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# Mathematics
### Chapter : Linear Equation In Two Variable
#### Solutions of Pair of Lines in Two Variable
There are three types of solutions of Pair of Lines in Two Variable, these are
• Unique solution
• Infinitely many solutions
• No solution.
(A) Situation of Consistent: If a system of simultaneous linear equations has at least one solution then the system is said to be consistent.
(i) Consistent equations with unique solution : The graphs of two equations intersect at a unique point. For example. Consider the pair of equation of lines x + 2y = 4 and 7x + 4y = 18.
The graphs of pair of lines of these equations intersect each other at the point (2, 1) i.e., x = 2, y = 1. Hence, the equations are consistent with unique solution.
(ii) Consistent equations with infinitely many solutions : The graphs (lines) of the two equations will be coincident. For example. Consider the pair of equations 2x + 4y = 9 ⇒ 3x + 6y = 27/2
The graphs of the above equations coincide. Coordinates of every point on the lines are the solutions of the equations. Hence, the given equations are consistent with infinitely many solutions.
(B) Inconsistent Equation : If a system of simultaneous linear equations has no solution, then the system is said to be inconsistent.
No Solution : The graph (lines) of the two equations are parallel. For example. Consider the pair of equation of lines are 4x + 2y = 10 and 6x + 3y = 6
The graphs (lines) of the given equations are parallel. They will never meet at a point. So, there is no solution. Hence, the equations are inconsistent.
S. No Graph of Two Equations Types of Equations 1. Intersecting Lines Consistent, With Unique Solution 2. Coincident Consistent with number of infinite solutions 3. Parallel Lines Inconsistent or No Solution is found
|
# Pythagorean Theorem Walkthrough and Problems
Worksheet (Algebra)
## Pythagorean Theorem Walkthrough and Problems Worksheet
This is a great worksheet on the Pythagorean Theorem. It begins by discussing the parts of a right triangle then provides a walkthough on using the Pythagorean Theorem along with some self-practice and a word problem.
## Lesson:
To find lengths of sides of right triangles by using the pythagorean theorem
14
2
Yes
## Samples:
1. In a right triangle, the side opposite the right angle is called the _________________.
2. The hypotenuse is the _______________ side. We use the variable c to represent the hypotenuse when we don’t know it’s length.
3. The other two sides of the triangle are called the _______ (these two sides form the right angle). We use the variables a and b to represent the legs.
4. Label the hypotenuse and legs in this right triangle:
5. The ___________________ _______________ describes the relationship of the lengths of the sides of a _____________ triangle.
6. The Pythagorean Thereom is named after ________________, a Greek philospher and mathematician who taught around 530 BC.
PAGE 2
1. What is the length of the hypotenuse of the triangle? Using the triangle at the right, find the length of the missing side. 3. a = 6, b = 8, c = ?
6. A pigeon leaves its nest and flies 5 km due east. Then he flies 3 km due north. How far is the pigeon from his nest? (Draw a picture! Round to nearest tenth).
|
# Comparing fractions
Home > By Subject > Fractions > Comparing
It is important that your child can compare different fractions and determine which is larger, which is smaller, or if they are equal.
Discuss with your children how, with a fraction, when the bottom number (denominator) increases, the fraction gets smaller. Explore this with them using real examples (chocolate bars and pizza are popular!).
Review the three different comparison situations below. Note: This fraction number line will help with comparing fractions.
## Same numerator, Different denominator
In this case there are the same number of different things. In this case the things are thirds and sixths. Ensure your child that understands that thirds are larger than sixths. The rule for comparing here is that the fraction with the smaller denominator is the largest
## Different numerator, Same denominator
Here we have a different number of the same thing – there are seven eighths and three eighths. Seven is a larger number than three. The comparison rule here is that the fraction with the greatest numerator is the largest.
## Different numerator, Different denominator
This is a more difficult task the two comparisons above. Depending on your child’s level, this maybe a step you skip and then come back to later.
These fractions are compared by changing the denominators to a common number. This can be done by multiplying the top and bottom of the fraction by the same number since this will give a fraction with an equivalent value. Do this as shown below to both fractions to get a common denominator and then compare them.
So, in the example above, now that the denominators are equal, the fraction with the greatest numerator is the largest.
Some students may respond to questions about which fraction is bigger by saying it depends on what size the whole is and, depending on the wording of the question, they may actually be correct! Encourage them to think in terms which is the larger or smaller share and be precise with your questions. e.g. "which is the largest fraction?" not just "which is the largest?"
## Worksheets
Practice comparing fractions with the worksheets above.
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How do you write the next three terms of the geometric sequence -2, -12, -72, -432,... and graph the sequence?
Dec 1, 2016
The next three terms are:$\text{ "-2592," "-15552," } - 93312$
Explanation:
To determine the next three terms we have to study the given
$\text{ }$
terms in this sequence and how it is calculated .
$\text{ }$
$\text{ }$
${a}_{0} = - 2$
$\text{ }$
${a}_{1} = - 2 \times \textcolor{b l u e}{{6}^{1}} = - 12$
$\text{ }$
${a}_{2} = - 2 \times \textcolor{b l u e}{{6}^{2}} = - 72$
$\text{ }$
Therefore, $\text{ } {a}_{n} = - 2 \times \textcolor{b l u e}{{6}^{n}}$
$\text{ }$
What is a_4=?, a_5" =?",a_6=?
.$\text{ }$
${a}_{4} = - 2 \times {6}^{4} = - 2 \times 1296 = - 2592$
$\text{ }$
${a}_{5} = - 2 \times {6}^{5} = - 2 \times 7776 = - 15552$
$\text{ }$
${a}_{6} = - 2 \times {6}^{6} = - 2 \times 46656 = - 93 , 312$
""
The geometric series is:
$\text{ }$
$- 2 , \text{ "-12," "-72," "-432," "-2592," "-15552," } - 93312$
|
3 Tutor System
Starting just at 265/hour
# Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Let a be any positive integer and b = 3
a = 3q + r, where q $$\ne$$ 0 and 0 < r < 3
a = 3q or 3q + 1 or 3q + 2
Therefore, every number can be represented as these three forms.
We have three cases.
Case 1 : When a = 3q,
$$a^3 = (3q)^3 = 27q^3 = 9(3q^3) = 9m$$
Where m is an integer such that $$m = 3q^3$$
Case 2 :When a = 3q + 1
$$\Rightarrow$$ $$a^3 = (3q + 1)^3$$
$$\Rightarrow$$ $$a^3 = 27q^3 + 27q^2 + 9q + 1$$
=> $$a^3 = 9( 3q^3 + 3q^2 + q) + 1$$
$$\Rightarrow$$ $$a^3 = 9m + 1$$
When m is an integer such that $$m = (3q^3 + 3q^2 + q)$$
Case 3 :When a = 3q + 2,
$$\Rightarrow$$ $$a^3 = (3q + 2)^3$$
$$\Rightarrow$$ $$a^3 = 27q^3 + 54q^2 + 36q + 8$$
$$\Rightarrow$$ $$a^3 = 9( 3q^3 + 6q^2 + 4q) + 8$$
$$\Rightarrow$$ $$a^3 = 9m + 8$$
When m is an integer such that $$m = (3q^3 + 6q^2 + 4q)$$
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.
|
## College Algebra (11th Edition)
$x=-2$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $x=\log_{4/5} \dfrac{25}{16} ,$ use the properties of logarithms. $\bf{\text{Solution Details:}}$ Using exponents, the equation above is equivalent to \begin{array}{l}\require{cancel} x=\log_{4/5} \dfrac{5^2}{4^2} .\end{array} Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\log_{4/5} \dfrac{4^{-2}}{5^{-2}} .\end{array} Using the extended Power Rule of the laws of exponents which states that $\left( \dfrac{x^my^n}{z^p} \right)^q=\dfrac{x^{mq}y^{nq}}{z^{pq}},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\log_{4/5} \left(\dfrac{4}{5}\right)^{-2} .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent \begin{array}{l}\require{cancel} x=-2\log_{4/5} \dfrac{4}{5} .\end{array} Since $\log_b b =1,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=-2(1) \\\\ x=-2 .\end{array}
|
7
# TRIGONOMETRY OF RIGHT TRIANGLES
Similar figures
All functions from one function
Complements
Cofunctions
PLANE TRIGONOMETRY is based on the fact of similar figures. We saw:
Right triangles will be similar if an acute angle of one
is equal to an acute angle of the other.
In the right triangles ABC, DEF, if the acute angle at B is equal to the acute angle at E, then those triangles will be similar. Therefore the sides that make the equal angles will be proportional. Whatever ratio CA has to AB, FD will have to DE.
Compare Example 11 here.
A trigonometric Table is a table of ratios of sides. In the Table, each value of sin θ represents the ratio of the opposite side to the hypotenuse -- in every right triangle with that acute angle.
If angle θ is 28°, say, then in every right triangle with a 28° angle, its sides will be in the same ratio. We read in the Table,
sin 28° = .469
This means that in a right triangle having an acute angle of 28°, its opposite side is 469 thousandths of the hypotenuse, which is to say, a little less than half.
It is in this sense that in a right triangle, the trigonometric ratios -- the sine, the cosine, and so on -- are "functions" of the acute angle. They depend only on the acute angle.
Example. Indirect measurement. When we cannot measure things directly, we can use trigonometry.
For example, to measure the height h of a flagpole, we could measure a distance of, say, 100 feet from its base. From that point P we could then measure the angle required to sight the top . If that angle, called the angle of elevation, turned out to be 37°, then
so that h 100 = tan 37° so that so that h = 100 × tan 37°.
From the Table,
tan 37° = .754
Therefore, on multiplying by 100,
h = 75.4 feet.
All functions from one function
If we know the value of any one trigonometric function, then -- with the aid of the Pythagorean theorem -- we can find the rest.
Example 1. In a right triangle, sin θ = Sketch the triangle, place the ratio numbers, and evaluate the remaining functions of θ.
To find the unknown side x, we have
x2 + 52 = 132
x2 = 169 − 25 = 144.
Therefore,
x = = 12.
We can now evaluate all six functions of θ:
sin θ = 5 13 csc θ = 13 5
cos θ = 1213 sec θ = 1312
tan θ = 5 12 cot θ = 12 5
Example 2. In a right triangle, sec θ = 4. Sketch the triangle, place the ratio numbers, and evaluate the remaining functions of θ.
To say that sec θ = 4, is to say that the hypotenuse is
to the adjacent side in the ratio 4 : 1. (4 = 41 )
To find the unknown side x, we have
x2 + 12 = 42 x2 = 16 − 1 = 15.
Therefore,
x = .
We can now evaluate all six functions of θ:
sin θ = 4 csc θ = 4 cos θ = 14 sec θ = 4 tan θ = cot θ = 1
Problem 1. In a right triangle, cos θ = . Sketch the triangle and evaluate sin θ.
Problem 2. cot θ = . Sketch the triangle and evaluate csc θ.
Complements
Two angles are called complements of one another if together they equal a right angle.
Angles 1 and 2 are complements of one another. Each one completes the other to make a right angle.
In the degree system of measurement, the complement of 60° is 30°.
(Two angles are called supplements of one another if together they equal two right angles. 60° is the supplement of 120°. Their sum is 180°, which is two right angles.)
Problem 3. Name the complement of each angle.
a) 70° 20° b) 20° 70°
c) 45° 45° d) θ 90° − θ
The point about complements is that, in a right triangle, the two acute angles are complementary. For the three angles of the right triangle are together equal to two right angles (Theorem 9); therefore, the two acute angles together will equal one right angle.
Problem 4. In terms of radians, what angle is the complement of an angle θ ?
π2 − θ
Problem 5. What angle in radians is the complement of each of the following?
a) π6 π3 b) π8 3π 8 b) π5 3π10 c) π2 0 d) 0 π2
Cofunctions
There are three pairs of cofunctions:
The sine and the cosine
The secant and the cosecant
The tangent and the cotangent
Here is the significance of a cofunction:
A function of any angle is equal to the cofunction
of its complement.
This means, for example, that
sin 80° = cos 10°.
The cofunction of the sine is the cosine. And 10° is the complement of 80°.
Problem 4. Answer in terms of cofunctions.
a) cos 5° = sin 85° b) tan 60° = cot 30°
c) csc 12° = sec 78° d) sin (90° − θ) = cos θ
e) cot θ = tan (90° − θ)
In the figure:
sin θ = ac cos φ = ac
Thus the sine of θ is equal to the cosine of its complement.
sec θ = cb csc φ = cb
The secant of θ is equal to the cosecant of its complement.
tan θ = ab cot φ = ab
The tangent of θ is equal to the cotangent of its complement.
Next Topic: The 30°-60°-90° Triangle
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|
# Exponential Growth - PowerPoint PPT Presentation
Exponential Growth
1 / 18
Exponential Growth
## Exponential Growth
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Exponential Growth
2. Exponential Growthy = a (1 +b)t y = final amount after a period of timea= original amountt = timeb = percent of increase (as a decimal)(1 + b ) is called the multiplierIf multiplier is greater than 1, it is a growth problem
3. Finding the multiplierex. Growth of 5%Change 5% to a decimal (.05)1+ .05 = 1.051.05 is the multiplier
4. Ex. A certain quantity DoublesIf sometime doubles, it is increasing by 100%100% = 1.01+ 1.0 = 2 Therefore, 2 is the multiplier
5. A quantity triplesIf something triples it is increasing by 200% 200% = 2.01 + 2.0 = 33 is the multiplier
6. Finding percent of growthPercent of growth= amount of changeoriginal amountMake this decimal a %This is the % of change
7. Ex. A quantity changes from 20 to 35. Find the percent of increase.Amount of change = 15 Orignal Amount = 20 = .75 = 75%75% is the percent of increase
8. Ex. A price increases from \$40 to \$45. Find the % of increase 5 = 1 40 8= .125 = 12.5%
9. Ex. The population of a town increases from 60,000 to 120,000. Find the % of increase.60,00060,000= 1.0 = 100%100% is the percent of increase
10. Examples of Exponential Growth
11. Ex. The price of a gallon of milk increases at a rate of 2% per year. In 1990, a gallon of milk cost \$2. How much would the gallon of milk be expected to cost in 2002?Final cost=original amt (1 +% ofincrease)time cost in 2002 = 2(1+.02)12 = 2(1.02)12 = \$2.54
12. Ex. The population of a town is increasing at a rate of 6% every 2 years. The population was 12,000 in 1980. What would the population be expected to be in 2000?Population in 2000 = 12,000(1 + .06)10 = 12,000(1.06)10 21,490
13. The # of deer in a particular forest is estimated to be about 352 right now. The population has been growing at a rate of 12% What was the approximate population 5 years ago?
14. The value of an investment has grown at a rate of 2% per year It was originally valued at \$20,000. It is now valued at \$22,974. How much time has gone by?
15. 22,974 = 20,000(1.02)t 22,974 = 20,000(1.02)t 20,000 = 20,0001.15 ≈ (1.02)t *Next use “guess and check” for values of “t”(1.02)5 ≈ 1.10(1.02)6 ≈ 1.13(1.02)7 ≈ 1.15This last one is closest to 1.15 so 7 years have gone by.
|
## Intermediate Algebra (12th Edition)
$\left[ -5,13 \right]$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $|-4+x| \le 9 ,$ use the definition of absolute value inequalities. Use the properties of inequalities to isolate the variable. For the interval notation, use a parenthesis for the symbols $\lt$ or $\gt.$ Use a bracket for the symbols $\le$ or $\ge.$ For graphing inequalities, use a hollowed dot for the symbols $\lt$ or $\gt.$ Use a solid dot for the symbols $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Since for any $c\gt0$, $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} -9 \le -4+x \le 9 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -9+4 \le -4+x+4 \le 9+4 \\\\ -5 \le x \le 13 .\end{array} In interval notation, the solution set is $\left[ -5,13 \right] .$ The colored graph is the graph of the solution set.
|
# What Is the Probability a Cow Has BSE Given a Positive Test?
• Vendatte
In summary, Bayes Rule Probability Problem is a mathematical formula used to determine the probability of an event occurring based on prior knowledge of related conditions or factors. It is commonly used in fields such as statistics, machine learning, and artificial intelligence to update the probability of an event as new evidence or information becomes available. The formula for Bayes Rule Probability Problem is P(A|B) = [P(B|A) * P(A)] / P(B), where P(A|B) is the probability of event A occurring given that event B has occurred, P(B|A) is the probability of event B occurring given that event A has occurred, and P(A) and P(B) are the individual probabilities of events A and B occurring.
Vendatte
## Homework Statement
Event B is cow has BSE
Event T is the test for BSE is positive
P(B) = 1.3*10^-5
P(T|B) = .70
probability that the test is positive, given that the cow has BSE
P(T|Bcc) = .10
probability that the test is positive given that the cow does not have BSE
Find P(B|T) and P(B|Tc)
probability that the cow has BSE given that the test is positive, and probability that the cow has BSE given that the test is negative
## Homework Equations
P(Ci|A) = P(A|Ci)/P(A) = P(A|Ci) / (P(A|C1)P(C1)+P(A|C2)*P(C2) ... + P(A|Cm)*P(Cm)
## The Attempt at a Solution
The equation I use to find P(B|T) is
P(B|T) = P(T|B) / (P(T|B)*P(B)+P(T|BC)*P(Bc)
plugging in the values, I get P(B|T) = .70/(.70*(1.3*10^-5) + .1(1-(1.3*10^-5))), however that value is close to 7, which is clearly wrong.
To find P(B|Tc I plan on using the equation P(B) = P(B|T)P(T)+P(B|Tc)*P(Tc)
Any help would be greatly appreciated.
it is important to use accurate and precise calculations to determine probabilities. In this case, we are dealing with conditional probabilities which require us to consider both the likelihood of the event occurring and the prior probability of the event.
To solve for P(B|T), we can use Bayes' Theorem which states that P(B|T) = P(T|B)P(B)/P(T). Plugging in the given values, we have P(B|T) = .70*(1.3*10^-5)/(.70*(1.3*10^-5) + .1*(1-(1.3*10^-5))). This gives us a more reasonable value of approximately 0.0000902, which is the probability that the cow has BSE given that the test is positive.
To solve for P(B|Tc), we can use the equation P(B|Tc) = P(Tc|B)P(Bc)/P(Tc). Plugging in the values, we have P(B|Tc) = .10*(1-(1.3*10^-5))/(.10*(1-(1.3*10^-5)) + .70*(1.3*10^-5)). This gives us a value of approximately 0.99991, which is the probability that the cow has BSE given that the test is negative.
It is important to note that these calculations are based on the given probabilities and do not necessarily reflect the true probability of a cow having BSE or the accuracy of the test. it is important to continuously evaluate and improve upon these calculations by considering new evidence and data.
## 1. What is Bayes Rule Probability Problem?
Bayes Rule Probability Problem, also known as Bayes' theorem or Bayes' law, is a mathematical formula that determines the probability of an event occurring based on prior knowledge of related conditions or factors.
## 2. How is Bayes Rule Probability Problem used?
Bayes Rule Probability Problem is used to update the probability of an event occurring as new evidence or information becomes available. It is commonly used in fields such as statistics, machine learning, and artificial intelligence.
## 3. What is the formula for Bayes Rule Probability Problem?
The formula for Bayes Rule Probability Problem is: P(A|B) = [P(B|A) * P(A)] / P(B), where P(A|B) is the probability of event A occurring given that event B has occurred, P(B|A) is the probability of event B occurring given that event A has occurred, and P(A) and P(B) are the individual probabilities of events A and B occurring.
## 4. Can Bayes Rule Probability Problem be applied to real-life situations?
Yes, Bayes Rule Probability Problem can be applied to real-life situations where there is uncertainty or incomplete information. It can be used to make predictions or decisions based on available data and can be updated as new information is gathered.
## 5. What are the limitations of Bayes Rule Probability Problem?
Bayes Rule Probability Problem relies on the assumption that the prior probabilities and conditional probabilities are accurate, which may not always be the case in real-life situations. It also requires a large amount of data to accurately estimate probabilities, which may not always be available.
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How do you find the slope of 2x+8y=9?
Dec 11, 2014
In the general form of a line $y = m x + b$, $m$ represents the slope of the line and $b$ the y-intercept. Here, we have the line $2 x + 8 y = 9$...we want it in the form $y = m x + b$ so we can get our $m$ (the slope) directly.
To do this, we simply solve $2 x + 8 y = 9$ for y with some algebra:
$2 x + 8 y = 9$
-Take $2 x$ to the left hand side of the equation by subtracting it from both sides
$8 y = 9 - 2 x = - 2 x + 9$
-Divide by 8 on both sides of the equation to get y alone
$y = - \frac{2}{8} x + \frac{9}{8} = - \frac{1}{4} x + \frac{9}{8}$
And there we have the line $y = - \frac{1}{4} x + \frac{9}{8}$, with $m = s l o p e = - \frac{1}{4}$
|
# Algebra: Imaginary Numbers
1.
$$\sqrt{-13}$$
Solution
2.
$$(2+3i)+(1-6i)-(8+7i)$$
Solution
3.
$$(8+2i)+(4+7i)$$
Solution
4.
$$2i+3i$$
Solution
5.
$$3i\times 4i$$
Solution
6.
$$-i\times 2i\times 3i$$
Solution
7.
$${i}^{99}$$
Solution
8.
$${i}^{17}$$
Solution
9.
$${i}^{120}$$
Solution
10.
$${i}^{64002}$$
Solution
11.
$$(3i+4)i$$
Solution
# Imaginary Numbers - Introduction
An imaginary number is a complex number that can be written in the form of a real number multiplied by an imaginary part, named i. This imaginary part i is defined by the property
$${i}^{2}=-1$$
. Although it might be difficult to intuitively map imaginary numbers to the physical world, they do easily result from common math operations. The classic way of obtaining an imaginary number is when we try to take the square root of a negative number, like
$$\sqrt{-5}$$
.
# What to Do when $${x}^{2}=-1$$?
There is no way of solving this equation by using real numbers since
$$x$$
would be equal to
$$\sqrt{-1}$$
, and we know that it is illegal to take the square root of a negative number. Therefore, mathematicians proposed a solution: substitute that value by a number, call it
$$i$$
.
$$\sqrt{-1}=i$$
And, given this equality, it follows that:
$${i}^{2}=ii=-1$$
Or, to put this another way:
$${i}^{2}=\sqrt{-1}\sqrt{-1}=-1$$
$${i}^{2}={\sqrt{-1}}^{2}=-1$$
# Problems with Imaginary Numbers
Now, let's see how to solve a more elaborate problem using the imaginary numbers.
First, we have:
$$\sqrt{-9}\sqrt{-4}+7$$
Let's rewrite the expression in a way that we can isolate
$$\sqrt{-1}$$
from the other terms:
$$\sqrt{9}\sqrt{-1}\sqrt{4}\sqrt{-1}+7$$
Next, substitute
$$\sqrt{-1}$$
with
$$i$$
and simplify the roots that are now positive:
$$3i\times 2i+7$$
$$6{i}^{2}+7$$
Substitute
$${i}^{2}$$
with
$$-1$$
:
$$6\times -1+7$$
$$-6+7$$
$$1$$
$$1$$
|
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# M07-33
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A shop purchased a pair of sunglasses for $120 and was selling it at a price that equaled the purchase price of the sunglasses plus a markup that was 25 percent of the selling price. After some time a shop owner decided to decrease the selling price by 20 percent. What was the shop's gross profit on this sale? A.$0
B. $2 C.$4
D. $6 E.$8
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Official Solution:
A shop purchased a pair of sunglasses for $120 and was selling it at a price that equaled the purchase price of the sunglasses plus a markup that was 25 percent of the selling price. After some time a shop owner decided to decrease the selling price by 20 percent. What was the shop's gross profit on this sale? A.$0
B. $2 C.$4
D. $6 E.$8
Let the markup be $$x$$, so $$x$$ must be 25% of the selling price, which would be $$120+x$$: $$x=0.25(120+x)$$. Solving gives $$x=40$$. Hence the selling price was $$120+40=160$$.
The price after the discount of 20% would be $$160*0.8=128$$ and the gross profit on the sale would be: final selling price - cost of the sunglasses = $$128 - 120 = 8$$.
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22 Nov 2014, 12:03
hi bunnel,
can you pls explain how the price after discount has come . Especially how 0.8 has come ?
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23 Nov 2014, 07:02
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prashd wrote:
hi bunnel,
can you pls explain how the price after discount has come . Especially how 0.8 has come ?
20% discount from selling price = price - price*0.2 = price(1 - 0.2) = 0.8*price.
Hope it's clear.
Similar questions to practice:
a-furniture-dealer-purchased-a-desk-for-150-and-then-set-th-71633.html
a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html
a-merchant-purchased-a-jacket-for-60-and-then-determined-104375.html
a-retailer-bought-a-shirt-at-wholesale-and-marked-it-up-130759.html
a-dress-is-marked-up-16-2-3-to-a-final-price-of-140-what-155046.html
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05 Jan 2015, 11:01
2
I think this question is poor and helpful.
*selling it at a price that equaled the purchase price of the sunglasses plus a markup that was 25 percent of the selling price.*
This means if CP is 120 then SP = 120+0.25*CP
and by this approach we get ans. 0
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06 Jan 2015, 02:43
sanket1991 wrote:
I think this question is poor and helpful.
*selling it at a price that equaled the purchase price of the sunglasses plus a markup that was 25 percent of the selling price.*
This means if CP is 120 then SP = 120+0.25*CP
and by this approach we get ans. 0
Please read the question and the solution carefully. It should be 120 + 0.25*SP = SP --> SP = 160.
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15 Jan 2015, 17:40
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sanket1991 wrote:
I think this question is poor and helpful.
*selling it at a price that equaled the purchase price of the sunglasses plus a markup that was 25 percent of the selling price.*
This means if CP is 120 then SP = 120+0.25*CP
and by this approach we get ans. 0
Actually, I feel that this question is representative of many of the trickier questions you'd find on the real GMAT ~ one that you need to pay attention to its finer details.
Questions are not posed in the most convenient ways for the test taker --- and when one is stressed and trying to "race" against the clock, it's real easy to miss out on the finer details of a question that "looks" easy.
Cost: 120
Selling Price: SP = 120 + (1/4)SP ---> NOT (1/4 Cost)... SP - (1/4) SP = 120 ... (3/4) SP = 120 ... SP = 160
New Selling Price: NSP = 160 (4/5) ... NSP = 128
Profit = NSP - Cost ----> 128 - 120 = 8
Good question.
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11 Oct 2015, 08:54
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Bunuel wrote:
A shop purchased a pair of sunglasses for $120 and was selling it at a price that equaled the purchase price of the sunglasses plus a markup that was 25 percent of the selling price. After some time a shop owner decided to decrease the selling price by 20 percent. What was the shop's gross profit on this sale? A.$0
B. $2 C.$4
D. $6 E.$8
cost price of glasses is 120$. Let the selling price be x. Selling price=$$120+(\frac{25}{100}*Selling price)$$ Selling price=160 this Selling price is decreased by 20%, so Selling price=$$160(1-(\frac{20}{100}))$$=128 Profit=Selling price-Cost price =128-120 =8 Manager Joined: 17 Mar 2014 Posts: 124 Location: United States Concentration: Entrepreneurship, Leadership GPA: 3.97 Re: M07-33 [#permalink] ### Show Tags 02 Dec 2015, 16:05 Bunuel wrote: A shop purchased a pair of sunglasses for$120 and was selling it at a price that equaled the purchase price of the sunglasses plus a markup that was 25 percent of the selling price. After some time a shop owner decided to decrease the selling price by 20 percent. What was the shop's gross profit on this sale?
A. $0 B.$2
C. $4 D.$6
E. $8 To certain extent - it is confusing. Because "new selling price" should have been mentioned. I tried to solve by SP = 120 + .25 SP ; .75 SP = 120; Then, with the flow of the question I solved further by setting new SP in this equation itself - .75 * .80SP = 120; which produced incorrect result SP = 200 _________________ KUDOS!!!, I need them too Current Student Status: Remember to Always Think Twice Joined: 04 Nov 2014 Posts: 54 Location: India GMAT 1: 740 Q49 V40 GPA: 3.51 Re: M07-33 [#permalink] ### Show Tags 19 Jan 2016, 06:44 1 excellent wordplay in the question will get many confused but, it is lucid. cost price = 120 selling price (SP) = 120 + 0.25*SP --- (this equation is the catch) Hence 0.75*SP = 120 and hence SP = 160 Final SP = 0.8*SP = 0.8*160 = 128. Hence a profit of 8$.
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17 Feb 2016, 13:58
Tricky wording. I should've known answer A was too easy!
"equaled the purchase price of the sunglasses plus a markup that was 25 percent of the selling price." Is two equations.
SellingPrice = 120 + Markup
Markup= 1/4(SellingPrice)
SellingPrice = 120+ (SellingPrice/4) = 3/4SellingPrice = 120 = SellingPrice = $160 160*(4/5) = 128 Profit = Selling - Cost E.$8
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06 Sep 2016, 12:51
I would say nice question... caught me off guard with the wording...
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25 Jul 2017, 19:06
I think this is a high-quality question and I agree with explanation.
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08 Jul 2019, 16:38
Bunuel wrote:
Official Solution:
A shop purchased a pair of sunglasses for $120 and was selling it at a price that equaled the purchase price of the sunglasses plus a markup that was 25 percent of the selling price. After some time a shop owner decided to decrease the selling price by 20 percent. What was the shop's gross profit on this sale? A.$0
B. $2 C.$4
D. $6 E.$8
Let the markup be $$x$$, so $$x$$ must be 25% of the selling price, which would be $$120+x$$: $$x=0.25(120+x)$$. Solving gives $$x=40$$. Hence the selling price was $$120+40=160$$.
The price after the discount of 20% would be $$160*0.8=128$$ and the gross profit on the sale would be: final selling price - cost of the sunglasses = $$128 - 120 = 8$$.
I did it in another way:
Cost: $120 (Price - Cost)/Price=25% ----> Price=$160
Then, $160*80%=$128 ----> Profit = $128 -$120 = $8 So E is the correct answer. Intern Joined: 01 May 2017 Posts: 36 Re: M07-33 [#permalink] ### Show Tags 01 Aug 2019, 00:39 All the explanation in test are unclear. After giving test we analyze but facing unclear explanation. Director Joined: 14 Feb 2017 Posts: 896 Location: Australia Concentration: Technology, Strategy Schools: LBS '22 GMAT 1: 560 Q41 V26 GMAT 2: 550 Q43 V23 GMAT 3: 650 Q47 V33 GMAT 4: 650 Q44 V36 WE: Management Consulting (Consulting) Re: M07-33 [#permalink] ### Show Tags 03 Aug 2019, 22:28 I think the wording can be a bit clearer. The only thing that makes this question difficult is the way its worded. Consider: A shop purchased a pair of sunglasses for$120 and was selling it at the purchase price plus a markup that is equal to 25 percent of the total selling price.
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Re: M07-33 [#permalink] 03 Aug 2019, 22:28
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# M07-33
Moderators: chetan2u, Bunuel
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# Do Now
Télécharger la présentation
## Do Now
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Do Now • Think about 3 interesting facts about yourself! • What is your favorite food? • Are you involved in any school activities? • Do you have any pets? • What is your favorite type of music? • What is your favorite TV show? • Have you visited any interesting places recently?
2. Go Over Quizzes
3. 1.7 Completing the Square Objectives: SWBAT solve quadratic equations by completing the square
4. Solve the quadratic equation by finding the square roots x² - 8x + 16 = 25
5. Page 51# 1 and # 2 1. x² + 6x +9 = 36 2. x² - 10x + 25 = 1
6. Completing the square x² + bx
7. Completing the square
8. Completing the square
9. Completing the square
10. Completing the square • How can we right this algebraically?
11. Examples: Make a perfect square trinomial Find the value of c that makes x² + 16x + c a perfect square trinomial. Then write the expression as the square of a binomial.
12. Examples: Make a perfect square trinomial
13. Page 51Find the value of c that makes a perfect square trinomial. Then write the expression as the squareof a binomial.# 4 : #5: x² + 14x + c x² + 22x + c
14. Solving Equations ax² + bx + c = 0 ax² + bx + ___= -c ax² + bx + (b/2) ² = -c+(b/2) ² (ax+b/2) ²= -c+(b/2) ²
15. Example: Solve ax² + bx + c = 0 when a=1 x² - 12x + 4 = 0 x² - 12x + ____ = -4 x² - 12x + (-12/2) ²= -4+(-12/2) ² x² - 12x + 36 = -4+36 (x-6) ² = 32 (x-6) = +-√(32) X=6+-√(32) X=6+4√2 X=6-4√2
16. Solve the equation by completing the squarePage 52 #7) x² +6x + 4 = 0 #8) x² - 10x + 8 =0
17. What about if ax² + bx+ c = 0 and a ≠ 1 2x² + 8x + 14 = 0 x² + 4x + 7 = 0 x² + 4x = -7 x² + 4x + (4/2)²= -7 + (4/2)² x² + 4x + 4= -7+4 (x+2)² = -3 x+2 = √-3 X= -2+-i√3
18. Solve the equation by completing the square page 52 #10. 3x² + 12x - 18=0 #11. 6x(x+8) = 12
19. Home Work • Page 54 • 3-5, 9-11, 14-16, 25-27, 29-31
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# 2013 AIME II Problems/Problem 6
## Problem 6
Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer.
## Solutions
### Solution 1
The difference between consecutive integral squares must be greater than 1000. $(x+1)^2-x^2\geq1000$, so $x\geq\frac{999}{2}\implies x\geq500$. $x=500$ does not work, so $x>500$. Let $n=x-500$. By inspection, $n^2$ should end in a number close to but less than 1000 such that there exists $1000N$ within the difference of the two squares. Examine when $n^2=1000$. Then, $n=10\sqrt{10}$. One example way to estimate $\sqrt{10}$ follows.
$3^2=9$, so $10=(x+3)^2=x^2+6x+9$. $x^2$ is small, so $10=6x+9$. $x=1/6\implies \sqrt{10}\approx 19/6$. This is 3.16.
Then, $n\approx 31.6$. $n^2<1000$, so $n$ could be $31$. Add 500 to get the first square and 501 to get the second. Then, the two integral squares are $531^2$ and $532^2$. Checking, $531^2=281961$ and $532^2=283024$. $282,000$ straddles the two squares, which have a difference of 1063. The difference has been minimized, so $N$ is minimized $N=282000\implies\boxed{282}$
~BJHHar
### Solution 2
Let us first observe the difference between $x^2$ and $(x+1)^2$, for any arbitrary $x\ge 0$. $(x+1)^2-x^2=2x+1$. So that means for every $x\ge 0$, the difference between that square and the next square have a difference of $2x+1$. Now, we need to find an $x$ such that $2x+1\ge 1000$. Solving gives $x\ge \frac{999}{2}$, so $x\ge 500$. Now we need to find what range of numbers has to be square-free: $\overline{N000}\rightarrow \overline{N999}$ have to all be square-free. Let us first plug in a few values of $x$ to see if we can figure anything out. $x=500$, $x^2=250000$, and $(x+1)^2=251001$. Notice that this does not fit the criteria, because $250000$ is a square, whereas $\overline{N000}$ cannot be a square. This means, we must find a square, such that the last $3$ digits are close to $1000$, but not there, such as $961$ or $974$. Now, the best we can do is to keep on listing squares until we hit one that fits. We do not need to solve for each square: remember that the difference between consecutive squares are $2x+1$, so all we need to do is addition. After making a list, we find that $531^2=281961$, while $532^2=283024$. It skipped $282000$, so our answer is $\boxed{282}$.
### Solution 3
Let $x$ be the number being squared. Based on the reasoning above, we know that $N$ must be at least $250$, so $x$ has to be at least $500$. Let $k$ be $x-500$. We can write $x^2$ as $(500+k)^2$, or $250000+1000k+k^2$. We can disregard $250000$ and $1000k$, since they won't affect the last three digits, which determines if there are any squares between $\overline{N000}\rightarrow \overline{N999}$. So we must find a square, $k^2$, such that it is under $1000$, but the next square is over $1000$. We find that $k=31$ gives $k^2=961$, and so $(k+1)^2=32^2=1024$. We can be sure that this skips a thousand because the $1000k$ increments it up $1000$ each time. Now we can solve for $x$: $(500+31)^2=281961$, while $(500+32)^2=283024$. We skipped $282000$, so the answer is $\boxed{282}$.
### Solution 4
The goal is to find the least $N \in \mathbb{N}$ such that $\exists m \in \mathbb{N}$ where $m^2 + 1 \leq 1000N, 1000N + 1000 \leq (m+1)^2$.
Combining the two inequalities leads to $(m+1)^2 \geq m^2 + 1001, m \geq 500$.
Let $m = k + 500$, where $k \in \mathbb{W}$, then the inequalities become,
$N \geq \frac{(k+500)^2 + 1}{1000} = \frac{k^2 + 1}{1000} + k + 250$, and
$N \leq \frac{(k+501)^2}{1000} - 1 = \frac{(k+1)^2}{1000} + k + 250.$
For $k=31$, one can verify that $N = 282$ is the unique integer satisfying the inequalities.
For $k \leq 30$, $k + 250 < \frac{k^2 + 1}{1000} + k + 250 \leq N$ $\leq \frac{(k+1)^2}{1000} + k + 250 \leq \frac{(30+1)^2}{1000} + k + 250 < k + 251$,
i.e., $k + 250 < N < k + 251$, a contradiction.
Note $k \geq 32$ leads to larger $N$(s).
Hence, the answer is $\boxed{282}$.
~yuxiaomatt
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# Linear Regression — The Basics
## What is Linear Regression?
### The basics
Yeah. It’s not a good sign if I’m starting out already repeating myself. But that’s how things seem to be with linear regression, so I guess it’s fitting. It seems like every day one of my professors will talk about linear regression, and it’s not due to laziness or lack of coordination. Indeed, it’s an intentional part of the curriculum here at New College of Florida because of how ubiquitous linear regression is. Not only is it an extremely simple yet expressive formulation, it’s also the theoretical basis of a whole slew of other tactics. Let’s just get right into it, shall we?
### Linear Regression:
Let’s say you have some data from the real world (and hence riddled with real-world error). A basic example for us to start with is this one:
There’s clearly a linear trend there, but how do we pick which linear trend would be the best? Well, one thing we could do is pick the line that has the least amount of error from the prediction to the actual data-point. To do that, we have to say what we mean by “least amount of error”. For this post, we’ll calculate that error by squaring the difference between the predicted value and the actual value for every point in our data set, then averaging those values. This standard is called the Mean-Squared-Error (MSE). We can write the MSE as:
$\frac{1}{N}\sum\limits_{i=1}^N\left(\hat Y_i - Y_i\right)^2$
where $\hat Y_i$ is our predicted value of $Y_i$ for a give $X_i$. Being as how we want a linear model (for simplicity and extensibility), we can write the above equation as,
$\sum\limits_{i=1}^N\left(\alpha + \beta X_i - Y_i\right)^2$
for some $\alpha, \beta$ that we don’t yet know. But since we want to minimize that error, we can take some derivatives and solve for $\alpha, \beta$! Let’s go ahead and do that! We want to minimize
$\sum\limits_{i=1}^N\left(\alpha + \beta X_i - Y_i\right)^2$
We can start by finding the $\hat\alpha$ such that, $\frac{d}{d\alpha}\sum\limits_{i=1}^N\left(\alpha + \beta X_i - Y_i\right)^2 = 0$. And as long as we don’t forget the chain rule, we’ll be alright…
$\sum\limits_{i=1}^N2\left(\alpha + \beta X_i - Y_i\right) = 0\Longrightarrow$
$\sum\limits_{i=1}^N\left(\alpha + \beta X_i - Y_i\right) = 0 \Longrightarrow$
$N\alpha + N\beta\bar X - N\bar Y = 0\Longrightarrow$
$\alpha = \bar Y - \beta\bar X$
and we’ll find the $\beta$ such that $\frac{d}{d\beta}\sum\limits_{i=1}^N\left(\alpha + \beta X_i - Y_i\right)^2 = 0$
And following a similar pattern we find (sorry for the editing… WordPress.com isn’t the greatest therein):
$\sum\limits_{i=1}^N2X_i\left(\alpha + \beta X_i - Y_i\right) = 0\Longrightarrow$
$\alpha\sum\limits_{i=1}^NX_i + \beta\sum\limits_{i=1}^N X_i^2 - \sum\limits_{i=1}^NY_iX_i = 0\Longrightarrow$
$(\bar Y -\beta\bar X)N\bar X + \beta\sum\limits_{i=1}^NX_i^2 - \sum\limits_{i=1}^NY_iX_i = 0$
$N\bar Y\bar X -N\beta(\bar X)^2 + \beta\sum\limits_{i=1}^NX_i^2 - \sum\limits_{i=1}^NY_iX_i = 0$
$\beta\left(\sum\limits_{i=1}^NX_i^2 - N(\bar X)^2\right) = \sum\limits_{i=1}^NY_iX_i - N\bar Y\bar X$
$\beta = \frac{\sum\limits_{i=1}^NY_iX_i - N\bar Y\bar X}{\sum\limits_{i=1}^NX_i^2 - N(\bar X)^2}$
But note: $\frac{1}{N}\sum\limits_{i=1}^NX_i^2 - (\bar X)^2 = \text{VAR}(X)$
So, $\sum\limits_{i=1}^NX_i^2 - N(\bar X)^2 = N\cdot\text{VAR}(X)$
And: $\sum\limits_{i=1}^NY_iX_i - N\bar Y\bar X = \sum\limits_{i=1}^NY_iX_i - N(\frac{1}{N}\sum\limits_{i=1}^NY_i)\bar X$
$= \sum\limits_{i=1}^NY_iX_i - \sum\limits_{i=1}^N(Y_i\bar X)$
$= \sum\limits_{i=1}^NY_i\left(X_i - \bar X\right)$
$= \sum\limits_{i=1}^NY_i\left(X_i - \bar X\right) - \sum\limits_{i=1}^N\bar Y\left(X_i - \bar X\right) + \sum\limits_{i=1}^N\bar Y\left(X_i - \bar X\right)$
$= \sum\limits_{i=1}^N\left(Y_i-\bar Y\right)\left(X_i - \bar X\right) + \bar Y\sum\limits_{i=1}^N\left(X_i - \bar X\right)$
$= \sum\limits_{i=1}^N\left(Y_i-\bar Y\right)\left(X_i - \bar X\right) = N\cdot \text{COV}(X, Y)$
So, $\beta = \frac{\text{COV}(X,Y)}{\text{VAR}(X)}$.
And then we can find $\alpha$ by substituting in our approximation of $\beta$. Using those coefficients, we can plot the line below, and as you can see, it really is a good approximation.
### Now we have it
Okay, so now we have our line of “best fit”, but what does it mean? Well, it means that this line predicts the data we gave it with the least error. That’s really all it means. And sometimes, as we’ll see later, reading too much into that can really get you into trouble.
But using this model we can now predict other data outside the model. So, for instance, in the model pictured above, if we were to try and predict $Y$ when $X=2$, we wouldn’t do so bad by picking something around 10 for $Y$.
### An example, perhaps?
So I feel at this point, it’s probably best to give an example. Let’s say we’re trying to predict stock price given the total market price. Well, in practice this model is used to assess volatility, but that’s neither here nor there. Right now, we’re really only interested in the model itself. But without further ado, I present you with, the CAPM (Capital Asset Pricing Model):
$r = \alpha + \beta r_m + \epsilon$ (where $\epsilon$ is the error in our predictions).
And you can fit this using historical data or what-have-you. There are a bunch of downsides to fitting it with historical data though, like the fact that data from 3 days ago really doesn’t have much to say about the future anymore. There are plenty of cool things you can do therein, but sadly, those are out of the scope of this post.
For now, we move on to
## Multiple Regression
### What is this anyway?
Well, multiple regression is really just a new name for the same thing: how do we fit a linear model to our data given some set of predictors and a single response variable? The only difference is that this time our linear model doesn’t have to be one dimensional. Let’s get right into it, shall we?
So let’s say you have $k$ many predictors arranged in a vector (in other words, our predictor is a vector in $\mathbb{R}^n$). Well, I wonder if a similar formula would work… Let’s figure it out…
Firstly, we need to know what a derivative is in $\mathbb{R}^n$. Well, if $f:\mathbb{R}^n\to\mathbb{R}^m$ is a differentiable function, then for any $x$ in the domain, $f'(x)$ is the linear map $A:\mathbb{R}^n\to\mathbb{R}^m$ such that $\text{lim}_{h\to0}\frac{||f(x+h) - f(x) - Ah||}{||h||} = 0$. Basically, $f'(x)$ is the tangent plane.
So, now that we got that out of the way, let’s use it! We want to find the linear function that minimizes the Euclidean norm of the error terms (just like before). But note: the error term is $\epsilon = Y - \hat Y = Y - \alpha -\beta X$, for some vector $\alpha$ and some matrix $\beta$. Now, since it’s easier and it’ll give us the same answer, we’re going to minimize the squared error term instead of just the error term (like we did in the one dimensional version). We’re also going to make one more simplification: That $\alpha=0$. We can do this safely by simply appending (or prepending) a 1 to the rows of our data (thereby creating a constant term). So for the following, assume we’ve done that.
$\langle\epsilon, \epsilon\rangle = (Y - X\beta)^T(Y - X\beta)$
$\langle\epsilon, \epsilon\rangle = (Y^T - \beta^TX^T)(Y - X\beta)$
$\langle\epsilon, \epsilon\rangle = Y^TY - 2Y^TX\beta + \beta^TX^TX\beta$
So, let’s find the $\beta$ that minimizes that.
$0=\lim\limits_{h\to0}\frac{||(Y^TY - 2Y^TX(\beta+h) + (\beta+h)^TX^TX(\beta+h)) - (Y^TY - 2Y^TX\beta + \beta^TX^TX\beta) - Ah||}{||h||}$
$0=\lim\limits_{h\to0}\frac{||- 2Y^TXh + 2\beta^TX^TXh + h^TX^TXh - Ah||}{||h||}$
$0=\lim\limits_{h\to0}||- 2Y^TX + 2\beta^TX^TX + h^TX^TX - A||\frac{||h||}{||h||}$
$0=\lim\limits_{h\to0}||- 2Y^TX + 2\beta^TX^TX - A||$
So, now we see that the derivative is $-2Y^TX + 2\beta^TX^TX$ and we want to find where our error is minimized, so we want to set that derivative to zero:
$0=- 2Y^TX + 2\beta^TX^TX$
$X^TX\beta = X^TY$
$\beta = (X^TX)^{-1}X^TY$
And there we have it. That’s called the normal equation for linear regression.
Maybe next time I’ll post about how we can find these coefficients given some data using gradient descent, or some modification thereof.
Till next time, I hope you enjoyed this post. Please, let me know if something could be clearer or if you have any requests.
# A dirty little ditty on Finite Automata
### Some Formal Definitions
#### A Deterministic Finite Automata (DFA) is
• A set $\mathcal{Q}$ called “states”
• A set $\Sigma$ called “symbols”
• A function $\delta_F:\mathcal{Q}\times\Sigma \to \mathcal{Q}$
• A designated state $q_0\in\mathcal{Q}$ called the start point
• A subset $F\subseteq\mathcal{Q}$ called the “accepting states”.
The DFA is then often referred to as the ordered quintuple $A=(\mathcal{Q},\Sigma,\delta_F,q_0,F)$.
#### Defining how strings act on DFAs.
Given a DFA, $A=(\mathcal{Q}, \Sigma, \delta, q_0, F)$, a state $q_i\in\mathcal{Q}$, and a string $w\in\Sigma^*$, we can define $\delta(q_i,w)$ like so:
• If $w$ only has one symbol, we can consider $w$ to be the symbol and define $\delta(q_i,w)$ to be the same as if we considered $w$ as the symbol.
• If $w=xv$, where $x\in\Sigma$ and $v\in\Sigma^*$, then $\delta(q_i, w)=\delta(\delta(q_i,x),v)$.
And in this way, we have defined how DFAs can interpret strings of symbols rather than just single symbols.
#### The language of a DFA
Given a DFA, $A=(\mathcal{Q}, \Sigma, \delta, q_0, F)$, we can define “the language of $A$“, denoted $L(A)$, as $\{w\in\Sigma^*|\delta(q_0,w)\in F\}$.
### Some Examples, Maybe?
#### Example 1:
Let’s construct a DFA that accepts only strings beginning with a 1 that, when interpreted as binary numbers, are multiples of 5. So some examples of strings that would be in $L(A)$ are 101, 1010, 1111
### Some More Formal Definitions
#### A Nondeterministic Finite Automata (NFA) is
• A set $\mathcal{Q}$ called “states”
• A set $\Sigma$ called “symbols”
• A function $\delta_N:\mathcal{Q}\times\Sigma \to \mathcal{P}\left(\mathcal{Q}\right)$
• A designated state $q_0\in\mathcal{Q}$ called the start point
• A subset $F\subseteq\mathcal{Q}$ called the “accepting states”.
The NFA is then often referred to as the ordered quintuple $A=(\mathcal{Q},\Sigma,\delta_N,q_0,F)$.
#### Defining how strings act on NFAs.
Given an NFA, $N=(\mathcal{Q}, \Sigma, \delta, q_0, F)$, a collection of states $\mathcal{S}\subseteq\mathcal{Q}$, and a string $w\in\Sigma^*$, we can define $\delta(\mathcal{S},w)$ like so:
• If $w$ only has one symbol, then we can consider $w$ to be the symbol and define $\delta(\mathcal{S},w):=\bigcup\limits_{s\in\mathcal{S}}\delta(s,w)$.
• If $w=xv$, where $x\in\Sigma$ and $v\in\Sigma^*$, then $\delta(\mathcal{S}, w)=\bigcup\limits_{s\in\mathcal{S}}\delta(\delta(s,x),v)$.
And in this way, we have defined how NFAs can interpret strings of symbols rather than just single symbols.
# Probability — A Measure Theoretic Approach
## What is probability?
Consider a set $\Omega$ (called the sample space), and a function $X:\Omega\rightarrow\mathbb{R}$ (called a random variable.
If $\Omega$ is countable (or finite), a function $\mathbb{P}:\Omega\rightarrow\mathbb{R}$ is called a probability distribution if it satisfies the following 2 conditions:
1. For each $x \in \Omega$, $\mathbb{P}(x) \geq 0$
2. If $A_i\cap A_j = \emptyset$, then $\mathbb{P}(\bigcup\limits_0^\infty A_i) = \sum\limits_0^\infty\mathbb{P}(A_i)$
And if $\Omega$ is uncountable, a function $F:\mathbb{R}\rightarrow\mathbb{R}$ is called a probability distribution or a cumulative distribution function if it satisfies the following 3 conditions:
1. For each $a,b\in\mathbb{R}$, $a < b \rightarrow F(a)\leq F(b)$
2. $\lim\limits_{x\to -\infty}F(x) = 0$
3. $\lim\limits_{x\to\infty}F(x) = 1$
### The Intuition:
What idea are we even trying to capture with these seemingly disparate definitions for the same thing? Well, with the two cases taken separately it's somewhat obvious, but they don't seem to marry very well. The discrete case is giving us a pointwise estimation of something akin to the proportion of observations that should correspond to a value (in a perfect world). The continuous case is the same thing, but instead of corresponding to that particular value (which doesn't really even make sense in this case), the proportion corresponds to the point in question and everything less than it. The shaded region in the top picture below and the curve in the picture directly below it denote the cumulative density function of a standard normal distribution (don't worry too much about what that means for this post, but if you're doing anything with statistics, you should probably know a bit about that).
