities for the thirteenth card (because it can be anything except the ♦7, ♠K, ♣4, or any of the nine other cards already dealt). That means that the number of possible hands is 52×51×50×···×42×41×40 = 3954242643911239680000.
## How many hands include the ace of spades?
There are (524) ways to choose the ace of spade. After we choose the ace of spade, there are total of 51 cards left in the deck with 3 aces. There are (513) ways to choose the ace of hearts. Then we add them both.
## What is 13th hand?
There are special names for specific types of hands. A ten, jack, queen, king, or ace is called an “honor.” Getting the three top cards (ace, king, and queen) of three suits and the ace, king, and queen, and jack of the remaining suit is called 13 top honors. Getting all cards of the same suit is called a 13-card suit.
## How many different 13-card bridge hands are there that contain all four aces?
to select 9 of the 48 remaining cards. Thus there are 1,677,106,640 ways to select a 13-card bridge hand with all four aces.
## How many different 5 card hands are possible?
Probability of a Full House
First, count the number of five-card hands that can be dealt from a standard deck of 52 cards. We did this in the previous section, and found that there are 2,598,960 distinct poker hands. Next, count the number of ways that five cards can be dealt to produce a full house.
## How many different 5 card hands which contain at least one ace are possible from a standard deck of 52 cards?
If you want exactly one ace, then your answer is correct. (525) is the number of 5-card hands in the deck, and you have 4 choices for which ace to include (hence, (41)), and 48 choose 4 choices for the other 4 cards (hence, (484)).
## How many different 5 card hands are possible that contain exactly three aces?
How to show that there are 4704 ways to choose 5 cards with exactly 3 Ace’s from a deck of 52 cards? – Mathematics Stack Exchange.
## What is the total number of possible hands that can be dealt if the hand contains exactly one heart?
Since there are a total of 13 hearts in a standard deck, we need to select exactly one heart from the 13 hearts and then from the remaining 39 cards, we need to select 4 cards. Hence, the total number of possible hands in the given case is 1,069,263.
## What is a 13-card bridge hand?
A Bridge hand consists of 13 random cards taken from a deck that holds 52 cards. The total number of possible Bridge hands is thus: COMBIN(52, 13) = 635,013,559,600.
## How many 13-card bridge hands are there?
There are 40 possibilities for the thirteenth card (because it can be anything except the ♦7, ♠K, ♣4, or any of the nine other cards already dealt). That means that the number of possible hands is 52×51×50×···×42×41×40 = 3954242643911239680000.
## What is the probability that a bridge hand of 13 cards dealt at random from a standard deck of cards contains exactly one ace and exactly two kings?
There is a 44% chance of getting exactly one ace.
## What is the perfect bridge hand?
Every bridge player fantasizes about the perfect hand – being dealt the 13 cards of one suit – and the perfect game, in which each of the four players receives all 13 cards of one suit. … It contains the cards which dealt one perfect suit to each player.
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# Learn how to subtract ordinary fractions: theory, steps and a practical example. Fractions with different (like) or unlike denominators
## How to: Subtracting ordinary (simple, common) math fractions. Steps.
There are two cases regarding the denominators when we subtract ordinary fractions:
• A. the fractions have like denominators;
• B. the fractions have unlike denominators.
### A. How to subtract ordinary fractions that have like denominators?
• Simply subtract the numerators of the fractions.
• The denominator of the resulting fraction will be the common denominator of the fractions.
• Reduce the resulting fraction.
### An example of subtracting ordinary fractions that have like denominators, with explanations
• #### 3/18 + 4/18 - 5/18 = (3 + 4 - 5)/18 = 2/18;
• We simply subtracted the numerators of the fractions: 3 + 4 - 5 = 2;
• The denominator of the resulting fraction is: 18;
### B. To subtract fractions with different denominators (unlike denominators), build up the fractions to the same denominator. How is it done?
• #### 1. Reduce the fractions to the lowest terms (simplify them).
• Factor the numerator and the denominator of each fraction, break them down to prime factors (run their prime factorization).
• #### Factor numbers online, break them down to their prime factors.
• Calculate GCF, the greatest common factor of the numerator and of the denominator of each fraction.
• GCF is the product of all the unique common prime factors of the numerator and of the denominator, multiplied by the lowest exponents.
• #### Calculate the greatest common factor, GCF, online.
• Divide the numerator and the denominator of each fraction by their GCF - after this operation the fraction is reduced to its lowest terms equivalent.
• #### 2. Calculate the least common multiple, LCM, of all the fractions' new denominators:
• LCM is going to be the common denominator of the added fractions, also called the lowest common denominator (the least common denominator).
• Factor all the new denominators of the reduced fractions (run the prime factorization).
• The least common multiple, LCM, is the product of all the unique prime factors of the denominators, multiplied by the largest exponents.
• #### 3. Calculate each fraction's expanding number:
• The expanding number is the non-zero number that will be used to multiply both the numerator and the denominator of each fraction, in order to build all the fractions up to the same common denominator.
• Divide the least common multiple, LCM, calculated above, by the denominator of each fraction, in order to calculate each fraction's expanding number.
• #### 4. Expand each fraction:
• Multiply each fraction's both numerator and denominator by the expanding number.
• At this point, fractions are built up to the same denominator.
• #### 5. Subtract the fractions:
• In order to subtract all the fractions simply subtract all the fractions' numerators.
• The end fraction will have as a denominator the least common multiple, LCM, calculated above.
### An example of subtracting fractions that have different denominators (unlike denominators), step by step explanations
• #### 2. Calculate the least common multiple, LCM, of all the fractions' new denominators
• Factor all the denominators, break them down to their prime factors, then multiply all these prime factors, uniquely, by the largest exponents.
• #### 3. Calculate each fraction's expanding number:
• Divide the least common multiple, LCM, by the denominator of each fraction.
• #### 4. Expand each fraction:
• Multiply both the numerator and the denominator of each fraction by their expanding number.
• #### 5. Subtract the fractions:
• Simply subtract the numerators of the fractions. The denominator = LCM.
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# Ex.4.3 Q1 Quadratic Equations Solutions - NCERT Maths Class 10
Go back to 'Ex.4.3'
## Question
Find the roots of the following quadratic equations, if they exist, by the method of completing the square.
(i) $$2x^2-7x+3=0$$
(ii) $$2x^2+x-4=0$$
(iii) $$4x^2+ 4\sqrt {\left( 3 \right)} x +3=0$$
(iv) $$2x^{2}+x+4=0$$
Video Solution
Ex 4.3 | Question 1
## Text Solution
What is the unknown?
Reasoning:
Steps required to solve a quadratic equation by applying the ‘completing the square’ method are given below:
Let the given quadratic equation be: $$ax{}^\text{2}+bx+c=0$$
(i) Divide all the terms by $$a$$
\begin{align}\frac{{a{x^2}}}{a} + \frac{{bx}}{a} + \frac{c}{a} &= 0\\{x^2} + \frac{{bx}}{a} + c &= 0\end{align}
(ii) Move the constant term \begin{align}\frac{c}{a}\end{align} to the right side of the equation:
${x^2} + \frac{b}{a}x = - \frac{c}{a}$
(iii) Complete the square on the left side of the equation by adding \begin{align} \frac{{{b^2}}}{{4{a^2}}}.\end{align} Balance this by adding the same value to the right side of the equation.
Steps:
(i) $$2x^2-7x+3=0$$
Divide $$2x{}^\text{2}-7x+3=0$$ by $$2:$$
\begin{align}{x^2} - \frac{7}{2}x + \frac{3}{2}& = 0\\{x^2} - \frac{7}{2}x &= - \frac{3}{2}\\\end{align}
Since \begin{align}\frac{7}{2} \div 2 = \frac{7}{4} \end{align}, \begin{align}{\left( {\frac{7}{4}} \right)^2} \end{align} should be added to both sides of the equation:
\begin{align}{x^2} - \frac{7}{2}x + {\left( {\frac{7}{4}} \right)^2} &= \frac{{ - 3}}{2} + {\left( {\frac{7}{4}} \right)^2}\\{\left( {x - \frac{7}{4}} \right)^2} &= \frac{{ - 3}}{2} + \frac{{49}}{{16}}\\ &= \frac{{ - 24 + 49}}{{16}}\\{\left( {x - \frac{7}{4}} \right)^2} & = \frac{{25}}{{16}}\\{\left( {x - \frac{7}{4}} \right)^2} &= {\left( { \pm \frac{5}{4}} \right)^2}\\x - \frac{7}{4} = \frac{5}{4} &\qquad \!\!x - \frac{7}{4} = \frac{{ - 5}}{4}\\x = \frac{5}{4} + \frac{7}{4} &\qquad \!\!x = \frac{{ - 5}}{4} + \frac{7}{4}\\ x = \frac{{12}}{4} &\qquad x = \frac{2}{4}\\x = 3&\qquad x = \frac{1}{2}\end{align}
Roots are \begin{align}3, \;\frac{1}{2}.\end{align}
(ii) $$2x^2+x-4=0$$
\begin{align}{x^2} + \frac{x}{2} - 2 &= 0\\{x^2} + \frac{x}{2} &= 2\\{x^2} + \frac{x}{2} &= 2\end{align}
Since, \begin{align} \frac{{\rm{1}}}{{\rm{2}}} \div 2 = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4},\; {\left( {\frac{1}{4}} \right)^2} \end{align} should be added on both sides
\begin{align}{x^2} + \frac{x}{2} + {\left( {\frac{1}{4}} \right)^2} &= 2 + {\left( {\frac{1}{4}} \right)^2}\\ {\left( {{x^2} + \frac{1}{4}} \right)^2} &= 2 + \frac{1}{{16}}\\{\left( {{x^2} + \frac{1}{4}} \right)^2} &= \frac{{32 + 1}}{{16}}\\{\left( {{x^2} + \frac{1}{4}} \right)^2} &= \frac{{33}}{{16}}\\\left( {x + \frac{1}{4}} \right) &= \pm \frac{{\sqrt {33} }}{4}\end{align}
\begin{align} \left( {x + \frac{1}{4}} \right) \!\!= \frac{{\sqrt {33} }}{4} ,& \quad \!\!\!\!\!\!\left( \!{x \!+\! \frac{1}{4}} \!\!\right) \!\!= - \frac{{\sqrt {33} }}{4}\\ x = \frac{{\sqrt {33} }}{4} - \!\! \frac{1}{4}, &\quad \!\!x \!\!= - \!\! \frac{{\sqrt {33} }}{4} - \! \frac{1}{4}\\x = \frac{{\sqrt {33} - 1}}{4}, &\quad \!\!x \! = \frac{{ - \sqrt {33} - 1}}{4}\end{align}
Roots are \begin{align} \frac{{\sqrt {33} - 1}}{4},\frac{{ - \sqrt {33} - 1}}{4}\end{align}
iii) $$4x+ 4\sqrt {\left( 3 \right)} x +3=0$$
\begin{align}x^2+ \sqrt 3 x +\frac{3}{4}&= 0\\{x^2} + \sqrt 3 x& = - \frac{3}{4}\\ x^2 \!\! + \!\! \sqrt 3 x \!\!+ \!\! {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} & \!\! = - \frac{3}{4} \!\!+\!\! {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \end{align}
\begin{align}{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \end{align} is added on both sides
\begin{align}{{\left( {x + \frac{{\sqrt 3 }}{2}} \right)}^2} &= - \frac{3}{4} + \frac{3}{4}\\{\left( {x + \frac{{\sqrt 3 }}{2}} \right)^2} &= 0\\x = - \frac{{\sqrt 3 }}{2} &\quad x= - \frac{{\sqrt 3 }}{2}\end{align}
Roots are \begin{align} - \frac{{\sqrt 3 }}{2},\, - \frac{{\sqrt 3 }}{2}\end{align}
iv) $$2x^{2}+x+4=0$$
\begin{align}{x^2} + \frac{x}{2} + 2 &= 0\\{x^2} + \frac{x}{2}& = - 2\\{x^2} + \frac{x}{2} + {\left( {\frac{1}{4}} \right)^2} &= - 2 + {\left( {\frac{1}{4}} \right)^2}\\{\left( {x + \frac{1}{4}} \right)^2} &= - 2 + \frac{1}{{16}}\\{\left( {x + \frac{1}{4}} \right)^2} &= \frac{{ - 32 + 1}}{{16}}\\{\left( {x + \frac{1}{4}} \right)^2} &= \frac{{ - 31}}{{16}} < 0\end{align}
Square of any real number can’t be negative.
$$\therefore\;$$Real roots don’t exist.
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# Maharashtra Board 10th Class Maths Part 1 Practice Set 1.4 Solutions Chapter 1 Linear Equations in Two Variables
## Maharashtra Board 10th Class Maths Part 1 Practice Set 1.4 Solutions
Question 1.
Solve the following simultaneous equations.
Solution:
i. The given simultaneous equations are
∴ Equations (i) and (ii) become
2p – 3q = 15 …(iii)
8p + 5q = 77 …(iv)
Multiplying equation (iii) by 4, we get
8p – 12q = 60 …(v)
Subtracting equation (v) from (iv), we get
ii. The given simultaneous equations are
Substituting x = 3 in equation (vi), we get
3 + y = 5
∴ y = 5 – 3 = 2
∴ (x, y) = (3, 2) is the solution of the given simultaneous equations.
iii. The given simultaneous equations are
∴ Equations (i) and (ii) become
27p + 31q = 85 …(iii)
31p + 27q = 89 …(iv)
Adding equations (iii) and (iv), we get
iv. The given simultaneous equations are
Substituting x = 1 in equation (vi), we get
3(1) + y = 4
∴ 3 + y = 4
∴ y = 4 – 3 = 1
∴ (x, y) = (1, 1) is the solution of the given simultaneous equations.
Question 1.
Complete the following table. (Textbook pg. no. 16)
Solution:
Question 2.
In the above table the equations are not linear. Can you convert the equations into linear equations? (Textbook pg. no. 17)
Yes, the above given simultaneous equations can be converted to a pair of linear equations by making suitable substitutions.
Steps for solving equations reducible to a pair of linear equations.
• Step 1: Select suitable variables other than those which are in the equations.
• Step 2: Replace the given variables with new variables such that the given equations become linear equations in two variables.
• Step 3: Solve the new simultaneous equations and find the values of the new variables.
• Step 4: By resubstituting the value(s) of the new variables, find the replaced variables which are to be determined.
Question 3.
To solve given equations fill the below boxes suitably. (Text book pg.no. 19)
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# Free math homework help
Here, we will be discussing about Free math homework help. We will also look at some example problems and how to approach them.
## The Best Free math homework help
There are a lot of Free math homework help that are available online. this analysis method is called time domain analysis method. If the time variable is transformed into other variables in order to solve the equation by hand, it is called transformation domain analysis method accordingly. This chapter mainly explains the time-domain analysis method, which is also the focus of the postgraduate entrance examination. When we are in contact with ordinary differential equations, we can only solve some special forms of equations, such as first-order linear differential equations, differential equations with separable variables, Bernoulli differential equations, etc.
In the play, it seems that Lin Chaoxi traversed the parallel world with Ji Jiang because of the inspiration of a photo. In order to return to the strawberry world, he completed the task of winning the cup in the photo. His love of mathematics has become a task-based journey. Lin Zhaoxi went through three times to regain mathematics, save Lao Lin and harvest friendship. In the TV series, Lin Chaoxi and star Ji Jiang crossed the parallel world together to complete the task of winning the cup in order to return to the strawberry world.
Since the distances from the discrete points on the building contour to each crane layout point have been put into the list, the distances of each crane layout point are summed separately. The lowest index value among all distances is the nearest distance between each crane hoisting area and the crane. Summary: as can be seen from the above examples, there are various areas of the shadow part of the circle, and there are also many ways to solve it. However, as long as the shadow in the thinking is removed through appropriate transformation and flexible processing according to the characteristics of the figure, it will certainly bring a bright future to the solution of the problem.
However, even if no analytical solution is found, some properties of the solution can still be confirmed. When the analytical solution cannot be obtained, the numerical solution can be found by numerical analysis and computer. Dynamic system theory emphasizes the quantitative analysis of differential equation system, and many numerical methods can calculate the numerical solution of differential equation with certain accuracy. Is it true that the sooner you master algebra and equations, the better? no, it isn't.
When it comes to the second-order differential equation, there are fewer equations that can be solved, and many special functions are defined by the solution of the second-order differential equation, such as hypergeometric functions, Legendre functions, Bessel functions, Airy functions... We mentioned earlier that K (s, t) in the integral equation is the kernel function of the integral equation, so we guess that the difficulty of solving the integral equation is probably related to this kernel, and the more special the kernel, the easier it will be. In this section, I will begin to introduce the solution of Fredholm equation of the second kind. The reason why we don't start with other equations is that these equations are easier to solve than other equations. Explain the neural network as a discrete format for solving differential equations? The field of numerical solution will pay attention to the numerical convergence of discrete schemes, but what is the connection between this and differential equations? How to map the input-output mapping of the network connection to the infinite dimensional mapping of differential equations? Using the knowledge of dynamic system to analyze the properties of neural network? Different body tissues (such as bones, muscles, blood, etc.) have different absorption intensities for X-rays, and CT machines use this characteristic in combination with the principle of solving linear equations to characterize the internal structure of the human body.
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Learn the basics of exactly how to calculate percentages of quantities in this easy lesson! To find a percentage of any type of number, usage this generic reminder of TRANSLATION: readjust the percentage into a decimal, and also the word "of" into multiplication. See countless examples below.
You are watching: 80 of what number is 160
The concepts and ideas of this class are also explained in this video:
You have learned that to find 1% of a number way finding 1/100 the it. Similarly, recognize 60% that a number method finding 60/100 (or 6/10) of it.In this expressions, words “of” translates into multiplication:1% the 90 → 1% × 90&60% the \$700 → 60% × \$700.We can additionally write those percentages together decimals:1% that 90 → 0.01 × 9060% that \$700 → 0.6 × \$700.This gives us another method to calculation the portion of a number (or percent of part quantity): To calculation a percent of some number, change the percentage into a decimal, and the native "of" into multiplication. Example 1. Find 70% the 80.Following the shortcut, we compose this as 0.7 × 80.Remember that in decimal multiplication, friend multiply together if there were no decimal points, and the price will have actually as many “decimal digits” come the appropriate of the decimal suggest as the total number of decimal digits of every one of the factors. So as soon as you main point 0.7 × 80, think of multiply 7 × 80 = 560. Because 0.7 has one decimal digit, and 80 has none, the answer has one decimal digit: 56.0 Thus, 0.7 × 80 = 56.You can additionally use “common sense” to reason it through logically: 0.7 × 80 should be much less than 80, yet much more than 1/2 the 80, which is 40. Because 7 × 8 = 56, you understand that the answer must be 56—not 5.6 or 560. Example 2. Find 3% that \$4,000.First compose it as 0.03 × \$4,000. Climate multiply 3 × \$4,000 = \$12,000. Finally put the decimal point where it gives the answer 2 decimal digits: \$120.00. Example 3. Find 23% of 5,500km.Write the expression together 0.23×5,500km and also use a calculator to calculation the product. The answer is 1,265km. This answer provides sense because 10% of 5,500km is 550km, for this reason 20% is 1,100km. Hence 1,265km together 23% the 5,500km is a reasonable answer.
1. "Translate" the expressions into multiplication through a decimal. Calculate.
a. 20% the 70______ × ______ = ______ b. 90% of 50______ × ______ = ______ c. 9% the 3,000______ × ______ = ______
2. "Translate" the other way: compose the multiplications together expressions of "percentage the the number".
a. 0.6 × 50_____% the ______ = ______ b. 0.03 × \$400_____% of ______ = ______ c. 0.08 × 6_____% the ______ = ______
3. Usage a calculator to discover percentages of this quantities.
a. 17% the \$4500
b. 67% of 27 m
4. Use mental math to uncover percentages of these quantities.
a. 25% the 240 mi
5. a. A lake has a 30-km lengthy shoreline. 6% of that is sandy beach. What percentage of the coastline is no sandy beach?
6. Twenty percent that a university’s 4,000 students have actually a scholarship.
a. What percent that the students do not have a scholarship?
b. How countless students have actually a scholarship?
9. Discover the expressions v the same value as 20% of \$620.
0.02 × \$620 \$620 ÷ 5 \$620 ÷ 10 × 2 2 × \$62 15 × \$620
0.2 × \$62020 × \$620\$620 ÷ 411. The table listed below shows Andy’s consumption of time in one day.
a. Calculate the moment he invested in every activity. round the minute to the nearest minute.
b.
See more: Is 85 A Prime Or Composite Number ? Is 85 A Prime Or Composite Number
label the sections in the circle graph v the name of each activity.
Andy’s usage of Time
Activity Percent Minutes Hours/minutes Sleep 38% School 21% Soccer 10% 144 2 h 24 min Play 11% Eating 9% Chores 9% Hygiene 2% TOTAL 100% 1440 24 hours
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# 1950 AHSME Problems/Problem 24
## Problem
The equation $x + \sqrt{x-2} = 4$ has:
$\textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root}$
## Solutions
### Solution 1
$x + \sqrt{x-2} = 4$ Original Equation
$\sqrt{x-2} = 4 - x$ Subtract x from both sides
$x-2 = 16 - 8x + x^2$ Square both sides
$x^2 - 9x + 18 = 0$ Get all terms on one side
$(x-6)(x-3) = 0$ Factor
$x = \{6, 3\}$
If you put down A as your answer, it's wrong. You need to check for extraneous roots.
$6 + \sqrt{6 - 2} = 6 + \sqrt{4} = 6 + 2 = 8 \ne 4$
$3 + \sqrt{3-2} = 3 + \sqrt{1} = 3 + 1 = 4 \checkmark$
There is $\boxed{\textbf{(E)} \text{1 real root}}$
### Solution 2
It's not hard to note that $x=3$ simply works, as $3 + \sqrt{1} = 4$. But, $x$ is increasing, and $\sqrt{x-2}$ is increasing, so $3$ is the only root. If $x < 3$, $x + \sqrt{x-2} < 4$, and similarly if $x > 3$, then $x + \sqrt{x-2} > 4$. Imaginary roots don't have to be considered because squaring the equation gives a quadratic, and quadratics either have only imaginary or only real roots.
### Solution 3
We can create symmetry in the equation: $$x+\sqrt{x-2} = 4$$ $$x-2+\sqrt{x-2} = 2.$$ Let $y = \sqrt{x-2}$, then we have $$y^2+y-2 = 0$$ $$(y+2)(y-1) = 0$$ The two roots are $\sqrt{x-2} = -2, 1$.
Notice, that the first root is extraneous as the range for the square root function is always the non-negative numbers (remember, negative numbers in square roots give imaginary numbers - imaginary numbers in square roots don't give negative numbers); thus, the only real root for $x$ occurs for the second root; squaring both sides and solving for $x$ gives $x=3 \Rightarrow \boxed{\textbf{(E)} \text{1 real root}}$.
~Vndom
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This lesson will focus on analyzing and solving different trigonometric equations. For that purpose, inverse trigonometric functions and trigonometric identities will be used.
### Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
## Riding a Ferris Wheel
On Sunday, Magdalena and her younger sister Paulina went to the amusement park Adventurally with their father. They all had a lot of fun going on numerous rides, including a Ferris Wheel. When they first saw it close up, the girls were so amazed by its size that they asked one of the workers for more details about it.
The worker replied that it has a diameter of meters and it turns at a rate of revolutions per minute. When Magdalena's and Paulina's father heard this, he said that the height of their seat above the ground in meters after minutes can be modeled by the following function.
Then he asked his daughters two questions.
a How long will they have to wait until their seat is meters above the ground for the first time? Round the answer to the first decimal if needed.
b What height will the girls be at after minutes of riding the Ferris wheel? Round the answer to one decimal if needed.
Discussion
## Presenting Inverse Trigonometric Functions
There are functions that undo the trigonometric functions, so to speak. These functions are called inverse trigonometric functions.
Concept
## Inverse Trigonometric Functions
The inverse trigonometric functions are the inverse functions of the trigonometric functions. For example, the inverse sine is the inverse function of the sine function. The main inverse trigonometric functions are shown in the table below.
Trigonometric Function Inverse Trigonometric Function
The inverse trigonometric functions relate an input, which represents the ratio of two sides of a right triangle, to the measure of one of its two acute angles. The output angles are measured in radians.
The domain of the corresponding trigonometric function must be restricted in order for its inverse to be defined as a function.
The properties of the main inverse trigonometric functions are summarized in the following table.
Inverse Trigonometric Function Domain Range
All real numbers
Note that the inverse trigonometric functions are also called and respectively. These can also be written as and respectively.
The inverse trigonometric functions are especially useful when solving equations involving trigonometric functions.
Discussion
## Trigonometric Equations and Their Solutions
Contrary to trigonometric identities — which are true for all values of the variable for which both sides are defined — some equations involving trigonometric functions are true only for certain values of the variable. Now such equations will be presented.
Concept
## Trigonometric Equation
A trigonometric equation is an equation that includes one or more trigonometric functions. Consider the following example trigonometric equations.
If a trigonometric equation consists only trigonometric functions and constants, the solution is found by looking for the values of the argument that make the equation true. Solving trigonometric equations is similar to solving algebraic equations.
If possible, it is useful to rewrite a trigonometric equation such that they have the same trigonometric function on its both sides.
However, in cases where it is not possible, the inverse function to the trigonometric function that appears in the equation can be used.
Since the functions on the left-hand side are inverses, they cancel each other out and can be simplified to The value of the right-hand side, on the other hand, can be found by using a calculator.
Note that because trigonometric functions are periodic, they can have numerous angles corresponding to the same trigonometric value. This point can be illustrated by the following graph that shows at least four different solutions of the given equation.
To sum up, the following facts can be used to solve trigonometric equations. Note that and written below are integers.
Equation Solutions
It can be concluded that trigonometric identities are a special case of trigonometric equation. Next, it will be shown how to solve different trigonometric equations.
Example
## Decoding the Message with the Solutions to Trigonometric Equations
Today Magdalena’s math class started learning how to solve trigonometric equations. To encourage students to practice solving some more equations at home, Magdalena's teacher gave the students a fun task. A secret message has been encoded. The only way to read the message is to solve two equations to decode it.
Input the answers to the first equation first. The smaller number in each pair should be written first. Write the numbers one after the other without any commas or spaces. Students who decode the message will get a bonus point on the upcoming test. The teacher said that
a
b
### Hint
a Gather all the trigonometric functions and constants on different sides of the equation and simplify.
b Use the Double-Angle Identity for cosine to write in terms of Then gather all the terms on one side of the equation and rewrite them as a product of two expressions.
### Solution
a To solve the given equation, start by rewriting it so that all the trigonometric function appear on one side of the equation and the constants on the other side.
Now, try to find the angle for which the value of equals The table of the trigonometric values of notable angles can be useful here.
It can be concluded that this equation has two possible solutions.
b Start by analyzing the given equation.
There are two different trigonometric functions involved in the equation, sine and cosine. Therefore, the first step is to rewrite one side of the equation in terms of the other. To do this, the Double-Angle Identity for cosine can be used.
Next, to solve this equation, rewrite the left-hand side as a product of two expressions.
Factor
By the Zero Product Property, in order for the equation to be true, at least one of the parentheses should be equal to
Note that adding to both sides of the Equation results in However, this is not possible, as the values of sine are always between and Therefore, this equation has no solution. Now consider Equation To find the value of or arcsine of think about the angles for which the value of sine is According to the table of the trigonometric ratios of common angles, there are two such angles in the interval between and Therefore, the given equation has two possible solutions.
Now that all the solutions have been found, the code mentioned at the beginning can be determined. In Part A, the solutions and were found, while the solutions from Part B were and The code can be found by combining these numbers in the order directed at the start.
Try it out to see the hidden message!
Example
## Math Challenge With a Gift
Magdalena really liked the math joke that her teacher encoded with the solutions to the exercise. She and her friend Davontay decided to give each other two more equations to solve. If they solve them correctly, they will gift each other an envelope with another math joke they came up with or heard somewhere.
Magdalena received the following two trigonometric equations from Davontay. He told her to write all the possible solutions in the form of an equation where is an integer number.
a
b
### Hint
a Use the Double-Angle Identity for cosine to rewrite in terms of
b Apply one of the Pythagorean Identities and simplify the equation.
### Solution
a In the first equation, the same trigonometric function appears two times but it has different arguments. Therefore, use the Double-Angle Identity for cosine to rewrite the equation.
Apply this identity and then simplify the equation.
In order to find the values of that satisfy this equation, think about the angles for which the cosine equals According to the table of the trigonometric ratios of common angles, there is at least one such angle.
Since cosine is a periodic function with a period of , for every integer the following angles are the solutions to the equation.
b Start by analyzing the given equation.
There are two trigonometric functions in the equation, tangent and secant. Recall that one of the Pythagorean Identities relates these two functions.
Substitute for into the equation and simplify it. Next, look for the angles for which tangent is equal to or
To find the general equation for the solutions, note that all these angles can be expressed in the following way.
Therefore, all the solutions to the equation can be given by the following equation where is an integer number.
After correctly solving both equations, Magdalena can finally read the joke that Davontay found for her.
Example
## Making a Bet on Equations
The teacher gave the class a couple of exercises to solve for homework. She also warned that one of the equations has extraneous solutions and that the students should identify them. To make things interesting, Magdalena and Davontay decided to make a bet about which equation has extraneous solutions.
After each chose an equation, they started solving them to see who guessed correctly. The winner will get the last piece of cake left in the fridge. To solve the equations, they must write all the solutions in radians such that
a
b
Who guessed correctly?
### Hint
a Use the Pythagorean Identity to rewrite in terms of Then factor the equation and apply the Zero Product Property.
b Move to the right side of the equation and then square both sides. Check the solutions by substituting them into the original equation and seeing if it remains true.
### Solution
a Start by examining the first equation.
Since there are two different trigonometric functions, one should be rewritten in terms of the other. To do so, use the Pythagorean Identity, but first rewrite the equation so that is isolated on one side of the equation.
Substitute for and simplify the equation.
Next, factor the expression on the left-hand side into two parentheses.
By the Zero Product Property, at least one of the parentheses must be equal to Two equations can be formed by applying this property. To solve these two equations, try to find the angles for which the values of sine are and For the first equation, there are two such angles — and — while for the second there is only one —
Finally, the solutions should be checked whether there are any extraneous solutions. To do that, substitute each of them into the original equation and see if it is true.
Simplify left-hand side
The other two solutions can be checked in a similar fashion.
Solution Substitute Evaluate True or False
Therefore, there are three solutions to the equation and none of them are extraneous.
b Again, begin by analyzing the given equation.
First, move to the right-hand side of the equation. Then, square each side of the equation.
Now, to rewrite one trigonometric function in terms of the other, use the Pythagorean Identity. Similarly to Part A, substitute for To find the solutions of this equation, look for the angles for which the sine equals The table with the trigonometric ratios of notable angles can be useful here.
Finally, substitute each of these solutions into the original equation in order to determine whether any of them is extraneous.
Solution Substitute Evaluate True or False
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# Euler-Lagrange Equation and How to Use it in 5 Easy Steps
Level 3 (with higher mathematics)
Level 3 requires the basics of vector calculus, differential and integral calculus. Suitable for undergraduates and high school students.
Updated by Alexander Fufaev on
## Video - Euler-Lagrange Equation
Let us consider a particle in the gravitational field which is thrown vertically upwards from the height $$h_1 = 0$$ at the time $$t_1 = 0$$. It moves upwards in a straight line and arrives at the ground at the same location $$h_2 = 0$$ at time $$t_2$$.
In a position-time diagram, the particle has started at height $$h_1$$ at time $$t_1$$. Let us denote this starting point with $$\text{A}$$. And the particle has arrived at the end point $$\text{B}$$ at height $$h_2$$ at time $$t_2$$.
The connection between $$\text{A}$$ and $$\text{B}$$ must be a parabola in this problem. But why is this path $$h(t)$$ a parabola and not any other path? Why does nature choose exactly this path? And not for any other path as shown in illustration 2?
## Action Functional
To be able to answer this question, we need a quantity called action, which is abbreviated with the letter $$S$$. We can assign an action value to each of these conceivable paths. The action $$S$$ takes a whole function $$h$$ in the argument and spits out a number S[h], namely the value of the action for the corresponding function.
Such a mathematical object that eats a function $$h$$ and spits out a number is called a functional. Since the functional spits out the action, $$S[h]$$ is also called an action functional. To distinguish the action functional from usual functions that take just a number in the argument, we use square brackets.
The action has the unit Joulesecond ($$\mathrm{Js}$$). For example, one path might have the value $$3.5 \, \mathrm{Js}$$, another path might have the value $$5.6 \, \mathrm{Js}$$, and the parabolic path might have $$2 \, \mathrm{Js}$$.
So now back to the question: Why is the path a parabola? Experience shows that:
That means, if we calculate the action for all possible paths $$h_1(t)$$, $$h_2(t)$$, $$h_3(t)$$ and so on between $$A$$ and $$B$$, then nature takes the action value which is either maximum, minimum or a saddle point. Nature chooses one of these paths. Which of these extremal paths nature takes concretely depends on the considered problem.
In the case of a particle thrown upwards in the gravitational field, the path has the smallest action, so a minimum. Since the parabolic path has the smallest action, the particle chooses this path in the position vs. time diagram.
But how do we calculate these values of the action? For this we need the Lagrange function $$L$$. It depends on the time $$t$$, on the function value $$h(t)$$ and on the velocity $$\dot{h}(t)$$. The Lagrange function has the unit of energy, that is, joule ($$\mathrm{J}$$). If we integrate the Lagrange function over the time t between t1 and t2, we get a quantity that has the unit joules seconds. This is exactly the action we need. Thus we can calculate concretely the value of the action for each possible path "h", if we specify the Lagrange function.
If we integrate the Lagrange function over the time $$t$$ between $$t_1$$ and $$t_2$$, we get a quantity that has the unit joulesecond:
Action Functional Expressed with the Height Function
Formula anchor
This is exactly the action we need. Thus we can calculate concretely the value of the action for each possible path $$h$$, if we specify the Lagrange function.
Mostly you use the letter $$q$$ instead of $$h$$ and $$\dot{q}$$ instead of $$\dot{h}$$ and call $$q$$ as generalized coordinate and the derivate $$\dot{q}$$ as generalized velocity. What is meant by "generalized" you will learn in another lessen. For us it is only important to know that $$q$$ can be for example a height above the ground or an angle or any other quantity which can depend on the time $$t$$.
Action functional as integral of the Lagrange function
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## Euler-Lagrange equation to determine the path
Of course, it is totally cumbersome to calculate the integral for all possible paths and to take the path that results in the smallest value of the integral. To avoid such a huge task, the so-called Euler-Lagrange equation comes into play:
Euler-Lagrange equation
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The derivation of the Euler-Lagrange equation is based on the definition of action 2 and the stationary-action principle. In this lesson, we do not want to know how to obtain the Euler-Lagrange equation, but how to use it to determine the path $$q(t)$$ we are looking for.
The Euler-Lagrange equation contains the partial derivative of the Lagrangian function $$L$$ with respect to the genaralized velocity $$\dot{q}$$: $$\frac{\partial L}{\partial \dot{q}}$$. This derivative $$\frac{\partial L}{\partial \dot{q}}$$ is also called generalized momentum and abbreviated with $$p$$. The generalized momentum is then differentiated with respect to $$t$$.
Euler-Lagrange equation using momentum
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The second term in the Euler-Lagrange equation is the derivative of the Lagrangian function $$L$$ with respect to the generalized coordinate $$q$$: $$\frac{\partial L}{\partial q}$$.
If we bring the time derivative of the momentum to the other side, we can read from the Euler-Lagrange equation whether the momentum is conserved.
Conservation of momentum in the Euler-Lagrange equation
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For this, the time derivative of the momentum must be zero. So we only have to calculate if $$\frac{\partial L}{\partial q}$$ is equal zero.
## Lagrange function as ingredient
The Lagrange function $$L$$ is a scalar function that cannot be derived, but can only be guessed or motivated. If you think you have discovered a suitable Lagrangian for a problem, be it from quantum mechanics, classical mechanics or relativity, you can easily check whether the Lagrangian you found describes your problem correctly or not by using the Euler-Lagrange equation.
In most cases of classical mechanics, the Lagrange function is the difference between the kinetic energy $$W_{\text{kin}}$$and the potential energy $$W_{\text{pot}}$$ of a particle:
Lagrangian function of classical mechanics
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Thus, if we know the kinetic and potential energies of a particle, we can determine the Lagrange function $$L$$ and then use $$L$$ in the Euler-Lagrange equation 3.
## Example: How to use Euler-Lagrange equation
Let's look at our example and how we can calculate the parabola from the Lagrange function and the Euler-Lagrange equation. For this we have to do 5 steps.
### Step #1: Choose generalized coordinates
First, we need to know what $$q$$ and $$\dot{q}$$ represent. In our example, $$q = h$$ and $$\dot{q} = v$$, where $$v(t)$$ is the velocity of the thrown particle. Velocity here is nothing else than the time derivative of the path: $$\dot{q} = \dot{h}$$.
### Step #2: Set up Lagrange function
Next, we need to specify the Lagrange function 6 as a function of $$h$$ and $$\dot{h}$$. The kinetic energy $$W_{\text{kin}}(t, h, \dot{h})$$ of the thrown particle is given by:
Kinetic energy for the Lagrangian
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The potential energy $$W_{\text{pot}}(t, h, \dot{h})$$ of the particle in the gravitational field is given by:
Potential energy for the Lagrangian
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Thus, the Lagrange function is $$L(t, h, \dot{h})$$ for our problem:
Lagrange function for a particle in the gravitational field
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### Step #3: Calculate derivatives in the Euler-Lagrange equation
Now we can use the specified Lagrangian function 9 to calculate the derivatives occurring in the Euler-Lagrange equation 3. We first write the Euler-Lagrange equation using $$h$$ and $$\dot{h}$$:
Euler-Lagrange equation for a particle in the gravitational field
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• Differentiate the Lagrange function 9 with respect to $$h$$:
Lagrange function differentiated with respect to h
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• Differentiate the Lagrangian function 9 with respect to $$\dot{h}$$:
Lagrange function differentiated with respect to the velocity
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• Differentiate the momentum $$\frac{\partial L}{\partial \dot{q}} = m \, \dot{h}$$ with respect to time $$t$$:
Differentiation of the momentum with respect to time
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Let us substitute the calculated derivatives into the Euler-Lagrange equation:
Euler-Lagrange equation calculated
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Let's cancel the mass $$m$$ and bring $$\ddot{h}$$ to the other side:
Set up differential equation
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What we get is a differential equation for the function we are looking for $$h$$. Here you can see the meaning of the Euler-Lagrange equation! It tells us which differential equation we have to solve to find out the time behavior of $$h$$.
Note that our example is a one-dimensional problem, so we only got one differential equation out of it. For more complex multidimensional problems, we get several differential equations.
### Step #4: Solve the differential equation
Now we have to solve the differential equation 15 set up with the help of the Euler-Lagrange equation. To do this, we integrate both sides over time $$t$$:
Integrate both sides of the differential equation
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Then we get:
Result of the first integration
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This results in two integration constants. We combine these to one integration constant $$C_1$$. Then we integrate both sides again over time to eliminate the time derivative of $$h$$:
Integrate both sides of the differential equation a second time
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Then we get:
Solution of the differential equation
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Here $$C_2$$ is a combined integration constant. Now we have solved the differential equation for $$h$$!
### Step #5: Use boundary conditions
As a last step, we have to insert the boundary conditions of the considered problem into the solution of the differential equation and determine the unknown constants $$C_1$$ and $$C_2$$.
In our problem, we threw the particle from the height $$h_1 = 0$$ at the time $$t_1 = 0$$. So the first constraint is: $$h(0) = 0$$. Let's insert it into the result 19:
Determine second integration constant
Formula anchor
Thus we have figured out the integration constant $$C_2 = 0$$. Let's insert it into the Eq. 19:
Second constant inserted into the differential equation
Formula anchor
To find out the constant $$C_1$$, we need a second boundary condition. We know that the path of $$h(t)$$ ends at the point $$\text{B}$$. The point $$\text{B}$$ corresponds to the time $$t_2$$ at which the particle landed on the ground at $$h_2 = 0$$. The second boundary condition is therefore: $$h(t_2) = 0$$. Insert it into the differential equation:
Determine first integration constant
Formula anchor
Insert $$C_1$$ into the Eq. 21:
Final result: Function $$h$$
Formula anchor
If you plot the result in the position-time diagram, you get a parabola. The Euler-Lagrange equation delivers exactly the result we expected!
The Euler-Lagrange equation, together with the Lagrange function, helps us to set up differential equations for a concrete problem. The solution of these differential equations yields the shape of the function that nature allows.
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# Party Preparations: How Many Ways Can Susan Assign Tasks to Her Helpful Friends?
When it comes to party preparations, assigning tasks to friends can be a tricky business. It’s not just about who can do what, but also about how many different ways tasks can be assigned. This is a question that Susan, who is planning a party, is grappling with. She has five friends – Vicki, Luis, Maddy, Hank, and Rob – who have volunteered to help. She needs someone to buy beverages, someone to arrange for food, and someone to send the invitations. But how many ways can she assign these tasks? Let’s delve into this interesting problem and find out.
## Understanding the Problem
The problem at hand is essentially a permutation problem. In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order. In Susan’s case, she has five friends and three tasks, and she needs to figure out how many different ways she can assign these tasks.
## Calculating the Permutations
In mathematics, the number of permutations of n distinct objects taken r at a time is given by the formula nPr = n! / (n-r)!. Here, n! denotes the factorial of n, which is the product of all positive integers less than or equal to n. In Susan’s case, n is 5 (the number of her friends) and r is 3 (the number of tasks), so the number of permutations is 5P3 = 5! / (5-3)! = 60.
## Interpreting the Result
This means that there are 60 different ways Susan can assign the tasks of buying beverages, arranging for food, and sending the invitations to her five friends. This includes all possible combinations, such as Vicki buying beverages, Luis arranging for food, and Maddy sending the invitations, or Hank buying beverages, Rob arranging for food, and Vicki sending the invitations, and so on.
## Conclusion
So, Susan has plenty of options when it comes to assigning tasks for her party preparations. This mathematical approach not only provides a precise answer to her question, but also illustrates the power of permutations in solving real-world problems. Whether you’re planning a party or tackling a complex project, understanding permutations can help you see all the possible ways you can assign tasks and make informed decisions.
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# GRE Subject Test: Math : Factorials
## Example Questions
← Previous 1
### Example Question #1 : Factorials
Simplify the following expression:
Explanation:
Recall that ! means facotrial in math. This means we multiply the number by all positive integers less than itself. In other words, this...
Becomes
This is a great job for a calculator, which yields:
### Example Question #2 : Factorials
Evaluate:
Explanation:
Step 1: We need to define factorial (!). The operation (!) means that you multiply the number next to the ! by the next consecutive number down to 1.
Step 2: Expand the equation:
Step 3: Evaluate the expression:
The answer to the expression is .
### Example Question #3 : Factorials
Explanation:
is the product of consecutive numbers 1 through
### Example Question #4 : Factorials
Explanation:
is the product of consecutive numbers 1 through
### Example Question #5 : Factorials
Evaluate:
Explanation:
Step 1: Rewrite the equation by using rule of factorials...
Rule of factorials: Multiply the number in front of factorial and each number one lower than the rest until I hit .
Step 2: Reduce any common numbers (colored in blue):
We get !
Step 3: Evaluate what's left
The value of the fraction is .
### Example Question #6 : Factorials
What is equal to?
Explanation:
depicts the permutation of n total objects with k objects being arranged.
Thus,
The upper factorial is the upper index, and the lower factorial is the difference of the indices.
The or will cancel out.
### Example Question #7 : Factorials
Twelve horses run a race. In how many ways can 3 horses finish in 1st, 2nd, and 3rd place, in any order?
Explanation:
This particular question is asking one to find the combination. This is because the order of the 1st, 2nd, and 3rd place finishers is not important.
Thus evaluate .
Because and both consist of , the will cancel out leaving:
### Example Question #8 : Factorials
Evaluate:
Explanation:
Step 1: Evaluate
Step 2: Evaluate
Step 3: Evaluate
Step 4: Evaluate
### Example Question #9 : Factorials
Evaluate:
Explanation:
Step 1: Do the division of the factorials
Step 2: Simplify by cancelling out any terms on both top and bottom:
Step 3: Multiply the result of the division by 3..
The value of that expression is .
### Example Question #10 : Factorials
Evaluate:
Explanation:
Step 1: Find
Step 2: Find
Step 3: Subtract the values from step 1 and step 2:
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# Eureka Math Geometry Module 1 Lesson 2 Answer Key
## Engage NY Eureka Math Geometry Module 1 Lesson 2 Answer Key
### Eureka Math Geometry Module 1 Lesson 2 Exercise Answer Key
Opening Exercise
You need a compass, a straightedge, and another student’s Problem Set.
Directions:
Follow the directions from another student’s Problem Set write-up to construct an equilateral triangle.
→ What kinds of problems did you have as you followed your classmate’s directions?
→ Think about ways to avoid these problems. What criteria or expectations for writing steps in constructions should be included in a rubric for evaluating your writing? List at least three criteria.
### Eureka Math Geometry Module 1 Lesson 2 Exploratory Challenge Answer Key
Exploratory Challenge 1.
You need a compass and a straightedge.
Using the skills you have practiced, construct three equilateral triangles, where the first and second triangles share a common side and the second and third triangles share a common side. Clearly and precisely list the steps needed to accomplish this construction.
Switch your list of steps with a partner, and complete the construction according to your partner’s steps. Revise your drawing and list of steps as needed.
Construct three equilateral triangles here:
1. Draw a segment AB.
2. Draw circle A: center A, radius AB.
3. Draw circle B: center B, radius BA.
4. Label one intersection as C; label the other intersection as D.
5. Draw circle C: center C, radius CA.
6. Label the intersection of circle C with circle A (or the intersection of circle C with circle B) as E.
7. Draw all segments that are congruent to $$\overline{A B}$$ between the labeled points.
There are many ways to address Step 7; students should be careful to avoid making a blanket statement that would allow segment BE or segment CD.
Exploratory Challenge 2.
On a separate piece of paper, use the skills you have developed in this lesson to construct a regular hexagon. Clearly and precisely list the steps needed to accomplish this construction. Compare your results with a partner, and revise your drawing and list of steps as needed.
1. Draw circle K: center K, any radius.
2. Pick a point on the circle; label this point A.
3. Draw circle A: center A, radius AK.
4. Label the intersections of circle A with circle K as B and F.
5. Draw circle B: center B, radius BK.
6. Label the intersection of circle B with circle K as C.
7. Continue to treat the intersection of each new circle with circle K as the center of a new circle until the next circle to be drawn is circle A.
8. Draw $$\overline{A B}$$, $$\overline{B C}$$, $$\overline{C D}$$, $$\overline{D E}$$, $$\overline{E F}$$, $$\overline{F A}$$.
Can you repeat the construction of a hexagon until the entire sheet is covered in hexagons (except the edges, which are partial hexagons)?
Yes, this result resembles wallpaper, tile patterns, etc.
### Eureka Math Geometry Module 1 Lesson 2 Problem Set Answer Key
Why are circles so important to these constructions? Write out a concise explanation of the importance of circles in creating equilateral triangles. Why did Euclid use circles to create his equilateral triangles in Proposition 1? How does construction of a circle ensure that all relevant segments are of equal length?
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Home » Math Theory » Operations of Numbers » Addition Of Numbers Within 100
# Addition Of Numbers Within 100
## Introduction
Addition is one of the most important arithmetic operations used in mathematics. It is the method of calculating the total of two or more numbers to know the sum of the numbers. We use addition not just while studying mathematics but in solving our day to day problems as well. Before we learn how to add numbers let us understand the definition of addition.
Addition, in mathematics, can be defined as the process of combining two or more numbers together to make a new total or sum. The numbers to be added together are called addends and the result thus obtained is called the sum.
## Addition of a Single Digit Number with a Single Digit Number
We know that the single digit numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. This is the order in which we count the numbers where 2 comes after 1, 3 comes after 2 and so on. A single digit has the place value of a unit in the place value system of the numbers.
Let us consider a situation where you have 3 chocolates. Your brother gives you 4 more chocolates. How many chocolates do you have now?
In order to add 4 chocolates to the 3 chocolates that you already have, starting from 3, you will move 4 places to the right of the counting of numbers. So, 4 points after 3, in the number system is 7. Hence, 3 + 4 = 7
Let us take another example.
Let us add 2 to 4. Again if we want to add 2 to 4 we will have to move 2 numbers to the right of 4. So, moving 2 points ahead of 4, we will get, 6 which means that 4 + 2 = 6.
Now, that we have understood how to add a single digit number to another single digit number, let us learn how to add a single digit number to a two digit number.
## Addition of a Single Digit Number with a Two Digit Number
In the counting of numbers, two digit numbers start from 10. Then we have 11, 12, 13, 14, 15, ….. and so on. It is important to note here that two digit number “ab” can be written in the form 10 x a + b, where a has the ten’s place value and b has the unit value in the number system.
### Addition with No Carrying Forward
For adding a Single Digit with a Two Digit Number, the digit at the one’s place value of first number will be added to the digit at the one’s place of the second number.
Let us understand it using an example.
Suppose we want to add 5 and 13.
5 – 5 is at the unit’s place in the number.
13 – 1 is the ten’s place while 3 is the unit’s place.
So, we add the unit’s digit of both the number as shown below –
Here we can see that 5 + 3 = 8 < 10, so the result obtained was a single digit number at the unit’s place. But if we were to add the numbers 5 + 16, we would have got 5 + 6 = 11, then how would we have written 11 in the answer. This is where the concept of carrying forward comes into force.
Let us consider two numbers 5 and 16. How will we add these two numbers? Let us find out.
5 = 0 tens + 5 ones
16 = 1 tens + 6 ones.
Therefore,
5 + 16 =
0 tens + 5 ones
+1 tens + 6 ones
——————–
1 tens + 11 ones
——————–
Now, 11 ones = 1 tens + 1 ones
Therefore, we have 5 + 16 = 1 tens +1 tens + 1 ones = 2 tens + 1 ones.
Hence, 5 + 16 = 21
## Addition of a Two Digit Number with a Two Digit Number
Let us now learn how to add two numbers of two digits. We have just discussed that in a two digit number, we have two digits, one at the ten’s place and the other at the one’s place. Let us see how to add these numbers.
### Addition with No Carrying Forward
Consider two numbers 12 and 16. Suppose we want to add these two numbers. How will we find their sum? Let us find out.
Recall that,
12 = 1 tens + 2 ones
16 = 1 tens + 6 ones
Therefore,
The sum of 12 and 16 will be given by adding the respective one’s digits with each other and the ten’s digits with each other. We will have,
12 +16 = =
1 tens + 2 ones
+ 1 tens + 6 ones
———————
2 tens + 8 ones
———————
Hence, 12 + 16 = 28
Here we can see that 2 + 6 = 8 < 10, so the result obtained was a single digit number at the unit’s place. But if we were to add the numbers such as 8 + 9, we would have got 8 + 9 = 17, then how would we have written 17 in the answer. This is the same situation, we discussed while understanding two to add a single digit and a two digit number. Here as well, we will proceed in the same manner, i.e. by carrying forward.
Let us consider two numbers 28 and 16. How will we add these two numbers? Let us find out.
1. We know that 28 = 2 tens + 8 ones and 16 = 1 tens + 6 ones
2. Now, 8 ones + 6 ones = 14 ones. As the sum of the digits at the one’s place exceeds 9, you must carry ones into tens.
3. 14 ones = 10 ones + 4 ones = 1 tens + 4 ones
4. Write 4 under ones column and carry
6. 1 ten ( that was carried over ) + 2 tens + 1 ten = 4 tens.
7. Thus 28 + 16 = 44
## Addition Using a Number Line
Let us now understand the concept of adding one or digit numbers using the number line.
We move to the right on our number line for adding two numbers. The following example shall help you understand how to add two numbers to a number line.
For example, you want to add 3 + 2
Steps involved –
1. First draw a number line having both negative as well as positive numbers.
2. Mark the first number 3 on this number line.
3. Move 2 points right as we have to add the numbers.
4. The number we have our point on is our answer, i.e. 5
Hence 3 + 2 = 5
## Addition Using the Hundred Grid
Another way of adding two digits numbers is by using the hundred grid. A hundred grid is where you mark numbers from 1 to 100 in the following manner –
Let us use the above grid to add two numbers. For this let us have two numbers 57 and 16 and we want to find their sum.
We shall use the following steps to add these two numbers using the hundred grid –
1. Mark the bigger number on the hundred grid. In this case, the bigger number is 57. So we ill mark it on the number grid. We will get,
1. If the number to be added is more than 9, break it into tens and ones. In our case, we have the number 16. So, breaking it into tens and ones, we have, 16 = 1 tens + 6 ones.
2. Jump as many tens as in the second number. The number 16 has 1 tens, so we will jump 1 tens to reach the number 67 and mark it on the grid as shown below –
1. Now, we will move forward as many ones as in the second number. 16 has 6 ones, so, we will move 6 numbers forward from 67 to reach 73 and show it on the grid as below –
1. The number that we have reached is our answer. So, 57 + 16 = 73
The understanding of the addition table is very important for better learning of the addition of small numbers. The addition table is a table that shows the results of the addition of a single digits number with another single number.
How do we use the above table to add two single digit numbers? Let us find out.
Suppose we want to add the number 6 and 9.
In order to use this table, of the two given numbers, we will mark one number of the horizontal ( rows ) top of the table and the other in the vertical (columns ) headings of the table. For instance, we will mark 6 on the vertical side and 8 on the side of the row to get –
Now we will check the intersection of these two numbers on the matrix table. In this case, we can see that they intersect at the number 14. Hence, 8 + 16 = 14
## Some Important Results
1. Zero added to any number gives the same number, for example, 0 + 7 = 7.
2. When you add 1 to a number, the answer is the number “ just after” example, 5 + 1 = 6.
3. Changing order of numbers while adding does not change the answer, for example 4 + 3 = 3 + 4 = 7.
4. “ + “ is the sign used for addition.
5. 5 + 3 = 8 is read as “ five plus three is equal to eight”.
## Key Notes and Summary
1. Addition, in mathematics, can be defined as the process of combining two or more numbers together to make a new total or sum. The numbers to be added together are called addends and the result thus obtained is called the sum.
2. A single digit has the place value of a unit in the place value system of the numbers.
3. For adding a Single Digit with a Two Digit Number, the digit at the one’s place value of first number will be added to the digit at the one’s place of the second number.
4. The addition table is a table that shows the results of the addition of a single digits number with another single number.
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# 9.3 Double-angle, half-angle, and reduction formulas (Page 4/8)
Page 4 / 8
Given that $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =-\frac{4}{5}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ lies in quadrant IV, find the exact value of $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\left(\frac{\alpha }{2}\right).$
$-\frac{2}{\sqrt{5}}$
## Finding the measurement of a half angle
Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{5}{3}\text{\hspace{0.17em}}$ for higher-level competition, what is the measurement of the angle for novice competition?
Since the angle for novice competition measures half the steepness of the angle for the high level competition, and $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{5}{3}\text{\hspace{0.17em}}$ for high competition, we can find $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ from the right triangle and the Pythagorean theorem so that we can use the half-angle identities. See [link] .
$\begin{array}{ccc}\hfill {3}^{2}+{5}^{2}& =& 34\hfill \\ \hfill c& =& \sqrt{34}\hfill \end{array}$
We see that $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{3}{\sqrt{34}}=\frac{3\sqrt{34}}{34}.\text{\hspace{0.17em}}$ We can use the half-angle formula for tangent: $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\frac{\theta }{2}=\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\theta }{1+\mathrm{cos}\text{\hspace{0.17em}}\theta }}.\text{\hspace{0.17em}}$ Since $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in the first quadrant, so is $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\frac{\theta }{2}.\text{\hspace{0.17em}}$
$\begin{array}{ccc}\hfill \mathrm{tan}\text{\hspace{0.17em}}\frac{\theta }{2}& =& \sqrt{\frac{1-\frac{3\sqrt{34}}{34}}{1+\frac{3\sqrt{34}}{34}}}\hfill \\ & =& \sqrt{\frac{\frac{34-3\sqrt{34}}{34}}{\frac{34+3\sqrt{34}}{34}}}\hfill \\ & =& \sqrt{\frac{34-3\sqrt{34}}{34+3\sqrt{34}}}\hfill \\ & \approx & 0.57\hfill \end{array}$
We can take the inverse tangent to find the angle: $\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}\left(0.57\right)\approx 29.7°.\text{\hspace{0.17em}}$ So the angle of the ramp for novice competition is $\text{\hspace{0.17em}}\approx 29.7°.$
Access these online resources for additional instruction and practice with double-angle, half-angle, and reduction formulas.
## Key equations
Double-angle formulas $\begin{array}{ccc}\hfill \mathrm{sin}\left(2\theta \right)& =& 2\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta \hfill \\ \hfill \mathrm{cos}\left(2\theta \right)& =& {\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \hfill \\ & =& 1-2{\mathrm{sin}}^{2}\theta \hfill \\ & =& 2{\mathrm{cos}}^{2}\theta -1\hfill \\ \hfill \mathrm{tan}\left(2\theta \right)& =& \frac{2\mathrm{tan}\text{\hspace{0.17em}}\theta }{1-{\mathrm{tan}}^{2}\theta }\hfill \end{array}$ Reduction formulas $\begin{array}{ccc}\hfill {\mathrm{sin}}^{2}\theta & =& \frac{1-\mathrm{cos}\left(2\theta \right)}{2}\hfill \\ \hfill {\mathrm{cos}}^{2}\theta & =& \frac{1+\mathrm{cos}\left(2\theta \right)}{2}\hfill \\ \hfill {\mathrm{tan}}^{2}\theta & =& \frac{1-\mathrm{cos}\left(2\theta \right)}{1+\mathrm{cos}\left(2\theta \right)}\hfill \end{array}$ Half-angle formulas $\begin{array}{ccc}\hfill \mathrm{sin}\text{\hspace{0.17em}}\frac{\alpha }{2}& =& ±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \\ \hfill \mathrm{cos}\text{\hspace{0.17em}}\frac{\alpha }{2}& =& ±\sqrt{\frac{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \\ \hfill \mathrm{tan}\text{\hspace{0.17em}}\frac{\alpha }{2}& =& ±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }}\hfill \\ & =& \frac{\mathrm{sin}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }\hfill \\ & =& \frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{\mathrm{sin}\text{\hspace{0.17em}}\alpha }\hfill \end{array}$
## Key concepts
• Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent. See [link] , [link] , [link] , and [link] .
• Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term. See [link] and [link] .
• Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not. See [link] , [link] , and [link] .
## Verbal
Explain how to determine the reduction identities from the double-angle identity $\text{\hspace{0.17em}}\mathrm{cos}\left(2x\right)={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x.$
Use the Pythagorean identities and isolate the squared term.
Explain how to determine the double-angle formula for $\text{\hspace{0.17em}}\mathrm{tan}\left(2x\right)\text{\hspace{0.17em}}$ using the double-angle formulas for $\text{\hspace{0.17em}}\mathrm{cos}\left(2x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{sin}\left(2x\right).$
We can determine the half-angle formula for $\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{x}{2}\right)=\frac{\sqrt{1-\mathrm{cos}\text{\hspace{0.17em}}x}}{\sqrt{1+\mathrm{cos}\text{\hspace{0.17em}}x}}\text{\hspace{0.17em}}$ by dividing the formula for $\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{x}{2}\right)\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{x}{2}\right).\text{\hspace{0.17em}}$ Explain how to determine two formulas for $\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{x}{2}\right)\text{\hspace{0.17em}}$ that do not involve any square roots.
$\text{\hspace{0.17em}}\frac{1-\mathrm{cos}\text{\hspace{0.17em}}x}{\mathrm{sin}\text{\hspace{0.17em}}x},\frac{\mathrm{sin}\text{\hspace{0.17em}}x}{1+\mathrm{cos}\text{\hspace{0.17em}}x},$ multiplying the top and bottom by $\text{\hspace{0.17em}}\sqrt{1-\mathrm{cos}\text{\hspace{0.17em}}x}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\sqrt{1+\mathrm{cos}\text{\hspace{0.17em}}x},$ respectively.
For the half-angle formula given in the previous exercise for $\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{x}{2}\right),$ explain why dividing by 0 is not a concern. (Hint: examine the values of $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ necessary for the denominator to be 0.)
## Algebraic
For the following exercises, find the exact values of a) $\text{\hspace{0.17em}}\mathrm{sin}\left(2x\right),$ b) $\text{\hspace{0.17em}}\mathrm{cos}\left(2x\right),$ and c) $\text{\hspace{0.17em}}\mathrm{tan}\left(2x\right)\text{\hspace{0.17em}}$ without solving for $\text{\hspace{0.17em}}x.$
If $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x=\frac{1}{8},$ and $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in quadrant I.
a) $\text{\hspace{0.17em}}\frac{3\sqrt{7}}{32}\text{\hspace{0.17em}}$ b) $\text{\hspace{0.17em}}\frac{31}{32}\text{\hspace{0.17em}}$ c) $\text{\hspace{0.17em}}\frac{3\sqrt{7}}{31}$
If $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x=\frac{2}{3},$ and $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in quadrant I.
If $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x=-\frac{1}{2},$ and $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in quadrant III.
a) $\text{\hspace{0.17em}}\frac{\sqrt{3}}{2}\text{\hspace{0.17em}}$ b) $\text{\hspace{0.17em}}-\frac{1}{2}\text{\hspace{0.17em}}$ c) $\text{\hspace{0.17em}}-\sqrt{3}\text{\hspace{0.17em}}$
x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0
x exposent4+4x exposent3+8x exposent2+4x+1=0
HERVE
How can I solve for a domain and a codomains in a given function?
ranges
EDWIN
Thank you I mean range sir.
Oliver
proof for set theory
don't you know?
Inkoom
find to nearest one decimal place of centimeter the length of an arc of circle of radius length 12.5cm and subtending of centeral angle 1.6rad
factoring polynomial
find general solution of the Tanx=-1/root3,secx=2/root3
find general solution of the following equation
Nani
the value of 2 sin square 60 Cos 60
0.75
Lynne
0.75
Inkoom
when can I use sin, cos tan in a giving question
depending on the question
Nicholas
I am a carpenter and I have to cut and assemble a conventional roof line for a new home. The dimensions are: width 30'6" length 40'6". I want a 6 and 12 pitch. The roof is a full hip construction. Give me the L,W and height of rafters for the hip, hip jacks also the length of common jacks.
John
I want to learn the calculations
where can I get indices
I need matrices
Nasasira
hi
Raihany
Hi
Solomon
need help
Raihany
maybe provide us videos
Nasasira
Raihany
Hello
Cromwell
a
Amie
What do you mean by a
Cromwell
nothing. I accidentally press it
Amie
you guys know any app with matrices?
Khay
Ok
Cromwell
Solve the x? x=18+(24-3)=72
x-39=72 x=111
Suraj
Solve the formula for the indicated variable P=b+4a+2c, for b
Need help with this question please
b=-4ac-2c+P
Denisse
b=p-4a-2c
Suddhen
b= p - 4a - 2c
Snr
p=2(2a+C)+b
Suraj
b=p-2(2a+c)
Tapiwa
P=4a+b+2C
COLEMAN
b=P-4a-2c
COLEMAN
like Deadra, show me the step by step order of operation to alive for b
John
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has
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# RD Sharma Chapter 1 Class 10 Maths Exercise 1.4 Solutions
RD Sharma Chapter 1 Class 10 Maths Exercise 1.1 Solutions -Real Number is written by expert mathematics teachers. The solutions in the chapter are written according to RD Sharma’s latest Class 10 Mathematics Solutions course. This chapter will introduce you to real numbers. Here you will learn about Euclid’s division lemma and Euclid’s division algorithm. This chapter also introduces you to the fundamental principle of arithmetic, the fundamental factor, the method of application of LCM, HCF, and HCF and LCM in real-world problems.
RD Sharma Chapter 1 Class 10 Maths Exercise 1.4 Solutions: There are 16 questions in total. In which, 1 question out of 7 asks you to find the LCM and HCF of a given pair of integers. Questions 8, 9, 10, and 14 ask you to find the smallest number that can be divided by the given numbers. Questions 11, 15 and 16 are scenario based questions, where they are based on real world HCF and LCM problems. Questions 12 and 13 ask you to find the largest number, which is divisible by the given numbers.
## Download RD Sharma Chapter 1 Class 10 Maths Exercise 1.4 Solutions
RD Sharma Solutions Class 10 Maths Chapter 1 Ex 1.4
## Important Definition for RD Sharma Chapter 1 Class 10 Maths Exercise 1.4 Solutions
• Euclid’s Division Lemma
This section also teaches you about Euclid’s Division Lemma. It states that the given two integers a and b, there always exists a unique pair of integers which are named as q and r such that a=b×q+r and 0≤r<b.