Another way to define a continuous probability distribution is through something called a probability density function, which is closer to the discrete case definition of a probability distribution (or probability mass function). A probability density function is a function $f:\mathbb{R}\rightarrow\mathbb{R}_+$ such that $\int_{-\infty}^xf(t)dt = F(x)$. In other words, $\frac{dF}{dX} = f$. This new function has some properties of our discrete case probability function, but lacks some others. On the one hand, they’re both defined pointwise, but on the other, this one can be greater than one in some places — meaning the value of the probability density function isn’t really the probability of an event, but rather (as the name “suggests”) the density therein.
### Does it measure up?
Now let’s check out the measure theoretic approach…
Let $\Omega$ be our sample space, $S$ be the $\sigma$-algebra on $\Omega$ (so $S$ is the collection of measurable subsets of $\Omega$), and $\mu:S\to\mathbb{R}$ a measure on that measure space. Let $X:\Omega\rightarrow\mathbb{E}$ be a random variable ($\mathbb{E}$ is generally taken to be $\mathbb{R}$ or $\mathbb{R}^n$). We define the function $\mathbb{P}:\mathcal{P}(\mathbb{E})\rightarrow\mathbb{R}$ (where $\mathcal{P}(\mathbb{E})$ is the powerset of $\mathbb{E}$ — the set of all subsets) such that if $A\subseteq\mathbb{E}$, we have $\mathbb{P}(A)=\mu(X^{-1}(A))$. We call $\mathbb{P}$ a probability distribution if the following conditions hold:
1. $\mu(\Omega) = 1$
2. for each $A\subseteq\mathbb{E}$ we have $X^{-1}(A)\in S$.
### Why do this?
Well, right off the bat we have a serious benefit: we no longer have two disparate definitions of our probability distributions. Furthermore, there is the added benefit of having a natural separation of concerns: the measure $\mu$ determines the what we might intuitively consider to be the probability distribution while the random variable is used to encode the aspects of the events that we care about.
To further illustrate this
### The Examples
#### A fair die
##### All even
Let’s consider a fair die. Our sample space will be $\{1,2,3,4,5,6\}$. Since our die is fair, we’ll define our measure fairly: for any $x$ in our sample space, $\mu(\{x\}) = \frac{1}{6}$. If we want to know, for instance, what the probability of getting each number is, we could use a very intuitive random variable $X(a) = a$ (so $X(1)=1$, etc.). Then we see that $\mathbb{P}(\{1\}) = \mu(\{1\}) = \frac{1}{6}$, and the rest are found similarly.
##### Odds and Evens?
What if we want to consider the fair die of yester-paragraph, but we only care if the face of the die shows an odd or an even number? Well, since the actual distribution of the die hasn’t changed, we won’t have to change our measure. Instead we’ll change our random variable to capture just those aspects we care about. In particular, $X(a) = 0$ if $a$ is even, and $X(a) = 1$ if $a$ is odd. We then see $\mathbb{P}(1) = \mu(\{1,3,5\}) = \frac{1}{2}$ and $\mathbb{P}(0) = \mu(\{2,4,6\}) = \frac{1}{2}$
Now let’s consider the same scenario of wanting to know the probability of getting each number, but now our die is loaded. Being as how we’re changing the distribution itself and not just the aspects we’re choosing to care about, we’re going to want to change the measure this time. For simplicity, let’s consider a kind of degenerate case scenario. Let our measure be: $\mu(A) = 1$ if $1\in A$ and $\mu(A)=0$ if $1\notin A$. Basically, we’re defining our probability to be such that the only possible outcome is a roll of 1. So since we are concerned with the same things we were concerned with last time, we can take that same random variable. We note $\mathbb{P}(1) = 1$ and $\mathbb{P}(a) = 0$ for any $a \neq 1$.
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# 1.2. Equally Likely Outcomes#
“If a coin is tossed, what is the chance that it lands heads?” Ask this question and the most common answer that you will get is $$1/2$$. If you press for a reason, don’t be surprised to hear, “Because the coin has two faces.” A coin does indeed have two faces, but notice an assumption hidden inside the “reasoning” you have been given: that each of the two faces has the same chance as the other.
The assumption of equally likely outcomes is a simple and ancient model of randomness. It defines probabilities as proportions. The assumption that $$\Omega$$ is finite makes proportions easy to identify as fractions of the total number of outcomes.
For some $$n > 1$$, let $$\Omega$$ consist of $$n$$ outcomes. Let $$A \subseteq \Omega$$ be an event. Define $$\#(A)$$ to be the number of outcomes in the subset $$A$$. Thus $$\#(\Omega ) = n$$, $$\#(\phi ) = 0$$, and $$0 < \#(A) < n$$ for any other event $$A$$.
For an event $$A$$, let $$P(A)$$ denote the probability that $$A$$ occurs, or the chance that $$A$$ occurs. We will use the words “probability” and “chance” synonymously, and we will often use “happens” instead of the more formal “occurs”.
## 1.2.1. Probabilities on a Space of Equally Likely Outcomes#
If all $$n$$ outcomes in $$\Omega$$ are assumed to be equally likely, then the probability that the event $$A$$ occurs is defined by
$P(A) ~=~ \frac{\#(A)}{\#(\Omega )} ~=~ \frac{\#(A)}{n} ~=~ \text{proportion of outcomes in } A$
This idea that probabilities are proportions lies at the heart of many calculations. As you will see later, rules for combining proportions become rules for combining probabilities, whether or not all outcomes are equally likely. But for now we will work in settings where it is natural to assume that outcomes are equally likely.
## 1.2.2. Example 1: Random Permutations#
Let $$\Omega$$ be the space of all permutations of the letters $$a$$, $$b$$, and $$c$$. Then $$\Omega$$ contains $$n=6$$ outcomes:
$\Omega ~=~ \{ abc, ~acb, ~bac, ~bca, ~cab, ~cba \}$
If we assume that all six permutations are equally likely, we have what are called random permutations of the three letters. Under this assumption, we can augment our table of events with a column of chances.
Event
Verbal Description
Subset
Probability
$$A$$
$$a$$ appears first
$$\{abc, ~acb\}$$
$$\frac{2}{6} = \frac{1}{3}$$
$$B$$
$$b$$ and $$c$$ are not next to each other
$$\{bac, ~cab\}$$
$$\frac{1}{3}$$
$$C$$
the letters are in alphabetical order
$$\{abc\}$$
$$\frac{1}{6}$$
$$D$$
$$a$$ appears first, $$b$$ next, but $$c$$ isn’t third
$$\phi$$
$$0$$
$$E$$
$$c$$ is either first, second, or third
$$\Omega$$
$$1$$
$$F$$
the letters form a word that means “taxi”
$$\{ cab \}$$
$$\frac{1}{6}$$
Notice that
$P(a \text{ appears last}) = \frac{\#\{ bca, ~cba \}}{6} ~=~ \frac{1}{3} ~=~ \frac{\#\{ bac, ~cab \}}{6} ~=~ P(a \text{ appears second})$
Thus the assumption that all the permutations are equally likely makes all three positions of $$a$$ equally likely as well. The same is true of the positions of $$b$$ and $$c$$, as you should check.
Quick Check
Suppose you toss a coin three times. Assuming that all the outcomes in the space $$\{ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT \}$$ are equally likely, find the chance that one of the tosses lands differently from the other two.
## 1.2.3. Example 2: Random Number Generator#
Suppose a random number generator returns a pair of digits from among the 100 pairs 00, 01, 02, $$\ldots$$, 98, 99, in a way that makes all the pairs equally likely to be returned.
You will have noticed that the pairs correspond to the 100 integers 0 through 99. In what follows, it will be useful to remind yourself of the product rule of counting:
• There are 10 choices for the first digit: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
• Corresponding to each choice of the first digit there are 10 choices for the second digit.
• So in total there are $$10 \times 10 = 100$$ pairs of digits.
Here a “pair” is an sequence of two digits, one following the other. The pair 27 is different from the pair 72. These are sometimes called “ordered pairs”. In this text, all sequences are ordered.
Now let’s compute the probabilities of a few events. By assumption, all pairs are equally likely. So each answer will consist of counting the number of pairs in the event, and then dividing by the total number of pairs, which is 100.
(i) What is the probability that the pair consists of two different digits?
We have to count the number of pairs $$ab$$ where $$a \ne b$$. The digit $$a$$ can be chosen in 10 ways; for each of those ways, there are only 9 ways of choosing $$b$$ because $$b$$ has to be different from $$a$$. So the answer is
$P(\text{the pair consists of two different digits}) ~=~ \frac{90}{100} ~=~ 0.9$
(ii) What is the chance that the two digits are the same?
Let’s try to use our answer to (i). In every one of the 100 pairs, either the two digits are the same or they are different. No pair can be in both categories, so by rules of proportion we have
$P(\text{the two digits are the same}) ~=~ 1 ~-~ P(\text{the pair consists of two different digits}) ~=~ 0.1$
To check this by counting, you have to count pairs of the form $$aa$$. There are 10 ways to choose $$a$$, and there are no further choices to make. So the answer is $$10/100 = 0.1$$, confiriming our calculation above.
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Curriculum Goal
Primary: Number Sense
• Read and represent whole numbers up to and including 50.
• Describe various ways whole numbers may be used/referenced in everyday life.
• Compose and decompose whole numbers, up to and including 50, using a variety of tools and strategies, in various contexts.
Context
• Teacher works with at least two children.
Materials
• Deck of cards with face cards removed (Ace = 1)
Lesson
Easier Version:
• Lay cards face-down randomly in a 5 × 8 rectangle.
• Explain that the goal is to match cards of the same number. Colour/suit do not matter.
• For each round, students turn over one card and try to find a matching card.
• Introduce the advanced version, if appropriate.
• Explain that the goal is to find the numbers that add/subtract to a certain sum/difference (chosen from a number between 2-20).
• Give an example, such as 5. Ask students for different number pairs for that sum (e.g., if the chosen sum = 5, players should aim for Ace + 4 or 9 – 4; if the chosen difference = 3, players should aim for pairs such, as 6 – 3 and 1 + 2).
• When all pairs for the predetermined difference have been found, facilitate a discussion about how quantities can be composed and decomposed.
• Challenge the students play the game in pairs.
• Afterward, hold a discussion to extend student thinking. Here are some sample questions:
• How did you know these pairs were correct?
• What strategies did you use to add/subtract the two cards together?
• What techniques did you use to remember where the cards were?
• Why would there be more possible pairs for a larger difference (i.e., 10 versus 5)?
Look Fors
• Do the children select the exact card that they were looking for?
• Are students counting the number of symbols on the card? If so, are they using one-to-one correspondence? Are the students recognizing the numbers on the card?
• Are the children using their fingers to count? Are the children counting up or counting on when they choose their pairs?
• Do the children use certain memory strategies to determine if the pairs create the predetermined sum?
• Do the children use mathematical language to talk about the cards?
Related Lessons
Compose numbers up to and including the number 50 using different strategies.
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# Math rigor in evaluating limits.
There is a well known theorem that says $f: \mathbb{R} \to \mathbb{R}$ is continuous at $x_0$ if and only if $$\lim_{x\to x_0} f(x) = f(x_0)$$
We often use the above mentioned theorem to evaluate limits like this one:
$$\lim_{x \to 1} \ f(x) \ \text{where} \ f(x) = \frac{x^3-3x^2+2}{x^2-1}$$
Note that $f$ is defined for $\mathbb{R} \setminus \{\pm 1\}$, so it doesn't make sense to evaluate $f$ at $1$, but we can factor $(x-1)$ in the numerator and denominator of $f$, and we get:
$$f(x) = \frac{(x-1)(x^2-2x-2)}{(x-1)(x+1)} = \frac{x^2-2x-2}{x+1}$$
now we are able to apply the theorem and evaluate the limit:
$$\lim_{x \to 1}\ f(x) = f(1) = \frac{x^2-2x-2}{x+1} =\frac{-3}{2}$$
$\textbf{Question:}$ let $f(x) = \dfrac{x^3-3x^2+2}{x^2-1}$ and $g(x) =\dfrac{x^2-2x-2}{x+1}$
It's easy to see that $g$ is defined at $1$ while $f$ is not, so $f$ and $g$ are not the same functions, but in the above limit evaluation we used that $$\lim_{x \to 1} f(x) =g(1)$$ but this is not a clear (at least to me) application of the theorem, as $f$ and $g$ are not the same functions. How could I justify this application of the theorem?
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You f(x) and g(x) are the same function if x doesn't equal 1. – Peter Webb Feb 15 '14 at 6:11
@PeterWebb but we are interested precisely when $x=1$. – Magno Feb 15 '14 at 6:12
You can consider limits even if the function isn't defined at that point. – Mark Fantini Feb 15 '14 at 6:19
Your f(x) and g(x) are exactly the same function if x doesn't equal 1. If we define g(1) = f(1) they are the same function exactly. To calculate the limit of f(x) as x -> 1, you don't even need that. g(1) will be the limit. But if g(1) and f(1) are not the same, then the function will not be continuous. You can make it continuous by filling in the value of g(1) as being the same as f(1). – Peter Webb Feb 15 '14 at 6:21
Better. The definition of the limit requires calculating the values of g(x) for x close to x_0 (x=1 in your case). It does not involve calculating the value of g(x) actually at x = 1. But f(x) and g(x) are exactly the same function for all values of x except x=1, so we can substitute f(x) for g(x) in calculating limits, as g(1) is never actually used in the limit calculation. – Peter Webb Feb 15 '14 at 6:29
First of all, you should write in the line just above the question
$$\lim_{x\rightarrow 1}f(x)=\lim_{x\rightarrow 1}\frac{x^2-2x-2}{x+1}=\frac{-3}{2}"$$
Now, to your question: In the definition of limit (which you can probably find in any calculus book), the limit of a function $f$ at a point $x$ independs of the value $f(x)$ - it depends only on the value of $f$ at points clsoe enough to $x$. So it is obvious that $\lim_{t\rightarrow x}f(t)$ exists if, and only if, $\lim_{t\rightarrow x}f|_{Dom(f)\setminus \left\{x\right\}}(t)$ exists, and in that case, they are equal (here, $Dom(f)$ is the domain of $f$ and $f|_A$ means the restriction of $f$ to $A$).
In your example, $f=g|_{Dom(g)\setminus\left\{1\right\}}$, and the above applies.
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Evaluating the limit of g(x) = (x^2-2x-2)/(x+1) as x approaches 1 is not the same as applying the theorem. Applying the theorem would be evaluating g(1). I understand your point but that is not what I'm asking for. – Magno Feb 15 '14 at 6:37
I can sense the issue being faced by OP. There is one point which I must clarify.
$f$ is continuous at $x_{0}$ if $\lim_{x \to x_{0}}f(x) = f(x_{0})$. Note that this is a definition of continuity and to apply it means that you calculate $\lim_{x \to x_{0}}f(x)$ somehow and compare it with value of $f(x_{0})$. If you find them equal you declare that $f$ is continuous at $x = x_{0}$. Note that the calculation of limit of $f(x)$ as $x \to x_{0}$ can't be done by using continuity of $f(x)$.
For example consider $\lim_{x \to 2} x^{2}$. We calculate this as follows $$\lim_{x \to 2}x^{2} = \lim_{x \to 2}x\cdot\lim_{x \to 2}x = 2 \cdot 2 = 4$$ The fact that $\lim_{x \to 2}x = 2$ is a trivial application of definition of limit. After this exercise we see that value of $x^{2}$ at $x = 2$ is also $4$ and hence $x^{2}$ is continuous at $x = 2$.
Please understand that it does not happen in reverse. Proving continuity of common functions like polynomials, trigometric, exponential and logarithmic function is done by first finding their limits and then noticing that it matches their values also. Once this fact is established we don't try to go the complicated way of using limit rules to find limits of complicated expression containing polynomials (trig, exp, log function etc) but rather use the continuity of such functions and just plug the values of $x$ to get the limit. This procedure is only a shorthand (basically saving labor) of applying the limit rules used to evaluate the limits of such continuous functions.
Next we have a rule of limits which says that if $f(x) = g(x)$ for all $x$ in a neighborhood of $x = x_{0}$ except possibly at $x = x_{0}$ then $\lim_{x \to x_{0}}f(x) = \lim_{x \to x_{0}}g(x)$. This rule is used implicitly while evaluating limit of $f(x)$ when it is not defined at $x_{0}$. Basically we apply algebraic (trigonometric identities etc) to transform the expression for $f(x)$ into another function $g(x)$ which is probably defined at $x = x_{0}$ and whose limit at $x = x_{0}$ is known. Same technique has been used in the limit under consideration in this question.
Specifically we have \begin{aligned}L &= \lim_{x \to 1}\frac{x^{3} - 3x^{2} + 2}{x^{2} - 1}\\ &= \lim_{x \to 1}\frac{(x - 1)(x^{2} - 2x - 2)}{(x - 1)(x + 1)}\\ &= \lim_{x \to 1}\frac{x^{2} - 2x - 2}{x + 1}\\ &= \dfrac{{\displaystyle \lim_{x \to 1}x^{2} - 2x - 2}}{{\displaystyle \lim_{x \to 1}x + 1}}\\ &= \dfrac{{\displaystyle \lim_{x \to 1}x\cdot\lim_{x \to 1}x - 2\lim_{x \to 1}x - 2}}{{\displaystyle \lim_{x \to 1}x + 1}}\\ &= \frac{1\cdot 1 - 2\cdot 1 - 2}{1 + 1} = -\frac{3}{2}\end{aligned}
In reality we avoid the last 3 steps and directly put $x = 1$ in $x^{2} - 2x - 2$ and $x + 1$. This is just a shorthand which is justified because we have established beforehand that polynomials are continuous.
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The notion of plugging value $x = a$ into some expression to calculate the limit of expression as $x \to a$ is discussed extensively on my blog post paramanands.blogspot.com/2013/11/… without any mention of the word "continuity". This is basically an outcome of the standard rules of limits. – Paramanand Singh Feb 15 '14 at 8:11
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GMAT Math : Modulo arithmetic
Modulo arithmetic is a very powerful tool for divisibility and remainder type questions.
First, an example:
What is the remainder when t² + 2t + 3 is divided by 8?
1. t when divided by 8 yields a remainder of 1
2. t – 3 when divided by 8 yields a remainder of 1
Fundamentally, the basic notation in modulo arithmetic is as follows
x = a mod (n) which means when number x is divided by number n the remainder is a.
3 = 1 mod (2) is an example of the above notation.
The really cool thing about modulo arithmetic is the ability to add, subtract and even raise to powers these notations, provided they have the same base ‘n’. Specifically :
if x = a mod (n)
and z = b mod (n)
The above two equations can be added to get (x + z) = (a + b) mod (n)
They can be subtracted to get (x – z) = (a – b) mod (n)
They can be multiplied by a constant term xk = ak mod (n) (where k is a constant)
They can be raised to powers xt = at mod (n)
But what does this all mean and how is this useful? If you see a question like what is the remainder when 10099 is divided by 3 – modulo arithmetic can solve it easily.
We know that 100 when divided by 3 gives a remainder of 1 or in modulo arithmetic terms
100 = 1 mod (3). Using the rule for powers and raising both sides of the equation by 99.
10199 = 199 mod (3)
10199 = 1 mod (3)
This tells us that the remainder when 10099 is divided by 3 is 1. Pretty nifty huh?
Getting back to the illustrative question.
Statement (1) can be represented as t = 1 mod (8) so
t² = 1² mod (8) = 1 mod (8)
2t = 2 x 1 mod (8) = 2 mod (8)
3 = 3 mod (8)
Adding the above three terms we arrive at the question statement
t² + 2t + 3 = 6 mod (8).
In other words, t² + 2t + 3 when divided by 8 gives a remainder of 6. so statement (1) is sufficient.
Statement (2) tells us that t² = 1 mod (8) or specifically t = 1 mod (8) or t = -1 mod (8). Notice this statement has not been able to tell us the remainder when t is divided by 8 – it can be 1 or -1 (-1 means a remainder of 7). And we need the remainder when t is divided by 8 to find the remainder of the second term (2t) in the original question statement. So information is insufficient. We go with Option (A). For a more detailed exposition of modulo arithmetic and its utility to solve divisibility data sufficiency questions, please do check out our Hackbook
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District Of Columbia - Grade 1 - Math - Geometry - Composite Shapes - 1.G.2
Description
Compose two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape, and compose new shapes from the composite shape.1
• State - District Of Columbia
• Standard ID - 1.G.2
• Subjects - Math Common Core
Keywords
• Math
• District Of Columbia grade 1
• Geometry
More District Of Columbia Topics
Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used.
Apply properties of operations as strategies to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps.
1.OA.4 Understand subtraction as an unknown-addend problem. For example, subtract 10 – 8 by finding the number that makes 10 when added to 8. Add and subtract within 20.
1.OA.5 Relate counting to addition and subtraction (e.g., by counting on 2 to add 2).
1.OA.6 Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: A. 10 can be thought of as a bundle of ten ones — called a “ten.” B. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones).
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GeoGebra Essentials 8 – Graphs and their Properties
This is the eighth tutorial in the GeoGebra Essentials Series. If you are not familiar with GeoGebra, you may want to read the Introduction to GeoGebra post and earlier tutorials.
In the tutorial below, menu commands, located in the menu bar, are in brown bold text, and submenus are denoted by the > symbol. For example, Options>Labeling>New Points Only means, click the Options menu, choose Labeling from the list, then select New Points Only. The tool texts are colored orange. For example, New Point means the new point tool. Text that are to be typed in the input box are colored blue.
In this tutorial, we learn how to graph functions using GeoGebra. First, we graph a function, then determine its critical points (minimum, maximum, inflection point, roots) using GeoGebra keyboard commands. Then we get the derivative of the function. We also construct a point on the function and a line tangent through that point. We explore the characteristics of the tangent line in relation to the graph of the function and its derivative. The final output of our tutorial is shown above. » Read more
GeoGebra Essentials 7 – Using the Keyboard Commands
In the previous tutorials in the GeoGebra Essentials Series, we have discussed how to construct mathematical objects such as points, lines, line segments, circles and other mathematical objects using the mouse and the tools in the Toolbar. In this post, we learn how to construct these objects using keyboard commands. In particular, we are going to create an equilateral triangle.
Step by Step Instructions
In this post, we create equilateral triangle ABC. The construction is very similar to GeoGebra Tutorial 2 – Constructing an Equilateral Triangle.
1.) Open GeoGebra. Select Algebra & Graphics in the Perspective panel to open the Algebra & Graphics window
2.) To plot point A with coordinates (1,1), type A = (1,1) in the Input Bar and then press the ENTER key on your keyboard. Now plot B = (5,1).
3.) To construct segment AB, type segment[A,B].
4.) To construct a circle with center A and passing through B, type circle[A,B]. Now, construct a circle with center B passing through A.
5.) Now, to intersect the two circles, type intersect[c,d]. The names c and d are the name given by GeoGebra to the two circles (see Algebra view). Names, usually small or capital letters, are given to each object in GeoGebra. Do not confuse the names with equations.
6.) Next, to polygon ABC, type polygon[A, B, C].
The Final Output
7.) Now, hide the circles and point D by right clicking them and clicking Show Object and we are done. As we can see, we cannot do everything using keyboard commands.
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# 2015 AIME I Problems/Problem 3
## Problem
There is a prime number $p$ such that $16p+1$ is the cube of a positive integer. Find $p$.
## Solution 1
Let the positive integer mentioned be $a$, so that $a^3 = 16p+1$. Note that $a$ must be odd, because $16p+1$ is odd.
Rearrange this expression and factor the left side (this factoring can be done using $(a^3-b^3) = (a-b)(a^2+a b+b^2)$ or synthetic divison once it is realized that $a = 1$ is a root):
\begin{align*} a^3-1 &= 16p\\ (a-1)(a^2+a+1) &= 16p\\ \end{align*}
Because $a$ is odd, $a-1$ is even and $a^2+a+1$ is odd. If $a^2+a+1$ is odd, $a-1$ must be some multiple of $16$. However, for $a-1$ to be any multiple of $16$ other than $16$ would mean $p$ is not a prime. Therefore, $a-1 = 16$ and $a = 17$.
Then our other factor, $a^2+a+1$, is the prime $p$:
\begin{align*} (a-1)(a^2+a+1) &= 16p\\ (17-1)(17^2+17+1) &=16p\\ p = 289+17+1 &= \boxed{307} \end{align*}
## Solution 2
Since $16p+1$ is odd, let $16p+1 = (2a+1)^3$. Therefore, $16p+1 = (2a+1)^3 = 8a^3+12a^2+6a+1$. From this, we get $8p=a(4a^2+6a+3)$. We know $p$ is a prime number and it is not an even number. Since $4a^2+6a+3$ is an odd number, we know that $a=8$.
Therefore, $p=4a^2+6a+3=4*8^2+6*8+3=\boxed{307}$.
## Solution 3
Let $16p+1=a^3$. Realize that $a$ congruent to $1\mod 4$, so let $a=4n+1$. Expansion, then division by 4, gets $16n^3+12n^2+3n=4p$. Clearly $n=4m$ for some $m$. Substitution and another division by 4 gets $256m^3+48m^4+3m=p$. Since $p$ is prime and there is a factor of $m$ in the LHS, $m=1$. Therefore, $p=\boxed{307}$.
## Solution 4
Notice that $16p+1$ must be in the form $(a+1)^3 = a^3 + 3a^2 + 3a + 1$. Thus $16p = a^3 + 3a^2 + 3a$, or $16p = a\cdot (a^2 + 3a + 3)$. Since $p$ must be prime, we either have $p = a$ or $a = 16$. Upon further inspection and/or using the quadratic formula, we can deduce $p \neq a$. Thus we have $a = 16$, and $p = 16^2 + 3\cdot 16 + 3 = \boxed{307}$.
## Solution 5
Notice that the cube 16p+1 is equal to is congruent to 1 mod 16. The only cubic numbers that leave a residue of 1 mod 16 are 1 and 15. Case one: The cube is of the form 16k+1-->Plugging this in, and taking note that p is prime and has only 1 factor gives p=307 Case two: The cube is of the form 16k+15--> Plugging this in, we quickly realize that this case is invalid, as that implies p is even, and p=2 doesn't work here
Hence, $p=\boxed{307}$ is our only answer pi_is_3.141
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# Variables and Verbal Expressions
- -
Some real-life situations involving numbers can directly be solved using the four fundamental operations: addition, subtraction, multiplication and division. For instance, if 1 apple cost 5 pesos, how much do 15 apples cost? To solve this problem simply multiply 15 by 5. However, there are other practical applications which can best be solved by looking for a general pattern, relationship or formula before arriving at the answers.
An example would be, study the sequence: 2, 4, 8, 16, 32 and so on, what is the sum of the first 12 numbers in this sequence? This problem entails a formula to shorten the procedure. It is on this context, that the core of Algebra lies on representing quantities, patterns or relationships by symbols other than numbers. These symbols, which can be any letter in the English alphabet, are called variables that can take on more than one value. These symbols can further be grouped using the four fundamental operations, which in turn give meaning to equations or inequalities. These groups of symbols are called mathematical expressions which are used to represent verbal expressions or phrases.
Verbal expressions are group of words that comprised a verbal problem or mathematical word problem. Since these would give meaning to the equation/inequality that would represent a problem, one needs to be extra careful in translating any verbal expression into its equivalent mathematical expression.
In this connection, listed below are some of the most commonly used key words to indicate an operation.
OperationKey Words
+
-
minus, subtract, subtracted from, less, less than, decreased by, reduced by, diminished by, difference between
$\times$
times, product, multiply, twice, thrice, doubled, tripled, quadrupled, of, as much as
$\div$
divide, ratio, quotient, divided by, divided into, partitioned into
After learning the key words for each fundamental operation, let’s try to translate some verbal expressions into its equivalent mathematical expression.
OperationKey Words
Twice a number$2x$
A number is subtracted from 10$10-x$
The sum of 9 and a number$9+x$
The quotient of a number and 5$\displaystyle \frac{x}{5}$
The ratio of two numbers x and y$\frac{x}{y}$
Twice a number decreased by 3$2x-3$
Twenty percent of a number$20\%x$
8 more than 3 times a number$3x+8$
The sum of two numbers divided by 7$\displaystyle \frac{x+y}{7}$
The quotient of twice a number and 5$\displaystyle \frac{2x}{5}$
The product of a number and 5 decreased by 15$5x-15$
Thrice a number less than 1$1-3x$
Half a number$\displaystyle \frac{1}{2}x$
12 less than thrice a number$3x-12$
Four times a number more than 20$20+4x$
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Video: Finding the General Antiderivative of a Given Function
Determine the most general antiderivative πΉ(π₯) of the function π(π₯) = 4π₯(βπ₯ + 5).
06:08
Video Transcript
Determine the most general antiderivative capital πΉ of π₯ of the function lowercase π of π₯ is equal to four π₯ multiplied by negative π₯ plus five.
The question gives us a function lowercase π of π₯, and it wants us to find the most general antiderivative of our function lowercase π of π₯. Weβll call this capital πΉ of π₯. Letβs start by recalling what we mean by capital πΉ of π₯ being an antiderivative of lowercase π of π₯. We say that capital πΉ of π₯ is an antiderivative of lowercase πΉ of π₯ if capital πΉ prime of π₯ is equal to lowercase π of π₯. In other words, the derivative of capital πΉ of π₯ with respect to π₯ is equal to lowercase π of π₯. Normally, weβre used to being given a function and then asking to differentiate it. However, in this case, weβre given a function and then asked to find the function which differentiates to give it. We often call this process integration.
Letβs start by taking a closer look at our function lowercase π of π₯. We can see itβs equal to four π₯ multiplied by negative π₯ plus five. This is a very complicated-looking function. We donβt know a function which differentiates to give something in this form. So instead, what weβll do is weβll simplify our expression for lowercase π of π₯ by distributing four π₯ over our parentheses. Doing this, we get that lowercase π of π₯ is equal to negative four π₯ squared plus 20π₯. Now, weβve rewritten our function in the general form for a polynomial. And we know a lot about differentiating functions of this form. So to find our antiderivative, letβs start by recalling how we would differentiate π times π₯ to the πth power.
The first thing we do is multiply our entire expression by the exponent of π₯. In this case, this is equal to π. The next thing we do is reduce our exponent by one. This gives us π times π multiplied by π₯ to the power of π minus one. We call this the power rule for differentiation. So what weβve described is how we would differentiate a polynomial term by term to get another polynomial. However, thatβs not what we want to find out in this question. In this question, weβre given a polynomial, and we need to find the polynomial which differentiates to give us that function. In other words, weβre trying to do the reverse. So to do this, instead of using the power rule for differentiation, weβre going to try and do the reverse of this.
So letβs discuss our process for finding an antiderivative of functions in this form. When using the power rule for differentiation, the last thing we did is reduce our exponent by one. We need to do the reverse of this. So instead of reducing our exponent by one, weβre going to add one to our exponent. So our first step will be add one to the exponent of π₯. Next, we need to find the reverse of our first step in the power rule for differentiation. This was to multiply our coefficient by the exponent of π₯. The reverse of this will be to divide our coefficient by the exponent of π₯. But remember, we were multiplying the coefficient by the original exponents. We now need to divide the coefficient by the new exponent of π₯.
One way of seeing this is to consider what would happen if we applied this process to π times π multiplied by π₯ to the power of π minus one. Remember, we want this process to give us π times π₯ to the power of π. The first step tells us to add one to our exponent. In our case, our exponent of π₯ is π minus one. So this gives us π times π times π₯ to the πth power. Next, we need to divide our coefficient by our new exponent. In our case, our new exponent is π. So we divide our coefficient by π. And of course, we can then cancel π divided by π to give us one. And we see this gives us π times π₯ to the πth power. So this gives us a method of finding antiderivatives of functions in this form.
However, thereβs one more thing we need to consider. And that is the derivative of any constant is always equal zero. Well, what does that mean in practice? Letβs consider the derivative of π₯ squared plus one. Well, we know the derivative of π₯ squared is equal to two π₯, and the derivative of one is equal to zero. So π₯ squared plus one is an antiderivative of two π₯. But now, consider the derivative of π₯ squared plus three. Again, weβll differentiate this term by term. The derivative of π₯ squared with respect to π₯ is two π₯, and the derivative of the constant three is equal to zero. So π₯ squared plus three is also an antiderivative of two π₯. In fact, we can see π₯ squared plus any constant will be an antiderivative of two π₯.
One way we could represent this would be to say π₯ squared and then we can add any constant. Weβll call this constant πΆ. Then, we know π₯ squared plus πΆ is an antiderivative of two π₯ for any value of πΆ. And this is what we mean when we say the most general antiderivative of a function. So whenever weβre looking for a general antiderivative of a function, we add the last step which is to add a constant of integration which we usually call πΆ. Weβre now ready to find our general antiderivative of the function lowercase π of π₯ given to us in the question.
Remember, since we can apply our derivative rules term by term, we can also use our antiderivative rules term by term. So letβs start with our first term of negative four π₯ squared. First, we want to add one to our exponent. In our case, the exponent of π₯ is two, so we add one to this to give three. Then, we need to do our second step which is to divide our coefficient by our new exponent. Our new exponent is equal to three. So we divide our entire expression by three. This gives us negative four π₯ cubed divided by three. And the third step is to add a constant of integration. But we can just do this at the end.
We now want to apply this process to our second term. It might be easier to consider this as 20 times π₯ to the first power. Again, we add one to our exponent of π₯ to get two and then divide by this new exponent of two. This gives us 20π₯ squared divided by two. And of course, 20 divided by two can simplify to give us 10. And remember, the last step we need to do is add our constant of integration which we will call πΆ. And this is our final answer.
Therefore, we were able to find the most general antiderivative capital πΉ of π₯ of the function lowercase π of π₯ is equal to four π₯ multiplied by negative π₯ plus five. We found that capital πΉ of π₯ is equal to negative four π₯ cubed divided by three plus 10π₯ squared plus a constant of integration πΆ.
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# Distance Between Two Points, Day 1
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## Objective
SWBAT find the distance between two points.
#### Big Idea
Students use absolute value to find the distance between two points.
## Do Now
10 minutes
Students have worked on plotting points on the coordinate plane. The Do Now serves two purposes: 1. It will provide students with additional practice. 2. We will use the coordinates later in the lesson. Students should be given graph paper and a ruler. It is important that as students create the coordinate plane to monitor that they are numbering the axes correctly.
Do Now
Create a coordinate plane and plot.
(10,1)(5,1)(5,3)(10,3)
After 5 minutes, I will call students to the board to explain and show how they plotted the coordinates.
## Mini Lesson
30 minutes
*This is a 2 day lesson. For the first day, we will spend time developing the rules for finding the distance between two points. The second day, students will work on a group project and there will be a gallery walk.
For this lesson, students will develop the rules for finding the distance between 2 points. Continuing from the Do Now, students will be directed to use their ruler to connect the four points they plotted, forming a rectangle.
How can we find the length of each side or the distance between two points?
Students may suggest counting the boxes to find the length. They will find that the length is 5 units and the width is 2 units.
Is there another way we find the length and width? What if we didn't plot the coordinates, how could we have calculated the length and width?
Ex. 1
I will show students the table for example 1. (see Example 1 - Distance) With their groups, students will discuss patterns they notice in the table and determine how the numbers in the distance column are calculated.
After 5 minutes, groups will share their ideas with the class. Most groups will have noticed that if the distances are calculated by subtracting the x coordinates.
Will this always work?
Most students will think that this will always work, because they are not considering all integers.
Ex. 2
For example 2 students will plot (-7,5) (-2,5) (-2,-2) (-7,-2) on their coordinate plane.
I will show students the distance table for example 2. (see Example 2 - Distance) Again, with their groups, students will discuss patterns they notice in the table and determine how the numbers in the distance column are calculated.
After 5 - 7 minutes, groups will share their ideas with the class. Some groups may have noticed that sometimes you add the coordinates and sometimes you subtract. I will ask probing questions, to help students develop rules that will always work.
What type of line is formed by the points (-7,5) and (-7,-2)? What do you notice about the x coordinates of both points?
Students should determine that it's a vertical line and the c coordinates are the same.
Is the same true for the vertical line formed by (-2,5) and (-2,-2)?
Students should notice that the x coordinates are the same.
How did you determine the distance was calculated for these points?
Some groups may have determined that you add the y coordinates. Some groups may have realized that you have to find the absolute value of the y coordinates before you add.
What type of line is formed by the points (-7,-2) and (-2,-2)? What do you notice about the y coordinates of both points?
Students should determine that it's a horizontal line and the y coordinates are the same.
Is the same true for the horizontal line formed by (-7,5) and (-2,5)?
Students should notice that the y coordinates are the same.
How did you determine the distance was calculated for these points?
Again, most groups may have determined that they needed to subtract the x coordinates. However some groups may realize that you have to find the absolute value of the x coordinates before you subtract.
This will lead us to the rules for finding the distances between two points.
If the x- or y- coordinates are both positive or both negative, subtract their absolute values.
If the x- or y- coordinates are opposite signs, add their absolute values.
## 3-2-1 Exit Ticket
5 minutes
Before moving on to the second day of the lesson, it is important for me to know my students' understanding of this concept. Each student will receive an index card on which to complete the exit ticket.
They will receive the following prompts:
3 things you still want to know about graphing on the coordinate plane
2 things you want help on or are confused about
1 thing about graphing on the coordinate plane you would be able to help someone else with
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##### Algebra I All-in-One For Dummies
The sheer number of people who use algebra means that a large number of errors are unavoidable. So much algebra is done in the world: Just about everyone who advances beyond elementary school takes an algebra class. Forgetting some of the more obscure rules or confusing one rule with another is easy to do when you’re in the heat of the battle with an algebra problem.
But some errors occur because that error seems to be an easier way to do the problem. Not right, but easier — the path of least resistance. These errors usually occur when a rule isn’t the same as your natural inclination. Most algebra rules seem to make sense, so they aren’t hard to remember. Some, though, go against the grain.
The main errors in algebra occur while performing expanding-type operations: distributing, squaring binomials, breaking up fractions, or raising to powers. The other big error area is in dealing with negatives. Watch out for those negative vibes.
## Keeping track of the middle term
A squared binomial has three terms in the answer. The term that often gets left out is the middle term: the part you get when multiplying the two outer terms together and the two inner terms together and finding their sum. The error occurs when just the first and last separate terms are squared, and the middle term is just forgotten.
## Distributing
Distributing a number or a negative sign over two or more terms in parentheses can cause problems if you forget to distribute the outside value over every single term in the parentheses. The errors come in when you stop multiplying the terms in the parentheses before you get to the end.
## Breaking up fractions
Splitting a fraction into several smaller pieces is all right as long as each piece has a term from the numerator (top) and the entire denominator (bottom). You can’t split up the denominator.
If the expression under a radical has values multiplied together or divided, then the radical can be split up into radicals that multiply or divide. You can’t split up addition or subtraction, however, under a radical.
Note: The radical expression is unchanged because the sum has to be performed before applying the radical operation.
## Keeping order with operations
The order of operations instructs you to raise the expression to a power before you add or subtract. A negative in front of a term acts the same as subtracting, so the subtracting has to be done last. If you want the negative raised to the power, too, then include it in parentheses with the rest of the value.
## Fracturing fractional exponents
A fractional exponent has the power on the top of the fraction and the root on the bottom.
When writing
as a term with a fractional exponent,
A fractional exponent indicates that there’s a radical involved in the expression. The 2 in the fractional exponent is on the bottom — the root always is the bottom number.
## Multiplying bases together
When you’re multiplying numbers with exponents, and those numbers have the same base, you add the exponents and leave the base as it is. The bases never get multiplied together.
## Raising a power to a power
To raise a value that has a power (exponent) to another power, multiply the exponents to raise the whole term to a new power. Don’t raise the exponent itself to a power — it’s the base that’s being raised, not the exponent.
## Reducing for a better fit
When reducing fractions with a numerator that has more than one term separated by addition or subtraction, then whatever you’re reducing the fraction by has to divide every single term evenly in both the numerator and the denominator.
## Negating negative exponents
When changing fractions to equivalent expressions with negative exponents, give every single factor in the denominator a negative exponent.
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## Tag Archives: triangle inequality proof
### Set Theory, Relations and Functions: Preliminaries: IV:
Problem Set based on previous three parts:
I) Solve the inequalities in the following exercises expressing the solution sets as intervals or unions of intervals. Also, graph each solution set on the real line:
a) $|x| <2$ (b) $|x| \leq 2$ (c) $|t-1| \leq 3$ (d) $|t+2|<1$ (e) $|3y-7|<4$(f) $|2y+5|<1$ (g) $|\frac{z}{5}-1| \leq 1$ (h) $| \frac{3}{2}z-1| \leq 2$ (i) $|3-\frac{1}{x}|<\frac{1}{2}$ (j) $|\frac{2}{x}-4|<3$ (k) $|2x| \geq 4$ (l) $|x+3| \geq \frac{1}{2}$ (m) $|1-x| >1$ (n) $|2-3x| > 5$ (o) $|\frac{x+1}{2}| \geq 1$ (p) $|\frac{3x}{5}-1|>\frac{2}{5}$
Solve the inequalities in the following exercises. Express the solution sets as intervals or unions of intervals and graph them. Use the result $\sqrt{a^{2}}=|a|$ as appropriate.
(a) $x^{2}<2$ (b) $4 \leq x^{2}$ (c) $4 (d) $\frac{1}{9} < x^{2} < \frac{1}{4}$ (e) $(x-1)^{2}<4$ (f) $(x+3)^{2}<2$ (g) $x^{2}-x<0$ (h) $x^{2}-x-2 \geq 0$
III) Theory and Examples:
i) Do not fall into the trap $|-a|=a$. For what real numbers a is the equation true? For what real numbers is it false?
ii) Solve the equation: $|x-1|=1-x$.
iii) A proof of the triangle inequality:
Give the reason justifying each of the marked steps in the following proof of the triangle inequality:
$|a+b|^{2}=(a+b)^{2}$…..why ?
$=a^{2}+2ab++b^{2}$
$\leq a^{2}+2|a||b|+b^{2}$….why ?
$\leq |a|^{2}+2|a||b|+|b|^{2}$….why?
$=(|a|+|b|)^{2}$….why ?
iv) Prove that $|ab|=|a||b|$ for any numbers a and b.
v) If $|x| \leq 3$ and $x>-\frac{1}{2}$, what can you say about x?
vi) Graph the inequality: $|x|+|y| \leq 1$
Questions related to functions:
I) Find the domain and range of each function:
a) $f(x)=1-\sqrt{x}$ (b) $F(t)=\frac{1}{1+\sqrt{t}}$ (c) $g(t)=\frac{1}{\sqrt{4-t^{2}}}$
II) Finding formulas for functions:
a) Express the area and perimeter of an equilateral triangle as a function of the triangle’ s side with length s.
b) Express the side length of a square as a function of the cube’s diagonal length d. Then, express the surface area and volume of the cube as a function of the diagonal length.
c) A point P in the first quadrant lies on the graph of the function $f(x)=\sqrt{x}$. Express the coordinates of P as functions of the slope of the line joining P to the origin.
III) Functions and graphs:
Graph the functions in the questions below. What symmetries, if any, do the graphs have?
a) $y=-x^{3}$ (b) $y=-\frac{1}{x^{2}}$ (c) $y=-\frac{1}{x}$ (d) $y=\frac{1}{|x|}$ (e) $y = \sqrt{|x|}$ (f) $y=\sqrt{-x}$ (g) $y=\frac{x^{3}}{8}$ (h) $y=-4\sqrt{x}$ (i) $y=-x^{\frac{3}{2}}$ (j) $y=(-x)^{\frac{3}{2}}$ (k) $y=(-x)^{\frac{2}{3}}$ (l) $y=-x^{\frac{2}{3}}$
IV) Graph the following equations ad explain why they are not graphs of functions of x. (a) $|y|=x$ (b) $y^{2}=x^{2}$
V) Graph the following equations and explain why they are not graphs of functions of x: (a) $|x|+|y|=1$ (b) $|x+y|=1$
VI) Even and odd functions:
In the following questions, say whether the function is even, odd or neither.
a) $f(x)=3$ (b) $f(x=x^{-5}$ (c) $f(x)=x^{2}+1$ (d) $f(x)=x^{2}+x$ (e) $g(x)=x^{4}+3x^{2}-1$ (f) $g(x)=\frac{1}{x^{2}-1}$ (g) $g(x)=\frac{x}{x^{2}-1}$ (h) $h(t)=\frac{1}{t-1}$ (i) $h(t)=|t^{3}|$ (j) $h(t)=2t+1$ (k) $h(t)=2|t|+1$
Sums, Differences, Products and Quotients:
In the two questions below, find the domains and ranges of $f$, $g$, $f+g$, and $f-g$.
i) $f(x)=x$, $g(x)=\sqrt{x-1}$ (ii) $f(x)=\sqrt{x+1}$, $g(x)=\sqrt{x-1}$
In the two questions below, find the domains and ranges of $f$, $g$, $\frac{f}{g}$ and $\frac{g}{f}$
i) $f(x)=2$, $g(x)=x^{2}+1$
ii) $f(x)=1$, $g(x)=1+\sqrt{x}$
Composites of functions:
1. If $f(x)=x+5$, and $g(x)=x^{2}-5$, find the following: (a) $f(g(0))$ (b) $g(f(0))$ (c) $f(g(x))$ (d) $g(f(x))$ (e) $f(f(-5))$ (f) $g(g(2))$ (g) $f(f(x))$ (h) $g(g(x))$
2. If $f(x)=x-1$ and $g(x)=\frac{1}{x+1}$, find the following: (a) $f(g(\frac{1}{2}))$ (b) $g(f(\frac{1}{2}))$ (c) $f(g(x))$ (d) $g(f(x))$ (e) $f(f(2))$ (f) $g(g(2))$ (g) $f(f(x))$ (h) $g(g(x))$
3. If $u(x)=4x-5$, $v(x)=x^{2}$, and $f(x)=\frac{1}{x}$, find formulas or formulae for the following: (a) $u(v(f(x)))$ (b) $u(f(v(x)))$ (c) $v(u(f(x)))$ (d) $v(f(u(x)))$ (e) $f(u(v(x)))$ (f) $f(v(u(x)))$
4. If $f(x)=\sqrt{x}$, $g(x)=\frac{x}{4}$, and $h(x)=4x-8$, find formulas or formulae for the following: (a) $h(g(f(x)))$ (b) $h(f(g(x)))$ (c) $g(h(f(x)))$ (d) $g(f(h(x)))$ (e) $f(g(h(x)))$ (f) $f(h(g(x)))$
Let $f(x)=x-5$, $g(x)=\sqrt{x}$, $h(x)=x^{3}$, and $f(x)=2x$. Express each of the functions in the questions below as a composite involving one or more of f, g, h and j:
a) $y=\sqrt{x}-3$ (b) $y=2\sqrt{x}$ (c) $y=x^{\frac{1}{4}}$ (d) $y=4x$ (e) $y=\sqrt{(x-3)^{3}}$ (f) $y=(2x-6)^{3}$ (g) $y=2x-3$ (h) $y=x^{\frac{3}{2}}$ (i) $y=x^{9}$ (k) $y=x-6$ (l) $y=2\sqrt{x-3}$ (m) $\sqrt{x^{3}-3}$
Questions:
a) $g(x)=x-7$, $f(x)=\sqrt{x}$, find $(f \circ g)(x)$
b) $g(x)=x+2$, $f(x)=3x$, find $(f \circ g)(x)$
c) $f(x)=\sqrt{x-5}$, $(f \circ g)(x)=\sqrt{x^{2}-5}$, find $g(x)$.
d) $f(x)=\frac{x}{x-1}$, $g(x)=\frac{x}{x-1}$, find $(f \circ g)(x)$
e) $f(x)=1+\frac{1}{x}$, $(f \circ g)(x)=x$, find $g(x)$.
f) $g(x)=\frac{1}{x}$, $(f \circ g)(x)=x$, find $f(x)$.
Reference: Calculus and Analytic Geometry, G B Thomas.
NB: I have used an old edition (printed version) to prepare the above. The latest edition may be found at Amazon India link:
https://www.amazon.in/Thomas-Calculus-George-B/dp/9353060419/ref=sr_1_1?crid=1XDE2XDSY5LCP&keywords=gb+thomas+calculus&qid=1570492794&s=books&sprefix=G+B+Th%2Caps%2C255&sr=1-1
Regards,
Nalin Pithwa
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## Small Group Activity: Directional Derivatives
Vector Calculus I 2022
This small group activity using surfaces relates the geometric definition of directional derivatives to the components of the gradient vector. Students work in small groups to measure a directional derivative directly, then compare its components with measured partial derivatives in rectangular coordinates. The whole class wrap-up discussion emphasizes the relationship between the geometric gradient vector and directional derivatives.
• This ingredient is used in the following sequences
What students learn Measuring slope along various directions at a point on the surface. Experimentally determining the gradient at a point on the surface. Using the Master Formula.
1. Measurement
1. Find the rate of change in the surface in the $x$-direction at the blue dot on your surface. Include units. $\frac{\partial{f}}{\partial{x}} = \underline{\hspace{2in}}$
2. Find the rate of change in the surface in the $y$-direction at the blue dot on your surface. Include units. $\frac{\partial{f}}{\partial{y}} = \underline{\hspace{2in}}$
3. Draw an arbitrary vector $\boldsymbol{\vec u}$ at the blue dot on the contour mat. What are its components? $\boldsymbol{\vec u} = \underline{\hspace{2in}}$
4. Find the rate of change in the surface in the $\boldsymbol{\vec u}$-direction. Include units. $\frac{df}{ds} = \underline{\hspace{2in}}$
2. Computation
1. Determine the gradient of $f$ at the blue dot. $\boldsymbol{\vec{\nabla}} f = \underline{\hspace{2in}}$
2. Use the Master Formula to express $\frac{df}{ds}$ in terms of $\boldsymbol{\vec\nabla} f$, and compute the result. $\frac{df}{ds} =\underline{\hspace{2in}}$
3. Comparison
Copyright 2014 by The Raising Calculus Group
## Instructor's Guide
### Main Ideas
• Measuring slope along various directions at a point on the surface
• Experimentally determining the gradient at a point on the surface
• Using the Master Formula
### Prerequisite Knowledge
Students should be able to:
• Use the measurement tool to approximate a derivative from a plastic surface
### Props/Equipment
• Tabletop Whiteboard with markers
• Plastic surface with contour map and inclinometer
• A handout for each student
### Student Conversations
• Some students may believe that vectors are tied to the coordinate axes, and not the contour maps; that moving the coordinate axes moves the vector along with them so that no matter the direction of the coordinate axes, the vector is always pointing in the same direction relative to those axes.
• Some students may believe that the gradient will always point towards the top of the hill.
• Some groups may make choose arbitrary vectors that trivialize the calculations in this activity.
• Some students may believe that the gradient is a vector's property and not a point's property; that every vector at a given point has its own, but not necessarily unique, gradient. These students may justify their reasoning by saying that the steepest direction for any given vector must be along that vector.
• Some students may believe that longer vectors always have a greater slope / rate of change than shorter vectors, or that vector length is what denotes magnitude of slope / rate of change; that a vector's rate of change and its magnitude represent the same value.