• Euclid’s Division Algorithm
This part of Class 10 Maths Real Numbers teaches you about Euclid’s algorithm theorem. It tells you how Euclid’s Division Algorithm is based on Euclid’s Division Lemma.
Here you will also come to know how to find the HCF of the two given numbers with Euclid’s Division Algorithm say p and g, where p>g. Now, if we are applying Euclid’s theorem to this, then p=g×q+r and 0≤r<g.
Suppose, if we consider r=0, with the HCF as g, now, if we apply Euclid’s division Lemma to g(the divisor) and r (the remainder). With this, we get another pair of remainder and quotient. This method is repeated until a remainder 1 is attained. The divisor that step is the HCF of the given set of numbers.
• The Fundamental Theorem of Arithmetic
This section of RD Sharma Solutions for Class 10 Maths Chapter 1-Real Numbers, introduces you to prime factorisation and how it is the method of representing a natural number through the multiplication of prime numbers. For example, 36=2 x 2 x3 x 3 is the prime factorisation of 36.
It also explains the Fundamental Theorem of Arithmetic which states that the prime factorisation of the given number should be unique if the arrangement of prime factors is ignored.
For Example 36= 2 x 2 x 3 x 3 or 36 = 2 x 3 x 2 x 3.
In this section, you’ll also learn the method of finding LCM, method of finding HCF, the product of two numbers which is HCF X LCM of the two numbers and application of HCF and LCM in real-world problems.
• Revisiting Irrational Numbers
This element of Class 10 Maths Real Numbers, explains about the irrational numbers. Basically, these are the numbers which cannot be expressed in the form of a fraction (p/q). Examples of Irrational numbers are √2,π, and e.
• Revisiting Rational Numbers and Their Decimal Expansions
This element of RD Sharma Solutions for Class 10 Maths Chapter 1-Real Numbers, teaches you about the terminating and non-terminating decimals.
1. When a number stops at a certain point, then it is called terminating decimal.
Example: 8.42, 7.58. 1.23, and so on.
2. When a decimal number doesn’t terminate after a certain point, then it is considered as non-terminating decimals.
Know more at the official website.
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# Random variables: How would you explain it to a beginner?
Different types of random variables: (discrete) Binomial, hypergeometric, geometric, Poisson (continuous) Uniform, normal, exponential
Random variables are very useful tools when solving simple and complex problems related to probability. They're used in diverse situations in many different forms, so how should you, for instance, describe in very general terms what they are to a student who is just starting to learn about mathematics?
Not really looking for a formal definition here, but more of a "here is how it's relevant" to your studies and your life kind of 101-deal. Something that even a middle school or high schooler could understand.
"Let $X$ be the number of times you can a sum of $7$ when you throw three dice. Then the probability distribution of $X$ is given by $\Pr(X=0)=\text{whatever}$, $\Pr(X=1)=\text{whatever}$, $\Pr(X=2)=\text{whatever}$, $\Pr(X=3)=\text{whatever}$." Etc. Then $X$ is an example of a "random variable". For continuous distributions, speak of the probability that $X$ is between two numbers, rather than the probability that $X$ is equal to some number. In other words, I would not start by stating a precise definition of the concept of "random variable".
• Strongly agree. Students can learn to use (correctly!) random variables in sentences before being exposed to formal definitions. – André Nicolas Nov 22 '11 at 2:07
Suppose you have a space $\Omega$ where you have a probability defined. The possible outcomes of tossing a coin, for example. Now, imagine you are gambling... for each possible outcome it is defined the amount of money you will get (or loose). This is a random variable!!
The catch is that with a function $f: \Omega \to \mathbb{R}$, you can transport the probability defined for $\Omega$ to a probability in $\mathbb{R}$. So, if for instance you get $10$ bucks when you get heads, but looses $5$ for tails, then you can talk about the probability of loosing $5$ bucks when you toss the coin...
Notice that it does not make much sense to ask the expected "value" for heads or tails. If instead of heads and tails you get a coin with faces coloured green and blue, it is probably meaningless to say that in mean, the expected colour is cyan... On the other hand, if you are talking about loosing and gaining money, it makes sense to talk about the expected amount of money you will gain or loose. And this is the expectation.
It is important to emphasise that a "random variable" is NOT a function that gives randomly different values for the same "input". The amount you get for heads or tails is always the same for a fixed $f$! It is just a way to transport the probability in $\Omega$ to a probability in "amounts" (real numbers), so you can talk about expectation.
Usually one is not interested in the random variable itself... people talk about random variable when they just want to talk about the probability they induce in $\mathbb{R}$. This induced probability is the distribution of the random variable.
Two random variables $f$ and $g$ are independent, for example, when knowing or not the outcome of $f$ (in terms of events: $f \in A \subset \mathbb{R}$), makes no difference in determining the probability of $g$'s outcome.
• My example also shows that you would enjoy very much gambling with me!! :-) – André Caldas Nov 22 '11 at 2:57
• Plus one because of the example :) – Leandro Nov 23 '11 at 2:08
It seems to me that an intuitive application where random variables play an important role may help motivate the concept.
Consider the manager of a customer support center who has to decide how many customer support personnel to hire to man the telephone lines. The number of support personnel is dependent on the number of calls that come in; a number that is likely to vary depending on the time of the day, day of the week etc.
Thus, it seems reasonable to assume that the number of calls in any given time period is uncertain with a range of plausible values (say, the number of calls to arrive at the center per minute can range anywhere between 5 to 10). One way to capture the above scenario is to let $N$ be the number of calls that arrive per unit time period.
In the above scenario, we would call $N$ a random variable as we do now know for sure the value we would observe apriori (i.e., we do not know how many calls would come per minute). Then, we can (depending on the situation) assume that $P(N=5) = 0.2$, $P(N=6) = 0.3$ and so on to capture our uncertainty.
The advantage to the above approach is that a student immediately appreciates the practical application and utility of the concept of random variables. The disadvantage, however, is that it requires a more elaborate explanation as the application needs to be sufficiently realistic.
• This is one example where one is talking about the distribution and not the random variable itself. Notice that there is even no mention of $N$'s domain of definition. It is difficult to have a realistic example because the set $\Omega$, where $N$ is defined is not "realistic"... – André Caldas Nov 22 '11 at 3:01
• @Andre I thought we were talking about beginners and motivating them to see why random variables are important. We can always polish up the example as much as desired. – tards Nov 22 '11 at 3:04
• Sorry, @tards... I was not complaining. Well, maybe a little... I think we are used to complicating things too much. In your example, you do exactly what the author of the question does. You are not at all out of scope. I just think that people do not realize they are not talking about a random variable... they are just talking about a distribution over the reals. I just find it too complicated to put random variables when you do not need them. If you do not need (or do not care about) the domain of definition, you do not need the random variable. – André Caldas Nov 22 '11 at 3:20
A random variable is another word for a function. Its input is the outcome of a random event, and it returns some aspect of the outcome.
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## What is the formula to solve for circumference?
C = 2πr
The circumference is the distance around a circle. In other words, it’s the perimeter of the circle. And we find the circumference by using the formula C = 2πr.
## How do you find the radius with the circumference?
To calculate the circumference, you need the radius of the circle:
1. Multiply the radius by 2 to get the diameter.
2. Multiply the result by π, or 3.14 for an estimation.
3. That’s it; you found the circumference of the circle.
What are the two formulas for circumference of a circle?
To calculate the circumference of a circle, multiply the diameter of the circle with π (pi). The circumference can also be calculated by multiplying 2×radius with pi (π=3.14).
Is circumference twice the diameter?
This means that the circumference (C) is always about 3.14 (π) times the diameter (d). The formula below allows you to easily calculate the circumference of a circle when you know its diameter. To find the circumference of a circle, multiply π times the diameter of the circle.
### Is the radius half the circumference?
The distance around a rectangle or a square is as you might remember called the perimeter. The distance around a circle on the other hand is called the circumference (c). Half of the diameter, or the distance from the midpoint to the circle border, is called the radius of the circle (r).
### What is a circumference in math?
The circumference is the length of any great circle, the intersection of the sphere with any plane passing through its centre. A meridian is any great circle passing through a point designated a pole.
Why is circumference 2pir?
Basically this is a definition thing. π is defined to be the ratio of the circumference of a circle over its diameter (or 2 times its radius). This ratio is a constant since all circles are geometrically similar and linear proportions between any similar geometric figures are constant.
What is perimeter formula?
The perimeter is equal to the boundary of the rectangle which can be calculated with the formula: Perimeter = Length + Length + Width + Width = 2(Length + Width).
#### What is circumference diameter?
A circle is a geometric form of which every point on the outside of the circle is the same distance away from the center. The distance around the edge of the circle is called the circumference. The distance from one side of the circle to the other, going through the center of the circle, is the diameter.
#### Is the circumference 3 times the diameter?
The circumference is about 3 times the diameter of the circle. The ratio of circumference to diameter (C ÷ d) is always π. This means that the circumference (C) is always about 3.14 (π) times the diameter (d).
What is the formula for the circumference of a circle?
Formula for circumference. The following equation describes the relation between the circumference and the radius R of a circle: C = 2πR. Where π is a constant approximately equal to 3.14159…
What is the ratio of circumference to diameter?
Circumference to diameter. You have probably noticed that, since diameter is twice the radius, the proportion between the circumference and the diameter is equal to π: C/D = 2πR / 2R = π. This proportion (circumference to diameter) is the definition of the constant pi.
## What is the purpose of a circumference calculator?
It is a tool specifically created to find the diameter, circumference and area of any circle. Read on to learn:
## What is the chemical formula for sucrose sugar?
Sucrose PubChem CID 5988 Structure Find Similar Structures Chemical Safety Laboratory Chemical Safety Summary (LCSS Molecular Formula C12H22O11 Synonyms sucrose 57-50-1 saccharose Cane sugar Ta
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# 5.4: The Exponential Distribution
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The exponential distribution is often concerned with the amount of time until some specific event occurs. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length, in minutes, of long distance business telephone calls, and the amount of time, in months, a car battery lasts. It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution.
Values for an exponential random variable occur in the following way. There are fewer large values and more small values. For example, the amount of money customers spend in one trip to the supermarket follows an exponential distribution. There are more people who spend small amounts of money and fewer people who spend large amounts of money.
The exponential distribution is widely used in the field of reliability. Reliability deals with the amount of time a product lasts.
##### Example $$\PageIndex{1}$$
Let $$X$$ = amount of time (in minutes) a postal clerk spends with his or her customer. The time is known to have an exponential distribution with the average amount of time equal to four minutes.
$$X$$ is a continuous random variable since time is measured. It is given that $$\mu = 4$$ minutes. To do any calculations, you must know $$m$$, the decay parameter.
$$m = \dfrac{1}{\mu}$$. Therefore, $$m = \dfrac{1}{4} = 0.25$$.
The standard deviation, $$\sigma$$, is the same as the mean. $$\mu = \sigma$$
The distribution notation is $$X \sim Exp(m)$$. Therefore, $$X \sim Exp(0.25)$$.
The probability density function is $$f(x) = me^{-mx}$$. The number $$e = 2.71828182846$$... It is a number that is used often in mathematics. Scientific calculators have the key "$$e^{x}$$." If you enter one for $$x$$, the calculator will display the value $$e$$.
The curve is:
$$f(x) = 0.25e^{-0.25x}$$ where $$x$$ is at least zero and $$m = 0.25$$.
For example, $$f(5) = 0.25e^{-(0.25)(5)} = 0.072$$. The value 0.072 is the height of the curve when x = 5. In Example $$\PageIndex{2}$$ below, you will learn how to find probabilities using the decay parameter.
The graph is as follows:
Notice the graph is a declining curve. When $$x = 0$$,
$$f(x) = 0.25e^{(-0.25)(0)} = (0.25)(1) = 0.25 = m$$. The maximum value on the y-axis is m.
##### Exercise $$\PageIndex{1}$$
The amount of time spouses shop for anniversary cards can be modeled by an exponential distribution with the average amount of time equal to eight minutes. Write the distribution, state the probability density function, and graph the distribution.
$$X \sim Exp(0.125)$$;
$$f(x) = 0.125e^{-0.125x}$$;
##### Example $$\PageIndex{2}$$
1. Using the information in the Example, find the probability that a clerk spends four to five minutes with a randomly selected customer.
2. Half of all customers are finished within how long? (Find the 50th percentile)
3. Which is larger, the mean or the median?
a. Find $$P(4 < x < 5)$$.
The cumulative distribution function (CDF) gives the area to the left.
$P(x < x) = 1 – e^{-mx}$
$P(x < 5) = 1 – e(–0.25)(5) = 0.7135$
and
$(P(x < 4) = 1 – e^{(-0.25)(4)} = 0.6321$
You can do these calculations easily on a calculator.
The probability that a postal clerk spends four to five minutes with a randomly selected customer is
$P(4 < x < 5) = P(x < 5) – P(x < 4) = 0.7135 − 0.6321 = 0.0814.$
On the home screen, enter (1 – e^(–0.25*5)) – (1 – e^(–0.25*4)) or enter e^(–0.25*4) – e^(–0.25*5).
b. Find the 50th percentile.
$$P(x < k) = 0.50$$, $$k = 2.8$$ minutes (calculator or computer)
Half of all customers are finished within 2.8 minutes.
You can also do the calculation as follows:
$P(x < k) = 0.50$
and
$P(x < k) = 1 – e^{-0.25k}$
Therefore,
$0.50 = 1 − e^{-0.25k}$
and
$e^{-0.25k} = 1 − 0.50 = 0.5$
Take natural logs:
$\ln(e^{-0.25k}) = \ln(0.50).$
So,
$-0.25k = ln(0.50).$
Solve for $$k: k = \dfrac{ln(0.50)}{-0.25} = 0.28$$ minutes. The calculator simplifies the calculation for percentile k. See the following two notes.
A formula for the percentile $$k$$ is $$k = ln(1 − \text{Area To The Left}) - mk = ln(1 - \text{Area To The Left}) - m$$ where $$ln$$ is the natural log.
c. From part b, the median or 50th percentile is 2.8 minutes. The theoretical mean is four minutes. The mean is larger.
##### Exercise $$\PageIndex{2}$$
The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days. Find the probability that a traveler will purchase a ticket fewer than ten days in advance. How many days do half of all travelers wait?
$$P(x < 10) = 0.4866$$
50th percentile = 10.40
##### Collaborative Exercise
Have each class member count the change he or she has in his or her pocket or purse. Your instructor will record the amounts in dollars and cents. Construct a histogram of the data taken by the class. Use five intervals. Draw a smooth curve through the bars. The graph should look approximately exponential. Then calculate the mean.
Let $$X =$$ the amount of money a student in your class has in his or her pocket or purse.
The distribution for $$X$$ is approximately exponential with mean, $$\mu =$$ _______ and $$m =$$ _______. The standard deviation, $$\sigma =$$ ________.
Draw the appropriate exponential graph. You should label the x– and y–axes, the decay rate, and the mean. Shade the area that represents the probability that one student has less than \$.40 in his or her pocket or purse. (Shade $$P(x < 0.40)$$).
##### Example $$\PageIndex{3}$$
On the average, a certain computer part lasts ten years. The length of time the computer part lasts is exponentially distributed.
1. What is the probability that a computer part lasts more than 7 years?
2. On the average, how long would five computer parts last if they are used one after another?
3. Eighty percent of computer parts last at most how long?
4. What is the probability that a computer part lasts between nine and 11 years?
a. Let $$x =$$ the amount of time (in years) a computer part lasts.
$\mu = 10$
so
$m = \dfrac{1}{\mu} = \dfrac{1}{10} = 0.1$
Find $$P(x > 7)$$. Draw the graph.
$P(x > 7) = 1 – P(x < 7).$
Since $$P(X < x) = 1 – e^{-mx}$$ then
$P(X > x) = 1 –(1 –e^{-mx}) = e^{-mx}$
$P(x > 7) = e^{(–0.1)(7)} = 0.4966.$
The probability that a computer part lasts more than seven years is 0.4966.
On the home screen, enter e^(-.1*7).
b. On the average, one computer part lasts ten years. Therefore, five computer parts, if they are used one right after the other would last, on the average, (5)(10) = 50 years.
c. Find the 80th percentile. Draw the graph. Let k = the 80th percentile.
Solve for $$k: k = \dfrac{ln(1-0.80)}{-0.1} = 16.1$$ years
Eighty percent of the computer parts last at most 16.1 years.
On the home screen, enter $$\dfrac{ln(1-0.80)}{-0.1}$$
d. Find $$P(9 < x < 11)$$. Draw the graph.
$P(9 < x < 11) = P(x < 11) - P(x < 9) = (1 - e^{(–0.1)(11)}) - (1 - e^{(–0.1)(9)}) = 0.6671 - 0.5934 = 0.0737.$
The probability that a computer part lasts between nine and 11 years is 0.0737.
On the home screen, enter e^(–0.1*9) – e^(–0.1*11).
##### Exercise $$\PageIndex{3}$$
On average, a pair of running shoes can last 18 months if used every day. The length of time running shoes last is exponentially distributed. What is the probability that a pair of running shoes last more than 15 months? On average, how long would six pairs of running shoes last if they are used one after the other? Eighty percent of running shoes last at most how long if used every day?
$$P(x > 15) = 0.4346$$
Six pairs of running shoes would last 108 months on average.
80th percentile = 28.97 months
##### Example $$\PageIndex{4}$$
Suppose that the length of a phone call, in minutes, is an exponential random variable with decay parameter = $$\dfrac{1}{12}$$. If another person arrives at a public telephone just before you, find the probability that you will have to wait more than five minutes. Let $$X$$ = the length of a phone call, in minutes.
What is $$m$$, $$\mu$$, and $$\sigma$$? The probability that you must wait more than five minutes is _______ .
• $$m = \dfrac{1}{12}$$
• $$\mu = 12$$
• $$\sigma = 12$$
$$P(x > 5) = 0.6592$$
##### Exercise $$\PageIndex{4}$$
Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay parameter $$\dfrac{1}{20}$$. Let $$S =$$ the distance people are willing to commute in miles. What is $$m$$, $$\mu$$, and $$\sigma$$? What is the probability that a person is willing to commute more than 25 miles?
$$m = \dfrac{1}{20}$$; $$\mu = 20$$; $$\sigma = 20$$; $$P(x > 25) = 0.2865$$
##### Example $$\PageIndex{5}$$
The time spent waiting between events is often modeled using the exponential distribution. For example, suppose that an average of 30 customers per hour arrive at a store and the time between arrivals is exponentially distributed.
1. On average, how many minutes elapse between two successive arrivals?
2. When the store first opens, how long on average does it take for three customers to arrive?
3. After a customer arrives, find the probability that it takes less than one minute for the next customer to arrive.
4. After a customer arrives, find the probability that it takes more than five minutes for the next customer to arrive.
5. Seventy percent of the customers arrive within how many minutes of the previous customer?
6. Is an exponential distribution reasonable for this situation?
1. Since we expect 30 customers to arrive per hour (60 minutes), we expect on average one customer to arrive every two minutes on average.
2. Since one customer arrives every two minutes on average, it will take six minutes on average for three customers to arrive.
3. Let $$X =$$ the time between arrivals, in minutes. By part a, $$\mu = 2$$, so $$m = \dfrac{1}{2} = 0.5$$.
Therefore, $$X \sim Exp(0.5)$$.
The cumulative distribution function is $$P(X < x) = 1 – e(–0.5x)^{e}$$.
Therefore $$P(X < 1) = 1 - e^{(–0.5)(1)} \approx 0.3935$$.
$$1 - e^(–0.5) \approx 0.3935$$
4. $$P(X > 5) = 1 - P(X < 5) = 1 - (1 - e^{(-5)(0.5)}) = e^{-2.5} \approx 0.0821$$.
Figure $$\PageIndex{9}$$.
$$1 - (1 - e^{( – 5*0.5)}$$ or $$e^{(-5*0.5)}$$
5. We want to solve $$0.70 = P(X < x)$$ for $$x$$.
Substituting in the cumulative distribution function gives $$0.70 = 1 – e^{–0.5x}$$, so that $$e^{–0.5x} = 0.30$$. Converting this to logarithmic form gives $$-0.5x = ln(0.30)$$, or $$x = \dfrac{ln(0.30)}{-0.5} \approx 2.41$$ minutes.
Thus, seventy percent of customers arrive within 2.41 minutes of the previous customer.
You are finding the 70th percentile $$k$$ so you can use the formula $$k = \dfrac{ln(1-\text{Area To The left Of k})}{-m}$$
$$k = \dfrac{ln(1-0.70)}{(-0.5)} \approx 2.41$$ minutes
Figure $$\PageIndex{10}$$.
This model assumes that a single customer arrives at a time, which may not be reasonable since people might shop in groups, leading to several customers arriving at the same time. It also assumes that the flow of customers does not change throughout the day, which is not valid if some times of the day are busier than others.
##### Exercise $$\PageIndex{5}$$
Suppose that on a certain stretch of highway, cars pass at an average rate of five cars per minute. Assume that the duration of time between successive cars follows the exponential distribution.
1. On average, how many seconds elapse between two successive cars?
2. After a car passes by, how long on average will it take for another seven cars to pass by?
3. Find the probability that after a car passes by, the next car will pass within the next 20 seconds.
4. Find the probability that after a car passes by, the next car will not pass for at least another 15 seconds.
1. At a rate of five cars per minute, we expect $$\dfrac{60}{5} = 12$$ seconds to pass between successive cars on average.
2. Using the answer from part a, we see that it takes $$(12)(7) = 84$$ seconds for the next seven cars to pass by.
3. Let $$T =$$ the time (in seconds) between successive cars.
The mean of $$T$$ is 12 seconds, so the decay parameter is $$\dfrac{1}{12}$$ and $$T \sim Exp\dfrac{1}{12}$$. The cumulative distribution function of $$T$$ is $$P(T < t) = 1 – e^{−\dfrac{t}{12}}$$. Then $$P(T < 20) = 1 –e^{−\dfrac{20}{12}} \approx 0.8111$$.
Figure $$\PageIndex{11}$$.
$$P(T > 15) = 1 – P(T < 15) = 1 – (1 – e^{−\dfrac{15}{12}}) = e^{−\dfrac{15}{12}} \approx 0.2865$$.
## Memorylessness of the Exponential Distribution
In Example recall that the amount of time between customers is exponentially distributed with a mean of two minutes ($$X \sim Exp (0.5)$$). Suppose that five minutes have elapsed since the last customer arrived. Since an unusually long amount of time has now elapsed, it would seem to be more likely for a customer to arrive within the next minute. With the exponential distribution, this is not the case–the additional time spent waiting for the next customer does not depend on how much time has already elapsed since the last customer. This is referred to as the memoryless property. Specifically, the memoryless property says that
$P(X > r + t | X > r) = P(X > t)$
for all $$r \geq 0$$ and $$t \geq 0$$.
For example, if five minutes has elapsed since the last customer arrived, then the probability that more than one minute will elapse before the next customer arrives is computed by using $$r = 5$$ and $$t = 1$$ in the foregoing equation.
$$P(X > 5 + 1 | X > 5) = P(X > 1) = e(–0.5)(1) e(–0.5)(1) \approx 0.6065$$.
This is the same probability as that of waiting more than one minute for a customer to arrive after the previous arrival.
The exponential distribution is often used to model the longevity of an electrical or mechanical device. In Example, the lifetime of a certain computer part has the exponential distribution with a mean of ten years ($$X \sim Exp(0.1)$$). The memoryless property says that knowledge of what has occurred in the past has no effect on future probabilities. In this case it means that an old part is not any more likely to break down at any particular time than a brand new part. In other words, the part stays as good as new until it suddenly breaks. For example, if the part has already lasted ten years, then the probability that it lasts another seven years is $$P(X > 17 | X > 10) = P(X > 7) = 0.4966$$.
##### Example $$\PageIndex{6}$$
Refer to Example where the time a postal clerk spends with his or her customer has an exponential distribution with a mean of four minutes. Suppose a customer has spent four minutes with a postal clerk. What is the probability that he or she will spend at least an additional three minutes with the postal clerk?
The decay parameter of $$X$$ is $$m = \dfrac{1}{4} = 0.25$$, so $$X \sim Exp(0.25)$$.
The cumulative distribution function is $$P(X < x) = 1 - e^{–0.25x}$$.
We want to find $$P(X > 7 | X > 4)$$. The memoryless property says that $$P(X > 7 | X > 4) = P(X > 3)$$, so we just need to find the probability that a customer spends more than three minutes with a postal clerk.
This is $$P(X > 3) = 1 - P(X < 3) = 1 - (1 - e^{-0.25 \cdot 3}) = e^{–0.75} \approx 0.4724$$.
$$1 - (1 - e^(-0.25*2)) = e^(-0.25*2)$$.
##### Exercise $$\PageIndex{6}$$
Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. If a bulb has already lasted 12 years, find the probability that it will last a total of over 19 years.
Let $$T =$$ the lifetime of the light bulb. Then $$T \sim Exp\left(\dfrac{1}{8}\right)$$.
The cumulative distribution function is $$P(T < t) = 1 − e^{-\dfrac{t}{8}}$$
We need to find $$P(T > 19 | T = 12)$$. By the memoryless property,
$$P(T > 19 | T = 12) = P(T > 7) = 1 - P(T < 7) = 1 - (1 - e^{-7/8}) = e^{-7/8} \approx 0.4169$$.
1 - (1 – e^(–7/8)) = e^(–7/8).
## Relationship between the Poisson and the Exponential Distribution
There is an interesting relationship between the exponential distribution and the Poisson distribution. Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of $$\mu$$ units of time. Also assume that these times are independent, meaning that the time between events is not affected by the times between previous events. If these assumptions hold, then the number of events per unit time follows a Poisson distribution with mean $$\lambda = \dfrac{1}{\mu}$$. Recall from the chapter on Discrete Random Variables that if $$X$$ has the Poisson distribution with mean $$\lambda$$, then $$P(X = k) = \dfrac{\lambda^{k}e^{-\lambda}}{k!}$$. Conversely, if the number of events per unit time follows a Poisson distribution, then the amount of time between events follows the exponential distribution. $$(k! = k*(k-1*)(k - 2)*(k - 3) \dotsc 3*2*1)$$
Suppose $$X$$ has the Poisson distribution with mean $$\lambda$$. Compute $$P(X = k)$$ by entering 2nd, VARS(DISTR), C: poissonpdf$$(\lambda, k$$). To compute $$P(X \leq k$$), enter 2nd, VARS (DISTR), D:poissoncdf($$\lambda, k$$).
##### Example $$\PageIndex{7}$$
At a police station in a large city, calls come in at an average rate of four calls per minute. Assume that the time that elapses from one call to the next has the exponential distribution. Take note that we are concerned only with the rate at which calls come in, and we are ignoring the time spent on the phone. We must also assume that the times spent between calls are independent. This means that a particularly long delay between two calls does not mean that there will be a shorter waiting period for the next call. We may then deduce that the total number of calls received during a time period has the Poisson distribution.
1. Find the average time between two successive calls.
2. Find the probability that after a call is received, the next call occurs in less than ten seconds.
3. Find the probability that exactly five calls occur within a minute.
4. Find the probability that less than five calls occur within a minute.
5. Find the probability that more than 40 calls occur in an eight-minute period.
1. On average there are four calls occur per minute, so 15 seconds, or $$\dfrac{15}{60} = 0.25$$ minutes occur between successive calls on average.
2. Let $$T =$$ time elapsed between calls. From part a, $$\mu = 0.25$$, so $$m = \dfrac{1}{0.25} = 4$$. Thus, $$T \sim Exp(4)$$.
The cumulative distribution function is $$P(T < t) = 1 - e^{–4t}$$.
The probability that the next call occurs in less than ten seconds (ten seconds $$= \dfrac{1}{6}$$ minute) is $$P(T < \dfrac{1}{6}) = 1 - e^{-4 (\dfrac{1}{6})} \approx 0.4866)$$.
Figure $$\PageIndex{13}$$
3. Let $$X =$$ the number of calls per minute. As previously stated, the number of calls per minute has a Poisson distribution, with a mean of four calls per minute.
Therefore, $$X \sim Poisson(4)$$, and so $$P(X = 5) = \dfrac{4^{5}e^{-4}}{5!} \approx 0.1563$$. ($$5! = (5)(4)(3)(2)(1)$$)
$$\text{poissonpdf}(4, 5) = 0.1563$$.
4. Keep in mind that $$X$$ must be a whole number, so $$P(X < 5) = P(X \leq 4)$$.
To compute this, we could take $$P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$$.
Using technology, we see that $$P(X \approx 4) = 0.6288$$.
$$\text{poisssoncdf}(4, 4) = 0.6288$$
5. Let $$Y =$$ the number of calls that occur during an eight minute period.
Since there is an average of four calls per minute, there is an average of $$(8)(4) = 32$$ calls during each eight minute period.
Hence, $$Y \sim Poisson(32)$$. Therefore, $$P(Y > 40) = 1 - P(Y \leq 40) = 1 - 0.9294 = 0.0707$$.
$$1 - \text{poissoncdf}(32, 40). = 0.0707$$
##### Exercise $$\PageIndex{7}$$
In a small city, the number of automobile accidents occur with a Poisson distribution at an average of three per week.
1. Calculate the probability that there are at most 2 accidents occur in any given week.
2. What is the probability that there is at least two weeks between any 2 accidents?
1. Let $$X =$$ the number of accidents per week, so that $$X \sim Poisson(3)$$. We need to find $$P(X \leq 2) \approx 0.4232$$
$$\text{poissoncdf}(3, 2)$$
2. Let $$T =$$ the time (in weeks) between successive accidents.
Since the number of accidents occurs with a Poisson distribution, the time between accidents follows the exponential distribution.
If there are an average of three per week, then on average there is $$\mu = \dfrac{1}{3}$$ of a week between accidents, and the decay parameter is $$m = \dfrac{1}{\left(\dfrac{1}{3}\right)} = 3$$.
To find the probability that there are at least two weeks between two accidents, $$P(T > 2) = 1 - P(T < 2) = 1 – (1 – e(–3)(2)) = e^{–6} \approx 0.0025$$.
e^(-3*2).
## Review
If $$X$$ has an exponential distribution with mean $$\mu$$, then the decay parameter is $$m = \dfrac{1}{\mu}$$, and we write $$X \sim Exp(m)$$ where $$x \geq 0$$ and $$m > 0$$. The probability density function of $$X$$ is $$f(x) = me^{-mx}$$ (or equivalently $$f(x) = \dfrac{1}{\mu}e^{-\dfrac{x}{\mu}}$$). The cumulative distribution function of $$X$$ is $$P(X \leq X) = 1 - e^{-mx}$$.
The exponential distribution has the memoryless property, which says that future probabilities do not depend on any past information. Mathematically, it says that $$P(X > x + k | X > x) = P(X > k)$$.
If $$T$$ represents the waiting time between events, and if $$T \sim Exp(\lambda)$$, then the number of events $$X$$ per unit time follows the Poisson distribution with mean $$\lambda$$. The probability density function of $$PX$$ is $$(X = k) = \dfrac{\lambda^{k}e^{-k}}{k!}$$. This may be computed using a TI-83, 83+, 84, 84+ calculator with the command $$\text{poissonpdf}(\lambda, k)$$. The cumulative distribution function $$P(X \leq k)$$ may be computed using the TI-83, 83+,84, 84+ calculator with the command $$\text{poissoncdf}(\lambda, k)$$.
## Formula Review
Exponential: $$X \sim Exp(m)$$ where $$m =$$ the decay parameter
• pdf: $$f(x) = me^{(–mx)}$$ where $$x \geq 0$$ and $$m > 0$$
• cdf: $$P(X \leq x) = 1 - e^{(–mx)}$$
• mean $$\mu = \dfrac{1}{m}$$
• standard deviation $$\sigma = \mu$$
• percentile $$k: k = \dfrac{ln(1 - \text{Area To The Left Of k})}{-m}$$
• $$P(X > x) = e^{(–mx)}$$
• $$P(a < X < b) = e^{(–ma)} - e^{(–mb)}$$
• Memoryless Property: $$P(X > x + k | X > x) = P(X > k)$$
• Poisson probability: $$P(X = k) = \dfrac{\lambda^{k}e^{k}}{k!}$$ with mean $$\lambda$$
• $$k! = k*(k - 1)*(k - 2)*(k - 3) \dotsc 3*2*1$$
## References
1. Data from the United States Census Bureau.
2. Data from World Earthquakes, 2013. Available online at http://www.world-earthquakes.com/ (accessed June 11, 2013).
3. “No-hitter.” Baseball-Reference.com, 2013. Available online at http://www.baseball-reference.com/bullpen/No-hitter (accessed June 11, 2013).
4. Zhou, Rick. “Exponential Distribution lecture slides.” Available online at www.public.iastate.edu/~riczw/stat330s11/lecture/lec13.pdf (accessed June 11, 2013).
## Glossary
decay parameter
The decay parameter describes the rate at which probabilities decay to zero for increasing values of $$x$$. It is the value $$m$$ in the probability density function $$f(x) = me^{(-mx)}$$ of an exponential random variable. It is also equal to $$m = \dfrac{1}{\mu}$$, where $$\mu$$ is the mean of the random variable.
memoryless property
For an exponential random variable $$X$$, the memoryless property is the statement that knowledge of what has occurred in the past has no effect on future probabilities. This means that the probability that $$X$$ exceeds $$x + k$$, given that it has exceeded $$x$$, is the same as the probability that $$X$$ would exceed $$k$$ if we had no knowledge about it. In symbols we say that $$P(X > x + k | X > x) = P(X > k)$$
Poisson distribution
If there is a known average of $$\lambda$$ events occurring per unit time, and these events are independent of each other, then the number of events $$X$$ occurring in one unit of time has the Poisson distribution. The probability of k events occurring in one unit time is equal to $$P(X = k) = \dfrac{\lambda^{k}e^{-\lambda}}{k!}$$.
This page titled 5.4: The Exponential Distribution is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.
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How to Find the Mode of Grouped Data (With Examples)
Often we may want to calculate the mode of data that is grouped in some way.
Recall that the mode represents the value that occurs most frequently.
For example, suppose we have the following grouped data:
While it’s not possible to calculate the exact mode since we don’t know the raw data values, it is possible to estimate the mode using the following formula:
Mode of Grouped Data = L + W[(Fm – F1)/( (Fm-F1) + (Fm – F2) )]
where:
• L: Lower limit of modal class
• W: Width of modal class
• Fm: Frequency of modal class
• F1: Frequency of class immediately before modal class
• F2: Frequency of class immediately after modal class
Note: The modal class is simply the class with the highest frequency. In the example above, the modal class would be 21-30 since it has the highest frequency.
The following examples show how to calculate the mode of grouped data in different scenarios.
Example 1: Calculate the Mode of Grouped Data
Suppose we have the following frequency distribution that shows the exam scored receive by 40 students in a certain class:
In this example, the modal class is 71-80.
Knowing this, we can calculate the following values:
• L: Lower limit of modal class: 71
• W: Width of modal class: 9
• Fm: Frequency of modal class: 15
• F1: Frequency of class immediately before modal class: 8
• F2: Frequency of class immediately after modal class: 8
We can plug these values into the formula to calculate the mode of the distribution:
• Mode = L + W[(Fm – F1)/( (Fm-F1) + (Fm – F2) )]
• Mode = 71 + 9[(15-8) / ( (15-8) + (15-8) )]
• Mode = 75.5
We estimate that the modal exam score is 75.5.
Example 2: Calculate the Mode of Grouped Data
Suppose we have the following frequency distribution that shows the number of points scored per game by 60 basketball players:
In this example, the modal class is 11-20.
Knowing this, we can calculate the following values:
• L: Lower limit of modal class: 11
• W: Width of modal class: 9
• Fm: Frequency of modal class: 25
• F1: Frequency of class immediately before modal class: 8
• F2: Frequency of class immediately after modal class: 14
We can plug these values into the formula to calculate the mode of the distribution:
• Mode = L + W[(Fm – F1)/( (Fm-F1) + (Fm – F2) )]
• Mode = 11 + 9[(25-8) / ( (25-8) + (25-14) )]
• Mode = 16.46
We estimate that the modal points scored is 16.46.
The following tutorials explain how to perform other common operations with grouped data:
4 Replies to “How to Find the Mode of Grouped Data (With Examples)”
1. Judah Muthaura says:
In my class the teacher always ignores the h, which is here defined as the width of the modal class. Is it ok to ignore it you’re using a different formula here?
2. Gonzalo Edwards says:
What happens if the modal class is the first or the last class? Do we just plug in a 0 in the formula?
3. Henry says:
It is really helpful I didn’t think so but the graphics need some improvement
4. Brima Kamara says:
Good.
I want to join you in your email.
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# Vectors
## Describing translations using vectors
Vectors are used to describe movement. We often use them to describe the movement of shapes on or off of graphs.
We also write vectors like this
which refers to vector AB, this can be a line or two points, point A and point B,. It can also be referred to as a or as below:
When it comes to adding and subtracting vectors its quite simple, add the x values and add the y values together.
When vectors are added together the arrows follow the same direction eg.
Eg. Subtracting:
a-b =
## Multiplying by a scalar
Vectors have a direction and a magnitude. So for example the vector3a is three times as large asa.
To multiply by a scalar, simply multiply the column vector by the scale factor, or multiply the length by the scalar.
For example:
## Diagramatic Vectors
Often, we have vectors put together to describe a shape or a diagram.
For example:
In this diagram we can describe the different lengths in terms of their vectors.
Eg. OP=a PQ=b QR=-a RO=-b and RP =-b + a
You may often need to find vectors that describe how to get from one point of the diagram to the other.
For example:
Given that:
And point C is on the line DE such that EC: DC is in the ratio 1:3 find an expression for the line the vector
.
Firstly we need to think about the line DE and how we can express that as a vector.
DE =b-a
Now we need to think about the ratio. We know that the line DE is split in the ratio of 1:3. Thereforeb-a = the whole. When we have a ratio we need to split a whole into equal parts, in this case 1 part to 3 parts.
Therefore
EC: DC
1/4 (b-a) : 3/4 (b-a)
Now we can work out the whole vector
To go from F to D =a
Then we will need to add the vector
which we know is 3/4 (b-a)
a + 3/4 (b-a) =
a + 3/4 b- 3/4 a = 1/4 a +1/4 b
## Vectors for proofs
Often you will be required to prove a vector can be written in a certain way or that one vector relates to another in a certain way.
Common things that you may need to prove:
1. That two lines are parallel
2. That a point on a line is a midpoint.
For these specific proofs remember
1. When two lines are parallel they have the same gradient. Therefore if two vectors have a different magnitude but the same letter eg. 4v and2v they are parallel.
2. A midpoint is half of a line, so the scalar of the vector should be ½ of the given vector.
For example to prove that VY is parallel to ZX , given:
Firstly, put all the information you have about the vectors onto the diagram.
To prove that the two lines are parallel we are going to need to find the vectors for both lines.
We can get from Z to X by adding the vectors
= a + 3b + -a+b
= 4b
Now we just need to work out the vector for the line VY
Which we can get to by adding the vectors
= 1/3(a+3b) + -1/3 a
= 1/3 a +b - 1/3 a
= b
Therefore because b is a factor of 4b the lines must be parallel!
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# Question 88ca9
Oct 20, 2017
$c = 3$
#### Explanation:
$0.2 \left(10 - 5 c\right) = 5 c - 16$
In order to solve this, we need to isolate $c$, which means get it by itself. So, let's start by getting rid of those parentheses and distributing that $0.2$.
$0.2 \left(10 - 5 c\right) = 5 c - 16$
$2 - 1 \cdot c = 5 c - 16$
Now let's get all the $\textcolor{\mathmr{and} a n \ge}{c o n s t a n t s}$ (numbers) on one side
subtract 2 on both sides
$- c = 5 c \textcolor{\mathmr{and} a n \ge}{- 16 - 2}$
Now let's get the color(blue)(variab l es_# on one side
subtract $5 c$ from both sides
$\textcolor{b l u e}{- c - 5 c} = - 16 - 2$
Let's combine everything and see what we've got
$- 6 c = - 18$
We're almost there, but $c$ still isn't by itself. We need to get rid of that pesky $- 6$
divide by $- 6$ on both sides
$c = \frac{- 18}{-} 6$
$c = 3$
$. . . . . . . . . . . . . . . . . . . . . . . .$
To check our work, let's plug in $3$ for $c$ and make sure both sides are equal:
$0.2 \left(10 - 5 \textcolor{red}{c}\right) = 5 \textcolor{red}{c} - 16$
$0.2 \left(10 - \left(5 \cdot 3\right)\right) = \left(5 \cdot 3\right) - 16$
$0.2 \left(10 - 15\right) = 15 - 16$
$0.2 \left(- 5\right) = - 1$
$\textcolor{g r e e n}{- 1 = - 1}$
We were right! Great job
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# Cpm mathematics homework help
In addition, Cpm mathematics homework help can also help you to check your homework. Our website can help me with math work.
## The Best Cpm mathematics homework help
Best of all, Cpm mathematics homework help is free to use, so there's no sense not to give it a try! Solving a Rubik's cube is usually a matter of determining the shortest path between two corners. If, for example, the corner on the left is U-1 and the corner on the right is U-5, then the shortest route to the center must be U-2, U-4 and U-6. The shortest route is usually not the easiest route; in fact, it may be quite difficult to determine. However, this process can be simplified by determining a general solution for a given configuration that can then be used as a guide as to how to solve any other configuration. The most common approach to solving a Rubik's Cube is solving one side at a time. To do so, turn the cube over so that it is shaking in its frame. Each side will independently move in the frame and create one of four possible positions: solid yellow, solid red, solid blue or solids green and orange. When each side has been moved into position, you have determined your final position relative to the center of the cube (your "target" or "goal"). Once you know how to move each side individually, you will have solved half of your cube. Now you need to combine all of your individual solutions into one solution that shows all six faces solved. For our example above, you would need to perform six operations: Movement 1: -U-
Solving exponential equations can be a bit tricky. Most of the time you will need to use an inverse function to get from one number to the other. However, it is possible to solve some equations without using such techniques. Here are some examples: One way to solve an exponential equation is to use a logarithm table. For example, if you have an equation of the form y = 4x^2 + 32, then you would use the logarithm table found here. Then, you would find that log(y) = -log(4) = -2 and log(32) = 2. These values would be used in the original equation to obtain the solution: 4*y = -2*4 + 32 = -16 + 32 = 16. This value is the desired answer for y in this problem. Another way to solve an exponential equation is by using a combination of substitution and elimination. You can start by putting x into both sides of the equation and simplifying: ax + b c where a c if and only if b c/a . Then, once this is done, you can eliminate b from each side (using square roots or taking logs if necessary) to obtain a single solution that does not involve x . c if and only if , then you can substitute for y in both sides, thus eliminating x
One important thing to remember about solving absolute value equations is that you can only use addition and subtraction operations when solving them. You can’t use multiplication or division to solve absolute value equations because those operations change the number in the equation rather than just finding its absolute value. To solve absolute value equations, all you have to do is add or subtract one number from both sides of the equation until you get 0 on one side and then subtract that number from both sides again until you get 0 on both sides. Example: Find the absolute value of 6 + 4 = 10 Subtracting 4 from both sides gives us 2 math>egin{equation} ext{Absolute Value} end{equation} The absolute value of a number x is the distance between 0 and x, or egin{equation}label{eq:absv} ext{x}} Therefore, egin
Expanded form is the usual way you might see it in an equation: To solve an exponential equation, expand both sides and then factor out a common factor. Each side will have one number multiplied by another specific number raised to a power. Then take that power and multiply it by itself (to get one number squared). That’s your answer! Base form is used for when we’re given just the base (or “base-rate”) value of something: To solve a base-rate problem, first find the base rate (number of events per unit time), then subtract that from 1. Finally, multiply the result by the event rate (also called “per unit time”).
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# Amarge the following in descending order.(i) $\frac{2}{9}, \frac{2}{3} \cdot \frac{8}{21}$(ii) $\frac{1}{5}, \frac{3}{7}+\frac{7}{10}$In a magic square"
Given:
$(i)\ \frac{2}{9}, \frac{2}{3} \times \frac{8}{21}$
$(ii)\ \frac{1}{5}, \frac{3}{7},\ \frac{7}{10}$
Solution::
$(i)\ \frac{2}{9}, \frac{2}{3} \times \frac{8}{21}$
LCM of 3, 9, 21 is 63
Given fractions can be rewritten as
$\frac{14}{63}$, $\frac{42}{63}$, $\frac{24}{63}$
So descending order would be $\frac{42}{63}$, $\frac{24}{63}$, $\frac{14}{63}$
or $\frac{2}{3}$, $\frac{8}{21}$, $\frac{2}{9}$ or
$\frac{2}{3}> \frac{8}{21}> \frac{2}{9}$
$(ii)\ \frac{1}{5}, \frac{3}{7},\ \frac{7}{10}$
LCM of 5, 7, 10 is 70
Given fractions can be rewritten as $\frac{14}{70}$, $\frac{30}{70}$, \frac{49}{70}$Descending order is$\frac{49}{70} > \frac{30}{70} > \frac{14}{70}$or$\frac{7}{10} > \frac{3}{7} > \frac{1}{5}\$
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Updated on: 10-Oct-2022
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# Factors of 25
The number 25 is a composite number, meaning it has factors other than 1 and itself. The factors of 25 are the numbers that divide 25 evenly, leaving no remainder.
The factors of 25 are: 1, 5, 25.
Here’s a breakdown of the factors:
1: 1 is a factor of every number.
5: 5 divides 25 evenly, as 25 ÷ 5 = 5.
25: 25 is the number itself and divides evenly, as 25 ÷ 25 = 1.
These factors are important in various mathematical concepts, such as prime factorization, finding common factors, and determining divisibility. Additionally, the factors of a number can be used in solving equations, simplifying fractions, and exploring the properties of numbers.
## How many Factors of 25
The number 25 has a total of 3 factors. These factors are: 1, 5, and 25.
#### Factor Pairs of 25
Factor pairs of 25 are pairs of numbers that, when multiplied together, result in 25. The factor pairs of 25 are:
1 and 25 (1 * 25 = 25)
5 and 5 (5 * 5 = 25)
So, the factor pairs of 25 are (1, 25) and (5, 5).
#### All Factors of 25 using Multiplication
To find all the factors of 25 using multiplication, we can start with 1 and multiply it by various numbers until we reach 25. The factors of 25 are:
1 * 25 = 25
5 * 5 = 25
So, the factors of 25 obtained through multiplication are 1, 25, 5, and 5.
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# Applications of Systems of Linear Equations Example 1: Steve invested \$12,000 for one year in two different accounts, one at 3.5% and the other at 4%.
## Presentation on theme: "Applications of Systems of Linear Equations Example 1: Steve invested \$12,000 for one year in two different accounts, one at 3.5% and the other at 4%."— Presentation transcript:
Applications of Systems of Linear Equations Example 1: Steve invested \$12,000 for one year in two different accounts, one at 3.5% and the other at 4%. The combined interest on the two accounts was \$445. Use a system of equations to determine the amount in each account. Make a drawing to illustrate the situation.
Part at 3.5% Part at 4% 1) Variable declarations Let x be the amount at 3.5%. Let y be the amount at 4%.
2) Write the equations. Since this is a system with two variables, we need two equations. Steve invested \$12,000 for one year in two different accounts, one at 3.5% and the other at 4%. The combined interest on the two accounts was \$445. Use a system of equations to determine the amount in each account. \$12,000 is the amount in the combined accounts. (amount at 3.5%) + (amount at 4%) = (total amount)
Use the interest on the accounts to write the second equation. Steve invested \$12,000 for one year in two different accounts, one at 3.5% and the other at 4%. The combined interest on the two accounts was \$445. Use a system of equations to determine the amount in each account. 3.5% Account: 4% Account:
Steve invested \$12,000 for one year in two different accounts, one at 3.5% and the other at 4%. The combined interest on the two accounts was \$445. Use a system of equations to determine the amount in each account. 3.5% Account: 4% Account: (interest on 3.5%) + (interest on 4%) = (total interest)
The two equations are … 3) Solve the system: Multiply every term in the second equation by 1000 to eliminate decimals.
Multiply the top equation by ( - 35).
Add the equations: Substitute into the first of the original equations:
4) Write an answer in words, explaining the meaning in light of the application Part at 3.5% Part at 4% Steve invested 7000 at 3.5% and 5000 at 4%.
Example 2: Sam leaves a gas station in Denver at 11:00am heading west. At the same time, Mary leaves the same station heading east. Since Sam is driving in the mountains, his average rate is 10 mph slower than Mary’s. At 3:00pm they are 480 miles apart. Determine Sam’s rate. Applications of Systems of Linear Equations
Drt Sam Mary Sam leaves a gas station in Denver at 11:00am heading west. At the same time, Mary leaves the same station heading east. Since Sam is driving in the mountains, his average rate is 10 mph slower than Mary’s. At 3:00pm they are 480 miles apart. Determine Sam’s rate. Let x represent Sam’s rate. 1) Variable declaration: Let y represent Mary’s rate
Drt Sam Mary Both Sam and Mary were traveling the same amount of time, from 11:00am to 3:00pm, which is 4 hours. Sam leaves a gas station in Denver at 11:00am heading west. At the same time, Mary leaves the same station heading east. Since Sam is driving in the mountains, his average rate is 10 mph slower than Mary’s. At 3:00pm they are 480 miles apart. Determine Sam’s rate.
Drt Sam Mary Sam leaves a gas station in Denver at 11:00am heading west. At the same time, Mary leaves the same station heading east. Since Sam is driving in the mountains, his average rate is 10 mph slower than Mary’s. At 3:00pm they are 480 miles apart. Determine Sam’s rate. Since distance = rate × time, Sam’s distance is … … and Mary’s distance is…
Drt Sam Mary Sam leaves a gas station in Denver at 11:00am heading west. At the same time, Mary leaves the same station heading east. Since Sam is driving in the mountains, his average rate is 10 mph slower than Mary’s. At 3:00pm they are 480 miles apart. Determine Sam’s rate. (Sam’s distance) + (Mary’s distance) = 480 miles 2) Write the first equation
Drt Sam Mary Sam leaves a gas station in Denver at 11:00am heading west. At the same time, Mary leaves the same station heading east. Since Sam is driving in the mountains, his average rate is 10 mph slower than Mary’s. At 3:00pm they are 480 miles apart. Determine Sam’s rate. (Sam’s rate) = (Mary’s rate) - 10 Write the second equation
3) Solve the system: Substitute the expression in the second equation for x in the first equation.
Solve the equation:
4) Write an answer in words, explaining the meaning in light of the application What was asked for in the application Sam leaves a gas station in Denver at 11:00am heading west. At the same time, Mary leaves the same station heading east. Since Sam is driving in the mountains, his average rate is 10 mph slower than Mary’s. At 3:00pm they are 480 miles apart. Determine Sam’s rate.
y =Mary’s rate Mary’s rate was 65 mph. Sam’s rate was x. Sam’s rate was 55 mph.
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Assignment 4: Activity- Exploration of Medians
Many of us know that the medians of a given triangle can be found by connecting opposite vertices to the midpoint of each side of the triangle. This process can be achieved by using a compass, ruler, and pencil, but can also be achieved by using the mathematics program Geometers Sketchpad (GSP). Let's first begin our exploration of medians by walking through the steps necessary to construct the medians of any given triangle using GSP.
1. Place three points anywhere on the plane. These will be the vertices of your triangle.
2. Label these three points A, B, and C to identify your vertices.
3. Connect all three vertices to form the sides of your triangle, AB, BC, and AC.
4. Using the midpoint tool, under the Construct tab, find the midpoints of all three sides.
5. Now, connect all three midpoints to the opposite vertices of the triangle.
6. You can also label the three midpoints, using the text tool.
Through these easy steps, we have created the three medians of a given triangle. Looking at the last image, we can see that the three medians intersect at one given point inside the triangle. This point is called the Centroid and is usually labeled as G.
An exploration that would be extremely useful for students would be to figure out if the Centroid, the intersection of the triangle's three medians, can always be found inside the triangle. Students should be encouraged to explore all different kinds of triangles such as right triangles, isosceles triangles, and even equilateral triangles. Through this mini-exploration students will be able to discover more about the Centroid of triangles.
After allowing students to explore this question on their own or in groups, teachers could present the class with an illustration of a varying triangle and the location of the centroid as the triangle changes. On GSP teachers can easily vary the vertex A along a line to illustrate for students in one image the location of the Centroid as the triangle changes. Taking your already existing triangle ABC, teachers can merge the vertex A onto a line, l, and animate the vertex. This will automatically cause the vertex A to follow along the line, l, changing the shape of the triangle ABC. Students will be able to study the location of the Centroid, G, as the point varies. Screen shots of this animation can be seen below.
From the above snapshots of the animation on GSP students will be able to conclude that no matter the shape of the triangle, the Centroid will always be inside the triangle. As an additional exploration students could work on animating all three vertices on three different lines, l, m, and n to fully comprehend this mathematical principle about the Centroid. Merging vertex B onto a line m, and merging vertex C onto a line n, students could animate all three vertices. This exploration could be done after the teachers animation is shown, encouraging the students to try it on their own. Snapshots of what students might create can be seen below.
This activity/exploration is a great way to get students working hands-on to learn the mathematical concept of medians and the Centroid of a triangle. By working in small groups or pairs, students would even be able to learn from one another. In addition, through this exploration students would be required to discuss their ideas using proper mathematical terms, thus developing their mathematical vocabulary.
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### High School Mathematics - 210.2 Heights and Distances
Introduction One of the most important applications of trigonometry is the application of measurement of heights and distances which cannot be measured directly. It is also extensively used by astronomers in determining the distance of the heavenly bodies like the sun and moon and stars. Two angles are very often used in the practical applications of trigonometry . Line of sight The line of object is the line from our eyes to object. Angle of Elevation If the object is above the horizontal level of our eyes, we have to turn our head upwards to view the object. This is called the angle of elevation. Angle of Depression If the object is below the horizontal level of our eyes, we have to bend our head to view it. This is called the angle of elevation. Example 1: The angle of elevation of the top of a tower at a distance of 100 metres from its foot on a horizontal plane is found to be 60o. Find the height of the tower. Let CA be the tower equal to h metres in length and B point at a distance of 100 metres from its foot C. It is given that angle ABC = 60o From the right angle triangle ABC we have h/100 = tan 60 Solving the above we get h = 173.2 metres Answer: Height of the tower is 173.2 metres Example 2: From the top of a cliff, 200 metres high, the angle of depression of the top and bottom of a tower are observed to be observed to be 30o and 60o, find the height of the tower. Solution: Let AB represent the tower and P the top of cliff MP. If PX be the horizontal line through P, then angle XPA = 30o and angle XPB = 60o. Let the height of the tower be h metres. From A draw AL perpendicular to PM. ML = AB = h=> LP = (200-h) Again, ang PBM = ang XPB = 60 ang PAL = ang XPA = 30 (alternate angles ) From the right triangle PMB, BM/200 = cot 60 BM = 200 cot 60 = 200/3 From the right triangle PLA, AL /AP = cot 30 =>AL = LP cot 30 = =Ö(200-h)3 But AL = BM = =Ö(200-h)3 =200/3 Answer: h = 133 1/3 metres Directions: Answer the following questions. Also write at least 10 examples of your own.
Q 1: What is the angle of elevation of the sun when the length of the shadow of a pole is 31/2 times the height of the pole?90o30o60o Q 2: The angular elevation of a tower from a point is 30 degrees at a point in a horizontal line to the foot of the tower and 100 metres near it is 60 degrees, find the distance of the first point from the tower. 100 metres150 metres86.6 metres Q 3: The angle of elevation of the top of a tower from a point 60m from its foot is 30o. Find the height of the tower.2031/2 m100 m20 m Q 4: A person standing on the bank of a river, observes that the angle subtended by a tree on the opposite bank is 60 degrees, when he retires 40 metres from the bank, he finds the angle to be 30 degrees. Find the height of the tree.64 metres20 metres34 metres Q 5: The angle of elevation of the top of a tower which is yet incomplete at a point 120 metres from its base is 45o. How much higher should it be raised so that the elevation at the same point may become 60o?100 metres36.85 metres87.84 metres Q 6: The elevation of a tower at a point 60 metres from it is cot-13/5 Obtain the height of the tower.45 metres100 metres110 metres Q 7: A vertical flagstaff stands on a horizontal plane from a point distant 150 metres from its foot, the angle of elevation of its top is found to be 30o, find the height of the flagstaff.100.4 metres113.4 metres100 metres Q 8: From the light house the angles of depression of two ships on opposite sides of the light house are observed to be 30o and 45o. If the height of the light house is 300 metres, find the distance between the ships if the line joining them passes through foot of the light house.809 metres800 metres819.6 metres Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!
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# Disjoint sets Information
(Redirected from Disjoint set)
https://en.wikipedia.org/wiki/Disjoint_set
Two disjoint sets
In mathematics, two sets are said to be disjoint sets if they have no element in common. Equivalently, two disjoint sets are sets whose intersection is the empty set. [1] For example, {1, 2, 3} and {4, 5, 6} are disjoint sets, while {1, 2, 3} and {3, 4, 5} are not disjoint. A collection of two or more sets is called disjoint if any two distinct sets of the collection are disjoint.
## Generalizations
A disjoint collection of sets
This definition of disjoint sets can be extended to a family of sets ${\displaystyle \left(A_{i}\right)_{i\in I}}$: the family is pairwise disjoint, or mutually disjoint if ${\displaystyle A_{i}\cap A_{j}=\varnothing }$ whenever ${\displaystyle i\neq j}$. Alternatively, some authors use the term disjoint to refer to this notion as well.
For families the notion of pairwise disjoint or mutually disjoint is sometimes defined in a subtly different manner, in that repeated identical members are allowed: the family is pairwise disjoint if ${\displaystyle A_{i}\cap A_{j}=\varnothing }$ whenever ${\displaystyle A_{i}\neq A_{j}}$ (every two distinct sets in the family are disjoint). [2] For example, the collection of sets { {0, 1, 2}, {3, 4, 5}, {6, 7, 8}, ... } is disjoint, as is the set { {..., −2, 0, 2, 4, ...}, {..., −3, −1, 1, 3, 5} } of the two parity classes of integers; the family ${\displaystyle (\{n+2k\mid k\in \mathbb {Z} \})_{n\in \{0,1,\ldots ,9\}}}$ with 10 members is not disjoint (because the classes of even and odd numbers are each present five times), but it is pairwise disjoint according to this definition (since one only gets a non-empty intersection of two members when the two are the same class).
Two sets are said to be almost disjoint sets if their intersection is small in some sense. For instance, two infinite sets whose intersection is a finite set may be said to be almost disjoint. [3]
In topology, there are various notions of separated sets with more strict conditions than disjointness. For instance, two sets may be considered to be separated when they have disjoint closures or disjoint neighborhoods. Similarly, in a metric space, positively separated sets are sets separated by a nonzero distance. [4]
## Intersections
Disjointness of two sets, or of a family of sets, may be expressed in terms of intersections of pairs of them.
Two sets A and B are disjoint if and only if their intersection ${\displaystyle A\cap B}$ is the empty set. [1] It follows from this definition that every set is disjoint from the empty set, and that the empty set is the only set that is disjoint from itself. [5]
If a collection contains at least two sets, the condition that the collection is disjoint implies that the intersection of the whole collection is empty. However, a collection of sets may have an empty intersection without being disjoint. Additionally, while a collection of less than two sets is trivially disjoint, as there are no pairs to compare, the intersection of a collection of one set is equal to that set, which may be non-empty. [2] For instance, the three sets { {1, 2}, {2, 3}, {1, 3} } have an empty intersection but are not disjoint. In fact, there are no two disjoint sets in this collection. Also the empty family of sets is pairwise disjoint. [6]
A Helly family is a system of sets within which the only subfamilies with empty intersections are the ones that are pairwise disjoint. For instance, the closed intervals of the real numbers form a Helly family: if a family of closed intervals has an empty intersection and is minimal (i.e. no subfamily of the family has an empty intersection), it must be pairwise disjoint. [7]
## Disjoint unions and partitions
A partition of a set X is any collection of mutually disjoint non-empty sets whose union is X. [8] Every partition can equivalently be described by an equivalence relation, a binary relation that describes whether two elements belong to the same set in the partition. [8] Disjoint-set data structures [9] and partition refinement [10] are two techniques in computer science for efficiently maintaining partitions of a set subject to, respectively, union operations that merge two sets or refinement operations that split one set into two.
A disjoint union may mean one of two things. Most simply, it may mean the union of sets that are disjoint. [11] But if two or more sets are not already disjoint, their disjoint union may be formed by modifying the sets to make them disjoint before forming the union of the modified sets. [12] For instance two sets may be made disjoint by replacing each element by an ordered pair of the element and a binary value indicating whether it belongs to the first or second set. [13] For families of more than two sets, one may similarly replace each element by an ordered pair of the element and the index of the set that contains it. [14]
## References
1. ^ a b Halmos, P. R. (1960), Naive Set Theory, Undergraduate Texts in Mathematics, Springer, p. 15, ISBN 9780387900926.
2. ^ a b Smith, Douglas; Eggen, Maurice; St. Andre, Richard (2010), A Transition to Advanced Mathematics, Cengage Learning, p. 95, ISBN 978-0-495-56202-3.
3. ^ Halbeisen, Lorenz J. (2011), Combinatorial Set Theory: With a Gentle Introduction to Forcing, Springer monographs in mathematics, Springer, p. 184, ISBN 9781447121732.
4. ^ Copson, Edward Thomas (1988), Metric Spaces, Cambridge Tracts in Mathematics, vol. 57, Cambridge University Press, p. 62, ISBN 9780521357326.
5. ^ Oberste-Vorth, Ralph W.; Mouzakitis, Aristides; Lawrence, Bonita A. (2012), Bridge to Abstract Mathematics, MAA textbooks, Mathematical Association of America, p. 59, ISBN 9780883857793.
6. ^ See answers to the question ″Is the empty family of sets pairwise disjoint?″
7. ^ Bollobás, Béla (1986), Combinatorics: Set Systems, Hypergraphs, Families of Vectors, and Combinatorial Probability, Cambridge University Press, p. 82, ISBN 9780521337038.
8. ^ a b Halmos (1960), p. 28.
9. ^ Cormen, Thomas H.; Leiserson, Charles E.; Rivest, Ronald L.; Stein, Clifford (2001), "Chapter 21: Data structures for Disjoint Sets", Introduction to Algorithms (Second ed.), MIT Press, pp. 498–524, ISBN 0-262-03293-7.
10. ^ Paige, Robert; Tarjan, Robert E. (1987), "Three partition refinement algorithms", SIAM Journal on Computing, 16 (6): 973–989, doi: 10.1137/0216062, MR 0917035, S2CID 33265037.
11. ^ Ferland, Kevin (2008), Discrete Mathematics: An Introduction to Proofs and Combinatorics, Cengage Learning, p. 45, ISBN 9780618415380.
12. ^ Arbib, Michael A.; Kfoury, A. J.; Moll, Robert N. (1981), A Basis for Theoretical Computer Science, The AKM series in Theoretical Computer Science: Texts and monographs in computer science, Springer-Verlag, p. 9, ISBN 9783540905738.
13. ^ Monin, Jean François; Hinchey, Michael Gerard (2003), Understanding Formal Methods, Springer, p. 21, ISBN 9781852332471.
14. ^ Lee, John M. (2010), Introduction to Topological Manifolds, Graduate Texts in Mathematics, vol. 202 (2nd ed.), Springer, p. 64, ISBN 9781441979407.
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Question Video: Finding the Range of a Given Sine Function | Nagwa Question Video: Finding the Range of a Given Sine Function | Nagwa
# Question Video: Finding the Range of a Given Sine Function Mathematics • First Year of Secondary School
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Find the range of the function π(π) = 3 sin (π) + 4.
02:18
### Video Transcript
Find the range of the function π of π equals three sin π plus four.
Letβs recall first some key properties of the sine function. First, itβs a periodic function with a period of two π radians or 360 degrees. The graph oscillates between negative one and one, taking every value in between. This tells us that the range of the sine function is the closed interval from negative one to positive one. We can use this information to determine the range of the function π of π.
Remember, the range is the set of all possible values of the function itself or all possible output values. So weβre looking for all possible values of the expression three sin π plus four when π is any real number. Letβs consider the function transformations that have been applied to sin π to give three sin π plus four.
Multiplying sin π by three stretches the graph of the function vertically by a factor of three. This has the effect of multiplying each value in the range by three. So, after multiplying by three, the new range will be the closed interval from negative three to positive three.