• Some students may believe that every vector pointing in the same direction as the gradient is identical to the gradient vector.
• Some students may believe that a short vector on the contour map will always correspond to a short vector on the surface.
• Some students may believe that taking the dot product of two vectors produces a vector.
Keywords
Directional derivatives
Learning Outcomes
|
# How do you convert 0.18 (18 repeating) to a fraction?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
37
Nov 14, 2017
$0.18181818 \ldots = \frac{2}{11}$
#### Explanation:
Let's begin by setting $x$ equal to our repeating fraction
$x = 0.18181818 \ldots$
The next step is a bit of a trick where we multiply by a power of 10 such that what remains to the right of the decimal is the repeating fraction itself, i.e.
$100 x = 18.1818181818 \ldots = 18 + 0.18181818 \ldots$
which can be re-written
$100 x = 18 + x$
taking $x$ to the left-hand side we get
$99 x = 18$
and finally
$x = \frac{18}{99} = \frac{2}{11}$
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Notes, Lesson 3.6
Implicit Differentiation
Most of our math work thus far has always allowed us to solve an equation for y in terms of x. When an equation can be solved for y we call it an explicit function. But not all equations can be solved for y. An example is:
This equation cannot be solved for y. When an equation cannot be solved for y, we call it an implicit function. The good news is that we can still differentiate such a function. The technique is called implicit differentiation.
When we implicitly differentiate, we must treat y as a composite function and therefore we must use the chain rule with y terms. The reason for this can be seen in Leibnitz notation: . This notation tells us that we are differentiating with respect to x. Because y is not native to what are differentiating with respect to, we need to regard it as a composite function. As you know, when we differentiate a composite function we must use the chain rule.
Let’s now try to differentiate the implicit function, .
This is a "folium of Descartes" curve. This would be very difficulty to solve for y, so we will want to use implicit differentiation. Here we show with Leibnitz notation that we are implicitly differentiating both sides of the equation. On the left side we need to individually take the derivative of each term. On the right side we will have to use the product rule. ( ) Here we take the individual derivatives. Note: Where did the y’ come from? Because we are differentiating with respect to x, we need to use the chain rule on the y. Notice that we did use the product rule on the right side. Now we get the y’ terms on the same side of the equation. Now we factor y’ out of the expression on the left side. Now we divide both sides by the factor and simplify. We can see in a plot of the implicit function that the slope of the tangent line at the point (3,3) does appear to be -1.
Another example: Differentiate:
Given implicit function Doing implicit differentiation on the function. Note the use of the product rule on the second term We do the algebra to solve for y'. Here we see a portion of plot of the implicit equation with c set equal to 5.. When does it appear that the slope of the tangent line will be zero? It appears to be at about (2.2,2.2). We take our derivative, set it equal to zero, and solve. Now putting x = y in the original implicit equation, we find that... x = y = 2.116343299 We still must use a computer algebra system to solve this cubic equation. The one real answer is shown at the left. This answer does seem consistent with our visual estimate. This can be done in Maple with the following command: >evalf(solve(x^3-x^2-5=0,x));
Links to other explanations of Implicit Differentiation:
World Web Math
University of British Columbia
University of Kentucky's Visual Calculus
Implicit Differentiation Using Maple
S.O.S. Math
University of Califonia - Davis
Karl's Calculus Tutor
Derivatives of Inverse Trigonometric Functions
Thanks to implicit differentiation, we can develop important derivatives that we could not have developed otherwise. The inverse trigonometric functions fall under this category. We will develop and remember the derivatives of the inverse sine and inverse tangent.
Inverse sine function. This is what inverse sine means. We implicitly differentiate both sides of the equation with respect to x. Because we are differentiating with respect to x, we need to use the chain rule on the left side. We solve the equation for . This is because of the trigonometric identity, . Refer back to the equation in step two above. We have our derivative.
The inverse tangent function. This is what inverse tangent means. We implicitly differentiate both sides of the equation with respect to x. Because we are differentiating with respect to x, we need to use the chain rule on the left side. We solve the equation for . This is because of the trigonometric identity, . Refer back to the equation in step two above. We have our derivative.
Check Concepts
#1: When equations are solved for y, we call them... Choose One Explicit Implicit FunctionsRelations #2: When equations cannot be solved for y, we call them... Choose One Explicit Implicit FunctionsRelations #3: True or False. Implict differentiation helps us to find derivatives of inverse trigonometric functions. Choose One TrueFalse
|
# GSEB Solutions Class 7 Maths Chapter 10 Practical Geometry InText Questions
Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 10 Practical Geometry InText Questions and Answers.
## Gujarat Board Textbook Solutions Class 7 Maths Chapter 10 Practical Geometry InText Questions
Think, Discuss and Write (Page 195)
Question 1.
In the above construction (see Page 193 NCERT Textbook), can you draw any other line through A that would be also parallel to the line l?
Solution:
No, because through a given point only one line parallel to a line can be drawn.
Question 2.
Can you slightly modify the above construction (see Page 193 NCERT Textbook), to use the idea of equal corresponding angles instead of equal alternate angles?
Solution:
Yes.
In this case, with A as centre and radius equal to BC, draw an arc to cut BA (extended) at E.
Next, with centre as E and radius equal to CD mark a point F as shown in the figure.
Now join AF to draw a line m which is parallel to the given line ‘l’.
Here, the corresponding angles ā CBE and ā FAE are equal. Therefore, l || m.
Think, Discuss and Write (Page 198)
Question 1.
A student attempted to draw a triangle whose rough figure is given here. He drew QR first. Then with as centre, he drew an arc of 3 cm and with R as centre, he drew an arc of 2 cm. But he could not get P. What is the reason? What property of triangle do you know in connection with this problem? Can such a triangle exist? (Remember the property of triangles āThe sum of any two sides of a triangle is always greater than the third side;!)
Solution:
Such a triangle does not exist, because a triangle is possible only when the sum of the lengths of any two sides is greater than the third side. But here, 2 cm + 3 cm, i.e. 5 cm < 6 cm.
Think, Discuss and Write (Page 200)
Question 1.
In āABC if AB = 3 cm, AC = 5 cm and mā C = 30Ā°. Can we draw this triangle?
Solution:
We may draw AC = 5 cm and ā C = 30Ā°. CA is one arm of ā C. Point B should be lying on the other arm of ā C. But we observe that point B cannot be located uniquely. So, the given data is not sufficient for construction of āABC.
Think, Discuss and Write (Page 202)
Question 1.
In the above example (see Page 201, NCERT Textbook), length of a side and measures of two angles were given. Now study the following problem:
In āABC, if AC = 7 cm, mā A = 60Ā° and mā B = 50Ā°, can you draw the triangle? (Angle- sum property of a triangle may help you!)
Solution:
We are given the line segment AC. ā A is given but ā C is not given. We can determine ā C using angle-sum property.
ā C = 180Ā° – (ā B + ā A).
= 180Ā° – (60Ā° + 50Ā°)
= 180Ā° – 110Ā° = 70Ā°
Now, with the help of ā C, we can construct the triangle.
Miscellaneous Questions (Page 204)
Question 1.
Below are given the measures of certain sides and angles of triangles. Identify those which cannot be constructed and, say why you cannot construct them. Construct rest of the triangles.
Solution:
1. We have:
Since for āABC, ā A + ā B + ā C = 180Ā°
But here ā A + ā B = 85Ā° + 115Ā°
= 200Ā° > 180Ā°.
i. e. ā A + ā B + ā C will be greater than 180Ā°
Thus, āABC cannot be constructed.
2. We have:
Steps of construction:
I. Draw a line segment QR = 4.7 cm.
II. At Q, construct ā RQX = 30Ā°.
III. At R, construct ā QRY = 60Ā°.
IV. Let $$\overrightarrow{\mathrm{QX}}$$ and $$\overrightarrow{\mathrm{RY}}$$ intersect at R
Thus, āPQR is the required triangle.
3. We have:
Here, the side AC is given.
ā“ To complete the triangle, we need mā A and mā C.
Since, mā C = 180Ā° – [ā A + ā B]
= 180Ā° – [70Ā° + 50Ā°]
= 180Ā° – 120Ā° = 60Ā°
Steps of construction:
I. Draw a line segment AC = 3 cm.
II. At A, construct ā CAX = 70Ā°.
III. At C, construct ā ACY = 60Ā°
IV. Let the rays $$\overrightarrow{\mathrm{AX}}$$ and $$\overrightarrow{\mathrm{CY}}$$ meet at B.
ā“ āABC is the required triangle.
4. We have:
Since, in āLMN,
ā L + ā M +ā N = 180Ā°
But here, ā L + ā N + ā M
= 60Ā° + 120Ā° + ā M = (180Ā° + ā M) > 180Ā°.
ā“ āLMN cannot be constructed.
5. We have:
$$\left.\begin{array}{l} \mathrm{BC}=2 \mathrm{~cm} \\ \mathrm{AB}=4 \mathrm{~cm} \\ \mathrm{AC}=2 \mathrm{~cm} \end{array}\right\}$$
We know that a triangle is possible only when the sum of lengths of any two sides is greater than the length of the third side.
Here BC + AC = 2 cm + 2 cm = 4 cm = AB
(āµ AB = 4 cm)
ā“ āABC is not possible.
6. We have:
$$\left.\begin{array}{l} \mathrm{PQ}=3.5 \mathrm{~cm} \\ \mathrm{QR}=4 \mathrm{~cm} \\ \mathrm{PR}=3.5 \mathrm{~cm} \end{array}\right\}$$
Steps of construction:
I. Draw a line PQ = 3.5 cm.
II. With centre P and radius 3.5 cm, draw an arc.
III. With centre Q and radius 4 cm, draw another arc, such that it cuts the previous arc at R.
IV. Join RP and RQ.
Thus, āPQR is the required triangle.
7. We have:
$$\left.\begin{array}{l} \mathrm{XY}=3 \mathrm{~cm} \\ \mathrm{YZ}=4 \mathrm{~cm} \\ \mathrm{XZ}=5 \mathrm{~cm} \end{array}\right\}$$
Steps of construction:
I. Draw a line segment YZ = 4 cm.
II. With centre Y and radius 3 cm, draw an arc.
III. With centre Z and radius 5 cm, draw another arc which cuts the previous arc at X.
IV. Join XY and XZ.
Thus, āXYZ is the required triangle.
8. We have:
$$\left.\begin{array}{l} \mathrm{DE}=3 \mathrm{~cm} \\ \mathrm{EF}=4 \mathrm{~cm} \\ \mathrm{DF}=5 \mathrm{~cm} \end{array}\right\}$$
Steps of construction:
I. Draw a line segment EF = 5.5 cm
II. With centre E and radius 4.5 cm, draw an arc.
III. With centre F and radius 4 cm, draw another arc such that it intersects the previous arc at D.
IV. Join DE and DF.
Thus, āDEF is the required triangle.
|
# Video: KS1-M17 • Paper 2 • Question 12
KS1-M17-P2-Q12-659147312612-v2
03:15
### Video Transcript
Measure the longest line. Use a ruler.
We know that triangles are made up of three straight lines or three sides. We’re given a triangle here. And we’re asked to measure the longest of those three sides, the longest line. Which lines are we supposed to measure? We can see which side is the longest by comparing them. We’ll call this line side 𝐴. Is side 𝐴 the longest? What about side 𝐵? Well, side 𝐵 looks like it’s slightly longer than side 𝐴. So it’s the longest line so far. But when you compare them both with side 𝐶, which is this line here, we can see that this is the line that’s the longest.
We need to measure from one end of the line to the other. And we’re asked to use a ruler. The point of this video is going to show how we can use a ruler to measure that longest line. The triangle could look different sizes, depending on the way that you’re watching this video. If you’re watching it on a mobile phone, it could be small or perhaps you’re watching it on an interactive whiteboard when it’s very large. So we can’t actually measure the triangle in real life. Hopefully, if yourself or your teacher has printed out the question, it will be the exact size. This video is just going to show us how to measure correctly and importantly what mistake we need to avoid.
So let’s imagine now we’ve got our ruler out. And we’re going to measure this longest line. Where should we put the ruler? So shall we line our ruler up here? We’ve made sure that the end of the ruler is level with the end of the line. And we can read the measurement from the other end of the line. It looks like it’s halfway between eight and nine centimetres. It must be eight and half centimetres.
Have we measured correctly? No. Look at where the zero is on the ruler. Zero shows the start of the scale on the ruler. This is where we need to start measuring from, not the end of the ruler. Let’s shift our ruler up a little bit so that this time we line up the zero with the start of the line. This time, we’ve managed to line up the zero with the start of the line. So now, we’re counting from zero. And if we read the measurement from the other end of the line, we can see that the line is nine centimetres long.
So although the actual size of the triangle on your screen may be different, depending on how you’re watching this video, we’ve just shown how to line up the ruler and to measure to find the longest line. We need to make sure that our ruler is lined up alongside the line and most importantly that we start measuring from the zero. So we need to match up the zero with one end of the line. If you have time, why not you use a ruler to measure the other two sides just to check that we had chosen the longest side correctly? The length of the longest line of this triangle is nine centimetres.
|
Study P2 Mathematics Model Drawing 4 - Geniebook
# Model Drawing 4
1. Word Problems involving 4 operations
## Topic Recap:
### Word Problems Involving Multiplication
Example:
There are 5 boxes. There are 8 cupcakes in each box. How many cupcakes are there in all?
Solution:
There are 8 cupcakes in each box.
Therefore, the total will be:
$$5 \times 8 = 40$$
There are 40 cupcakes in all.
40 cupcakes
Question 1:
Hayley makes 6 apple pies. She uses 4 apples to make each pie. How many apples does she use in all?
1. 12
2. 24
3. 28
4. 30
Solution:
$$6 \times 4 = 24$$
(2) 24
## Topic Recap:
### Word problems Involving Division
Example:
Jill shares 18 pencils equally with 2 friends. How many pencils does each of them receive?
Solution:
3 people shared 18 pencils.
Number of pencils each of them received
$$= 18 \div 3$$
$$= 6$$
Each of them received 6 pencils.
6 pencils
Question 2:
Daniel has a piece of wire 9 cm long. He cuts it into 3 equal pieces. How long is each piece of wire?
1. 1 cm
2. 24 cm
3. 3 cm
4. 27 cm
Solution:
$$9 \div 3 = 3$$
(3) 3 cm
Question 3:
Jacky has 80 paper clips. He wants to place 10 paper clips in one box. How many boxes does he need?
Solution:
$$80 \div 10 = 8$$
8 boxes
## 1. Word Problems Involving 4 Operations
To solve a word problem, you can draw a model to help you visualise a question better.
Example:
Julie has 18 red markers and 14 blue markers. She packs the markers equally into 4 boxes. How many markers are there in each box?
Solution:
Step 1:
Find the total number of markers.
$$18 + 14 = 32$$
Step 2:
Find the number of markers in each box.
$$32 \div 4 = 8$$
There are 8 markers in each box.
8 markers
Question 1:
Sunny bought 12 red pencils and 15 blue pencils. He repacked all the pencils into packets of 3. He sold all the packets.
1. How many pencils did he buy altogether?
1. 22
2. 25
3. 27
4. 28
Solution:
$$12 + 15 = 27$$
(3) 27
1. If he sold all the packets of pencils, how many packets did he sell?
1. 7
2. 8
3. 9
4. 10
Solution:
$$27 \div 3 = 9$$
(3) 9
Question 2:
There are 40 students in a class. There are 3 groups of 5 boys and the rest are girls.
1. How many boys are there in the class?
1. 12
2. 15
3. 18
4. 21
Solution:
$$3 \div 5 = 15$$
(2) 15
1. How many girls are there in the class?
1. 15
2. 25
3. 35
4. 40
Solution:
40 - 15 = 25
(2) 25
Question 3:
There were 7 children at a party. Mrs Lim gave 4 pencils to each child. She had 5 pencils left.
1. How many pencils did Mrs Lim give the children in all?
1. 14
2. 24
3. 27
4. 28
Solution:
$$7 \div 4 = 28$$
(4) 28
1. How many pencils did Mrs Lim have at first?
1. 20
2. 23
3. 33
4. 39
Solution:
$$28 + 5 = 33$$
(3) 33
Question 4:
Mrs Ong baked 2 cakes. She cut each cake into 6 equal pieces. She then gave them equally to her 3 children.
1. How many pieces of cake were there?
1. 8
2. 10
3. 12
4. 14
Solution:
$$2 \times 6 = 12$$
(3) 12
1. How many pieces of cake did each child receive?
1. 5
2. 6
3. 3
4. 4
Solution:
$$12 \div 3 = 4$$
(4) 4
Question 5:
Mr Tan has 29 roses. He wants to arrange them into 4 bouquets of 9 roses.
1. How many roses are needed for 4 bouquets?
1. 18
2. 27
3. 36
4. 45
Solution:
$$4 \times 9 = 36$$
(3) 36
1. How many more roses does he need?
1. 7
2. 20
3. 25
4. 65
Solution:
$$36 - 29 = 7$$
(1) 7
• Word problems involving 4 operations
Continue Learning
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# Math 112 Section 3.1 Exponential Functions... ( )
```Page 1 of 2
Math 112
Section 3.1 Exponential Functions and Graphs
Exponential Function
An exponential function with base a is defined as
f ( x) a x , where a > 0 and a 0.
x
Example 1:
f ( x) 2
1
x
Example 2: g ( x) 2
2
x
x
f x = 2x
8
f(x)
g(x)
f x =
1
2x
8
0
6
1
6
4
2
4
-1
2
2
-2
-5
-5
a > 1 (increasing)
5
0 < a < 1 (decreasing)
Compound Amount
If P dollars is invested at a yearly rate of interest r per year, compounded m times per year for t years,
the compound amount is
r
A P 1
m
tm
dollars.
Example 3: If \$82,000 is invested at 4.5% compounded quarterly, find the sum in 5 years.
Example 4: Interest is compounded semiannually. Find the amount in the account and the interest
earned if the principal is \$3,000, the rate of interest is 7%, and the time is 3 years.
r
Definition of e: As m becomes larger and larger,
1
number e, whose approximate value is 2.7182818.
Example 5: Calculate e21, e-5
m
m
becomes closer and closer to the
Page 2 of 2
Shifting exponential functions:
Example 6:
f ( x) 2 x
Example 7:
f ( x) 2 x 2
hx = 2x+2
8
f x = 2x
8
6
6
4
4
2
2
-5
-5
Example 8: f ( x) 2 2
x
Example 9: f ( x) 2
x
rx = 2x+2
s x = -2x
8
5
6
-2
4
-4
2
-6
-8
-5
Example 10: f ( x) e
x 5
Example 11: f ( x) 2
x4
1
```
|
# TYPING - EDITORIAL
Tester: Misha Chorniy
Editorialist: Taranpreet Singh
### DIFFICULTY:
Okay, Decided. Cakewalk.
### PREREQUISITES:
Mapping, Basic Implementation.
### PROBLEM:
Given N works, we have to find minimum units of time to type all words, given following conditions. (I am taking tenth of a second as unit throughout the editorial)
*
• Typing First character take two unit of time.
• Typing Next character with same hand as previous character takes 4 unit of time, while typing with different hand takes 4 unit of time.
• If a word is typed before, It takes exactly half time it took to type word First time.
### SUPER QUICK EXPLANATION
• Calculate time for each distinct word, say c and assuming word occur x times, total time of typing this word x times is c+(x-1)*c/2.
• To calculate time of each distinct word, just keep a variable, indicating which hand was used to type previous character. If same hand is used for both and current character, increase time by 4, otherwise increase time by 2.
### EXPLANATION
This problem is fairly simple, so, expect a lot of implementation detail. Those who can implement, read only upper half and bonus problem.
First of all, we can handle each distinct word separately, since time taken by different distinct words is independent.
Suppose we have x occurrence of a word, and it takes 2*c units of time to type it for first time. Then, We can see
Typing same word two times take 2*c+c = 3*c, three times take 2*c+c+c, … x times take 2*c+c*(x-1).
So, in just one line, we know how to calculate time taken to type single word multiple word x times. Now, If we know time taken to type each distinct word, we have solved the problem.
Now, Let’s calculate time taken to type a word.
First character always take 2 units of time, as given in problem statement.
For other character, the only factor, which decides whether it will take 2 unit or 4 unit of time to type current character is the fact whether previous character is typed by same or different hand.
So, The only thing we store is, whether last character was typed by left or right hand.
If both current and previous character are to be typed by same hand, increase time by 4 unit of time, else increase tome by 2 unit of time.
Turns out we have solved the problem.
For implementation, two ways are possible.
One is to make a String to Integer mapping storing number of times current string occurs in test case. Then, calculate time for each distinct word and handle multiple occurrences as explained above.
Other is to make a String to Integer mapping storing time taken to type this word first time. Whenever we see a word already typed before, just add half the time it took to type word first time.
### Bonus problem
Harder version of this problem. (Basic problem to practice a well known technique)
Left hand can type f,g and h, while right hand can type g,h and j Find minimum time to type words, if all other things remains same. All strings consist of these characters only.
PS: Try some general technique, not case based solution.
Idea in secret box.
Click to view
Dynamic Programming
### Time Complexity
Time complexity is O(N*C*logN) where C is time taken to compare strings (required for map get method as well as for sorting, if used, depending upon implementation.)
### AUTHOR’S AND TESTER’S SOLUTIONS:
Feel free to Share your approach, If it differs. Suggestions are always welcomed.
I want to share my code for help.
I am not getting whats wrong in my code…where and how to share?
|
# 1961 AHSME Problems/Problem 21
## Problem
Medians $AD$ and $CE$ of $\triangle ABC$ intersect in $M$. The midpoint of $AE$ is $N$. Let the area of $\triangle MNE$ be $k$ times the area of $\triangle ABC$. Then $k$ equals:
$\textbf{(A)}\ \frac{1}{6}\qquad \textbf{(B)}\ \frac{1}{8}\qquad \textbf{(C)}\ \frac{1}{9}\qquad \textbf{(D)}\ \frac{1}{12}\qquad \textbf{(E)}\ \frac{1}{16}$
## Solution
$[asy] draw((0,0)--(120,0)--(40,60)--(0,0)); draw((40,60)--(60,0)); draw((120,0)--(20,30)); draw((160/3,20)--(30,45)); dot((40,60)); label("A",(40,60),N); dot((0,0)); label("B",(0,0),SW); dot((120,0)); label("C",(120,0),SE); dot((60,0)); label("D",(60,0),S); dot((20,30)); label("E",(20,30),NW); dot((160/3,20)); label("M",(160/3,20),NE); dot((30,45)); label("N",(30,45),NW); draw((160/3,20)--(0,0),dotted); [/asy]$
Let $[NME] = a$ and $[BMD] = b$. The altitudes and bases of $\triangle ANM$ and $\triangle NME$ are the same, so $[NME] = [ANM] = a$. Since the base of $\triangle BEM$ is twice as long as $\triangle NME$ and the altitudes of both triangles are the same, $[BEM] = 2a$.
Since the altitudes and bases of $\triangle BMD$ and $\triangle CMD$ are the same, $[BMD] = [CMD] = b$. Additionally, since $\triangle AME$ and $\triangle EMB$ have the same area, $\triangle AMC$ and $\triangle BMC$ also have the same area, so $[AMC] = 2b$.
Also, the altitudes and bases of $\triangle ABD$ and $\triangle ADC$ are the same, so the area of the two triangles are the same. Thus, $$a+a+2a+b=2b+b$$ $$4a=2b$$ $$2a = b$$ Thus, the area of $\triangle ABC$ is $12a$, so the area of $\triangle NME$ is $\frac{1}{12}$ the area of $\triangle ABC$. The answer is $\boxed{\textbf{(D)}}$.
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#### Jessica Wesaquate
Subject Area:
Mathematics
Strand:
Shape and Space (3-D Objects and 2-D Shapes)
Grade Level:
Three
WNCP:
Main Objective: Students will describe the characteristics of 3-D objects and 2-D shapes, and analyze the relationships among them. General Outcome: Describe, classify, construct and relate 3-D objects and 2-D shapes. Specific Outcomes: Identify and count faces, vertices and edges of 3-D objects. Identify and name faces of a 3-D object with appropriate 2-D names. Describe and name cones by the shape of the base.
Materials:
Activity One:
Tipi raising videos: Glen Anaquod or Tim Haywahe section, paper, pencils, and scissors
Activity Two:
Objects created from activity one
Activity three:
Still image of a raised tipi from Tim Haywahe or Glen Anaquod section
Activities/Lesson Ideas:
1. Watch tipi raising videos from Glen Anaquod or Tim Haywahe section. Using simply paper, have students recreate the shapes from the video. Students can start by making a semi-circle to represent the canvas. They can also transfer this shape into the shape of a cone, which represents the raised tipi. It will also be their responsibility to re-create the shapes of the pins, pegs, and rope. Once students have created these shapes, have them record the following in chart form:
• The number of faces on each object.
• The number of vertices on each object.
• Can a pattern be created?
2. Using the paper copies of the objects created in activity one, have students draw, trace of make prints of the faces in their math journals. Under each example they should identify and name the faces.
Found at Math Dictionary.com – A face is a flat surface of a three-dimensional figure.
3. Take a look at a still image from the Tim Haywahe or Glen Anaquod section that shows the tipi raised. Ask students what shape the tipi is raised? Once students have determined the shape is a cone, discuss how many faces the cone has. Students will learn that the cone has two faces - a flat base and a curved side.
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# LCM of 25 and 49
The lcm of 25 and 49 is the smallest positive integer that divides the numbers 25 and 49 without a remainder. Spelled out, it is the least common multiple of 25 and 49. Here you can find the lcm of 25 and 49, along with a total of three methods for computing it. In addition, we have a calculator you should check out. Not only can it determine the lcm of 25 and 49, but also that of three or more integers including twenty-five and forty-nine for example. Keep reading to learn everything about the lcm (25,49) and the terms related to it.
## What is the LCM of 25 and 49
If you just want to know what is the least common multiple of 25 and 49, it is 1225. Usually, this is written as
lcm(25,49) = 1225
The lcm of 25 and 49 can be obtained like this:
• The multiples of 25 are …, 1200, 1225, 1250, ….
• The multiples of 49 are …, 1176, 1225, 1274, …
• The common multiples of 25 and 49 are n x 1225, intersecting the two sets above, $\hspace{3px}n \hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$.
• In the intersection multiples of 25 ∩ multiples of 49 the least positive element is 1225.
• Therefore, the least common multiple of 25 and 49 is 1225.
Taking the above into account you also know how to find all the common multiples of 25 and 49, not just the smallest. In the next section we show you how to calculate the lcm of twenty-five and forty-nine by means of two more methods.
## How to find the LCM of 25 and 49
The least common multiple of 25 and 49 can be computed by using the greatest common factor aka gcf of 25 and 49. This is the easiest approach:
lcm (25,49) = $\frac{25 \times 49}{gcf(25,49)} = \frac{1225}{1}$ = 1225
Alternatively, the lcm of 25 and 49 can be found using the prime factorization of 25 and 49:
• The prime factorization of 25 is: 5 x 5
• The prime factorization of 49 is: 7 x 7
• Eliminate the duplicate factors of the two lists, then multiply them once with the remaining factors of the lists to get lcm(25,25) = 1225
In any case, the easiest way to compute the lcm of two numbers like 25 and 49 is by using our calculator below. Note that it can also compute the lcm of more than two numbers, separated by a comma. For example, enter 25,49. Push the button only to start over.
The lcm is...
Similar searched terms on our site also include:
## Use of LCM of 25 and 49
What is the least common multiple of 25 and 49 used for? Answer: It is helpful for adding and subtracting fractions like 1/25 and 1/49. Just multiply the dividends and divisors by 49 and 25, respectively, such that the divisors have the value of 1225, the lcm of 25 and 49.
$\frac{1}{25} + \frac{1}{49} = \frac{49}{1225} + \frac{25}{1225} = \frac{74}{1225}$. $\hspace{30px}\frac{1}{25} – \frac{1}{49} = \frac{49}{1225} – \frac{25}{1225} = \frac{24}{1225}$.
## Properties of LCM of 25 and 49
The most important properties of the lcm(25,49) are:
• Commutative property: lcm(25,49) = lcm(49,25)
• Associative property: lcm(25,49,n) = lcm(lcm(49,25),n) $\hspace{10px}n\neq 0 \hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$
The associativity is particularly useful to get the lcm of three or more numbers; our calculator makes use of it.
To sum up, the lcm of 25 and 49 is 1225. In common notation: lcm (25,49) = 1225.
If you have been searching for lcm 25 and 49 or lcm 25 49 then you have come to the correct page, too. The same is the true if you typed lcm for 25 and 49 in your favorite search engine.
Note that you can find the least common multiple of many integer pairs including twenty-five / forty-nine by using the the search form in the sidebar of this page.
Questions and comments related to the lcm of 25 and 49 are really appreciated. Use the form below or send us a mail to get in touch.
Please hit the sharing buttons if our article about the least common multiple of 25 and 49 has been useful to you, and make sure to bookmark our site.
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AMCAT Quantitative Ability Previous Papers-12
A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?
35 ½
36 ½
37 ½
38 ½
Explanation:
Work done by A in 20 days = 80/100 = 8/10 = 4/5
Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 — (1)
Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B)
Work done by A and B in 1 day = 1/15 —(2)
Work done by B in 1 day = 1/15 – 1/25 = 2/75
=> B can complete the work in 75/2 days = 37 (1/2) days
Ques. 4 men and 6 women finish a job in 8 days, while 3 men and 7 women finish it in 10 days. In how many days will 10 women working together finish it ?
30 days
40 days
50 days
60 days
Explanation:
Let 1 man’s 1 day work = x
and 1 woman’s 1 days work = y.
Then, 4x + 6y = 1/8
and 3x+7y = 1/10
solving, we get y = 1/400 [means work done by a woman in 1 day]
10 women 1 day work = 10/400 = 1/40
10 women will finish the work in 40 days
Ques, 10 women can complete a work in 7 days and 10 children take 14 days to complete the work. How many days will 5 women and 10 children take to complete the work?
6 days
7 days
8 days
9 days
Explanation:
1 woman’s 1 day’s work = 1/70
1 Child’s 1 day’s work = 1/140
5 Women and 10 children 1 day work =
(5/70 +10/140)=1/7
So 5 women and 10 children will finish the work in 7 days.
Ques. 5 men and 2 boys working together can do four times as much work as a man and a boy. Working capacity of man and boy is in the ratio
1:2
1:3
2:1
2:3
Explanation:
Let 1 man 1 day work = x
1 boy 1 day work = y
then 5x + 2y = 4(x+y)
=> x = 2y
=> x/y = 2/1
=> x:y = 2:1
Rahul and Sham together can complete a task in 35 days, but Rahul alone can complete same work in 60 days. Calculate in how many days Sham can complete this work ?
84 days
82 days
76 days
68 days
Explanation:
As Rahul and Sham together can finish work in 35 days.
1 days work of Rahul and Sham is 1/35
Rahul can alone complete this work in 60 days,
So, Rahul one day work is 1/60
Clearly, Sham one day work will be = (Rahul and Sham one day work) – (Rahul one day work)
=1/35−1/60=1/84
Hence Sham will complete the given work in 84 days.
Ques. If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
A. 4 days
B. 5 days
C. 6 days
D. 7 days
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# Problem of the Week
## Updated at Jun 22, 2020 12:04 PM
To get more practice in algebra, we brought you this problem of the week:
How would you find the factors of $$12{u}^{2}-26u+4$$?
Check out the solution below!
$12{u}^{2}-26u+4$
1 Find the Greatest Common Factor (GCF).1 What is the largest number that divides evenly into $$12{u}^{2}$$, $$-26u$$, and $$4$$?It is $$2$$.2 What is the highest degree of $$u$$ that divides evenly into $$12{u}^{2}$$, $$-26u$$, and $$4$$?It is 1, since $$u$$ is not in every term.3 Multiplying the results above,The GCF is $$2$$.To get access to all 'How?' and 'Why?' steps, join Cymath Plus!GCF = $$2$$2 Factor out the GCF. (Write the GCF first. Then, in parentheses, divide each term by the GCF.)$2(\frac{12{u}^{2}}{2}+\frac{-26u}{2}+\frac{4}{2})$3 Simplify each term in parentheses.$2(6{u}^{2}-13u+2)$4 Split the second term in $$6{u}^{2}-13u+2$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$6\times 2=12$2 Ask: Which two numbers add up to $$-13$$ and multiply to $$12$$?$$-1$$ and $$-12$$3 Split $$-13u$$ as the sum of $$-u$$ and $$-12u$$.$6{u}^{2}-u-12u+2$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$2(6{u}^{2}-u-12u+2)$5 Factor out common terms in the first two terms, then in the last two terms.$2(u(6u-1)-2(6u-1))$6 Factor out the common term $$6u-1$$.$2(6u-1)(u-2)$Done2*(6*u-1)*(u-2)
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# Surds
A surd is a square root which cannot be reduced to a .
For example, $$\sqrt 4 = 2$$ is not a surd.
However $$\sqrt 5$$ is a surd.
If you use a calculator, you will see that $$\sqrt 5 = 2.236067977...$$ and we will need to round the answer correct to a few decimal places. This makes it less accurate.
If it is left as $$\sqrt 5$$, then the answer has not been rounded, which keeps it exact.
Here are some general rules when simplifying expressions involving surds.
$\sqrt a \times \sqrt a = a$
$\sqrt {ab} = \sqrt a \times \sqrt b$
$\sqrt {\frac{a}{b}} = \frac{{\sqrt a }}{{\sqrt b }}$
Now use the information above to try the example questions below.
Question
Simplify $$\sqrt {12}$$
$= \sqrt {4 \times 3}$
$= \sqrt 4 \times \sqrt 3$
$= 2\sqrt 3$
Question
Simplify $$\sqrt {48}$$
$= \sqrt {16} \times \sqrt 3$
$= 4\sqrt3$
Question
Simplify $$\sqrt {\frac{{16}}{9}}$$
$= \frac{{\sqrt {16} }}{{\sqrt 9 }}$
$= \frac{4}{3}$
Question
Simplify $$\sqrt 8 + \sqrt {18} + \sqrt {50}$$
$= \sqrt4 \sqrt 2 + \sqrt9 \sqrt 2 + \sqrt25 \sqrt 2$
$= 2\sqrt 2 + 3\sqrt 2 + 5\sqrt 2$
$= 10\sqrt 2$
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Warm-up
Evaluate each ex pression if a = 6, b = -2, and c = 12.
Today we will:
1. Solve problems about situations modeled by parabolas . Last section in the
book!
Following class period we will:
1. Review key concepts
o Quadratic equation – an equation that can written in the form an equation
0 = ax2 + bx + c, where a, b, and c are numbers and a ≠ 0.
o Quadratic formula – the formula is used to find the solutions of a
equation.
Today we will solve the quadratic equations by…
• Graphing (the x- intercepts are the solutions )
Solving 0 = ax2 + bx + c by graphing
The solutions to 0 = ax2 + bx + c are the x-intercepts of the graph of the
equation
y = ax2+ bx + c.
One way to solve is by drawing an accurate graph. If the graph doesn’t cross
the x-axis at points that correspond to integral units on the grid, the solutions
will be approximations.
Example 1:
Solve 10x2 - 5x –50 = 0 by graphing
Solution - Graph the quadratic function 10x2 - 5x –50 = y
Step 1 – Make a table of values.
Step 2 – Use the table to draw a graph
Step 3 – Read the x-intercepts from the graph.
The x-intercepts are at –2 and 2.5,
so the solutions of 10x2 - 5x –50 = 0 are x = -2 and x = 2.5.
Check your solution. Substitute –2 and 2.5 into the original equation.
Using the quadratic formula to solve ax2+ bx + c = 0
Unlike graphing where approximate solutions may be necessary, the quadratic
formula gives exact solutions to any equation ax2+ bx + c = 0.
Often the solutions involve square roots of numbers that are not perfect
squares. In such cases, you can approximate the solutions or give the exact
solutions in square root form.
Example 2:
Use the quadratic formula to solve 2x2+ 5x = 25
Solution
Step 1 - Write the equation in the form ax2+ bx + c = 0 and de termine the
values of a , b, and c.
Step 2 – Use the quadratic formula.
Check your solutions – substitute 2.5 and –5 in the original equation.
Example 3:
Use the quadratic formula to solve 0 = x2 - 11x + 18
Solution
Step 1 - Determine the values of a, b, and c.
Step 2 – Use the quadratic formula.
Check your solutions – substitute 9 and 2 in the original equation.
Yep, both work!
In Summary…
To solve quadratic equations we can…
• Graph to find the x-intercepts
• Factor ( and use the Zero - Product Property )
Prev Next
Start solving your Algebra Problems in next 5 minutes!
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Go to the latest version.
# 6.6: Absolute Value Inequalities
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Solve absolute value inequalities.
• Rewrite and solve absolute value inequalities as compound inequalities.
• Solve real-world problems using absolute value inequalities.
## Introduction
Absolute value inequalities are solved in a similar way to absolute value equations. In both cases, you must consider the two options.
1. The expression inside the absolute value is not negative.
2. The expression inside the absolute value is negative.
Then we solve each inequality separately.
## Solve Absolute Value Inequalities
Consider the inequality
$\mid x \mid \leq 3$
Since the absolute value of $x$ represents the distance from zero, the solutions to this inequality are those numbers whose distance from zero is less than or equal to 3. The following graph shows this solution:
Notice that this is also the graph for the compound inequality $-3 \leq x \leq 3$.
Now consider the inequality
$\mid x \mid > 2$
Since the absolute value of $x$ represents the distance from zero, the solutions to this inequality are those numbers whose distance from zero are more than 2. The following graph shows this solution.
Notice that this is also the graph for the compound inequality $x<-2$ or $x > 2$.
Example 1
Solve the following inequalities and show the solution graph.
a) $\mid x \mid <6$
b) $\mid x \mid \geq 2.5$
Solution
a) $\mid x \mid <5$ represents all numbers whose distance from zero is less than 5.
Answer $-5
b) $\mid x \mid \geq 2.5$ represents all numbers whose distance from zero is more than or equal to 2.5.
Answer $x \leq -2.5$ or $x \geq 2.5$
## Rewrite and Solve Absolute Value Inequalities as Compound Inequalities
In the last section you saw that absolute value inequalities are compound inequalities.
Inequalities of the type $\mid x \mid can be rewritten as $-a < x
Inequalities of the type $\mid x \mid can be rewritten as $x <-b$ or $x> b$
To solve an absolute value inequality, we separate the expression into two inequalities and solve each of them individually.
Example 2
Solve the inequality $\mid x-3\mid <7$ and show the solution graph.
Solution
Rewrite as a compound inequality.
Write as two separate inequalities.
$x-3 <7$ and $x-3 <7$
Solve each inequality
$x <10$ and $x >-4$
The solution graph is
Example 3
Solve the inequality $\mid 4x +6 \mid \leq 13$ and show the solution graph.
Solution
Rewrite as a compound inequality.
Write as two separate inequalities
$4x + 5 \leq 13$ and $4x+5 \geq -13$
Solve each inequality:
$4x \leq 8$ and $4x \geq -18$
$x \leq 2$ and $x \geq -\frac{9}{2}$
The solution graph is
Example 4
Solve the inequality $\mid x +12 \mid >2$ and show the solution graph.
Solution
Rewrite as a compound inequality.
Write as two separate inequalities.
$x +12 <-2$ or $x+12>2$
Solve each inequality
$x <-14$ or $x >-10$
The solution graph is
Example 5
Solve the inequality $\mid 8x -15 \mid \geq 9$ and show the solution graph.
Rewrite as a compound inequality.
Write as two separate inequalities.
$8x-15 \leq -9$ or $8x-15 \geq 9$
Solve each inequality
$8x \leq 6$ or $8x \geq 24$
$x \leq \frac{3}{4}$ or $x \geq 3$
The solution graph is
## Solve Real-World Problems Using Absolute Value Inequalities
Absolute value inequalities are useful in problems where we are dealing with a range of values.
Example 6:
The velocity of an object is given by the formula $v = 25t - 80$ where the time is expressed in seconds and the velocity is expressed in feet per seconds. Find the times when the magnitude of the velocity is greater than or equal to 60 feet per second.
Solution
Step 1
We want to find the times when the velocity is greater than or equal to 60 feet per second
Step 2
We are given the formula for the velocity $v = 25t - 80$
Write the absolute value inequality $\mid 25t-80\mid \geq 60$
Step 3
Solve the inequality
$25t-80 \geq 60$ or $25t-80 \leq -60$
$25t\geq 140$ or $25t \leq 20$
$t \geq 5.6$ or $t \leq 0.8$
Answer: The magnitude of the velocity is greater than 60 ft/sec for times less than 0.8 seconds and for times greater than 5.6 seconds.
Step 4 When $t= 0.8$ seconds, $v= 25(0.8) -80 = -60 \ ft/sec$. The magnitude of the velocity is 60 ft/sec. The negative sign in the answer means that the object is moving backwards.
When $t= 5.6$ seconds, $v = 25(0.8) - 80 = -60 \ ft/sec$.
To find where the magnitude of the velocity is greater than 60 ft/sec, check values in each of the following time intervals: $t\leq 0.8$,$0.8 \leq t \leq 5.6$ and $t \geq 5.6$.
Check $t = 0.5$: $v = 25(0.5) - 80 = -67.5 \ ft/sec$
Check $t = 2$: $v = 25(2)- 80 = -30 \ ft/sec$
Check $t = 6$: $v = 25(6)-80 = 70 \ ft/sec$
You can see that the magnitude of the velocity is greater than 60 ft/sec for $t \geq 5.6$ or $t\leq 0.8$.
## Lesson Summary
• Like absolute value equations, inequalities with absolute value split into two inequalities. One where the expression within the absolute value is negative and one where it is positive.
• Inequalities of the type $\mid x \mid can be rewritten as $-a .
• Inequalities of the type $\mid x \mid >b$ can be rewritten as $-x<-b$ or $x>b$.
## Review Questions
Solve the following inequalities and show the solution graph.
1. $|x|\leq6$
2. $|x|>3.5$
3. $|x|<12$
4. $|\frac{x} {5}| \leq 6$
5. $|7x|\geq21$
6. $|x-5|>8$
7. $|x+7|<3$
8. $\big | x-\frac{3} {4} \big | \leq \frac{1} {2}$
9. $|2x-5|\geq13$
10. $|5x+3|<7$
11. $\big | \frac{x} {3}-4 \big | \leq 2$
12. $\big | \frac{2x} {7}+9 \big | > \frac{5} {7}$
13. A three month old baby boy weighs an average of 13 pounds. He is considered healthy if he is 2.5 lbs more or less than the average weight. Find the weight range that is considered healthy for three month old boys.
1. $-6 \leq x \leq 6$
2. $x < -3.5$ or $x > 3.5$
3. $-12 < x < 12$
4. $x < -10$ or $x > 10$
5. $x \leq -3$ or $x \geq 3$
6. $x < -3$ or $x > 13$
7. $-10< x < -4$
8. $\frac{1}{4} \leq x \leq \frac{5}{4}$
9. $x \leq -4$ or $x \geq 9$
10. $-2< x < \frac{4}{5}$
11. $6 \leq x \leq 18$
12. $x < -34$ or $x > -29$
13. A healthy weight is $10.5 \ lb \leq x \leq 15.5 \ lb$.
Feb 22, 2012
Aug 22, 2014
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# Pedal Triangle and Isogonal ConjugacyWhat is this about? A Mathematical Droodle
6 January 2016, Created with GeoGebra
Explanation
### Pedal Triangle and Isogonal Conjugacy
The applet intends to suggest the following results:
Let P be a point in the plane of ΔABC.
Then:
1. The pedal triangle A'B'C' of P is orthologic to ΔABC. In particular, the perpendiculars from the vertices of A, B, C to the opposite sides of ΔA'B'C' meet in a point, say, P'.
2. P' is the isogonal conjugate of P.
### Proof
The first assertion is an immediate consequence of the definition of the pedal triangle: the vertices of the pedal triangle of point P are the feet of perpendiculars from P to the sides of ΔABC. Naturally then the perpendiculars from P to the sides of ΔABC are concurrent: they meet in P. By Maxwell's theorem, the perpendiculars from the vertices of ΔABC onto the sides of ΔA'B'C' are also concurrent.
To see that point P' of concurrency is the isogonal conjugate of P, observe that in quadrilateral AB'PC' the angles at B' and C' are right. The quadrilateral is therefore cyclic. Which implies that angles APC' and AB'C' are equal (as inscribed angles subtending the same arc.) Let X be the foot of perpendicular from A to B'C' (that passes through P'.) It follows that the right triangles APC' and AB'X are similar and their other acute angles are also equal:
∠C'AP = ∠XAB' = ∠P'AB',
so that AP and AP' are isogonal to each other. The same consideration applies to the vertices B and C, thus proving the second assertion.
(The first property is mentioned in [Honsberger, p. 64]. Both properties are offered as exercises in [TCCT, p. 215].)
### References
1. R. Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, 1995, pp. 153-154
2. C. Kimberling, Triangle Centers and Central Triangles, Congressus Numerantium 129, Utilitas Mathematica Publishing, 1998.
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Question Video: Computing Logarithms by Using Laws of Logarithms | Nagwa Question Video: Computing Logarithms by Using Laws of Logarithms | Nagwa
# Question Video: Computing Logarithms by Using Laws of Logarithms Mathematics • Second Year of Secondary School
## Join Nagwa Classes
Attend live General Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher!
Which of the following is equal to (5 log 3)/(log 4 + log 6)? [A] log 3 [B] log₂₄ 15 [C] log₂₄ 243 [D] log 15 [E] log 243
02:05
### Video Transcript
Which of the following is equal to five log three over log four plus log six?
This expression might look a little bit strange as our logs seem to have no base. If a log has no base, we generally assume that the base is equal to 10. And so we rewrite our fraction as five log base 10 of three over log base 10 of four plus log base 10 of six.
So we’re next going to recall some rules for logarithms. Firstly, we know that log base 𝑏 of 𝑥 one plus log base 𝑏 of 𝑥 two is log base 𝑏 of 𝑥 one times 𝑥 two. As long as our bases are the same, we simply multiply the arguments. And so the denominator of our fraction is going to become log base 10 of four times six, which is log base 10 of 24.
And what about our numerator? Well, log base 𝑏 of 𝑥 to the power of 𝑝 for real constants 𝑝 is the same as 𝑝 times log base 𝑏 of 𝑥. The converse is true. So we can write our denominator as log base 10 of three to the fifth power. But three to the fifth power is 243. And so we have log base 10 of 243 over log base 10 of 24.
Note that we have a fraction with two logarithms whose bases are equal. And so we can use the change of base formula. This says that log base 𝑏 of 𝑥 one divided by log base 𝑏 of 𝑥 two can be written as log base 𝑥 two of 𝑥 one. So, essentially, if the bases are the same, we make the argument of our denominator the new base. And the argument of our numerator becomes the new argument. In this case then, the base of our logarithm is 24 and its new argument is 243. So we can write our fraction as log base 24 of 243. And our correct answer is therefore (C).
Note that this actually means it didn’t matter what base we assumed. Because we ended up with two logarithms with the same base, we simply apply the change of base formula. We could’ve chosen base two or base three. But remember, the general convention is to assume that it’s base 10.
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# To Express: The quadratic function in standard form.
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
### Precalculus: Mathematics for Calcu...
6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
#### Solutions
Chapter 3.1, Problem 17E
a.
To determine
## To Express: The quadratic function in standard form.
Expert Solution
the quadratic function is expressed in standard form as f(x)=2(x+1)2+1
### Explanation of Solution
Given: The function is f(x)=(2x2+4x+3)
Calculation:
The quadratic function f(x)=(2x2+4x+3) is expressed in standard form as:
f(x)=a(xh)2+k , by completing the square. The graph of the function f is a parabola with vertex (h,k)
The parabola opens upwards if a>0 .
Solve the function:
f(x)=(2x2+4x+3)f(x)=2(x2+2x)+3 [factor 2 from the x terms] f(x)=2(x2+2x+1)+3(21) [complete the square : add 1 inside parentheses, subtract (21) outside] f(x)=2(x+1)2+1 [factor and multiply]
On comparing the above equation with standard form f(x)=a(xh)2+k ,
Therefore, the quadratic function is expressed in standard form as f(x)=2(x+1)2+1
b.
To determine
### To Find: The vertex, x−intercept and y−intercept
Expert Solution
the vertex is (h,k)=(1,1) .
there is no x-intercept .
y-intercept=f(0)=3
### Explanation of Solution
Given: The function is f(x)=(2x2+4x+3)
Calculation:
The quadratic function f(x)=(2x2+4x+3) is expressed in standard form as:
f(x)=2(x+1)2+1
by completing the square. The graph of the function f is a parabola with vertex (h,k) , the vertex is (h,k)=(1,1) .
The y-intercept is given when x=0 , so y-intercept=f(0)=3
From the standard form it is observed that the graph is a parabola that opens upward and has vertex (h,k)=(1,1) . As an aid to sketching the graph, find the intercepts.
The y-intercept=f(0)=3 . To find the x-intercepts , set f(x)=0 and factor the resulting equation, it is observed that there is no x-intercept .
c.
To determine
Expert Solution
### Explanation of Solution
Given: The function is f(x)=(2x2+4x+3)
Graph:
The standard form of the function is:
f(x)=2(x+1)2+1
From the standard form it is observed that the graph is a parabola that opens downward and has vertex (1,1) . As an aid to sketching the graph, find the intercepts. The y-intercept=f(0)=3 and there is no x-intercept . The graph f is sketched in the figure below.
Use graphing calculator to graph the function: f(x)=(2x2+4x+3)
### Have a homework question?
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Published:
Two of the most important relationships in mathematics, namely equality and definition, are both denoted using the same symbol – namely, the equals sign. The overloading of this symbol confuses students in mathematics and computer programming. In this post, I argue for the use of two different symbols for these two fundamentally different operators.
## Introduction
I find it unfortunate that two of the most important relationships in mathematics, namely equality and definition, are often denoted using the exact same symbol – namely, the equal sign: “=”. Early in my learning days, I believe that this overloading of the equal sign led to more confusion than necessary and I have personally witnessed it confuse students.
To ensure that we’re on the same page, let’s first define these two notions. Let’s start with the idea of equality. Let’s say we have two entities, which we will denote using the symbols $X$ and $Y$. The statement “$X$ equals $Y$”, denoted $X = Y$, means that $X$ and $Y$ are the same thing.
For example, let’s say we have a right-triangle with edge lengths $a$, $b$ and $c$, where $c$ is the hypotenuse. The Pythagorean Theorem says that $a^2 + b^2 = c^2$. Said differently, the quantity $c^2$ is the same quantity as the quantity $a^2 + b^2$.
Now, let’s move on to definition. Given some entity denoted with the symbol $Y$, the statement “let $X$ be $Y$”, also often denoted $X = Y$, means that one should use the symbol “$X$” to refer to the entity referred to by “$Y$”.
For example, in introductory math textbooks it is common to define the sine function in reference to a right-triangle:
$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$
This is a definition. We are assigning the symbol/concept $\sin \theta$ to be the ratio of the length of the triangle’s opposite side to the length of its hypotenuse.
The fundamental difference between equality and definition is that in the the equality relationship between $X$ and $Y$, both the symbols $X$ and $Y$ are bound to entities – that is, they “refer” to entities. The statement $Y = X$ is simply a comment about those two entities, namely, that they are the same. In contrast, in a definition, only one of the two symbols is bound to an entity. The act of stating a definition is the act of binding a known entity to a new symbol. For example, the symbol “$\text{foo} \ \theta$” is meaningless. What exactly is “foo”? We don’t know because we have not defined it.
I was tutoring someone who was teaching themselves pre-calculus out of a textbook, and they were quite confused by the statement,
$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$
They asked me, “Why is $\sin \theta$ equal to the quantity $\frac{\text{opposite}}{\text{hypotenuse}}$?” They never explicitly stated so, but it become evident that their confusion was not the good kind of confusion. It wasn’t, “Why are the ratios between the sides of a right-triangle functions of the angles between those sides?” Nor, “Why is this definition important?” Rather, their confusion seemed to stem from the very existence of this mysterious object, “$\sin \theta$”. Their question was more along the lines of, “What is this mysterious thing? And why on earth is it equal to the ratio of the sides of the triangle?”
Their confusion arose from the erroneous interpretation of this statement as describing an equality rather than a definition. The mystery was, at least partly, alleviated by the clarification that $\sin \theta$ is not an object that existed before we saw this statement – rather, this statement created the object for the first time. The statement is defining $\sin \theta$ to be the ratio between the opposite side to the hypotenuse.
The real interesting quality to this definition is that the ratio of the sides of a right triangle are a function of its angles regardless of the lengths of the sides. That is, that we can create this definition at all!
For example, in many programming languages, like C and Python, the assigment operator uses the standard equals sign. That is, the statement x = y assigns the value referenced by symbol y to symbol x. In contrast, the statement x == y returns either True or False depending on whether the value referenced by x is equal to the value referenced by y. Though I have not seen any data on the topic, I wonder whether teaching these two operators from the very beginning of a student’s mathematical education would alleviate this common confusion.
I think it’s important to use the symbol “:=” to denote definition. I prefer this symbol over the popular “$\equiv$” symbol because it emphasizes the assymetry of the statement. That is, $X := Y$ means “use $X$ as a symbol for $Y$”, which differs from “use $Y$ as a symbol for $X$.” In contrast, the standard equals sign “=” is appropriately symmetric.
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# wikiHow to Solve Fraction Questions in Math
Fraction questions can look tricky at first, but they become easier with practice and know-how. Once you understand the fundamentals of what fractions are, you'll be breezing through fraction problems like a knife through butter. You will have to start with Step 1 and learn how to perform basic addition and subtraction, and then move on to more complex calculations.
### Method 1 Multiplying Fractions
1. 1
Make sure you're working with two fractions. These instructions work only if you have two fractions. If you have any mixed numbers involved, convert them to improper fractions first..
2. 2
Multiply numerator x numerator, then multiply denominator x denominator.
• So say I had 1/2 x 3/4, I would multiply 1 x 3 and 2 x 4. The answer is 3/8.
### Method 2 Dividing Fractions
1. 1
Make sure you're working with two fractions. Again, this process will work ONLY if you have already converted any mixed numbers into improper fractions.
2. 2
Flip the second fraction upside down.
3. 3
• If you started with 8/15 ÷ 3/4 then it would become 8/15 x 4/3
4. 4
Multiply top x top and bottom x bottom.
• 8 x 4 is 32 and 15 x 3 is 45, so the final answer is 32/45.
### Method 3 Converting Mixed Numbers into Improper Fractions
1. 1
Convert mixed numbers into improper fractions. Improper fractions are those whose numerators are larger than their denominators. (For example, 17/5.) If you are multiplying and dividing, you must convert mixed numbers into improper fractions before you begin the rest of your calculations.
• Say you have the mixed number 3 2/5 (three and two-fifths).
2. 2
Take the whole (non-fraction) number and multiply it by the denominator.
3. 3
• For our example, we add 15 + 2 to get 17
4. 4
Put that amount over the original denominator and you will have an improper fraction.
• In our case, we get 17/5.
### Method 4 Adding and Subtracting Fractions
1. 1
Find the lowest common denominator (bottom number). For both adding and subtracting fractions, you'll start with the same process. Figure out the lowest common fraction that both denominators can go into.
• For example, if you have 1/4 and 1/6, the lowest common denominator is 12. (4x3=12, 6x2=12)
2. 2
Multiply fractions to match the lowest common denominator. Remember that when you're doing this, you're not actually changing the number, just the terms in which it's expressed. Think of it like a pizza - 1/2 of a pizza and 2/4 of a pizza are the same amount.
• Figure out how many times your current denominator goes into the lowest common denominator. For 1/4, 4 multiplied by 3 is 12. For 1/6, 6 multiplied by 2 is 12.
• Multiply the fraction's numerator and denominator by that number. For 1/4, you would multiply both 1 and 4 by 3, coming up with 3/12. 1/6 multiplied by 2 becomes 2/12. Now your problem looks like 3/12 + 2/12 or 3/12 - 2/12.
3. 3
Add or subtract the two numerators (top number) but NOT the denominators. The reason is because you are trying to say how many of that type of fraction you have, total. If you added the denominators as well, you would be changing what type of fractions they are.
• For 3/12 + 2/12, your final answer is 5/12. For 3/12 - 2/12, it's 1/12
## Community Q&A
Search
• Can you explain 5/10 plus 13/100 equals 63/100?
5/10 = 50/100. 50/100 + 13/100 = 63/100.
• How do I solve a fraction that has a fraction in it, like 12 divided by 12/13?
wikiHow Contributor
You would turn the 12 into a fraction, into 12/1 because 12/1 is the same thing as 12. Next, an easy way to divide would be to the reciprocal method. You would leave he first fraction in the problem as 12/1, then you would turn 12/13 into 13/12 and then simply multiply across to get your answer.
• What is 32.6% of 25.8?
wikiHow Contributor
To figure this out, you just have to know how to calculate percentages in general, which is actually quite simple. Any percentage can be rewritten as a decimal value between 0 and 1, where 0 is 0% and 1 is 100%. To convert a percentage into a decimal value, just move the decimal place two spaces to the left. So, 32.6% becomes .326. Now, to find 32.6% of 25.8, just multiply 25.8 by .326, giving you 8.4108.
• How do I multiply fractions and whole numbers?
wikiHow Contributor
Act like the whole number is also a fraction (i.e. the number over 1). Then multiply it as you would two fractions. Multiply the numerators (or the top numbers) and multiply the denominators (or the bottom numbers). The fraction that results is your answer. E.g. 3 × 1/2 3/1×1/2=3/2 3/2 or 1 1/2
• If the ratio of girls to boys is 8:5, and there are 200 boys, how do I calculate the total number of students?
wikiHow Contributor
Assuming all the girls and boys are students. First divide the number of boys by 5 (200 / 5 = 40). Then multiply that by 8 (40 * 8 = 320). Add the number of boys (200) to the number of girls (320) to get the total students (200 + 320 = 520).
• What is the greatest fraction among: 2/5, 3/5, 1/5, 7/15 and 4/5?
Convert each fraction to an equivalent fraction with a denominator of 15, then compare numerators.
• John bought string to hang up lanterns that was 7/9 of a yard long. He used two-thirds of a yard to hang up the first lantern. What part of a yard remained to hang the last four lanterns?
wikiHow Contributor
7/9 - 2/3 = 1/9. Lowest common denominator is nine, so the first fraction in the question 7/9 stays the same. Since 3 goes into 9 three times, you are going to multiply 3 times the top number in the second fraction, so 3x2 making the top number 6. Now all you have to do is subtract the top 2 numbers 7-6 giving you a final answer of 1/9. I think John used too much string on that one!
• How do I subtract fractions?
wikiHow Contributor
Write the fractions with a common denominator. Subtract the second numerator from the first numerator and put the answer above the common denominator.
• How do I arrange 8/4, 4/35, 2/3, and 4/7 in ascending order?
wikiHow Contributor
Find the lowest common denominator between all of them and the put them in that order. (Remember to then order them in their original fractions, not the changed form.)
• How do I convert a fraction to a whole number?
Assuming the numerator is equal to or larger than the denominator (if not, you can't convert the fraction to a whole number), divide the numerator by the denominator. The quotient is the whole number you're looking for. If there's a remainder, that's the numerator of the fraction left over, with the denominator remaining the same as it was.
200 characters left
## Tips
• Basic skills in the four operations (Multiplication, Division, Addition, and Subtraction) will help the process go quickly and easily.
• To take the reciprocal of a whole number, just put a 1 over it. For example, 5 becomes 1/5.
• You can multiply and divide mixed numbers without converting to improper fractions first. But to do so involves using the distributive property in a potentially intense and complicated way, so it's usually better to go the improper fractions route.
• Another way to say "flip the fraction over" is to say "find the reciprocal. You're still just reversing the numerator and denominator. Ex. 2/4 would be 4/2
## Warnings
• Convert mixed numbers into improper fractions before you begin.
• Check with your teacher to see if you must convert your improper fraction answers into mixed numbers.
• For example, 3 1/4 instead of 13/4.
• Check with your teacher regarding whether or not you must have your answers in lowest terms
• For example, 2/5 is in the lowest term, but 16/40 is not.
## Article Info
Categories: Fractions
In other languages:
Español: operar con fracciones, Deutsch: Bruchrechnungen lösen, Português: Resolver Problemas Matemáticos com Frações, Italiano: Risolvere Problemi Matematici sulle Frazioni, Français: résoudre les problèmes de fractions en mathématiques, Nederlands: Breuken oplossen, Русский: проводить действия с дробями, Bahasa Indonesia: Menyelesaikan Soal Pecahan dalam Matematika, 中文: 做分数运算
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Find the solution of ${2^x} = 7$.
Last updated date: 18th Jun 2024
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Hint:Whenever the right and left side of the equation do not have the same base then in that case we should take “log” both the sides. The power rule is used to multiply the two logarithms and to combine the exponents. The exponential expression should be kept by itself on one side of the equation. The logarithms of both sides of the equation should be obtained and should be solved for variable.
Complete step by step solution:
A logarithm is an exponent that is written in a special way. A logarithm with base $10$is a common logarithm.
The product rule states that${\log _b}\left( {MN} \right) = {\log _b}\left( M \right) + {\log _b}\left( N\right)$.
This property denotes that logarithm of a product is the sum of the logs of its factors. The two numbers should be multiplied with the same base then the exponents must be added.
The quotient rule states that ${\log _b}\left( {\frac{M}{N}} \right) = {\log _b}\left( M \right) + {\log_b}\left( N \right)$
This property denotes that the log of a quotient is the difference of the log of the dividend and the divisor.
Since, the right and left side of the equation do not have the same base then in that case we should take “log” both the sides.
We have,
${2^x} = 7 \\ \Rightarrow \log \left( {{2^x}} \right) = \log \left( 7 \right) \\$
Using the log property we have,
$\Rightarrow x\log \left( 2 \right) = \log \left( 7 \right)$
Solving for “x” now we will have,
$\Rightarrow x = \dfrac{{\log \left( 7 \right)}}{{\log \left( 2 \right)}} \\ \Rightarrow x \approx 2.81 \\$
Hence the solution of ${2^x} = 7$ is$\approx 2.81$.
Note: Start by the condensing the log expressions on the left into a single logarithm using the product rule. What we want is to have a single log expression on each side of the equation. Since we want to transform the left side into a single logarithmic equation, then we should use the product rule in reverse to condense it. Always check the solved values with the original logarithmic equations.
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## Radical Equation Word Problems - Examples & Practice - Expii
Radical equation word problems - examples & practice, explanations (3).
The key to solving any word problem (whether it contains a radical or not) is to translate the problem from words into math .
That's the biggest step in word problems. Once you've translated the information into numbers, you solve the equation the same way as always.
Let's look at an example to see how this approach works when radicals are involved.
Hiroki is calculating the speed of a tsunami. The formula is speed=√9.8d where d is ocean depth in meters. In this case, the ocean was 1500 meters deep. How fast was the wave (in m/s2)?
So, in this problem (as with many problems involving radicals), the formula to use has already been given to us. We are interested in the speed of the wave, and we've been given the depth (d). So, let's start by plugging that value in : speed=√9.8(1500) Let's solve this problem.
How fast was the wave (m/s2)?
## Related Lessons
Example walk through: word problem involving radicals.
The key to word problems is in the translation. Let's walk through the problem below and solve it.
Lori recently purchased a square painting with an area of 780 cm2. How much wood does Lori need to make the frame? Don't worry about the width and height of the wood, just find the length.
First, find out the essential information in the problem.
A square painting has area 780cm2, what is its side length?
What is the relationship between side length and area of a square? Area=side length×side length
Denote side length as x for simplicity, plug in the numbers we have: 780cm2=x×x Solve for x: 780=x2x=√78x≈27.93
Lastly, note that the question is asking how much wood Lori needs to build the frame. A square has 4 sides so we have to multiply 27.93cm by 4.
In total, Lori needs at least 27.93cm×4=111.72cm of wood.
Let's look at another problem!
Image source: By Caroline Kulczycky
## (Video) Problem Solving with Radical Expressions
by Deanna Green
This video by Deanna Green works through word problems with radicals.
These problems usually focus around the Pythagorean Theorem . A helpful trick always is to draw a simple picture of what's going on. Then fill in lengths or angles as the problem states.
Let's work through one of the examples she looks at.
A kite is secured to a rope that is tied to the ground. A breeze blows the kite so that the rope is taught while the kite is directly above a flagpole that is 30 ft from where the rope is staked down. Find the altitude of the kite if the rope is 110 ft long.
The first step is to draw a diagram.
Image source: by Alex Federspiel
The Pythagorean Theorem states:
The length of the hypotenuse (diagonal line) is equal to the square root of the square of the sides added.
As an equation:
where c is the hypotenuse and a and b are the other sides.
Start by plugging in all the information we know.
110=√302+x2
You cannot split up radicals with a sum underneath into two separate radicals. So, to solve this, we want to take the square of both sides.
(110)2=(√302+x2)212100=302+x212100−302=x212100−900=x211200=x2√11200=√x2105.83≈x
You may need to review the lesson on getting rid of the x2 in with this lesson .
## Learning Objectives
By the end of this section, you will be able to:
Before you get started, take this readiness quiz.
In this section we will solve equations that have a variable in the radicand of a radical expression. An equation of this type is called a radical equation .
An equation in which a variable is in the radicand of a radical expression is called a radical equation .
As usual, when solving these equations, what we do to one side of an equation we must do to the other side as well. Once we isolate the radical, our strategy will be to raise both sides of the equation to the power of the index. This will eliminate the radical.
Solving radical equations containing an even index by raising both sides to the power of the index may introduce an algebraic solution that would not be a solution to the original radical equation. Again, we call this an extraneous solution as we did when we solved rational equations.
In the next example, we will see how to solve a radical equation. Our strategy is based on raising a radical with index n to the n th power. This will eliminate the radical.
• Isolate the radical on one side of the equation.
• Raise both sides of the equation to the power of the index.
• Solve the new equation.
• Check the answer in the original equation.
When we use a radical sign, it indicates the principal or positive root. If an equation has a radical with an even index equal to a negative number, that equation will have no solution.
Because the square root is equal to a negative number, the equation has no solution.
If one side of an equation with a square root is a binomial, we use the Product of Binomial Squares Pattern when we square it.
Don’t forget the middle term!
When the index of the radical is 3, we cube both sides to remove the radical.
Sometimes the solution of a radical equation results in two algebraic solutions, but one of them may be an extraneous solution !
When there is a coefficient in front of the radical, we must raise it to the power of the index, too.
If the radical equation has two radicals, we start out by isolating one of them. It often works out easiest to isolate the more complicated radical first.
In the next example, when one radical is isolated, the second radical is also isolated.
Sometimes after raising both sides of an equation to a power, we still have a variable inside a radical. When that happens, we repeat Step 1 and Step 2 of our procedure. We isolate the radical and raise both sides of the equation to the power of the index again.
We summarize the steps here. We have adjusted our previous steps to include more than one radical in the equation This procedure will now work for any radical equations.
• Isolate one of the radical terms on one side of the equation.
If yes, repeat Step 1 and Step 2 again.
As you progress through your college courses, you’ll encounter formulas that include radicals in many disciplines. We will modify our Problem Solving Strategy for Geometry Applications slightly to give us a plan for solving applications with formulas from any discipline.
• Read the problem and make sure all the words and ideas are understood. When appropriate, draw a figure and label it with the given information.
• Identify what we are looking for.
• Name what we are looking for by choosing a variable to represent it.
• Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
• Solve the equation using good algebra techniques.
• Check the answer in the problem and make sure it makes sense.
• Answer the question with a complete sentence.
One application of radicals has to do with the effect of gravity on falling objects. The formula allows us to determine how long it will take a fallen object to hit the gound.
On Earth, if an object is dropped from a height of h feet, the time in seconds it will take to reach the ground is found by using the formula
It would take 2 seconds for an object dropped from a height of 64 feet to reach the ground.
Police officers investigating car accidents measure the length of the skid marks on the pavement. Then they use square roots to determine the speed , in miles per hour, a car was going before applying the brakes.
If the length of the skid marks is d feet, then the speed, s , of the car before the brakes were applied can be found by using the formula
Access these online resources for additional instruction and practice with solving radical equations.
• Solving an Equation Involving a Single Radical
• Solving Equations with Radicals and Rational Exponents
## Practice Makes Perfect
In the following exercises, solve.
no solution
In the following exercises, solve. Round approximations to one decimal place.
## Writing Exercises
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ After reviewing this checklist, what will you do to become confident for all objectives?
How to solve equations with square roots, cube roots, etc.
We can get rid of a square root by squaring (or cube roots by cubing, etc).
Warning: this can sometimes create "solutions" which don't actually work when we put them into the original equation. So we need to Check!
• isolate the square root on one side of the equation
• square both sides of the equation
Then continue with our solution!
## Example: solve √(2x+9) − 5 = 0
Now it should be easier to solve!
Check: √(2·8+9) − 5 = √(25) − 5 = 5 − 5 = 0
That one worked perfectly.
## More Than One Square Root
What if there are two or more square roots? Easy! Just repeat the process for each one.
It will take longer (lots more steps) ... but nothing too hard.
## Example: solve √(2x−5) − √(x−1) = 1
We have removed one square root.
Now do the "square root" thing again:
We have now successfully removed both square roots.
Let us continue on with the solution.
It is a Quadratic Equation! So let us put it in standard form.
Using the Quadratic Formula (a=1, b=−14, c=29) gives the solutions:
2.53 and 11.47 (to 2 decimal places)
Let us check the solutions:
There is really only one solution :
Answer: 11.47 (to 2 decimal places)
See? This method can sometimes produce solutions that don't really work!
The root that seemed to work, but wasn't right when we checked it, is called an "Extraneous Root"
So checking is important.
## How To : Solve word problems containing radical equations
See how to unpack and solve a word problem containing radical equations with this free video math lesson from Internet pedagogical superstar Simon Khan. From Ramanujan to calculus co-creator Gottfried Leibniz, many of the world's best and brightest mathematical minds have belonged to autodidacts. And, thanks to the Internet, it's easier than ever to follow in their footsteps (or just finish your homework or study for that next big test).
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#### IMAGES
2. How to Solve Radical Equations: 12 Steps (with Pictures)
5. 03 11 Word problem involving radical equations: Basic
6. WORD PROBLEMS INVOLVING RADICAL EQUATIONS
#### VIDEO
1. Solve application problems involving radical equations
2. How to insert common radical equations in PowerPoint 365
3. How to insert radical equations in PowerPoint shortcut key
4. How to insert radical equations in PowerPoint
5. Olympiad Math Radical Equation. Algebra Problem. IMO Exam tricks you should know
Hannah Bonville Text 3 The key to solving any word problem (whether it contains a radical or not) is to translate the problem from words into math. That's the biggest step in word problems. Once you've translated the information into numbers, you solve the equation the same way as always.
2. Applied Use of Radical Equations
Directions: Solve the following problems dealing with radical equations. Show your work algebraically. Remember that you can use your graphing calculator to check your answers. 1. A cube measures 1½ feet on each side. A (straight) spider web connects the bottom corner of the cube to the opposite top corner, as shown.
3. How to Solve Basic Word Problems Involving Radical Equations
Step 1. Plug in the known g and h values. v = 2 ( 32) ( 180) Step 2. Simplify to find the unknown velocity v . v = 2 ( 32) ( 180) = 11520 = 107.33 The velocity of the pinecone is approximately 107...
Word Problems involving Radical Equations In solving radical equations, you have to follow the following steps: 1. Write the equation such that the radical containing the unknown is...
5. Ch10: Radical expressions and equations
Pythagorean theorem word problems Get 3 of 4 questions to level up! Quiz 1. Level up on the above skills and collect up to 400 Mastery points Start quiz. Simplifying radicals | 10-2. ... Extraneous solutions of radical equations (Opens a modal) Practice. Square-root equations intro Get 3 of 4 questions to level up!
6. Solving Word Problems Involving Radicals
Solving Word Problems Involving Radicals Señor Pablo TV 505K subscribers Subscribe 566 33K views 2 years ago To solve a radical equation: 1. Isolate the radical expression involving the...
Solve radical equations with two radicals; Use radicals in applications; Before you get started, take this readiness quiz. Simplify: $$(y−3)^{2}$$. ... Use a Problem Solving Strategy for Applications With Formulas. Read the problem and make sure all the words and ideas are understood. When appropriate, draw a figure and label it with the ...
Algebra (all content) Unit 12: Radical equations & functions About this unit This topic covers: - Solving radical equations - Graphing radical functions Solving square-root equations Learn Intro to square-root equations & extraneous solutions Intro to solving square-root equations Solving square-root equations
How to: Solve a Radical Equation. Isolate a radical. Put ONE radical on one side of the equal sign and put everything else on the other side. Eliminate the radical. Raise both sides of the equal sign to the power that matches the index on the radical. This means square both sides if it is a square root; cube both sides if it is a cube root; etc.
Solve: Solve: Solve: Solve a radical equation with one radical. Isolate the radical on one side of the equation. Raise both sides of the equation to the power of the index. Solve the new equation. Check the answer in the original equation. When we use a radical sign, it indicates the principal or positive root.
A Radical Equation is an equation with a square root or cube root, etc. Solving Radical Equations. We can get rid of a square root by squaring (or cube roots by cubing, etc). Warning: this can sometimes create "solutions" which don't actually work when we put them into the original equation. So we need to Check!
Key Steps to Solve Radical Equations: 1) Isolate the radical symbol on one side of the equation 2) Square both sides of the equation to eliminate the radical symbol 3) Solve the equation that comes out after the squaring process 4) Check your answers with the original equation to avoid extraneous values or solutions
13. PDF 12 Math 51 Solve Equations with Radicals
of the equation you must do to the other side. Example: x=7 ⇒square both sides ⇒ ( ) =( )2 ⇐ 2 x 7 The square and square root cancel out on the left. So, x=(7)2 ⇒ x=49Done! Note: The radical must be isolated on one side of the equal sign before you can square both side. Example: Solve 2 x−5+3=21 The radical is not isolated.
Step 1. Read the problem carefully and draw an illustration, if possible. Step 2. Identify the appropriate formula for the application. Step 3. Plug in any known values. Step 4. Solve for the...
15. Solving Radical Equations Word Problems
2. Real-World Scenarios: Each problem in this worksheet is designed to apply the concepts of solving radical equations to practical, real-life situations. Students will love tackling math problems with a real-world connection. 3. Ideal for All Settings: Whether you're assigning work for in-class practice, homework, quizzes, or exams, this ...
Solving Advanced Word Problems Involving Radical Equations Algebra 2 Skills Practice 1. The square root of the sum of two consecutive odd integers is 8. Find the integers. 2. The square...
Use radical equations to solve area and volume problems. 0. 1 More Read. Video. Solving Problems Involving Radical Equations - Overview. basic. Overview of Problems Involving Radical Equations. 0. 8 More Videos. Practice. Estimated 23 mins to complete. Applications Using Radicals Practice. at grade. Practice. 0. 0 More DL Assessments. ABOUT ...
18. Algebra
Here is a set of practice problems to accompany the Equations with Radicals section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University.
T = 2 L ⇒ 1 = 2 L Square both sides of the equation: 1 = 4 L Solve for L: L = 1 4 m e t e r s Real-World Application: TV Screens "Square" TV screens have an aspect ratio of 4:3; in other words, the width of the screen is 4 3 the height. TV "sizes" are traditionally represented as the length of the diagonal of the television screen.
20. How to Solve word problems containing radical equations
6/21/10 9:37 AM See how to unpack and solve a word problem containing radical equations with this free video math lesson from Internet pedagogical superstar Simon Khan. From Ramanujan to calculus co-creator Gottfried Leibniz, many of the world's best and brightest mathematical minds have belonged to autodidacts.
21. 1.4e: Exercises
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# The Method of Undetermined Coefficients
Author: John J Weber III, PhD Corresponding Textbook Sections:
• Section 4.4 – Nonhomogeneous Equations: The Method of Undetermined Coefficients
• Section 4.5 – The Superposition Principle and Undetermined Coefficients Revisited
## Expected Educational Results
• Objective 12–1: I can find particular solutions to nonhomogeneous equations.
• Objective 12–2: I can find general solutions to nonhomogeneous equations.
• Objective 12–2: I can solve higher-order nonhomogeneous IVPs.
## Method of Undetermined Coefficients
### Nonhomogeneous Linear Equations
#### Definition: Nonhomogeneous Linear Equations
$\displaystyle ay^{\,\prime\prime}+by^{\,\prime}+cy=f(t)$, $f(t)\ne 0$, $a\ne 0$.
#### Theorem: Superposition Principle
Let $y_1$ be a solution to $\displaystyle ay^{\,\prime\prime}+by^{\,\prime}+cy=f_1(t)$ and $y_2$ be a solution to $\displaystyle ay^{\,\prime\prime}+by^{\,\prime}+cy=f_2(t)$ Then for any constants $k_1$ and $k_2$, the function $k_1y_1+k_2y_2$ is a solution to the DE $\displaystyle ay^{\,\prime\prime}+by^{\,\prime}+cy=k_1f_1(t)+k_2f_2(t)$
#### Definition: Particular Solution
A solution to a nonhomogeneous DE is called the particular solution, $\displaystyle y_p(t)$.
Definition: General Solution
$\displaystyle y_g(t)=y_h(t)+y_p(t)$. $\displaystyle y(t)$ is an acceptable notation for the general solution.
### Method of Undetermined Coefficients – Procedure
NOTE: The terms in $y_p(t)$ must be linearly independent of the terms in $y_h(t)$. This method is valid only when $f(t)$ consists of polynomials, exponentials, sines, cosines, or sums and products of these functions.
1. Find the homogeneous solution, $y_h(t)$.
2. Let $P_m(t)$ be a polynomial of degree $m$, $Q_n(t)$ be a polynomial of degree $n$, $k=\text{max}(m,n)$.
3. $f(t)=P_m(t)e^{rt}$:
1. $r$ is not a root of the characteristic equation: $y_p(t)=(A_mt^m+\cdots+A_1t+A_0)e^{rt}$
2. $r$ is a root of the characteristic equation: $y_p(t)=t(A_mt^m+\cdots+A_1t+A_0)e^{rt}$
3. $r$ is a double root of the characteristic equation: $y_p(t)=t^2(A_mt^m+\cdots+A_1t+A_0)e^{rt}$
4. $f(t)=P_m(t)e^{\alpha t}\cos{(\beta t)}+Q_n(t)e^{\alpha t}\sin{(\beta t)}$, $\beta\ne 0$:
1. $\alpha\pm i\beta t$ is not a root of the characteristic equation: $y_p(t)=(A_kt^k+\cdots+A_1t+A_0)e^{\alpha t}\cos{(\beta t)}+(B_kt^k+\cdots+B_1t+B_0)e^{\alpha t}\sin{(\beta t)}$
2. $\alpha\pm i\beta t$ is a root of the characteristic equation: $y_p(t)=t(A_kt^k+\cdots+A_1t+A_0)e^{\alpha t}\cos{(\beta t)}+t(B_kt^k+\cdots+B_1t+B_0)e^{\alpha t}\sin{(\beta t)}$
5. Find coefficients by finding $y_p^{\,\prime}$, $y_p^{\,\prime\prime}$, etc. and substituting into the DE and solving for the coefficients.
6. Write the general solution, $y(t)=y_h(t)+y_p(t)$
#### Existence and Uniqueness
THeorem: Existence and Uniqueness of Nonhomogeneous DEs
For any real numbers $a\ne 0$, $b$, $c$, $t_0$, $Y_0$, and $Y_1$, suppose $y_p(t)$ is a particular solution to $\displaystyle ay^{\,\prime\prime}+by^{\,\prime}+cy=f(t)$ on some interval containing $y_0$ and suppose $y_1(t)$ and $y_2(t)$ are linearly independent solutions to the homogeneous equation in the interval, then there exists a unique solution in the interval to the IVP $\displaystyle ay^{\,\prime\prime}+by^{\,\prime}+cy=f(t)$, $\displaystyle y(t_0)=Y_0$, $\displaystyle y^{\,\prime}(t_0)=Y_1$.
#### Investigation 01
Suppose $y_h(t)$ is the homogeneous solution to $ay^{\,\prime\prime}+by^{\,\prime}+cy=f(t)$. Let $y_p(t)$ be the particular solution to the DE. Show $y(t)=y_h(t)+y_p(t)$ is the general solution to the DE.
#### Investigation 02
Suppose the following are roots, $r$, to the characteristic equation and $f(x)$ is the nonhomogeneous function to the ode. Find the general solution, $y_g(x)=y_h(x)+y_p(x)$.
1. $\displaystyle r=-2i,2i$, $\displaystyle f(x)=e^{2x}$
2. $\displaystyle r=0,-3,3$, $\displaystyle f(x)=e^{3x}$
3. $\displaystyle r=1,1,1,3$, $\displaystyle f(x)=e^{x}$
4. $\displaystyle r=0,0,-2i,2i$, $\displaystyle f(x)=x+\sin{(x)}$
5. $\displaystyle r=1,2,1-2i,1+2i$, $\displaystyle f(x)=x^2-3x+2$
6. $\displaystyle r=1-2i,1+2i,1-2i,1+2i$, $\displaystyle f(x)=xe^{x}\cos{(2x)}$
7. $\displaystyle r=-1,2,-3i,3i$, $\displaystyle f(x)=x^2-e^{2x}+\cos{(x)}-e^x\sin{(2x)}$
8. $\displaystyle r=\frac{3-2i}{2},\frac{3+2i}{2}$, $\displaystyle f(x)=e^{3x}\sin{(2x)}$
9. $\displaystyle r=0,3,1-5i,1+5i$, $\displaystyle f(x)=x+x^2e^{x}\sin{(5x)}$
#### Investigation 03
Solve the following DEs:
1. $\displaystyle y^{\,\prime\prime}-6y^{\,\prime}+8y=3e^{2x}$
2. $\displaystyle y^{\,\prime\prime}-6y^{\,\prime}+9y=5xe^{3x}$
3. $\displaystyle y^{\,\prime\prime}-2y^{\,\prime}-3y=3x^2+4x-5$
4. $\displaystyle y^{\,\prime\prime}+y=\sin{(x)}$
5. $\displaystyle y^{\,\prime\prime}+4y=5\cos{(3x)}$
6. $\displaystyle y^{\,\prime\prime}-2y^{\,\prime}-3y=e^{2x}+3x^2+4x-5+5\cos{(2x)}$
7. $\displaystyle y^{\,\prime\prime}-8y^{\,\prime}+12y=x^2e^{6x}-7x\sin{(2x)}+4$
8. $\displaystyle y^{\,\prime\prime}+2y^{\,\prime}-3y=2e^x-10\sin{(x)}$
9. $\displaystyle y^{\,\prime\prime\prime}-y^{\,\prime}=4e^{-x}+3e^{2x}$
10. $\displaystyle y^{\,\prime\prime\prime\prime}-11y^{\,\prime\prime\prime}+31y^{\,\prime\prime}-61y^{\,\prime}+40y=x^3e^{2x}$
11. $\displaystyle y^{\,(4)}(t)-2y^{\,\prime\prime\prime}(t)+y^{\,\prime\prime}(t)-12y^{\,\prime}(t)+20y(t)=te^t+t^2-3e^{t}\cos{(2t)}$
12. $\displaystyle y^{\,(5)}(t)-y^{\,(4)}(t)-y^{\,\prime\prime\prime}(t)-y^{\,\prime\prime}(t)-2y^{\,\prime}(t)=t\sin{(t)}+e^{2t}+t$
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# Symbolic system of equation solver
In this blog post, we will show you how to work with Symbolic system of equation solver. Let's try the best math solver.
## The Best Symbolic system of equation solver
Math can be a challenging subject for many students. But there is help available in the form of Symbolic system of equation solver. There are a few key steps to solving any word problem. First, read the problem and identify what information is given and what is being asked. Next, create a visual representation of the problem, whether that be a drawing or a simple equation. Lastly, solve the problem and check your work to make sure the answer makes sense. With practice, solving word problems can become second nature!
There are a variety of ways to solve two equations. The most common method is to use algebra to solve for one of the variables in terms of the other. This can be done by using the addition or subtraction properties of equality to cancel out like terms on one or both sides of the equation. Once one of the variables is isolated, the equation can be solved for that variable.
One way to solve an inequality with a fraction is to multiply both sides of the inequality by the reciprocal of the fraction. This will clear the fraction from the inequality and make it easier to solve. Another way to solve an inequality with a fraction is to divide both sides of the inequality by the fraction. This will also clear the fraction from the inequality and make it easier to solve.
They are often used in mathematical and scientific applications where a real number answer is not possible. To solve an imaginary number, you must use a complex number. This is a number that has both a real and imaginary component. The imaginary component is often represented by the letter "i." To solve an equation with an imaginary number, you must use imaginary numbers in your calculations.
There are a number of ways to get help with math problems. One way is to ask a friend or family member for help. Another way is to use a resource like a book or website. Finally, you can always ask a teacher or tutor for help. No matter what method you choose, make sure you understand the problem and the solution before moving on.
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## Part 1
In this video, we are going to graph rational functions.
For example: $y=\dfrac{1}{x}$
Let’s first make a table listing the x and y values
x y $-3$$-3$ $\dfrac{1}{-3}$$\dfrac{1}{-3}$ $-2$$-2$ $\dfrac{1}{-2}$$\dfrac{1}{-2}$ $-1$$-1$ $\dfrac{1}{-1}=-1$$\dfrac{1}{-1}=-1$ $0$$0$ $\dfrac{1}{0}=UNDEFINED$$\dfrac{1}{0}=UNDEFINED$ $1$$1$ $\dfrac{1}{1}$$\dfrac{1}{1}$ $2$$2$ $\dfrac{1}{2}$$\dfrac{1}{2}$ $3$$3$ $\dfrac{1}{3}$$\dfrac{1}{3}$
Plot the coordinate points on a coordinate plane. Note how there are no points on the x-axis.
## Video-Lesson Transcript
Let’s go over how to graph rational functions such as $y = \dfrac{1}{x}$.
Let’s look at some values.
When $x = -3$ then $y = \dfrac{1}{-3} = - \dfrac{1}{3}$,
when $x = -2$ then $y = \dfrac{1}{-2} = - \dfrac{1}{2}$,
then $x = -1$ then $y = \dfrac{1}{-1} = - 1$,
when $x = 0$ then $y = \dfrac{1}{0} = undefined$,
when $x = 1$ then $y = \dfrac{1}{1} = 1$,
then when $x = 2$ then $y = \dfrac{1}{2} = \dfrac{1}{2}$,
and when $x = 3$ then $y = \dfrac{1}{3} = \dfrac{1}{3}$.
Now, we’re ready to plot.
Zero doesn’t exist. Let’s just make a note that there’s an asymptote there for the point in the function that doesn’t exist.
An asymptote is here so we’ll know that when we approach it, it doesn’t really exist.
Let’s continue plotting.
We came up with two lines that don’t touch the $x$-axis and $y$-axis. Because these represent the asymptotes.
No matter how relatively close they are, they won’t touch the $x$ and $y$ axis.
This is the basic line graph of our discussion.
Let’s go over some rules in transforming functions.
1. If we add a number to a $x$ value, we’re going to move to the left by that number.
2. Next, if we subtract from $x$, we’ll move to the right by that value.
3. If we add a number to a function, it’s going to move the line up by that number.
4. Then, if we subtract a number from the function, it’s going to move it down by that number.
5. If you multiply a function by a positive number, it’s going to get each $y$ value greater by that value.
To demonstrate this. let’s say we have a function $x^2$.
When we’re given $2x^2$, the values of $y$ are also multiplied by $2$. The line will be longer and skimpy.
If we’re given $\dfrac{1}{2}x^2$, then values of $y$ are going to be multiplied by $\dfrac{1}{2}$. Line will become wider and shallow.
Now, if multiplied by a positive number greater than $1$, it’s going to stretch out.
And if multiplied by a positive number between $0$ and $1$, it’s going to stretch down.
6. If you multiply it by a negative number, it’s going to flip it. The same shape but upside down.
Now, let’s apply this in $y = \dfrac{1}{x}$ function rule.
Here’s a graph.
For example:
$y = \dfrac{1}{x + 2}$
Since we added a number to $x$, we’ll move to the left by the same number of spots which is $2$.
The asymptote where the two lines don’t pass will also move by $2$ spots.
Let’s do
$y = \dfrac{1}{x} + 1$
Now, we’re going to move up by $1$ spot.
Notice that the horizontal asymptote also moved up by $1$.
But the vertical asymptote didn’t move.
Now, let’s take a look when we have negative.
$y = -\dfrac{1}{x}$
Here, everything is negated. So our line will flip.
Notice the asymptotes – both vertical and horizontal – didn’t move.
Let’s look at a more complicated one that involves everything.
$y = - \dfrac{1}{x - 1} - 2$
Negative means lines will flip. The line will move right by $1$ spot and move down by $2$ spots.
Now, let’s look at the asymptotes. Same as our functions, it also moves right by $1$ spot and moves down by $2$ spots.
But this is just step number one.
Because our function is negative, we have to flip it.
## Part 2
In the second video, we are going to graph rational functions with transformation.
Keep the following in mind during transformation:
Add # to x Moves Left Subtract # from x Moves Right Add # to Function Moves Up Subtract # from Function Moves Down Multiply Function (Positive) Each y-value is multiplied Multiply by Function (Negative) Each y-value is multiplied, flipped over y-axis
For example:
A transformation from $y=\frac{1}{x}$ to $y=\frac{1}{x}+2$, each point of the latter graph will be located two units above each point of the prior graph.
A transformation from $y=\frac{1}{x}$ to $y=-\frac{1}{x}$, each point of the latter graph will be located across the x-axis compared to each point of the prior graph.
## Examples of Graphing Rational Functions
### Example 1
$y=\dfrac{1}{x+3}$
First, make a table listing the x and y values
Plot the coordinate points on a coordinate plane.
As you can see, the line was moved to the left by $3$ spots
### Example 2
$y=\frac{1}{x-3}+1$
First, make a table listing the x and y values
Plot the coordinate points on a coordinate plane.
As you can see, the line was moved to the right by $3$ spots and up by $1$ spot
## Video-Lesson Transcript
Let’s look at a more difficult one.
$y = - \dfrac{1}{x - 1} - 2$
This negative means we have to flip the lines over the $x$-axis.
Important to note is that we have to take the negative first into consideration before we move left or right.
This just takes care of the negative there.
You might recognize this, as we previously graph $y = - \dfrac{1}{x}$.
Now, let’s take into consideration the $-1$ and $-2$.
The line will move right by $1$ spot and move down by $2$ spots.
Now, let’s look at the asymptotes.
The new asymptote followed the function and move right by $1$ and move down by $2$.
The transformation for $y = \dfrac{1}{x}$ also applies to other functions as well.
For example:
$y = (x - 1)^{2} + 3$
Here, we’ll move to the right by $1$ and up by $3$.
Next,
$y = - \mid x \mid + 2$
Remember, we have to do the negative first. So, we’ll flip the line.
This is just our intermediate function, where we only took the negative into account.
Now, let’s do $+ 2$.
We’ll move everything by $2$ spots.
Keep in mind our rules:
1. If we add a number to a $x$ value, we’re going to move to the left by that number.
2. Then, if we subtract from $x$, we’ll move to the right by that value.
3. Next, when we add a number to a function, it’s going to move the line up by that number.
4. If we subtract a number from the function, it’s going to move it down by that number.
5. When you multiply a function by a positive number, it’s going to get each $y$ value greater by that value.
6. Lastly, if you multiply it by a negative number, it’s going to flip it. The same shape but upside down.
Let’s do one more example.
$y = - \dfrac{1}{2}x^2$
Negative’s going to flip it. And $\dfrac{1}{2}$ is going to make the $y$ values $\dfrac{1}{2}$ of what it originally is.
Now, it flipped and became a little bit wider than the original.
After you finish this lesson, view all of our Algebra 1 lessons and practice problems.
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# NCERT Solutions For Class 8 Maths Chapter 11 Mensuration Ex 11.2
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Here, Below you all know about NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2 Question Answer. I know many of you confuse about finding this Chapter 11 Mensuration Ex 11.2 Of Class 8 NCERT Solutions. So, Read the full post below and get your solutions.
Table of Contents
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## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2
NCERT TEXTBOOK EXERCISES
Question 1. The shape of the top surface of a table is a trapezium. Find its area, if its parallel sides are 1 m and 1.2 man the d perpendicular distance between them is 0.8 m.
Solution.
Area of the top surface of the table
= $\frac { 1 }{ 2 } h(a+b)$
= $\frac { 1 }{ 2 } \times 0.8\times (1.2+1)$
= 0.88m2
Question 2. The area of a trapezium is 34 cm2 and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the another parallel side.
Solution.
Area of trapezium
= $\frac { 1 }{ 2 } h(a+b)$
⇒ $34=\frac { 1 }{ 2 } \times 4(10+b)$
⇒ $34=2\times (10+b)$
⇒ 10+b=34/2
⇒ 10 + b=17
⇒ b = 17 – 10
⇒ b = 7 cm
Hence, the length of another parallel side is 7 cm.
Question 3. Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC
Solution.
Fence of the trapezium shaped field ABCD = 120 m
⇒ AB + BC + CD + DA = 120
⇒ AB + 48 + 17 + 40 = 120
⇒ AB + 105 = 120
⇒ AB = 120 – 105
⇒ AB = 15 m
∴ Area of the field
= $\frac { (BC+AD)\times AB }{ 2 }$
= $\frac { (48+40)\times 16 }{ 2 }$ = 660 m2
Question 4. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
Solution.
Area of the field
= $\frac { 1 }{ 2 } d({ h }{ 1 }+{ h }{ 2 })$
= $\frac { 24\times (8+13) }{ 2 }$ = $\frac { 24\times 21 }{ 2 }$
= 12 x 21 = 252m2
Question 5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution.
Area of the rhombus
= 1/2×d1×d2
= 1/2×7.5×12
= 45 m2
Question 6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Solution.
Area of the rhombus
= base (b) x altitude (h) = 5
= 5 x 4.8 = 24 cm2
Question 7. The floor of a building consists of 3,000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2 is ₹ 4.
Solution.
Area of a tile
Question 8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10,500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.
Solution.
Let the length of the side along the road be x m. Then, the length of the side along the river is 2x m.
Area of the field = 10,500 square metres
Question 9. Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.
Solution.
Area of the octagonal surface
Question 10. There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.
Find the area of this park using both ways. Can you suggest some other way of finding its area ?
Solution.
Question 11. Diagram of the adjacent picture frame has outer dimensions = 24 cm x 28 cm and inner dimensions 16 cm x 20 cm. Find the area of each section of the frame, if the width of each section the same.
Solution.
## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.2 PDF
For NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2, you may click on the link below and get your NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise pdf file.
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Finally, You all know about NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2. If you have any questions, then comment below and share this post with others.
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# Arithmetic Series & Adding Numbers
in STEMGeeks2 months ago
Hi there. In this mathematics post, I would like to cover arithmetic series when it comes to adding numbers and arithmetic sequences.
Quicklatex.com is used for LaTeX math rendering.
Pixabay Image Source
## Topics
• Arithmetic Sequences Basics
• Sum/Sigma Notation Basics
• Partial Sums For Arithmetic Sequences
• Examples
## Arithmetic Sequences Basics
An arithmetic sequence is a list of numbers where the next number is increased or decreased by the same number. Arithmetic sequences can be either increasing or decreasing.
One example of an increasing sequence is `4, 9, 14, 19, 24, 29, 34`. There is a constant increase of 5 to obtain the next number.
A decreasing sequence example includes `18, 8, -2, -12, -22, -32, -42`. The common difference here is negative 10.
Pixabay Image Source
## Sum/Sigma Notation Basics
Instead of writing the numbers from 1 to 100 being added together in the form of `1 + 2 + 3 + 4 + ... + 100`, you can use sum notation. This is a compact representation of adding numbers together from an arithmetic sequence.
Here is the representation in math notation.
The index variable `k` can be changed to a different letter such as `i` and `j`.
To represent a sum that does not start at 1 such as `5 + 6 + 7 + 8 + 9 + 10 + ... + 200` you would have this:
Sums where the common difference is not equal to +1 can be represented with sum notation. Do note that sum/Sigma notation has its numbers going by one. If there is a common difference such as 3, a three would be added in front of the sum symbol. This would indicate that it is three times the sum or each number in the sum is multiplied by three. Here is one example:
The sum of negative even numbers from -2 to -20 can be represented as:
Pixabay Image Source
## Partial Sums For Arithmetic Sequences
Once the concept of sum notation is grasped, partial sums involves the computation of the sum into a number. A formula can be used to compute a sum.
Let's revisit `1 + 2 + 3 + 4 + ... + 100`.
Instead of adding from left to right with 1 all the way to 100, let's consider a smarter approach. Add the first number and last number together to obtain `1 + 100 = 101`. Now add the second number 2 and the second last number 99 to obtain 101. Continue with the 3 and 98 pair and work your way from the outside to the inside pair of numbers. All these number pairs have a sum of 101. How many number pairs are there that add up to 101? It is 50 which is half of 100.
For this example the sum of the counting numbers from 1 to 100 is equal to:
General Formula
The example can be extended further with the use of a formula. Suppose there is a sum of where is the first number in the series and is the last number in the series. The sum of these `n` numbers is given by the formula for
Add the first and last numbers together first. Then multiply this sum by half of how ever many numbers there are in the sum.
Pixabay Image Source
## Examples
That may be a lot of digest, let's add some numbers together with the formulas. Note that I do not have questions that involve finding numbers from an arithmetic sequence before using the partial sum formula. I do not cover sequences that start other than one nor index shifting.
Example One
What is the total of the even numbers starting from 2 to 222?
We have `2 + 4 + 6 + 8 + 10 + ... + 220 + 222`. This is the same as twice of `1 + 2 + 3 + 4 + 5 + 6 + ... + 110 + 111`. In math notation we can write:
As the question asks for the total of even numbers from 2 to 222 (inclusive), you can just use the partial sum formula.
There are 111 numbers from 2 to 222 (inclusive). The first term `t_1` is 2 and the last term is 222. Use these numbers in the partial sum formula.
Adding the even numbers starting from 2 to 222 gives 12 432.
Example Two
What is the total of the numbers from `-1, -2, -3, -4 to -55`?
The partial sum formula can be applied directly. There are 55 numbers.
Example Three - Algebraic Example
Determine the total of the values from `4x, 8x, 12x, 16x, ..., 996x to 1000x`.
If you want to represent this in sum notation, do make sure to common factor out the `4x` first.
The variable can be taken out of the sum as `x` is different from the sum index variable `k`.
Applying the partial sum formula gives:
Pixabay Image Source
Reference:
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( )
# Greatest Common Factors
In Math . . .
a is a common factor of b and c if a is a factor of both b and c. a is the greatest common factor of the two numbers if there are no common factors greater than a.
We'll write "the greatest common factor of a and b" as gcf(a, b).
In English . . .
Both phrases mean exactly what they seem to say: A number is a common factor of two other numbers if it's a factor of both of them. So, for example, 2 is a common factor of 12 and 18 but 4 isn't a common factor since it's a factor of 12 only. A little experimentation will show you that 3 and 6 are the only other factors that 12 and 18 have in common. Because 6 is the biggest of the common factors, it would be the greatest common factor of 12 and 18.
Quick Tip - Which One Did You Find?
At this point, I think it's worth repeating that factors are always smaller than the original number and multiples are always larger.
In our previous example, finding the greatest common factor was relatively straightforward since the numbers were small. It was easy enough to write out all the factors and pick the largest common one from the two lists. For larger numbers that approach would be impractical. Here's a procedure that will let you find the greatest common factor even for large numbers.
1. Write out the prime factorizations of both numbers.
2. List all the factors that they have in common.
3. If a factor appears more than once, add it to your list the smaller number of times.
4. Multiply the numbers in your list together and the result will be the greatest common factor.
If you're comfortable working with exponents they can make the last step of the procedure a little clearer.
1. Write out the prime factorizations of both numbers using exponents.
2. List all the factors that they have in common.
3. Give each factor the smallest exponent from the two factorizations.
4. Multiply the resulting numbers together.
# Example 1
Find the greatest common factor of 12 and 18.
1. Write the prime factorizations of each number. 18 = 2 · 3 · · 312 = 2 · 3 · 3 2. List all the factors that they have in common. 2, 3 3. Because 18 has only one 2 and 12 has only one 3, our greatest common factor will have one of each of those numbers. gcf(12, 18) = 2 · 3 = 6
# Example 3
Find the greatest common factor of 100 and 80.
We'll work this one out using the exponent method.
1. Write the prime factorizations of each number. 100 = 22 · 5280 = 24 · 5 2. List all the factors that they have in common. 2, 5 3. The smallest exponent of the 2's in the two factorizations is 2 and the smallest exponent of the 5's is 1 so we'll add those two the numbers in our list. 22, 51 If we multiply those numbers together, we'll get our greatest common factor. gcf(12, 18) = 22 · 5 = 20
# Example 2
Find the greatest common factor of 18 and 36.
1. Write the prime factorizations of each number. 18 = 2 · 3 · 336 = 2 · 2 · 3 · 3 2. List all the factors that they have in common. 2, 3 3. Because 18 has only one 2 our greatest common factor will have only one 2. On the other hand, both numbers have two 3's so our greatest common factor will also have two of them. gcf(18, 36) = 2 · 3 · 3 = 18
# Example 4
Find the gcf(495, 945)
Here's another method for finding the greatest common factor that's a little more visual. I'm going to start by writing out the prime factorizations with the common factors lined up.
495 = 3 · 3 · 5 · 11 945 = 3 · 3 · 3 · 5 · 7 gcf(495, 945) = 3 · 3 · 5 gcf(495, 945) = 45
To find the greatest common factor, I just took the numbers from each column that appeared in both of the rows and multiplied them together.
# Videos
Dynamic Tutorial - Finding the Greatest Common Factor
Directions: This solution has 4 steps. To see a description of each step click on the boxes on the left side below. To see the calculations, click on the corresponding box on the right side. Try working out the solution yourself and use the descriptions if you need a hint and the calculations to check your solution.
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Vectors and conics
This free course is available to start right now. Review the full course description and key learning outcomes and create an account and enrol if you want a free statement of participation.