Adding four to the function shifts the graph vertically upwards by four units. This has the effect of increasing each value in the range by four. So, in particular, negative three becomes one and three becomes seven. The range of the function π of π equals three sin π plus four is therefore the closed interval from one to seven.
We could also show this by using inequalities. If sin π is greater than or equal to negative one and less than or equal to positive one, then three sin π is greater than or equal to negative three and less than or equal to positive three. And three sin π plus four is greater than or equal to one and less than or equal to seven.
Using two methods then, weβve shown that the range of the function π of π equals three sin π plus four is the closed interval from one to seven.
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# 1989 AIME Problems/Problem 4
## Problem
If $a are consecutive positive integers such that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube, what is the smallest possible value of $c$?
## Solution
Since the middle term of an arithmetic progression with an odd number of terms is the average of the series, we know $b + c + d = 3c$ and $a + b + c + d + e = 5c$. Thus, $c$ must be in the form of $3 \cdot x^2$ based upon the first part and in the form of $5^2 \cdot y^3$ based upon the second part, with $x$ and $y$ denoting an integers. $c$ is minimized if it’s prime factorization contains only $3,5$, and since there is a cubed term in $5^2 \cdot y^3$, $3^3$ must be a factor of $c$. $3^35^2 = \boxed{675}$, which works as the solution.
## Solution 2
Let $b$, $c$, $d$, and $e$ equal $a+1$, $a+2$, $a+3$, and $a+4$, respectively. Call the square and cube $k^2$ and $m^3$, where both k and m are integers. Then:
$5a + 10 = m^3$
Now we know $m^3$ is a multiple of 125 and $m$ is a multiple of 5. The lower $m$ is, the lower the value of $c$ will be. Start from 5 and add 5 each time.
$m = 5$ gives no solution for k
$m = 10$ gives no solution for k
$m = 15$ gives a solution for k.
$10 + 5a = 15^3$
$2 + a = 675$
$c = \boxed{675}$
-jackshi2006
## Solution 3
Let the numbers be $a,a+1,a+2,a+3,a+4.$ When then know $3a+6$ is a perfect cube and $5a+10$ is perfect cube. Since $5a+10$ is divisible by $5$ we know that $5a+10 = (5k)^3$ since otherwise we get a contradiction. This means $a = 25k^3 - 2$ in which plugging into the other expression we know $3(25k^3 - 2) + 6 = 75k^3$ is a perfect square. We know $75 = 5^2 \cdot 3$ so we let $k = 3$ to obtain the perfect square. This means that $c = a+2 = (25 \cdot 27 - 2)+2 = 25 \cdot 27 = \boxed{675}.$
## Solution 4
(This is literally a combination of 1 and 3)
Since $a$, $b$, $c$, $d$, and $e$ are consecutive, $a = c-2$, $b = c-1$, $c=c$, $d = c+1$, and $e = c+2$.
Because $b+c+d = 3c$ is a perfect square, and $a+b+c+d+e = 5c$ is a perfect cube, we can express $c$ as $c = 3^{n} \cdot 5^{k}$.
Now, by the problem's given information,
$k \equiv 0 \text{(mod 2)}$
$n \equiv 0 \text{(mod 3)}$
and because ALL exponents have to be cubes/squares,
$k \equiv 2 \text{(mod 3)}$
$n \equiv 1 \text{(mod 2)}$
Therefore,
$k = 2$
$n = 3$
$c = 3^3 \cdot 5^2 = \boxed{675}$.
~ethanhansummerfun
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# Algebra II : Solving Equations
## Example Questions
### Example Question #1 : Linear Systems With Three Variables
Solve this system of equations.
, ,
, ,
, ,
, ,
, ,
, ,
Explanation:
Equation 1:
Equation 2:
Equation 3:
Adding the terms of the first and second equations together will yield .
Then, add that to the third equation so that the y and z terms are eliminated. You will get .
This tells us that x = 1. Plug this x = 1 back into the systems of equations.
Now, we can do the rest of the problem by using the substitution method. We'll take the third equation and use it to solve for y.
Plug this y-equation into the first equation (or second equation; it doesn't matter) to solve for z.
We can use this z value to find y
So the solution set is x = 1, y = 2, and z = –5/3.
### Example Question #2 : Solving Equations
Solve for
Explanation:
To solve this problem we can first add to each side of the equation yielding
Then we take the square root of both sides to get
Then we calculate the square root of which is .
### Example Question #3 : Solving Equations
Solve this system of equations for :
Explanation:
Multiply the top equation by 3 on both sides, then add the second equation to eliminate the terms:
### Example Question #91 : Factoring Polynomials
Solve for .
Explanation:
Multiply both sides by 3:
Distribute:
Subtract from both sides:
Add the terms together, and subtract from both sides:
Divide both sides by :
Simplify:
### Example Question #1 : Solving Equations
Solve for :
Explanation:
Distribute the x through the parentheses:
x2 –2x = x2 – 8
Subtract x2 from both sides:
–2x = –8
Divide both sides by –2:
x = 4
### Example Question #1 : Solving Equations
Solve for :.
Explanation:
First factor the expression by pulling out :
Factor the expression in parentheses by recognizing that it is a difference of squares:
Set each term equal to 0 and solve for the x values:
### Example Question #1 : Systems Of Equations
Solve the system of equations.
None of the other answers are correct.
Explanation:
Isolate in the first equation.
Plug into the second equation to solve for .
Plug into the first equation to solve for .
Now we have both the and values and can express them as a point: .
### Example Question #1 : Solving Equations
Solve for and .
Cannot be determined.
Explanation:
1st equation:
2nd equation:
Subtract the 2nd equation from the 1st equation to eliminate the "2y" from both equations and get an answer for x:
Plug the value of into either equation and solve for :
### Example Question #9 : Solving Equations
What is a solution to this system of equations:
Explanation:
Step 1: Multiply first equation by 2 and add the result to the second equation. The result is:
Step 2: Multiply first equation by 3 and add the result to the third equation. The result is:
Step 3: Multiply second equation by 23 and add the result to the third equation. The result is:
Step 4: solve for z.
Step 5: solve for y.
Step 6: solve for x by substituting y=2 and z=1 into the first equation.
### Example Question #10 : Solving Equations
What is a solution to this system of equations?
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Review question
# Can we calculate the total time taken for this car journey? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R7250
## Solution
The diagram shows a speed-time graph of a car.
1. Calculate the acceleration of the car during the first $20$ seconds.
The acceleration at any point is given by the gradient of the line, so the acceleration is constant on each section.
The gradient for the first section is $\dfrac{15}{20}=0.75$, and so the acceleration of the car during the first $20$ seconds is $\quantity{0.75} {m/s^2}$.
1. Calculate the distance the car travels from rest before it begins to decelerate.
The distance the car travels is given by the area below the graph up to $t = 70s$.
We’ll split this area into two pieces, the left hand side triangle (the distance the car travels during the first 20 seconds), and the rectangle for $20s \leq t \leq 70s$.
The base of the triangle is $20$ and the height is $15$, and so the area of the triangle is given by $\dfrac{20 \times 15}{2} = 150$.
The rectangle has sides of lengths $15$ and $70-20 = 50$ and so has an area of $15 \times 50 = 750.$
Therefore, the total distance the car travels before it begins to decelerate is $150 + 750 = \quantity{900}{m}$.
1. Given that the car decelerates at $\quantity{0.5} {m/s^2}$, calculate the total time taken for the journey.
We’ll describe the deceleration of the car as a linear function, or the equation of a straight line, and then compute the $t$-intercept.
The general form for a linear function is $y = ax + b$, where $a$ is the slope of the function and $b$ is the $y$-intercept.
We already know that the gradient of the graph is $-0.5$, so the function is of the form $y = -0.5x + b.$
The graph passes through the point $(70, 15)$, so the function has to satisfy $15 = -0.5 \times 70 + b$, which yields $b = 50$.
The function describing the graph for $t \geq 70$ is therefore given by $v = -0.5t + 50.$
The $t$-intercept is the solution to $0 = -0.5t + 50$, which gives $t = 100$. Therefore, the total time taken for the journey is $\quantity{100}{s}$.
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# 10.3: Surface Area of Prisms
Difficulty Level: At Grade Created by: CK-12
## Introduction
The Doll House
On her first day wrapping boxes, Candice had a customer come up to the counter with a huge box in her arms.
“This is a doll house for my niece,” the woman said. “Can you please wrap it?”
“Certainly,” Candice said looking at the box.
The box was . Candice looked at the different rolls of wrapping paper. She was sure that she was going to need the largest roll of wrapping paper that she could find, and even then she was sure that it was going to take two pieces of paper. The wrapping station has a huge roll of paper on a big roller. This one was brand new.
“How big is this roll?” Candice asked Mrs. Scott.
“It is feet long," Mrs. Scott said.
“That should be big enough,” Candice thought.
Candice isn’t sure that the roll of paper will be enough for the doll house. This is a problem that can be solved using surface area. If Candice can figure out the surface area of the box, then she will know how much of the wrapping paper she will need to wrap the doll house.
This lesson will teach you all about surface area. Pay attention and at the end of the lesson you will know how to figure out the wrapping paper dilemma.
What You Will Learn
By the end of this lesson, you will have an understanding of the following skills.
• Recognize the surface area of prisms as the sum of the areas of the faces using nets.
• Find surface areas of rectangular prisms using formulas.
• Find surface areas of triangular prisms using formulas.
• Solve real-world problems involving surface area of prisms.
Teaching Time
I. Recognize Surface Area of Prisms as the Sum of the Areas of the Faces using Nets
In this lesson, we will look at prisms with more detail. Remember that a prism is a three-dimensional object with two congruent parallel bases. The shape of the base names the prism and there are rectangles for the sides of the prism.
When we worked with two-dimensional figures, we measured the area of those figures. The area is the space that is contained in a two-dimensional figure. Now we are going to look at the area of three-dimensional figures. Only this isn’t called simply area anymore, it is called surface area.
What is surface area?
The surface area is the covering of a three-dimensional figure. Imagine you could wrap a three dimensional figure in wrapping paper, like a present. The amount of wrapping paper needed to cover the figure represents its surface area.
To find the surface area, we must be able to calculate the area of each face and then add these areas together.
One way to do this is to use a net. Remember that a net is a two-dimensional representation of a three-dimensional solid. A net is a stretched out, or unfolded version of a solid.
If we find the area of each section of the net and then add up those measurements, then we will know the measurement of the “cover” of the figure.
We can do this with prisms of all different kinds. Let’s look at a net for a rectangular prism.
Now we can find the area of each part of the rectangular prism. Notice that the rectangular prism is made up of rectangles. To find the area of a rectangle, we use this formula.
Next we can find the area of each part of the prism. Remember that there are six faces that we need to measure!
The answer is 282 sq. inches.
We found the area of each rectangular face and then added all of these areas together. The total surface area of the rectangular prism is 282 square inches. Using a net helped us to locate all of the faces and find the measurements of each side.
Example
What is the surface area of the figure below?
The first thing we need to do is draw a net. Get ready to exercise your imagination! It may help to color the top and bottom faces to keep you on track. Begin by drawing the bottom face. It is a triangle. Each side of the face is connected to a side face. What shape is each side face? They are rectangles, so we draw rectangles along each side of the triangular base. Lastly, we draw the top face, which can be connected to any of the side faces.
Next let’s fill in the measurements for the sides of each face so that we can calculate their area. Be careful! This time two of the faces are triangles. Remember, we calculate the area of triangles with the formula . We need to know the height of the triangles, look at the diagram to find it. We are going to find the areas of two triangles and three rectangles.
Now that we have the measurements of all the faces, let’s calculate the area of each. Remember to use the correct area formula.
We used the formula to find the area of the top and bottom faces. We used the formula to find the area of the three side faces.
When we add these together, we get a surface area of 524 square centimeters for this triangular prism.
II. Find Surface Area of Rectangular Prisms Using Formulas
Nets let us see each face so that we can calculate the area. However, we can also use a formula to represent the faces as we find their area. Let’s look again at our calculations for the rectangular prism we dealt with.
Notice that our calculations repeat in pairs. This is because every face in a rectangular prism is opposite a face that is congruent. In other words, the top and bottom faces have the same measurements, the two long side faces have the same measurements, and the two short side faces have the same measurements. Therefore we can create a formula for surface area that gives us a short cut. We simply double each calculation to represent a pair of faces.
The formula looks like this.
In this rectangular prism, inches, inches, and inches. We simply put these numbers into the formula and solve for surface area. Let’s try it.
As we have already seen, the surface area of this prism is 282 square inches. This formula just saves us a little time by allowing us to calculate the area of pairs of faces at a time. Let’s try another example.
Example
What is the surface area of the figure below?
All of the faces of this prism are rectangles, so we can use our formula for finding the surface area of rectangular prisms. We simply put the measurements into the formula and solve for , surface area.
This rectangular prism has a surface area of 938 square centimeters. If you’re not sure which measurements go with which side of the prism, try drawing a net.
Let’s practice using this formula.
10D. Lesson Exercises
Use the formula to find the surface area of each rectangular prism.
1. Length of 8 in, width of 4 inches, height of 6 inches
2. Length of 5 ft, width of 4 ft, height of 2 ft
Write this formula for finding the surface area of a rectangular prism down in your notebook.
Now let’s see how we can use a formula to find the surface area of a triangular prism.
III. Find Surface Areas of Triangular Prisms using Formulas
Triangular prisms have a different formula for finding surface area because they have two triangular faces opposite each other. Remember, the formula for the area of triangles is not the same as the area formula for rectangles, so we’ll have to proceed differently here to find a formula for surface area.
First, we know that we need to find the area of the two triangular faces. Each face will have an area of . Remember, we can use a formula to calculate the area of a pair of faces. Therefore we can double this formula to find the area of both triangular faces at once. This gives us . The numbers cancel each other out, and we’re left with . That part was easy!
Next, we need to calculate the area of each side face. The length of each rectangle is the same as the height of the prism, so we’ll call this . The width of each rectangle is actually the same as each side of the triangular base. Instead of multiplying the length and width for each rectangle, we can combine this information. We can multiply the perimeter of the triangular base, since it is the sum of each “width” of a rectangular side, by the height of the prism, .
If we put these pieces together—the area of the bases and the area of the side faces—we get this formula.
To use this formula, we fill in the base and height of the prism’s triangular base, the lengths of the base’s sides, and the height of the prism. Don’t confuse the height of the triangular base with the height of the prism! Let’s try out the formula.
Example
What is the surface area of the figure below?
We have all of the measurements we need. Let’s put them into the formula and solve for surface area, .
This triangular prism has a surface area of 222 square centimeters.
The biggest thing that we need to watch out for with this formula is that we put the correct measurement in the correct spot.
IV. Solve Real-World Problems Involving Surface Area of Prisms
We have learned two ways to find surface area: drawing a net or using a formula.
Write both of these formulas down in your notebooks. Then continue.
We can use either of these methods to solve word problems involving surface area. Nets may be especially useful if the problem does not provide an image of the prism. If you choose to use a formula, be sure you know whether the problem deals with a rectangular or a triangular prism. Let’s practice using what we have learned.
Example
Crystal is wrapping the box below in wrapping paper for her brother’s birthday. How much wrapping paper will she need?
First of all, is this a rectangular or triangular prism? All of the faces are rectangles, so it is a rectangular prism. The picture clearly shows us what its length, width, and height are, so let’s use the formula for finding the surface area of rectangular prisms. Simply put the measurements in for the appropriate variables in the formula.
Crystal will need 468 square inches of wrapping paper in order to cover the present.
Now let’s go back to the problem from the introduction and use what we have learned to figure out the surface area.
## Real–Life Example Completed
The Doll House
Here is the original problem once again. Reread the problem and underline the important information.
On her first day wrapping boxes, Candice had a customer come up to the counter with a huge box in her arms.
“This is a doll house for my niece,” the woman said. “Can you please wrap it?”
“Certainly,” Candice said looking at the box.
The box was . Candice looked at the different rolls of wrapping paper. She was sure that she was going to need the largest roll of wrapping paper that she could find, and even then she was sure that it was going to take two pieces of paper. The wrapping station has a huge roll of paper on a big roller. This one was brand new.
“How big is this roll?” Candice asked Mrs. Scott.
“It is feet long," Mrs. Scott said.
“That should be big enough,” Candice thought.
To figure out if the wrapping paper on the roll will be enough to wrap the doll house, Candice has to figure out the surface area of the box. She knows the dimensions, so she can use these dimensions to help solve the problem.
The box is .
Since a box is a rectangular prism, and we know the length and width, we can use this formula for the surface area of the box.
Next, we can substitute the given measurements into the formula.
Let’s change that into square feet by dividing by 144.
We can round up to 41 square feet just to be sure.
Now the measurement of the wrapping paper was in inches and feet.
, so let’s change it to feet
The wrapping paper will be enough to cover the box of the doll house.
## Vocabulary
Here are the vocabulary words that are found in this lesson.
Prism
a three-dimensional solid with two congruent parallel bases.
Area
the space enclosed inside a two-dimensional figure.
Surface Area
the covering of a three dimensional solid.
Net
a two-dimensional representation of a three-dimensional solid.
Rectangular Prism
a prism with rectangles as bases and faces.
Triangular Prism
a prism with triangles as bases and rectangles as faces.
## Technology Integration
1. http://www.mathplayground.com/mv_surface_area_prisms.html – This is a Brightstorm video on how to find the surface area of a prism.
## Time to Practice
Directions: Use the formula for surface area to find the surface area of each rectangular prism.
1. A rectangular prism with a length of 10 in, width of 8 in and height of 5 inches.
2. A rectangular prism with a length of 8 in, width of 8 in and height of 7 inches.
3. A rectangular prism with a length of 12 m, width of 4 m and height of 6 meters.
4. A rectangular prism with a length of 10 in, a width of 6 in and a height of 7 inches.
5. A rectangular prism with a length of 12 m, a width of 8 m and a height of 5 meters.
6. A rectangular prism with a length of 9 ft, a width of 7 feet and a height of 6 feet.
7. A rectangular prism with a length of 10 m, a width of 8 m and a height of 2 m.
8. A rectangular prism with a length of 6 ft, a width of 5 feet and a height of 3 feet.
9. A rectangular prism with a length of 3 feet, a width of 6 feet and a height of 2 feet.
10. A rectangular prism with a length of 4 feet, a width of 4 feet and a height of 4 feet.
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# Parallelograms
## Quadrilateral with two pairs of parallel sides.
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Parallelograms
### Parallelograms
A parallelogram is a quadrilateral with two pairs of parallel sides.
Notice that each pair of sides is marked parallel (for the last two shapes, remember that when two lines are perpendicular to the same line then they are parallel). Parallelograms have a lot of interesting properties.
1. Opposite Sides Theorem: If a quadrilateral is a parallelogram, then both pairs of opposite sides are congruent.
If then
2. Opposite Angles Theorem: If a quadrilateral is a parallelogram, then both pairs of opposite angles are congruent.
If then
3. Consecutive Angles Theorem: If a quadrilateral is a parallelogram, then all pairs of consecutive angles are supplementary.
If then
\begin{align*}m \angle A + m \angle D = 180^\circ\\ m \angle A + m \angle B = 180^\circ\\ m \angle B + m \angle C = 180^\circ\\ m \angle C + m \angle D = 180^\circ\end{align*}
4. Parallelogram Diagonals Theorem: If a quadrilateral is a parallelogram, then the diagonals bisect each other.
If then
What if you were told that \begin{align*}FGHI\end{align*} is a parallelogram and you are given the length of \begin{align*}FG\end{align*} and the measure of \begin{align*}\angle F \end{align*}? What can you determine about \begin{align*}HI\end{align*}, \begin{align*}\angle H \end{align*}, \begin{align*}\angle G \end{align*}, and \begin{align*}\angle I \end{align*}?
### Examples
#### Example 1
Show that the diagonals of \begin{align*}FGHJ\end{align*} bisect each other.
Find the midpoint of each diagonal.
\begin{align*}&\text{Midpoint of} \ \overline{FH}: \qquad \left ( \frac{-4 + 6 }{2}, \frac{5 - 4}{2} \right ) = (1, 0.5)\\ &\text{Midpoint of} \ \overline{GJ}: \qquad \left ( \frac{3 - 1}{2}, \frac{3 - 2}{2} \right ) = (1,0.5)\end{align*}
Because they are the same point, the diagonals intersect at each other’s midpoint. This means they bisect each other.
#### Example 2
Find the measures of \begin{align*}a\end{align*} and \begin{align*}b\end{align*} in the parallelogram below:
Consecutive angles are supplementary so \begin{align*}127^\circ +m\angle b=180^\circ\end{align*} which means that \begin{align*}m\angle b=53^\circ\end{align*}. \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are alternate interior angles and since the lines are parallel (since its a parallelogram), that means that \begin{align*}m\angle a =m\angle b=53^\circ\end{align*}.
#### Example 3
\begin{align*}ABCD\end{align*} is a parallelogram. If \begin{align*}m \angle A = 56^\circ\end{align*}, find the measure of the other angles.
First draw a picture. When labeling the vertices, the letters are listed, in order.
If \begin{align*}m \angle A = 56^\circ\end{align*}, then \begin{align*}m \angle C = 56^\circ\end{align*} by the Opposite Angles Theorem.
\begin{align*}m \angle A + m \angle B & = 180^\circ \quad \text{by the Consecutive Angles Theorem.}\\ 56^\circ + m \angle B & = 180^\circ\\ m \angle B & = 124^\circ \quad m \angle D = 124^\circ \quad \text{because it is an opposite angle to} \ \angle B.\end{align*}
#### Example 4
Find the values of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
Remember that opposite sides of a parallelogram are congruent. Set up equations and solve.
\begin{align*}6x - 7 & = 2x + 9 && y + 3 = 12\\ 4x & = 16 && \qquad y = 9\\ x & = 4\end{align*}
#### Example 5
Prove the Opposite Sides Theorem.
Given: \begin{align*}ABCD\end{align*} is a parallelogram with diagonal \begin{align*}\overline{BD}\end{align*}
Prove: \begin{align*}\overline{AB} \cong \overline{DC},\overline{AD} \cong \overline{BC}\end{align*}
Statement Reason
1. \begin{align*}ABCD\end{align*} is a parallelogram with diagonal \begin{align*} \overline{BD}\end{align*} 1. Given
2. \begin{align*}\overline{AB} \| \overline{DC}, \overline{AD} \| \overline{BC}\end{align*} 2. Definition of a parallelogram
3. \begin{align*}\angle ABD \cong \angle BDC, \angle ADB \cong \angle DBC\end{align*} 3. Alternate Interior Angles Theorem
4. \begin{align*}\overline{DB} \cong \overline{DB}\end{align*} 4. Reflexive PoC
5. \begin{align*}\triangle ABD \cong \triangle CDB\end{align*} 5. ASA
6. \begin{align*}\overline{AB} \cong \overline{DC}, \overline{AD} \cong \overline{BC}\end{align*} 6. CPCTC
The proof of the Opposite Angles Theorem is almost identical.
### Review
\begin{align*}ABCD\end{align*} is a parallelogram. Fill in the blanks below.
1. If \begin{align*}AB = 6\end{align*}, then \begin{align*}CD =\end{align*} ______.
2. If \begin{align*}AE = 4\end{align*}, then \begin{align*}AC =\end{align*} ______.
3. If \begin{align*}m \angle ADC = 80^\circ, m \angle DAB\end{align*} = ______.
4. If \begin{align*}m \angle BAC = 45^\circ, m \angle ACD\end{align*} = ______.
5. If \begin{align*}m \angle CBD = 62^\circ, m \angle ADB\end{align*} = ______.
6. If \begin{align*}DB = 16\end{align*}, then \begin{align*}DE\end{align*} = ______.
7. If \begin{align*}m \angle B = 72^\circ\end{align*} in parallelogram \begin{align*}ABCD\end{align*}, find the other three angles.
8. If \begin{align*}m \angle S = 143^\circ\end{align*} in parallelogram \begin{align*}PQRS\end{align*}, find the other three angles.
9. If \begin{align*}\overline{AB} \perp \overline{BC}\end{align*} in parallelogram \begin{align*}ABCD\end{align*}, find the measure of all four angles.
10. If \begin{align*}m \angle F = x^\circ\end{align*} in parallelogram \begin{align*}EFGH\end{align*}, find the other three angles.
For questions 11-18, find the values of the variable(s). All the figures below are parallelograms.
Use the parallelogram \begin{align*}WAVE\end{align*} to find:
1. \begin{align*}m \angle AWE\end{align*}
2. \begin{align*}m \angle ESV\end{align*}
3. \begin{align*}m \angle WEA\end{align*}
4. \begin{align*}m \angle AVW\end{align*}
Find the point of intersection of the diagonals to see if \begin{align*}EFGH\end{align*} is a parallelogram.
1. \begin{align*}E(-1, 3), F(3, 4), G(5, -1), H(1, -2)\end{align*}
2. \begin{align*}E(3, -2), F(7, 0), G(9, -4), H(5, -4)\end{align*}
3. \begin{align*}E(-6, 3), F(2, 5), G(6, -3), H(-4, -5)\end{align*}
4. \begin{align*}E(-2, -2), F(-4, -6), G(-6, -4), H(-4, 0)\end{align*}
Fill in the blanks in the proofs below.
1. Opposite Angles Theorem
Given: \begin{align*}ABCD\end{align*} is a parallelogram with diagonal \begin{align*}\overline{BD}\end{align*}
Prove: \begin{align*}\angle A \cong \angle C\end{align*}
Statement Reason
1. 1. Given
2. \begin{align*}\overline{AB} \| \overline{DC},\overline{AD} \| \overline{BC}\end{align*} 2.
3. 3. Alternate Interior Angles Theorem
4. 4. Reflexive PoC
5. \begin{align*}\triangle ABD \cong \triangle CDB\end{align*} 5.
6. \begin{align*}\angle A \cong \angle C\end{align*} 6.
1. Parallelogram Diagonals Theorem
Given: \begin{align*}ABCD\end{align*} is a parallelogram with diagonals \begin{align*}\overline{BD}\end{align*} and \begin{align*}\overline{AC}\end{align*}
Prove: \begin{align*}\overline{AE} \cong \overline{EC}, \overline{DE} \cong \overline{EB}\end{align*}
Statement Reason
1. 1.
2. 2. Definition of a parallelogram
3. 3. Alternate Interior Angles Theorem
4. \begin{align*}\overline{AB} \cong \overline{DC}\end{align*} 4.
5. 5.
6. \begin{align*}\overline{AE} \cong \overline{EC}, \overline{DE} \cong \overline{EB}\end{align*} 6.
1. Find \begin{align*}x, y^\circ,\end{align*} and \begin{align*}z^\circ\end{align*}. (The two quadrilaterals with the same side are parallelograms.)
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### Vocabulary Language: English Spanish
TermDefinition
parallelogram A quadrilateral with two pairs of parallel sides. A parallelogram may be a rectangle, a rhombus, or a square, but need not be any of the three.
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# Lesson 11
Introducing the Number $i$
• Let’s meet $$i$$.
### 11.1: Math Talk: Squared
Find the value of each expression mentally.
$$\left(2\sqrt{3}\right)^2$$
$$\left(\frac12 \sqrt{3}\right)^2$$
$$\left(2\sqrt{\text- 1}\right)^2$$
$$\left(\frac12 \sqrt{\text- 1}\right)^2$$
### 11.2: It is $i$
Find the solutions to these equations, then plot the solutions to each equation on the imaginary or real number line.
1. $$a^2 = 16$$
2. $$b^2 = \text- 9$$
3. $$c^2 = \text- 5$$
### 11.3: The $i$’s Have It
Write these imaginary numbers using the number $$i$$.
1. $$\sqrt{\text- 36}$$
2. $$\sqrt{\text- 10}$$
3. $$\text- \sqrt{\text- 100}$$
4. $$\text- \sqrt{\text- 17}$$
### 11.4: Complex Numbers
1. Label at least 8 different imaginary numbers on the imaginary number line.
2. When we add a real number and an imaginary number, we get a complex number. The diagram shows where $$2 + i$$ is in the complex number plane. What complex number is represented by point $$A$$?
3. Plot these complex numbers in the complex number plane and label them.
1. $$\text- 2 - i$$
2. $$\text- 6 + 3i$$
3. $$5+4i$$
4. $$1 - 3i$$
Diego says that all real numbers and all imaginary numbers are complex numbers but not all complex numbers are imaginary or real. Do you agree with Diego? Explain your reasoning.
### Summary
A square root of a number $$a$$ is a number whose square is $$a$$. In other words, it is a solution to the equation $$x^2 = a$$. Every positive real number has two real square roots. For example, look at the number 35. Its square roots are $$\sqrt{35}$$ and $$\text-\sqrt{35}$$, because those are the two numbers that square to make 35 (remember, the $$\sqrt{}$$ symbol is defined to indicate the positive square root). In other words, $$\left(\sqrt{35}\right)^2=35$$ and $$\left(\text-\sqrt{35}\right)^2=35$$.
Similarly, every negative real number has two imaginary square roots. The two square roots of -1 are written $$i$$ and $$\text- i$$. That means that
$$\displaystyle i^2 = \text-1$$
and
$$\displaystyle (\text- i)^2 = \text-1$$
Another example would be the number -17. Its square roots are $$i\sqrt{17}$$ and $$\text-i\sqrt{17}$$, because
$$\displaystyle \begin{array}{} \left(i\sqrt{17}\right)^2 &= 17i^2 \\ &=\text-17 \end{array}$$
and
$$\displaystyle \begin{array}{} \left(\text-i\sqrt{17}\right)^2 &=17(\text-i)^2 \\ &= 17i^2 \\ & =\text-17 \end{array}$$
In general, if $$a$$ is a positive real number, then the square roots of $$\text- a$$ are $$i \sqrt{a}$$ and $$\text- i \sqrt{a}$$.
Rarely, we might see something like $$\sqrt{\text-17}$$. It’s not immediately clear which of the two square roots it is supposed to represent. By convention, $$\sqrt{\text-17}$$ is defined to indicate the square root on the positive imaginary axis, so $$\sqrt{\text- 17}=i\sqrt{17}$$.
When we add a real number and an imaginary number, we get a complex number. Together, the real number line and the imaginary number line form a coordinate system that can be used to represent any complex number as a point in the complex plane. For example, the point shown represents the complex number $$\text- 3 + 2i$$.
In this context, people call the real number line the real axis and the imaginary number line the imaginary axis. This is different than the coordinate plane you have seen before because those points were pairs of real numbers, like $$(\text- 3,2)$$, but in the complex plane, each point represents a single complex number. Note that since the real number line is part of the complex plane, real numbers are a special type of complex number. For example, the real number 5 can be described as the point $$5+0i$$ in the complex plane.
### Glossary Entries
• complex number
A number in the complex plane. It can be written as $$a + bi$$, where $$a$$ and $$b$$ are real numbers and $$i^2 = \text-1$$.
• imaginary number
A number on the imaginary number line. It can be written as $$bi$$, where $$b$$ is a real number and $$i^2 = \text-1$$.
• real number
A number on the number line.
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A linear pair is a pair of angles that lie next to each other on a line and whose measures add to equal 180 degrees. Linear pair: Two adjacent angles are said to be linear pair if their sum is equal to 180°. S(1) is an exception, but S(2) is clearly true because 2 is a prime number. If we apply a function to every element in a set, the answer is still a set. Proof: ∵ ABC is an isosceles triangle If we apply a function to every element in a set, the answer is still a set. Proof by Induction is a technique which can be used to prove that a certain statement is true for all natural numbers 1, 2, 3, … The “statement” is usually an equation or formula which includes a variable n which could be any natural number. Proof for complementary case is similar. We need to show that given a linear pair … Sets are built up from simpler sets, meaning that every (non-empty) set has a minimal member. If we want to prove a statement S, we assume that S wasn’t true. If you start with different axioms, you will get a different kind of mathematics, but the logical arguments will be the same. Some theorems can’t quite be proved using induction – we have to use a slightly modified version called Strong Induction. Now let us assume that S(1), S(2), …, S(k) are all true, for some integer k. We know that k + 1 is either a prime number or has factors less than k + 1. This works for any initial group of people, meaning that any group of k + 1 also has the same hair colour. It turns out that the principle of weak induction and the principle of strong induction are equivalent: each implies the other one. The first step, proving that S(1) is true, starts the infinite chain reaction. PAIR-SET AXIOM Similarly, ∠GON and ∠HON form a linear pair and so on. There is a set with no members, written as {} or ∅. Proofs are what make mathematics different from all other sciences, because once we have proven something we are absolutely certain that it is and will always be true. Imagine that we place several points on the circumference of a circle and connect every point with each other. In the above example, we could count the number of intersections in the inside of the circle. D-2 For all points A and B, AB ‚ 0, with equality only when A = B. D-3: For all points A and B, AB = BA. It is really just a question of whether you are happy to live in a world where you can make two spheres from one…. ∠AOC + ∠BOC = 180° Axiom 6.1: If a ray stands on a line, then the sum of two adjacent angles so formed is 180°. ∠AOC + ∠BOC = 180° Axiom 6.1: If a ray stands on a line, then the sum of two adjacent angles so formed is 180°. This property is called as the linear pair axiom Not all points lie on the same line. Linear pair: Two adjacent angles are said to be linear pair if their sum is equal to 180°. However this is not as problematic as it may seem, because axioms are either definitions or clearly obvious, and there are only very few axioms. We could now try to prove it for every value of x using “induction”, a technique explained below. Then if we have k + 1 disks: In total we need (2k – 1) + 1 + (2k – 1) = 2(k+1) – 1 steps. We have a pair of adjacent angles, and this pair is a linear pair, which means that the sum of the (measures of the) two angles will be 180 0. We first check the equation for small values of n: Next, we assume that the result is true for k, i.e. 4 Thinking carefully about the relationship between the number of intersections, lines and regions will eventually lead us to a different equation for the number of regions when there are x = V.Axi points on the circle: Number of regions = x4 – 6 x3 + 23 x2 – 18 x + 2424 = (Math.pow(V.Axi,4) - 6*Math.pow(V.Axi,3) + 23*Math.pow(V.Axi,2) - 18*V.Axi + 24)/24. In Axiom 6.1, it is given that 'a ray stands on a line'. Can you find the mistake? Find the axiom or theorem from a high school book that corresponds to the Supplement Postulate. AXIOM-1 : If a ray stands on a line, then the sum of two adjacent angles so formed is 180°. Using this assumption, we try to deduce that S(. AXIOM OF INFINITY AXIOM OF REPLACEMENT Exercise 2.42. Proof of vertically opposite angles theorem. In fig 6.15,angle pqr=angle prq, then prove thatangle pqs=angle prt - 4480658 Start Over The Fundamental Theorem of Arithmetic states that every integer greater than 1 is either a prime number, or it can be written as the product of prime numbers in an essentially unique way. 2 1 5 from the axiom of parallel lines corresponding angles. But the fact that the Axiom of Choice can be used to construct these impossible cuts is quite concerning. Here, ∠BOC + ∠COA = 180°, so they form linear pair. One example is the Continuum Hypothesis, which is about the size of infinite sets. Towards the end of his life, Kurt Gödel developed severe mental problems and he died of self-starvation in 1978. Using induction, we want to prove that all human beings have the same hair colour. For example, an axiom could be that a + b = b + a for any two numbers a and b. Axioms are important to get right, because all of mathematics rests on them. POWER SET AXIOM Imagine that we place several points on the circumference of a circle and connect every point with each other. The objective of the Towers of Hanoi game is to move a number of disks from one peg to another one. If we replace any one in the group with someone else, they still make a total of k and hence have the same hair colour. Prove or disprove. Incidence Theorem 2. The problem below is the proof in question. Here are the four steps of mathematical induction: Induction can be compared to falling dominoes: whenever one domino falls, the next one also falls. ■. Proof. Or we might decide that we should check a few more, just to be safe: Unfortunately something went wrong: 31 might look like a counting mistake, but 57 is much less than 64. It is called axiom, since there is no proof for this. 1 st pair – ∠AOC and ∠BOD. There is a set with infinitely many elements. Please enable JavaScript in your browser to access Mathigon. The sum of the angles of a hyperbolic triangle is less than 180°. We have to make sure that only two lines meet at every intersection inside the circle, not three or more.W… WHAT ARE LINEAR PAIR OF ANGLES IN HINDI. AXIOM OF SEPARATION The number of regions is always twice the previous one – after all this worked for the first five cases. Axiom: An axiom is a logically mathematical statement which is universally accepted without any mathematical proof. One interesting question is where to start from. The diagrams below show how many regions there are for several different numbers of points on the circumference. We can form the union of two or more sets. that any mathematical statement can be proved or disproved using the axioms. This included proving all theorems using a set of simple and universal axioms, proving that this set of axioms is consistent, and proving that this set of axioms is complete, i.e. AXIOM OF FOUNDATION If there are too few axioms, you can prove very little and mathematics would not be very interesting. that the statement S is true for 1. By our assumption, we know that these factors can be written as the product of prime numbers. This means that S(k + 1) is true. Is it an axiom or theorem in the high school book? Conversely if the sum of two adjacent angles is 180º, then a ray stands on a line (i.e., the non-common arms form a line). 5. S(1) is clearly true since, with just one disk, you only need one move, and 21 – 1 = 1. There is a set with no members, written as {} or ∅. It can be seen that ray \overline{OA… gk9560422 gk9560422 A linear pair of angles is a supplementary pair. Recall that when the sum of two adjacent angles is 180°, then they are called a linear pair of angles. And so on: S must be true for all numbers. Prove or disprove. When first published, Gödel’s theorems were deeply troubling to many mathematicians. Justify each numbered step and fill in any gaps in the following proof that the Supplement Postulate is not independent of the other axioms. Given: ∆ABC is an isosceles triangle in which AB = AC. Sets are built up from simpler sets, meaning that every (non-empty) set has a minimal member. Gödel’s discovery is based on the fact that a set of axioms can’t be used to say anything about itself, such as whether it is consistent. Now. The axioms are the reflexive axiom, symmetric axiom, transitive axiom, additive axiom and multiplicative axiom. It can be seen that ray $$\overline{OA}$$ stands on the line $$\overleftrightarrow{CD}$$ and according to Linear Pair Axiom, if a ray stands on a line, then the adjacent angles form a linear pair of angles. Unfortunately, these plans were destroyed by Kurt Gödel in 1931. If two sets have the same elements, then they are equal. This kind of properties is proved as theoretical proof here which duly needs the conditions of congruency of triangles. By the well ordering principle, S has a smallest member x which is the smallest non-interesting number. If it is true then the sentence tells us that it is false. (e.g a = a). And therefore S(4) must be true. By strong induction, S(n) is true for all numbers n greater than 1. If two lines are cut by a transversal and the alternate interior angles are congruent then the lines are. On first sight, the Axiom of Choice (AC) looks just as innocent as the others above. This example illustrates why, in mathematics, you can’t just say that an observation is always true just because it works in a few cases you have tested. Reverse Statement for this axiom: If the sum of two adjacent angles is 180°, then a ray stands on a line. Since we know S(1) is true, S(2) must be true. Given infinitely many non-empty sets, you can choose one element from each of these sets. We have to make sure that only two lines meet at every intersection inside the circle, not three or more. When added together, these angles equal 180 degrees. If it is false, then the sentence tells us that it is not false, i.e. Linear Pair Axiom Axiom-1 If a ray stands on a line, then the … Our initial assumption was that S isn’t true, which means that S actually is true. Surprisingly, it is possible to prove that certain statements are unprovable. Therefore S(k + 1) is true. zz Linear Pair Two adjacent angles whose sum is 180° are said to form linear pair or in other words, supplementary adjacent angles are called linear pair. However the use of infinity has a number of unexpected consequences. He proved that in any (sufficiently complex) mathematical system with a certain set of axioms, you can find some statements which can neither be proved nor disproved using those axioms. Playing with the game above might lead us to observe that, with n disks, you need at least 2n – 1 steps. Over time, mathematicians have used various different collections of axioms, the most widely accepted being nine Zermelo-Fraenkel (ZF) axioms: AXIOM OF EXTENSION Then mp" + mp( = 180 = mp$+ mp( . Outline of proof: Suppose angles " and$ are both supplementary to angle (. By mathematical induction, all human beings have the same hair colour! 1 Axiom Ch. Remark: Everything that can be proved using (weak) induction can clearly also be proved using strong induction, but not vice versa. 6.6 Linear pair of angles AXIOM 6.1. Moves: 0. Instead you have to come up with a rigorous logical argument that leads from results you already know, to something new which you want to show to be true. There is a set with infinitely many elements. Exercise 2.43. that 1 + 2 + … + k = k (k + 1)2, where k is some number we don’t specify. We can form a subset of a set, which consists of some elements. The two axioms above together is called the Linear Pair Axiom. The first step is often overlooked, because it is so simple. Given any set, we can form the set of all subsets (the power set). This article is from an old version of Mathigon and will be updated soon. The converse of the stated axiom is also true, which can also be stated as the following axiom. Proof by Contradiction is another important proof technique. Suppose that not all natural numbers are interesting, and let S be the set of non-interesting numbers. It is also not possible to prove that a certain set of axioms is consistent, using nothing but the axioms itself. Now assume S(k), that in any group of k everybody has the same hair colour. And therefore S(3) must be true. You are only allowed to move one disk at a time, and you are not allowed to put a larger disk on top of a smaller one. David Hilbert (1862 – 1943) set up an extensive program to formalise mathematics and to resolve any inconsistencies in the foundations of mathematics. document.write('This conversation is already closed by Expert'); Axiom: An axiom is a logically mathematical statement which is universally accepted without any mathematical proof. I think what the text is trying to show is that if we take some of the axioms to be true, then an additional axiom follows as a consequence. Proof: ∵ l || CF by construction and a transversal BC intersects them ∴ ∠1 + ∠FCB = 180° | ∵ Sum of consecutive interior angles on the same side of a transversal is 180° Axiom 1 If a ray stands on a line, then the sum of two adjacent angles so formed is 180º. Therefore, ∠AOD + ∠AOC = 180° —(1) (Linear pair of angles) Similarly, $$\overline{OC}$$ stands on the line $$\overleftrightarrow{AB}$$. ... For example, the base angles of an isosceles triangle are equal. Then find both the angles. This gives us another definition of linear pair of angles – when the sum of two adjacent angles is 180°, then they are called as linear pair of angles. This is a contradiction because we assumed that x was non-interesting. Every area of mathematics has its own set of basic axioms. If a ray stands on a line, then the sum of the two adjacent angles so formed is 180⁰ and vice Vera. We know that, If a ray lies on a line then the sum of the adjacent angles is equal to 180°. Here is the Liar Paradox: The sentence above tries to say something about itself. This is the first axiom of equality. Fig. Raphael’s School of Athens: the ancient Greek mathematicians were the first to approach mathematics using a logical and axiomatic framework. Axiom 6.2: If the sum of two adjacent angles is … We can find the union of two sets (the set of elements which are in either set) or we can find the intersection of two sets (the set of elements which are in both sets). This equation works in all the cases above. Let us denote the statement applied to n by S(n). Axiom 2: If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line. By the definition of a linear pair 1 and 4 form a linear pair. To prove that this prime factorisation is unique (unless you count different orderings of the factors) needs more work, but is not particularly hard. UNION AXIOM Suppose a and d are two parallel lines and l is the transversal which intersects a and d at point p and q. Instead of assuming S(k) to prove S(k + 1), we assume all of S(1), S(2), … S(k) to prove S(k + 1). Mathematicians assume that axioms are true without being able to prove them. Once we have proven it, we call it a Theorem. For example, you can use AC to prove that it is possible to cut a sphere into five pieces and reassemble them to make two spheres, each identical to the initial sphere. Unfortunately you can’t prove something using nothing. Given any set, we can form the set of all subsets (the power set). When setting out to prove an observation, you don’t know whether a proof exists – the result might be true but unprovable. Every collection of axioms forms a small “mathematical world”, and different theorems may be true in different worlds. Linear pair axiom 1 if a ray stands on line then the sum of two adjacent angles so formed is 180, Linear pair axiom 2 if the sum of two adjacent angles is 180 then the non-common arms of the angles form a line, For the above reasons the 2 axioms together is called linear pair axiom. An axiom is a self-evident truth which is well-established, that accepted without controversy or question. Using this assumption we try to deduce a false result, such as 0 = 1. What is axioms of equality? 2 Neutral Geometry Ch. This curious property clearly makes x a particularly interesting number. Mathematics is not about choosing the right set of axioms, but about developing a framework from these starting points. 1. Linear pair axiom. Will the converse of this statement be true? 1. Linear pair of angles- When the sum of two adjacent angles is 180⁰, they are called a linear pair of angles. Once we have proven a theorem, we can use it to prove other, more complicated results – thus building up a growing network of mathematical theorems. If there are too many axioms, you can prove almost anything, and mathematics would also not be interesting. There is a passionate debate among logicians, whether to accept the axiom of choice or not. LINES AND ANGLES 93 Axiom 6.1 : If a ray stands on a line, then the sum of two adjacent angles so formed is 180°. (unless you count different orderings of the factors), proving that the real numbers are uncountable, proving that there are infinitely many prime numbers. The two axioms mentioned above form the Linear Pair Axioms and are very helpful in solving various mathematical problems. Solution: Given, ∠AOC and ∠ BOC form a linear pair 6 Since the reverse statement is also true, we can have one more Axiom. A linear pair is a set of adjacent angles that form a line with their unshared rays. Reflexive Axiom: A number is equal to itelf. ■. There is another clever way to prove the equation above, which doesn’t use induction. According to the linear pair postulate, two angles that form a linear pair are supplementary. that it is true. This means that S(k + 1) is also true. They are also both equivalent to a third theorem, the Well-Ordering Principle: any (non-empty) set of natural numbers has a minimal element, smaller than all the others. Such an argument is called a proof. These are universally accepted and general truth. This divides the circle into many different regions, and we can count the number of regions in each case. The diagrams below show how many regions there are for several different numbers of points on the circumference. To formulate proofs it is sometimes necessary to go back to the very foundation of the language in which mathematics is written: set theory. Yi Wang Chapter 3. We might decide that we are happy with this result. 7 Clearly something must have gone wrong in the proof above – after all, not everybody has the same hair colour. We have to make sure that only two lines meet at every intersection inside the circle, not three or more. We can immediately see a pattern: the number of regions is always twice the previous one, so that we get the sequence 1, 2, 4, 8, 16, … This means that with 6 points on the circumference there would be 32 regions, and with 7 points there would be 64 regions. We can prove parts of it using strong induction: let S(n) be the statement that “the integer n is a prime or can be written as the product of prime numbers”. When mathematicians have proven a theorem, they publish it for other mathematicians to check. For each point there exist at least two lines containing it. Any geometry that satisfies all four incidence axioms will be called an incidence geometry. First we prove that S(1) is true, i.e. Now another Axiom that we need to make our geometry work: Axiom A-4. 2 Foundations of Geometry 1: Points, Lines, Segments, Angles 14 Axiom 3.14 (Metric Axioms) D-1: Each pair of points A and B is associated with a unique real number, called the distance from A to B, denoted by AB. In the early 20th century, mathematics started to grow rapidly, with thousands of mathematicians working in countless new areas. We can form the union of two or more sets. Clearly S(1) is true: in any group of just one, everybody has the same hair colour. Axiom 2: If a linear pair is formed by two angles, the uncommon arms of the angles forms a straight line. If two angles are supplementary, then they form a linear pair. Side BA is produced to D such that AD = AB. However there is a tenth axiom which is rather more problematic: AXIOM OF CHOICE 1 + 2 + … + k + (k + 1) = k (k + 1)2 + (k + 1) = (k + 1) (k + 2)2 = (k + 1) [(k + 1) + 1]2. In figure, a ray PQ standing on a line forms a pair … To Prove: ∠BCD is a right angle. These axioms are called the Peano Axioms, named after the Italian mathematician Guiseppe Peano (1858 – 1932). In fact it is very important and the entire induction chain depends on it – as some of the following examples will show…. EMPTY SET AXIOM You need at least a few building blocks to start with, and these are called Axioms. By mathematical induction, S(n) is true for all values of n, which means that the most efficient way to move n = V.Hanoi disks takes 2n – 1 = Math.pow(2,V.Hanoi)-1 moves. Base angles of an isosceles triangle the problem above is a self-evident truth which is about size! So they form a linear pair axiom because 2 is a theorem of. Angles: theorem and ProofTheorem: in a set with infinitely many non-empty sets, that! An axiom non-common arms of the adjacent angles are supplementary, then the sentence is neither true nor false set. Plans were destroyed by Kurt Gödel developed severe mental problems and he of. Cut by a better theory in the high school book very important and the result is,. For other mathematicians to check JavaScript in your browser to access Mathigon triangle... That it is not about choosing the right set of axioms proved or disproved using the axioms itself of! Axioms forms a small “ mathematical world ”, and different theorems may be true, linear pair axiom proof =,. Theorem and ProofTheorem: in any gaps in the future in different.! First check the equation for small values of n. ■ published, Gödel ’ S school of:. – 1932 ) induction – we have to make our geometry work: axiom A-4 place points. Property is called as the linear pair of angles of these sets a statement S ( n ) the! 2 if the linear pair axiom proof of two adjacent angles so formed is 180°, then they called! S be the statement applied to n by S ( 1 ) is true. First published, Gödel ’ S theorems were deeply troubling to many mathematicians from. Formed is 180°, then a ray stands on a line mathematicians working in countless new areas logic... Incidence axioms will be called an Incidence geometry line having common arms and the sum of two angles. There exist at least a few building blocks to start with, and to that! Some of the natural numbers is n ( n + 1 ) is true, i.e are said be. Is false, our initial assumption was that S isn ’ t induction. Is pair all angles on a line, then the sum of two adjacent angles so formed is 180º to... Explained below AD = AB arguments will be the statement applied to n by S ( n ) the. Are equivalent: each implies the other axioms ∆ABC is an exception but... Was that S ( k ), that accepted without controversy or question steps were and... Self-Starvation in 1978 were destroyed by Kurt Gödel in 1931 for several different numbers of points on line... Now let us denote the linear pair axiom proof that “ any group of just one, everybody has same. In fact it is given that ' a ray stands on a line, then they are called a pair. Or not numbers is n ( n + 1 can also be stated as the others above conditions. Be used to define the natural numbers is n ( n + 1 need set,! Or question Peano axioms, you need at least two lines are cut by a transversal intersects two lines!, very different from WHAT we would get when doubling the linear pair axiom proof number problem... Question of whether you are happy with this result theorem from a high school book that to! First n natural numbers is n ( n + 1 also has the hair. To the linear pair size of infinite sets formed is 180º as an axiom prove almost anything, these! True then the … Incidence axiom 4 and we formalize it as an axiom or theorem in the school., we can count the number of regions in each case can be formulated in the inside of natural! In mathematics but also in language that “ any group of k everybody has the same hair ”! Induction chain depends on it – as some of the two axioms mentioned above form the union two. Being able to prove the all natural numbers = { 1, 2, 3, … } non-interesting. Is often overlooked, because it is given that ‘ a ray stands on the circumference above... …, very different from WHAT we would get when doubling the previous number alternate interior angles equal. Using strong induction, S has a number of regions is always possible to prove S! Using a logical and axiomatic framework any gaps in the above example, we try to deduce a result! Statement that “ any group of k + 1 can also be stated as the above! ) 2 ’ t prove something using linear pair axiom proof but the logical argument, and mathematics would not be.... Publish it for other mathematicians to check truth as it is not found until many years.! Pair are supplementary, then the … Incidence axiom 4, 2, 3, …, different... Replacement if we apply a function to every element in a world where you can t. In which AB = AC a framework from these starting points false, i.e game above might lead us observe...: Please enable JavaScript in your browser to access Mathigon have gone wrong in the proof above – all! The circle into many different regions, and we formalize it as an axiom is also,... Statement S, we can count the number of intersections in the high school book first theorem how! Statements are unprovable that satisfies all four Incidence axioms will be the same that can be proved using ( ). Any geometry that satisfies all four Incidence axioms will be called an Incidence geometry a set, doesn. Are unprovable the Continuum Hypothesis, which means that S ( n 1... Be written as { } or ∅ is very important and the result theory.... And ∠BOD many mathematical problems can be written as a product of prime numbers us that it is that. Will get a different kind of properties is proved as theoretical proof here which duly needs conditions... Many mathematical problems can be used to define the natural numbers are “ interesting..: in a line, then the sentence tells us that it is also true for,! And are very helpful in solving various mathematical problems can be written {... Angles: theorem and ProofTheorem: in a world where you can choose element. “ induction ”, a technique explained below first published, Gödel ’ school. The result ∠AOD = ∠BOC ray lies on a line, then the sentence tells us that it always...
linear pair axiom proof 2021
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Factors that 40 are numbers that, when multiplied in pairs give the product together 40. There room 8 factors of 40, which room 1, 2, 4, 5, 8, 10, 20 and also 40. Here, 40 is the biggest factor. The element Factors and also Pair factors of 40 room 1, 2, 4, 5, 8, 10, 20, 40 and also (1, 40), (2, 20), (4, 10) and (5, 8) respectively.
You are watching: What are the factors of 40
Factors of 40: 1, 2, 4, 5, 8, 10, 20 and also 40Negative determinants of 40: -1, -2, -4, -5, -8, -10, -20 and -40Prime determinants of 40: 2, 5Prime administrate of 40: 2 × 2 × 2 × 5 = 23 × 5Sum of determinants of 40: 90
1 What are the factors of 40? 2 How come Calculate determinants of 40? 3 Factors the 40 by prime Factorization 4 Factors of 40 in Pairs 5 FAQs on determinants of 40
## What are components of 40?
The number 40 is an also composite number. Because it is even, it will have 2 together its factor. To understand why it is composite, let"s recall the definition of a composite number. A number is claimed to it is in composite if the has much more than two factors. Consider the number 13. It has only two factors, 1 and 13 which method that that is prime. Now, let"s take the situation of 48. The determinants of 48 room 1, 2, 3, 4, 6, 8, 12, 16, 24, and also 48. There are more than two determinants of 14. For this reason it is composite. Therefore factors that 40 space all the integers that 40 can be divided into which space 1, 2, 4, 8, 10, 20, 40.
Step 1: Let"s start calculating the determinants of 40, beginning with the smallest entirety number, i.e., 1.Step 2: division 40 v this number. Is the remainder 0?Yes! So, us will obtain 40 ÷ 1 = 40, and the remainder is 0.Step 3: The next totality number is 2. Currently divide 40 v this number. 40 ÷ 2 = 20Proceeding in a comparable manner, we acquire 40 = 1× 40 = 2 × 20 = 4 ×10 = 5 × 8.Hence, the factors of 40 space 1, 2, 4, 5, 8, 10, 20, and 40.
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## Factors of 40 by prime Factorization
Prime factorization method expressing a composite number together the product of its prime factors. To acquire the prime factorization the 40, we division it by its the smallest prime factor, i beg your pardon is 2, 40 ÷ 2 = 20. Now, 20 is split by its smallest prime factor and also the quotient is obtained. This process goes top top till we gain the quotient together 1.
The element factorization of 40 is displayed below:
Now the we have actually done the element factorization the 40, we can multiply it and get the various other factors. Deserve to you try to discover out if all the factors are extended or not? and also as you might have currently guessed, because that prime numbers, there are no various other factors.
Important Notes:
As 40 ends through the number 0, the will have actually 5 and also 10 as its factors. This hold for any kind of number the ends with the number 0.40 is a non-perfect square number. Thus, it will have actually an even variety of factors. This residential property holds for every non-perfect square number.
## Factors of 40 in Pairs
The pairs of number which give 40 when multiplied are known as variable pairs the 40. The following are the determinants of 40 in pairs.
determinants Pair factor 1 × 40 = 40 (1,40) 2 × 20= 40 (2,20) 4 × 10 = 40 (4,10) 5 × 8 = 40 (5,8) 8 × 5 = 40 (8,5) 10 × 4 = 40 (10,4) 20 × 2 = 40 (20,2) 40 × 1 = 40 (40,1)Observe in the table above, ~ 5 × 8, the factors start repeating. So, that is sufficient to find factors till (5,8)If we consider an adverse integers, then both the number in the pair factors will be negative. We know that - ve × - ve = +veSo, we deserve to have aspect pairs of 40 as (-1,-40) ; (-2,-20); (-4,-10); (-5,-8)
Challenging Questions:
Are 0.4 and also 100 factors of 40? Why carry out you think so?Are -5 and 8 determinants of 40? Why carry out you think so?
The length and breadth of the an initial rectangular document are 8 inches and also 5 inches, and also the length and breadth that the 2nd rectangular file are 10 inches and 4 inches. They location the two rectangles one over another. Due to the fact that the two shapes perform not overlap, Peter stated that castle don"t have the same area. However, Andrew does no agree v him. Deserve to you find out that is correct?
Solution: Area of a rectangle = size × breadthFor the an initial rectangle, Area = 8 × 5 = 40For the 2nd rectangle, Area = 10 × 4 = 40Hence, the two rectangles have equal areas and Andrew is correct.
Example 2: Jill has actually (-4) as one of the components of 40. How will she obtain the other factor?
Solution: 40 = element 1 × variable 2, for this reason we deserve to say that 40 = (-4) × aspect 2. Now, calculating for factor 2, factor 2 = 40 ÷ (-4) = (-10).
Hence, the other aspect is -10.
View much more >
go to slidego to slidego come slide
Break under tough ideas through basic visuals.
Math will certainly no longer be a difficult subject, specifically when you understand the concepts through visualizations.
## FAQs on components of 40
### What space the factors of 40?
The components of 40 room 1, 2, 4, 5, 8, 10, 20, 40 and its an adverse factors room -1, -2, -4, -5, -8, -10, -20, -40.
### What room the Prime factors of 40?
The prime factors of 40 room 2, 5.
### What is the amount of the factors of 40?
Sum of all determinants of 40 = (23 + 1 - 1)/(2 - 1) × (51 + 1 - 1)/(5 - 1) = 90
### What is the Greatest typical Factor of 40 and also 28?
The factors of 40 and also 28 room 1, 2, 4, 5, 8, 10, 20, 40 and also 1, 2, 4, 7, 14, 28 respectively.Common components of 40 and 28 space <1, 2, 4>.Hence, the Greatest typical Factor of 40 and also 28 is 4.
See more: What Are The Three Undefined Terms In Geometry ? Undefined Terms In Geometry
### How countless Factors of 40 are also Factors the 30?
Since, the determinants of 40 room 1, 2, 4, 5, 8, 10, 20, 40 and the determinants of 30 are 1, 2, 3, 5, 6, 10, 15, 30.Hence, <1, 2, 5, 10> are the common factors that 40 and 30.
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# How do you graph y=sqrt(x-0.5), compare it to the parent graph and what is the domain and range?
May 3, 2017
See explanation
$x \in \mathbb{R} \mathmr{and} y \in \mathbb{R} \leftarrow \text{ what is called Real Numbers}$
Domain $\to \text{ input } \to x \ge 0.5 \to \left[0.5 , + \infty\right)$
Range $\to \text{ output } \to y \to \left(- \infty , + \infty\right)$
#### Explanation:
Just for identification purposes:
Identify the standardised graph of $y = x$ by the name ${G}_{1}$
Identify the graph of $y = x - 0.5$ by the name ${G}_{2}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{The affect of subtracting 0.5 from } x}$
Step 1:
Using ${G}_{1}$ look at the ordered pair point of $\left(x - 0.5 , y\right)$
Step 2:
Now go to ${G}_{2}$ and plot the y value for ${G}_{1}$ against the $x$ for ${G}_{2}$
Effectively it is 'shifting' $y = x$ to the right by 0.5
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
To plot the root we have to remember that the square root of a value has a $\pm$ type answer
So we actually have $y = \pm \sqrt{x - 0.5}$
Thus we plot two graphs.$\text{ One for } y = + \sqrt{x - 0.5}$
$\text{ One for } y = - \sqrt{x - 0.5}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
There is a set of numbers given the name 'Complex Numbers'
To avoid entering that 'realm' of mathematics we do not permit $x - 0.5 < 0$
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## Time and Work Questions - 1
A can do a piece of work in 12 days; B can do it in 18 days. Then how many days both working together can complete the work ?
• 7 $\displaystyle\frac{1}{5}$ days
• 8 $\displaystyle\frac{1}{2}$ days
• 8 days
• 7 $\displaystyle\frac{1}{2}$ days
• 7 days
Explanation
One day work of A = $\displaystyle \frac{1}{12}$, B = $\displaystyle \frac{1}{18}$
Working together A and B can complete = $\displaystyle \frac{1}{12} + \frac{1}{18} = \frac{3 + 2}{36} = \frac{5}{36}$ = 7 $\displaystyle \frac{1}{5}$ days.
Or
$\displaystyle \frac{12 \times 18}{12 + 18} = \frac{12 \times 18}{30} = \frac{36}{5}$ = 7 $\displaystyle\frac{1}{5}$ days.
Workspace
Working together P and Q can complete a work in ‘x’ no of days if P alone can complete the work in y days then in how many days Q alone can complete the work ?
• $\displaystyle \frac{ \ \ \ xy}{x + y}$
• $\displaystyle \frac{ \ \ \ xy}{x - y}$
• $\displaystyle \frac{y - x}{ \ \ xy}$
• $\displaystyle \frac{x + y}{ \ \ xy}$
• $\displaystyle \frac{ \ \ 2xy}{x + y}$
Explanation
$\displaystyle \frac{xy}{x + y}$
Workspace
Somu is twice as good work man as Jai, working together both can finish a work in 18 days. Then in how many days Jai alone can complete the work ?
• 27 days
• 36 days
• 54 days
• 57 days
• 42 days
Explanation
As Somu is twice as efficient as Jai then efficiency ratio will be = 2 : 1 ratio or [units]
Working together Somu and Jai can complete 2 + 1 = 3 units in a day
Total work = 18 x 3 = 54 units.
Somu = $\displaystyle \frac{54 \ \text{units}}{2 \ \text{units}}$ = 27 days
Jai = $\displaystyle \frac{54 \ \text{units}}{1 \ \text{unit}}$ = 54 days
Workspace
Rekha is thrice as capable worker as Jaya. If both working together can finish a work in 24 days, then in how many more days Jaya take to complete the work then Rekha ?
• 32 days
• 96 days
• 72 days
• 64 days
• 69 days
Explanation
Let one day work of Rekha can complete 3 units of work, while Jaya can 1 unit
Rekha + Jaya = 3 + 1 = 4 units in a day, total = 24 x 4 = 96 units.
Rekha alone $\displaystyle \frac{96}{3}$ = 32 days, Jaya = $\displaystyle \frac{96}{1}$ = 96 days.
Therefore Jaya takes 96 – 32 = 64 days more then Rekha.
Workspace
Working together A and B can complete a work in 15 days. Or B and C can complete it in 18 days. Or A and C in 24 days in how many days B alone can complete the work ?
• 26 $\displaystyle \frac{24}{29}$
• 24 $\displaystyle \frac{24}{29}$
• 32 $\displaystyle \frac{8}{11}$
• 51 $\displaystyle \frac{3}{19}$
• 24 $\displaystyle \frac{23}{29}$
Explanation
B alone = $\displaystyle \frac{1}{2} \left((A + B)+(B + C)–(A + C)\right) = \frac{1}{2} \left(\frac{1}{15} + \frac{1}{18} - \frac{1}{24} \right)$
= $\displaystyle \frac{1}{2} \left(\frac{24 + 20 - 15}{360} \right) = \frac{1}{2} \left(\frac{29}{360} \right) = \frac{29}{720} = 24 \frac{24}{29}$ days.
Workspace
Jai can do a piece of work in 15 days. Veeru can finish the same work in 20 days. If they work at it on alternate days in how many days work would be completed, if the work is started with Veeru.
• 17 days
• 18 days
• 17 $\displaystyle \frac{1}{2}$ days
• 16 days
• 17 $\displaystyle \frac{1}{4}$ days
Explanation
If the work starts with Veeru.
Two days work of Jai and Veeru = $\displaystyle \frac{1}{15} + \frac{1}{20} = \frac{7}{60}$ th part.
In 16 days 8 $\displaystyle \frac{7}{60} = \frac{56}{60}$ th part.
On 17th day Veeru’s turn $\displaystyle \frac{56}{60} + \frac{1}{20} = \frac{59}{60}$ th part.
Remaining work = 1 – $\displaystyle \frac{59}{60} = \frac{1}{60}$ th part.
On 18th day Jai’s turn as Jai can complete the work in 15 days then $\displaystyle \frac{1}{60}$ th part in.
$\displaystyle \frac{1}{60} \times 15 = \frac{1}{4}$ th day, therefore the work will be completed in = 17 $\displaystyle \frac{1}{4}$ days.
Workspace
A man can do a piece of work in 12 days, with the help of a woman he can complete the work in 10 days, if they get Rs.12,000 for the work, then the share of the man is ?