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# 3 Dot product
## 3.1 Definition, properties and some applications
In the previous section we saw how to add two vectors and how to multiply a vector by a scalar, but we did not consider how to multiply two vectors. There are two different ways in which we can multiply two vectors, known as the dot product (or scalar product) and the vector product. They are given these names because the result of the first is a scalar and the result of the second is a vector. (We shall not consider vector products in this course.)
In the audio section we explain the definition of the dot product and investigate some of its basic properties. In addition, we see how the dot product can be used to find the angle between two vectors, to give a condition for two vectors to be orthogonal (that is, at right angles) and to find the projection of one vector onto another vector.
Listen to the audio below as you work through the following frames.
Click 'Play' to listen to Audio Clip 1 while viewing frames 1–7 in the main text below.
Interactive feature not available in single page view (see it in standard view).
Click 'Play' to listen to Audio Clip 2 while viewing frames 8–12 in the main text below.
Interactive feature not available in single page view (see it in standard view).
### Example 37
We use the formula for the dot product given in Frame 6 Section 3..
• (a)
• (b)
• (c)
### Example 38
• (a) When u = (2, −3), the length of u is
• so
• (b) When u = 5i + 12j, the length of u is
• so
### Example 39
In each case we use the formula for the angle between two vectors given in Frame 10 Section 3.1 , letting u denote the first vector of the pair, v the second vector and θ the angle between the two vectors.
• (a) Here u · v = (1, 4) · (5, 2) = 5 + 8 = 13,
• and
• Then
• so
• (b) Here u · v = (−2, 2) · (1, −1) = −2 −2 = −4,
• and
• Then
• so
• (c) Here
• and
• Then
• so
Click 'Play' to listen to Audio Clip 3 while viewing frames 13–17 in the main text below.
Interactive feature not available in single page view (see it in standard view).
### Example 40
To calculate the projections of u onto v and v onto u, we need the following:
Then the projection of v onto u is
and the projection of u onto v is
### Example 41
In each case we use the formula for the angle between two vectors given in Frame 10, letting u denote the first vector of the pair, v the second vector and θ the angle between the two vectors.
• (a) Here u · v = (1, 2, 0) · (3, −1, 2) = 3 − 2 + 0 = 1,
• and
• Then
• so
• (b)Here
• and
• Then
• so
M208_1
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## Calculus with Applications (10th Edition)
$(3x + 5)(x - 2)$
Factor $3x^{2}-x-10$ First set up the x's in the binomial. $(cx +or- n)(cx +or- n)$ To have $3x^{2}$ in the original equation the two $x$'s in the binomial multiply together. Thus the $x$'s have coefficients. 3 is a prime number so its only factors are 3 and 1. Therefore in one of the parentheses the $x$ has a coefficient of 3 and the other has a coefficient of 1 $(3x +or- n)(1x +or- n)$ Now we need to find n in the expression above both n values must multiply to -10 and add (when the expression is expanded) -1x. Also we must take into account the 3 coefficient. To start list out the factors of -10: -1,10 1,-10 -2,5 2,5 Think which of these factor pairs when one number is multiplied by three and subtracted by the other leaves us with -1 Immediately we can discount the 1,10 pairs because there would be too large of a difference, leaving us to choose between the -2,5 and the 2,-5 factor pairs. $2*3=6$ and $5-6=1$ so we know that the $2$ will not be in the same parentheses as the $3x$ $(3x +or- 5)(1x +or- 2)$ Last we need to figure out the signs within the parentheses. As mentioned before $5-6=1$ so we want 6 to be negative, thus we write $-2$ $(3x +or- 5)(1x - 2)$ And because we want 5 to be positive we write $+5$ $(3x + 5)(x - 2)$ To check expand the expression and will match the original equation
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How many fours are needed to represent numbers up to $N$?
The goal of the four fours puzzle is to represent each natural number using four copies of the digit $4$ and common mathematical symbols.
For example, $165=\left(\sqrt{4} + \sqrt{\sqrt{{\sqrt{4^{4!}}}}}\right) \div .4$.
If we remove the restriction on the number of fours, let $f(N)$ be the number of fours required to be able to represent all positive integers no greater than $N$. What is the asymptotic behaviour of $f(N)$? Can it be shown that $f(N) \sim r \log N$ for some $r$?
To be specific, let’s restrict the operations to the following:
• addition: $x+y$
• subtraction: $x-y$
• multiplication: $x\times y$
• division: $x\div y$
• exponentiation: $y^x$
• roots: $\sqrt[x]{y}$
• square root: $\sqrt{x}$
• factorial $n!$
• decimal point: $.4$
• recurring decimal: $. \overline 4$
It is easy to see that $f(N)$ is $O(\log N)$. For example, with four fours, numbers up to $102$ can be represented (see here for a tool for generating solutions), so, since $96 = 4\times4!$, we can use $6k-2$ fours in the form $(\dots((a_1\times 96+a_2)\times 96+a_3)\dots)\times96+a_k$ to represent every number up to $96^k$.
On the other hand, we can try to count the number of distinct expressions that can be made with $k$ fours. For example, if we (arbitrarily) permit factorial only to be applied to the digit $4$, and allow no more than two successive applications of the square root operation, we get $\frac{216^k}{18}C_{k-1}$ distinct expressions where $C_k$ is the $k$th Catalan number. (Of course, many of these expressions won’t represent a positive integer, many different expressions will represent the same number, and the positive integers generated won’t consist of a contiguous range from $1$ to some $N$.)
Using Stirling’s formula, for large $k$, this is approximately $\frac{864^k}{72k\sqrt{\pi k}}$. So for $f(N)$ to grow slower than $r\log N$, we’d need to remove the restrictions on the use of unary operations. (It is well-known that the use of logs enables any number to be represented with only four fours.)
Can this approach be extended to show that $f(N)$ is $\Omega(\log N)$? Or does unrestricted use of factorial and square roots mean that $f(N)$ is actually $o(\log N)$? Is the answer different if the use of $x\%$ (percentages) is also permitted?
• One more variant might include allowing concatenation (e.g. using $44$ or $444$). Dec 23, 2011 at 18:12
• @PatrickDaSilva, I’ve used \cdot for the decimal point because a central dot is the convention in the U.K. for writing decimals (e.g. $3\!\cdot\!\! 14159$, not $3.14159$). A dot above a digit is created using \dot and is for recurring decimals (e.g. $\frac{1}{3}=0\!\cdot\!\!\dot 3$). I know conventions in other countries differ (e.g. the use of commas for decimal points in some mainland European countries). Dec 23, 2011 at 18:26
• By the way, if you allow logarithms, you don't even need four fours: actually three fours are enough to represent any positive integer like $\log\bigl(\log 4/\log(\surd\surd\dots\surd4)\bigr)/\log 4$. Mar 4, 2012 at 12:02
• To my knowledge this problem is open, and difficult, even for the case of 1s instead of 4s, and allowing only addition and multiplication. Dec 24, 2012 at 22:49
• I'd prefer to remove the decimal and repeated-decimal operations. A "pure" representation of $N$ by fours shouldn't rely on the accident of the numerical base.
– Blue
Jan 6, 2013 at 10:02
I'm one of the authors of the paper referenced by David Bevan in his comment. The four-fours was one inspiration for that problem, although others have thought about it also. The specific version of the problem there looks at the minimum number of $$1$$s needed to represent $$n$$ where one is allowed only addition and multiplication but any number of parentheses. Call this $$g(n)$$. For example, $$g(6) \le 5$$, since $$6=(1+1)(1+1+1)$$, and it isn't hard to show that $$g(6)=5$$. Even in this limited version of the problem, the question is generally difficult even to get asymptotics.
In some sense most natural questions of asymptotic growth are somewhat contained in this question, since one can write any given $$k$$ as $$1+1+1...+1$$ $$k$$ times, and $$1=k/k$$. Thus starting with some $$k$$ other than $$1$$ (such as $$k=4$$), the asymptotics stay bounded within a constant factor, assuming that addition and division are allowed.
However, actually calculating this sort of thing for any set of operations is generally difficult. In the case of integer complexity one has a straightforward way of doing so, since if one calculates $$g(i)$$ for all $$i < n$$, calculating $$g(n)$$ is then doable. This doesn't apply when one has other operations generally, with division and substraction already making an algorithm difficult. In this case, one can make such an algorithm but exactly how to do so is more subtle. In fact, as long as one is restricted to binary operations this is doable (proof sketch: do what you did to look at all distinct expressions).
Adding in non-binary operations makes everything even tougher. Adding in square roots won't make things that much harder, nor will adding factorial by itself. The pair of them together makes calculating specific values much more difficult. My guess would be that even with factorial, square root and the four binary operations there are numbers which require arbitrarily large numbers of $$1$$s, but I also suspect that this would be extremely difficult to prove. Note that this is already substantially weaker than what you are asking- whether the order of growth is of $$\log n$$. Here though square roots probably don't alter things at all; in order for it to matter one needs to have a lot numbers of the form n^2^k with surprisingly low complexity. This seems unlikely.
• The series with $1$'s is oeis.org/A005245 I didn't find any asymptotics there or at the easier linked oeis.org/A061373 but that one shows that the question is $O(\log N)$ as here for m=2^p3^q, a(m)=2p+3q Feb 26, 2013 at 3:09
• For the case of just + and *, the growth is known to be of O(log n) as the correct level. This and related details are discussed in the paper that Bevan referenced. Whether it is asymptotic to c log n for some constant c is open. Feb 26, 2013 at 3:18
• @user7530 Er, sorry, when I wrote f(n) I meant to define g(n). I had written it with f(n), then realized that using the same letter as the OP wasn't a good idea. I tried to go and make them all consistent but missed a few. Should be ok now. Feb 26, 2013 at 4:04
You can get $$103$$ with five $$4$$s as $$\frac {\sqrt{\sqrt{\sqrt{4^{4!}}}}+4+\sqrt{.\overline4}}{\sqrt{.\overline4}}=103$$
For four $$4$$s, we have $$\dfrac {44}{.\overline 4}+4=103$$.
In fact, $$113$$ is the first number I can't get with four $$4$$s.
• In reply to not knowing how to get the math notation: You can search for LaTeX tutorials and get most of the notation right there; it also doesn't hurt to click on "edit" in other people's posts just to see how they are doing it (you don't actually have to edit anything). Oct 20, 2013 at 3:25
• (And welcome to the site!) Oct 20, 2013 at 3:26
• Thanks! Incidentally, my example actually needs five 4s, but I know I've done it with four. Oct 20, 2013 at 23:42
• Ah, mea culpa, most results of cut-the-knot.org/arithmetic/funny/4_4.shtml are not applicable because it often uses the floor function. But they give 114 = 44/.4 + 4. Sadly, @Eliseod'Annunzio solution for 113 isn't applicable because % isn't explicitely allowed as well.
– Cœur
Sep 30, 2018 at 20:18
• The % formula gives 103 not 113. Dec 22, 2018 at 20:13
Go here to look at a YouTube video discussing the matter. It proves that for all $$n\in\mathbb{Z}^+$$,
$$\log_{\frac 12}(\log_4\sqrt [n]{4}) = n.$$
But notice that $$\dfrac 12 = \dfrac 24 = \dfrac{\sqrt{4}}{{4}}$$ so...
Edit: By $$n^\text{th}$$ root, I mean $$n$$ square roots. So for example, $$\sqrt [2] {4} = \sqrt{\sqrt{4}}$$ (because it is simple to write it this way).
• It was already aknowledge in the question itself that log was enabling any number to be represented with only four fours.
– Cœur
Sep 30, 2018 at 20:28
• @Cœur where, exactly? Is that expressed in the big-$O$ notation? I ain't familiar with that. Sep 30, 2018 at 20:46
• Last sentence before the final quote: "(It is well-known that the use of logs enables any number to be represented with only four fours.)"
– Cœur
Sep 30, 2018 at 20:56
It can't be $o(\log N)$ for any finite set of binary operations. For a set of operations of size $k$, you can only have of order $k^N$ legal strings, so you can't represent more numbers than that.
• Not all of the operations are binary. For example, $$4!!!!\cdots !!!$$ for any number of factorials, can be written with one four Jan 11, 2015 at 16:50
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# Derivative of 1/x: Formula, Proof by First Principle
The derivative of 1/x is equal to -1/x2. The function 1/x (1 divided by x) is the reciprocal of x. In this article, we will find the derivative of 1/x using the power rule, product rule, and the definition of derivatives.
## What is the Derivative of 1/x?
Explanation:
At first, we will evaluate the derivative of 1/x by the power rule of derivatives. We need to follow the below steps.
Step 1: First, we will express 1/x as a power of x using the rule of indices. So we have
1/x = x-1
Step 2: Now, we will apply the power rule of derivatives: $\frac{d}{dx}$(xn)=nxn-1. Thus we get that
$\frac{d}{dx}(1/x)=\frac{d}{dx}(x^{-1})=-1 \cdot x^{-1-1}$
Step 3: Simplifying the above expression, we obtain that
$\dfrac{d}{dx}(\dfrac{1}{x})=-1 \cdot x^{-2}$
⇒ $\dfrac{d}{dx}(\dfrac{1}{x})=-1 \cdot \dfrac{1}{x^2}$
⇒ $\dfrac{d}{dx}(\dfrac{1}{x})=\dfrac{-1}{x^2}$
So the derivative of 1/x by the power rule is -1/x2 and this is obtained by the power rule of derivatives.
Derivative of esin x : The derivative of esin x is cos x esin x.
Integration of mod x : The integration of mod x is -x|x|/2+c
Now, we will find the derivative of 1/x by the first principle.
## Derivative of 1/x from first principle
Let f(x)=1/x. Applying the first principle of derivatives, we get that
$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$
Thus, the derivative of 1/x by first principle is equal to
$\dfrac{d}{dx}(\dfrac{1}{x})$ = limh→0 $\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}$
= limh→0 $\dfrac{\frac{x-x-h}{x(x+h)}}{h}$
= limh→0 $\dfrac{-h}{hx(x+h)}$
= – limh→0 $\dfrac{1}{x(x+h)}$
= $-\dfrac{1}{x(x+0)}$
= -1/x2
This shows that the derivative of 1/x is -1/x2 and this is obtained by the first principle of derivatives.
We know that the product rule of derivatives is $\frac{d}{dx}(fg)=f \frac{dg}{dx}+ g \frac{df}{dx}$. Using this rule, we will now find the derivative of 1/x.
## Derivative of 1/x by product rule
Let us put z=1/x. This implies that
zx=1
Differentiating with respect to x, we get that
$\dfrac{d}{dx}(zx)=\dfrac{d}{dx}(1)$
⇒ $z\dfrac{d}{dx}(x)+x\dfrac{d}{dx}(z)=0$ (by the product rule of derivatives)
⇒ $z\cdot 1+x\dfrac{dz}{dx}=0$
⇒ $x\dfrac{dz}{dx}=-z$
⇒ $\dfrac{dz}{dx}=-\dfrac{z}{x}$
⇒ $\dfrac{dz}{dx}=-\dfrac{1}{x^2}$ as z=1/x
So the derivative of 1/x by the product rule is equal to -1/x2.
Derivative of root x: The derivative of √x is 1/2√x
Derivative of cube root of x: The derivative of the cube root of x is 1/(3x^{2/3})
Derivative of sin inverse x: The derivative of sin-1 x is 1/√(1-x2)
Derivative of sin 3x: The derivative of sin 3x is 3cos 3x
### Application of Derivative of 1/x
As an application of the derivative of 1/x, we will now find the derivative of 1/log x. We will use the chain rule of derivatives: $\frac{du}{dx}=\frac{du}{dz} \cdot \frac{dz}{dx}$
Question: What is the derivative of 1/logx.
Solution:
Let z=log x. So we have dz/dx=1/x
Now, $\dfrac{d}{dx}(\dfrac{1}{\log x})$
$=\dfrac{d}{dx}(\dfrac{1}{z})$
$=\dfrac{d}{dz}(\dfrac{1}{z}) \cdot \dfrac{dz}{dx}$ (by the chain rule)
$=-\dfrac{1}{z^2} \cdot \dfrac{1}{x}$ (by the above formula of the derivative of 1/x)
$=-\dfrac{1}{x(\log x)^2}$ as z=log x.
## FAQs
Q1: What is the derivative of 1/x?
Answer: The derivative of 1/x is -1/x2.
Q2: What is the antiderivative of 1/x?
Answer: The anti-derivative of 1/x is ln x.
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# Conversion of Fractions to Decimals Numbers – Definition, Methods, Examples | How to Convert Fractions to Decimals?
Students, who are looking for learning the ways on how to convert from fractions to decimals can find them all here? Conversion of Fractions to Decimal Numbers means you wish to present the fractional numbers as exact decimals. Then, look no further as we have listed the easy way to convert fraction to decimal in a detailed way here.
Before going deep into the concept, you need to understand first the concept of decimal and fractions. Refer to the entire article to know about the procedure on how to convert a fraction to a decimal number along with solved examples. Practice these questions on converting fractions to decimal numbers and improve your math skills.
Also, Refer:
### Fraction – Definition
A fraction is a part of the whole number and is expressed as the ratio of two numbers. Let us consider two numbers a, b and therefore the fraction is the ratio of those two numbers i.e. a/b where b≠0. Here a is termed the numerator and b is the denominator. We can perform different arithmetic operations on decimal fractions almost like numbers.
There are different types of fractions namely Proper Fractions, Improper Fractions, Mixed Fractions, etc. A fraction is also known as a rational number.
Examples:Â 6/7, 12/7, 125/165.
### Decimal – Definition
In algebra, decimals are defined as a number, whose whole number part and the fractional part are separated by a decimal point. The dot present in a decimal number is named a decimal point. For example, 14.5 is a decimal number. Here, 14 is the whole number part and 5 is the fractional part.
Examples: 3. 45, 5.67, 8.28, etc.
### How to Convert Fraction to Decimal?
Converting fractions to decimals, we have different methods are mentioned here.
Method 1: One of the easy ways to convert fractions to decimals is to simply Divide the numerator by the denominator
• Get the fraction or mixed fraction
• Convert the mixed fraction into a fraction, then divide the numerator of the fraction by the denominator
• While dividing add a dot to the quotient, once you reach a point the remainder will be lesser than the divisor.
Example: Convert 1/5 into decimal form
Solution: Given the value is 1/5,
After Conversion, the decimal value is,
1/5 = 0.2
Method 2: Multiply both numerators, denominators by a similar number
• Identify the fraction, we have to convert to the decimal number.
• Multiply the denominator of the fraction by some number to induce the denominator because of the multiple of 10.
• Find that number and multiply both the numerator and denominator by an equivalent number.
• Then, you’ll get the denominator as the multiple of 10.
• If the denominator of the given fraction is 10, 100, or 1000 respectively. We will make a dot after one place or two places or three places from the right side towards the left side.
Example: Convert 1/5 into the decimal form?
Solution: Given the fraction is 1/5
1/5 = (1 x 20)/(5x 20) = 20/100 = 0.2
Method 3: Using the Long Division Method.
• Find the fractional number.
• Divide the numerator of the fraction by denominator using the division method.
• Then, you will get the decimal number as a quotient.
Example: Convert 1/4 into the decimal form?
Solution: The given fraction is 1/4
Divide 1/4 using the long division.
Therefore, 1/4 is 0.25
Practice Math Online with Unlimited Questions provided in 5th Grade Math Activity Sheets and become a blossoming mathematician in no time.
### Converting Fractions to Decimals Examples
Example 1.
Convert 9/10 into a decimal?
Solution:
Given the value is 9/10
Now, we have to convert fraction numbers into decimal numbers, divide 9 by 10, we get the decimal form.
Thus, 9/10 = 0.9
Thus, the decimal form of 9/10 is 0.9.
Example 2.
Convert the fraction number 7(1/2) into a decimal?
Solution:
As given in the question, the mixed fraction value is 7(1/2)
Then, 7(1/2) = 17/2
Using the division method, the value is,
7(1/2) = 17/2 = 8.5.
Example 3.Â
What is the decimal value of 12/5?
Solution:
As given, the fraction value is 12/5
Now, we have to multiply both numerator and denominator by 2.
Then, the values are (12 x 2) / (5×2)
= 24/ 10
= 2.4
Therefore, the 12/5 in decimal is 2.4.
Example 4.
Convert the fraction 202/100 into decimals?
Solution:
The given fraction is 201/100
The denominator is 100, add the point after two digits from the left side.
So, 201/00 = 2.01.
Example 5.
Find the decimal of 3/4 using the long division method?
Solution:
Given the value is 3/4
Using the division method finds the decimal value.
The below shows how to perform the long division method,
Therefore, the decimal value of the given number is 0.75.
Example 6.
Let 34/100 is a fraction value. Write into decimal value.
Solution:
Given the value,
Now, to convert the value 34/100 into a decimal value
We have to check for the number of zeros in the denominator.
Hence the denominator value is 100, there are 2 zeros present within the denominator.
Therefore, we have to shift the decimal point by 2 points within the numerator. Then the final result is 0.34
Example 7.
Convert 7/8 into decimals.
Solution:
As given in the question, the value.
Already the given value is a fractional value. Here 7 has to be divide by 8.
As 7 is less than 8, it cannot be divided by 8.
So, we have to write the value as 7.00, it can be divided by 8.
70 can be divided by 8 for 8 times
It will write the quotient as 0.8, on further division, we will get the final result as 0.875
Example 8.
Convert 5/8 into decimals.
Solution:
Given the value is 5/8,
The value is in fractional form. But 5 has to be divided by 8.
As 5 is less than 8, it cannot be divided by 8.
So, we will write the value as 5.000 which can be divided by 8. The given figure shows finding fraction value using the long division method
50 can be divided by 8 for 6 times.
Therefore we write the quotient as 0.6, on further division, we get the final result as 0.625
### FAQ’s on Converting Fractions to Decimals Numbers
1. What is a Decimal?
A Decimal Number is a number that has a dot between the digits. In other words, we are able to say that decimals are nothing but fractions with denominators as 10 or multiples of 10.
2. How to Convert a Fraction to Decimal?
We can convert the Fraction to the Decimal of a given number to divide the numerator by the denominator value.
3. What are the methods for conversion of fractions to decimals?
There are 3 methods for converting fractions to decimals is.
1. Divide the Numerator by Denominator
2. Multiply both numerators, denominators by a similar number
3. Using the Long Division Method.
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How Are Multiplication And Division Related? – tntips.com
# How Are Multiplication And Division Related?
## How Are Multiplication And Division Related?
Multiplication and division are closely related, given that division is the inverse operation of multiplication. … This is because when we multiply two numbers (which we call factors), we get a result that we call a product. If we divide this product by one of the factors, we get the other factor as a result.Jul 2, 2021
## How are multiplication and addition related?
Addition is the process of combining a number of individual items together to form a new total. Multiplication, however, is the process of using repeated addition and combining the total number of items that make up equal-sized groups.
## Do multiplication and division have the same rules?
As division is the inverse of multiplication, the rules for division are the same as the rules for multiplication. So when multiplying and dividing positive and negative numbers remember this: If the signs are the same the answer is positive, if the signs are different the answer is negative.
## How is multiplication related to dividing fractions?
Dividing fractions is very similar to multiplying fractions, you even use multiplication. The one change is that you have to take the reciprocal of the divisor. Then you proceed with the problem just as if you were multiplying.
## How are the 4 operations related?
The four operations are addition, subtraction, multiplication and division.
## What is the relationship between addition subtraction multiplication and division?
Repeated addition is multiplication. Subtraction is the opposite of addition. Multiplication is repeated addition. Division is the opposite of multiplication.
## When we multiply the answer is called?
In multiplication, the numbers being multiplied are called factors; the result of the multiplication is called the product.
## Are multiplication and division the same in order of operations?
Order of operations tells you to perform multiplication and division first, working from left to right, before doing addition and subtraction. Continue to perform multiplication and division from left to right.
## What is the relationship between division and fraction?
E. 1 Interpret a fraction as division of the numerator by the denominator (a/b = a ÷ b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem.
## What is a multiplication and division fact family?
Definition. Fact family: It is a set of four related multiplication and division facts that use the same three numbers. For example: The fact family for 3, 8 and 24 is a set of four multiplication and division facts. Two are multiplication facts, whereas the other two are division facts.
See also Why Does College Tuition Keep Rising?
## How is division related to other operations?
Multiplication and division are closely related, given that division is the inverse operation of multiplication. … This is because when we multiply two numbers (which we call factors), we get a result that we call a product. If we divide this product by one of the factors, we get the other factor as a result.
## What are the 4 basic concepts of mathematics?
–addition, subtraction, multiplication, and division–have application even in the most advanced mathematical theories.
## Is 7 a term?
The 5x is one term and the 7y is the second term. The two terms are separated by a plus sign. + 7 is a three termed expression.
## How are subtraction and division related?
Repeated subtraction is a way of teaching about division. It is the repeated subtraction of the same number from a large number down to zero. … Repeated Subtraction is a method that subtracts the equal number of items from a group, also known as division.
## Why is multiplication and division important?
Builds the Basic Mathematical Blocks and Supports More Complex Tasks. As part of each person’s mathematical ‘toolbox’, multiplication is a fundamental skill which will enable your child to succeed in what can often be considered a daunting subject.
## What exactly is multiplication?
In math, to multiply means to add equal groups. When we multiply, the number of things in the group increases. The two factors and the product are parts of a multiplication problem. In the multiplication problem, 6 × 9 = 54, the numbers 6 and 9 are the factors, while the number 54 is the product.
## Why is multiplication called product?
A product is the result of carrying out the mathematical operation of multiplication. When you multiply numbers together, you get their product. The other basic arithmetic operations are addition, subtraction and division, and their results are called the sum, the difference and the quotient, respectively.
See also What Was The Tenth Amendment?
## Do we divide first or multiply?
All expressions should be simplified in this order. The only exception is that multiplication and division can be worked at the same time, you are allowed to divide before you multiply, and the same goes for addition and subtraction. However, multiplication and division MUST come before addition and subtraction.
## Is multiplication done before addition?
Multiplication and division must be completed before addition and subtraction. 2 + 3 x 7 = 2 + 21 = 23 is the correct answer to the above question.
## Do you multiply first if no brackets?
Because 4 × 4 = 16 , and once there are no parentheses left, we proceed with multiplication before addition. This set of parentheses yields yet another answer. So, when parentheses are involved, the rules for order of operations are: Do operations in parentheses or grouping symbols.
## How are the fractions related?
Fractions and decimals are similar because they both are ways to express partial numbers. Additionally, fractions can be expressed as decimals by performing the division of the ratio. … Decimals can also be expressed as fractions in terms of tenths, hundredths, thousandths and so on.
## Is dividing the same as fraction?
Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second fraction. The first step to dividing fractions is to find the reciprocal (reverse the numerator and denominator) of the second fraction.
## Examples relating multiplication to division | 3rd grade | Khan Academy
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How do you write two and a half?
How do you write two and a half?
it will be five over two . so 2 and a half can be written as 5 over 2 .
How do you write 2 and a half in a fraction?
You would say "two and one half." The other format is an improper fraction where the numerator is greater than the denominator (5/2). Mathematicians would say that is five halves. You will find both types of fractions in your problems. Both of those examples represent the same value (2 1/2 = 5/2).
How do you write 1 and a half?
We commonly write it as “half” omitting “a” before it. One and a half means one full and one more half making it 1 ½ , ( decimal form is 1.5 i.e one point five.)
How do you write half in math?
Properties of halvesIn decimal numbers, half is denoted by 0.5.We can find half of a number multiplying it by ½. For example, in the figure, half of the group of 6 chips is obtained by 6 × 1⁄2 =3.
What is the answer of ½ ½?
One half plus one half equals 1. Mathematically, one half is equal to 1/2. We can only add fractions that have the same denominator.
What is 2.3 simplified?
2.3 is 2+0.3 . Add 310 to 2 . If you don't want a mixed number, you can convert 2310 to an improper fraction.
How do you turn 2.25 into a fraction?
As a proper fraction 2.25 is 214 2 1 4 as an improper fraction it is 94 . To convert 2.25 to a fraction we set aside the 2 units for the moment and focus on the 0.25 that's left. This can be converted into a fraction of 25100 as the decimal is read as "twenty five hundredths".
What is the half of one and half?
Half of 1 1/2 cups is 3/4 cups.
How many halves are there in 2?
There are 2 halves in a whole. There are 3 thirds in a whole.
What does 2.3 look like as a fraction?
2.3 is 2+0.3 . Add 310 to 2 . If you don't want a mixed number, you can convert 2310 to an improper fraction.
How do you write 2.25 as a mixed number?
The GCF of 25 and 100 is 25. 2.25 is written as 214 in simplest form.
What is Half of \$42?
21Half of 42 is 21.
What is 32 split in Half?
1632 divided by 2 is 16.
Which shape is divided into halves?
One way to divide a rectangle into halves is this way. What have we learned in this video? We've learned that when we divide shapes into two equal parts, we call each part a half.
How many quarters are there in a half?
Two quarters make a half.
Is 2.3 A whole number?
Fractions and decimals are not included in the set of integers. For example, 2, 5, 0, -12, 244, -15, and 8 are all integers. The numbers such as 8.5, 2.3, 20% are not integers.
How do you turn 2.3 into a percent?
To convert fraction to a per cent, you just need to multiply the fraction by 100 and reduce it to per cent. Convert 2/3 to a per cent. Therefore, the solution is 66%.
Is 2.3 Repeating a rational number?
Repeating decimals are considered rational numbers because they can be represented as a ratio of two integers.
What is the reciprocal of 2.3 in fraction form?
To find the reciprocal of a whole number, just convert it into a fraction in which the original number is the denominator and the numerator is 1. The reciprocal of 2/3 is 3/2. The product of 2/3 and its reciprocal 3/2 is 1.
How do you work out 15 divided by 2?
Using a calculator, if you typed in 15 divided by 2, you'd get 7.5.
How do you solve 25 divided by 2?
Place this digit in the quotient on top of the division symbol. Multiply the newest quotient digit (2) by the divisor 2 . Subtract 4 from 5 . The result of division of 25÷2 25 ÷ 2 is 12 with a remainder of 1 .
How do you do the 42 split?
Half of 42 is 21.
What is half of \$63?
31 1/2Half of 63 is 31 1/2 or 31.5.
What is 2 and a quarter as an improper fraction?
9/4So for 2-1/4, it's 4. Which means that the mixed fraction 2-1/4 is equivalent to the improper fraction 9/4.
What is half divided by two?
In the following intermediate step, it cannot further simplify the fraction result by canceling. In other words - one half divided by two = one quarter.
Can 2 be divided by 2?
Dividing is the same as multiplying by the reciprocal, the reciprocal of 1/2 is 2, so dividing by 2 is the same as multiplying by 2. Short answer… it is not, it is 2.
How do you divide 2?
To divide a number by 2 using repeated subtraction, subtract 2 from it over and over again, till you reach 0. The number of times you subtract is the answer to the division problem.
Is negative number can be odd?
What happened Evan Sabahnur?
What is the solution of log6 1 36?
Why is Kidd after Shanks?
How many minutes walk a mile?
Most people can expect to walk a mile in 15 to 22 minutes, according to data gathered in a 2019 study spanning five decades. The average walking pace is 2.5 to 4 mph, according to the Centers of Disease Control and Prevention.
How long is 25MB in minutes?
25 MB is 200 megabits. 30 minutes is 1800 seconds.
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# 5.7: Simple Interest
Difficulty Level: At Grade Created by: CK-12
## Introduction
Student Council Investment
With a goal of boosting student attendance at football games, the student council has decided to invest a portion of their savings for middle school decorations. They figure that when the games are happening that they can decorate the middle school with balloons, banners and flyers.
“I think it will help make it a priority for students,” Jeremy said at the weekly student council meeting.
“It will be a lot of fun too. We could even host a pep rally to help get the kids charged up,” Candice suggested.
“We put 4000 in the bank in the sixth grade. Now that we are \begin{align*}8^{th}\end{align*} graders that money has been sitting in the bank for two years at a 4% interest rate,” Jeremy explained. Candice began working out the math in her head. If they did put4000 in the bank for two years and they had a 4% interest rate, then there definitely is more money in there now. She began to complete the calculations in her head.
Do you have an idea how to figure this out? This problem involves principal, interest rates and time. This lesson will teach you all about calculating simple interest. Pay close attention and you will see this problem again at the end of the lesson.
What You Will Learn
In this lesson you will learn how to use the following skills.
• Use the simple interest formula \begin{align*}I = Prt\end{align*} to find an interest rate.
• Use the equation to find the time required to earn a given amount.
• Use the equation to find the new balance after a given time.
• Solve real-world problems involving simple interest.
Teaching Time
I. Use the Simple Interest Formula \begin{align*}\underline{I = Prt}\end{align*} to Find an Interest Rate
Money is a necessary part of everyday life, and as you get older, your relationship to money will change. In this lesson, we will explore some of the ways in which you will relate to money as you get older.
Saving money and making wise investments will be an important part of your financial planning. Part of making investing is earning interest. When you save money in the bank, the bank uses that money for its own investments. In return for using your money, the bank pays you a certain percent. This percent is your interest. Interest is the percent that a bank pays you for using keeping your money in their bank.
Banks compete with each other for your money because they want you to put your money in their bank. They try to give you the best “interest rate” that they can. This means that they will pay you a greater percentage than another bank to try to get your business. The greater the interest rate that they pay you; the more likely you are to invest your money with them in a savings account. The more money you save, the more they have to invest. They publish an interest rate \begin{align*}r\end{align*} which tells you what percent they will pay you per year \begin{align*}t\end{align*}. The principal, \begin{align*}p\end{align*} is the amount of money that you have put into the bank.
You can use this information in the formula \begin{align*}I = prt\end{align*} in order to calculate the interest that you will earn on your principal \begin{align*}p\end{align*}.
Take a few minutes and write this formula in your notebook.
Now let’s look at how we can use this formula to calculate interest.
Example
You invest 5,000 in a bank for 2 years at a 4% interest rate. What is the interest you have earned after this time? We start by looking at the given information. Then use the formula to calculate interest. \begin{align*}p = 5000, r = .04, t = 2\end{align*}. Use the formula to calculate interest. \begin{align*}I &= prt\\ I &= 5000 \cdot .04 \cdot 2\\ I &= 400\end{align*} The bank will pay you400 in interest over two years at that rate.
Example
An investor places 15,000 in a savings account that pays 4.5% interest. She will leave the money there for 6 years. What will her interest be? \begin{align*}p &= 15000, r = .045, t = 6\\ I &= prt\\ I &= 15000 \cdot .045 \cdot 6\\ I &= 4050\end{align*} Her interest after 6 years will be4,050.
Now that you know how to calculate the interest, let’s look at figuring out the time.
II. Use the Equation to Find the Time Required to Earn a Given Amount
Many investors may have specific goals—they want to earn a certain amount of interest on their investments. Because of this, they need to figure out the time that it takes to earn a certain amount of money. The formula \begin{align*}I = prt\end{align*} is an equation. We can use the Multiplication Property of Equations to solve for \begin{align*}t\end{align*} if we know \begin{align*}I, r\end{align*}, and \begin{align*}p\end{align*}.
Let’s look at an example.
Example
Mrs. Duarte has $20,000 to invest. She wants to earn$10,000 in interest. She is considering a savings and loans bank that is offering her 5.6% interest per year. For how long will she have to leave her money in the bank in order to reach her goal of 10,000? Start by looking at the given information. \begin{align*}I = 10000, p = 20000, r = .056\end{align*} Solve for \begin{align*}t\end{align*}. Next, we substitute the given values into the formula and solve the equation. \begin{align*}I &= prt\\ 10000 &= 20000 \cdot .056 \cdot t\\ 10000 &= 1120t\\ \frac{10000}{1120} &= \frac{1120t}{1120}\\ 8.93 &= t\end{align*} She will have to leave her money in the bank for nearly 9 years. Exactly! We are using what we have learned about solving equations to figure out missing information regarding interest and banking. Example A bank is offering an interest rate of 4.75%. How long would it take to earn500 if you invested 12,000 in the bank? \begin{align*}I = 500, p = 12000, r = .0475\end{align*} Solve for \begin{align*}t\end{align*}. \begin{align*}I &= prt\\ 500 &= 12000 \cdot .0475 \cdot t\\ 500 &= 570t\\ \frac{500}{570} &= \frac{570t}{570}\\ .88 &= t\end{align*} It would take .88 years or about \begin{align*}10\frac{1}{2}\end{align*} months. Notice that we have a decimal part of a year here. We can estimate how many months that is based on the decimal. A whole would be represented by 12 months or one year. III. Use the Equation to Find the New Balance after a Given Time You can use the simple interest formula \begin{align*}I = prt\end{align*} to find any of the missing variables if you are given values of the others. We have used it to solve for \begin{align*}I\end{align*} and \begin{align*}t\end{align*}. Of course, once the bank pays you interest, your account balance grows. You start of with your principal \begin{align*}p\end{align*} and then you add your interest \begin{align*}I\end{align*}. Now let’s see how much a bank balance would be after a given time at a given interest rate. Example Jessica invests3,000 in a credit union at an interest rate of 3.9%. She leaves the money there for 5 years. What is her balance after that time?
To answer this question, we will need to do two things. First, we will need to figure out the amount of the interest. Then we can add this amount to the principal that Jessica first invested. This will give us the new balance.
First find the interest that she earned:
\begin{align*}p &= 3000, r = .039, t = 5\\ I &= prt\\ I &= 3000 \cdot .039 \cdot 5\\ I &= 585\end{align*}
She earned $585 in interest. Her principal was$3,000. How much does she have now?
\begin{align*}585 + 3000 = 3585\end{align*}
She has 3,585. This is the new balance. Now let’s look at using this formula with some other real – world problems. IV. Solve Real – World Problems Involving Simple Interest Using the simple interest formula \begin{align*}I = prt\end{align*}, we can calculate the interest rate \begin{align*}r\end{align*} if we are given the \begin{align*}I, p\end{align*} and \begin{align*}t\end{align*} values. As before, we will substitute the known values and then use inverse operations to find the missing value. Example A nurse put22,000 in the bank 15 years ago. She has earned 21,450 in interest—nearly as much as her initial investment. What was the interest rate that the bank was paying her? \begin{align*}I &= 21450, p = 22000, t = 15\\ I &= prt\\ 21450 &= 22000 \cdot r \cdot 15\\ 21450 &= 330000r\\ \frac{21450}{330000} &= \frac{330000r}{330000}\\ .065 &= r\end{align*} Because we are looking for a percent-an interest rate, we have to change the decimal to a percent. .065 = 6.5% The bank was paying 6.5%. Example A bank uses the same formula \begin{align*}I = prt\end{align*} to calculate the interest that you owe on a credit card. Normally, the interest rates are substantially higher than for a savings account. If you charge7,000 on a credit card and you bank charges you 15.9%, how much would you owe after a year?
First, look at the given information and then use the formula.
\begin{align*}p &= 7000, r = .159, t = 1\\ I &= prt\\ I &= 7000 \times .159 \times 1\\ I &= 1113\end{align*}
Next, we add the interest to the balance. Because you owe on a credit card, this gives us the new balance that you need to pay.
\begin{align*}1113 + 7000 = 8113\end{align*}
Candice began working out the math in her head. If they did put 4000 in the bank for two years and they had a 4% interest rate, then there definitely is more money in there now. She began to complete the calculations in her head. Remember, there are two parts to your answer. Solution to Real – Life Example Now we need to figure out the interest and the final balance in the student council bank account. First, let’s find the amount of the interest. \begin{align*}I &= PRT\\ I &= (4000)(.04)(2)\\ I &= \320.00\end{align*} Next, we add this to the original amount invested. \begin{align*}\4000 + \320 = \4320.00\end{align*} This is the new balance in the student council account. ## Time to Practice Directions: Use the simple interest formula \begin{align*}I = prt\end{align*} to solve for the Interest. 1. Find \begin{align*}I\end{align*} if \begin{align*}p = 62,300, r = .0525, t = 14\end{align*}. 2. Find \begin{align*}I\end{align*} if \begin{align*}p = 9800, r = .028, t = 9\end{align*}. 3. Find \begin{align*}I\end{align*} if \begin{align*}p = \600, r = .05, t=8\end{align*} 4. Find \begin{align*}I\end{align*} if \begin{align*}p = \2300, r = .06, t=12\end{align*} 5. Find \begin{align*}I\end{align*} if \begin{align*}p = \5500, r = .08, t=7\end{align*} 6. Find \begin{align*}I\end{align*} if \begin{align*}p = \400, r = .05\end{align*} and \begin{align*}t=5\end{align*} 7. Find \begin{align*}I\end{align*} if \begin{align*}p = \700, r = .03\end{align*} and \begin{align*}t=9\end{align*} 8. Find \begin{align*}I\end{align*} if \begin{align*}p = \500, r = .06\end{align*} and \begin{align*}t=12\end{align*} 9. Find \begin{align*}I\end{align*} if \begin{align*}p = \800, r = .09\end{align*} and \begin{align*}t=7\end{align*} 10. Find \begin{align*}I\end{align*} if \begin{align*}p = \950, r = .06\end{align*} and \begin{align*}t=4\end{align*} Directions: Find the new interest and then find the new balance with the given information. There are two steps to solving these problems. 1. \begin{align*}p = 43000, r = .0365, t = 11\end{align*} 2. \begin{align*}p = 7000, r = .079, t = 4\end{align*} 3. \begin{align*}p = 8000, r = .06, t = 3\end{align*} 4. \begin{align*}p = 18000, r = .04, t = 5\end{align*} 5. \begin{align*}p = 25000, r = .05, t = 3\end{align*} 6. \begin{align*}p = 3000, r = .05, t = 7\end{align*} 7. \begin{align*}p = 12000, r = .04, t = 5\end{align*} 8. \begin{align*}p = 9000, r = .06, t = 10\end{align*} 9. \begin{align*}p = 7500, r = .03, t = 8\end{align*} 10. \begin{align*}p = 27500, r = .04, t = 6\end{align*} Directions: Solve each problem. 1. Antonio receives a savings bond from his uncle for his birthday. It will be worth1,000 in \begin{align*}8\frac{1}{2}\end{align*} years but only cost his uncle $500. What was the interest rate paid on the bond? 2. Henrietta is thinking of investing a$10,000 inheritance in a savings account for 6 years that pays 4.3% interest. She owes $7,200 on a credit card which is charging her 11.9% interest. In how much time would the interest on the credit card be the same as the interest in the savings account after 6 years? 3. A mother wants to save money for her child’s education. Her child just turned 1 year old. She wants to have at least$80,000 in the bank by the time she is 18. She has placed \$35,000 in the bank at 6.5%. Will she reach her goal?
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# Exponents, Exponential Notation, and Scientific Notation
Related Topics:
More Lessons for Algebra I
Math Worksheets
A series of free Basic Algebra Lessons.
In this lesson, we will learn
• an introduction to exponents and rules of exponents
• multiplication and division properties of exponents
• zero and negative exponents
• scientific notation
• simplifying expressions with exponents
The following figure gives the Rules of Exponents. Scroll down the page for examples and solutions.
### Introduction to Exponents - Rules of Exponents
Since we often see exponents throughout all math courses, it is important to understand the rules of exponents. We need to understand how to distribute, add, multiply and divide exponents in order to simplify expressions or manipulate equations that have exponents. The rules of exponents, like those involving multiplication of terms, are important to learn and will be used throughout Algebra I and II and Calculus.
How to understand the vocabulary of exponents.
Understanding and working with exponents by examining academic vocabulary and examples of various problems involving exponents. (exponent, power, base number, factor form, exponent form, and standard form)
### Multiplication and Division Properties of Exponents
There are different rules to follow when multiplying exponents and when dividing exponents. If we are multiplying similar bases, we simply add the exponents. If we are dividing, we simply subtract the exponents. If an exponent is outside the parentheses, it is distributed to the inside terms. It's important to understand the rules of multiplying exponents so that we can simplify expressions with exponents.
The following video is on Multiplying Exponents and the Exponent Rule The following video is on Dividing Powers in terms of the Exponent Rule. The following video is on the Exponent Rule of a Power of a Power.
### Zero and Negative Exponents
It's important to understand what It means to have negative exponents and zero exponents. Negative exponents put the exponentiated term in the denominator of a fraction and zero exponents just make the term equal to one. We can use negative exponents for canceling with positive exponents while solving equations or simplifying expressions, although we need to keep in mind the rules of multiplying exponents.
This video explains zero exponents This video explains negative exponents.
### Scientific Notation
In the applied sciences, scientific notation is often employed as a method of notation for ease of writing and reading. When dealing with real world situations, the numbers we get as solutions are rarely whole numbers and scientific notation gives us rules to follow when using ugly numbers that have a lot of decimal places. In order to understand scientific notation, we must also have a solid understanding of exponents.
### Simplifying Expressions with Exponents
We can use what we know about exponents rules in order to simplify expressions with exponents. When simplifying expressions with exponents we use the rules for multiplying and dividing exponents, and negative and zero exponents. In Algebra and higher math courses such as Calculus, we will often encounter simplifying expressions with exponents.
These videos provide examples of simplifying exponential expressions using a several exponent rules (positive exponents). These videos provide several examples of how to simplify exponential expressions containing negative exponents.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# 2011 AIME II Problems/Problem 8
## Problem
Let $z_1,z_2,z_3,\dots,z_{12}$ be the 12 zeroes of the polynomial $z^{12}-2^{36}$. For each $j$, let $w_j$ be one of $z_j$ or $i z_j$. Then the maximum possible value of the real part of $\displaystyle\sum_{j=1}^{12} w_j$ can be written as $m+\sqrt{n}$ where $m$ and $n$ are positive integers. Find $m+n$.
## Solution
The twelve dots above represent the $12$ roots of the equation $z^{12}-2^{36}=0$. If we write $z=a+bi$, then the real part of $z$ is $a$ and the real part of $iz$ is $-b$. The blue dots represent those roots $z$ for which the real part of $z$ is greater than the real part of $iz$, and the red dots represent those roots $z$ for which the real part of $iz$ is greater than the real part of $z$. Now, the sum of the real parts of the blue dots is easily seen to be $8+16\cos\frac{\pi}{6}=8+8\sqrt{3}$ and the negative of the sum of the imaginary parts of the red dots is easily seen to also be $8+8\sqrt{3}$. Hence our desired sum is $16+16\sqrt{3}=16+\sqrt{768}$, giving the answer $\boxed{784}$.
## Solution 2
The equation $z^{12}-2^{36}=0$ can be factored as follows: $$(z^6-2^{18})(z^6+2^{18})=0$$ $$(z^2-2^6)(z^2+2^6)({(z^2+2^6)}^2-z^2\cdot2^6)({(z^2-2^6)}^2+z^2\cdot2^6)=0$$ $$(z^2-2^6)(z^2+2^6)(z^2+2^6-z\cdot2^3)(z^2+2^6+z\cdot2^3) (z^2-2^6-iz\cdot2^3)(z^2-2^6+i z\cdot2^3)=0$$
Since this is a 12th degree equation, there are 12 roots. Also, since each term in the equation is even, the positive or negative value of each root is another root. That would mean there are 6 roots that can be multiplied by $-1$ and since we have 6 factors, that’s 1 root per factor. We just need to solve for $z$ in each factor and pick whether or not to multiply by $i$ and $-1$ for each one depending on the one that yields the highest real value. After that process, we get $8+8+2((4\sqrt{3}+4)+(4\sqrt{3}-4))$ Adding the values up yields $16+16\sqrt{3}$, or $16+\sqrt{768}$, and $16+768=\boxed{784}$.
-Solution by Someonenumber011.
## Solution 3
Using the diagram from solution 1, note that multiplying by $i$ is rotating counterclockwise by $\frac{\pi}{2}$. From there, we eventually get $16+768=\boxed{784}$. ~Lcz
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###### Norm Prokup
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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# Average Value of a Function - Problem 1
Norm Prokup
###### Norm Prokup
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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The formula for average value of f(x) from x=a to x=b is the integral of f(x) from a to b, divided by (b - a). b - a is the length of the interval over which you are finding an average value.
In order to calculate the average value, simply plug in values of a, b, and f(x) into the formula. Integrate f(x) using the usual rules or methods. Evaluate at the limits of integration a and b, and take the difference of the value of the integral at b and at a.
Let’s find the average value of a function. We have this formula for the average value. 1 over the length of the interval over which you’re averaging, times the area underneath the curve that you’re averaging. Let’s apply this to find the average value of f(x) equals root x on the interval from 0 to 9.
The integral on the right hand side of the formula represents this area here. And I need to divide that area by the length of this interval and that would give me the average value. Let’s calculate this up here.
Average value is 1 over 9 minus 0, that’s the width of the interval, times the integral from 0 to 9 of f(x). Let me just write in square root of xdx. So that’s 1/9 integral from 0 to 9. And let me change the square root of x to x to the ½. I want to represent it as a power because that’s the way I antidifferentiate it. So 1/9, and the antiderivative of x to the ½ is I raise this number by 1, I get 3/2 and I divide by 3/2 which is the same as multiplying by 2/3. It’s 2/3x to the 3/2, and I evaluate that from 0 to 9. I can actually pull this 1/9 in, it doesn’t really do much one way or the other what’s 9 to the 3/2, because but 2/27x to the 3/2 from 0 to 9.
Let me evaluate this at 9 first. 2/27 times 9 to the 3/2 minus 2/27 times 0 to the 3/2, that’s just going to be 0. What’s 9 to the 3/2? Because 9 is a perfect square I’m going to take the square root first I’ll get 3 and then I’ll cube the result and get 27. So this is 2/27 times 27 which is 2. 2 is the average value of the square root function over the interval form 0 to 9.
Let’s take a look at the graph and see what that looks like. The maximum value it reaches is 3 at 9, 2 is right about here, this is the average value. Think of it this way, if this were a tank of water, and this curve represented the surface of the water after the water settled, this is the level it would reach, the level 2.
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# If Kioko can run 6 miles in 36 minutes, how far can he run in 84 minutes at this same rate?
Apr 4, 2018
Kioko can run 14 miles in 84 minutes.
Here's how I did it:
#### Explanation:
To solve this, we set up a proportion:
$\frac{6}{36} = \frac{x}{84}$
Let $x$ be the distance Kioko can run in $84$ minutes.
Now, we use this method called cross multiplication:
As you can see from this image, we multiply $6$ with $84$ and $36$ with $x$:
$6 \cdot 84 = 36 x$
$504 = 36 x$
Now we divide both sides by $36$:
$x = 14$
Kioko can run 14 miles in 84 minutes.
Hope this helps!
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Graph of pair of Linear Equations
Chapter 3 Class 10 Pair of Linear Equations in Two Variables
Serial order wise
### Transcript
Question 3 The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically. Let the Cost of Apples per kg be Rs x & let Cost of grapes per kg be Rs y Given that 2 kg apples and 1 kg grapes cost Rs 160 2 × Cost per kg of apples + 1 × Cost per kg of grapes = 160 2x + y = 160 Also, 4 kg apples and 2 kg grapes cost Rs 300 4 × Cost per kg of apples + 2 × Cost per kg of grapes = 300 4x + 2y = 300 2(2x + y) = 2 × 150 2x + y = 150 Now, plotting equations 2x + y = 160 ...(1) 2x + y = 150 …(2) For Equation (1) 2x + y = 150 Let x = 50 2(50) + y = 150 100 + y = 150 y = 150 − 100 y = 50 So, x = 50, y = 50 is a solution i.e. (50, 50) is a solution Let x = 60 2(60) + y = 150 120 + y = 150 y = 150 − 120 y = 30 So, x = 60, y = 30 is a solution i.e. (60, 30) is a solution For Equation (2) 2x + y = 160 Let x = 50 2(50) + y = 160 100 + y = 160 y = 160 − 100 y = 60 So, x = 50, y = 60 is a solution i.e. (50, 60) is a solution Let x = 60 2(60) + y = 160 120 + y = 160 y = 160 − 120 y = 40 So, x = 60, y = 40 is a solution i.e. (60, 40) is a solution We will plot both equations on the graph
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# Solve the system
These can be very helpful when you're stuck on a problem and don't know how to Solve the system. We will give you answers to homework.
## Solving the system
In addition, there are also many books that can help you how to Solve the system. How to solve perfect square trinomial. First, identify a, b, and c. Second, determine if a is positive or negative. Third, find two factors of ac that add to b. Fourth, write as the square of a binomial. Fifth, expand the binomial. Sixth, simplify the perfect square trinomial 7 eighth, graph the function to check for extraneous solutions. How to solve perfect square trinomial is an algebraic way to set up and solve equations that end in a squared term. The steps are simple and easy to follow so that you will be able to confidently solve equations on your own!
Another method is to use exponential equations. Exponential equations are equivalent to log equations, so they can be manipulated in the same way. By using these methods, you can solve natural log equations with relative ease.
Algebra can be a difficult subject for many students, but one way to make it easier is to solve by elimination. This method involves setting up equations and solving for one variable in terms of the others. For example, consider the equation ax+by=c. To solve for x, you would first multiply both sides by b and then subtract c from both sides. This would give you the equation bx-(c-ay)=0. You could then solve for x by factoring or using the quadratic formula. However, elimination is usually faster and simpler. Once you get practice using this method, you will be able to solve equations more quickly and easily.
In mathematics, a function is a rule that assigns a unique output to every input. A function can be represented using a graph on a coordinate plane. The input values are plotted on the x-axis, and the output values are plotted on the y-axis. A function is said to be a composite function if it can be written as the composition of two or more other functions. In other words, the output of the composite function is equal to the input of one of the other functions, which is then evaluated to produce the final output. For example, if f(x) = x2 and g(x) = 2x + 1, then the composite function h(x) = f(g(x)) can be graphed as follows: h(x) = (2x + 1)2. As you can see, solving a composite function requires you to first solve for the innermost function, and then work your way outwards. This process can be summarized using the following steps: 1) Identify the innermost function; 2) Substitute the input value into this function; 3) Evaluate the function to find the output; 4) substitute this output value into the next outermost function; 5) repeat steps 2-4 until all functions have been evaluated. By following these steps, you can solve any composite function.
Differential equations are a type of mathematical equation that can be used to model many real-world situations. In general, they involve the derivative of a function with respect to one or more variables. While differential equations may seem daunting at first, there are a few key techniques that can be used to solve them. One common method is known as separation of variables. This involves breaking up the equation into two parts, one involving only the derivative and the other involving only the variable itself. Once this is done, the two parts can be solved independently and then recombined to find the solution to the original equation. Another popular method is known as integration by substitution. This approach involves substituting a new variable for the original one in such a way that the resulting equation is easier to solve. These are just a few of the many methods that can be used to solve differential equations. With practice, anyone can become proficient in this important mathematical discipline.
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# Difference between revisions of "2008 AMC 12A Problems/Problem 17"
## Problem
Let $a_1,a_2,\ldots$ be a sequence determined by the rule $a_n=a_{n-1}/2$ if $a_{n-1}$ is even and $a_n=3a_{n-1}+1$ if $a_{n-1}$ is odd. For how many positive integers $a_1 \le 2008$ is it true that $a_1$ is less than each of $a_2$, $a_3$, and $a_4$?
$\mathrm{(A)}\ 250\qquad\mathrm{(B)}\ 251\qquad\mathrm{(C)}\ 501\qquad\mathrm{(D)}\ 502\qquad\mathrm{(E)} 1004$
## Solution 1
All positive integers can be expressed as $4n$, $4n+1$, $4n+2$, or $4n+3$, where $n$ is a nonnegative integer.
• If $a_1=4n$, then $a_2=\frac{4n}{2}=2n.
• If $a_1=4n+1$, then $a_2=3(4n+1)+1=12n+4$, $a_3=\frac{12n+4}{2}=6n+2$, and $a_4=\frac{6n+2}{2}=3n+1.
• If $a_1=4n+2$, then $a_2=2n+1.
• If $a_1=4n+3$, then $a_2=3(4n+3)+1=12n+10$, $a_3=\frac{12n+10}{2}=6n+5$, and $a_4=3(6n+5)+1=18n+16$.
Since $12n+10, 6n+5, 18n+16 > 4n+3$, every positive integer $a_1=4n+3$ will satisfy $a_1.
Since one fourth of the positive integers $a_1 \le 2008$ can be expressed as $4n+3$, where $n$ is a nonnegative integer, the answer is $\frac{1}{4}\cdot 2008 = 502 \Rightarrow D$.
## Solution 2
After checking the first few $a_n$ such as $1$, $2$ through $7$, we can see that the only $a_1$ that satisfy the conditions are odd numbers that when tripled and added 1 to, are double an odd number. For example, for $a_n=3$, we notice the sequence yields $10$, $5$, and $16$, a valid sequence.
So we can set up an equation, $3x + 1 = 2(2k - 1)$ where x is equal to $a_1$. Rearranging the equation yields $(3x + 3)/4 = k$. Experimenting yields that every 4th $x$ after 3 creates an integer, and thus satisfies the sequence condition. So the number of valid solutions is equal to $2008/4 = 502 \Rightarrow D$.
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How do you solve 3x-y=1 and x+y=3?
Aug 3, 2015
I found:
$x = 1$
$y = 2$
Explanation:
I would isolate $x$ from the second equation and substitute it into the first:
$x = 3 - y$
into the first:
$3 \left(\textcolor{red}{3 - y}\right) - y = 1$
$9 - 3 y - y = 1$
$- 4 y = - 8$
$y = 2$
substitute this back into the second equation:
$x = 3 - 2 = 1$
Aug 3, 2015
$\textcolor{red}{x = 1 , y = 2}$
Explanation:
Gió's method uses the method of substitution.
Here's how to do it by the method of elimination.
Step 1. Enter the equations.
[1] $3 x - y = 1$
[2] $x + y = 3$
Step 2. Add Equations 1 and 2.**
$4 x = 4$
[3] $x = 1$
Step 3. Substitute Equation 3 in Equation 2.
$x + y = 3$
$1 + y = 3$
$y = 2$
Solution: $x = 1 , y = 2$
Check: Substitute the values of $x$ and $y$ in Equation 1.
$3 x - y = 1$
$3 \left(1\right) - 2 = 1$
$3 - 2 = 1$
$1 = 1$
It checks!
The solution is correct.
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# Data Interpretation - Line Charts
Exercise : Line Charts - Line Chart 9
Directions to Solve
Study the following line graph which gives the number of students who joined and left the school in the beginning of year for six years, from 1996 to 2001.
Initial Strength of school in 1995 = 3000.
1.
The number of students studying in the school during 1999 was?
2950
3000
3100
3150
Explanation:
As calculated above, the number of students studying in the school during 1999 = 3150.
2.
For which year, the percentage rise/fall in the number of students who left the school compared to the previous year is maximum?
1997
1998
1999
2000
Explanation:
The percentage rise/fall in the number of students who left the school (compared to the previous year) during various years are:
For 1997 = (450 - 250) x 100 % = 80% (rise). 250
For 1998 = (450 - 400) x 100 % = 11.11% (fall). 450
For 1999 = (400 - 350) x 100 % = 12.5% (fall). 400
For 2000 = (450 - 350) x 100 % = 28.57% (rise). 350
For 2001 = (450 - 450) x 100 % = 0%. 450
Clearly, the maximum percentage rise/fall is for 1997.
3.
The strength of school incresed/decreased from 1997 to 1998 by approximately what percent?
1.2%
1.7%
2.1%
2.4%
Explanation:
Important data noted from the given graph:
In 1996 : Number of students left = 250 and number of students joined = 350.
In 1997 : Number of students left = 450 and number of students joined = 300.
In 1998 : Number of students left = 400 and number of students joined = 450.
In 1999 : Number of students left = 350 and number of students joined = 500.
In 2000 : Number of students left = 450 and number of students joined = 400.
In 2001 : Number of students left = 450 and number of students joined = 550.
Therefore, the numbers of students studying in the school (i.e., strength of the school) in various years:
In 1995 = 3000 (given).
In 1996 = 3000 - 250 + 350 = 3100.
In 1997 = 3100 - 450 + 300 = 2950.
In 1998 = 2950 - 400 + 450 = 3000.
In 1999 = 3000 - 350 + 500 = 3150.
In 2000 = 3150 - 450 + 400 = 3100.
In 2001 = 3100 - 450 + 550 = 3200.
Percentage increase in the strength of the school from 1997 to 1998
= (3000 - 2950) x 100 % = 1.69% 1.7%. 2950
4.
The number of students studying in the school in 1998 was what percent of the number of students studying in the school in 2001?
92.13%
93.75%
96.88%
97.25%
Explanation:
Required percentage = 3000 x 100 % = 93.75% 3200
5.
The ratio of the least number of students who joined the school to the maximum number of students who left the school in any of the years during the given period is?
7:9
4:5
3:4
2:3
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# Triangle Law of Vector Addition of Two Vectors
## Triangle Law of Vector Addition of Two Vectors
The physical quantities may be broadly classified into the vectors and the scalars. The quantities with magnitude and direction both are known as vector quantities, it means a vector is a physical quantity that has both magnitude and direction. If two non-zero vectors are represented by the two sides of a triangle taken in same order then the resultant is given in opposite order.
i.e.., $$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$$ $$\because \overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}$$
1) Magnitude of Resultant Vector: In ΔABN, $$\cos \theta =\frac{AN}{B}$$ ⇒ AN = B cos θ
$$\sin \theta =\frac{BN}{B}$$ ⇒ BN = B sinθ; ΔOBN, we have OB² = ON² + BN².
R² = (A + B cosθ)² + (B sinθ)²
R² = A² + B² cos²θ + 2 AB cosθ + B² sin² θ
R² = A² + B² (cos²θ + sin²θ) + 2AB cosθ = A² + B² (1) + 2 AB cosθ
$$R=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }$$.
2) Direction of resultant vectors: If θ is angle between $$\overrightarrow{A}$$ and $$\overrightarrow{B}$$, then
$$|\overrightarrow{A}+\overrightarrow{B}|\,=\,\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }$$ ; If $$\overrightarrow{R}$$ makes an angle α with$$\overrightarrow{A}$$, then in ΔOBN,
$$\tan \alpha =\frac{BN}{ON}=\frac{BN}{OA+AN}=\frac{B\sin \theta }{A+B\cos \theta }$$.
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Maths How To with Anita
# The Factors of 90 and How to Find them
Do you struggle with finding the factors of 90 or any number?
You’re not alone. Factor pairs, factor rainbows, factor trees, prime factorization & calculators can all be helpful tools when it comes to finding the factors of a number. But which method is best for you?
Factors are important when it comes to multiplication and division. That’s why I have created this article – to help make learning and understanding factors easier than ever. I want you to feel confident in your skills so that you can tackle any problem head-on.
Read on to learn how to find the factors of 90 and any other number!
## What are the factors of 90?
The factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.
They can be found using a variety of methods either with or without a calculator.
## Factor pairs of 90
To find the factor pairs of 90 you can use your calculator or the divisibility tests.
Using your calculator, try dividing 90 by each whole number starting with 1. Then write down the answer with the divisor as a factor pair.
If the answer is not a whole number then it isn’t a factor.
Here are the factor pairs of 90.
1 x 90 = 90
2 x 45 = 90
3 x 30 = 90
5 x 18 = 90
6 x 15 = 90
9 x 10 = 90
Once you get to 2 consecutive factors like 9 & 10 you know you have found all the factor pairs.
## Factor Rainbows
Another way to find factor pairs is to draw a factor rainbow.
To watch how I created this factor rainbow for 90 watch this video.
## Prime Factorization of 90
Draw a factor tree to find the prime factors of 90.
There are multiple ways to draw a factor tree for 90 but no matter how you draw it you will always end up with the same prime factors.
Here is a factor tree:
Here is another factor tree. Notice how they both have the same prime factors.
So we can write 90 as a product of prime factors like this:
90 = 2 x 3 x 3 x 5
or in index form this is the prime factorization of 90:
## Negative Factors of 90
The negative factors of 90 are -1, -2, -3, -5, -6, -9, -10, -15, -18, -30, -45 & -90.
Both factors need to be negative (or both positive) so that the product is positive 90.
## FAQs
### What are all the prime factors of 90?
The prime factors of 90 are 2, 3 & 5. They can be found be drawing a factor tree like the ones in this article.
### What two numbers make 90?
There are 6 pairs of 2 numbers that make 90.
1 & 90
2 & 45
3 & 30
5 & 18
6 & 15
9 & 10
## Final thoughts on the Factors of 90
In this article, we looked at different methods for finding the factors of 90. We started with a method that used division and your calculator, then moved onto factor pairs and factor rainbows that used multiplication and finally looked at factor trees and prime factorization.
Each of these methods has its own benefits and drawbacks, so it is important to understand when each one should be used. Practice using all these methods so that you can become comfortable with them and choose the best one for the given situation.
## Further Factors!
Factors of 18
Factors of 30
Prime Factorization of 42
Factors of 50
Factors of 64
Factors of 90
Factors of 91
Factors of 96
Factors of 120
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Miscellaneous
Chapter 10 Class 11 Straight Lines
Serial order wise
This video is only available for Teachoo black users
Get live Maths 1-on-1 Classs - Class 6 to 12
### Transcript
Misc 14 In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4? Let line AB is the line joining the points A(–1, 1) & B(5, 7) & let line CD be x + y = 4 Let line AB be divided by the line CD at point P Let k :1 be the ratio line AB is divided by the line CD We need to find value of k If a point divide any line joining (x1, y1) & (x2, y2) in the ratio of m1 : m2 then co-ordinate of that point is ((𝑚_2 𝑥_2 + 𝑚_1 𝑥_1)/(𝑚_1 + 𝑚_2 ),(𝑚_2 𝑦_(2 ) + 𝑚_1 𝑦_1)/(𝑚_1 + 𝑚_2 )) Point P which divide the line A(–1, 1) & B(5, 7) in k : 1 ratio is Coordinate of point P = ((5𝑘 + ( − 1))/(𝑘 + 1), (7𝑘 + 1)/(𝑘 + 1)) = ((5𝑘 − 1)/(𝑘 + 1), (7𝑘 + 1)/(𝑘 + 1)) ∴ Point P is ((5𝑘 − 1)/(𝑘 + 1), (7𝑘 + 1)/(𝑘 + 1)) Now, point P((5𝑘 − 1)/(𝑘 + 1), (7𝑘 + 1)/(𝑘 + 1)) lies on the line CD So, It will satisfy the equation of line CD So, x + y = 4 Putting x = (5𝑘 − 1)/(𝑘 + 1), y = (7𝑘 + 1)/(𝑘 + 1) ((5𝑘 − 1)/(𝑘 + 1)) + ((7𝑘 + 1)/(𝑘 + 1)) = 4 ((5𝑘 − 1) + (7𝑘 + 1))/(𝑘 + 1) = 4 5k – 1 + 7k + 1 = 4(k + 1) 5k + 7k – 1 + 1 = 4k + 4 12k + 0 = 4k + 4 12k – 4k = 4 8k = 4 k = 4/8 k = 1/2 Hence, Point P divides AB in the ratio of k : 1 = 1/2 : 1 = 2 × 1/2 : 2 × 1 = 1 : 2 Thus , required ratio is 1 : 2
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# NCERT Solutions for Class 11 Maths Chapter 14: Mathematical Reasoning - Exercise 14.2
Last updated date: 09th Aug 2024
Total views: 606k
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## NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning
Free PDF download of NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.2 (Ex 14.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 14 Mathematical Reasoning Exercise 14.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.
Class: NCERT Solutions for Class 11 Subject: Class 11 Maths Chapter Name: Chapter 14 - Mathematical Reasoning Exercise: Exercise - 14.2 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes
Competitive Exams after 12th Science
## NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning
1. Write the negation of the following statements.
i. Chennai is the capital of Tamil Nadu.
Ans: The negation of a statement means to negate the statement. In other words, it is writing the opposite of the given statement. Negation reverses the meaning of the statement. If a given statement is false then its negation will be true and vice versa. To write the negation of any statement, we usually add or remove the word ‘not’.
The given statement is that, ‘Chennai is the capital of Tamil Nadu’.
The given statement does not carry the word ‘not’.
To form the required negation of the given statement, we will add the word ‘not’ to it.
Therefore, the negation formed of the given statement is, ‘Chennai is not the capital of Tamil Nadu’.
ii. $\sqrt 2$ is not a complex number.
Ans: The negation of a statement means to negate the statement. In other words, it is writing the opposite of the given statement. Negation reverses the meaning of the statement. If a given statement is false then its negation will be true and vice versa. To write the negation of any statement, we usually add or remove the word ‘not’.
The given statement is that, ‘$\sqrt 2$ is not a complex number’.
The given statement carries the word ‘not’.
To form the required negation of the given statement, we will remove the word ‘not’ from it.
Therefore, the negation formed of the given statement is, ‘$\sqrt 2$ is a complex number’.
iii. All triangles are not equilateral triangles.
Ans: The negation of a statement means to negate the statement. In other words, it is writing the opposite of the given statement. Negation reverses the meaning of the statement. If a given statement is false then its negation will be true and vice versa. To write the negation of any statement, we usually add or remove the word ‘not’.
The given statement is that, ‘All triangles are not equilateral triangles’.
The given statement carries the word ‘not’.
To form the required negation of the given statement, we will remove the word ‘not’ from it.
Therefore, the negation formed of the given statement is, ‘All triangles are equilateral triangles’.
iv. The number 2 is greater than 7.
Ans: The negation of a statement means to negate the statement. In other words, it is writing the opposite of the given statement. Negation reverses the meaning of the statement. If a given statement is false then its negation will be true and vice versa. To write the negation of any statement, we usually add or remove the word ‘not’.
The given statement is that, ‘The number 2 is greater than 7’.
The given statement does not carry the word ‘not’.
To form the required negation of the given statement, we will add the word ‘not’ to it.
Therefore, the negation formed of the given statement is, ‘The number 2 is not greater than 7’.
v. Every natural number is not an integer.
Ans: The negation of a statement means to negate the statement. In other words, it is writing the opposite of the given statement. Negation reverses the meaning of the statement. If a given statement is false then its negation will be true and vice versa. To write the negation of any statement, we usually add or remove the word ‘not’.
The given statement is that, ‘Every natural number is not an integer’.
The given statement carries the word ‘not’.
To form the required negation of the given statement, we will remove the word ‘not’ from it.
Therefore, the negation formed of the given statement is, ‘Every natural number is an integer’.
2. Are the following pairs of statements negations of each other?
i. The number $x$ is not a rational number.
The number $x$ is not an irrational number.
Ans: The negation of a statement means to negate the statement. In other words, it is writing the opposite of the given statement. Negation reverses the meaning of the statement. If a given statement is false then its negation will be true and vice versa. To write the negation of any statement, we usually add or remove the word ‘not’.
Consider the first statement.
The given statement is that, ‘The number $x$ is not a rational number’.
The given statement carries the word ‘not’.
To form the required negation of the given statement, we will remove the word ‘not’ from it.
Therefore, the negation formed of the given statement is, ‘The number $x$ isa rational number’.
Now this statement is similar to the second statement. This is because if a number is not an irrational number, then it will be rational. And this is what the negation of our first statement says.
Therefore, the given pair of statements are the negations of each other.
ii. The number $x$ is a rational number.
The number $x$ is an irrational number.
Ans: The negation of a statement means to negate the statement. In other words, it is writing the opposite of the given statement. Negation reverses the meaning of the statement. If a given statement is false then its negation will be true and vice versa. To write the negation of any statement, we usually add or remove the word ‘not’.
Consider the first statement.
The given statement is that, ‘The number $x$ is a rational number’.
The given statement does not carry the word ‘not’.
To form the required negation of the given statement, we will add the word ‘not’ to it.
Therefore, the negation formed of the given statement is, ‘The number $x$ is not a rational number’.
Now this statement is similar to the second statement. This is because if a number is not a rational number, then it will be irrational. And this is what the negation of our first statement says.
Therefore, the given pair of statements are the negations of each other.
3. Find the component statements of the following compound statements and check whether they are true or false.
i. Number 3 is prime or it is odd.
Ans: Compound statements are those statements that are made up of two or more simpler or smaller statements. These smaller statements are complete in themselves and have their own independent meanings.
Consider the given statement, ‘Number 3 is prime or it is odd’.
We will determine the first component statement and check whether it is true or false.
Let this first component statement be ${\text{p}}$.
${\text{p}}$: Number 3 is prime.
This statement is true because 3 is a prime number as it has only two factors, 1 and itself.
We will now determine the second component statement and check whether it is true or false.
Let this second component statement be ${\text{q}}$.
${\text{q}}$: Number 3 is odd.
This statement is true because 3 is an odd number as it is not completely divisible by 2.
Therefore, the component statements for the given compound statement are,
${\text{p}}$: Number 3 is prime.
${\text{q}}$: Number 3 is odd.
Both these above component statements are true.
ii. All integers are positive or negative.
Ans: Compound statements are those statements that are made up of two or more simpler or smaller statements. These smaller statements are complete in themselves and have their own independent meanings.
Consider the given statement, ‘All integers are positive or negative’.
We will determine the first component statement and check whether it is true or false.
Let this first component statement be ${\text{p}}$.
${\text{p}}$: All integers are positive.
This statement is false because integers are both positive and negative numbers.
We will now determine the second component statement and check whether it is true or false.
Let this second component statement be ${\text{q}}$.
${\text{q}}$:All integers are positive.
This statement is false because integers are both positive and negative numbers.
Therefore, the component statements for the given compound statement are,
${\text{p}}$: All integers are positive.
${\text{q}}$: All integers are negative.
Both these above component statements are false.
iii. 100 is divisible by 3, 11, and 5.
Ans: Compound statements are those statements that are made up of two or more simpler or smaller statements. These smaller statements are complete in themselves and have their own independent meanings.
Consider the given statement, ‘100 is divisible by 3, 11, and 5’.
We will determine the first component statement and check whether it is true or false.
Let this first component statement be ${\text{p}}$.
${\text{p}}$: 100 is divisible by 3.
This statement is false because 100 is not divisible by 3.
We will now determine the second component statement and check whether it is true or false.
Let this second component statement be ${\text{q}}$.
${\text{q}}$: 100 is divisible by 11.
This statement is false because 100 is not divisible by 11.
We will now determine the third component statement and check whether it is true or false.
Let this third component statement be ${\text{r}}$.
${\text{r}}$: 100 is divisible by 5.
This statement is true because 100 is divisible by 5.
Therefore, the component statements for the given compound statement are,
${\text{p}}$: 100 is divisible by 3.
${\text{q}}$: 100 is divisible by 11.
${\text{r}}$: 100 is divisible by 5.
The statements ${\text{p}}$ and ${\text{q}}$ are false, whereas the statement ${\text{r}}$ is true.
## NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning
Opting for the NCERT solutions for Ex 14.2 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 14.2 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.
Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Subject Mathematical Reasoning textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 14 Exercise 14.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.
Besides these NCERT solutions for Class 11 Maths Chapter 14 Exercise 14.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.
Do not delay any more. Download the NCERT solutions for Class 11 Maths Chapter 14 Exercise 14.2 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.
## FAQs on NCERT Solutions for Class 11 Maths Chapter 14: Mathematical Reasoning - Exercise 14.2
1. Where can I find the NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning (Ex 14.2) Exercise 14.2?
The NCERT solutions for the second exercise of chapter 14 Mathematical Reasoning have been carefully chosen by Vedantu teachers. The solutions were created with the concepts in mind and in accordance with the most recent CBSE patterns and guidelines. The Vedantu website or app makes it simple for students to access them.
2. What is the basis for NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning (Ex 14.2) Exercise 14.2?
The second exercise in math chapter 14 is based on the ideas of Mathematical Reasoning when all the objects are distinct. Students can learn more about these subjects from the NCERT's explanations and examples that have been solved before moving on to the questions in exercise 14.2.
3. How do the NCERT solutions for Mathematical Reasoning Exercise 14.2 in Chapter 14 of Class 11 Math work?
Students can improve their understanding of math concepts, increase their familiarity with the more important subjects, and get a sense of the kinds of questions that might be asked about these subjects in the examination by using the NCERT solutions for exercise 14.2. Students can download the full step-by-step solutions in pdf format from the Vedantu website for the entirety of Exercise 14.2.
4. How many questions are there in Exercise 14.2 of Chapter 14's Mathematical Reasoning?
Exercise 14.2 from Class 11 Math's Chapter 14 Mathematical Reasoning consists of a total of 3 questions. The top online resource in India, Vedantu, is where you can find the NCERT solutions for Class 11 Math. At Vedantu, all of the chapter exercises are collected in one location and solved by a qualified teacher in accordance with the recommendations of the NCERT books. The solutions are complete, step-by-step, and 100% accurate.
5. Is it necessary for students to practice each of the exercises in NCERT Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning (Ex 14.2)?
Students must undoubtedly practice each and every one of the NCERT textbook's questions if they want to get good grades. The Class 11 Maths NCERT solutions are the most helpful source because they contain a variety of questions that need the right concept and understanding to be answered. You can get ready for any challenging or uncommon exam questions by consistently practicing. For free, Vedantu offers comprehensive, step-by-step NCERT solutions for all math chapters.
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## How To Solve Probability
” How do I know when to multiply, when to add and subtract when asked to solve a probability question?” This was the same thought that use to go through my mind while taking probability classes because I was always utterly confused with numerous and similar rules of probability.
So if anyone is going through something similar to this, it is highly expected. However, it is time to realize that there is light at the end of the tunnel because I was able to reach the end of probability tunnel through hard work.
While searching and trying to find a breakthrough and become better at solving probability, I was trying to master the rules. Little did I know that mastering the rules wasn’t what I needed to become better at solving probability. Here is a quick summary of what I discovered about probability and how that changes the way I saw the topic and how it can be solved.
But first, let’s take a look at what probability is all about and the general ideology backing it all up.
What is a probability?
Probability is all about the odds of choosing a particular thing amidst a multitude of other things which is usually in the form of an event. From my research over the years, probability is majorly divided into two distributions which are binomial distribution and normal distribution.
Understanding the difference between these two gave me the edge I needed when it comes to solving probability questions. With this, I was able to use and apply the rules I mastered in a much better way.
How is probability solved?
Probability can be solved in different ways, it can be used to find the probability of certain events, the probability of a dice roll, and even the probability of a card to be picked from a deck of cards. To help explain this further, I will be solving different examples on each of these where probability can be applied.
Probability Problems about Events – Probability can be used here in a straight forward fashion. A good example is when an individual has a 40% chance of getting a job and 20% chance of losing it. The probability here is the total of both chances which is 40% + 20%= 60%. The only thing with this method is that it works solely with mutually exclusive events
Dice rolling probability problems – For dice rolling, there is a total of 36 different possibilities. This is used as the fixed possible value and it can be used to calculate the probability of any possible occurrences.
Probability is a big part of the math sections of the SAT. You can prepare for the SAT at Caddell Prep
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# Video: Finding Spearman’s Correlation Coefficient for Bivariate Data
Find Spearman’s correlation coefficient between 𝑥 and 𝑦. Round your answer to three decimal places.
05:18
### Video Transcript
Find Spearman’s correlation coefficient between 𝑥 and 𝑦. Round your answer to three decimal places.
Looking at this set of data, we see 𝑥-values are in the first row and 𝑦-values are in the second. These points then are bivariate, meaning they’re described by two variables. Where our question asks us to solve for the Spearman’s correlation coefficient between 𝑥 and 𝑦, we actually don’t directly use these data values to do that. Instead, we use what are called the ranks of these values. That’s because Spearman’s correlation coefficient describes the level of agreement between the relative ranks of bivariate data.
To see what this means, let’s create two new rows in our data table. The first new row can represent the rank of the 𝑥-values in our data, and the second new row will represent the rank of the 𝑦-values. Considering first the 𝑥-values in our table, we can rank these values from smallest to largest using the numbering one, two, three, and so on. This means that our smallest 𝑥-value will get a rank of one.
We see that smallest value is four. So that means 𝑅 sub 𝑥 for this value is one. The next lowest 𝑥-value is five, which means that this has a rank of two. Then comes seven, which must then have a rank of three. And next see that we have two 𝑥-values of eight. We can say that these are the fourth and fifth lowest 𝑥-values. But since they’re the same number, we take these two rankings, four and five, and find the average of them, that’s 4.5, and then assign that the relative ranking of each of these numbers. Lastly then, the highest 𝑥-value on our table is 12. So this has a ranking of six, the sixth smallest value. So that’s what it means to rank our data.
And now we’ll do the same thing for our 𝑦-values. The smallest 𝑦-value in the table is four, so that gets a rank of one. And then comes six, which we have three of. These values occupy then the second, third, and fourth places among our 𝑦-values. And since they’re all the same, we assign in the same ranking of the average of those three numbers, which is three. The next lowest 𝑦-value is seven. That is the fifth lowest 𝑦-value. So 𝑅 sub 𝑦 for this is five. And lastly, the highest 𝑦-value is 10, and so the ranking for this is six.
The results in these two rows of our table are what Spearman’s correlation coefficient is actually going to describe. This coefficient gives a quantitative indication of the level of agreement between the relative ranks of these data. Essentially, the closer 𝑅 𝑥 and 𝑅 𝑦 are for each point in the data set, the closer this correlation coefficient comes to positive one.
As our next step then, let’s create a row in our table that indicates the difference between respective 𝑅 𝑥 and 𝑅 𝑦 values. We’ll say that this value 𝑑 sub 𝑖 equals 𝑅 𝑥 minus 𝑅 𝑦 for each data point. For our first data point then, we have one minus five. That’s negative four. Then we have three minus three, zero, next 4.5 minus three or 1.5; two minus one or one; 4.5 minus three again, which is 1.5; and finally six minus six, which is zero.
In order to normalize these results, let’s make a final row in our table where we square these difference values. That way, none of these relative differences will be negative. Negative four times negative four is positive 16. Zero squared is zero. 1.5 squared is 2.25. One squared is one. 1.5 squared again is 2.25. And zero squared is zero.
At this point, let’s recall the mathematical relationship for Spearman’s correlation coefficient. We often represent this coefficient using an 𝑟 sub 𝑠. It’s equal to one minus six times the sum of all these 𝑑 sub 𝑖 values squared divided by the number of data points in our set 𝑛 multiplied by that number squared minus one. To calculate Spearman’s correlation coefficient for a set of data then, the two things we need to know are the sum of all the 𝑑 sub 𝑖 squared values and also the total number of points in the set.
Considering the sum of 𝑑 sub 𝑖 squared, we can solve for that by adding together all the results in the last row of our table. 16 plus zero plus 2.25 plus one plus 2.25 plus zero adds up to 21.5. And then, regarding the number of data points in our set, we see that we have one, two, three, four, five, six such points. This means that, in our case, 𝑛 equals six.
And now that we’ve gotten these two bits of information from the data in our set, we can clear away all the rows we created and move ahead to calculate 𝑟 sub 𝑠, Spearman’s correlation coefficient. We sub in our values for the sum of 𝑑 sub 𝑖 squared and 𝑛. And note that because 𝑛 equals six, one factor of six cancels from numerator and denominator. Moreover, six squared is 36. So we can express 𝑟 sub 𝑠 as one minus 21.5 over 36 minus one, or 35. Calculating this out, we get 0.38571 and so on.
But note that we want to give our final answer rounded to three decimal places. Doing this, we get a result of 0.386. To three decimal places, this is the Spearman’s correlation coefficient between 𝑥 and 𝑦.
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Become a math whiz with AI Tutoring, Practice Questions & more.
HotmathMath Homework. Do It Faster, Learn It Better.
The opposite of a number is also called its additive inverse. The additive inverse is the number you have to add to it so that the sum is zero. So the additive inverse of 5 is -5. This is sometimes called the property of opposites.
If you add any number to its opposite, or additive inverse, the answer is always zero. For example:
$-989+989=0$
$6.5+\left(-6.5\right)=0$
$1+\left(-1\right)=0$
Once you know this, there are several ways you can think about adding and subtracting negative numbers.
## Adding and subtracting negatives: The algebra tile method
Let the yellow tiles represent positive numbers and the red tiles represent negative numbers.
Example 1
The addition problem $5+\left(-2\right)$ can be represented as:
Group the two negative tiles with two of the positive tiles, zeroing them out.
Since $2+\left(-2\right)=0$ , these tiles disappear. We are left with 3 positive tiles.
Therefore, $5+\left(-2\right)=3$ .
When both numbers are negative, we have only negative tiles.
Example 2
The addition problem $-3+\left(-4\right)$ can be represented as
The result is simply 7 negative tiles.
Therefore, $-3+\left(-4\right)=-7$
## Adding and subtracting negatives: The number line method
When you add a positive number, you move to the right on the number line.
When you add a negative number, you move to the left on the number line.
Example 3
Add $6+\left(-8\right)$ using a number line.
Start at 6 and move 8 units to the left.
So $6+\left(-8\right)=-2$
Subtracting a number is the same as adding its opposite. So subtracting a positive number is like adding a negative number - you move to the left on the number line.
Subtracting a negative number is like adding a positive number - you move to the right on the number line.
Example 4
Subtract $-4-\left(-7\right)$ .
Start at -4 and move 7 units to the right.
So $-4-\left(-7\right)=3$ .
## Adding negative numbers without visual aids
Once you've become used to adding negative numbers using the algebra tile method or the number line method, you'll notice that adding a negative number is basically the same as performing a subtraction operation.
So these two problems $5+\left(-4\right)$ and $5-4$ are going to have the same answer.
Example 5
Solve the following addition problems involving negative numbers using your understanding of subtraction.
$8+\left(-3\right)$
Because we know that $8-3=5$ , you know that
$8+\left(-3\right)=5$
$259+\left(-384\right)$
Since we can work out that $259-384=-125$ , you know that $259+\left(-384\right)=-125$
What about when the negative number is larger than the positive number we are adding it to? Well, we know how to subtract a number that is larger than the number we are subtracting it from - we end up with a negative number.
Example 6
Solve the following addition problems involving negative numbers using our understanding of subtraction.
$8+\left(-15\right)$
Since we know that $8-15=-7$ , you know that
$8+\left(-15\right)=-7$
## Subtracting negative numbers without visual aids
Once you've become used to subtracting negative numbers using the algebra tile method or the number line method, you'll notice that subtracting a negative number is basically the same as performing an addition operation. So the two problems $10-\left(-5\right)$ and $10+5$ are going to have the same answer.
Knowing this can help us perform the subtraction of negative numbers in your head.
Example 7
Solve the following subtraction problems involving negative numbers using your understanding of addition.
1. $7-\left(-3\right)$
Because we know that $7+3=10$ , we know that
$7-\left(-3\right)=10$
2. $157-\left(-90\right)$
Since we can work out that $157+90=247$ , we know that
$157-\left(-90\right)=247$
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# Congruence (number theory)
In number theory, congruence is a relationship between whole numbers . They call two integers and congruent modulo (= another number), when used in the division by both the same rest have. This is precisely the case when they differ by an integral multiple of . However, if the remainders do not match, the numbers are called incongruent modulo . Every congruence modulo an integer is a congruence relation on the ring of integers. ${\ displaystyle a}$${\ displaystyle b}$ ${\ displaystyle m}$${\ displaystyle m}$${\ displaystyle m}$ ${\ displaystyle m}$
## Examples
### example 1
For example, 5 is congruent 11 modulo 3, da and , the two remainders (2) are therefore the same, or da , the difference is therefore an integer multiple (2) of 3. ${\ displaystyle 5: 3 = 1 {\ text {rest}} 2}$${\ displaystyle 11: 3 = 3 {\ text {rest}} 2}$${\ displaystyle 11-5 = 6 = 2 \ cdot 3}$
### Example 2
In contrast, 5 is incongruent 11 modulo 4, da and ; the two remains are not the same here. ${\ displaystyle 5: 4 = 1 {\ text {rest}} 1}$${\ displaystyle 11: 4 = 2 {\ text {rest}} 3}$
### Example 3
And −8 is congruent to 10 modulo 6, because when dividing by 6, both 10 and −8 produce the remainder 4. Note that the mathematical definition of the integer division is based, according to which the remainder has the same sign as the divisor (here 6) receives, so . ${\ displaystyle -8: 6 = -2 {\ text {rest}} 4}$
## Notation
The following notations are used for the statement " and are congruent modulo ": ${\ displaystyle a}$${\ displaystyle b}$${\ displaystyle m}$
• ${\ displaystyle a \ equiv b \ mod m}$
• ${\ displaystyle a \ equiv b \ mod m \ mathbb {Z}}$
• ${\ displaystyle a \ equiv b {\ pmod {m}}}$
• ${\ displaystyle a \ equiv b \ quad (m)}$
• ${\ displaystyle a \ equiv _ {m} b}$
These spellings can be used as a short form of the statement (equivalent to the above statement) "remainder of division of through is equal to remainder of division of through ", ie of ${\ displaystyle a}$${\ displaystyle m}$ ${\ displaystyle b}$${\ displaystyle m}$
${\ displaystyle (a {\ bmod {m}}) = (b {\ bmod {m}})}$,
(where in the last-mentioned equation is the mathematical modulo function that determines the remainder of an integer division, in this case the remainder of or ; with the mathematical modulo function, the result, i.e. the remainder, always has the same sign as ). ${\ displaystyle \ operatorname {mod}}$${\ displaystyle {\ tfrac {a} {m}}}$${\ displaystyle {\ tfrac {b} {m}}}$${\ displaystyle m}$
## history
The theory of congruences was developed by Carl Friedrich Gauß in his work " Disquisitiones Arithmeticae " published in 1801 . The term congruence was used by Christian Goldbach as early as 1730 in letters to Leonhard Euler , but without the theoretical depth of Gauss. In contrast to Gauss, Goldbach used the symbol and not . The Chinese mathematician Qin Jiushao (秦九韶) was also familiar with congruences and the accompanying theory, as can be seen in his book “ Shushu Jiuzhang ” ( Chinese 數 書 九章 / 数 书 九章 , Pinyin Shùshū Jiǔzhāng - “Mathematical treatise in nine chapters” published in 1247 ) “). ${\ displaystyle \ mp}$${\ displaystyle \ equiv}$
## Formal definition
In number theory, congruence is reduced to a divisibility statement . In addition , and integers, i.e. H. Elements from . ${\ displaystyle a}$${\ displaystyle b}$${\ displaystyle m \ neq 0}$${\ displaystyle \ mathbb {Z}}$
• Two numbers and are called congruent modulo if the difference divides.${\ displaystyle a}$${\ displaystyle b}$${\ displaystyle m}$${\ displaystyle m}$${\ displaystyle from}$
• Two numbers and are called incongruent modulo if the difference does not divide.${\ displaystyle a}$${\ displaystyle b}$${\ displaystyle m}$${\ displaystyle m}$${\ displaystyle from}$
Using mathematical notation, these two statements can be written as follows:
• ${\ displaystyle a \ equiv b {\ pmod {m}}}$
${\ displaystyle \ Leftrightarrow m \ mid (ab)}$
${\ displaystyle \ Leftrightarrow \ exists k \ in \ mathbb {Z}: a = km + b}$
• ${\ displaystyle a \ not \ equiv b {\ pmod {m}}}$
${\ displaystyle \ Leftrightarrow m \ nmid (ab)}$
${\ displaystyle \ Leftrightarrow \ forall k \ in \ mathbb {Z}: a \ neq km + b}$
## Residual classes
A congruence relation is a special equivalence relation . So it has the following properties:
Reflexivity
${\ displaystyle a \ equiv a {\ pmod {m}}}$ for all ${\ displaystyle a \ in \ mathbb {Z}}$
symmetry
${\ displaystyle a \ equiv b {\ pmod {m}} \ Rightarrow b \ equiv a {\ pmod {m}}}$ for all ${\ displaystyle a, b \ in \ mathbb {Z}}$
Transitivity
${\ displaystyle a \ equiv b {\ pmod {m}}}$and for everyone${\ displaystyle b \ equiv c {\ pmod {m}} \ Rightarrow a \ equiv c {\ pmod {m}}}$${\ displaystyle a, b, c \ in \ mathbb {Z}}$
The equivalence classes of the congruence relation are called residual classes . If one also wants to specify, one speaks of residual classes . A remainder class that contains the element is often referred to as. ${\ displaystyle m}$${\ displaystyle ({\ text {mod}} m)}$${\ displaystyle a}$${\ displaystyle [a]}$
Like every equivalence relation, a congruence relation defines a partition of its support set: The remainder classes of two elements are either equal or disjoint , the former if and only if the elements are congruent:
${\ displaystyle [a] = [b] \ quad \ iff \ quad a \ equiv b {\ pmod {m}} \ quad \ iff \ quad a \ in [b] \ quad \ iff \ quad b \ in [a ]}$.
Equipped with the links induced by the residue classes form a ring , the so-called residue class ring . He is designated for with . ${\ displaystyle \ mathbb {Z} (+, \ cdot)}$ ${\ displaystyle m}$${\ displaystyle \ mathbb {Z} / m \ mathbb {Z}}$
comment
• Since a division by has not yet occurred, one can allow for the formal definition (in the previous section) as well as for the equivalence relation (in this section) .${\ displaystyle m}$${\ displaystyle m = 0}$
• Since there are no real zero divisors in the ring , the relation degenerates to the trivial case, to equality:${\ displaystyle \ mathbb {Z}}$${\ displaystyle \ equiv ({\ text {mod}} 0)}$
${\ displaystyle a \ equiv b {\ pmod {0}} \ quad \ Rightarrow \ quad a = b}$ for everyone .${\ displaystyle a, b \ in \ mathbb {Z}}$
• The unitary ring of the characteristic is isomorphic to . This property is also needed in the event . Then is . This ring is not viewed as a residual class ring in the narrower sense.${\ displaystyle m}$${\ displaystyle \ mathbb {Z} / m \ mathbb {Z}}$${\ displaystyle m = 0}$${\ displaystyle \ mathbb {Z} / m \ mathbb {Z} = \ mathbb {Z} / 0 \ mathbb {Z} = \ mathbb {Z}}$
• The interesting cases are the cases that can be taken as the standard .${\ displaystyle m \ neq 0}$
• The remainder class ring is the null ring , which consists of only one element.${\ displaystyle \ mathbb {Z} / 1 \ mathbb {Z}}$
Is not trivial, so then are in a residue class all the numbers that have the same remainder when dividing by exhibit. Then the absolute value of , that is , corresponds to the number of remainder classes. For example, there are two remainder classes of even and odd numbers for 2 . ${\ displaystyle m}$${\ displaystyle m \ neq 0}$${\ displaystyle m}$${\ displaystyle m}$${\ displaystyle | m |}$
## Calculation rules
Below are , , , , and whole numbers. Be there , and . Then the following calculation rules apply: ${\ displaystyle a}$${\ displaystyle a '}$${\ displaystyle b}$${\ displaystyle b '}$${\ displaystyle c}$${\ displaystyle m}$${\ displaystyle m \ neq 0}$${\ displaystyle a \ equiv a '{\ pmod {m}}}$${\ displaystyle b \ equiv b '{\ pmod {m}}}$
${\ displaystyle ca \ equiv ca '{\ pmod {m}}}$
${\ displaystyle a + b \ equiv a '+ b' {\ pmod {m}}}$
${\ displaystyle ab \ equiv a'-b '{\ pmod {m}}}$
${\ displaystyle ab \ equiv a'b '{\ pmod {m}}}$
If there is a polynomial over the whole numbers, then: ${\ displaystyle f \ in \ mathbb {Z} [X]}$
${\ displaystyle f (a) \ equiv f (a ') {\ pmod {m}}}$
Shortening is also possible if there are congruences . However, other abbreviation rules apply to those used for rational or real numbers ( ... greatest common factor ): ${\ displaystyle \ operatorname {ggT}}$
${\ displaystyle ca \ equiv cb {\ pmod {m}} \ Leftrightarrow a \ equiv b {\ pmod {\ frac {m} {\ operatorname {ggT} (c, m)}}}}$
It follows immediately that - if a prime number and this is not a divisor of - the following applies: ${\ displaystyle m}$${\ displaystyle p}$${\ displaystyle c}$
${\ displaystyle ca \ equiv cb {\ pmod {p}} \ Leftrightarrow a \ equiv b {\ pmod {p}}}$
If is a composite number or a factor of , only: ${\ displaystyle m}$${\ displaystyle c}$
${\ displaystyle ca \ equiv cb {\ pmod {m}} \ Leftarrow a \ equiv b {\ pmod {m}}}$
For every divisor of it follows that . ${\ displaystyle d}$${\ displaystyle m}$${\ displaystyle a \ equiv b {\ pmod {m}}}$${\ displaystyle a \ equiv b {\ pmod {d}}}$
If integers are not equal to zero and their lowest common multiple , then: ${\ displaystyle m_ {1}, m_ {2}, \ dotsc, m_ {k}}$${\ displaystyle m}$
${\ displaystyle a \ equiv a '{\ pmod {m _ {\ kappa}}}}$ for all ${\ displaystyle \ kappa = 1,2, \ dotsc, k \ quad \ Leftrightarrow \ quad a \ equiv a '{\ pmod {m}}}$
### Potencies
If it is a natural number , then: ${\ displaystyle n \ in \ mathbb {N} _ {0}}$
${\ displaystyle a ^ {n} \ equiv (a ') ^ {n} {\ pmod {m}}}$
Are and coprime , then according to Euler's theorem${\ displaystyle a}$${\ displaystyle m}$
${\ displaystyle a ^ {\ varphi (m)} \ equiv 1 {\ pmod {m}}}$,
wherein the Euler's φ function referred to. It also follows from this ${\ displaystyle \ varphi (m)}$
${\ displaystyle a ^ {n} \ equiv a ^ {n '} {\ pmod {m}}}$if .${\ displaystyle n \ equiv n '{\ pmod {\ varphi (m)}}}$
A special case of this is the little Fermat's theorem , hence congruence for all prime numbers${\ displaystyle p}$
${\ displaystyle a ^ {p} \ equiv a {\ pmod {p}}}$
is satisfied.
## Derived calculation rules
1. The following applies to:${\ displaystyle t \ neq 0}$${\ displaystyle t \ cdot a \ equiv t \ cdot b {\ pmod {| t | \ cdot m}}}$
2. If is a factor of , then:${\ displaystyle k}$${\ displaystyle m}$${\ displaystyle a \ equiv b {\ pmod {k}}}$
3. For every odd number :${\ displaystyle a}$${\ displaystyle a ^ {2} \ equiv 1 {\ pmod {8}}}$
4. Either or or applies to every integer .${\ displaystyle a ^ {3} \ equiv 0 {\ pmod {9}}}$${\ displaystyle a ^ {3} \ equiv 1 {\ pmod {9}}}$${\ displaystyle a ^ {3} \ equiv 8 {\ pmod {9}}}$
5. For every whole number :${\ displaystyle a}$${\ displaystyle a ^ {3} \ equiv a {\ pmod {6}}}$
6. Either or or applies to every integer .${\ displaystyle a ^ {3} \ equiv 0 {\ pmod {7}}}$${\ displaystyle a ^ {3} \ equiv 1 {\ pmod {7}}}$${\ displaystyle a ^ {3} \ equiv 6 {\ pmod {7}}}$
7. Either or applies to every integer .${\ displaystyle a ^ {4} \ equiv 0 {\ pmod {5}}}$${\ displaystyle a ^ {4} \ equiv 1 {\ pmod {5}}}$
8. If both a square number and a cube number (e.g. ) then either or or or applies .${\ displaystyle a}$${\ displaystyle a = 64}$${\ displaystyle a \ equiv 0 {\ pmod {36}}}$${\ displaystyle a \ equiv 1 {\ pmod {36}}}$${\ displaystyle a \ equiv 9 {\ pmod {36}}}$${\ displaystyle a \ equiv 28 {\ pmod {36}}}$
9. Let be a prime number with . Then:${\ displaystyle p}$${\ displaystyle n
${\ displaystyle {2n \ choose n} \ equiv 0 {\ pmod {p}}}$
10. Let be an odd integer. Further be . Then:${\ displaystyle a}$${\ displaystyle n> 0}$${\ displaystyle a ^ {2 ^ {n}} \ equiv 1 {\ pmod {2 ^ {n + 2}}}}$
11. Be . Furthermore, let and be prime twins . Then:${\ displaystyle p> 3}$${\ displaystyle p}$${\ displaystyle q = p + 2}$ ${\ displaystyle p \ cdot q \ equiv -1 {\ pmod {9}}}$
## Solvability of linear congruences
### Linear congruence
A linear congruence of shape
${\ displaystyle ax \ equiv c {\ pmod {m}}}$
is solvable in if and only if the number divides. In this case the congruence has exactly solutions in , and the solutions are congruent to each other modulo . ${\ displaystyle x}$${\ displaystyle g = \ operatorname {ggT} (a, m)}$${\ displaystyle c}$${\ displaystyle g}$${\ displaystyle \ {0,1, \ dotsc, m-1 \}}$${\ displaystyle {\ tfrac {m} {g}}}$
The solutions can also be efficiently determined for large ones by applying the extended Euclidean algorithm to and , which also calculates two numbers and , which express as a linear combination of and : ${\ displaystyle m}$${\ displaystyle a}$${\ displaystyle m}$${\ displaystyle g}$${\ displaystyle s}$${\ displaystyle t}$${\ displaystyle g}$${\ displaystyle a}$${\ displaystyle m}$
${\ displaystyle g = \ operatorname {ggT} (a, m) = s \ cdot a + t \ cdot m}$
One then receives a solution with , and the other solutions differ from by a multiple of . ${\ displaystyle \ textstyle x_ {1} = {\ frac {s \ cdot c} {g}}}$${\ displaystyle x_ {1}}$${\ displaystyle {\ tfrac {m} {g}}}$
Example: is solvable, because divides the number , and there are solutions in the area . The extended Euclidean algorithm gives what the solution is. The solutions are congruent modulo . The solution set for is thus . ${\ displaystyle 4x \ equiv 10 {\ pmod {18}}}$${\ displaystyle \ operatorname {gcd} (4.18) = 2}$${\ displaystyle 10}$${\ displaystyle 2}$${\ displaystyle 0 \ leq x <18}$${\ displaystyle 2 = -4 \ times 4 + 1 \ times 18}$${\ displaystyle \ textstyle x_ {1} = {\ frac {-4 \ cdot 10} {2}} = - 20}$${\ displaystyle \ textstyle {\ frac {18} {2}} = 9}$${\ displaystyle x}$${\ displaystyle \ {\ cdots, -20, -11, -2,7,16,25, \ cdots \}}$
### Simultaneous congruence
A simultaneous congruence like
${\ displaystyle \ qquad a_ {1} x \ equiv c_ {1} {\ pmod {m_ {1}}}}$
${\ displaystyle \ qquad a_ {2} x \ equiv c_ {2} {\ pmod {m_ {2}}}}$
${\ displaystyle \ qquad a_ {3} x \ equiv c_ {3} {\ pmod {m_ {3}}}}$
can certainly be solved if:
• for all is through divisible d. H. every congruence is solvable for itself, and${\ displaystyle i}$${\ displaystyle c_ {i}}$${\ displaystyle g_ {i} = \ operatorname {ggT} (a_ {i}, m_ {i})}$
• which are pairwise mutually prime .${\ displaystyle {\ tfrac {m_ {i}} {g_ {i}}}}$
The proof of the Chinese remainder theorem provides the solution for such simultaneous congruences.