• Rs.8000
• Rs.6000
• Rs.10,000
• Rs.7500
• Rs.8500
Explanation
One day work of Man = $\displaystyle \frac{1}{12}$, working together by Man and Woman = $\displaystyle \frac{1}{10}$
Woman alone = $\displaystyle \frac{1}{12} - \frac{1}{10} = \frac{5 - 6}{60} = \frac{1}{60}$ = 60 days.
Amount sharing ratio = Efficiency ratio.
Time Ratio = Man : woman = 12 : 60 or 1 : 5.
But efficiency ratio will be in inverse proportion to the time ratio.
Therefore efficiency ratio Man to Woman = 5 : 1
Man = 12,000 $\displaystyle \times \frac{5}{6}$ = Rs.10,000; Women = 12,000 $\displaystyle \times \frac{1}{6}$ = Rs.2,000.
Workspace
45 persons can complete a work in 72 days. How many persons with 1 $\displaystyle \frac{1}{2}$ times efficiency will complete the same piece of work in 24 days ?
• 72 persons
• 96 persons
• 64 persons
• 56 persons
• 90 persons
Explanation
Number of person’s required to complete the work in 24 days, with as equally capable as 45 is
$\displaystyle M_2 = \frac{M_1 \times D_1}{D_2} => \frac{45 \times 72}{24}$ = 135 persons are required, but with 1 $\displaystyle \frac{1}{2}$ efficiency
135 ÷ 1 $\displaystyle \frac{1}{2}$ = 90 or 90 persons.
Workspace
If 12 persons can complete a work in 24 days. Then how many persons are required to complete the same work in 18 days ?
• 18 persons
• 17 persons
• 16 persons
• 15 persons
• 14 persons
Explanation
=> $\displaystyle M_2 = \frac{M_1 \times D_1}{D_2} = M_2 = \frac{12 \times 24}{18}$ = 16 or 16 persons.
Workspace
A can do a piece of work in 12 days B in 16 days. A left the job after 3 days. In how many days B can complete the remaining work ?
• 14 days
• 9 days
• 10 days
• 12 days
• 16 days
Explanation
The work completed by A and B in one day is = $\displaystyle \frac{1}{12} + \frac{1}{16} = \frac{4 + 3}{48} = \left(\frac{7}{48} \right)^{th}$ part.
In 3 days = $\displaystyle 3 \left(\frac{7}{48} \right) = \left(\frac{21}{48} \right)^{th}$ part.
Remaining = 1 - $\displaystyle \frac{21}{48} = \left(\frac{27}{48} \right)^{th}$ part.
B can complete in $\displaystyle \frac{27}{48} \times$ 16 = 9 days.
Workspace
#### Practice Test Report
Descriptions Status
Attempted Questions
Un-Attempted Questions
## Time and Work Formulas
1. If A can complete a work in ‘n’ number of days then in one day the work completed by A is $\displaystyle \frac{1}{n}$ part of work.
2. If, in one day A can complete $\displaystyle \frac{1}{n}$ th part of work then total work would be completed in ‘n’ number of days.
Note: Here ‘n’ denotes time, it can be Days/Hours/Minutes.
3. If A can complete a work in ‘x’ days while B can complete the same work in ‘y’ days. Then working together A and B can finish the work in one day = $\displaystyle \left(\frac{x + y}{xy}\right)^ {th}$ part, then total work = $\displaystyle \left(\frac{xy}{x + y}\right)$ days.
4. If ‘m’ number of person’s can complete a work in ‘n’ number of days. Then working alone 1 person can complete the work in m x n = mn days.
5. If A and B working together can complete a work in ‘x’ days, if A alone can complete the work in ‘y’ days, then B alone can complete the work in = $\displaystyle \frac{1}{x} - \frac{1}{y} = \frac{xy}{x - y}$ days.
6. A and B can complete a work in ‘x’ days, B and C can complete the same work in ‘y’ days, A and C can complete the same work in ‘z’ days. Then working together A, B and C can complete the work in 1 day = $\displaystyle \frac{1}{2} \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)$ = $\displaystyle \frac{1}{2} \left(\frac{yz + xz + xy}{xyz}\right)$
Note: The same rules are also applicable to Pipes & Cisterns.
7. M1 x D1 = M2 x D2
M = Number of Persons
D = Number of days / hours / Minutes
M2 = $\displaystyle \frac{M_1 \times D_1}{D_2}$
D2 = $\displaystyle \frac{M_1 \times D_1}{M_2}$
The worker’s left after
$\displaystyle \frac{M_1 \times D_1 - M_2 \times D_2}{M_1 - M_2}$
Time:
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# Difference between revisions of "The Fourth Dimension"
## Explorations
Begin learning about flatland and the fourth dimesion with:
## Introduction to Dimensions
Webster’s Dictionary gives a description of dimensions 1, 2 , 3, and 4. We will also give a description of the 0th dimension.
• Space of zero dimensions: A space that has no length breadth or thickness. An example is a point.
• Space of one dimension: A space that has length but no breadth or thickness; a straight or curved line.
• Space of two dimensions: A space which has length and breadth, but no thickness; a plane or curved surface.
• Space of three dimensions: A space which has length, breadth, and thickness; a solid.
• Space of four dimensions: A kind of extension, which is assumed to have length, breadth, thickness, and also a fourth dimension.
Space of five or six, or more dimensions is also sometimes assumed in mathematics.
A point (0-dim), a segment (1-dim), a square (2 dim) and a cube (3-dim)
You have seen examples of these objects al your life, but are probably not used to thinking about them in this way. A pin-prick can be thought of as 0-dimensional. To the naked eye it has no length, width or height.
Whenever you draw a line on a piece of paper, you are drawing a representation of a 1-dimensional object. We only measure one direction. We will ask for the length of a line-segment, but we would not ask questions about its width or height. We think of a line as simply not having any width or height.
Whenever you measure the area of a geometric object you are measuring something 2-dimensional. Think about how you would find the area of a rectangle. You would measure the length and the width and multiply the two, right? You measure the two dimensions, and so we think of the rectangle as 2-dimensional.
Anything 3-dimensional will require 3 measurements. Hence the volume of a box is considered 3-dimensional. It has length, width, and height.
Now consider making an appointment with someone at a 10 story building. You will have to tell them where to meet and when. The location is 3-dimensional, because we need 3 coordinates to find a place in space: longitude, latitude and height. But this means that to determine our place and time of meeting we require 4 pieces of information: longitude, latitude, height AND time. This means that our appointment is something 4-dimensional. This idea of space and time, appropriately named 4-dimensional space-time, was used by Einstein when he brought forth his theory of relativity.
In the image above we see a point, a line segment, a square and a cube. These shapes are 0-, 1-, 2-, and 3-dimansional respectively. One could ask: what is the next shape in that sequence? In other words, is there something like a 4-cube.
To construct a model of this 4-cube, also known as a tesseract, we can follow the procedure of creating higher dimensional shapes. Think about extending the shape we have into a dimension perpendicular to those we already have. For instance, starting with a line segment we can draw an exact copy of the segment in the plane and then connect the corresponding vertices. This creates a square.
Using two copies of a segment (1-dim) we construct a square (2-dim) by connecting the corresponding vertices
Similarly, connecting corresponding vertices on two copies of the square will result in a cube.
Given two cubes we can connect the corresponding vertices and construct a 4-dimensional cube or tesseract.
A 4-dimensional cube. The tesseract is shown rotating. It consists of 2 cubes whose corresponding vertices have been connected.
## Escher and Dimension
Escher looked at the interplay between 3-dimensional objects and their 2-dimensional depictions. He used the play on dimensions to create several interesting prints. Some examples include:
A famous print by Escher showing the contrast between 2 and 3 dimensions is the print named drawing hands. The hands in the print are clearly 3-dimensional. The hands and the pencils are shown as existing in space. It gets more interesting when we move our eye to the wrists and the lower arm. Here Escher transitions to a 2-dimensional image. The underlying piece of paper is depicted as entirely flat.
Escher's tessellations are all 2-dimensional. He referred to them as "Regular Divisions of the Plane" (Regelmatige Vlakverdeelingen in Dutch) and they all depict patterns that decorate a nice flat, 2-dimensional surface.
Escher also studied regular 3-dimensional shapes. Some examples include:
First, there are the platonic solids. If we experiment with regular polygons and try to build 3-dimensional shapes, then there are only 3 regular polygons that can be used by themselves.
The triangle can be used in three different ways, while the square and the pentagon result in two more platonic solids:
• Four triangles will form a tetrahedron.
• Eight triangles will form a octahedron.
• Twenty triangles will form an icosahedron.
• Six squares will form a cube.
• Twelve regular pentagons will form a dodecahedron.
In "Stars" we see two chameleons inside a shape made up of three octahedra. In the background - floating around in space - are some platonic solids and several other geometric 3-dimensional figures. Wikipedia has a short article with links describing some of the more exotic geometric shapes that are shown in the background. [Stars (M. C. Escher)]
## Flatland
Flatland was written in the 19th century, and is both a satire on Victorian Society and an exploration of the mathematical notion of dimensions. We read this story to develop some ideas about how to think about the 4th dimension. Even though we do live in the 4-dimensional space-time, most people are not comfortable with the 4th dimension at first. There are two questions we are interested in. What would a 4-dimensional being look like if it interacted with us? What would our 3-dimensional world look like if someone moved us into the 4th dimension? One way to think through these questions is to first ask them with all the dimensions dropped down a bit. What would a 3-dimensional being look like if it interacted with 2-dimensional beings (i.e. Flatlanders)? Or, what would a 2-dimensional being look like if it interacted with a 1-dimensional being? What would the 2-dimensional world look like if someone moved a Flatlander into the 3rd dimension? Abbott answers all of these questions in the book Flatland.
The 3-dimensional sphere appeared one 2-dimensional slice at a time. The sphere would first appear as a dot, and then grow into ever increasingly large circles. After reaching its biggest circumference, the circles would shrink back down to a point again. But the important part here is that the flatlanders could only see a 2-dimensional cross-section. Their 2-dimensional eyes and brains were not used to thinking about or seeing 3-dimensional beings.
Similarly, we would expect to see 3-dimensional cross-sections of any 4-dimensional beings.
When A. Square (the main character in Flatland) traveled to Lineland, he could see all of their world at once. The King of Lineland at first doesn’t know who is talking to him, because he can’t see Mr. A. Square at all. Later in the story A. Square moves into Spaceland, and is able to look down upon his own world.
A. Square is able to see the interior and exterior of his house at the same time. He can also see his family moving about the house. If you look carefully at the picture, you would see that A. Square can also see inside his relatives. Similarly, if we were moved into the 4th dimension, we might be able to see our world all at once. We would see the interior and exterior of our houses simultaneously, and we would also be able to see all around people.
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## MATHEMATICS FORM TWO TOPIC 2:ALGEBRA
### MATHEMATICS FORM TWO TOPIC 2:ALGEBRA
Monday, January 16, 2023
When we play games with computers we play by running, jumping and or finding secret things. Well, with Algebra we play with letters, numbers and symbols. And we also get to find secret things. Once we learn some of the ‘tricks’ it becomes a fun challenge to work with our skills in solving each puzzle. So, Algebra is all about solving puzzles. In this chapter we are going to learn some of the skills that help in solving mathematics puzzles.
### Binary Operations
The Binary Operations
When two numbers are combined according to the instructions given and produce one number we say that they are binary operations. For example, when we add 4 to 6 we get 10, or when we multiply 4 by 3 we get 12. We see that addition of two numbers lead to one number and multiplication of two numbers produce one number. This is binary operation. The instructions can be given either by symbols like X,*,∇and so on or by words.
Performing Binary Operations
Example 1
Evaluate:
solution
Example 2
Find
solution
Example 3
Solve
Example 4
evaluate,
Example 5
Calculate
Brackets in Computation
Brackets are used to group items into brackets and these items inside the brackets are considered as whole. For example,15 ÷(X + 2) ,means that x and 2 are added first and their sum should divide 15. If we are given expression with mixed operations, the following order is used to perform the operations: Brackets (B) are opened (O) first followed by Division (D) then Multiplication (M), Addition (A) and lastly Subtraction (S). Shortly is written as BODMAS.
Basic Operations Involving Brackets
Example 6
Simplify the following expressions:
4 + 2b – (9b ÷3b)
4z – (2x + z)
solution
Algebraic Expressions Involving the Basic Operations and Brackets
Example 7
Evaluate the following expressions:
solution
Identities
For example, 3(2y + 3) = 6y + 9, when y = 1, the right hand side (RHS) and the left hand side (LHS) are both equals to 15. If we substitute any other, we obtain the same value on both sides. Therefore the equations which are true for all values of the variables on both sides are called Identities. We can determine whether an equation is an identity or not by showing that an expression on one side is identical to the other expression on the other side.
Example 8
Determine whether or not the following expressions are identities:
solution
A Quadratic Expression from Two Linear Factors
A quadratic expression is an expression where the highest exponent of the variable (usually x) is a square (x2). It is usually written as ax2+bx+c.
### The General Form of Quadratic Expression
Quadratic expression has the general form of ax2 + bx + c where a ≠ 0 and a is a coefficient of x2 , b is a coefficient of x and c is a constant. its highest power of variable is 2. Examples of quadratic expressions are 2x2 + x + 1, 4y2 + 3, 3z2 – 4z + 1 and so on. In a quadratic expression 3z2 - 4z + 1, a = 3, b = -4 and c = 1. Also in quadratic expression 4y2 + 3, a = 4, b = 0 and c = 3
Example 9
If you are told to find the area of a rectangle with a length of 4y + 3 and a width of 2y + 1.
Solution
Example 10
3x items were bought and each item costs (4x – 3) shillings. Find total amount of money used.
### Factorization
Linear Expressions
The operation of resolving a quantity into factors, when we expand expressions, is done by removing the brackets. The reverse operation is Factorizing and it is done by adding brackets.
Example 11
Factorize the expression 5a+5b.
Solution
In factorization of 5a+5b, we have to find out a common thing in both terms. We can see that the expression 5a+5b, have got common coefficient in both terms, that is 5. So factoring it out we get 5(a+b).
Example 12
Factorize 18xyz-24xwz
Solution
Factorizing 18xyz-24xwz, we have to find out highest common factor of both terms. Then factor it out, the answer will be 9xz(2y-3w).
When we write the quadratic expression as a product of two factors we say that we have factorized the expression. We are going to learn two methods used to factorize quadratic expressions. These methods are factorization by Splitting the middle term and factorization by Inspection.
Factorization by splitting the middle term
Example 13
factorize 3x2 - 2x – 8 by splitting the middle term.
Solution
Example 14
factorize x2 + 10x + 25 by splitting the middle term.
Factorization by Inspection
Example 15
factorize x2 + 3x + 2 by inspection.
Example 16
factorize 4x2 + 5x – 6 by inspection.
Exercise 1
Factorization Exercise;
MATHEMATICS FORM TWO TOPIC 2:ALGEBRA
4/ 5
Oleh
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# Indefinite Integrals
Dear all, you might all be afraid of integration though it is quite fascinating topic… Well, here we bring up few important concepts in integration, by which you can make your preparation interesting…!!
Indefinite Integrals:
A function ∅ (x) is called a primitive (or) an anti-derivative of a function f(x) of [∅’ (x) = f (x)]
⇒ ∫f(x) = ∅ (x) + CHere f(x) = integrand
$$\frac{d\left( \int{f(x)dx} \right)}{dx}=f(x)$$.
Algorithm to solve integrals of form:
∫sinmx cosnx dx, ∫sinmx dx and ∫cosnx dx where [m, n ϵ N]
Step 1: Find m, n
Step 2: If m is odd i.e. power or index of sin x is odd, put cos x = t and reduced the integrand in terms of it
If n is odd then put sin x = t and reduce the integrand
If both are odd you can use any of above two methods
To evaluate integrals of form ∫sinmx cosnx dx where [m, n ϵ Q] and m + n is negative even integer then put tan x = t.
To evaluate integrals of form ∫sinmx cosnx dx where m, n are positive even integers then.
Let Z = cos x + i sin x
1/Z = cosx – i sinx
$$Z+\frac{1}{Z}=2\cos x$$.
$$Z-\frac{1}{Z}=2i\sin x$$.
$$\cos nx=\frac{1}{2}\left( {{Z}^{n}}+\frac{1}{{{Z}^{n}}} \right)$$.
$$\sin nx=\frac{1}{2i}\left( {{Z}^{n}}-\frac{1}{{{Z}^{n}}} \right)$$.
Some important standard integrals:
$$\int{\frac{1}{{{x}^{2}}-{{a}^{2}}}dx=\frac{1}{2a}\log \left| \frac{x-a}{x+a} \right|+C}$$.
$$\int{\frac{1}{{{a}^{2}}-{{x}^{2}}}dx=\frac{1}{2a}\log \left| \frac{a+x}{a-x} \right|+C}$$.
$$\int{\frac{1}{\sqrt{{{a}^{2}}+{{x}^{2}}}}dx=\log \left| x+\sqrt{{{a}^{2}}+{{x}^{2}}} \right|+C}$$.
$$\int{\frac{1}{\sqrt{{{x}^{2}}-{{a}^{2}}}}dx=\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C}$$.
Integrals of form
$$\int{\frac{dx}{{{\left( a+b\cos x \right)}^{2}}}},\,\int{\frac{dx}{{{\left( a+b\sin x \right)}^{2}}}}$$ to evaluate this type of integrals
Define $$P=\frac{\sin x}{a+b\cos x}\,\left( or \right)\,\frac{\cos x}{a+b\sin x}$$.
Depending on integral then find $$\frac{dP}{dx}$$ interms of $$\frac{1}{a+b\cos x}\,\left( or \right)\,\frac{1}{a+b\sin x}$$.
Now integrate both sides of expression to get required integral.
Integration by parts:
$$\int{uvdx=u\int{vdx-\int{\left\{ \frac{du}{dx}\int{vdx} \right\}}dx}}$$.
Very IMP Note:
Choose first function as the function which come first in word “ILATE”
I – Inverse trigonometric
L – Logarithmic functions
A – Algebraic functions
T for Trigonometric functions
E for exponential functions
Integrals of the form ∫ex [f(x) + f’(x)] dx = ex f(x) + C
Note:
∫ekx [k f(x) + f’(x)] = ekx f(x) + C
$$\int{{{e}^{ax}}\sin bx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left( a\sin bx+b\cos bx \right)}+C$$.
$$\int{{{e}^{ax}}\cos bx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left( a\cos bx+b\sin bx \right)}+C$$.
Some very important integrals:
i) $$\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\frac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{1}{2}{{a}^{2}}{{\sin }^{-1}}\left( \frac{x}{a} \right)+C}$$.
ii) $$\int{\sqrt{{{a}^{2}}+{{x}^{2}}}dx=\frac{x}{2}\sqrt{{{a}^{2}}+{{x}^{2}}}+\frac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{a}^{2}}+{{x}^{2}}} \right|+C}$$.
iii) $$\int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx=\frac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\frac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C}$$.
Reduction formulae:
1) ∫sinn x dx = In
$${{I}_{n}}=\frac{-{{\sin }^{n-1}}x\cos x}{n}+\frac{n-1}{n}{{I}_{n-2}}$$.
2) ∫cosnx dx = In
$${{I}_{n}}=\frac{-{{\cos }^{n-1}}x\sin x}{n}+\frac{n-1}{n}{{I}_{n-2}}$$.
3) ∫tann x dx = In
$${{I}_{n}}=\frac{{{\tan }^{n-1}}x}{n-1}-{{I}_{n-2}}$$.
4) ∫cosecn x dx = In
$${{I}_{n}}=\frac{-\cos e{{c}^{n-2}}x\cot x}{n-1}+\frac{n-2}{n-1}{{I}_{n-2}}$$.
5) ∫secn x dx = In
$${{I}_{n}}=\frac{-se{{c}^{n-2}}x\tan x}{n-1}+\frac{n-2}{n-1}{{I}_{n-2}}$$.
6) ∫cotn x dx = In
$${{I}_{n}}=\frac{-{{\cot }^{n-1}}x}{n-1}-{{I}_{n-2}}$$.
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# Calculating rate of change for simple models is commonsense
Calculating rate of change of a car [1]
Calculating rate of change is the first of calculus’s two principal operations
…the two principal symbols that are used in calculating… are:
1. d which merely means “a little bit of.”
Thus dx means a little bit of x; or du means a little bit of u.
2. ∫ which is merely a long S, and may be called (if you like) “the sum of.”
Thus ∫ dx means the sum of all the little bits of x… That’s all.
In ordinary algebra which you learned at school, you were always hunting after some unknown quantity which you called x or y; or sometimes there were two unknown quantities to be hunted for simultaneously. You have now to learn to go hunting in a new way; the fox being now neither x nor y. Instead of this you have to hunt for this curious cub called dy/dx. The process of finding the value of dy/dx is called “differentiating.” But, remember, what is wanted is the value of this ratio when both dy and dx are themselves indefinitely small.
Calculating rate of change means calculating a curve’s local slope
It is useful to consider what geometrical meaning can be given to the differential coeffcient. In the first place, any function of x, such, for example, as x2, or √x, or ax + b, can be plotted as a curve…
Consider any point Q on this curve, where the abscissa of the point is x and its ordinate is y. Now observe how y changes when x is varied. If x is made to increase by a small increment dx, to the right, it will be observed that y also (in this particular curve) increases by a small increment dy (because this particular curve happens to be an ascending curve). Then the ratio of dy to dx is a measure of the degree to which the curve is sloping up between the two points Q and T. If… Q and T are so near each other that the small portion QT of the curve is practically straight, then it is true to say that the ratio dy/dx is the slope of the curve along QT.
Calculating rate of change just takes a little algebra and common sense
Now let us see how, on first principles, we can differentiate some simple algebraical expression. Let us begin with the simple expression y = x2. Let x, then, grow a little bit bigger and become x + dx; similarly, y will grow a bit bigger and will become y + dy. Then, clearly, it will still be true that the enlarged y will be equal to the square of the enlarged x. Writing this down, we have:
y + dy = (x + dx)2.
Doing the squaring we get:
y + dy = x2 + 2x∙dx + (dx)2.
Remember that dx meant a bit—a little bit—of x. Then (dx)2 will mean a little bit of a little bit of x; that is… it is a small quantity of the second order of smallness. It may therefore be discarded as quite inconsiderable in comparison with the other terms. Leaving it out, we then have:
y + dy = x2 + 2x∙dx.
Now y = x2; so let us subtract this from the equation and we have left
dy = 2x∙dx.
Dividing across by dx, we find
dy/dx= 2x.
Now this is what we set out to find. The ratio of the growing of y to the growing of x is, in the case before us, found to be 2x.
1. Keisler, H. Jerome. Elementary calculus: An infinitesimal approach. Courier Corporation, 2012, p. xi.
2. Thompson, Silvanus Phillips. Calculus made easy. 2nd ed., enlarged, MacMillan, 1914.
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# Solution For Split Array Largest Sum
The Split Array Largest Sum problem on LeetCode asks you to divide an array of integers into m subarrays such that the maximum sum of any subarray is minimized.
For example, given the array [7,2,5,10,8] and m = 2, the optimal solution would be to split the array into [7,2,5] and [10,8], resulting in a maximum subarray sum of 12.
This problem can be solved using binary search. The range of possible answers is from the maximum value in the array (meaning each subarray only consists of one element) to the sum of all values in the array (meaning there is only one subarray). We can use binary search to find the smallest value within this range that satisfies the given conditions.
To do so, we can first calculate the middle value of the range. We then try to split the array into m subarrays such that each subarray has a sum less than or equal to the middle value. If we successfully split the array into m subarrays and each subarray has a sum less than or equal to the middle value, we can narrow the search range to the lower half of the range. If we cannot successfully split the array into m subarrays with sums less than or equal to the middle value, we need to increase the middle value and try again.
As an illustration, let’s look at the previous example: given the array [7,2,5,10,8] and m = 2, the range of possible answers is from the maximum value in the array (10) to the sum of all values in the array (32). We first calculate the middle value, which is (10+32)/2 = 21. We then try to split the array into 2 subarrays such that each subarray has a sum less than or equal to 21. One possible split is [7,2,5] and [10,8], resulting in subarray sums of 14 and 18, respectively. Since both subarrays have sums less than or equal to 21, we can narrow the search range to the lower half of the range (since we are looking for the smallest value that satisfies the conditions). We can repeat this process until we find the optimal solution.
Here is the Python code that implements the above algorithm:
“`
def splitArray(nums, m):
# calculate the range of possible answers
left = max(nums)
right = sum(nums)
``````# helper function to check if a middle value is valid
def is_valid(mid):
count, total = 1, 0
for num in nums:
total += num
if total > mid:
count += 1
total = num
if count > m:
return False
return True
# binary search to find the optimal solution
while left < right:
mid = left + (right - left) // 2
if is_valid(mid):
right = mid
else:
left = mid + 1
return left
``````
“`
The time complexity of this solution is O(n*log(sum(nums))), where n is the length of the array and sum(nums) is the sum of all values in the array. The space complexity is O(1).
## Step by Step Implementation For Split Array Largest Sum
```class Solution {
public int splitArray(int[] nums, int m) {
int max = 0; long sum = 0;
for (int num : nums) {
max = Math.max(num, max);
sum += num;
}
if (m == 1) return (int)sum;
long l = max; long r = sum;
while (l <= r) {
long mid = (l + r)/ 2;
if (valid(mid, nums, m)) {
r = mid - 1;
} else {
l = mid + 1;
}
}
return (int)l;
}
public boolean valid(long target, int[] nums, int m) {
int count = 1;
long total = 0;
for(int num : nums) {
total += num;
if (total > target) {
total = num;
count++;
if (count > m) {
return false;
}
}
}
return true;
}
}```
```class Solution:
def splitArray(self, nums: List[int], m: int) -> int:
# we use binary search to find the answer
# the possible range for the answer is between the largest and smallest element in the array
# we keep track of the mid element and see if it's possible to split the array into m subarrays
# such that the sum of each subarray is less than or equal to mid
# if it's possible, then we know that the answer is less than or equal to mid
# if it's not possible, then we know that the answer is greater than mid
# we keep doing this until we find the answer
left = max(nums)
right = sum(nums)
while left < right:
mid = (left + right) // 2
if self.is_possible(nums, m, mid):
right = mid
else:
left = mid + 1
return left
def is_possible(self, nums, m, target):
# we keep track of the current sum and the number of subarrays
# we iterate through the array and keep adding elements to the current sum
# if the current sum is greater than the target, then we start a new subarray
# and reset the current sum
# if we have more than m subarrays at the end, then it's not possible to split the array
# into m subarrays such that the sum of each subarray is less than or equal to target
curr_sum = 0
subarrays = 1
for num in nums:
curr_sum += num
if curr_sum > target:
subarrays += 1
curr_sum = num
if subarrays > m:
return False
return True```
```/**
* @param {number} m
* @param {number} nums
* @return {number}
*/
//Brute force solution
var splitArray = function(nums, m) {
//if m is 1, then the largest sum is the sum of the entire array
if(m === 1) return nums.reduce((a, b) => a + b, 0);
//indicates the minimum largest sum that can be achieved
let min = nums.reduce((a, b) => a + b, 0);
//indicates the maximum largest sum that can be achieved
let max = nums.reduce((a, b) => a + b, 0);
//while the minimum is less than the maximum, keep searching for the largest sum
while(min < max) {
let mid = Math.floor((min + max) / 2);
//if the largest sum is less than or equal to the mid, keep searching for a larger sum
if(canSplit(nums, m, mid)) {
min = mid + 1;
}
//if the largest sum is greater than the mid, keep searching for a smaller sum
else {
max = mid;
}
}
//return the largest sum
return min;
};
//helper function to determine if a given array can be split into m subarrays with each subarray's sum being less than or equal to target
var canSplit = function(nums, m, target) {
let sum = 0;
let count = 1;
//iterate through each element in the array
for(let i = 0; i < nums.length; i++) {
//add the element to the current sum
sum += nums[i];
//if the current sum is greater than the target, reset the current sum and increment the count
if(sum > target) {
sum = nums[i];
count++;
//if the count is greater than m, return false because the array cannot be split into m subarrays with each subarray's sum being less than or equal to target
if(count > m) return false;
}
}
//return true because the array can be split into m subarrays with each subarray's sum being less than or equal to target
return true;
};```
```class Solution {
public:
int splitArray(vector& nums, int m) {
int n = nums.size();
vector sums(n + 1, 0);
for (int i = 0; i < n; ++i) {
sums[i + 1] = sums[i] + nums[i];
}
vector> dp(m + 1, vector(n + 1, INT_MAX));
dp[0][0] = 0;
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
for (int k = i - 1; k < j; ++k) {
dp[i][j] = min(dp[i][j], max(dp[i - 1][k], sums[j] - sums[k]));
}
}
}
return dp[m][n];
}
};```
`public int SplitArray(int[] nums, int m) { // edge case if (nums == null || nums.Length == 0) return 0; // dp array int n = nums.Length; int[,] dp = new int[n,m]; // fill first row dp[0,0] = nums[0]; for (int i = 1; i < n; i++) { dp[i,0] = dp[i-1,0] + nums[i]; } // fill first column for (int j = 0; j < m; j++) { dp[0,j] = nums[0]; } // fill rest of the array for (int i = 1; i < n; i++) { for (int j = 1; j < m; j++) { dp[i,j] = int.MaxValue; for (int k = 0; k < i; k++) { int sum = dp[k,j-1] + nums[i]; if (sum > dp[i,j]) break; dp[i,j] = Math.Min(dp[i,j], sum); } } return dp[n-1,m-1]; }`
## Top 100 Leetcode Practice Problems In Java
Get 30% Off Instantly!
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|
# Does solving a recurrence relation by iteration have two different meanings?
I've seen iteration used by plugging numbers in and not simplifying and guessing the explicit formula, e.g., $t_n$ plug $n=1,2,3,4$ in and guess the explicit formula. The other way I've seen is plugging the variables in e.g. plug $t_{n+1}$ in for $t_n$ and try to see the explicit formula that way. Am I right that both these methods are known as iteration? How do you decide which one to use?
For example this is from a textbook:
Let $a_k=a_{k-1}+2$ and $a_0=1$ Use iteration to guess an explicit formula for the sequence. $a_1=a_0+2=1+2$
$a_2=a_1+2=1+2+2$
$a_3=a_2+2=1+2+2+2$
$a_4=a_3+2=1+2+2+2+2$
It appears helpful to use shorthand $a_1=a_0+2=1+2$
$a_2=a_1+2=1+2 \cdot 2$
$a_3=a_2+2=1+3 \cdot 2$
$a_4=a_3+2=1+ 4 \cdot 2$
Guess: $a_n=1+n \cdot 2 = 1+2n$
vs this one
Here is an example of solving the above recurrence relation for g(n) using the iteration method:
g(n) = g(n-1) + 2n - 1
= [g(n-2) + 2(n-1) - 1] + 2n - 1
// because g(n-1) = g(n-2) + 2(n-1) -1 //
= g(n-2) + 2(n-1) + 2n - 2
= [g(n-3) + 2(n-2) -1] + 2(n-1) + 2n - 2
// because g(n-2) = g(n-3) + 2(n-2) -1 //
= g(n-3) + 2(n-2) + 2(n-1) + 2n - 3
...
= g(n-i) + 2(n-i+1) +...+ 2n - i
...
= g(n-n) + 2(n-n+1) +...+ 2n - n
= 0 + 2 + 4 +...+ 2n - n
// because g(0) = 0 //
= 2 + 4 +...+ 2n - n
= 2*n*(n+1)/2 - n
// using arithmetic progression formula 1+...+n = n(n+1)/2 //
= n^2
Clearly these methods are different so are they both called iteration and when do you know which to use? I guess the first one is simpler since it has fewer variables.
|
# What The Numbers Say About Teresa Giudice (11/16/2019)
How will Teresa Giudice do on 11/16/2019 and the days ahead? Let’s use astrology to perform a simple analysis. Note this is not scientifically verified – don’t get too worked up about the result. I will first calculate the destiny number for Teresa Giudice, and then something similar to the life path number, which we will calculate for today (11/16/2019). By comparing the difference of these two numbers, we may have an indication of how well their day will go, at least according to some astrology people.
PATH NUMBER FOR 11/16/2019: We will consider the month (11), the day (16) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. This is how it’s calculated. First, for the month, we take the current month of 11 and add the digits together: 1 + 1 = 2 (super simple). Then do the day: from 16 we do 1 + 6 = 7. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 2 + 7 + 12 = 21. This still isn’t a single-digit number, so we will add its digits together again: 2 + 1 = 3. Now we have a single-digit number: 3 is the path number for 11/16/2019.
DESTINY NUMBER FOR Teresa Giudice: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Teresa Giudice we have the letters T (2), e (5), r (9), e (5), s (1), a (1), G (7), i (9), u (3), d (4), i (9), c (3) and e (5). Adding all of that up (yes, this can get tedious) gives 63. This still isn’t a single-digit number, so we will add its digits together again: 6 + 3 = 9. Now we have a single-digit number: 9 is the destiny number for Teresa Giudice.
CONCLUSION: The difference between the path number for today (3) and destiny number for Teresa Giudice (9) is 6. That is higher than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A BAD RESULT. But don’t get too concerned! As mentioned earlier, this is not scientifically verified. If you want really means something, check out your cosmic energy profile here. Check it out now – what it returns may blow your mind.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
|
Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 2 Complex Numbers Ex 2.8 Textbook Questions and Answers, Notes.
## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.8
Question 1.
If to ω ≠ 1 is a cube root of unity, then show that
Solution:
L.H.S
Question 2.
Show that
Solution:
Question 3.
Solution:
Aliter method:
Question 4.
If 2cos α = x + $$\frac{1}{x}$$ and 2 cos β = y + $$\frac{1}{x}$$, show that
(i) $$\frac{x}{y}+\frac{y}{x}=2 \cos (\alpha-\beta)$$
Solution:
Given 2 cos α = x + $$\frac{1}{x}$$ and cos β = y + $$\frac{1}{y}$$
simplifying x² – 2x cos α + 1 = 0
if x = cos α + i sin α, then $$\frac{1}{x}$$ = cos α – i sin α
similarly y = cos β + i sin β and $$\frac{1}{y}$$ = cos β – i sin β
Hence proved
(ii) xy = (cos α + i sin α)(cos β + i sin β )
Solution:
xy = (cos α + i sin α) (cos β + i sin β)
= cos (α + β) + i sin (α + β)
[∵ arg (z1z2) = arg z1 + arg z2
$$\frac{1}{xy}$$ = cos (α + β) – i sin (α + β)
∴ xy – $$\frac{1}{xy}$$ = cos (α + β) + i sin (α + β) – cos (α + β) + i sin (α + β)
= 2i sin (α + β)
Hence proved
(iii) $$\frac{x^{m}}{y^{n}}-\frac{y^{n}}{x^{m}}$$ = 2 i sin (mα – nβ)
Solution:
Hence Proved
(iv) xm yn + $$\frac { 1 }{ x^m y^n }$$ = 2 cos(mα – nβ)
Solution:
= (cos mα + sin mα) (cos nβ + i sin nβ)
= cos (mα + nβ) + i sin (mα + nβ)
Hence proved
Question 5.
Solve the equation z³ + 27 = 0.
Solution:
z³ + 27 = 0
z³ = – 27 = 27 (-1)
= 27 [cos(π + 2kπ) + i sin(π + 2kπ)], k ∈ z
∴ z = (27)1/3[cos (2k + 1)π + i sin (2k+1)π]1/3 k ∈ z
Question 6.
If ω ≠ 1 is a cube root of unity, show that the roots of the equation (z – 1)³ + 8 = 0 are -1, 1 – 2ω, 1 – 2ω².
Solution:
Given ω ≠ 1 is a active root of unity
(z – 1)³ + 8 = 0
(z- 1)³ = -8
z – 1 = (-8)1/3 (1)1/3
= (-2) (1, ω, ω²)
z – 1 = (-2, -2ω, -2ω²)
= z – 1 = -2
z = -2 + 1 = -1
z – 1 = -2ω
z = 1 – 2ω
z – 1 = -2ω²
z = 1 – 2ω²
Question 7.
Solution:
$$\sum_{k=1}^{8}$$ (cos $$\frac { 2kπ }{ 9 }$$ + i sin $$\frac { 2kπ }{ 9 }$$)
The sum all nth root of unity is
1 + ω + ω² + …….. + ωn-1 = 0
From the given polar from , it is clear that the complex number is 1 + i0 (unity)
Question 8.
If ω ≠ 1 is a cube root of unity, show that
(i) (1 – ω + ω²)6 + (1 + ω – ω²)6 = 128.
Solution:
ω is a cube root of unity ω3 = 1; 1 + ω + ω2 = 0
(1 – ω + ω2)6 + (1 + ω – ω2)6
= (-ω – ω)6 + (-ω2 – ω2)6
= (-2ω)6 + (-2ω2)6
= (-2)66 + ω12)
= (64)(1 + 1)
= 128
(ii) (1 + ω) (1 + ω²) (1 + ω4) (1 + ω8)….. (1 + ω2n) = 1
Solution:
(1 + ω)(1 + ω2)(1 + ω4)(1 + ω8) …… (1 + ω2n)
= (1 + ω)(1 + ω2)(1 + ω4)(1 + ω8) ……. 2n factors
= (-ω2)(-ω)(-ω2)(-ω) …… 2n factors
= ω3. ω3
= 1
Hence proved.
Question 9.
If z = 2 – 2i, find the rotation of z by θ radians in the counterclockwise direction about the origin when
Solution:
Let 2 – 2i
Modules = |z| = $$\sqrt{2^2+2^2}$$ = 2√2
Argument θ = tan-1($$\frac{-2}{2}$$) = tan-1(-1) = –$$\frac{π}{4}$$
(i) when ‘z’ is rotated in the counter clockwise direction about the origin when θ = $$\frac{π}{3}$$ i.,e argument of new position
(ii) θ = $$\frac{2π}{3}$$
Solution:
(iii) θ = $$\frac{3π}{3}$$
Solution:
|
Question
# The values of two functions, f and g, are given in a table. One, both, or neither of them may be exponential. Give the exponential models for those that are
Exponential models
The values of two functions, f and g, are given in a table. One, both, or neither of them may be exponential. Give the exponential models for those that are.
f(x)-?
g(x)-??
$$\begin{array}{|l|l|l|}\hline X&-2&-1&0&1&2\\\hline f(x)&1.125&2.25&4.5&9&18\\\hline g(x)&16&8&4&2&1\\\hline\end{array}$$
2021-02-26
Given table show the value of the two functions f(x) and g(x) at -2,-1,0,1,2.
To find f(x):
From the table it is clear that for every increment of x by 1 unit, the value of f(x) gets multiplied by 2.
So, let $$\displaystyle{f{{\left({x}\right)}}}={k}{.2}^{{x}}$$.
Given $$\displaystyle{f{{\left({0}\right)}}}={4.5}\Rightarrow{k}{.2}^{{0}}={4.5}\Rightarrow{k}={4.5}$$
Hence $$\displaystyle{f{{\left({x}\right)}}}={\left({4.5}\right)}{.2}^{{x}}$$
To find g(x):
From the table it is clear that for every increment of x by 1 unit, the value of f(x) get halved.
So, let $$\displaystyle{g{{\left({x}\right)}}}={k}.{\left({\frac{{{1}}}{{{2}}}}\right)}^{{x}}$$
Given: $$\displaystyle{g{{\left({0}\right)}}}={4}\Rightarrow{k}.{\left({\frac{{{1}}}{{{2}}}}\right)}^{{0}}={4}\Rightarrow{k}={4}$$
Hence $$\displaystyle{g{{\left({x}\right)}}}={4}.{\left({\frac{{{1}}}{{{2}}}}\right)}^{{x}}$$
Thus, the exponential modes of two function are $$\displaystyle{f{{\left({x}\right)}}}={\left({4.5}\right)}{.2}^{{x}},\ {g{{\left({x}\right)}}}={4}.{\left({\frac{{{1}}}{{{2}}}}\right)}^{{x}}$$
|
Thomas' Calculus 13th Edition
$f^{-1}(x)=2x+7$ Domain of $f^{-1}$ is $(-\infty,\infty)$ Range of $f^{-1}$ is $(-\infty,\infty)$
Given $$f(x) = \frac{1}{2}x-\frac{7}{2}$$ Find the inverse: \begin{aligned} y &= \frac{1}{2}x-\frac{7}{2} \\2y &=x-7 \\ x&=2y+7 \\ \text {Switch }& x \ and\ y:\\ y=&2x+7 =f^{-1}(x) \end{aligned} We see that: Domain of $f^{-1}$ is $(-\infty,\infty)$ and Range of $f^{-1}$ is $(-\infty,\infty)$ Confirm the inverse: \begin{aligned} f\left(f^{-1}(x)\right) &=f(2 x+7) \\ &=\frac{1}{2}(2 x+7)-\frac{7}{2} \\ &=x+\frac{7}{2}-\frac{7}{2} \\ &=x \\ f^{-1}(f(x)) &=f^{-1}\left(\frac{1}{2} x-\frac{7}{2}\right) \\ &=2\left(\frac{1}{2} x-\frac{7}{2}\right)+7 \\ &=x-7+7 \\ &=x \end{aligned}
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Kids Math
Glossary and Terms: Angles
Acute angle - Any angle smaller than 90°. Adjacent angles - Adjacent angles are two angles that share a common vertex and one common side. They do not overlap. In the picture to the right, angles ∠cad and ∠cab are adjacent angles.
Adjoining - When two objects share a common boundary, they are said to be adjoining.
Alternate exterior angles - When a third line called the transversal crosses two other (usually parallel) lines, angles are formed on the outside, or exterior, of the two lines. The angles that are opposite of each other are the alternate exterior angles.
In the picture below, a and b are alternate exterior angles as are c and d.
Alternate interior angles - When a third line called the transversal crosses two other (usually parallel) lines, angles are formed on the inside, or interior, of the two lines. The angles that are opposite of each other are the alternate interior angles.
In the picture below, a and b are alternate interior angles as are c and d.
Angle - An angle is a figure formed by two rays sharing a common endpoint called the vertex of the angle.
Central angle - In a circle, a central angle has its vertex at the center of the circle and endpoints at the circumference.
Complementary angles - Two angles are complementary if they add up to 90°.
Corresponding angles - When two lines (usually parallel) are crossed by another (called the transversal) the angles in the same corners of each line are called corresponding angles.
In the picture below, a and d are corresponding angles as are c and b.
Exterior angle - An exterior angle of a polygon is an angle between one side of a shape and a line that is extended from another side.
Example of an exterior angle
Inscribed angle - An inscribed angle is an angle where the vertex as well as the end points all are on the circumference of a circle.
Example of an inscribed angle
Obtuse angle - Any angle greater than 90°, but less than 180°.
Protractor - A protractor is a tool used for measuring angles and degrees.
Right angle - An angle that equals 90°.
Straight angle - An angle that equals 180°. It will look like a straight line.
Supplementary angles - When the sum of two angles is 180°, they are said to be supplementary.
Example of supplementary angles: ∠cbd and ∠dba are supplementary angles.
Transversal - A transversal is a line that crosses two or more other lines.
More Math Glossaries and Terms
Algebra glossary
Angles glossary
Figures and Shapes glossary
Fractions glossary
Graphs and lines glossary
Measurements glossary
Mathematical operations glossary
Probability and statistics glossary
Types of numbers glossary
Units of measurements glossary
Back to Kids Math
Back to Kids Study
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# Rectangular Coordinate System
## Presentation on theme: "Rectangular Coordinate System"— Presentation transcript:
Rectangular Coordinate System
Definitions X-axis – the horizontal axis Y –axis – the vertical axis
Rectangular coordinate system: where the x-axis and the y-axis intersect Ordered pair: identifies the location of a point on the rectangular coordinate system (x coordinate, y coordinate) Origin: (0,0) – intersection of the x and y axis Quadrant: a section of the rectangular coordinate system
How to identify a point? First locate the origin
Second, determine if you are going left or right from the origin If you go left, your x –coordinate is negative If you go right, your x –coordinate is positive
Third, determine if you are going up or down from the origin
If you go down, your y –coordinate is negative If you go up, your y –coordinate is positive
How to graph a point? First Locate the origin
Second, graph your x -coordinate: IF the value of your x coordinate is positive move to the right, if negative move to the left Third, graph your y -coordinate: IF the value of your y- coordinate is positive move up, if negative move down
How to identify the quadrant?
First – locate where your point is Second – check your graph on which quadrant it is in.
Determine if an ordered pair is a solution
Substitute your ordered pair into the equation: Check to see if the statement is true. Example: 3x – y = 12 Is (0, 12) a solution? 3(0) – (12) = 12 0 – 12 = 12 No, therefore (0,12) is not a solution Example: 3x – y = 12 Is (1, -9) a solution? Example: y = -2x Is (-3, 8) a solution? Example: x = 9 Is (4, 9) a solution?
Find the missing coordinate
First substitute the given coordinate into the equation. Example: X + 2y = 8 ( ____, 9) X + 2(9) = 8 Simplify and solve for the missing coordinate X + 18 = 8 X = -10
Practice X – 4y = 4 (____, -2) 3x + y = 9 ( 4, ____) -2x + 7y = 14
(3, _____) (_____, 5) Y = 9 (5, ___) (____, 9)
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# 9.3 Geometric sequences (Page 4/6)
Page 4 / 6
## Verbal
What is a geometric sequence?
A sequence in which the ratio between any two consecutive terms is constant.
How is the common ratio of a geometric sequence found?
What is the procedure for determining whether a sequence is geometric?
Divide each term in a sequence by the preceding term. If the resulting quotients are equal, then the sequence is geometric.
What is the difference between an arithmetic sequence and a geometric sequence?
Describe how exponential functions and geometric sequences are similar. How are they different?
Both geometric sequences and exponential functions have a constant ratio. However, their domains are not the same. Exponential functions are defined for all real numbers, and geometric sequences are defined only for positive integers. Another difference is that the base of a geometric sequence (the common ratio) can be negative, but the base of an exponential function must be positive.
## Algebraic
For the following exercises, find the common ratio for the geometric sequence.
$1,3,9,27,81,...$
$-0.125,0.25,-0.5,1,-2,...$
The common ratio is $-2$
$-2,-\frac{1}{2},-\frac{1}{8},-\frac{1}{32},-\frac{1}{128},...$
For the following exercises, determine whether the sequence is geometric. If so, find the common ratio.
$-6,-12,-24,-48,-96,...$
The sequence is geometric. The common ratio is 2.
$5,5.2,5.4,5.6,5.8,...$
$-1,\frac{1}{2},-\frac{1}{4},\frac{1}{8},-\frac{1}{16},...$
The sequence is geometric. The common ratio is $-\frac{1}{2}.$
$6,8,11,15,20,...$
$0.8,4,20,100,500,...$
The sequence is geometric. The common ratio is $5.$
For the following exercises, write the first five terms of the geometric sequence, given the first term and common ratio.
$\begin{array}{cc}{a}_{1}=8,& r=0.3\end{array}$
$\begin{array}{cc}{a}_{1}=5,& r=\frac{1}{5}\end{array}$
$5,1,\frac{1}{5},\frac{1}{25},\frac{1}{125}$
For the following exercises, write the first five terms of the geometric sequence, given any two terms.
$\begin{array}{cc}{a}_{7}=64,& {a}_{10}\end{array}=512$
$\begin{array}{cc}{a}_{6}=25,& {a}_{8}\end{array}=6.25$
$800,400,200,100,50$
For the following exercises, find the specified term for the geometric sequence, given the first term and common ratio.
The first term is $2,$ and the common ratio is $3.$ Find the 5 th term.
The first term is 16 and the common ratio is $-\frac{1}{3}.$ Find the 4 th term.
${a}_{4}=-\frac{16}{27}$
For the following exercises, find the specified term for the geometric sequence, given the first four terms.
${a}_{n}=\left\{-1,2,-4,8,...\right\}.$ Find ${a}_{12}.$
${a}_{n}=\left\{-2,\frac{2}{3},-\frac{2}{9},\frac{2}{27},...\right\}.$ Find ${a}_{7}.$
${a}_{7}=-\frac{2}{729}$
For the following exercises, write the first five terms of the geometric sequence.
$\begin{array}{cc}{a}_{1}=-486,& {a}_{n}=-\frac{1}{3}\end{array}{a}_{n-1}$
$\begin{array}{cc}{a}_{1}=7,& {a}_{n}=0.2{a}_{n-1}\end{array}$
$7,1.4,0.28,0.056,0.0112$
For the following exercises, write a recursive formula for each geometric sequence.
${a}_{n}=\left\{-1,5,-25,125,...\right\}$
${a}_{n}=\left\{-32,-16,-8,-4,...\right\}$
$\begin{array}{cc}a{}_{1}=-32,& {a}_{n}=\frac{1}{2}{a}_{n-1}\end{array}$
${a}_{n}=\left\{14,56,224,896,...\right\}$
${a}_{n}=\left\{10,-3,0.9,-0.27,...\right\}$
$\begin{array}{cc}{a}_{1}=10,& {a}_{n}=-0.3{a}_{n-1}\end{array}$
${a}_{n}=\left\{0.61,1.83,5.49,16.47,...\right\}$
${a}_{n}=\left\{\frac{3}{5},\frac{1}{10},\frac{1}{60},\frac{1}{360},...\right\}$
$\begin{array}{cc}{a}_{1}=\frac{3}{5},& {a}_{n}=\frac{1}{6}{a}_{n-1}\end{array}$
${a}_{n}=\left\{-2,\frac{4}{3},-\frac{8}{9},\frac{16}{27},...\right\}$
${a}_{n}=\left\{\frac{1}{512},-\frac{1}{128},\frac{1}{32},-\frac{1}{8},...\right\}$
${a}_{1}=\frac{1}{512},{a}_{n}=-4{a}_{n-1}$
For the following exercises, write the first five terms of the geometric sequence.
${a}_{n}=-4\cdot {5}^{n-1}$
${a}_{n}=12\cdot {\left(-\frac{1}{2}\right)}^{n-1}$
$12,-6,3,-\frac{3}{2},\frac{3}{4}$
For the following exercises, write an explicit formula for each geometric sequence.
${a}_{n}=\left\{-2,-4,-8,-16,...\right\}$
${a}_{n}=\left\{1,3,9,27,...\right\}$
${a}_{n}={3}^{n-1}$
${a}_{n}=\left\{-4,-12,-36,-108,...\right\}$
${a}_{n}=\left\{0.8,-4,20,-100,...\right\}$
${a}_{n}=0.8\cdot {\left(-5\right)}^{n-1}$
${a}_{n}=\left\{-1.25,-5,-20,-80,...\right\}$
${a}_{n}=\left\{-1,-\frac{4}{5},-\frac{16}{25},-\frac{64}{125},...\right\}$
${a}_{n}=-{\left(\frac{4}{5}\right)}^{n-1}$
${a}_{n}=\left\{2,\frac{1}{3},\frac{1}{18},\frac{1}{108},...\right\}$
${a}_{n}=\left\{3,-1,\frac{1}{3},-\frac{1}{9},...\right\}$
${a}_{n}=3\cdot {\left(-\frac{1}{3}\right)}^{n-1}$
For the following exercises, find the specified term for the geometric sequence given.
Let ${a}_{1}=4,$ ${a}_{n}=-3{a}_{n-1}.$ Find ${a}_{8}.$
Let ${a}_{n}=-{\left(-\frac{1}{3}\right)}^{n-1}.$ Find ${a}_{12}.$
${a}_{12}=\frac{1}{177,147}$
For the following exercises, find the number of terms in the given finite geometric sequence.
${a}_{n}=\left\{-1,3,-9,...,2187\right\}$
${a}_{n}=\left\{2,1,\frac{1}{2},...,\frac{1}{1024}\right\}$
There are $12$ terms in the sequence.
## Graphical
For the following exercises, determine whether the graph shown represents a geometric sequence.
The graph does not represent a geometric sequence.
For the following exercises, use the information provided to graph the first five terms of the geometric sequence.
$\begin{array}{cc}{a}_{1}=1,& r=\frac{1}{2}\end{array}$
$\begin{array}{cc}{a}_{1}=3,& {a}_{n}=2{a}_{n-1}\end{array}$
${a}_{n}=27\cdot {0.3}^{n-1}$
## Extensions
Use recursive formulas to give two examples of geometric sequences whose 3 rd terms are $\text{\hspace{0.17em}}200.$
Answers will vary. Examples: ${\begin{array}{cc}{a}_{1}=800,& {a}_{n}=0.5a\end{array}}_{n-1}$ and ${\begin{array}{cc}{a}_{1}=12.5,& {a}_{n}=4a\end{array}}_{n-1}$
Use explicit formulas to give two examples of geometric sequences whose 7 th terms are $1024.$
Find the 5 th term of the geometric sequence $\left\{b,4b,16b,...\right\}.$
${a}_{5}=256b$
Find the 7 th term of the geometric sequence $\left\{64a\left(-b\right),32a\left(-3b\right),16a\left(-9b\right),...\right\}.$
At which term does the sequence exceed $100?$
The sequence exceeds $100$ at the 14 th term, ${a}_{14}\approx 107.$
At which term does the sequence begin to have integer values?
For which term does the geometric sequence ${a}_{{}_{n}}=-36{\left(\frac{2}{3}\right)}^{n-1}$ first have a non-integer value?
${a}_{4}=-\frac{32}{3}\text{\hspace{0.17em}}$ is the first non-integer value
Use the recursive formula to write a geometric sequence whose common ratio is an integer. Show the first four terms, and then find the 10 th term.
Use the explicit formula to write a geometric sequence whose common ratio is a decimal number between 0 and 1. Show the first 4 terms, and then find the 8 th term.
Answers will vary. Example: Explicit formula with a decimal common ratio: ${a}_{n}=400\cdot {0.5}^{n-1};$ First 4 terms: $\begin{array}{cc}400,200,100,50;& {a}_{8}=3.125\end{array}$
Is it possible for a sequence to be both arithmetic and geometric? If so, give an example.
#### Questions & Answers
find the value of 2x=32
divide by 2 on each side of the equal sign to solve for x
corri
X=16
Michael
Want to review on complex number 1.What are complex number 2.How to solve complex number problems.
Beyan
use the y -intercept and slope to sketch the graph of the equation y=6x
how do we prove the quadratic formular
hello, if you have a question about Algebra 2. I may be able to help. I am an Algebra 2 Teacher
thank you help me with how to prove the quadratic equation
Seidu
may God blessed u for that. Please I want u to help me in sets.
Opoku
what is math number
4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
hi vedant can u help me with some assignments
Solomon
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
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Find the square root of 7+4sqr...
# Find the square root of 7+4sqrt(3). <br> (b) Find the square root of 10+sqrt(24)+sqrt(60)+sqrt(40)
Khareedo DN Pro and dekho sari videos bina kisi ad ki rukaavat ke!
Updated On: 27-06-2022
Text Solution
Solution : Let sqrt(7+4sqrt(3))=sqrt(x)+sqrt(y) <br> Squaring both the sides, <br> 7+4sqrt(3)=x+y+2sqrt(xy) <br> implies x+y=7 and sqrt(xy)=2sqrt(3) implies xy 12 <br> By solving , we get x=4 and y=3 <br> sqrt(x)+sqrt(y)=sqrt(4)+sqrt(3)=2+sqrt(3) <br> (b) Let the given expression be equal to a+sqrt(x)+sqrt(y)+sqrt(z) <br> As per the method discussed , a=10, b=24, c=60 and d=40 <br> x=(1)/(2)sqrt((bd)/(c))=(1)/(2)sqrt((24xx40)/(60))=2 <br> y=(1)/(2)sqrt((bc)/(d))=(1)/(2)sqrt((24xx60)/(40))=3 <br> z=(1)/(2)sqrt((cd)/(b))=(1)/(2)sqrt((60xx40)/(24))=5 <br> therefore sqrt(x)+sqrt(y)+sqrt(z)=sqrt(2)+sqrt(3)+sqrt(5) <br> Alternative method: <br> =sqrt(10+sqrt(24)+sqrt(60)+sqrt(40)) <br> =sqrt(10+2sqrt(6)+2sqrt(15)+2sqrt(10)) <br> =sqrt((2+3+5)+2sqrt(2(3))+2sqrt(3(5))=2sqrt(2(5))) <br> =sqrt((2)+sqrt(3)+sqrt(5))^(2) <br> =sqrt(2)+sqrt(3)+sqrt(5)
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Level-3 Mixed Graph Practice Test -1 Hard
Directions : Study the following pie chart and line chart carefully and answer the questions given beside.
The pie chart shows the number of students study in five different schools as a percentage of the total number of students studies in all five schools A, B, C, D, and E. All the students of the schools are divided into two houses Red and Green. The total number of students in all five schools is 15000.
Percentage of students
The line graph shows the number of girls and number of boys who are in red house in each of the five schools.
1 If ratio of number of girls in green house to red house in school B is 16:17 then find difference between number of boys and girls in green house.
A 140
B 180
C 120
D 170
E 210
Correct Option: D
Total students = 20% of 15000 = 3000
Total girls = 1650
Total boys = 3000 – 1650 = 1350
Boys in red house = 720
Boys in green house = 1350 – 720 = 630
Girls in green house = 1650 × 16/33 = 800
Difference = 800 – 630 = 170
Hence, option D is correct.
2 What is the ratio of girls in red house to green house in school C if total number of students in green house of school C is 1083?
A 18 : 11
B 11 : 15
C 20 : 13
D 24 : 19
E 10 : 3
Correct Option: C
For school C:
Total students = 15% of 15000 = 2250
Total girls = 990
Total boys = 2250 – 900 = 1260
Boys in red house = 567
Boys in green house = 1260 – 567 = 693
Girls in green house = 1083 – 690 = 390
Girls in red house = 990 – 390 = 600
Ratio = 600 : 390 = 20 : 13
Hence, option C is correct.
3 Number of boys in green house of school A is what percent of number of girls in green house of school A if total students in red house of school A is 750?
A 110%
B 120%
C 90%
D 70%
E 150%
Correct Option: A
For school A:
Total students = 12% of 15000 = 1800
Total girls = 810
Total boys = 1800 – 810 = 990
Boys in red house = 440
Boys in green house = 990 – 440 = 550
Girls in red house = 750 – 440 = 310
Girls in green house = 810 – 310 = 500
Percentage = 550 × 100/500 = 110%
Hence, option A is correct.
4 What is the total number of students in red house of school E if number of girls in green house of school E is 1155?
A 2245
B 2205
C 2285
D 2175
E 2255
Correct Option: B
For school E:
Total students = 28% of 15000 = 4200
Total girls = 2100
Total boys = 4200 – 2100 = 2100
Boys in red house = 1260
Boys in green house = 2100 – 1260 = 840
Girls in red house = 2100 – 1155 = 945
Total students in red house = 1260 + 945 = 2205
Hence, option B is correct.
5 What percent of girls are in green house out of total girls of school D if number of girls in red house of school D is 305 more than the boys in same house?
A 40%
B 50%
C 20%
D 60%
E 80%
Correct Option: B
For school D:
Total students = 25% of 15000 = 3750
Total girls = 2250
Total boys = 3750 – 2250 = 1500
Boys in red house = 820
Boys in green house = 1500 – 820 = 680
Girls in red house = 820 + 305 = 1125
Girls in green house = 2250 – 1125
Percentage = 1125 × 100/2250 = 50%
Hence, option B is correct.
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# Constructions Class 10 Maths Formulas
For those looking for help on Constructions Class 10 Math Concepts can find all of them here provided in a comprehensive manner. To make it easy for you we have jotted the Class 10 Constructions Maths Formulae List all at one place. You can find Formulas for all the topics lying within the Constructions Class 10 Constructions in detail and get a good grip on them. Revise the entire concepts in a smart way taking help of the Maths Formulas for Class 10 Constructions.
## Maths Formulas for Class 10 Constructions
The List of Important Formulas for Class 10 Constructions is provided on this page. We have everything covered right from basic to advanced concepts in Constructions. Make the most out of the Maths Formulas for Class 10 prepared by subject experts and take your preparation to the next level. Access the Formula Sheet of Constructions Class 10 covering numerous concepts and use them to solve your Problems effortlessly.
Determining a Point Dividing a given Line Segment, Internally in the given Ratio M : N
Let AB be the given line segment of length x cm. We are required to determine a point P dividing it internally in the ratio m : n.
Steps of Construction:
• Draw a line segment AB = x cm.
• Make an acute ∠BAX at the end A of AB.
• Use a compass of any radius and mark off arcs. Take (m + n) points A1, A2, … Am, Am+1, …, Am+n along AX such that AA1 = A1A2 = … = Am+n-1 , Am+n
• Join Am+nB.
• Passing through Am, draw a line AmP || Am+nB to intersect AB at P. The point P so obtained is the A required point which divides AB internally in the ratio m : n.
Construction of a Tangent at a Point on a Circle to the Circle when its Centre is Known
Steps of Construction:
• Draw a circle with centre O of the given radius.
• Take a given point P on the circle.
• Join OP.
• Construct ∠OPT = 90°.
• Produce TP to T’ to get TPT’ as the required tangent.
Construction of a Tangent at a Point on a Circle to the Circle when its Centre is not Known
If the centre of the circle is not known, then we first find the centre of the circle by drawing two non-parallel chords of the circle. The point of intersection of perpendicular bisectors of these chords gives the centre of the circle. Then we can proceed as above.
Construction of a Tangents from an External Point to a Circle when its Centre is Known
Steps of Construction:
• Draw a circle with centre O.
• Join the centre O to the given external point P.
• Draw a right bisector of OP to intersect OP at Q.
• Taking Q as the centre and OQ = PQ as radius, draw a circle to intersect the given circle at T and T’.
• Join PT and PT’ to get the required tangents as PT and PT’.
Construction of a Tangents from an External Point to a Circle when its Centre is not Known
If the centre of the circle is not known, then we first find the centre of the circle by drawing two non-parallel chords of a circle. The point of intersection of perpendicular bisectors of the chords gives the centre of the circle. Then we can proceed as above.
Construction of a Triangle Similar to a given Triangle as per given Scale Factor $$\frac { m }{ n }$$ , m < n.
Let ΔABC be the given triangle. To construct a ΔA’B’C’ such that each of its sides is $$\frac { m }{ n }$$ (m < n) of the corresponding sides of ΔABC.
Steps of Construction:
• Construct a triangle ABC by using the given data.
• Make an acute angle ∠BAX, below the base AB.
• Along AX, mark n points A1, A2 …, An, such that AA1 = A1A2 = … = Am-1 Am = … An-1 An.
• Join AnB.
• From Am, draw AmB’ parallel to AnB, meeting AB at B’.
• From B’, draw B’C’ parallel to BC, meeting AC at C’.
Triangle AB’C’ is the required triangle, each of whose sides is $$\frac { m }{ n }$$ (m < n) of the corresponding sides of ΔABC.
Construction of a Triangle Similar to a given Triangle as per given Scale Factor $$\frac { m }{ n }$$ , m > n.
Let ΔABC be the given triangle and we want to construct a ΔAB’C’, such that each of its sides is $$\frac { m }{ n }$$ (m > n) of the corresponding side of ΔABC.
Steps of Construction:
• Construct a ΔABC by using the given data.
• Make an acute angle ∠BAX, below the base AB. Extend AB to AY and AC to AZ.
• Along AX, mark m points A1, A2 …, An, ..Am, such that AA1 = A1A2 = A2A3 = … = An-1 An = … = Am-1 Am
• Join AnB.
• From Am, draw AmB’ parallel to AnB, meeting AY produced at B’.
• From B’, draw B’C’ parallel to BC, meeting AZ produced at C’.
• Triangle AB’C’ is the required triangle, each of whose sides is ($$\frac { m }{ n }$$) (m > n) of the corresponding sides of ΔABC.
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# 5.2 Unit circle: sine and cosine functions
Page 1 / 12
In this section, you will:
• Find function values for the sine and cosine of and $\text{\hspace{0.17em}}{60°}^{}\text{or}\left(\frac{\pi }{3}\right).$
• Identify the domain and range of sine and cosine functions.
• Use reference angles to evaluate trigonometric functions.
Looking for a thrill? Then consider a ride on the Singapore Flyer, the world’s tallest Ferris wheel. Located in Singapore, the Ferris wheel soars to a height of 541 feet—a little more than a tenth of a mile! Described as an observation wheel, riders enjoy spectacular views as they travel from the ground to the peak and down again in a repeating pattern. In this section, we will examine this type of revolving motion around a circle. To do so, we need to define the type of circle first, and then place that circle on a coordinate system. Then we can discuss circular motion in terms of the coordinate pairs.