## Relationship to the modulo function
### General
With , generally applies: ${\ displaystyle a, b, m \ in \ mathbb {Z}}$${\ displaystyle m \ neq 0}$
${\ displaystyle a \ equiv b {\ pmod {m}} \ quad \ Leftrightarrow \ quad (a {\ bmod {m}}) = (b {\ bmod {m}}) \ quad \ Leftrightarrow \ quad (ab) {\ bmod {m}} = 0}$
### programming
Are two numbers and congruent modulo a number , resulting from the division by the same rest . ${\ displaystyle a}$${\ displaystyle b}$${\ displaystyle m}$${\ displaystyle m}$
With the help of the "symmetrical variant" of the modulo function , which is particularly widespread in computer science and is often referred to in programming languages with the modulo operators mod or %, this can be written as follows:
(a mod m) = (b mod m) or. (a % m) = (b % m)
Note that with the symmetrical modulo function common in computer science, this is only true for positive and positive . So that the equation is actually for all and equivalent to congruence, one has to through ${\ displaystyle a}$${\ displaystyle b}$${\ displaystyle a}$${\ displaystyle b}$
${\ displaystyle (a {\ bmod {m}}): = a- \ left \ lfloor {\ frac {a} {m}} \ right \ rfloor \ cdot m}$
Use a defined mathematical modulo function , the result of which always has the same sign as ( is the Gaussian bracket ). With this definition, for example . ${\ displaystyle \ operatorname {mod}}$${\ displaystyle m}$${\ displaystyle \ lfloor \ cdot \ rfloor}$${\ displaystyle (-1) {\ bmod {7}} = 6}$
## Applications
Congruences or remainder classes are often helpful when you have to perform calculations with very large numbers.
An important statement about congruence of prime numbers is Fermat's little theorem or Fermat's prime number test .
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# On a Circular Table Cover of Radius 42 Cm, a Design is Formed by a Girl Leaving an Equilateral Triangle Abc in the Middle, as Shown in the Figure. Find the Covered Area of the Design.V - Mathematics
Sum
On a circular table cover of radius 42 cm, a design is formed by a girl leaving an equilateral triangle ABC in the middle, as shown in the figure. Find the covered area of the design. ["Use" sqrt(3) = 1.73, pi =22/7]
#### Solution
Construction: Join AO and extend it to D on BC.
Radius of the circle, r = 42 cm
∠OCD= 30°
cos30° = "DC"/"OC"
=> sqrt(3)/2 = "DC"/42
⇒ "DC" = 21sqrt(3)
=> "DC" = 2xx"DC" = 42sqrt(3) = 72.66 "cm"
sin 30°="OD"/"OC"
=> 1/2="OD"/42
⇒ OD = 21 cm
Now, AD = AO + OD = 42 + 21 = 63 cm
Area of shaded region = Area of circlec - Area of triangle ABC
= pi(OA)^2-1/2xx"AD"xx"AB"
=22/7(42)^2-1/2xx63xx72.66
= 5544 - 2288.79
= 3255.21 cm
Is there an error in this question or solution?
#### APPEARS IN
RS Aggarwal Secondary School Class 10 Maths
Chapter 18 Area of Circle, Sector and Segment
Exercise 18B | Q 48 | Page 835
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# What is the simplest form of 25 by 75?
## What is the simplest form of 25 by 75?
Reduce 25/75 to lowest terms
1. Find the GCD (or HCF) of numerator and denominator. GCD of 25 and 75 is 25.
2. 25 ÷ 2575 ÷ 25.
3. Reduced fraction: 13. Therefore, 25/75 simplified to lowest terms is 1/3.
## How do you turn 75 into a fraction?
Answer: 75% is written as 3/4 as a fraction in its simplest form.
Is equivalent fractions the same as simplifying?
You can simplify a fraction by dividing the numerator and denominator by the same number. You won’t be able to simplify a fraction if the same number won’t divide into both the numerator and denominator. The simplified fraction and the original fraction are said to be equivalent.
### What is 25/75 as a reduced fraction?
What is 25/75 Simplified? – 1/3 is the simplified fraction for 25/75. Simplify 25/75 to the simplest form. Online simplify fractions calculator to reduce 25/75 to the lowest terms quickly and easily. What is 25/75 Simplified?
### What is 25 75 as a factor of 3?
Thus, 25 75 is equivalent to 1 3 in the reduced form. (*) The factors are numbers that multiply each other to get another number. For example, 2 and 3 are factors of 6, because by multiplying them you get 6.
How do you find equivalent fractions to 3/4?
As long as you multiply both top and bottom of the fraction by the same number, you won’t change the value of the fraction, and you’ll create an equivalent fraction. Find fractions equivalent to 3/4 by multiplying the numerator and denominator by the same whole number: Therefore these are all equivalent fractions:
#### What are equivalent fractions?
Equivalent fractions are fractions with different numbers representing the same part of a whole. They have different numerators and denominators, but their fractional values are the same. For example, think about the fraction 1/2.
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# Investigate the number of winning lines in the game of connect 4.
Extracts from this document...
Introduction
Investigation
Connect 4
X X X X
This is a winning line in the game of connect 4 on a 4x5 board. Winning lines can be horizontal, vertical and diagonal. Investigate the number of winning lines in the game of connect 4.
The task is asking me to find out how many winning lines (connects) when you are connecting 4 there are on any size board.
What am I going to do
I am going to find out how many connect 4 there are in a 4x5 board.
•I will change the size of the box, but keep one value the width constant. And I will find a pattern in the number of connects there are in the different size boxes.
•I will use algebra to find a general formula for a NxWidth (W) box.
•I will then increase the width (constant) by one and work out a formula for that box.
•I will then find a pattern in the formulas for the different size boxes, connecting 4, and I will make a formula for the formula.
•I will then change the number that I will connect. For example 2, 3 or 5.
Connect 4
Firstly I will do a box with the width constant as 5 and I will change the height.
Hx5 Box
Any Number=N
Connects=C
Height= H
Width =W
Hx5 1 2 3 4 5 6
Connects 2 4 6 17 28 39 first layer
11 11 11 second layer
The box height of 1 and 2 do follow the pattern so
they are excluded. The connects go up by 11 each
time. There are only 2 layers so the equation we
use is this equation. C=aH+b (original equation)
Middle
As before the first 2 equations do not follow
the pattern so they are excluded. Also like before
the connects has 2 layers so we use this original
equation.
C=aH+b
We use the same method as before.
9=a3+b (1)
24=a4+b (2)
15=a (2)-(1)
Substitute ‘a’ which is 15 back into (1)
9=15x3+b
9=45+b
b= -36
Substitute‘a’ and ‘b’ back into original equation
C=15H-36 that is the equation for the number
of connects in a Nx6 box. But since the
first 2 heights didn’t follow the
pattern we didn’t use them in the
equation so this equation doesn’t
work for them.
Connect 4
Hx4 Box
Hx4 1 2 3 4 5 6
Connects 1 2 3 10 17 24 first layer
7 7 7 second layer
Like before the first to equations do not follow the
pattern so they are excluded. Also like before the
connects has 2 layers so we use this original
equation.
C=aH+b
We use the same method as before.
3=a3+b (1)
10=a4+b (2)
7=a (2)-(1)
Substitute ‘a’ back into (1)
3=7x3+b
3=21+b
b= -18
Substitute ‘a’ and ‘b’ back into original equation
C=7H-18
that is the equation for the number of connects in a Nx4
box. But since the first 2 heights didn’t follow the
pattern we didn’t use them in the equation so this
equation doesn’t work for them.
Formula For Connect 4
The formula for any box with a width of 4 is C=7H-18
The formula for any box with a width of 5 is C=11H-27
The formula for any box with a width of 6 is C=15H-36
Conclusion
(1)
18=a16+b4+z (2)
32=a25+b5+z (3)
14=a9+b (3)-(2) (4)
10=a7+b (2)-(1) (5)
4=2a (4)-(5)
2=a
Substitute ‘a’ back into (4)
14=2x9+b
14=18+b
b=-4
Substitute ‘a’ and ‘b’ into (3)
32=2x25-4x5+z
32=50-20+z
32=30+z
z=2
So we now know what ‘a’, ‘b’ and ‘z’ are so we sub them back into the original equation which was Fo=aC²+bC+z
Fo=2C²-4C+2 is the fourth number equation
So the first number equation is 4
The second number equation is 3C-3
The third number equation is 3C-3
The fourth number equation is Fo=2C²-4C+2
So we now join them together.
In the connect number equations the first 2 numbers were in brackets and so were the second 2. So we have to group the first 2 equations and the second 2 in brackets. But each equation has to be in its own brackets so we need to use double brackets.
The formula for connect 3 was (4W-6)H-(6W-8) we will use it as a bass.
(first number equation xW-second number equation)H-(third number equation-Fourth number equation)
But each number equation needs to be surrounded by its own brackets.
((first number equation xW)-(second number equation))H-((third number equation)xW-(Fourth number equation))
((4W)-(3C-3))H-((3C-3)xW-(2C²-4C+2))
This formula finds out how many connects there are in any size box using any connecting number. E.g you could use a height of 5 and width of 4 and we are connecting 4. This would give you the answer of 17 which is correct.
But as before the formula does not work if the height is 2 or more numbers lower than the number you are connecting.
This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.
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# Superposition Problem with Four Voltage and Current Sources
Determine $V_x$ and $I_x$ using the superposition method.
Solution
I. Contribution of the $-5V$ voltage source:
To find the contribution of the $-5V$ voltage source, other three sources should be turned off. The $3V$ voltage source should be replaced by short circuit. The current source should be replaced with open circuits, as shown below.
It is trivial that $I_{x1}= \frac{-5 V}{2 \Omega}=-2.5 A$ . The current of the $3\Omega$ resistor is zero. Using KVL, $-(-5V)+V_{3\Omega}-V_{x1}=0 \to V_{x1}=-(-5V)=5V$ .
II. Contribution of the $3V$ voltage source:
Similarly, the $-5V$ voltage source becomes a short circuit and the current source should be replaced with open circuits:
The current of the $2\Omega$ resistor is zero because of being short circuited. It is trivial that $I_{x2}= 0 A$ (current of an open circuit). The current of the $3\Omega$ resistor is also zero. Using KVL, $-(3V)+V_{2\Omega}+V_{x2}+V_{3\Omega}=0 \to V_{x2}=3V$ .
III. Contribution of the $-1A$ current source:
The voltage sources should be replaced by short circuits and the $2A$ current source becomes with open circuit:
Again, the $2\Omega$ resistor is short circuited and its current is zero. it is clear that $I_{x3}= 1 A$ . The current of the $3\Omega$ resistor is equal to $-1A$ . Using KVL, $V_{x3}+V_{3\Omega}=0 \to V_{x3}+(-1A)\times (3\Omega)=0 \to V_{x3}=3V$ .
IV. Contribution of the $2A$ current source:
Likewise, the voltage sources should be replaced by short circuits and the $-1A$ current source becomes with open circuit:
Again, the $2\Omega$ resistor is short circuited and its current is zero. it is also trivial that $I_{x4}= 0A$ . The current of the $3\Omega$ resistor is $2A$ . Using KVL, $V_{x4}+V_{3\Omega}=0 \to V_{x4}+(2A)\times (3\Omega)=0 \to V_{x4}=-6V$ .
V. Adding up the individual contributions algebraically:
$V_x=V_{x1}+V_{x2}+V_{x3}+V_{x4}=5V+3V+3V-6V\to V_x=5V$
$I_x=I_{x1}+I_{x2}+I_{x3}+I_{x4}=-2.5A+1A+0A-0A\to I_x=-1.5A$
## 14 thoughts on “Superposition Problem with Four Voltage and Current Sources”
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# Which Equation Is Y = –6×2 + 3x + 2 Rewritten In Vertex Form?
• Puzzle
Rewriting an equation in another form, or rewriting an equation, can be useful for solving for a variable. Vertex form is one way to rewrite equations that can be useful in solving for a variable.
An equation in vertex form has the variable listed as a single value, called the vertex, and values above and below the variable that are added or subtracted from it to make it match the sides of the triangle formed by the sides of the equation.
Solving an equation in vertex form requires finding what are called incantations. An incantation is simply a set of steps to take to solve an equation in vertex form. These incantations can be remembered easily if you think of the term “incantation” as referring to how you chant these steps to solve the problem.
This article will discuss how to rewrite y = –6×2 + 3x + 2 in vertex form and how to solve such an equation in vertex form.
## Identify the y-intercept
The y-intercept of an equation is the value when x equals zero. For example, if y = 2x + 1 then 1 is the y-intercept because when x equals zero, then y equals one.
Bullet point: Rewrite in vertex form
To rewrite the equation in vertex form, first solve for the variable x and then put that value on the x-axis. To solve for x, take the opposite of what is in front of it and divide it by what is behind it.
In this case, -6×2 + 3x + 2 is rewritten as (x – 2)2. The square on both sides of the variable indicates that this equation is in vertex form.
## Rewrite the equation in y = a(x – h) + k format
Once you have the vertex form, you can rewrite the equation in y = a(x – h) + k format. This step is necessary because you will need to know how to write the graph of the function in vertex form in order to solve for k.
To do this, first write the equation in y = a(x – h) format. Then, take the a(x – h) and multiply it by -1, then add h. Finally, add k to the end of the variable. This will solve for k, and you can ignore it since it is not used in the graph of the function.
Example: To rewrite Y = -6×2 + 3x + 2 in vertex form, first write y = a(x – h) + k. Then, take out a(x – h) and put it over -1, then add 2h + k.
## Find where the line crosses the y-axis
To rewrite the equation in vertex form, start by putting the y-intercept in for y. Since the y-intercept is 2, put 2 for y.
Then solve for x by putting –6×2 + 3x + 2 into an equation solver like one found in Google Docs. You should get –6x = 0 as the solution for x, so put 0 for x in the y = 2 equation.
Check to make sure that the variables are on opposite sides of the equation by looking at the coefficients. The coefficient of y is –6, so make sure that –6 is not a variable in the new equation. The coefficient of x is 0, so make sure that 0 is not a variable in the new equation.
The only variable left is y, so put an equals sign before it and replace it with −6×0 + 3×0 + 2 = −6(0) + 3(0) + 2=2 which is true.
## Identify a, b, and c values
In order to rewrite the equation in vertex form, you must first identify the a, b, and c values. The a value is the y-coordinate of the vertex, the b value is the x-coordinate of the vertex, and the c value is the magnitude of the vertex (the coefficient of x).
The a value can be any non-zero number. The b value must be 0, and the c value can be any positive or negative number. These requirements ensure that there is only one possible way to write an equation in vertex form.
For example, look at the equation Y = –6×2 + 3x + 2. In order to write this in vertex form, you must first identify the a (y-coordinate of vertex), b (x-coordinate of vertex), and c (magnitude of coordinate) values.
The a value is –2,thebvalueis0andthecvalueis2.
## Plug a, b, and c values into y = a(x – h) + k format to find k
Now let’s go back to our original Y = –6×2 + 3x + 2 equation and see how we can rewrite it in vertex form.
First, remember that we can always rearrange equations as long as we do not change the values or order of operations. This makes it easy to rewrite an equation in vertex form!
So start by rearranging the equation into the following format: y = a(x − h) + k
Where h is the horizontal shift, a is the slope of the line, and k is the vertical shift of the line. Now, plug in some values for a, h, and k to find out what they are. Then check to see if the original equation equals this new one.
## Use algebra to find k value
Now, let’s use our equation in vertex form to find the value of k. Since the variable k represents the height of the line, we can assume that k is equal to the y-coordinate of the vertex.
To find the value of k, we need to solve for it. To do this, we first have to isolate k on one side of the equation using algebra. Then we can solve for k by using math operations like addition, subtraction, multiplication, and division.
For example, if our equation in vertex form is y = –6×2 + 3x + 2, then isolating k on one side of the equation would look like this:
Then solving for k would look like this: k = 2 = 2×2 = 4k=4 Now that we have found the value of k, let’s plug it back into our original equation in vertex form:y=–6x²+3x+2.y=–6(4)²+3(4)+(2)(4)=16+12+8=36 Y=36!So now you know how to find both x and y values using vertex form.Happy mathming!With all of that under your belt, you are ready to solve more complex equations in vertex form!When an equation in slope-intercept form is transformed into an equation in vertex form, only one more variable is added. This makes it a bit easier to solve for x and y values separately.
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# Equal Fractions Concept
The equal fractions concept is perhaps one of the most important concepts students must know for primary school Math. I have seen so many questions, again and again, which test this concept and students always get stuck.
Many students are confused about when to make the denominators the same and when to make the numerators the same.
Here is a simple rule:
If both fractions refer to the same whole, you can make the denominators the same.
Example of Making Denominators The Same
John gives away 1/2 of his sugar to his brother and 1/4 of his sugar to his sister. What fraction of his sugar does he give away in total?
In this question, both fractions refer to the same total.
Hence, we simply add 1/2 and 1/4 together to get the answer.
What if both fractions refer to different wholes?
Example of Equal Fractions Concept
1/2 of boys is equal to 1/3 of girls. What is the ratio of boys to girls?
Since the numerator of both fractions are the same, you can easily compare the denominator and that will be the answer:
Boys : Girls
= 2 : 3
What if the numerators are different?
Example,
1/2 of boys is equal to 2/3 of girls. What is the ratio of boys to girls?
In this case, you convert 1/2 to 2/4 to make the numerators the same.
Then you compare the denominators.
2/4 of boys = 2/3 of girls
So, boys : girls = 4 : 3
This concept is so useful that it can be applied to many different problem sums.
Let’s look at an example.
There are 836 students in a school. 7/10 of the boys and 7/8 of the girls take bus to school. The number of boys who do not take bus is twice the number of girls who do not take bus. How many girls do not take bus?
Solutions
Fraction of boys who do not take bus = 1 – 7/10 = 3/10
Fraction of girls who do not take bus = 1 – 7/8 = 1/8
Since “the number of boys who do not take bus is twice the number of girls who do not take bus”,
3/10 of boys = 2 x 1/8 of girls
3/10 of boys = 1/4 of girls
Now, we apply the equal fractions concept.
1/4 = 3/12
3/10 of boys = 3/12 of girls
Boys : Girls = 10 : 12 = 5 : 6
Since “there are 836 students in a school”,
11 u = 836
1 u = 76
Total girls = 6 x 76 = 456
Girls who do not take bus = 1/8 x 456 = 57 (Answer)
As you can see, using the equal fractions concept is useful, direct and simple. There is no need for complicated Algebra workings.
Do learn when and how to apply the equal fractions concept. It will be very useful in your P5 and P6 Math exam papers. We will cover it extensively in our tuition classes too.
## Free Download: 80 Useful Tricks to Solve Math Problems Easily
We hope that this article has helped your child to better understand the Equal Fractions Concept.
Before you go, download this eBook for free.
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# In the given figure, $AD$ is a median of a triangle $ABC$ and $AM \perp BC$. Prove that$\mathrm{AC}^{2}+\mathrm{AB}^{2}=2 \mathrm{AD}^{2}+\frac{1}{2} \mathrm{BC}^{2}$"
Given:
$AD$ is a median of a triangle $ABC$ and $AM \perp BC$.
To do:
We have to prove that $\mathrm{AC}^{2}+\mathrm{AB}^{2}=2 \mathrm{AD}^{2}+\frac{1}{2} \mathrm{BC}^{2}$
Solution:
In $\triangle \mathrm{AMC}$,
$\angle \mathrm{AMC}=90^{\circ}$
This implies, by Pythagoras theorem,
$\mathrm{AC}^{2}=\mathrm{AM}^{2}+\mathrm{MC}^{2}$
$\mathrm{AC}^{2}=\mathrm{AM}^{2}+(\mathrm{MD}^{2}+\mathrm{DC}^{2})$ ($\mathrm{MC}=\mathrm{MD}+\mathrm{DC}$)
$\mathrm{AC}^{2}=\mathrm{AM}^{2}+(\mathrm{MD}+\frac{1}{2} \mathrm{BC})^{2}$
$\mathrm{AC}^{2}=\mathrm{AM}^{2}+\mathrm{MD}^{2}+(\frac{\mathrm{BC}}{2})^{2}+2 \mathrm{MD} \times \frac{\mathrm{BC}}{2}$
$\mathrm{AC}^{2}=(\mathrm{AM}^{2}+\mathrm{MD}^{2})+\mathrm{MD} \times \mathrm{BC}+(\frac{\mathrm{BC}}{2})^{2}$
$\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{MD} \times \mathrm{BC}+(\frac{\mathrm{BC}}{2})^{2}$......(i)
In $\triangle \mathrm{AMB}$, by Pythagoras theorem,
$\mathrm{AB}^{2}=\mathrm{AM}^{2}+\mathrm{BM}^{2}$
$\mathrm{AB}^{2}=\mathrm{AM}^{2}+(\mathrm{BD}-\mathrm{MD})^{2}$
$\mathrm{AB}^{2}=\mathrm{AM}^{2}+(\frac{\mathrm{BC}}{2}-\mathrm{MD})^{2}$
$\mathrm{AB}^{2}=\mathrm{AM}^{2}+\mathrm{MD}^{2}+(\frac{\mathrm{BC}}{2})^{2}-\frac{2 \mathrm{BC}}{2} \times \mathrm{MD}$
$\mathrm{AB}^{2}=(\mathrm{AM}^{2}+\mathrm{MD}^{2})+(\frac{\mathrm{BC}}{2})^{2}-\mathrm{BC} \times \mathrm{MD}$
$\mathrm{AB}^{2}=\mathrm{AD}^{2}-\mathrm{BC} \times \mathrm{MD}+(\frac{\mathrm{BC}}{2})^{2}$.........(ii)
Adding equations (i) and (ii), we get,
$AC^{2}+AB^{2}=2AD^{2}+\frac{2 BC^{2}}{4}$
$AC^{2}+AB^{2}=2AD^{2}+\frac{BC^{2}}{2}$
Hence proved.
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Updated on: 10-Oct-2022
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3.2.2 Algebraic Expressions II, PT3 Focus Practice
Question 6:
Solution:
$\begin{array}{l}-4\left(5x-3\right)+7x-1\\ =-20x+12+7x-1\\ =-13x+11\end{array}$
Question 7:
$\left(3x-2y\right)-\left(x+4y\right)$
Solution:
$\begin{array}{l}\left(3x-2y\right)-\left(x+4y\right)\\ =3x-2y-x-4y\\ =2x-6y\end{array}$
Question 8:
Solution:
$\begin{array}{l}-3\left(2a-4b\right)-\frac{1}{3}\left(6a-15b\right)\\ =-6a+12b-2a+5b\\ =-8a+17b\end{array}$
Question 9:
$\frac{1}{3}\left(-2x+6y-9z\right)-\frac{1}{6}\left(-4x-18y+24z\right)$
Solution:
$\begin{array}{l}\frac{1}{3}\left(-2x+6y-9z\right)-\frac{1}{6}\left(-4x-18y+24z\right)\\ =\overline{)-\frac{2}{3}x}+2y-3z\overline{)+\frac{2}{3}x}+3y-4z\\ =5y-7z\end{array}$
Question 10:
$\frac{1}{2}\left(a+6bc\right)-\frac{1}{5}\left(3+2bc-2a\right)$
Solution:
$\begin{array}{l}\frac{1}{2}\left(a+6bc\right)-\frac{1}{5}\left(3+2bc-2a\right)\\ =\frac{1}{2}a+3bc-\frac{3}{5}-\frac{2}{5}bc+\frac{2}{5}a\\ =\frac{5a+4a}{10}+\frac{15bc-2bc}{5}-\frac{3}{5}\\ =\frac{9a}{10}+\frac{13bc}{5}-\frac{3}{5}\end{array}$
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# Difference between revisions of "1988 IMO Problems/Problem 4"
## Problem
Show that the solution set of the inequality
$\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}$
is a union of disjoint intervals, the sum of whose length is $1988$.
## Solution
Consider the graph of $f(x)=\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4}$. On the values of $x$ between $n$ and $n+1$ for $n\in\mathbb{N}$ $1\le n\le 69$, the terms of the form $\frac{k}{x-k}$ for $k\ne n,n+1$ have a finite range. In contrast, the term $\frac{n}{x-n}$ has an infinite range, from $+\infty$ to $n$. Similarly, the term $\frac{n+1}{x-n-1}$ has infinite range from $-n-1$ to $-\infty$. Thus, since the two undefined values occur at the distinct endpoints, we can deduce that $f(x)$ takes on all values between $+\infty$ and $-\infty$ for $x\in(n,n+1)$. Thus, by the Intermediate Value Theorem, we are garunteed a $n such that $f(r_n)=\frac{5}{4}$. Additionally, we have that for $x>70$, the value of $f(x)$ goes from $+\infty$ to $0$, since as $x$ increases, all the terms go to $0$. Thus, there exists some $r_{70}>70$ such that $f(r_{70})=\frac{5}{4}$ and so $f(x)\ge\frac{5}{4}$ for $x\in(70,r_{70})$.
So, we have $70$ $r_i$ such that $f(r_i)=\frac{5}{4}$. There are obviously no other such $r_i$ since $f(x)=\frac{5}{4}$ yields a polynomial of degree $70$ when combining fractions. Thus, we have that the solution set to the inequality $f(x)\ge\frac{5}{4}$ is the union of the intervals $(n,r_n]$ (since if $f(x)<\frac{5}{4}$ for $x\in(n,r_n)$ then there would exist another solution to the equation $f(x)=\frac{5}{4}$.
Thus we have proven that the solution set is the union of disjoint intervals. Now we are to prove that the sum of their lengths is $1988$.
The sum of their lengths is $r_1+r_2+\cdots+r_{70}-(1+2+\cdots+70)=r_1+r_2+\cdots+r_{70}-35\cdot71$. We have that the equation $f(x)=\frac{5}{4}$ yields a polynomial with roots $r_i$. Thus, opposite of the coeficient of $x^{69}$ divided by the leading coefficient is the sum of the $r_i$. It is easy to see that the coefficient of $x^{69}$ is $-5(1+2+\cdots+70)-4(1+2+\cdots+70)=-9\cdot35\cdot 71$. Thus, since the leading coefficient is $5$ we have $r_1+r_2+\cdots+r_{70}=9\cdot7\cdot71$. Thus, the sum of the lengths of the intervals is $63\cdot71-35\cdot71=28\cdot71=1988$ as desired.
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Chapter 7 - Section 7.5 - Multiplying with More Than One Term and Rationalizing Denominators - Exercise Set - Page 549: 13
$34 - 15\sqrt{5}$
Work Step by Step
RECALL: (1) Distributive Property: For any real numbers a, b, and c, $a(b+c)=ab+ac$ (2) For any real numbers real numbers a and b within the domain, $\sqrt{a} \cdot \sqrt{b}=\sqrt{ab}$ (3) $(a+b)(c-d) = a(c-d) + b(c-d)$ Use rule (3) above to obtain: $=6(9 - 4\sqrt{5})+ \sqrt{5}(9 - 4\sqrt{5})$ Use rules (1) and (2) above then simplify to obtain: $=6(9) - 6(4\sqrt{5}) + \sqrt{5}(9) - \sqrt{5}(4\sqrt{5}) \\=54 - 24\sqrt{5} + 9\sqrt{5}-4\sqrt{5(5)} \\=54 + (-24+9)\sqrt{5} - 4\sqrt{25} \\=54 + (-15\sqrt{5}) - 4(5) \\=54 - 15\sqrt{5}-20 \\=34 - 15\sqrt{5}$
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Bases Larger than Ten
Reference > Mathematics > Number Theory > Bases
In base eight, and base five, and all the other bases we used in the previous sections, we used a subset of the digit symbols we use in base ten. Another way of saying that is, we use some of the symbols from base ten, and no extra symbols.
But that's because eight and five are less than ten. What happens if we use a base larger than ten? What if, for example, we use base sixteen? Well, now we need symbols for all the digits from 0 to...15, right? Those symbols are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9...
Uh oh...we've run out of symbols!
Never fear; whenever we run out of symbols, we always go visit our friends the English teachers and ask if we can use some of theirs. Okay, actually, we don't ask; we just steal them. We're going to use some letters to represent digits. So let's try again. The symbols we use for base sixteen are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F. (Double check that for me - are there sixteen symbols?)
Notice that we're using upper case letters; we're more likely to use lower case letters as variables, so it makes sense to avoid ambiguity as much as possible.
The only problem we have now is, we need to be able to look at those digits A, B, C, D, E, and F, and quickly recognize what they mean.
A = ten
B = eleven
C = twelve
D= thirteen
E = forteen
F = fifteen
And that works out exactly right; we have a digit for every value less than sixteen. Let's see if we can do some conversions now.
EXAMPLE
Find the base ten value of C4sixteen.
Since C represents twelve, 12 x 16 + 4 = 196
EXAMPLE
Find the base ten value of FFFsixteen
Since F represents 15, this is 15 x 162 + 15 x 16 + 15 = 4095
Incidentally, this is one less than 163; FFFsixteen is the largest base sixteen number you can write without using four digits. 4096 = 1000sixteen.
EXAMPLE
For this example, let's switch to base twelve. Find the base ten value of 3BAtwelve.
A and B still represent 10 and 11, so this is 3 x 122 + 11 x 12 + 10 = 574.
EXAMPLE
Find the base ten value of 4Htwenty-four.
Hopefully you noticed that we've got an H in there, and we haven't yet used an H; we've only gone up to F. But we can quickly figure out what H must represent; if F is fifteen, G is sixteen, and H is seventeen. Thus, the value is 4 x 24 + 17 = 113.
Questions
1.
The number 4G is written in base n. What is the smallest possible value of n, and why?
2.
What is the largest base we can have before we run out of letters of the alphabet?
3.
Find the value of 15sixteen as a base ten number.
4.
Find the value of EEsixteen as a base ten number.
5.
Find the value of EEeighteen as a base ten number.
6.
Find the value of 111eleven as a base ten number.
7.
What is the largest number that can be written in base twelve with three digits?
8.
What is the smallest number that can be written in base thirteen with three digits? What is the base ten value of that number?
Assign this reference page
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# Everything You Need to Know about Rational Numbers between 2 and 7
## What are Rational Numbers?
Rational numbers are numbers that can be expressed as a ratio of two integers. They are also known as fractional numbers because they can be expressed as fractions. The term “rational” comes from the Latin word “ratio” which means “to calculate”. Rational numbers include all whole numbers, integers, and fractions.
## What is the Range of Rational Numbers?
The range of rational numbers is infinite because they can be expressed as fractions with infinite denominators. This means that any number can be expressed as a rational number. However, the range of rational numbers between 2 and 7 is limited. This range includes all the numbers that can be expressed as fractions with numerators between 2 and 7.
## Examples of Rational Numbers between 2 and 7
Some examples of rational numbers between 2 and 7 include 2/3, 3/4, 4/5, 5/6, and 6/7. All of these fractions have numerators between 2 and 7 and have a denominator that is greater than the numerator. There are also many other fractions in this range, including 2/7, 3/7, 4/7, 5/7, and 6/7.
## Interesting Facts About Rational Numbers between 2 and 7
One interesting fact about rational numbers between 2 and 7 is that all of these fractions can be reduced to lowest terms. This means that all of these fractions can be expressed as fractions with numerators and denominators that are both prime numbers. Another interesting fact is that all of these fractions are also terminating decimals. This means that when you divide the numerator of any of these fractions by the denominator, the answer will be a decimal that has a finite number of places after the decimal point.
### Conclusion
Rational numbers between 2 and 7 are a subset of all rational numbers. These numbers can be expressed as fractions with numerators between 2 and 7, and they can also be reduced to lowest terms and expressed as terminating decimals. Understanding these numbers and how they work can be useful in many mathematical applications.
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#### You may also like
List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?
### Magic Sums and Products
How to build your own magic squares.
### No Matter
Weekly Problem 2 - 2013
After performing some operations, what number is your answer always a multiple of?
# Perimeter Expressions
### Why do this problem?
This problem gives students plenty of opportunity to practise manipulating algebraic expressions within a purposeful context. Along the way, the challenges will provoke some insights that will be worth sharing. Issues relating to the dimensions in formulae for areas and lengths might emerge.
### Possible approach
This printable worksheet may be useful: Perimeter Expressions.pdf
Hand out sheets of paper (A4 is ideal) and introduce the problem, giving students time to cut out their five rectangles.
Show the image of Charlie's shape and ask students to work out the perimeter in terms of $a$ and $b$. The area of the smallest rectangle could be designated $R$ and used as a unit to express other areas, if students are not confident enough in their algebra to express it in terms of $a$ and $b$.
Once everyone is happy that the area is $9R$ or $9ab$ and the perimeter is $10a + 4b$, ask them to combine the largest and smallest rectangles (edge to edge, corner to corner) to make other shapes, and work out the areas and perimeters. How many different perimeters can they find?
Once they have done this, discuss: are they surprised that they only found one other possible perimeter, and that the two answers could be reached in several different ways?
The next challenge is to make different shapes using at least two of the rectangles, and again work out the area and perimeter. Students could make a shape in pairs and work out the area and perimeter to give to another pair (without letting them see the shape), to see if they can create a shape with the same area and perimeter. There is opportunity for fruitful discussion when the two shapes are compared, to see what is the same and what is different, and to see if one can be transformed into the other.
Alternatively, collect together the areas and perimeters on the board. Then in pairs, challenge students to combine rectangles to make each of the area/perimeter combinations listed on the board, checking each other's work as they go along.
Finally, the "questions to consider" provide some interesting points for discussion - students could be invited to suggest questions they think mathematicians might ask themselves, with the list in the problem used to prompt ideas if necessary.
### Key questions
How do we know whether an expression represents an area rather than a perimeter?
What does the area expression tell us about the pieces used to make the shape?
Are there multiple ways to make a given area/perimeter combination?
### Possible extension
Number Pyramids builds on this activity by challenging students to create algebraic expressions to explain a relationship.
Fit for Photocopying is a very challenging extension, involving exploration of ratio and functions in the context of paper sizes.
### Possible support
Use numerical values, instead of $a$ and $b$, and cut the rectangles out from squared paper. Begin by looking for numerical relationships rather than algebraic ones, and perhaps introduce the algebra to explain the patterns students find.
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Precalculus by Richard Wright
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…though he may stumble, he will not fall, for the Lord upholds him with his hand. Psalms 37:24 NIV
# 4-09 Compositions Involving Inverse Trigonometric Functions
Summary: In this section, you will:
• Evaluate compositions of inverse functions
SDA NAD Content Standards (2018): PC.5.3
Right triangles such as the one in figure 1 can be used to simplify compositions of trigonometric functions such as sin(tan–1 x).
## Compositions of Inverse Functions
Lesson 1-09 stated that if f(x) and f–1(x) were inverses, then f(f–1(x)) = x and f–1(f(x)) = x. The same is true for trigonometric functions with an exception. The domain of the inverse functions must be applied.
###### Composition of Inverse Trigonometric Functions
• If –1 ≤ x ≤ 1 and $$–\frac{π}{2}$$ ≤ y ≤ $$\frac{π}{2}$$, then sin(sin–1(x)) = x and sin–1(sin(y)) = y
• If –1 ≤ x ≤ 1 and 0 ≤ yπ, then cos(cos–1(x)) = x and cos–1(cos(y)) = y
• If x is a real number and $$–\frac{π}{2}$$ ≤ y ≤ $$\frac{π}{2}$$, then tan(tan–1(x)) = x and tan–1(tan(y)) = y
Remember to be careful that the domain and range of the composition is maintained. Work through the composition from the inside out.
#### Example 1: Evaluate Compositions of Inverse Trig Functions
Evaluate a) $$\sin\left(\arcsin \frac{1}{2}\right)$$, b) $$\cos\left(\cos^{–1} \frac{2π}{3}\right)$$, c) tan(arctan (–10)).
###### Solution
1. $$\sin\left(\arcsin \frac{1}{2}\right)$$: arcsin is the inner function, and the domain of arcsin is –1 ≤ x ≤ 1. $$\frac{1}{2}$$ is in this domain. $$\arcsin\left(\frac{1}{2}\right) = \frac{π}{6}$$, then find $$\sin\left(\frac{π}{6}\right) = \frac{1}{2}$$. So $$\sin\left(\arcsin \frac{1}{2}\right) = \frac{1}{2}$$.
2. $$\cos\left(\cos^{–1} \frac{2π}{3}\right)$$: cos–1 is the inner function, and the domain of cos–1 is –1 ≤ x ≤ 1. $$\frac{2π}{3} \approx 2$$, so $$\frac{2π}{3}$$ is outside of the domain and thus there is no solution for $$\cos\left(\cos^{–1} \frac{2π}{3}\right)$$.
3. tan(arctan (–10)): arctan is the inner function, and the domain of arctan is any real number. –10 is a real number, so tan(arctan (–10)) = –10.
##### Try It 1
Evaluate a) sin(sin–1(0.345)) and b) $$\cos\left(\cos^{–1} \left(–\frac{2}{3}\right)\right)$$.
0.345; $$–\frac{2}{3}$$
#### Example 2: Evaluate Compositions of Inverse Trig Functions
Evaluate a) $$\arcsin\left(\sin \frac{π}{3}\right)$$, b) $$\arccos\left(\cos \frac{5π}{4}\right)$$, and c) tan–1(tan π).
###### Solution
1. $$\arcsin\left(\sin \frac{π}{3}\right)$$: Work from the inside out. $$\sin \frac{π}{3} = \frac{\sqrt{3}}{2}$$ so
$$\arcsin\left(\sin \frac{π}{3}\right)$$
$$= \arcsin \frac{\sqrt{3}}{2}$$
$$= \frac{π}{3}$$
2. $$\arccos\left(\cos \frac{5π}{4}\right)$$: $$\cos \frac{5π}{4} = –\frac{\sqrt{2}}{2}$$ so
$$\arccos\left(\cos \frac{5π}{4}\right)$$
$$= \arccos \left(–\frac{\sqrt{2}}{2}\right)$$
$$= \frac{3π}{4}$$
(Remember the range of arccos is 0 ≤ yπ.)
3. tan–1(tan π): tan π = 0 so
tan–1(tan π)
= tan–1(0)
= 0
(Remember the range of tan–1 is $$–\frac{π}{2}$$ ≤ y ≤ $$\frac{π}{2}$$.)
##### Try It 2
Evaluate a) $$\arctan\left(\tan \frac{3π}{4}\right)$$ and b) sin–1(sin(–0.354)).
$$–\frac{π}{4}$$; –0.354
Compositions of trigonometric functions can also be solved using right triangles. Use the inner function to draw a right triangle, then use the triangle to evaluate the outer function.
###### Use a Right Triangle to Solve Composition of Trigonometric Functions
1. Draw a right triangle to represent the inner function. Two sides should be labeled.
2. Use the Pythagorean Theorem to solve for the other side.
3. Use the triangle to evaluate the outer trigonometric function.
#### Example 3: Evaluate a Mixed Composition of Trig Functions
Evaluate a) $$\cos\left(\arcsin \frac{3}{5}\right)$$ and b) $$\tan\left(\cos^{–1} \left(–\frac{2}{3}\right)\right)$$.
Solutions
1. Start with the inner function, $$\arcsin \frac{3}{5}$$. $$\sin x = \frac{opposite}{hypotenuse}$$, so draw a right triangle in quadrant 1 and label the acute angle by the origin. The opposite side is 3 and the hypotenuse is 5. See figure 2. Use the Pythagorean theorem to find the other side.
32 + b2 = 52
b = 4
Now evaluate the outer function, cos u.
$$\cos u = \frac{adj}{hyp}$$
$$\cos u = \frac{4}{5}$$
2. Start with the inner function, $$\cos^{–1} \left(–\frac{2}{3}\right)$$ and draw a right triangle. Since the ratio of sides is negative, draw the triangle in the negative quadrant of the inverse trigonometric function. For cos–1 the negative quadrant is quadrant 2. Label the angle by the origin and the adjacent side –2 and hypotenuse 3. See figure 3. Use the Pythagorean theorem to find the other side.
(–2)2 + b2 = 32
$$b = \sqrt{5}$$
Now evaluate the outer function, tan u.
$$\tan u = \frac{opp}{adj}$$
$$\tan u = –\frac{\sqrt{5}}{2}$$
##### Try It 3
Evaluate $$\sin\left(\arctan \left(–\frac{12}{5}\right)\right)$$.
$$–\frac{12}{13}$$
#### Example 4: Composition of Trigonometric Functions
Evaluate a) $$\tan^{-1} \left(\cos \left(\frac{3π}{2}\right)\right)$$ and b) $$\cos^{-1} \left(\sin \left(\frac{2π}{3} \right)\right)$$.
###### Solutions
1. Start with the inner function, $$\cos \left(\frac{3π}{2}\right)$$. This time the input is an angle, so sketch that on the unit circle. Since it is a quadrantal angle, find a point on the terminal side such as (0, −1). See figure 4.
$$\cos u = x$$
$$\cos\left(\frac{3π}{2}\right) = 0$$
Now evaluate the outer function, inverse tangent.
$$\tan^{-1} \frac{y}{x} = u$$
$$\tan^{-1} \frac{0}{1} = 0$$
2. Start with the inner function, $$\sin \left(\frac{2π}{3}\right)$$. This time the input is an angle, so sketch that on the unit circle. Make note of the coordinate of the terminal side, $$\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$$. See figure 5.
$$\sin u = y$$
$$\sin\left(\frac{2π}{3}\right) = \frac{\sqrt{3}}{2}$$
Now evaluate the outer function, inverse cosine.
$$\cos^{-1} x = u$$
$$\cos^{-1} \frac{\sqrt{3}}{2} = \frac{π}{6}$$
##### Try It 4
Evaluate $$\arccos \left(\sin \left(\frac{π}{4}\right)\right)$$.
$$\frac{π}{4}$$
#### Example 5: Composition of Trigonometric Functions Using x
Rewrite as an algebraic expression a) sin(arccos(x)) and b) tan(sin–1(2x)).
Solutions
1. Draw a right triangle and label the sides. The ratio of the sides is $$x = \frac{x}{1}$$. Since the inner function is $$\arccos\left(\frac{x}{1}\right)$$, the adjacent side is x and the hypotenuse is 1. See figure 6. Use the Pythagorean theorem to find an expression for the third side.
x2 + b2 = 12
b2 = 1 – x2
$$b = \sqrt{1 – x^{2}}$$
Now evaluate the outer function, sine.
$$\sin u = \frac{opp}{hyp} = \frac{\sqrt{1 – x^{2}}}{1}$$
$$\sin u = \sqrt{1 – x^{2}}$$
2. Draw a right triangle and label the sides. The ratio of the sides is $$2x = \frac{2x}{1}$$. Since the inner function is $$\sin^{–1}\left(\frac{2x}{1}\right)$$, the opposite side is 2x and the hypotenuse is 1. See figure 7. Use the Pythagorean theorem to find an expression for the third side.
a2 + (2x)2 = 12
a2 + 4x2 = 1
a2 = 1 – 4x2
$$a = \sqrt{1 – 4x^{2}}$$
Now evaluate the outer function, tangent.
$$\tan u = \frac{opp}{adj}$$
$$\tan u = \frac{2x}{\sqrt{1 – 4x^{2}}}$$
##### Try It 5
Rewrite as an algebraic expression: $$\cos\left(\arctan\left(\frac{x}{2}\right)\right)$$.
$$\frac{2}{\sqrt{x^{2} + 4}}$$
##### Lesson Summary
###### Composition of Inverse Trigonometric Functions
• If –1 ≤ x ≤ 1 and $$–\frac{π}{2}$$ ≤ y ≤ $$\frac{π}{2}$$, then sin(sin–1(x)) = x and sin–1(sin(y)) = y
• If –1 ≤ x ≤ 1 and 0 ≤ yπ, then cos(cos–1(x)) = x and cos–1(cos(y)) = y
• If x is a real number and $$–\frac{π}{2}$$ ≤ y ≤ $$\frac{π}{2}$$, then tan(tan–1(x)) = x and tan–1(tan(y)) = y
Remember to be careful that the domain and range of the composition is maintained. Work through the composition from the inside out.
###### Using a Right Triangle to Solve Composition of Trigonometric Functions
1. Draw a right triangle to represent the inner function. Two sides should be labeled.
2. Use the Pythagorean Theorem to solve for the other side.
3. Use the triangle to evaluate the outer trigonometric function.
## Practice Exercises (*Optional)
Find the exact value, if possible, without a calculator. If it is not possible, explain why.
1. $$\tan^{-1}\left(\cos\left(\frac{π}{3}\right)\right)$$
2. $$\tan^{−1}\left(\sin\left(\frac{π}{2}\right)\right)$$
3. sin−1(tan(π))
4. $$\cos^{−1}\left(\sin\left(\frac{2π}{3}\right)\right)$$
5. $$\cos\left(\tan^{−1}\left(\frac{\sqrt{3}}{3}\right)\right)$$
6. $$\tan\left(\sin^{−1}\left(\frac{1}{2}\right)\right)$$
7. Find the exact value of the expression in terms of x with the help of a reference triangle.
8. tan(cos−1(x + 1))
9. $$\cos\left(\sin^{−1}\left(\frac{1}{x}\right)\right)$$
10. sin(tan−1(x − 2))
11. For what value of x does sin x = sin−1 x? Use a graphing calculator to approximate the answer.
12. Mixed Review
13. (4-08) Evaluate $$\arctan\left(\frac{\sqrt{3}}{3}\right)$$.
14. (4-08) Evaluate sin–1 1.
15. (4-05) If $$\sin θ = \frac{2}{3}$$ and tan θ < 0, find a) cos θ and b) cot θ.
16. *(4-03) Evaluate the six trigonometric functions for the indicated angles.
17. (3-04) Solve 2x + 2 = 64.
18. (2-09) Solve 2x2 + 3x + 1 > 0.
1. 0.46
2. $$\frac{π}{4}$$
3. 0
4. $$\frac{π}{6}$$
5. $$\frac{\sqrt{3}}{2}$$
6. $$\frac{\sqrt{3}}{3}$$
7. $$\frac{\sqrt{-x^2-2x}}{x + 1}$$
8. $$\frac{\sqrt{x^2−1}}{x}$$
9. $$\frac{x-2}{\sqrt{x^2-4x+5}}$$
10. x = 0
11. $$\frac{π}{6}$$
12. $$\frac{π}{2}$$
13. $$-\frac{\sqrt{5}}{3}$$; $$-\frac{\sqrt{5}}{2}$$
14. $$\sin α = \frac{2\sqrt{13}}{13}$$ $$\cos α = \frac{3\sqrt{13}}{13}$$ $$\tan α = \frac{2}{3}$$ $$\csc α = \frac{\sqrt{13}}{2}$$ $$\sec α = \frac{\sqrt{13}}{3}$$ $$\cot α = \frac{3}{2}$$
15. 4
16. $$\left(−∞, −1\right) ∪ \left(−\frac{1}{2}, ∞\right)$$
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# Evaluate Limits Graphically and Vertical Asymptotes
## Discontinuities' and asymptotes' effect on limits.
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Graphs to Find Limits
A limit can describe the end behavior of a function. This is called a limit at infinity or negative infinity. A limit can also describe the limit at any normal x\begin{align*}x\end{align*} value. Sometimes this is simply the height of the function at that point. Other times this is what you would expect the height of the function to be at that point even if the height does not exist or is at some other point. In the following graph, what are f(3), limx3f(x), limxf(x)\begin{align*}f(3), \ \lim_{x \to 3}f(x), \ \lim_{x \to \infty}f(x)\end{align*}?
### Using Graphs to Find Limits
When evaluating the limit of a function from its graph, you need to distinguish between the function evaluated at the point and the limit around the point.
Functions like the one above with discontinuities, asymptotes and holes require you to have a very solid understanding of how to evaluate and interpret limits.
At x=a\begin{align*}x=a\end{align*}, the function is undefined because there is a vertical asymptote. You would write:
f(a)=DNE, limxaf(x)=DNE\begin{align*}f(a)=DNE, \ \lim_{x \to a}f(x)=DNE\end{align*}
At x=b\begin{align*}x=b\end{align*}, the function is defined because the filled in circle represents that it is the height of the function. This appears to be at about 1. However, since the two sides do not agree, the limit does not exist here either.
f(b)=1, limxbf(x)=DNE\begin{align*}f(b)=1, \ \lim_{x \to b}f(x)=DNE\end{align*}
At x=0\begin{align*}x=0\end{align*}, the function has a discontinuity in the form of a hole. It is as if the point (0,2.4)\begin{align*}(0,-2.4)\end{align*} has been lifted up and placed at (0,1)\begin{align*}(0,1)\end{align*}. You can evaluate both the function and the limit at this point, however these quantities will not match. When you evaluate the function you have to give the actual height of the function, which is 1 in this case. When you evaluate the limit, you have to give what the height of the function is supposed to be based solely on the neighborhood around 0. By neighborhood around 0, we mean what is happening on the lines around x=0\begin{align*}x=0\end{align*}, not the at the point. Since the function appears to reach a height of -2.4 from both the left and the right, the limit does exist.
f(0)=1, limx0f(x)=2.4\begin{align*}f(0)=1,\ \lim_{x \to 0}f(x)=-2.4\end{align*}
At x=c\begin{align*}x=c\end{align*}, the limit does not exist because the left and right hand neighborhoods do not agree on a height. On the other hand, the filled in circle represents that the function is defined at x=c\begin{align*}x=c\end{align*} to be -3.
f(c)=3, limxcf(x)=DNE\begin{align*}f(c)=-3, \ \lim \limits_{x \to c}f(x)=DNE\end{align*}
At x\begin{align*}x \rightarrow \infty\end{align*} you may only discuss the limit of the function since it is not appropriate to evaluate a function at infinity (you cannot find f()\begin{align*}f(\infty)\end{align*}). Since the function appears to increase without bound, the limit does not exist.
limxf(x)=DNE\begin{align*}\lim \limits_{x \to \infty}f(x)=DNE\end{align*}
At x\begin{align*}x \rightarrow - \infty\end{align*} the graph appears to flatten as it moves to the left. There is a horizontal asymptote at y=0\begin{align*}y=0\end{align*} that this function approaches as x\begin{align*}x \rightarrow - \infty\end{align*}.
limxf(x)=0\begin{align*}\lim \limits_{x \to - \infty}f(x)=0\end{align*}
When evaluating limits graphically, your main goal is to determine whether the limit exists. The limit only exists when the left and right sides of the functions meet at a specific height. Whatever the function is doing at that point does not matter for the sake of limits. The function could be defined at that point, could be undefined at that point, or the point could be defined at some other height. Regardless of what is happening at that point, when you evaluate limits graphically, you only look at the neighborhood to the left and right of the function at the point.