## Finding function values for the sine and cosine
To define our trigonometric functions, we begin by drawing a unit circle, a circle centered at the origin with radius 1, as shown in [link] . The angle (in radians) that $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ intercepts forms an arc of length $\text{\hspace{0.17em}}s.\text{\hspace{0.17em}}$ Using the formula $\text{\hspace{0.17em}}s=rt,$ and knowing that $r=1,$ we see that for a unit circle , $\text{\hspace{0.17em}}s=t.\text{\hspace{0.17em}}$
Recall that the x- and y- axes divide the coordinate plane into four quarters called quadrants. We label these quadrants to mimic the direction a positive angle would sweep. The four quadrants are labeled I, II, III, and IV.
For any angle $\text{\hspace{0.17em}}t,$ we can label the intersection of the terminal side and the unit circle as by its coordinates, $\text{\hspace{0.17em}}\left(x,y\right).\text{\hspace{0.17em}}$ The coordinates $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ will be the outputs of the trigonometric functions $\text{\hspace{0.17em}}f\left(t\right)=\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(t\right)=\mathrm{sin}\text{\hspace{0.17em}}t,$ respectively. This means $\text{\hspace{0.17em}}x=\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$
## Unit circle
A unit circle has a center at $\text{\hspace{0.17em}}\left(0,0\right)\text{\hspace{0.17em}}$ and radius $\text{\hspace{0.17em}}1\text{\hspace{0.17em}}$ . In a unit circle, the length of the intercepted arc is equal to the radian measure of the central angle $\text{\hspace{0.17em}}1.$
Let $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ be the endpoint on the unit circle of an arc of arc length $\text{\hspace{0.17em}}s.\text{\hspace{0.17em}}$ The $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates of this point can be described as functions of the angle.
## Defining sine and cosine functions
Now that we have our unit circle labeled, we can learn how the $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates relate to the arc length and angle . The sine function relates a real number $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ to the y -coordinate of the point where the corresponding angle intercepts the unit circle. More precisely, the sine of an angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ equals the y -value of the endpoint on the unit circle of an arc of length $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ In [link] , the sine is equal to $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ Like all functions, the sine function has an input and an output. Its input is the measure of the angle; its output is the y -coordinate of the corresponding point on the unit circle.
The cosine function of an angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ equals the x -value of the endpoint on the unit circle of an arc of length $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ In [link] , the cosine is equal to $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$
Because it is understood that sine and cosine are functions, we do not always need to write them with parentheses: $\mathrm{sin}\text{\hspace{0.17em}}t$ is the same as $\mathrm{sin}\left(t\right)$ and $\mathrm{cos}\text{\hspace{0.17em}}t$ is the same as $\mathrm{cos}\left(t\right).$ Likewise, $\text{\hspace{0.17em}}{\mathrm{cos}}^{2}t\text{\hspace{0.17em}}$ is a commonly used shorthand notation for ${\left(\mathrm{cos}\left(t\right)\right)}^{2}.\text{\hspace{0.17em}}$ Be aware that many calculators and computers do not recognize the shorthand notation. When in doubt, use the extra parentheses when entering calculations into a calculator or computer.
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the  that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
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# Lehmer's GCD algorithm
Lehmer's GCD algorithm, named after Derrick Henry Lehmer, is a fast GCD algorithm, an improvement on the simpler but slower Euclidean algorithm. It is mainly used for big integers that have a representation as a string of digits relative to some chosen numeral system base, say β = 1000 or β = 232.
## Algorithm
Lehmer noted that most of the quotients from each step of the division part of the standard algorithm are small. (For example, Knuth observed that the quotients 1, 2, and 3 comprise 67.7% of all quotients.[1]) Those small quotients can be identified from only a few leading digits. Thus the algorithm starts by splitting off those leading digits and computing the sequence of quotients as long as it is correct.
Say we want to obtain the GCD of the two integers a and b. Let a ≥ b.
• If b contains only one digit (in the chosen base, say β = 1000 or β = 232), use some other method, such as the Euclidean algorithm, to obtain the result.
• If a and b differ in the length of digits, perform a division so that a and b are equal in length, with length equal to m.
• Outer loop: Iterate until one of a or b is zero:
• Decrease m by one. Let x be the leading (most significant) digit in a, x = a div β m and y the leading digit in b, y = b div β m.
• Initialize a 2 by 3 matrix
${\displaystyle \textstyle {\begin{bmatrix}A&B&x\\C&D&y\end{bmatrix}}}$ to an extended identity matrix ${\displaystyle \textstyle {\begin{bmatrix}1&0&x\\0&1&y\end{bmatrix}},}$
and perform the euclidean algorithm simultaneously on the pairs (x + Ay + C) and (x + By + D), until the quotients differ. That is, iterate as an inner loop:
• Compute the quotients w1 of the long divisions of (x + A) by (y + C) and w2 of (x + B) by (y + D) respectively. Also let w be the (not computed) quotient from the current long division in the chain of long divisions of the euclidean algorithm.
• If w1w2, then break out of the inner iteration. Else set w to w1 (or w2).
• Replace the current matrix
${\displaystyle \textstyle {\begin{bmatrix}A&B&x\\C&D&y\end{bmatrix}}}$
with the matrix product
${\displaystyle \textstyle {\begin{bmatrix}0&1\\1&-w\end{bmatrix}}\cdot {\begin{bmatrix}A&B&x\\C&D&y\end{bmatrix}}={\begin{bmatrix}C&D&y\\A-wC&B-wD&x-wy\end{bmatrix}}}$
according to the matrix formulation of the extended euclidean algorithm.
• If B ≠ 0, go to the start of the inner loop.
• If B = 0, we have reached a deadlock; perform a normal step of the euclidean algorithm with a and b, and restart the outer loop.
• Set a to aA + bB and b to Ca + Db (again simultaneously). This applies the steps of the euclidean algorithm that were performed on the leading digits in compressed form to the long integers a and b. If b ≠ 0 go to the start of the outer loop.
## References
1. ^ Knuth, The Art of Computer Programming vol 2 "Seminumerical algorithms", chapter 4.5.3 Theorem E.
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29
Q:
# Two trains A and B start simultaneously in the opposite direction from two points P and Q and arrive at their destinations 16 and 9 hours respectively after their meeting each other. At what speed does the second train B travel if the first train travels at 120 km/h
A) 90 km/h B) 160 km/h C) 67.5 km/h D) None of these
Explanation:
$\frac{{s}_{1}}{{s}_{2}}=\sqrt{\frac{{t}_{2}}{{t}_{1}}}$
$⇒\frac{120}{{s}_{2}}=\sqrt{\frac{9}{16}}=\frac{3}{4}$
$⇒$${s}_{2}=160km/h$
Q:
Ashwin fires two bullets from the same place at an interval of 15 minutes but Rahul sitting in a bus approaching the place hears the second sound 14 minutes 30 seconds after the first. If sound travels at the speed of 330 meter per second, what is the approximate speed of bus?
A) 330/29 m/s B) 330 x 30 m/s C) 330/14 m/s D) 330/900 m/s
Explanation:
Second gun shot take 30 sec to reach rahul imples distance between two.
given speed of sound = 330 m/s
Now, distance = 330 m/s x 30 sec
Hence, speed of the bus = d/t = 330x30/(14x60 + 30) = 330/29 m/s.
0 27
Q:
Important Time and Distance formulas with examples.
1. How to find Speed(s) if distance(d) & time(t) is given:
Ex: Find speed if a person travels 4 kms in 2 hrs?
Speed = D/T = 4/2 = 2 kmph.
2. Similarly, we can find distance (d) if speed (s) & time (t) is given by
Distance (D) = Speed (S) x Time (T)
Ex : Find distance if a person with a speed of 2 kmph in 2 hrs?
Distance D = S X T = 2 x 2 = 4 kms.
3. Similarly, we can find time (t) if speed (s) & distance (d) is given by
Time (T) =
Ex : Find in what time a person travels 4 kms with a speed of 2 kmph?
Time T = D/S = 4/2 = 2 hrs.
4. How to convert km/hr into m/sec :
Ex : Convert 36 kmph into m/sec?
36 kmph = 36 x 5/18 = 10 m/sec
5. How to convert m/sec into km/hr :
.
Ex : Convert 10 m/sec into km/hr?
10 m/sec = 10 x 18/5 = 36 kmph.
6. If the ratio of the speeds of A and B is a:b, then the ratio of the times taken by them to cover the same distance is b : a.
7. Suppose a man covers a certain distance at x km/ hr and an equal distance at y km/hr . Then, the average speed during the whole journey is $\frac{\mathbf{2}\mathbf{xy}}{\mathbf{x}\mathbf{+}\mathbf{y}}$ km/hr.
86
Q:
Two trains are running with speeds 30 kmph and 58 kmph respectively in the same direction. A man in the slower train passes the faster train in 18 seconds. Find the length of the faster train?
A) 105 mts B) 115 mts C) 120 mts D) 140 mts
Explanation:
Speeds of two trains = 30 kmph and 58 kmph
=> Relative speed = 58 - 30 = 28 kmph = 28 x 5/18 m/s = 70/9 m/s
Given a man takes time to cross length of faster train = 18 sec
Now, required Length of faster train = speed x time = 70/9 x 18 = 140 mts.
0 100
Q:
A train-A passes a stationary train B and a pole in 24 sec and 9 sec respectively. If the speed of train A is 48 kmph, what is the length of train B?
A) 200 mts B) 180 mts C) 160 mts D) 145 mts
Explanation:
Length of train A = 48 x 9 x 5/18 = 120 mts
Length of train B = 48 x 24 x 5/18 - 120
=> 320 - 120 = 200 mts.
1 250
Q:
Tilak rides on a cycle to a place at speed of 22 kmph and comes back at a speed of 20 kmph. If the time taken by him in the second case is 36 min. more than that of the first case, what is the total distance travelled by him (in km)?
A) 132 km B) 264 km C) 134 km D) 236 km
Explanation:
Let the distance travelled by Tilak in first case or second case = d kms
Now, from the given data,
d/20 = d/22 + 36 min
=> d/20 = d/22 + 3/5 hrs
=> d = 132 km.
Hence, the total distance travelled by him = d + d = 132 + 132 = 264 kms.
0 184
Q:
A train covers 180 km distance in 5 hours. Another train covers the same distance in 1 hour less. What is the difference in the distances covered by these trains in one hour if they are moving in the same direction?
A) 15 kms B) 9 kms C) 6 kms D) 18 kms
Explanation:
The first train covers 180 kms in 5 hrs
=> Speed = 180/5 = 36 kmph
Now the second train covers the same distance in 1 hour less than the first train => 4 hrs
=> Speed of the second train = 180/4 = 45 kmph
Now, required difference in distance in 1 hour = 45 - 36 = 9 kms.
4 364
Q:
Chennai express left Hyderabad for Chennai at 14 : 30 hours, travelling at a speed of 60 kmph and Charminar Express left Hyderabad for Chennai on the same day at 16 : 30 hours, travelling at a speed of 80 kmph. How far away from Hyderabad will the two trains meet?
A) 360 kms B) 480 kms C) 520 kms D) 240 kms
Explanation:
Now, the distance covered by Chennai express in 2 hrs = 60 x 2 = 120 kms
Let the Charminar Express takes 't' hrs to catch Chennai express
=> 80 x t = 60 x (2 + t)
=> 80 t = 120 + 60t
=> t = 6 hrs
Therefore, the distance away from Hyderabad the two trains meet = 80 x 6 = 480 kms.
6 301
Q:
A man covers half of his journey by train at 60 km/hr, half of the remaining by bus at 30 km/hr and the rest by cycle at 10 km/hr. Find his average speed during the entire journey?
A) 32 kmph B) 20 kmph C) 18 kmph D) 24 kmph
|
# CLASS-8SIMULTANEOUS LINEAR EQUATION - ELIMINATION METHOD
SIMULTANEOUS LINEAR EQUATION -
Elimination Method –
This method is also called the addition subtraction method
Step.1) Decide which variable will be easier to eliminate, try to avoid fractions
Step.2) Multiply one or both the equations by suitable numbers to ensure that the coefficients of the variable to be eliminated are the same in both the equations.
Step.3) Add or subtract the resulting equations to eliminate the variable
Step.4) Solve the resulting equation in one variable
Step.5) Substitute the value of the variable obtained in step.4in either of the given equations.
Step.6) Solve the resulting equation.
Step.7) Verify the correctness of the solution by substituting the values of the variables in the given equations.
There are some example are given below for your better understanding –
Example.1) Solve the equation 8x + 5y = 10 and 7x + 9y = 15
Ans.) That has observed that the given equations are 8x + 5y = 10 ………………………………. (1)
7x + 9y = 15 …………………………………….. (2)
From the equation (1), 8x + 5y = 10 or x = (10 – 5y)/8
Now, we would like to substitute the expression (10 – 5y)/8 for x in equation (2)
7x + 9y = 15
7 (10 – 5y)
So, --------------- + 9y = 15
8
or, 7 (10 – 5y) + 9y . 8 = 15 X 8
or, 70 – 35y + 72y = 120
or, 37y = 120 – 70
or, 37y = 50
or, y = 50/37
substituting the value of y = 50/37 in equation (1) we get
so, 8x + 5y = 10
or, 8x+ 5 (50/37) = 10
or, 8x + 250/37 = 10
or, 8x X 37 + 250 = 10 X 37
or, 296x = 370 – 250
or, x = 120/296 = 30/74 = 15/37
so, the required value is x = 15/37 and y = 50/37 (Ans.)
Example.2) Solve the equation 3y – 2x = 1 and 3x + 4y = 24
Ans.) That has observed that the given equations are 3y – 2x = 1 ………………………………. (1)
3x + 4y = 24 …………………………………….. (2)
From the equation (1), 3y – 2x = 1 or x = (3y – 1)/2
Now, we would like to substitute the expression (3y – 1)/2 for x in equation (2)
3x + 4y = 24
3 (3y – 1)
So, -------------- + 4y = 24
2
or, 3 (3y – 1) + 4y . 2 = 24 X 2
or, 9y – 3 + 8y = 48
or, 17y = 48 + 3
or, 17y = 51
or, y = 3
substituting the value of y = 3 in equation (1) we get
so, 3y – 2x = 1
or, 3 X 3 – 2x = 1
or, 2x = 9 - 1
or, 2x = 8
or, x = 4
so, the required value is x = 4 and y = 3 (Ans.)
Example.3) Solve the equation 2/x + 3/y = - 1 and 3/x + 5/y = - 2
1 1
Ans.) Let, --------- = p and --------- = q
x y
so, now we get the equations 2p + 3q = - 1 …………………………….(1)
and 3p + 5q = - 2 …………………………………… (2)
so, From the equation (1), 2p + 3q = - 1 or p = (- 3q – 1)/2
Now, we would like to substitute the expression (- 3q – 1)/2 for p in equation (2)
3p + 5q = - 2
3 (- 3q – 1)
So, ---------------- + 5q = - 2
2
or, 3 (- 3q – 1) + 5q . 2 = (-2) X 2
or, - 9q – 3 + 10q = - 4
or, 10q – 9q = 3 – 4
or, q = - 1
substituting the value of q = -1 in equation (1) we get
So, 2p + 3q = - 1
so, 2p + 3 (- 1) = - 1
or, 2p – 3 = - 1
or, 2p = 3 - 1
or, 2p = 2
or, p = 1
now to put the value of p & q we find –
1
p = 1 , --------- = 1, or x = 1
x
1
and q = - 1, or --------- = - 1 or y = -1
y
so, the required value is x = 1 and y = -1 (Ans.)
|
Exponential Functions
You can move the red dots to see the graph of different exponential functions.
Exponential functions are related to geometric progressions:
Geometric sequences graphic representations. Sum of terms of a geometric sequence and geometric series.
Exponential functions can be seen as extensions of geometric succesions to the Real Numbers. But it is not so easy to make this extension.
You can think that it is easy to understand the meaning of a function like 2x
(or 3x or 10x) because we can calculate thinks like
But, the meaning of
There is no simple way to defining 2x for irrational x.
In general, this kind of functions are called Exponential Functions
b is called the base of the exponential function.
We expect that this function satisfy the fundamental equation
And
Some of these functions are increasing (when b < 1)
And some are decreasing (when b > 1)
When an exponential function is increasing, 'at the end' it increases very rapidly even if the base is only a little bigger than 1.
We can see that all exponential functions have a very similar shape. But there is a exponential function very special. Its base is a number but it seems odd that we call number e (Euler was the first to use this notation for this number).
Tangents at 0. Existence of number e
If we look at the tangent at x=0 of such functions there is a base for which the slope of the tangent of the function at x=0 is 1. We are going to call such number as e
"The existence of a number e having the above property can be motivated as follows. Let a, b be numbers > 1 and suppose that a < b. Then for all numbers x, we have
If b is very large, then the curve y=bx will have a very steep slope at x=0.
It is plausible that as a increases from numbers close to 1 (and > 1) to very large numbers, the slope of ax at x=0 increases continuously from values close to 0 to large values, and therefore for some value of a, which we call e, this slope is precisely equal to 1. Thus in this naive approach, e is the number such that the slope of ex at x=0 is equal to 1. " (Serge Lang)
We shall find out eventually how to compute e. Its value is
Using limits, the slope of the tangent at x= 0 is (if this limit exits)
Notice that, if this limit exists, only depends on b.
Then we are saying that there is a number e such as
Thinking about the derivative of an exponential function, if that limit exits and using the property of the exponentials, we can write
This is to say, the derivative of an exponential function is a multiple of the function (we do not know yet the value of this factor)
And the number e has a very nice property
INVESTIGATE
Moving the red point that it is not over the vertical axis you can see different exponential functions. You can see the relation between the derivative of a function in a point and the tangent line. You can see a right triangle and that the hypotenuse is parallel to the tangent.
What happens when the base is e?.
If you move the two red points you can see the graph the more general exponential functions:
An exponential function is fixed when we know two points.
In this page we played with exponential functions but we have problems with the definition of such functions. Remember that we do not know the meaning of
The idea is to introduce logarithms first and then use logarithms to define exponential functions.
REFERENCES
A. I. Markushevich, Areas and Logarithms, D.C. Heath and Company, 1963.
Serge Lang, A First Course in Calculus, Third Edition, Addison-Wesley Publishing Company.
Tom M. Apostol, Calculus, Second Edition, John Willey and Sons, Inc.
Michael Spivak, Calculus, Third Edition, Publish-or-Perish, Inc.
Otto Toeplitz, The Calculus, a genetic approach, The University of Chicago Press, 1963.
Kenneth A. Ross, Elementary Analysis: The Theory of Calculus, Springer-Verlag New York Inc., 1980.
The natural logaritm can be defined using the integral of the rectangular hiperbola. In this page we are going to see an important property of this integral. Using this property you can justify that the logarithm of a product is the sum of the logarithms.
The main property of a logarithm function is that the logarithm of a product is the sum of the logarithms of the individual factors.
The logarithm of the number e is equal to 1. Using this definition of the number e we can approximate its value.
Constant e is the number whose natural logarithm is 1. It can be defined as a limit of a sequence related with the compound interest. Both definitions for e are equivalent.
After the definition of the natural logarithm function as an integral you can define the exponential function as the inverse function of the logarithm.
Different hyperbolas allow us to define different logarithms functions and their inversas, exponentials functions.
Mercator published his famous series for the Logarithm Function in 1668. Euler discovered a practical series to calculate.
By increasing the degree, Taylor polynomial approximates the exponential function more and more.
The Complex Exponential Function extends the Real Exponential Function to the complex plane.
The complex exponential function is periodic. His power series converges everywhere in the complex plane.
The Fundamental Theorem of Calculus tell us that every continuous function has an antiderivative and shows how to construct one using the integral.
The Second Fundamental Theorem of Calculus is a powerful tool for evaluating definite integral (if we know an antiderivative of the function).
As an introduction to Piecewise Linear Functions we study linear functions restricted to an open interval: their graphs are like segments.
A piecewise function is a function that is defined by several subfunctions. If each piece is a constant function then the piecewise function is called Piecewise constant function or Step function.
A continuous piecewise linear function is defined by several segments or rays connected, without jumps between them.
The integral of power functions was know by Cavalieri from n=1 to n=9. Fermat was able to solve this problem using geometric progressions.
If the derivative of F(x) is f(x), then we say that an indefinite integral of f(x) with respect to x is F(x). We also say that F is an antiderivative or a primitive function of f.
The integral concept is associate to the concept of area. We began considering the area limited by the graph of a function and the x-axis between two vertical lines.
Monotonic functions in a closed interval are integrable. In these cases we can bound the error we make when approximating the integral using rectangles.
If we consider the lower limit of integration a as fixed and if we can calculate the integral for different values of the upper limit of integration b then we can define a new function: an indefinite integral of f.
We can consider the polynomial function that passes through a series of points of the plane. This is an interpolation problem that is solved here using the Lagrange interpolating polynomial.
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Basic Math | Fractions and Decimals | Prealgebra | Workbooks | Glossary | Standards | Site Map | Help
# Pick-a-Bar: Finding Values in the Tenths Place
This little decimal activity has you looking for values in specific places. Start by remembering that whole numbers are found on the left side of the decimal point and values smaller than one are on the right side. As you move further and further to the right, the values get smaller even though you still use the same ten numbers. A five (5) in the tenths position is much larger than a five (5) in the ten-thousandths position. Here is a key to help you with the labels:
0.123456789
"0" is a whole number in the ones position.
"1" is the value in the tenths position.
"2" is the value in the hundredths position.
"3" is the value in the thousandths position.
"4" is the value in the ten-thousandths position.
"5" is the value in the hundred-thousandths position.
"6" is the value in the millionths position.
They go on like that. Remember that the positions that end with a "th" are decimal values less than one. This activity will have you looking for the values in the tenths position. That's the one right next to the decimal point.
Good luck and have fun.
## Directions
This is one of the NumberNut four-bar activities. Your question will be displayed at the top of the screen. Under that question, you will see four (4) possible answers. Just click on the bar with the correct answer. The next screen will show you the correct answer.
You get a happy face for every correct answer and a sad face for every wrong answer. The quiz is over after ten (10) questions. Take the quiz again and again because all of the questions are random. Chances are, you'll get a new quiz every time. It's good practice to learn these basic arithmetic operations.
RELATED LINKS LESSONS: - NumberNut.com: Decimals ACTIVITIES: - Pick-a-Bar: Naming Values with Tenths - Number Lines: Adding Tenths - Number Lines: Subtracting Tenths - QuickQuiz: Adding Tenth Values - QuickQuiz: Subtracting Tenth Values - More or Less: Adding Tenths - More or Less: Subtracting Tenths - Pick-a-Bar: Naming Values with Hundredths - Number Lines: Adding Hundredths - Number Lines: Subtracting Hundredths - QuickQuiz: Adding Hundredth Values - QuickQuiz: Subtracting Hundredth Values - More or Less: Adding Hundredths Quiz - More or Less: Subtracting Hundredths Quiz - Pick-a-Bar: Naming with Thousandths
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Polar Coordinates
# Polar Coordinates - Polar Coordinates Up to this point weve...
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Polar Coordinates Up to this point we’ve dealt exclusively with the Cartesian (or Rectangular, or x-y ) coordinate system. However, as we will see, this is not always the easiest coordinate system to work in. So, in this section we will start looking at the polar coordinate system. Coordinate systems are really nothing more than a way to define a point in space. For instance in the Cartesian coordinate system at point is given the coordinates ( x,y ) and we use this to define the point by starting at the origin and then moving x units horizontally followed by y units vertically. This is shown in the sketch below. This is not, however, the only way to define a point in two dimensional space. Instead of moving vertically and horizontally from the origin to get to the point we could instead go straight out of the origin until we hit the point and then determine the angle this line makes with the positive x -axis. We could then use the distance of the point from the origin and the amount we needed to rotate from the positive x -axis as the coordinates of the point. This is shown in the sketch below. Coordinates in this form are called polar coordinates .
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The above discussion may lead one to think that r must be a positive number. However, we also allow r to be negative. Below is a sketch of the two points and . From this sketch we can see that if r is positive the point will be in the same quadrant as θ. On the other hand if r is negative the point will end up in the quadrant exactly opposite θ. Notice as well that the coordinates describe the same point as the coordinates do. The coordinates tells us to rotate an angle of from the positive x -axis, this would put us on the dashed line in the sketch above, and then move out a distance of 2. This leads to an important difference between Cartesian coordinates and polar coordinates. In Cartesian coordinates there is exactly one set of coordinates for any given point. With polar coordinates this isn’t true. In polar coordinates there is literally an infinite number of coordinates for a given point. For instance, the following four points are all coordinates for the same point.
Here is a sketch of the angles used in these four sets of coordinates. In the second coordinate pair we rotated in a clock-wise direction to get to the point. We shouldn’t forget about rotating in the clock-wise direction. Sometimes it’s what
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## This note was uploaded on 11/10/2011 for the course MATH 136 taught by Professor Prellis during the Fall '08 term at Rutgers.
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Polar Coordinates - Polar Coordinates Up to this point weve...
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# If A= <8 ,-5 ,6 > and B= <7 ,1 ,0 >, what is A*B -||A|| ||B||?
Mar 14, 2018
$\boldsymbol{\underline{A}} \cdot \boldsymbol{\underline{B}} - | | \boldsymbol{\underline{A}} | | \setminus | | \boldsymbol{\underline{B}} | | = 51 - 25 \sqrt{2}$
#### Explanation:
We start with the vectors $A$ and $B$
$\boldsymbol{\underline{A}} = \left\langle8 , - 5 , 6\right\rangle$
$\boldsymbol{\underline{B}} = \left\langle7 , 1 , 0\right\rangle$
The dot (or scalar) product is given by:
$\boldsymbol{\underline{A}} \cdot \boldsymbol{\underline{B}} = \left\langle8 , - 5 , 6\right\rangle \cdot \left\langle7 , 1 , 0\right\rangle$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(8\right) \left(7\right) + \left(- 5\right) \left(1\right) + \left(6\right) \left(0\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 56 - 5 + 0$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 51$
And the norms are given by:
$| | \boldsymbol{\underline{A}} | | = | | \left\langle8 , - 5 , 6\right\rangle | |$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{{8}^{2} + {\left(- 5\right)}^{2} + {6}^{2}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{64 + 25 + 36}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{125}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 5$
$| | \boldsymbol{\underline{B}} | | = | | \left\langle7 , 1 , 0\right\rangle | |$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{{7}^{2} + {1}^{2} + {0}^{2}}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{49 + 1 + 0}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{50}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 5 \sqrt{2}$
So then:
$\boldsymbol{\underline{A}} \cdot \boldsymbol{\underline{B}} - | | \boldsymbol{\underline{A}} | | \setminus | | \boldsymbol{\underline{B}} | | = 51 - 25 \sqrt{2}$
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# Division and Multiplication Relationship
Do you need extra help for EL students? Try the Multiplication and Division: What's the Connection?Pre-lesson.
EL Adjustments
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Do you need extra help for EL students? Try the Multiplication and Division: What's the Connection?Pre-lesson.
Students will be able to show understanding of the inverse relationship between multiplication and division.
The adjustment to the whole group lesson is a modification to differentiate for children who are English learners.
EL adjustments
(5 minutes)
• Draw a picture of 7 groups of 12On the board and ask students to turn and share in pairs what they see and notice about the drawing. Listen for key terms such as "groups of," "multiplied," "divide," "repeated addition," etc.
• Ask students to share their ideas and write them on the board. If they haven't given an equation, ask them to write on the board as many equations they can think of that relate to the picture (e.g., 7 x 12 = ____, 12 + 12 + 12 + 12 + 12 + 12 + 12 = ____). Ask students to solve for the blanks in partners if they hadn't done so in the sharing portion.
• Circle the multiplication and division equations and rewrite them on the board stacked on top of each other. Explain that today they'll review the inverse relationship of multiplication and division to help solve future word problems.
(8 minutes)
• Define Inverse operationAs an operation that reverses the effect of another operation. With multiplication and division, if you multiply to get a product, you can use division to reverse the operation by dividing the product, and vice versa. The ProductIs the answer when two or more numbers are multiplied together.
• Provide a simple multiplication and division problem using the same numbers. Model how you can change a division problem into a multiplication problem to make the division problem easier to solve.
• Highlight that converting multiplication equations to division equations is a strategy to divide by focusing on memorized or familiar multiplication facts. Check your answer using the picture representation of your choice (e.g., arrays, equal groups, tape diagrams, etc.).
(20 minutes)
• Distribute the worksheet The Inverse Relationship of Division and read the directions. Tell students they'll work in pairs to answer the questions and find the inverse of each of the equations.
• Conduct a multiplication fluency game in which there are two teams placed in two lines perpendicular to the board. Project the Division Facts to 100 with One-Digit Divisors exercise and have one student from each team compete to quickly convert the division equation to a multiplication problem and provide the answer. The students who get the correct answer first win a point for their team. Allow them to use whiteboards as necessary.
(10 minutes)
• Distribute the Multiplication and Division Review worksheet and ask students to complete the top section on their own.
• Allow students to meet in partners to share their answers and correct misconceptions.
• Choose students to share any corrections they made and their process to get the right answer with the class.
Support:
• Provide a pre-lesson with simple multiplication and division problems with manipulatives and a review of vocabulary terms and their meanings.
• Allow students to practise converting equations with a common factor (e.g., 8 x 3 = 24, 9 x 3 = 27, etc.), then transition to other factors. Use a worksheet like the optional Division Facts: 9s worksheet for assistance.
Enrichment:
• Allow students additional practise with the inverse operations of multiplication and division with the maths Crossword Puzzle worksheet. Additionally, allow them to practise division with the Division with One-Digit Divisors and Missing Factors exercise.
• Ask them to complete the word problems in the Multiplication and Division Review worksheet and show their method and equations, or create their own word problems.
(7 minutes)
• Write the following numbers on the board: 72, 9, 8. Distribute the index cards and ask students to write a multiplication and division equation using those numbers. Then, ask them to write how they know their answers are correct.
• Allow them to read their explanations to their elbow partners.
(5 minutes)
• Choose a student to answer the following question: “How are multiplication and division inverse operations?”
• Have students turn to their elbow partner and have them discuss why it is important to understand the relationship between multiplication and division (e.g., solve unfamiliar division problems with multiplication and vice versa).
• Review some of the ideas students shared in partners with the whole class.
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Applying Reasoning and Proof
Reasoning About Fractions | Reasoning About Fractions in Action | Classroom Practice | Reasoning and Proof in Action | Classroom Checklist | Your Journal
Think about the Fraction Tracks game and consider the following questions. Select "Show Answer" to reveal our commentary.
Question: How might the fraction pieces help students clarify their thinking and develop some conjectures about equivalent fractions? Show Answer
Our Answer: For students who do not see some of the possible combinations, the fraction pieces allow them to find other combinations they can use in the game. The pieces also help students clarify their understanding and give them a visual model of the concept.
Question: How does having students share strategies support the development of reasoning and proof? Show Answer
Our Answer: Students bring the activity to understanding by using their own language, they expand on ideas by using their own examples, and they question and challenge one another in a different way to bring meaning and understanding to the mathematics of the activity.
Question: How might this activity be extended for students who show an understanding of equivalent fractions? Show Answer
Our Answer: There are many ways to extend this activity. One might be to extend the board to 2. Also, tenths could be removed or twelfths could be added to the board.
Question: In what ways does this task encourage reasoning and proof? Show Answer
Our Answer: This task expects students to apply and extend their previous experience with equivalent fractions to a new task. The expectation is that students will share their reasoning and will need to prove that their reasoning is accurate. The task also expects students to refine their thinking to help them develop understanding of the concept of equivalent fractions.
Question: How does Ms. Paster encourage students to share their reasoning and thinking with one another? Show Answer
Our Answer: Her questions lead students to include an explanation of their reasoning so that their partners also understand what they are thinking.
Question: What characteristics of a classroom that encourages reasoning and proof are present in this lesson? Show Answer
Our Answer: Students work in groups to play the game, talk about their strategies, explain their reasoning to one another, and then write down their ideas. The expectation is for students to be able to explain and justify their reasoning.
Question: How does this activity show how each of the process standards can be integrated into a single activity? Show Answer
Our Answer: Problem solving is enveloped in the task. Communication, in the form of the students writing and talking about their reasoning and justifying their thinking, is apparent in all parts of the class. Students connect their previous experience with equivalent fractions to a new situation. Students represent mathematical ideas through the use of models (fraction pieces and game boards).
Teaching Math Home | Grades 6-8 | Reasoning and Proof | Site Map | © |
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# NCERT Class 8 Maths Chapter 7 – Cubes and Cube roots solutions
### Access Answers of Maths NCERT Class 8 Chapter 7 – Cubes and Cube roots
Exercise 7.1 Page: 114
1. Which of the following numbers are not perfect cubes?
(i) 216
Solution:
By resolving 216 into prime factor,
216 = 2×2×2×3×3×3
By grouping the factors in triplets of equal factors, 216 = (2×2×2)×(3×3×3)
Here, 216 can be grouped into triplets of equal factors,
∴ 216 = (2×3) = 6
Hence, 216 is cube of 6.
(ii) 128
Solution:
By resolving 128 into prime factor,
128 = 2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors, 128 = (2×2×2)×(2×2×2)×2
Here, 128 cannot be grouped into triplets of equal factors, we are left of with one factors 2 .
∴ 128 is not a perfect cube.
(iii) 1000 Solution:
By resolving 1000 into prime factor,
1000 = 2×2×2×5×5×5
By grouping the factors in triplets of equal factors, 1000 = (2×2×2)×(5×5×5)
Here, 1000 can be grouped into triplets of equal factors,
∴ 1000 = (2×5) = 10
Hence, 1000 is cube of 10.
(iv) 100 Solution:
By resolving 100 into prime factor,
100 = 2×2×5×5
Here, 100 cannot be grouped into triplets of equal factors.
∴ 100 is not a perfect cube.
(v) 46656
Solution:
By resolving 46656 into prime factor,
46656 = 2×2×2×2×2×2×3×3×3×3×3×3
By grouping the factors in triplets of equal factors, 46656 = (2×2×2)×(2×2×2)×(3×3×3)×(3×3×3)
Here, 46656 can be grouped into triplets of equal factors,
∴ 46656 = (2×2×3×3) = 36
Hence, 46656 is cube of 36.
2. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243
Solution:
By resolving 243 into prime factor,
243 = 3×3×3×3×3
By grouping the factors in triplets of equal factors, 243 = (3×3×3)×3×3
Here, 3 cannot be grouped into triplets of equal factors.
∴ We will multiply 243 by 3 to get perfect cube.
(ii) 256 Solution:
By resolving 256 into prime factor,
256 = 2×2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors, 256 = (2×2×2)×(2×2×2)×2×2
Here, 2 cannot be grouped into triplets of equal factors.
∴ We will multiply 256 by 2 to get perfect cube.
(iii) 72
Solution:
By resolving 72 into prime factor,
72 = 2×2×2×3×3
By grouping the factors in triplets of equal factors, 72 = (2×2×2)×3×3
Here, 3 cannot be grouped into triplets of equal factors.
∴ We will multiply 72 by 3 to get perfect cube.
(iv) 675 Solution:
By resolving 675 into prime factor,
675 = 3×3×3×5×5
By grouping the factors in triplets of equal factors, 675 = (3×3×3)×5×5
Here, 5 cannot be grouped into triplets of equal factors.
∴ We will multiply 675 by 5 to get perfect cube.
(v) 100 Solution:
By resolving 100 into prime factor,
100 = 2×2×5×5
Here, 2 and 5 cannot be grouped into triplets of equal factors.
∴ We will multiply 100 by (2×5) 10 to get perfect cube.
3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81
Solution:
By resolving 81 into prime factor,
81 = 3×3×3×3
By grouping the factors in triplets of equal factors, 81 = (3×3×3)×3
Here, 3 cannot be grouped into triplets of equal factors.
∴ We will divide 81 by 3 to get perfect cube.
(ii) 128 Solution:
By resolving 128 into prime factor,
128 = 2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors, 128 = (2×2×2)×(2×2×2)×2
Here, 2 cannot be grouped into triplets of equal factors.
∴ We will divide 128 by 2 to get perfect cube.
(iii) 135 Solution:
By resolving 135 into prime factor,
135 = 3×3×3×5
By grouping the factors in triplets of equal factors, 135 = (3×3×3)×5
Here, 5 cannot be grouped into triplets of equal factors.
∴ We will divide 135 by 5 to get perfect cube.
(iv) 192 Solution:
By resolving 192 into prime factor,
192 = 2×2×2×2×2×2×3
By grouping the factors in triplets of equal factors, 192 = (2×2×2)×(2×2×2)×3
Here, 3 cannot be grouped into triplets of equal factors.
∴ We will divide 192 by 3 to get perfect cube.
(v) 704 Solution:
By resolving 704 into prime factor,
704 = 2×2×2×2×2×2×11
By grouping the factors in triplets of equal factors, 704 = (2×2×2)×(2×2×2)×11
Here, 11 cannot be grouped into triplets of equal factors.
∴ We will divide 704 by 11 to get perfect cube.
4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
Solution:
Given, side of cube is 5 cm, 2 cm and 5 cm.
∴ Volume of cube = 5×2×5 = 50
50 = 2×5×5
Here, 2 , 5 and 5 cannot be grouped into triplets of equal factors.
∴ We will multiply 50 by (2×2×5) 20 to get perfect cube. Hence, 20 cuboid is needed.
## Exercise 7.2 Page: 116
1. Find the cube root of each of the following numbers by prime factorisation method.
(i) 64
Solution:
64 = 2×2×2×2×2×2
By grouping the factors in triplets of equal factors, 64 = (2×2×2)×(2×2×2)
Here, 64 can be grouped into triplets of equal factors,
∴ 64 = 2×2 = 4
Hence, 4 is cube root of 64.
(ii) 512
Solution:
512 = 2×2×2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors, 512 = (2×2×2)×(2×2×2)×(2×2×2)
Here, 512 can be grouped into triplets of equal factors,
∴ 512 = 2×2×2 = 8
Hence, 8 is cube root of 512.
(iii) 10648
Solution:
10648 = 2×2×2×11×11×11
By grouping the factors in triplets of equal factors, 10648 = (2×2×2)×(11×11×11)
Here, 10648 can be grouped into triplets of equal factors,
∴ 10648 = 2 ×11 = 22
Hence, 22 is cube root of 10648.
(iv) 27000
Solution:
27000 = 2×2×2×3×3×3×3×5×5×5
By grouping the factors in triplets of equal factors, 27000 = (2×2×2)×(3×3×3)×(5×5×5)
Here, 27000 can be grouped into triplets of equal factors,
∴ 27000 = (2×3×5) = 30
Hence, 30 is cube root of 27000.
(v) 15625
Solution:
15625 = 5×5×5×5×5×5
By grouping the factors in triplets of equal factors, 15625 = (5×5×5)×(5×5×5)
Here, 15625 can be grouped into triplets of equal factors,
∴ 15625 = (5×5) = 25
Hence, 25 is cube root of 15625.
(vi) 13824
Solution:
13824 = 2×2×2×2×2×2×2×2×2×3×3×3
By grouping the factors in triplets of equal factors,
13824 = (2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)
Here, 13824 can be grouped into triplets of equal factors,
∴ 13824 = (2×2× 2×3) = 24
Hence, 24 is cube root of 13824.
(vii) 110592
Solution:
110592 = 2×2×2×2×2×2×2×2×2×2×2×2×3×3×3
By grouping the factors in triplets of equal factors,
110592 = (2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)
Here, 110592 can be grouped into triplets of equal factors,
∴ 110592 = (2×2×2×2 × 3) = 48
Hence, 48 is cube root of 110592.
(viii) 46656
Solution:
46656 = 2×2×2×2×2×2×3×3×3×3×3×3
By grouping the factors in triplets of equal factors,
46656 = (2×2×2)×(2×2×2)×(3×3×3)×(3×3×3)
Here, 46656 can be grouped into triplets of equal factors,
∴ 46656 = (2×2×3×3) = 36
Hence, 36 is cube root of 46656.
(ix) 175616
Solution:
175616 = 2×2×2×2×2×2×2×2×2×7×7×7
By grouping the factors in triplets of equal factors,
175616 = (2×2×2)×(2×2×2)×(2×2×2)×(7×7×7)
Here, 175616 can be grouped into triplets of equal factors,
∴ 175616 = (2×2×2×7) = 56
Hence, 56 is cube root of 175616.
(x) 91125
Solution:
91125 = 3×3×3×3×3×3×3×5×5×5
By grouping the factors in triplets of equal factors, 91125 = (3×3×3)×(3×3×3)×(5×5×5)
Here, 91125 can be grouped into triplets of equal factors,
∴ 91125 = (3×3×5) = 45
Hence, 45 is cube root of 91125.
2. State true or false.
(i) Cube of any odd number is even.
Solution:
False
(ii) A perfect cube does not end with two zeros.
Solution:
True
(iii) If cube of a number ends with 5, then its cube ends with 25.
Solution:
False
(iv) There is no perfect cube which ends with 8.
Solution:
False
(v) The cube of a two digit number may be a three digit number.
Solution:
False
(vi) The cube of a two digit number may have seven or more digits.
Solution:
False
(vii) The cube of a single digit number may be a single digit number.
Solution:
True
3. You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
Solution:
(i) By grouping the digits, we get 1 and 331
We know that, since, the unit digit of cube is 1, the unit digit of cube root is 1.
∴ We get 1 as unit digit of the cube root of 1331.
The cube of 1 matches with the number of second group.
∴ The ten’s digit of our cube root is taken as the unit place of smallest number.
We know that, the unit’s digit of the cube of a number having digit as unit’s place 1 is 1.
∴ ∛1331 = 11
(ii) By grouping the digits, we get 4 and 913
We know that, since, the unit digit of cube is 3, the unit digit of cube root is 7.
∴ we get 7 as unit digit of the cube root of 4913. We know 13 = 1 and 23 = 8 , 1 > 4 > 8.
Thus, 1 is taken as ten digit of cube root.
∴ ∛4913 = 17
(iii) By grouping the digits, we get 12 and 167.
We know that, since, the unit digit of cube is 7, the unit digit of cube root is 3.
∴ 3 is the unit digit of the cube root of 12167 We know 23 = 8 and 33 = 27 , 8 > 12 > 27.
Thus, 2 is taken as ten digit of cube root.
∴ ∛12167= 23
(iv) By grouping the digits, we get 32 and 768.
We know that, since, the unit digit of cube is 8, the unit digit of cube root is 2.
∴ 2 is the unit digit of the cube root of 32768. We know 33 = 27 and 43 = 64 , 27 > 32 > 64.
Thus, 3 is taken as ten digit of cube root.
∴ ∛32768= 32
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# Brief Introduction to Vectors and Matrices
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1 CHAPTER 1 Brief Introduction to Vectors and Matrices In this chapter, we will discuss some needed concepts found in introductory course in linear algebra. We will introduce matrix, vector, vector-valued function, and linear independency of a group of vectors and vector-valued functions. 1. Vectors and Matrices A matrix is a group of numbers(elements) that are arranged in rows and columns. In general, an m n matrix is a rectangular array of mn numbers (or elements) arranged in m rows and n columns. If m n the matrix is called a square matrix. For example a 2 2 matrix is a11 a 12 and an 3 3 matrix is a 21 a 22 a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 Generally, we use bold phase letter, like A, to denote a matrix, and lower case letters with subscripts, like a ij, to denote element of a matrix. Here a ij would be the element at i th row and j th column. So a 11 is an element at 1 st row and column. Sometime we use the abbreviation A (a ij ) for a matrix with elements a ij Special matrices. 0 denotes the zero matrix whose elements are all zeroes. So 2 2 and 3 3 zero matrices are 0 0 and Another special matrix is the identity matrix, denoted by I, a identity matrix is an matrix whose main diagonal elements are 1, and all 1
2 2 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES other elements are 0. So 2 2 and 3 3 zero matrices are 1 0 and A vector is a matrix with one row or one column. In this chapter, a vector is always a matrix with one column as x1 x 2 for a two-dimensional vector and x 1 x 2 x 3 for a three dimensional vector. Here the element has only one index that denotes the row position (Sometimes we use different variable to denote number in different position such as using x y for a 2-dimensional vector). denote a vector. We use bold lower case, such as v, to 1.2. Operations on Matrices. Arrange number in rectangular fashion, as a matrix, itself is not something terribly interesting. The most important advantage from that kind arrangement is that we can define matrix addition, multiplication, and scalar multiplication. Definition 1.1. (i) Equality: Two matrix A (a ij ) and B (b ij ) are equal if corresponding elements are equal, i.e. a ij b ij. (ii) Addition: If A (a ij ) and B (b ij ) and the sum of Aand B is A + B (c ij ) a ij + b ij. (iii) Scalar Product: If A (a ij ) is matrix and k is number(scalar), the ka (ka ij ) is product of k and A. From the above definition, we see that, to multiply a matrix by a number k, we simply multiply each of its entries by k; to add two matrices we just add their corresponding entries; A B A+( 1)B. Example 1.1. Let A
3 and 1. VECTORS AND MATRICES 3 B find (a) A + B, (b) 3A, (c) 4A B. Solution (a) (b) (c) 4A B, A + B ( 4) A The following fact lists all properties of matrix addition and scalar product. Theorem 1.1. Let A, bb, and C be matrices. Let a, b be scalars (numbers). We have (1) A A A, A A 0; (2) A + B B + A (commutativity); (3) A + (B + C) (A + B) + C, (ab)a a(ba) (associativity); (4) a(a+b) aa+ab, (a+b)a aa+ba (distributivity) When we have a row vector and a column vector with the same number of elements, we can define the dot product as Definition 1.2. Dot Product: in 2-dimension: Let x y1 x 1 x 2 and y, the y 2 dot product of x and y is, x y x 1 y 1 + x 2 y 2
4 4 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES in 3-dimension: Let x x 1 x 2 x 3 and y the dot product of x and y is, y 1 y 2 y 3, x y x 1 y 1 + x 2 y 2 + x 3 y 3 Definition 1.3. Matrix product Let A (a ij ) and B (b ij ), if the number of columns of A is the same as number of rows of B, then the product of A and B is given by AB (c ij ) where c ij is dot product of i th row of A with j th column of B. and Example 1.2. Let find AB A B , Solution AB (0) + 3 (3) ( 4) ( 1) (0) ( 4) Notice, the first element of AB is 2 (0) + 3 (3) which is the dot product of first row of A, and first column of B, 3 The following fact gives properties of matrix product, Theorem 1.2. Let A, B, C be three matrices and r be a scalar, we have A(BC) (AB)C, r(ab) A(rB) (associativity) A(B + C) AB + AC (distributivity) Notice, in general AB BA, that is for most of the times, AB is not equal to BA. Using the matrix notation and matrix product, we can write the following system of equations { ax1 + bx 2 y 1 cx 1 + dx 2 y 2
5 as Ax y with x x1 x 2, and y 1. VECTORS AND MATRICES 5 a b A c d. y 2 y1 Definition 1.4. A square (ex. 2 2 or 3 3) matrix A is invertible if there is a matrix A -1 such that AA -1 A -1 A I. Theorem 1.3. Let, a b A c d be a 2 2 matrix, if A is invertible, we have A -1 1 d b ad bc c d So if A is invertible, to solve Ax y, we need to simply multiply both sides with A -1, that is x x A -1 y. Example 1.3. Solve the system of equation { 3x1 4x 2 2 2x 1 + 5x 2 7 Solution x1 equation is The equation can be rewrite Ax y with 3 4 A, x 2, and y Now the inverse of A is so the solution is A -1 x A -1 y 1 7. So in matrix form the system of x1 2 x (5) ( 2)( 4) ,
6 6 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES x Example 1.4. Solve the system of equation 3x 1 4x 2 + 5x 3 2 2x 1 + 5x 2 7 x 1 5x 2 + 8x 3 1 Solution x 1 x 2 x 3 The equation can be rewrite Ax y with A , 1 5 8, and y equation, Ax y, is So in matrix form the system of x 1 x 2 x It is a little harder to compute the inverse of a 3 3 matrix, we will use Mathcad to solve the equation. Here is how to do it, Type A:CtrlM at a blank area to bring up the matrix definition screen, put 3 in the both input boxes and click OK, you will get a 3 3 matrix place holder like A :. Fill the entries of A in the corresponding position, using Tab key to navigate among the place holders(or just click each one). Type b:ctrlm in another blank area, the matrix definition screen is up again. This time put 3 in the number of row box, and 1 in the number of column box and click OK. You will get b: put the values of y in the corresponding position. Type A^-1 *b you will get the solution, which is,
7 1. VECTORS AND MATRICES 7 Notice, by default, Mathcad will display the results as decimal, you can double click on the result vector to change it to fraction, after you double click the result you will have a popup menu such as Figure 1. Format Result The next example shows how we can determine the unknown constants typically found in the initial value problems of system of differential equations. Example 1.5. Let x 1 (t) C 1 e t + C 2 e 2t and x 2 2C 1 e t C 2 e 2t. If x 1 (0) 2, x 2 (0) 3, find C 1 and C 2 Solution From x 1 (t) C 1 e t + C 2 e 2t, set t 0 we have x 1 (0) C 1 e 0 + C 2 e 2(0) C 1 + C 2. Similarly, x 2 (t) 2C 1 e t C 2 e 2t, gives x 2 (0) 2C 1 e ( 0) C 2 e 2(0) 2C 1 C 2. Together with x 1 (0) 2, x 2 (0) 3 we have the following system of equations, { C1 + C 2 1 2C 1 C 2 3 Rewrite the equation in matrix form 1 1 C C 2 3 and using Mathcad we find the solution is 4 C1 3 C 2 1 3,
8 8 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES So x 1 (t) 4 3 et 1 3 e 2t and x et e 2t. They are solution of the following system of differential equations, { x 1 (t) x 1(t) + x 2 (t) x 2 (t) 2x 1 a b 1.3. Eigenvalues and Eigenvectors. If A the de- c d termined of A is defined as A a b c d ad bc. For a 3 3 matrix a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 we can compute the matrix as a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 a 11 a 22 a 23 a 32 a 33 a 12 a 21 a 23 a 31 a 33 +a 13 a 21 a 22 a 31 a 32 In Mathcad, type the vertical bar to bring up the absolute evaluator, put the matrix in the place holder and press to compute the determinant. The following screen shot shows an example,. Figure 2. Compute determinant in Mathcad The concepts of eigenvalue and eigenvector play an important role in find solutions to system of differential equations. Definition 1.5. We say λ is an eigenvalue of a matrix A (2 2 or 3 3) if the determinant A λi 0. An nonzero vector v is an eigenvector associated with λ if Remark 1.1. Av λv.
9 1. VECTORS AND MATRICES 9 - The above definition of eigenvector and eigenvalue is valid for any square matrix with n rows and columns. - p(λ) A λi is a polynomial of degree n for n n matrix A, which is called the characteristic polynomial of A. - If we view A as an transform that maps a vector x to Ax, an eigenvector v defines a straight line passing origin that is invariant under A. - If v is an eigenvector then for and number s 0, sv is also an eigenvector. This is especially useful when using Mathcad to get eigenvectors, the result of Mathcad might look bad, you might need to remove the common factor of the component of the vector to make it better. Computing eigenvalues and eigenvectors of a given matrix is quite tedious, Mathcad provides two functions eigenvals() and eigenvecs() to compute eigenvalues and eigenvectors of a matrix. In Mathcad, eigenvecs(m) Returns a matrix containing the eigenvectors. The nth column of the matrix returned is an eigenvector corresponding to the nth eigenvalue returned by eigenvals. The results of these functions by default is in decimal, you can change it by using simplify key word as shown in the following diagram. (a) Find eigenvalue (b) Find eigenvector Figure 3. Compute eigenvalue and eigenvector in Mathcad Notice, in the diagram, the eigenvalues are listed as vector and the eigenvectors are listed in a matrix 1+ 3 (8+2 3) (8+2 3) 1 2 -( 3-1) (8-2 3) (8-2 3) 1 2, 3 3
10 10 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES each column represents a eigenvector. Since multiplying an eigenvector by a nonzero constant you still get an eigenvector, so we can simplify the eigenvectors as v 1, and v Here is how to use Mathcad, Define the matrix by type A:CtrlM and specify the row and column number, fill the entries. type eigenvals(, you will get eigenvals( ) and in the place holder type A. Click at end of the eigenvals(a) and press ShiftCtrl., you will get eigenvals(a). In the place holder type in key word simple. And click any area outside the box to get result. Using the same procedure for find eigenvector using eigenvecs() function. 2. Vector-valued functions A vector-valued function over a, b is a function whose value is a vector or matrix. For example the following functions are vector-valued functions, t Example 2.1. (1) v(t) t 2 (2) x 1 t 2 (3) A(t) e t 1 t 3 4t sin(t) 2.1. Arithmetics of vector-valued function. To add two vector-valued function is to add their corresponding components. To multiply a vector-valued function by a scalar function to to multiply each entry by the scalar function. To multiply a vector(matrix) valued function to another vectorvalued function is same as multiply a matrix with a vector. The following example illustrate how to add/subtract two vector-valued functions and how to multiply a vector-valued function by a scalar function and how to apply a vector-valued function that is matrix to a vector value function.
11 2. VECTOR-VALUED FUNCTIONS 11 Example 2.2. Suppose v(t) A(t) t t 2 t sin(t) 1 t3 4t (a) Find v(t) + x(t); (b) Let f(t) e t, find f(t)x(t); (c) Find A(t)x(t), x 1 t 2 e t., and Solution (a) v(t) + x(t) (b) (c) t t 2 t t 2 e t f(t)x(t) e t 1 t 2 e t et t 2 e t e 2t A(t)x(t) 1 t3 4t sin(t) t2 (t 3 4t + 5) + e t 2t 2 + e t sin(t) 2 + sin(t) t + 1 2t 2 t e t ; 1 t 2 e t ; 2.2. derivative and integrations of vector-valued functions. A vector-valued function is continuous if each of its entries are continuous. A vector-valued function is differentiable if each of its entries are differentiable. If v(t) is an vector-valued function, then the derivative dv(t) dt v (t) of v(t) is a vector-valued function whose entries are the derivative of corresponding entries of v(t). That is to find derivative of a vector-valued function we just need to find derivative of each of its component.
12 12 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES The antiderivative v(t) dt of an vector-valued function v(t) is a vector-valued function whose entries are the antiderivative of corresponding entries of v(t). 3t Example 2.3. Find derivative of x(t) 2 5 sin(t) Solution x (t) dx(t) dt d dt 3t 2 5 sin(t) d dt (3t2 5) d (sin(t)) dt Example 2.4. Find antiderivative of x(t) 3t 2 5 sin(t) 6t cos(t) Solution 3t 2 5 x(t) dt sin(t) t 3 5t + C 1 cos(t) + C 2 (3t dt 2 5) dt sin(t) dt C1 + C 2 t 3 5t cos(t) Theorem 2.1. Suppose v(t), x(t), A(t) are differentiable vectorvalued functions (A(t) is matrix), and f(t) is differentiable scalar function. We have, (1) Sum and Difference rule: - v(t) ± x(t) v (t) ± x (t), - v(t) ± x(t) dt v(t) dt ± x(t) dt. (2) Product rule: - f(t)v(t) f (t)v(t) + f(t)v (t), - A(t)x(t) A (t)x(t) + A(t)x (t), Using Mathcad to find derivative or antiderivative of a vectorvalued function using Mathcad, you need to find derivative or antiderivative component wise as shown in the following screen shot,
13 2. VECTOR-VALUED FUNCTIONS 13 Figure 4. Differentiate and integrate vector-valued function
14 14 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES Notice: - Press Shift/to get the derivative operator and press CtrlI to get the antiderivative operator. - To get dx(t)simplif y you type dx(t) and press ShiftCtrl. and type the key word simplify in the place holder before. - To execute symbolically ( operator), just press Ctrl. 3. Linearly independency 3.1. Linearly independency of vectors. Let x 1, x 2,, x n be n vectors, C 1, C 2,, C n are n scalars(numbers), the expression C 1 x 1 + C 2 x C n x n is called a linear combination of vectors x 1, x 2,, x n. if Definition 3.1. n vectors x 1, x 2,, x n is linearly independent C 1 x 1 + C 2 x C n x n 0 leads to C 1 0, C 2 0,, C n 0. A set of vectors are linearly dependent if they are not linearly independent. If 0 is one of x 1, x 2,, x n, then they linearly dependent. Two nonzero vectors x and y are linearly dependent if and only if x sy for some s 0. n nonzero vectors are linearly independent if one can be represented as linear combination of the others. Any three or more 2-dimensional vectors (vectors with two entries) are linear dependent. Any four or more 3-dimensional(vectors with three entries) vectors are linear dependent. To determine if a given set of vectors are linearly independent, create a matrix so that the row of the matrix are given vectors. Using Mathcad function rref( ) to find the reduced echelon form of the matrix, if the result contains one or more rows that are entirely zero the vectors are linearly dependent, otherwise the vectors are linearly independent.
15 3. LINEARLY INDEPENDENCY 15 Example 3.1. For x , x , and x , we can form a matrix, A apply rref(type rref and in the place holder type A, then press ), rref(a) We see that the vectors are linearly dependent as the last row is entirely zero Linearly independency of functions. We can also define linearly independency for a group of functions over an given interval a, b. Let f 1, f 2,, f n be n functions defined over a, b, C 1, C 2,, C n are n scalars(numbers), the expression, C 1 f 1 + C 2 f C n f n is called a linear combination of functions f 1, f 2,, f n. Definition 3.2. n functions f 1, f 2,, f n is linearly independent over a, bif (1) C 1 f 1 + C 2 f C n f n 0 for all a t b leads to C 1 0, C 2 0,, C n 0. A set of function are linearly dependent if they are not linearly independent. If 0 function is one of f 1, f 2,, f n, then they linearly dependent. Two nonzero functions f(t) and g(t) are linearly dependent over a, b if and only if f(t) sg(t) for a constant s 0 and all a t b, for example, f(t) t and g(t) 4t are linearly dependent but f(t) t and g(t) 4t 2 are not, even f(0) 4g(0) and f(1) 4g(1). There are exists infinite many functions that are linearly independent. For example the set {1, t, t 2, t 3,, t n, } is a linearly independent set.
16 16 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES Here are some sets of linearly independent functions that we encounter in solving a system of differential equations, assume k 1, k 2,, k n are different numbers, - {t k 1, t k 2,, t k n }. - {e k 1t, e k 1t,, e k 1t }. - {sin(k 1 t), sin(k 2 t),, sin(k n t)}. - {cos(k 1 t), cos(k 2 t),, cos(k n t)}. - The mixing of above sets. - For each above set, when multiplying each element by an common nonzero factor, we get another linearly independent set. The following screen shot displays a heuristic Mathcad function that tries to determine if a given set of functions are linearly independent. Figure 5. Calculus tool bar One warning, the result of the program is not very reliable, the user should check the result manually to confirm the result. To manually check if an set of functions are linearly independent on a, b, one need to show that the only solutions are C 1 0, C 2 0,, C n 0. if equation (1) holds for all t in a, b, which requires strong algebraic skill. One method is to choose n different numbers {t 1, t 2,, t n } from a, b and using the functions to create an matrix, the compute the determinant of the matrix A (f i (t j )), if the determinant is not zero, the functions are linearly independent, but if the determinant is zero, it is inconclusive(most likely are linearly dependent). Example 3.2. Determine if f 1 (t) t 2 2t + 3, f 2 (t) 2t 2 5t 6, and f 3 (t) 5t 2 11t + 4 are linearly independent.
17 3. LINEARLY INDEPENDENCY 17 Solution Choose t 1 1, t 2 0, and t 3 1, f 1(t 1 ) f 1 (t 2 ) f 1 (t 3 ) f 2 (t 1 ) f 2 (t 2 ) f 2 (t 3 ) f 3 (t 1 ) f 3 (t 2 ) f 3 (t 3 ) Compute the determinant, , 2 so the functions are linearly independent. Project At beginning you should enter: Project title, your name, ss#, and due date in the following format Project One: Define and Graph Functions John Doe SS# Due: Mon. Nov. 23rd, 2003 You should format the text region so that the color of text is different than math expression. You can choose color for text from Format >Style select normal and click modify, then change the settings for font. You can do this for headings etc. (1) Independent of functions as vectors Goal: Familiar your self with the concept of linearly independency. Use the Mathcad code provided at at the website mzhan/linear to check if given set of functions are linearly independent or not. {sin(x), sin(2x), sin(3x)} {t 2, 2t 2 2t + 4, 3t, 6} {e t, te t, t 3 e t } {e 2t, e t, e 3t } Using algebraic arguments or reasoning to verify the conclusion of the Mathcad code.
18 18 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES (2) Condition Number In solving Ax b, one number is very important, it is called the condition number, which can be defined as C(A) s, where λ s is the eigenvalue with smallest l absolute value and lambda l is the eigenvalue with largest absolute value, if C(A) is too large or too small, a little change in b will result in a large in the solution x. We say the system Ax b is not stable. Now if A Find all eigenvalues, all eigenvectors, and C(A). Find solution of Ax b if b Change b a little to b 1 1 we get different solution, 1.1 which component of the new solution change most? The change of the third component if 10% what is the percentage change of the most changed component? Note: Our definition of condition number is not accurate, the true definition is C(A) where is a 1 A A -1 given norm (metric). Mathcad provides three functions cond1(a), cond2(a) and condi(a) in compute condition number for A in different metric
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# How do you prove sec(x) + 1 + ((1-tan^2(x)) / (sec(x)-1)) = cos(x)/(1-cos(x))?
Jul 10, 2016
Do some conjugate multiplication, make use of trig identities, and simplify. See below.
#### Explanation:
Recall the Pythagorean Identity ${\sin}^{2} x + {\cos}^{2} x = 1$. Divide both sides by ${\cos}^{2} x$:
$\frac{{\sin}^{2} x + {\cos}^{2} x}{\cos} ^ 2 x = \frac{1}{\cos} ^ 2 x$
$\to {\tan}^{2} x + 1 = {\sec}^{2} x$
We will be making use of this important identity.
Let's focus on this expression:
$\sec x + 1$
Note that this is equivalent to $\frac{\sec x + 1}{1}$. Multiply the top and bottom by $\sec x - 1$ (this technique is known as conjugate multiplication):
$\frac{\sec x + 1}{1} \cdot \frac{\sec x - 1}{\sec x - 1}$
$\to \frac{\left(\sec x + 1\right) \left(\sec x - 1\right)}{\sec x - 1}$
$\to \frac{{\sec}^{2} x - 1}{\sec x - 1}$
From ${\tan}^{2} x + 1 = {\sec}^{2} x$, we see that ${\tan}^{2} x = {\sec}^{2} x - 1$. Therefore, we can replace the numerator with ${\tan}^{2} x$:
$\frac{{\tan}^{2} x}{\sec x - 1}$
$\frac{{\tan}^{2} x}{\sec x - 1} + \frac{1 - {\tan}^{2} x}{\sec x - 1} = \cos \frac{x}{1 - \cos x}$
We have a common denominator, so we can add the fractions on the left hand side:
$\frac{{\tan}^{2} x}{\sec x - 1} + \frac{1 - {\tan}^{2} x}{\sec x - 1} = \cos \frac{x}{1 - \cos x}$
$\to \frac{{\tan}^{2} x + 1 - {\tan}^{2} x}{\sec x - 1} = \cos \frac{x}{1 - \cos x}$
The tangents cancel:
$\frac{\cancel{{\tan}^{2} x} + 1 - \cancel{{\tan}^{2} x}}{\sec x - 1} = \cos \frac{x}{1 - \cos x}$
Leaving us with:
$\frac{1}{\sec x - 1} = \cos \frac{x}{1 - \cos x}$
Since $\sec x = \frac{1}{\cos} x$, we can rewrite this as:
$\frac{1}{\frac{1}{\cos} x - 1} = \cos \frac{x}{1 - \cos x}$
Adding fractions in the denominator, we see:
$\frac{1}{\frac{1}{\cos} x - 1} = \cos \frac{x}{1 - \cos x}$
$\to \frac{1}{\frac{1}{\cos} x - \frac{\cos x}{\cos x}} = \cos \frac{x}{1 - \cos x}$
$\to \frac{1}{\frac{1 - \cos x}{\cos} x} = \cos \frac{x}{1 - \cos x}$
Using the property $\frac{1}{\frac{a}{b}} = \frac{b}{a}$, we have:
$\cos \frac{x}{1 - \cos x} = \cos \frac{x}{1 - \cos x}$
And that completes the proof.
Jul 11, 2016
$L H S = \left(\sec x + 1\right) + \frac{1 - {\tan}^{2} x}{\sec x - 1}$
$= \frac{\left(\sec x + 1\right) \left(\sec x - 1\right) + 1 - {\tan}^{2} x}{\sec x - 1}$
$= \frac{{\sec}^{2} x - 1 + 1 - {\tan}^{2} x}{\sec x - 1}$
$= \cos \frac{x}{\cos} x \cdot \frac{\left({\sec}^{2} x - {\tan}^{2} x\right)}{\left(\sec x - 1\right)}$
$\textcolor{red}{\text{putting} , {\sec}^{2} x - {\tan}^{2} x = 1}$
$= \cos \frac{x}{\cos x \sec x - \cos x}$
$\textcolor{red}{\text{putting} , \cos x \sec x = 1}$
$= \cos \frac{x}{1 - \cos x} = R H S$
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### 2: Quadratic Relations of the Form y = ax² + bx + c
#### 2.1: Overall Expectations
2.1.1: determine the basic properties of quadratic relations;
2.1.2: relate transformations of the graph of y = x² to the algebraic representation y = a(x - h)² + k;
2.1.4: solve problems involving quadratic relations.
#### 2.2: Investigating the Basic Properties of Quadratic Relations
2.2.1: collect data that can be represented as a quadratic relation, from experiments using appropriate equipment and technology (e.g., concrete materials, scientific probes, graphing calculators), or from secondary sources (e.g., the Internet, Statistics Canada); graph the data and draw a curve of best fit, if appropriate, with or without the use of technology (Sample problem: Make a 1 m ramp that makes a 15° angle with the floor. Place a can 30 cm up the ramp. Record the time it takes for the can to roll to the bottom. Repeat by placing the can 40 cm, 50 cm, and 60 cm up the ramp, and so on. Graph the data and draw the curve of best fit.);
2.2.2: determine, through investigation with and without the use of technology, that a quadratic relation of the form y = ax² + bx + c (a "not equal to" 0) can be graphically represented as a parabola, and that the table of values yields a constant second difference (Sample problem: Graph the relation y = x² - 4x by developing a table of values and plotting points. Observe the shape of the graph. Calculate first and second differences. Repeat for different quadratic relations. Describe your observations and make conclusions, using the appropriate terminology.);
2.2.3: identify the key features of a graph of a parabola (i.e., the equation of the axis of symmetry, the coordinates of the vertex, the y-intercept, the zeros, and the maximum or minimum value), and use the appropriate terminology to describe them;
#### 2.3: Relating the Graph of y = x² and Its Transformations
2.3.1: identify, through investigation using technology, the effect on the graph of y = x² of transformations (i.e., translations, reflections in the x-axis, vertical stretches or compressions) by considering separately each parameter a, h, and k [i.e., investigate the effect on the graph of y = x² of a, h, and k in y = x² + k, y = (x - h)², and y = ax²];
2.3.2: explain the roles of a, h, and k in y = a(x - h)² + k, using the appropriate terminology to describe the transformations, and identify the vertex and the equation of the axis of symmetry;
2.3.3: sketch, by hand, the graph of y = a(x - h)² + k by applying transformations to the graph of y = x² [Sample problem: Sketch the graph of y =- 1/2(x - 3)² + 4, and verify using technology.];
2.3.4: determine the equation, in the form y = a(x - h)² + k, of a given graph of a parabola.
2.4.1: expand and simplify second-degree polynomial expressions [e.g., (2x + 5)², (2x - y)(x + 3y)], using a variety of tools (e.g., algebra tiles, diagrams, computer algebra systems, paper and pencil) and strategies (e.g., patterning);
2.4.2: factor polynomial expressions involving common factors, trinomials, and differences of squares [e.g., 2x² + 4x, 2x - 2y + ax - ay, x² - x - 6, 2a² + 11a + 5, 4x² - 25], using a variety of tools (e.g., concrete materials, computer algebra systems, paper and pencil) and strategies (e.g., patterning);
2.4.3: determine, through investigation, and describe the connection between the factors of a quadratic expression and the x-intercepts (i.e., the zeros) of the graph of the corresponding quadratic relation, expressed in the form y = a(x - r)(x - s);
2.4.4: interpret real and non-real roots of quadratic equations, through investigation using graphing technology, and relate the roots to the x-intercepts of the corresponding relations;
2.4.5: express y = ax² + bx + c in the form y = a(x - h)² + k by completing the square in situations involving no fractions, using a variety of tools (e.g. concrete materials, diagrams, paper and pencil);
2.4.6: sketch or graph a quadratic relation whose equation is given in the form y = ax² + bx + c, using a variety of methods (e.g., sketching y = x² - 2x - 8 using intercepts and symmetry; sketching y = 3x² - 12x + 1 by completing the square and applying transformations; graphing h = -4.9t² + 50t + 1.5 using technology);
2.4.7: explore the algebraic development of the quadratic formula (e.g., given the algebraic development, connect the steps to a numerical example; follow a demonstration of the algebraic development [student reproduction of the development of the general case is not required]);
2.4.8: solve quadratic equations that have real roots, using a variety of methods (i.e., factoring, using the quadratic formula, graphing) (Sample problem: Solve x² + 10x + 16 = 0 by factoring, and verify algebraically. Solve x² + x - 4 = 0 using the quadratic formula, and verify graphically using technology. Solve -4.9t² + 50t + 1.5 = 0 by graphing h = -4.9t² + 50t + 1.5 using technology.).
#### 2.5: Solving Problems Involving Quadratic Relations
2.5.1: determine the zeros and the maximum or minimum value of a quadratic relation from its graph (i.e., using graphing calculators or graphing software) or from its defining equation (i.e., by applying algebraic techniques);
2.5.2: solve problems arising from a realistic situation represented by a graph or an equation of a quadratic relation, with and without the use of technology (e.g., given the graph or the equation of a quadratic relation representing the height of a ball over elapsed time, answer questions such as the following: What is the maximum height of the ball? After what length of time will the ball hit the ground? Over what time interval is the height of the ball greater than 3 m?).
### 3: Analytic Geometry
#### 3.1: Overall Expectations
3.1.3: verify geometric properties of triangles and quadrilaterals, using analytic geometry.
#### 3.2: Using Linear Systems to Solve Problems
3.2.1: solve systems of two linear equations involving two variables, using the algebraic method of substitution or elimination (Sample problem: Solve y = 1/2x - 5, 3x + 2y = -2 for x and y algebraically, and verify algebraically and graphically);
3.2.2: solve problems that arise from realistic situations described in words or represented by linear systems of two equations involving two variables, by choosing an appropriate algebraic or graphical method (Sample problem: The Robotics Club raised \$5000 to build a robot for a future competition. The club invested part of the money in an account that paid 4% annual interest, and the rest in a government bond that paid 3.5% simple interest per year. After one year, the club earned a total of \$190 in interest. How much was invested at each rate? Verify your result.).
#### 3.3: Solving Problems Involving Properties of Line Segments
3.3.2: develop the formula for the length of a line segment, and use this formula to solve problems (e.g., determine the lengths of the line segments joining the midpoints of the sides of a triangle, given the coordinates of the vertices of the triangle, and verify using dynamic geometry software);
3.3.3: develop the equation for a circle with centre (0, 0) and radius r, by applying the formula for the length of a line segment;
3.3.4: Determine the radius of a circle with centre (0, 0), given its equation; write the equation of a circle with centre (0, 0), given the radius; and sketch the circle, given the equation in the form x² + y² = r²;
3.3.5: solve problems involving the slope, length, and midpoint of a line segment (e.g., determine the equation of the right bisector of a line segment, given the coordinates of the endpoints; determine the distance from a given point to a line whose equation is given, and verify using dynamic geometry software).
#### 3.4: Using Analytic Geometry to Verify Geometric Properties
3.4.1: determine, through investigation (e.g., using dynamic geometry software, by paper folding), some characteristics and properties of geometric figures (e.g., medians in a triangle, similar figures constructed on the sides of a right triangle);
3.4.2: verify, using algebraic techniques and analytic geometry, some characteristics of geometric figures (e.g., verify that two lines are perpendicular, given the coordinates of two points on each line; verify, by determining side length, that a triangle is equilateral, given the coordinates of the vertices);
### 4: Trigonometry
#### 4.1: Overall Expectations
4.1.1: use their knowledge of ratio and proportion to investigate similar triangles and solve problems related to similarity;
4.1.2: solve problems involving right triangles, using the primary trigonometric ratios and the Pythagorean theorem;
#### 4.2: Investigating Similarity and Solving Problems Involving Similar Triangles
4.2.1: verify, through investigation (e.g., using dynamic geometry software, concrete materials), the properties of similar triangles (e.g., given similar triangles, verify the equality of corresponding angles and the proportionality of corresponding sides);
4.2.2: describe and compare the concepts of similarity and congruence;
4.2.3: solve problems involving similar triangles in realistic situations (e.g., shadows, reflections, scale models, surveying) (Sample problem: Use a metre stick to determine the height of a tree, by means of the similar triangles formed by the tree, the metre stick, and their shadows.).
#### 4.3: Solving Problems Involving the Trigonometry of Right Triangles
4.3.1: determine, through investigation (e.g., using dynamic geometry software, concrete materials), the relationship between the ratio of two sides in a right triangle and the ratio of the two corresponding sides in a similar right triangle, and define the sine, cosine, and tangent ratios (e.g., sin A = opposite/hypotenuse);
4.3.2: determine the measures of the sides and angles in right triangles, using the primary trigonometric ratios and the Pythagorean theorem;
4.3.3: solve problems involving the measures of sides and angles in right triangles in reallife applications (e.g., in surveying, in navigating, in determining the height of an inaccessible object around the school), using the primary trigonometric ratios and the Pythagorean theorem.
#### 4.4: Solving Problems Involving the Trigonometry of Acute Triangles
4.4.2: explore the development of the cosine law within acute triangles (e.g., use dynamic geometry software to verify the cosine law; follow the algebraic development of the cosine law and identify its relationship to the Pythagorean theorem and the cosine ratio [student reproduction of the development of the formula is not required]);
4.4.4: solve problems involving the measures of sides and angles in acute triangles.
Correlation last revised: 8/18/2015
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## Staging content lifeboat
### Course: Staging content lifeboat>Unit 14
Lesson 4: Fractions and whole numbers
# Whole numbers as fractions
Let's explore how to represent whole numbers as fractions. We shade circles to demonstrate 3/1 equals 3, and discuss how the fraction symbol represents division. We then illustrate the concept on a number line, connecting fractions to whole numbers. Created by Sal Khan.
## Want to join the conversation?
• What if I have 1 whole, and I have nothing shaded in? Will that just be 0/1?
• Yes. The whole represents the denominator. There are 0 parts, so 0 as the numerator. 0 over 1. 0/1. So it is definitely 0/1. Thanks for asking!
• Then how about 5/1 ? It is read as five one or five ones ?
• Five ones. This fraction means 5 wholes.
• can fractions have negative or decimal numbers inside?
• Yes! Fractions are essentially a form of division, so you can have a decimal or another fraction as the numerator or denominator.
• can a whole number be written as a fraction with a denominator of 1?
• Yes, all whole numbers can be written as a fraction with a denominator of 1.
• Seriously, where would this be used? Is the point something other than math. I understand whats happening here but how would this be used in the real world?
I'm a paranoid person and this question is a mental health tool.
• Oh I PROMISE YOU, you need to be able to do fractions
So they haven't explained it yet but fractions are actually just a different way of expressing division in equations
so the fraction 10/2 is literally 10 divided by 2
And if you did the division section you'll know that division helps in separating groups equally.
If you need to spilt 10k between 10 people how many will received money?
If I have 3 pencils and 3 classes how many pencils can I have for each class
Then when the number doesn't spilt evenly fractions are there to explain what part of a whole part is needed to be given
Once again like with money: If my business made \$47 today and I need to split it among 3 workers how much money does each worker get?
\$15.66 (not rounded because we don't have extra money to give to all three)
(1 vote)
• what if the number im trying to make into a fraction is 7 what would the answer be
• Any number is the same as itself over 1. So 7 is the same as 7/1.
Have a blessed, wonderful day!
• how do you shade in more then the denominator
(1 vote)
• Would it be one whole if it is 6/6 or 8/8 for an example?
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Lesson 3
# Lesson 3 - by an even number from this maximum So f x = 3 x...
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Finding Roots of Polynomials Let’s start with a quadratic: p q x 2 + r s x + t u = 0 with p,q,r,s,t,u all integers Then find a common denominator and a common factor and divide them out: commonfactor commondenominator ( ax 2 + x + c ) = 0 So we call f ( x ) = ax 2 + bx + c This gives a new problem, find roots of f ( x ) = 0 Now we use Descartes Rule of Signs : The number of sign changes in the coefficients of f ( x ) (meaning we list the coefficients from first to last and count how many times they change from positive to negative) tells us the maximum number of positive roots the polynomial has, and the number of sign changes in the coefficients of f ( - x ) gives us the maximum number of negative roots the polynomial has. Furthermore, the actual number of positive or negative roots will differ
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Unformatted text preview: by an even number from this maximum. So, f ( x ) = 3 x 2 + 2 x + 5 has no positive roots and has 2 or 0 negative roots. Vieta’s Formula says that real roots of f ( x ) must have a numerator which is a divisor of the last term c and a denominator which is a divisor of the first term a . Last but not least, all of the roots must add up to-b/a . This week, we will practice using these rules where the maximum number of roots is the number of roots and all roots happen once Test for a Root f ( x ) has a root a when f ( x ) = ( x-a ) g ( x ) or f ( a ) = 0 Then use long division to remove the root from the polynomial....
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## This note was uploaded on 03/22/2010 for the course MATH 1104 taught by Professor Unknown during the Fall '10 term at Carleton.
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# What is the power of a quotient property?
Dec 25, 2014
The Power of a Quotient Rule states that the power of a quotient is equal to the quotient obtained when the numerator and denominator are each raised to the indicated power separately, before the division is performed.
i.e.: ${\left(\frac{a}{b}\right)}^{n} = {a}^{n} / {b}^{n}$
For example:
${\left(\frac{3}{2}\right)}^{2} = {3}^{2} / {2}^{2} = \frac{9}{4}$
You can test this rule by using numbers that are easy to manipulate:
Consider: $\frac{4}{2}$ (ok it is equal to $2$ but for the moment let it stay as a fraction), and let us calculate it with our rule first:
${\left(\frac{4}{2}\right)}^{2} = {4}^{2} / {2}^{2} = \frac{16}{4} = 4$
Let us, now, solve the fraction first and then raise to the power of $2$:
${\left(\frac{4}{2}\right)}^{2} = {\left(2\right)}^{2} = 4$
This rule is particularly useful if you have more difficult problems such as an algebraic expression (with letters):
Consider: ${\left(\frac{x + 1}{4 x}\right)}^{2}$
You can now write:
${\left(\frac{x + 1}{4 x}\right)}^{2} = {\left(x + 1\right)}^{2} / {\left(4 x\right)}^{2} = \frac{{x}^{2} + 2 x + 1}{16 {x}^{2}}$
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# 20 Is What Percent of 52? Round Your Answer to the Nearest Tenth.
20 Is What Percent of 52? Round Your Answer to the Nearest Tenth.
When faced with the question “20 is what percent of 52?” it might seem daunting at first, especially if you haven’t dealt with percentages in a while. However, with a little bit of knowledge and some simple calculations, finding the answer becomes a breeze. In this article, we will break down the process of determining what percent 20 is of 52 and provide you with a clear answer. Additionally, we will address some frequently asked questions to further enhance your understanding of percentages.
Calculating the Percentage:
To find out what percent 20 is of 52, we need to divide 20 by 52 and then multiply the result by 100. This will give us the percentage value we are looking for. Let’s go through the steps:
Step 1: Divide 20 by 52
20 ÷ 52 = 0.3846 (rounded to four decimal places)
Step 2: Multiply the result by 100
0.3846 × 100 = 38.46 (rounded to two decimal places)
Therefore, 20 is approximately 38.5% of 52.
1. What is a percentage?
A percentage is a way of expressing a fraction or a ratio as a portion of 100. It represents a proportion or a part out of a whole.
2. How do I calculate a percentage?
To calculate a percentage, divide the part (20) by the whole (52) and multiply the result by 100. This will give you the percentage value.
3. How do I round to the nearest tenth?
To round to the nearest tenth, look at the digit in the hundredth place. If it is 5 or greater, round up; if it is less than 5, round down. In the case of 38.46, rounding to the nearest tenth gives us 38.5.
4. Can I use a calculator to find percentages?
Absolutely! Calculators can be a helpful tool when dealing with complex calculations. However, it’s always beneficial to understand the underlying concepts and be able to solve problems manually as well.
5. What are some real-life applications of percentages?
Percentages are widely used in various fields, including finance, statistics, and everyday situations. They are commonly used to calculate discounts, interest rates, tax rates, and analyze data in research or surveys.
6. How do I calculate a percentage increase or decrease?
To calculate a percentage increase or decrease, subtract the original value from the new value, divide the result by the original value, and multiply by 100. If the result is positive, it represents an increase, and if it is negative, it represents a decrease.
7. Are percentages always written as whole numbers?
No, percentages can be expressed as decimals or fractions as well. For example, 50% can also be written as 0.5 or 1/2.
Conclusion:
Determining what percent 20 is of 52 requires dividing 20 by 52 and then multiplying the result by 100. The answer, rounded to the nearest tenth, is approximately 38.5%. Percentages are an essential concept in various aspects of life, including finance, statistics, and everyday calculations. By understanding the fundamentals of percentages, you can confidently tackle similar problems and apply this knowledge to real-life situations.
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# Topic 1.1: Modulus Functions
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Category: Education
## Presentation Description
Absolute value, Solving modulus equations, modulus inequalities, quadratic modulus inequalities.
## Presentation Transcript
### Modulus functions :
Modulus functions |4| = |-4| = |a| = b => |x - a| = b => |a| = |b| => |x - a| = |b| => 4 4 a = b (x – a) = b
### Modulus inequalities :
Modulus inequalities |x| < 1 => |x| > 1 => |a| = b => + a = b -1 < x < 1 x < -1 or x > 1
### Quadratic inequalities :
Quadratic inequalities Quadratic inequality (x - a)(x – b) < 0 => where a is smaller than b (x – a)(x – b) > 0 = > a < x < b, x > b or x < a. < >
### Solve |x – 3| < 7 :
Solve |x – 3| < 7 +(x – 3) < 7 (or) –(x – 3) < 7 x < 10 -4 10 –x + 3 < 7 -x < 4 x > -4 Answer is -4 < x < 10.
### Solve |2x – 1| > 7 :
Solve |2x – 1| > 7 (or) +(2x – 1) > 7 2x > 7 + 1 2x > 8 x > 4 -(2x – 1) > 7 -2x + 1 > 7 -2x > 7 - 1 -2x > 6 x < -3 -3 4 Answer is x < -3 (or) x > 4
### Solve 3x + 1 > |x – 5| :
Solve 3x + 1 > |x – 5| (or) 3x + 1 > - (x – 5) 3x + 1 > - x + 5 4x > 4 x > 1 3x + 1 > +(x – 5) 3x + 1 > +x – 5 2x > - 6 x > - 3 0 2 CHECK FOR x = 0 3x + 1 > |x – 5| 3(0) + 1 > |0 – 5| 1 > 5 (FALSE) Since it’s wrong, therefore Reject x > -3 Therefore answer is x > 1
### Solve |x – 3| > 2x + 1 :
Solve |x – 3| > 2x + 1 +(x – 3) > 2x + 1 + x – 3 > 2x + 1 -x > 4 x < -4 (or) -(x – 3) > 2x + 1 - x + 3 > 2x + 1 -3x > - 2 x < -4 2 3 0 -5 CHECK FOR x = 0 |x – 3| > 2x + 1 |0 – 3| > 2(0) + 1 3 > 1 (TRUE) Since it’s correct, therefore x < 2 3
### Solve |2x - 3| < |x + 1| :
Solve |2x - 3| < |x + 1| |a| = |b| => <
### Solve |2x - 3| > |x + 1| :
Solve |2x - 3| > |x + 1| |a| = |b| => >
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Home | Teacher | Parents | Glossary | About Us
Study these two triangles. How are they alike? If we rotated the pink triangle and slid it on top of the green one, the two would fit on top of each other exactly. All the sides and all the angles are equal, so these two triangles are congruent.
If two figures are congruent, then their corresponding parts are congruent. Let's look at the corresponding parts of triangles ABC and DFE.
A triangle has three sides and three angles. If two triangles are congruent, then the sides and angles that match are called corresponding parts. Here, angle A corresponds to angle D, angle B corresponds to angle F, and angle C corresponds to angle E. Side AB corresponds to side DF, side BC corresponds to side FE, and side CA corresponds to side ED.
Congruent figures are named in the order of their corresponding parts. For these triangles, we say "triangle ABC is congruent to triangle DFE," not "triangle DEF," because vertex A corresponds to vertex D, vertex B corresponds to vertex F, and vertex C corresponds to vertex E.
The lines crossing the sides (sometimes called "hash marks") are another way to keep track of which side corresponds to another. The side with one hash mark in triangle ABC corresponds to the side with one hash mark in triangle DFE.
How can we tell if two triangles are congruent? Just "eyeballing" them is not good enough, because our eyes can play tricks on us. One way is to trace around one of the triangles and place it over the other. If the angles and sides match exactly, then the two are congruent. But there is another way to tell. If two triangles have congruent corresponding parts, then the triangles are congruent triangles.
You can measure the corresponding parts. If they are all congruent, then the triangles are congruent.
These principles and rules of congruence hold true for any polygon.
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Lost Car Key Puzzle
# Lost Car Key Puzzle
I drive to work, and put my car keys in my pocket. Whenever I sit down in a meeting there is a 20% chance that my car keys might fall out of my pocket (unnoticed). At lunchtime, I reach for my keys and find out they are missing. If I’ve had three sit-down meetings that morning, what is the chance that the car keys fell out in the first meeting location? (You can assume the only place the keys could have fallen out is in one of those three locations).
What is the chance the car keys fell out in the first meeting?
### Solution
A quick knee-jerk reaction might be to say “20%”, but that’s not correct. Yes, there is a 20% chance that they could fall out in any location, but we have the additional information that we know they have fallen out. What we need to calculate is that, knowing they have fallen out, the chance this happened in the first meeting.
There are lots of different ways of looking at this, but probably the easiest to understand is to draw a tree diagram showing all the possible states.
We know with 100% certainty that the keys were in my pocket when I arrived, and with 100% certainty they were missing at lunchtime. There is only one of three locations they could have been lost.
At the first meeting, there is a 1/5 chance (20%) they could have fallen out. This means there is a 4/5 chance they could still be in my pocket at the start of the second meeting.
At the second meeting, there is a the same 1/5 chance, so the probability they fell out there is 4/5 × 1/5 (They survived the first meeting, and then were lost in the second).
The chance they fell out in the third meeting is 1/5, multiplied by the chance they were still around at the start of the third meeting (4/5)2.
The three possible ways they could have fallen out are shown in yellow: 1/5, 4/25 and 16/125.
Putting these into a common denominator we get: 25/125, 20/125 and 16/125, and we know it's one of those. Put another way, if we repeated this day 125 times, we'd expect that I'd lose my keys 25+20+16 times, which is 61 times.
Of these 61 times, 25 of them would be from the first meeting, so the chance of me having lost them then is 25/(25+20+16) = 25/61 ≈ 41%.
Image: Matthew Ragan
The chance they fell out in the first meeting is ≈ 41%!
This is more than twice the chance that the naïve answer of 20% might suggest. I know it's a cliché that you end up finding something in the last place you look, but of course that's because you stop looking when you've found something!
What this math suggests, however, is that if you are looking for a lost item, rather than following the advice of 'retracing your steps' and going backwards from where you currently are, a better strategy is to go back to the place you last know you defintely had the item and go forwards from there.
### Bayes' Theorem
If you are familiar with Bayes' Theorem, an equivalent way of looking at this is to say that it's the probability of it falling out and it being in meeting room one out of (divided by) the probability of it falling out somewhere = (1/5) / (1-64/125).
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# The Holistic Approach to Absolute Values – Part V
We will continue our holistic approach to absolute values and add more complications to these types of questions. This article should set you up for any question of this kind. Note that this is a 750+ level concept, so if you are targeting a lower score, it may not be necessary for you to know.
(Before you continue reading, be sure to check out Part I, Part II, Part III and Part IV of this lesson.)
Let’s look at the following GMAT question:
For how many integer values of x, is |x – 6| > |3x + 6|?
(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite
In this question, we are given the inequality |x – 6| > 3*|x + 2|
Using the same logic as we did in the previous two posts, we will word the inequality like this: the distance from 6 should be more than three times the distance from -2.
At x = -2, the distance from 6 is 8 and the distance from -2 is 0. This means the distance from 6 is more than three times the distance from -2.
At x = -1, the distance from 6 is 7 and the distance from -2 is 1. Three times the distance from -2 is 3. This means the distance from 6 is more than three times the distance from -2.
At some point on the right of -1, the distance from 6 will be equal to three times the distance from -2. The distance between -2 and 6 is 8. If we split this 8 into 4 equal parts to get to x = 0, the distance from 6 will be equal to three times the distance from -2.
Now for every point to the right of 0, the distance from 6 will be less than three times the distance from -2.
Let’s try to go to the left of -2 instead. Will there be a point to the left of -2 where the distance from 6 will be equal to three times the distance from -2? Say that point is “a” units away from -2. -2 must then be 2a units away from 6 to ensure that 6 is a total of 3a units away from that point.
The distance between -2 and 6 is 8 – this 8 needs to be equal to 2a, so “a” must be 4 units.
The point where the distance from 6 will be equal to three times the distance from -2 will be 4 units to the left of -2, i.e. at -6. So at points to the right of -6 (but left of 0), the distance from 6 will be more than three times the distance from -2.
Note that for all values to the left of -6, the distance from 6 will be less than three times the distance from -2.
Hence, our x will lie in the range from -6 to 0.
-6 < x < 0
With these parameters, we will have 5 integer solutions: -5, -4, -3, -2 and -1. Hence, our answer is C.
Let’s look at a second question:
For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?
(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite
Now the true value of this method is visible, as we have three or more terms. The arduous algebra involved in this given inequality makes our logical approach much more attractive.
First note that we have the term |5 – x|. This is the same as |x – 5| because |x| = |-x|.
We will word the inequality like this: the distance from 5 + the distance from 8 should be greater than the distance from -7.
Let’s find the point where the sum of the distance from 5 and the distance from 8 is equal to distance from -7. Say that point is “a” units to the left of 5.
a + a + 3 = 12 – a
a = 3
So the point is 3 units to the left of 5, which means it is at 2. For all points to the left of 2, the sum of the distance from 5 and the distance from 8 will be greater than the distance from -7.
How about the points that are to the right of 8? Say there is a point “b” units away from 8 where the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.
3 + b + b = 15 + b
b = 12
So if we go 12 units to the right of 8, i.e. at x = 20, the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.
For all points to the right of 20, the sum of the distance from 5 and the distance from 8 is greater than the distance from -7, so there will be infinite points for which the sum of the distance from 5 and the distance from 8 is greater than the distance from -7. Therefore, our answer is E.
Using this concept, try to answer the following question on your own: For how many integer values of x, is |x – 6| – |3x + 6| > 0?
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# The Holistic Approach to Absolute Values – Part IV
Last week, we looked at some absolute value questions involving inequalities. Today, we’ll continue this discussion by adding some more complications to our questions. Consider the question: What is the minimum value of the expression |x – 3| + |x + 1| + |x|? Technically, |x – 3| + |x + 1| + |x| is the sum of “the distance of x from 3,” “the distance of x from -1” and “the distance of x from 0.” To make solving such questions simpler, we’ll often use a parallel situation:
Imagine that there are 3 friends with houses at points -1, 0 and 3 in a straight line. They decide to meet at the point x.
• |x – 3| will be the distance covered by the friend at 3 to reach x.
• |x + 1| will be the distance covered by the friend at -1 to reach x.
• |x| will be the distance covered by the friend at 0 to reach x.
So, the total distance the friends will cover to meet at x will be |x – 3| + |x + 1| + |x|.
Now we can choose to minimize this total distance, bring it to some particular value or make it more or less than some particular value.
If we want to minimize the total distance, we just make the friends meet at the second guy’s house, i.e. at the point 0. The friend at 3 and the friend at -1 need to travel 4 units total to meet anyway, so there’s no point in making the guy at 0 travel any distance at all. So the minimum total distance would be 4, which would then be the minimum value of |x – 3| + |x + 1| + |x|. This minimum value is given by the expression at x = 0.
With this in mind, when we move to the right or to the left of x = 0, the total distance will increase and, hence, the value of the expression |x – 3| + |x + 1| + |x| will also increase.
Thereafter, it is easy to solve for |x – 3| + |x + 1| + |x| = 10 or |x – 3| + |x + 1| + |x| < 10, etc., as seen in our previous post.
Today, let’s look at how to solve a more advanced GMAT question using the same logic:
For how many integer values of x, is |x – 5| + |x + 1| + |x| + |x – 7| < 15?
(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite
In our parallel situation of friends and houses, we now have 4 friends with houses at points -1, 0, 5 and 7.
The friends at -1 and 7 are 8 units apart, so they will need to cover at least this total distance together to meet. It doesn’t matter where they meet between -1 and 7 (inclusive), they will need to cover exactly 8 units.
The friends at 0 and 5 will need to travel a minimum distance of 5 to meet. They can meet anywhere between 0 and 5 (inclusive) and the distance they will cover will still be 5.
So, all four friends can meet anywhere between 0 and 5 (inclusive) and the total distance covered will be 8 + 5 = 13. This would be the minimum total distance, and hence, the minimum value of the expression |x – 5| + |x + 1| + |x| + |x – 7|.
When we move to the left of 0 or to the right of 5, the total distance covered will be more than 13. At any point between -1 and 7, the total distance covered by the friends at -1 and 7 will be only 8. When we move 1 unit to the left of 0 and reach -1, the total distance covered by the friends at 0 and 5 will be 1 + 6 = 7. So to meet at -1, the total distance traveled by all friends together will be 8 + 7 = 15.
Similarly, when we move 1 unit to the right of 5 and reach 6, the total distance covered by the four friends will be again 8 + 7 = 15. So at points x = -1 and x = 6, the value of the expression will be 15. Between these two points (excluding the points themselves), the value of the expression will be less than 15.
So now we know -1 < x < 6. With these parameters, x can take 6 integer values: 0, 1, 2, 3, 4, 5. Therefore, the answer is D.
Note that when we had 3 points on the number line, the minimum total distance was found at the second point. Now when we have 4 points on the number line, the minimum total distance has been found to be in the range between second and third points.
Let’s look at another question:
For how many integer values of x, is |2x – 5| + |x + 1| + |x| < 10?
(A) 1
(B) 2
(C) 4
(D) 5
(E) Infinite
|2x – 5| + |x + 1| + |x| < 10
2*|x – 5/2| + |x + 1| + |x| < 10
In this sum, now the distance from 5/2 is added twice.
In our parallel situation, this is equivalent to two friends living at 5/2, one living at 0 and one living at -1. Now note that the expression may not take the minimum value of x = 0 because there are 2 people who will need to travel from 5/2.
We have four friends in all, so we can expect to get a range in which we will get the minimum value of the expression. The second and third people are at 0 and 5/2, respectively.
The total distance at x = 0 will be 1 + 2*(5/2) = 6. The total distance at x = 5/2 will be 7/2 + 5/2 = 6.
So if we move to the left of 0 or to the right of 5/2, the total distance will increase. If we move 1 unit to the right of 5/2 and reach 7/2, the total distance covered by the four friends will be 9/2 + 7/2 + 2 = 10. If we move 1 unit to the left of 0 and reach -1, the total distance covered by the four friends will be 0 + 1 + 2*(7/2) = 8. Now all four friends are at -1. To cover a distance of another 2, they should move another 0.5 units to the left of -1 to reach -1.5.
Now the total distance covered by the four friends will be 0.5 + 1.5 + 2*4 = 10, so the total distance when x lies between the points -1.5 and 3.5 (excluding the points themselves) will be less than 10.
Now we know -1.5 < x < 3.5. With these parameters, x can take 5 integer values: -1, 0, 1, 2 and 3. Therefore, the answer is D.
Now use these concepts to solve the following question: For how many integer values of x, is |3x – 3| + |2x + 8| < 15?
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# The Patterns to Solve GMAT Questions with Reversed-Digit Numbers – Part II
, I wrote about the GMAT’s tendency to ask questions regarding the number properties of two two-digit numbers whose tens and units digits have been reversed.
The biggest takeaways from that post were:
1. Anytime we add two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 11.
2. Anytime we take the difference of two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 9.
For the hardest GMAT questions, we’re typically mixing and matching different types of number properties and strategies, so it can be instructive to see how the above axioms might be incorporated into such problems.
Take this challenging Data Sufficiency question, for instance:
When the digits of two-digit, positive integer M are reversed, the result is the two-digit, positive integer N. If M > N, what is the value of M?
(1) The integer (M –N) has 12 unique factors.
(2) The integer (M –N) is a multiple of 9.
The average test-taker looks at Statement 1, sees that it will be very difficult to simply pick numbers that satisfy this condition, and concludes that this can’t possibly be enough information. Well, the average test-taker also scores in the mid-500’s, so that’s not how we want to think.
First, let’s concede that Statement 1 is a challenging one to evaluate and look at Statement 2 first. Notice that Statement 2 tells us something we already know – as we saw above, anytime you have two two-digit numbers whose tens and units digits are reversed, the difference will be a multiple of 9. If Statement 2 is useless, we can immediately prune our decision tree of possible correct answers. Either Statement 1 alone is sufficient, or the statements together are not sufficient, as Statement 2 will contribute nothing. So right off the bat, the only possible correct answers are A and E.
If we had to guess, and we recognize that the average test-taker would likely conclude that Statement 1 couldn’t be sufficient, we’d want to go in the opposite direction – this question is significantly more difficult (and interesting) if it turns out that Statement 1 gives us considerably more information than it initially seems.
In order to evaluate Statement 1, it’s helpful to understand the following shortcut for how to determine the total number of factors for a given number. Say, for example, that we wished to determine how many factors 1000 has. We could, if we were sufficiently masochistic, simply list them out (1 and 1000, 2 and 500, etc.). But you can see that this process would be very difficult and time-consuming.
Alternatively, we could do the following. First, take the prime factorization of 1000. 1000 = 10^3, so the prime factorization is 2^3 * 5^3. Next, we take the exponent of each prime base and add one to it. Last, we multiply the results. (3+1)*(3+1) = 16, so 1000 has 16 total factors. More abstractly, if your number is x^a * y^b, where x and y are prime numbers, you can find the total number of factors by multiplying (a+1)(b+1).
Now let’s apply this process to Statement 1. Imagine that the difference of M and N comes out to some two-digit number that can be expressed as x^a * y^b. If we have a total of 12 factors, then we know that (a+1)(b+1) = 12. So, for example, it would work if a = 3 and b = 2, as a + 1 = 4 and b + 1 = 3, and 4*3 =12. But it would also work if, say, a = 5 and b = 1, as a + 1 = 6 and b + 1 = 2, and 6*2 = 12. So, let’s list out some numbers that have 12 factors:
1. 2^3 * 3^2 (3+1)(2+1) = 12
2. 2^5 * 3^1 (5+1)(1+1) = 12
3. 2^2 * 3^3 (2+1)(3+1) = 12
Now remember that M – N, by definition, is a multiple of 9, which will have at least 3^2 in its prime factorization. So the second option is no longer a candidate, as its prime factorization contains only one 3. Also recall that we’re talking about the difference of two two-digit numbers. 2^2 * 3^3 is 4*27 or 108. But the difference between two positive two-digit numbers can’t possibly be a three-digit number! So the third option is also out.
The only possibility is the first option. If we know that the difference of the two numbers is 2^3 * 3^2, or 8*9 = 72, then only 91 and 19 will work. So Statement 1 alone is sufficient to answer this question, and the answer is A.
Algebraically, if M = 10x + y, then N = 10y + x.
M – N = (10x + y) – (10y + x) = 9x – 9y = 9(x – y).
If 9(x – y) = 72, then x – y = 8. If the difference between the tens and units digits is 8, the numbers must be 91 and 19.
Takeaway: the hardest GMAT questions will require a balance of strategy and knowledge. In this case, we want to remember the following:
• Anytime we take the difference of two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 9.
• If one statement is easier to evaluate than the other, tackle the easier one first. If it’s the case that one statement gives you absolutely nothing, and the other is complex, there is a general tendency for the complex statement alone to be sufficient.
• For the number x^a * y^b, where x and y are prime numbers, you can find the total number of factors by multiplying (a+1)(b+1).
By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.
# The Holistic Approach to Absolute Values – Part III
A while back, we discussed some holistic approaches to answering absolute value questions. Today, we will enhance our understanding of absolute values with some variations that you might see on the GMAT.
Instead of looking at how to solve equations, like we did in our previous post, we will look at how to solve inequalities using the same concept.
A quick review:
• |x| = The distance of x from 0 on the number line. For example, if |x| = 4, x is 4 away from 0. So x can be 4 or -4.
• |x – 1| = The distance of x from 1 on the number line. For example, if |x – 1| = 4, x is 4 away from 1. so x can be 5 or -3.
• |x| + |x – 1| = The sum of distance of x from 0 and distance of x from 1 on the number line. for example, if x = 5, the distance of x from 0 is 5 and the distance of x from 1 is 4. The sum of the distances is 5 + 4 = 9. So |x| + |x – 1| = 5 + 4 = 9.
Let’s move ahead now and see how we can use these concepts to solve inequalities:
For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?
(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite
In the previous post, we saw the a similar question, except it involved an equation rather than an inequality. For that problem, we found that the two points where the total distance is equal to 10 are -2.667 and 4:
What will be the total distance at any value of x between these two points?
Say, x = 0
|x – 3| + |x + 1| + |x|
= 3 + 1 + 0
= 4
Say, x = 3
|x – 3| + |x + 1| + |x|
0 + 4 + 3
= 7
In both cases, we see that the total distance covered is less than 10. Note that the minimum distance covered will be 4 at x = 0 (discussed in the previous post) so by moving to the right of 0 or to the left of 0 on the number line, we get to the points where the distance increases to 10. So for every point in between, the total distance will be less than 10 (the entire red region).
Hence, at integer points x = -2, -1, 0, 1, 2 and 3 (which are all between -2.667 and 4), the total distance will be less than 10. The total distance will be less than 10 for all non-integer points lying between -2.667 and 4 too, but the question only asks for the integer values, so that is all we need to focus on. (Of course, there are infinite non-integer points between any two distinct points on the number line.) Hence, the answer will be 6 points, or D.
Along the same lines, consider a slight variation of this question:
For how many integer values of x, is |x – 3| + |x + 1| + |x| > 10?
(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite
What will the answer be here? We hope you immediately jumped to answer choice E – for every integer value of x to the right of 4 or to the left of -2.667, the total distance will be more than 10 (the blue regions). So there will be infinite such integer points (all integers greater than 4 or less than -2.667). Thus, the answer is E.
We hope this logic is clear. We will look at some other variations of this concept next week!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# Data Sufficiency Questions: How to Know When Both Statements Together Are Not Sufficient
Today we will discuss a problem we sometimes face while attempting to solve Data Sufficiency questions for which the answer is actually E (when both statements together are not sufficient to answer the question). Ideally, we would like to find two possible answers to the question asked so that we know that the data of both statements is not sufficient to give us a unique answer. But what happens when it is not very intuitive or easy to get these two distinct cases?
Let’s try to answer these questions in today’s post using using one of our own Data Sufficiency questions.
A certain car rental agency rented 25 vehicles yesterday, each of which was either a compact car or a luxury car. How many compact cars did the agency rent yesterday?
(1) The daily rental rate for a luxury car was \$15 higher than the rate for a compact car.
(2) The total rental rates for luxury cars was \$105 higher than the total rental rates for compact cars yesterday
We know from the question stem that the total number of cars rented is 25. Now we must find how many compact cars were rented.
There are four variables to consider here:
1. Number of compact cars rented (this is what we need to find)
2. Number of luxury cars rented
3. Daily rental rate of compact cars
4. Daily rental rate of luxury cars
Let’s examine the information given to us by the statements:
Statement 1: The daily rental rate for a luxury car was \$15 higher than the rate for a compact car.
This statement gives us the difference in the daily rental rates of a luxury car vs. a compact car. Other than that, we still only know that a total of 25 cars were rented. We have no data points to calculate the number of compact cars rented, thus, this statement alone is not sufficient. Let’s look at Statement 2:
Statement 2: The total rental rates for luxury cars was \$105 higher than the total rental rates for compact cars yesterday.
This statement gives us the difference in the total rental rates of luxury cars vs. compact cars (we do not know the daily rental rates). Again, we have no data points to calculate the number of compact cars rented, thus, this statement alone is also not sufficient.
Now, let’s try to tackle both statements together:
The daily rate for luxury cars is \$15 higher than it is for compact cars, and the total rental rates for luxury cars is \$105 higher than it is for compact cars. What constitutes this \$105? It is the higher rental cost of each luxury car (the extra \$15) plus adjustments for the rent of extra/fewer luxury cars hired. That is, if n compact cars were rented and n luxury cars were rented, the extra total rental will be 15n. But if more luxury cars were rented, 105 would account for the \$15 higher rent of each luxury car and also for the rent of the extra luxury cars.
Event with this information, we still should not be able to find the number of compact cars rented. Let’s find 2 cases to ensure that answer to this question is indeed E – the first one is quite easy.
The total extra money collected by renting luxury cars is \$105.
105/15 = 7
Say out of 25 cars, 7 are luxury cars and 18 are compact cars. If the rent of compact cars is \$0 (theoretically), the rent of luxury cars is \$15 and the extra rent charged will be \$105 (7*15 = 105) – this is a valid case.
Now how do we get the second case? Think about it before you read on – it will help you realize why the second case is more of a challenge.
Let’s make a slight change to our current numbers to see if they still fit:
Say out of 25 cars, 8 are luxury cars and 17 are compact cars. If the rent of compact cars is \$0 and the rent of luxury cars is \$15, the extra rent charged should be \$15*8 = \$120, but notice, 9 morecompact cars were rented than luxury cars. In reality, the extra total rent collected is \$105 – the \$15 reduction is because of the 9 additional compact cars. Hence, the daily rental rate of each compact car would be \$15/9 = \$5/3.
This would mean that the daily rental rate of each luxury car is \$5/3 + \$15 = \$50/3
The total rental cost of luxury cars in this case would be 8 * \$50/3 = \$400/3
The total rental cost of compact cars in this case would be 17 * \$5/3 = \$85/3
The difference between the two total rental costs is \$400/3 – \$85/3 = 315/3 = \$105
Everything checks out, so we know that there is no unique answer to this question – for any number of compact cars you use, you will come up with the same answer. Thus, Statements 1 and 2 together are not sufficient.
The strategy we used to find this second case to test is that we tweaked the numbers we were given a little and then looked for a solution. Another strategy is to try plugging in some easy numbers. For example:
Instead of using such difficult numbers, we could have tried an easier split of the cars. Say out of 25 cars, 10 are luxury and 15 are compact. If the rent of compact cars is \$0 and the rent of luxury cars is \$15, the extra rent charged should be 10*\$15 = \$150 extra, but it is actually only \$105 extra, a difference of \$45, due to the 5 additional compact cars. The daily rental rent of 5 extra compact cars would be \$45/5 = \$9. Using these numbers in the calculations above, you will see that the difference between the rental costs is, again, \$105. This is a valid case, too.
Hence, there are two strategies we saw in action today:
• Tweak the numbers slightly to see if you will get the same results
• Go for the easy split when choosing numbers to plug in
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# GMAT Tip of the Week: Taking The GMAT Like It’s Nintendo Switch
The non-election trending story of the day is the announcement of the forthcoming Nintendo Switch gaming system, a system that promises to help you take the utmost advantage of your leisure time…but that may help you maximize the value of your GMAT experience, too.
How?
The main feature of the Switch (and the driving factor behind its name) is its flexibility. It can be an in-home gaming system attached to a fixed TV set, but then immediately Switch to a hand-held portable system that allows you to continue your game on the go. Nintendo’s business plan is primarily based on offering flexibility…and on the GMAT, your plan should be to prove to business schools that you can offer the same.
The GMAT, of course, tests algebra skills and critical thinking skills and grammar skills, but beneath the surface it also has a preference for testing flexibility. Many problems will punish those with pure tunnel vision, but reward those who can identify that their first course of action isn’t working and who can then Switch to another plan. This often manifests itself in:
• Math problems that seem to require algebra…but halfway through beg to be back-solved using answer choices.
• Sentence Correction problems that seem to ask you to make a decision about one major difference…but for which the natural choices leave you with clearer-cut errors elsewhere.
• Critical Reasoning answer choices that seem out of scope at first, but reward those who read farther and then see their relevance.
• Data Sufficiency problems for which you’ve made a clear, confident decision on one statement…but then the other statement shows you something you hadn’t considered before and forces you to reconsider.
• The overall concept that if you’re a one-trick pony – you’re a master of plugging in answer choices, for example – you’ll find questions that just won’t reward that strategy and will force you to do something else.
Flexibility matters on the GMAT! As an example, consider the following Data Sufficiency question:
Is x/y > 3?
1) 3x > 9y
2) y > 3y
If you’re like many, you’ll confidently address the algebra in Statement 1, divide both sides by 3 to get x > 3y, and then see that if you divide both sides by y, you can make it look exactly like the question stem: x/y > 3. And you may very well say, “Statement 1 is sufficient!” and confidently move on to Statement 2.
But when you look at Statement 2 – either conceptually or algebraically – something should stand out. For one, there’s no way that it’s sufficient because it doesn’t help you determine anything about x. And secondly, it brings up the point that “y is negative” (algebraically you’d subtract y from both sides to get 0 > 2y, then divide by 2 to get 0 > y). And here’s where, if it hasn’t already, your mind should Switch to “positive/negative number properties” mode. If you weren’t thinking about positive vs. negative properties when you considered Statement 1, this one gives you a chance to Switch your thinking and reconsider – what if y were negative? Algebraically, you’d then have to flip the sign when you divide both sides by y:
3x > 9y : Divide both sides by 3
x > 3y : Now divide both sides by y, but remember that if y is positive you keep the sign (x/y > 3), and if y is negative you flip the sign (x/y < 3).
With this in mind, Statement 1 doesn’t really tell you anything. x/y can be greater than 3 or less than 3, so all Statement 1 does is eliminate that x/y could be exactly 3. Now you have the evidence to Switch your answer. If you initially thought Statement 1 was sufficient, Statement 2 has given you a chance to reassess (thereby demonstrating flexibility in thinking) and realize that it’s not, until you know whether y is negative or positive.
Statement 2 supplies that missing piece, and the answer is thus C. But more important is the lesson – because the GMAT so values mental flexibility, it will often provide you with clues that can help you change your mind if you’re paying attention. So on the GMAT, take a lesson from Nintendo Switch: flexibility is an incredibly marketable skill, so look for clues and opportunities to Switch your line of thinking and save yourself from trap answers.
By Brian Galvin.
# How to Solve “Unsolvable” Equations on the GMAT
The moment we see an equation involving the variable x, we have a habit of jumping right into attempting to solve it. But what happens when we are not able to solve it? Let’s say, for example, we have an equation such as x^2 + 1 = 0. How would we solve for x here? We can’t because x has no real value. Note that x^2 is non-negative – it would be either 0 or positive. 1, we know, is positive. So together, a positive number and a non-negative number cannot add up to 0.
In this example, it relatively easy to see that the equation has no real solution. In others, it may not be so obvious, so we will need to use other strategies.
We know how to solve third degree equations. The first solution is found by trial and error – we try simple values such as -2, -1, 0, 1, 2, etc. and are usually able to find the first solution. Then the equation of third degree is split into two factors, including a quadratic. We know how to solve a quadratic, and that is how we get all three solutions, if it has any.
But what if we are unable to find the first solution to a third degree equation by trial and error? Then we should force ourselves to wonder if we even need to solve the equation at all. Let’s take a look at a sample question to better understand this idea:
Is x < 0?
(1) x^3 + x^2 + x + 2 = 0
(2) x^2 – x – 2 < 0
In this problem, x can be any real number – we have no constraints on it. Now, is x negative?
Statement 1: x^3 + x^2 + x + 2 = 0
If we try to solve this equation as we are used to doing, look at what happens:
If you plug in x = 2, you get 16 = 0
If you plug in x = 1, you get 5 = 0
If you plug in x = 0, you get 2 = 0
If you plug in x = -1, you get 1 = 0
If you plug in x = -2, you get -4 = 0
We did not find any root for the equation. What should we do now? Note that when x goes from -1 to -2, the value on the left hand side changes from 1 to -4, i.e. from a positive to a negative. So, in between -1 and -2 there will be some value of x for which the left hand side will become 0. That value of x will not be an integer, but some decimal value such as -1.3 or -1.4, etc.
Even after we find the first root, making the quadratic will be very tricky and then solving it will be another uphill task. So we should ask ourselves whether we even need to solve this equation.
Think about it – can x be positive? If x is indeed positive, x^3, x^2 and x all will be positive. Then, if we add four positive numbers (x^3, x^2, x and 2) we will get a positive sum – we cannot get 0. Obviously x cannot be 0 since that will give us 2 = 0.
This means the value of x must be negative, but what it is exactly doesn’t matter. We know that x has to be negative, and that is sufficient to answer the question.
Statement 2: x^2 – x – 2 < 0
This, we can easily solve:
x^2 – 2x + x – 2 < 0
(x – 2)*(x + 1) < 0
We know how to solve this inequality using the method discussed here.
This this will give us -1 < x < 2.
Since x can be a non-integer value too, x can be negative, 0, or positive. This statement alone is not sufficient,and therefore, the answer is A.
To evaluate Statement 1, we didn’t need to solve the equation at all. We figured out everything we wanted to know by simply using some logic.
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# Quarter Wit, Quarter Wisdom: A GMAT Quant Question That Troubles Many!
What determines whether or not a question can be considered a GMAT question? We know that GMAT questions that are based on seemingly basic concepts can be camouflaged such that they may “appear” to be very hard. Is it true that a question requiring a lot of intricate calculations will not be tested in GMAT? Yes, however it is certainly possible that a question may “appear” to involve a lot of calculations, but can actually be solved without any!
In the same way, it is possible that a question may appear to be testing very obscure concepts, while it is really solvable by using only basic ones.
This happens with one of our own practice questions – we have often heard students exclaim that this problem isn’t relevant to the GMAT since it “tests an obscure number property”. It is a question that troubles many people, so we decided to tackle it in today’s post.
We can easily solve this problem with just some algebraic manipulation, without needing to know any obscure properties! Let’s take a look:
† and ¥ represent non-zero digits, and (†¥)² – (¥†)² is a perfect square. What is that perfect square?
(A) 121
(B) 361
(C) 576
(D) 961
(E) 1,089
The symbols † and ¥ are confusing to work with, so the first thing we will do is replace them with the variables A and B.
The question then becomes: A and B represent non-zero digits, and (AB)² – (BA)² is a perfect square. What is that perfect square?
As I mentioned before, we have heard students complain that this question isn’t relevant to the GMAT because it “uses an obscure number property”. Now here’s the thing – most advanced number property questions CAN be solved in a jiffy using some obscure number property such as, “If you multiply a positive integer by its 22nd multiple, the product will be divisible by …” etc. However, those questions are not actually about recalling these so-called “properties” – they are about figuring out the properties using some generic technique, such as pattern recognition.
For this question, the complaint is often that is that the question tests the property, “(x + y)*(x – y) (where x and y are two digit mirror image positive integers) is a multiple of 11 and 9.” It doesn’t! Here is how we should solve this problem, instead:
Given the term (AB)^2, where A and B are digits, how will you square this while keeping the variables A and B?
Let’s convert (AB)^2 to (10A + B)^2, because A is simply the placeholder for the tens digit of the number. If you are not sure about this, consider the following:
58 = 50 + 8 = 10*5 + 8
27 = 20 + 7 = 10*2 + 7
…etc.
Along those same lines:
AB = 10A + B
BA = 10B + A
Going back to our original question:
(AB)^2 – (BA)^2
= (10A + B)^2 – (10B + A)^2
= (10A)^2 + B^2 + 2*10A*B – (10B)^2 – A^2 – 2*10B*A
= 99A^2 – 99B^2
= 9*11*(A^2 – B^2)
We know now that the expression is a multiple of 9 and 11. We would not have known this beforehand. Now we’ll just use the answer choices to figure out the solution. Only 1,089 is a multiple of both 9 and 11, so the answer must be E.
We hope you see that this question is not as hard as it seems. Don’t get bogged down by unknown symbols – just focus on the next logical step at each stage of the problem.
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# GMAT Tip of the Week: As You Debate Over Answer Choices… Just Answer The Freaking Question!
If you’re like many – to the dismay of the NFL and the advertising industry – you’re planning to watch another presidential debate this coming Sunday. And just like Trump-Clinton I and Pence-Kaine earlier this week, this debate will provide plenty of opportunities to be annoyed, frustrated, and disappointed…but it will also provide an ever-important lesson about the GMAT.
It’s no surprise that candidate approval ratings are low for the same reason that far too many GMAT scores are lower than candidates would hope. Why?
People don’t directly answer the question.
This is incredibly common in the debates, where the poor moderators are helpless against the talking points and stump speeches of the candidates. The public then suffers because people cannot get direct answers to the questions that matter. This is also very common on the GMAT, where students will invest the time in critical thought and calculation, and then levy an answer that just doesn’t hit the mark. Consider the example:
Donald has \$520,000 in campaign money available to spend on advertising for the month of October, and his advisers are telling him that he should spend a minimum of \$360,000 in the battleground states of Ohio, Florida, Virginia, and North Carolina. If he plans to spend the minimum amount in battleground states to appease his advisers, plus impress his friends by a big ad spend specific to New York City (and then he will skip advertising in the rest of the country), how much money will he have remaining if he wants 20% of his ad spend to take place in New York City?
(A) \$45,000
(B) \$52,000
(C) \$70,000
(D) \$90,000
(E) \$104,000
As people begin to calculate, it’s common to try to determine all of the facets of Donald’s ad spend. If he’s spending only the \$360,000 in battleground states plus the 20% he’ll spend in New York City, then \$360,000 will represent 80% of his total ad spend. If \$360,000 = 0.8(Total), then the total will be \$450,000. That means that he’ll spend \$90,000 in New York City. Which is answer choice D…but that’s not the question!
The question asked for how much of his campaign money would be left over, so the calculation you need to focus on is the \$520,000 he started with minus the \$450,000 he spent for a total of \$70,000, answer choice C. And in a larger context, you can learn a major lesson from Wharton’s most famous alumnus: it’s not enough for your answer to be related to the question. On the GMAT, you must answer the question directly! So make sure that you:
1. Double check which portion of a word problem the question asked for. Don’t be relieved when your algebra spits out “a” number. Make sure it’s “the” number.
2. Be careful with Strengthen/Weaken Critical Reasoning problems. A well-written Strengthen problem will likely have a good Weaken answer choice, and vice-versa.
3. In algebra problems, make sure to identify the proper variable (or combination of variables if they ask, for example “What is 6x – y?”).
4. With Data Sufficiency problems, pay attention to the exact values being asked for. One of the most common mistakes that people make is saying that a statement is insufficient because they’re looking to fill in all variables, when actually it is sufficient to answer the exact combination that the test asked for.
As you watch the debate this weekend, notice (How could you not?) how absurd it is that the candidates just about never directly answer the question…and then vow to not make the same mistake on your GMAT exam.
By Brian Galvin.
# How to Use the Pythagorean Theorem With a Circle
It does not surprise anyone when they learn that the properties of circles are tested on the GMAT. Most test-takers will nod and rattle off the relevant equations by rote: Area = Π*radius^2; Circumference = 2Π* radius; etc. However, many of my students are caught off guard to learn that the equation for a circle on the coordinate plane is our good friend the Pythagorean theorem. Why on earth would an equation for a right triangle describe a circle?
Remember: the GMAT loves to test shapes in combination: a circle inscribed in a square, for example, or the diagonal of a rectangle dividing it into two right triangles. So you should expect that triangles will appear just about anywhere – including in circles. Especially in coordinate geometry questions, where the coordinate grid allows for right angles everywhere, you should bring the Pythagorean Theorem with you to just about every GMAT geometry problem you see, even if the triangle isn’t immediately apparent. Let’s talk about how the Pythagorean Theorem can present itself in circle problems – “Pythagorean circle problems” if you will. (And note that the Pythagorean Theorem doesn’t have to “announce itself” by telling you you’re dealing with a right triangle! Very often it’s on you to determine that it applies.)
Take a look at the following diagram in which a circle is centered on the origin (0,0) in the coordinate plane:
Designate a random point on the circle (x,y). If we draw a line from the center of the circle to x,y, that line is a radius of the circle. Call it r. If we drop a line down from (x,y) to the x-axis, we’ll have a right triangle (and an opportunity to therefore apply the Pythagorean Theorem to this circle):
Note that the base of the triangle is x, and the height of the triangle is y. So now we have our Pythagorean Theorem equation: x^2 + y^2 = r^2. This is also the equation for a circle centered on the origin on the coordinate plane. [The more general equation for a circle with a center (a,b) is (x-a)^2 + (y-b)^2 = r^2. When a circle is centered on the origin, (a,b) is simply (0,0.)]
This Pythagorean equation of a circle ends up being an immensely useful tool to use on the GMAT. Take the following Data Sufficiency question, for example:
A certain circle in the xy-plane has its center at the origin. If P is a point on the circle, what is the sum of the squares of the coordinates of P?
(1) The radius of the circle is 4
(2) The sum of the coordinates of P is 0
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient
C. Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
D. EACH statement ALONE is sufficient
E. Statements (1) and (2) TOGETHER are NOT sufficient
So let’s draw this, designating P as (x,y):
Now we draw our trust right triangle by dropping a line down from P to the x-axis, which will give us this:
We’re looking for x^2 + y^2. Hopefully, at this point, you notice what the question is going for – because we have a right triangle, x^2 + y^2 = r^2, meaning that all we need is the radius!
Statement 1 is pretty straightforward – if r = 4, we can insert this into our equation of x^2 + y^2 = r^2 to get x^2 + y^2 = 4^2. So x^2 + y^2 = 16. Clearly, this is sufficient.
Now look at Statement 2. If the sum of x and y is 0, we can say x = 1 and y = -1 or x = 2 and y = -2 or x = 100 and y = -100, etc. Each of these will yield a different value for x^2 + y^2, so this statement alone is clearly not sufficient. Our answer is A.
Takeaway: any shape can appear on the coordinate plane, and given the right angles galore in the coordinate grid you should be on the lookout for right triangles, specifically. If the shape in question is a circle, remember to use the Pythagorean theorem as your equation for the circle, and what would have been a challenging question becomes a tasty piece of baklava. (We are talking about principles elucidated by the ancient Greeks, after all.)
And a larger takeaway: it’s easy to memorize formulas for each shape, so what does the GMAT like to do? See if you can apply knowledge about one shape to a problem about another (for example, applying Pythagorean Theorem to a circle). For this reason it’s important to know the “usual suspects” of how shapes get tested together. Triangles and circles work well together, for example:
-If a triangle is formed with two radii of a circle, that triangle is therefore isosceles since those radii necessarily have the same measure.
-If a triangle is formed by the diameter of a circle and two chords connecting to a point on the circle, that triangle is a right triangle with the diameter as the hypotenuse (another way that the GMAT can combine Pythagorean Theorem with a circle).
-When a circle appears in the coordinate plane, you can use Pythagorean Theorem with that circle to find the length of the radius (which then opens you up to diameter, circumference, and area).
In general, whenever you’re stuck on a geometry problem on the GMAT a great next step is to look for (or draw) a diagonal line that you can use to form a right triangle, and then that triangle lets you use Pythagorean Theorem. Whether you’re dealing wit a rectangle, square, triangle, or yes circle, Pythagorean Theorem has a way of proving extremely useful on almost any GMAT geometry problem, so be ready to apply it even to situations that didn’t seem to call for Pythagorean Theorem in the first place.
By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.
# Quarter Wit, Quarter Wisdom: Try to Answer This GMAT Challenge Question!
Today, we will give you a GMAT challenge question. The challenge of reviewing this question is not that the question is hard to understand – it is that you will need to solve this official question within a minute using minimum calculations.
Let’s take a look at the question stem:
Date of Transaction Type of Transaction June 11 Withdrawal of \$350 June 16 Withdrawal of \$500 June 21 Deposit of x dollars
For a certain savings account, the table shows the three transactions for the month of June. The daily balance for the account was recorded at the end of each of the 30 days in June. If the daily balance was \$1,000 on June 1 and if the average (arithmetic mean) of the daily balances for June was \$1,000, what was the amount of the deposit on June 21?
(A) \$1,000
(B) \$1,150
(C) \$1,200
(D) \$1,450
(E) \$1,600
The average of daily balances = (Balance at the end of June 1 + Balance at the end of June 2 + … + Balance at the end of June 30) / 30 = 1000
Now we have been given the only three transactions that took place:
• A withdrawal of \$350 on June 11 – so on June 11, the account balance goes down to \$650.
• A withdrawal of \$500 on June 16 – so on June 16, the account balance goes down to \$150.
• A deposit of \$x on June 21 – So on June 21, the account balance goes up to 150 + x.
Now we can plug in these numbers to say the average of daily balances = [1000 + 1000 + …(for 10 days, from June 1 to June 10) + 650 + 650 + … (for 5 days, from June 11 to June 15) + 150 + … (for 5 days, from June 16 to June 20) + (150 + x) + (150 + x) + … (for 10 days, from June 21 to June 30)] / 30 = 1000
One might then end up doing this calculation to find the value of x:
[(1000 * 10) + (650 * 5) + (150 * 5) + ((150 + x) * 10)] / 30 = 1000
x = \$1,450
But this calculation is rather tedious and time consuming. Can’t we use the deviation method ? After all, we are dealing with large values here! How?
Note that we are talking about the average of certain data values. Also, we know the deviations from those data values:
• The amount from June 11 to June 30 is 350 less.
• The amount from June 16 to June 30 is another 500 less.
• The amount from June 21 to June 30 is x in excess.
Through the deviation method, we can see the shortfall = the excess:
350 * 20 + 500 * 15 = x * 10
x = 1,450 (D)
This simplifies our calculation dramatically! Though saving only one minute on a question like this may not seem like a very big deal, saving a minute on every question by using a more efficient method could be the difference between a good Quant score and a great Quant score!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# Quarter Wit, Quarter Wisdom: Evaluating Nasty GMAT Answer Choices
In some Quant questions, we are given big nasty numbers in the answer choices and little else in the question stem. Often in such cases, the starting point is difficult for the test-taker to find, so today, we will discuss how to handle such questions.
The first and only rule with these types of problems is that familiarity helps. Evaluate the answer choices that make sense to you first.
Let’s look at a few questions to understand how to do that:
Which of the following is NOT prime?
(A) 1,556,551
(B) 2,442,113
(C) 3,893,257
(D) 3,999,991
(E) 9,999,991
The first thing that comes to mind when we consider how to find prime numbers should be to “check the number N for divisibility by all prime factors until we get to the √N.” But note that here, we have four numbers that are prime and one number that is not. Also, the numbers are absolutely enormous and, hence, will be very difficult to work with. So, let’s slide down to a number that seems a bit more sane: 3,999,991 (it is very close to a number with lots of 0’s).
3,999,991 = 4,000,000 – 9
= (2000)^2 – 3^2
This is something we recognise! It’s a difference of squares, which can be written as:
= (2000 + 3) * (2000 – 3)
= 2003 * 1997
Hence, we see that 3,999,991 is a product of two factors other than 1 and itself, so it is not a prime number. We have our answer! The answer is D.
Let’s try another problem:
Which of the following is a perfect square?
(A) 649
(B) 961
(C) 1,664
(D) 2,509
(E) 100,000
Here, start by looking at the answer choices. The first one that should stand out is option E, 100,000, since multiples of 10 are always easy to handle. However, we have an odd number of zeroes here, so we know this cannot be a perfect square.
Next, we look at the answer choices that are close to the perfect squares that we intuitively know, such as 30^2 = 900, 40^2 = 1600, 50^2 = 2500. The only possible number whose perfect square could be 961 is 31 – 31^2 will end with a 1 and will be a bit greater than 900 (32^2 will end with a 4, so that cannot be the square root of 961, and the perfect squares of other greater numbers will be much greater than 900).
31^2 = (30 + 1)^2 = 900 + 1 + 2*30*1 = 961
So, we found that 961 is a perfect square and is, hence, the answer!
In case 961 were not a perfect square, we would have tried 1,664 since it is just 64 greater than 1,600. It could be the perfect square of 42, as the perfect square of 42 will end in a 4.
If 1,664 were also not a perfect square (it is not), we would have looked at 2,509. We would have known immediately that 2,509 cannot be a perfect square because it is too close to 2,500. 2,509 ends in a 9, so we may have considered 53 to be its square root, but the difference between consecutive perfect squares increases as we get to greater numbers.
(4^2 is 16 while 5^2 is 25 – the difference between them is 9. The difference between 5^2 and 6^2 will be greater than 9, and so will the difference between the perfect squares of any pair of consecutive integers greater than 6. Hence, the difference between the squares of 50 and 53 certainly cannot be 9.)