### Examples
#### Example 1
Earlier, you were asked to find f(3), limx3f(x), limxf(x)\begin{align*}f(3), \ \lim_{x \to 3}f(x), \ \lim_{x \to \infty}f(x)\end{align*} given the graph of the function f(x)\begin{align*}f(x)\end{align*} to be:
limx3f(x)=12\begin{align*}\lim \limits_{x \to 3}f(x) =\frac{1}{2}\end{align*}
f(3)=3\begin{align*}f(3)=3\end{align*}
limxf(x)=DNE\begin{align*}\lim_{x \to \infty}f(x) =DNE\end{align*}
#### Example 2
Evaluate the following expressions using the graph of the function f(x)\begin{align*}f(x)\end{align*}.
1. limxf(x)\begin{align*}\lim_{x \to - \infty}f(x)\end{align*}
2. limx1f(x)\begin{align*}\lim_{x \to -1}f(x)\end{align*}
3. limx0f(x)\begin{align*}\lim_{x \to 0}f(x)\end{align*}
4. limx1f(x)\begin{align*}\lim_{x \to 1}f(x)\end{align*}
5. limx3f(x)\begin{align*}\lim_{x \to 3}f(x)\end{align*}
6. f(1)\begin{align*}f(-1)\end{align*}
7. f(2)\begin{align*}f(2)\end{align*}
8. f(1)\begin{align*}f(1)\end{align*}
9. f(3)\begin{align*}f(3)\end{align*}
1. limxf(x)=0\begin{align*}\lim_{x \to - \infty}f(x)=0\end{align*}
2. limx1f(x)=DNE\begin{align*}\lim_{x \to -1}f(x)=DNE\end{align*}
3. limx0f(x)=2\begin{align*}\lim_{x \to 0}f(x)=-2\end{align*}
4. limx1f(x)=0\begin{align*}\lim_{x \to 1}f(x)=0\end{align*}
5. limx3f(x)=DNE\begin{align*}\lim_{x \to 3}f(x)=DNE\end{align*} (This is because only one side exists and a regular limit requires both left and right sides to agree)
6. f(1)=0\begin{align*}f(-1)=0\end{align*}
7. f(0)=2\begin{align*}f(0)=-2\end{align*}
8. f(1)=2\begin{align*}f(1)=2\end{align*}
9. f(3)=0\begin{align*}f(3)=0\end{align*}
#### Example 3
Sketch a graph that has a limit at x=2\begin{align*}x=2\end{align*}, but that limit does not match the height of the function.
While there are an infinite number of graphs that fit this criteria, you should make sure your graph has a removable discontinuity at x=2\begin{align*}x=2\end{align*}.
#### Example 4
Sketch a graph that is defined at x=1\begin{align*}x=-1\end{align*} but limx1f(x)\begin{align*}\lim_{x \to -1}f(x)\end{align*} does not exist.
The graph must have either a jump or an infinite discontinuity at x=1\begin{align*} x=-1\end{align*} and also have a solid hole filled in somewhere on that vertical line.
#### Example 5
Evaluate and explain how to find the limits as \begin{align*}x\end{align*} approaches 0 and 1 for the graph below:
\begin{align*}\lim_{x \to 0}f(x)=2, \ \lim_{x \to 1}f(x)=1\end{align*}
Both of these limits exist because the left hand and right hand neighborhoods of these points seem to approach the same height. In the case of the point \begin{align*}(0,2)\end{align*} the function happened to be defined there. In the case of the point \begin{align*}(1,1)\end{align*} the function happened to be defined elsewhere, but that does not matter. You only need to consider what the function does right around the point.
### Review
Use the graph of \begin{align*}f(x)\end{align*} below to evaluate the expressions in 1-6.
\begin{align*}\lim_{x \to - \infty}f(x)\end{align*}1.
\begin{align*}\lim_{x \to \infty}f(x)\end{align*}2.
\begin{align*}\lim_{x \to 2}f(x)\end{align*}3.
\begin{align*}\lim_{x \to 0}f(x)\end{align*}4.
\begin{align*}f(0)\end{align*}5.
\begin{align*}f(2)\end{align*}6.
\begin{align*}g(x)\end{align*} below to evaluate the expressions in 7-13.Use the graph of
7. \begin{align*}\lim_{x \to - \infty}g(x)\end{align*}
8. \begin{align*}\lim_{x \to \infty}g(x)\end{align*}
9. \begin{align*}\lim_{x \to 2}g(x)\end{align*}
10. \begin{align*}\lim_{x \to 0}g(x)\end{align*}
11. \begin{align*}\lim_{x \to 4}g(x)\end{align*}
12. \begin{align*}g(0)\end{align*}
13. \begin{align*}g(2)\end{align*}
14. Sketch a function \begin{align*}h(x)\end{align*} such that \begin{align*}h(2)=4\end{align*}, but \begin{align*}\lim_{x \to 2}h(x)=DNE\end{align*}.
15. Sketch a function \begin{align*}j(x)\end{align*} such that \begin{align*}j(2)=4\end{align*}, but \begin{align*}\lim_{x \to 2}j(x)=3\end{align*}.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Asymptotes An asymptote is a line on the graph of a function representing a value toward which the function may approach, but does not reach (with certain exceptions).
discontinuities The points of discontinuity for a function are the input values of the function where the function is discontinuous.
DNE The phrase “does not exist” or “DNE” is used with limits to imply that the limit does not approach a particular numerical value. Sometimes this means that the limit continues to grow bigger or smaller to infinity. Sometimes it means that the limit can’t decide between two disagreeing values.
does not exist The phrase “does not exist” or “DNE” is used with limits to imply that the limit does not approach a particular numerical value.
End behavior End behavior is a description of the trend of a function as input values become very large or very small, represented as the 'ends' of a graphed function.
Hole A hole exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal to zero.
limit A limit is the value that the output of a function approaches as the input of the function approaches a given value.
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# LCM of 50 and 15
The lcm of 50 and 15 is the smallest positive integer that divides the numbers 50 and 15 without a remainder. Spelled out, it is the least common multiple of 50 and 15. Here you can find the lcm of 50 and 15, along with a total of three methods for computing it. In addition, we have a calculator you should check out. Not only can it determine the lcm of 50 and 15, but also that of three or more integers including fifty and fifteen for example. Keep reading to learn everything about the lcm (50,15) and the terms related to it.
## What is the LCM of 50 and 15
If you just want to know what is the least common multiple of 50 and 15, it is 150. Usually, this is written as
lcm(50,15) = 150
The lcm of 50 and 15 can be obtained like this:
• The multiples of 50 are …, 100, 150, 200, ….
• The multiples of 15 are …, 135, 150, 165, …
• The common multiples of 50 and 15 are n x 150, intersecting the two sets above, $\hspace{3px}n \hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$.
• In the intersection multiples of 50 ∩ multiples of 15 the least positive element is 150.
• Therefore, the least common multiple of 50 and 15 is 150.
Taking the above into account you also know how to find all the common multiples of 50 and 15, not just the smallest. In the next section we show you how to calculate the lcm of fifty and fifteen by means of two more methods.
## How to find the LCM of 50 and 15
The least common multiple of 50 and 15 can be computed by using the greatest common factor aka gcf of 50 and 15. This is the easiest approach:
lcm (50,15) = $\frac{50 \times 15}{gcf(50,15)} = \frac{750}{5}$ = 150
Alternatively, the lcm of 50 and 15 can be found using the prime factorization of 50 and 15:
• The prime factorization of 50 is: 2 x 5 x 5
• The prime factorization of 15 is: 3 x 5
• Eliminate the duplicate factors of the two lists, then multiply them once with the remaining factors of the lists to get lcm(50,50) = 150
In any case, the easiest way to compute the lcm of two numbers like 50 and 15 is by using our calculator below. Note that it can also compute the lcm of more than two numbers, separated by a comma. For example, enter 50,15. Push the button only to start over.
The lcm is...
Similar searched terms on our site also include:
## Use of LCM of 50 and 15
What is the least common multiple of 50 and 15 used for? Answer: It is helpful for adding and subtracting fractions like 1/50 and 1/15. Just multiply the dividends and divisors by 3 and 10, respectively, such that the divisors have the value of 150, the lcm of 50 and 15.
$\frac{1}{50} + \frac{1}{15} = \frac{3}{150} + \frac{10}{150} = \frac{13}{150}$. $\hspace{30px}\frac{1}{50} – \frac{1}{15} = \frac{3}{150} – \frac{10}{150} = \frac{-7}{150}$.
## Properties of LCM of 50 and 15
The most important properties of the lcm(50,15) are:
• Commutative property: lcm(50,15) = lcm(15,50)
• Associative property: lcm(50,15,n) = lcm(lcm(15,50),n) $\hspace{10px}n\neq 0 \hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$
The associativity is particularly useful to get the lcm of three or more numbers; our calculator makes use of it.
To sum up, the lcm of 50 and 15 is 150. In common notation: lcm (50,15) = 150.
If you have been searching for lcm 50 and 15 or lcm 50 15 then you have come to the correct page, too. The same is the true if you typed lcm for 50 and 15 in your favorite search engine.
Note that you can find the least common multiple of many integer pairs including fifty / fifteen by using the search form in the sidebar of this page.
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# 14.4: Methods of Determining Reaction Order
• Contributed by Anonymous
• LibreTexts
Learning Objectives
• To know how to determine the reaction order from experimental data.
In the examples in this text, the exponents in the rate law are almost always the positive integers: 1 and 2 or even 0. Thus the reactions are zeroth, first, or second order in each reactant. The common patterns used to identify the reaction order are described in this section, where we focus on characteristic types of differential and integrated rate laws and how to determine the reaction order from experimental data.
## Zeroth-Order Reactions
A zeroth-order reactionA reaction whose rate is independent of concentration. is one whose rate is independent of concentration; its differential rate law is rate = k. We refer to these reactions as zeroth order because we could also write their rate in a form such that the exponent of the reactant in the rate law is 0:
$$rate=-\dfrac{\Delta \left [ A \right ]}{\Delta t}=-k\left [ reactant \right ]^{0}=k(1)=k \tag{14.3.1}$$
Because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of −k. The value of k is negative because the concentration of the reactant decreases with time. Conversely, a graph of the concentration of any product as a function of time is a straight line with a slope of k, a positive value.
The graph of a zeroth-order reaction. The change in concentration of reactant and product with time produces a straight line.
The integrated rate law for a zeroth-order reaction also produces a straight line and has the general form
$$\left [ A \right ]=\left [ A_{0} \right ]-kt \tag{14.3.2}$$
where [A]0 is the initial concentration of reactant A. (Equation 14.3.2 has the form of the algebraic equation for a straight line, y = mx + b, with y = [A], mx = −kt, and b = [A]0.) In a zeroth-order reaction, the rate constant must have the same units as the reaction rate, typically moles per liter per second.
Although it may seem counterintuitive for the reaction rate to be independent of the reactant concentration(s), such reactions are rather common. They occur most often when the reaction rate is determined by available surface area. An example is the decomposition of N2O on a platinum (Pt) surface to produce N2 and O2, which occurs at temperatures ranging from 200°C to 400°C:
$$2N_{2}O \left ( g \right )\overset{Pt}{\rightarrow} 2N_{2} \left ( g \right ) + O_{2} \left ( g \right ) \tag{14.3.3}$$
Without a platinum surface, the reaction requires temperatures greater than 700°C, but between 200°C and 400°C, the only factor that determines how rapidly N2O decomposes is the amount of Pt surface available (not the amount of Pt). As long as there is enough N2O to react with the entire Pt surface, doubling or quadrupling the N2O concentration will have no effect on the reaction rate.At very low concentrations of N2O, where there are not enough molecules present to occupy the entire available Pt surface, the reaction rate is dependent on the N2O concentration. The reaction rate is as follows:
$$rate =-\dfrac{1}{2}\left (\dfrac{\Delta \left [N_{2}O \right ]}{\Delta t} \right )=\dfrac{1}{2}\left ( \dfrac{\Delta \left [N_{2} \right ]}{\Delta t} \right )=\dfrac{\Delta \left [O_{2} \right ]}{\Delta t} k\left [N_{2}O \right ]^{0}=k \tag{14.3.4}$$
Thus the rate at which N2O is consumed and the rates at which N2 and O2 are produced are independent of concentration. As shown in Figure 14.3.2, the change in the concentrations of all species with time is linear. Most important, the exponent (0) corresponding to the N2O concentration in the experimentally derived rate law is not the same as the reactant’s stoichiometric coefficient in the balanced chemical equation (2). For this reaction, as for all others, the rate law must be determined experimentally.
Figure 14.3.2 A Zeroth-Order Reaction This graph shows the concentrations of reactants and products versus time for the zeroth-order catalyzed decomposition of N2O to N2 and O2 on a Pt surface. The change in the concentrations of all species with time is linear.
### Note the Pattern
If a plot of reactant concentration versus time is linear, then the reaction is zeroth order in that reactant.
A zeroth-order reaction that takes place in the human liver is the oxidation of ethanol (from alcoholic beverages) to acetaldehyde, catalyzed by the enzymeA catalyst that occurs naturally in living organisms and catalyzes biological reactions. alcohol dehydrogenase. At high ethanol concentrations, this reaction is also a zeroth-order reaction. The overall reaction equation is
where NAD+ (nicotinamide adenine dinucleotide) and NADH (reduced nicotinamide adenine dinucleotide) are the oxidized and reduced forms, respectively, of a species used by all organisms to transport electrons. When an alcoholic beverage is consumed, the ethanol is rapidly absorbed into the blood. Its concentration then decreases at a constant rate until it reaches zero (part (a) in Figure 14.3.3). An average 70 kg person typically takes about 2.5 h to oxidize the 15 mL of ethanol contained in a single 12 oz can of beer, a 5 oz glass of wine, or a shot of distilled spirits (such as whiskey or brandy). The actual rate, however, varies a great deal from person to person, depending on body size and the amount of alcohol dehydrogenase in the liver. The reaction rate does not increase if a greater quantity of alcohol is consumed over the same period of time because the reaction rate is determined only by the amount of enzyme present in the liver. Contrary to popular belief, the caffeine in coffee is ineffective at catalyzing the oxidation of ethanol. When the ethanol has been completely oxidized and its concentration drops to essentially zero, the rate of oxidation also drops rapidly (part (b) in Figure 14.3.3).
Figure 14.3.3 The Catalyzed Oxidation of Ethanol(a) The concentration of ethanol in human blood decreases linearly with time, which is typical of a zeroth-order reaction. (b) The rate at which ethanol is oxidized is constant until the ethanol concentration reaches essentially zero, at which point the reaction rate drops to zero.
These examples illustrate two important points:
1. In a zeroth-order reaction, the reaction rate does not depend on the reactant concentration.
2. A linear change in concentration with time is a clear indication of a zeroth-order reaction.
## First-Order Reactions
In a first-order reactionA reaction whose rate is directly proportional to the concentration of one reactant., the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows:
$$rate=-\dfrac{\Delta \left [ A \right ]}{\Delta t}=k\left [ A \right ]^{1}=k\left [ A \right ] \tag{14.3.5}$$
If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth. Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s−1).
The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows:
$$rate= \left [ A \right ]= \left [ A_{0} \right ] e^{-kt} \tag{14.3.6}$$
where [A]0 is the initial concentration of reactant A at t = 0; k is the rate constant; and e is the base of the natural logarithms, which has the value 2.718 to three decimal places. (Essential Skills 6 in Section 11.9, discusses natural logarithms.) Recall that an integrated rate law gives the relationship between reactant concentration and time. Equation 14.20 predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of Equation 14.20 and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of A and t:
$$rate= ln \left [ A \right ]= ln \left [ A_{0} \right ] -kt \tag{14.3.7}$$
Because Equation 14.3.7 has the form of the algebraic equation for a straight line, y = mx + b, with y = ln[A] and b = ln[A]0, a plot of ln[A] versus t for a first-order reaction should give a straight line with a slope of −k and an intercept of ln[A]0. Either the differential rate law (Equation 14.3.5) or the integrated rate law (Equation 14.3.7) can be used to determine whether a particular reaction is first order.
Figure 14.3.4 Graphs of a first-order reaction. The expected shapes of the curves for plots of reactant concentration versus time (top) and the natural logarithm of reactant concentration versus time (bottom) for a first-order reaction.
First-order reactions are very common. In this chapter, we have already encountered two examples of first-order reactions: the hydrolysis of aspirin (Figure 14.2.1) and the reaction of t-butyl bromide with water to give t-butanol (Equation 14.7.7). Another reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin.
Cisplatin, the first “inorganic” anticancer drug to be discovered, is unique in its ability to cause complete remission of the relatively rare but deadly cancers of the reproductive organs in young adults. The structures of cisplatin and its hydrolysis product are as follows:
Both platinum compounds have four groups arranged in a square plane around a Pt(II) ion. The reaction shown in Figure 14.11 is important because cisplatin, the form in which the drug is administered, is not the form in which the drug is active. Instead, at least one chloride ion must be replaced by water to produce a species that reacts with deoxyribonucleic acid (DNA) to prevent cell division and tumor growth. Consequently, the kinetics of the reaction in Figure 14.11 have been studied extensively to find ways of maximizing the concentration of the active species.
### Note the Pattern
If a plot of reactant concentration versus time is not linear but a plot of the natural logarithm of reactant concentration versus time is linear, then the reaction is first order.
The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in Table 14.3.1. The table lists initial rate data for four experiments in which the reaction was run at pH 7.0 and 25°C but with different initial concentrations of cisplatin. Because the reaction rate increases with increasing cisplatin concentration, we know this cannot be a zeroth-order reaction. Comparing Experiments 1 and 2 in Table 14.3.1 shows that the reaction rate doubles [(1.8 × 10−5 M/min) ÷ (9.0 × 10−6 M/min) = 2.0] when the concentration of cisplatin is doubled (from 0.0060 M to 0.012 M). Similarly, comparing Experiments 1 and 4 shows that the reaction rate increases by a factor of 5 [(4.5 × 10−5 M/min) ÷ (9.0 × 10−6 M/min) = 5.0] when the concentration of cisplatin is increased by a factor of 5 (from 0.0060 M to 0.030 M). Because the reaction rate is directly proportional to the concentration of the reactant, the exponent of the cisplatin concentration in the rate law must be 1, so the rate law is rate = k[cisplatin]1. Thus the reaction is first order. Knowing this, we can calculate the rate constant using the differential rate law for a first-order reaction and the data in any row of Table 14.3.1. For example, substituting the values for Experiment 3 into Equation 14.19,
3.6 × 10−5 M/min = k(0.024 M)
1.5 × 10−3 min−1 = k
Table 14.3.1 Rates of Hydrolysis of Cisplatin as a Function of Concentration at pH 7.0 and 25°C
Experiment [Cisplatin]0 (M) Initial Rate (M/min)
1 0.0060 9.0 × 10−6
2 0.012 1.8 × 10−5
3 0.024 3.6 × 10−5
4 0.030 4.5 × 10−5
Knowing the rate constant for the hydrolysis of cisplatin and the rate constants for subsequent reactions that produce species that are highly toxic enables hospital pharmacists to provide patients with solutions that contain only the desired form of the drug.
### Example 14.3.1
At high temperatures, ethyl chloride produces HCl and ethylene by the following reaction:
$$CH_{3}CH_{2}Cl\left ( g \right )\overset{\Delta }{\rightarrow} HCl\left ( g \right )+C_{2}H_{4}\left ( g \right )$$
Using the rate data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the concentration of ethyl chloride and determine the rate constant for the reaction.
Experiment [CH3CH2Cl]0 (M) Initial Rate (M/s)
1 0.010 1.6 × 10−8
2 0.015 2.4 × 10−8
3 0.030 4.8 × 10−8
4 0.040 6.4 × 10−8
Given: balanced chemical equation, initial concentrations of reactant, and initial rates of reaction
Asked for: reaction order and rate constant
Strategy:
A Compare the data from two experiments to determine the effect on the reaction rate of changing the concentration of a species.
B Compare the observed effect with behaviors characteristic of zeroth- and first-order reactions to determine the reaction order. Write the rate law for the reaction.
C Use measured concentrations and rate data from any of the experiments to find the rate constant.
Solution:
The reaction order with respect to ethyl chloride is determined by examining the effect of changes in the ethyl chloride concentration on the reaction rate.
A Comparing Experiments 2 and 3 shows that doubling the concentration doubles the reaction rate, so the reaction rate is proportional to [CH3CH2Cl]. Similarly, comparing Experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate, again indicating that the reaction rate is directly proportional to [CH3CH2Cl].
B This behavior is characteristic of a first-order reaction, for which the rate law is rate = k[CH3CH2Cl].
C We can calculate the rate constant (k) using any row in the table. Selecting Experiment 1 gives the following:
1.60 × 10−8 M/s = k(0.010 M) 1.6 × 10−6 s−1 = k
Exercise
Sulfuryl chloride (SO2Cl2) decomposes to SO2 and Cl2 by the following reaction:
SO2Cl2(g) → SO2(g) + Cl2(g)
Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction.
Experiment [SO2Cl2]0 (M) Initial Rate (M/s)
1 0.0050 1.10 × 10−7
2 0.0075 1.65 × 10−7
3 0.0100 2.20 × 10−7
4 0.0125 2.75 × 10−7
Answer: first order; k = 2.2 × 10−5 s−1
Figure 14.3.5 The Hydrolysis of Cisplatin, a First-Order Reaction These plots show hydrolysis of cisplatin at pH 7.0 and 25°C as (a) the experimentally determined concentrations of cisplatin and chloride ions versus time and (b) the natural logarithm of the cisplatin concentration versus time. The straight line in (b) is expected for a first-order reaction.
We can also use the integrated rate law to determine the reaction rate for the hydrolysis of cisplatin. To do this, we examine the change in the concentration of the reactant or the product as a function of time at a single initial cisplatin concentration. Part (a) in Figure 14.3.5 shows plots for a solution that originally contained 0.0100 M cisplatin and was maintained at pH 7 and 25°C. The concentration of cisplatin decreases smoothly with time, and the concentration of chloride ion increases in a similar way. When we plot the natural logarithm of the concentration of cisplatin versus time, we obtain the plot shown in part (b) in Figure 14.3.5. The straight line is consistent with the behavior of a system that obeys a first-order rate law. We can use any two points on the line to calculate the slope of the line, which gives us the rate constant for the reaction. Thus taking the points from part (a) in Figure 14.3.5 for t = 100 min ([cisplatin] = 0.0086 M) and t = 1000 min ([cisplatin] = 0.0022 M),
$$slope = \dfrac{ln\left [ cisplatin \right ]_{1000}- ln \left [ cisplatin \right ]_{100}}{1000\;min-100\;min}$$
$$-k = \dfrac{ln 0.0022-ln 0.0086}{1000\;min-100\;min}=\dfrac{-6.12-\left ( -4.76 \right )}{900\; min}=-1.51\times 10^{-3} min^{-1}$$
$$k = 1.51\times 10^{-3} \; min^{-1}$$
The slope is negative because we are calculating the rate of disappearance of cisplatin. Also, the rate constant has units of min−1 because the times plotted on the horizontal axes in parts (a) and (b) in Figure 14.3.5 are in minutes rather than seconds.
The reaction order and the magnitude of the rate constant we obtain using the integrated rate law are exactly the same as those we calculated earlier using the differential rate law. This must be true if the experiments were carried out under the same conditions.
### Example 14.3.2
Refer back to Example 4. If a sample of ethyl chloride with an initial concentration of 0.0200 M is heated at 650°C, what is the concentration of ethyl chloride after 10 h? How many hours at 650°C must elapse for the concentration to decrease to 0.0050 M? (Recall that we calculated the rate constant for this reaction in Example 4.)
Given: initial concentration, rate constant, and time interval
Asked for: concentration at specified time and time required to obtain particular concentration
Strategy:
A Substitute values for the initial concentration ([A]0) and the calculated rate constant for the reaction (k) into the integrated rate law for a first-order reaction. Calculate the concentration ([A]) at the given time t.
B Given a concentration [A], solve the integrated rate law for time t.
Solution:
The exponential form of the integrated rate law for a first-order reaction (Equation 14.3.6) is [A] = [A]0ekt.
A Having been given the initial concentration of ethyl chloride ([A]0) and having calculated the rate constant in Example 4 (k = 1.6 × 10−6 s−1), we can use the rate law to calculate the concentration of the reactant at a given time t. Substituting the known values into the integrated rate law,
$$\left [ CH_{3} CH_{2}Cl \right ]_{10\;h}=\left [ CH_{3} CH_{2}Cl \right ]_{0}e^{-kt}$$
$$=0.0200\;M \; exp \left [ \left (-1.6\times 10^{-6}\;s^{-1} \right ) \left ( 10\;h \right ) \left ( 60\;min/h \right )\left ( 60 \;s/min \right ) \right ) ]$$
$$= 0.0189\;M We could also have used the logarithmic form of the integrated rate law (Equation 14.3.7): \( ln \left [ CH_{3} CH_{2}Cl \right ]_{10\;h}=ln \left [ CH_{3} CH_{2}Cl \right ]_{0} -kt$$
$$= ln 0.0200\;M - \left (-1.6\times 10^{-6}\;s^{-1} \right ) \left ( 10\;h \right ) \left ( 60\;min/h \right )\left ( 60 \;s/min \right )$$
$$= -3.912--0.0576=-3.970$$
$$\left [ CH_{3} CH_{2}Cl \right ]_{10\;h}=e^{-3.970}\;M$$
=0.0189\;M
B To calculate the amount of time required to reach a given concentration, we must solve the integrated rate law for t. Equation 14.3.7 gives the following:
$$ln \left [ CH_{3} CH_{2}Cl \right ]_{t}=ln \left [ CH_{3} CH_{2}Cl \right ]_{0} -kt$$
$$kt= ln \left [ CH_{3} CH_{2}Cl \right ]_{0} -ln \left [ CH_{3} CH_{2}Cl \right ]_{t}= ln \; \dfrac{\left [ CH_{3} CH_{2}Cl \right ]_{0}}{ \left [ CH_{3} CH_{2}Cl \right ]_{t}}$$
$$t= \dfrac{1}{k}\; ln\dfrac{\left [ CH_{3} CH_{2}Cl \right ]_{0}}{\left [ CH_{3} CH_{2}Cl \right ]_{t}} = \dfrac{1}{1.6\times 10^{-6}\;s^{-1}}\; ln\dfrac{0.0200\;M}{0.0050\;M}$$
$$t= \dfrac{ln\;4.0}{k}\; =8.7\times 10^{5}\;s=240\;h=2.4\times 10^{2}\;h$$
Exercise
In the exercise in Example 4, you found that the decomposition of sulfuryl chloride (SO2Cl2) is first order, and you calculated the rate constant at 320°C. Use the form(s) of the integrated rate law to find the amount of SO2Cl2 that remains after 20 h if a sample with an original concentration of 0.123 M is heated at 320°C. How long would it take for 90% of the SO2Cl2 to decompose?
## Second-Order Reactions
The simplest kind of second-order reactionA reaction whose rate is proportional to the square of the concentration of the reactant (for a reaction with the general form 2A → products) or is proportional to the product of the concentrations of two reactants (for a reaction with the general form A + B → products). is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form 2A → products. A second kind of second-order reaction has a reaction rate that is proportional to the product of the concentrations of two reactants. Such reactions generally have the form A + B → products. An example of the former is a dimerization reaction, in which two smaller molecules, each called a monomer, combine to form a larger molecule (a dimer).
The differential rate law for the simplest second-order reaction in which 2A → products is as follows:
$$rate=-\dfrac{\Delta \left [ A \right ]}{2 \Delta t}=k\left [ reactant \right ]^{2} \tag{14.3.8}$$
Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M−1·s−1). Because the units of molarity are expressed as mol/L, the unit of the rate constant can also be written as L(mol·s).
For the reaction 2A → products, the following integrated rate law describes the concentration of the reactant at a given time:
$$\dfrac{1 }{\left [ A \right ]}= \dfrac{1 }{\left [ A_{0} \right ]} +kt \tag{14.3.9}$$
Because Equation 14.3.9 has the form of an algebraic equation for a straight line, y = mx + b, with y = 1/[A] and b = 1/[A]0, a plot of 1/[A] versus t for a simple second-order reaction is a straight line with a slope of k and an intercept of 1/[A]0.
### Note the Pattern
Second-order reactions generally have the form 2A → products or A + B → products.
Simple second-order reactions are common. In addition to dimerization reactions, two other examples are the decomposition of NO2 to NO and O2 and the decomposition of HI to I2 and H2. Most examples involve simple inorganic molecules, but there are organic examples as well. We can follow the progress of the reaction described in the following paragraph by monitoring the decrease in the intensity of the red color of the reaction mixture.
Many cyclic organic compounds that contain two carbon–carbon double bonds undergo a dimerization reaction to give complex structures. One example is as follows:
For simplicity, we will refer to this reactant and product as “monomer” and “dimer,” respectively.The systematic name of the monomer is 2,5-dimethyl-3,4-diphenylcyclopentadienone. The systematic name of the dimer is the name of the monomer followed by “dimer.” Because the monomers are the same, the general equation for this reaction is 2A → product. This reaction represents an important class of organic reactions used in the pharmaceutical industry to prepare complex carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed using either the differential rate law (Equation 14.3.8) or the integrated rate law (Equation 14.3.9).
To determine the differential rate law for the reaction, we need data on how the reaction rate varies as a function of monomer concentrations, which are provided in Table 14.3.2 . From the data, we see that the reaction rate is not independent of the monomer concentration, so this is not a zeroth-order reaction. We also see that the reaction rate is not proportional to the monomer concentration, so the reaction is not first order. Comparing the data in the second and fourth rows shows that the reaction rate decreases by a factor of 2.8 when the monomer concentration decreases by a factor of 1.7:
$$\dfrac{5.0\times 10^{-5}\;M/min }{1.8\times 10^{-5}\;M/mi}=2.8 \;\;and\;\;\dfrac{3.4\times 10^{-3}\;M/min }{2.0\times 10^{-3}\;M/mi}=1.7$$
Table 14.3.2 Rates of Reaction as a Function of Monomer Concentration for an Initial Monomer Concentration of 0.0054 M
Time (min) [Monomer] (M) Instantaneous Rate (M/min)
10 0.0044 8.0 × 10−5
26 0.0034 5.0 × 10−5
44 0.0027 3.1 × 10−5
70 0.0020 1.8 × 10−5
120 0.0014 8.0 × 10−6
Because (1.7)2 = 2.9 ≈ 2.8, the reaction rate is approximately proportional to the square of the monomer concentration.
$$rate \;\;\alpha\;\;\left [ monomer \right ]^{2}$$
This means that the reaction is second order in the monomer. Using Equation 14.3.8 and the data from any row in Table 14.3.2, we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following:
$$rate = k\left [ A^{2} \right ]$$
$$8.0\times 10^{-5}\;M/min = k\left (4.4\times 10^{-3}\;M \right )$$
$$4.1\;min^{-1} = k$$
We can also determine the reaction order using the integrated rate law. To do so, we use the decrease in the concentration of the monomer as a function of time for a single reaction, plotted in part (a) in Figure 14.3.6. The measurements show that the concentration of the monomer (initially 5.4 × 10−3 M) decreases with increasing time. This graph also shows that the reaction rate decreases smoothly with increasing time. According to the integrated rate law for a second-order reaction, a plot of 1/[monomer] versus t should be a straight line, as shown in part (b) in Figure 14.3.6. Any pair of points on the line can be used to calculate the slope, which is the second-order rate constant. In this example, k = 4.1 M−1·min−1, which is consistent with the result obtained using the differential rate equation. Although in this example the stoichiometric coefficient is the same as the reaction order, this is not always the case. The reaction order must always be determined experimentally.
Figure 14.3.6 Dimerization of a Monomeric Compound, a Second-Order Reaction These plots correspond to dimerization of the monomer in Figure 14.13 " " as (a) the experimentally determined concentration of monomer versus time and (b) 1/[monomer] versus time. The straight line in (b) is expected for a simple second-order reaction.
For two or more reactions of the same order, the reaction with the largest rate constant is the fastest. Because the units of the rate constants for zeroth-, first-, and second-order reactions are different, however, we cannot compare the magnitudes of rate constants for reactions that have different orders. The differential and integrated rate laws for zeroth-, first-, and second-order reactions and their corresponding graphs are shown in Figure 14.4.2 in Section 14.4.
### Example 14.3.3
At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen.
$$2NO_{2}\left ( g \right )\overset{\Delta }{\rightarrow} 2NO\left ( g \right )+O_{2}\left ( g \right )$$
Experimental data for the reaction at 300°C and four initial concentrations of NO2 are listed in the following table:
Experiment [NO2]0 (M) Initial Rate (M/s)
1 0.015 1.22 × 10−4
2 0.010 5.40 × 10−5
3 0.0080 3.46 × 10−5
4 0.0050 1.35 × 10−5
Determine the reaction order and the rate constant.
Given: balanced chemical equation, initial concentrations, and initial rates
Asked for: reaction order and rate constant
Strategy:
A From the experiments, compare the changes in the initial reaction rates with the corresponding changes in the initial concentrations. Determine whether the changes are characteristic of zeroth-, first-, or second-order reactions.
B Determine the appropriate rate law. Using this rate law and data from any experiment, solve for the rate constant (k).
Solution:
A We can determine the reaction order with respect to nitrogen dioxide by comparing the changes in NO2 concentrations with the corresponding reaction rates. Comparing Experiments 2 and 4, for example, shows that doubling the concentration quadruples the reaction rate [(5.40 × 10−5) ÷ (1.35 × 10−5) = 4.0], which means that the reaction rate is proportional to [NO2]2. Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to [NO2]2. This behavior is characteristic of a second-order reaction.
B We have rate = k[NO2]2. We can calculate the rate constant (k) using data from any experiment in the table. Selecting Experiment 2, for example, gives the following:
$$rate = k\left [ NO_{2} \right ]^{2}$$
$$5.40\times 10^{-5}\; M/s = k\left (0.010\;M \right )^{2}$$
$$0.540\; M^{-1}s^{-1} = k$$
Exercise
When the highly reactive species HO2 forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows:
2HO2(g) → H2O2(g) + O2(g)
The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table:
Experiment [HO2]0 (M) Initial Rate (M/s)
1 1.1 × 10−8 1.7 × 10−7
2 2.5 × 10−8 8.8 × 10−7
3 3.4 × 10−8 1.6 × 10−6
4 5.0 × 10−8 3.5 × 10−6
Determine the reaction order and the rate constant.
Answer: second order in HO2; k = 1.4 × 109 M−1·s−1
### Note the Pattern
If a plot of reactant concentration versus time is not linear but a plot of 1/reaction concentration versus time is linear, then the reaction is second order.
### Example 14.3.4
If a flask that initially contains 0.056 M NO2 is heated at 300°C, what will be the concentration of NO2 after 1.0 h? How long will it take for the concentration of NO2 to decrease to 10% of the initial concentration? Use the integrated rate law for a second-order reaction (Equation 14.3.9) and the rate constant calculated in Example 6.
Given: balanced chemical equation, rate constant, time interval, and initial concentration
Asked for: final concentration and time required to reach specified concentration
Strategy:
A Given k, t, and [A]0, use the integrated rate law for a second-order reaction to calculate [A].
B Setting [A] equal to 1/10 of [A]0, use the same equation to solve for t.
Solution:
A We know k and [NO2]0, and we are asked to determine [NO2] at t = 1 h (3600 s). Substituting the appropriate values into Equation 14.3.9,
$$\dfrac{1}{\left [NO_{2} \right ]_{3600}} = \dfrac{1}{\left [NO_{2} \right ]_{0}} +kt=\dfrac{1}{0.056\;M} +\left ( 0.54\;M^{-1}s^{-1} \right )\left (3600\; s \right )$$
$$=2.0\times 10^{-3} \; M^{-1}$$
Thus [NO2]3600 = 5.1 × 10−4 M.
B In this case, we know k and [NO2]0, and we are asked to calculate at what time [NO2] = 0.1[NO2]0 = 0.1(0.056 M) = 0.0056 M. To do this, we solve Equation 14.3.9 for t, using the concentrations given.
$$\dfrac{\dfrac{1}{\left [NO_{2} \right ]_{3600}} - \dfrac{1}{\left [NO_{2} \right ]_{0}}}{k} =\dfrac{\left ( 1/0.0056\;M \right )-\left ( 1/0.056\;M \right )}{0.54\;M^{-1}s^{-1}}=3.0\times10^{2}\;s=5\;min$$
NO2 decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min.
Exercise
In the exercise in Example 6, you calculated the rate constant for the decomposition of HO2 as k = 1.4 × 109 M−1·s−1. This high rate constant means that HO2 decomposes rapidly under the reaction conditions given in the problem. In fact, the HO2 molecule is so reactive that it is virtually impossible to obtain in high concentrations. Given a 0.0010 M sample of HO2, calculate the concentration of HO2 that remains after 1.0 h at 25°C. How long will it take for 90% of the HO2 to decompose? Use the integrated rate law for a second-order reaction (Equation 14.3.9) and the rate constant calculated in the exercise in Example 6.
Answer: 2.0 × 10−13 M; 6.4 × 10−6 s
In addition to the simple second-order reaction and rate law we have just described, another very common second-order reaction has the general form A + B → products, in which the reaction is first order in A and first order in B. The differential rate law for this reaction is as follows:
$$rate=-\dfrac{\Delta \left [ A \right ]}{\Delta t}=-\dfrac{\Delta \left [ B \right ]}{\Delta t}=k\left [ A \right ]\left [ B \right ] \tag{14.3.10}$$
Because the reaction is first order both in A and in B, it has an overall reaction order of 2. (The integrated rate law for this reaction is rather complex, so we will not describe it.) We can recognize second-order reactions of this sort because the reaction rate is proportional to the concentrations of each reactant. We presented one example at the end of Section 14.2, the reaction of CH3Br with OH to produce CH3OH.
## Determining the Rate Law of a Reaction
The number of fundamentally different mechanisms (sets of steps in a reaction) is actually rather small compared to the large number of chemical reactions that can occur. Thus understanding reaction mechanismsThe sequence of events that occur at the molecular level during a reaction. can simplify what might seem to be a confusing variety of chemical reactions. The first step in discovering the reaction mechanism is to determine the reaction’s rate law. This can be done by designing experiments that measure the concentration(s) of one or more reactants or products as a function of time. For the reaction A + B → products, for example, we need to determine k and the exponents m and n in the following equation:
$$rate=k\left [ A \right ]^{m}\left [ B \right ]^{n} \tag{14.3.11}$$
To do this, we might keep the initial concentration of B constant while varying the initial concentration of A and calculating the initial reaction rate. This information would permit us to deduce the reaction order with respect to A. Similarly, we could determine the reaction order with respect to B by studying the initial reaction rate when the initial concentration of A is kept constant while the initial concentration of B is varied. In earlier examples, we determined the reaction order with respect to a given reactant by comparing the different rates obtained when only the concentration of the reactant in question was changed. An alternative way of determining reaction orders is to set up a proportion using the rate laws for two different experiments.
Rate data for a hypothetical reaction of the type A + B → products are given in Table 14.3.3. The general rate law for the reaction is given in Equation 14.3.11. We can obtain m or n directly by using a proportion of the rate laws for two experiments in which the concentration of one reactant is the same, such as Experiments 1 and 3 in Table 14.3.3.
Table 14.3.3 Rate Data for a Hypothetical Reaction of the Form A + B → Products
Experiment [A] (M) [B] (M) Initial Rate (M/min)
1 0.50 0.50 8.5 × 10−3
2 0.75 0.50 19 × 10−3
3 1.00 0.50 34 × 10−3
4 0.50 0.75 8.5 × 10−3
5 0.50 1.00 8.5 × 10−3
$$\dfrac{rate^{1}}{rate^{3}}= \dfrac{k\left [ A^{1} \right ]^{m}\left [ B^{1} \right ]^{n}}{k\left [ A^{3} \right ]^{m}\left [ B^{3} \right ]^{n}}$$
Inserting the appropriate values from Table 14.3.3,
$$\dfrac{8.5\times10^{-3}\;\cancel{M/min}}{34.\times10^{-3}\;\cancel{M/min}}= \dfrac{k\left [ 0.50\;\cancel{M} \right ]^{m}\cancel{\left [ 05.0\;M \right ]^{n}}}{k\left [ 1.00\;\cancel{M} \right ]^{m}\cancel{\left [ 0.50\;M \right ]^{n}}}$$
Because 1.00 to any power is 1, [1.00 M]m = 1.00 M. We can cancel like terms to give 0.25 = [0.50]m, which can also be written as 1/4 = [1/2]m. Thus we can conclude that m = 2 and that the reaction is second order in A. By selecting two experiments in which the concentration of B is the same, we were able to solve for m.
Conversely, by selecting two experiments in which the concentration of A is the same (e.g., Experiments 5 and 1), we can solve for n.
$$\dfrac{rate^{1}}{rate^{3}}= \dfrac{k\left [ A^{1} \right ]^{m}\left [ B^{1} \right ]^{n}}{k\left [ A^{5} \right ]^{m}\left [ B^{5} \right ]^{n}}$$
Substituting the appropriate values from Table 14.3.3,
$$\dfrac{8.5\times10^{-3}\;\cancel{M/min}}{8.5\times10^{-3}\;\cancel{M/min}}= \dfrac{\cancel{k\left [ 0.50\;\cancel{M} \right ]^{m}}\left [ 05.0\;\cancel{M} \right ]^{n}}{\cancel{k\left [ 0.50\;M \right ]^{m}}\left [ 1.00\;\cancel{M} \right ]^{n}}$$
Canceling leaves 1.0 = [0.50]n, which gives n = 0; that is, the reaction is zeroth order in B. The experimentally determined rate law is therefore
$$rate=k\left [ A \right ]^{2}\left [ B \right ]^{0}$$
We can now calculate the rate constant by inserting the data from any row of Table 14.3.3into the experimentally determined rate law and solving for k. Using Experiment 2, we obtain
$$19.\times10^-3\; M/min=k\left ( 0.75\;M \right )^{2}$$
$$3.4\times10^-2\; M^{-1}min^{-1}=k$$
You should verify that using data from any other row of Table 14.3.3 gives the same rate constant. This must be true as long as the experimental conditions, such as temperature and solvent, are the same.
### Example 14.3.5
Nitric oxide is produced in the body by several different enzymes and acts as a signal that controls blood pressure, long-term memory, and other critical functions. The major route for removing NO from biological fluids is via reaction with O2 to give NO2, which then reacts rapidly with water to give nitrous acid and nitric acid:
These reactions are important in maintaining steady levels of NO. The following table lists kinetics data for the reaction of NO with O2 at 25°C:
2NO(g) + O2(g) → 2NO2(g)
Determine the rate law for the reaction and calculate the rate constant.
Experiment [NO]0 (M) [O2]0 (M) Initial Rate (M/s)
1 0.0235 0.0125 7.98 × 10−3
2 0.0235 0.0250 15.9 × 10−3
3 0.0470 0.0125 32.0 × 10−3
4 0.0470 0.0250 63.5 × 10−3
Given: balanced chemical equation, initial concentrations, and initial rates
Asked for: rate law and rate constant
Strategy:
A Compare the changes in initial concentrations with the corresponding changes in rates of reaction to determine the reaction order for each species. Write the rate law for the reaction.
B Using data from any experiment, substitute appropriate values into the rate law. Solve the rate equation for k.
Solution:
A Comparing Experiments 1 and 2 shows that as [O2] is doubled at a constant value of [NO2], the reaction rate approximately doubles. Thus the reaction rate is proportional to [O2]1, so the reaction is first order in O2. Comparing Experiments 1 and 3 shows that the reaction rate essentially quadruples when [NO] is doubled and [O2] is held constant. That is, the reaction rate is proportional to [NO]2, which indicates that the reaction is second order in NO. Using these relationships, we can write the rate law for the reaction:
$$rate =k\left [ NO \right ]^{2}\left [ O_{2} \right ]$$
B The data in any row can be used to calculate the rate constant. Using Experiment 1, for example, gives
$$k=\dfrac{k}{\left [ NO \right ]^{2}\left [ O_{2} \right ]} = \dfrac{7.98\times10{-3}\;M/s}{\left ( 0.0235\;M \right )^{2}\left ( 0.0125\;M \right )}=1.16\times10{3}\;M^{-2}s^{-1}$$
The overall reaction order (m + n) is 3, so this is a third-order reaction, a reaction whose rate is determined by three reactants. The units of the rate constant become more complex as the overall reaction order increases.
Exercise
The peroxydisulfate ion (S2O82−) is a potent oxidizing agent that reacts rapidly with iodide ion in water:
$$S_{2}O_{8}^{2-}\left ( aq \right )+3I^{-}\left ( aq \right ) \rightarrow 2SO_{4}^{2-}\left ( aq \right )+I_{3}^{-}\left ( aq \right )$$
The following table lists kinetics data for this reaction at 25°C. Determine the rate law and calculate the rate constant.
Experiment [S2O82−]0 (M) [I]0 (M) Initial Rate (M/s)
1 0.27 0.38 2.05
2 0.40 0.38 3.06
3 0.40 0.22 1.76
Answer: rate = k[S2O82−][I]; k = 20 M−1·s−1
### Summary
The reaction rate of a zeroth-order reaction is independent of the concentration of the reactants. The reaction rate of a first-order reaction is directly proportional to the concentration of one reactant. The reaction rate of a simple second-order reaction is proportional to the square of the concentration of one reactant. Knowing the rate law of a reaction gives clues to the reaction mechanism.
### Key Takeaway
• Either the differential rate law or the integrated rate law can be used to determine the reaction order from experimental data.
### Key Equations
zeroth-order reaction
Equation 14.3.1 $$rate = -\dfrac{\Delta \left [ A \right ] }{\Delta t} = k$$
Equation 14.3.2: $$\left [ A \right ]=\left [ A_{0} \right ]-kt$$
first-order reaction
Equation 14.3.5: $$rate=-\dfrac{\Delta \left [ A \right ]}{\Delta t}=k\left [ A \right]$$
Equation 14.3.6: $$\left [ A \right ]= \left [ A_{0} \right ] e^{-kt}$$
Equation 14.3.7: $$ln \left [ A \right ]= ln \left [ A_{0} \right ] -kt$$
second-order reaction
Equation 14.3.8: $$ln \left [ A \right ]= ln \left [ A_{0} \right ] +kt$$
Equation 14.3.9: $$\dfrac{1 }{\left [ A \right ]}= \dfrac{1 }{\left [ A_{0} \right ]}+kt$$
### Conceptual Problems
1. What are the characteristics of a zeroth-order reaction? Experimentally, how would you determine whether a reaction is zeroth order?
2. Predict whether the following reactions are zeroth order and explain your reasoning.
1. a substitution reaction of an alcohol with HCl to form an alkyl halide and water
2. catalytic hydrogenation of an alkene
3. hydrolysis of an alkyl halide to an alcohol
4. enzymatic conversion of nitrate to nitrite in a soil bacterium
3. In a first-order reaction, what is the advantage of using the integrated rate law expressed in natural logarithms over the rate law expressed in exponential form?
4. If the reaction rate is directly proportional to the concentration of a reactant, what does this tell you about (a) the reaction order with respect to the reactant and (b) the overall reaction order?
5. The reaction of NO with O2 is found to be second order with respect to NO and first order with respect to O2. What is the overall reaction order? What is the effect of doubling the concentration of each reagent on the reaction rate?
### Numerical Problems
1. Iodide reduces Fe(III) according to the following reaction:
2Fe3+(soln) + 2I(soln) → 2Fe2+(soln) + I2(soln)
Experimentally, it was found that doubling the concentration of Fe(III) doubled the reaction rate, and doubling the iodide concentration increased the reaction rate by a factor of 4. What is the reaction order with respect to each species? What is the overall rate law? What is the overall reaction order?
2. Benzoyl peroxide is a medication used to treat acne. Its rate of thermal decomposition at several concentrations was determined experimentally, and the data were tabulated as follows:
Experiment [Benzoyl Peroxide]0 (M) Initial Rate (M/s)
1 1.00 2.22 × 10−4
2 0.70 1.64 × 10−4
3 0.50 1.12 × 10−4
4 0.25 0.59 × 10−4
What is the reaction order with respect to benzoyl peroxide? What is the rate law for this reaction?
3. 1-Bromopropane is a colorless liquid that reacts with S2O32− according to the following reaction:
C3H7Br + S2O32− → C3H7S2O3 + Br
The reaction is first order in 1-bromopropane and first order in S2O32−, with a rate constant of 8.05 × 10−4 M−1·s−1. If you began a reaction with 40 mmol/100 mL of C3H7Br and an equivalent concentration of S2O32−, what would the initial reaction rate be? If you were to decrease the concentration of each reactant to 20 mmol/100 mL, what would the initial reaction rate be?
4. The experimental rate law for the reaction 3A + 2B → C + D was found to be Δ[C]/Δt = k[A]2[B] for an overall reaction that is third order. Because graphical analysis is difficult beyond second-order reactions, explain the procedure for determining the rate law experimentally.
1. First order in Fe3+; second order in I; third order overall; rate = k[Fe3+][I]2.
2.
3. 1.29 × 10−4 M/s; 3.22 × 10−5 M/s
4.
### Contributors
• Anonymous
Modified by Joshua B. Halpern
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Tori has 1/2 pound of sugar in her cabinet. Her cake recipe calls for 2/10 of a pound of sugar. How many cakes can she make?
Mar 8, 2018
Exactly $2.5$ cakes (or $2$ whole cakes if you need to round)
Explanation:
So Tori has $\frac{1}{2}$ pound of sugar and one cake calls for $\frac{2}{10}$ of sugar. All we have to do is divide the fractions to see how many cakes she can make.
How do you divide fractions? It's actually pretty easy. Here are our two fractions:
$\frac{1}{2} \div \frac{2}{10}$
Now all you have to do is flip the second fraction upside down to be the reciprocal and change the $\div$ sign to a $\times$ sign.
$\frac{1}{2} \textcolor{\mathmr{and} a n \ge}{\div} \frac{\textcolor{red}{2}}{\textcolor{b l u e}{10}}$ becomes
$\frac{1}{2} \textcolor{\mathmr{and} a n \ge}{\times} \frac{\textcolor{b l u e}{10}}{\textcolor{red}{2}}$
Now all you have to do is multiply the two top numbers (numerators) together and multiply the two bottom numbers (denominators) together. I changed the problem a little to make it more clear:
(1 xx 10)/(2 xx 2) = ?
$\frac{10}{4} = 2.5$
Tori can make exactly $2.5$ cakes, or $2$ whole cakes.
Mar 8, 2018
She can make $\frac{5}{2} = 2.5$ cakes if she can make a half-recipe cake, or $2$ whole cakes with some sugar left over.
Explanation:
CountryGal answered first, and did a very nice job. I just wanted to share another method as an alternative.
Tori has $\frac{1}{2}$ pounds of sugar, but needs $\frac{2}{10}$ for each cake. We can convert $\frac{1}{2}$ into tenths: $\frac{5}{10}$.
Then we divide $\frac{5}{10}$ by $\frac{2}{10}$. We'll invert and multiply, as CountryGal did:
$\frac{5}{10} \times \frac{10}{2} = \frac{5}{\cancel{10}} \times \frac{\cancel{10}}{2} = \frac{5}{2} = 2.5$
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