Therefore, our answer is B. Let’s try one more question:
When a certain perfect square is increased by 148, the result is another perfect square. What is the value of the original perfect square?
(A) 1,296
(B) 1,369
(C) 1,681
(D) 1,764
(E) 2,500
This question is, again, on perfect squares. We can use the same concepts here, too.
30^2 = 900
31^2 = 961 (=(30+1)^2 = 900 + 1 + 2*30)
40^2= 1,600
41^2 = 1,681 (=(40+1)^2 = 1,600 + 1 + 2*40)
50^2 = 2,500
51^2 = 2,601 (=(50+1)^2 = 2,500 + 1 + 2*50)
We know that the difference between consecutive squares increases as we go to greater numbers: going from 30^2 to 31^2 is a difference of 61, while jumping from 40^2 to 41^2 is a difference of 81.
All the answer choices lie in the range from 900 to 2500. In this range, the difference between consecutive squares is between 60 and 100. So, when you add 148 to a perfect square to get another perfect square in this range, we can say that the numbers must be 2 apart, such as 33 and 35 or 42 and 44, etc. Also, the numbers must lie between 30 and 40 because twice 61 is 122 and twice 81 is 162 – 148 lies somewhere in between 122 and 162.
A and B are the only two possible options.
Consider option A – it ends in a 6, so the square root must end in a 6, too. If you add 148, then it will end with a 4 (the perfect square of a number ending in 8 will end in 4). So this answer choice works.
Consider option B – it ends in a 9, so the square root must end in a 3 or a 7. When you add 148, it ends in 7. No perfect square ends in 7, so this option is out. Our answer is, therefore, A.
We hope you see how a close evaluation of the answer choices can help you solve questions of this type. Go get ’em!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# GMAT Tip of the Week: Gary Johnson, Aleppo, and What To Do When Your Mind Goes Blank
Arguably the biggest news story this week was presidential hopeful Gary Johnson’s reply to a foreign policy question. “What is Aleppo?” is what Johnson responded, his mind evidently blanking on the epicenter of Syrian civil war and its resulting refugee crisis. And regardless of your opinion of Johnson’s fitness to be the architect of American foreign policy, there’s one major lesson there for your GMAT aspirations:
In pressure situations, it’s not uncommon for your brain to fail you as you “blank” on a concept you know (or should know). So it’s important to have strategies ready for that moment that very well may come to you. To paraphrase the Morning Joe question to Johnson:
What would you do about “Aleppo?”
Meaning: what would you do if your mind were to go blank on an important GMAT rule or formula?
There are four major strategies that should be in your toolkit for such a situation:
1) Test Small Numbers
You should absolutely know formulas like exponent rules or relationships like that between dividend, divisor, and remainder in division, but sometimes your mind just goes blank. In those cases, remember that math rules are logically-derived, not arbitrarily ordained! Math rules will hold for all possible values, so if you’re unsure, test numbers. For example, if you’re forced to solve something like:
(x^15)(x^9) =
And you’ve blanked on what to do with exponents, try testing small numbers like (2^2)(2^3). Here, that’s (4)(8) = 32, which is 2^5. So if you’re unsure, “Do I add or multiply the exponents?” you should see from the small example that you definitely don’t multiply, and that your hunch that, “Maybe I add?” works in this case, so you can more confidently make that decision.
When integer y is divided by integer z, the quotient is equal to x. Which of the following represents the remainder in terms of x, y, and z?
(A) x – yz
(B) zy – x
(C) y – zx
(D) zy – x
(E) zx – y
Many students memorize equations to organize dividend, divisor, quotient, and remainder, but in the fog of war on test day it can even be difficult to remember which element of the division problem is the dividend (it’s the number you start with) and which is the divisor (it’s the one you divide by). So if your mind has blanked on any part of the equation or on which element is which, just test it with small numbers to remind yourself how the concept works:
11 divided by 4 is 2 with a remainder of 3. How do you get to the remainder? You take the 11 you started with and subtract the 8 that you get from taking the divisor of 4 and multiplying it by the quotient of 2. So the answer is y (what you started with) minus zx (the divisor times the quotient), or answer choice C.
Simply put, if you blank on a rule or concept, you can test small numbers to remind yourself how it works.
2) Use Process of Elimination and Work Backwards From the Answer Choices
One beautiful thing about the GMAT is that, while in “the real world” if you need to know the Pythagorean Theorem and blank on it, you’re out of luck (well, unless you have a Google-enabled Smartphone in your pocket which you almost certainly do…), on the GMAT you have answer choices as assets. So if your own work stalls in progress, you can look to the answer choices to eliminate options you know for sure you wouldn’t get with that math:
What is x^5 + x^6? You know you don’t add or multiply those exponents, so even if you don’t see to factor out the common x^5, you could eliminate answer choices like x^11 and x^30.
Or you can look to the answer choices to see if they help you determine how you’d apply a rule. For example, if a problem forces you to employ the side ratios for a 45-45-90 triangle and you’ve forgotten them, the presence of some square roots of 2 in the answer choices can help you remember. The square root of 2 is greater than 1, and two sides must match, so if someone spots you “the rule includes a square root of 2” the only thing it can really be is the ratio x : x : x(√2)
Gary Johnson should have been so lucky – had the question been posed as, “What would you do about Aleppo, which is either a DJ on the new Drake album; the epicenter of the Syrian crisis; or a new restaurant in the Garment District?” he would get that question right every single time. Answer choices are your friends…when you blank, consult them!
3) Think Logically
Similar to that 45-45-90 “what else could it be?” logic, many times when you blank on a rule, you can work your way to either the rule itself or just to the answer by thinking logically about it. For example, if you end up with math that includes a radical sign in the denominator and can’t quite remember the steps for rationalizing the denominator:
What is 1/(1 – √2)?
(A) √2
(B) 1 – √2
(C) 1 + √2
(D) -1 – √2
(E) √2 – 1
Not all is lost! Sure, algebraically you should multiply the numerator and the denominator by the conjugate (1 + 2) but you can also logically work with this one. The numerator is 1, and the denominator is 1 – the square root of 2. You know that 2 is between 1 and 2, so what do you know about the denominator? It’s negative, and it’s a fraction (or decimal), so once you’ve taken 1 divided by that, your answer must be a negative number to the left of -1 – only answer choice D would work. So, yeah, you blanked on the steps, but you can still employ logic to back into the answer.
4) Write Down Everything You Know
Blanking is particularly troublesome because it’s that moment of panic. You’re trying to retrace your mental steps and the answer is elusive; it’s a moment you’re not in control of at that point. So take control! The more you’re actively working – jotting down other related formulas or facts you know, working on other facets of the diagram or problem and saving that step for last, etc. – the more you’re controlling, or at least actively managing, the situation.
Gary Johnson couldn’t get away with a “Who Wants to Be a Millionaire?” style talk-through-it (“Um, I know it’s not the name of any congressmen; it’s not Zika, it’s not…”) without looking dumb, but no one is going to audit your scratchwork and release it to Huffington Post, so you’re free to jot down half-baked thoughts and trial calculations to your heart’s content. Actively manage the situation, and you can work your way through that dreaded “my mind is blank” moment.
So learn from Gary Johnson. No matter how much you’ve prepared for your GMAT, there’s a chance that your mind will go blank on something you know that you know, but just can’t recall in the moment. But you have options, so heed the wisdom above, and let Trump or Clinton handle the gaffes for the day while you move on confidently to the next question.
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And as always, be sure to follow us on Facebook, YouTubeand Twitter!
By Brian Galvin.
# How to Simplify Complicated Combination and Permutation Questions on the GMAT
When test-takers first learn how to tackle combination and permutation questions, there’s typically a moment of euphoria when the proper approach really clicks.
If, for example, there are 10 people in a class, and you wish to find the number of ways you can form a cabinet consisting of a president, a vice president, and a treasurer, all you need to do is recognize that if you have 10 options for the president, you’ll have 9 left for the vice president, and 8 remaining for the treasurer, and the answer is 10*9*8. Easy, right?
But on the GMAT, as in life, anything that seems too good to be true probably is. An easy question can be tackled with the type of mechanical thinking illustrated above. A harder question will require a more sophisticated approach in which we consider disparate scenarios and perform calculations for each.
Take this question, for example:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?
A) 84
B) 91
C) 100
D) 105
E) 243
It’s natural to see this problem and think, “All I have to do is reason out how many options I have for each digit. So for the hundreds digit, I have 3 options (7, 8, or 9); the tens digit has to be different from the hundreds digit, and it must be non-zero, so I’ll have 8 options here; then the last digit has to be odd, so…”
Here’s where the trouble starts. The number of eligible numbers in the 700’s will not be the same as the number of eligible numbers in the 800’s -if the digits must all be different, then a number in the 700’s can’t end in 7, but a number in the 800’s could. So, we need to break this problem into separate cases:
First Case: Numbers in the 700’s
If we’re dealing with numbers in the 700’s, then we’re calculating how many ways we can select a tens digit and a units digit. 7___ ___.
Let’s start with the units digit. Well, we know that this number needs to be odd. And we know that it must be different from the hundreds and the tens digits. This leaves us the following options, as we’ve already used 7 for the hundreds digit: 1, 3, 5, 9. So there are 4 options remaining for the units digit.
Now the tens digit must be a non-zero number that’s different from the hundreds and units digit. There are 9 non-zero digits. We’re using one of those for the hundreds place and one of those for the units place, leaving us 7 options remaining for the tens digit. If there are 4 ways we can select the units digit and 7 ways we can select the tens digit, there are 4*7 = 28 options in the 700’s.
Second Case: Numbers in the 800’s
Same logic: 8 ___ ___. Again, this number must be odd, but now we have 5 options for the units digit, as every odd number will obviously be different from the hundreds digit, which is even (1, 3, 5, 7, or 9). The tens digit logic is the same – 9 non-zero digits total, but it must be different from the hundreds and the units digit, leaving us 7 options remaining. If there are 5 ways we can select the units digit and 7 ways we can select the tens digit, there are 5*7 = 35 options in the 800’s.
Third Case: Numbers in the 900’s
This calculation will be identical to the 700’s scenario: 9___ ___. For the units digit, we want an odd number that is different from the hundreds digit, giving us (1, 3, 5, 7), or 4 options. We’ll have 7 options again for the tens digit, for the same reasons that we’ll have 7 options for the tens digit in our other cases. If there are 4 ways we can select the units digit and 7 ways we can select the tens digit, then there are 4*7 = 28 options in the 900’s.
To summarize, there are 28 options in the 700’s, 35 options in the 800’s, and 28 options in the 900’s. 28 + 35 + 28 = 91. Therefore, B is the correct answer.
Takeaway: for a simpler permutation question, it’s fine to simply set up your slots and multiply. For a more complicated problem, we’ll need to work case-by-case, bearing in mind that each individual case is, on its own, actually not nearly as hard as it looks, sort of like the GMAT itself.
By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.
# Quarter Wit, Quarter Wisdom: Using the Deviation Method for Weighted Averages
We have discussed how to use the deviation method to find the arithmetic mean of numbers. It is very useful in cases where the numbers are huge, as it considerably brings down the calculation time.
The same method can be applied to weighted averages, as well. Let’s look at an example very similar to the one we examined when we were working on deviations in the case of arithmetic means:
What is the average of 452, 452, 453, 460, 467, 480, 499, 499, 504?
What would you say the average is here? Perhaps, around 470?
Shortfall:
We have two 452s – 452 is 18 less than 470.
453 is 17 less than 470.
460 is 10 less than 470.
467 is 3 less than 470.
Overall, the numbers less than 470 are (2*18) + 17 + 10 + 3 = 66 less than 470.
Excess:
480 is 10 more than 470.
We have two 499s – 499 is 29 more than 470.
504 is 34 more than 470.
Overall, the numbers more than 470 are 10 + (2*29) + 34 = 102 more than 470.
The shortfall is not balanced by the excess; there is an excess of 102-66 = 36.
So what is the average? If we assume that the average of these 9 numbers is 470, there will be an excess of 36. We need to distribute this excess evenly among all of the numbers, and hence, the average will increase by 36/9 = 4.
Therefore, the required mean is 470 + 4 = 474. (If we had assumed the mean to be 474, the shortfall would have balanced the excess.)
This method is used in exactly the same way when we have a simple average as when we have a weighted average. The reason we are reviewing it is that it can be very handy in weighted average questions involving more than two quantities.
We often deal with questions on weighted averages involving two quantities using the scale method. Let’s see how to use the deviation method for more than 2 quantities on an official GMAT question:
Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x+y+z gallons of a 1.5 percent grade, what is x in terms of y and z?
(A) y + 3z
(B) (y +z) / 4
(C) 2y + 3z
(D) 3y + z
(E) 3y + 4.5z
Grade 1 milk contains 1% fat. Grade 2 milk contains 2% fat. Grade 3 milk contains 3% fat. The mixture of all three contains 1.5% fat. So, grade 1 milk provides the shortfall and grades 2 and 3 milk provide the excess.
Shortfall = x*(1.5 – 1)
Excess = y*(2 – 1.5) + z*(3 – 1.5)
Since 1.5 is the actual average, the shortfall = the excess.
x*(1.5 – 1) = y*(2 – 1.5) + z*(3 – 1.5)
x/2 = y/2 + 3z/2
x = y + 3z
And there you have it – the answer is A.
We easily used deviations here to arrive at the relation. It’s good to have this method – useful for both simple averages and weighted averages – in your GMAT toolkit.
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# GMAT Tip of the Week: 6 Reasons That Your Test Day Won’t Be A Labor Day
As the northern hemisphere drifts toward autumn, two events have become just about synonymous: Labor Day and Back to School. If you’re spending this Labor Day weekend getting yourself ready to go back to graduate school, you may well labor over GMAT study materials in between barbecues and college football games. And if you do, make sure you heed this wisdom: GMAT test day should not be Labor Day!
What does that mean?
On a timed test like the GMAT, one of the biggest drains on your score can be a combination of undue time and undue energy spent on problems that could be done much simpler. “The long way is the wrong way” as a famous GMAT instructor puts it – those seconds you waste, those extra steps that could lead to error or distraction, they’ll add up over the test and pull your score much lower than you’d like it to be. With that in mind, here are six ways to help you avoid too much labor on test day:
QUANTITATIVE SECTION
1) Do the math in your order, only when necessary.
Because the GMAT doesn’t allow a calculator, it heavily rewards candidates who can find efficient ways to avoid the kind of math for which you’d need a calculator. Very frequently this means that the GMAT will tempt you with calculations that you’d ordinarily just plug-and-chug with a calculator, but that can be horribly time-consuming once you start.
For example, a question might require you to take an initial number like 15, then multiply by 51, then divide by 17. On a calculator or in Excel, you’d do exactly that. But on the GMAT, that calculation gets messy. 15*51 = 765 – a calculation that isn’t awful but that will take most people a few steps and maybe 20 seconds. But then you have to do some long division with 17 going into 765. Or do you? If you’re comfortable using factors, multiples, and reducing fractions, you can see those two steps (multiply by 51, divide by 17) as one: multiply by 51/17, and since 51/17 reduces to 3, then you’re really just doing the calculation 15*3, which is easily 45.
The lesson? For one, don’t start doing ugly math until you absolutely know you have to perform that step. Save ugly math for later, because the GMAT is notorious for “rescuing” those who are patient enough to wait for future steps that will simplify the process. And, secondly, get really, really comfortable with factors and divisibility. Quickly recognizing how to break a number into its factors (51 = 3*17; 65 = 5*13; etc.) allows you to streamline calculations and do much of the GMAT math in your head. Getting to that level of comfort may take some labor, but it will save you plenty of workload on test day.
2) Recognize that “Answers Are Assets.”
Another way to avoid or shortcut messy math is to look at the answer choices first. Some problems might look like they involve messy algebra, but can be made much easier by plugging in answer choices and doing the simpler arithmetic. Other times, the answer choices will lead themselves to process of elimination, whether because some choices do not have the proper units digit, or are clearly too small.
Still others will provide you with clues as to how you have to attack the math. For example, if the answer choices are something like: A) 0.0024; B) 0.0246; C) 0.246; D) 2.46; E) 24.6, they’re not really testing you on your ability to arrive at the digits 246, but rather on where the decimal point should go (how many times should that number be multiplied/divided by 10). You can then set your sights on the number of decimal places while not stressing other details of the calculation.
Whatever you do, always scan the answer choices first to see if there are easier ways to do the problem than to simply slog through the math. The answers are assets – they’re there for a reason, and often, they’ll provide you with clues that will help you save valuable time.
3) Question the Question – Know where the game is being played.
Very often, particularly in Data Sufficiency, the GMAT Testmaker will subtly provide a clue as to what’s really being tested. And those who recognize that can very quickly focus on what matters and not get lost in other elements of the problem.
For example, if the question stem includes an inequality with zero (x > 0 or xy < 0), there’s a very high likelihood that you’re being tested on positive/negative number properties. So, when a statement then says something like “1) x^3 = 1331”, you can hold off on trying to take the cube root of 1331 and simply say, “Odd exponent = positive value, so I know that x is positive,” and see if that helps you answer the question without much calculation. Or if the problem asks for the value of 6x – y, you can say to yourself, “I may not be able to solve for x and y individually, but if not, let’s try to isolate exactly that 6x – y term,” and set up your algebra accordingly so that you’re efficiently working toward that specific goal.
Good test-takers tend to see “where the game is being played” by recognizing what the Testmaker is testing. When you can see that a question is about number properties (and not exact values) or a combination of values (and not the individual values themselves) or a comparison of values (again, not the actual values themselves), you can structure your work to directly attack the question and not fall victim to a slog of unnecessary calculations.
VERBAL SECTION
4) Focus on keywords in Critical Reasoning conclusions.
The Verbal section simply looks time-consuming because there’s so much to read, so it pays to know where to spend your time and focus. The single most efficient place to spend time (and the most disastrous if you don’t) is in the conclusion of a Strengthen or Weaken question. To your advantage, noticing a crucial detail in a conclusion can tell you exactly “where the game is being played” (Oh, it’s not how much iron, it’s iron PER CALORIE; it’s not that Company X needs to reduce costs overall, it’s that it needs to reduce SHIPPING costs; etc.) and help you quickly search for the answer choices that deal with that particular gap in logic.
On the downside, if you don’t spend time emphasizing the conclusion, you’re in trouble – burying a conclusion-limiting word or phrase (like “per calorie” or “shipping”) in a long paragraph can be like hiding a needle in a haystack. The Testmaker knows that the untrained are likely to miss these details, and have created trap answers (and just the opportunity to waste time re-reading things that don’t really matter) for those who fall in that group.
5) Scan the Sentence Correction answer choices before you dive into the sentence.
Much like “Answers are Assets” above, a huge help on Sentence Correction problems is to scan the answer choices quickly to see if you can determine where the game is being played (Are they testing pronouns? Verb tenses?). Simply reading a sentence about a strange topic (old excavation sites, a kind of tree that only grows on the leeward slopes of certain mountains…) and looking for anything that strikes you as odd or ungrammatical, that takes time and saps your focus and energy.
However, the GMAT primarily tests a handful of concepts over and over, so if you recognize what is being tested, you can read proactively and look for the words/phrases that directly control that decision you’re being asked to make. Do different answers have different verb tenses? Look for words that signal time (before, since, etc.). Do they involve different pronouns? Read to identify the noun in question and determine which pronoun it needs. You’re not really being tasked with “editing the sentence” as much as your job is to make the proper decision with the choices they’ve already given you. They’ve already narrowed the scope of items you can edit, so identify that scope before you take out the red marking pen across the whole sentence.
As the Veritas Prep Reading Comprehension lesson teaches, stop at the end of each paragraph of a reading passage to ask yourself whether you understand Scope, Tone, Organization, and Purpose. The top two time-killers on Reading Comprehension passages/problems are re-reading (you get to the end and realize you don’t really know what you just read) and over-reading (you took several minutes absorbing a lot of details, but now the clock is ticking louder and you haven’t looked at the questions yet).
STOP will help you avoid re-reading (if you weren’t locked in on the first paragraph, you can reread that in 30 seconds and not wait to the end to realize you need to reread the whole thing) and will give you a quick checklist of, “Do I understand just enough to move on?” Details are only important if you’re asked about them, so focus on the major themes (Do you know what the paragraph was about – a quick 5-7 word synopsis is perfect – and why it was written? Good.) and save the details for later.
It may seem ironic that the GMAT is set up to punish hard-workers, but in business, efficiency is everything – the test needs to reward those who work smarter and not just harder, so an effective test day simply cannot be a Labor Day. Use this Labor Day weekend to study effectively so that test day is one on which you prioritize efficiency, not labor.
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And as always, be sure to follow us on Facebook, YouTubeand Twitter!
By Brian Galvin.
# GMAT Geometry Practice Questions and Problems
Would you call yourself a math person? If so, you’ll be glad to know that there are plenty of algebra, geometry, arithmetic, and other types of math problems on the GMAT. Perhaps you like math but need a little review when it comes to the topic of geometry. If so, learn some valuable tips on how to prep for GMAT geometry problems before you get started studying for the exam.
Learn and Practice the Basic Geometry Formulas
Knowing some basic formulas in geometry is an essential step to mastering these questions on the GMAT. One formula you should know is the Pythagorean Theorem, which is a^2 + b^2 = c^2, where c stands for the longest side of a right triangle, while a and b represent the other two sides.
Another formula to remember is the area of a triangle, which is A = 1/2bh, where A is the area, b is the length of the base, and h is the height. The formula for finding the area of a rectangle is l*w = A (length times width equals the area). Once you learn these and other basic geometry formulas for the GMAT, the next step is to put them into practice so you know how to use them when they’re called for on the exam.
Complete Practice Quizzes and Questions
Reviewing problems and their answers and completing GMAT geometry practice questions are two ways to sharpen your skills for this section of the test. This sort of practice also helps you become accustomed to the timing when it comes to GMAT geometry questions. These questions are found within the Quantitative section of the GMAT.
You are given just 75 minutes to finish 37 questions in this section. Of course, not all 37 questions involve geometry – GMAT questions in the Quantitative section also include algebra, arithmetic, and word problems – but working on completing each geometry problem as quickly as possible will help you finish the section within the time limit. In fact, you should work on establishing a rhythm for each section of the GMAT so you don’t have to worry about watching the time.
Use Simple Study Tools to Review Problems
Another way to prepare for GMAT geometry questions is to use study tools such as flashcards to strengthen your skills. Some flashcards are virtual and can be accessed as easily as taking your smartphone out of your pocket. If you prefer traditional paper flashcards, they can also be carried around easily so you can review them during any free moments throughout the day. Not surprisingly, a tremendous amount of review can be accomplished at odd moments during a single day.
In addition, playing geometry games online can help you hone your skills and add some fun to the process at the same time. You could try to beat your previous score on an online geometry game or even compete against others who have played the same game. Challenging another person to a geometry game can sometimes make your performance even better.
Study With a Capable Tutor
Preparing with a tutor can help you to master geometry for GMAT questions. A tutor can offer you encouragement and guide you in your studies. All of our instructors at Veritas Prep have taken the GMAT and earned scores that have put them in the 99th percentile of test-takers. When you study with one of our tutors, you are learning from an experienced instructor as well as someone who has been where you are in the GMAT preparation process.
Our prep courses instruct you on how to approach geometry questions along with every other topic on the GMAT. We know that memorizing facts is not enough: You must apply higher-order thinking to every question, including those that involve geometry. GMAT creators have designed the questions to test some of the skills you will need in the business world.
Taking a practice GMAT gives you an idea of what skills you’ve mastered and which you need to improve. Our staff invites you to take a practice GMAT for free. We’ll give you a score report and a performance analysis so you have a clear picture of what you need to focus on. Then, whether you want help with geometry or another subject on the GMAT, our team of professional instructors is here for you.
# How to Solve “Hidden” Factor Problems on the GMAT
One of the interesting things to note about newer GMAC Quant questions is that, while many of these questions test our knowledge of multiples and factors, the phrasing of these questions is often more subtle than earlier versions you might have seen. For example, if I ask you to find the least common multiple of 6 and 9, I’m not being terribly artful about what topic I’m testing you on – the word “multiple” is in the question itself.
But if tell you that I have a certain number of cupcakes and, were I so inclined, I could distribute the same number of cupcakes to each of 6 students with none left over or to each of 9 students with none left over, it’s the same concept, but I’m not telegraphing the subject in the same conspicuous manner as the previous question.
This kind of recognition comes in handy for questions like this one:
All boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over. After 60 additional boxes arrived and no boxes were removed, all the boxes in the warehouse were arranged in stacks of 14 boxes each, with no boxes left over. How many boxes were in the warehouse before the 60 additional boxes arrived?
(1) There were fewer than 110 boxes in the warehouse before the 60 additional arrived.
(2) There were fewer than 120 boxes in the warehouse after the 60 additional arrived.
Initially, we have stacks of 12 boxes with no boxes left over, meaning we could have 12 boxes or 24 boxes or 36 boxes, etc. This is when you want to recognize that we’re dealing with a multiple/factor question. That first sentence tells you that the number of boxes is a multiple of 12. After 60 more boxes were added, the boxes were arranged in stacks of 14 with none left over – after this change, the number of boxes is a multiple of 14.
Because 60 is, itself, a multiple of 12, the new number must remain a multiple of 12, as well. [If we called the old number of boxes 12x, the new number would be 12x + 60. We could then factor out a 12 and call this number 12(x + 5.) This number is clearly a multiple of 12.] Therefore the new number, after 60 boxes are added, is a multiple of both 12 and 14. Now we can find the least common multiple of 12 and 14 to ensure that we don’t miss any possibilities.
The prime factorization of 12: 2^2 * 3
The prime factorization of 14: 2 * 7
The least common multiple of 12 and 14: 2^2 * 3 * 7 = 84.
We now know that, after 60 boxes were added, the total number of boxes was a multiple of 84. There could have been 84 boxes or 168 boxes, etc. And before the 60 boxes were added, there could have been 84-60 = 24 boxes or 168-60 = 108 boxes, etc.
A brief summary:
After 60 boxes were added: 84, 168, 252….
Before 60 boxes were added: 24, 108, 192….
That feels like a lot of work to do before even glancing at the statements, but now look at how much easier they are to evaluate!
Statement 1 tells us that there were fewer than 110 boxes before the 60 boxes were added, meaning there could have been 24 boxes to start (and 84 once 60 were added), or there could have been 108 boxes to start (and 168 once 60 were added). Because there are multiple potential solutions here, Statement 1 alone is not sufficient to answer the question.
Statement 2 tells us that there were fewer than 120 boxes after 60 boxes were added. This means there could have been 84 boxes – that’s the only possibility, as the next number, 168, already exceeds 120. So we know for a fact that there are 84 boxes after 60 were added, and 24 boxes before they were added. Statement 2 alone is sufficient, and the answer is B.
Takeaway: questions that look strange or funky are always testing concepts that have been tested in the past – otherwise, the exam wouldn’t be standardized. By making these connections, and recognizing that a verbal clue such as “none left over” really means that we’re talking about multiples and factors, we can recognize even the most abstract patterns on the toughest of GMAT questions.
By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.
# GMAT Probability Practice: Questions and Answers
The Quantitative portion of the GMAT contains questions on a variety of math topics. One of those topics is probability. GMAT questions of this sort ask you to look for the likelihood that something will occur. Probability is not as familiar to many as Algebra, Geometry, and other topics on the test. This is why some test-takers hesitate when they see the word “probability” on a summary of the GMAT. However, this is just another topic that can be mastered with study and practice.
You may already know that there are certain formulas that can help solve GMAT probability questions, but there is more to these problems than teasing out the right answers. Take a look at some advice on how to tackle GMAT probability questions to calm your fears about the test:
Probability Formulas
As you work through GMAT probability practice questions, you will need to know a few formulas. One key formula to remember is that the probability equals the number of desired outcomes divided by the number of possible outcomes. Another formula deals with discrete events and probability – that formula is P(A and B) = P(A)*P(B). Figuring out the probability of an event not occurring is one minus the probability that the event will occur. Putting these formulas into practice is the most effective way to remember them.
Is it Enough to Know the Basic Formulas for Probability?
Some test-takers believe that once you know the formulas related to probability for GMAT questions, then you have the keys to success on this portion of the test. Unfortunately, that is not always the case. The creators of the GMAT are not just looking at your ability to plug numbers into formulas – you must understand what each question is asking and why you arrived at a particular answer. Successful business executives use reason and logic to arrive at the decisions they make. The creators of the GMAT want to see how good you are at using these same tools to solve problems.
The Value of Practice Exams
Taking a practice GMAT can help you determine your skill level when it comes to probability questions and problems on every other section of the test. Also, a practice exam gives you the chance to become accustomed to the amount of time you’ll have to finish the various sections of the test.
At Veritas Prep, we have one free GMAT practice test available to anyone who wants to get an idea of how prepared they are for the test. After you take the practice test, you will receive a score report and thorough performance analysis that lets you know how you fared on each section. Your performance analysis can prove to be one of the most valuable resources you have when starting to prepare for the GMAT. Follow-up practice tests can be just as valuable as the first one you take. These tests reveal your progress on probability problems and other skills on the GMAT. The results can guide you on how to adjust your study schedule to focus more time on the subjects that need it.
Getting the Right Kind of Instruction
When it comes to probability questions, GMAT creators have been known to set subtle traps for test-takers. In some cases, you may happen upon a question with an answer option that jumps out at you as the right choice. This could be a trap.
If you study for the GMAT with Veritas Prep, we can teach you how to spot and avoid those sorts of traps. Our talented instructors have not only taken the GMAT; they have mastered it. Each of our tutors received a score that placed them in the 99th percentile. Consequently, if you study with Veritas Prep, you’ll benefit from the experience and knowledge of tutors who have conquered the GMAT. When it comes to probability questions, GMAT tutors at Veritas Prep have you covered!
In addition to providing you with effective GMAT strategies, tips, and top-quality instruction, we also give you choices regarding the format of your courses. We have prep classes that are given online and in person – learn your lessons where you want, and when you want. You may want to go with our private tutoring option and get a GMAT study plan that is tailored to your needs. Contact Veritas Prep today and dive into your GMAT studies!
# Quarter Wit, Quarter Wisdom: The Power of Deduction on GMAT Data Sufficiency Questions
In a previous post, we have discussed how to find the total number of factors of a number. What does the total number of factors a number has tell us about that number? One might guess, “Not a lot,” but it actually does tell us quite a bit! If the total number of factors is odd, you know the number must be a perfect square. If the total number of factors is even, you know the number is not a perfect square.
We know that the total number of factors of a number A (prime factorised as X^p * Y^q *…) is given by (p+1)*(q+1)… etc.
So, if we know that a number has, say, 6 total factors, what can we say about the number?
6 = (p+1)*(q+1) = 2*3, so p = 1 and q = 2 or vice versa.
A = X^1 * Y^2 where X and Y are distinct prime numbers.
Today, we will look at a data sufficiency question in which we can use factors to deduce much more information than what we might first guess:
When the digits of a two-digit, positive integer M are reversed, the result is the two-digit, positive integer N. If M > N, what is the value of M?
Statement 1: The integer (M – N) has 12 unique factors.
Statement 2: The integer (M – N) is a multiple of 9.
With this question, we are told that M is a two-digit integer and N is obtained by reversing it. So if M = 21, then N = 12; if M = 83, then N = 38 (keeping in mind that M must be greater than N). In the generic form:
M = 10a + b and N =10b + a (where a and b are single-digit numbers from 1 to 9. Neither can be 0 or greater than 9 since both M and N are two-digit numbers.)
We also know that no matter what M and N are, M > N. Therefore:
10a + b > 10b + a
9a > 9b
a > b
Let’s examine both of our given statements:
Statement 1: The integer (M – N) has 12 unique factors.
First, let’s figure out what M – N is:
M – N = (10a + b) – (10b + a) = 9a – 9b
Say M – N = A. This would mean A = 9(a-b) = 3^2 * (a-b)
The total number of factors of A where A = X^p * Y^q *… can be calculated using the formula (p+1)*(q+1)* …
We know that A has 3^2 as a factor, so X = 3 and p = 2. Therefore, the total number of factors would be (2+1)*(q+1)*… = 3*(q+1)*… = 12, so (q+1)*… must be 4.
Case 1:
This means q may be 3 so that (q+1) is 4. Since a-b must be less than or equal to 9 and must also be the cube of a number, (a-b) must be 8. (Note that a-b cannot be 1 because then the total number of factors of A would only be 3.)
So, a must be 9 and b must be 1 in this case (since a > b). The integers will be 91 and 19, and since M > N, M = 91.
Case 2:
Another possibility is that (a-b) is a product of two prime factors (other than 3), both with the power of 1. In that case, the total number of factors = (2+1)*(1+1)*(1+1) = 12
Note, however, that the two prime factors (other than 3) with the smallest product is 2*5 = 10, but the difference of two single-digit positive integers cannot be 10. This means that only Case 1 can be true, therefore, Statement 1 alone is sufficient. This is certainly not what we expected to find from just the total number of factors!
Statement 2: The integer (M – N) is a multiple of 9.
M – N = (10a + b) – (10b + a) = 9a – 9b, so M – N = 9 (a-b) . This is already a multiple of 9.
We get no new information with this statement; (a-b) can be any integer, such as 2 (a = 5, b = 3 or a = 7, b = 5), etc. This statement alone is insufficient, therefore our answer is A.
Don’t take the given data of a GMAT question at face value, especially if you are expecting questions from the 700+ range. Ensure that you have deduced everything that you can from it before coming to a conclusion.
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# GMAT Math Cheat Sheet: Formulas and Tips for Success
An individual who is creating a study plan for the GMAT knows that math must be a part of the equation. Though many people love all sorts of math, there are some who become worried about the Quantitative portion of the exam.
If you’re concerned about the math questions on the GMAT, it can be useful to become more familiar with the specific content in this section. Find out about the types of problems in the Quantitative section and consider some GMAT geometry formulas. Also, check out a gathering of tips on how to prep in an effective way:
What is in the Quantitative Section?
Data Sufficiency and Problem-Solving are the two types of questions in the Quantitative section. The Problem-Solving questions are multiple-choice and test your skills in algebra, basic arithmetic, and geometry. The basic arithmetic questions involve decimals, positive and negative integers, fractions, percentages, and averages. The problems you find in this section are on par with the level of material taught in high school math classes. Though many of the questions on the exam involve basic arithmetic, it’s helpful to have a GMAT formula sheet to refer to when preparing for algebra and geometry problems.
GMAT Formulas for the Math Section
Your GMAT math formulas cheat sheet should include the Pythagorean Theorem. This formula helps you to find the measurement of the third side of a right triangle when given the measurements of the other two sides. Another item on your GMAT math cheat sheet should be A = 1/2 bh, which is the formula for finding the area of a triangle. Distance = rate*time is a very helpful formula to know, too. Find the area of a rectangle in fast fashion by using the formula A = lw. The formula A = s2 will help you discover the area of a square.
Moving Beyond Memorization
A GMAT math formulas cheat sheet is an effective study tool, but it’s equally important to know which formula to apply to a problem, so you should spend time practicing problems that employ each of those formulas. This way, on test day, you’ll be familiar with the formulas and feel comfortable using them. The easiest way to do this, of course, is to let us help you.
The expert instructors at Veritas Prep partner with students to help them learn and to practice these formulas for the Quantitative section. We hire tutors who have excellent teaching skills as well as GMAT scores in the 99th percentile. When you study with us, you know you’re learning from the best! Our instructors work through practice math problems with you to ensure that you understand how to solve them in the most efficient way.
Get the Timing Right
Test-takers are given 75 minutes to tackle the 37 questions in the Quantitative section. This sounds like a long time, but if you get hung up on one question for several minutes, you could end up running out of time for this section. In order to avoid this, you should take timed practice tests. Taking timed tests allows you to establish a rhythm for solving problems and answering questions. Once you establish a rhythm, you don’t have to be so concerned about running out of time before you finish all of the problems.
More Tips for Mastering the Quantitative Section
Studying with a GMAT math cheat sheet is one way to prepare for the test. Another way to save test time and make questions more manageable is to eliminate answer options that are clearly wrong – this allows your mind to focus only on the legitimate choices. Estimating the answer to a problem as you read through it is another way to save test time and arrive at answers more quickly.
Our GMAT curriculum teaches you how to approach questions on the separate math topics within the Quantitative section. Our strategies give you the tools you need to problem-solve like a business professional! We are proud to provide both online and in-person courses that prepare you for the GMAT. Veritas Prep instructors offer solid instruction as well as encouragement to individuals with the goal of acing the GMAT and getting into a preferred business school. Let us partner with you on the road to GMAT success! Contact us to talk with one of our course advisers today.
# When to Pick Your Own Numbers on GMAT Quant Questions
The other day, while working with a tutoring student, I was enumerating the virtues of various test-taking strategies when the student sheepishly interrupted my eloquent paean to picking numbers. She’d read somewhere that these strategies were fine for easy to moderate questions, but that for the toughest questions, you just had to bear down and solve the problem formally. Clearly, she is not a regular reader of our fine blog.
As luck would have it, on her previous practice exam she’d received the following problem, which both illustrates the value of picking numbers and demonstrates why this approach works so well.
A total of 30 percent of the geese included in a certain migration study were male. If some of the geese migrated during the study and 20 percent of the migrating geese were male, what was the ratio of the migration rate for the male geese to the migration rate for the female geese?
[Migration rate for geese of a certain sex = (number of geese of that sex migrating) / (total number of geese of that sex)]
A) 1/4
B) 7/12
C) 2/3
D) 7/8
E) 8/7
This is a perfect opportunity to break out two of my favorite GMAT tools: picking numbers and making charts. So, let’s say there are 100 geese in our population. That means that if 30% are male, we’ll have 30 male geese and 70 females geese, giving us the following chart:
Male Female Total Migrating Not-Migrating Total 30 70 100
Now, let’s say 10 geese were migrating. That means that 90 were not migrating. Moreover, if 20 percent of the migrating geese were male, we know that we’ll have 2 migrating males and 8 migrating females, giving us the following:
Male Female Total Migrating 2 8 10 Not-Migrating Total 30 70 100
(Note that if we wanted to, we could fill out the rest of the chart, but there’s no reason to, especially when we’re trying to save as much time as possible.)
Our migration rate for the male geese is 2/30 or 1/15. Our migration rate for the female geese is 8/70 or 4/35. Ultimately, we want the ratio of the male migration rate (1/15) to the female migration rate (4/35), so we need to simplify (1/15)/(4/35), or (1*35)/(15*4) = 35/60 = 7/12. And we’re done – B is our answer.
My student was skeptical. How did we know that 10 geese were migrating? What if 20 geese were migrating? Or 50? Shouldn’t that change the result? This is the beauty of picking numbers – it doesn’t matter what number we pick (so long as we don’t end up with an illogical scenario in which, say, the number of migrating male geese is greater than the number of total male geese). To see why, watch what happens when we do this algebraically:
Say that we have a total of “t” geese. If 30% are male, we’ll have 0.30t male geese and 0.70t females geese. Now, let’s call the migrating geese “m.” If 20% are male, we’ll have 0.20m migrating males and 0.80m migrating females. Now our chart will look like this:
Male Female Total Migrating 0.20m 0.80m m Not-Migrating Total 0.30t 0.70t t
The migration rate for the male geese is 0.20m/0.30t or 2m/3t. The migration rate for the female geese is 0.80m/0.70t or 8m/7t. We want the ratio of the male migration rate (2m/3t) to the female migration rate (8m/7t), so we need to simplify (2m/3t)/(8m/7t) = (2m*7t)/(3t * 8m) = 14mt/24mt = 7mt/12mt = 7/12. It’s clear now why the numbers we picked for m and t don’t matter – they cancel out in the end.
Takeaway: We cannot say this enough: the GMAT is not testing your ability to do formal algebra. It’s testing your ability to make good decisions in a stressful environment. So your goal, when preparing for this test, isn’t to become a virtuoso mathematician, even for the toughest questions. It’s to practice the kind of simple creative thinking that will get you to your answer with the smallest investment of your time.
By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.
# Quarter Wit, Quarter Wisdom: Know Your Subtraction for the GMAT!
Your first reaction to the title of this post is probably, “I already know my subtraction!” No surprise there. But what is surprising is that our statistics tell us that the following GMAT question – which is nothing extraordinary, but does involve some tricky subtraction – is a 700-level question. That made us decide to write this post. We will discuss this concept along with the question:
The last digit of 12^12 + 13^13 – 14^14 × 15^15 =
(A) 0
(B) 1
(C) 5
(D) 8
(E) 9
This is a simple question based on the cyclicity of units digits. There are 3 terms here: 12^12, 13^13 and (14^14)*(15^15). Let’s find the last digit of each of these terms:
12^12
The units digit of 12 is 2.
2 has a cyclicity of 2 – 4 – 8 – 6.
The cycles end at the powers 4, 8, 12 … etc. So, twelve 2’s will end in a units digit of 6.
13^13
The units digit of 13 is 3.
3 has a cyclicity of 3 – 9 – 7 – 1.
A new cycle starts at the powers 1, 5, 9, 13 … etc. So, thirteen 3’s will end in a units digit of 3.
(14^14)*(15^15)
This term is actually the most simple to manage in the case of its units digit – an even number multiplied by a multiple of 5 will end in 0. Also, note that this will be a huge term compared to the other two terms.
This is what our expression looks like when we consider just the units digits of these terms:
(A number ending in 6) + (A number ending in 3) – (A much greater number ending in 0)
Looking at our most basic options, a number ending in 6 added to a number ending in 3 will give us a number ending in 9 (as 3 + 6 = 9). So, the expression now looks like this:
(A number ending in 9) – (A much greater number ending in 0)
It is at this point that many people mess up. They deduce that 9-0 will end in a 9, and hence, the answer will be E. All their effort goes to waste when they do this. Let’s see why:
How do you subtract one number out of another? Take, for example, 10-7 = 3
This can also be written as 7-10 = -3. (Here, you are still subtracting the number with a lower absolute value from the number with a greater absolute value, but giving it a negative sign.)
Let’s try to look at this in tabular form. The number with the greater absolute value goes on the top and the number with the smaller absolute value goes under it. You then subtract and the result gets the sign of the number with the greater absolute value.
(i) 100-29
100
-29
071
(ii) 29-100
100
-29
071
(But since the sign of 100 is negative, your answer is actually -71.)
So, the number with greater absolute value is always on top. Going back to our original question now, (A number ending in 9) – (A much greater number ending in 0) will look like:
abcd0
– pq9
ghjk1
Ignoring the letter variables (these are simply placeholders), note that the greater number ending in 0 will be on the top and the smaller one ending in 9 will be below it. This means the answer will be a negative number ending in a units digit of 1. Therefore, our answer is B.
As we learn more advanced concepts, make sure you are not taking your basic principles for granted!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# GMAT Tip of the Week: What Simone Biles and the Final Five Can Teach You About GMAT Math
On this Friday, ending the first week of the Rio Olympics, your office has undoubtedly said the name “Simone” exponentially more than ever before. Michael Phelps’ blowout win – his 4th straight – in the 200 IM was incredible, but last night belonged to two Texans named Simone.
Swimmer Simone Manuel and gymnast Simone Biles each won historic gold medals, and if you’re at all inspired to pursue your own “go for the gold” success in business school (maybe Stanford like Manuel, or UCLA like Biles), you can learn a lot from the Olympic experience. Two lessons, in particular, stand out from the performance of Biles and her “Final Five” teammates:
There’s no way to watch Olympic gymnastics and not be overwhelmingly impressed by the skills that each gymnast brings to competition. So at times it’s frustrating and saddening to hear the TV announcers discuss deduction after deduction; shouldn’t everyone at all times just be yelling, “Wow!!!!” at the otherworldly talents of each athlete?
Much like the GMAT, though, Olympic gymnastics is not about the sheer possession of these skills – at that level, everyone has them. It’s more about the ability to execute them and, as becomes evident from the expert commentary of Tim Dagget and Nastia Liukin, to connect them. It’s not the uneven bars handstand or release itself that wins the gold, it’s the ability to connect skill after skill as part of a routine. The line, “She was supposed to connect that skill to another…” is always followed by, “That will be a deduction” – both in Olympic gymnastics and on the GMAT.
How does that affect you?
By test day, you had better have all of the necessary skills to compete on the GMAT Quant Section. Area of a triangle, Pythagorean Theorem, Difference of Squares…if you don’t know these rules, you’re absolutely sunk. But to do really well, you need to quickly connect skill to skill, and connect items in the problems to the skills necessary to work with them. For example:
If a problem includes a term x^4 – 1, you should immediately be thinking, “That connects really well to the Difference of Squares rule: a^2 – b^2 = (a + b)(a – b), and since x^4 is a square [it’s (x^2)^2] and 1 is a square (it’s 1^2), I can write that as (x^2 + 1)(x^2 – 1), and for good measure I could apply Difference of Squares to the (x^2 – 1) term too.” The GMAT won’t ever specifically tell you, “Use the Difference of Squares,” so it’s your job to immediately connect the symptoms of Difference of Squares (an even exponent, a subtraction sign, a square of some kind, even if it’s 1) to the opportunity to use it.
If you see a right triangle, you should recognize that Area and Pythagorean Theorem easily connect. In a^2 + b^2 = c^2, sides a and b are perpendicular and allow you to use them as the base and height in the area formula. And the Pythagorean Theorem includes three squares with the opportunity to create subtraction [you could write it as a^2 = c^2 – b^2, allowing you to say that a^2 = (c + b)(c – b)…], so you could connect yet another skill to it to help solve for variables.
Similarly, if you see a square or rectangle, its diagonal is the hypotenuse of a right triangle, allowing you to use the sides as a and b in the Pythagorean setup, which could also connect to Difference of Squares…etc.
When you initially learned most of these skills in high school (much like when Biles, Aly Raisman, Gabby Douglas, etc. learned handstands and cartwheels in Gymboree), you learned them as individual, isolated skills. “Here’s the formula, and here are 10 questions that test it.” On the GMAT – as in the Olympics – you’re being tested more on your ability to connect them, to see opportunities to use a skill that’s not obvious at first (“Well, I’m not sure what to do but I do have multiple squared terms so let me try to apply Difference of Squares…or maybe I can use a and b in the Area calculation.”), but that helps you build more knowledge of the problem.
So as you study, don’t just learn individual skills. Look for opportunities to connect them, and look for signals that will tell you that a connection is possible. A rectangle problem with a square root of 3 in the answer choices should tell you “the diagonal of this rectangle may very well be connected to a 30-60-90 triangle, since those have the 1, √3, 2 side ratio…” The GMAT is about connections more so than just skills, so study accordingly.
Stick the Landing
If you’re like most in the “every four years I love gymnastics for exactly one week” camp, the single most important thing you look for on any apparatus is, “Did he/she stick the landing?” A hop or a step on the landing is the most noticeable deduction on a gymnastics routine…and the same holds true for the GMAT.
Again, the GMAT is testing you on how well you connect a variety of skills, so naturally there are places for you to finish the problem a step short. A problem that requires you to leverage the Pythagorean Theorem and the Area of a Triangle may ask for the sum of sides A and B, for example, but if you’ve solved for the sides individually first, you might see a particular value (A = 6) on your noteboard and in the answer choices and choose it without double checking that you answered the proper question.
That is a horrible and unnecessary “deduction” on your GMAT score: you did all the work right, all the hard part right (akin to the flip-and-two-twists in the air on your vault or the dazzling array of jumps and handstands on the tiny beam) and then botched the landing.
On problems that include more than one variable, circle the variable that the test is looking for and then make sure that you submit the proper answer for that variable. If a problem asks for a combination of variables (a + b, for example), write that down at the top of your scratchwork and go back to it after you’ve calculated. Take active steps to ensure that you stick the landing, because nothing is worse than doing all the work right and then still getting the problem wrong.
In summary, recognize that there are plenty of similarities between the GMAT and GyMnAsTics [the scoring system is too complex for the layman to worry about, the “Final Five” are more important than you think (hint: the test can’t really use the last five questions of a section for research purposes since so many people are rushing and guessing), etc.]. So take a lesson from Simone Biles and her gold-medal-winning teammates: connect your skills, stick the landing, and you’ll see your score vault to Olympian heights.
By Brian Galvin.
# Quarter Wit, Quarter Wisdom: Linear Relations in GMAT Questions
We have covered the concepts of direct, inverse and variation in previous posts. Today, we will discuss what we mean by “linearly related”. A linear relation is one which, when plotted on a graph, is a straight line. In linear relationships, any given change in an independent variable will produce a corresponding change in the dependent variable, just like a change in the x-coordinate produces a corresponding change in the y-coordinate on a line.
We know the equation of a line: it is y = mx + c, where m is the slope and c is a constant.
Let’s illustrate this concept with a GMAT question. This question may not seem like a geometry question, but using the concept of linear relations can make it easy to find the answer:
A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?
(A) 20
(B) 36
(C) 48
(D) 60
(E) 84
Let’s think of the two scales R and S as x- and y-coordinates. We can get two equations for the line that depicts their relationship:
30 = 6m + c ……. (I)
60 = 24m + c ……(II)
(II) – (I)
30 = 18m
m = 30/18 = 5/3
Plugging m = 5/3 in (I), we get:
30 = 6*(5/3) + c
c = 20
Therefore, the equation is S = (5/3)R + 20. Let’s plug in S = 100 to get the value of R:
100 = (5/3)R + 20
R = 48
Alternatively, we have discussed the concept of slope and how to deal with it without any equations in this post. Think of each corresponding pair of R and S as points lying on a line – (6, 30) and (24, 60) are points on a line, so what will (r, 100) be on the same line?
We see that an increase of 18 in the x-coordinate (from 6 to 24) causes an increase of 30 in the y-coordinate (from 30 to 60).
So, the y-coordinate increases by 30/18 = 5/3 for every 1 point increase in the x-coordinate (this is the concept of slope).
From 60 to 100, the increase in the y-coordinate is 40, so the x-coordinate will also increase from 24 to 24 + 40*(3/5) = 48. Again, C is our answer.
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# GMAT Math Help: Understanding and Solving Combinatorics Problems
Students who are taking the GMAT are going to encounter combinatorics problems. If you are a little rusty on your math topics, you may be asking, “What is combinatorics?” Combinatorics has to do with counting and evaluating the possibilities within a scenario that involve various amounts of people or things. Learn more about GMAT combinatorics questions and how to arrive at the right answers to be better prepared for the test.
Permutations
Picture a certain number of people or objects. Permutations are the possible arrangements that those people or objects can be in. One of the things you have to decide when looking at combinatorics problems is whether order is an important factor. If order is important in a problem, then the answer has to do with permutations. If order is not important in a problem, then the answer deals with combinations.
For example, say you line up five postcards from different cities on a tabletop. You may wonder how many different orders you can put these postcards in. Another way to say that would be, “How many different permutations can I make with these five postcards?” To figure out this problem, you would need the help of an equation: 5! = (5) (4) (3) (2) (1) = 120. The exclamation point in the formula is a symbol that means “factorial.”
Combinations
When working on combinatorics questions that deal with combinations, the order/arrangement of items is not important. For example, say that you have eight books and you want to know how many ways you can group three of those books on a library shelf. You could plug numbers into the three places in this formula to figure out the answer: (8) (7) (6) = 336 ways. This is the slot method of solving a combination problem.
Combinations With a Large Amount of Numbers
You will quickly find yourself needing combinatorics help if you try to count up a lot of numbers in one combination problem on the GMAT. Furthermore, you’ll use a lot of valuable test time with this counting method. Knowing the formula for combinations can help you to find the solution to a problem in a much shorter amount of time. The formula is nCr = n!/r!(n-r)! Here, n is the total number of options, r is the number of options chosen, and ! is the symbol for factorial.
Preparing for Applied Combinatorics Questions on the GMAT
One of the most effective ways of preparing for applied combinatorics questions is to take practice tests and review the various steps of problems. You want to get into the habit of approaching a problem by asking yourself whether order is a factor in a problem. This will help you determine whether a problem deals with permutations or combinations. Then, you can start to attack a problem from the right angle.
In addition, it’s important to time yourself when taking a practice Quantitative test. Though there are not many of these problems on the test, you have to get into the habit of spending only a certain amount of minutes on each problem so you don’t run out of test time before finishing.
We have a program of study at Veritas Prep that prepares you for questions on combinatorics as well as all of the other problems in the Quantitative section. We instruct you on how to approach test questions instead of just coaching you on how to memorize facts. Pair up with one of our skilled instructors at Veritas Prep and you will be studying with someone who scored in the 99th percentile on the GMAT. We believe that in order to perform at your best on the GMAT, you have to learn from a first-rate instructor! Our instructors can work through a combinatorics tutorial with you to determine what your strengths and weaknesses are in this branch of math. Then, we give you strategies that help you to improve.
For your convenience, we offer both in-person and online GMAT prep courses. We recognize that professionals in the business world have busy schedules, so we provide several study options to fit your life. When it comes to the topic of combinatorics, GMAT tips, instruction, and encouragement, we are your test prep experts. Contact us today and let us know how we can help you achieve your top GMAT score!
# Probability and Combinations: What You’ll Need to Know for the GMAT
If you’ve been paying attention to the exciting world of GMAT prep, you know thatfairly recently. I’d mentioned in a previous post that I was going to write about any conspicuous trends I noted, and one unmistakable pattern I’ve seen with my students is that probability questions seem to be cropping up with greater and greater frequency.
While these questions don’t seem fundamentally different from what we’ve seen in the past, there does seem to be a greater emphasis on probability questions for which a strong command of combinations and permutations will prove indispensable.
First, recall that the probability of x is the number ways x can occur/number of total possible outcomes (or p(x) = # desired/ # total). Another way to think about this equation is to see it as a ratio of two combinations or permutations. The number of ways x can occur is one combination (or permutation), and the total number of possible outcomes is another.
Keeping this in mind, let’s tackle this new official prompt:
From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?
(A) 3/7
(B) 5/12
(C) 27/70
(D) 2/7
(E) 9/35
Typically, I’ll start by calculating the total number of possible outcomes, as this calculation tends to be the more straightforward one. We’ve got 8 volunteers, and we want to know the number of total ways we can select 4 people from these 8 volunteers. Note, also, that the order does not matter – group of Tiffany, Mike, Louis, and Amy is the same as a group of Louis, Amy, Mike, and Tiffany. We’re not assigning titles or putting anyone in seats, so this is a combination.
If we use our combination formula N!/[(K!*(N-K)!] then N, our total pool of candidates, is 8, and K, the number we’re selecting, is 4. We get 8!/(4!*4!), which comes out to 70. At this point, we know that the denominator must be a factor of 70, so anything that doesn’t meet this criterion is out. In this case, this only allows us to eliminate B.
Now we want our desired outcomes, in which Andrew is selected and Karen is not. Imagine that you’re responsible for assembling this group of four from a total pool of eight people. You plan on putting your group of four in a conference room. Your supervisor tells you that Andrew must be in and Karen must not be, so you take Andrew and put him in the conference room. Now you’ve got three more spots to fill and seven people remaining. But remember that Karen cannot be part of this group. That means you only have 6 people to choose from to fill those other 3 spots in the room.
Put another way, think of the combination as the number of choices you have. Andrew and Karen are not choices – you’ve been ordered to include one of them and not the other. Of the 4 spots in the conference room, you only get to choose 3. And you’re only selecting from the other 6 people for those spots. Now N = 6 and K = 3. Plugging these into our trusty combination formula, we get 6!/(3!*3!), which comes out to 20.
Summarizing, we know that there are 20 ways to create our desired group of 4, and 70 total ways to select 4 people from a pool of 8, giving us a probability of 20/70, or 2/7, so the correct answer is D.
Takeaway: Probability questions can be viewed as ratios of combinations or permutations, so when you brush up on combinatorics, you’re also bolstering your probability fundamentals. Anytime you’re stuck on a complex probability question, break your calculation down into its component parts – find the total number of possible outcomes first, then find the total number of desired outcomes. Like virtually every hard question on the GMAT, probability questions are never as hard as they first seem.
Plan on taking the GMAT soon? We have GMAT prep courses starting all the time. And be sure to follow us on FacebookYouTube,and Twitter!
By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.
# Advanced Number Properties on the GMAT – Part VI
Most people feel that the topic of number properties is hard or at least a little tricky. The reason is that no matter how much effort you put into it, you will still come across new concepts every time you sit with some 700+ level problems of this topic. There will be some concepts you don’t know and will need to “figure out” during the actual test. I came across one such question the other day. It brought forth a concept I hadn’t thought about before so I decided to share it today:
Say you have N consecutive integers (starting from any integer). What can you say about their sum? What can you say about their product?
Say N = 3
The numbers are 5, 6, 7 (any three consecutive numbers)
Their sum is 5 + 6 + 7 = 18
Their product is 5*6*7 = 210
Note that both the sum and the product are divisible by 3 (i.e. N).
Say N = 5
The numbers are 2, 3, 4, 5, 6 (any five consecutive numbers)
Their sum is 2 + 3 + 4 + 5 + 6 = 20
Their product is 2*3*4*5*6 = 720
Again, note that both the sum and the product are divisible by 5 (i.e. N)
Say N = 4
The numbers are 3, 4, 5, 6 (any five consecutive numbers)
Their sum is 3 + 4 + 5 + 6 = 18
Their product is 3*4*5*6 = 360
Now note that the sum is not divisible by 4, but the product is divisible by 4.
If N is odd then the sum of N consecutive integers is divisible by N, but this is not so if N is even.
Why is this so? Let’s try to generalize – if we have N consecutive numbers, they will be written in the form:
(Multiple of N),
(Multiple of N) +1,
(Multiple of N) + 2,
… ,
(Multiple of N) + (N-2),
(Multiple of N) + (N-1)
In our examples above, when N = 3, the numbers we picked were 5, 6, 7. They would be written in the form:
(Multiple of 3) + 2 = 5
(Multiple of 3) = 6
(Multiple of 3) + 1 = 7
In our examples above, when N = 4, the numbers we picked were 3, 4, 5, 6. They would be written in the form:
(Multiple of 4) + 3 = 3
(Multiple of 4) = 4
(Multiple of 4) + 1 = 5
(Multiple of 4) + 2 = 6
etc.
What happens in case of odd integers? We have a multiple of N and an even number of other integers. The other integers are 1, 2, 3, … (N-2) and (N-1) more than a multiple of N.
Note that these extras will always add up in pairs to give the sum of N:
1 + (N – 1) = N
2 + (N – 2) = N
3 + (N – 3) = N
So when you add up all the integers, you will get a multiple of N.
What happens in case of even integers? You have a multiple of N and an odd number of other integers. The other integers are 1, 2, 3, … (N-2) and (N-1) more than a multiple of N.
Note that these extras will add up to give integers of N but one will be leftover:
1 + (N – 1) = N
2 + (N – 2) = N
3 + (N – 3) = N
The middle number will not have a pair to add up with to give N. So when you add up all the integers, the sum will not be a multiple of N.
For example, let’s reconsider the previous example in which we had four consecutive integers:
(Multiple of 4) = 4
(Multiple of 4) + 1 = 5
(Multiple of 4) + 2 = 6
(Multiple of 4) + 3 = 3
1 and 3 add up to give 4 but we still have a 2 extra. So the sum of four consecutive integers will not be a multiple of 4.
Let’s now consider the product of N consecutive integers.
In any N consecutive integers, there will be a multiple of N. Hence, the product will always be a multiple of N.
Now take a quick look at the GMAT question that brought this concept into focus:
Which of the following must be true?
1) The sum of N consecutive integers is always divisible by N.
2) If N is even then the sum of N consecutive integers is divisible by N.
3) If N is odd then the sum of N consecutive integers is divisible by N.
4) The Product of K consecutive integers is divisible by K.
5) The product of K consecutive integers is divisible by K!
(A) 1, 4, 5
(B) 3, 4, 5
(C) 4 and 5
(D) 1, 2, 3, 4
(E) only 4
Let’s start with the first three statements this question gives us. We can see that out of Statements 1, 2 and 3, only Statement 3 will be true for all acceptable values of N. Therefore, all the answer choices that include Statements 1 and 2 are out, i.e. options A and D are out. The answer choices that don’t have Statement 3 are also out, i.e. options C and E are out. This leaves us with only answer choice B, and therefore, B is our answer.
This question is a direct application of what we learned above so it doesn’t add much value to our learning as such, but it does have an interesting point. By establishing that B is the answer, we are saying that Statement 5 must be true.
5) The product of K consecutive integers is divisible by K!
We will leave it to you to try to prove this!
(For more advanced number properties on the GMAT, check out Parts I, II, III, IV and V of this series.)
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# Quarter Wit, Quarter Wisdom: Divisibility by Powers of 2
We know the divisibility rules of 2, 4 and 8:
For 2 – If the last digit of the number is divisible by 2 (is even), then the number is divisible by 2.
For 4 – If the number formed by last two digits of the number is divisible by 4, then the number is divisible by 4.
For 8 – If the number formed by last three digits of the number is divisible by 8, then the number is divisible by 8.
A similar rule applies to all powers of 2:
For 16 – If the number formed by last four digits of the number is divisible by 16, then the number is divisible by 16.
For 32 – If the number formed by last five digits of the number is divisible by 32, then the number is divisible by 32.
and so on…
Let’s figure out why:
The generic rule can be written like this: A number M is divisible by 2^n if the last n digits of M are divisible by 2^n.
Take, for example, a division by 8 (= 2^3), where M = 65748048 and n = 3.
Our digits of interest are the last three digits, 048.
48 is completely divisible by 8, so we conclude that 65748048 is also divisible by 8.
A valid question here is, “What about the remaining five digits? Why do we ignore them?”
Breaking down M, we can see that 65748048 = 65748000 + 048 (we’ve separated the last three digits).
Now note that 65748000 = 65748 * 1000. Since 1000 has three 0s, it is made up of three 2s and three 5s. Because 1000 it has three 2s as factor, it also has 8 as a factor. This means 65748000 has 8 as a factor by virtue of its three 0s.
All we need to worry about now is the last three digits, 048. If this is divisible by 8, 65748048 will also be divisible by 8. If it is not, 65748048 will not be divisible by 8.
In case the last three digits are not divisible by 8, you can still find the remainder of the number. Whatever remainder you get after dividing the last three digits by 8 will be the remainder when you divide the entire number by 8. This should not be a surprise to you now – 65748000 won’t have a remainder when divided by 8 since it is divisible by 8, so whatever the remainder is when the last 3 digits are divided by 8 will be the remainder when the entire number is divided by 8.
In the generic case, the number M will be split into a number with n zeroes and another number with n digits. The number with n zeroes will be divisible by 2^n because it has n 2s as factors. We just need to see the divisibility of the number with n digits.
We hope you have understood this concept. Let’s take look at a quick GMAT question to see this in action:
What is the remainder when 1990990900034 is divided by 32 ?
(A) 16
(B) 8
(C) 4
(D) 2
(E) 0
Breaking down our given number, 1990990900034 = 1990990900000 + 00034.
1990990900000 ends in five 0’s so it is divisible by 32. 34, when divided by 32, gives us a remainder of 2. Hence, when 1990990900034 is divided by 32, the remainder will be 2. Our answer is D.
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# How to Use Units Digits to Avoid Doing Painful Calculations on the GMAT
During the first session of each new class I teach, we do a quick primer on the utility of units digits. Imagine I want to solve 130,467 * 367,569. Without a calculator, we are surely entering a world of hurt. But we can see almost instantaneously what the units digit of this product would be.
The units digit of 130,467 * 367,569 would be the same as the units digit of 7*9, as only the units digits of the larger numbers are relevant in such a calculation. 7*9 = 63, so the units digit of 130,467 * 367,569 is 3. This is one of those concepts that is so simple and elegant that it seems too good to be true.
And yet, this simple, elegant rule comes into play on the GMAT with surprising frequency.
Take this question for example:
If n is a positive integer, how many of the ten digits from 0 through 9 could be the units digit of n^3?
A) three
B) four
C) six
D) nine
E) ten
Surely, you think, the solution to this question can’t be as simple as cubing the easiest possible numbers to see how many different units digits result. And yet that’s exactly what we’d do here.
1^3 = 1
2^3 = 8
3^3 = 27 à units 7
4^3 = 64 à units 4
5^3 = ends in 5 (Fun fact: 5 raised to any positive integer will end in 5.)
6^3 = ends in 6 (Fun fact: 6 raised to any positive integer will end in 6.)
7^3 = ends in 3 (Well 7*7 = 49. 49*7 isn’t that hard to calculate, but only the units digit matters, and 9*7 is 63, so 7^3 will end in 3.)
8^3 = ends in 2 (Well, 8*8 = 64, and 4*8 = 32, so 8^3 will end in 2.)
9^3 = ends in 9 (9*9 = 81 and 1 * 9 = 9, so 9^3 will end in 9.)
10^3 = ends in 0
Amazingly, when I cube all the integers from 1 to 10 inclusive, I get 10 different units digits. Pretty neat. The answer is E.
Of course, this question specifically invoked the term “units digit.” What are the odds of that happening? Maybe not terribly high, but any time there’s a painful calculation, you’d want to consider thinking about the units digits.
Take this question, for example:
A certain stock exchange designates each stock with a one, two or three letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be replaced and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?
A) 2,951
B) 8,125
C) 15,600
D) 16,302
E) 18,278
Conceptually, this one doesn’t seem that bad.
If I wanted to make a one-letter code, there’d be 26 ways I could do so.
If I wanted to make a two-letter code, there’d be 26*26 or 26^2 ways I could do so.
If I wanted to make a three-letter code, there’d be 26*26*26, or 26^3 ways I could so.
So the total number of codes I could make, given the conditions of the problem, would be 26 + 26^2 + 26^3. Hopefully, at this point, you notice two things. First, this arithmetic will be deeply unpleasant to do. Second, all of the answer choices have different units digits!
Now remember that 6 raised to any positive integer will always end in 6. So the units digit of 26 is 6, and the units digit of 26^2 is 6 and the units digit of 26^3 is also 6. Therefore, the units digit of 26 + 26^2 + 26^3 will be the same as the units digit of 6 + 6 + 6. Because 6 + 6 + 6 = 18, our answer will end in an 8. The only possibility here is E. Pretty nifty.
Takeaway: Painful arithmetic can always be avoided on the GMAT. When calculating large numbers, note that we can quickly find the units digit with minimal effort. If all the answer choices have different units digits, the question writer is blatantly telegraphing how to approach this problem.
By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.
# Quarter Wit Quarter Wisdom: What is Your Favorite Number?
Fans of The Big Bang Theory will remember Sheldon Cooper’s quote from an old episode on his favorite number:
The best number is 73. Why? 73 is the 21st prime number. Its mirror, 37, is the 12th and its mirror, 21, is the product of multiplying 7 and 3… and in binary 73 is a palindrome, 1001001, which backwards is 1001001.”
Though Sheldon’s logic is infallible, my favorite number is 1001 because it has a special role in standardized tests.
1001 is 1 more than 1000 and hence, is sometimes split as (1000 + 1). It sometimes appears in the a^2 – b^2 format such as 1001^2 – 1, and its factors are 7, 11 and 13 (not the factors we usually work with).
Due to its unusual factors and its convenient location (right next to 1000), it could be a part of some tough-looking GMAT questions and should be remembered as a “special” number. Let’s look at a question to understand how to work with this number.
Which of the following is a factor of 1001^(32) – 1 ?
(A) 768
(B) 819
(C) 826
(D) 858
(E) 924
Note that 1001 is raised to the power 32. This is not an exponent we can easily handle. If we try to use a binomial here and split 1001 into (1000 + 1), all we will achieve is that upon expanding the given expression, 1 will be cancelled out by -1 and all other terms will have 1000 in common. None of the answer choices are factors of 1000, however, so we must look for some other factor of 1001^(32) – 1.
Without a calculator, it is not possible for us to find the factors of 1001^(32) – 1, but we do know the prime factors of 1001 and hence, the prime factors of 1001^32. We may not be able to say which numbers are factors of 1001^(32) – 1, but we will be able to say which numbers are certainly not factors of this!
Let me explain:
1001 = 7 * 11 * 13 (Try dividing 1001 by 7 and you’ll get 143. 143 is divisible by 11, giving you 13.)
1001^32 = 7^32 * 11^32 * 13^32
Now, what can we say about the prime factors of 1001^(32) – 1? Whatever they are, they are certainly not 7, 11 or 13 – two consecutive integers cannot have any common prime factor (discussed here and continued here).
Now look at the answer choices and try dividing each by 7:
(A) 768 – Not divisible by 7
(B) 819 – Divisible by 7
(C) 826 – Divisible by 7
(D) 858 – Not divisible by 7
(E) 924 – Divisible by 7
Options B, C and E are eliminated. They certainly cannot be factors of 1001^(32) – 1 since they have 7 as a prime factor, and we know 1001^(32) – 1 cannot have 7 as a prime factor.
Now try dividing the remaining options by 11:
(A) 768 – Not divisible by 11
(D) 858 – Divisible by 11
D can also be eliminated now because it has 11 as a factor. By process of elimination, the answer is A; it must be a factor of 1001^(32) – 1.
I hope you see how easily we used the factors of 1001 to help us solve this difficult-looking question. And yes, another attractive feature of 1001 – it is a palindrome in the decimal representation itself!
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# GMAT Tip of the Week: How to Avoid GMAT (and Pokemon Go) Traps
In seemingly the most important development in world history since humans learned to create fire, Pokemon Go has arrived and is taking the world by storm. Rivaling Twitter and Facebook for mobile phone attention and battling the omnipresent selfie as a means of death-by-mobile-phone, Pokemon Go is everywhere you want to be…and often in places you don’t.
And that is why Pokemon Go is responsible for an ever-important GMAT lesson.
Perhaps most newsworthy about Pokemon Go these days is the dangerous and improper places that it has led its avid users. On the improper side, such solemn and dignified places as the national Holocaust Museum and Arlington National Cemetery have had to actively prohibit gamers from descending upon mourners/commemorators while playing the game. And as for danger, there have been several instances of thieves luring gamers into traps and therefore robbing them of valuable (if you’re playing the game, you definitely have a smartphone) items.
And the GMAT can and will do the same thing.
How?
If you’re reading this on our GMAT blog, you’ve undoubtedly already learned that, on Data Sufficiency problems, you cannot assume that a variable is positive, or that it is an integer. But think about what makes Pokemon Go users so vulnerable to being lured into a robbery or to losing track of basic human decency. They’re so invested in the game that they lose track of the situations they’re being lured into.
Similarly, the most dangerous GMAT traps are those for which you should absolutely know better, but the testmaker has gotten your mind so invested in another “game” that you lose track of something basic. Consider the example:
If y is an odd integer and the product of x and y equals 222, what is the value of x?
(1) x is a prime number
(2) y is a 3 digit number
Statement 1 is clearly sufficient. Since y is odd, and an integer, and the product of integers x and y is an even integer, that means that x must be even. And since x also has to be prime (which is how you know it’s an integer, too), the only even prime is 2, making x = 2.
From there your mind is fixated on the game. You can quickly see that in that case y = 111 and x = 2. Which you then have to forget about as you attack Statement 2. But here’s the reason that less than 25% of users in the Veritas Prep Question Bank get this right, while nearly half incorrectly choose D. Statement 1 has gotten your mind fixated on the even/odd/prime game, meaning that you may only be thinking about integers (and positive integers at that) at this point.
That y is a 3-digit number DOES NOT mean that it has to be 111. It could be -111 (making x = -2) or 333 (making x = 2/3). So only Statement 1 alone is sufficient, but the larger lesson is more important. Just like Pokemon Go has the potential to pollute your mind and have you see the real world through its “enhanced reality” lens, so does a statement that satisfies your intellect (“Ah, 2 is the only even prime number!”) give you just enough tunnel vision that you make poor decisions and fall for traps.
The secret here is that almost no one scoring above a 500 carries over all of Statement 1 (“Oh, well I already know that x = 2!”) – a total rookie mistake. It’s that Statement 1 got you fixated on definitions of types of integers (prime, even, odd) and therefore got your mind looking through the “enhanced reality” of integers-only.
The lesson? Much like Pokemon Go, the GMAT has tools to get you so invested in a particular facet of a game that you lose your universal awareness of your surroundings. Know that going in – that you have to consciously step back from that enhanced reality you’ve gained after Statement 1 and look at the whole picture. So take a lesson from Pokemon Go and know when to stop and step back.
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And as always, be sure to follow us on Facebook, YouTubeand Twitter!
By Brian Galvin.
# How to Approach Difficult GMAT Problems
My students have a hard time understanding what makes a difficult GMAT question difficult. They assume that the tougher questions are either testing something they don’t know, or that these problems involve a dizzying level of complexity that requires an algebraic proficiency that’s simply beyond them.
One of my main goals in teaching a class is to persuade everyone that this is not, in fact, how hard questions work on this test. Hard questions don’t ask you do to something you don’t know how to do. Rather, they’re cleverly designed to provoke an anxiety response that makes it difficult to realize that you do know exactly how to solve the problem.
Take this official question, for example:
Let a, b, c and d be nonzero real numbers. If the quadratic equation ax(cx + d) = -b(cx +d) is solved for x, which of the following is a possible ratio of the 2 solutions?
A) –ab/cd
B) –ac/bd
D) ab/cd
Most students see this and panic. Often, they’ll start by multiplying out the left side of the equation, see that the expression is horrible (acx^2 + adx), and take this as evidence that this question is beyond their skill level. And, of course, the question was designed to elicit precisely this response. So when I do this problem in class, I always start by telling my students, much to their surprise, that every one of them already knows how to do this. They’ve just succumbed to the question writer’s attempt to convince them otherwise.
So let’s start simple. I’ll write the following on the board: xy = 0. Then I’ll ask what we know about x or y. And my students shrug and say x or y (or both) is equal to 0. They’ll also wonder what on earth such a simple identity has to do with the algebraic mess of the question they’d been struggling with.
I’ll then write this: zx + zy = 0. Again, I’ll ask what we know about the variables. Most will quickly see that we can factor out a “z” and get z(x+y) = 0. And again, applying the same logic, we see that one of the two components of the product must equal zero – either z = 0 or x + y = 0.
Next, I’ll ask if they would approach the problem any differently if I’d given them zx = -zy – they wouldn’t.
Now it clicks. We can take our initial equation in the aforementioned problem: ax(cx +d) = -b(cx+d), and see that we have a ‘cx + d’ on both sides of the equation, just as we’d had a “z” on both sides of the previous example. If I’m able to get everything on one side of the equation, I can factor out the common term.
Now ax(cx +d) = -b(cx+d) becomes ax(cx +d) + b(cx+d) = 0.
Just as we factored out a “z” in the previous example, we can factor out “cx + d” in this one.
Now we have (cx + d)(ax + b) = 0.
Again, if we multiply two expressions to get a product of zero, we know that at least one of those expressions must equal 0. Either cx + d = 0 or ax + b = 0.
If cx + d = 0, then x = -d/c.
If ax + b = 0, then x = -b/a.
Therefore, our two possible solutions for x are –d/c and –b/a. So, the ratio of the two would simply be (-d/c)/(-b/a). Recall that dividing by a fraction is the equivalent of multiplying by the reciprocal, so we’re ultimately solving for (-d/c)(-a/b). Multiplying two negatives gives us a positive, and we end up with da/cb, which is equivalent to answer choice E.
Takeaway: Anytime you see something on the GMAT that you think you don’t know how to do, remind yourself that the question was designed to create this false impression. You know how to do it – don’t hesitate to dive in and search for how to apply this knowledge.
By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.
# How to Go From a 48 to 51 in GMAT Quant – Part VII
Both a test-taker at the 48 level and one at the 51 level in the GMAT Quant section, are conceptually strong – given an unlimited time frame, both will be able to solve most GMAT questions correctly. The difference lies in the two things a test-taker at the 51 level does skillfully:
1. Uses holistic, big-picture methods to solve Quant questions.
2. Handles questions he or she finds difficult in a timely manner.
We have been discussing holistic methods on this blog for a long time now and will continue discussing them. (Before you continue reading, be sure to check out parts I, II, III, IVV and VI of this series.)
Today we will focus on “handling the hard questions in a timely manner.” Note that we do not say “solving the hard questions in a timely manner.” Occasionally, one might be required to make a quick call and choose to guess and move on – but again, that is not the focus of this post. We are actually going to talk about the “lightbulb” moment that helps us save on time. There are many such moments for the 51 level test-taker – in fact, the 51 scorers often have time left over after attempting all these questions.
Test takers at the 48 level will also eventually reach the same conclusions but might need much more time. That will put pressure on them the next time they look at the ticking clock, and once their cool is lost, “silly errors” will start creeping in. So it isn’t about just that one question – one can end up botching many other questions too.
There are many steps that can be easily avoided by a lightbulb moment early on. This is especially true for Data Sufficiency questions.
Let’s take an official example:
Pam owns an inventory of unopened packages of corn and rice, which she has purchased for \$17 and \$13 per package, respectively. How many packages of corn does she have ?
Statement 1: She has \$282 worth of packages.
Statement 2: She has twice as many packages of corn as of rice.
A high scorer will easily recognize that this question is based on the concept of “integral solutions to an equation in two variables.” Since, in such real world examples, x and y cannot be negative or fractional, these equations usually have a finite number of solutions.
After we find one solution, we will quickly know how many solutions the equation has, but getting the first set of values that satisfy the equation requires a little bit of brute force.
The good thing here is that this is a Data Sufficiency question – you don’t need to find the actual solution. The only thing we need is to establish that there is a single solution only. (Obviously, there has to be a solution since Pam does own \$282 worth of packages.)
So, the test-taker will start working on finding the first solution (using the method discussed in this post). We are told:
Price of a packet of corn = \$17
Price of a packet of rice = \$13
Say Pam has “x” packets of corn and “y” packets of rice.
Statement 1: She has \$282 worth of packages
Using Statement 1, we know that 17x + 13y = 282.
We are looking for the integer values of x and y.
If x = 0, y will be 21.something (not an integer)
If x = 1, y = 20.something
If x = 2, y = 19.something
If x = 3, y = 17.something
This is where the 51 level scorer stops because they never lose sight of the big picture. The “lightbulb” switches on, and now he or she knows that there will be only one set of values that can satisfy this equation. Why? Because y will be less than 17 in the first set of values that satisfies this equation. So if we want to get the next set that satisfies, we will need to subtract y by 17 (and add 13 to x), which will make y negative.
So in any case, there will be a unique solution to this equation. We don’t actually need to find the solution and hence, nothing will be gained by continuing these calculations. Statement 1 is sufficient.
Statement 2: She has twice as many packages of corn as of rice.
Statement 2 gives us no information on the total number of packages or the total amount spent. Hence, we cannot find the total number of packages of corn using this information alone. Therefore, our answer is A.
I hope you see how you can be alert to what you want to handle these Quant questions in a timely manner.
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
# GMAT Tip of the Week: The Overly Specific Question Stem
For most of our lives, we ask and answer relatively generic questions: “How’s it going?” “What are you up to this weekend?” “What time do the Cubs play tonight?”
And think about it, what if those questions were more specific: “Are you in a melancholy mood today?” “Are you and Josh going to dinner at Don Antonio’s tonight and ordering table-side guacamole?” “Do the Cubs play at 7:05 tonight on WGN?” If someone is asking those questions instead, you’re probably a bit suspicious. Why so specific? What’s your angle?
The same is true on the GMAT. Most of the question stems you see are relatively generic: “What is the value of x?” “Which of the following would most weaken the author’s argument?” So when the question stem get a little too specific, you should become a bit suspicious. What’s the test going for there? Why so specific?
The overly-specific Critical Reasoning question stem is a great example. Consider the problem:
Raisins are made by drying grapes in the sun. Although some of the sugar in the grapes is caramelized in the process, nothing is added.
Moreover, the only thing removed from the grapes is the water that evaporates during the drying, and water contains no calories or nutrients.
The fact that raisins contain more iron per food calorie than grapes do is thus puzzling.
Which one of the following, if true, most helps to explain why raisins contain more iron per calorie than do grapes?
(A) Since grapes are bigger than raisins, it takes several bunches of grapes to provide the same amount of iron as a handful of raisins does.
(B) Caramelized sugar cannot be digested, so its calories do not count toward the food calorie content of raisins.
(C) The body can absorb iron and other nutrients more quickly from grapes than from raisins because of the relatively high water content of grapes.
(D) Raisins, but not grapes, are available year-round, so many people get a greater share of their yearly iron intake from raisins than from grapes.
(E) Raisins are often eaten in combination with other iron-containing foods, while grapes are usually eaten by themselves.
Look at that question stem: a quick scan naturally shows you that you need to explain/resolve a paradox, but the question goes into even more detail for you. It reaffirms the exact nature of the paradox – it’s not about “iron,” but instead that that raisins contain more iron per calorie than grapes do. By adding that extra description into the question stem, the testmaker is practically yelling at you, “Make sure you consider calories…don’t just focus on iron!” And therefore, you should be prepared for the correct answer B, the only one that addresses calories, and deftly avoid answers A, C, D, and E, which all focus only on iron (and do so tangentially to the paradox).
Strategically speaking, if a Critical Reasoning question stem gets overly specific, you should pay particular attention to the specificity there…it’s most likely directing you to the operative portion of the argument.
Overly specific questions are most helpful in Data Sufficiency questions (and that same logic will help on Problem Solving too, as you’ll see). The testmaker knows that you’ve trained your entire algebraic life to solve for individual variables. So how can a question author use that lifetime of repetition against you? By asking you to solve for a specific combination that doesn’t require you to find the individual values. Consider this example, which appears courtesy the Official Guide for GMAT Quantitative Review:
If x^2 + y^2 = 29, what is the value of (x – y)^2?
(1) xy = 10
(2) x = 5
Two major clues should stand out to you that you need to Leverage Assets on this problem. For one, using both statements together (answer choice C) is dead easy. If xy = 10 and x = 5 then y = 2 and you can solve for any combination of x and y that anyone could ever ask for. But secondly and more subtly, the question stem should jump out as a classic way-too-specific, Leverage Assets question stem. They asked for a really, really specific value: (x – y)^2.
Now, immediately upon seeing that specificity you should be thinking, “That’s too specific…there’s probably a way to solve for that exact value without getting x and y individually.” That thought process alone tells you where to spend your time – you want to really leverage Statement 1 to try to make it work alone.
And if you’re still unconvinced, consider what the specificity does: the “squared” portion removes the question of negative vs. positive from the debate, removing one of the most common reasons that a seemingly-sufficient statement just won’t work. And, furthermore, the common quadratic (x – y)^2 shares an awful lot in common with the x^2 and y^2 elsewhere in the question stem. If you expand the parentheses, you have “What is x^2 – 2xy + y^2?” meaning that you’re already 2/3 of the way there (so to speak), since they’ve spotted you the sum x^2 + y^2.
The important strategy here is that the overly-specific question stem should scream “LEVERAGE ASSETS” and “You don’t need to solve for x and y…there’s probably a way to solve directly for that exact combination.” Since you know that you’re solving for the expanded x^2 – 2xy + y^2, and you already know that x^2 + y^2 = 29, you’re really solving for 29 – 2xy. Since you know from Statement 1 that xy = 20, then 29 – 2xy will be 29 – 2(10), which is 9.
Statement 1 alone is sufficient, even though you don’t know what x and y are individually. And one of the major signals that you should recognize to help you get there is the presence of an overly specific question stem.
So remember, in a world of generic questions, the oddly specific question should arouse a bit of suspicion: the interrogator is up to something! On the GMAT, you can use that to your advantage – an overly specific Critical Reasoning question usually tells you exactly which keywords are the most important, and an overly specific Data Sufficiency question stem begs for you to leverage assets and find a way to get the most out of each statement.
By Brian Galvin.
# Don’t Swim Against the Arithmetic Currents on the GMAT Quant Section
When I was a child, I was terrified of riptides. Partially, this was a function of having been raised by unusually neurotic parents who painstakingly instilled this fear in me, and partially this was a function of having inherited a set of genes that seems to have predisposed me towards neuroticism. (The point, of course, is that my parents are to blame for everything. Perhaps there is a better venue for discussing these issues.)
If there’s a benefit to fears, it’s that they serve as potent motivators to find solutions to the troubling predicaments that prompt them. The solution to dealing with riptides is to avoid struggling against the current. The water is more powerful than you are, so a fight is a losing proposition – rather, you want to wait for an opportunity to swim with the current and allow the surf to bring you back to shore. There’s a profound wisdom here that translates to many domains, including the GMAT.
In class, whenever we review a strategy, my students are usually comfortable applying it almost immediately. Their deeper concern is about when to apply the strategy, as they’ll invariably find that different approaches work with different levels of efficacy on different problems. Moreover, even if one has a good strategy in mind, the way the strategy is best applied is often context-dependent. When we’re picking numbers, we can say that x = 2 or x = 100 or x = 10,000; the key is not to go in with a single approach in mind. Put another way, don’t swim against the arithmetic currents.
Let’s look at some questions to see this approach in action:
At a picnic there were 3 times as many adults as children and twice as many women as men. If there was a total of x men, women, and children at the picnic, how many men were there, in terms of x?
A) x/2
B) x/3
C) x/4
D) x/5
E) x/6
The moment we see “x,” we can consider picking numbers. The key here is contemplating how complicated the number should be. Swim with the current – let the question tell you. A quick look at the answer choices reveals that x could be something simple. Ultimately, we’re just dividing this value by 2, 3, 4, 5, or 6.
Keeping this in mind, let’s think about the first line of the question. If there are 3 times as many adults as children, and we’re keeping things simple, we can say that there are 3 adults and 1 child, for a total of 4 people. So, x = 4.
Now, we know that among our 3 adults, there are twice as many women as men. So let’s say there are 2 women and 1 man. Easy enough. In sum, we have 2 women, 1 man, and 1 child at this picnic, and a total of 4 people. The question is how many men are there? There’s just 1! So now we plug x = 4 into the answers and keep going until we find x = 1. Clearly x/4 will work, so C is our answer. The key was to let the question dictate our approach rather than trying to impose an approach on the question.
Let’s try another one:
Last year, sales at Company X were 10% greater in February than in January, 15% less in March than in February, 20% greater in April than in March, 10% less in May than in April, and 5% greater in June than in May. On which month were sales closes to the sales in January?
A) February
B) March
C) April
D) May
E) June
Great, you say. It’s a percent question. So you know that picking 100 is often a good idea. So, let’s say sales in January were 100. If we want the month when sales were closest to January’s level, we want the month when sales were closest to 100, Sales in February were 10% greater, so February sales were 110. (Remember that if sales increase by 10%, we can multiply the original number by 1.1. If they decrease by 10% we could multiply by 0.9, and so forth.)
So far so good. Sales in March were 15% less than in February. Well, if sales in Feb were 110, then the sales in March must be 110*(0.85). Hmm… A little tougher, but not insurmountable. Now, sales in April were 20% greater than they were in March, meaning that April sales would be 110*(0.85)*1.2. Uh oh. Once you see that sales are 10% less in May than they were in April, we know that sales will be 110*(0.85)*1.2*0.9.
Now you need to stop. Don’t swim against the current. The arithmetic is getting hard and is going to become time-consuming. The question asks which month is closest to 100, so we don’t have to calculate precise values. We can estimate a bit. Let’s double back and try to simplify month by month, keeping things as simple as possible.
Our February sales were simple: 110. March sales were 110*0.85 – an unpleasant number. So, let’s try thinking about this a little differently. 100*0.85 = 85. 10*0.85 = 8.5. Add them together and we get 85 + 8.5 = 93.5. Let’s make life easier on ourselves – we’ll round up, and call this number 94.
April sales are 20% more than March sales. Well, 20% of 100 is clearly 20, so 20% of 94 will be a little less than that. Say it’s 18. Now sales are up to 94 + 18 = 112. Still not close to 100, so we’ll keep going.
May sales are 10% less than April sales. 10% of 112 is about 11. Subtract 11 from 112, and you get 101. We’re looking for the number closest to 100, so we’ve got our answer – it’s D, May.
Takeaway: Don’t try to impose your will on GMAT questions. Use the structural clues of the problems to dictate how you implement your strategy, and be prepared to adjust midstream. The goal is never to conquer the ocean, but rather, to ride the waves to calmer waters.
By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.
# How to Simplify Percent Questions on the GMAT
One of the most confounding aspects of the GMAT is its tendency to make simple concepts seem far more complex than they are in reality. Percent questions are an excellent example of this.
When I introduce this topic, I’ll typically start by asking my class the following question: If you’ve completed 10% of a project how much is left to do? I have never, in all my years of teaching, had a class that was unable to tell me that 90% of the project remains. It’s more likely that they’ll react as though I’m insulting their collective intelligence. And yet, when test-takers see this concept under pressure, they’ll often fail to recognize it.
Take the following question, for example:
Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session. When the readout indicated 24 min 18 sec, she had completed 10% of her exercise session. The readout indicated which of the following when she had completed 40% of her exercise session.
(A) 10 min. 48 sec.
(B) 14 min. 52 sec.
(C) 14 min. 58 sec.
(D) 16 min. 6 sec.
(E) 16 min. 12 sec.
Hopefully, you’ve noticed that this question is testing the same simple concept that I use when introducing percent problems to my class. And yet, in my experience, a solid majority of students are stumped by this problem. The reason, I suspect, is twofold. First, that figure – 24 min. 18 sec. – is decidedly unfriendly. Painful math often lends itself to careless mistakes and can easily trigger a panic response. Second, anxiety causes us to work faster, and when we work faster, we’re often unable to recognize patterns that would be clearer to us if we were calm.
There’s interesting research on this. Psychologists, knowing that the color red prompts an anxiety response and that the color blue has a calming effect, conducted a study in which test-takers had to answer math questions – the questions were given to some subjects on paper with a red background and to other subjects on paper with a blue background. (The control group had questions on standard white paper.) The red anxiety-producing background noticeably lowered scores and the calming blue background boosted scores.
Now, the GMAT doesn’t give you a red background, but it does give you unfriendly-seeming numbers that likely have the same effect. So, this question is as much about psychology as it is about mathematical proficiency. Our job is to take a deep breath or two and rein in our anxiety before we proceed.
If Dara has completed 10% of her workout, we know she has 90% of her workout remaining. So, that 24 min. 18 sec. presents 90% of her total workout. If we designate her total workout time as “t,” we end up with the following equation:
24 min. 18 sec. = 0.90t
Let’s work with fractions to solve. 18 seconds is 18/60 minutes, which simplifies to 3/10 minutes. 0.9 is 9/10, so we can rewrite our equation as:
24 + 3/10 = (9/10)t
(243/10) = (9/10)t
(243/10)*(10/9) = t
27 = t
Not so bad. Dara’s full workout is 27 minutes long.
We want to know how much time is remaining when Dara has completed 40% of her workout. Well, if she’s completed 40% of her workout, we know she has 60% of her workout remaining. If her full workout is 27 minutes, then 60% of this value is 0.60*27 = (3/5)*27 = 81/5 = 16 + 1/5, or 16 minutes 12 seconds. And we’ve got our answer: E.
Now, let’s say you get this problem with 20 seconds remaining on the clock and you simply don’t have time to solve it properly. Let’s estimate.
Say, instead of 24 min 18 seconds remaining, Dara had 24 minutes remaining (so we know we’re going to underestimate the answer). If that’s 90% of her workout time, 24 = (9/10)t, or 240/9 = t.
We want 60% of this, so we want (240/9)*(3/5).
Because 240/5 = 48 and 9/3 = 3, (240/9)*(3/5) = 48/3 = 16.
We know that the correct answer is over 16 minutes and that we’ve significantly underestimated – makes sense to go with E.
Takeaway: Don’t let the question-writer trip you up with figures concocted to make you nervous. Take a breath, and remember that the concepts being tested are the same ones that, when boiled down to their essence, are a breeze when we’re calm.
By David Goldstein, a Veritas Prep GMAT instructor based in Boston. You can find more articles written by him here.
# GMAT Tip of the Week: The Least Helpful Waze To Study
If you drive in a large city, chances are you’re at least familiar with Waze, a navigation app that leverages user data to suggest time-saving routes that avoid traffic and construction and that shave off seconds and minutes with shortcuts on lesser-used streets.
And chances are that you’ve also, at some point or another, been inconvenienced by Waze, whether by a devout user cutting blindly across several lanes to make a suggested turn, by the app requiring you to cut through smaller streets and alleys to save a minute, or by Waze users turning your once-quiet side street into the Talladega Superspeedway.
To its credit, Waze is correcting one of its most common user that it often leads users into harrowing and time-consuming left turns. But another major concern still looms, and it’s one that could damage both your fender and your chances on the GMAT:
Beware the shortcuts and “crutches” that save you a few seconds, but in doing so completely remove all reasoning and awareness.
With Waze, we’ve all seen it happen: someone so beholden to, “I must turn left on 9th Street because the app told me to!” will often barrel through two lanes of traffic – with no turn signal – to make that turn…not realizing that the trip would have taken the exact same amount of time, with much less risk to the driver and everyone else on the road, had he waited a block or two to safely merge left and turn on 10th or 11th. By focusing so intently on the app’s “don’t worry about paying attention…we’ll tell you when to turn” features, the driver was unaware of other cars and of earlier opportunities to safely make the merge in the desired direction.
The GMAT offers similar pitfalls when examinees rely too heavily on “turn your brain off” tricks and techniques. As you learn and practice them, strategies like the “plumber butt” for rates and averages may seem quick, easy, and “turn your brain off” painless. But the last thing you want to do on a higher-order thinking test like the GMAT is completely turn your brain off. For example, a “turn your brain off” rate problem might say:
John drives at an average rate of 45 miles per hour. How many miles will he drive in 2.5 hours?
And using a Waze-style crutch, you could remember that to get distance you multiply time by rate so you’d get 112.5 miles. That may be a few seconds faster than performing the algebra by thinking “Rate = Distance over Time”; 45 = D/2.5; 45(2.5) = D; D = 112.5.
But where a shortcut crutch saves you time on easier problems, it can leave you helpless on longer problems that are designed to make you think. Consider this Data Sufficiency example:
A factory has three types of machines – A, B, and C – each of which works at its own constant rate. How many widgets could one machine A, one Machine B, and one Machine C produce in one 8-hour day?
(1) 7 Machine As and 11 Machine Bs can produce 250 widgets per hour
(2) 8 Machine As and 22 Machine Cs can produce 600 widgets per hour
Here, simply trying to plug the information into a simple diagram will lead you directly to choice E. You simply cannot separate the rate of A from the rate of B, or the rate of B from the rate of C. It will not fit into the classic “rate pie / plumber’s butt” diagram that many test-takers use as their “I hate rates so I’ll just do this trick instead” crutch.
However, those who have their critical thinking mind turned on will notice two things: that choice E is kind of obvious (the algebra doesn’t get you very close to solving for any one machine’s rate) so it’s worth pressing the issue for the “reward” answer of C, and that if you simply arrange the algebra there are similarities between the number of B and of C:
7(Rate A) + 11(Rate B) = 250
8(Rate A) + 22(Rate C) = 600
Since 11 is half of 22, one way to play with this is to double the first equation so that you at least have the same number of Bs as Cs (and remember…those are the only two machines that you don’t have “together” in either statement, so relating one to the other may help). If you do, then you have:
14(A) + 22(B) = 500
8(A) + 22(C) = 600
Then if you sum the questions (Where does the third 22 come from? Oh, 14 + 8, the coefficients for A.), you have:
22A + 22B + 22C = 1100
So, A + B + C = 50, and now you know the rate for one of each machine. The two statements together are sufficient, but the road to get there comes from awareness and algebra, not from reliance on a trick designed to make easy problems even easier.
The lesson? Much like Waze, which can lead to lack-of-awareness accidents and to shortcuts that dramatically up the degree of difficulty for a minimal time savings, you should take caution when deciding to memorize and rely upon a knee-jerk trick in your GMAT preparation.
Many are willing (or just unaware that this is the decision) to sacrifice mindfulness and awareness to save 10 seconds here or there, but then fall for trap answers because they weren’t paying attention or become lost when problems are more involved because they weren’t prepared.
So, be choosy in the tricks and shortcuts you decide to adopt! If a shortcut saves you a or two of calculations, it’s worth the time it takes to learn and master it (but probably never worth completely avoiding the “long way” or knowing the general concept). But if its time savings are minimal and its grand reward is that, “Hey, you don’t have to understand math to do this!” you should be wary of how well it will serve your aspirations of scores above around 600.
Don’t let these slick shortcut waze of avoiding math drive you straight into an accident. Unless the time savings are game-changing, you shouldn’t make a trade that gains you a few seconds of efficiency on select, easier problems in exchange for your awareness and understanding.
Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And as always, be sure to follow us on, YouTubeGoogle+ and Twitter!
By Brian Galvin.
# GMAT Tip of the Week: The Curry Twos Remind You To Keep The GMAT Simple
Happy Friday from Veritas Prep headquarters, where we’re actively monitoring the way that Twitter is reacting to UnderArmour’s release of the new Steph Curry shoes. What’s the problem with the Curry Twos? Essentially they’re too plain and buttoned up – much more Mickelson than Michael, son.
OK, so what? The Curry 2s are more like the Curry 401(k)s. Why should that matter for your GMAT score?
Because on the GMAT, you want to be as simple and predictable as a Steph Curry sneaker.
What does that mean? One of the biggest study mistakes that people make is that once they’ve mastered a core topic like “factoring” or “verb tenses,” they move on to more obscure topics and spend their valuable study time on those.
There are two major problems with this: 1) the core topics appear much more often and are much more repeatable, and 2) in chasing the obscure topics later in their study regimen, people spend the most valuable study time – that coming right before the test – feverishly memorizing things they probably won’t see or use at the expense of practicing the skills and strategies that they’ll need to use several times on test day.
Consider an example: much like Twitter is clowning the Curry Twos, a handful of Veritas Prep GMAT instructors were laughing this time last week about an explanation in a practice test (by a company that shall remain nameless…) for a problem similar to:
Two interconnected, circular gears travel at the same circumferential rate. If Gear A has a diameter of 30 centimeters and Gear B has a diameter of 50 centimeters, what is the ratio of the number of revolutions that Gear A makes per minute to the number of revolutions that Gear B makes per minute?
(A) 3:5
(B) 9:25
(C) 5:3
(D) 25:9
(E) Cannot be determined from the information provided
Now, the “Curry Two” approach – the tried and true, “don’t-overcomplicate-this-for-the-sake-of-overcomplicating-it” method – is to recognize that the distance around any circle (a wheel, a gear, etc.) is its circumference. And circumference is pi * diameter. So, if each gear travels the same circumferential distance, that distance for any given period of time is “circumference * number of revolutions.” That then means that the circumference of A times the number of revolutions of A is equal to the circumference of B times the number of revolutions for B, and you know that’s:
30π * A = 50π * B (where A = # of revolutions for A, and B = # of revolutions for B). Since you want the ratio of A:B, divide both sides by B and by 30, and you have A/B = 50/30, or A:B = 5:3 (answer choice C).
Why were our instructors laughing? The explanation began, “There is a simple rule for interconnected gears…” Which is great to know if you see a gear-based question on the test or become CEO of a pulley factory, but since the GMAT officially tests “geometry,” you’re much better off recognizing the relationship between circles, circumferences, and revolutions (for questions that might deal with gears, wheels, windmills, or any other type of spinning circles) than you are memorizing a single-use rule about gears.
Problems like this offer the “Curry Two” students a fantastic opportunity to reinforce their knowledge of circles, their ability to think spatially about shapes, etc. But, naturally, there are students who will add “gear formula” to their deck of flashcards and study that single-use rule (which 99.9% of GMAT examinees will never have the opportunity to use) with the same amount of time/effort/intensity as they revisit the Pythagorean Theorem (which almost everyone will use at least twice).
Hey, the Curry Twos are plain, boring, and predictable, as are the core rules and skills that you’ll use on the GMAT. But simple, predictable, and repeatable are what win on this test, so heed this lesson. As 73 regular season opponents learned this basketball season, Curry Twos lead to countless Curry 3s, and on the GMAT, “Curry Two” strategies will help you curry favor with admissions committees by leading to Curry 700+ scores.
By Brian Galvin.
# How to Go From a 48 to 51 in GMAT Quant – Part VI
Today’s post the next part in our “How to Go From 48 to 51 in Quant” series. Again, we will learn a technique that can be employed by the test-taker at an advanced stage of preparation – requiring one to understand the situations in which one can use this simplifying technique.
(Before you continue reading, be sure to check out parts I, II, III, IV, and V of this series.)
We all love to use the plug-in method on GMAT Quant questions. We have an equation given, and if the answer choices are the possible values of x, we just plug in these values to find the one that satisfies the equation.
But what if the answer choices are all complicated values? What if it seems that five times the calculation (in the worst case) will be far more time consuming than actually solving the given equation? Then one is torn between using the favorite plug-in method and using algebra. Let’s take an example to review the methods we can use to solve the question and learn how to simplify the plug-in process by approximating the five available options:
If |4x−4|=|2x+30|, which of the following could be a value of x?
(A) –35/3
(B) −21/2
(C) −13/3
(D) 11/5
(E) 47/5
This question is an ideal candidate for the “plug-in” method. Here, you have the absolute value equation with the potential values of x given in the answer choices. The problem is that the values of x given are fractional. Of course, if we do plan to solve the equation rather than “plug-in”, we can still solve it using our holistic approach rather than pure algebra. Let’s take a look at that now, and later we will discuss the trick to making the answer choices easier for us to plug in.
Method 1:
|4x – 4| = |2x + 30|
4 * |x – 1| = 2 * |x + 15|
2 * |x – 1| = |x + 15|
This is how we rephrase the equation in our words: twice the distance of x from 1 should be equal to the distance of x from -15.
——————(-15) —————————————————(0)——(1)——————
There are two ways to find the value of x:
Case 1: x could be between -15 and 1 such that the distance between them is split in the ratio 2:1.
or
Case 2: x could be to the right of 1 such that the distance between x and -15 is twice the distance between x and 1.
Let’s examine both of these cases in further detail:
Case 1: The distance from -15 to 1 is of 16 units – this can be split into 3 sections of 16/3 units each. So, the distance of x from 1 should be 16/3, which would make the distance of x from -15 two times 16/3, i.e. 32/3.
So, x should be at a point 16/3 away from 1 toward the left.
x = 1 – 16/3 = -13/3
This is one of our answer choices and, hence, the correct answer. Normally, we would just move on to the next question at this point, but had we not found -13/3 in the answer options, we would have moved on to Case 2:
Case 2: The distance between -15 and 1 is 16 units. x should be an additional 16 units to the right of 1, so the distance between x and 1 is 16 and the distance between x and -15 is two times 16, i.e. 32. This means that x should be 16 units to the right of 1, i.e. x = 17. If you would not have found -13/3 in the answer choices, then you would have found 17.
Now let’s move on to see how we can make the plug-in method work for us in this case by examining each answer choice we are given:
Method 2:
|4x – 4| = |2x + 30|
2 * |x – 1| = |x + 15|
(A) -35/3
It is difficult to solve for x = -35/3 to see if both sides match. Instead, let’s solve for the closest integer, -12.
2 * |-12 – 1| = |-12 + 15|
On the left-hand side, you will get 26, but on the right-hand side, you will get 3.
These values are far away from each other, so x cannot be -35/3. As the value of x approaches the point where the equation holds – i.e. where the two sides are equal to each other – the gap between the value of the two sides keeps reducing. With such a huge gap between the value of the two sides in this case, it is unlikely that a small adjustment of -35/3 from -12 will bring the two sides to be equal.
(B) -21/2
For this answer choice, let’s solve for the nearest integer, x = -10.
2 * |-10 – 1| = |-10 + 15|
On the left-hand side, you will get 22; on the right-hand side, you will get 5.
Once again, these values are far away from each other and, hence, x will not be -21/2.
(C) -13/3
For this answer choice, let’s solve for x = -4.
2 * |-4 -1| = |-4 + 15|
On the left-hand side, you will get 10; on the right-hand side, you will get 11.
Here, there is a possibility that x can equal -13/3, as the two sides are so close to one another – plug in the actual value of -13/3 and you will see that the left-hand side of the equation does, in fact, equal the right-hand side. Therefore, C is the correct answer.
Basically, we approximated the answer choices we were given and shortlisted the one that gave us very close values. We checked for that and found that it is the answer.
We can also solve this question using pure algebra (taking positive and negative signs of absolute values) but in my opinion, the holistic Method 1 is almost always better than that. Out of the two methods discussed above, you can pick the one you like better, but note that Method 2 does have limited applications – only if you are given the actual values of x, can you use it. Method 1 is far more generic for absolute value questions.
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# Math help from the Learning Centre
This guide provides useful resources for a wide variety of math topics. It is targeted at students enrolled in a math course or any other Centennial course that requires math knowledge and skills.
## Introduction to Annuities
What is an annuity? It is a series of equal payments which occurs at equal time intervals - usually monthly, quarterly, semi-annually or annually. There are two important terms of annuities - ordinary and simple.
Ordinary Annuity : Has payments at the end of each time interval. There is no payment at the start of the term of the annuity.
Simple Annuity : Frequency of the payment is the same as the frequency of the compound interest.
Below are the definitions of the terms we use for ordinary simple annuities:
$$S_n \ or \ FV =$$ future value at $$n$$ payment
$$\ \ \ \ \ \ \ \ \ \ \ \ \ R =$$ regular payment amount
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i = \frac{j}{m} =$$ periodic rate of interest
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ j =$$ nominal annual rate of interest
$$\ \ \ \ \ \ \ \ \ \ \ \ \ m =$$ compounded frequency daily: $$m = 365$$; monthly: $$m = 12$$; quarterly $$m = 4$$
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ n = (length \ of \ period) \times m =$$ time
$$A_n \ or \ PV =$$ present value at $$n$$ payment
## Future Value and Present Value of a Simple Ordinary Annuity
The future value ($$FV$$ or $$S_{n}$$) of an annuity is the value of a current asset at a future date based on an assumed rate of growth. The future value is important to investors and financial planners as they use it to estimate how much an investment made today will be worth in the future.
$S_{n} = \frac{R[(1+i)^n - 1]}{i}$
Example 1
Suppose you invested $1000 per quarter over a 15 year period. If money earns an annual rate of 6.5% compounded quarterly, what would be the value at the end of the time period? Solution $$\Longrightarrow S_{n} = ?$$ | $$R = 1000$$ | $$i = \frac{6.5\%}{4} = 0.01625$$ | $$n = 15*4 = 60$$ $$\Longrightarrow S_{60} = \frac{1000[(1+0.01625)^{60} - 1]}{0.01625} = 10, 033.67$$ The present value of an annuity is the amount of cash today equivalent in value to a payment, or to a stream of payments, to be received in the future. It is calculated using the following formula: $A_{n} = \frac{R[1- (1+i)^{-n}]}{i}$ Example 2 Suppose you with to be able to withdraw$3000 at the end of each month for two years. How much money must you deposit now at 2.75% interest compounded monthly?
Solution
$$\Longrightarrow A_{n} = ?$$ | $$R = 3000$$ | $$i = \frac{2.75\%}{12} = 0.002291666$$ | $$n = 2*12 = 24$$
$$\Longrightarrow A_{24} = \frac{3000[1-(1+0.002291666)^{-24}]}{0.002291666} = 69, 977.66$$
## Calculating the Payment of a Simple Ordinary Annuity
Since there are different formulas for the present and future values of annuities. The first step is to determine whether the given value of the question is the future or present value. If we rearranged the formulas to solve for the payment, we have the following.
$R=\frac{i\cdot S_n}{\left(1+i\right)^n-1}$
$R=\frac{i\cdot A_n}{1-\left(1+i\right)^{-n}}$
Let's try the following example.
Example
How much would you have to pay into an account at the end of every 6 months to accumulate $10,000 in eight years if interest is 3% compounded semi-annually? Solution To solve any annuity, we need to pick out the important pieces of information from the question. • Payment is at the end of the period which implies this is an ordinary annuity. • Payment is every 6 months and compounding is semi-annual. Since it is the same this is a simple annuity. • The nominal interest rate, j, is 3% compounded semi-annually. • Thus, the periodic interest rate is $$i=\frac{0.03}{2}=0.015$$. • There are payments every 6 months for eight years, so $$n=16$$. • We are trying to accumulate to$10,000. Since this is a value after all the payments, this is the future value.
\begin{align}R&=\frac{i\cdot S_n}{\left(1+i\right)^n-1}\\&=\frac{0.015\cdot 10,000}{\left(1+0.015\right)^{16}-1}\\&=557.65\end{align}
Therefore, a regular payment of $557.65 every 6 months for eight years will be required to accumulate to$10,000.
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## Lesson: Ordering Greater Numbers Introducing the Concept
Before students order greater numbers, use this lesson to reinforce their understanding of place value of greater numbers.
Materials: transparency made from Learning Tool 4 in the Learning Tools Folder
Preparation: Write the value of each place on Learning Tool 4 and then make a transparency.
Prerequisite Skills and Background: Students should know place value to hundreds.
Write 330 on the place-value chart.
• Ask: What digit is in the ones place? (0) What is the value of the digit 0? (0)
Repeat these questions for the digits in the tens and hundreds places. (tens place: 3, value: 30; hundreds place: 3, value: 300)
• Ask: What is the number?
Have the class read the number aloud. (three hundred thirty)
Point to the thousands, ten thousands, and hundred thousands places.
• Say: These are the thousands places. This is the thousands place. Its value is 1,000. This is the ten thousands place. Its value is 10,000. This is the hundred thousands place. Its value is 100,000.
Write the digit 9 in the thousands place.
• Ask: What is the value of the digit 9? (9,000)
• Say: Read the number aloud. (nine thousand, three hundred thirty)
Write the digit 8 in the ten thousands place.
• Ask: What is the value of the digit 8? (80,000) What is the value of the digit 9? (9,000) So what is the total value of the thousands places? (89,000)
• Say: Read the number aloud. (eighty-nine thousand, three hundred thirty)
Write the digit 6 in the hundred thousands place.
• Ask: What is the value of the digit 6 in the hundred thousands place? (600,000)
• Ask: What is the total value of the thousands places? (689,000)
• Say: Read the number aloud. (six hundred eighty-nine thousand, three hundred thirty)
• Ask: Which digit has the greatest value? (6) How do you know?
Elicit from students that the greatest place is the hundred thousands place, so 6 has the greatest value.
• Ask: Is the value of the digit 9 greater than or less than the value of the digit 8?
Students should realize that 9,000 is less than 80,000, so the value of the digit 9 is less than the value of the digit 8.
• Ask: Which places have the same digit? (the hundreds and tens places) Do the digits have the same value?
Students should realize that the digit 3 in the hundreds place has a value of 300 and the digit 3 in the tens place has a value of 30.
• Ask: Since the value of the digit in the ones place is 0, could we just drop the 0? Why or why not?
Lead students to discover that without the 0, the number would be 68,933.
Replace 689,330 with 455,072. Ask questions similar to those above.
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1975 AHSME Problems/Problem 28
Problem 28
In $\triangle ABC$ shown in the adjoining figure, $M$ is the midpoint of side $BC, AB=12$ and $AC=16$. Points $E$ and $F$ are taken on $AC$ and $AB$, respectively, and lines $EF$ and $AM$ intersect at $G$. If $AE=2AF$ then $\frac{EG}{GF}$ equals
$[asy] draw((0,0)--(12,0)--(14,7.75)--(0,0)); draw((0,0)--(13,3.875)); draw((5,0)--(8.75,4.84)); label("A", (0,0), S); label("B", (12,0), S); label("C", (14,7.75), E); label("E", (8.75,4.84), N); label("F", (5,0), S); label("M", (13,3.875), E); label("G", (7,1)); [/asy]$
$\textbf{(A)}\ \frac{3}{2} \qquad \textbf{(B)}\ \frac{4}{3} \qquad \textbf{(C)}\ \frac{5}{4} \qquad \textbf{(D)}\ \frac{6}{5}\\ \qquad \textbf{(E)}\ \text{not enough information to solve the problem}$
Solution
Here, we use Mass Points. Let $AF = x$. We then have $AE = 2x$, $EC = 16-2x$, and $FB = 12 - x$ Let $B$ have a mass of $2$. Since $M$ is the midpoint, $C$ also has a mass of $2$. Looking at segment $AB$, we have $$2 \cdot (12-x) = \text{m}A_{AB} \cdot x$$ So $$\text{m}A_{AB} = \frac{24-2x}{x}$$ Looking at segment $AC$,we have $$2 \cdot (16-2x) = \text{m}A_{AC} \cdot 2x$$ So $$\text{m}A_{AC} = \frac{16-2x}{x}$$ From this, we get $$\text{m}E = \frac{16-2x}{x} + 2 \Rrightarrow \text{m}E = \frac{16}{x}$$ and $$\text{m}F = \frac{24-2x}{x} + 2 \Rrightarrow \text{m}F = \frac{24}{x}$$ We want the value of $\frac{EG}{GF}$. This can be written as $$\frac{EG}{GF} = \frac{\text{m}F}{\text{m}E}$$ Thus $$\frac{\text{m}F}{\text{m}E} = \frac {\frac{24}{x}}{\frac{16}{x}} = \frac{3}{2}$$ $\boxed{A}$
~JustinLee2017
Solution 2
Since we only care about a ratio $EG/GF$, and since we are given $M$ being the midpoint of $\overline{BC}$, we realize we can conveniently also choose $E$ to be the midpoint of $\overline{AC}$. (we're free to choose any point $E$ on $\overline{AC}$ as long as $AE$ is twice $AF$, the constraint given in the problem). This means $AE = 16/2 = 8$, and $AF = 4$. We then connect $ME$ which creates similar triangles $\triangle EMC$ and $\triangle ABC$ by SAS, and thus generates parallel lines $\overline{EM}$ and $\overline{AB}$. This also immediately gives us similar triangles $\triangle AFG \sim \triangle MEG, => EG/FG = ME/AF = 6/4 = \boxed{3/2}$ (note that $ME = 6$ because $ME/AB$ is in $1:2$ ratio).
~afroromanian
Solution 3
In order to find $\frac{EG}{FG}$, we can apply the law of sine to this model. Let:
$\angle MAC=\alpha, \; \angle MAB=\beta, \; \angle AGE=\gamma,\; \angle AMC=\delta$
Then, in the $\triangle AMC$ and $\triangle AMB$:
$\frac{MC}{sin\alpha}=\frac{AC}{sin\delta};$ $\frac{MB}{sin\beta}=\frac{AB}{sin\delta}=\frac{MC}{sin\alpha}\cdot\frac{3}{4};$ $\frac{sin\alpha}{sin\beta}=\frac{3}{4}$
In the $\triangle AGE$ and $\triangle AGF$:
$\frac{FG}{sin\beta}=\frac{AF}{sin\gamma};$ $\frac{EG}{sin\alpha}=\frac{AE}{sin\gamma};$ $\frac{2FG}{sin\beta}=\frac{EG}{sin\alpha};$ $\frac{EG}{FG}=\frac{2sin\alpha}{sin\beta}=\frac{3}{2}$
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# Lesson Video: Arc Length by Integration Mathematics • Higher Education
In this video, we will learn how to use integration to find the length of a curve.
14:26
### Video Transcript
In this video, weβll learn how to use integration to find the length of a curve. By this stage, you might be aware of some of the key applications of integration, areas, volumes, and average values. But did you know that integration can also be used to find the length of a curve? In this video, weβll derive the formula for arc length using integration and then look at some of the key applications of this formula.
Suppose that a curve π is defined by the equation π¦ equals π of π₯, where π is a continuous function over the close interval π to π.
That might look a little something like this. We can then find an estimate to the length of π by dividing our interval into subintervals and then working out the length of the straight line that joins the function at these intervals as shown. Now, imagine the number of subintervals π gets ever larger. What will happen to our approximation? As the number of subintervals increase, each straight line gets smaller, and our approximation will get closer and closer to the exact value of the length of π. So weβre going to define the length of the curve πΏ with an equation π¦ equals π of π₯ as the limit of the length of these inscribed polygons, assuming that limit exists.
The problem is this isnβt a hugely useful formula for us. But it does allow us to derive an integral formula for πΏ given that π is continuous and differentiable over that closed interval π to π. But what we can do is work out the length of the line segments by using the distance formula or the Pythagorean theorem by defining Ξπ¦ subscript π as the difference between π¦ subscript π and π¦ subscript π minus one, which is, of course, equal to the difference between π π₯ subscript π minus π of π₯ subscript π minus one. We can say that the length of the line segments are given by the square root of π₯ π minus π₯ π minus one squared plus π¦ π minus π¦ π minus one square. And we can redefine that as the square to of Ξπ₯ π squared plus Ξπ¦ π squared.
Then by applying the mean value theorem to our function π on the close interval π₯ π minus one π₯ π, we find that thereβs this number π₯ π star between π₯ π minus one and π₯ π such that π of π₯ π minus π of π₯ π minus one is equal to the derivative of the function evaluated at this number π₯ star π times π₯ π minus π₯ π minus one. And then we redefine this using the notation from before. We find that Ξπ¦ π is equal to π prime of π₯ star π times Ξπ₯. We replaced Ξπ¦ π in our expression for the length of the line segments. And then we factor by the square root of Ξπ₯ squared. And we find that the length of our line segments is equal to Ξπ₯ times the square root of one plus π prime of π₯ star π squared.
We substitute this into our original limit. And we find that πΏ is equal to the limit as π₯ tends to infinity of the sum between π equals one and π of Ξπ₯ times the square root of one plus π prime of π₯ star π squared. By definition, we can say that this is equal to the integral between π and π of the square root of one of π prime of π₯ squared with respect to π₯. And weβve obtained the ultimate formula.
This says that if π prime is continuous on the close interval π to π. Then the length of the curve given by π¦ equals π of π₯, given that π₯ is greater than or equal to zero and less than or equal to π, is πΏ equals the definite integral evaluated between π and π of the square root of one plus π prime of π₯ squared with respect to π₯. Using Libenieβs notation, we can write this as the integral between π and π of the square root of one plus dπ¦ by dπ₯ squared evaluated with respect to π₯. Weβre now going to have a look at the application of this formula.
Calculate the arc length of the curve of π¦ equals the square root of four minus π₯ squared between π₯ equals zero and π₯ equals two, giving your answer to five decimal places.
Using Libenieβs notation, the formula for the arc length of a curve is given by the definite integral evaluated between π and π of the square root of one plus dπ¦ by dπ₯ squared with respect to π₯. We know that π¦ is equal to the square root of four minus π₯ squared, so weβll begin by working out dπ¦ by dπ₯. If we write π¦ as four minus π₯ squared to the power of one-half, then we can use the general power rule to find the derivative of this function with respect to π₯. Itβs a half times four minus π₯ squared to the power of negative half multiplied by the derivative of the function that sits inside the brackets. Thatβs negative two π₯.
Dividing through by two and rewriting dπ¦ by dπ₯, we obtain negative π₯ over the square root of four minus π₯ squared. We let π be equal to zero and π be equal to two. And this means the length of the curve that weβre interested in is equal to the definite integral between zero and two of the square root of one plus negative π₯ over the square of four minus π₯ squared squared with respect to π₯. And this simplifies quite nicely. The integrand becomes the square root of one plus π₯ squared over four minus π₯ squared. We can simplify the expression inside the square root by multiplying both the numerator and denominator of the number one by four minus π₯ squared. And when we add that, we get four minus π₯ squared plus π₯ squared over four minus π₯ squared. Negative π₯ squared plus π₯ squared is zero. And so our integrand becomes the square root of four over four minus π₯ squared.
This can be simplified even further. If we take out our factor of four, we see that the integrand could be written as the square root of four times one over the square of four minus π₯ squared. The square to four is two. And weβre allowed state constant factors outside of the integral. So we have that πΏ is equal to two times the definite integral evaluated between zero and two of one over the square of four minus π₯ squared. Now, this might look really nasty. But we can evaluate this integral by using a substitution. We recall that the derivative of arc sine of π₯ is one over the square root of one minus π₯ squared. So we rewrite our integrand slowly.
This time we take out our factor four on the denominator. And we see that two times one over the square of four cancels. We can also write π₯ squared over four as π₯ over two squared. We choose π’ then for our substitution to be equal to π₯ over two. Then dπ’ by dπ₯ is equal to one-half. And we can say this is the equivalent to saying two dπ’ equals dπ₯. Weβre also going to need to change the limits on our integral. When π₯ is equal to two, π’ is equal to two divided by two, which is one. And when π₯ is equal to zero, π’ is zero divided by two, which is zero.
So we now see that πΏ is equal to the definite integral between zero and one of one over the square root of one minus π’ squared times two dπ’. Once again, weβll take out our common factor of two. And we can now evaluate the integral of one over the square root of one minus π’ squared. Itβs simply arc sin of π’. And weβre looking to evaluate two times arc sine of π’ between one and zero. Thatβs two times arc sin of one minus arc sin of zero, which is simply π. And we see that correct to five decimal places, the arc length of the curve πΏ is 3.14159.
Weβre now going to look at how to find the arc length of a curve defined as π₯ in terms of π¦. For a curve with equation π₯ equals π of π¦, where π of π¦ is continuous and has a continuous derivative on the close interval π to π, the length of the curve between π¦ equals π and π¦ equals π is given by the definite integral evaluated between π and π of the square root of one plus dπ₯ by dπ¦ squared dπ¦. The application of this formula is just the same as the application of the previous formula we used. Letβs see what this might look like.
Find the length of the length of the arc of the curve π¦ squared equals π₯ between π¦ equals zero and π¦ equals four.
Remember, for a curve with equation π₯ equals π of π¦, where π of π¦ is continuous and has a continuous derivative on the close interval π to π, the arc length of the curve between π¦ equals π and π¦ equals π is given by the definite integral evaluated between π and π of the square root of one plus dπ₯ by dπ¦ squared dπ¦. Letβs compare this formula to our question.
Our function π₯ in terms of π¦ is π₯ equals π¦ squared. So we can see weβre going to need to differentiate π₯ with respect to π¦. The derivative of π₯ with respect to π¦ is two π¦. Weβre also told that we need to find the length of the arc between π¦ equals zero and π¦ equals four. So weβre going to say that π is equal to zero and π is equal to four. Substituting everything we know into this formula and we obtain πΏ to be equal to the definite integral between zero and four of the square root of one plus two π¦ squared, evaluated with respect to π¦. Well, two π¦ all squared is the same as four π¦ squared. So to find the length of the arc weβre interested in, we need to evaluate the definite integral of the square root of one plus four π¦ squared between the limits of zero and four. Weβre going to use our graphical calculator to do so. And when we do, we obtain that the length of the arc is equal to 16.81863 which is equal to 16.8 units correct to three significant figures.
We saw that the processes in the two examples weβve looked at are almost identical. We can, therefore, redefine and formalize. So we have a single formula. Arc length is given by πΏ equals the integral dπ where dπ is equal to the square root of one plus dπ¦ by dπ₯ squared dπ₯ if π¦ is equal to π of π₯ over some closed interval π to π. And dπ equals the square root of one plus dπ₯ by dπ¦ squared dπ¦ if π₯ is equal to π of π¦ for some closed interval π to π. In our final example, weβre going to look how to define an arc length function π of π₯ from a point given.
Find the arc length function for the curve π¦ equals four π₯ to the power of three over two, taking the point with coordinates one, four as the starting point.
Remember, the arc length formula for a function π¦ equals π of π₯ is given by πΏ equals the integral dπ , where dπ is equal to the square root of one plus dπ¦ by dπ₯ squared dπ₯. In our case, π¦ is equal to four π₯ to the power of three over two. So weβll begin by simply working out what dπ¦ by dπ₯ is. Remember to differentiate the function of this sort, we multiply by the power and then reduce that power by one. So dπ¦ by dπ₯ is equal to three over two times four π₯ to the power of one-half. Four divided by two is two. So we see that dπ¦ by dπ₯ is equal to six π₯ to the power of one-half. And this means we can substitute what we know so far into our formula for dπ . We get the square root of one plus six π₯ to the power of one-half squared dπ₯.
Well, this simplifies quite nicely; we end up with the square root of one plus 36π₯. And this means that πΏ is going to be equal to the integral of the square root of one plus 36π₯ evaluated with respect to π₯. But what are our limits? Well, weβre working in terms of π₯. And we know we have a starting point at π₯ equals one. So thatβs the lower limit. Weβre looking for a general function, so the upper limit is simply going to be π₯. And we can see that the argument function will be given by the integral evaluated between one and π₯ of the square root of one plus 36π₯ dπ₯. So how do we evaluate this Integral? We cannot rely on calculator, and, in fact, weβre going to need to use a substitution.
We have a composite function. Weβre going to let π’ be equal to the function inside the square root; thatβs one plus 36π₯. We can say that the derivative of π’ with respect π₯ is, therefore, 36. And we can also equivalently say that one over 36 dπ’ equals dπ₯. Before we can make any substitution though, weβre going to need to change the limits. We use our substitution to do so. And the upper limit is π₯, so when π₯ is equal to π₯, π’ is just one plus 36π₯. And the lower limit is one, so when π₯ is equal to one, π’ is one plus 36 times one which is 37. We now replace one plus 36 with π’. Iβve changed the square root to be the power of one-half. And we replace dπ₯ with 36 dπ’. So all thatβs left is to evaluate this integral between the limits given.
The integral of π’ to the power of one-half is π’ to the power of three over two divided by three over two. So this means that the integral of π’ to the power of one-half over 36 is equal to π’ to the power of three over two over 36 divided by three over two. And, of course, dividing by three over two is the same as multiplying by two-thirds. We can simplify this a little. And we see that our arc length function is now one over 54 times π’ to the power of three over two. Evaluated between 37 and one plus 36π₯. Thatβs one over 54 times one plus 36π₯ to the power of three over two minus one over 54 times 37 to the power of three over two. And weβre done.
We found the arc length function for the curve π¦ equals four π₯ to the power of three over two, taking the point with coordinates one, four as the starting point. Itβs useful to know that now we have this arc length function. We can find the length of the curve between the point π₯ equals one and any other point by substituting that value of π₯ into this formula.
In this video, weβve seen that we can use integration to help us find the arc length of curves given in the form π¦ equals π of π₯ and π₯ equals π of π¦. We saw that the arc length formula is given by πΏ is equal to the integral dπ , where dπ is equal to the square root of one plus dπ¦ by dπ₯ squared dπ₯ if π¦ is a function in π₯ over some close interval π to π. And itβs equal to the square root of one plus dπ₯ dπ¦ squared dπ¦ if π₯ is a function in π¦ over some closed interval π to π. We also saw that we can use this formula to find the general equation for the arc length for a function from a given starting point by changing the upper limit of our integral to π₯ or π¦, depending on the context.
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Gauthmath
# How to Rationalize the Denominator
## How to Rationalize the Denominator
#### Context of this video
Gauthmath helps with your math. Welcome to Gauthmath.
In this video, we're going to introduce a key skill called the rationalization of denominator.
What does rationalization mean?
We know that the correct output of radical simplifications shouldn't have any radicals in the denominator.
To rationalize denominator means to eliminate any radical expressions in the denominator under the premise that the value of the expression remains unchanged.
By doing this, we convert the denominator into rational numbers or integrals.
So how do we achieve that?
For example, in this expression, we've multiplied the numerator and the denominator by a square root of five at the same time.
The denominator turned from irrational square root of 5 into rational number, 5.
This is rationalization of the denominator.
The key point of this whole process is to find a number or an expression that can turn the irrational denominator into a rational form after multiplication.
In this case, the denominator and numerator were both multiplied with the square root of 5 so that the radical sign in the denominator was eliminated.
The value of the original expression is unchanged and therefore the rationalization of denominator is completed!
For a simple denominator like the square root of five, all we have to do is to multiply the irrational number by itself.
However, for an irrational expression like this one, the denominator is the difference between the square root of 5 and the square root of 3. Which multiplier should we use then?
No matter which number we choose from A,B,C, the denominator result from the multiplication will still be radicals.
Especially option C, after we square the denominator, though the square root of 5 and square root of 3 turn into integers 5 and 3,the expansion of the difference of squares contains a radical term
Are there any multiplication formulas that can result in square terms only, with no radical products?
Oh yea, the formula of the difference of squares satisfies this requirement. The product contains two squares with no radical terms.
By using the formula, we should multiply the denominator by square root of 5 plus square root of 3
In this way, the denominator turns into the difference of 5 and 3 , which is 2. The radical sign disappeared, hooray!
Since both the numerator and denominator contain 2, we can simplify the fraction. Thus the final result is square root of 5 plus square root of 3.
To summarize, when the denominator is the difference or sum of two radicals, there are only 3 steps we need to take to rationalize it.
First, based on the formula of the difference of squares, we need to find the other factor of the denominator.
Then we multiply both the numerator and denominator by this factor.
Finally we simplify the fraction by cancelling common factors.
Let's try this question! Emm...Which one is the correct answer?
Options A,B,C are all wrong, the answer is D.
The denominator is square root of 2 minus square root of 5
The factor in pair with the denominator is square root of 2 plus square root of 5
Multiply it with both the numerator and denominator, then calculate the new denominator.
Attention! Options A,B,C were wrong due to this step.
According to the formula of the difference of squares, the denominator equals the square of square root of 2 minus the square of square root of 5
2 minus 5 is negative 3
After simplifying the fraction, we get the final answer, negative 2 times the sum of the square root of 2 and the square root of 5
Never let down your guard in any step.
Have questions about rationalizing denominators? Find an expert to help you on our app Gauthmath! Homework solver, step by step.
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